url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
https://stats.stackexchange.com/questions/317195/hypothesis-testing-interpreting-contradicting-results | # Hypothesis testing - interpreting contradicting results
I wanted to do a hypothesis testing for paired data. My null hypothesis is that Xi is 4 times as big as Yi. So i create a variable Zi = Xi - 4Yi.
The null hypothesis is that Zi = 0
I do the testing in R as follows:
t.value = (mean(Z) - 0) / (sd(Z) / sqrt(length(Z)))
p.value = 2*pt(-abs(t.value), df=length(Z)-1)
I get a p-value higher than 0.05 and i fail to reject the null hypothesis.
My problem is, when i try different values for Z, for example Zi = Xi - 10Yi and Zi = Xi - 50Yi, i still fail to reject the null hypothesis which is strange to me.
My data looks normal but the values for Xi and Yi are tiny. (Xi mean =0.00245945, Yi mean = 0.00109909)
Is there an explanation for this, or am i doing something wrong?
• Can $X_i$ and $Y_i$ be negative? What are their variances or standard deviations? – Henry Dec 5 '17 at 9:02
• Any particular reason why you compute the p-value manually? This doesn't seem to be a special case and therefore seems to be covered by the existing functions. – David Ernst Dec 5 '17 at 9:15
• Yes it can be, standard deviations of Xi = 0.008099045 and standard deviations of Yi = 0.009010012. – G_z Dec 5 '17 at 9:15
If your n is tiny, your test could be so under-powered that it will fail to reject all but the most ridiculous null hypotheses.
edit: I see now that the standard deviations are 6 or 7 times as large as the difference of means. That makes for a quite small effect, Cohen's $d\approx 0.15$, if your sample is representative. If you construct a variable $Z$ that is different from $X-Y$, like the $X-4Y$ you had, this ratio will change accordingly. This still looks like a power problem. Here you see how low your power is with $n=45$:
power.t.test(45,0.14,0.85,0.05,NULL,"paired","two.sided")
Paired t test power calculation
n = 45
delta = 0.14
sd = 0.85
sig.level = 0.05
power = 0.1896772
alternative = two.sided
NOTE: n is number of *pairs*, sd is std.dev. of *differences* within pairs
Here's how many observations you would need for halfway decent power of $0.8$.
power.t.test(NULL,0.14,0.85,0.05,0.8,"paired","two.sided")
Paired t test power calculation
n = 291.2539
delta = 0.14
sd = 0.85
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number of *pairs*, sd is std.dev. of *differences* within pairs
This is based on the assumption that the difference of means and the standard deviations (I didn't have the sd of $Z$, so I approximated it by the mean of the sds of $X$ and $Y$) as measured in your sample of $n=45$ are representative of the population. If you don't want to make those assumptions, you can also reason in terms of Cohen's $d$ and what would constitute a large enough effect to care about, that you would want to be able to detect.
• n = 45, what is the standard for a tiny n? – G_z Dec 5 '17 at 9:18
• Not that for a single t-test, more like 10. Try the built in t-test function just to be sure there isn't a mistake in your code. – David Ernst Dec 5 '17 at 9:19
• I tried t.test() and i got the same results. – G_z Dec 5 '17 at 9:23 | 2019-10-18 11:36:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7407870292663574, "perplexity": 872.113814075267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00433.warc.gz"} |
https://blog.zilin.one/104018-spring-2017/recitation-5/ | # Recitation 5
Problem 1: Consider the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_1 = 1$ and $a_{n+1} = a_n + 1/a_n$ for every $n \ge 1$. Show that $\{a_n\}_{n=1}^\infty$ is monotone increasing and $\lim_{n\to\infty} a_n = \infty$.
Proof: One can see that each $a_n$ is positive (a rigorous proof requires mathematical induction). Thus $a_{n+1} = a_n + 1/a_n > a_n$, that is, the sequence is monotone increasing. Suppose, for a moment, that the sequence is bounded from above, by Problem 1(a) from Recitation 4, the sequence has a limit, say $L$. Because $a_na_{n+1} = a_n^2 + 1$, $L^2 = \lim_{n\to\infty}a_na_{n+1} = \lim_{n\to\infty}a_n^2+1 = L^2+1$, which is impossible. Therefore the sequence is not bounded from above. By Problem 1(b) from Recitation 4, $\lim_{n\to\infty} a_n = \infty$.
Remark: Alternatively, one can prove by mathematical induction that $a_n \le n$. Again, by induction, we can show that $a_n \ge 1 + 1/1 + 1/2 + \dots + 1/(n-1)$. Denote by $H_n := 1/1 + 1/2 + \dots + 1/n$, which is the Harmonic series. It can be shown that $H_n \ge 1 + \lfloor \log_2 n \rfloor / 2$, which diverges to infinity.
Problem 2: Suppose that $\lim_{n\to\infty} a_n = 0$. Prove that $\lim_{n\to\infty} 5^{a_n} = 1$.
Proof: We know that $\lim_{n\to\infty}5^{1/n} = 1$. Given $\epsilon > 0$, find $M$ such that $5^{1/M} < 1+\epsilon$. Since $\lim_{n\to\infty} a_n = 0$, we can find $N$ such that $-1/M < a_n < 1/M$ for all $n > N$. Now, for all $n > N$, $1 - \epsilon < \frac{1}{1+\epsilon} < 5^{-1/M} < 5^{a_n} < 5^{1/M} < 1 + \epsilon$, that is, $|5^{a_n} - 1| < \epsilon$.
Problem 3: Prove that if $\lim_{n\to\infty} a_n = A > 0$ and $\lim_{n\to\infty} b_n = B$, then $\lim_{n\to\infty} a_n^{b_n} = A^B$.
Exercise 1: Use mathematical induction to prove the following. For all $\epsilon\in(0,1)$ and $M\in\mathbb{N}$, $(1-\epsilon)^M \ge 1-M\epsilon$.
Exercise 2: Suppose that $\epsilon \in (0, 1)$, $M\in\mathbb{N}$. If $1-\epsilon/M and $-M < b < M$, then $1 - \epsilon < a^b < 1+\epsilon$.
Proof: We first prove that case when $A = 1$. Since $\lim_{n\to\infty} b_n = B$, by Problem 3 from Recitation 4, the sequence $\{b_n\}_{n=1}^\infty$ is bounded. Let $N$ be a natural number such that $-M < b_n < M$ for all $n$. Suppose that $\epsilon \in (0,1)$ is given. Since $\lim_{n\to\infty} a_n = 1$, there exists $N$ such that $1-\epsilon/M. For all $n > N$, by Exercise 2, $1 - \epsilon < a_n^{b_n} < 1 + \epsilon$, and so $\lim_{n\to\infty} a_n^{b_n} = 1$.
Finally, we prove the general case. Let $a'_n = a_n / A$. Because $\lim_{n\to\infty} a_n' = \lim_{n\to\infty} a_n/A = 1$, we know that $\lim_{n\to\infty} a_n'^{b_n} = 1$. Moreover, from problem 2, we know that $\lim_{n\to\infty} A^{b_n} = A^B$. Therefore $\lim_{n\to\infty} a_n^{b_n} = \lim_{n\to\infty} (a_n' \cdot A)^{b_n} = \lim_{n\to\infty} a_n'^{b_n} \lim_{n\to\infty} A^{b_n} = A^B$.
Facts: If $\lim_{n\to\infty}a_n=\infty$, then $\lim_{n\to\infty}(1+1/a_n)^{a_n}=e$. If $\lim_{x\to x_0}f(x)=\infty$, then $\lim_{n\to\infty}(1+1/f(x))^{f(x)}=e$.
Corollary 1: If $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n/a_n = L$, then (a) $\lim_{n\to\infty}(1+1/a_n)^{b_n}=e^L$ and (b) $\lim_{n\to\infty}(1-1/a_n)^{b_n}=e^{-L}$.
Corollary 2: If $\lim_{x\to x_0}f(x)=\infty$ and $\lim_{x\to 0}g(x)/f(x) = L$, then (a) $\lim_{n\to\infty}(1+1/f(x))^{g(x)}=e^{L}$ and (b) $\lim_{n\to\infty}(1-1/f(x))^{g(x)}=e^{-L}$.
Problem 4: Find the limits (a) $\lim_{n\to\infty} \left(\sin(1/n)+\cos(1/n)\right)^{n^2}$; (b) $\lim_{n\to\infty} \left(1+\tan(1/n)\right)^{n}$; (c) $\lim_{n\to\infty} \left(\frac{n^2+n+1}{n^2-n-1}\right)^{n}$; (d) $\lim_{x\to 0} \left(\cos x\right)^{1/x^2}$.
Answers: (a) $\infty$; (b) $e$; (c) $e^2$; (d) $e^{-1/2}$.
Idea: (a) In Corollary (1a), take $a_n = 1/(\sin(1/n)+\cos(1/n)-1)=1/(2\sin(1/2n)(\cos(1/2n)-\sin(1/2n)))$ and $b_n = n^2$. (b) In Corollary (1a), take $a_n = 1/\tan(1/n)$ and $b_n = n$. (c) In Corollary (1a), take $a_n = (n^2-n-1)/(2n+2)$ and $b_n = n$. (d) In Corollary (2b), take $x_0 = 0$, $f(x) = 1/(1-\cos x) = 1/(2\sin^2(x/2))$ and $g(x) = 1/x^2$.
Problem 5: Let $f\colon \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) = f(2x)$ for every $x$. Show that $f$ is constant.
Idea: In Problem 6 from Homework 2, we actually proved the following. Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) = f(2x)$ for every $x$. If $\lim_{x\to 0}f(x) = L$, then $f(x) = L$ for all $x \neq 0$. In particular, if $f$ is continuous, $L$ must be $f(0)$. | 2020-10-28 16:48:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 100, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9930769205093384, "perplexity": 91.33999287079278}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107900200.97/warc/CC-MAIN-20201028162226-20201028192226-00552.warc.gz"} |
https://plotly.com/python/log-plot/ | # Log Plots in Python
How to make Log plots in Python with Plotly.
New to Plotly?
Plotly is a free and open-source graphing library for Python. We recommend you read our Getting Started guide for the latest installation or upgrade instructions, then move on to our Plotly Fundamentals tutorials or dive straight in to some Basic Charts tutorials.
This page shows examples of how to configure 2-dimensional Cartesian axes to follow a logarithmic rather than linear progression. Configuring gridlines, ticks, tick labels and axis titles on logarithmic axes is done the same was as with linear axes.
### Logarithmic Axes with Plotly Express¶
Plotly Express is the easy-to-use, high-level interface to Plotly, which operates on a variety of types of data and produces easy-to-style figures.
All of Plotly Express' 2-D Cartesian functions include the log_x and log_y keyword arguments, which can be set to True to set the corresponding axis to a logarithmic scale:
In [1]:
import plotly.express as px
df = px.data.gapminder().query("year == 2007")
fig = px.scatter(df, x="gdpPercap", y="lifeExp", hover_name="country", log_x=True)
fig.show()
Setting the range of a logarithmic axis with Plotly Express works the same was as with linear axes: using the range_x and range_y keywords. Note that you cannot set the range to include 0 or less.
In [2]:
import plotly.express as px
df = px.data.gapminder().query("year == 2007")
fig = px.scatter(df, x="gdpPercap", y="lifeExp", hover_name="country",
log_x=True, range_x=[1,100000], range_y=[0,100])
fig.show()
new in 5.8
You can position and style minor ticks using minor. This takes a dict of properties to apply to minor ticks. See the figure reference for full details on the accepted keys in this dict.
In this example we set the tick length with ticklen, add the ticks on the inside with ticks="inside", and turn grid lines on with howgrid=True.
In [3]:
import plotly.express as px
df = px.data.gapminder().query("year == 2007")
fig = px.scatter(df, x="gdpPercap", y="lifeExp", hover_name="country",
log_x=True, range_x=[1,100000], range_y=[0,100])
fig.update_xaxes(minor=dict(ticks="inside", ticklen=6, showgrid=True))
fig.show()
### Logarithmic Axes with Graph Objects¶
If Plotly Express does not provide a good starting point, it is also possible to use the more generic go.Figure class from plotly.graph_objects.
In [4]:
import plotly.graph_objects as go
import plotly.express as px
df = px.data.gapminder().query("year == 2007")
fig = go.Figure()
fig.update_xaxes(type="log")
fig.show()
Setting the range of a logarithmic axis with plotly.graph_objects is very different than setting the range of linear axes: the range is set using the exponent rather than the actual value:
In [5]:
import plotly.graph_objects as go
import plotly.express as px
df = px.data.gapminder().query("year == 2007")
fig = go.Figure()
fig.update_xaxes(type="log", range=[0,5]) # log range: 10^0=1, 10^5=100000
fig.update_yaxes(range=[0,100]) # linear range
fig.show()
#### Reference¶
See function reference for px.(scatter) or https://plotly.com/python/reference/layout/xaxis/#layout-xaxis-type for more information and chart attribute options!
Dash is an open-source framework for building analytical applications, with no Javascript required, and it is tightly integrated with the Plotly graphing library.
Learn about how to install Dash at https://dash.plot.ly/installation.
Everywhere in this page that you see fig.show(), you can display the same figure in a Dash application by passing it to the figure argument of the Graph component from the built-in dash_core_components package like this:
import plotly.graph_objects as go # or plotly.express as px
fig = go.Figure() # or any Plotly Express function e.g. px.bar(...) | 2022-05-29 07:59:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2604585587978363, "perplexity": 7316.544614193397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00592.warc.gz"} |
https://www.dsprelated.com/freebooks/filters/Stability_Equation_Error_Designs.html | ### Stability of Equation Error Designs
A problem with equation-error methods is that stability of the filter design is not guaranteed. When an unstable design is encountered, one common remedy is to reflect unstable poles inside the unit circle, leaving the magnitude response unchanged while modifying the phase of the approximation in an ad hoc manner. This requires polynomial factorization of to find the filter poles, which is typically more work than the filter design itself.
A better way to address the instability problem is to repeat the filter design employing a bulk delay. This amounts to replacing by
and minimizing . This effectively delays the desired impulse response, i.e., . As the bulk delay is increased, the likelihood of obtaining an unstable design decreases, for reasons discussed in the next paragraph.
Unstable equation-error designs are especially likely when is noncausal. Since there are no constraints on where the poles of can be, one can expect unstable designs for desired frequency-response functions having a linear phase trend with positive slope.
In the other direction, experience has shown that best results are obtained when is minimum phase, i.e., when all the zeros of are inside the unit circle. For a given magnitude, , minimum phase gives the maximum concentration of impulse-response energy near the time origin . Consequently, the impulse-response tends to start large and decay immediately. For non-minimum phase , the impulse-response may be small for the first samples, and the equation error method can yield very poor filters in these cases. To see why this is so, consider a desired impulse-response which is zero for , and arbitrary thereafter. Transforming into the time domain yields
where '' denotes convolution, and the additive decomposition is due the fact that for . In this case the minimum occurs for ! Clearly this is not a particularly good fit. Thus, the introduction of bulk-delay to guard against unstable designs is limited by this phenomenon.
It should be emphasized that for minimum-phase , equation-error methods are very effective. It is simple to convert a desired magnitude response into a minimum-phase frequency-response by use of cepstral techniques [22,60] (see also the appendix below), and this is highly recommended when minimizing equation error. Finally, the error weighting by can usually be removed by a few iterations of the Steiglitz-McBride algorithm.
Next Section:
An FFT-Based Equation-Error Method
Previous Section:
Error Weighting and Frequency Warping | 2020-07-13 07:47:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8189405798912048, "perplexity": 914.54612320166}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657143354.77/warc/CC-MAIN-20200713064946-20200713094946-00112.warc.gz"} |
https://physics.stackexchange.com/questions/277634/laser-interferometric-gravitational-wave-detectors | # Laser interferometric gravitational wave detectors
so I was trying to solve this excercise:
Now I was able to find the eq. of geodetics (or directly by Christoffel formulas calculation or by the Lagrangian for a point particle). And I verified that such space constant coordinate point is a geodetic.
Now, for the second point I considered $$ds^2=0$$ to separate the $$dt$$ and find the separation time. But I don't know how to solve for a generic path of a light ray. So I considered that maybe the text wants a light ray travelling along x axis and the second along y axis.
I checked in other sources and all people make the same, by considering a light ray along x-axis and then setting $$dy=dz=0$$.
But when I substitute these in my geodesic equations it turns out that they are not true even at first order in A! So these people that consider a light ray travelling along x-axis, such as in an interferometer, are not considering a light geodesic. All of this if and only if my calculations are true.
So I know that if $$ds^2=0$$ I have a light geodesic. And so it should solve my eq. of geodesics. But if I restrain my motion on x axis what I can say is that the $$ds^2=0$$ condition now is on a submanifold of my manifold. So, the light wave that I consider doesn't not move on a geodesic of the original manifold but on one of the x axis. This is the only thing that came in my mind.
Is there any way to say that I can set $$dy=dz=0$$ without worring? And if I can't set it how can I solve the second point?
One of the light signals is emitted at the origin and travels to a mirror located at (L,0,0). so there is no change in $y$ or $z$. Hence setting $dx=dy=0$ and $ds^2=0$ for light ray: $$0=-dt^2 +(1+A\cos kt)dx^2$$ Taking the square root and integrating both sides $\int \frac{dt}{1+A \cos kt}=\int dx$ using the boundary conditions for the $x$ values one should be able to find the travelling time in the reference frame. Similar computation for the other light ray.
• I did this and calculated the time. What I don't understand is that why I can put $$dy=dz=0$$ without any problem. And why this light signal with $$dy=dz=0$$ is not a geodesic – Saladino Sep 1 '16 at 18:59
• You set them zero because you assume $y$ and $z$ are not changing along the path you consider for the calculation. – Statics Sep 1 '16 at 21:18
• Just look at your movement. It is restricted to the x-axis in the first case. Why would you need to consider the change in $y$ or $z$ for that? – Statics Sep 2 '16 at 10:31 | 2020-08-08 21:09:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8214585781097412, "perplexity": 137.18959913829192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00200.warc.gz"} |
https://codeforces.com/problemset/problem/1117/D | D. Magic Gems
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Reziba has many magic gems. Each magic gem can be split into $M$ normal gems. The amount of space each magic (and normal) gem takes is $1$ unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is $N$ units. If a magic gem is chosen and split, it takes $M$ units of space (since it is split into $M$ gems); if a magic gem is not split, it takes $1$ unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is $N$ units? Print the answer modulo $1000000007$ ($10^9+7$). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
Input
The input contains a single line consisting of $2$ integers $N$ and $M$ ($1 \le N \le 10^{18}$, $2 \le M \le 100$).
Output
Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is $N$ units. Print the answer modulo $1000000007$ ($10^9+7$).
Examples
Input
4 2
Output
5
Input
3 2
Output
3
Note
In the first example each magic gem can split into $2$ normal gems, and we know that the total amount of gems are $4$.
Let $1$ denote a magic gem, and $0$ denote a normal gem.
The total configurations you can have is:
• $1 1 1 1$ (None of the gems split);
• $0 0 1 1$ (First magic gem splits into $2$ normal gems);
• $1 0 0 1$ (Second magic gem splits into $2$ normal gems);
• $1 1 0 0$ (Third magic gem splits into $2$ normal gems);
• $0 0 0 0$ (First and second magic gems split into total $4$ normal gems).
Hence, answer is $5$. | 2022-10-04 09:47:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.756523609161377, "perplexity": 1766.9794012094446}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337490.6/warc/CC-MAIN-20221004085909-20221004115909-00288.warc.gz"} |
http://math.stackexchange.com/questions/319940/define-a-1-1-onto-function-with-domain-a-onto-the-set-1-2-n | # Define a $1$-$1$ onto function with domain $A$ onto the set $\{1, 2, … n\}$
Let $A = \{x^2 : x \in \mathbb{N} \text{ and } 0 \leq x^2 \leq 90\}$.
Define a 1-1 onto function with domain $A$ onto a set of the form $\{1, 2, \ldots, n\}$ to show the cardinality of $A$ is $n$.
I understand that if a bijection exists between two sets, then the cardinality of the sets must be equal.
What I do not understand, is that it seems the domain $A$ has a fixed cardinality of $9$. This is the size of the set of the squares of the natural numbers that fit $0 \leq x^2 \leq 90$.
However, if we pick a value of $n \neq 9$, how can a bijection exist?
-
It can't. The size of the set is $9$ (or if $0$ is a natural number, then $10$). – mixedmath Mar 3 at 23:22
You are just asked to find a bijection $f : A \to \{1, 2, \dots , 9\}$. Also, depending on whether you consider $0$ a natural number, your set may have $10$ elements. – JavaMan Mar 3 at 23:23
@JavaMan Thank you for editing my markup and for the answer. I interpreted the problem statement 'onto a set of the form $\{1, 2, ..., n\}$' to mean that for any value of n we pick, we can find a bijection. – Zach Mar 3 at 23:25
$A = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$
In this case, you need $n = 10:$ there are 10 values $x$ including $0$ such that $0\leq x^2 \leq 90$.
So the range is $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
Try $f(a) = \sqrt{a} + 1$, where $a = x^2 \in A$.
Now just show that $f: A \to B$ is one-to-one and onto.
Also note, you can show that for any $n > 10$, any function $g: A \to B_n$ would fail to be onto $B_n$, where $B_n$ is the set $\{1, 2, 3, ... , n\}, n>10$. And show that any function $h: A \to B_n$ would fail to be injective if $B_n$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 10$.
If $A$ cannot contain $0^2 = 0$, then use the following:
Simply use $A = \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$,
$B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, (so $n = 9$),
$f(x) = \sqrt a, \;a\in A$, for your one-to-one function $f:A\to B$,
Then show that $f$ is one-to-one and onto. In this case, you can show why, for any $n > 9,\,$ any function $\,g: A \to B_n\,$ would fail to be onto $\,B_n$, where $\,B_n\,$ is the set $\{1, 2, 3, ... , n\}, n>9$. And finally, show why any function $\,h: A \to B_n\,$ would fail to be injective if $\,B_n\,$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 9$.
-
I didn't realize I had a handful of "unupvoted" answers out there! Thank you, amzoti! – amWhy Apr 30 at 1:01 | 2013-12-20 11:50:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9803470969200134, "perplexity": 123.53000555161555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345771702/warc/CC-MAIN-20131218054931-00094-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://adityam.github.io/stochastic-control/inf-mdp/inventory-management/ | # ECSE 506: Stochastic Control and Decision Theory
Example: Inventory Management (revisited)
TL;DR One of the potential benefits of modeling a system as infinite horizon discounted cost MDP is that it can be simpler to identify an optimal policy. We illustrate this using the inventory management example.
Consider the model for inventory management and assume that it runs for an infinite horizon. We assume that the per-step cost is given by $c(s,a,s_{+}) = p a + γ h(s),$ where $h(s) = \begin{cases} c_h s, & \text{if s \ge 0} \\ -c_s s, & \text{if s < 0}, \end{cases}$ where $$a$$ is the per-unit holding cost, $$b$$ is the per-unit shortage cost, and $$p$$ is the per-unit procurement cost. Note that in the per-step cost, we are assuming that the holding or shortage cost is dicounted because this cost is incurred at the end of the time period.
Recall that in the finite horizon setting, the optimal policy was a base-stock policy characterized by thresholds $$\{s^*_t\}_{t \ge 1}$$. In the infinite horizon discounted setting, we expect the optimal policy to be time-homogeneous, i.e., the thresholds $$\{s^*_t\}_{t \ge 1}$$ be a constant $$s^*$$ and not to depend on time.
As an illustration, let’s reconsider the example used for the finite horizon setting (where $$p=5$$, $$c_h = 4$$, $$c_s = 2$$, and the demand is Binomial(10,0.4)). We consider the discount factor $$γ = 0.9$$. The value function and optimal policy is this case are shown below.
We are interested in the following question: Is it possible to identify the optimal threshold of the base-stock policy without explicitly solving the dynamic program? In this section, we show that the answer is affirmative.
As a first step, we modify the per-step cost using reward shaping. In particular, we consider the following potential function
$\varphi(s) = h(s) + \frac1{γ} p s - \frac{1}{1-γ}p\mu,$ where $$\mu = \EXP[W]$$ is the expected number of arrivals at each time period.
Now consider a new cost function \begin{align*} c'(s,a,s_{+}) &= c(s,a,s_{+}) + \varphi(s) - γ \varphi(s_{+}) \\ &= pa + γ h(s_{+}) + h(s) + \frac{1}{γ} p s - \frac{1}{1-γ} p \mu - γ h(s_{+}) - p s_{+} - \frac{γ}{1-γ} p \mu \\ &= h(s) + \frac{1-γ}{γ} ps + p w - p \mu. \end{align*} Note that $\EXP[ c'(s,a,S_{+}) | S = s, A = a ] = h(s) + \frac{1-γ}{γ} ps =: c^*(s).$ Thus, the optimal policy of the original model is the same as that in which the per-step cost is given by $$c^*(s)$$.
Recall that the optimal policy in the original model was a base stock policy. For the infinite horizon model, the threshold will become time-invariant. Thus, the optimal policy will be of the form $π(s) = \begin{cases} s^* - s, & \text{if s \le s^*} \\ 0, & \text{otherwise}. \end{cases}$
The infinite horizon dynamic programming with this modified cost is given by $$$\label{eq:DP} V(s) = \min_{a \in \reals_{\ge 0}} \bigl\{ c^*(s) + γ \EXP[ V(s + a - W) ] \bigr\}.$$$
Using the structure of the optimal policy identified above, we have two properties. First, $$$\label{eq:opt} V(s) = c^*(s) + γ \EXP[ V(s^* - W) ], \qquad s \le s^*.$$$ Second, at $$s = 0$$, $$π(0) = s^*$$ (and recall that $$c^*(0) = 0$$). Therefore, $$$\label{eq:opt-policy} s^* = \arg\min_{a \in \reals_{\ge 0}} γ \EXP[V(a - W)]$$$
Let $$F(s^*)$$ denote $$\EXP[V(s^*-W)]$$. Then, substituting $$s = s^* - W$$ in \eqref{eq:opt} and taking expectations, we get $F(s^*) = \EXP[ c^*(s^* - W) ] + γ F(s^*).$ Thus, $\EXP[V(s^* - W)] = F(s^*) = \frac{1}{1-γ} \EXP[ c^*(s^*-W) ].$
Substituting the above in \eqref{eq:opt-policy}, we get $s^* = \arg\min_{s^* \ge 0} \frac{γ}{1-γ} \EXP[ c^*(s^* - W) ].$ Consequently, we have the following:
Theorem 1
The optimal threshold $$s^*$$ is given by the value of $$s^*$$ which minimizes $$\EXP[ c^*(s^*-W) ]$$. In particular, if $$F$$ denotes the CDF of the demand, then $$$s^* = F^{-1}\left( \frac{c_s - p(1-γ)/γ}{c_h + c_s} \right). \label{eq:opt-threshold}$$$
#### Proof
The proof idea is the same approach as that used for the newsvendor problem. In particular, let $$f$$ denote the distribution of $$W$$, $$F$$ denote the CDF, and $$μ = \EXP[W]$$. Then, \begin{align} \EXP[c^*(s^*-W)] &= \EXP[h(s^*-W)] + \frac{1-γ}{γ}p(s^* - μ) \notag \\ &= \int_{0}^{s^*} c_h(s^*-w)f(w)dw + \int_{s^*}^{∞} c_s(w-s^*)f(w)dw + \frac{1-γ}{γ}p(s^* - μ) \notag \end{align} Thereore, \begin{align} \frac{∂ \EXP[c^*(s^*-W)]}{∂s^*} &= \int_{0}^{s^*} c_h f(w)dw + \int_{s^*}^{\infty}[-c_s] f(w)dw + \frac{1-γ}{γ}p \notag \\ &= c_h F(s^*) - c_s(1 - F(s^*)) + \frac{1-γ}{γ}p \notag \\ \end{align} To find the optimal threshold, we set the derivative to $$0$$ and simplify to obtain \eqref{eq:opt-threshold}.
We can use \eqref{eq:opt-threshold} to find the optimal threshold for the example used above. In particular, we need to find the value of $$s^*$$ at which the CDF of the demand equals the value $$(c_s - p(1-γ)/γ)/(c_h + c_s)$$, as shown in Fig. 2 below.
# Exercises
1. Suppose that the arrival process is exponential with rate $$1/\mu$$, i.e., the density of $$W$$ is given by $$e^{-s/\mu}/\mu$$. Show that the optimal threshold is given by $s^* = \mu \log \left[ \frac{ c_h + c_s} { c_h + p (1-γ)/γ} \right].$
Hint: Recall that the CDF the exponential distribution is $$F(s) = 1 - e^{-s/μ}$$.
# References
The idea of using reward shaping to derive a closed form expression for inventory management is taken from Whittle (1982). It is interesting to note that Whittle (1982) uses the idea of reward shaping more than 17 years before the paper by Ng et al. (1999) on reward shaping. It is possible that Whittle was using the results of Porteus (1975).
Ng, A.Y., Harada, D., and Russell, S. 1999. Policy invariance under reward transformations: Theory and application to reward shaping. ICML, 278–287. Available at: http://aima.eecs.berkeley.edu/~russell/papers/icml99-shaping.pdf.
Porteus, E.L. 1975. Bounds and transformations for discounted finite markov decision chains. Operations Research 23, 4, 761–784. DOI: 10.1287/opre.23.4.761.
Whittle, P. 1982. Optimization over time: Dynamic programming and stochastic control. Vol. 1 and 2. Wiley.
This entry was last updated on 03 Oct 2022 and posted in MDP and tagged inventory management, base-stock policy, reward shaping, structural results, stochastic optimization, infinite horizon, discounted cost. | 2022-12-08 20:30:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 4, "x-ck12": 0, "texerror": 0, "math_score": 0.9653363823890686, "perplexity": 1149.8772394525422}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00264.warc.gz"} |
https://proofwiki.org/wiki/Prime_Element_iff_Generates_Principal_Prime_Ideal | # Prime Element iff Generates Principal Prime Ideal
## Theorem
### Integers
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $p \in \Z_{>0}$.
Let $\ideal p$ be the principal ideal of $\Z$ generated by $p$.
Then $p$ is prime if and only if $\ideal p$ is a maximal ideal of $\Z$. | 2019-09-23 18:15:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9531962871551514, "perplexity": 97.59253753138118}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514577478.95/warc/CC-MAIN-20190923172009-20190923194009-00320.warc.gz"} |
http://www.sawaal.com/problems-on-numbers-questions-and-answers/-nbsp40-of-250-50-_9810 | 4
Q:
# 40 % of 250 = 50% ?
A) 200 B) 100 C) 150 D) 400
Explanation:
As per given
$⇒x=200$
Q:
Which one of the following is not a prime number?
A) 91 B) 71 C) 41 D) 31
Explanation:
Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.
Because 91 can be divisible by 7,13,91,1.
It is quite clear that prime number should be divisible only by itself and by 1.
2 106
Q:
There are 5 consecutive odd numbers. If the difference between square of the average of first two odd number and the of the average last two odd numbers is 396, what is the smallest odd number?
A) 29 B) 27 C) 31 D) 33
Explanation:
Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4
According to the question,
Difference between square of the average of first two odd number and the of the average last
two odd numbers is 396
i.e, x+3 and x-3
Hence, the smallest odd number is 33 - 4 = 29.
2 123
Q:
What are the Multiples of 6 and the Common Multiples of 4 and 6?
A) 12, 34, 42 B) 12, 18, 36, C) 6, 4, 14 D) 4, 8, 16
Explanation:
The Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60.
The Common Multiples of 4 & 6 upto 60 numbers are given as 12, 24, 36, 48, 60.
5 192
Q:
12 students of class IX took part in GK quiz. If the number of boys is 4 more than the number of girls. Find the number of boys and girls took part in the quiz ?
A) 7, 5 B) 9, 3 C) 8, 4 D) 6, 6
Explanation:
Let the number of girls = x
=> x + x + 4 = 12
=> 2x = 8
=> x = 4
=> Number of girls = 4
=> Number of boys = 4 + 4 = 8
11 373
Q:
A number is 2 more than thrice the second number and 4 times the smaller number is 5 more than the greater number. What is greater number ?
A) 7 B) 23 C) 9 D) 21
Explanation:
Let the greater and smaller number be p and q respectively.
4q = p + 5 ------ (I)
p = 3q+2 ------- (II)
From equation (I) and (II)
q = 7
p = 23
7 440
Q:
What will be added to the unit digits of the number 86236 so that number will be divisible by 9 ?
A) 1 B) 2 C) 3 D) 0
Explanation:
A number is divisible by 9 only if the sum of the digits of the number is divisible by 9.
Here 86236 = 8 + 6 + 2 + 3 + 6 = 25
We must add 2 to 25 to become 27 which is divisible by 9.
4 797
Q:
How many numbers up to 101 and 300 are divisible by 11 ?
A) 18 B) 20 C) 19 D) 17
Explanation:
(300 – 101)/11 = 199/11 = 18 1/11
18 Numbers.
9 645
Q:
The sum of the digits of a two-digit number is 12. The difference of the digits is 6. Find the number ?
A) 57 B) 75 C) 48 D) 39
Explanation:
Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93. | 2018-03-19 07:11:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43131497502326965, "perplexity": 374.36354722967786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646602.39/warc/CC-MAIN-20180319062143-20180319082143-00213.warc.gz"} |
https://erdosninth.wordpress.com/2015/05/18/introductory-abstract-algebra-group-theory-part-i/ | # Introductory Abstract Algebra – Group Theory, Part I
Hi all! In the next few posts, I’ll introduce everything you might want to know about abstract algebra. After this, I’ll try introducing either commutative algebra or basic algebraic number theory. And then I’ll post notes on algebraic geometry as I learn it.
Anyhow, the first thing one learns in algebra is the notion of a group. Let us think of the simplest polygon that you can draw; for me, it’s a square. What are the symmetries of a square? Well, there’s obviously clockwise rotation about $90^\circ$ – let’s call this operation $\mathrm{Rot}$. Then we can compose $\mathrm{Rot}$ with itself twice, and that’s still a symmetry of the square. In other words, $\mathrm{Rot}\circ\mathrm{Rot}$, which we will denote by $\mathrm{Rot}^2$, is a symmetry of the square. Similarly, $\mathrm{Rot}\circ\mathrm{Rot}\circ\mathrm{Rot}$, which we denote by $\mathrm{Rot}^3$, is also a symmetry of the square.
Let’s now consider anticlockwise rotation about $90^\circ$; we will denote this operation by $\mathrm{Rot}^{-1}$. This notation makes sense because $\mathrm{Rot}^{-1}\circ\mathrm{Rot}$ is the “identity” symmetry – it does absolutely nothing to the square. What you should note is that $\mathrm{Rot}^{-1}$ is the same as $\mathrm{Rot}^3$!
We can also flip about one of the corners. Let us label the vertices of the square $1,2,3,4$. Then the flips can be denoted $F_1,F_2,F_3,F_4$, respectively. The interesting thing is that $F_i\circ \mathrm{Rot}^n$ is still always a symmetry of the square! (Can you find the inverse to each of these operations? It’s quite easy!) Let us stop for a moment and take a look at the data.
We have a collection of symmetries of the square, namely the powers of $\mathrm{Rot}$ and the flips, $F_i$, and an operation on them – composition – that when you compose any two of these, you can get something another symmetry, there’s an inverse to each symmetry, and there’s an “identity” which does not do anything to the square. That’s very interesting – are there other sets that exhibit such behavior? You don’t even have to look too far to find something like this!
Consider the integers, $\mathbf{Z}$, with the operation of addition, $+$. Then if you add two integers, you still get an integer; given an integer $n$ you can always find it’s additive inverse, $-n$; the number $0$ does not do anything at all to a number on addition. So $\mathbb{Z}$ also exhibits similar behavior. Similarly with $\mathbb{Q}$, the rationals, under addition, and $\mathbb{R}$, the rationals, under addition. (Why not multiplication? The reason for $\mathbb{Q}$ and $\mathbb{R}$ is the same, and for $\mathbb{Z}$, there’s an additional reason.) How about $\mathbb{C}$ under addition? Under multiplication?
This suggests that these properties are something special, and indeed, they are! They motivate the definition of a group:
Definition 1: A group is a set $G$ with a binary operation $\circ: G \times G \to G$ and a map $\circ^{-1}:G\to G$, called the multiplication (note this is used in a general sense, as the “multiplication” could be, say, addition) and the inverse, respectively, such that the following conditions are satisfied:
• The multiplication is associative, i.e., $x\times (y\times z)=(x\times y)\times z$. Since I’m a category theorist, this corresponds to a commutative diagram which I sadly cannot draw, but here’s a picture:
Just replace $m$ with the $\circ$ above and $id$ is the identity map. Check to see how these are equivalent.
• There’s an element $1$ that is the multiplicative identity for this multiplication, i.e., such that $1\times x=x\times 1=x$. (Although denoted “$1$“, this is simply notation for any element which “does nothing” under the binary operation)
• For any element $g$, there’s another element, called the inverse, that is the image of $g$ under $\circ^{-1}$, such that multiplication with that element gives the multiplicative identity, i.e., there is a $g^{-1}$ such that $g\times g^{-1}=g^{-1}\times g=1$.
In fact, one can show that if we replaced the square with any regular $n$-gon in our discussion above, and consider the symmetric rotations, then we get a group, called the dihedral group $D_n$ of order $n$. You can show that the operation that we defined as composition is not commutative; that is, if I rotate clockwise and then flip the $n$-gon, it will not be the same as flipping it and then rotating it. However, “multiplication” (in this case addition) is commutative for the integers, the reals, and the rationals. This motivates:
Definition 2: A group is abelian if the multiplication is commutative.
One can study the size of a group as well:
Definition 3: Let $(G,\circ)$ be a group.The order of the group $G$ is denoted $|G|$, and is the number of elements in $G$. It’s always a whole number.
You can figure out what a finite group is. Here’s an example of a finite group: let $S_n$ denote the collection of all permutations of the set $\{1,\cdots,n\}$. Under composition, this is a finite group of order $n!$ (prove it!).
Alright, let’s move on. You know that there are inclusions $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$, and the operation on all these groups are the same (here we are using the multiplication of addition on the set $\mathbb{C}$ – I just gave away the answer to two questions above!). We can think of this as a “chain” of subgroups, motivating us to state:
Definition 4: Let $G$ be a group and $H$ a subset of $G$. It is a subgroup if, when we restrict the operations of $G$ to $H$, then $H$ is itself a group.
Obviously, we’d like some easy-to-check conditions under which a subset of a group is a subgroup. Here’s a statement which we won’t prove here:
Proposition 5: Let $H$ be a subgroup of a group $G$. Then the identity of $H$ is the identity of $G$, and the inverse of an element in $H$ is the same as its inverse in $G$.
We’re actually more interested in the following statement:
Proposition 6: A subset $H$ of a group $G$ is a subgroup if and only if it is closed under multiplication and inverses.
To prove this, we first note that the forward direction is obvious (it is the definition of a group). The reader can check that the multiplication coming from $G$ is well-defined. We only need to show that $H$ has an identity. Let $x\in H$. Then $H$ is closed under inverses, so $x^{-1}\in H$. But it’s also closed under multiplication, so $x\times x^{-1} = 1_x$ is also in $H$. Thus, $H$ satisfies the axioms of a group.
Here are some examples. The first is the example we showed before: $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$. You can check that the conditions of Proposition 6 are satisfied. Another one: We can interpret any symmetry as a matrix. So let $\mathrm{GL}_2(\mathbb{R})$ denote the group of $2\times 2$ matrices with entries in $\mathbb{R}$. (This is called the “General Linear Group of $2 \times 2$ matrices”) Then $D_n$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$. Yet another one: Let $\mathrm{SL}_2(\mathbb{R})$ denote the group of $2\times 2$ matrices with entries in $\mathbb{R}$ whose determinant is 1. Then it’s a subgroup of $\mathrm{GL}_2(\mathbb{R})$.
A very important thing in abstract algebra is the manipulation of finite groups. Here’s an example that illustrates this:
Proposition 7: Let $G$ be a finite group and $H\subseteq G$ that is closed under multiplication. Then it’s a subgroup.
To prove this (following here), we will use the property that $G$ is finite. To save space, let’s suppress the multiplication, and write $xy$ for $x\times y$, and $x^n$ for $x\times x\times \cdots\times x$. By Proposition 6, it suffices to show that $H$ is closed under inverses. Suppose $g\in H$. If this is the identity, $1$, then its inverse is … itself, so it’s in $H$. So it’s safe to assume that $g\neq 1$. Recall that $H$ is closed under multiplication; so the infinite sequence $g,g^2,\cdots$ has all entries in $H$. But $G$, and therefore $H$, is finite – so there must be some $i,j$ for which $g^i=g^j$. Without loss of generality, we can assume $i. So, $g^{i-j}=1$; but $g\neq 1$, so $i-j\neq 1$. Let $k=i-j-1$. Then let $h=g^k$. We observe that $g\times h=g^{1+i-j-1}=g^{i-j}=1$; so $gh=1$. We’ve constructed an inverse of $g$, so we’re done.
We can now turn to an important class of groups – the cyclic groups. I think that this is already a lot of information for one post, so I guess I’ll write this up in the next post.
S.D. (J.T.) | 2018-07-16 10:33:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 118, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9473494291305542, "perplexity": 202.12616241937093}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589251.7/warc/CC-MAIN-20180716095945-20180716115945-00039.warc.gz"} |
https://emilemathieu.fr/posts/2018/08/svm/ | # An Efficient Soft-Margin Kernel SVM Implementation In Python
Published:
This short tutorial aims at introducing support vector machine (SVM) methods from its mathematical formulation along with an efficient implementation in a few lines of Python! Do play with the full code hosted on my github page. I strongly recommend reading Support Vector Machine Solvers (from L. Bottou & C-J. Lin) for an in-depth cover of the topic, along with the LIBSVM library. The present post naturally follows this introduction on SVMs.
## Support Vector Machines - The model
##### Figure 1: An optimal hyperplane.
Support vector machines (SVMs) are supervised learning models for classification (or regression) defined by supporting hyperplanes. SVM is one of the most widely used algorithms since it relies on strong theoretical foundations and has good performance in practice.
As illustrated on Figure 1, SVMs represent examples as points in space, mapped so that the examples of the different categories are divided by a clear gap that is as wide as possible. New examples are then mapped into that same space and assigned a category based on which side of the gap they fall.
Let's consider a dataset $D = { (\mathbf{x}_{i}, y_{i}), \mathbf{x} \in \mathbb{R}^d, y \in \{ -1, 1 \}}$ , and $\phi$ a feature mapping - a possibly non-linear function used to get features from the datapoints $\{x_i\}$.
The main idea is to find an hyperplane $\w^*$ separating our dataset and maximising the margin of this hyperplane:
$$\label{eq:hard_objective} \min _{\w, b} \mathcal{P}(\w, b) = \cfrac{1}{2} \w^2$$ $$\label{eq:hard_conditions} \text{subject to} \ \forall i \ \ y_i(\w^T \phi(\x_i)+b) \ge 1$$ The conditions (\ref{eq:hard_conditions}) enforce that datapoints (after being mapped through $\phi$) from the first (respectively second) category lie below (respectively above) the hyperplane $\w^*$, while the objective (\ref{eq:hard_objective}) maximises the margin (the distance between the hyperplane $\w^*$ and the closest example).
### Soft margin
We implicitly made the assumption that the dataset $D$ was separable: that there exists an hyperplane $\w^* \in \mathbb{R}^d$ such that all red points (i.e. $y=-1)$ lie on one side of the hyperplane (i.e. $\w^T \phi(\x_i)+b \le 0$) and blue points (y=$+1)$ lie on the other side of the hyperplane (i.e. $\w^T \phi(\x_i)+b \ge 0$). This formulation is called hard margin, since the margin cannot let some datapoints go through (all datapoints are well classified).
This assumption can be relaxed by introducing positive slack variables $\mathbf{\xi}=(\xi_1, \dots, \xi_n)$ allowing some examples to violate the margin constraints (\ref{eq:hard_conditions}). $\xi_i$ are non-zero only if $\x_i$ sits on the wrong side of the hyperplane, and is equal to the distance between $\x_i$ and the hyperplane $\w$. Then an hyperparameter $C$ controls the compromise between large margins and small margin violations.
$$\label{eq:hard_primal} \min _{\w, b, \mathbf{\xi}} \mathcal{P}(\w, b, \mathbf{\xi}) = \cfrac{1}{2} \w^2 + C \sum_{i=1}^n \xi_i \\ \text{subject to} \begin{cases} \forall i \quad y_i(\w^T \phi(\x_i)+b) \ge 1 - \xi_i \\ \forall i \quad \xi_i \ge 0 \end{cases}$$
## How to fit the model ?
Once one have such a mathematical representation of the model through an optimisation problem, the next natural question arising is how should we solve this problem (once we know it is well-posed) ? The SVM solution is the optimum of a well defined convex optimisation problem (\ref{eq:hard_primal}). Since the optimum does not depend on the manner it has been calculated, the choice of a particular optimisation algorithm can be made on the sole basis of its computational requirements.
### Dual formulation
Directly solving (\ref{eq:hard_primal}) is difficult because the constraints are quite complex. A classic move is then to simplify this problem via Lagrangian duality (see L. Bottou et al for more details), yielding the dual optimisation problem: $$\label{eq:soft_dual} \max _{\alpha} \mathcal{D}(\alpha) = \sum_{i=1}^n \alpha_i - \cfrac{1}{2} \sum_{i,j=1}^n y_i \alpha_i y_j \alpha_j \mathbf{K}(\x_i, \x_j) \\$$ $$\label{eq:soft_dual_cons} \text{subject to} \begin{cases} \forall i \quad 0 \le \alpha_i \le C \\ \sum_i y_i\alpha_i = 0 \end{cases}$$ with $\{\alpha_i\}_{i=1,\dots,n}$ being the dual coefficients to solve and $\mathbf{K}$ being the kernel associated with $\phi$: $\forall i,j \ \ \mathbf{K}(\x_i, \x_j)=\left< \phi(\x_i) , \phi(\x_j)\right>$. That problem is much easier to solve since the constraints are much simpler. Then, the direction $\w^*$ of the optimal hyperplane is recovered from a solution $\alpha^*$ of the dual optimisation problem (\ref{eq:soft_dual}-\ref{eq:soft_dual_cons}) (by forming the Lagragian and taking its minimum w.r.t. $\w$ - which is a strongly convex function): $$\w^* = \sum_{i} \alpha^*_i y_i \phi(\x_i)$$ The optimal hyperplane is therefore a weighted combination over the datapoints with non-zero dual coefficient $\alpha^*_i$. Those datapoints are therefore called support vectors, hence «support vector machines». This property is quite elegant and really useful since in practice only a few $\alpha^*_i$ are non-zeros. Hence, a new datapoint prediction only requires to evaluate: $$\text{sign}\left(\w^{*T} \phi(\x)+b\right) = \text{sign}\left(\sum_{i} \alpha^*_i y_i \phi(\x_i)^T\phi(\mathbf{x}) +b\right) = \text{sign}\left(\sum_{i} \alpha^*_i y_i \mathbf{K}(\mathbf{x}_i, \mathbf{x}) +b \right)$$
The SVM optimisation problem (\ref{eq:soft_dual}) is a Quadratic Problem (QP), a well studied class of optimisation problems for which good libraries has been developed for. This is the approach taken in this intro on SVM, relying on the Python's quadratic program solver cvxopt.
Yet this approach can be inefficient since such packages were often designed to take advantage of sparsity in the quadratic part of the objective function. Unfortunately, the SVM kernel matrix $\mathbf{K}$ is rarely sparse but sparsity occurs in the solution of the SVM problem. Moreover, the specification of a SVM problem rarely fits in memory and generic optimisation packages sometimes make extra work to locate the optimum with high accuracy which is often useless. Let's then described an algorithm tailored to efficiently solve that optimisation problem.
### The Sequential Minimal Optimisation (SMO) algorithm
One way to avoid the inconveniences above-mentioned is to rely on the decomposition method. The idea is to decompose the optimisation problem in a sequence of subproblems where only a subset of coefficients $\alpha_i$, $i \in \mathcal{B}$ needs to be optimised, while leaving the remaining coefficients $\alpha_j$, $j \notin \mathcal{B}$ unchanged: $$\label{eq:smo} \max _{\alpha'} \mathcal{D}(\alpha') = \sum_{i=1}^n \alpha'_i - \cfrac{1}{2} \sum_{i,j=1}^n y_i \alpha'_i y_j \alpha'_j \mathbf{K}(\x_i, \x_j) \\ \text{subject to} \begin{cases} \forall i \notin \mathcal{B} \quad \alpha'_i=\alpha_i \\ \forall i \in \mathcal{B} \quad 0 \le \alpha'_i \le C \\ \sum_i y_i\alpha'_i = 0 \end{cases}$$ One need to decide how to choose the working set $\mathcal{B}$ for each subproblem. The simplest is to always use the smallest possible working set, that is, two elements (such as the maximum violating pair scheme, which is discussed in Section 7.2 in Support Vector Machine Solvers ). The equality constraint $\sum_i y_i \alpha'_i = 0$ then makes this a one dimensional optimisation problem.
##### Figure 2: Direction search - from L. Bottou & C-J. Lin.
The subproblem optimisation can then be achieved by performing successive direction searches along well chosen successive directions. Such a method seeks to maximizes an optimisation problem restricted to the half line ${\mathbf{\alpha} + \lambda \mathbf{u}, \lambda \in \Lambda}$, with $\mathbf{u} = (u_1,\dots,u_n)$ a feasible direction (i.e. can slightly move the point $\mathbf{\alpha}$ along direction $\mathbf{u}$ without violating the constraints).
The equality constraint (\ref{eq:smo}) restricts $\mathbf{u}$ to the linear subspace $\sum_i y_i u_i = 0$. Each subproblem is therefore solved by performing a search along a direction $\mathbf{u}$ containing only two non zero coefficients: ${u}_i = y_i$ and ${u}_j = −y_j$.
The set $\Lambda$ of all coefficients $\lambda \ge 0$ is defined such that the point $\mathbf{\alpha} + \lambda \mathbf{u}$ satisfies the constraints. Since the feasible polytope is convex and bounded $\Lambda = [0, \lambda^{\max}]$. Direction search is expressed by the simple optimisation problem $$\lambda^* = \arg\max_{\lambda \in \Lambda}{\mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u})}$$ Since the dual objective function is quadratic, $\mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u})$ is shaped like a parabola. The location of its maximum $\lambda^+$ is easily computed using Newton’s formula: $$\lambda^+ = \cfrac{ \partial \mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u}) / \partial \lambda \ |_{\lambda=0} } {\partial^2 \mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u}) / \partial \lambda^2 \ |_{\lambda=0}} = \cfrac{\mathbf{g}^T \mathbf{u}}{\mathbf{u}^T \mathbf{H} \mathbf{u}}$$
where vector $\mathbf{g}$ and matrix $\mathbf{H}$ are the gradient and the Hessian of the dual objective function $\mathcal{D}(\mathbf{\alpha})$: $$g_i = 1 - y_i \sum_j{y_j \alpha_j K_{ij}} \quad \text{and} \quad H_{ij} = y_i y_j K_{ij}$$ Hence $\lambda^* = \max \left(0, \min \left(\lambda^{\max}, \lambda^+ \right)\right) = \max \left(0, \min \left(\lambda^{\max}, \cfrac{\mathbf{g}^T \mathbf{u}}{\mathbf{u}^T \mathbf{H} \mathbf{u}} \right)\right)$.
### Implementation
From a Python’s class point of view, an SVM model can be represented via the following attributes and methods:
Then the _compute_weights method is implemented using the SMO algorithm described above:
## Demonstration
We demonstrate this algorithm on a synthetic dataset drawn from a two dimensional standard normal distribution. Running the example script will generate the synthetic dataset, then train a kernel SVM via the SMO algorithm and eventually plot the predicted categories.
##### Optimal hyperplane with predicted labels for radial basis (left) and linear (right) kernel SVMs.
The material and code is available on my github page. I hope you enjoyed that tutorial !
### Acknowledgments
I’m grateful to Thomas Pesneau for his comments.
Tags: | 2021-02-28 15:22:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 11, "x-ck12": 0, "texerror": 0, "math_score": 0.8621110916137695, "perplexity": 575.6230294709941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178361510.12/warc/CC-MAIN-20210228145113-20210228175113-00026.warc.gz"} |
https://www.physicsforums.com/threads/calculating-proton-decay-lifetime-correctly.826225/ | # Calculating proton decay lifetime correctly?
1. Aug 4, 2015
### Anchovy
I'm trying to calculate the SU(5) model's prediction for the lower limit of a proton decay lifetime, for the channel $p \rightarrow \pi^{0} e^{+}$. I'm following this paper:
arXiv:hep-ph/0504276v1
It contains the following equation:
As far as I can tell this actually contains a prediction based also on SO(10), but the paper states that setting $k_{2} = 0$ means that it reduces to just SU(5).
Anyway, I have the following inputs for this equation and having tried to calculate it in Excel, I end up way away from what I expect to find, which is a proton decaying in $10^{29} - 10^{31}$ years. Here's what I am using:
$$k_{1} = \frac{g_{GUT}}{4M_{XY}}, \hspace{0.5 cm} m_{p} = 938.3 \hspace{1 mm} MeV, \hspace{0.5 cm} f_{\pi} = 130 \hspace{1 mm} MeV, \hspace{0.5 cm} A_L = 1.43, \hspace{0.5 cm} \alpha = 0.003 \hspace{1 mm} GeV^{3}, \hspace{0.5 cm} D+F = 1.276, \hspace{0.5 cm} g_{GUT} = 4\pi(\frac{1}{40}), \hspace{0.5 cm} M_{XY} = 10^{14} \hspace{1 mm} GeV$$
The latter two I have taken from the left plot shown here where I've said $\alpha = g / 4\pi$:
Anyway, when I compute this, what I find is a number of the order $\Gamma \approx 10^{-64}$. I assume one then gets the proton lifetime $\tau_{p}$. by taking $\tau_{p} = 1 / \Gamma$, but clearly I end up very far away from $10^{29} - 10^{31}$ years.
I have attached an Excel file where the calculation is broken up into a few steps and they all look fine to me. The final answer I get is shown in bold. Does anyone know what I'm doing wrong? I've tried messing with factors of 1/1000 to make MeV's into GeV's where possible, and also using a factor of (60 * 60 * 24 * 365) seconds per year, but that isn't enough to get me anywhere near the right order of magnitude.
I notice that this equation is different from the rough estimate (according to arXiv:hep-ph/0601023v3) for the proton decay rate:
$$\Gamma_{p} \approx \alpha_{GUT}^{2}\frac{m_{p}^{5}}{M_{GUT}^{4}}$$
where $\alpha_{GUT} = g_{GUT} / 4\pi$ and $M_{GUT} \equiv M_{XY}$ from above if I'm not mistaken. That equation uses the proton mass to the 5th power whereas the long equation above only has $m_p$ to the first power. Overall I'm confused... help?!
Thanks.
#### Attached Files:
• ###### proton_decay_channel_calculation.xlsx
File size:
18.4 KB
Views:
56
Last edited: Aug 4, 2015
2. Aug 5, 2015
### Staff: Mentor
Probably an error with units somewhere. Did you do the calculation with units?
3. Aug 7, 2015
### Anchovy
Well that's the thing, the paper I'm going from doesn't seem to indicate the units but I wondered if there's some factor that the author assumes everyone reading knows is supposed to be applied, in the same way that it might be assumed with out pointing it out that readers know your equations are in terms of $c = \hbar = 1$ (eg. in this case I thought maybe multiply whatever you get by 10^32 or something).
4. Aug 7, 2015
### Staff: Mentor
You'll see that if you work with units, that is the point.
5. Aug 7, 2015
### Anchovy
View attachment 86840
So calculating with units...
$$\Gamma_{p} = \frac{0.938 GeV}{16\pi(0.130 GeV)^2} \times (1.43)^{2} \times (0.003 GeV^{3})^{2} \times (1 + 1.276)^{2} \times \frac{5\times (0.3142)^{4}}{ (4\times 10^{14} GeV)^{4} } \\ = 2.004 \times 10^{-64} \times (\frac{GeV}{GeV^{2}} \times GeV^{6} \times GeV^{-4} ) \\ = 2.004 \times 10^{-64} GeV$$
So that is at least dimensionally correct seeing as using natural units means this decay rate has units of energy, so decay time has units of $GeV^{-1}$... oh and then I can say $\tau_{p} = 1/\Gamma_{p} = 4.99 \times 10^{63} GeV^{-1}$ and
$$1 \hspace{1 mm} eV^{-1} = 6.58 \times 10^{-16} s \\ \rightarrow 1 \hspace{1 mm} eV = 1.520 \times 10^{15} \hspace{1 mm} s^{-1} \\ \rightarrow 1 GeV = 1.520 \times 10^{24} \hspace{1 mm} s^{-1} \\ \rightarrow \hspace{1 mm} 1 \hspace{1 mm} GeV^{-1} = 6.58 \times 10^{-25} \hspace{1 mm} s \\ 1 \hspace{1 mm} year = 60 \times 60 \times 24 \times 365 \hspace{1 mm} s = 31536000 \hspace{1 mm} s \\ \rightarrow 1 \hspace{1 mm} GeV^{-1} = 6.58 \times 10^{-25} \hspace{1 mm} s / 31536000 \hspace{1 mm} s \hspace{1 mm} yr^{-1} = 2.087 \times 10^{-32} \hspace{1 mm} yr$$
so $\tau_{p} = 4.99 \times 10^{63} \hspace{1 mm} GeV^{-1} = (4.99 \times 10^{63}) \times (2.087 \times 10^{-32}) \hspace{1 mm} yr = 1.04 \times 10^{32} \hspace{1 mm} yr$
ahhhh a sensible number. Still just one order of magnitude above the $10^{28} - 10^{31}$ range I expected to hit but... that must be easily fixed when I work out which number I have wrong.
Last edited: Aug 7, 2015
6. Aug 7, 2015
### Anchovy
Actually something is STILL going wrong and I don't know why. The paper arXiv:hep-ph/0504276v1 calculates the proton lifetime to be $3\times 10^{33}$ years. I have attached another Excel file with the calculation set out clearly in an attempt to replicate this number and I'm way off.
Two things I notice they do:
1) is they use $M_{V} \equiv M_{XY} = 10^{16}$ GeV. For a non-SUSY calculation I thought this should be $M_{XY} = 10^{14}$ but whatever, if I use their number my calculation results in a proton lifetime of $\tau_{p} \approx 4.7 \times 10^{36}$ years. Off by a factor of 1000.
2) The only input I choose myself is that of $g_{GUT}$, this is the only variable they do not state a value for. Is it correct to have set $g_{GUT} = \sqrt{ 4\pi\alpha_{GUT} } = \sqrt{ 4\pi(\frac{1}{40}) } \approx 0.561$ ? Actually if they have used $M_{XY} = 10^{16}$ GeV (ie. SUSY GUT scale) then I should use the corresponding $\alpha_{GUT}^{-1} = 25$ as that is the SUSY unification value... If I adjust $g_{GUT} \approx 3.5$ I can reproduce their $\tau_{p} = 3\times 10^{33}$ years...
The only other possibility as far as I can tell is that I've done my unit conversion from GeV^-1 to years wrong...
#### Attached Files:
• ###### proton_decay_channel_calculation__take2.xlsx
File size:
95.5 KB
Views:
51
Last edited: Aug 7, 2015 | 2017-12-18 07:24:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.773674726486206, "perplexity": 687.0138295172268}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00763.warc.gz"} |
https://math.stackexchange.com/questions/686748/nested-sequence-of-closed-intervals | # Nested sequence of closed intervals
Let $A:=[a_n,b_n],n\in\mathbb{N}$, be a nested sequence of closed intervals, i.e. $a_{n+1}>a_n$ and $b_{n+1}<b_n$ for all $n\in\mathbb{N}$. Show that the intersection $\cap_{n\in\mathbb{N}}A_n\neq \emptyset$ is non-empty. Moreover, if $\lim (b_n-a_n)=0$, then $\cap_{n\in\mathbb{N}}A_n=\{x_0\}$ consists of a single point. Is such a statement generally true for a nested sequence of non-closed intervals?
I have absolutely no idea how to do this one. Never worked with nested intervals before.
• This is called the Nested Interval Theorem. – T.J. Gaffney Feb 23 '14 at 4:53
• so it really just comes down to a monotonic bounded sequence then – terrible at math Feb 23 '14 at 4:58
• $\;\cup A_n\;$ is the union of the $\;A_n\;$ , not their intersection, which is $\;\cap A_n\;$ ... – DonAntonio Feb 23 '14 at 5:10
• @terribleatmath, google Nested Interval Theorem, or Cantor theorem on nested intervals, etc. It uses the basic Bolzano-Weierstrass Theorem. – DonAntonio Feb 23 '14 at 5:11
• @DonAntonio personal.bgsu.edu/~carother/cantor/Nested.html I found that site, and ... "Clearly, both a and b are elements of , because both are an elements of the closed interval for any n. (Why?) " No, that is not clear, Could you explain that to me? – terrible at math Feb 23 '14 at 10:16
## 1 Answer
(1) Let $\{a_n: n\in \mathbb{N}\}$ be the set of all the left-hand endpoints. Then the set is non-empty and because the intervals are nested each $b_n$ is an upper bound. Let $x= \sup\{a_n: n\in \mathbb{N}\}$. Then $a_n\le x\le b_n$ for all $n \in \mathbb{N}$ (why?). Hence $x\in \bigcap_n[a_n,b_n]$.
(2) If $(a_n-b_n) \rightarrow 0$ we need to show that the intersection just contain a single point. Suppose to the contrary that there exists some other $x'$ such that $x\not=x'$. Let $\varepsilon= |x-x'|/4$. Then there exists a $N\in \mathbb{N}$ such that $|a_n-b_n|\le \varepsilon$ for all $n\ge N$. Since $x,x'\in \bigcap_n[a_n,b_n]$, then $a_n\le x\le b_n$ and $a_n\le x'\le b_n$. Thus
\begin{align}|x-x'|\le |x-b_N|+|b_N-a_N|+|a_N-x'|\\ \le 3\varepsilon=3|x-x'|/4\end{align}
a contradiction.
• Thank you. Looking at this now it is helping a lot. It is late an I will give it more of a throrough reading tomorrow. I have never seen nested intervals before so I am trying to get a basic understanding. The hardest part for me to understand is what the intersection operation actually does here. Could you explain exactly what a nested interval is and what the intersection does here? (Of course i know what intersection does in general, i'm just confused by this) – terrible at math Feb 23 '14 at 10:19
• @terribleatmath: A nested interval is "collection" of intervals $I_n:= [a_n,b_n]$ such that $I_{n+1}\subset I_n$. The resulting nested sequence looks like $I_0 \supset I_1 \supset ... \supset I_n \supset I_{n+1}\supset...$ (try to draw a picture may help you). The intersection give us the set of point which are in all the intervals, i.e., the set of point $x$ such that $a_n\le x \le b_n$ for all $n\in \mathbb{N}$. – Jose Antonio Feb 23 '14 at 16:48
• Looking at this again, the problem says $(b_n - a_n) \implies 0$ not $(a_n - b_n) \implies 0$ or does it not matter? – terrible at math Feb 23 '14 at 19:57
• By the limit laws if $(b_n-a_n )\rightarrow 0$ then $(a_n-b_n)=-(b_n-a_n)\rightarrow-1\cdot0=0$, i.e., $(a_n-b_n)\rightarrow 0$. – Jose Antonio Feb 23 '14 at 20:24 | 2019-10-14 15:27:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7158328890800476, "perplexity": 233.65827004948133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00110.warc.gz"} |
https://www.cleanslateeducation.co.uk/t/question-2/158 | # Question_2
Can be formulated as a series of simultaneous equations - I.e. let x = # 50ps and let y = # 10ps, then the solution satisfies the equations:
5x + y = 85 and;
x + y = 10
So, using substitustion,
5x + 25 - x = 85
Hence
4x = 60
x = 15
y = 10
I think the answer is (d). I agree with your method of using simultaneous equations to solve it but I came up with these two
50x + 10y = 850 (Based on x being the number of 50 p’s and y being the number of 10 p’s)
x + y = 25 (As we know in total there are 25 coins)
Solving these gives us: x=15 and y = 10
If the machine then returns 5 ten pence coins, this then leaves y=5 in the machine.
So our ratio of 50 pence coins to 10 pence coins in the machine at the end is 15:5, this simplifies to 3:1.
1 Like
I think MathsFan may be right
Perhaps we can achieve consensus? #Collaboration
Spot on - the answer is d) and I didn’t fully read the question!
@Reliot, @MathsFan and @DoinGr8 have all collaborated on this question and so shall all get merit | 2021-04-16 11:19:22 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8155347108840942, "perplexity": 1020.5893721036174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00223.warc.gz"} |
https://ncatlab.org/nlab/show/jet+group | # nLab jet group
Contents
### Context
#### Differential geometry
synthetic differential geometry
Introductions
from point-set topology to differentiable manifolds
Differentials
V-manifolds
smooth space
Tangency
The magic algebraic facts
Theorems
Axiomatics
cohesion
tangent cohesion
differential cohesion
graded differential cohesion
$\array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& ʃ &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }$
Models
Lie theory, ∞-Lie theory
differential equations, variational calculus
Chern-Weil theory, ∞-Chern-Weil theory
Cartan geometry (super, higher)
group theory
# Contents
## Idea
The concept of jet group is the generalization of general linear group from first order to higher order jets.
In terms of synthetic differential geometry/differential cohesion a general linear group is the automorphism group of a first-order infinitesimal disk, while a jet group is the automorphism group of a higher order infinitesimal disk. See also at differential cohesion – Frame bundles.
## Properties
### Homotopy type
For all $k \in \mathbb{N}$, the homotopy type of the orientation preserving jet group $GL^k_p(n)$ is that of the ordinary orientation-preserving general linear group $GL(n)$, and the canonical projection
$GL^k_+(n) \longrightarrow GL_+(n)$
is, on the level of the underlying topological spaces, a homotopy equivalence, indeed it preserves the maximal compact subgroup, which is the special orthogonal group $SO(n)$ on both sides (recalled e.g. in Dartnell 94, section 1).
### Group homology
The canonical projection $GL^k_+(n) \longrightarrow GL_+(n)$ also induces an isomorphism on group homology with constant integer coefficients
$H_\bullet^{grp}(GL^k_+(n), \mathbb{Z}) \stackrel{\simeq}{\longrightarrow} H_\bullet^{grp}(GL_+(n),\mathbb{Z}) \,.$
## References
Original discussion (in the context of integrability of G-structures) is due to
• Victor Guillemin, section 3 of The integrability problem for $G$-structures, Trans. Amer. Math. Soc. 116 (1965), 544–560. (JSTOR)
Textbook accounts and lecture notes include
• C.L. Terng, Natural vector bundles and natural differential operators, Amer. J. Math. 100 (1978) 775-828.
• Demeter Krupka, Josef Janyška, Lectures on differential invariants, Univerzita JEP, Brno, 1990.
See also
Discussion of the group homology of jet groups includes
• Pablo Dartnell, On the homology of groups of jets, Journal of Pure and Applied Algebra Volume 92, Issue 2, 7 March 1994, Pages 109–121 (publisher)
• Dror Farjoun, Jekel, Suciu, Homology of jet groups (pdf)
Last revised on January 16, 2015 at 20:09:28. See the history of this page for a list of all contributions to it. | 2019-04-23 14:26:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.834918200969696, "perplexity": 2530.2714679865394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578605510.53/warc/CC-MAIN-20190423134850-20190423160850-00455.warc.gz"} |
https://pos.sissa.it/334/044/ | Volume 334 - The 36th Annual International Symposium on Lattice Field Theory (LATTICE2018) - Applications beyond QCD
U(1) vacuum, Chern-Simons diffusion and real-time simulations
A. Florio
Full text: pdf
Published on: 2019 May 29
Abstract
Despite its importance for subjects ranging from cosmology to plasma physics, first principle simulations of the dynamics associated with a $U(1)$ chiral anomaly have been started only recently. In this work, we report on the current status of these investigations. We discuss a possible set-up and highlight some results. We present a determination of the Chern-Simons diffusion rate, which shows some discrepancy with the usual effective description which is anomalous magnetohydrodynamic. We also present some exploratory results on the behaviour of an initial chiral chemical potential.
DOI: https://doi.org/10.22323/1.334.0044
Open Access | 2020-05-25 11:35:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4689007103443146, "perplexity": 1593.077339929747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347388427.15/warc/CC-MAIN-20200525095005-20200525125005-00241.warc.gz"} |
https://math.stackexchange.com/questions/750669/inequality-with-moments-mf3-le-mf2-mf | # Inequality with moments $m(f^3) \le m(f^2) m(f)$
Let $m$ a probability measure, $f$ a positive measurable function (one can assume it is bounded, the existence of the moments is not a problem here).
Is $m(f^3) \le m(f^2) m(f)$?
No. Consider $([0,1],\mathcal{B}([0,1]),\lambda|_{[0,1]})$ and $f(x) := 1+x$. Then $$\int_0^1 (1+x)^n \, dx = \frac{1}{n+1} (1+x)^{n+1} \bigg|_{x=0}^1 = \frac{2^{n+1}-1}{n+1}$$
for any $n \in \mathbb{N}$. Hence,
$$\frac{15}{4} = \int_0^1 (1+x)^3 \, dx > \left( \int_0^1 (1+x)^2 \, dx \right) \cdot \left( \int_0^1 (1+x) \, dx \right) = \frac{7}{3} \cdot \frac{3}{2} = \frac{7}{2}.$$
No. Actually, for every probability measure $m$ and nonnegative function $f$, $$m(f^3)\geqslant m(f^2)\cdot m(f),$$ with equality if and only if $f$ is ($m$-almost surely) constant.
• Because $f^2$ and $f$ are increasing functions of $f$ hence they are positively correlated. I might have already explained this several times on the site. – Did Apr 13 '14 at 18:42 | 2019-10-15 18:59:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9310114979743958, "perplexity": 171.62979400598874}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00172.warc.gz"} |
https://gamedev.stackexchange.com/questions/48608/3d-to-2d-collision | 3d to 2d Collision
I have a 3d model which rotates around my screen, it's always on the Z axis = 0 where everything of my 2d images are. My 2d images fly out of the center and get bigger to give the view as getting closer. I'm trying to detect a collision so I can disable my 2d images to load new ones into the game.
My collision detection isn't working at all and I'm wondering if anyone can shed any light on this.
foreach (coins coin in coins)
{
coin.Update(_graphics.GraphicsDevice);
if ((coin.getCoinXPosition() + scale < getModelXPosition() - scale) &&
(coin.getCoinXPosition() - scale > getModelXPosition() + scale) &&
(coin.getCoinYPosition() + scale < getModelYPosition() - scale) &&
(coin.getCoinYPosition() - scale > getModelYPosition() + scale))
{
coin.coinVisible = false;
score++;
//coins.Remove(coin);
}
}
• "Isn't working at all"? What have you tried? – Anko Feb 5 '13 at 12:56 | 2019-08-24 09:18:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17927636206150055, "perplexity": 2145.1693765222435}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00320.warc.gz"} |
https://www.zbmath.org/?q=an%3A0848.03005 | # zbMATH — the first resource for mathematics
A complete many-valued logic with product-conjunction. (English) Zbl 0848.03005
Studies in the field of fuzzy sets related to triangular norms have shown that the infinitely many-valued logic over the real unit interval with 1 as the only designated truth degree and a product-based conjunction connective has not been discussed much till now, despite some representation theorems of $$t$$-norms which refer to the product as a kind of basic $$t$$-norm.
Choosing the product conjunction, a corresponding implication connective via residuation (i.e., both of them as an adjoint pair), and defining negation formally from implication and the truth degree constant 0 like in intuitionistic logic, the authors constitute the system they investigate.
The completeness proof is given via algebraic studies essentially like the corresponding proof for Łukasiewicz’s infinite-valued logic, but now introducing and investigating product algebras for this product logic instead of the MV-algebras for the Łukasiewicz case.
##### MSC:
03B50 Many-valued logic 03B52 Fuzzy logic; logic of vagueness 03G25 Other algebras related to logic
Full Text:
##### References:
[1] Alsina, C., Trillas, E., Valverde L.: On some logical connectives for fuzzy set theory. J. Math. Anal. Appl.93, 15–26 (1983) · Zbl 0522.03012 [2] Balbes, R., Dwinger, P.: Distributive lattices. Missouri: Univ. Missouri Press 1974 · Zbl 0321.06012 [3] Birkhoff, G.: Lattice theory, vol. 25. New York: 1948, Amer. Math. Soc. Colloquium Publ. · Zbl 0033.10103 [4] Bouchon, B.: Fuzzy inferences and conditional possibility distributions Fuzzy Sets Syst.23, 33–41 (1978) · Zbl 0633.68100 [5] Fuchs, L.: Partially ordered algebraical systems. New York: Pergamon Press 1963 · Zbl 0137.02001 [6] Gödel, K.: Zum intuitionistischen Aussagenkalkül. Anz. Akad. Wissensch. Wien, Math.-naturwissensch. Klasse69, 65–66 (1932). Erg. math. Kolloqu.4, 40 (1933) · JFM 58.1001.03 [7] Gottwald, S.: Mehrwertige Logik. Berlin: Akademie-Verlag 1988 [8] Grätzer, G.: Universal Algebra. Berlin Heidelberg New York: Springer 1979 · Zbl 0412.08001 [9] Gurevich, Y., Kokorin, A.I.: Universal equivalence of ordered Abelian groups (in Russian). Algebra i Logika2.1, 37–39 (1963) [10] Hájek, P.: Fuzzy logic and arithmetical hierarchy. Fuzzy Sets Syst.73, 359–363 (1995) · Zbl 0857.03011 [11] Hájek, P.: Fuzzy logic and arithmetical hierarchy, vol. II. Submitted · Zbl 0857.03011 [12] Hájek, P., Havránek, T., Jiroušek, R.: Uncertain information processing in expert systems. CRC Press 1992 [13] Hájek, P., Valdés, J.J.: Algebraic foundations of uncertainty processing in rule-based expert systems I. Comp. Artif. Intell.9, 325–334 (1990) [14] Höhle, U.: Commutative residuated monoids. In: Höhle, U., Klement, P., (eds) Non-classical logics and their applications to fuzzy subsets (A handbook of the mathematical foundations of the fuzzy set theory). Dordrecht. Kluwer 1995 [15] Ling, C.H.: Representation of associative functions Publ. Math. Debrecen12, 182–212 (1965) [16] Łukasiewicz, J.: Selected works. Amsterdam: North-Holland 1970 · Zbl 0212.00902 [17] Novák, V.: On the syntactico-semantical completeness of first-order fuzzy logic I, II Kybernetika26, 47–26, 134–152 (1990) · Zbl 0705.03009 [18] Paris, J.B.: The uncertain reasoner’s companion – a mathematical perspective. Cambridge: Cambridge University Press 1994 · Zbl 0838.68104 [19] Pavelka, J.: On fuzzy logic I, II, III. Z. Math. Logik Grundl. Math.25, 45–52, 119–134, 447–464 (1979) · Zbl 0435.03020 [20] Rose, A., Rosser, J.B.: Fragments of many-valued statement calculi. Trans. A.M.S.87, 1–53 (1958) · Zbl 0085.24303 [21] Scarpelini, B.: Die Nichtaxiomatisierbarkeit des unendlichwertigen Prädikatenkalküls von łukasiewicz. J. Symb. Log.27, 159–170 (1962) · Zbl 0112.24503 [22] Schweizer, B., Sklar, A.: Associative functions and abstract semi-groups. Publ. Math. Debrecen10, 69–81 (1963) · Zbl 0119.14001 [23] Schweizer, B., Sklar, A.: Probabilistic metric spaces. Amsterdam: North Holland 1983 · Zbl 0546.60010 [24] Takeuti, G., Titani S.: Fuzzy Logic and fuzzy set theory. Anal. Math. Logic32, 1–32 (1992) · Zbl 0786.03039 [25] Zadeh, L.: Fuzzy logic. IEEE Comput.1, 83 (1988) · Zbl 0634.03020
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-08-01 04:39:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.761099636554718, "perplexity": 14442.706528194754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00317.warc.gz"} |
https://msp.org/ant/2014/8-3/p03.xhtml | #### Vol. 8, No. 3, 2014
Recent Issues
The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-7833 (e-only) ISSN: 1937-0652 (print) Author Index To Appear Other MSP Journals
The algebraic dynamics of generic endomorphisms of $\mathbb{P}^n$
### Najmuddin Fakhruddin
Vol. 8 (2014), No. 3, 587–608
##### Abstract
We investigate some general questions in algebraic dynamics in the case of generic endomorphisms of projective spaces over a field of characteristic zero. The main results that we prove are that a generic endomorphism has no nontrivial preperiodic subvarieties, any infinite set of preperiodic points is Zariski-dense and any infinite subset of a single orbit is also Zariski-dense, thereby verifying the dynamical “Manin–Mumford” conjecture of Zhang and the dynamical “Mordell–Lang” conjecture of Denis and Ghioca and Tucker in this case.
##### Keywords
generic endomorphisms, projective space
Primary: 37P55
Secondary: 37F10 | 2022-05-27 21:10:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5220004916191101, "perplexity": 4175.331077136304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00080.warc.gz"} |
https://gmatclub.com/forum/everyone-shakes-hands-with-everyone-else-in-a-room-total-number-of-ha-102740.html?fl=similar | It is currently 17 Oct 2017, 03:21
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Events & Promotions
Events & Promotions in June
Open Detailed Calendar
Everyone shakes hands with everyone else in a room. Total number of ha
Author Message
TAGS:
Hide Tags
Senior Manager
Joined: 13 Aug 2010
Posts: 281
Kudos [?]: 35 [0], given: 1
Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]
Show Tags
12 Oct 2010, 21:27
00:00
Difficulty:
(N/A)
Question Stats:
92% (00:30) correct 8% (00:00) wrong based on 13 sessions
HideShow timer Statistics
Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?
a. 14
b. 12
c. 11
d. 15
e. 16
[Reveal] Spoiler: OA
Kudos [?]: 35 [0], given: 1
Retired Moderator
Joined: 02 Sep 2010
Posts: 793
Kudos [?]: 1184 [0], given: 25
Location: London
Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]
Show Tags
12 Oct 2010, 22:06
prab wrote:
Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?
a.14
b.12
c.11
d.15
e.16
In a room of n people, the number of possible handshakes is C(n,2) or n(n-1)/2
So n(n-1)/2 = 66 OR n(n-1)=132 OR n=12
_________________
Kudos [?]: 1184 [0], given: 25
Senior Manager
Joined: 13 Aug 2010
Posts: 281
Kudos [?]: 35 [0], given: 1
Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]
Show Tags
12 Oct 2010, 22:19
can you please explain why are we using n(n-1)/2, m not able to grab the concept.
Kudos [?]: 35 [0], given: 1
Retired Moderator
Joined: 02 Sep 2010
Posts: 793
Kudos [?]: 1184 [0], given: 25
Location: London
Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]
Show Tags
12 Oct 2010, 22:34
Number of handshakes will be the number of ways to choose 2 people out of n people. For every choice of two people, there is a handshake.
This number is C(n,2) = $$\frac{n!}{(n-2)!2!}$$
_________________
Kudos [?]: 1184 [0], given: 25
Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink] 12 Oct 2010, 22:34
Display posts from previous: Sort by | 2017-10-17 10:21:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1739700436592102, "perplexity": 5664.634332189446}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00026.warc.gz"} |
https://math.stackexchange.com/questions/3297316/sum-of-a-sum-of-a-squared-difference | Sum of a Sum of a Squared Difference
How did the author jump from the second equation to the third equation? I suspect there’s a rule I’m forgetting that allows for this, any help is appreciated.
• I'm more interested in how they jumped from the first line to the second. Where do the 8s and 7s come from, what happened to good old $\mu$? – Vincent Jul 19 at 10:01
Note that
\begin{align} -\left(\sum_{i=1}^{160} (x_i - 8)^2 - \sum_{i=1}^{160} (x_i - 7)^2\right) & = -\left(\sum_{i=1}^{160} (x_i^2 - 16x_i + 64) - (x_i^2 - 14x_i + 49)\right) \\ & = -\left(\sum_{i=1}^{160} (-2x_i + 15)\right) \\ & = 2\sum_{i=1}^{160} x_i - 2400 \tag{1}\label{eq1} \end{align}
As you can see, the $$15$$ is a constant repeating $$160$$ times for a total of $$15 \times 160 = 2400$$. Also, the author used the minus sign in front to remove the first minus sign for the $$2x_i$$, moved the $$2$$ outside the summation and changed the plus to a minus for the sum of $$2400$$.
• Brilliant, thanks for breaking that down for me. – Seraphim Jul 19 at 1:36
• @Seraphim You're welcome. I'm glad I could be of some help. – John Omielan Jul 19 at 1:37
A simpler way, using $$a^2-b^2 = (a-b)(a+b)$$, is
$$\begin{array}\\ \sum_{i=1}^{160} (x_i - 8)^2 - \sum_{i=1}^{160} (x_i - 7)^2) &=\sum_{i=1}^{160} ((x_i - 8)^2 -(x_i - 7)^2)\\ &=\sum_{i=1}^{160} ((x_i - 8) -(x_i - 7))((x_i - 8) +(x_i - 7))\\ &=\sum_{i=1}^{160} (-1)(2x_i -15)\\ &=160\cdot 15-2\sum_{i=1}^{160} x_i\\ \end{array}$$
• Simpler? Debatable. Cooler? Yes. – Hendrix Jul 19 at 2:22
• I think it's simple because you don't have to expand the squares. – marty cohen Jul 19 at 2:30 | 2019-08-24 00:57:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9803899526596069, "perplexity": 783.2505935017631}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00336.warc.gz"} |
https://couryes.com/biological-statistic-analysis-dai-xie-biol220/ | ## 统计代写|生物统计分析代写Biological statistic analysis代考|BIOL220
2022年10月11日
couryes-lab™ 为您的留学生涯保驾护航 在代写生物统计分析Biological statistic analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写生物统计分析Biological statistic analysis代写方面经验极为丰富，各种生物统计分析Biological statistic analysis相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
couryes™为您提供可以保分的包课服务
## 统计代写|生物统计分析代写Biological statistic analysis代考|An Example with Sub-sampling
In biological experimentation, experimental units are frequently sub-sampled, and the data contain several response values for each experimental unit. In our example, we might still randomize the drug treatments on the mice, but take four blood samples instead of one from each mouse and measure them independently. Then, the mice are still the experimental units for the treatment, but the blood samples now provide the response units. The Hasse diagrams in Fig. $4.5$ illustrate this design.
The treatment structure is identical to our previous example, and contains Drug as its only relevant factor. The unit structure now contains a new factor (Sample) with 128 levels, one for each measured enzyme level. It is the response factor that provides the observations. Since each sample belongs to one mouse, and each mouse has several samples, the factor (Sample) is nested in (Mouse). The observations are then partitioned first into 32 groups-one per mouse-and further into 128-one per sample per mouse. For the experiment structure, we randomize Drug on (Mouse), and arrive at the diagram in Fig. 4.5C.
The $F$-test for the drug effect again uses the mean squares for Drug on 3 degrees of freedom. Using our rule, we find that (Mouse) – and not (Sample) – is the experimental unit factor that provides the estimate of the variance for the $F$-denominator on 28 degrees of freedom. As far as this test is concerned, the 128 samples are technical replicates or pseudo-replicates. They do not reflect the biological variation against which we need to test the differences in enzyme levels for the four drugs, since drugs are randomized on mice and not on samples.
## 统计代写|生物统计分析代写Biological statistic analysis代考|The Linear Model
For a completely randomized design with $k$ treatment groups, we can write each datum $y_{i j}$ explicitly as the corresponding treatment group mean and a random deviation from this mean:
$$y_{i j}=\mu_i+e_{i j}=\mu+\alpha_i+e_{i j} .$$
The first model is called a cell means model, while the second, equivalent, model is a parametric model. If the treatments had no effect, then all $\alpha_i-\mu_i-\mu$ are zero and the data are fully described by the grand mean $\mu$ and the residuals $e_{i j}$. Thus, the parameters $\alpha_i$ measure the systematic difference of each treatment from the grand mean and are independent of the experimental units.
It is crucial for an analysis that the linear model fully reflects the structure of the experiment. The Hasse diagrams allow us to derive an appropriate model for any experimental design with comparative ease. For our example, the diagram in Fig.4.4C has three factors: M, Drug, and (Mouse), and these are reflected in the three sets of parameters $\mu, \alpha_i$, and $e_{i j}$. Note that there are four parameters $\alpha_i$ to produce the four group means, but given three and the grand mean $\mu$, the fourth parameter can be calculated; thus, there are four parameters $\alpha_i$, but only three can be independently estimated given $\mu$, as reflected by the three degrees of freedom for Drug. Further, the $e_{i j}$ are 32 random variables, and this is reflected in the fact that (Mouse) is a random factor. Given estimates for $\mu$ and $\alpha_i$, the $e_{i j}$ in each of the four groups must sum to zero and only 28 values are independent.
For the sub-sampling example in Fig.4.5, the linear model is
$$y_{i j k}=\mu+\alpha_i+m_{i j}+e_{i j k} \text {, }$$ where $m_{i j}$ is the average deviation of measurements of mouse $j$ in treatment group $i$ from the treatment group mean, and $e_{i j k}$ are the deviations of individual measurements of a mouse to its average. These terms correspond exactly to $\mathbf{M}$, Drug, (Mouse), and (Sample).
# 生物统计分析代考
## 统计代写|生物统计分析代写生物统计分析代考|线性模型
$$y_{i j}=\mu_i+e_{i j}=\mu+\alpha_i+e_{i j} .$$
$$y_{i j k}=\mu+\alpha_i+m_{i j}+e_{i j k} \text {, }$$，其中$m_{i j}$为处理组$i$中小鼠$j$的测量值与处理组平均值的平均偏差，$e_{i j k}$为单个小鼠测量值与其平均值的偏差。这些术语对应$\mathbf{M}$，药物，(老鼠)和(样本)。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 2023-03-26 06:49:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6892815828323364, "perplexity": 1475.5918556353759}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00531.warc.gz"} |
http://clay6.com/qa/37390/find-the-derivative-of-large-frac-for-some-constant-a | Browse Questions
# Find the derivative of $\large\frac{x^n-a^n}{x-a}$ for some constant $a$
Let $f(x) = \large\frac{x^n-a^n}{x-a}$ | 2017-01-22 12:24:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9981812238693237, "perplexity": 963.245506850778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281424.85/warc/CC-MAIN-20170116095121-00158-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/200481-factoring-stuff-out-limits.html | # Thread: Factoring stuff out of limits
1. ## Factoring stuff out of limits
If I do this:
$\lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\frac{2\alpha}{5}} = \lim_{\alpha\rightarrow 0} \frac{5sin\alpha}{2\alpha} = \lim_{\alpha\rightarrow 0} \frac{5}{2}\cdot \frac{sin\alpha}{\alpha}$
Am I allowed to do this?
$\frac{5}{2} \cdot \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\alpha} = \frac{5}{2} \cdot 1 = \frac{5}{2}$
Or did I do something in the very first steps incorrectly?
2. ## Re: Factoring stuff out of limits
Yes you can factor out constants
3. ## Re: Factoring stuff out of limits
Your derivation is correct. The theorem about the limit of product says that if the two limits on the right exist, then
$\lim\limits_{x \to p} & (f(x)\cdot g(x)) = \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x)$
Since $\lim_{\alpha\to0}\frac{5}{2}$ and $\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}$ exist, your application of the theorem is valid.
4. ## Re: Factoring stuff out of limits
Also I didn't want to create another thread since this is kind of relevant:
If I have:
$\lim_{\alpha \rightarrow 0} \frac{\sin ^{2}\alpha }{\alpha^{2}} = \lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha})^{2}$
Can I do this?
$\lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha} \cdot \frac{\sin\alpha }{\alpha}) = (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) \cdot (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) = 1 \cdot 1 = 1$
But this wouldn't work with a variable, only constants?
5. ## Re: Factoring stuff out of limits
The theorem is applicable to the product of any two functions; one of them does not have to be a constant. | 2017-06-28 11:03:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5413350462913513, "perplexity": 1222.706388579771}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00339.warc.gz"} |
https://zbmath.org/?q=ai%3Ayamada.kosaku+st%3As | # zbMATH — the first resource for mathematics
## Found 2 Documents (Results 1–2)
Berkes, I. (ed.) et al., Limit theorems in probability and statistics. Fourth Hungarian colloquium on limit theorems in probability and statistics, Balatonlelle, Hungary, June 28-July 2, 1999. Vol. II. Budapest: János Bolyai Mathematical Society. 553-573 (2002).
MSC: 60G52 60G51 | 2020-10-27 23:56:58 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8809189796447754, "perplexity": 5392.54935214107}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00388.warc.gz"} |
https://goodboychan.github.io/python/datacamp/time_series_analysis/2020/06/09/01-TSA-Putting-It-All-Together.html | import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
plt.rcParams['figure.figsize'] = (10, 5)
## Cointegration Models
• What is Cointegration?
• Two series, $P_t$ and $Q_t$ can be random walks
• But the linear combination $P_t - c Q_t$ may not be a random walk!
• If that's true
• $P_t - c Q_t$ is forecastable
• $P_t$ and $Q_t$ are said to be cointegrated
• Analogy: Dog on a Leash
• $P_t =$ Owner
• $Q_t =$ Dog
• Both series look like a random walk
• Difference of distance between them, looks mean reverting
• If dog falls too far behind, it gets pulled forward
• If dog gets too far ahead, it gets pulled back
• What types of series are cointegrated?
• Economic substitutes
• Heating Oil and Natural Gas
• Corn and Wheat
• Corn and Sugar
• $\dots$
• Bitcoin and Ethereum
• Two steps to test for Cointegration
• Regress $P_t$ on$Q_t$ and get slope $c$
• Run Augmented Dickey-Fuller test on $P_t - c Q_t$ to test for random walk
• Alternatively, can use coint function in statsmodels that combines both steps
### A Dog on a Leash? (Part 1)
The Heating Oil and Natural Gas prices are pre-loaded in DataFrames HO and NG. First, plot both price series, which look like random walks. Then plot the difference between the two series, which should look more like a mean reverting series (to put the two series in the same units, we multiply the heating oil prices, in $/gallon, by 7.25, which converts it to$/millionBTU, which is the same units as Natural Gas).
The data for continuous futures (each contract has to be spliced together in a continuous series as contracts expire) was obtained from Quandl.
• Preprocess
HO = pd.read_csv('./dataset/CME_HO1.csv', index_col=0)
HO.index = pd.to_datetime(HO.index, format='%m/%d/%Y')
NG.index = pd.to_datetime(NG.index, format='%m/%d/%Y')
HO = HO.sort_index()
NG = NG.sort_index()
HO.head()
Close
Date
1991-01-02 0.7330
1991-01-03 0.7024
1991-01-04 0.6830
1991-01-07 0.7617
1991-01-08 0.7430
NG.head()
Close
Date
1991-01-02 1.832
1991-01-03 1.782
1991-01-04 1.743
1991-01-07 1.785
1991-01-08 1.786
plt.subplot(2, 1, 1)
plt.plot(7.25 * HO, label='Heating Oil');
plt.plot(NG, label='Natural Gas');
plt.legend(loc='best', fontsize='small');
plt.subplot(2, 1, 2)
plt.plot(7.25 * HO - NG, label='Spread');
plt.legend(loc='best', fontsize='small');
plt.axhline(y=0, linestyle='--', color='k');
### A Dog on a Leash? (Part 2)
To verify that Heating Oil and Natural Gas prices are cointegrated, First apply the Dickey-Fuller test separately to show they are random walks. Then apply the test to the difference, which should strongly reject the random walk hypothesis.
from statsmodels.tsa.stattools import adfuller
# Compute the ADF for HO and NG
print("The p-value for the ADF test on HO is ", result_HO[1])
print("The p-value for the ADF test on NG is ", result_NG[1])
The p-value for the ADF test on HO is 0.956710878501786
The p-value for the ADF test on NG is 0.9008747444676729
### Are Bitcoin and Ethereum Cointegrated?
Cointegration involves two steps: regressing one time series on the other to get the cointegration vector, and then perform an ADF test on the residuals of the regression. In the last example, there was no need to perform the first step since we implicitly assumed the cointegration vector was (1,−1). In other words, we took the difference between the two series (after doing a units conversion). Here, you will do both steps.
You will regress the value of one cryptocurrency, bitcoin (BTC), on another cryptocurrency, ethereum (ETH). If we call the regression coefficient b, then the cointegration vector is simply (1,−b). Then perform the ADF test on BTC −b ETH.
• Preprocess
BTC = pd.read_csv('./dataset/BTC.csv', index_col=0)
BTC.index = pd.to_datetime(BTC.index, format='%Y-%m-%d')
ETH.index = pd.to_datetime(ETH.index, format='%Y-%m-%d')
import statsmodels.api as sm
# Regress BTC on ETH
result = sm.OLS(BTC, ETH).fit()
b = result.params[1]
The p-value for the ADF test is 0.02336900232347285
## Case Study: Climate Change
### Is Temperature a Random Walk (with Drift)?
An ARMA model is a simplistic approach to forecasting climate changes, but it illustrates many of the topics covered in this class.
The DataFrame temp_NY contains the average annual temperature in Central Park, NY from 1870-2016 (the data was downloaded from the NOAA here). Plot the data and test whether it follows a random walk (with drift).
• Preprocess
temp_NY = pd.read_csv('./dataset/NOAA_TAVG.csv', index_col=0)
temp_NY.index = pd.to_datetime(temp_NY.index, format='%Y')
TAVG
DATE
1870-01-01 53.8
1871-01-01 51.3
1872-01-01 51.3
1873-01-01 50.9
1874-01-01 51.3
temp_NY.plot();
# Compute and print ADF p-value
print("The p-value for the ADF test is ", result[1])
The p-value for the ADF test is 0.583293898787112
### Getting "Warmed" Up: Look at Autocorrelations
Since the temperature series, temp_NY, is a random walk with drift, take first differences to make it stationary. Then compute the sample ACF and PACF. This will provide some guidance on the order of the model.
from statsmodels.graphics.tsaplots import plot_acf, plot_pacf
# Take first difference of the temperature Series
chg_temp = temp_NY.diff()
chg_temp = chg_temp.dropna()
# Plot the ACF and PACF on the same page
fig, axes = plt.subplots(2, 1)
# Plot the ACF
plot_acf(chg_temp, lags=20, ax=axes[0]);
# Plot the PACF
plot_pacf(chg_temp, lags=20, ax=axes[1]);
plt.tight_layout()
### Which ARMA Model is Best?
Recall from Chapter 3 that the Akaike Information Criterion (AIC) can be used to compare models with different numbers of parameters. It measures goodness-of-fit, but places a penalty on models with more parameters to discourage overfitting. Lower AIC scores are better.
Fit the temperature data to an AR(1), AR(2), and ARMA(1,1) and see which model is the best fit, using the AIC criterion. The AR(2) and ARMA(1,1) models have one more parameter than the AR(1) has.
from statsmodels.tsa.arima_model import ARMA
# Fit the data to an AR(1) model and print AIC:
mod_ar1 = ARMA(chg_temp, order=(1, 0))
res_ar1 = mod_ar1.fit()
print("The AIC for an AR(1) is: ", res_ar1.aic)
# Fit the data to an AR(2) model and print AIC
mod_ar2 = ARMA(chg_temp, order=(2, 0))
res_ar2 = mod_ar2.fit()
print("The AIC for an AR(2) is: ", res_ar2.aic)
# fit the data to an ARMA(1, 1) model and print AIC
mod_arma11 = ARMA(chg_temp, order=(1, 1))
res_arma11 = mod_arma11.fit()
print("The AIC for an ARMA(1,1) is: ", res_arma11.aic)
/home/chanseok/anaconda3/lib/python3.7/site-packages/statsmodels/tsa/base/tsa_model.py:162: ValueWarning: No frequency information was provided, so inferred frequency AS-JAN will be used.
% freq, ValueWarning)
/home/chanseok/anaconda3/lib/python3.7/site-packages/statsmodels/tsa/base/tsa_model.py:162: ValueWarning: No frequency information was provided, so inferred frequency AS-JAN will be used.
% freq, ValueWarning)
/home/chanseok/anaconda3/lib/python3.7/site-packages/statsmodels/tsa/base/tsa_model.py:162: ValueWarning: No frequency information was provided, so inferred frequency AS-JAN will be used.
% freq, ValueWarning)
The AIC for an AR(1) is: 510.5346898313909
The AIC for an AR(2) is: 501.9274123160227
The AIC for an ARMA(1,1) is: 469.0729133043228
### Don't Throw Out That Winter Coat Yet
Finally, you will forecast the temperature over the next 30 years using an ARMA(1,1) model, including confidence bands around that estimate. Keep in mind that the estimate of the drift will have a much bigger impact on long range forecasts than the ARMA parameters.
Earlier, you determined that the temperature data follows a random walk and you looked at first differencing the data. In this exercise, you will use the ARIMA module on the temperature data (before differencing), which is identical to using the ARMA module on changes in temperature, followed by taking cumulative sums of these changes to get the temperature forecast.
from statsmodels.tsa.arima_model import ARIMA
# Forecast temperatures using an ARIMA(1,1,1) model
mod = ARIMA(temp_NY, order=(1,1,1))
res = mod.fit()
# Plot the original series and the forecasted series
res.plot_predict(start='1872-01-01', end='2046-01-01');
plt.savefig('../images/tavg_predict.png')
/home/chanseok/anaconda3/lib/python3.7/site-packages/statsmodels/tsa/base/tsa_model.py:162: ValueWarning: No frequency information was provided, so inferred frequency AS-JAN will be used.
% freq, ValueWarning)
/home/chanseok/anaconda3/lib/python3.7/site-packages/statsmodels/tsa/base/tsa_model.py:162: ValueWarning: No frequency information was provided, so inferred frequency AS-JAN will be used.
% freq, ValueWarning) | 2022-06-29 03:03:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5088691115379333, "perplexity": 6891.394102522832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00441.warc.gz"} |
https://www.physics-in-a-nutshell.com/article/1/overview-and-classification | Physics in a nutshell
$\renewcommand{\D}[2][]{\,\text{d}^{#1} {#2}}$ $\DeclareMathOperator{\Tr}{Tr}$
Stop the war — peace for Ukraine!
# Overview and Classification
## Definition of a Solid
Solid State Physics deals - as the name already implies - with the physical properties of solid materials. A material is referred to as solid if it is composed of a larger number ($\propto 10^{23}$) of smallest constituents (atoms, molecules, …) which are in fixed positions and very tightly packed with a strong mutual attraction.[1] Ideal solids are incompressible just as liquids but in contrast to liquids solids show a strong resistance to shear stress which can be regarded as the main distinctive feature.[2] Due to these properties usually strong forces/torques are required to change the shape of a solid.
## Classification of Solids
The most common classification of solids regards the extend of inner order on a large scale. Solids which exhibit a very regular, periodic structure are referred to as crystals while solids with no such positional large scale order are called amorphous.[3][4][5]
Many solids have their origin in a molten substance that is cooled down. When this cooling process is slow enough, the constituents have enough time to arrange themselves in the positions of lowest energy which usually produces a very regular pattern. Materials that only consist of single atoms or very simple molecules therefore often form crystals. An exception is glass which has a highly unordered (amorphous) structure (due to rather fast cooling processes). Large molecules as for instance in organic materials or plastics usually also form amorphous solids.[6]
## References
[1] W. Demtröder Experimentalphysik 1 Springer 2013 (ch. 6.1) [2] E. N. Economou The Physics of Solids Springer 2010 (ch. 3.1) [3] S. Hunklinger Festkörperphysik De Gruyter 2014 (ch. 1) [4] E. N. Economou The Physics of Solids Springer 2010 (ch. 3.1.1) [5] J. S. Blakemore Solid-State Physics Cambridge University Press 2004 (pp. 4-5) [6] E. N. Economou The Physics of Solids Springer 2010 (3.1.4) [7] R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 5.2.3)
Your browser does not support all features of this website! more | 2023-03-24 18:59:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5946852564811707, "perplexity": 2028.9250596801207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945288.47/warc/CC-MAIN-20230324180032-20230324210032-00007.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/6/lesson/6.4.2/problem/6-161 | ### Home > PC > Chapter 6 > Lesson 6.4.2 > Problem6-161
6-161.
Suppose $y$ varies inversely as $x + 6$ and $y = 1$ when $x = 1$.
1. Express $y$ as a function of $x$, $y = f\left(x\right)$.
1. Write the inverse proportion. Don't forget '$k$'.
2. Substitute a point and solve for '$k$'.
3. Write the function.
2. Find $f\left(−3\right)$, $f\left(0\right)$, $f( \frac { 1 } { 3 } )$, and $f( \frac { 1 } { a } )$.
Use the function found in part (a) above. | 2021-10-25 03:42:22 | {"extraction_info": {"found_math": true, "script_math_tex": 13, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9915282130241394, "perplexity": 1218.1357440441318}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587623.1/warc/CC-MAIN-20211025030510-20211025060510-00619.warc.gz"} |
https://randommathgenerator.com/2021/09/18/the-work-of-alice-chang/ | # The work of Alice Chang
Sun-Yung Alice Chang is one of the pre-eminent scholars of modern geometry, and I wanted to understand the nature of her work. Although I’ve read papers by her before, this article written by her students Matthew Gursky and Yi Wang serves as a solid introduction to her work.
Given below are my notes as I try to understand the article. I first attach a screenshot from the article, and then below that write my notes and questions and such. I often make speculations which are either verified or proved wrong in the best next screenshot. I learned a lot from writing these notes. These can hopefully be useful to anyone working on PDE theory or Riemannian geometry.
Note: This article is unlikely to be helpful to anyone trying to learn about the topic. It is perhaps more useful as a record for myself of how I try and understand Mathematics (or anything at all). One of the reasons that I’ve started blogging less frequently on research papers is that I now try and write detailed notes on every paper before I even think of blogging about it, and that often leaves me too little time to devote to my day job.
• Why is it important to bound the norm of the function by that of its derivative? Why can’t we just measure both separately without having to compare both of them? I don’t know. It seems that we want to make the deduction that $W^{1,p}\subset L^{p*}$, where $1/p+1/p*=1/n$. Why would we want to make such a deduction though? Imagine that you’re trying to find out about an object. You see what it looks like. You hear what it sounds like. You taste what it tastes like. You throw it around and see how it flies in the air. It is the accumulation of all of these infinite observations that tells you all that object is and can be. Seeing if a function belongs to a certain $W^{k,p}$ or an $L^q$ is like that. We want to study the membership (or non-membership) of functions in all of these infinite sets to get a sense of what that function is like.
• Things like Sobolev’s inequality tell you that if a ball flies quickly through the air, it must be aerodynamically shaped, etc. Measuring one property tells you another. It gives you embedding theorems.
• Why do we need to consider weak derivatives? What is wrong with regular derivatives? Well it turns out that the mathematics of proving that smooth or differentiable functions indeed solve PDEs does not exist. Smooth functions do not form a nice enough algebra for which functional analysis can prove the existence of solutions. However, if we enlarge the set a little bit, functional analysis can prove that an element within this set satisfies this PDE. But in order to be able to access all elements of this set, we will need to weaken the question itself; ie weaken the definition of the PDE. It should now incorporate weak derivatives. After we have found the solution, we can prove by some limiting methods that the solution is indeed smooth. Hence, we have found a solution for the original question.
• Was this the only method possible? I am not sure. Maybe a new branch of mathematics would have rendered all of this obsolete. However, if we want to use functional analysis, we have to use this method.
• What is happening here? It turns out that the extremal function is not just one function. It is a family of functions. Moreover, by changing the parameters $a,b$, we can get a function almost completely concentrated at a point. Hence, the Sobolev constant is independent of the domain.
• We choose a cutoff function, and then let it remain fixed. Now we change $a,b$ such that as soon as the domain is inside the domain of the cutoff function, the Sobolev constant is realized.
• Why is the Sobolev constant not attained? I thought that it is attained for extremal functions. Let us walk back a little bit. In the second image, we are only considering compactly supported functions. We are able to reach the same upper bound with the much smaller set of compactly supported functions. However, the extremal function defined above is not compact supported, and it doesn’t become compactly supported for any values of $a,b$ (even though there is pointwise convergence). Hence, the Sobolev constant is not attained.
• Let us walk back a little bit. In the second image, we are only considering compactly supported functions. We are able to reach the same upper bound with the much smaller set of compactly supported functions. However, the extremal function defined above is not compact supported, and it doesn’t become compactly supported for any values of $a,b$ (even though there is pointwise convergence). Hence, the Sobolev constant is not attained.
• The assumption in the formula is that $n\neq p$. However, when $n=p$ we see the inequality shown above. This looks like the $\log$ formulae we generally see in extremal cases.
• Why are we restricting ourselves only to compactly supported functions? Probably because we don’t have the concept of an extremal function here. Hence, we want to better understand this space.
• What does the optimal value of $\beta$ mean? I think they mean an explicit value, which depends only on the dimension and not the domain.
• The surprising thing about this theorem is that they were able to find an extremal function. But I thought extremals don’t exist…? But that’s only for $p\neq n$.
• Why should volume forms have anything to do with the metric? Why not just be completely coordinate dependent? I suppose we could have that. However, most manifolds don’t have a fixed coordinate system. Hence, we have a notion of integral that is invariant with respect to coordinate charts. That is why we include the metric in the definition of the volume form.
• I think that in the theorems given above, we were assuming that the dimension of $\Omega$ was the dimension of the space. Hence, when we have domains of lower dimensions, we need to modify the inequality, which is precisely what we have done on the sphere.
• Why should we care about these inequalities at all? I think we are making an observation about these functions. We are establishing a property of these functions.
• Why do we linearize differential operators? Probably because they are easier to solve, and help us figure out important properties of the actual operator. But the solutions of these linearized operators only give us a hint as to the nature of the actual solutions. A major fact in mathematics is that these linearized operators give us MUCH more information than we’d expect.
• Why must questions about geometry lead to PDE’s? Because the very basic notions of geometry, like curvature, are differential operators of the metric. Hence, differentiation is built into the very fabric of geometry.
• What is a variational approach? It is an approach in which you find the answer by minimizing some functional, because differentiation is easy! Hence, the hard part of this approach is finding the right functional.
• A conformal map of the sphere also changes the metric conformally. Why do we care about this? Because it turns out that these conformal maps generate an infinite family of extremal points for every extremal point that you can construct by yourself.
• What does it mean to say that the conformal group of the sphere is noncompact? Well it is a noncompact set in the space of all maps of the sphere to itself. But how does it matter? I don’t know. But it probably helps prove that some manifolds are not the sphere, or something.
• What does bubbling mean? Is there a bubble being formed somewhere? We are constructing a family of conformal maps on the sphere such that the conformal factors progressively accumulate at the north pole. Well….so what? $J_1$ is still minimized for all of these factors right? Yes, they are. However, this just shows us that $J_1$ doesn’t satisfy the Palais-Smale condition. But how does that matter? First of all, the round metric minimizes $J_1$. Hence, all of these metrics also minimize $J_1$. Who cares about the Palais-Smale condition? I think the point is that extremal points other than these would be difficult to find.
• Does bubbling happen only when we start with the round metric, and then construct the family described above? I think so. What is the point though? Why do we do it? Apparently it is the only thing that “messes up” the geometry of the sphere. We will probably have occasion to talk about it later.
• What does this theorem intend to say? I don’t think it is a generalization of the previous problem, because $\Delta_0 K\neq 0$. However, as long as there are at least two maximum points and all the saddle points are positive, there exists a metric that is conformal to the round metric such that its Gaussian curvature is given by this function.
• Note that we are not just concerned with any metric here. This metric here has to be conformal to the round metric. Hence, this is not like the Yamabe problem, where we start with a round metric and then try to get a conformal metric whose scalar curvature is constant.
• Is a saddle point also a critical point? Yes. Why do we want to find saddle points? Are those the points that satisfy the $\text{Gauss curvature}=K$ condition? I don’t understand why this has to be the case. How does $J_K$ come in? Well, we just insert the desired $K$ somewhere into the formula. Critical points will have exactly that Gaussian curvature. Why a saddle point though? Well maybe all of these critical points have to be saddle points. Critical point does not mean minima
• What is the conformal invariance of this problem? It just means that the pullback of the metric after a conformal change is also a critical point. Is this what conformal invariance means in general? Yes, conformal invariance means that a conformal change in the metric implies no change or very manageable change
• What does conformal invariance have to do with this? I think that through conformal invariance, we construct a saddle point of $J_K$ which is not even bounded in $W^{1,2}$. So what? Why is that a problem, as long as the $J_K$ value is finite? So remember that we are trying to find solutions of the original PDE, and hence the solution must lie inside $W^{1,2}$. Therefore, even though we may find a saddle point, it might not be the solution of the PDE. Therefore, we have not solved the problem we set out to solve. The key step in the proof is showing that at the saddle points where this might happen, $\Delta_0 K<0$. Hence, we don’t care about those solutions
• Why is the spectrum of the laplacian important? In some sense, it is the skeleton of the linear operator. If there is an eigenbasis, then we know exactly how the laplacian acts on the whole space. Note that this condition doesn’t come up naturally in Physics. We have equations like $\Delta \text{potential}=\text{charge}$ and $\Delta Q=\partial Q/\partial t$. The need for eigenvalues of the laplacian probably comes up somewhere else. Does it give us information about the metric or the manifold? It gives us (partial) information about the metric. Note that if the spectra of two linear operators (counter with multiplicities) are the same, then they are the same (provided they are of the same rank). However, this is not the same with metrics with the same spectra of the laplacian. Why did we have to choose the laplacian though? Why not a different linear transformation? That’s a good question. Maybe we can construct a linear transformation which satisfies this property? I’m not sure.
• What is the heat kernel?
• The heat kernel is not the solution to $\text{Heat PDE}=\delta$. It is the solution to $\text{Heat PDE}=0$, with $\text{temperature}=\delta(y)$. Hence, it is an actual solution to the heat equation, and not an attempt to form a general solution to $\text{Heat PDE}=f$
• Why do we need to form an equation over $\Bbb{R}^{d}\times \Bbb{R}^d\times \Bbb{R}_+$, instead of just $\Bbb{R}^d\times \Bbb{R}_+$? I don’t know. I would have just done the latter and said $\text{temperature}=\delta(a)$ at some $a\in \Bbb{R}^d$. I think that we want $y$ to be a variable because we want to be able to assess what happens when $y$ is varied. Hence, we might be able to solve a more general formulation of the problem.
• What does $\delta(x-y)$ mean? I interpret $y$ to be constant. However, if this were a function on $\Bbb{R}^{2d}$, this would be a function that is infinite on the diagonal. Is the integral of the $\delta$ function still $1$? I think so.
• I think that we are trying to form a solution for a generalized boundary condition, and not a generalized heat PDE.
• What does $\delta_x(y)$ mean? Well $\delta(x-y)$ is kind of symmetric. $\delta_x(y)$ makes it clear that $x$ is the independent variable and $y$ is fixed for the purposes of this computation.
• How is the spectrum of $\Delta$ related to the heat PDE? It turns out that the kernel of the heat PDE can be written in terms of the spectra and eigenbasis of the $\Delta$ operator. That is how the spectra is relevant to Physics. A lot of PDE theory is about reducing solving a differential equation problem to a Linear algebra problem. That is exactly what we’re trying to do here as well.
• What is the trace of the heat kernel? Assuming an orthonormal eigenbasis, maybe the trace is $\sum e^{-\lambda_n t}\phi^n(x)\phi_n(x)$? Yes, that would be the only way that this would make sense, as traces of functions don’t really exist.
• What did Brooks-Perry-Yang show? They showed that if all the isospectral metrics lay in the same conformal class, then this set would be compact. Alice showed that we don’t need to assume that all these metrics lie in the same conformal class. I think that the conformal class needs to contain a metric of negative curvature so that bubbling doesn’t occur (which destroys compactness)
• What does modulo conformal group mean? Well the action of the conformal group creates sequences that don’t have a limit inside the space. Thereby compactness is lost. But if all conformally equivalent metrics are considered to be one equivalence class, then the set is compact.
• Why do we define the laplacian to be $-\Delta$? Probably because $\Delta u +\lambda u=0$ implies $\text{laplacian }u=\lambda u$. We want those $\lambda$‘s to be our eigenvalues.
• Why do we have the $\zeta$ function for eigenvalues? Well we have them for $\{1,2,3,\dots\}$. Why not for eigenvalues? But that doesn’t really answer the question as to why this could be important. I think it’s possible that $\zeta(g_1)=\zeta(g_2)$ tells us something about the similarities between the metrics involves, although it is weaker than being isospectral.
• Why the $n/2$ part? A function is defined if we have finite values. An analytic function is defined if it’s equal to its Taylor series. We see that if $Re(s)>n/2$, we have convergence, as $\sum 1/n^k<\infty$ for $k>1$. The complex exponents don’t matter for convergence, which is all we care about right now. Why analytic though? How do we get powers of $s$ in the expansion? I don’t know. Maybe there is an extended proof for this.
• How do meromorphic functions make sense? They’re infinite! Well, they give us information on the nature of the poles, and that can help us determine important properties of the function in a neighborhood of the poles. Meromorphic functions are also important in Physics. Consider the electrostatic potential at the position of a charge. We still want to be able to study this potential, although it clearly has poles.
• The determinant of the laplacian is essentially the product of all eigenvalues. Why would this be an important property? Again, this is a weakening of the notion of isospectral matrices.
• What does a global invariant mean in this context? It doesn’t need to be measured at a point. “Niceness” is a local property of humans, because it needs to be measured within each human. The population of humans is not a local property. It’s a global property. In what sense is it an invariant? I’m not sure
• What does equation (11) say? It says that there’s an upper bound to the fraction on the right, if you start with the round metric on the circle, and that it is $0$ if the conformal metric is a pullback of a conformal map! Hence, not all conformal metrics are pullbacks of conformal maps. Therefore, because we have a $\log$ on the left, we have invariance under action of the conformal group of the determinant as long as we start with the round metric.
• The round metric only maximizes the determinant as long as we’re in the same conformal class.
• How does formula (10) tell you anything about compactness? Maybe the boundedness of the determinant tells us about the compactness of the metrics themselves? That’s possible. But wouldn’t isospectral metrics have the same determinant as well? Yes. I don’t understand how the proof works.
• What are the changes we make to the laplacian to make it conformally invariant? Well, we change the functional space we act on completely, and we also add a scalar curvature term. The good thing about this is what it matches the good old laplacian on Euclidean space, and hence it may be thought of as a “correct generalization” of the laplacian. As long as things “are the same” on Euclidean space, we’re fine.
• Why does the laplacian change the weights of the weight classes that it acts on? Doesn’t a laplacian generally preserve the rank of the differential forms it acts on? Well in this case, taking $\tilde{g}^{ij}$ brings along with it the factor $\frac{1}{t^2}$. This causes the weight to change.
• Aren’t we “cheating” when we act on weight classes instead of considering $\hat{g}=e^{2w}g$? Well, it’s like studying a line bundle on $\Bbb{R}$ instead of studying the space on all smooth functions on $\Bbb{R}$. We are saying that we can study $f:C^\infty (M)\to C^\infty (M)$ by just studying $\text{line bundle}\to \text{line bundle}$ because of the extraordinary homogeneity restrictions on the former map. Is that what is happening here? I think it is, because we are only consider the “extraordinarily homogeneous maps”- the conformally invariant maps.
• Why are Paneitz operators important? Because they are conformally invariant. But so what? There are several conformally invariant operators! Maybe this is a new operator that cannot be generated by previously known conformally invariant operators, that arose in a specific situation. Moreover, it seems to be the natural generalization of the laplacian to four dimensions. Wait. Why only four dimensions? Moreover, isn’t the correct generalization the conformal laplacian? Well the conformal laplacian is a generalization. We can include other terms as well that become $0$ in the Euclidean case, and get a valid generalization (of course, we need to have conformal invariance as well). If $X_{\hat{g}}=e^{nw}X_g$, is $X$ the correct generalization of the laplacian in $n$ dimensions? Shouldn’t it act on weighted classes? Shouldn’t it change the weight by $2$? Moreover, shouldn’t we look for a generalization that works in all dimensions (like the conformal laplacian)?
• Well first we generalize the notion of Gaussian curvature to $Q$, and then say that the Paneitz operator is the correct generalization of the laplacian. So a generalization of the laplacian doesn’t need to be equal to the usual laplcian in Euclidean space. It just needs to satisfy the same kind of PDE as the laplacian. Of course, equation (15) is only true in four dimensions.
• What is the $Q$ curvature? It is a geometric quantity whose integral is a topological invariant in four dimensions. Is this the only geometric quantity with this property? It is possible that it is not. However, it comes up naturally in the generalization of Gauss-Bonnet, which is called Chern-Gauss-Bonnet. But was this really needed? What does this tell us about the manifold? I think it mostly tells us what is not possible, given the topology of the manifold, because we can have really wild metrics on the manifold. Of course all of this is independent of the embedding.
• Both $K$ and $Q$ are completely determined by the metric.
• What is the point of this generalization though? It essentially tells us what a $4$-dimensional manifold given a certain topology can “look like”. It restricts the class of metrics, and hence the ways that this manifold can be embedded inside Euclidean space isometrically. Generally, topology tells us very little about what a manifold can look like. This takes us towards that goal, but eliminating what a manifold cannot look like.
• Why is scale invariance important? Well it’s sort of like conformal invariance, but weaker. Hence, when trying to construct conformally invariant quantities, we may mod out by scaling stuff, and then tweak the right things to get conformal invariance. I think of it as a stepping stone.
• What is happening here? It is interesting that for all kinds of operators $A_g$, $F_A$ can be represented as a sum of three functionals with different coefficients. We sort of have a vector space with a basis thing going on here.
• How do we associate a Lagrangian to a geometric problem? Well the functional that we try to find the extremal point of is generally the lagrangian.
• They’re not saying that $I$ is the Euler-Lagrange equation given on the side. They’re just discussing the nature of critical points and the corresponding Euler-Lagrange equation on the right. Hence, if you want to make $|W_g|^2, Q_g$ or $\Delta_g R_g$ constant, you’ll have to extremize the functionals I, II or III
• What is the theorem that they proved? They proved that the round metric extremizes every normalized functional of the form $F_A$ on $S^4$. This is a pretty strong result. I think that we don’t have the tools to study $S^{2n}$ in general.
• So we are specifically considering $A_g=L_g$
• How is the dimension being $4$ relevant here? I think we are using $Q$ curvature and Paneitz operator here. Note that II has $Q_g=c$ its its Euler-Lagrange equation. It is possible that we extremize all three functionals I, II and III, and we get the Paneitz operator from II.
• I think that $R$ affects the definition of $L$
• How might the proof go? The positive scalar curvature might make $L$ “nice”, leading to a relatively simple fourth order PDE that we’re able to solve. The $\chi\leq 0$ probably makes the PDE even easier to solve. Hence, we’re affecting both the metric and $L$ to make things “easier”. Why do we need to be four dimensions though? Well we have to prove that II can be extremized. It’s probably difficult to do in other dimensions, but gives us a solvable 4th order PDE in four dimensions.
• I find it difficult to understand theorems like these. How are these conditions intuitive? How do you know how many to apply and when to stop? Are these the weakest conditions that can be imposed? Let’s take an analogy. We know that compact sets are closed. OK. But is this the weakest condition possible? No. There are lots of non-compact sets that are closed. However, this is still useful information. There are probably some important metrics that satisfy the conditions given above. And it is good to know that this large class of metrics satisfy this property. We don’t always want to construct the largest class of objects that satisfy a given property. We mostly want to understand individual objects, or classes of objects. And sometimes specifying this awkward class of objects takes awkward wording and seemingly arbitrary conditions, like those stated above. These conditions may in fact encode very easy to see geometric properties that we are missing. Much like the formal definition of compactness hides the intuitive notion of “closed and bounded”. But how can we be sure that we are including all “nice” metrics that we want to study? I think that we are definitely including a large subclass. $R>0$ and $\chi\leq 0$ aren’t that hard to visualize.
• How do we have a parallel for surfaces with high genus and four dimensional manifolds with $\chi\leq 0$? Well a high genus implies $\chi\leq 0$. The only difference is that we can discard the $R>0$ assumption in surfaces. I think the article suggests that we may be able to remove that assumption in four dimensional manifolds as well.
• When you minimize II, you get $Q_g=c$. How do we get $c=0$? It has something to do with $P_g$ having a trivial kernel. Maybe $c$ is the dimension about the kernel, assuming $P_g\geq 0$? I don’t know.
• Note that all our minimization and extremization is occuring in the same confomal class.
• Why couldn’t we do this for the round metric? Well we know that the round metric maximizes everything in its conformal class, and also has constant $Q$ curvature. Alice has studied this for all other conformal classes.
• We no longer start with a functional that is the $\log$ of determinants. We play around with coefficients $\gamma_i$ such that we get an “easy” Euler-Lagrange equation.
• If we find critical points of $F_A$ with positive scalar curvature, then the Ricci curvature will also be positive. So? Well we want to construct metrics with positive Ricci curvature, and this is hard in general. However, playing around with values of $\gamma_2,\gamma_3$ so that we get the “right” second order PDE is easy. And that is what we do here
• What does it mean to say that the conformal change of the curvature tensor is determined by the Schouten tensor? Well if the Schouten tensor is conformally invariant (which it is not), the curvature tensor will also be conformally invariant. However, if the Schouten tensor has a different transformation law, that will determine the transformation law of the curvature tensor.
• Keeping $\text{volume}=1$ is equivalent to normalizing the relevant functionals so as to have scale invariance.
• Why are $\sigma_k$ curvatures important? They are an attempt to generalize the $\int R$ functional. We want to minimize this functional, so as to get $\sigma_k=c$ manifolds.
• Why don’t we want to attain manifolds with $K=c$ or $Q_g=c$? Why $\sigma_k=c$? I think we want to solve both problems. Of course the former is dependent on the embedding, while the latter isn’t.
• Note the following interesting fact: $\int K$ when $n=2$ and $\int Q$ when $n=4$ are topological invariants. However, we can solve PDE’s so as to find $K=c$ or $Q=c$ solutions. For $\int \sigma_1$ when $n=2$ and $\int \sigma_2$ when $n=4$ on the other hand, we have conformal invariance (at least for $n=4$) instead of topological invariance. There are no PDE’s here, and obviously no minimization. But we should still be able to somehow make $\sigma_2=c$ for $n=4$ right, even though the integral doesn’t change? I’m not sure.
• It seems that $Q_g$ and $\sigma_2$ are related. Are $K$ and $R$ related in the same way? I’m not sure. Maybe they are in two dimensions
• I don’t understand why the conformal invariance of $\int \sigma_2$ precludes the possibility that $\sigma_2=c$
• Note that $K,Q$ are found to be constant only in dimensions $2,4$ respectively. On the other hand, $\sigma_1,\sigma_2$ are found to be constant in all dimensions except $2,4$
• We assume local conformal flatness. Do we have the same assumption in the Yamabe problem? No. I suppose it’s needed for higher $k$. Yes, that’s exactly what is stated.
• What is happening with $\sigma_1$? I think that we can find it to be constant in all dimensions. Hence, the problem starts only with $k\geq 2$. Therefore, the analogy with $K,Q$ isn’t exactly clear. $\int R$ may not even be a conformal invariant. In fact it clearly isn’t. We minimize it to find the solution of $R=c$
• So I was right. In $4$ dimensions, we can still have $\sigma_2=c$. Of course, the assumption is that $\sigma_1>0$. I don’t think we had to assume that $K>0$ to prove $Q=c$.
• What does it mean to say that the conditions are conformally invariant? Is $R>0$ a conformally invariant condition? I’m not sure.
• What are quermassintegral inequalities? They’re easier to describe in convex geometry, and compare volumes of the form $\int \sigma_k(L)d\mu$, where $L$ is the second fundamental form. These inequalities are not expected to hold in nonconvex domains. However, Alice and her collaborators could prove such an inequality for $(k+1)$-convex domains.
• But $(k+1)$-convex domains are also convex right? Where does the non-convex part come in? I’m not sure of this point. I think the implication is that the non-convex domains that lie inside this $(k+1)$-convex domain satisfy the inequality. Again, I’m not sure.
• Why is this space called conformally compact Einstein? Well it is conformally compact because it is conformally equivalent to an almost compact manifold, which is the disc. Note that the Poincare disc is not the same as the Euclidean disc. It is the Euclidean disc that is almost compact. The Poincare disc has to have its metric conformally changed to make it the Euclidean disc. It is conformally Einstein because $\text{Ric}(g_+)=-ng_+$. It is actually Einstein, and no conformal factors are needed to make it Einstein
• I think this is where the Einstein part of Poincare-Einstein comes in. $g_+$ doesn’t even have to be in normal form for the Einstein condition to hold.
• I think $g_+$ is always fixed. Moreover, the conformal class $[g]$ on the boundary is also fixed.
• Why do we need a conformal class on the boundary at all? We want to construct quantities that are conformally invariant, so that we may use them in Physical laws.
• I think that the volume is in terms of the metric $g_+$, and not $\rho^2 g_+$
• What does $o(1)$ mean? It refers to a quantity $a(n)$ which satisfies $\lim\limits_{n\to\infty} a(n)=0$. Why do we need to use this notation? Why should we compare functions to powers of $n$? Moreover, why should we care about behaviour as $n\to\infty$? Computational costs depend only on $n\to\infty$. But why? This is because $n$ often does get very large. And we want a good feeling for the computational costs involved. $O(n^3)$ might be manageable, but $O(n^{3.1})$ might become completely impossible. We could also have had things like $O(2^n)$ right? Well those problems are completely intractable, with impossible computational costs. We want to restrain ourselves to polynomials and monomials. I still don’t understand why knowing something is $O(n^3)$ instead of $O(n^4)$ would help. Well, suppose we want a result for $n=100$ in finite time. The former will take $10^6$ seconds, which is approximately 9 days, while the latter will take $900$ days, which is about three years. Hence, if we are already aware of the value of $n$ needed in our computation, and it is high, then knowing the exponent of $n$ is pivotal.
• The coefficients are not conformal invariants, and depend upon curvature terms of the boundary metric. Hence, a defining function has already been defined.
• I’m always surprised by inequalities leading to homeomorphism or diffeomorphism theorems. How do they work? Can’t we make the volume arbitrarily small? Not when we have a blowup near the boundary.
• How is this volume renormalized? We are not talking about the whole volume expansion. We are only talking about the conformally invariant constant term $V$
• What was the Anderson formula? It relates $V$ to $\chi$, although it need to say $V>\frac{1}{2}\frac{4\pi^2}{3}\chi$ (which implies homeomorphism to $B^4$)
• It turns that this is true only when $n=3$. There is a generalization of this to higher dimensions, which involves a conformally invariant integral.
• Have I constructed a fractional power laplacian in my paper? I have proven its existence. However, I haven’t factorized a fractional power laplacian. I have constructed GJMS operators anyway whole symbols would be fractional power laplacians.
• Is the Paneitz operator a GJMS operator for $k=2$? Yes. That is correct.
• How can Fourier analysis be used to construct fractional power laplacians? Well differential operators are analogous to multiplication by a variable. Hence, fractional differential operators are probably multiplication with fractional powers of polynomials, which are not that bad. We can then take the inverse Fourier transform.
• What is an elliptic extension problem? I think it is just an elliptic PDE with a certain boundary condition.
• How can the Cafarelli-Silvestre construction be interpreted as acting on the hyperbolic half space? Maybe we modify the elliptic PDE on Euclidean space to become the scattering operator on hyperbolic PDE? Yes, that is possible.
• How does making the space a hyperbolic half plane give us the extended interval $\gamma\in (0,n/2)$? This is because of the usual ambient restrictions. But what if $n$ is odd, which I think is the assumption here? I’m not sure. However, what I learned is that $w=-\frac{n}{2}+k$ is always negative. As $f\in \xi[w]\implies \tilde{f}=t^{-w/2}f$, this implies that $t$ will be raised to positive powers. | 2021-10-25 02:02:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 197, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8723928928375244, "perplexity": 201.43824185653682}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587608.86/warc/CC-MAIN-20211024235512-20211025025512-00541.warc.gz"} |
http://openstudy.com/updates/55c27e06e4b06d80932e46b2 | ## anonymous one year ago What statement correctly describes the key features of the graph of f(x) = 4(1/2)^x + 1 − 3?
1. anonymous
@phi
2. anonymous
Y-intercept of (0, −1), starts up on the left, gets closer to y = −3 on the right Y-intercept of (0, −1), starts down on the left, gets closer to y = −3 on the right Y-intercept of (0, 1), starts up on the left, gets closer to y = −3 on the right Y-intercept of (0, 1), starts down on the left, gets closer to y = −3 on the right
3. phi
is this $y = \left(\frac{1}{2}\right)^{x+1} -3$?
4. anonymous
you forgot the 4 in front of the 1/2
5. anonymous
other than that, you were right
6. phi
$y = 4\left(\frac{1}{2}\right)^{x+1} -3$
7. anonymous
yes! :)
8. phi
the y-intercept is the y value when x is 0 any idea what y is when x=0?
9. anonymous
1?
10. phi
you erase the x and put 0 in its place in the formula what do you get ?
11. anonymous
ok hold on:)
12. anonymous
0+1=1 right?
13. anonymous
-1! sorry
14. phi
the exponent x+1 becomes 0+1 or just 1 now do (1/2)^ 1 which is just 1/2 then do 1/2 * 4 - 3 or 2 - 3 or -1
15. phi
follow?
16. anonymous
yes okay so now we can elimate some answer choices and yes i do
17. phi
yes, it's A or B
18. phi
what do you get for x= -2 any idea?
19. anonymous
hold on:)
20. anonymous
-5?
21. phi
x+1 becomes -2 + 1 = -1 now do (1/2)^(-1) the negative exponent means you can "flip" the fraction and make the exponent positive so you get 2^1 or just 2 now do 4*2-3 = 8-3 = 5
22. anonymous
Oh! I get it :)
23. phi
at x=2, y=5 at x=0, y= -1 starts high on the left and gets smaller (asymptotes to -3)
24. phi
**x=-2 , y=5
25. anonymous
I see, I see
26. anonymous
so would it be B?
27. phi
yes, I think B. The language is not very clear. (whoever wrote it flunked basic English)
28. anonymous
hahah! I totally agree<3 :D | 2016-10-24 22:11:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8486019372940063, "perplexity": 4354.249422895143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719784.62/warc/CC-MAIN-20161020183839-00007-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://docs.pymc.io/api/distributions/multivariate.html | # Multivariate¶
MvNormal(name, *args, **kwargs) Multivariate normal log-likelihood. MatrixNormal(name, *args, **kwargs) Matrix-valued normal log-likelihood. KroneckerNormal(name, *args, **kwargs) Multivariate normal log-likelihood with Kronecker-structured covariance. MvStudentT(name, *args, **kwargs) Multivariate Student-T log-likelihood. Wishart(name, *args, **kwargs) Wishart log-likelihood. LKJCholeskyCov(name, eta, n, sd_dist[, …]) Wrapper function for covariance matrix with LKJ distributed correlations. LKJCorr(name, *args, **kwargs) The LKJ (Lewandowski, Kurowicka and Joe) log-likelihood. Multinomial(name, *args, **kwargs) Multinomial log-likelihood. Dirichlet(name, *args, **kwargs) Dirichlet log-likelihood. DirichletMultinomial(name, *args, **kwargs) Dirichlet Multinomial log-likelihood.
class pymc3.distributions.multivariate.Dirichlet(name, *args, **kwargs)
Dirichlet log-likelihood.
$f(\mathbf{x}|\mathbf{a}) = \frac{\Gamma(\sum_{i=1}^k a_i)}{\prod_{i=1}^k \Gamma(a_i)} \prod_{i=1}^k x_i^{a_i - 1}$
Support $$x_i \in (0, 1)$$ for $$i \in \{1, \ldots, K\}$$ such that $$\sum x_i = 1$$ Mean $$\dfrac{a_i}{\sum a_i}$$ Variance $$\dfrac{a_i - \sum a_0}{a_0^2 (a_0 + 1)}$$ where $$a_0 = \sum a_i$$
Parameters
a: array
Concentration parameters (a > 0).
logp(value)
Calculate log-probability of Dirichlet distribution at specified value.
Parameters
value: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Dirichlet distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.DirichletMultinomial(name, *args, **kwargs)
Dirichlet Multinomial log-likelihood.
Dirichlet mixture of Multinomials distribution, with a marginalized PMF.
f(x mid n, a) = frac{Gamma(n + 1)Gamma(sum a_k)}
{Gamma(n + sum a_k)}
prod_{k=1}^K frac{Gamma(x_k + a_k)}
{Gamma(x_k + 1)Gamma(a_k)}
Support $$x \in \{0, 1, \ldots, n\}$$ such that $$\sum x_i = n$$ Mean $$n \frac{a_i}{\sum{a_k}}$$
Parameters
nint or array
Total counts in each replicate. If n is an array its shape must be (N,) with N = a.shape[0]
aone- or two-dimensional array
Dirichlet parameter. Elements must be strictly positive. The number of categories is given by the length of the last axis.
shapeinteger tuple
Describes shape of distribution. For example if n=array([5, 10]), and a=array([1, 1, 1]), shape should be (2, 3).
logp(value)
Calculate log-probability of DirichletMultinomial distribution at specified value.
Parameters
value: integer array
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Dirichlet-Multinomial distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.KroneckerNormal(name, *args, **kwargs)
Multivariate normal log-likelihood with Kronecker-structured covariance.
$f(x \mid \mu, K) = \frac{1}{(2\pi |K|)^{1/2}} \exp\left\{ -\frac{1}{2} (x-\mu)^{\prime} K^{-1} (x-\mu) \right\}$
Support $$x \in \mathbb{R}^N$$ Mean $$\mu$$ Variance $$K = \bigotimes K_i$$ + sigma^2 I_N
Parameters
mu: array
Vector of means, just as in MvNormal.
covs: list of arrays
The set of covariance matrices $$[K_1, K_2, ...]$$ to be Kroneckered in the order provided $$\bigotimes K_i$$.
chols: list of arrays
The set of lower cholesky matrices $$[L_1, L_2, ...]$$ such that $$K_i = L_i L_i'$$.
evds: list of tuples
The set of eigenvalue-vector, eigenvector-matrix pairs $$[(v_1, Q_1), (v_2, Q_2), ...]$$ such that $$K_i = Q_i \text{diag}(v_i) Q_i'$$. For example:
v_i, Q_i = tt.nlinalg.eigh(K_i)
sigma: scalar, variable
Standard deviation of the Gaussian white noise.
References
1
Saatchi, Y. (2011). “Scalable inference for structured Gaussian process models”
Examples
Define a multivariate normal variable with a covariance $$K = K_1 \otimes K_2$$
K1 = np.array([[1., 0.5], [0.5, 2]])
K2 = np.array([[1., 0.4, 0.2], [0.4, 2, 0.3], [0.2, 0.3, 1]])
covs = [K1, K2]
N = 6
mu = np.zeros(N)
with pm.Model() as model:
vals = pm.KroneckerNormal('vals', mu=mu, covs=covs, shape=N)
Effeciency gains are made by cholesky decomposing $$K_1$$ and $$K_2$$ individually rather than the larger $$K$$ matrix. Although only two matrices $$K_1$$ and $$K_2$$ are shown here, an arbitrary number of submatrices can be combined in this way. Choleskys and eigendecompositions can be provided instead
chols = [np.linalg.cholesky(Ki) for Ki in covs]
evds = [np.linalg.eigh(Ki) for Ki in covs]
with pm.Model() as model:
vals2 = pm.KroneckerNormal('vals2', mu=mu, chols=chols, shape=N)
# or
vals3 = pm.KroneckerNormal('vals3', mu=mu, evds=evds, shape=N)
neither of which will be converted. Diagonal noise can also be added to the covariance matrix, $$K = K_1 \otimes K_2 + \sigma^2 I_N$$. Despite the noise removing the overall Kronecker structure of the matrix, KroneckerNormal can continue to make efficient calculations by utilizing eigendecompositons of the submatrices behind the scenes [1]. Thus,
sigma = 0.1
with pm.Model() as noise_model:
vals = pm.KroneckerNormal('vals', mu=mu, covs=covs, sigma=sigma, shape=N)
vals2 = pm.KroneckerNormal('vals2', mu=mu, chols=chols, sigma=sigma, shape=N)
vals3 = pm.KroneckerNormal('vals3', mu=mu, evds=evds, sigma=sigma, shape=N)
are identical, with covs and chols each converted to eigendecompositions.
logp(value)
Calculate log-probability of Multivariate Normal distribution with Kronecker-structured covariance at specified value.
Parameters
value: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Multivariate Normal distribution with Kronecker-structured covariance.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
pymc3.distributions.multivariate.LKJCholeskyCov(name, eta, n, sd_dist, compute_corr=False, store_in_trace=True, *args, **kwargs)
Wrapper function for covariance matrix with LKJ distributed correlations.
This defines a distribution over Cholesky decomposed covariance matrices, such that the underlying correlation matrices follow an LKJ distribution [1] and the standard deviations follow an arbitray distribution specified by the user.
Parameters
name: str
The name given to the variable in the model.
eta: float
The shape parameter (eta > 0) of the LKJ distribution. eta = 1 implies a uniform distribution of the correlation matrices; larger values put more weight on matrices with few correlations.
n: int
Dimension of the covariance matrix (n > 1).
sd_dist: pm.Distribution
A distribution for the standard deviations.
compute_corr: bool, default=False
If True, returns three values: the Cholesky decomposition, the correlations and the standard deviations of the covariance matrix. Otherwise, only returns the packed Cholesky decomposition. Defaults to False to ensure backwards compatibility.
store_in_trace: bool, default=True
Whether to store the correlations and standard deviations of the covariance matrix in the posterior trace. If True, they will automatically be named as {name}_corr and {name}_stds respectively. Effective only when compute_corr=True.
Returns
packed_chol: TensorVariable
If compute_corr=False (default). The packed Cholesky covariance decomposition.
chol: TensorVariable
If compute_corr=True. The unpacked Cholesky covariance decomposition.
corr: TensorVariable
If compute_corr=True. The correlations of the covariance matrix.
stds: TensorVariable
If compute_corr=True. The standard deviations of the covariance matrix.
Notes
Since the Cholesky factor is a lower triangular matrix, we use packed storage for the matrix: We store the values of the lower triangular matrix in a one-dimensional array, numbered by row:
[[0 - - -]
[1 2 - -]
[3 4 5 -]
[6 7 8 9]]
The unpacked Cholesky covariance matrix is automatically computed and returned when you specify compute_corr=True in pm.LKJCholeskyCov (see example below). Otherwise, you can use pm.expand_packed_triangular(packed_cov, lower=True) to convert the packed Cholesky matrix to a regular two-dimensional array.
References
1
Lewandowski, D., Kurowicka, D. and Joe, H. (2009). “Generating random correlation matrices based on vines and extended onion method.” Journal of multivariate analysis, 100(9), pp.1989-2001.
2
J. M. isn’t a mathematician (http://math.stackexchange.com/users/498/ j-m-isnt-a-mathematician), Different approaches to evaluate this determinant, URL (version: 2012-04-14): http://math.stackexchange.com/q/130026
Examples
with pm.Model() as model:
# Note that we access the distribution for the standard
# deviations, and do not create a new random variable.
sd_dist = pm.Exponential.dist(1.0)
chol, corr, sigmas = pm.LKJCholeskyCov('chol_cov', eta=4, n=10,
sd_dist=sd_dist, compute_corr=True)
# if you only want the packed Cholesky (default behavior):
# packed_chol = pm.LKJCholeskyCov('chol_cov', eta=4, n=10, sd_dist=sd_dist)
# chol = pm.expand_packed_triangular(10, packed_chol, lower=True)
# Define a new MvNormal with the given covariance
vals = pm.MvNormal('vals', mu=np.zeros(10), chol=chol, shape=10)
# Or transform an uncorrelated normal:
vals_raw = pm.Normal('vals_raw', mu=0, sigma=1, shape=10)
vals = tt.dot(chol, vals_raw)
# Or compute the covariance matrix
cov = tt.dot(chol, chol.T)
Implementation In the unconstrained space all values of the cholesky factor are stored untransformed, except for the diagonal entries, where we use a log-transform to restrict them to positive values.
To correctly compute log-likelihoods for the standard deviations and the correlation matrix seperatly, we need to consider a second transformation: Given a cholesky factorization $$LL^T = \Sigma$$ of a covariance matrix we can recover the standard deviations $$\sigma$$ as the euclidean lengths of the rows of $$L$$, and the cholesky factor of the correlation matrix as $$U = \text{diag}(\sigma)^{-1}L$$. Since each row of $$U$$ has length 1, we do not need to store the diagonal. We define a transformation $$\phi$$ such that $$\phi(L)$$ is the lower triangular matrix containing the standard deviations $$\sigma$$ on the diagonal and the correlation matrix $$U$$ below. In this form we can easily compute the different likelihoods separately, as the likelihood of the correlation matrix only depends on the values below the diagonal, and the likelihood of the standard deviation depends only on the diagonal values.
We still need the determinant of the jacobian of $$\phi^{-1}$$. If we think of $$\phi$$ as an automorphism on $$\mathbb{R}^{\tfrac{n(n+1)}{2}}$$, where we order the dimensions as described in the notes above, the jacobian is a block-diagonal matrix, where each block corresponds to one row of $$U$$. Each block has arrowhead shape, and we can compute the determinant of that as described in [2]. Since the determinant of a block-diagonal matrix is the product of the determinants of the blocks, we get
$\text{det}(J_{\phi^{-1}}(U)) = \left[ \prod_{i=2}^N u_{ii}^{i - 1} L_{ii} \right]^{-1}$
class pymc3.distributions.multivariate.LKJCorr(name, *args, **kwargs)
The LKJ (Lewandowski, Kurowicka and Joe) log-likelihood.
The LKJ distribution is a prior distribution for correlation matrices. If eta = 1 this corresponds to the uniform distribution over correlation matrices. For eta -> oo the LKJ prior approaches the identity matrix.
Support Upper triangular matrix with values in [-1, 1]
Parameters
n: int
Dimension of the covariance matrix (n > 1).
eta: float
The shape parameter (eta > 0) of the LKJ distribution. eta = 1 implies a uniform distribution of the correlation matrices; larger values put more weight on matrices with few correlations.
Notes
This implementation only returns the values of the upper triangular matrix excluding the diagonal. Here is a schematic for n = 5, showing the indexes of the elements:
[[- 0 1 2 3]
[- - 4 5 6]
[- - - 7 8]
[- - - - 9]
[- - - - -]]
References
LKJ2009
Lewandowski, D., Kurowicka, D. and Joe, H. (2009). “Generating random correlation matrices based on vines and extended onion method.” Journal of multivariate analysis, 100(9), pp.1989-2001.
logp(x)
Calculate log-probability of LKJ distribution at specified value.
Parameters
x: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from LKJ distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.MatrixNormal(name, *args, **kwargs)
Matrix-valued normal log-likelihood.
$f(x \mid \mu, U, V) = \frac{1}{(2\pi^{m n} |U|^n |V|^m)^{1/2}} \exp\left\{ -\frac{1}{2} \mathrm{Tr}[ V^{-1} (x-\mu)^{\prime} U^{-1} (x-\mu)] \right\}$
Support $$x \in \mathbb{R}^{m \times n}$$ Mean $$\mu$$ Row Variance $$U$$ Column Variance $$V$$
Parameters
mu: array
Array of means. Must be broadcastable with the random variable X such that the shape of mu + X is (m,n).
rowcov: mxm array
Among-row covariance matrix. Defines variance within columns. Exactly one of rowcov or rowchol is needed.
rowchol: mxm array
Cholesky decomposition of among-row covariance matrix. Exactly one of rowcov or rowchol is needed.
colcov: nxn array
Among-column covariance matrix. If rowcov is the identity matrix, this functions as cov in MvNormal. Exactly one of colcov or colchol is needed.
colchol: nxn array
Cholesky decomposition of among-column covariance matrix. Exactly one of colcov or colchol is needed.
Examples
Define a matrixvariate normal variable for given row and column covariance matrices:
colcov = np.array([[1., 0.5], [0.5, 2]])
rowcov = np.array([[1, 0, 0], [0, 4, 0], [0, 0, 16]])
m = rowcov.shape[0]
n = colcov.shape[0]
mu = np.zeros((m, n))
vals = pm.MatrixNormal('vals', mu=mu, colcov=colcov,
rowcov=rowcov, shape=(m, n))
Above, the ith row in vals has a variance that is scaled by 4^i. Alternatively, row or column cholesky matrices could be substituted for either covariance matrix. The MatrixNormal is quicker way compute MvNormal(mu, np.kron(rowcov, colcov)) that takes advantage of kronecker product properties for inversion. For example, if draws from MvNormal had the same covariance structure, but were scaled by different powers of an unknown constant, both the covariance and scaling could be learned as follows (see the docstring of LKJCholeskyCov for more information about this)
# Setup data
true_colcov = np.array([[1.0, 0.5, 0.1],
[0.5, 1.0, 0.2],
[0.1, 0.2, 1.0]])
m = 3
n = true_colcov.shape[0]
true_scale = 3
true_rowcov = np.diag([true_scale**(2*i) for i in range(m)])
mu = np.zeros((m, n))
true_kron = np.kron(true_rowcov, true_colcov)
data = np.random.multivariate_normal(mu.flatten(), true_kron)
data = data.reshape(m, n)
with pm.Model() as model:
# Setup right cholesky matrix
sd_dist = pm.HalfCauchy.dist(beta=2.5, shape=3)
colchol_packed = pm.LKJCholeskyCov('colcholpacked', n=3, eta=2,
sd_dist=sd_dist)
colchol = pm.expand_packed_triangular(3, colchol_packed)
# Setup left covariance matrix
scale = pm.Lognormal('scale', mu=np.log(true_scale), sigma=0.5)
rowcov = tt.nlinalg.diag([scale**(2*i) for i in range(m)])
vals = pm.MatrixNormal('vals', mu=mu, colchol=colchol, rowcov=rowcov,
observed=data, shape=(m, n))
logp(value)
Calculate log-probability of Matrix-valued Normal distribution at specified value.
Parameters
value: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Matrix-valued Normal distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.Multinomial(name, *args, **kwargs)
Multinomial log-likelihood.
Generalizes binomial distribution, but instead of each trial resulting in “success” or “failure”, each one results in exactly one of some fixed finite number k of possible outcomes over n independent trials. ‘x[i]’ indicates the number of times outcome number i was observed over the n trials.
$f(x \mid n, p) = \frac{n!}{\prod_{i=1}^k x_i!} \prod_{i=1}^k p_i^{x_i}$
Support $$x \in \{0, 1, \ldots, n\}$$ such that $$\sum x_i = n$$ Mean $$n p_i$$ Variance $$n p_i (1 - p_i)$$ Covariance $$-n p_i p_j$$ for $$i \ne j$$
Parameters
n: int or array
Number of trials (n > 0). If n is an array its shape must be (N,) with N = p.shape[0]
p: one- or two-dimensional array
Probability of each one of the different outcomes. Elements must be non-negative and sum to 1 along the last axis. They will be automatically rescaled otherwise.
logp(x)
Calculate log-probability of Multinomial distribution at specified value.
Parameters
x: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Multinomial distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.MvNormal(name, *args, **kwargs)
Multivariate normal log-likelihood.
$f(x \mid \pi, T) = \frac{|T|^{1/2}}{(2\pi)^{k/2}} \exp\left\{ -\frac{1}{2} (x-\mu)^{\prime} T (x-\mu) \right\}$
Support $$x \in \mathbb{R}^k$$ Mean $$\mu$$ Variance $$T^{-1}$$
Parameters
mu: array
Vector of means.
cov: array
Covariance matrix. Exactly one of cov, tau, or chol is needed.
tau: array
Precision matrix. Exactly one of cov, tau, or chol is needed.
chol: array
Cholesky decomposition of covariance matrix. Exactly one of cov, tau, or chol is needed.
lower: bool, default=True
Whether chol is the lower tridiagonal cholesky factor.
Examples
Define a multivariate normal variable for a given covariance matrix:
cov = np.array([[1., 0.5], [0.5, 2]])
mu = np.zeros(2)
vals = pm.MvNormal('vals', mu=mu, cov=cov, shape=(5, 2))
Most of the time it is preferable to specify the cholesky factor of the covariance instead. For example, we could fit a multivariate outcome like this (see the docstring of LKJCholeskyCov for more information about this):
mu = np.zeros(3)
true_cov = np.array([[1.0, 0.5, 0.1],
[0.5, 2.0, 0.2],
[0.1, 0.2, 1.0]])
data = np.random.multivariate_normal(mu, true_cov, 10)
sd_dist = pm.Exponential.dist(1.0, shape=3)
chol, corr, stds = pm.LKJCholeskyCov('chol_cov', n=3, eta=2,
sd_dist=sd_dist, compute_corr=True)
vals = pm.MvNormal('vals', mu=mu, chol=chol, observed=data)
For unobserved values it can be better to use a non-centered parametrization:
sd_dist = pm.Exponential.dist(1.0, shape=3)
chol, _, _ = pm.LKJCholeskyCov('chol_cov', n=3, eta=2,
sd_dist=sd_dist, compute_corr=True)
vals_raw = pm.Normal('vals_raw', mu=0, sigma=1, shape=(5, 3))
vals = pm.Deterministic('vals', tt.dot(chol, vals_raw.T).T)
logp(value)
Calculate log-probability of Multivariate Normal distribution at specified value.
Parameters
value: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Multivariate Normal distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.MvStudentT(name, *args, **kwargs)
Multivariate Student-T log-likelihood.
$f(\mathbf{x}| \nu,\mu,\Sigma) = \frac {\Gamma\left[(\nu+p)/2\right]} {\Gamma(\nu/2)\nu^{p/2}\pi^{p/2} \left|{\Sigma}\right|^{1/2} \left[ 1+\frac{1}{\nu} ({\mathbf x}-{\mu})^T {\Sigma}^{-1}({\mathbf x}-{\mu}) \right]^{-(\nu+p)/2}}$
Support $$x \in \mathbb{R}^p$$ Mean $$\mu$$ if $$\nu > 1$$ else undefined Variance $$\frac{\nu}{\mu-2}\Sigma$$if $$\nu>2$$ else undefined
Parameters
nu: int
Degrees of freedom.
Sigma: matrix
Covariance matrix. Use cov in new code.
mu: array
Vector of means.
cov: matrix
The covariance matrix.
tau: matrix
The precision matrix.
chol: matrix
The cholesky factor of the covariance matrix.
lower: bool, default=True
Whether the cholesky fatcor is given as a lower triangular matrix.
logp(value)
Calculate log-probability of Multivariate Student’s T distribution at specified value.
Parameters
value: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Multivariate Student’s T distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
class pymc3.distributions.multivariate.Wishart(name, *args, **kwargs)
Wishart log-likelihood.
The Wishart distribution is the probability distribution of the maximum-likelihood estimator (MLE) of the precision matrix of a multivariate normal distribution. If V=1, the distribution is identical to the chi-square distribution with nu degrees of freedom.
$f(X \mid nu, T) = \frac{{\mid T \mid}^{nu/2}{\mid X \mid}^{(nu-k-1)/2}}{2^{nu k/2} \Gamma_p(nu/2)} \exp\left\{ -\frac{1}{2} Tr(TX) \right\}$
where $$k$$ is the rank of $$X$$.
Support $$X(p x p)$$ positive definite matrix Mean $$nu V$$ Variance $$nu (v_{ij}^2 + v_{ii} v_{jj})$$
Parameters
nu: int
Degrees of freedom, > 0.
V: array
p x p positive definite matrix.
Notes
This distribution is unusable in a PyMC3 model. You should instead use LKJCholeskyCov or LKJCorr.
logp(X)
Calculate log-probability of Wishart distribution at specified value.
Parameters
X: numeric
Value for which log-probability is calculated.
Returns
TensorVariable
random(point=None, size=None)
Draw random values from Wishart distribution.
Parameters
point: dict, optional
Dict of variable values on which random values are to be conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not specified).
Returns
array
pymc3.distributions.multivariate.WishartBartlett(name, S, nu, is_cholesky=False, return_cholesky=False, testval=None)
Bartlett decomposition of the Wishart distribution. As the Wishart distribution requires the matrix to be symmetric positive semi-definite it is impossible for MCMC to ever propose acceptable matrices.
Instead, we can use the Barlett decomposition which samples a lower diagonal matrix. Specifically:
\begin{align}\begin{aligned}\begin{split}\text{If} L \sim \begin{pmatrix} \sqrt{c_1} & 0 & 0 \\ z_{21} & \sqrt{c_2} & 0 \\ z_{31} & z_{32} & \sqrt{c_3} \end{pmatrix}\end{split}\\\begin{split}\text{with} c_i \sim \chi^2(n-i+1) \text{ and } n_{ij} \sim \mathcal{N}(0, 1), \text{then} \\ L \times A \times A.T \times L.T \sim \text{Wishart}(L \times L.T, \nu)\end{split}\end{aligned}\end{align}
Parameters
S: ndarray
p x p positive definite matrix Or: p x p lower-triangular matrix that is the Cholesky factor of the covariance matrix.
nu: int
Degrees of freedom, > dim(S).
is_cholesky: bool (default=False)
Input matrix S is already Cholesky decomposed as S.T * S
return_cholesky: bool (default=False)
Only return the Cholesky decomposed matrix.
testval: ndarray
p x p positive definite matrix used to initialize
Notes
This is not a standard Distribution class but follows a similar interface. Besides the Wishart distribution, it will add RVs name_c and name_z to your model which make up the matrix.
This distribution is usually a bad idea to use as a prior for multivariate normal. You should instead use LKJCholeskyCov or LKJCorr. | 2021-04-12 11:42:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7702949643135071, "perplexity": 6232.286696553248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00401.warc.gz"} |
https://math.stackexchange.com/questions/1561202/convergence-of-measures-on-mathbbt/1561336 | # Convergence of measures on $\mathbb{T}$
Denote by $M(\mathbb T)$ the set of complex-values measures on the circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$.Prove that $D(T)$, the set of discrete measures on $\mathbb{T}$ is:
• closed in $M(\mathbb{T})$ under norm convergence.
• dense in $M(\mathbb{T})$ under weak convergence.
I had problem formulating both of these questions:
About the norm convergence, suppose we take a sequence $\{\mu_n\}_n\subset D(\mathbb T)$, we can assume it convergences to infintite sum of delta measures (otherwise the sum is finite and the claim is trivial) $$\mu_n\to\mu=\sum_{k=1}^\infty c_k\delta_{x_k},\quad x_k\in\mathbb{T}.$$ How can I formulate that $$\Vert\mu -\mu_n\Vert =\Vert \sum_{k=1}^\infty c_k\delta {x_k} -\sum_{j=1}^n c_j\delta_{x_j}\Vert \to 0?$$
About the second point: Let $\nu\in M(\mathbb{T})$. I thought taking a sequence of discrete measures $$\mu_n=\sum_{j=1}^n d_j\delta_{e^{\frac{2\pi i jt}{n}}}$$ but I don't know how to choose the coefficients exactly such that $\mu_n\to\nu$ for this $\nu\in M(\mathbb T)$?
How do I prove the convergence in norm and find the coefficients ?
For the first part: Let $\mu_i = \sum_j a_j \delta_{x_{ij}}$ Then consider the measurable set $A = \mathbb{T} - \{x_{ij}\}$ By definition of uniform convergence, fo$$B \in A is measurable, then by the definition of uniform convergence there exists N s.t \forall n > N$$|\mu_n(B) - \mu(B)| < \epsilon$$But because B \in A \mu_n(B) = 0, hence \mu is 0 except on a countable subset of \mathbb{T} and the result follows. Note that we didn't need to find the coefficients, as the proof just becomes messier. For the second part: I couldn't if mu = \nu_1 + \nu_2 where \nu_1 is the absolutely continuous part of \mu and \nu_2 is the singular part of \mu with respect to the Lebesgue measure, then we can use the Radon-Nikodym theorem to conclude \nu_1(E) = \int_E f(x)dx for some f measurable and finite a.e. By a famous theorem in measure theory, f can be uniforlmy (!) approximated by finite linear combinations of elementary functions f_k. Given E : measurable$$\nu_1(E) = \int_E f(x)dx = lim_{k \to \infty} \int_E f_k(x)dx For each $f_k$ we can easily find a corresponding discrete measure which coincides with $f_k$ so this part is done (note the choice of $f_k$ can be made independent of the set $E$ Also note that the interchanging of limit and integral is justified as the convergence is uniform: say by Dominated convergence theorem).
We are left with $\nu_2$ which is zero outside a set of Lebesgue measure 0. I'm not sure how to proceed if it's not a discrete set: I will update the post if I find a solution.
• He may have a definition of the type: consider the measures as a bounded linear functional on $C_0(\mathbb{T})$, the compactly supported continuous functions on the torus. They form a normed space with respect to ... norm (I have seen only the uniform norm on measures here, but who knows) – Milen Ivanov Dec 5 '15 at 21:55 | 2020-01-19 13:35:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9695500731468201, "perplexity": 195.07696925840833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00071.warc.gz"} |
https://www.ncatlab.org/nlab/show/classical+limit | Contents
# Contents
## Idea
To a large extent quantum physics is a deformation of classical physics parameterized by Planck's constant $\hbar$. For a given quantum system, the limit $\hbar \to 0$ (in as far as that can be made sense of), is called its classical limit.
One imagines that from a quantum field theory one can derive a classical field theory of which the original quantum field theory may be regarded as a quantization. Whereas quantization may be haphazard, since quantum physics seems to be a more fundamental description of reality that need not be derivable from anything classical, we expect taking the classical limit to be more systematic, since the appearance of a classical world in some regimes should be explicable by fundamental quantum theories. However, it may not be as systematic as some people would like, as in this blog discussion.
What precisely this means and to which degree precisely it makes sense depends on which precise formulation of these concepts one considers. For instance when regarding deformation quantization as definition the quantum theory then taking the classical limit is an obvious straightforward operation. But in more complete contexts, such as when the quantum field theory is realized as an FQFT or AQFT, the passage to the classical limit may be subtle or not make sense at all.
Most prescriptions for taking classical limits are formulated in algebraic settings. To fit for instance a 1-dimensional FQFT into the discussion one is likely to find in the literature, one will have to encode it equivalently in its spectral triple (as described there) and then proceed with that.
## Formalizations
In various formalizations of (aspects of) quantum physics, the classical limit has a corresponding formalization. The following list discusses a few.
### Relations to coherent states
One method for producing classical mechanics from a quantum theory is by looking at coherent states? of the quantum theory. The standard (Glauber) coherent states have a localized probability distribution in classical phase space whose center follows the classical equations of motion when the Hamiltonian is quadratic in positions and momenta.
(For nonquadratic Hamiltonians, this only holds approximately over short times. For example, for the 2-body problem with a $1/r^2$ interaction, Glauber coherent states are not preserved by the dynamics. In this particular case, there are, however, alternative $SO(2,4)$-based coherent states that are preserved by the dynamics, smeared over Kepler-like orbits. The reason is that the Kepler 2-body problem – and its quantum version, the hydrogen atom – are superintegrable systems with the large dynamical symmetry group $SO(2,4)$.)
In general, roughly, coherent states form a nice orbit of unit vectors of a Hilbert space $H$ under a dynamical symmetry group $G$ with a triangular decomposition, such that the linear combinations of coherent states are dense in $H$, and the inner product $\phi^*\psi$ of coherent states $\phi$ and $\psi$ can be calculated explicitly in terms of the highest weight representation theory of $G$. The diagonal of the $N$-th tensor power of $H$ has coherent states $\psi_N$ (labelled by the same classical phase space as the original coherent states, and corresponding to the $N$-fold highest weight) with inner product
$\phi_N^* \psi_N = (\phi^* \psi)^N$
and for $N\to \infty$, one gets a good classical limit. For the Heisenberg group, $\phi^*\psi$ is a $1/\hbar$-th power, and the $N$-th power corresponds to replacing $\hbar$ by $\hbar/N$. Thus one gets the standard classical limit.
### Formulation in geometric quantization
In the context of geometric quantization the classical limit corresponds to taking high tensor powers of the prequantum line bundle with itself. See at Planck's constant the section Planck’s constant - In geometric quantization.
## References
Basic literature on relations between coherent states and the classical limit, based on irreducible unitary representations of Lie groups includes the book
• A. M. Perelomov, Generalized Coherent States and Their Applications, Springer-Verlag, Berlin, 1986.
and the paper
• L. Yaffe, Large $N$ limits as classical mechanics, Rev. Mod. Phys. 54, 407–435 (1982)
Both references assume that the Lie group is finite-dimensional and semisimple. This excludes the Heisenberg group, in terms of which the standard (Glauber) coherent states are usually defined. However, the Heisenberg group has a triangular decomposition, and this suffices to apply Perelomov’s theory in spirit.
The online book
contains a general discussion of the relations between classical mechanics and quantum mechanics, and discusses in Chapter 16 the concept of a triangular decomposition of Lie algebras and a summary of the associated representation theory (though in its present version not the general relation to coherent states).
For other relevant approaches to a rigorous classical limit, see the online sources
• <http://www.projecteuclid.org/Dienst/Repository/1.0/Disseminate/euclid.cmp/1103859040/body/pdf>
• <http://www.univie.ac.at/nuhag-php/bibtex/open_files/si80_SIMON!!!.pdf>
• <http://arxiv.org/abs/quant-ph/9504016>
• <http://arxiv.org/pdf/math-ph/9807027>
Last revised on September 1, 2013 at 14:59:27. See the history of this page for a list of all contributions to it. | 2020-04-09 00:37:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.873066782951355, "perplexity": 376.8904052494269}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371826355.84/warc/CC-MAIN-20200408233313-20200409023813-00208.warc.gz"} |
https://scipost.org/SciPostPhys.9.5.077 | ## Pseudoscalar pair production via off-shell Higgs in composite Higgs models
Diogo Buarque Franzosi, Gabriele Ferretti, Li Huang, Jing Shu
SciPost Phys. 9, 077 (2020) · published 20 November 2020
### Abstract
We propose a new type of search for a pseudoscalar particle $\eta$ pair produced via an off-shell Higgs, $pp\to h^*\to \eta\eta$. The search is motivated by a composite Higgs model in which the $\eta$ is extremely narrow and decays almost exclusively into $Z\gamma$ in the mass range $65$ GeV$\lesssim m_\eta \lesssim 160$ GeV. We devise an analysis strategy to observe the novel $Z\gamma Z\gamma$ channel and estimate potential bounds on the Higgs-$\eta$ coupling. The experimental sensitivity to the signatures depends on the power to identify fake photons and on the ability to predict large photon multiplicities. This search allows us to exclude large values of the compositeness scale $f$, being thus complementary to other typical processes.
### Authors / Affiliations: mappings to Contributors and Organizations
See all Organizations.
Funders for the research work leading to this publication | 2023-03-20 14:59:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31492409110069275, "perplexity": 2692.4400755303504}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00431.warc.gz"} |
http://programming-coursework-help.co.uk/castanet-kamloops-yrc/charge-of-uranium-nucleus-481fd1 | charge of uranium nucleus
By | December 30, 2020
1.6022×10⁻¹⁹C = 1.474×10⁻¹⁷C. The nuclear density for a typical nucleus can be approximately calculated from the size of the nucleus and from its mass. what is the difference between her potential energy at the top of the slope and the bottom of theslope?water, in the given figure find charge on capacitor at steady state, Which is a possible verbal sign of a person contemplating suicide?A. …, 23. The heaviest natural element is Uranium-238, which contains 92 protons for 146 neutrons. Example: For example, natural uranium consists primarily of isotope 238 U (99.28%), therefore the atomic mass of uranium element is close to the atomic mass of 238 U isotope (238.03u). The inner core s electrons move faster, and are drawn in to the heavy nucleus, shielding it … According to Coulomb's law magnitude of the electrical field at distance r to the nucleus is: Which of these elements might you therefore expect to have the smaller critical mass? The fission of one atom of uranium-235 releases 202.5 MeV (3.24×10 J) inside the reactor. The radius of the uranium nucleus is approximately 7.4*10^-15m . The copper atom has a number 29 in Mendeleyev's periodic table of elements. The radius of the uranium nucleus is approximately 7.4*10^-15m . In symmetric fission, the nucleus of uranium (238 U) splits into two nuclei of palladium (119 Pd). Moreover, if we lose 4 nuclear particles of the original 235, there are 231 remaining. A neutral atom contains equal numbers of protons and electrons. Ask Question. A) number of protons : 92 , number of neutrons 143. (The left subscript on the element symbol denotes Z, while the superscript denotes A.) + X What is X? There are 92 protons in the nucleus, so the overall positive charge within the nucleus is overall positive. A nucleus with greater binding energy has a lower total energy, and therefore a lower mass according to Einstein’s mass-energy equivalence relation E = mc 2. (ii) How many neutrons are there in the nucleus? 1.47×10-17 C You Are Correct. (A)What is the total charge of the uranium nucleus? The three nuclides, U-233, U-235, and U-238, are isotopes of uranium because they have the same number of protons per atom and. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. +100. true A nuclear process in which small nuclei are combined into larger nuclei, with an accompanying release of energy is called _____. After decay, the daughter nucleus is Ni, which has Z = 28, and there is an electron, so that the total charge is also 28 + (–1) or 27. Favorite Answer. )II) What is the total change in the climber's potential energy as she climbs down the mountain to fetch her fallenbottle? The height of the well (E1 minus E2) is 6 MeV, so the nucleus has not left the well, but tunnelling and hence nuclear fission are Now we will go above Radon and look at Uranium. They are thus the densest part of an atom. A transverse wave is described by the equationy = Asin 2rent - x2). Its radius of this nucleus will be: As a result, Z is used to calculate a nucleus’s charge. Uranium-235 releases an average of 2.5 neutrons per fission, while plutonium-239 releases an average of 2.7 neutrons per fission. The reason the Adam divided by two to the power of one third. If our uranium nucleus loses 2 protons, there are 90 protons remaining, identifying the element as thorium. the same number of electrons per atom ... An electron has a positive charge and is located in the nucleus. Uranium Nuclei Calculation: The calculation for the two uranium nuclei goes like this: The electrical charge of each nucleus is 92 times the charge of each proton, or 92(1.602x10 -19 coulomb) = 1.47x10 … This is referred to as an unchanged charge distribution. The nuclei of large atoms, such as uranium, with 920protons, can be modeled as spherically symmetric spheres of charge. The unstable nucleus of uranium-236 can be regarded as a uniformly charged sphere of charge Q = + 92 e and radius R = 7.4 × 10 − 15 m. In nuclear fission, this can divide into two smaller nuclei, each with half the charge and half the volume of the original uranium-236 nucleus. Big stuff like Uranium usually undergoes alpha particle decay. An electron has a negative charge and is located outside the nucleus. Thus, we use subtraction to identify the isotope of the thorium atom—in this case, $$\ce{^{231}_{90}Th}$$. Which answer choice best explains why uranium-238 can be used in power plants to generate electricity? (ii) How many neutrons are there in the nucleus? Nuclear Physics Multiple Choice Questions 1. Alpha particles, also called alpha rays or alpha radiation, consist of two protons and two neutrons bound together into a particle identical to a helium-4 nucleus.They are generally produced in the process of alpha decay, but may also be produced in other ways.Alpha particles are named after the first letter in the Greek alphabet, α.The symbol for the alpha particle is α or α 2+. Nos partenaires et nous-mêmes stockerons et/ou utiliserons des informations concernant votre appareil, par l’intermédiaire de cookies et de technologies similaires, afin d’afficher des annonces et des contenus personnalisés, de mesurer les audiences et les contenus, d’obtenir des informations sur les audiences et à des fins de développement de produit. a) What is the electric field this nucleus produces just outside its surface? In symmetric fission, the nucleus of uranium (238 U) splits into two nuclei of palladium (119 Pd). Solution for A uranium nucleus emits an α particle. Question: Nuclear Charge, Electric Field And Electric Force: What Is The Total Charge Of The Uranium Nucleus? An alpha particle is a helium nucleus with 2 protons and 2 neutrons. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. (The Neutral Uranium Atom Has 92 Electrons.) The nucleus (plural, nuclei) is defined as the dense, central part of an atom, consisting of two subatomic particles, namely protons and neutrons. The radius of the uranium nucleus is approximately 7.4 x 10^-15 meters What is the electric field this nucleus produces just outside its surface? 145 C. 235 D. Same number E. Different for different isotopes 2. Yahoo fait partie de Verizon Media. The nucleus of uranium-92 contains 92 protons, each with charge +e=1.6x10^-19C. The radius of the uranium nucleus is approximately 7.4×10−15m . The 92 protons of the uranium nucleus must be conserved, and complementary fission-product pairs—such as krypton-36 with barium-56, rubidium-37 with cesium-55, or strontium-38 with xenon-54—would be possible. The nucleus of a certain type of uranium atom contains 92 protons and 143 neutrons. U-238 has 92 protons and (238-92 = 146 neutrons). withdrawn from activities. Vous pouvez modifier vos choix à tout moment dans vos paramètres de vie privée. For example, in 60 Co decay, total charge is 27 before decay, since cobalt has Z = 27. Click here to get an answer to your question ️ the charge on uranium nucleus is 1.5×10^-17c and charge on alpha particle is 3.2×10^-19c what is electrostati… 94 B. Please make the explanation idiot proof as all the other forums ive looked at i just dont understand However, forces in the nucleus counteract this repulsion and hold the nucleus together. Another 8.8 MeV escapes the reactor as anti-neutrinos. (i) Explain what is meant by isotopes. What Is The Magnitude Of Its Electric Field At A Distance Of 6.39×10-10 M From The Nucleus? writing poems in which death is the main ideaB. 1 3 2 3 ... 19 What is the approximate mass of a nucleus of uranium? For all nuclei naturally present in the Universe, Z varies from 1 to 92, and A from 1 to 238. Calculate the specific charge of nucleus /ckg^-1 . Informations sur votre appareil et sur votre connexion Internet, y compris votre adresse IP, Navigation et recherche lors de l’utilisation des sites Web et applications Verizon Media. Each proton has a positive charge of 1.6*10-19 C. Therefore, the total charge of a copper's nucleus is 1-)What is the electric field this nucleus produces just outside its surface? What must be the charge of X? The word ‘nucleus’ means ‘kernel of a nut’. How many up quarks and down quarks must a proton contain? We can see below that uranium-238 still has 92 protons but it now has 146 neutrons so its nucleon number is now 238. If you divide the charge (Q) of a particle or atom by it’s mass (m) then you will have found the specific charge in coulombs per kilogram (C kg-1). The nucleus, therefore, has a combined mass of 238 nucleons, and a charge … Assuming charge is conserved, the resulting nucleus must be a. thorium b. plutonium c. radium d. curium So that means Uranium is going to lose two protons and 2 neutrons. If the nucleus is unstably large, it will emit a 'package' of two protons and two neutrons called an alpha particle. Express your … Please help with this i can never calculate specific charge. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. If we model the uranium nucleus as a uniformly charged sphere of radius 7.75 fm, what is the volume charge density of the uranium nucleus? The maximum particlevelocity is equal to 3 times the wave velocity ifπΑ21A(1) I just published a long paper on building the elements. A Uranium nucleus (mass 238 u, charge 92e) decays, emitting an alpha particle (mass 4 u, charge 2e) and leaving a Thorium nucleus (mass 234 u, charge 90e). Pour autoriser Verizon Media et nos partenaires à traiter vos données personnelles, sélectionnez 'J'accepte' ou 'Gérer les paramètres' pour obtenir plus d’informations et pour gérer vos choix. These two nuclei will repel each other and should gain a total kinetic energy of c. 200 Mev., as calculated from nuclear radius and charge. The uranium nucleus is spherical with a radius of 7.4 x 10-15 m. Assume that the two palladium nuclei adopt a spherical shape immediately after fission; at this instant, the configuration is as shown in Figure 26.6. Answered: The nuclei of large atoms, such as… | bartleby. Express your answer using two significant figures. We see that charge is conserved in β − decay, since the total charge is Z before and after the decay. The nucleus, that dense central core of the atom, contains both protons and neutrons. Please make the explanation idiot proof as all the other forums ive looked at i just dont understand I.e. So it is fine to ignore charge in balancing nuclear reactions and concentration on balancing mass and atomic numbers only. As a result, Z is used to calculate a nucleus’s charge. (b) An unstable isotope of uranium may split into a caesium nucleus, a rubidium nucleus and four neutrons in the following process. The nucleus, therefore, has a combined mass of 238 nucleons, and a charge … A 10–15 kg B 10 –20 kg C 10 –25 kg D 10 –30 kg 238 Q1) what is the specific charge of a proton at rest. Themass of the climber is 60 kg and her water bottle has a mass of 500 gm .I if the bottle starts from rest, how fast is it travelling by the time it reaches the bottom of the slope? With this information, the positive charge of actinium given in Coulombs is 89 x (+ 1.6 x 10-19). Nuclides of uranium-238 decay over time. The nucleus of uranium-92 contains 92 protons, each with charge +e=1.6x10^-19C. It means that its nucleus has 29 protons. Its nucleus is so full of protons and neutrons that it draws its core electron shells in close. This amount of energy may actually be expected to be available from the difference in packing fraction between uranium and the elements in the middle of the periodic system. The nuclei of large atoms, such as uranium, with 920protons, can be modeled as spherically symmetric spheres of charge. Discuss the scope of Physical Education in career making., what is the order of magnitude of 5 x 10^-7, A mountain climber who is climbing a mountain during winter, by mistake drops her water bottle which thenslides 100 m down the side of a steep icy slo The nucleus of the uranium atom contains 92 protons. The diameter of the nucleus is in the range of 1.7566 fm (1.7566 × 10 −15 m) for hydrogen (the diameter of a single proton) to about 11.7142 fm for uranium. The radius of the uranium nucleus is approximately 7.4 $\times$ 10$^{-15}$ m. (a) What is the electric field this nucleus produces just outside its surface? (iii) Calculate the charge to mass ratio, in C kg –1, for the nucleus. baryon number 4 & charge +2 ? (b) An unstable isotope of uranium may split into a caesium nucleus, a rubidium nucleus and four neutrons in the following process. With uranium, when a neutron hits a U-238 nucleus and sticks to it, a U-239 nucleus is formed, in an excited state (higher up the well). Solution for Electric Fields in an Atom. What is the number of a neutron in the nucleus of uranium: A. Neutron Number and Mass Number of Uranium. The charge of an electron is -1, the charge of a proton is +1. Chemically, uranium is fascinating. Alpha particle. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge.… See also: Mass Number. Prompt neutrons in fission. …, pe to a point which is 10 m lower than the climber's initial position. That corresponds to 19.54 TJ/mol, or 83.14 TJ/kg. 1-)What is the electric field this nucleus produces just outside its surface? What is the charge of… the uranium nucleus? To be useful in an atomic bomb, the uranium in uranium-235 must be enriched to 70% or more. The average number of neutrons emitted per fission (represented by the symbol v̄) varies with the fissioning nucleus. Please help with this i can never calculate specific charge. In it, I explained the structure and radioactivity of Technetium and Radon. Charge of the nucleus=atomic mass (ie no of protons )×(+e) Where +e=1.6×10^-19 C Join Yahoo Answers and get 100 points today. (iii) Calculate the charge to mass ratio, in C kg –1, for the nucleus. The protons of an atom are all crammed together inside the nucleus. So the charge of the nucleus is +92. The total number of neutrons in the nucleus of an atom is called the neutron number of the atom and is given the symbol N.Neutron number plus atomic number equals atomic mass number: N+Z=A.The difference between the neutron number and the atomic number is known as the … Specific charge. Express your … charge of 92 protons = 92 x (+1.6 x 10^-19) = 1.47 x 10^-17 C. 10. You can specify conditions of storing and accessing cookies in your browser, The charge on uranium nucleus is 1.5×10^-17c and charge on alpha particle is 3.2×10^-19c what is electrostatic force between a uranium nucleus and an alpha particle separated by 1.0×10^-13m? ✌. 94 B. express your … Découvrez comment nous utilisons vos informations dans notre Politique relative à la vie privée et notre Politique relative aux cookies. One atom of uranium-235 releases 202.5 MeV ( from a slow neutron ) explained the structure and of..., each with charge +e=1.6x10^-19C of uranium-92 contains 92 protons units in the nucleus Adam divided two... Mass numbers of protons: 92, number of electrons per atom... an electron -1. For more than 99.9 % of an electron is -1, the uranium,! You therefore expect to have the smaller critical mass with charge +e=1.6x10^-19C the maximum particlevelocity is equal 3... 86 electrons. Cu the atomic mass is less than 63 so this must be enriched 70... Structure and radioactivity of Technetium and Radon nucleus with 2 protons and neutrons. Have a positive charge, and like charges repel each other Adam divided by two the... Of an electron has a negative charge and is located in the missing particle an unchanged charge distribution now will. ) calculate the charge to mass ratio, in 60 Co decay, since cobalt Z. Its radius of the uranium nucleus, that dense central core of the electrical field at Distance r to power... Atom has 92 electrons. superscript denotes a. Distance of 6.39×10-10 from! Are combined into larger nuclei, with 92 protons, can be modeled as spherically symmetric spheres of.. Spheres of charge Big stuff like uranium usually forms privée et notre Politique relative cookies! 2Rent - x2 ) at rest is equal to 3 times the charge of the of!... an electron in uranium-235 must be enriched to 70 % or more negatively called! With 920protons, can be modeled as spherically symmetric spheres of charge electrons., identifying the symbol. And atomic numbers only Different isotopes 2 since cobalt has Z = 27 privée et Politique! Go above Radon and look at uranium a nucleus accounts for more than 99.9 % of an atom of! Dans vos paramètres de vie privée element symbol denotes Z, while plutonium-239 releases an average of neutrons... Tout moment dans vos paramètres de vie privée is Z before and the! ’ s mass but is 100,000 times smaller than it in size 10-19 ) nucleus of?. In C kg –1, for the nucleus of uranium: a. we see that charge is in. In stars TJ/mol, or 83.14 TJ/kg and 2 neutrons, it 's uranium, with 92 protons can! 83.14 TJ/kg the power of one atom of uranium-235 releases an average of 2.7 neutrons per (... Charge and is located in the nucleus, shielding it … alpha particle is.! Modifier vos choix à tout moment dans vos paramètres de vie privée i can calculate! Natural element is uranium-238, which is plus 92 times the wave velocity ifπΑ21A ( 1 …... Over one proton is +1.6 x 10-19 ) J ) inside the reactor, has negative... I just published a long paper on building the elements that occur naturally in/on the earth, it emit... To be useful in an atom ’ s mass but is 100,000 times smaller than it in.! That is found in small quantities in uranium ore. Big stuff like uranium forms! Writing poems in which small nuclei are combined into larger nuclei, with 92 protons for 146 neutrons.. As uranium, with 920protons, can be modeled as spherically symmetric spheres charge... Than 99.9 % of an electron is -1, the positive charge within the nucleus is so full of:... Technetium and Radon each proton carries a positive charge within the nucleus, therefore, has a charge! Chemically, uranium is fascinating means relativistic effects come into play that affect the electron orbital energies i the... We lose 4 nuclear particles of the nucleus power of one third u-238 has 92 protons is described by equationy! Many neutrons are neutral is found in small quantities in uranium ore. Big like! The elements, has a positive charge of the nucleus of the in. 'S uranium, with charge of uranium nucleus accompanying release of energy as she climbs the. Charge in balancing nuclear reactions and concentration on balancing mass and atomic only! Is called _____ 227 atomic mass is less than 63 so this must be to... The size of the original 235, there are 90 protons remaining, identifying the element symbol Z... Nucleus ’ s mass but is 100,000 times smaller than it in size 146 neutrons vos informations charge of uranium nucleus! Different isotopes 2 to lose two protons and electrons. charge of uranium nucleus about 2.0 for the nucleus gains some 4.8 (... The structure and radioactivity of Technetium and Radon ' of two protons electrons! And electrons have a negative charge % or more for 146 neutrons ) orbital energies central of. Fused in stars part charge of uranium nucleus an atom ’ s charge negatively chargedparticles called electrons )... Fields in an atom nucleus accounts for more than 99.9 % of an has. Of electrons per atom... an electron has a combined mass of a neutron in nucleus! Nucleussurrounded by one or more a from 1 to 238 10-19 ) enriched to 70 % or more break. Heavy nucleus, showing How it is fine to ignore charge in balancing nuclear reactions concentration. Bomb, the positive charge within the nucleus is approximately 7.4 x 10^-15 meters what the... The positive charge of an atom are all crammed together inside the?! Nucleussurrounded by one or more negatively chargedparticles called electrons. M charge of uranium nucleus the size of uranium. A what is the number of electrons per atom... an electron, the. 2.0 for the spontaneous fission of uranium-238 and 4.0 for that of fermium-257: Chemically, uranium is to! Dominant factor … alpha particle decay be modeled as spherically symmetric spheres of.... The positive charge within the nucleus of uranium is equal to 3 times charge. For that of fermium-257 have no charge, and a from 1 92... Nuclei of palladium ( 119 Pd ) a proton at rest symbol Z... | 2021-09-17 03:35:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5706948041915894, "perplexity": 1522.8352063636967}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00467.warc.gz"} |
https://drc.bmj.com/content/7/1/e000621 | Article Text
Association of socioeconomic status and DKA readmission in adults with type 1 diabetes: analysis of the US National Readmission Database
1. Estelle Everett,
2. Nestoras Nicolas Mathioudakis
1. Endocrinology, Johns Hopkins University, Baltimore, Maryland, USA
1. Correspondence to Dr Nestoras Nicolas Mathioudakis; nmathio1{at}jhmi.edu
## Abstract
Objective To identify patient and hospital predictors of recurrent diabetic ketoacidosis (DKA) admissions in adults in the USA with type 1 diabetes, focusing on socioeconomic indicators.
Research design and methods This cross-sectional study used the National Readmission Database to identify adult patients with type 1 diabetes admitted for DKA between 2010 and 2015. The index DKA admission was defined as the first admission within the calendar year and the primary outcome was recurrent DKA admission(s) within the same calendar year. Multivariable logistic regression analysis was performed using covariates of patient and hospital factors at the index admission to determine the odds of DKA readmission(s).
Results Among 181 284 index DKA admissions, 39 693 (22%) had at least one readmission within the calendar year, of which 33 931 (86%) and 5762 (14%) had 1–3 and ≥4 DKA readmissions, respectively. When compared with the highest income quartile, patients in the first and second income quartiles had 46% (95% CI 30% to 64%) and 34% (95% CI 19% to 51%) higher odds of four or more DKA readmissions, respectively. Medicaid and Medicare insurance were both associated with a 3.3-fold adjusted risk (95% CI 3.0 to 3.7) for ≥4 readmissions compared with private insurance, respectively. Younger age, female sex, and discharge against medical advice were also predictive.
Conclusions Lower socioeconomic status and Medicaid insurance are strong predictors of DKA readmissions in adults with type 1 diabetes in the USA. Further studies are needed to understand the mediators of this association to inform multilevel interventions for this high-risk population.
Significance of the study The association of socioeconomic status (SES) and hospital admission for DKA has been studied in pediatrics with type 1 diabetes, but the data in adults are limited, and studies evaluating recurrent DKA admissions are scarcer. To our knowledge, this is the first study to describe predictors of recurrent DKA admissions in adults with type 1 diabetes on a national level in the USA. We found that those at highest risk of recurrent DKA are young women with low SES who had Medicaid or Medicare insurance. These findings should prompt further studies to explore the mediators of these disparities in patients with type 1 diabetes, as recurrent DKA results in high healthcare utilization and increased risk of long-term complications.
• type 1 diabetes
• diabetes ketoacidosis
• socioeconomic status
This is an open access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited, appropriate credit is given, any changes made indicated, and the use is non-commercial. See: http://creativecommons.org/licenses/by-nc/4.0/.
View Full Text
## Statistics from Altmetric.com
### Significance of this study
• Diabetic ketoacidosis (DKA) is a serious acute complication of type 1 diabetes (T1D) that is associated with poor clinical outcomes.
• Studies evaluating the association of socioeconomic status and recurrent DKA are limited in adult populations of T1D, especially in the USA.
#### What are the new findings?
• In this nationally representative sample of adults with T1D, we found that being from the lowest income quartile was associated with a near 50% increased odds of >4 DKA readmissions in a calendar year.
• Other factors associated with recurrent DKA admission included younger age, female sex, Medicaid, Medicare, no insurance, and discharge against medical advice.
#### How might these results change the focus of research or clinical practice?
• Further research is needed to understand the drivers of these disparities so that we can intervene in this high-risk population.
## Introduction
Diabetic ketoacidosis (DKA) is a serious acute complication of type 1 diabetes and the leading cause of death in children and young adults in this population. A study in the UK found that while a single episode of DKA has been associated with a 5.2% risk of death, that risk increases by 6-fold with five or more admissions.1 In the USA, despite recent advances in medical care of type 1 diabetes (eg, closed-loop insulin pumps, continuous glucose monitoring), the number of hospitalizations for DKA in 2009 had increased by 40% from the previous decade.2 Medical expenditures related to these DKA hospitalizations are estimated to be over US$2.4 billion.3 Inferences regarding the socioeconomic factors linked to readmission for DKA among adults in the USA are primarily derived from limited studies conducted in children with type 1 diabetes4–8 or adults with type 1 diabetes in European countries.2 9 10 Most of the studies conducted in the USA were primarily restricted to single hospitals, hospital systems or registries, limiting their generalizability.2 9 11 The Type 1 Diabetes Exchange Registry evaluated the role of socioeconomic factors on admission for DKA, but this study focused on predictors of at least one admission as the investigators could not reliably measure the number of DKA admissions for each patient, which were largely self-reported.12 Although there have been studies exploring predictors of hospital readmission in adults with type 2 diabetes, these studies evaluated all-cause admissions, not DKA admissions specifically.13 Several measures are considered independent surrogates of socioeconomic status (SES), including education, occupation, income, and area level measures.14 Area level measures are especially good indicators of SES in the USA as social structures often segregate persons by SES, thus making where one resides a very useful indicator of their SES.14 Using the National Readmission Database, which strategically samples US hospitals to allow generalizability to the entire country, we sought to evaluate the role of SES and other patient and hospital factors on DKA readmission in adults with type 1 diabetes in the USA. The primary objective of this study was to determine whether known area level median income, a well-established proxy for SES, is independently associated with readmission for DKA. We hypothesized that lower median income would be associated with higher likelihood of readmission. A secondary objective was to explore whether other patient and hospital factors, some of which are also established social determinants of health, are associated with this outcome. Despite the higher risk of DKA for adults with type 1 diabetes compared with other diabetes types, to our knowledge no previous study has focused specifically on identifying predictors of DKA readmission on a national level in this patient population. ## Research design and methods ### Study population This study used the Healthcare Cost and Utilization Project National Readmission Database (NRD), developed by the US Agency for Healthcare Research and Quality. The NRD is a publicly available deidentified database containing admission data from hospitals in 27 geographically dispersed states, accounting for 58% of the total US resident population and 57% of all US hospitalizations. We selected all patients aged 18 or older with an admission diagnosis of type 1 diabetes who had at least one admission for DKA between January 1, 2010 and December 31, 2015. Patient admissions were identified through International Classification of Disease (ICD)-9 codes reported in the NRD as a primary diagnosis, defined by the NRD as the condition at the time of discharge thought to be chiefly responsible for the admission to the hospital. We used four ICD-9 codes (250.11, 250.13, 250.21, and 250.23), which code for either ketoacidosis or hyperosmolarity in either controlled or uncontrolled type 1 diabetes. Hyperosmolarity ICD-9 codes (250.21 and 250.23) accounted for the primary diagnosis for only 2% of admissions. These admissions were included in the study as approximately 40% of participants that had an admission with a primary admission diagnosis of hyperosmolality also had either a secondary diagnosis of DKA or had other admissions with a primary diagnosis was DKA. There was also a subset of admissions with a primary DKA diagnosis but also had a secondary diagnosis of hyperosmolality. Considering the prevalence of mixed DKA with hyperosmolality in type 1 diabetes15–19 and the lack of convention for coding these patients,15 the very low number of patients (~1%) with only hyperosmolality ICD codes, were considered to have negligible influence on the generalizability of the findings to patients with type 1 diabetes and DKA. ### Predictor variables NRD variables are reported by admission and include hospital-level and patient-level data. Given that we had an interest in characteristics that predict readmission, the covariates analyzed in this study represent characteristics from the index admission, which we defined as the first hospitalization for DKA in the calendar year. Patient-level variables included age, sex, urban–rural classification, whether the patient resided in the same state in which they were hospitalized, median income, admission payer, length of stay (LOS), number of chronic illnesses (defined as conditions lasting 12 months or longer requiring ongoing medical intervention or self-care), and number of diagnoses. Race and ethnicity data are not available in the NRD. Number of diagnoses and chronic diseases were determined by ICD-9 codes on file at the time of discharge. Residential status was reported per the 2010 United States urban–rural classification, which defines an urbanized area as 50 000 residents or more and rural as less than 50 000 residents. Median household income was reported as quartiles by the NRD and was established based on median income data per zip code for the calendar year. Admission payer included Medicare, Medicaid, private payers (private health maintenance organizations or prefeered provider organization), and other (worker’s compensation, title V, Civilian Health and Medical Health and Medical Program of the Department of Veterans Affairs or other government programs). Hospital-level variables included rural–urban classification, size, ownership, and teaching status. Hospital ownership was described as either government, private/non-profit, and private/investment. A teaching hospital was defined as having an American Medical Association–approved residency program, by being a member of the Council of Teaching Hospitals or having a ratio of full-time equivalent interns and residents to beds of 0.25 or higher. The size of a hospital was characterized as small, medium, or large, based on several factors including bed size adjusting for US region, rural or urban location, and teaching status. Other predictor variables included were weekday versus weekend admission, LOS, and disposition location. ### Outcome variable The primary outcome was readmission for DKA occurring after but within the same calendar year as the index admission. Since each admission receives a unique identifier that is consistent at the patient level for a calendar year and there is no patient-specific identifier across calendar years, admissions could only be tracked for a given patient within a calendar year and a given patient could have multiple index admissions in different calendar years. Consequently, the follow-up time interval was variable depending on the time of year of the index admission (ie, only 1 month of follow-up data available for an admission occurring on December 1, 2015 vs 6 months of follow-up data for an admission occurring on July 1, 2015). We evaluated DKA readmissions as a categorical outcome with three levels: no readmission, 1–3 readmissions, and 4 or more readmissions. The rationale for selecting these categories was based on a previous study that demonstrated a substantial increase in mortality for patients with 5 or more admissions (ie, 4 or more readmissions). ### Statistical analysis Descriptive statistics were used to characterize the overall study population and by DKA readmission outcome category at the level of patient admissions. Since all continuous variables were non-normally distributed (as assessed by the Skewness and Kurtosis test), medians and IQRs are reported and counts and frequencies are reported for categorical data. Statistical differences were calculated for each of the readmission groups relative to the no readmission group. For continuous measures, the Wilcoxon rank-sum test was used to analyze differences given the non-parametric distribution. The χ2 test was used to evaluate differences in proportions of categorical variables. Multivariable logistic regression was used to evaluate the association of DKA readmission categories and predictor variables. Given the large sample size of the study, we expected most predictors to be statistically significant even if the magnitude of effect was small; to select the most parsimonious model, we relied on the best subsets achieved by Aikaike information criterion rather than stepwise regression.20 A variance inflation factor <10 was considered to indicate collinearity,21 and no variables were found to be collinear. The patient’s urban–rural status was not included in the model due to a large proportion of missing data. Other than this variable, there was near complete ascertainment in the NRD for most predictors. The following variables had very little (<2%) missing data: payer, median income, patient urban/rural status, and LOS. Considering the nearly complete outcome ascertainment and lack of significant differences in the proportion of missing data by outcome group, we excluded admissions with any missing data from the regression analyses. Two regression models were built with respect to the DKA outcome: 1–3 readmissions (model 1) and ≥4 readmissions (model 2), with the reference group being no readmissions for each model. For model 1, the following variables were included: age, sex, median income, insurance type, whether patient is a resident of the state in which they were hospitalized, number of chronic illnesses, number of diagnoses, hospital bed size, LOS, and disposition. Model 2 additionally contained hospital urban/rural status as a covariate. Discharge-level weights provided by the NRD were used to inflate effect estimates to national estimates, accounting for the stratified single-stage cluster sampling design. All analyses were performed with Stata statistical software, V.14.2 (StataCorp). A two-sided p value of <0.05 was considered statistically significant. ## Results Among 264 948 DKA admissions of patients with type 1 diabetes, 181 284 were identified as index admissions. Among these 181 284 index DKA admissions, 39 693 (22%) had at least one readmission, of which 33 931 (86%) and 5762 (14%) had 1–3 and ≥4 DKA readmissions, respectively, in the calendar year. Overall, in a calendar year, the total number of DKA readmissions ranged from 1 to 19. In the 1–3 readmission group, the median time to the first readmission was 73 days. Approximately 25% had their first readmission within 30 days, 43% within 60 days, and 58% within 90 days. In the ≥4 readmission group, the median time to the first readmission was 33 days, with 47%, 72%, and 86% of the first readmissions occurring within 30, 60, and 90 days, respectively. Baseline characteristics of the entire cohort and readmission subgroups are reported in table 1. Table 1 Baseline characteristics at index admission Overall, this study cohort consisted of admissions of relatively young (median age 35) patients with type 1 diabetes from low-income areas (62% below the 50th percentile for median income) admitted to large (58%), urban (88%), private (84%), non-teaching (88%) hospitals. When comparing the two groups with readmissions with the group without readmissions, we observed several differences. An inverse association was noted with age and women were disproportionally represented in the DKA readmission groups. There was an inverse association with median income and category of DKA readmissions: the proportion of patients in the lowest income quartile increased with increasing DKA admission category. The proportion of admissions covered by private insurance declined with increasing DKA readmission category, with nearly 50% fewer admissions covered by private insurance in those with ≥4 DKA readmissions compared with those with none. Similarly, the proportion of admissions covered by Medicaid increased by 60% in the highest DKA readmission group. There were no notable differences observed in number of chronic illnesses and number of diagnoses, median length of stay, weekday versus weekend admission, hospital size, and hospital ownership. Following discharge, transfer to another hospital/facility or home health services was associated with lower prevalence of DKA readmission. Conversely, leaving the hospital against medical advice (AMA) was associated with higher prevalence of DKA readmission. The proportions of patients admitted to teaching hospitals increased with DKA readmission categories. The results of the fully adjusted regression models are shown in figure 1 and table 2. Table 2 Predictors of DKA readmission in multivariable logistic regression models Figure 1 Risk factors for diabetic ketoacidosis (DKA) readmissions comparing those without DKA readmissions with those with 1–3 and ≥4, respectively. Reference for categorical variables are fourth income quartile for median income, private insurance for payer, large hospital size, routine discharge for type of discharge. AMA, against medical advice; LOS, length of stay. With respect to SES, there was a 19% and 46% increase, respectively, in the adjusted odds of 1–3 and ≥4 DKA readmissions for those in the lowest income quartile compared with those in the highest. With respect to insurance status, the adjusted odds of readmission was 2.08 (95% CI 1.97 to 2.19) and 3.33 (95% CI 2.94 to 4.379) for those with Medicare in the 1–3 and ≥4 DKA readmissions, respectively, compared with those with private insurance. For Medicaid patients, these odds were 1.89 (95% CI 1.82 to 1.97) and 3.33 (95% CI 3.02 to 3.67), respectively. After adjustment, increasing age was associated with lower odds of DKA readmission, with increasing effect size by higher DKA readmission category. Women had a significantly higher odds of DKA readmission, with a 40% increase in odds of ≥4 readmissions. Although length of stay was not associated with DKA readmission, hospital transfer, facility transfer, or discharge with home health services each reduced the odds of 1–3 readmissions by approximately 10%–15%, with increasing effect size in the higher DKA readmission category. Conversely, patients who were discharged against medical advice had approximately a 41% increased odds of 1–3 admissions and 70% increased odds for ≥4 readmissions. Hospital size, hospital teaching status, and hospital ownership were only weakly associated with DKA readmission. ## Discussion In this study, we found that lower SES, using the surrogate of median area income, was strongly associated with DKA readmission. Similarly, insurance status, an established social determinant of health that is closely linked to SES, was the strongest predictor of DKA readmission. To place these findings in context, a patient from an area with the lowest income quartile would be expected to have a 46% increase in the odds of four or more DKA readmissions in a given calendar year, while a patient with Medicare insurance would have over a 3-fold increased odds of this outcome. The inverse association with median income and DKA readmission in this study expands on previous findings from the Type 1 Diabetes Exchange Registry, which showed nearly 2-fold increased odds of at least one DKA admission in those with a household income of US$35 000 or less compared with those with US\$75 000 or more.12 Unlike that study, which was conducted in an ambulatory population with type 1 diabetes where the outcome was a single DKA admission (most of which were patient reported), our study was focused on confirmed DKA readmissions, which represents a higher risk outcome. Additionally, in that study, extremes of income were evaluated, while the present study used a greater number of income categories, which are normalized to the general US population over time and allow inferences to be drawn regarding the magnitude and strength of the association. Our findings also align with those obtained in a UK population which showed a 2-fold risk of DKA admission in patients from more deprived areas.10
With respect to insurance status, only one other study has evaluated the association of DKA readmission and payer in adults with diabetes; however, this study included a mix of adult patients with type 1 and type 2 diabetes limited to the Chicago area and only reported proportions of patients with different payers without any adjusted measure of association. In that study, among patients with 2–3 DKA admissions, 19% were insured by Medicaid and 22% by Medicare, while only 18% had private insurance. In comparison, among the equivalent group in our study (1–3 readmissions), 34% were insured by Medicaid, 17% by Medicare, and 23% by private insurance. Thus, compared with the Chicago study, there was a much larger proportion of patients insured by Medicaid nationally who had a moderate number of DKA readmissions.
Other notable predictors of DKA readmissions identified in this study were age, female sex, and leaving the hospital AMA. Each 1-year increase in age was associated with a 7% decrease in the odds of four or more DKA readmissions. The inverse association of age and DKA admission has been previously described, although the effect sizes were not as marked as in this study,10 24 likely owing to the difference in outcome as we selected for a higher risk group. Female sex has generally been shown to be a risk factor for DKA admissions.10 24 The reasons for this are not readily apparent. Several possible explanations include (1) men are less likely to access medical services, (2) women may intentionally avoid glycemic control for weight gain, and 3) hormonal effects on glucose regulation. Lastly, the caregiver role of many women can sometime inadvertently result in self-neglect.25
In conclusion, lower median income and Medicare and Medicaid insurance were strong predictors of recurrent DKA admissions in adults with type 1 diabetes. Further studies are needed to understand how these socioeconomic factors mediate this association and to identify strategies to overcome disparities in this potentially life-threatening outcome.
1. 1.
2. 2.
3. 3.
4. 4.
5. 5.
6. 6.
7. 7.
8. 8.
9. 9.
10. 10.
11. 11.
12. 12.
13. 13.
14. 14.
15. 15.
16. 16.
17. 17.
18. 18.
19. 19.
20. 20.
21. 21.
22. 22.
23. 23.
24. 24.
25. 25.
26. 26.
27. 27.
28. 28.
29. 29.
View Abstract
## Footnotes
• Contributors EE and NNM designed the study, analyzed and interpreted the data, and wrote the manuscript.
• Funding EE was supported by the Clinical Research and Epidemiology in Diabetes and Endocrinology Training Grant of the NIDDK through grant number T32 DK062707.
• Competing interests None declared.
• Patient consent for publication Not required.
• Provenance and peer review Not commissioned; externally peer reviewed.
• Data availability statement Data are available in a public, open-access repository.
## Request Permissions
If you wish to reuse any or all of this article please use the link below which will take you to the Copyright Clearance Center’s RightsLink service. You will be able to get a quick price and instant permission to reuse the content in many different ways. | 2019-07-16 18:49:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2008250653743744, "perplexity": 6095.146277322698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524685.42/warc/CC-MAIN-20190716180842-20190716202842-00472.warc.gz"} |
https://www.physicsforums.com/threads/proof-using-rolles-theorem.297483/ | # Proof Using Rolle's Theorem
1. Mar 5, 2009
### killpoppop
1. The problem statement, all variables and given/known data
Let f be continuous on [a,b] and differentiable on (a,b) Suppose that:
f2(b) f2(a) = b2 - a2:
Prove (using Rolle's theorem) that:
( exists x belonging to (a, b) ) ( f'(x)f(x) = x )
I just don't know where to start i've done basic proofs with the theorem but only when f(a) = f(b). Any help would be greatly appreciated.
Thanks
2. Mar 5, 2009
### Dick
I think you mean f(b)^2-f(a)^2=b^2-a^2. Define g(x)=f(x)^2-x^2. Can you apply Rolle's theorem to g(x) on the interval [a,b]?
3. Mar 5, 2009
### killpoppop
No the notation i used is what is displayed in the question. I've never seen it before. Is it just the second derivative?
4. Mar 5, 2009
### Dick
No. '^2' just means squared. E.g. f(b)^2=f(b)*f(b). Which is the same as $f^2(b)$. The problem I'm having is that what you wrote looks like f(b)^2*f(a)^2=b^2-a^2. I really think it should be f(b)^2-f(a)^2. With a minus sign between the two squares, not a product of them.
5. Mar 5, 2009
### killpoppop
Yep. So sorry your spot on =] | 2019-03-19 18:49:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7746511101722717, "perplexity": 1481.1477268898266}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202125.41/warc/CC-MAIN-20190319183735-20190319205735-00190.warc.gz"} |
https://testbook.com/objective-questions/mcq-on-pulse-modulation--5eea6a0f39140f30f369e5ce | # Quantizing noise occurs in
1. PCM
2. TDM
3. FDM
4. PPM
Option 1 : PCM
## Detailed Solution
Quantization
• It is the process through which a range of continuous analog values are quantized or rounded off to a single value, thereby forming samples of a discrete digital signal.
• Quantization Error occurs when there is a difference between an input value and it’s quantized value.
• Quantization occurs when an analog signal is converted into it’s digital form, thus it occurs in Pulse Code modulation (PCM).
PCM:
• PCM stands for Pulse Code Modulation.
• It is a technique by which an analog signal gets converted into digital form to have signal transmission through a digital network.
• The major steps involved in PCM are sampling, quantizing, and encoding.
• With PCM, the amplitude of the analog signal is sampled at regular intervals and translated into a binary number.
• The difference between the original signal and the translated digital signal is called the quantizing error.
Some Advantages associated with PCM are:
• Immunity to transmission noise and interference.
• It is possible to regenerate the coded signal along the transmission path.
• The Quantization Noise depends on the number of quantization levels and not on the number of samples produced per second.
• The storage of Coded signals is easy.
• Requires larger Bandwidth.
• Need synchronization
• Not compatible with analog systems.
# The number of bits used in a 4096 level PCM system is:
1. 12
2. 16
3. 20
4. 10
Option 1 : 12
## Detailed Solution
Concept:
The number of levels for an n-bit PCM system is given by:
L = 2n
We can also state that the number of bits for a given quantization level will be:
n = log2 L
Calculation:
The number of levels given = 4096, i.e. L = 4096
The number of bits used will be:
n = log2 (4096)
= log2 (212)
= 12 log2 (2)
n = 1
The bandwidth of PCM is given by:
$$BW=n{f_s}$$
n = number of bits to encode
fs = sampling frequency
# The main advantage of PCM is:
1. Less bandwidth
2. Better performance in presence of a noise
3. Possibility of multiplexing
4. Less power
Option 2 : Better performance in presence of a noise
## Detailed Solution
• PCM (Pulse Code Modulation) is a digital scheme for transmitting analog data.
• The amplitude of an analog signal can take any value over a continuous range, i.e. it can take on infinite values.
• But, digital signal amplitude can take on finite values.
• Analog signals can be converted into digital by sampling and quantizing.
• Encoding is possible in PCM.
• Very high noise immunity, i.e. better performance in the presence of noise.
• Convenient for long-distance communication.
• Good signal to noise ratio.
• The circuitry is complex.
• It requires a large bandwidth.
• Synchronization is required between transmitter and receiver.
# In TDM systems, channel separation is done with the use of
1. AND gates
2. Bandpass filter
3. Dfferentiator circuit
4. Integrator circuit
Option 1 : AND gates
## Detailed Solution
Time Division Multiplexing:
• Time-division multiplexing (TDM) is a method of putting multiple data streams in a single signal by separating the signal into many segments, each having a very short duration. Each individual data stream is reassembled at the receiving end based on the timing.
• In Time Division Multiplexing (TDM), the time frame is divided into slots. This technique is used to transmit a signal over a single communication channel, by allotting one slot for each message.
• For separating channels in TDM, it is necessary to use Time slots. For this purpose AND gates are used.
• Time Division Multiplexing (TDM) can be classified into Synchronous TDM and Asynchronous TDM.
• Synchronous TDM
• In Synchronous TDM, the input is connected to a frame. If there are ‘n’ number of connections, then the frame is divided into ‘n’ time slots. One slot is allocated for each input line.
• In this technique, the sampling rate is common for all signals and hence the same clock input is given. The MUX allocates the same slot to each device at all times.
• Asynchronous TDM
• In Asynchronous TDM, the sampling rate is different for each of the signals and a common clock is not required.
• If the allotted device for a time slot transmits nothing and sits idle, then that slot can be allotted to another device, unlike synchronous
• This type of TDM is used in Asynchronous transfer mode networks.
# The bandwidth required in DPCM is less than that of PCM because
1. the number of bits per code is reduced resulting in a reduced bit rate
2. the difference signal is larger in amplitude than actual signal
3. more quantization levels are needed
4. the successive samples of signal often differ in amplitude
Option 1 : the number of bits per code is reduced resulting in a reduced bit rate
## Detailed Solution
PCM:
• PCM stands for Pulse Code Modulation.
• With PCM, the amplitude of the analog signal is sampled at regular intervals and translated into a binary number.
• The difference between the original signal and the translated digital signal is called the quantizing error.
DPCM:
• DPCM stands for Differential Pulse Code modulation.
• It is a signal encoder that uses PCM but adds some functionalities based on the prediction of the samples of the signal.
• The input can be an analog signal or a digital signal.
• The number of encoded bits is smaller than the PCM system. This results in fewer bandwidth requirements by the DPCM system.
A comparison of different modulation schemes is as shown in the table below:
Parameter PCM Delta Modulation (DM) DPCM Number of bits It can use 4, 8 or 16 bits per sample It uses only one bit for one sample Bits can be more than one but are less than PCM Level/Step size Step size is fixed Step size is fixed and cannot be varied Fixed number of levels are used. Quantization error or Distortion Quantization error depends on the number of levels used Slope overload distortion and granular noise is present Slope overload distortion and quantization noise is present Bandwidth of the transmission channel Highest bandwidth is required since the number of bits are high Lowest bandwidth is required The bandwidth required is lower than PCM Signal to Noise ratio Good Poor Fair Area of Application Audio and Video Telephony Speech and images Speech and video
# To avoid slope overload error in delta modulation, the maximum amplitude of the input signal is
1. A ≤ 2πfm
2. A ≤ som 2πfm
3. $$A \le \frac{{2\pi {f_m}}}{{{\rm{\Delta }}{f_s}}}$$
4. $$A \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m}}}$$
Option 4 : $$A \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m}}}$$
## Detailed Solution
Concept:
Two types of distortions (Errors) occurs in Delta modulation system i.e.
2) Granular Error
This is explained with the help of the given figure:
To avoid the slope overload error, the optimum or desired condition is:
$$\frac{{\rm{\Delta }}}{{{T_s}}} = \frac{{d\;m\left( t \right)}}{{dt}}$$
i.e. the step rise of the quantized output must follow the input
$$\frac{{\rm{\Delta }}}{{{T_s}}} < \frac{{d\;m\left( t \right)}}{{dt}}$$
To prevent/avoid slope overload error, the condition that shall be satisfied is:
$$\frac{{d\;m\left( t \right)}}{{dt}} \le \frac{{\rm{\Delta }}}{{{T_s}}}$$
m(t) is a sinusoidal waveform given as:
m(t) = Am sin ωm t
The condition to avoid slope overload, therefore, becomes:
$$\frac{d}{{dt}}({A_m}\sin {\omega _m}t) \le \frac{{\rm{\Delta }}}{{{T_s}}}$$
$${A_m}\cos {\omega _m}t\;\left( {{\omega _m}} \right) \le \frac{{\rm{\Delta }}}{{{T_s}}}$$
Am cos (2π (mt)).2πfm ≤ Δ⋅fs
$${A_m} \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m} \cdot \cos \left( {2\pi {f_m}f} \right)}}$$
For maximum Amplitude, the above condition becomes:
$${A_m} \le \frac{{{\rm{\Delta }} \cdot {f_s}}}{{2\pi {f_m}}}$$
# The bandwidth requirement of a telephone channel is
1. 3 kHz
2. 5 kHz
3. 10 kHz
4. 15 kHz
Option 1 : 3 kHz
## Detailed Solution
The bandwidth requirement of a telephone channel is approximately 3 kHz.
• The internationally accepted frequency range of standard telephone channel is 300 – 3400 Hz.
• The range of audible human frequency varies from 20 Hz to about 20 kHz which is much higher than the bandwidth of the telephone channel.
• Clearly, the telephone channel's bandwidth is much lower than the human ear's audible bandwidth. The main reasons for it are:
• Acceptable Voice Quality when reproduced at the receiving end.
• Low Cost.
• Savings in Bandwidth.
• Hence, if we provide about 4 kHz of BW (including a guard band) per telephone channel, it would be sufficient for voice quality.
# In PCM, if the number of quantization levels is increased from 4 to 64, then the bandwidth requirement will approximately be:
1. 4 times
2. 16 times
3. 8 times
4. 3 times
Option 4 : 3 times
## Detailed Solution
Concept:
The number of levels for an n-bit PCM system is given by:
L = 2n
We can also state that the number of bits for a given quantization level will be:
n = log2 L
Also, the bandwidth of PCM is given by:
$$BW=n{f_s}$$
n = number of bits to encode
fs = sampling frequency
Calculation:
For L = 4 quantization levels, the number of bits n = log2 4 = 2 bits. The bandwidth is, therefore:
B.W. = 2 fs
Similarly, For L = 64 quantization levels, the number of bits n = log2 64 = 6 bits. The bandwidth is, therefore:
B.W. = 6 fs
Clearly, the Bandwidth is increased by 3 times.
# Limiter circuit is not needed in the following detector:
1. Foster - Seeley discriminator
2. Balanced slope
3. Ratio detector
4. None
Option 3 : Ratio detector
## Detailed Solution
Concept:
• Frequency Demodulators are often sensitive to amplitude variations. Therefore a limiter amplifier stage must be used before the detector, to remove amplitude variations in the signal which would be detected as noise.
• The limiter acts as a Class-A amplifier at lower amplitudes; at higher amplitudes, it becomes a saturated amplifier that clips off the peaks and limits the amplitude.
• The FM ratio detector was the more common because it offered a better level of amplitude modulation rejection of amplitude modulation.
• This enabled the circuit to provide a greater level of noise immunity as most noise is amplitude noise.
It also enables the FM detector to operate more effectively even with lower levels of limiting in the preceding IF stages of the receiver.
# The main advantage of frequency modulation over amplitude modulation is
1. That there will be no distortion
2. That the complete information is contained in the side bands
3. That is uses a wider band of frequencies
4. The elimination of noises
Option 4 : The elimination of noises
## Detailed Solution
• The Noise affects the amplitude of the modulated signal
• In the Amplitude Modulated system, the message is contained in the amplitude modulations of the carrier; hence the introduction of noise distorts the amplitude and hence the message contained in the signal
• In a frequency modulated system the message is contained in the frequency variations of the carrier, therefore, the introduction of noise does not affect the message contained in the message signal
• Since the amplitude of FM remains constant, the noise can be eliminated using an amplitude limiter at the demodulator.
Features: AM FM Noise immunity In AM, the message is stored in the form of variation in amplitude. Noise affects the amplitude of signal most so AM is less noise immune. In FM, the message is stored in the form of variation in frequency so it has better noise immunity. Bandwidth B.W. required in AM is = 2fm. Hence, less bandwidth is required in case of AM. B.W. required in FM is = 2(β+1)fm. Hence, more bandwidth is required in the case of FM. Transmitted power Power transmitted in AM is given by: $${P_T} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)$$ As the modulation index ‘μ’ increases power in AM increases. In FM, power transmitted is always equal to the total power of carrier before modulation. Hence, FM requires less power than AM.
# The plot of modulation index versus carrier amplitude for an amplitude modulated system yields
1. Horizontal line
2. Vertical line
3. Parabola
4. Hyperbola
Option 4 : Hyperbola
## Detailed Solution
Concept:
For an amplitude modulated system, the modulation index is defined as:
$$\mu=\frac{A_m}{A_c}$$
Am = Message signal amplitude
Ac = Carrier Amplitude
Observation: Since the modulation index is related to the carrier amplitude through an inverse relationship, we conclude that the plot of modulation index versus carrier amplitude will result in a hyperbola.
A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.
Amplitude Modulation: The amplitude of the carrier is varied according to the amplitude of the message signal.
Frequency Modulation: The frequency of the carrier is varied according to the amplitude of the message signal.
Phase Modulation: The Phase of the carrier is varied according to the amplitude of the message signal.
# In TV, an electrical disturbance (noise) affects
1. neither the video nor the audio signals
2. only the audio signals
3. both the video and audio signals
4. only the video signals
Option 4 : only the video signals
## Detailed Solution
In TV transmission:
• A video is Vestigial sideband modulated, which is a type of amplitude modulated waveform
• The Video signals are thus encoded in amplitude variations of the carriers
• The Audio signal is encoded in FM waveform
• Thus, the audio signals are encoded as frequency variations of the carrier
Noise is the signal that affects amplitude majorly. Thus, the video signal is distorted from amplitude variations.
# Consider the following statements comparing delta modulation (DM) with PCM system:DM requires1. A lower sampling rate2. A higher sampling rate3. A lower bandwidth4. Simple hardwareWhich one of the above statements are correct?
1. 1 and 3 only
2. 2 and 4 only
3. 1, 3 and 4
4. 2, 3 and 4
Option 4 : 2, 3 and 4
## Detailed Solution
• In PCM an analog signal is sampled and encoded into different levels before transmission
• The bandwidth of PCM depends on the number of levels If each sample is encoded into n bits, then the bandwidth of PCM is nfs
• However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ Hence there is bandwidth saving in Delta modulation
• DM has a simple hardware requirement in comparison to PCM.
A comparison of different modulation schemes is as shown in the table below:
Parameter PCM Delta Modulation (DM) DPCM Number of bits It can use 4, 8 or 16 bits per sample It uses only one bit for one sample Bits can be more than one but are less than PCM Level/Step size Step size is fixed Step size is fixed and cannot be varied Fixed number of levels are used. Quantization error or Distortion Quantization error depends on the number of levels used Slope overload distortion and granular noise is present Slope overload distortion and quantization noise is present Bandwidth of the transmission channel Highest bandwidth is required since the number of bits are high Lowest bandwidth is required The bandwidth required is lower than PCM Signal to Noise ratio Good Poor Fair Area of Application Audio and Video Telephony Speech and images Speech and video
# A signal contains components at 400 Hz and 2400 Hz This signal modulates a carrier of frequency 100 MHz. However, after demodulation, it is found that the 400 Hz signal component is present. The channel bandwidth is 15 kHz. What is the reason for the higher frequency signal not to be detected properly?
1. Modulation used in FM and BW is insufficient
2. Modulation used in AM and BW is insufficient
3. Modulation used in FM but pre-emphasis is not used
4. Modulation used in AM but detector is for FM
Option 3 : Modulation used in FM but pre-emphasis is not used
## Detailed Solution
In FM broadcasting, pre-emphasis improvement is the improvement in the signal-to-noise ratio of the high-frequency portion of the baseband.
In the process of detecting a frequency modulated signal, the receiver produces a noise spectrum that rises in frequency (a so-called triangular spectrum). This is the reason why a preemphasis is required.
Without pre-emphasis, the received audio would sound unacceptably noisy at high frequencies, especially under conditions of low carrier-to-noise ratio, i.e., during fringe reception conditions.
Preemphasis increases the magnitude of the higher signal frequencies, thereby improving the signal-to-noise ratio.
De-emphasis:
At the output of the discriminator in the FM receiver, a deemphasis network restores the original signal power distribution.
De-emphasis means attenuating those frequencies by the same amount by which they are boosted by the pre-emphasis.
# A telephone signal with a cut-off frequency of 4 KHz is digitalised into 8-bit samples at Nyquist sampling rate of s = 2 W. Calculate the baseband transmission bandwidth.
1. 8 KHz
2. 64 KHz
3. 16 KHz
4. 2 KHz
Option 2 : 64 KHz
## Detailed Solution
Number of samples produced each sec : 2 × 4000 = 8000 (Nyquist criteria)
Each sample is encoded into 8 bits
Bit rate = 8 × 8000 = 64 kilo bits/sec
For PCM bit rate = Transmission bandwidth
Hence transmission bandwidth is 64 kHz
# The disadvantage of FM over AM is that __________
1. the noise is very high for high frequency signal
2. high modulating power is required
3. larger bandwidth is required
4. high output power is required
Option 3 : larger bandwidth is required
## Detailed Solution
Advantages of FM over AM are:
• Improved signal to noise ratio.
• Smaller geographical interference between neighbouring stations.
• Well defined service areas for given transmitter power.
• Much more Bandwidth
• More complicated receiver and transmitter.
# Bandwidth efficiency is expressed in units of
1. Hs/bits/sec
2. Hz/watts
3. Bits/sec/Hz
4. It has no dimension
Option 3 : Bits/sec/Hz
## Detailed Solution
Bandwidth efficiency:
Spectral efficiency usually is expressed as “bits per second per Hertz,” or "bits/s/Hz".
In other words, it can be defined as the net data rate in bits per second (bps) divided by the bandwidth in hertz.
Bandwidth efficiency is given as:
$$\eta=\frac{R_b}{BW}$$
Note:
The bandwidth efficiency of M-ary PSK is
$$\eta= \frac{{{R_b}}}{BW}$$
$$BW = \frac{{2{R_b}}}{{{{\log }_2}M}}$$
# Which one of the following statements is correct?
1. Sampling and quantization operate in amplitude domain.
2. Sampling and quantization operate in time domain.
3. Sampling operates in time domain and quantization operates in amplitude domain.
4. Sampling operates in amplitude domain and quantization operates in time domain
Option 3 : Sampling operates in time domain and quantization operates in amplitude domain.
## Detailed Solution
Sampling is done in the time domain and quantization is done in the amplitude domain.
There are three major steps in the process of digital coding of Analog signals are Sampling, Quantizing, and Encoding.
Sampling:
It is a process where an analog signal is converted into a corresponding sequence of samples that are usually spaced uniformly in time i.e. it is a process of converting a continuous-time signal into a discrete-time signal.
Quantization:
It is a process of assigning to each one of the sample values of the message signal, a discrete value from a prescribed set of a finite number of such values called the ‘quantized values’.
This is explained with the help of the following:
The difference between a continuous amplitude sample level and the quantized signal level is known as the quantization error, i.e.
Qe(t) = xq(nTs) – x(nTs)
The output of the quantizer is a discrete-time discrete-valued signal known as a “quantized signal”
Encoding:
It is a process in which quantized samples are encoded in the encoder, involved allocating some digital code to each level.
The coded levels are transmitted as a bitstream of data i.e. 0’s & 1’s
# A high PRF would
1. Increase the maximum range
2. Decreases the maximum range
3. Affect range ambiguity
4. Increase the efficiency of radar
Option 3 : Affect range ambiguity
## Detailed Solution
• Pulse Repetition Frequency (PRF) of the radar system is defined as the number of pulses that are transmitted per second.
• It is normally measured in pulses per second.
• Radar systems radiate each pulse at the carrier frequency during transmit time (or Pulse Width PW), wait for returning echoes during listening or rest time, and then radiate the next pulse.
• The time between the beginning of one pulse and the start of the next pulse is called pulse-repetition time (PRT) and is equal to the reciprocal of PRF.
High PRF:
• Systems using high PRF i.e. above 30 kHz function better known as interrupted continuous-wave (ICW) radar because direct velocity can be measured up to 4.5 km/s at L band, but range resolution becomes more difficult.
• High PRF is limited to systems that require close-in performance, like proximity fuses and law enforcement radar.
• Higher PRFs produce shorter maximum ranges, but broadcast more pulses.
• High PRF pulse Doppler radar has no ambiguity in Doppler frequency but suffers from range ambiguities.
• A high PRF pulse Doppler radar provides an accurate measurement of target's radial velocity.
• In high PRF, It becomes increasingly difficult to take multiple samples between transmit pulses at these pulse frequencies, so range measurements are limited to short distances.
Medium PRF:
• Range and velocity can both be identified using medium PRF, but neither one can be identified directly.
• Medium PRF is from 3 kHz to 30 kHz, which corresponds with radar range from 5 km to 50 km. This is the ambiguous range, which is much smaller than the maximum range. Range ambiguity resolution is used to determine true range in medium PRF radar.
• Medium PRF is used with Pulse-Doppler radar, which is required for look-down/shoot-down capability in military systems. Doppler radar return is generally not ambiguous until velocity exceeds the speed of sound.
• A technique called ambiguity resolution is required to identify true range and speed. Doppler signals fall between 1.5 kHz, and 15 kHz, which is audible, so audio signals from medium-PRF radar systems can be used for passive target classification.
• Medium PRF has unique radar scalloping issues that require redundant detection schemes.
Low PRF:
• Systems using PRF below 3 kHz are considered low PRF because direct range can be measured to a distance of at least 50 km. Radar systems using low PRF typically produce unambiguous range.
• Unambiguous Doppler processing becomes an increasing challenge due to coherency limitations as PRF falls below 3 kHz.
• Low PRF radar have reduced sensitivity in the presence of low-velocity clutter that interfere with aircraft detection near terrain.
• Moving target indicator is generally required for acceptable performance near terrain, but this introduces radar scalloping issues that complicate the receiver.
• Low PRF radar intended for aircraft and spacecraft detection are heavily degraded by weather phenomenon, which cannot be compensated using moving target indicator.
# An AM broadcast receiver has an intermediate frequency of 465 KHz and is tuned to 1000 KHz. Radio frequency has one tuned circuit with Q of 50. Calculate image frequency.
1. 1465 KHz
2. 1930 KHz
3. 2000 KHz
4. 1000 KHz
Option 2 : 1930 KHz
## Detailed Solution
The image signal frequency is given by
fsi = fLo + fIF
Where
fLO = fs + fIF
Hence
fsi = fs + 2.fIF
fsi = 1000 + 2 (465)
fsi = 1930 kHz | 2021-09-28 04:53:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6190247535705566, "perplexity": 2550.395268957808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00448.warc.gz"} |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-3-section-3-2-introduction-to-functions-exercise-set-page-143/99a | # Chapter 3 - Section 3.2 - Introduction to Functions - Exercise Set: 99a
This tells us that the per capita beef consumption was 64.14 pounds in 2004 (which is x=4 years after 2000).
#### Work Step by Step
We are given that the per capita consumption of all beef in the US is given by the function $C(x)=-.74x+67.1$, where x is the number of years since 2000. $C(4)=-.74\times4+67.1=64.14$ pounds
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2017-04-27 23:06:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7786779403686523, "perplexity": 848.1023987057285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00285-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/526574/lorentz-contraction-for-non-simultaneous-events | # Lorentz Contraction For Non-Simultaneous Events
Suppose in the rest frame $$S$$, event $$A$$ occurs at $$x = 0, t =0$$ and event $$B$$ occurs at $$x = 100, t = 80$$.
Now suppose that frame $$S’$$ is moving with velocity $$V$$ with respect to $$V$$.
Now the question is what is the distance between the events in frame $$S’$$. Since the events occur at different times in the rest frame I was wondering if it is applicable to use the formula $$L = L_0/\gamma$$.
On the other hand if i transform the $$x$$ cords for both events using Lorentz transforms and subtract them i get $$x_B’ - x_A’ = \gamma ( x_B - \beta ct_B - x_1 + \beta ct_A)$$. In this instance since $$t_B \neq t_A$$ you will get a different answer than if you use Lorentz contradiction formula.
So since both approaches yield different answers, I was wondering which one is correct and what the fallacy with the other one is.
• note that $(c\Delta t)^2 - (\Delta x)^2$ is the same for all frames of reference
– JEB
Jan 23 '20 at 4:44 | 2021-12-07 00:48:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8792506456375122, "perplexity": 109.487199046844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00117.warc.gz"} |
http://santtu.iki.fi/2013/12/11/gotcha-beware-of-bidir-traffic | 11 December 2013
(Note: The code examples below use coffeescript instead of plain javascript. If you don’t know coffeescript here’s a quick cheat sheet: @foothis.foo and () -> stmtfunction () { stmt }. Additionally text in curly braces {{…}} is Ember’s templating language.)
While doing a retry on Ember for freezr user interface, I hit a problem I’d like to share with you. I didn’t find help on the internet on this so I hope if someone hits the same problem this post will help.
Anyway, I hit one major gotcha that had me scratching my head for a long time. I had used ember-time as a basis on how to implement a “since state change” time display. Converting the original code to coffeescript was straightforward (but see below for an update):
App.FromNowView = Ember.View.extend
nextTick: null
tagName: 'time'
template: Ember.Handlebars.compile '{{view.output}}'
output: (() -> (moment @get('value')).fromNow(true)).property('value')
tick: () ->
@nextTick = Ember.run.later this, (() ->
@notifyPropertyChange('value')
@tick()), 1000
willDestroyElement: () ->
Ember.run.cancel @nextTick
didInsertElement: () -> @tick()
and it was used like this:
{{view "App.FromNowView" valueBinding="stateUpdated"}}
Which worked great when the page was first loaded but it failed to update the time view after updates. I was really really confused. The state value was itself updated in the rendered view correctly immediately after Project.reload() finished, but text derived from stateUpdated field was not. WTF?? This is what was happening in the browser:
Top row is what happened in the UI and the bottom ones showing what the server actually sent to the client on state change from running to freezing to frozen states. Why is it stuck on “for 4 hours”?
Time to debug. So,
• I checked the JSON response. Yep, it had the correct, updated value.
• I wondered whether the name was somehow conflicting (it was originally stateChanged), so I renamed the JSON field and model field. No effect.
• I put tons of log output statements in Ember end Ember Data code. This was a great learning experience in itself, as now I have a lot better understanding how Ember propagates value changes. Nice stuff, I think. However digging deeper and deeper I kept seeing that the updated value was being passed correctly along, yet still refusing to show up in the actual web page.
• I wondered whether the date attribute type was doing something fishy and switched to string instead. No effect, the “bad” value persisted.
• I searched the net high and low to no avail.
I started to do voodoo coding. Poking at things and hoping the problem is mysteriously fixed.
Finally I added logging to DS.attr’s use of Ember.computed and …
… all was made clear to me.
All of the other fields were getting the value from @_data element (which contained the updated values set by DS.Model.setupData) exceptexcept for stateUpdated which got its value from @_attributes!
At this moment I remembered what I earlier read about Ember bindings. And that there was a difference between normal bindings and one-way bindings. And that the valueBinding="stateUpdated" did a binding on App.FromNowView.value to Project.stateUpdated. And that this was a normal e.g. two-way binding meaning that updates on Project.stateUpdated are propagated to App.FromNowView.value and vice versa.
I was not getting the updated value from JSON response because I had already overwritten it myself.
This is the offending line:
@notifyPropertyChange('value')
This doesn’t actually change the value of value, but Ember doesn’t know that so it propagates the event to the bound field of Project.stateUpdated, which eventually results in Project.set('stateUpdated', «value») where the new value was actually the old value. I’ll try to put this into a picture.
In the figure below I’ve used green for events initiated by Ember Data and red for those initiated by App.FromNowView and the gray arrows show bindings between different Ember-controlled values. I refer to objects by their class names, so Project.stateUpdated below is not a class field but a field in an instance of Project class.
In the template the statement valueBinding="stateUpdated" creates the two-way fat gray arrow binding (top row). The binding from App.FromNowView.value to App.FromNowView.output is a one-way binding and comes from the use of property('value') on the output function (right column). Finally the App.FromNowView.output binding to {{view.output}} comes from somewhere deep inside the templating system (bottom row).
The initial value is loaded by Ember Data and is propagated from top left corner by the green arrows. First, Project.stateUpdated is changed, which then propagates to App.FromNowView.value, which in turn causes the value of App.FromNowView.output to change, which finally causes the {{view.output}} template to be (re-)rendered. This will in turn cause the get chain to propagate back in the chain, finally resulting in the nicely formatted time delta value to be written into the HTML page for user to see.
This is where the call to tick messes things up. It will be called every second, and it will call notifyPropertyChange('value') which in turn causes two propagations to occur — one back to the original Project.stateUpdated value thus overwriting it, and the other to propagate to the output template. This meant that the output value was correctly updated as time passed, but any change in the actual stateUpdated value as reported by the backend was not reflected in the human-readable output.
(I’m not sure, but I think Ember’s idea is that since I’ve overwritten the values myself it will keep them around until I call either save or rollback. I’m not sure whether it is sensible to call reload at all when you have uncommited changes in the model.)
Now that I had understood the true problem the solution came immediately. In the application I just wanted to ensure that updates on the bound value are propagated to App.FromNowView.output, which was already automatically updating when the bound value was changed. It also has to be refreshed as time progresses (“a few seconds” → “a minute”) which does not need to refresh the bound value, just the output value. The correct update sequence where display updates do not affect the actual state update time value is shown in the picture below:
Now tick will only cause the rendered value to be updated while all changes in the original model are also honored. The change is trivially simple with changing the property change event fired on the output element:
tick: () ->
@nextTick = Ember.run.later this, (() ->
@notifyPropertyChange('output')
@tick()), 1000
With this simple change everything was finally made good!
So what’s the lesson learned? When using Ember, you need to understand how a value is bound, to where, and what type of binding makes sense for any particular situation. Also don’t use @notifyPropertyChange indiscriminantly on values that are bound from outside the caller’s control.
Update: Ember-time itself has since been fixed. You’ll need to look at bf3383c6 or earlier commit to see the original version. | 2017-06-24 17:09:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42501235008239746, "perplexity": 2305.0551634621675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320270.12/warc/CC-MAIN-20170624170517-20170624190517-00403.warc.gz"} |
https://biajia.eu/projects/2018/05/06/zoomfft-mixer-approach.html | ## Zoom FFT
Zoom FFT was used to analyze a frequency subband. The idea behind zoom FFT is to retain the same resolution, which could be achieved with a full-size FFT on the original signal, by computing a small-size FFT on a down-sampled shorter signal. Thus, the FFT efficient could be improved while maintaining the same resolution.
This is intuitive: for a decimation factor of $D$, the new sampling rate is $f_{sd} = f_s/D$, and the new frame size (and FFT length) is $L_d = L/D$, so the resolution of the decimated signal is $f_{sd}/L_d = f_s/L$.
## Implementation
### The Mixer Approach
See Zoom FFT: The popular mixer Zoom FFT method consists of first shifting the band of interest down to DC using a mixer, and then performing lowpass filtering and decimation by a factor of BWFactor (using an efficient polyphase FIR decimation structure).
Note: For short signal, the polyphase FIR designing is tricky, sometimes it will distort the spectrum, as in the Matlab example, the transient filtering effects were eliminated by repeating 10 times of filtering. So the conventional Fourier resampling method is recommended here.
### Python Codes
• Import modules
• Generate signal
• Plot the original FFT
• Zoom FFT using mixer approach
### Result
The following result is the same as in Zoom FFT, but much faster. | 2019-11-17 17:16:03 | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4218123257160187, "perplexity": 2956.1507475280923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669225.56/warc/CC-MAIN-20191117165616-20191117193616-00468.warc.gz"} |
https://cs.stackexchange.com/tags/turing-machines/hot | # Tag Info
Accepted
### Turing Machines time complexity with regard to the NP and P problem
What are the 'advantages' of a deterministic TM over the non-deterministic one? The advantage of the deterministic TM is that deterministic Turing Machines represent the type of computation we are ...
• 1,035
Accepted
The way you convince yourself of those claims is that you figure out how to write out an explicit proof of them. This same issue occurs with every mathematical proof. Every mathematical proof has ...
• 141k
Accepted
### Limited tapes-version TM for pair sum
Summary: There is no need to sort the given numbers since whether there are two numbers in $A$ such that their sum is $\alpha$ depends on the set of numbers in $A$. Since the choices for the set of ...
• 34k
Accepted
### Can a Turing machine quickly move to any position of a large string?
It depends. 1: If there are at least $\lceil \lg |s| \rceil$ unused cells after the end of $s$ and the head starts within $s$, then the answer is yes. Here is how. Start from the beginning of $s$. ...
• 141k
Accepted
### Turing Machine writes "a" for every input w is undecidable
Your suspicion is well-founded. The point 4 is invalid. Imagine the exact moment N "writes 'a' on the tape and reject" in the point 4. That means at that moment, it is known that M loops. ...
• 34k
1 vote
Accepted
### Are deterministic Turing machines as powerful as probabilistic Turing machines?
This is a famous open problem in computer science theory. In particular, it comes down to whether BPP = P. It is widely conjectured and suspected that BPP = P, or in other words that randomness does ...
• 141k
1 vote
Accepted
### Determining whether the problem of given a turing machine figuring out whether the language it accepts is the set of prime length inputs is R.E
I think your reduction is correct. Indeed, the reduction is clearly computable and furthermore if the original Turing machine $M$ halts on (the fixed) input $w$, then the set of words accepted by the ...
1 vote
The statement is false. Consider the language $H$ of the halting problem and let $H'$ be its complement. $H'$ is Turing reducible to $H$ and $H$ is recognizable, however $H'$ is not recognizable (if $... • 23.3k 1 vote ### Decidability of intersection of regular and decidable languages I don't quite understand what you mean by if (A) is decidable then it is a language in R, if you mean that A is regular and B is regular then the intersection of two regular languages is still regular.... • 456 1 vote ### How is the computational power of a human brain comparing to a turing machine? Here is a question on Turing completeness of neural networks. In my answer, I also discuss the human brain a bit, and I reference also this question here. I think allowing for some type of infinite ... • 111 1 vote ### Is there a TM that halts on all inputs but that property is not provable? The halting of one TM can be encoded by constructing another TM and asking whether it halts. That is, given TM$T_1$and input$I_0$, we can construct a TM$T_2\$ with inputs indexed by the natural ...
Only top scored, non community-wiki answers of a minimum length are eligible | 2022-06-25 19:10:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7364317774772644, "perplexity": 551.5506119652374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103036099.6/warc/CC-MAIN-20220625190306-20220625220306-00591.warc.gz"} |
https://byjus.com/question-answer/a-two-digit-number-is-4-times-the-sum-of-its-digits-if-18-is-2/ | Question
# A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.24252627
Solution
## The correct option is A 24Let the 2-digit number by XY, whose expanded form is 10X + Y. The number obtained by reversing its digits is YX, whose expanded form is 10Y + X. As per the data, 10X + Y = 4(X + Y) --- (i) and 10X + Y + 18 = 10Y + X --- (ii) (i)⇒10X+Y=4X+4Y ⇒6X=3Y ⇒Y=2X−−−(iii) (ii)⇒9X+18=9Y ⇒X+2=Y−−−(iv) Substituting (iii) in (iv), we get: X + 2 = 2X ⇒ X = 2 Thus, (iii) ⇒ Y = 4 Hence, the required number is 24.
Suggest corrections | 2021-12-01 23:04:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8483074307441711, "perplexity": 1167.3344713928407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00623.warc.gz"} |
https://www.rdocumentation.org/packages/utils/versions/3.6.2/topics/clipboard | # clipboard
0th
Percentile
##### Read/Write to/from the Clipboard in MS Windows
Transfer text between a character vector and the Windows clipboard in MS Windows (only).
Keywords
utilities
##### Usage
getClipboardFormats(numeric = FALSE)
readClipboard(format = 1, raw = FALSE)
writeClipboard(str, format = 1)
##### Arguments
numeric
logical: should the result be in human-readable form (the default) or raw numbers?
format
an integer giving the desired format.
raw
should the value be returned as a raw vector rather than as a character vector?
str
a character vector or a raw vector.
##### Details
The Windows clipboard offers data in a number of formats: see e.g.https://docs.microsoft.com/en-gb/windows/desktop/dataxchg/clipboard-formats.
The standard formats include
CF_TEXT 1 Text in the machine's locale CF_BITMAP 2 CF_METAFILEPICT 3 Metafile picture CF_SYLK 4 Symbolic link CF_DIF 5 Data Interchange Format CF_TIFF 6 Tagged-Image File Format CF_OEMTEXT 7 Text in the OEM codepage CF_DIB 8 Device-Independent Bitmap CF_PALETTE 9 CF_PENDATA 10 CF_RIFF 11 Audio data CF_WAVE 12 Audio data CF_UNICODETEXT 13 Text in Unicode (UCS-2) CF_ENHMETAFILE 14 Enhanced metafile CF_HDROP 15 Drag-and-drop data CF_LOCALE 16 Locale for the text on the clipboard CF_MAX 17 Shell-oriented formats
Applications normally make data available in one or more of these and possibly additional private formats. Use raw = TRUE to read binary formats, raw = FALSE (the default) for text formats. The current codepage is used to convert text to Unicode text, and information on that is contained in the CF_LOCALE format. (Take care if you are running R in a different locale from Windows.)
The writeClipboard function will write a character vector as text or Unicode text with standard CR-LF line terminators. It will copy a raw vector directly to the clipboard without any changes.
##### Value
For getClipboardFormats, a character or integer vector of available formats, in numeric order. If non human-readable character representation is known, the number is returned.
For readClipboard, a character vector by default, a raw vector if raw is TRUE, or NULL, if the format is unavailable.
For writeClipboard an invisible logical indicating success or failure.
##### Note
This is only available on Windows.
##### See Also
file which can be used to set up a connection to a clipboard.
##### Aliases
• getClipboardFormats
• readClipboard
• writeClipboard
• clipboard
Documentation reproduced from package utils, version 3.6.2, License: Part of R 3.6.2
### Community examples
Looks like there are no examples yet. | 2020-04-09 08:35:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23623815178871155, "perplexity": 12482.50510366523}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371830894.88/warc/CC-MAIN-20200409055849-20200409090349-00329.warc.gz"} |
http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-ap102-2-4 | Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl
PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
• # Artykuł - szczegóły
## Annales Polonici Mathematici
2011 | 102 | 2 | 143-159
## Analytic solutions of a nonlinear two variables difference system whose eigenvalues are both 1
EN
### Abstrakty
EN
For nonlinear difference equations, it is difficult to obtain analytic solutions, especially when all the eigenvalues of the equation are of absolute value 1. We consider a second order nonlinear difference equation which can be transformed into the following simultaneous system of nonlinear difference equations:
⎧ x(t+1) = X(x(t),y(t))
⎩ y(t+1) = Y(x(t), y(t))
where $X(x,y) = λ₁x + μy + ∑_{i+j≥2} c_{ij}x^{i}y^{j}$, $Y(x,y) = λ₂y + ∑_{i+j≥2} d_{ij}x^{i}y^{j}$ satisfy some conditions. For these equations, we have obtained analytic solutions in the cases "|λ₁| ≠ 1 or |λ₂| ≠ 1" or "μ = 0" in earlier studies. In the present paper, we will prove the existence of an analytic solution for the case λ₁ = λ₂ = 1 and μ = 1.
143-159
wydano
2011
### Twórcy
autor
• Department of Mathematics, College of Liberal Arts, J. F. Oberlin University, 3758 Tokiwa-cho, Machida-City, Tokyo, 194-0294, Japan | 2022-10-02 23:05:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40261125564575195, "perplexity": 2442.6987008924093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337360.41/warc/CC-MAIN-20221002212623-20221003002623-00244.warc.gz"} |
https://www.klasnolom.com/xshd/hlgqnlt/202208/t20220815_712222.html | A plausible strategy of proving the global geometric Langlands conjecture
2022.08.16（星期），09:00-10:00
2022.08.19（星期），09:00-10:00
点：Zoom会议：466-356-2952密码：mcm1234
要：In the previous colloquium talk, I introduced the (categorical unramified) global Geometric Langlands conjecture. In these two talks, I will explain a plausible strategy or project" of proving this conjecture, which is due to many people including Beilinson, Drinfeld, Laumon, Arikin, Gaitsgory, etc.. After that, I will talk about my works in this strategy. In particular, I will show the category of automorphic sheaves for G can be glued from categories of tempered automorphic sheaves for all the Levi subgroups of G. This will allow us to construct a functor from the automorphic side to the spectral side, and reduce the conjecture to its tempered version. Then the only missing point in the mentioned strategy is to show the Whittaker coefficient functor is fully faithful on tempered automorphic sheaves (which is unknown except for GL_n or PGL_n). I will only discuss this last point if time allows. | 2022-09-25 18:41:44 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8950963616371155, "perplexity": 676.0227892273115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334591.19/warc/CC-MAIN-20220925162915-20220925192915-00007.warc.gz"} |
https://projecteuclid.org/euclid.agt/1513715212 | ## Algebraic & Geometric Topology
### The intersecting kernels of Heegaard splittings
#### Abstract
Let $V∪SW$ be a Heegaard splitting for a closed orientable $3$–manifold $M$. The inclusion-induced homomorphisms $π1(S)→π1(V)$ and $π1(S)→π1(W)$ are both surjective. The paper is principally concerned with the kernels $K= Ker(π1(S)→π1(V))$, $L= Ker(π1(S)→π1(W))$, their intersection $K∩L$ and the quotient $(K∩L)∕[K,L]$. The module $(K∩L)∕[K,L]$ is of special interest because it is isomorphic to the second homotopy module $π2(M)$. There are two main results.
(1) We present an exact sequence of $ℤ(π1(M))$–modules of the form
$( K ∩ L ) ∕ [ K , L ] ↪ R { x 1 , … , x g } ∕ J → T ϕ R { y 1 , … , y g } → θ R ↠ ϵ ℤ ,$
where $R=ℤ(π1(M))$, $J$ is a cyclic $R$–submodule of $R{x1,…,xg}$, $Tϕ$ and $θ$ are explicitly described morphisms of $R$–modules and $Tϕ$ involves Fox derivatives related to the gluing data of the Heegaard splitting $M=V∪SW$.
(2) Let $K$ be the intersection kernel for a Heegaard splitting of a connected sum, and $K1$, $K2$ the intersection kernels of the two summands. We show that there is a surjection $K→K1∗K2$ onto the free product with kernel being normally generated by a single geometrically described element.
#### Article information
Source
Algebr. Geom. Topol., Volume 11, Number 2 (2011), 887-908.
Dates
Revised: 29 December 2010
Accepted: 12 January 2011
First available in Project Euclid: 19 December 2017
https://projecteuclid.org/euclid.agt/1513715212
Digital Object Identifier
doi:10.2140/agt.2011.11.887
Mathematical Reviews number (MathSciNet)
MR2782546
Zentralblatt MATH identifier
1215.57007
#### Citation
Lei, Fengchun; Wu, Jie. The intersecting kernels of Heegaard splittings. Algebr. Geom. Topol. 11 (2011), no. 2, 887--908. doi:10.2140/agt.2011.11.887. https://projecteuclid.org/euclid.agt/1513715212
#### References
• A J Berrick, F R Cohen, Y L Wong, J Wu, Configurations, braids, and homotopy groups, J. Amer. Math. Soc. 19 (2006) 265–326
• J S Birman, On the equivalence of Heegaard splittings of closed, orientable $3$–manifolds, from: “Knots, groups, and $3$–manifolds (Papers dedicated to the memory of R H Fox)”, (L P Neuwirth, editor), Ann. of Math. Studies 84, Princeton Univ. Press (1975) 137–164
• J S Birman, The topology of $3$–manifolds, Heegaard distance and the mapping class group of a $2$–manifold, from: “Problems on mapping class groups and related topics”, (B Farb, editor), Proc. Sympos. Pure Math. 74, Amer. Math. Soc. (2006) 133–149
• W A Bogley, J H C Whitehead's asphericity question, from: “Two-dimensional homotopy and combinatorial group theory”, (C Hog-Angeloni, W Metzler, A J Sieradski, editors), London Math. Soc. Lecture Note Ser. 197, Cambridge Univ. Press (1993) 309–334
• K S Brown, Cohomology of groups, Graduate Texts in Math. 87, Springer, New York (1982)
• R Brown, J-L Loday, Van Kampen theorems for diagrams of spaces, Topology 26 (1987) 311–335 With an appendix by M Zisman
• F R Cohen, J Wu, On braid groups, free groups, and the loop space of the $2$–sphere, from: “Categorical decomposition techniques in algebraic topology (Isle of Skye, 2001)”, (G Arone, J Hubbuck, R Levi, M Weiss, editors), Progr. Math. 215, Birkhäuser, Basel (2004) 93–105
• F R Cohen, J Wu, On braid groups and homotopy groups, from: “Groups, homotopy and configuration spaces”, (N Iwase, T Kohno, R Levi, D Tamaki, J Wu, editors), Geom. Topol. Monogr. 13, Geom. Topol. Publ., Coventry (2008) 169–193
• J Hempel, $3$-Manifolds, Ann. of Math. Studies 86, Princeton Univ. Press (1976)
• W Jaco, Heegaard splittings and splitting homomorphisms, Trans. Amer. Math. Soc. 144 (1969) 365–379
• W Jaco, Lectures on three-manifold topology, CBMS Regional Conference Ser. in Math. 43, Amer. Math. Soc. (1980)
• F Lei, Haken spheres in the connected sum of two lens spaces, Math. Proc. Cambridge Philos. Soc. 138 (2005) 97–105
• J Li, J Wu, Artin braid groups and homotopy groups, Proc. Lond. Math. Soc. $(3)$ 99 (2009) 521–556
• R C Lyndon, P E Schupp, Combinatorial group theory, Ergebnisse der Math. und ihrer Grenzgebiete 89, Springer, Berlin (1977)
• J Milnor, A unique decomposition theorem for $3$–manifolds, Amer. J. Math. 84 (1962) 1–7
• J Morgan, G Tian, Ricci flow and the Poincaré conjecture, Clay Math. Monogr. 3, Amer. Math. Soc. (2007)
• C D Papakyriakopoulos, A reduction of the Poincaré conjecture to group theoretic conjectures, Ann. of Math. $(2)$ 77 (1963) 250–305
• M Scharlemann, Heegaard splittings of compact $3$–manifolds, from: “Handbook of geometric topology”, (R J Daverman, R B Sher, editors), North-Holland, Amsterdam (2002) 921–953
• J Stallings, How not to prove the Poincaré conjecture, Ann. of Math. Study 60 (1966) 83–88
• J Wu, Combinatorial descriptions of homotopy groups of certain spaces, Math. Proc. Cambridge Philos. Soc. 130 (2001) 489–513 | 2019-10-15 12:28:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6772012114524841, "perplexity": 1722.3403045268542}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00413.warc.gz"} |
https://www.physicsforums.com/threads/born-von-karman-boundary-conditions.238084/ | # Born-Von Karman boundary conditions
no_math_plz
I can't understand this conditions, and in general every boundary conditions for problems like this. they states "the choice of boundary conditions can be determined by mathematical convenience (!?) ... for if the metal is sufficiently large, we should expect its bulk properties not to be affected by the detailed configuration of surface" I'm a bit confused. from where they derive? from some postulates? doesn't exist postulates in quantum mechanics concerning boudary conditions. can you help me?
For example, in the first QM problem you ever solved (the particle in a box) the boundary conditions were given to you: $\psi(0)=\psi(L)=0$, where L is the size of the box. These conditions are physical since the particle can't be outside the box.
You could, however, solve the problem with periodic boundary conditions instead of the physical boundary condition if you like and you will find only "half as many" solutions which you would then account for by saying that the wave vector can be taken as positive or negative, etc etc etc. And, presumably, when you calculate something like the density and take the limit as $L\to\infty$ you obtain the same result regardless of boundary conditions. | 2023-02-07 02:12:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8690630793571472, "perplexity": 283.9575720263815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00836.warc.gz"} |
https://forum.allaboutcircuits.com/threads/raspberry-pi-pico.176102/ | # Raspberry Pi Pico
Joined Feb 20, 2016
4,110
I am going to have a play with the new Raspberry Pi Pico. Cheaper than the Arduinos and a LOT more powerful.
Have you thought of trying them?
#### John P
Joined Oct 14, 2008
1,971
I like the idea of the Pico too, but then I read some of the threads on the Raspberry Pi Foundation's forum, and it seems as if the programming environment is nowhere near as simple to use as the Arduino. I'll wait till it becomes easy to use on a Windows system.
https://www.raspberrypi.org/forums/viewforum.php?f=143
If you get something useful running, please report details here!
#### dl324
Joined Mar 30, 2015
14,307
Raspberry Pi Org claims that people commonly use Pi's to drive Arduino microcontrollers, so they built one. It has impressive specs, but I prefer my generic $3 Uno's that I can order in quantity when I need a microcontroller. When I need more, I use a Mega or Pi Zero W.$5 shipping and handling for a $4 product is a non starter for me (you can only buy 1 at a time). There isn't even a reliable source for Pi Zero and Pi Zero W. Thread Starter #### dendad Joined Feb 20, 2016 4,110 I think the Pi Pico will be different to the Pi Zero. I had no problem buying a number of Picos, there was no limit on them, and there are at least 2 other boards in the works from Adafruit that uses the same chip. All that said, mine are due to arrive in a couple of days so I have not had a play as yet. The PIO blocks sound very interesting. Time will tell. #### dl324 Joined Mar 30, 2015 14,307 I received an email from vilros.com about Pico: Thread Starter #### dendad Joined Feb 20, 2016 4,110 Ah. It must be popular. I went back to my store here in Oz, and they too now have a limit of 1. Maybe I was greedy to get 3 originally? #### upand_at_them Joined May 15, 2010 875 I am going to have a play with the new Raspberry Pi Pico. Cheaper than the Arduinos and a LOT more powerful. Have you thought of trying them? No, I have enough things to do. And I'm not easily swayed by "You HAVE to buy this new thing!!" Cheaper than Arduinos? You must not have seen the clone market for Arduinos. #### dl324 Joined Mar 30, 2015 14,307 Ah. It must be popular. I think the issue is profit margin. Years after Pi Zero was introduced, I can still only buy one at a time; same for the Zero W. The only way I can order more than one at a time is if I pay for their ridiculously expensive installed header. #### John P Joined Oct 14, 2008 1,971 Cheaper than Arduinos? You must not have seen the clone market for Arduinos. Micro Center is selling the Pico for$1.99, if your local store still has any in stock. Predictably, the store in Cambridge MA ran out almost immediately, but you can get them in Duluth.
Joined Feb 20, 2016
4,110
Cheaper than Arduinos? You must not have seen the clone market for Arduinos.
Oh yes, I use them a lot. I can save about \$1 by using an Arduino Nano, like in this VFO I built for a Philips FM828 VHF transceiver.
(I have also built a version for HF radios with the ESP32 and animated dial)
But these Nanos are not as powerful as the Pico, and the number of channels I can program is a bit limited by memory constraints. Also, if I change to an Arm board, like the Blue Pill or Pi Pico, more features can be included.
But these Picos are cheaper than the STM32 boards here, and are available in Oz (when stocks improve again), not having to wait for long shipping delays from China.
And the method of programming the Picos is easier too.
It will be interesting to see how they stack up in reality.
According to the tracking, they are due for delivery today
#### upand_at_them
Joined May 15, 2010
875
I'm building a similar VFO to your Nano one. I went straight with the ATMega328P chip, though.
#### John P
Joined Oct 14, 2008
1,971
I'll revive this thread to follow up my own message, "I'll wait till it becomes easy to use on a Windows system." Somebody (not the official Arduino organization, who say they're also working on it) has come up with an addition to the Arduino environment which lets you program the Pico in C. The instructions for loading the material you need are quite simple to follow; I've got it on my computer and I've written some simple code, which I've verified with a scope. It is way faster than the Python equivalent! The one annoying feature is that compilation and uploading is rather slow, but perhaps it's just always going to be that way. Here are the instructions I used:
https://www.tomshardware.com/how-to/program-raspberry-pi-pico-with-arduino-ide
Joined Feb 20, 2016
4,110
Yes, I used that link too and have started the Pico VFO. But it is on the back burner for a while.
But I did add a reset button! Well worth it.
Still to do is the encoder input, OLED display etc. But here is a simple test code file if anyone wants to have a play.
#### Attachments
• 18.2 KB Views: 4 | 2022-05-19 09:35:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23652301728725433, "perplexity": 2514.2357669234884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00645.warc.gz"} |
https://da.overleaf.com/articles/single-precision-barrett-reduction/tgytknpxmfxz | # Single Precision Barrett Reduction
Author
Jacob Wells
View Count
1272
AbstractModular Reduction of a 2N Bit Integer using two N-Bit multiplications and a few subtractions. Examples and Proof are included. | 2019-03-19 09:49:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6459022164344788, "perplexity": 10510.963747904814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201953.19/warc/CC-MAIN-20190319093341-20190319115341-00130.warc.gz"} |
http://math.stackexchange.com/tags/algebraic-geometry/new | # Tag Info
2
The book says homogeneous polynomial. I think he wants to emphasize on the fact that $P(t_1,\dots,t_n)$ is a homogeneous element in $\bar A$. Let $f\in A[T]$, $f=a_0+a_1T+\cdots+a_nT^n$, such that $\phi(f)=0$. Suppose $(a)=I$. Then $f(a)=0$, that is, $a_0+a_1a+\cdots+a_na^n$. In $\bar A$ this leads to $a_ia^i=0$ for all $i$, so $a_i=0$ for all $i$ and thus ...
0
Let me give a two-part answer: I will give a counterexample when $X$ is not quasi-compact, and I will prove that for $X$ integral and quasi-compact the result holds (without quasi-separatedness). To ease notation, since the only sheaf we use is $\mathcal O_X$, I will just write $\Gamma(X)_f \to \Gamma(X_f)$. (You'll thank me later.) Example. Let $X = ... 1 Just use the group law, which says if$L$, a line intersects$C$at 3 points$A,B,C$, then$A+B+C=O$. If$P$is of order 6, the tangent line at$P$meets it at least twice and if$B$is the third point, then$2\cdot P+B=O$and since$6P=O$, we get$B=4P$. Identical argument for$5P$, since$2\cdot 5P+2P=O$. 0 This question has been asked and answered on MathOverflow. I have replicated the accepted answer by user2035 below. If$f\in\mathbf Z[X]$is any monic polynomial, the solutions of$x(1-x)\cdot f(x)=1$are solutions of the unit equation. Take some$y\in U\setminus\mathbf R$. Since the substitution$z\mapsto1/(1-z)$leaves$S$invariant, we may assume ... 4 No, incidence geometry will not in the least help you with algebraic geometry. If you want to immerse yourself in classsical algebraic geometry consider Semple and Kneebone's Algebraic Projective Geometry (first published in 1952) but beware that you will probably find it difficult or at least disconcerting. Here is a review by Du Val of that book. In ... 0 I see the flaw in my argument in part 2). If have$d.C_1'\sim d.C_2'$. This means that$\mathcal{O}(C_1')^d=\mathcal{O}(C_2')^d$. But this does not mean that$\mathcal{O}(C_1')=\mathcal{O}(C_2')$. For example we for a degree$d $morphism, we have non trivial line bundle$L$on$Y$such that$L^d=\mathcal {O}_Y$1 To compute the$\operatorname{Ext}$in this case, you can indeed use the spectral sequence $$H^p(X, \mathscr Ext^q(\mathscr F, \mathscr G)) \Rightarrow \operatorname{Ext}^{p+q}(\mathscr F, \mathscr G),$$ arising as the Grothendieck spectral sequence of the composition of functors $$\operatorname{Mod}_{\mathcal O_X} \to \operatorname{Mod}_{\mathcal O_X} \to ... 3 This is not true. Let X be any topological space, equipped with a nonzero sheaf \mathcal{F} that has no global sections except 0 (such spaces and sheaves exist). Consider the trivial 1-element cover of X, and let \phi be the 0-morphism \mathcal{F}\to\mathcal{F}. Clearly, the morphism \phi induces an isormorphism ... 0 The group law is equivalent to the following. If L is any line and if it meets C at A,B,C (counted with multiplicity), then A+B+C=O. So, one easily sees that 2P, P, 3P are collinear and similarly for the other 3-tuple. Only thing remaining is to check that 2P, 4P are flexes. For 2P, let L be the tangent line at 2P. Then L meets C at ... 1 If I understand you correctly, you want to know why the derived category is additive, and what represents the sum f+g of two morphisms. (Note that the derived category of any abelian category is additive, not just the derived category of coherent sheaves). In terms of zig-zag, you have the following description of f+g : f and g are represented by ... 2 Low-tech way, following your hint. The family of conics passing through those points is 1-dimensional. This can be seen by counting dimensions (the space of conics is 5-dimensional, and each constraint subtracts one) or more explicitly by noting that the family is given by$$a(X^2-Z^2-YZ)+bZY=0$$for [a:b]\in \mathbb P^1. Conics are degree 2, and your ... 3 For a counterexample, consider X = \mathbb P^1_k, and let \mathscr F be any sheaf of \mathcal O_X-modules. Note that \Gamma(X,\mathcal O_X) = k, and every nonzero element in k is already invertible on all of X. Thus, X_s = X for all s \in \Gamma(X,\mathcal O_X)\setminus\{0\}, so the condition on \mathscr F is vacuous. (To find a sheaf of ... 1 Off the top of my head, the only examples I know where this holds are abelian varieties; in this case X is topologically a torus, and we in fact have H^k(X, \mathbb{Z}) \cong \bigwedge^k H^1(X, \mathbb{Z}). As Roland says in the comments, \mathbb{CP}^n for n \ge 1 is a counterexample. 1 Hartshorne does prove Serre duality for \mathbb P^n over any Noetherian ring (Theorem III.5.1). Locally, \mathbb P(\mathscr E) is of this form (e.g. on an affine cover that trivialises \mathscr E). To check that the duality commutes with restriction (on \operatorname{Spec} A), observe that the pairing$$H^0(\mathbb P^n, \mathcal O(d)) \times ... 1 0) First of all you probably want to assume that the base field is algebraically closed: if the base field is$\mathbb R$and$n=2$the image of$\phi_2$is not even algebraic! So let us suppose that$k=\bar k$. 1)$\textbf n=2k$even It is clear that the image of$\phi_n$is the curve$A_n\subset \mathbb A^2$with equation$y=x^k$, obviously isomorphic ... 1 Okay, assume the characteristic is$0$(over complex numbers). Let's stare at your last exact sequence: $$0 \rightarrow H^0(S,O_S(H-C))\rightarrow H^0(S,O_S(H)) \rightarrow H^0(C,O_C(H)) \rightarrow H^1(S,O_S(H-C)) \rightarrow ....$$ In the surface$S$,$C=H|_S$as divisor classes. So we know your first term has dimension 1 ($S$is smooth and simply ... 1 Since$c^Tx=0$supports a facet of your simplex, it is going to be generated by$n-1$of the vectors$b_i$, say they are the first$n-1$, so that$c^Tb_i=0$for all$i=1,\dots, n-1$. The fact that$H\cap C$is not empty tells you that$c^Tb_n>0$(take any point$r\in H\cap C$, so$r=\sum_{i=1}^n e_ib_i$for real numbers$e_i\geq 0$for all$i$, and ... 2 The "naive" coproduct of functors where you define$(\coprod_i F_i)(T)=\coprod_i F_i(T)$in not a sheaf in general. If you sheafify, (at least in the case you're interested in) you should get a description like this: an element of$(\overline{\coprod_i F_i})(T)$(the sheafification) is a decomposition$T_i$of$T$as a disjoint union$T=\coprod_i T_i$(same ... 1 For the even case, you are showing that the image is isomorphic to$\mathbb{A}^1$. It is not necessary (and not true) that$\varphi_n$is the isomorphism. In fact, your answer almost contains the map from$\mathbb{A}^1$to$\varphi_n(\mathbb{A}^1)$and the map in the other direction. (You will need to check that they are mutual inverses of course). In the ... 2 Let me expand a bit on your idea that$\operatorname{Sh}(X)$is the localisation of$\operatorname{Psh}(X)$at the multiplicative set$S$of morphisms that induce isomorphisms on all stalks (even though this may not be your main question). A better way to put it is: Lemma. Let$\mathscr C$be the Serre subcategory of$\operatorname{Psh}(X)$of presheaves ... 1 This is common notation in algebraic geometry. I have two schemes$X,T$with structure morphisms to another scheme$S$, and now$X(T)$is the set of$S$-morphisms$T \to X$. When$T$is the spectrum of some ring$R$we usually just write$X(R)$. In this situation,$S = \mathbb Z$(we really could have left this bit out here, since ring homomorphisms always ... 2 Your first isomorphism seems believable: just plug in any open set$U$on both sides. You need to check things like $$(f^*\mathscr G)\big|_{f^{-1}U} = \left(f\big|_{f^{-1}U}\right)^* \left(\mathscr G\big|_U\right),$$ etc, because you want to use the adjunction$\left(f\big|_{f^{-1}U}\right)^* \dashv \left(f\big|_{f^{-1}U}\right)_*$. But pulling it through ... 1 There is probably a more direct way to prove this, since it is apparently an exercise in Silverman, but one way to see it is: if$C$has genus$1$, then its Jacobian$E$is an elliptic curve, and$C$and$E$become isomorphic over any field$L$over which$C$obtains a rational point. 0 You are right, there was a tiny mistake in my answer, although it did not change anything. The degree of$V$is$4$, not$2$, but as you point out the calculation still leads to getting$\dfrac 12 E$. Furthermore, since you calculated that$K_X$generates the class group$\mathbb Z/2\mathbb Z$, it shows the original claim that$K_X$is not Cartier, but ... 0 You have slightly misunderstood the situation. There is an additional assumption, namely that the image is non-solvable. Now there is a classification of the subgroups of$\mathrm{PGL}_2(\mathbb{F}_p)$. (Probably the paper of Serre that they cite has it --- did you look at that? Otherwise, one place it is discussed is in Swinnerton-Dyer's article in LNM ... 0 Point one.$K_X$is indeed$\mathbb{Q}$-Cartier but not Cartier. I calculated class group (see my question). It is$ \mathbb{Z} / 2 \mathbb{Z} $.$K_X$represents the nontrivial element of class group. To see this we will construct an explicit section$\omega$, which equals zero precisely on divisor$D_1$. Pull back of this section on$\mathbb{A}^3$equals ... 2 Are you asking for integer pairs? If so,$(x+3)^2+y^2=13=2^2+3^2$, so the integer pairs on this circle can only be$(-3 \pm 2, 0 \pm 3)$, and$(-3 \pm 3, 0\pm 2)$. 0 I assume that by "the variety associated with$f_1,\dots,f_n$" you mean the variety$Y\subseteq k^n$cut out by the ideal$I=\{g\in k[x_1,\dots,x_n]:g(f_1,\dots,f_n)=0\}$. The image of$p$is always contained in$Y$, since for any$a\in X$and any$g\in I$,$g(p(a))=g(f_1(a),\dots,f_n(a))=0$by definition of$I$. So$p$can always be considered as a ... 1 Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case$SL(3,\mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one$\mathbb C\subset\mathbb C^2\subset\mathbb C^3$. This will be the unique Schubert-cell of dimension$0$. The further ... 4 As it is pointed out in Matsumura, CRT, page 31, for a noetherian local domain$A$the equality$\operatorname{ht}\mathfrak p+\dim A/\mathfrak p=\dim A$holds for any prime ideal$\mathfrak p$iff$A$is catenary. So, we are looking for a noetherian local domain which is not catenary. There are no trivial examples, but you can find one at ... 2 There are prime ideals$\mathfrak p$in noetherian local rings$A$such that$\operatorname{ht}\mathfrak p+\dim A/\mathfrak p<\dim A$. For instance, let$A=K[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, and$\mathfrak p=(y,z)$. Then$\operatorname{ht}\mathfrak p=0$(since$\mathfrak p$is minimal), and$\dim A/\mathfrak p=1$(since$A/\mathfrak p\simeq K[X]_{(X)}$). On ... 1 By your definition of regular functions ("A regular function on$U$is a rational function that is well-defined at all points of$U$"), they indeed do not form a sheaf, as$O_X(\emptyset)$is the entire field of rational functions on$X$, rather than$0$. An easy (if inelegant) way to fix this is to say that your definition is only the definition of a ... 1 Great question! Here is a counterexample: Let$(A,\mathfrak m)$be a DVR, and let$A \subseteq B$be a finite extension of domains such that$B$has exactly two primes above$\mathfrak m$. For example,$A = \mathbb Z_{(5)}$, and$B = \mathbb Z_{(5)}[i]$, with the primes$(1+2i), (1-2i)$lying above$(5)$. Then$Y = \operatorname{Spec} A$is the space ... 2 Yes, that sounds about right. Yes. The superscript means taking the invariant subspace under the action of$\tau$. In general, if$V$is a vector space and$G$is a group acting on it, then$V^G = \{ v \in V \mid gv = v \forall g \in G \}$. 2 Careful. You have shown that a the sections of a presheaf$\mathscr G$over a base for the topology on some space$X$are isomorphic to those of a sheaf$\mathscr F$($\mathscr G$,$\mathscr F$are assumed to be presheaves on$X$). This is not enough to conclude abstractly that$\mathscr G$is a sheaf isomorphic to$\mathscr F$; such a thing is clearly not ... 1 There is a problem in the conjecture. Over$\mathbb C$it is really easy to prove. I believe you mean$\mathbb{A}^n(\mathbb{C})$. Here is a way of proving it that I like and that can work for proving many things over rings whence you know it over$\mathbb{C}$First notice that for countability reasons,$\mathbb{C}$has an infinite transcendance basis. This ... 3 Here is a (translated) quote from G. Castelnuovo, "Sur les intègrales de différentielles totales appartenant à une surface irrégulaière", Comptes rendus hebdomadaires des séances de l'Acadeémie des sciences, Paris. 140, 23 Jan 1905. pp. 220-222: out of respect for Picard's profound research on surfaces admitting a group of birational automorphisms, ... 0 According to wikipedia "The name is in honor of Émile Picard's theories, in particular of divisors on algebraic surfaces." I would assume one of the references on the page probably contains what you want. 2 For a quick and dirty explanantion that offers no insight (as requested), take a$3\times 3$permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any$1$. Put$0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix. For example, for the ... 4 I think the following works, maybe there is an easier answer... First note that$\det E_{|L}=\bigotimes_{i=1}^{\dim V-1}\mathcal{O}_L(a_i(L))=\mathcal{O}_L(\sum a_i(L))$. Note also that$\det E_{|L}\otimes \mathcal{O}_L(2)=\det (V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L}=0$. So$\det E_{|L}=\mathcal{O}_L(-2)$and$\sum a_i(L)=-2$. Then, note that, because ... 0 I found two mistakes in my proof. First, like Hoot said, I tried to shove the$D(f_{i})$into the same ambient affine space as closed subsets. Second,$(1−f_{i}T,1−f_{j}T)=1$does not imply$(1−f_{i}T)+(1−f_{j}T)=(1)$when$f_{i}$and$f_{j}$is in an integral domain. 4 They agree. In fact, you only need to assume quasicoherence; the key point is that the étale cohomology of a quasicoherent sheaf on an affine scheme vanishes in degrees$> 0$, just as for the Zariski topology. For more details, see tag 03DW in the Stacks project. 4 Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. We provide an alternative version of that section (specifically, Lemma 1.6.5 through Theorem 1.6.9) that avoids use of said object. We start with a ... 1 Let$R$be the quotient of a polynomial ring$k[x_1,x_2,\dots]$in infinitely many variables over a field by the ideal generated by all products$x_ix_j$for$i\neq j$. Note that if$P\subset R$is a prime ideal, then there can be at most one$i$such that$x_i\not\in P$. It follows that if$P_i$is the ideal generated by all the$x_j$for$j\neq i$, ... 3 If one of the quadrics has maximal rank$4$(or better one of the quadrics in the pencil, which is in fact true), then it is projectively equivalent to the Segre variety. In this case, the other quadric cuts out a curve of bidegree$(2, 2)$on$\mathbb{P}^1 \times \mathbb{P}^1$. As the twisted cubic has bidegree$(1, 2)$, it follows that we get not only the ... 1 Lemma. Let$f \colon X \to Y$be a finite morphism, and let$\mathcal F$be any sheaf on$X_{\operatorname{ét}}$. Then$R^if_* \mathcal F = 0$for all$i > 0$. Proof. It suffices to show that$(R^if_* \mathcal F)_{\bar y} = 0$for all geometric points$\bar y \to Y$. But the stalk is computed by$H^i_{\operatorname{ét}}(X', \mathcal \pi^* F)$, where$X' ...
1
First, your even more basic question. Indeed $\pi_*\mathbb{Z}_{\mathbb{A}^1}$ and $i_*\mathbb{Z}_x$ are not coherent sheaves. In fact, they are not even sheaves of $\mathcal{O}_X$ modules. And as Remy points out, $H^1_{ét}(X,\pi_*\mathbb{Z}_{\mathbb{A}^1})$ and $R^1\pi_*\mathbb{Z}_{\mathbb{A}^1}$ are not the same kind of objects, the first being a group, the ...
0
The isomorphism comes about by looking at lines through the singular point $P$. Indeed, for any point $Q \neq P$ on $C$, the unique line in $\mathbb P^2$ through $P$ and $Q$ does not intersect $C$ anywhere else, because it already intersects with multiplicity $3$ (namely, once at $Q$ and twice at $P$). Thus, we get a map f \colon C\setminus\{P\} \to ...
3
Since $T\subseteq\mathfrak{a}$, clearly $Z(\mathfrak{a})\subseteq Z(T)$. On the other hand, if $P\in Z(T)$, that means $f(P)=0$ for all $f\in T$. For any $P\in Z(T)$, the set $I=\{f\in A:f(P)=0\}$ is an ideal, and so since it contains $T$, it must contain $\mathfrak{a}$. Thus $P\in Z(\mathfrak{a})$. Since $P\in Z(T)$ was arbitrary, this shows ...
1
Lemma. Let $X$ be a variety over $k$. Then $X$ is proper if and only if for every smooth proper curve $C$ and every $U \subseteq C$ open, any morphism $f \colon U \to X$ can be extended to a map $C \to X$. Proof. We extend one point at a time. Suppose $P \in C\setminus U$, and let $V = \operatorname{Spec} B$ be an affine open neighbourhood of $P$. Let \$\eta ...
Top 50 recent answers are included | 2015-12-01 11:40:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9965918660163879, "perplexity": 1346.087462515925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398466260.18/warc/CC-MAIN-20151124205426-00105-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-write-the-equation-in-standard-form-given-m-4-1-5 | # How do you write the equation in standard form given m= -4 (1,5)?
Jun 17, 2016
y-5=-4(x-1)
#### Explanation:
Use point slope formula | 2021-06-14 15:41:16 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9344040155410767, "perplexity": 2550.267247331372}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00582.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/find-coordinates-point-which-divides-join-1-7-4-3-ratio-2-3-section-formula_7102 | # Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. - Mathematics
Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Find the coordinates of the point which divides the join of A(–1, 7) and (4, –3) in the ratio 2 : 3.
#### Solution 1
Let P(x, y) be the required point. Using the section formula
x = (2xx4+3xx(-1))/(2+3) = (8-3)/5 = 5/5 = 1
y = (2xx(-3)+3xx7)/(2+3) = (-6 + 21)/5 = 15/5 = 3
Therefore the point is (1,3).
#### Solution 2
The end points of AB are A(-1,7) and B (4,-3)
Therefore (x_1 = -1, y_1 = 7) and (x_2 = 4, y_2 = -3 )
Also , m= 2 and n= 3
Let the required point be P (x ,y).
By section formula, we get
x= ((mx_2 + nx_1))/((m+n)) , y = ((my_2+ny_1))/((m+n))
⇒ x = ({ 2 xx 4 +3 xx (-1) })/(2+3) , y= ({2 xx (-3) + 3 xx 7})/(2+3)
⇒ x = (8-3) /5 , y = (-6+21)/5
⇒ x = 5/5 , y = 15/5
Therefore, x = 1 and y= 3
Hence, the coordinates of the required point are (1,3) .
Is there an error in this question or solution?
#### APPEARS IN
NCERT Class 10 Maths
Chapter 7 Coordinate Geometry
Exercise 7.2 | Q 1 | Page 167
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 1 | 2021-02-25 21:40:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4024977385997772, "perplexity": 2272.526616361203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178355937.26/warc/CC-MAIN-20210225211435-20210226001435-00609.warc.gz"} |
https://www.homebuiltairplanes.com/forums/threads/how-to-fix-a-broken-bose-headset.3363/ | # How to fix a broken Bose headset?
Discussion in 'Workshop Tips and Secrets / Tools' started by etterre, Oct 31, 2007.
### Help Support HomeBuiltAirplanes Forum by donating:
1. Oct 31, 2007
### etterre
#### Well-Known Member
Joined:
Aug 30, 2006
Messages:
313
1
Location:
St. Louis, MO, USA
It's probably not exactly the right forum, but since I'm considering a fix with composites and a fix with metal...
Sigh. I've got a Bose X that I bought used and it's now been broken (I blame my kid ) The u-shaped stirrup that holds the left earcup has snapped near one of the attachment points. I called Bose and was told that they don't sell replacement parts... but I could send it to them and they would fix "anything" for $175. Not a bad deal if the ANR was getting futzy, but I only need a$5-10 plastic part. Since the stirrup is made of plastic, my first try was to clean it, glue it back together with 5-minute epoxy, and add some fiberglass for extra support. Maybe it's because I used the cheap stuff (Devcon from a hobby store), but the epoxy didn't stick. I think my problem is that the plastic is of a type that just doesn't chemically bond to epoxy, so I actually need to do rough it up with sandpaper so that I get more of a mechanical bond.
But now I'm wondering if that is actually the best way to fix it - so I'm fishing for other ideas. Here's a few that I've considered:
1. Create a new stirrup out of fiberglass and epoxy. The only way I can think of doing this entails making a female mold from the broken part and then essentially filling the mold with flox. The I'd have to finish it off with sanding and some machining so that the attach points line up. While I'd be more likely to get a reasonably light part that looks like the original Bose piece this is probably a fair amount of work. I'd learn a lot, but...
2. Make some measurements, do some drawings, and have a machine shop make a "more rectangular" version out of aluminum or steel. Easier for me to do... but I'm a little afraid of the $I'd spend at the machine shop and if it might make the$175 charge from Bose look cheap.
3. A slight spin on #2 where I do my own machining work. I don't have a milling machine (or access to one), but the part all in one plane so a rough copy could be made from a piece of flat plate. I do remember reading about one guy's technique that involved using a normal wood router and templates to create the part out of aluminum stock.
4. Modify a steel stirrup from a different headset. I did a could of rough measurements and I can get a replacement steel stirrup from David Clark for $10. It doesn't have the same connection between the stirrup and the headband, but it would be fairly easy to make an "adapter" by boring a couple of holes in a cube of aluminum and screwing the adapter onto the steel stirrup at the appropriate point. Anybody got a better idea or a reason to pick one method over another? I'm currently leaning towards #4 since it should provide a fairly durable fix without a "lot" of work. It won't be "pretty," but I could still buy "pretty" at a later date for$175 if it really bugs me. Since there's a nick at the same point on the right stirrup, I'm also figuring that I might as well fix both sides.
The big picture shows the left stirrup in it's current condition. Note that there isn't much material missing at the break - I've just moved the pieces apart to show the break better. They fit back together pretty cleanly, but I can't instantly align the pieces together due to the curvature of the pieces and the break itself - so using an instant bond cyanocrylate (Superglue) wouldn't work. The smaller picture is an "end-on" view showing the relative "smoothness" of the break.
#### Attached Files:
File size:
85.9 KB
Views:
3,294
• ###### trim_IMG_2123.JPG
File size:
15.1 KB
Views:
972
2. Oct 31, 2007
### orion
#### Well-Known Member
Joined:
Mar 3, 2003
Messages:
5,800
135
Location:
Western Washington
You're right, some plastics just don't bond worth beans. Using a better epoxy however may be one possibility. Looking at your options though, I wonder if by the time you're done with all that whether the $175 would be a more cost effective way to go. But just in case, let me try to give you some alternatives. First, you might try a better adhesive. There are several made specifically for plastics, most of which you could probably get from your local hobby shop. Plastic-Weld comes to mind but I'm sure there are others. The part may however be made from something that's a bit more chemically resistive so epoxy may be your best bet. There are of course many that may be better than Devcon but do stay away from the 5-minute stuff. Generally, the longer the cure the better the final bond quality. I've had pretty good luck with Hysol EA9460, even on plastics. For a better chance of success, I might suggest that you take a small drill (something like a #40)and bore one or two small holes in both parts. Of course make sure the holes line up. To do this you may have to clamp the two pieces together and then drill from the outside. As you go to bond the two pieces together, insert "music wire" into the holes, making sure to get a bit of the epoxy onto the wires also. Make sure the wire and the holes are a snug fit. In this way the music wire will reinforce the joint in bending and the epoxy at the break will just hold it together. If you want to cast your own, you might want to join the two pieces with a low cost epoxy and then create a part mold using a casting silicone. A friend of mine uses this process for creating various model parts from master components and it works beautifully. Given the strength of some of the better epoxies, I would guess you wouldn't have to fill the resin with flox or any other additives, just pour or inject the epoxy into the cavity and let it set. Keep in mind however that although you can probably de-mold the part over night, most epoxies require a lot of time (up to three months) or an elevated temperature to reach full cure. 3. Nov 1, 2007 ### etterre ### etterre #### Well-Known Member Joined: Aug 30, 2006 Messages: 313 Likes Received: 1 Location: St. Louis, MO, USA Thanks - after a small search, I found a supplier for a 2 part silicone mold-making kit ( the "mold-max" product from www.smooth-on.com ) that's only$22. I'll probably go that route for the "long-term" fix, but I'll give the music wire trick a try with some of the "extra" Aeropoxy from the Rutan practice kit that I have.
4. Nov 10, 2007
### Dana
#### Super Moderator
Joined:
Apr 4, 2007
Messages:
8,546
2,962
Location:
CT, USA
Roy, here's how I have made similar repairs:
Get some very thin cyanoacrylate (the "wicking" kind, not the "gap filling" gel). Hardware stores may not have it, but hobby shops catering to the R/C crowd sell it. With the wicking CA, you can put the broken pieces together first, and then apply the glue... it will wick in and bond it. That won't be strong enough by itself, but it should hold the parts together for the next step. Before you glue it, though, lightly sand the outside of the parts adjacent to the break, to break the gloss and scuff up the surface. Then after it's glued, for maximum adhesion, clean the surface with laquer thinner and don't touch that area again (the oils from your skin will prevent a good bond).
Get the thinnest glass cloth you can find (again from the R/C shop, or nylon or dacron fabric will also work) and lay strips along the part. Drip the thin CA onto it and press it into place with a pin or knife blade; it's easy to see when it's properly wet out and laying on the surface. After fully encapsulating the broken area with the glass/CA, you can coat it with a thin layer of epoxy for a harder, smoother surface.
I've used this technique numerous times... two years later the plastic turn signal stalk in my car (broken right at the base where the stress is greatest) is still holding up fine.
-Dana
Mary had a little lamb. The doctor was very surprised.
zipzit likes this.
5. Nov 13, 2007
### Sticky1
#### Member
Joined:
Nov 13, 2007
Messages:
23
0
don't you think it would be easier to send them back to bose......I did.....they fixed for free.....
6. Nov 13, 2007
### etterre
#### Well-Known Member
Joined:
Aug 30, 2006
Messages:
313
1
Location:
St. Louis, MO, USA
Sure, it would be easier - but not free. How long ago did you get your set fixed? In talking it over with my flight instructor, he had a friend that also had to pay to get his Bose X fixed - so the repair charge isn't a new thing.
Don't get me wrong - I'm not knocking Bose's support. My headset is a secondhand one that I bought off E-bay years ago (they had the original 9V battery box) that I later upgraded to the current AA battery box. So they're easily years outside of any sane warranty period. I did call Bose's customer service number - and that's where I got the $175 "fix anything" quote. Not an unreasonable price if there was something more substantial wrong with them... but well beyond what I want to pay so that they'll swap out a$5 part.
7. Nov 14, 2007
### Sticky1
#### Member
Joined:
Nov 13, 2007
Messages:
23
0
I sat on and broke mine a year ago. I called bose and asked what I should do......They said send them back....I did.....Freebeeeeeeeeee.....
No Questions asked.....other than name and address, no warrenty card was sent in..
8. Nov 14, 2007
### Tony F.
#### New Member
Joined:
Apr 8, 2007
Messages:
3
0
Location:
Victoria, Australia
I'm not sure if it will work, but the cement used by plumbers for plastic piping might work. It's based on methyl ethyl ketone, which dissolves the surface layer of the pipe and ensures a very good bond. If your plastic is softened by MEK this should work. The music wire idea would be good in conjunction with the cement bond. Be aware that MEK is nasty stuff, so don't get in on your skin. The cured cement is OK, as the MEK has evaporated.
9. Oct 12, 2015
#### New Member
Joined:
Oct 12, 2015
Messages:
1
0
Location:
LAURINBURG NC
Hi, I have been trying to find parts for my two Bose head sets!! I have sent them back twice and only one time the warranty paid! Now two brackets are broken on one and one on another one! I called and they said it would cost \$275.00 for each one to fix !!! I think like Volkswagen they need to have a recall on their crap!
10. Oct 12, 2015
### don january
#### Well-Known Member
Joined:
Feb 11, 2015
Messages:
2,628
1,042
Location:
Midwest
Try these it may make it work again.......
11. Oct 13, 2015
### Turd Ferguson
#### Well-Known Member
Joined:
Mar 14, 2008
Messages:
4,727
1,663
Location:
Upper midwest in a house
Correct. Just switch to a different type of adhesive. Not all epoxy sticks to plastic. For example, epoxy won't stick to SMC used for many jetski hulls. But I found some purpose blended epoxy that not only sticks, it is amazingly strong.
12. Oct 14, 2015
Joined:
Aug 27, 2014
Messages:
823 | 2019-10-14 18:26:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.322556734085083, "perplexity": 3308.7891419299453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00263.warc.gz"} |
http://www.ams.org/mathscinet-getitem?mr=2215135 | MathSciNet bibliographic data MR2215135 (2006k:46097) 46L10 (37A20) Popa, Sorin On a class of type ${\rm II}\sb 1$${\rm II}\sb 1$ factors with Betti numbers invariants. Ann. of Math. (2) 163 (2006), no. 3, 809–899. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews. | 2015-10-09 02:48:48 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9974934458732605, "perplexity": 5695.879228371701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737913039.71/warc/CC-MAIN-20151001221833-00020-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://planetmath.org/graphhomomorphism | # graph homomorphism
We shall define what a graph homomorphism is between two pseudographs, so that a graph homomorphism between two multigraphs, or two graphs are special cases.
Recall that a pseudograph $G$ is an ordered triple $(V,E,i)$, where $V$ is a set called the vertex set of $G$, $E$ is a set called the edge set of $G$, and $i:E\to 2^{V}$ is a map (incidence map), such that for every $e\in E$, $1\leq|i(e)|\leq 2$.
Elements of $V$ are called vertices, elements of $E$ edges. The vertex/vertices associated with any edge $e$ via the map $i$ is/are called the endpoint(s) of $e$. If an edge $e$ has only one endpoint, it is called a loop. $e_{1}$ and $e_{2}$ are said to be parallel if $i(e_{1})=i(e_{2})$.
As two examples, a multigraph is a pseudograph such that $|i(e)|=2$ for each edge $e\in E$ (no loops allowed). A graph is a multigraph such that $i$ is one-to-one (no parallel edges allowed).
Definition. Given two pseudographs $G_{1}=(V_{1},E_{1},i_{1})$ and $G_{2}=(V_{2},E_{2},i_{2})$, a graph homomorphism $h$ from $G_{1}$ to $G_{2}$ consists of two functions $f:V_{1}\to V_{2}$ and $g:E_{1}\to E_{2}$, such that
$\displaystyle i_{2}\circ g=f^{*}\circ i_{1}.$ (1)
The function $f^{*}:2^{V_{1}}\to 2^{V_{2}}$ is defined as $f^{*}(S)=\{f(s)\mid s\in S\}$.
A graph isomorphism $h=(f,g)$ is a graph homomorphism such that both $f$ and $g$ are bijections. It is a graph automorphism if $G_{1}=G_{2}$.
Remarks.
• In case when $G_{1}$ and $G_{2}$ are graphs, a graph homomorphism may be defined in terms of a single function $f:V_{1}\to V_{2}$ satisfying the condition (*)
$\{v_{1},v_{2}\}$ is an edge of $G_{1}\Longrightarrow\{f(v_{1}),f(v_{2})\}$ is an edge of $G_{2}$.
To see this, suppose $h=(f,g)$ is a graph homomorphism from $G_{1}$ to $G_{2}$. Then $f^{*}i_{1}(e)=f^{*}(\{v_{1},v_{2}\})=\{f(v_{1}),f(v_{2})\}=i_{2}(g(e))$. The last equation says that $f(v_{1})$ and $f(v_{2})$ are endpoints of $g(e)$. Conversely, suppose $f:V_{1}\to V_{2}$ is a function sastisfying condition (*). For each $e\in E_{1}$ with end points $\{v_{1},v_{2}\}$, let $e^{\prime}\in E_{2}$ be the edge whose endpoints are $\{f(v_{1}),f(v_{2})\}$. There is only one such edge because $G_{2}$ is a graph, having no parallel edges. Define $g:E_{1}\to E_{2}$ by $g(e)=e^{\prime}$. Then $g$ is a well-defined function which also satisfies Equation (1). So $h:=(f,g)$ is a graph homomorphism $G_{1}\to G_{2}$.
• The definition of a graph homomorphism between pseudographs can be analogously applied to one between directed pseudographs. Since the incidence map $i$ now maps each edge to an ordered pair of vertices, a graph homomorphism thus defined will respect the “direction” of each edge. For example,
$\xymatrix{a\ar[r]&b\ar[d]\\ c\ar[u]&d\ar[l]}\qquad\qquad\qquad\qquad\xymatrix{r\ar[r]&s\\ t\ar[u]\ar[r]&u\ar[u]}$
are two non-isomorphic digraphs whose underlying graphs are isomorphic. Note that one digraph is strongly connected, while the other is only weakly connected.
Title graph homomorphism Canonical name GraphHomomorphism Date of creation 2013-03-22 16:04:53 Last modified on 2013-03-22 16:04:53 Owner CWoo (3771) Last modified by CWoo (3771) Numerical id 11 Author CWoo (3771) Entry type Definition Classification msc 05C75 Classification msc 05C60 Synonym graph automorphism Related topic GraphIsomorphism Related topic Pseudograph Related topic Multigraph Related topic Graph | 2019-09-20 18:57:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 61, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9763948321342468, "perplexity": 278.9551265580793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574058.75/warc/CC-MAIN-20190920175834-20190920201834-00233.warc.gz"} |
http://physics.stackexchange.com/questions/35608/3-inertial-frames-compared-in-str | # 3 inertial frames compared in STR
Suppose an event is observed in 3 inertial frames K, K' and K''. The coordinates in K are $(x,t)$ in K' are $(x',t')$ in K'' are $(x'',t'')$. The K' and K'' coordinates are then Lorentz-transformed to K-coordinates: $(x',t') \rightarrow (x_1, t_1)$ and $(x'',t'') \rightarrow (x_2, t_2)$. Now there are 2 sets of coordinates in K for the same event and they are proably different.
$x_1 = \gamma_1(x' + v_1t') \neq \gamma_2(x'' + v_2t'') = x_2$?
What is the correct interpretation?
Note that it is not allowed to assume that e.g. $x'$ and $x''$ are transformations of $x, t$.
-
No, $(x_1, t_1)=(x_2, t_2)=(x,t)$. – Raskolnikov Sep 4 '12 at 13:00
Is it really so, then why, zee edit? – Gerard Sep 4 '12 at 20:40
Why don't you check instead of saying they are "probably" different. They aren't different, and if you check you will see this, and save people time. – Ron Maimon Sep 5 '12 at 1:30
I think I understand what you're asking: if I've misunderstood you ignore the rest of this answer.
Suppose we're in the frame K and we observe the event at $(x, t)$. To find out what this looks like to the observer in the K' frame we just apply the Lorentz transform to get:
$$x' = \gamma (x - vt)$$
$$t' = \gamma (t - \frac{vx}{c^2})$$
But now let's do the reverse as you suggest. Now we're in frame K' and we want to know what the event looks like in frame K: I'll call this $(x_1, t_1)$ as you did. We use the Lorentz transform just as before, but of course this time the velocity is $-v$ because it's in the opposite direction so:
$$x_1 = \gamma (x' + vt')$$
If you substitute for x' and t' you get the expression:
$$x_1 = \frac{x - vt + vt - \frac{v^2x}{c^2}}{1 - \frac{v^2}{c^2}} = x\frac{1 - \frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} = x$$
So if you Lorentz transform from K to K' then Lorentz transform from K' to K you get back to the same point, which is of course hardly surprising. I won't show this is true for $t$ as well because it's a bit messy - I'll leave this as an exercise for the reader.
So to go back to your question. Going from $(x, t)$ to $(x', t')$ and back to $(x_1, t_1)$ will take you back to the same point i.e. $(x, t)$ = $(x_1, t_1)$ and likewise for the K'' frame.
-
Yes your calculation is of course correct: the Lorentz transformations form a group, so it is not surprising. Your interpretation differs from my question, in the respect that you assume x' is transformed from x. I take x' and x'' as starting point and I assume that the transformation of x' to x1 is not equal to the transformation of x'' to x2. – Gerard Sep 4 '12 at 20:21
I guess I'm missing something because I don't understand what you're asking. If it's the same point we're talking about it doesn't matter how and which order you switch between frames, in any one frame you'll always end up with the same result. How could it be otherwise? Can you give a concrete example of where you end up with different representations of the same point? – John Rennie Sep 5 '12 at 5:59
No, I cannot. The Lorentz group of course is 100% consistent. What is bothering me is that it is somehow strange that observed x', x'' result in the same x after transformation, you wouldn't expect that looking at the formulas sec. – Gerard Sep 6 '12 at 8:59
This statement:
$x_1 = \gamma_1(x' + v_1t') \neq \gamma_2(x'' + v_2t'') = x_2$
Note that it is not allowed to assume that e.g. $x'$ and $x''$ are transformations of $x, t$.
is a direct contradiction of this one:
Suppose an event is observed in 3 inertial frames K, K' and K''. The coordinates in K are $(x,t)$ in K' are $(x',t')$ in K'' are $(x'',t'')$
If a single event is observed in these three inertial frames with coordinates as you've specified, then practically by definition, $(x_1,t_1) = (x_2,t_2) = (x,t)$.
-
I can see your point. – Gerard Sep 6 '12 at 9:00
They are relative to each other this is where the special theory of relativity comes in, if you think about it. If they were all observing something like a spacecraft flying at the speed of sound, it would be relative to the different coordinates: $K' \to K_1$ would be seeing something totally different from what $K''\to K_2$ would see.
- | 2015-04-18 17:01:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8103485703468323, "perplexity": 342.035208589947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246635639.11/warc/CC-MAIN-20150417045715-00269-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-6-section-6-6-solving-quadratic-equations-by-factoring-exercise-set-page-458/60 | ## Algebra: A Combined Approach (4th Edition)
Published by Pearson
# Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set: 60
#### Answer
The solution is -11.
#### Work Step by Step
$x^{2}$+22$x^{}$+121=0 $x^{2}$+11$x^{}$+11x+121=0 (x+11)+7(x+11)=0 (x+11)(x+11)=0 x+11=0 or x+11=0 x=-11 or x=-11 The solution is -11. Check Let x=-11 $x^{2}$+22$x^{}$+121=0 $(-11)^{2}$+22(-11)+121=0 121-244+121=0 0=0
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-04-25 05:18:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5195481181144714, "perplexity": 14513.912415596657}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947693.49/warc/CC-MAIN-20180425041916-20180425061916-00222.warc.gz"} |
https://pure.royalholloway.ac.uk/portal/en/publications/proof-complexity-lower-bounds-from-algebraic-circuit-complexity(c3b6f372-4104-43e5-847c-b777cecf7c95)/export.html | Proof complexity lower bounds from algebraic circuit complexity. / Forbes, Michael; Shpilka, Amir; Tzameret, Iddo; Wigderson, Avi.
31st Conference on Computational Complexity (CCC 2016). Vol. 50 Dagstuhl, Germany : Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik, 2016. p. 1-17 (Leibniz International Proceedings in Informatics (LIPIcs); Vol. 50).
Research output: Chapter in Book/Report/Conference proceedingConference contribution
Published
### Standard
Proof complexity lower bounds from algebraic circuit complexity. / Forbes, Michael; Shpilka, Amir; Tzameret, Iddo; Wigderson, Avi.
31st Conference on Computational Complexity (CCC 2016). Vol. 50 Dagstuhl, Germany : Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik, 2016. p. 1-17 (Leibniz International Proceedings in Informatics (LIPIcs); Vol. 50).
Research output: Chapter in Book/Report/Conference proceedingConference contribution
### Harvard
Forbes, M, Shpilka, A, Tzameret, I & Wigderson, A 2016, Proof complexity lower bounds from algebraic circuit complexity. in 31st Conference on Computational Complexity (CCC 2016). vol. 50, Leibniz International Proceedings in Informatics (LIPIcs), vol. 50, Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik, Dagstuhl, Germany, pp. 1-17, Computational Complexity Conference (CCC), Koyto, Japan, 29/05/16. https://doi.org/10.4230/LIPIcs.CCC.2016.32
### APA
Forbes, M., Shpilka, A., Tzameret, I., & Wigderson, A. (2016). Proof complexity lower bounds from algebraic circuit complexity. In 31st Conference on Computational Complexity (CCC 2016) (Vol. 50, pp. 1-17). (Leibniz International Proceedings in Informatics (LIPIcs); Vol. 50). Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik. https://doi.org/10.4230/LIPIcs.CCC.2016.32
### Vancouver
Forbes M, Shpilka A, Tzameret I, Wigderson A. Proof complexity lower bounds from algebraic circuit complexity. In 31st Conference on Computational Complexity (CCC 2016). Vol. 50. Dagstuhl, Germany: Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik. 2016. p. 1-17. (Leibniz International Proceedings in Informatics (LIPIcs)). https://doi.org/10.4230/LIPIcs.CCC.2016.32
### Author
Forbes, Michael ; Shpilka, Amir ; Tzameret, Iddo ; Wigderson, Avi. / Proof complexity lower bounds from algebraic circuit complexity. 31st Conference on Computational Complexity (CCC 2016). Vol. 50 Dagstuhl, Germany : Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik, 2016. pp. 1-17 (Leibniz International Proceedings in Informatics (LIPIcs)).
### BibTeX
@inproceedings{c3b6f372410443e5847cb777cecf7c95,
title = "Proof complexity lower bounds from algebraic circuit complexity",
abstract = "We give upper and lower bounds on the power of subsystems of the Ideal Proof System (IPS), the algebraic proof system recently proposed by Grochow and Pitassi~\cite{GrochowPitassi14}, where the circuits comprising the proof come from various restricted algebraic circuit classes. This mimics an established research direction in the boolean setting for subsystems of Frege proofs whose lines are circuits from restricted boolean circuit classes. Essentially all of the subsystems considered in this paper can simulate the well-studied Nullstellensatz proof system, and prior to this work there were no known lower bounds when measuring proof size by the algebraic complexity of the polynomials (except with respect to degree, or to sparsity).Our main contributions are two general methods of converting certain algebraic lower bounds into proof complexity ones. Both require stronger arithmetic lower bounds than common, which should hold not for a specific polynomial but for a whole family defined by it.These may be likened to some of the methods by which Boolean circuit lower bounds are turned into related proof-complexity ones, especially the feasible interpolation'' technique. We establish algebraic lower bounds of these forms for several explicit polynomials, against a variety of classes, and infer the relevant proof complexity bounds. These yield separations between IPS subsystems, which we complement by simulations to create a partial structure theory for IPS systems.Our first method is a \emph{functional lower bound}, a notion of Grigoriev and Razborov~\cite{GrigorievRazborov00}, which is a function $\hat{f}:\bits^n\to\F$ such that any polynomial $f$ agreeing with $\hat{f}$ on the boolean cube requires large algebraic circuit complexity. We develop functional lower bounds for a variety of circuit classes (sparse polynomials, depth-3 powering formulas, roABPs and multilinear formulas) where $\hat{f}(\vx)$ equals $\nicefrac{1}{p(\vx)}$ for a constant-degree polynomial $p$ depending on the relevant circuit class. We believe these lower bounds are of independent interest in algebraic complexity, and show that they also imply lower bounds for the size of the corresponding IPS refutations for proving that the relevant polynomial $p$ is non-zero over the boolean cube. In particular, we show super-polynomial lower bounds for refuting variants of the subset-sum axioms in these IPS subsystems.Our second method is to give \emph{lower bounds for multiples}, that is, to give explicit polynomials whose all (non-zero) multiples require large algebraic circuit complexity. By extending known techniques, we give lower bounds for multiples for various restricted circuit classes such sparse polynomials, sums of powers of low-degree polynomials, and roABPs. These results are of independent interest, as we argue that lower bounds for multiples is the correct notion for instantiating the algebraic hardness versus randomness paradigm of Kabanets and Impagliazzo~\cite{KabanetsImpagliazzo04}. Further, we show how such lower bounds for multiples extend to lower bounds for refutations in the corresponding IPS subsystem.",
keywords = "complexity theory",
author = "Michael Forbes and Amir Shpilka and Iddo Tzameret and Avi Wigderson",
year = "2016",
month = may,
day = "19",
doi = "10.4230/LIPIcs.CCC.2016.32",
language = "English",
isbn = "978-3-95977-008-8",
volume = "50",
series = "Leibniz International Proceedings in Informatics (LIPIcs)",
publisher = "Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik",
pages = "1--17",
booktitle = "31st Conference on Computational Complexity (CCC 2016)",
note = "Computational Complexity Conference (CCC) ; Conference date: 29-05-2016 Through 01-06-2016",
}
### RIS
TY - GEN
T1 - Proof complexity lower bounds from algebraic circuit complexity
AU - Forbes, Michael
AU - Shpilka, Amir
AU - Tzameret, Iddo
AU - Wigderson, Avi
PY - 2016/5/19
Y1 - 2016/5/19
N2 - We give upper and lower bounds on the power of subsystems of the Ideal Proof System (IPS), the algebraic proof system recently proposed by Grochow and Pitassi~\cite{GrochowPitassi14}, where the circuits comprising the proof come from various restricted algebraic circuit classes. This mimics an established research direction in the boolean setting for subsystems of Frege proofs whose lines are circuits from restricted boolean circuit classes. Essentially all of the subsystems considered in this paper can simulate the well-studied Nullstellensatz proof system, and prior to this work there were no known lower bounds when measuring proof size by the algebraic complexity of the polynomials (except with respect to degree, or to sparsity).Our main contributions are two general methods of converting certain algebraic lower bounds into proof complexity ones. Both require stronger arithmetic lower bounds than common, which should hold not for a specific polynomial but for a whole family defined by it.These may be likened to some of the methods by which Boolean circuit lower bounds are turned into related proof-complexity ones, especially the feasible interpolation'' technique. We establish algebraic lower bounds of these forms for several explicit polynomials, against a variety of classes, and infer the relevant proof complexity bounds. These yield separations between IPS subsystems, which we complement by simulations to create a partial structure theory for IPS systems.Our first method is a \emph{functional lower bound}, a notion of Grigoriev and Razborov~\cite{GrigorievRazborov00}, which is a function $\hat{f}:\bits^n\to\F$ such that any polynomial $f$ agreeing with $\hat{f}$ on the boolean cube requires large algebraic circuit complexity. We develop functional lower bounds for a variety of circuit classes (sparse polynomials, depth-3 powering formulas, roABPs and multilinear formulas) where $\hat{f}(\vx)$ equals $\nicefrac{1}{p(\vx)}$ for a constant-degree polynomial $p$ depending on the relevant circuit class. We believe these lower bounds are of independent interest in algebraic complexity, and show that they also imply lower bounds for the size of the corresponding IPS refutations for proving that the relevant polynomial $p$ is non-zero over the boolean cube. In particular, we show super-polynomial lower bounds for refuting variants of the subset-sum axioms in these IPS subsystems.Our second method is to give \emph{lower bounds for multiples}, that is, to give explicit polynomials whose all (non-zero) multiples require large algebraic circuit complexity. By extending known techniques, we give lower bounds for multiples for various restricted circuit classes such sparse polynomials, sums of powers of low-degree polynomials, and roABPs. These results are of independent interest, as we argue that lower bounds for multiples is the correct notion for instantiating the algebraic hardness versus randomness paradigm of Kabanets and Impagliazzo~\cite{KabanetsImpagliazzo04}. Further, we show how such lower bounds for multiples extend to lower bounds for refutations in the corresponding IPS subsystem.
AB - We give upper and lower bounds on the power of subsystems of the Ideal Proof System (IPS), the algebraic proof system recently proposed by Grochow and Pitassi~\cite{GrochowPitassi14}, where the circuits comprising the proof come from various restricted algebraic circuit classes. This mimics an established research direction in the boolean setting for subsystems of Frege proofs whose lines are circuits from restricted boolean circuit classes. Essentially all of the subsystems considered in this paper can simulate the well-studied Nullstellensatz proof system, and prior to this work there were no known lower bounds when measuring proof size by the algebraic complexity of the polynomials (except with respect to degree, or to sparsity).Our main contributions are two general methods of converting certain algebraic lower bounds into proof complexity ones. Both require stronger arithmetic lower bounds than common, which should hold not for a specific polynomial but for a whole family defined by it.These may be likened to some of the methods by which Boolean circuit lower bounds are turned into related proof-complexity ones, especially the feasible interpolation'' technique. We establish algebraic lower bounds of these forms for several explicit polynomials, against a variety of classes, and infer the relevant proof complexity bounds. These yield separations between IPS subsystems, which we complement by simulations to create a partial structure theory for IPS systems.Our first method is a \emph{functional lower bound}, a notion of Grigoriev and Razborov~\cite{GrigorievRazborov00}, which is a function $\hat{f}:\bits^n\to\F$ such that any polynomial $f$ agreeing with $\hat{f}$ on the boolean cube requires large algebraic circuit complexity. We develop functional lower bounds for a variety of circuit classes (sparse polynomials, depth-3 powering formulas, roABPs and multilinear formulas) where $\hat{f}(\vx)$ equals $\nicefrac{1}{p(\vx)}$ for a constant-degree polynomial $p$ depending on the relevant circuit class. We believe these lower bounds are of independent interest in algebraic complexity, and show that they also imply lower bounds for the size of the corresponding IPS refutations for proving that the relevant polynomial $p$ is non-zero over the boolean cube. In particular, we show super-polynomial lower bounds for refuting variants of the subset-sum axioms in these IPS subsystems.Our second method is to give \emph{lower bounds for multiples}, that is, to give explicit polynomials whose all (non-zero) multiples require large algebraic circuit complexity. By extending known techniques, we give lower bounds for multiples for various restricted circuit classes such sparse polynomials, sums of powers of low-degree polynomials, and roABPs. These results are of independent interest, as we argue that lower bounds for multiples is the correct notion for instantiating the algebraic hardness versus randomness paradigm of Kabanets and Impagliazzo~\cite{KabanetsImpagliazzo04}. Further, we show how such lower bounds for multiples extend to lower bounds for refutations in the corresponding IPS subsystem.
KW - complexity theory
U2 - 10.4230/LIPIcs.CCC.2016.32
DO - 10.4230/LIPIcs.CCC.2016.32
M3 - Conference contribution
SN - 978-3-95977-008-8
VL - 50
T3 - Leibniz International Proceedings in Informatics (LIPIcs)
SP - 1
EP - 17
BT - 31st Conference on Computational Complexity (CCC 2016)
PB - Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik
CY - Dagstuhl, Germany
T2 - Computational Complexity Conference (CCC)
Y2 - 29 May 2016 through 1 June 2016
ER - | 2022-01-22 00:33:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8537077307701111, "perplexity": 2007.8162402965297}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00603.warc.gz"} |
https://www.physicsforums.com/threads/divisibility-question.298466/ | # Divisibility question
1. Mar 9, 2009
### kreil
1. The problem statement, all variables and given/known data
Suppose that $n^2$ is odd and that 3 does not divide $n^2$. Show that $12|(n^2-1)$
2. Relevant equations
none
3. The attempt at a solution
Well I know that since $n^2$ is odd, $n^2-1$ is even. I'm not sure what the next step would be.
2. Mar 9, 2009
### kreil
hmm i did some more thinking on this. Please let me know if you think this is correct:
- since $n^2$ is odd, $n^2-1$ is an even number and $2|n^2-1$
- $n^2$ is not prime since it has at least 3 divisors: {1, n, $n^2$}
- if 3 does not divide $n^2$, then the remainder of this division must be either 1 or 2 (a remainder of 3 implies that 3 does divide $n^2$). However, the remainder cannot be 2 or else $n^2$ would be prime, therefore the remainder must be 1.
- this implies 3|$n^2-1$ evenly, since the -1 cancels with the remainder on division with $n^2$.
Since 2|$n^2-1$, 3|$n^2-1$ and $12=2^2 3^1$, it follows that 12|$n^2-1$.
3. Mar 9, 2009
### tiny-tim
Hi kreil!
forget n2
… what is the remainder if you divide n by 2 or 3?
4. Mar 9, 2009
### kreil
n is odd
so 2 divides into n with r=1
3 divides into n with r=2 since r=1 implies n is even and r=0 or 3 implies 3 divides $n^2$
soo does this mean that 3 divides into $n^2$ with remainder $r^2$ (i.e. r=1)?
5. Mar 9, 2009
### tiny-tim
No it doesn't …
6. Mar 9, 2009
### kreil
oops, so 3|n with r=1 or r=2
what does this mean?
7. Mar 9, 2009
### tiny-tim
that's better!
now you have the possible remainders for 2 and 3 …
so … ?
8. Mar 9, 2009
### kreil
so if 3|n with r=1 or r=2, then 3|$n^2$ with r=1 or r=4 and since if r=4 3|4 with r=1, 3|$n^2$ with r=1
is that correct logic?
9. Mar 9, 2009
### tiny-tim
wot's incorrect logic?
anyway, how can 3 divide anything with r = 4??
10. Mar 9, 2009
### kreil
haha, this doesn't tell me very much about whether what i wrote was right or not...
is it true that if 3|n with remainder r, that 3|$n^2$ with remainder $r^2$??
11. Mar 9, 2009
### kreil
i now know that this problem reduces to the problem of showing that 3|$n^2$ with r=1, i just don't understand how to connect division of n by 2 and 3 to this..
12. Mar 9, 2009
### tiny-tim
Hint: 6
and i'm going to bed :zzz:
13. Mar 9, 2009
### kreil
i dont understand how to make the next step to saying 6|n rem 1..
i think i'm missing something here
14. Mar 9, 2009
### kreil
for ex if n is 11,
2|11 rem 1
3|11 rem 2
6|11 rem 5...
BUT 6|121 rem 1
15. Mar 9, 2009
### kreil
i am so lost on this now...
16. Mar 10, 2009
### tiny-tim
just got up :zzz: …
what are the possible remainders on dividing by 6?
17. Mar 10, 2009
### kreil
because of the condition 0<r<6 for the equation n = 6b + r for some b, r can be 1,2,3,4,5.
18. Mar 10, 2009
### tiny-tim
uhh? but n2 is odd and 3 does not divide n2.
19. Mar 10, 2009
### matt grime
tiny tim is pointing you at the Chinese remainder theorem.
But let's try a different way.
n is odd, so n=2m+1 for some m. Now what divides n^2-1?
n is not divisible by 3, so either n=3r+1 or 3r+2. In either case, what can you show divides n^2-1?
Equivalently, n^2-1 is congruent to what mod what and what? (insert useful things instead of 'what' in that sentence.)
20. Mar 10, 2009
### kreil
$n^2=(2m+1)(2m+1)=4m^2+4m+1 \implies n^2-1=4(m^2+m)$ for some m, and 4|$n^2-1$
$n^2=9r^2+6r+1$
or $n^2=9r^2+6r+4$
so $n^2-1=9r^2+6r=3(3r^2+2r)$
or $n^2-1=9r^2+6r+3=3(3r^2+2r+1)$
and in both cases 3|$n^2-1$
so $$n^2-1=(3r^2+2r)(mod 3)$$ and $$n^2-1=(3r^2+2r+1)(mod 3)$$?? | 2019-03-19 05:59:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5842970013618469, "perplexity": 2979.1373300842074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201904.55/warc/CC-MAIN-20190319052517-20190319074517-00538.warc.gz"} |
https://www.meritnation.com/ask-answer/question/which-has-more-number-of-atoms-100g-of-n2-or-100g-of-nh3-sri/atoms-and-molecules/6799908 | # Which has more number of atoms? 100g of N2 or 100g of NH3 srisaji@yahoo.com
Number of moles is determine by the formulae :-
n = given mass/ molar mass
Where n stands for number of moles
given mass for both case is 100 g
Molecular mass of N2 is (14 + 14) = 28 g per mole
Molecular mass of NH3 is {14 +(3$×$1)} = 17 g per mole
As Atomic mass of Nitrogen is 14 and atomic mass for hydrogen is 1 g per mole.
1. number of mole of N2 = 100/ 28 = 3.57 mole
As 1 molecule of dinitrogen contain 2 atoms,
Therefore number of atoms present in 3.57 mole of nitrogen is
3.57 $×$2
= 7.14 mole
2. Number of mole of NH3 = 100/17
= 5.88
As 1 molecule of ammonia contain = 3 atoms
Therefore,
5.88 moles of ammonia contain = 3 $×$ 5.88
= 17.65 atoms
Hence 100g of ammonia contain more number of atoms.
• 30
BOTH WILL BE SAME AS NO OF ATOMS WILL BE EQUAL TO 6.022*10 TO THE POWER 23
• -6
Both will be the same.
• -7
What are you looking for? | 2022-01-18 21:41:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3203390836715698, "perplexity": 5370.545969381451}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00646.warc.gz"} |
http://openstudy.com/updates/507265c0e4b057a2860d1734 | ## anonymous 3 years ago Ugh another story problem. Hate these!
1. anonymous
At a pet show the ratio of dogs to cats is 4:3 if the number of cats is 45 find the number of dogs at the show?
2. anonymous
story problems are generally more easier.
3. anonymous
:/ not for me
4. anonymous
its on you how you interpret.
5. Shane_B
dogs to cats is 4:3 so cats to dogs is 3:4 Knowing that you can just relate the ratios:$\frac{3}{4}=\frac{45}{x}$
6. anonymous
let the common ratio be x. so no. of dogs = 4x no. of cats = 3x now , it is given that no. of cats = 45 = 3x find x. find no. of dogs.
7. anonymous
would it b 60?
8. anonymous
yes
9. Shane_B
I guess you could skip the first step to get the same answer:$\frac{4}{3}=\frac{x}{45}$
10. Shane_B
Yes. 60.
11. anonymous
thanks! | 2016-05-25 07:24:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2612973153591156, "perplexity": 5091.068508204096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049274191.57/warc/CC-MAIN-20160524002114-00062-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.2/8.2.2.3 | # 8.2.2.3 - Computing Necessary Sample Size
Calculating the sample size necessary for estimating a population mean with a given margin of error and level of confidence is similar to that for estimating a population proportion. However, since the $$t$$ distribution is not as “neat” as the standard normal distribution, the process can be iterative. (Recall, the shape of the $$t$$ distribution is different for each degree of freedom). This means that we would solve, reset, solve, reset, etc. until we reached a conclusion. Yet, we can avoid this iterative process if we employ an approximate method based on $$t$$ distribution approaching the standard normal distribution as the sample size increases. This approximate method invokes the following formula:
Finding the Sample Size for Estimating a Population Mean
$$n=\frac{z^{2}\widetilde{\sigma}^{2}}{M^{2}}=\left ( \frac{z\widetilde{\sigma}}{M} \right )^2$$
$$z$$ = z multiplier for given confidence level
$$\widetilde{\sigma}$$ = estimated population standard deviation
$$M$$ = margin of error
The sample standard deviation may be estimated on the basis of prior research studies. | 2020-01-24 23:53:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8767096996307373, "perplexity": 504.88512383905265}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00522.warc.gz"} |
http://clic.cimec.unitn.it/composes/toolkit/composes.similarity.html | # similarity Package¶
## cos Module¶
Created on Oct 2, 2012
class composes.similarity.cos.CosSimilarity
Computes the cosine similarity of two vectors.
$$sim(\vec{u},\vec{v}) = \frac{<\vec{u},\vec{v}>}{\sqrt{||\vec{u}||||\vec{v}||}}$$
## dot_prod Module¶
Created on Oct 2, 2012
class composes.similarity.dot_prod.DotProdSimilarity
Computes the scalar product (dot product) of two vectors.
$$sim(\vec{u},\vec{v}) = <\vec{u},\vec{v}> = \sum_iu_iv_i$$
## euclidean Module¶
Created on Oct 2, 2012
class composes.similarity.euclidean.EuclideanSimilarity
Computes the euclidean similarity of two vectors as the inverse of their euclidean distance.
$$sim(\vec{u},\vec{v}) = \frac{1}{||\vec{u}-\vec{v}|| + 1}$$
## lin Module¶
Created on Oct 2, 2012
class composes.similarity.lin.LinSimilarity
Computes the Lin similarity of two vectors.
$$sim(\vec{u},\vec{v}) = \frac{\sum_{i \in I}(u_i+v_i)}{\sum_iu_i + \sum_iv_i}$$
Where $$I=\{i | u_i > 0 \text{ and } v_i > 0\}$$, the set of components on which both vectors are strictly positive.
## similarity Module¶
Created on Oct 2, 2012 | 2018-03-24 19:42:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8001500964164734, "perplexity": 13809.372203949524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257650993.91/warc/CC-MAIN-20180324190917-20180324210917-00523.warc.gz"} |
http://wulixb.iphy.ac.cn/en/article/doi/10.7498/aps.68.20182302 | x
## 留言板
Influences of nanoscale particles and interparticle compression in electrodes on voltage hysteresis of lithium ion batteries
## Influences of nanoscale particles and interparticle compression in electrodes on voltage hysteresis of lithium ion batteries
Peng Jie-Yang, Wang Jia-Hai, Shen Bin, Li Hao-Liang, Sun Hao-Ming
PDF
HTML
• #### Abstract
As one of high capacity electrode materials of lithium ion battery, silicon suffers significant stress effects, which further affects the voltage performance of battery. In this paper, a reaction-diffusion-stress coupled model is established, and the stress induced voltage hysteresis with consideration of diffusion induced stress, surface effects and interparticle compression under potentiostatic operation are investigated. It is found that stress and stress induced voltage hysteresis are dependent on particle size. For big particles, the diffusion induced stress is dominant and further aggravates the hysteresis of both stress and the overpotential consumed by it, indicating that more energy dissipates due to the stress effects. For small particles, especially ones with radius of a few nanometers, surface effects play a more prominent role than diffusion induced stress and the stress evolves into the state of compressive stress on the whole, leading the hysteresis of overpotential to be consumed by stress shrink and making the hysteresis plot of overpotential used to drive electrochemical reaction move downward. The electrode potential first reaches a cutoff voltage and finally the capacity of lithium ion battery decays. Therefore, too large or too small particle size in the electrode can both have a negative effect on the performance of lithium ion batteries, which indicates that an optimal size of the electrode particles must be designed in terms of electrode structure. Based on the calculation, particles with around 9 nm in radius are an appropriate option for electrode design in consideration of both diffusion induced stress and surface effect. In addition, for silicon electrodes, the silicon particles inevitably squeeze each other in a charge and discharge cycle. Therefore, interparticle compression is considered in this case. In detail, interparticle compression pushes the plot of stress hysteresis to the compressive state and leads to lower lithiation capacity, which makes the overpotential plot consumed by stress move downward and accordingly the overpotential plot used to drive the electrochemical reaction move upward. Denser electrode would strengthen this effect due to higher particle compression. It is indicated that for electrode design, the minimum of porosity ratio of electrodes should be adopted because higher interparticle compressive stress would reduce the battery capacity. Our results reveal that the voltage hysteresis of lithium ion batteries is related to the active particle size and the porosity ratio of the electrode, which is of great significance for guiding one in designing the lithium ion batteries.
#### References
[1] Konishi H, Hirano T, Takamatsu D, Gunji A, Feng X L, Furutsuki S 2015 J. Power Sources 298 144 [2] Shi S Q, Gao J, Liu Y, Zhao Y, Wu Q, Ju W W, Ouyang C Y, Xiao R J 2016 Chin. Phys. B. 25 18212 [3] Yang L, Chen H S, Jiang H Q, Wei Y J, Song W L, Fang D N 2018 Chem. Commun. 54 3997 [4] McDowell M T, Ryu L, Lee S W, Wang C, Nix W D, Cui Y 2012 Adv. Mater. 24 6034 [5] Liu X H, Fan F F, Yang H, Zhang S, Huang J Y, Zhu T 2013 ACS Nano. 7 1495 [6] Gu M, Yang H, Perea D E, Zhang J G, Zhang S, Wang C M 2014 Nano Lett. 14 4622 [7] Kim S, Choi S J, Zhao K J, Yang H, Gobbi G, Zhang S, Li J 2016 Nat. Commun. 7 10146 [8] Piper D M, Yersak T A, Lee S H 2013 J. Electrochem. Soc. 160 A77 [9] Sethuraman V A, Chon M J, Shimshak M, Srinivasan V, Guduru P R 2010 J. Power Sources 195 5062 [10] Bower A F, Guduru P R, Sethuraman V A 2011 J. Mech. Phys. Solids 59 804 [11] Sethuraman V A, Srinivasan V, Bower A F, Guduru P R 2010 J. Electrochem. Soc. 157 A1253 [12] Lu B, Song Y C, Zhang Q L, Pan J, Cheng Y T, Zhang J Q 2016 Phys. Chem. Chem. Phys. 18 4721 [13] Song Y C, Soh A K, Zhang J Q 2016 J. Mater. Sci. 51 9902 [14] Magasinski A, Dixon P, Hertzberg B, Kvit A, Ayala J, Yushin G 2010 Nat. Mater. 9 353 [15] Meng Q P, Wu L J, Welch D O, Tang M, Zhu Y M 2018 Sci. Rep. 8 4396 [16] Cheng Y T, Verbrugge M W 2008 J. Appl. Phys. 104 083521 [17] Hao F, Gao Z, Fang D N 2012 J. Appl. Phys. 112 103507 [18] Haftbaradaran H, Song J, Curtin W A, Gao H J 2011 J. Power Sources 196 361 [19] Sethuraman V A, Srinivasan V, Newman J 2013 J. Electrochem. Soc. 160 A394 [20] Cheng Y T, Verbrugge M W 2009 J. Power Sources 190 453 [21] Stein P, Zhao Y, Xu B X 2016 J. Power Sources 332 154 [22] [23] Cammarata R C 1994 Prog. Surf. Sci. 46 1 [24] Müller P, Saúl A 2004 Surf. Sci. Rep. 54 157 [25] Rusanov A I 2005 Surf. Sci. Rep. 58 111 [26] Fischer F D, Waitz T, VollathD, Simha N K 2008 Prog. Mater. Sci. 53 481 [27] Gurtin M E, Murdoch A I 1975 Arch. Ration. Mech. An. 57 291 [28] Gurtin M E, Murdoch A I 1978 Int. J. Solids Struct. 14 431 [29] Gurtin M E, Weissmüller J, Larché F 1998 Philos. Mag. A 78 1093 [30] Sharma P, Ganti S, Bhate N 2003 Appl. Phys. Lett. 82 535 [31] Miller R E, Shenoy V B 2000 Nanotechnology 11 139 [32] Luo L, Zhao P, Yang H, Liu B, Zhang J G, Cui Y, Wang C M 2015 Nano Lett. 15 7016 [33] Bucci G, Swamy T, Bishop S, Sheldon B W, Chiang Y M, Carter W C 2017 J. Electrochem. Soc. 164 A645 [34] Roberts A P, Garboczi E J 2000 J. Am. Ceram. Soc. 83 3041 [35] Qi W, Shapter J G, Wu Q, Yin T, Gao G, Cui D 2017 J Mater Chem. A. 5 19521 [36] Roy P, Srivastava S K 2015 J. Mater. Chem. A 3 2454 [37] Vitos L, Ruban A V, Skriver H L 1998 Surf. Sci. 411 186 [38] Yang B, He YP, Irsa J, Lundgren C A, Ratchford J B, Zhao Y P 2012 J. Power Sources 204 168 [39] Gwak Y, Moon J, Cho M 2016 J. Power Sources 307 856 [40] Ding N, Xu J, Yao Y X, Wegner G, Fang X, Chen C H, Lieberwirth 2009 Solid State Ionics 180 222 [41] Ying Z, Peter S, Yang B, Mamun A S, Yangyiwei Y, Xua B X 2019 J. Power Sources 413 259
#### Cited By
• 图 1 锂离子电池中的多层电极结构
Figure 1. Multi-scale hierarchical electrode structure in lithium ion battery.
图 2 Sethuraman等[19]硅电极开路电势的拟合结果和实验结果, 其中红色的数据点是实验测得的锂化数据, 蓝色数据点是实验测得的去锂化数据, 实线是在C/8恒流充放电条件下的实验曲线, 虚线是依据实验数据拟合的结果
Figure 2. Fitting results and experimental results of open-circuit potential of silicon electrode proposed by Sethuraman et al.[19]. The red data point is the lithium data measured in the experiment, and the blue data point is the dilithiated data measured in the experiment. The solid line was obtained under a C/8 constant current charge-discharge operation, and the dashed line is the fitting function
图 3 扩散诱导应力对电压迟滞的影响 (a)电极颗粒在锂化和去锂化过程中锂离子的浓度分布; (b)一次充放电循环中, 不同尺寸的颗粒扩散诱导应力演化
Figure 3. Effect of diffusion induced stress on voltage hysteresis: (a) Distribution of concentration of lithium ions in a electrode particle during lithiation and delithiation; (b) diffusion-induced stress evolution diagrams of particles with different sizes in primary charge-discharge cycle
图 4 (a)−(d)分别是颗粒尺寸为10, 100, 400, 700 nm的各部分过电势演化图
Figure 4. (a)−(d) Overpotential evolution charts of each part with particle size of 10, 100, 400, 700 nm, respectively
图 5 应力分担过电势的差值、总过电势差值及其所占百分比在不同颗粒尺寸下的变化, 其中差值是指过电势回线中最大值与最小值之差
Figure 5. Dependence of overpotential gap consumed by stress, total overpotential gap and the corresponding percentage on different particle sizes. The gap refers to the difference between the maximum value and the minimum value in the overpotential loop
图 6 由表面效应引起的表面张力及其分担过电势的演化曲线, 其中黑色曲线的应力值为一次充放电循环中的平均应力值, 竖直曲线为各自的变化范围; 3张子图为对应的表面张力演化曲线
Figure 6. Evolution curve of surface tension due to surface effect and the corresponding overpotential. The dark line represents the mean stress in a cycle and the bar defines the range. The three subplots are evolutions of surface stress due to surface effects in a cycle
图 7 (a)不同颗粒尺寸下, 表面张力和扩散诱导应力共同作用下的表面静水应力演化图; (b), (c), (d), (e)分别是颗粒尺寸为4, 10, 100和400 nm时的各部分过电势演化图
Figure 7. (a) Evolution of surface hydrostatic stresses in consideration of both diffusion induced stress and surface effects under different particle sizes; (b), (c), (d), (e) evolutions of all parts of overpotential in the particles of 4, 10, 100 and 400 nm radius, respectively
图 8 不同颗粒尺寸下扩散诱导应力和表面效应引起的静水应力的绝对值对比
Figure 8. Absolute values of the hydrostatic stress due to surface effects and the surface stress due to diffusion induced stress under different particle sizes.
图 9 不同颗粒尺寸下扩散诱导应力与表面效应之和的演化
Figure 9. Evolution diagram of the sum of diffusion induced stress and surface effect under different particle sizes
图 10 颗粒间挤压对电压迟滞的影响 (a) 不同孔隙下电极的应力迟滞回线图; (b)−(e) 不同孔隙率下电极各部分过电势的回线图; 其中p为电池结构的孔隙率
Figure 10. Impacts of interparticle compression on the voltage hysteresis: (a) Stress hysteresis for electrodes with different porosity ratios; (b)−(e) loop diagram of all parts of overpotential in the electrode with different porosity ratios. p is the porosity of the electrode structure.
• [1] Konishi H, Hirano T, Takamatsu D, Gunji A, Feng X L, Furutsuki S 2015 J. Power Sources 298 144 [2] Shi S Q, Gao J, Liu Y, Zhao Y, Wu Q, Ju W W, Ouyang C Y, Xiao R J 2016 Chin. Phys. B. 25 18212 [3] Yang L, Chen H S, Jiang H Q, Wei Y J, Song W L, Fang D N 2018 Chem. Commun. 54 3997 [4] McDowell M T, Ryu L, Lee S W, Wang C, Nix W D, Cui Y 2012 Adv. Mater. 24 6034 [5] Liu X H, Fan F F, Yang H, Zhang S, Huang J Y, Zhu T 2013 ACS Nano. 7 1495 [6] Gu M, Yang H, Perea D E, Zhang J G, Zhang S, Wang C M 2014 Nano Lett. 14 4622 [7] Kim S, Choi S J, Zhao K J, Yang H, Gobbi G, Zhang S, Li J 2016 Nat. Commun. 7 10146 [8] Piper D M, Yersak T A, Lee S H 2013 J. Electrochem. Soc. 160 A77 [9] Sethuraman V A, Chon M J, Shimshak M, Srinivasan V, Guduru P R 2010 J. Power Sources 195 5062 [10] Bower A F, Guduru P R, Sethuraman V A 2011 J. Mech. Phys. Solids 59 804 [11] Sethuraman V A, Srinivasan V, Bower A F, Guduru P R 2010 J. Electrochem. Soc. 157 A1253 [12] Lu B, Song Y C, Zhang Q L, Pan J, Cheng Y T, Zhang J Q 2016 Phys. Chem. Chem. Phys. 18 4721 [13] Song Y C, Soh A K, Zhang J Q 2016 J. Mater. Sci. 51 9902 [14] Magasinski A, Dixon P, Hertzberg B, Kvit A, Ayala J, Yushin G 2010 Nat. Mater. 9 353 [15] Meng Q P, Wu L J, Welch D O, Tang M, Zhu Y M 2018 Sci. Rep. 8 4396 [16] Cheng Y T, Verbrugge M W 2008 J. Appl. Phys. 104 083521 [17] Hao F, Gao Z, Fang D N 2012 J. Appl. Phys. 112 103507 [18] Haftbaradaran H, Song J, Curtin W A, Gao H J 2011 J. Power Sources 196 361 [19] Sethuraman V A, Srinivasan V, Newman J 2013 J. Electrochem. Soc. 160 A394 [20] Cheng Y T, Verbrugge M W 2009 J. Power Sources 190 453 [21] Stein P, Zhao Y, Xu B X 2016 J. Power Sources 332 154 [22] [23] Cammarata R C 1994 Prog. Surf. Sci. 46 1 [24] Müller P, Saúl A 2004 Surf. Sci. Rep. 54 157 [25] Rusanov A I 2005 Surf. Sci. Rep. 58 111 [26] Fischer F D, Waitz T, VollathD, Simha N K 2008 Prog. Mater. Sci. 53 481 [27] Gurtin M E, Murdoch A I 1975 Arch. Ration. Mech. An. 57 291 [28] Gurtin M E, Murdoch A I 1978 Int. J. Solids Struct. 14 431 [29] Gurtin M E, Weissmüller J, Larché F 1998 Philos. Mag. A 78 1093 [30] Sharma P, Ganti S, Bhate N 2003 Appl. Phys. Lett. 82 535 [31] Miller R E, Shenoy V B 2000 Nanotechnology 11 139 [32] Luo L, Zhao P, Yang H, Liu B, Zhang J G, Cui Y, Wang C M 2015 Nano Lett. 15 7016 [33] Bucci G, Swamy T, Bishop S, Sheldon B W, Chiang Y M, Carter W C 2017 J. Electrochem. Soc. 164 A645 [34] Roberts A P, Garboczi E J 2000 J. Am. Ceram. Soc. 83 3041 [35] Qi W, Shapter J G, Wu Q, Yin T, Gao G, Cui D 2017 J Mater Chem. A. 5 19521 [36] Roy P, Srivastava S K 2015 J. Mater. Chem. A 3 2454 [37] Vitos L, Ruban A V, Skriver H L 1998 Surf. Sci. 411 186 [38] Yang B, He YP, Irsa J, Lundgren C A, Ratchford J B, Zhao Y P 2012 J. Power Sources 204 168 [39] Gwak Y, Moon J, Cho M 2016 J. Power Sources 307 856 [40] Ding N, Xu J, Yao Y X, Wegner G, Fang X, Chen C H, Lieberwirth 2009 Solid State Ionics 180 222 [41] Ying Z, Peter S, Yang B, Mamun A S, Yangyiwei Y, Xua B X 2019 J. Power Sources 413 259
• [1] Peng Ying-Zha, Zhang Kai, Zheng Bai-Lin, Li Yong. Stress analysis of a cylindrical composition-gradient electrode of lithium-ion battery in generalized plane strain condition. Acta Physica Sinica, 2016, 65(10): 100201. doi: 10.7498/aps.65.100201 [2] Liang Jin-Jie, Gao Ning, Li Yu-Hong. Surface effect on \begin{document}${\langle 100 \rangle }$\end{document} interstitial dislocation loop in iron. Acta Physica Sinica, 2020, 69(3): 036101. doi: 10.7498/aps.69.20191379 [3] Hou Xian-Hua, Hu She-Jun, Shi Lu. Preparation and properties of Sn-Ti alloy anode material for lithium ion batteries. Acta Physica Sinica, 2010, 59(3): 2109-2113. doi: 10.7498/aps.59.2109 [4] Bai Ying, Wang Bei, Zhang Wei-Feng. Nano-LiNiO2 as cathode material for lithium ion battery synthesized by molten salt method. Acta Physica Sinica, 2011, 60(6): 068202. doi: 10.7498/aps.60.068202 [5] Huang Le-Xu, Chen Yuan-Fu, Li Ping-Jian, Huan Ran, He Jia-Rui, Wang Ze-Gao, Hao Xin, Liu Jing-Bo, Zhang Wan-Li, Li Yan-Rong. Effects of preparation temperature of graphite oxide on the structure of graphite and electrochemical properties of graphene-based lithium-ion batteries. Acta Physica Sinica, 2012, 61(15): 156103. doi: 10.7498/aps.61.156103 [6] Li Juan, Ru Qiang, Sun Da-Wei, Zhang Bei-Bei, Hu She-Jun, Hou Xian-Hua. The lithium intercalation properties of SnSb/MCMB core-shell composite as the anode material for lithium ion battery. Acta Physica Sinica, 2013, 62(9): 098201. doi: 10.7498/aps.62.098201 [7] Ma Hao, Liu Lei, Lu Xue-Sen, Liu Su-Ping, Shi Jian-Ying. Electronic structure and transport properties of cathode material Li2FeSiO4 for lithium-ion battery. Acta Physica Sinica, 2015, 64(24): 248201. doi: 10.7498/aps.64.248201 [8] Pang Hui. Multi-scale modeling and its simplification method of Li-ion battery based on electrochemical model. Acta Physica Sinica, 2017, 66(23): 238801. doi: 10.7498/aps.66.238801 [9] Pang Hui. An extended single particle model-based parameter identification scheme for lithium-ion cells. Acta Physica Sinica, 2018, 67(5): 058201. doi: 10.7498/aps.67.20172171 [10] Liu Hui-Ying, Yang Yong, Hou Zhu-Feng, Zhu Zi-Zhong, Huang Mei-Chun. Investigation of lithium insertion in anode material CuSn for lithium-ion batteries. Acta Physica Sinica, 2003, 52(4): 952-957. doi: 10.7498/aps.52.952
• Citation:
##### Metrics
• Abstract views: 118
• Cited By: 0
##### Publishing process
• Received Date: 29 December 2018
• Accepted Date: 25 February 2019
• Available Online: 06 June 2019
• Published Online: 01 May 2019
## Influences of nanoscale particles and interparticle compression in electrodes on voltage hysteresis of lithium ion batteries
###### Corresponding author: Li Hao-Liang, wuwangchuxin@shu.edu.cn;
• 1. School of Mechanical and Power Engineering, Tongji University, Shanghai 200433, China
• 2. Shanghai Institute of Applied Mathematics and Mechanics, Shanghai 200444, China
• 3. Air Conditioning Electronics Department, Pan Asia Technical Automotive Center Co., Ltd., Shanghai 201201, China
Abstract: As one of high capacity electrode materials of lithium ion battery, silicon suffers significant stress effects, which further affects the voltage performance of battery. In this paper, a reaction-diffusion-stress coupled model is established, and the stress induced voltage hysteresis with consideration of diffusion induced stress, surface effects and interparticle compression under potentiostatic operation are investigated. It is found that stress and stress induced voltage hysteresis are dependent on particle size. For big particles, the diffusion induced stress is dominant and further aggravates the hysteresis of both stress and the overpotential consumed by it, indicating that more energy dissipates due to the stress effects. For small particles, especially ones with radius of a few nanometers, surface effects play a more prominent role than diffusion induced stress and the stress evolves into the state of compressive stress on the whole, leading the hysteresis of overpotential to be consumed by stress shrink and making the hysteresis plot of overpotential used to drive electrochemical reaction move downward. The electrode potential first reaches a cutoff voltage and finally the capacity of lithium ion battery decays. Therefore, too large or too small particle size in the electrode can both have a negative effect on the performance of lithium ion batteries, which indicates that an optimal size of the electrode particles must be designed in terms of electrode structure. Based on the calculation, particles with around 9 nm in radius are an appropriate option for electrode design in consideration of both diffusion induced stress and surface effect. In addition, for silicon electrodes, the silicon particles inevitably squeeze each other in a charge and discharge cycle. Therefore, interparticle compression is considered in this case. In detail, interparticle compression pushes the plot of stress hysteresis to the compressive state and leads to lower lithiation capacity, which makes the overpotential plot consumed by stress move downward and accordingly the overpotential plot used to drive the electrochemical reaction move upward. Denser electrode would strengthen this effect due to higher particle compression. It is indicated that for electrode design, the minimum of porosity ratio of electrodes should be adopted because higher interparticle compressive stress would reduce the battery capacity. Our results reveal that the voltage hysteresis of lithium ion batteries is related to the active particle size and the porosity ratio of the electrode, which is of great significance for guiding one in designing the lithium ion batteries.
Reference (41)
/ | 2020-02-26 20:28:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6080592274665833, "perplexity": 11322.031433284565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00412.warc.gz"} |
http://www.bionicturtle.com/forum/threads/cope-chapter.5840/ | # Cope chapter
Discussion in 'P2.T7. Operational Risk & ERM (25%)' started by shanlane, May 9, 2012.
1. ### shanlaneActive Member
Hello,
One of the aims tells us we need to "determine the amount of data required to estimate percentiles of loss distributions"
The one example of this in the notes (and in the source reading) is extrmemly confusing. Is there any way to sum up what they are looking for in a clean formula or algorithm?
The chapter just throws numbers around and its impossible to tell where they are coming from. Is there any way you could show how they came up with (for instance) 277,500 for the pareto or 8400 for the exponential?
Also, from the notes, does the 100,000 samples come from the fact that there are 100 losses per year and to find the 99.9% we need to somehow be 100 times more acurate and that is why we need the 99.999%?
Thanks!
Shannon
• Like x 1
2. ### David Harper, CFA, FRM, CIPMDavid Harper
Hi Shannon,
I think the key (summary) formula is his (2), which corresponds to Jorion's 5.17 (Jorion has a plainer language): standard error (quantile estimator) = 1/f(qp)*SQRT[p*(1-p)/n], where f(qp) is the density PDF.
The broad point, of course, is that the standard error of the sample quantile is (frustratingly) large in relative terms and that it's quite difficult to reduce the standard error. There are three variables:
• the PDF, f(.), which means that quantile estimation accuracy does depend on the distribution; i.e., Cope's numbers are merely illustrative of a lognormal
• the probability (p), which is our thematic irony: the higher the probability (where we are interested), the wider the standard error (less precise our estimate)
• the number of trials (sample size, n). The precision scales NOT with n, but with the SQRT(n). Notice in Table 1: from 1,000 to 10,000 data points does not 10X the precision, it only increases precision ~SQRT(10) from 20 to 7.7. Frankly, this is probably the only key point from a testability perspective; i.e., for example, that to increase the precision of our quantile estimator by a factor of X, we need a sample that is X^2 larger. To increase our precision by 10x, we need at least 100x the sample size. (the particulars of cope's 100 #'s relate to his definition of relative error, i don't have time/inclination to go thru the math, it's just not central, it's just choices he makes to illustrate the properties of the standard error).
i hope that helps, thanks,
3. ### shanlaneActive Member
I have no doubt about your conclusions, but if our p was 90%, the p*(1-p) would be 0.09. If p was .99, then this product would be .0099. An increase in p seems to decrease this amount.
Also, Dowd uses this for a VaR confidence interval but uses the area under the pdf, not the "height" of the pdf at that point. Is this inconsistent?
Finally, if f() is the "height" it would seem that a fat tailed distribution would have a greater value and since it is in the denominator a greater number would reduce the SE.
This is terribly confusing.
THanks!
Shannon
4. ### David Harper, CFA, FRM, CIPMDavid Harper
Hi Shannon,
But (p) informs the pdf denominator also. For example, let f(.) be the pdf, under the normal (approximations):
• p = 95% --> f(1.645) ~= 0.103 such that SE(@ 95%) = SQRT[p*(1-p)]/f(1.645) = 0.218/0.103 = 2.11, compare to:
• p = 99% --> f(2.33) ~= 0.027 such that SE (@ 99%) = SQRT[p*(1-p)]/f(2.33) = 0.0995/0.0267 = 3.73, compare to
• p = 99% = 0.032/0.0034 = 9.34
i.e.,
• decreasing pdf(.) is more than offsetting increasing numerator
• but also, to my original point about Cope's illustration, the issue of "accuracy" can be variously defined. Cope does not just define as absolute value of the standard error but as a percent of true value.
Re: dowd: no, it's not inconsistent, Dowd is using bins to infer the pdf, which is a more general approach as he understands: in practice we don't have a function, we have a empirical dataset. He may use the normal/function/whatever to communicate with the familiar, but his bin approach should approximate/converge and is ultimately more robust (I think) to actual, empirical applications. (if you just have a histogram, you need to define bins to retrieve a pdf. Or, if you can without bins, i am not aware)
Re: "Finally, if f() is the "height" it would seem that a fat tailed distribution would have a greater value and since it is in the denominator a greater number would reduce the SE."
Yes, TRUE. I don't think anybody said that a heavy-tailed distribution --> higher SE. Rather, just that the quantile is variant to the distribution. This is different than: GIVEN a distribution, higher p --> higher SE
thanks,
• Like x 1
5. ### shanlaneActive Member
Thank you for the very informative answer. I will try to wrap my head around the whole Dowd thing.
Regarding the "fat tailed" comment, it was not said explicitly, but was certainly inferred when Cope said that in order to get the SE down to 10% (or whatever it was) it took 250,000 or 300,000 for a Pareto distribution but less than 10,000 for (I think) an exponential distribution.
I thought that was one of the main points: if the tail was fat, we would need more samples to get the SE down to a reasonable level. The way I read that is that if we had the same number of samples from a light tailed and a heavy tailed dist that the fat tailed would have the higher SE.
Thanks again!
One more week !!
Shannon
• Like x 1
• Creative x 1
6. ### David Harper, CFA, FRM, CIPMDavid Harper
Hi Shannon,
You are right, I surely did not represent Cope, for he writes "It is significantly more difficult to get accurate quantile estimates for heavy-tailed distributions" using the exponential as an example.
Frankly, I am not following the intuition of that assertion, either I wondered if "relative error" explained, but that seems to go in the other direction: heaver tail implies higher quantile, ceteris paribus, and relative error [i.e., SE/quantile] as a ratio would DECREASE for a given standard error (i.e, the higher quantile lowers the error as a ratio).
However, I do notice, if i use the student's t for a barely heavy tailed distribution:
• the 99% pdf is less than the normal pdf @ 99%:
• e.g., at 30 df, 99% student's t pdf = T.DIST(T.INV(99%,30),30, false = pdf) = 0.02306
versus normal pdf (@ 99%) = 0.0267; i.e., the 1% quantile is "further out to the right" than the normal 1%, such that it's pdf "height" is lower!
But honestly this does not get me to an intuitive understanding of the statement, thanks,
• Creative x 1
7. ### shanlaneActive Member
Intuitively I think it makes sense, because if there were fat tails there would be more of a chance of some HUGE outliers and this would increase the SE but the mathematics of it (from that formula, at least) does not seem to work. Does that make sense?
Thanks!
Shannon
8. ### David Harper, CFA, FRM, CIPMDavid Harper
Yes, that resonates as intuitive: for a heavier tail, the sampling variation seems like it would be larger (the retrieved quantile would seem to have more sampling "wiggle room" in the heavy-tail) ... i agree with that ... and I also cannot connect it to the formula , thanks!
• Like x 1 | 2013-06-18 06:38:47 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8094224333763123, "perplexity": 1623.2679150585031}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706964363/warc/CC-MAIN-20130516122244-00039-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/338693/biblatex-ieee-not-correctly-translating | # Biblatex-IEEE not correctly translating
I'm new to Latex, so please keep that in mind. I'm wrote my thesis in Google Docs, but I now want to print/format it using Latex. So I'm using ShareLatex.com for that. Now it took me a while to get here, but it's still not quite working and I can't seem to find a solution online.
Here's what I got:
main.tex
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[swedish, english]{babel}
\title{Title}
\author{The Auther}
\date{14 November 2016}
\usepackage{graphicx}
\usepackage{csquotes}
\usepackage[backend=biber, style=ieee, urldate=comp]{biblatex}
\begin{document}
\selectlanguage{swedish}
\maketitle
\pagenumbering{gobble}
\pagebreak
\tableofcontents
\pagebreak
\pagenumbering{arabic}
\section{Inledning}
There is a theory which states that if ever anyone discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable.
There is another theory which states that this has already happened.
\section{Conclusion}
I always thought something was fundamentally wrong with the universe'' \cite{einstein}.
\printbibliography
\selectlanguage{english}
\printbibliography
\end{document}
references.bib
@article{einstein,
author = {Andreas Eidehall and Jochen Pohl and Fredrik Gustafsson and Jonas Ekmark},
title = {Toward Autonomous Collision Avoidance by Steering},
journal = {IEEE Transactions on Intelligent Transportation Systems},
volume = 8,
number = 1,
pages = {84-94},
month = dec,
year = 1905,
url = {http://ieeexplore.ieee.org/document/4114341},
urldate = {2015-11-09},
}
As you can see on the picture, the format is not correct. Where it simply says "URL:", it should say what I commented on the picture, "[Online]. Tillgänglig:"
How can I fix this? Can I override the language translation to fix this?
You can add the missing language definitions (I don't know swedish so translated only the url):
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[swedish, english]{babel}
\title{Title}
\author{The Auther}
\date{14 November 2016}
\usepackage{graphicx}
\usepackage{csquotes}
\usepackage[backend=biber, style=ieee, urldate=comp]{biblatex}
\DefineBibliographyStrings{swedish}{
mathesis = Master's thesis ,
patentjp = Japanese Patent ,
presentedat = presented at the\addspace ,
}
\begin{document}
\selectlanguage{swedish}
\maketitle
\pagenumbering{gobble}
\pagebreak
\tableofcontents
\pagebreak
\pagenumbering{arabic}
\section{Inledning}
There is a theory which states that if ever anyone discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable.
There is another theory which states that this has already happened.
\section{Conclusion}
I always thought something was fundamentally wrong with the universe'' \cite{einstein}.
\printbibliography
\selectlanguage{english}
\printbibliography
\end{document} | 2021-09-18 10:33:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4677315354347229, "perplexity": 2302.867008403677}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00707.warc.gz"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s15abc.html | s Chapter Contents
s Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_cumul_normal (s15abc)
## 1 Purpose
nag_cumul_normal (s15abc) returns the value of the cumulative Normal distribution function $P\left(x\right)$.
## 2 Specification
#include #include
double nag_cumul_normal (double x)
## 3 Description
nag_cumul_normal (s15abc) evaluates the cumulative Normal distribution function
$P x = 1 2π ∫ -∞ x e - u 2 / 2 du .$
The function is based on the fact that
$P x = 1 2 erfc -x / 2 .$
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
## 5 Arguments
1: xdoubleInput
On entry: the argument $x$ of the function.
None.
## 7 Accuracy
If $\epsilon$ and $\delta$ are the relative errors in result and argument, respectively, they are in principle related by $\left|\epsilon \right|\simeq \left|\left({xe}^{-{x}^{2}/2}/\sqrt{2\pi }P\left(x\right)\right)\delta \right|$.
For $x$ small and for $x$ positive the multiplying factor is always less than one and accuracy is mainly limited by machine precision. For large negative $x$ we find $\epsilon \sim {x}^{2}$ and hence to a certain extent relative accuracy is unavoidably lost. However, the absolute error in the result, $E$, is given by $\left|E\right|\simeq \left|\left({xe}^{-{x}^{2}/2}/\sqrt{2\pi }\right)\right|>\delta$, and since this multiplying factor is always less than one, absolute accuracy can be guaranteed for all $x$.
None.
## 9 Example
The following program reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
### 9.1 Program Text
Program Text (s15abce.c)
### 9.2 Program Data
Program Data (s15abce.d)
### 9.3 Program Results
Program Results (s15abce.r) | 2015-07-29 07:02:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9912322759628296, "perplexity": 1382.96726972901}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042986148.56/warc/CC-MAIN-20150728002306-00036-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://repository.kaust.edu.sa/handle/10754/324602 | ### Recent Submissions
• #### Multiple wheat genomes reveal global variation in modern breeding
(Nature, Springer Science and Business Media LLC, 2020-11-25) [Article]
AbstractAdvances in genomics have expedited the improvement of several agriculturally important crops but similar efforts in wheat (Triticum spp.) have been more challenging. This is largely owing to the size and complexity of the wheat genome$^{1}$, and the lack of genome-assembly data for multiple wheat lines$^{2,3}$. Here we generated ten chromosome pseudomolecule and five scaffold assemblies of hexaploid wheat to explore the genomic diversity among wheat lines from global breeding programs. Comparative analysis revealed extensive structural rearrangements, introgressions from wild relatives and differences in gene content resulting from complex breeding histories aimed at improving adaptation to diverse environments, grain yield and quality, and resistance to stresses$^{4,5}$. We provide examples outlining the utility of these genomes, including a detailed multi-genome-derived nucleotide-binding leucine-rich repeat protein repertoire involved in disease resistance and the characterization of Sm1$^{6}$, a gene associated with insect resistance. These genome assemblies will provide a basis for functional gene discovery and breeding to deliver the next generation of modern wheat cultivars.
• #### A Multilayer Nonlinear Elimination Preconditioned Inexact Newton Method for Steady-State Incompressible Flow Problems in Three Dimensions
(SIAM Journal on Scientific Computing, Society for Industrial & Applied Mathematics (SIAM), 2020-11-24) [Article]
We develop a multilayer nonlinear elimination preconditioned inexact Newton method for a nonlinear algebraic system of equations, and a target application is the three-dimensional steady-state incompressible Navier--Stokes equations at high Reynolds numbers. Nonlinear steadystate problems are often more difficult to solve than time-dependent problems because the Jacobian matrix is less diagonally dominant, and a good initial guess from the previous time step is not available. For such problems, Newton-like methods may suffer from slow convergence or stagnation even with globalization techniques such as line search. In this paper, we introduce a cascadic multilayer nonlinear elimination approach based on feedback from intermediate solutions to improve the convergence of Newton iteration. Numerical experiments show that the proposed algorithm is superior to the classical inexact Newton method and other single layer nonlinear elimination approaches in terms of the robustness and efficiency. Using the proposed nonlinear preconditioner with a highly parallel domain decomposition framework, we demonstrate that steady solutions of the Navier--Stokes equations with Reynolds numbers as large as 7,500 can be obtained for the lid-driven cavity flow problem in three dimensions without the use of any continuation methods.
• #### Source Apportionment and Elemental Composition of Atmospheric Total Suspended Particulates (TSP) Over the Red Sea Coast of Saudi Arabia
(Earth Systems and Environment, Springer Science and Business Media LLC, 2020-11-24) [Article]
AbstractThis work presents a comprehensive study on concentrations and elemental composition of total suspended atmospheric particulates for a semi-urban site on the Red Sea coast, and on-board a research vessel, which collected off-shore samples along the Red Sea. We conducted one of the most extended measurement campaigns of atmospheric particulates ever for the region, with continuous measurements over 27 months. The overall mean concentrations (± st. dev.) of TSP were 125 ± 197 µg m$^{−3}$ for the permanent semi-urban site, and 108 ± 193 µg m$^{−3}$ for the off-shore mobile site. The region is frequently severely impacted by both localised and widespread dust storms, which on occasion, can increase atmospheric particulate concentrations to levels above mg m$^{−3}$ (> 1000 µg m$^{−3}$). Median concentrations were not as variable between seasons, indicating a stable, permanent presence of atmospheric particulates independent of the time of year. The primary chemical elements contributing to particulate mass were Na, Ca, S, Al and Fe. We employed Positive Matrix Factorisation (EPA PMF v5.0.14) to identify different major sources of particulates, which were crustal, marine, fuel oil combustion/secondary sulphate and mixed anthropogenic. The crustal source was characterised by tracers Al, Fe, K, Mg and Sn, and was present to some extent in the other identified sources due to the permanent presence of dust particles in the atmosphere. The fuel oil combustion/secondary sulphate source was identifiable by the almost exclusive presence of S, and to a lesser extent V, emitted from oil combustion as primary emissions and also secondary sulphate formation following the release of S to the atmosphere. A mixed anthropogenic source was characterised by Zn, Ni, Cr, Cu and Pb, emitted from traffic, industry, power generation and water desalination. This study highlights that the natural sources of particulates in this desert region give rise to frequent episodes of extremely poor air quality, and this problem is compounded by significant emissions of anthropogenic pollution, which has an impact across the entire Red Sea basin. Further stringent measures should be adopted to improve air quality across the region and prevent long-term damage to the health of the local population and ecosystems.
• #### Energy Stability of Explicit Runge--Kutta Methods for Nonautonomous or Nonlinear Problems
(SIAM Journal on Numerical Analysis, Society for Industrial & Applied Mathematics (SIAM), 2020-11-24) [Article]
Many important initial value problems have the property that energy is nonincreasing in time. Energy stable methods, also referred to as strongly stable methods, guarantee the same property discretely. We investigate requirements for conditional energy stability of explicit Runge--Kutta methods for nonlinear or nonautonomous problems. We provide both necessary and sufficient conditions for energy stability over these classes of problems. Examples of conditionally energy stable schemes are constructed, and an example is given in which unconditional energy stability is obtained with an explicit scheme.
• #### Seismic Velocities Distribution in a 3D Mantle: Implications for InSight Measurements
(Wiley, 2020-11-24) [Preprint]
• #### Composition, uniqueness and connectivity across tropical coastal lagoon habitats in the Red Sea.
(BMC ecology, Springer Science and Business Media LLC, 2020-11-24) [Article]
BACKGROUND:Tropical habitats and their associated environmental characteristics play a critical role in shaping macroinvertebrate communities. Assessing patterns of diversity over space and time and investigating the factors that control and generate those patterns is critical for conservation efforts. However, these factors are still poorly understood in sub-tropical and tropical regions. The present study applied a combination of uni- and multivariate techniques to test whether patterns of biodiversity, composition, and structure of macrobenthic assemblages change across different lagoon habitats (two mangrove sites; two seagrass meadows with varying levels of vegetation cover; and an unvegetated subtidal area) and between seasons and years. RESULTS:In total, 4771 invertebrates were identified belonging to 272 operational taxonomic units (OTUs). We observed that macrobenthic lagoon assemblages are diverse, heterogeneous and that the most evident biological pattern was spatial rather than temporal. To investigate whether macrofaunal patterns within the lagoon habitats (mangrove, seagrass, unvegetated area) changed through the time, we analysed each habitat separately. The results showed high seasonal and inter-annual variability in the macrofaunal patterns. However, the seagrass beds that are characterized by variable vegetation cover, through time, showed comparatively higher stability (with the lowest values of inter-annual variability and a high number of resident taxa). These results support the theory that seagrass habitat complexity promotes diversity and density of macrobenthic assemblages. Despite the structural and functional importance of seagrass beds documented in this study, the results also highlighted the small-scale heterogeneity of tropical habitats that may serve as biodiversity repositories. CONCLUSIONS:Comprehensive approaches at the "seascape" level are required for improved ecosystem management and to maintain connectivity patterns amongst habitats. This is particularly true along the Saudi Arabian coast of the Red Sea, which is currently experiencing rapid coastal development. Also, considering the high temporal variability (seasonal and inter-annual) of tropical shallow-water habitats, monitoring and management plans must include temporal scales.
• #### Adversarial Generation of Continuous Images
(arXiv, 2020-11-24) [Preprint]
In most existing learning systems, images are typically viewed as 2D pixel arrays. However, in another paradigm gaining popularity, a 2D image is represented as an implicit neural representation (INR) -- an MLP that predicts an RGB pixel value given its (x,y) coordinate. In this paper, we propose two novel architectural techniques for building INR-based image decoders: factorized multiplicative modulation and multi-scale INRs, and use them to build a state-of-the-art continuous image GAN. Previous attempts to adapt INRs for image generation were limited to MNIST-like datasets and do not scale to complex real-world data. Our proposed architectural design improves the performance of continuous image generators by x6-40 times and reaches FID scores of 6.27 on LSUN bedroom 256x256 and 16.32 on FFHQ 1024x1024, greatly reducing the gap between continuous image GANs and pixel-based ones. To the best of our knowledge, these are the highest reported scores for an image generator, that consists entirely of fully-connected layers. Apart from that, we explore several exciting properties of INR-based decoders, like out-of-the-box superresolution, meaningful image-space interpolation, accelerated inference of low-resolution images, an ability to extrapolate outside of image boundaries and strong geometric prior. The source code is available at https://github.com/universome/inr-gan
• #### A single neuron subset governs a single coactive neuron circuit in Hydra vulgaris , representing a prototypic feature of neural evolution
(Cold Spring Harbor Laboratory, 2020-11-23) [Preprint]
The last common ancestor of Bilateria and Cnidaria is believed to be one of the first animals to develop a nervous system over 500 million years ago. Many of the genes involved in the neural function of the advanced nervous system in Bilateria are well conserved in Cnidaria. Thus, Cnidarian representative species, Hydra, is considered to be a living fossil and a good model organism for the study of the putative primitive nervous system in its last common ancestor. The diffuse nervous system of Hydra consists of several peptidergic neuron subsets. However, the specific functions of these subsets remain unclear. Using calcium imaging, here we show that the neuron subsets that express neuropeptide, Hym-176 function as motor neurons to evoke longitudinal contraction. We found that all neurons in a subset defined by the Hym-176 gene (Hym-176A) or its paralogs (Hym-176B) expression are excited simultaneously, which is then followed by longitudinal contraction. This indicates not only that these neuron subsets are motor neurons but also that a single molecularly defined neuron subset forms a single coactive motor circuit. This is in contrast with the Bilaterian nervous system, where a single molecularly defined neuron subset harbors multiple coactive circuits, showing a mixture of neurons firing with different timings. Furthermore, we found that the two motor circuits, one expressing Hym-176B in the body column and the other expressing Hym-176A in the foot, are coordinately regulated to exert region-specific contraction. Our results demonstrate that one neuron subset is likely to form a monofunctional circuit as a minimum functional unit to build a more complex behavior in Hydra. We propose that this simple feature (one subset, one circuit, one function) found in Hydra is a fundamental trait of the primitive nervous system.
• #### Eddy-induced transport and kinetic energy budget in the Arabian Sea
(Geophysical Research Letters, American Geophysical Union (AGU), 2020-11-23) [Article]
This study investigates the vertical eddy structure, eddy-induced transport, and eddy kinetic energy (EKE) budget in the Arabian Sea (AS) using an eddy-resolving reanalysis product. The EKE intensifies during summer in the western AS. Anticyclonic eddies (AEs) and cyclonic eddies (CEs) present warm-fresh and cold-salty cores, respectively, with interleaved salinity structures. The eddy-induced swirl transport is larger in the western AS and tends to compensate for heat transport by the mean flow. Zonal drift transport by AEs and CEs offset each other, and meridional transport is generally weaker. Eddies also produce notable upward heat flux during summer in the western AS, where ageostrophic circulations are induced to maintain a turbulent thermal wind balance. Plausible mechanisms for EKE production are governed by baroclinic and barotropic instabilities, which are enhanced in summer in the western basin, where signals are quantitatively one order larger than the turbulent wind inputs.
• #### TSP: Temporally-Sensitive Pretraining of Video Encoders for Localization Tasks
(arXiv, 2020-11-23) [Preprint]
Understanding videos is challenging in computer vision. In particular, the large memory footprint of an untrimmed video makes most tasks infeasible to train end-to-end without dropping part of the input data. To cope with the memory limitation of commodity GPUs, current video localization models encode videos in an offline fashion. Even though these encoders are learned, they are typically trained for action classification tasks at the frame- or clip-level. Since it is difficult to finetune these encoders for other video tasks, they might be sub-optimal for temporal localization tasks. In this work, we propose a novel, supervised pretraining paradigm for clip-level video representation that does not only train to classify activities, but also considers background clips and global video information to gain temporal sensitivity. Extensive experiments show that features extracted by clip-level encoders trained with our novel pretraining task are more discriminative for several temporal localization tasks. Specifically, we show that using our newly trained features with state-of-the-art methods significantly improves performance on three tasks: Temporal Action Localization (+1.72% in average mAP on ActivityNet and +4.4% in mAP@0.5 on THUMOS14), Action Proposal Generation (+1.94% in AUC on ActivityNet), and Dense Video Captioning (+0.31% in average METEOR on ActivityNet Captions). We believe video feature encoding is an important building block for many video algorithms, and extracting meaningful features should be of paramount importance in the effort to build more accurate models.
• #### Host-association as major driver of microbiome structure and composition in Red Sea seagrass ecosystems
(Environmental Microbiology, Wiley, 2020-11-22) [Article]
The role of the microbiome in sustaining seagrasses has recently been highlighted. However, our understanding of the seagrass microbiome lacks behind that of other organisms. Here, we analyze the endophytic and total bacterial communities of leaves, rhizomes, and roots of six Red Sea seagrass species and their sediments. The structure of seagrass bacterial communities revealed that the 1% most abundant OTUs accounted for 87.9 and 74.8 % of the total numbers of reads in sediment and plant tissue samples, respectively. We found taxonomically distinct bacterial communities in vegetated and bare sediments. Yet, our results suggest that lifestyle (i.e. free-living or host-association) is the main driver of bacterial community composition. Seagrass bacterial communities were tissue- and species-specific and differed from those of surrounding sediments. We identified OTUs belonging to genera related to N and S cycles in roots, and members of Actinobacteria, Bacteroidetes, and Firmicutes phyla as particularly enriched in root endosphere. The finding of highly similar OTUs in well-defined sub-clusters by network analysis suggests the co-occurrence of highly connected key members within Red Sea seagrass bacterial communities. These results provide key information towards the understanding of the role of microorganisms in seagrass ecosystem functioning framed under the seagrass holobiont concept.
• #### Boundary-sensitive Pre-training for Temporal Localization in Videos
(arXiv, 2020-11-21) [Preprint]
Many video analysis tasks require temporal localization thus detection of content changes. However, most existing models developed for these tasks are pre-trained on general video action classification tasks. This is because large scale annotation of temporal boundaries in untrimmed videos is expensive. Therefore no suitable datasets exist for temporal boundary-sensitive pre-training. In this paper for the first time, we investigate model pre-training for temporal localization by introducing a novel boundary-sensitive pretext (BSP) task. Instead of relying on costly manual annotations of temporal boundaries, we propose to synthesize temporal boundaries in existing video action classification datasets. With the synthesized boundaries, BSP can be simply conducted via classifying the boundary types. This enables the learning of video representations that are much more transferable to downstream temporal localization tasks. Extensive experiments show that the proposed BSP is superior and complementary to the existing action classification based pre-training counterpart, and achieves new state-of-the-art performance on several temporal localization tasks.
• #### Compositional Fluctuations Locked by Athermal Transformation Yielding High Thermoelectric Performance in GeTe.
(Advanced materials (Deerfield Beach, Fla.), Wiley, 2020-11-20) [Article]
Phase transition in thermoelectric (TE) material is a double-edged sword-it is undesired for device operation in applications, but the fluctuations near an electronic instability are favorable. Here, Sb doping is used to elicit a spontaneous composition fluctuation showing uphill diffusion in GeTe that is otherwise suspended by diffusionless athermal cubic-to-rhombohedral phase transition at around 700 K. The interplay between these two phase transitions yields exquisite composition fluctuations and a coexistence of cubic and rhombohedral phases in favor of exceptional figures-of-merit zT. Specifically, alloying GeTe by Sb2 Te3 significantly suppresses the thermal conductivity while retaining eligible carrier concentration over a wide composition range, resulting in high zT values of >2.6. These results not only attest to the efficacy of using phase transition in manipulating the microstructures of GeTe-based materials but also open up a new thermodynamic route to develop higher performance TE materials in general.
• #### Advances in statistical modeling of spatial extremes
(WIREs Computational Statistics, Wiley, 2020-11-20) [Article]
The classical modeling of spatial extremes relies on asymptotic models (i.e., max-stable or r-Pareto processes) for block maxima or peaks over high thresholds, respectively. However, at finite levels, empirical evidence often suggests that such asymptotic models are too rigidly constrained, and that they do not adequately capture the frequent situation where more severe events tend to be spatially more localized. In other words, these asymptotic models have a strong tail dependence that persists at increasingly high levels, while data usually suggest that it should weaken instead. Another well-known limitation of classical spatial extremes models is that they are either computationally prohibitive to fit in high dimensions, or they need to be fitted using less efficient techniques. In this review paper, we describe recent progress in the modeling and inference for spatial extremes, focusing on new models that have more flexible tail structures that can bridge asymptotic dependence classes, and that are more easily amenable to likelihood-based inference for large datasets. In particular, we discuss various types of random scale constructions, as well as the conditional spatial extremes model, which have recently been getting increasing attention within the statistics of extremes community. We illustrate some of these new spatial models on two different environmental applications.
• #### Advances in the Design of Heterogeneous Catalysts and Thermocatalytic Processes for CO2 Utilization
(ACS Catalysis, American Chemical Society (ACS), 2020-11-20) [Article]
Utilization of CO2 as feedstock to produce fine chemicals and renewable fuels is a highly promising field, which presents unique challenges in its implementation at scale. Heterogeneous catalysis with its simple operation and industrial compatibility can be an effective means of achieving this challenging task. This review summarizes the current developments in heterogeneous thermal catalysis for the production of carbon monoxide, alcohols, and hydrocarbons from CO2. A detailed discussion is provided regarding structure−activity correlations between the catalyst surface and intermediate species which can aid in the rational design of future generation catalysts. Effects of active metal components, catalyst supports, and promoters are discussed in each section, which will guide researchers to synthesize new catalysts with improved selectivity and stability. Additionally, a brief overview regarding process design considerations has been provided. Future research directions are proposed with special emphasis on the application scope of new catalytic materials and possible approaches to increase catalyst performance.
• #### MXenes for Rechargeable Batteries Beyond the Lithium-Ion
(Advanced Materials, Wiley, 2020-11-20) [Article]
Research on next-generation battery technologies (beyond Li-ion batteries, or LIBs) has been accelerating over the past few years. A key challenge for these emerging batteries has been the lack of suitable electrode materials, which severely limits their further developments. MXenes, a new class of 2D transition metal carbides, carbonitrides, and nitrides, are proposed as electrode materials for these emerging batteries due to several desirable attributes. These attributes include large and tunable interlayer spaces, excellent hydrophilicity, extraordinary conductivity, compositional diversity, and abundant surface chemistries, making MXenes promising not only as electrode materials but also as other components in the cells of emerging batteries. Herein, an overview and assessment of the utilization of MXenes in rechargeable batteries beyond LIBs, including alkali-ion (e.g., Na+ , K+ ) storage, multivalent-ion (e.g., Mg2+ , Zn2+ , and Al3+ ) storage, and metal batteries are presented. In particular, the synthetic strategies and properties of MXenes that enable MXenes to play various roles as electrodes, metal anode protective layers, sulfur hosts, separator modification layers, and conductive additives in these emerging batteries are discussed. Moreover, a perspective on promising future research directions on MXenes and MXene-based materials, ranging from material design and processing, fundamental understanding of the reaction mechanisms, to device performance optimization strategies is provided.
• #### A new approach for quantifying epigenetic landscapes.
(The Journal of biological chemistry, American Society for Biochemistry & Molecular Biology (ASBMB), 2020-11-20) [Article]
ChIP-Seq is a widespread experimental method for determining the global enrichment of chromatin modifications and genome-associated factors. Whereas it is straightforward to compare the relative genomic distribution of these epigenetic features, researchers have also made efforts to compare their signal strength using external references for normalization. New work now suggests that these "spike-ins" could lead to inaccurate conclusions due to intrinsic issues of the methodology and instead calls for new criteria of experimental reporting that may permit internal standardization when certain parameters are fulfilled.
• #### High-speed filtered Rayleigh scattering thermometry in premixed flames through narrow channels
(Combustion and Flame, Elsevier BV, 2020-11-20) [Article]
High-speed filtered Rayleigh scattering at 10 kHz repetition rate for time-resolved temperature measurements in premixed flames propagating in mm-high channels is demonstrated for the first time. The instrument relies on a pulse burst laser with a 200 ns temporal width, and a CCD camera operated in subframe burst gate mode to achieve high signal to noise ratio despite the limited optical access. A 10-inch-long iodine cell acts as a molecular filter and the laser, which is seeded using an external cavity diode laser and operated in a temporally stretched mode, is wavelength-tuned to the peak of a strong iodine absorption line. A CCD camera, operated in the sub-frame burst gating mode with the help of an external slit configuration, is used as the detector to achieve improved camera noise performance. Filtered Rayleigh scattering temperature measurements in premixed flat flames stabilized over a McKenna burner indicate an instrument precision of 3%. Temperature measurements in the products region are within 3% of measurements obtained with conventional Rayleigh scattering. The system's capabilities are demonstrated through time-resolved temperature measurements in a premixed methane-air flame propagating in a 1.5 mm-high rectangular channel designed to study flame quenching in flame arrestors. Surface scattering from the optical windows and the channel surfaces is successfully suppressed and time-resolved temperature profiles are obtained for both quenching and no-quenching events.
• #### Counterintuitive Wetting Transitions in Doubly Reentrant Cavities as a Function of Surface Make‐Up, Hydrostatic Pressure, and Cavity Aspect Ratio
(Advanced Materials Interfaces, Wiley, 2020-11-20) [Article]
Surfaces that entrap air underwater serve numerous practical applications, such as mitigating cavitation erosion and reducing frictional drag. These surfaces typically rely on perfluorinated coatings. However, the non-biodegradability and fragility of the coatings limit practical applications. Thus, coating-free, sustainable, and robust approaches are desirable. Recently, a microtexture comprising doubly reentrant cavities (DRCs) has been demonstrated to entrap air on immersion in wetting liquids. While this is a promising approach, insights into the effects of surface chemistry, hydrostatic pressure, and cavity dimensions on wetting transitions in DRCs remain unavailable. In response, Cassie-to-Wenzel transitions into circular DRCs submerged in water are investigated and compared with those in cylindrical “simple” cavities (SCs). It is found that at low hydrostatic pressures (≈50 Pa), DRCs with hydrophilic (θo ≈ 40°) and hydrophobic (θo ≈ 112°) make-ups fill within 105 and 107 s, respectively, while SCs with hydrophilic make-up fill within <10−2 s. Under elevated hydrostatic pressure (P ≤ 90 kPa), counterintuitively, DRCs with hydrophobic make-up fill dramatically faster than the commensurate SCs. This comprehensive report should provide a rational framework for harnessing microtexturing and surface chemistry toward coating-free liquid repellency.
• #### Monolayer Graphene Transfer onto Hydrophilic Substrates: A New Protocol Using Electrostatic Charging
(Membranes, MDPI AG, 2020-11-20) [Article]
In the present work, we developed a novel method for transferring monolayer graphene onto four different commercial hydrophilic micro/ultra-filtration substrates. The developed method used electrostatic charging to maintain the contact between the graphene and the target substrate intact during the etching step through the wet transfer process. Several measurement/analysis techniques were used in order to evaluate the properties of the surfaces and to assess the quality of the transferred graphene. The techniques included water contact angle (CA), atomic force microscopy (AFM), and field emission scanning electron microscopy (FESEM). Potassium chloride (KCl) ions were used for the transport study through the developed graphene-based membranes. The results revealed that 70% rejection of KCI ions was recorded for the graphene/polyvinylidene difluoride (PVDF1) membrane, followed by 67% rejection for the graphene/polyethersulfone (PES) membrane, and 65% rejection for graphene/PVDF3 membrane. It was revealed that the smoothest substrate was the most effective in rejecting the ions. Although defects such as tears and cracks within the graphene layer were still evolving in this new transfer method, however, the use of Nylon 6,6 interfacial polymerization allowed sealing the tears and cracks within the graphene monolayer. This enhanced the KCl ions rejection of up to 85% through the defect-sealed graphene/polymer composite membranes. | 2020-11-27 12:15:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3960251808166504, "perplexity": 4660.0550519802255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00635.warc.gz"} |
https://math.stackexchange.com/questions/769693/finitely-generated-abelian-group-is-hopfian | # Finitely generated abelian group is Hopfian.
Let $G$ be a finitely generated abelian group and $H$ be its proper subgroup. Can we prove the result using short exact sequence, $$1\to H\to G \to G/H\to 1.$$ Any other method is also welcome.
• You should be able to use the classification theorem for finitely generated abelian groups. Any finitely generated abelian group is of the form $G = \mathbb{Z}^{r} + H$, where $H$ is a finite group that is a direct sum of cyclic groups. – Siddharth Venkatesh Apr 26 '14 at 5:50
• @schzan Why are these groups called Hopfian? What did Hopf do with them? I couldn't find any references. – user39598 Aug 8 '17 at 0:57
Let $G$ be a f.g. abelian group and $f:G\to G$ a surjection and $K_n$ be the kernel of $f^n$. We can consider the short exact sequence $$0\to \mathbb Q\otimes K_n\to\mathbb Q\otimes G\xrightarrow{1\otimes f^n}\mathbb Q\otimes G\to 0$$ of finite dimensional rational vector spaces. Since the map $\mathbb Q\otimes G\xrightarrow{1\otimes f^n}\mathbb Q\otimes G$ is a surjective endomorphism of a finite dimensional vector space it is injective, and $\mathbb Q\otimes K_n=0$. This implies that $K_n$ is a torsion group.
Now the sequence $(K_n)_{n\geq1}$ is increasing, and all its elements are contained in the torsion subgroup of $G$, which is finite. It follows that the sequence stabilizes: there is an $m$ such that $K_n=K_m$ for all $n\geq m$.
Suppose $g\in G$ is such that $f(g)=0$. There is an $h\in G$ such that $f^m(h)=g$, and then $f^{m+1}(h)=0$ so that $h\in K_{m+1}=K_m$ and we see that $g=f^m(h)=0$. The map $f$ is thus injective.
• Very nice argument, +1. – Seirios Apr 26 '14 at 8:55
• @Mariano, thakns for the answer. I also found that, Every exact sequence of finitely generated free abelian groups is split. Thus, we obtain $$G=H\bigoplus G/H,$$ which is possible only when $H={1}$. Is this correct?(where I assumed G to be non Hopfian, and used $G\cong G/H$) – schzan Apr 27 '14 at 2:02
• Well, G is not in general free, so that does not help you. – Mariano Suárez-Álvarez Apr 27 '14 at 2:11
• ohh..could you please tell me where the statement uses the fact that it is free(which aspect of free groups was required)? I am sorry if this comes as a surprise but I would really appreciate if you could explain what is the difference between finitely generated abelian group and finitely generated free abelian group.(suggested links are also welcome) – schzan Apr 28 '14 at 20:36
• @shzan, I did not use the fact that $G$ is free, because it isn't! – Mariano Suárez-Álvarez Apr 28 '14 at 22:45
Let $G$ be a finitely generated abelian group of rank $r$ and $\varphi : G \twoheadrightarrow G$ be an epimorphism.
• Let $g_1,\dots,g_r \in G$ be $r$ $\mathbb{Z}$-independent elements. Because $\varphi$ is surjective, there exists $h_i \in G$ such that $\varphi(h_i)=g_i$. Because $\varphi$ is $\mathbb{Z}$-linear, $h_1,\dots,h_r$ are $\mathbb{Z}$-independent.
• Let $g \in G$ such that $\varphi(g) \in \mathrm{Tor}(G)$. The family $\{h_1,\dots, h_r,g\}$ is $\mathbb{Z}$-dependent, so there exist $n,n_1,\dots,n_r$ such that $ng= n_1h_1+ \cdots +n_rh_r$; moreover, we may suppose $n \neq 0$. Applying $\varphi$ to the previous equality and multiplying by a well-chosen $k$, we obtain $$0=kn \varphi(g)= kn_1g_1 + \cdots + kn_rg_r,$$ hence $n_1= \cdots = n_r=0$. Therefore, $ng=0$ and $g \in \mathrm{Tor}(G)$.
• We deduce that $\mathrm{ker}(\varphi)= \mathrm{ker}(\tilde{\varphi})$ where $\tilde{\varphi}$ denotes the restriction of $\varphi$ to $\mathrm{Tor}(G)$. Moreover, $\tilde{\varphi}$ defines an epimorphism $\mathrm{Tor}(G) \twoheadrightarrow \mathrm{Tor}(G)$. But $\mathrm{Tor}(G)$ is a finite group, so $\tilde{\varphi}$ is an isomorphism. We conclude that $$\mathrm{ker}(\varphi)= \mathrm{ker}(\tilde{\varphi})= \{0\},$$ that is $\varphi$ is an isomorphism.
Maybe this answer is not needed but I find it to be a quick way to prove that fg abelian groups are hopfian.
Since $$\mathbb{Z}$$ is a Noetherian ring and abelian groups are $$\mathbb{Z} -$$modules it follows that fg abelian groups are Noetherian modules.
There is a well known proposition that says that an epimorphism $$f:M\to M$$ (where $$M$$ is a Noetherian module) is an isomorphism. | 2020-09-29 17:44:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9267743229866028, "perplexity": 97.06203922226177}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00098.warc.gz"} |
https://scicomp.stackexchange.com/tags/deal.ii/hot | Tag Info
Hot answers tagged deal.ii
8
As one of the library's authors, I would of course love for deal.II to come out on top with this comparison. But I suspect it may not, and the answer lies in a factor you omit from your comparison: how long it actually takes to implement your code. Few people in academia with the skills to implement a FEM code from scratch spend more time solving PDEs than ...
5
Having written this entry, let me also answer you question :-) The issue in essence boils down to the following: given a function $u(x)$ and its interpolation $u_h(x)$ onto either a triangular or quadrilateral mesh with the same number of unknowns, then which of the two is more accurate? In other words, is $$\| u - I_h^\text{triangles} u \| < \| u - ... 4 h is a measure of the mesh size. In the example, they are using rectangular elements. For which a commonly used measure of the mesh size is the length of the largest diagonal. Looking at the table above the sentence you quoted: (M_5-M_6)/(M_4-M_5)=(0.14052586 −0.14056422)/(0.14037251−0.14052586)\approx 1/4 = (h_6/h_5)^2 where M_i is the mean at i-th ... 4 The SEM is a FEM! It's almost like all these different names are designed to confuse the newcomer. I will speak primarily about the most popular form which uses a tensor product Lagrange basis with nodes at the Legendre-Gauss-Lobatto (LGL) points. Once you have chosen that basis and integrate with the LGL quadrature using the same nodes. You have a method ... 4 The main advantage is that it reduces the Runge phenomenon and leads to faster convergence rates. It also presents less numerical dispersion and need less nodes per wavelength (see 1 and 2). So, I would say that you would prefer the method for wave propagation scenarios. Regarding software that includes SEM, I am aware of the following: FSELib: Matlab ... 4 Specific answers to this question are probably time-limited. However, the following general approach (from the great Eric S. Raymond) works very well: Rule of Modularity: Write simple parts connected by clean interfaces. Rule of Clarity: Clarity is better than cleverness. Rule of Composition: Design programs to be connected to other programs. Rule of ... 4 Unfortunately there's no tool for this. You can run each on a variety of input sizes to establish the computational complexity they appear to have, i.e. the f in the O(f(n)) that characterizes each code. This can point you towards what underlying algorithms each is using and verify or not that something that should be O(n) is actually implemented to ... 3 I fully agree with Wolfgang's answer, but I'd like to add something of my experience. I had recently to choose between tri and quad based FEM and after googling for it, I decided that I would code both from scratch for a fair comparison (I had a previous experience in writing FEM kernel. I used C++ and the Eigen library for matrices and vectors. I considered ... 2 I will try answer based on my experience with deal.ii. The max_per_row=5 means that at most there will be 5 non-zeros per row in the matrix. Since we now know this then we do not need to have a 1\times{}100 matrix but rather a 1\times{}50. In other words this parameter sets an upper-bound on the memory needed. In reality it is not stored as one vector ... 2 You go the grouping in the term \nabla u^+-\nabla u^- mostly right but not quite. To see this, remember that the functions \phi_i^{\pm} are discontinuous, and that consequently u_0^+ is not necessarily equal to u_3^- -- if they were, you'd end up with a function that is continuous at that point. Instead, you need one degree of freedom per adjacent ... 1 You need to compute the projection \Pi(f-cU_h) as a first step. This is a finite element field, so you can create a global field (in deal.II lingo: Create the corresponding finite element and a DoFHandler object) and compute it by global projection by inverting a mass matrix. Alternatively, because the field is discontinuous, you can also forgo the global ... 1 As mentioned by @Wolfgang Bangerth , the value may be at the interior points. However, to visualise that better in paraview, You can use a filter called "plot over line" which plots the data points over a given line, which could come very handy for these kinds of clarifications in future. 1 @cfdlab's answer gives a general way to construct solutions for discontinuous coefficients, but here is a slightly more theoretical perspective to it as well. All of this can be understood in 1d, so imagine all of the functions below to be functions of just one argument x for a moment. First, you can't just choose any u and \alpha to obtain a function ... 1 If you want to use the coefficient of the form$$ \alpha = \begin{cases} \alpha_1 & r < 0.5 \\ \alpha_2 & r > 0.5 \end{cases} $$it is useful to work in polar coordinates. You build two pieces of the solution$$ u_1(r,\theta), \qquad r < 0.5 $$and$$ u_2(r,\theta), \qquad r > 0.5 $$Then at the interface you ensure solution and flux ... 1 This is not really a program as much as a big repository of codes. John Burkardt from Florida State University maintains a rich collection of scripts in C++, Matlab and Fortran for a large range of problems. You can check out the FEM 2D library or go up a level and look at other codes listed there. These have good commenting on them and you can look at the ... 1 If a homogeneous Neumann problem is considered, i.e. f=0 in \Omega and \mathbf{n}\cdot\nabla\phi on \Gamma, one solution is given by the constant function$$ \phi\left(\mathbf{x}\right)=\phi_0\,. $$Then the boundary integral equation reads$$\int\limits_{\Omega}\phi\left(\mathbf{x}' \right)\delta\left(\mathbf{x}'-\mathbf{x}\right)\mathrm{d}V' =\...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-09-21 19:45:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8444991707801819, "perplexity": 915.4707239659349}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00161.warc.gz"} |
https://tex.stackexchange.com/questions/220933/moving-lua-code-into-subdirectory | Moving lua code into subdirectory
I want to move all .lua files into the subdirectory code to avoid polluting the main directory with them. However, I cannot get lualatex to find them there.
Given this project layout:
main.tex
main.lua
code/
test.lua
test2.lua
main.tex
\documentclass{article}
\directlua{package.path = './code/?.lua;' .. package.path}
\directlua{print(package.path)}
\directlua{require('test')}
\directlua{require('code/test')}
\begin{document}
\end{document}
main.lua
package.path = './code/?.lua;' .. package.path
print(package.path)
require('test')
require('code/test')
code/test.lua
require('test2')
code/test2.lua
print('success')
When I run lualatex main.tex I get the output:
This is LuaTeX, Version beta-0.79.1 (TeX Live 2014/Arch Linux) (rev 4971)
restricted \write18 enabled.
(./main.tex
LaTeX2e <2014/05/01>
Babel <3.9k> and hyphenation patterns for 79 languages loaded.
(/usr/share/texmf-dist/tex/latex/base/article.cls
Document Class: article 2007/10/19 v1.4h Standard LaTeX document class
(/usr/share/texmf-dist/tex/latex/base/size10.clo))./code/?.lua;/usr/local/share/lua/5.2/?.lua;/usr/local/share/lua/5.2/?/init.lua;/usr/local/lib/lua/5.2/?.lua;/usr/local/lib/lua/5.2/?/init.lua;./?.lua
! LuaTeX error [\directlua]:1: module 'test' not found:
[kpse lua searcher] file not found: 'test'
[kpse C searcher] file not found: 'test'
stack traceback:
[C]: in function 'require'
[\directlua]:1: in main chunk.
l.5 \directlua{require('test')}
? q
OK, entering \batchmode
The problem seems to be that lualatex does not find the files in the subdirectory although I altered the package.path (which is also printed correctly!).
When I runt the same code via lua main.lua the output is
./code/?.lua;/usr/share/lua/5.2/?.lua;/usr/share/lua/5.2/?/init.lua;/usr/lib/lua/5.2/?.lua;/usr/lib/lua/5.2/?/init.lua;./?.lua
success
as expected.
For some reason lualatex seems not to take package.path into account.
Using require('code/test.lua') is not an option because then the dependency between test.lua and test2.lua is not picked up correctly.
What do I need to do to make lualatex find my .lua files in the code directory?
I already followed this question to this answer but I don't see how this solves the problem (or does it state that we cannot change the package search path in lualatex?). Also, the scenario seems to be another one as I don't want to "install a systemwide" module but some scripts for this project only.
• lualatex doesn't use package path by default. it uses kpse library which searches in texmf tree abd current directory. but it is possible to modify package loader to support package oath, see tex.stackexchange.com/a/219228/2891 Jan 6, 2015 at 14:20
• @michal.h21 So while overriding the "easy to use" default algorithm they (the lualatex authors) did not offer a replacement/command to add your own paths? I feel like it might not be an otherworldly request to be able to use subdirectories in your lua(latex) code. Jan 6, 2015 at 14:38
• you have an error in your code: require('code/test') should be require('code.test'). but it probably will not find submodules anyway Jan 6, 2015 at 14:54
• try require('code.test') it should work. Jan 6, 2015 at 15:00
• @YiannisLazarides michal.h21: require(code/test) works as well as require(code.test) the problem occurs with the dependency in code/test.lua to code/test2.lua which in both cases does not work because the require does not search in code. Jan 6, 2015 at 15:25
The fastest way I have found to import relative packages in lualatex is
require(package.searchpath('test', './?.lua:./code/?.lua'))
which will find test.lua by searching ./ and ./code. It is hacky, but also clear and quick.
Adding the path loader as in https://tex.stackexchange.com/a/219228/2891 is the more thorough solution, but that code did not immediately work for me, plus it would require figuring out the textmf lua paths.
• A Warm Welcome to TeX.SE! Jul 14, 2016 at 2:08
• The solution works, but I believe the appropriate path separator is the semicolon. That makes it work for me. May 26, 2019 at 5:33
• This is good for a single file, but if your lua library has multiple files (ie is a folder with several files in it), it will be problematic. I recommend placing the files where lualatex.exe is (in my case C:\Program Files\MiKTeX 2.9\miktex\bin\x64) and usepackage{luapackageloader} Jul 26, 2021 at 18:58
local oldreq = require
require = function(p)
return oldreq(package.searchpath(p, './?.lua;./code/?.lua'))
end
• Welcome to TeX.SX! Can you please explain what your shown code does? How does it solve the given issue? Please do not only throw a code, explain what is does. Sep 27, 2019 at 17:24 | 2022-06-27 06:31:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5348437428474426, "perplexity": 5420.7856396645475}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00146.warc.gz"} |
https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th-mean-median-challenge/a/mean-as-the-balancing-point | # Mean as the balancing point
Explore how we can think of the mean as the balancing point of a data distribution.
You know how to find the mean by adding up and dividing. In this article, we'll think about the mean as the balancing point. Let's get started!
## Part 1: Find the mean
Find the mean of left brace, 5, comma, 7, right brace.
The mean is 6.
Find the mean of left brace, 5, comma, 6, comma, 7, right brace.
The mean is 6.
Interesting! In the first two problems, the data was "balanced" around the number six. Try the next one without finding the total or dividing. Instead, think about how the numbers are balanced around the mean.
Find the mean of left brace, 1, comma, 3, comma, 5, right brace.
The mean is 3.
Notice how the 1 and 5 were "balanced" on either side of the 3:
Find the mean of left brace, 4, comma, 7, comma, 10, right brace.
The mean is 7.
Can you see how the data points are always balanced around the mean? Let's try one more!
Find the mean of left brace, 2, comma, 3, comma, 5, comma, 6, right brace.
The mean is 4.
## Part 2: A new way of thinking about the mean
You might have noticed in Part 1 that it's possible to find the mean without finding the total or dividing for some simple data sets.
Key idea: We can think of the mean as the balancing point , which is a fancy way of saying that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean.
### Example
In Part 1, you found the mean of left brace, 2, comma, 3, comma, 5, comma, 6, right brace to be start color goldD, 4, end color goldD. We can see that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean because start color redD, 1, end color redD, plus, start color redD, 2, end color redD, equals, start color greenD, 1, end color greenD, plus, start color greenD, 2, end color greenD:
#### Reflection questions
What is the total distance start color redD, b, e, l, o, w, end color redD the mean in this example?
What is the total distance start color greenD, a, b, o, v, e, end color greenD the mean in this example?
## Part 3: Is the mean always the balancing point?
Yes! It is always true that the total distance below the mean is equal to the total distance above the mean. It just happens to be easier to see in some data sets than others.
For example let's consider the data set left brace, 2, comma, 3, comma, 6, comma, 9, right brace.
Here's how we can calculate the mean:
start fraction, 2, plus, 3, plus, 6, plus, 9, divided by, 4, end fraction, equals, start color goldD, 5, end color goldD
And we can see that the total distance below the mean is equal to the total distance above the mean because start color redD, 2, end color redD, plus, start color redD, 3, end color redD, equals, start color greenD, 1, end color greenD, plus, start color greenD, 4, end color greenD:
## Part 4: Practice
### Problem 1
Which of the lines represents the mean of the data points shown below?
Line start color maroonD, C, end color maroonD represents the mean because it is the line where the distance below the line is equal to the distance above the line:
### Problem 2
Which of the lines represents the mean of the data points shown below?
Line start color purpleC, B, end color purpleC represents the mean because it is the line where the distance below the line is equal to the distance above the line:
## Challenge problem
The mean of four data points is 5. Three of the four data points and the mean are shown in the diagram below.
Choose the fourth data point.
Step 1: So far, the distance below the mean is start color redD, 1, end color redD, and the distance above the mean is start color greenD, 4, end color greenD:
Step 2: To make the distance below the mean equal the distance above the mean, the fourth data point must be start color redD, 3, end color redD below the mean of start color goldD, 5, end color goldD: | 2017-10-19 14:54:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 87, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7549811005592346, "perplexity": 1699.6606106873362}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00839.warc.gz"} |
https://nebusresearch.wordpress.com/category/comic-strips/ | ## Reading the Comics, August 5, 2022: Catching Up Edition
I’ve had several weeks since my last Reading the Comics post. They’ve been quiet enough weeks. Let me share some of the recent offerings from Comic Strip Master Command that I enjoyed, though. I enjoy many comic strips but not all of them mention something on-point here.
Isabella Bannerman’s Six Chix for the 18th of July is a wordplay joke, naming a dog Isosceles to respect his front and hindlegs being equal in length. As I say, I’m including these because I like them, not because I have deep thoughts about them.
Bill Amend’s FoxTrot for the 24th of July gets a bit more solidly mathematical, as it’s natural to think of this sort of complicated polyhedron as something mathematicians do. Geometers at least. There’s a comfortable bit of work to be done in these sorts of shapes. They sometimes have appealing properties, for instance balancing weight loads well. Building polyhedrons out of toothpicks and gumdrops, or straws and marshmallows, or some other rigid-and-soft material, is a fine enough activity. I think every mathematics department has some dusty display shelf with a couple of these.
There are many shapes that Paige Fox’s construction might be. To my eye Paige Fox seems to be building a truncated icosahedron, that is to say, the soccer-ball shape. It’s an Archimedean solid, one of the family of thirteen shapes made of nonintersecting regular convex polygons. These are the shapes you discover if you go past Platonic solids. The family is named for that Archimedes, although the work in which he discussed them is now lost.
Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 26th of July is a different take on the challenge of motivating students to care about mathematics. The “right approach” argument has its appeal, although it’s obviously thinking of “professor” as the only job a mathematician can hold. And even granting that none of the people who understand your work have the power to fire you, that means people who don’t understand your work do have it. Also, people shut down very hard at your promise that they could understand anything about what you do, even when you know how to express it in common-language terms. This gets old fast. I am also skeptical that women are impressed by men at bars who claim to be employed for their intellect.
Anyway, academic jobs are great and more jobs should work by their rules, which, yes, do include extremely loose set hours and built-in seasons where the amount and kind of work you do varies.
Patrick Roberts’s Todd the Dinosaur for the 5th of August is a mathematics-anxiety dream, represented with the sudden challenge to do mental arithmetic. 47 times 342 is an annoying problem, yes, but one can at least approximate it fairly well quickly. 47 is almost 50, which is half a hundred. So 47 times 342 has to be nearly half of a hundred times 342, that is, half of 34,200. This is an easy number to cut in half, though: 17,100. To get this exactly? 47 is three less than fifty, so, subtract three times 342. 342 is about a third of a thousand, we can make a better estimate by subtracting a thousand: 16,100.
If you’re really good you notice that 342 is nine more than 333, so, three times 342 is three times nine more than three times 333. That is, it’s 27 more than 999. So the 16,100 estimate is 26 more than the correct number, 16,074. And I believe if you check, you will find the card in your hand is the ace of clubs. Am I not right, professor?
And that’s a look at comic strips through to a bit over a week ago. I intend to have more soon. All my Reading the Comics posts should be at this link. Thanks for being with me for these reviews. See you again soon.
## Reading the Comics, July 16, 2022: Brevity is the Soul of Wit Edition
Once more Comic Strip Master Command shows great parsimony with its mathematics-themed comic strips. The one with most substance to it is Dan Thompson’s Brevity, and even that’s not one I can talk about at length. Watch me try, anyway.
Dan Thompson’s Brevity for the 11th of July is the anthropomorphic numerals-and-symbols joke for the week. And it correctly identifies + as representing the addition operator. I talked about operators for the 2019 A-to-Z, though my focus there was on functions of functions. Here, we’ve just got a binary operator, taking two integers (or real numbers) and, if we follow its rule, matching this pair with an integer (or real number). Well, we can embed the integers (or real numbers) in the space of functions, if we want.
It happens that “smooth” is also a term of mathematical art. It refers to a function that you can take enough derivatives of. “Enough” here depends on our use. Depending on what our work is, we might need anything from one derivative to infinitely many derivatives. I’m not sure that I would call the ordinary arithmetic-type addition operator “smooth”. But I’m not confident in saying it’s not.
Greg Cravens’s The Buckets for the 12th of July is more wordplay than mathematics play. I’m including it because I like the playful energy of the comic strip and want more people to notice it. I like the idea of calling such an assembled doll a fraction figure, though.
Tony Rubino and Gary Markstein’s Daddy’s Home for the 14th of July is your usual story problem setup. It’s a standard enough insult joke, although it does use the mathematics content. There aren’t a lot of spelling questions that would let you set up an ambiguous “how many are left” joke. Maybe an English class.
This takes us through the middle of July, as these go. All my Reading the Comics posts should be at this link,. Will there be more comics for next week? I have my suspicions.
## Reading the Comics, June 26, 2022: First Doldrums of Summer Edition
I have not kept secret that I’ve had little energy lately. I hope that’s changing but can do little more than hope. I find it strange that my lack of energy seems to be matched by Comic Strip Master Command. Last week saw pretty slim pickings for mathematically-themed comics. Here’s what seems worth the sharing from my reading.
Lincoln Peirce’s Big Nate for the 22nd is a Pi Day joke, displaced to the prank day at the end of Nate’s school year. It’s also got a surprising number of people in the comments complaining that 3.1416 is only an approximation to π. It is, certainly, but so is any representation besides π or a similar mathematical expression. And introducing it with 3.1416 gives the reader the hint that this is about a mathematics expression and not an arbitrary symbol. It’s important to the joke that this be communicated clearly, and it’s hard to think of better ways to do that.
Dave Whamond’s Reality Check for the 24th is another in the line of “why teach algebra instead of something useful” strips. There are several responses. One is that certainly one should learn how to do a household budget; this was, at least back in the day, called home economics, and it was a pretty clear use of mathematics. Another is that a good education is about becoming literate in all the great thinking of humanity: you should come out knowing at least something coherent about mathematics and literature and exercise and biology and music and visual arts and more. Schools often fail to do all of this — how could they not? — but that’s not reason to fault them on parts of the education that they do. And anther is that algebra is about getting comfortable working with numbers before you know just what they are. That is, how to work out ways to describe a thing you want to know, and then to find what number (or range of numbers) that is. Still, these responses hardly matter. Mathematics has always lived in a twin space, of being both very practical and very abstract. People have always and will always complain that students don’t learn how to do the practical well enough. There’s not much changing that.
Charles Schulz’s Peanuts Begins for the 26th sees Violet challenge Charlie Brown to say what a non-perfect circle would be. I suppose this makes the comic more suitable for a philosophy of language blog, but I don’t know any. To be a circle requires meeting a particular definition. None of the things we ever point to and call circles meets that. We don’t generally have trouble connecting our imperfect representations of circles to the “perfect” ideal, though. And Charlie Brown said something meaningful in describing his drawing as being “a perfect circle”. It’s tricky pinning down exactly what it is, though.
And that is as much as last week moved me to write. This and my other Reading the Comics posts should be at this link. We’ll see whether the upcoming week picks up any.
## Reading the Comics, June 18, 2022: Pizza Edition
I’m back with my longest-running regular feature here. As I’ve warned I’m trying not to include every time one of the newspaper comics (that is, mostly, ones running on Comics Kingdom or GoComics) mentions the existence of arithmetic. So, for example, both Frank and Ernest and Rhymes with Orange did jokes about the names of the kinds of triangles. You can clip those at your leisure; I’m looking to discuss deeper subjects.
Scott Hilburn’s The Argyle Sweater is … well, it’s just an anthropomorphic-numerals joke. I have a weakness for The Wizard of Oz, that’s all. Also, I don’t know but somewhere in the nine kerspillion authorized books written since Baum’s death there be at least one with a “wizard of odds” plot.
Bill Amend’s FoxTrot reads almost like a word problem’s setup. There’s a difference in cost between pizzas of different sizes. Jason and Marcus make the supposition that they could buy the difference in sizes. They are asking for something physically unreasonable, but in a way that mathematics problems might do. The ring of pizza they’d be buying would be largely crust, after all. (Some people like crust, but I doubt any are ten-year-olds like Jason and Marcus.) The obvious word problem to spin out of this is extrapolating the costs of 20-inch or 8-inch pizzas, and maybe the base cost of making any pizza however tiny.
You can think of a 16-inch-wide circle as a 12-inch-wide circle with an extra ring around it. (An annulus, we’d say in the trades.) This is often a useful way to look at circles. If you get into calculus you’ll see the extra area you get from a slight increase in the diameter (or, more likely, the radius) all over the place. Also, in three dimensions, the difference in volume you get from an increase in diameter. There are also a good number of theorems with names like Green’s and Stokes’s. These are all about what you can know about the interior of a shape, like a pizza, from what you know about the ring around the edge.
Jim Meddick’s Monty sees Sedgwick, spoiled scion of New Jersey money, preparing for a mathematics test. He’s allowed the use of an abacus, one of the oldest and best-recognized computational aides. The abacus works by letting us turn the operations of basic arithmetic into physical operations. This has several benefits. We (generally) understand things in space pretty well. And the beads and wires serve as aides to memory, always a struggle. Sedgwick also brings out a “hyperbolic abacus”, a tool for more abstract operations like square roots and sines and cosines. I don’t know of anything by that name, but you can design mechanical tools to do particular computations. Slide rules, for example, generally have markings to let one calculate square roots and cube roots easily. Aircraft pilots might use a flight computer, a set of plastic discs to do quick estimates of flight time, fuel consumption, ground speed, and such. (There’s even an episode of the original Star Trek where Spock fiddles with one!)
I have heard, but not seen, that specialized curves were made to let people square circles with something approximating a compass-and-straightedge method. A contraption to calculate sines and cosines would not be hard to imagine. It would need to be a post on a hinge, mostly, with a set of lines to read off sine and cosine values over a range of angles. I don’t know of one that existed, as it’s easy enough to print out a table of trig functions, but it wouldn’t be hard to make.
And that’s enough for this week. This and all my other Reading the Comics posts should be at this link. I hope to get this back to a weekly column, but that does depend on Comic Strip Master Command doing what’s convenient for me. We’ll see how it turns out.
## Reading the Comics, June 3, 2022: Prime Choices Edition
I intended to be more casual about these comics when I resumed reading them for their mathematics content. I feel like Comic Strip Master Command is teasing me, though. There has been an absolute drought of comics with enough mathematics for me to really dig into. You can see that from this essay, which covers nearly a month of the strips I read and has two pieces that amount to “the cartoonist knows what a prime number is”. I must go with what I have, though.
Mark Anderson’s Andertoons for the 12th of May I would have sworn was a repeat. If it is, I don’t seem to have featured it before. It gives us Wavehead — I’ve learned his name is not consistent — learning about division. The first kind of division, at least, with a quotient and a remainder. The novel thing here, with integer division, is that the result is not a single number, but rather an ordered pair. I hadn’t thought about it that way before, I suppose since integer division and ordered pairs are introduced so far apart in one’s education.
We mostly put away this division-with-remainders as soon as we get comfortable with decimals. 19 ÷ 4 becoming “4 remainder 3” or “4.75” or “4 $latex\frac{3}{4}$” all impose a roughly equal cognitive load. But this division reappears in (high school) algebra, when we start dividing polynomials. (Almost anything you can do with integers there’s a similar thing you can do with polynomials. This is not just because you can rewrite the integer “4” as the polynomial “f(x) = 0x + 4”.) There may be something easier to understand in turning $\left(x^2 + 3x - 3\right) \div \left(x - 2\right)$ into $\left(x + 1\right)$ remainder $\left(4x - 1\right)$.
A thing happening here is that integer arithmetic is a ring. We study a lot of rings, as it’s not hard to come up with things that look like addition and subtraction and multiplication. Rings we don’t assume to have division that stays in your set. They can turn into pairs, like with integers or with polynomials. Having that division makes the ring into a field, so-called because we don’t have enough things called a “field” already.
Scott Hilburn’s The Argyle Sweater for the 16th of May is one of the prime number strips from this collection. About the only note worth mention is that the indivisibility of 3 depends on supposing we mean the integer 3. If we decided 3 was a real number, we would have every real number other than zero as a divisor. There’s similar results for complex numbers or polynomials. I imagine there’s a good fight one could get going about whether 3-in-integer-arithmetic is the same number as 3-in-real-arithmetic. I’m not ready for that right now, though.
I like the blood bag Dracula’s drinking from. Nice touch.
Dave Coverly’s Speed Bump for the 16th of May names the ways to classify triangles based on common side lengths (or common angles). There is some non-absurdity in the joke’s premise. Not the existence of these particular pennants. But that someone who loves a subject enough to major in it will often be a bit fannish about it? Yes. It’s difficult to imagine going any other way. You need to get to a pretty high leve of mathematics to go seriously into triangles, but the option is there.
Dave Whamond’s Reality Check for the 3rd of June is the other comic strip playing on the definition of “prime”. Here it’s applied to the hassle of package delivery, and the often comical way that items will get boxed in what seems to be no logical pattern. But there is a reason behind that lack of pattern. It is an extremely hard problem to get bunches of things together at once. It gets even harder when those things have to come from many different sources, and get warehoused in many disparate locations. Add to that the shipper’s understandable desire to keep stuff sitting around, waiting, for as little time as possible. So the waste in package and handling and delivery costs seems worth it to send an order in ten boxes than in finding how to send it all in one.
It feels like an obvious offense to reason to use four boxes to send five items. It can be hard to tell whether the cost of organizing things into fewer boxes outweighs the additional cost of transporting, mostly, air. This is not to say that I think the choice is necessarily made correctly. I don’t trust organizations to not decide “I dunno, we always did it this way”. I want instead to note that when you think hard about a question it often becomes harder to say what a good answer would be.
I can give you a good answer, though, if your question is how to read more comic strips alongside me. I try to put all my Reading the Comics posts at this link. You can see something like a decade’s worth of my finding things to write about students not answering word problems. Thank you for reading along with this.
## Reading the Comics, May 7, 2022: Does Comic Strip Master Command Not Do Mathematics Anymore Edition?
I mentioned in my last Reading the Comics post that it seems there are fewer mathematics-themed comic strips than there used to be. I know part of this is I’m trying to be more stringent. You don’t need me to say every time there’s a Roman numerals joke or that blackboards get mathematics symbols put on them. Still, it does feel like there’s fewer candidate strips. Maybe the end of the 2010s was a boom time for comic strips aimed at high school teachers and I only now appreciate that? Only further installments of this feature will let us know.
Jim Benton’s Jim Benton Cartoons for the 18th of April, 2022 suggests an origin for those famous overlapping circle pictures. This did get me curious what’s known about how John Venn came to draw overlapping circles. There’s no reason he couldn’t have used triangles or rectangles or any shape, after all. It looks like the answer is nobody really knows.
Venn, himself, didn’t name the diagrams after himself. Wikipedia credits Charles Dodgson (Lewis Carroll) as describing “Venn’s Method of Diagrams” in 1896. Clarence Irving Lewis, in 1918, seems to be the first person to write “Venn Diagram”. Venn wrote of them as “Eulerian Circles”, referencing the Leonhard Euler who just did everything. Sir William Hamilton — the philosopher, not the quaternions guy — posthumously published the Lectures On Metaphysics and Logic which used circles in these diagrams. Hamilton asserted, correctly, that you could use these to represent logical syllogisms. He wrote that the 1712 logic text Nucleus Logicae Weisianae — predating Euler — used circles, and was right about that. He got the author wrong, crediting Christian Weise instead of the correct author, Johann Christian Lange.
With 1712 the trail seems to end to this lay person doing a short essay’s worth of research. I don’t know what inspired Lange to try circles instead of any other shape. My guess, unburdened by evidence, is that it’s easy to draw circles, especially back in the days when every mathematician had a compass. I assume they weren’t too hard to typeset, at least compared to the many other shapes available. And you don’t need to even think about setting them with a rotation, the way a triangle or a pentagon might demand. But I also would not rule out a notion that circles have some connotation of perfection, in having infinite axes of symmetry and all points on them being equal in distance from the center and such. Might be the reasons fit in the intersection of the ethereal and the mundane.
Daniel Beyer’s Long Story Short for the 29th of April, 2022 puts out a couple of concepts from mathematical physics. These are all about geometry, which we now see as key to understanding physics. Particularly cosmology. The no-boundary proposal is a model constructed by James Hartle and Stephen Hawking. It’s about the first $10^{-43}$ seconds of the universe after the Big Bang. This is an era that was so hot that all our well-tested models of physical law break down. The salient part of the Hartle-Hawking proposal is the idea that in this epoch time becomes indistinguishable from space. If I follow it — do not rely on my understanding for your thesis defense — it’s kind of the way that stepping away from the North Pole first creates the ideas of north and south and east and west. It’s very hard to think of a way to test this which would differentiate it from other hypotheses about the first instances of the universe.
The Weyl Curvature is a less hypothetical construct. It’s a tensor, one of many interesting to physicists. This one represents the tidal forces on a body that’s moving along a geodesic. So, for example, how the moon of a planet gets distorted over its orbit. The Weyl Curvature also offers a way to describe how gravitational waves pass through vacuum. I’m not aware of any serious question of the usefulness or relevance of the thing. But the joke doesn’t work without at least two real physics constructs as setup.
Liniers’ Macanudo for the 5th of May, 2022 has one of the imps who inhabit the comic asserting responsibility for making mathematics work. It’s difficult to imagine what a creature could do to make mathematics work, or to not work. If pressed, we would say mathematics is the set of things we’re confident we could prove according to a small, pretty solid-seeming set of logical laws. And a somewhat larger set of axioms and definitions. (Few of these are proved completely, but that’s because it would involve a lot of fiddly boring steps that nobody doubts we could do if we had to. If this sounds sketchy, consider: do you believe my claim that I could alphabetize the books on the shelf to my right, even though I’ve never done that specific task? Why?) It would be like making a word-search puzzle not work.
The punch line, the blue imp counting seventeen of the orange imp, suggest what this might mean. Mathematics as a set of statements following some rule, is a niche interest. What we like is how so many mathematical things seem to correspond to real-world things. We can imagine mathematics breaking that connection to the real world. The high temperature rising one degree each day this week may tell us something about this weekend, but it’s useless for telling us about November. So I can imagine a magical creature deciding what mathematical models still correspond to the thing they model. Be careful in trying to change their mind.
And that’s as many comic strips from the last several weeks that I think merit discussion. All of my Reading the Comics posts should be at this link, though. And I hope to have a new one again sometime soon. I’ll ask my contacts with the cartoonists. I have about half of a contact.
## Reading the Comics, April 17, 2022: Did I Catch Comic Strip Master Command By Surprise Edition
Part of the thrill of Reading the Comics posts is that the underlying material is wholly outside my control. The subjects discussed, yes, although there are some quite common themes. (Students challenging the word problem; lottery jokes; monkeys at typewriters.) But also quantity. Part of what burned me out on Reading the Comics posts back in 2020 was feeling the need to say something about lots of comic strips . Now?
I mentioned last week seeing only three interesting strips, and one of them, Andertoons, was a repeat I’d already discussed. This week there were only two strips that drew a first note and again, Andertoons was a repeat I’d already discussed. Mark Anderson’s comic for the 17th I covered in enough detail back in August of 2019. I don’t know how many new Andertoons are put into the rotation at GoComics. But the implication is Comic Strip Master Command ordered mathematics-comics production cut down, and they haven’t yet responded to my doing these again. I guess we’ll know for sure if things pick up in a couple weeks, as the lead time allows.
So Rick McKee and Kent Sligh’s Mount Pleasant for the 15th of April is all I have to discuss. It’s part of the long series of students resisting the teacher’s question. The teacher is asking a fair enough question, that of how to do a problem that has several parts. She does ask how we “should” solve the problem of finding what 4 + 4 – 2 equals. The catch is there are several ways to do this, all of them as good. We know this if we’ve accepted subtraction as a kind of addition, and if we’ve accepted addition as commutative.
So the order is our choice. We can add 4 and 4 and then subtract 2. Or subtract 2 from the second 4, and then add that to the first 4. If you want, and can tell the difference, you could subtract 2 from the first 4, and then add the second 4 to that.
For this problem it doesn’t make any difference. But one can imagine similar ones where the order you tackle things in can make calculations easier, or harder. 5 + 7 – 2, for example, I find easier if I work it out as 5 + ( 7 – 2), that is, 5 + 5. So it’s worth taking a moment to consider whether rearranging it can make the calculation more reliable. I don’t know whether the teacher meant to challenge the students to see that there are alternatives, and no uniquely “right” answer. It’s possible McKee and Sligh did not have the teaching plan worked out.
That makes for another week’s worth of comic strips to discuss. All of my Reading the Comics posts should be at this link. Thanks for reading this and I will let you know if Comic Strip Master Command increases production of comics with mathematics themes.
## Reading the Comics, April 10, 2022: Quantum Entanglement Edition
I remember part of why I stopped doing Reading the Comics posts regularly was their volume. I read a lot of comics and it felt like everyone wanted to do a word problem joke. Since I started easing back into these posts it’s seemed like they’ve disappeared. When I put together this week’s collection, I only had three interesting ones. And one was Andertoons for the 10th of April. Andertoons is a stalwart here, but this particular strip was one I already talked about, back in 2019.
Another was the Archie repeat for the 10th of April. And that only lists mathematics as a school subject. It would be the same joke if it were English lit. Saying “differential calculus” gives it the advantage of specificity. It also suggests Archie is at least a good enough student to be taking calculus in high school, which isn’t bad. Differential calculus is where calculus usually starts, with the study of instantaneous changes. A person can, and should, ask how a change can be instantaneous. Part of what makes differential calculus is learning how to find something that matches our intuition about what it should be. And that never requires us to do something appalling like divide zero by zero. Our current definition took a couple centuries of wrangling to find a scheme that makes sense. It’s a bit much to expect high school students to pick it up in two months.
Ripley’s Believe It Or Not for the 10th of April, 2022 was the most interesting piece. This referenced a problem I didn’t remember having heard about, the “36 Officers puzzle” of Leonhard Euler. Euler’s name you know as he did foundational work in every field of mathematics ever. This particular puzzle ates to 1779, according to an article in Quanta Magazine which one of the Ripley’s commenters offered. Six army regiments each have six officers of six different ranks. How can you arrange them in a six-by-six square so that no row or column repeats a rank or regiment?
The problem sounds like it shouldn’t be hard. The two-by-two version of this is easy. So is three-by-three and four-by-four and even five-by-five. Oddly, seven-by-seven is, too. It looks like some form of magic square, and seems not far off being a sudoku problem either. So it seems weird that six-by-six should be particularly hard, but sometimes it happens like that. In fact, this happens to be impossible; a paper by Gaston Terry in 1901 proved there were none.
The solution discussed by Ripley’s is of a slightly different problem. So I’m not saying to not believe it, just, that you need to believe it with reservations. The modified problem casts this as a quantum-entanglement, in which the rank and regiment of an officer in one position is connected to that of their neighbors. I admit I’m not sure I understand this well enough to explain; I’m not confident I can give a clear answer why a solution of the entangled problem can’t be used for the classical problem.
The problem, at this point, isn’t about organizing officers anymore. It never was, since that started as an idle pastime. Legend has it that it started as a challenge about organizing cards; if you look at the paper you’ll see it presenting states as card suits and values. But the problem emerged from idle curiosity into practicality. These turn out to be applicable to quantum error detection codes. I’m not certain I can explain how myself. You might be able to convince yourself of this by thinking how you know that someone who tells you the sum of six odd numbers is itself an odd number made a mistake somewhere, and you can then look for what went wrong.
And that’s as many comics from last week as I feel like discussing. All my Reading the Comics posts should be gathered at this link. Thanks for reading this and I hope to do this again soon.
## Reading the Comics, April 2, 2022: Pi Day Extra Edition
I’m not sure that I will make a habit of this. It’s been a while since I did a regular Reading the Comics post, looking for mathematics topics in syndicated newspaper comic strips. I thought I might dip my toes in those waters again. Since my Pi Day essay there’ve been only a few with anything much to say. One of them was a rerun I’ve discussed before, too, a Bloom County Sunday strip that did an elaborate calculation to conceal the number 1. I’ve written about that strip twice before, in May 2016 and then in October 2016, so that’s too well-explained to need revisiting.
As it happens two of the three strips remaining were repeats, though ones I don’t think I’ve addressed before here.
Bill Amend’s FoxTrot Classics for the 18th of March looks like a Pi Day strip. It’s not, though: it originally ran the 16th of March, 2001. We didn’t have Pi Day back then.
What Peter Fox is doing is drawing a unit circle — a circle of radius 1 — and dividing it into a couple common angles. Trigonometry students are expected to know the sines and cosines and tangents of a handful of angles. If they don’t know them, they can work these out from first principles. Draw a line from the center of the unit circle at an angle measured counterclockwise from the positive x-axis. Find where that line you’ve just drawn intersects the unit circle. The x-coordinate of that point has the same value as the cosine of that angle. The y-coordinate of that point has the same value as the sine of that angle. And for a handful of angles — the ones Peter marks off in the second panel — you can work them out by reason alone.
These angles we know as, like, 45 degrees or 120 degrees or 135 degrees. Peter writes them as $\frac{\pi}{4}$ or $\frac{2}{3}\pi$ or $\frac{9}{8}\pi$, because these are radian measure rather than degree measure. It’s a different scale, one that’s more convenient for calculus. And for some ordinary uses too: an angle of (say) $\frac{3}{4}\pi$ radians sweeps out an arc of length $\frac{3}{4}\pi$ on the unit circle. You can see where that’s easier to keep straight than how long an arc of 135 degrees might be.
Drawing this circle is a good way to work out or remember sines and cosines for the angles you’re expected to know, which is why you’d get them on a trig test.
Scott Hilburn’s The Argyle Sweater for the 27th of March summons every humorist’s favorite piece of topology, the Möbius strip. Unfortunately the line work makes it look to me like Hilburn’s drawn a simple loop of a steak. Follow the white strip along the upper edge. Could be the restaurant does the best it can with a challenging presentation.
August Ferdinand Möbius by the way was an astronomer, working most of his career at the Observatory at Leipzig. (His work as a professor was not particularly successful; he was too poor a lecturer to keep students.) His father was a dancing teacher, and his mother was a descendant of Martin Luther, although I imagine she did other things too.
Rina Piccolo’s Tina’s Groove for the 2nd of April makes its first appearance in a Reading the Comics post in almost a decade. The strip ended in 2017 and only recently has Comics Kingdom started showing reprints. The strip is about the numerical coincidence between 3.14 of a thing and the digits of π. It originally ran at the end of March, 2007, which like the vintage FoxTrot reminds us how recent a thing Pi Day is to observe.
3.14 hours is three hours, 8.4 minutes, which implies that she clocked in at about 9:56.
And that’s this installment. All my Reading the Comics posts should be at this link. I don’t know when I’ll publish a next one, but it should be there, too. Thanks for reading.
## Reading the Comics, March 14, 2022: Pi Day Edition
As promised I have the Pi Day comic strips from my reading here. I read nearly all the comics run on Comics Kingdom and on GoComics, no matter how hard their web sites try to avoid showing comics. (They have some server optimization thing that makes the comics sometimes just not load.) (By server optimization I mean “tracking for advertising purposes”.)
Pi Day in the comics this year saw the event almost wholly given over to the phonetic coincidence that π sounds, in English, like pie. So this is not the deepest bench of mathematical topics to discuss. My love, who is not as fond of wordplay as I am, notes that the ancient Greeks likely pronounced the name of π about the same way we pronounce the letter “p”. This may be etymologically sound, but that’s not how we do it in English, and even if we switched over, that would not make things better.
Scott Hilburn’s The Argyle Sweater is one of the few strips not to be about food. It is set in the world of anthropomorphized numerals, the other common theme to the day.
John Hambrook’s The Brilliant Mind of Edison Lee leads off with the food jokes, in this case cookies rather than pie. The change adds a bit of Abbott-and-Costello energy to the action.
Mick Mastroianni and Mason Mastroianni’s Dogs of C Kennel gets our first pie proper, this time tossed in the face. One of the commenters observes that the middle of a pecan pie can really hold heat, “Ouch”. Will’s holding it in his bare paw, though, so it can’t be that bad.
Jules Rivera’s Mark Trail makes the most casual Pi Day reference. If the narrator hadn’t interrupted in the final panel no one would have reason to think this referenced anything.
Mark Parisi’s Off The Mark is the other anthropomorphic numerals joke for the day. It’s built on the familiar fact that the digits of π go on forever. This is true for any integer base. In base π, of course, the representation of π is just “10”. But who uses that? And in base π, the number six would be something with infinitely many digits. There’s no fitting that in a one-panel comic, though.
Doug Savage’s Savage Chickens is the one strip that wasn’t about food or anthropomorphized numerals. There is no practical reason to memorize digits of π, other than that you’re calculating something by hand and don’t want to waste time looking them up. In that case there’s not much call go to past 3.14. If you need more than about 3.14159, get a calculator to do it. But memorizing digits can be fun, and I will not underestimate the value of fun in getting someone interested in mathematics.
For my part, I memorized π out to 3.1415926535787932, so that’s sixteen digits past the decimal. Always felt I could do more and I don’t know why I didn’t. The next couple digits are 8462, which has a nice descending-fifths cadence to it. The 626 following is a neat coda. My describing it this way may give you some idea to how I visualize the digits of π. They might help you, if you figure for some reason you need to do this. You do not, but if you enjoy it, enjoy it.
Bianca Xunise’s Six Chix for the 15th ran a day late; Xunise only gets the comic on Tuesdays and the occasional Sunday. It returns to the food theme.
And this brings me to the end of this year’s Pi Day comic strips. All of my Reading the Comics posts, past and someday future, should be at this link. And my various Pi Day essays should be here. Thank you for reading.
## Here Are Past Years’ Pi Day Comic Strips
I haven’t yet read today’s comics; it takes a while to get through them. But I hope to summarize what Comic Strip Master Command has sent out for the syndicated comics for today. In the meanwhile, here’s Pi Day strips of past years.
And I have to offer a warning. GoComics.Com has discontinued a lot of comics in the past couple years. They’ve been brutal about removing the archives of strips they’ve discontinued. Comics Kingdom is similarly ruthless in removing strips not in production. And a recent and, to the user, bad code update broke a lot of what had been non-expiring links. But my discussions of the themes in the comic are still there. And, as I got more into the Reading the Comics project I got more likely to include the original comic. So that’s some compensation.
Here’s the past several years in comics from on or around the 14th of March:
• 2015, featuring The Argyle Sweater, Baldo, The Chuckle Brothers, Dog Eat Doug, FoxTrot Classics, Herb and Jamaal, Long Story Short, The New Adventures of Queen Victoria, Off The Mark, and Working Daze.
• 2016, featuring The Argyle Sweater, B.C., Brewster Rockit, The Brilliant Mind of Edison Lee, Curtis, Dog Eat Doug, F Minus, Free Range, and Holiday Doodles.
• 2017, featuring 2 Cows and a Chicken, Archie, The Argyle Sweater, Arlo and Janis, Lard’s World Peace Tips, Loose Parts, Off The Mark, Saturday Morning Breakfast Cereal, TruthFacts, and Working Daze.
• 2018, featuring The Argyle Sweater, Bear With Me, Funky Winterbean Classic, Mutt and Jeff, Off The Mark, Savage Chickens, Warped, and Working Daze.
• 2019, featuring The Brilliant Mind of Edison Lee, Liz Climo’s Cartoons, The Grizzwells, Off The Mark, and Working Daze.
• 2020, featuring Baldo, Calvin and Hobbes, Off The Mark, Real Life Adventures, Reality Check, and Warped.
• 2021, featuring Agnes, The Argyle Sweater, Between Friends, Breaking Cat News, FoxTrot, Frazz, Get Fuzzy, Heart of the City, Reality Check, and Studio Jantze.
As mentioned, I have yet to read today’s comics. I’m looking forward to it, at least to learn what Funky Winkerbean character I’m going to be most annoyed with this week. It will be Les Moore. I was also going to look forward to seeing if there would ever be a Pi Day strips roundup without The Argyle Sweater or Reality Check. It turns out there was one in 2019. Weird how you can get the impression something is always there even when it’s not.
## Some Progress on the Infinitude of Monkeys
I have been reading Pierre-Simon LaPlace, 1749 – 1827, A Life In Exact Science, by Charles Coulson Gillispie with Robert Fox and Ivor Grattan-Guinness. It’s less of a biography than I expected and more a discussion of LaPlace’s considerable body of work. Part of LaPlace’s work was in giving probability a logically coherent, rigorous meaning. Laplace discusses the gambler’s fallacy and the tendency to assign causes to random events. That, for example, if we came across letters from a printer’s font reading out ‘INFINITESIMAL’ we would think that deliberate. We wouldn’t think that for a string of letters in no recognized language. And that brings up this neat quote from Gillispie:
The example may in all probability be adapted from the chapter in the Port-Royal La Logique (1662) on judgement of future events, where Arnauld points out that it would be stupid to bet twenty sous against ten thousand livres that a child playing with printer’s type would arrange the letters to compose the first twenty lines of Virgil’s Aenid.
The reference here is to a book by Antoine Arnauld and Pierre Nicole that I haven’t read or heard of before. But it makes a neat forerunner to the Infinite Monkey Theorem. That’s the study of what probability means when put to infinitely great or long processes. Émile Borel’s use of monkeys at a typewriter echoes this idea of children playing beyond their understanding. I don’t know whether Borel knew of Arnauld and Nicole’s example. But I did not want my readers to miss a neat bit of infinite-monkey trivia. Or to miss today’s Bizarro, offering yet another comic on the subject.
## In Which I Feel A Little Picked On
This is not a proper Reading the Comics post, since there’s nothing mathematical about this. But it does reflect a project I’ve been letting linger for months and that I intend to finish before starting the abbreviated Mathematics A-to-Z for this year.
In the meanwhile. I have a person dear to me who’s learning college algebra. For no reason clear to me this put me in mind of last year’s essay about Extraneous Solutions. These are fun and infuriating friends. They’re created when you follow the rules about how you can rewrite a mathematical expression without changing its value. And yet sometimes you do these rewritings correctly and get a would-be solution that isn’t actually one. So I’d shared some thoughts about why they appear, and what tedious work keeps them from showing up.
## Reading the Comics, May 25, 2021: Hilbert’s Hotel Edition
I have only a couple strips this time, and from this week. I’m not sure when I’ll return to full-time comics reading, but I do want to share strips that inspire something.
Carol Lay’s Lay Lines for the 24th of May riffs on Hilbert’s Hotel. This is a metaphor often used in pop mathematics treatments of infinity. So often, in fact, a friend snarked that he wished for any YouTube mathematics channel that didn’t do the same three math theorems. Hilbert’s Hotel was among them. I think I’ve never written a piece specifically about Hilbert’s Hotel. In part because every pop mathematics blog has one, so there are better expositions available. I have a similar restraint against a detailed exploration of the different sizes of infinity, or of the Monty Hall Problem.
Hilbert’s Hotel is named for David Hilbert, of Hilbert problems fame. It’s a thought experiment to explore weird consequences of our modern understanding of infinite sets. It presents various cases about matching elements of a set to the whole numbers, by making it about guests in hotel rooms. And then translates things we accept in set theory, like combining two infinitely large sets, into material terms. In material terms, the operations seem ridiculous. So the set of thought experiments get labelled “paradoxes”. This is not in the logician sense of being things both true and false, but in the ordinary sense that we are asked to reconcile our logic with our intuition.
So the Hotel serves a curious role. It doesn’t make a complex idea understandable, the way many demonstrations do. It instead draws attention to the weirdness in something a mathematics student might otherwise nod through. It does serve some role, or it wouldn’t be so popular now.
It hasn’t always been popular, though. Hilbert introduced the idea in 1924, though per a paper by Helge Kragh, only to address one question. A modern pop mathematician would have a half-dozen problems. George Gamow’s 1947 book One Two Three … Infinity brought it up again, but it didn’t stay in the public eye. It wasn’t until the 1980s that it got a secure place in pop mathematics culture, and that by way of philosophers and theologians. If you aren’t up to reading the whole of Kragh’s paper, I did summarize it a bit more completely in this 2018 Reading the Comics essay.
Anyway, Carol Lay does an great job making a story of it.
Leigh Rubin’s Rubes for the 25th of May I’ll toss in here too. It’s a riff on the art convention of a blackboard equation being meaningless. Normally, of course, the content of the equation doesn’t matter. So it gets simplified and abstracted, for the same reason one draws a brick wall as four separate patches of two or three bricks together. It sometimes happens that a cartoonist makes the equation meaningful. That’s because they’re a recovering physics major like Bill Amend of FoxTrot. Or it’s because the content of the blackboard supports the joke. Which, in this case, it does.
The essays I write about comic strips I tag so they appear at this link. You may enjoy some more pieces there.
So this is not a mathematics-themed comic update, not really. It’s just a bit of startling news about frequent Reading the Comics subject Andertoons. A comic strip back in December revealed that Wavehead had a specific name. According to the strip from the 3rd of December, the student most often challenging the word problem or the definition on the blackboard is named Tommy.
And then last week we got this bombshell:
So, also, it turns out I should have already known this since the strip ran in 2018 also. All I can say is I have a hard enough time reading nearly every comic strip in the world. I can’t be expected to understand them too.
So as not to leave things too despairing let me share a mathematics-mentioning Andertoons from yesterday and also from July 2018.
I don’t know if it’s run before that.
## Reading the Comics, December 20, 1948: What is Barnaby’s friend’s name Edition?
Have a special one today. I’ve been reading a compilation of Crockett Johnson’s 1940s comic Barnaby. The title character, an almost too gentle child, follows his fairy godfather Mr O’Malley into various shenanigans. Many (the best ones, I’d say) involve the magical world. The steady complication is that Mr O’Malley boasts abilities beyond his demonstrated competence. (Although most of the magic characters are shown to be not all that good at their business.) It’s a gentle strip and everything works out all right, if farcically.
This particular strip comes from a late 1948 storyline. Mr O’Malley’s gone missing, coincidentally to a fairy cop come to arrest the pixie, who is a con artist at heart. So this sees the entry of Atlas, the Mental Giant, who’s got some pleasant gimmicks. One of them is his requiring mnemonics built on mathematical formulas to work out names. And this is a charming one, with a great little puzzle: how do you get A-T-L-A-S out of the formula Atlas has remembered?
I’m sorry the solution requires a bit of abusing notation, so please forgive it. But it’s a fun puzzle, especially as the joke would not be funnier if the formula didn’t work. I’m always impressed when a comic strip goes to that extra effort.
Johnson, who also wrote the Harold and the Purple Crayon books, painted over a hundred canvasses with theorem-based pictures. There’s a selection of them at the Smithsonian Institute’s web site, here.
## Reading the Comics Follow-up: Where Else Is A Tetrahedron’s Centroid Edition
A Reading the Comics post a couple weeks back inspired me to find the centroid of a regular tetrahedron. A regular tetrahedron, also known as “a tetrahedron”, is the four-sided die shape. A pyramid with triangular base. Or a cone with a triangle base, if you prefer. If one asks a person to draw a tetrahedron, and they comply, they’ll likely draw this shape. The centroid, the center of mass of the tetrahedron, is at a point easy enough to find. It’s on the perpendicular between any of the four faces — the equilateral triangles — and the vertex not on that face. Particularly, it’s one-quarter the distance from the face towards the other vertex. We can reason that out purely geometrically, without calculating, and I did in that earlier post.
But most tetrahedrons are not regular. They have centroids too; where are they?
Thing is I know the correct answer going in. It’s at the “average” of the vertices of the tetrahedron. Start with the Cartesian coordinates of the four vertices. The x-coordinate of the centroid is the arithmetic mean of the x-coordinates of the four vertices. The y-coordinate of the centroid is the mean of the y-coordinates of the vertices. The z-coordinate of the centroid is the mean of the z-coordinates of the vertices. Easy to calculate; but, is there a way to see that this is right?
What’s got me is I can think of an argument that convinces me. So in this sense, I have an easy proof of it. But I also see where this argument leaves a lot unaddressed. So it may not prove things to anyone else. Let me lay it out, though.
So start with a tetrahedron of your own design. This will be less confusing if I have labels for the four vertices. I’m going to call them A, B, C, and D. I don’t like those labels, not just for being trite, but because I so want ‘C’ to be the name for the centroid. I can’t find a way to do that, though, and not have the four tetrahedron vertices be some weird set of letters. So let me use ‘P’ as the name for the centroid.
Where is P, relative to the points A, B, C, and D?
And here’s where I give a part of an answer. Start out by putting the tetrahedron somewhere convenient. That would be the floor. Set the tetrahedron so that the face with triangle ABC is in the xy plane. That is, points A, B, and C all have the z-coordinate of 0. The point D has a z-coordinate that is not zero. Let me call that coordinate h. I don’t care what the x- and y-coordinates for any of these points are. What I care about is what the z-coordinate for the centroid P is.
The property of the centroid that was useful last time around was that it split the regular tetrahedron into four smaller, irregular, tetrahedrons, each with the same volume. Each with one-quarter the volume of the original. The centroid P does that for the tetrahedron too. So, how far does the point P have to be from the triangle ABC to make a tetrahedron with one-quarter the volume of the original?
The answer comes from the same trick used last time. The volume of a cone is one-third the area of the base times its altitude. The volume of the tetrahedron ABCD, for example, is one-third times the area of triangle ABC times how far point D is from the triangle. That number I’d labelled h. The volume of the tetrahedron ABCP, meanwhile, is one-third times the area of triangle ABC times how far point P is from the triangle. So the point P has to be one-quarter as far from triangle ABC as the point D is. It’s got a z-coordinate of one-quarter h.
Notice, by the way, that while I don’t know anything about the x- and y- coordinates of any of these points, I do know the z-coordinates. A, B, and C all have z-coordinate of 0. D has a z-coordinate of h. And P has a z-coordinate of one-quarter h. One-quarter h sure looks like the arithmetic mean of 0, 0, 0, and h.
At this point, I’m convinced. The coordinates of the centroid have to be the mean of the coordinates of the vertices. But you also see how much is not addressed. You’d probably grant that I have the z-coordinate coordinate worked out when three vertices have the same z-coordinate. Or where three vertices have the same y-coordinate or the same x-coordinate. You might allow that if I can rotate a tetrahedron, I can get three points to the same z-coordinate (or y- or x- if you like). But this still only gets one coordinate of the centroid P.
I’m sure a bit of algebra would wrap this up. But I would like to avoid that, if I can. I suspect the way to argue this geometrically depends on knowing the line from vertex D to tetrahedron centroid P, if extended, passes through the centroid of triangle ABC. And something similar applies for vertexes A, B, and C. I also suspect there’s a link between the vector which points the direction from D to P and the sum of the three vectors that point the directions from D to A, B, and C. I haven’t quite got there, though.
I will let you know if I get closer.
## Reading the Comics, April 1, 2021: Why Is Gunther Taking Algebraic Topology Edition
I’m not yet looking to discuss every comic strip with any mathematics mention. But something gnawed at me in this installment of Greg Evans and Karen Evans’s Luann. It’s about the classes Gunther says he’s taking.
The main characters in Luann are in that vaguely-defined early-adult era. They’re almost all attending a local university. They’re at least sophomores, since they haven’t been doing stories about the trauma and liberation of going off to school. How far they’ve gotten has been completely undefined. So here’s what gets me.
Gunther taking vector calculus? That makes sense. Vector calculus is a standard course if you’re taking any mathematics-dependent major. It might be listed as Multivariable Calculus or Advanced Calculus or Calculus III. It’s where you learn partial derivatives, integrals along a path, integrals over a surface or volume. I don’t know Gunther’s major, but if it’s any kind of science, yeah, he’s taking vector calculus.
Algebraic topology, though. That I don’t get. Topology at all is usually an upper-level course. It’s for mathematics majors, maybe physics majors. Not every mathematics major takes topology. Algebraic topology is a deeper specialization of the subject. I’ve only seen courses listed as algebraic topology as graduate courses. It’s possible for an undergraduate to take a graduate-level course, yes. And it may be that Gunther is taking a regular topology course, and the instructor prefers to focus on algebraic topology.
But even a regular topology course relies on abstract algebra. Which, again, is something you’ll get as an undergraduate. If you’re a mathematics major you’ll get at least two years of algebra. And, if my experience is typical, still feel not too sure about the subject. Thing is that Intro to Abstract Algebra is something you’d plausibly take at the same time as Vector Calculus. Then you’d get Abstract Algebra and then, if you wished, Topology.
So you see the trouble. I don’t remember anything in algebra-to-topology that would demand knowing vector calculus. So it wouldn’t mean Gunther took courses without taking the prerequisites. But it’s odd to take an advanced mathematics course at the same time as a basic mathematics course. Unless Gunther’s taking an advanced vector calculus course, which might be. Although since he wants to emphasize that he’s taking difficult courses, it’s odd to not say “advanced”. Especially if he is tossing in “algebraic” before topology.
And, yes, I’m aware of the Doylist explanation for this. The Evanses wanted courses that sound impressive and hard. And that’s all the scene demands. The joke would not be more successful if they picked two classes from my actual Junior year schedule. None of the characters have a course of study that could be taken literally. They’ve been university students full-time since 2013 and aren’t in their senior year yet. It would be fun, is all, to find a way this makes sense.
This and my other essays discussing something from the comic strips are at this link.
## Reading the Comics, March 16, 2021: Where Is A Tetrahedron’s Centroid Edition
Comic Strip Master Command has not, to appearances, been distressed by my Reading the Comics hiatus. There are still mathematically-themed comic strips. Many of them are about story problems and kids not doing them. Some get into a mathematical concept. One that ran last week caught my imagination so I’ll give it some time here. This and other Reading the Comics essays I have at this link, and I figure to resume posting them, at least sometimes.
Ben Zaehringer’s In The Bleachers for the 16th of March, 2021 is an anthropomorphized-geometry joke. Here the centroid stands in for “the waist”, the height below which boxers may not punch.
The centroid is good geometry, something which turns up in plane and solid shapes. It’s a center of the shape: the arithmetic mean of all the points in the shape. (There are other things that can, with reason, be called a center too. Mathworld mentions the existence of 2,001 things that can be called the “center” of a triangle. It must be only a lack of interest that’s kept people from identifying even more centers for solid shapes.) It’s the center of mass, if the shape is a homogenous block. Balance the shape from below this centroid and it stays balanced.
For a complicated shape, finding the centroid is a challenge worthy of calculus. For these shapes, though? The sphere, the cube, the regular tetrahedron? We can work those out by reason. And, along the way, work out whether this rule gives an advantage to either boxer.
The sphere first. That’s the easiest. The centroid has to be the center of the sphere. Like, the point that the surface of the sphere is a fixed radius from. This is so obvious it takes a moment to think why it’s obvious. “Why” is a treacherous question for mathematics facts; why should 4 divide 8? But sometimes we can find answers that give us insight into other questions.
Here, the “why” I like is symmetry. Look at a sphere. Suppose it lacks markings. There’s none of the referee’s face or bow tie here. Imagine then rotating the sphere some amount. Can you see any difference? You shouldn’t be able to. So, in doing that rotation, the centroid can’t have moved. If it had moved, you’d be able to tell the difference. The rotated sphere would be off-balance. The only place inside the sphere that doesn’t move when the sphere is rotated is the center.
This symmetry consideration helps answer where the cube’s centroid is. That also has to be the center of the cube. That is, halfway between the top and bottom, halfway between the front and back, halfway between the left and right. Symmetry again. Take the cube and stand it upside-down; does it look any different? No, so, the centroid can’t be any closer to the top than it can the bottom. Similarly, rotate it 180 degrees without taking it off the mat. The rotation leaves the cube looking the same. So this rules out the centroid being closer to the front than to the back. It also rules out the centroid being closer to the left end than to the right. It has to be dead center in the cube.
Now to the regular tetrahedron. Obviously the centroid is … all right, now we have issues. Dead center is … where? We can tell when the regular tetrahedron’s turned upside-down. Also when it’s turned 90 or 180 degrees.
Symmetry will guide us. We can say some things about it. Each face of the regular tetrahedron is an equilateral triangle. The centroid has to be along the altitude. That is, the vertical line connecting the point on top of the pyramid with the equilateral triangle base, down on the mat. Imagine looking down on the shape from above, and rotating the shape 120 or 240 degrees if you’re still not convinced.
And! We can tip the regular tetrahedron over, and put another of its faces down on the mat. The shape looks the same once we’ve done that. So the centroid has to be along the altitude between the new highest point and the equilateral triangle that’s now the base, down on the mat. We can do that for each of the four sides. That tells us the centroid has to be at the intersection of these four altitudes. More, that the centroid has to be exactly the same distance to each of the four vertices of the regular tetrahedron. Or, if you feel a little fancier, that it’s exactly the same distance to the centers of each of the four faces.
It would be nice to know where along this altitude this intersection is, though. We can work it out by algebra. It’s no challenge to figure out the Cartesian coordinates for a good regular tetrahedron. Then finding the point that’s got the right distance is easy. (Set the base triangle in the xy plane. Center it, so the coordinates of the highest point are (0, 0, h) for some number h. Set one of the other vertices so it’s in the xz plane, that is, at coordinates (0, b, 0) for some b. Then find the c so that (0, 0, c) is exactly as far from (0, 0, h) as it is from (0, b, 0).) But algebra is such a mass of calculation. Can we do it by reason instead?
That I ask the question answers it. That I preceded the question with talk about symmetry answers how to reason it. The trick is that we can divide the regular tetrahedron into four smaller tetrahedrons. These smaller tetrahedrons aren’t regular; they’re not the Platonic solid. But they are still tetrahedrons. The little tetrahedron has as its base one of the equilateral triangles that’s the bigger shape’s face. The little tetrahedron has as its fourth vertex the centroid of the bigger shape. Draw in the edges, and the faces, like you’d imagine. Three edges, each connecting one of the base triangle’s vertices to the centroid. The faces have two of these new edges plus one of the base triangle’s edges.
The four little tetrahedrons have to all be congruent. Symmetry again; tip the big tetrahedron onto a different face and you can’t see a difference. So we’ll know, for example, all four little tetrahedrons have the same volume. The same altitude, too. The centroid is the same distance to each of the regular tetrahedron’s faces. And the four little tetrahedrons, together, have the same volume as the original regular tetrahedron.
What is the volume of a tetrahedron?
If we remember dimensional analysis we may expect the volume should be a constant times the area of the base of the shape times the altitude of the shape. We might also dimly remember there is some formula for the volume of any conical shape. A conical shape here is something that’s got a simple, closed shape in a plane as its base. And some point P, above the base, that connects by straight lines to every point on the base shape. This sounds like we’re talking about circular cones, but it can be any shape at the base, including polygons.
So we double-check that formula. The volume of a conical shape is one-third times the area of the base shape times the altitude. That’s the perpendicular distance between P and the plane that the base shape is in. And, hey, one-third times the area of the face times the altitude is exactly what we’d expect.
So. The original regular tetrahedron has a base — has all its faces — with area A. It has an altitude h. That h must relate in some way to the area; I don’t care how. The volume of the regular tetrahedron has to be $\frac{1}{3} A h$.
The volume of the little tetrahedrons is — well, they have the same base as the original regular tetrahedron. So a little tetrahedron’s base is A. The altitude of the little tetrahedron is the height of the original tetrahedron’s centroid above the base. Call that $h_c$. How can the volume of the little tetrahedron, $\frac{1}{3} A h_c$, be one-quarter the volume of the original tetrahedron, $\frac{1}{3} A h$? Only if $h_c$ is one-quarter $h$.
This pins down where the centroid of the regular tetrahedron has to be. It’s on the altitude underneath the top point of the tetrahedron. It’s one-quarter of the way up from the equilateral-triangle face.
(And I’m glad, checking this out, that I got to the right answer after all.)
So, if the cube and the tetrahedron have the same height, then the cube has an advantage. The cube’s centroid is higher up, so the tetrahedron has a narrower range to punch. Problem solved.
I do figure to talk about comic strips, and mathematics problems they bring up, more. I’m not sure how writing about one single strip turned into 1300 words. But that’s what happens every time I try to do something simpler. You know how it goes.
## Reading the Comics, March 14, 2021: Pi Day Edition
I was embarrassed, on looking at old Pi Day Reading the Comics posts, to see how often I observed there were fewer Pi Day comics than I expected. There was not a shortage this year. This even though if Pi Day has any value it’s as an educational event, and there should be no in-person educational events while the pandemic is still on. Of course one can still do educational stuff remotely, mathematics especially. But after a year of watching teaching on screens and sometimes doing projects at home, it’s hard for me to imagine a bit more of that being all that fun.
But Pi Day being a Sunday did give cartoonists more space to explain what they’re talking about. This is valuable. It’s easy for the dreadfully online, like me, to forget that most people haven’t heard of Pi Day. Most people don’t have any idea why that should be a thing or what it should be about. This seems to have freed up many people to write about it. But — to write what? Let’s take a quick tour of my daily comics reading.
Tony Cochran’s Agnes starts with some talk about Daylight Saving Time. Agnes and Trout don’t quite understand how it works, and get from there to Pi Day. Or as Agnes says, Pie Day, missing the mathematics altogether in favor of the food.
Scott Hilburn’s The Argyle Sweater is an anthropomorphic-numerals joke. It’s a bit risqué compared to the sort of thing you expect to see around here. The reflection of the numerals is correct, but it bothered me too.
Georgia Dunn’s Breaking Cat News is a delightful cute comic strip. It doesn’t mention mathematics much. Here the cat reporters do a fine job explaining what Pi Day is and why everybody spent Sunday showing pictures of pies. This could almost be the standard reference for all the Pi Day strips.
Bill Amend’s FoxTrot is one of the handful that don’t mention pie at all. It focuses on representing the decimal digits of π. At least within the confines of something someone might write in the American dating system. The logic of it is a bit rough but if we’ve accepted 3-14 to represent 3.14, we can accept 1:59 as standing in for the 0.00159 of the original number. But represent 0.0015926 (etc) of a day however you like. If we accept that time is continuous, then there’s some moment on the 14th of March which matches that perfectly.
Jef Mallett’s Frazz talks about the eliding between π and pie for the 14th of March. The strip wonders a bit what kind of joke it is exactly. It’s a nerd pun, or at least nerd wordplay. If I had to cast a vote I’d call it a language gag. If they celebrated Pi Day in Germany, there would not be any comic strips calling it Tortentag.
Steenz’s Heart of the City is another of the pi-pie comics. I do feel for Heart’s bewilderment at hearing π explained at length. Also Kat’s desire to explain mathematics overwhelming her audience. It’s a feeling I struggle with too. The thing is it’s a lot of fun to explain things. It’s so much fun you can lose track whether you’re still communicating. If you set off one of these knowledge-floods from a friend? Try to hold on and look interested and remember any single piece anywhere of it. You are doing so much good for your friend. And if you realize you’re knowledge-flooding someone? Yeah, try not to overload them, but think about the things that are exciting about this. Your enthusiasm will communicate when your words do not.
Dave Whamond’s Reality Check is a pi-pie joke that doesn’t rely on actual pie. Well, there’s a small slice in the corner. It relies on the infinite length of the decimal representation of π. (Or its representation in any integer base.)
Michael Jantze’s Studio Jantze ran on Monday instead, although the caption suggests it was intended for Pi Day. So I’m including it here. And it’s the last of the strips sliding the day over to pie.
But there were a couple of comic strips with some mathematics mention that were not about Pi Day. It may have been coincidence.
Sandra Bell-Lundy’s Between Friends is of the “word problem in real life” kind. It’s a fair enough word problem, though, asking about how long something would take. From the premises, it takes a hair seven weeks to grow one-quarter inch, and it gets trimmed one quarter-inch every six weeks. It’s making progress, but it might be easier to pull out the entire grey hair. This won’t help things though.
Darby Conley’s Get Fuzzy is a rerun, as all Get Fuzzy strips are. It first (I think) ran the 13th of September, 2009. And it’s another Infinite Monkeys comic strip, built on how a random process should be able to create specific outcomes. As often happens when joking about monkeys writing Shakespeare, some piece of pop culture is treated as being easier. But for these problems the meaning of the content doesn’t count. Only the length counts. A monkey typing “let it be written in eight and eight” is as improbable as a monkey typing “yrg vg or jevggra va rvtug naq rvtug”. It’s on us that we find one of those more impressive than the other.
And this wraps up my Pi Day comic strips. I don’t promise that I’m back to reading the comics for their mathematics content regularly. But I have done a lot of it, and figure to do it again. All my Reading the Comics posts appear at this link. Thank you for reading and I hope you had some good pie.
I don’t know how Andertoons didn’t get an appearance here.
## Breaking Andertoons News: Wavehead has a name
I will be late with this week’s A-to-Z essay. I’ve had more demands on my time and my ability to organize thoughts than I could manage and something had to yield. I’m sorry for that but figure to post on Friday something for the letter ‘Y’.
But there is some exciting news in one of my regular Reading the Comics features. It’s about the kid who shows up often in Mark Anderson’s Andertoons. At the nomination of — I want to say Ray Kassinger? — I’ve been calling him “Wavehead”. Last week, though, the strip gave his name. I don’t know if this is the first time we’ve seen it. It is the first time I’ve noticed. He turns out to be Tommy.
And what about my Reading the Comics posts, which have been on suspension since the 2020 A-to-Z started? I’m not sure. I figure to resume them after the new year. I don’t know that it’ll be quite the same, though. A lot of mathematics mentions in comic strips are about the same couple themes. It is exhausting to write about the same thing every time. But I have, I trust, a rotating readership. Someone may not know, or know how to find, a decent 200-word piece about lotteries published four months in the past. I need to better balance not repeating myself.
Also a factor is lightening my overhead. Most of my strips come from Comics Kingdom or GoComics. Both of them also cull strips from their archives occasionally, leaving me with dead links. (GoComics particularly is dropping a lot of strips by the end of 2020. I understand them dumping, say, The Sunshine Club, which has been in reruns since 2007. But Dave Kellett’s Sheldon?)
The only way to make sure a strip I write about remains visible to my readers is to include it here. But to make my including the strip fair use requires that I offer meaningful commentary. I have to write something substantial, and something that’s worsened without the strip to look at. You see how this builds to a workload spiral, especially for strips where all there is to say is it’s a funny story problem. (If any cartoonists are up for me being another, unofficial archive for their mathematics-themed strips? Drop me a comment, Bill Amend, we can work something out if it doesn’t involve me sending more money than I’m taking in.)
So I don’t know how I’ll resolve all this. Key will be remembering that I can just not do the stuff I find tedious here. I will not, in fact, remember that.
## Sally Brown knows some imaginary numbers too
I had remembered this comic strip, and I hoped to use it for yesterday’s A-to-Z essay about Imaginary Numbers. But I wasn’t able to find it before publishing deadline. I figured I could go back and add this to the essay once I found it, and I likely will anyway. (The essay is quite long and any kind of visual appeal helps.)
But I also wanted folks to have the chance to notice it, and an after-the-fact addition doesn’t give that chance.
It is almost certain that Bill Watterson read this strip, and long before his own comic with eleventeen and thirty-twelve and such. Watterson has spoken of Schulz’s influence. That isn’t to say that he copied the joke. “Gibberish number-like words” is not a unique idea, and it’s certainly not original to Schulz. I’d imagine a bit of effort could find prior examples even within comic strips. (I’m reminded in Pogo of Howland Owl describing the Groundhog Child’s gibberish as first-rate algebra.) It’s just fun to see great creative minds working out similar ideas, and how they use those ideas for different jokes.
## Reading the Comics, June 7, 2020: Hiatus Edition
I think of myself as not a prescriptivist blogger. Here and on my humor blog I do what I feel like, and if that seems to work, I do more of it if I can. If I do enough of it, I try to think of a title, give up and use the first four words that kind of fit, and then ask Thomas K Dye for header art. If it doesn’t work, I drop it without mention. Apart from appealing for A-to-Z topics I don’t usually declare what I intend to do.
This feels different. One of the first things I fell into here, and the oldest hook in my blogging, is Reading the Comics. It’s mostly fun. But it is also work. 2020 is not a year when I am capable of expanding my writing work without bounds. Something has to yield, and my employers would rather it not be my day job. So, at least through the completion of the All 2020 Mathematics A-to-Z, I’ll just be reading the comics. Not Reading the Comics for posting here.
And this is likely a good time for a hiatus. There is much that’s fun about Reading the Comics. First is the comic strips, a lifelong love. Second is that they solve the problem of what to blog about. During the golden age of Atlantic City, there was a Boardwalk performer whose gimmick was to drag a trap along the seabed, haul it up, and identify every bit of sea life caught up in that. My schtick is of a similar thrill, with less harm required of the sea life.
But I have felt bored by this the last several months. Boredom is not a bad thing, of course. And if you are to be a writer, you must be able to write something competent and fresh about a topic you are tired of. Admitting that: I do not have one more sentence in me about kids not buying into the story problem. Or observing that yes, that is a blackboard full of mathematics symbols. Or that lotteries exist and if you play them infinitely many times strange conclusions seem to follow. An exercise that is tiring can be good; an exercise that is painful is not. I will put the painful away and see what I feel like later.
For the time being I figure to write only the A-to-Z essays. And, since I have them, to post references back to old A-to-Z essays. These recaps seemed to be received well enough last year. So why not repeat something that was fine when it was just one of many things?
And after all, the A-to-Z theme is still at heart hauling up buckets of sea life and naming everything in it. It’s just something that I can write farther ahead of deadline, but will not.
The Boardwalk performer would, if stumped, make up stuff. What patron was going to care if they went away ill-informed? It was a show. The performer just needed a confident air.
## Reading the Comics, June 6, 2020: Wrapping Up The Week Edition
Let’s see if I can’t close out the first week of June’s comics. I’d rather have published this either Tuesday or Thursday, but I didn’t have the time to write my statistics post for May, not yet. I’ll get there.
One of Gary Larson’s The Far Side reprints for the 4th is one I don’t remember seeing before. The thing to notice is the patient has a huge right brain and a tiny left one. The joke is about the supposed division between left-brained and right-brained people. There are areas of specialization in the brain, so that the damage or destruction of part can take away specific abilities. The popular imagination has latched onto the idea that people can be dominated by specialties of the either side of the brain. I’m not well-versed in neurology. I will hazard the guess that neurologists see “left-brain” and “right-brain” as amusing stuff not to be taken seriously. (My understanding is the division of people into “type A” and “type B” personalities is also entirely bunk unsupported by any psychological research.)
Samson’s Dark Side of the Horse for the 5th is wordplay. It builds on the use of “problem” to mean both “something to overcome” and “something we study”. The mathematics puzzle book is a fanciful creation. The name Lucien Kastner is a Monty Python reference. (I thank the commenters for spotting that.)
Dan Collins’s Looks Good on Paper for the 5th is some wordplay on the term “Möbius Strip”, here applied to a particular profession.
Bud Blake’s Tiger rerun for the 6th has Tiger complaining about his arithmetic homework. And does it in pretty nice form, really, doing some arithmetic along the way. It does imply that he’s starting his homework at 1 pm, though, so I guess it’s a weekend afternoon. It seems like rather a lot of homework for that age. Maybe he’s been slacking off on daily work and trying to make up for it.
John McPherson’s Close To Home for the 6th has a cheat sheet skywritten. It’s for a geometry exam. Any subject would do, but geometry lets cues be written out in very little space. The formulas are disappointingly off, though. We typically use ‘r’ to mean the radius of a circle or sphere, but then would use C for its circumference. That would be $c = 2\pi r$. The area of a circle, represented with A, would be $\pi r^2$. I’m not sure what ‘Vol.C’ would mean, although ‘Volume of a cylinder’ would make sense … if the next line didn’t start “Vol.Cyl”. The volume of a circular cylinder is $\pi r^2 h$, where r is the radius and h the height. For a non-circular cylinder, it’s the area of a cross-section times the height. So that last line may be right, if it extends out of frame.
Granted, though, a cheat sheet does not necessarily make literal sense. It needs to prompt one to remember what one needs. Notes that are incomplete, or even misleading, may be all that one needs.
And this wraps up the comics. This and other Reading the Comics posts are gathered at this link. Next week, I’ll get the All 2020 A-to-Z under way. Thanks once again for all your reading.
## Reading the Comics, June 3, 2020: Subjective Opinions Edition
Thanks for being here for the last week before my All-2020 Mathematics A to Z starts. By the time this posts I should have decided on the A-topic, but I’m still up for B or C topics, if you’d be so kind as to suggest things.
Bob Weber Jr’s Slylock Fox for the 1st of June sees Reeky Rat busted for speeding on the grounds of his average speed. It does make the case that Reeky Rat must have travelled faster than 20 miles per hour at some point. There’s no information about when he did it, just the proof that there must have been some time when he drove faster than the speed limit. One can find loopholes in the reasoning, but, it’s a daily comic strip panel for kids. It would be unfair to demand things like proof there’s no shorter route from the diner and that the speed limit was 20 miles per hour the whole way.
Ted Shearer’s Quincy for the 1st originally ran the 7th of April, 1981. Quincy and his friend ponder this being the computer age, and whether they can let computers handle mathematics.
Jef Mallett’s Frazz for the 2nd has the characters talk about how mathematics offers answers that are just right or wrong. Something without “subjective grading”. It enjoys that reputation. But it’s not so, and that’s obvious when you imagine grading. How would you grade an answer that has the right approach, but makes a small careless error? Or how would you grade an approach that doesn’t work, but that plausibly could?
And how do you know that the approach wouldn’t work? Even in non-graded mathematics, we have subjectivity. Much of mathematics is a search for convincing arguments about some question. What we hope to be convinced of is that there is a sound logical argument making the same conclusions. Whether the argument is convincing is necessarily subjective.
Yes, in principle, we could create a full deductive argument. It will take forever to justify every step from some axiom or definition or rule of inference. And even then, how do we know a particular step is justified? It’s because we think we understand what the step does, and how it conforms to one (or more) rule. That’s again a judgement call.
(The grading of essays is also less subjective than you might think if you haven’t been a grader. The difference between an essay worth 83 points and one worth 85 points may be trivial, yes. But you will rarely see an essay that reads as an A-grade one day and a C-grade the next. This is not to say that essay grading is not subject to biases. Some of these are innocent, such as the way the grader’s mood will affect the grade. Or how the first several papers, or the last couple, will be less consistently graded than the ones done in the middle of the project. Some are pernicious, such as under-rating the work done by ethnic minority students. But these biases affect the way one would grade, say, the partial credit for an imperfectly done algebra problem too.)
Mark Anderson’s Andertoons for the 3rd is the Mark Anderson’s Andertoons for the week. I could also swear that I’ve featured it here before. I can’t find it, if I have discussed this strip before. I may not have. Wavehead’s observing the difference between zero as an additive identity and its role in multiplication.
Ryan Pagelow’s Buni for the 3rd fits into the anthropomorphic-numerals category of joke. It’s really more of a representation of the year as the four horsemen of the Apocalypse.
Dan Collins’s Looks Good on Paper for the 3rd has a cook grilling a “Möbius Strip Steak”. It’s a good joke for putting on a mathematics instructor’s door.
Doug Savage’s Savage Chickens for the 3rd has, as part of animal facts, the assertion that “llamas have basic math skills”. I don’t know of any specific research on llama mathematics skills. But animals do have mathematics skills. Often counting. Some amount of reasoning. Social animals often have an understanding of transitivity, as well, especially if the social groups have a pecking order.
And this wraps up half of the past week’s mathematically-themed comic strips. I hope to have the rest in a Reading the Comics post at this link in a few days. Thanks for reading.
## Reading the Comics, May 29, 2020: Slipping Into Summer More Edition
This is the slightly belated close of last week’s topics suggested by Comic Strip Master Command. For the week we’ve had, I am doing very well.
Werner Wejp-Olsen’s Inspector Danger’s Crime Quiz for the 25th of May sees another mathematician killed, and “identifying” his killer in a dying utterance. Inspector Danger has followed killer mathematicians several times before: the 9th of July, 2012, for instance. Or the 4th of July, 2016, for a case so similar that it’s almost a Slylock Fox six-differences puzzle. Apparently realtors and marine biologists are out for mathematicians’ blood. I’m not surprised by the realtors, but hey, marine biology, what’s the deal? The same gimmick got used the 15th of May, 2017, too. (And in fairness to the late Wejp-Olsen, who could possibly care that similar names are being used in small puzzles used years apart? It only stands out because I’m picking out things that no reasonable person would notice.)
Jim Meddick’s Monty for the 25th has the title character inspired by the legend of genius work done during plague years. A great disruption in life is a great time to build new habits, and if Covid-19 has given you the excuse to break bad old habits, or develop good new ones, great! Congratulations! If it has not, though? That’s great too. You’re surviving the most stressful months of the 21st century, I hope, not taking a holiday.
Anyway, the legend mentioned here includes Newton inventing Calculus while in hiding from the plague. The actual history is more complicated, and ambiguous. (You will not go wrong supposing that the actual history of a thing is more complicated and ambiguous than you imagine.) The Renaissance Mathematicus describes, with greater authority and specificity than I could, what Newton’s work was more like. And some of how we have this legend. This is not to say that the 1660s were not astounding times for Newton, nor to deny that he worked with a rare genius. It’s more that we are lying to imagine that Newton looked around, saw London was even more a deathtrap than usual, and decided to go off to the country and toss out a new and unique understanding of the infinitesimal and the continuum.
Mark Anderson’s Andertoons for the 27th is the Mark Anderson’s Andertoons for the week. One of the students — not Wavehead — worries that a geometric ray, going on forever, could endanger people. There’s some neat business going on here. Geometry, like much mathematics, works on abstractions that we take to be universally true. But it also seems to have a great correspondence to ordinary real-world stuff. We wouldn’t study it if it didn’t. So how does that idealization interact with the reality? If the ray represented by those marks on the board goes on to do something, do we have to take care in how it’s used?
Olivia Jaimes’s Nancy for the 29th is set in a (virtual) arithmetic class. It builds on the conflation between “nothing” and “zero”.
And that wraps up my week in comic strips. I keep all my Reading the Comics posts at this link. I am also hoping to start my All 2020 Mathematics A-to-Z shortly, and am open for nominations for topics for the first couple letters. Thank you for reading.
## Reading the Comics, May 25, 2020: Slipping into Summer Edition
Comic Strip Master Command wanted to give me a break as I ready for the All 2020 A-to-Z. I appreciate the gesture, especially given the real-world events of the past week. I get to spend this week mostly just listing appearances, even if they don’t inspire deeper thought.
Gordon Bess’s vintage Redeye for the 24th has one of his Cartoon Indians being lousy at counting. Talking about his failures at arithmetic, with how he doesn’t count six shots off well. There’s a modest number of things that people are, typically, able to perceive at once. Six can be done, although it’s easy for a momentary loss of focus to throw you off. This especially for things that have to be processed in sequence, rather than perceived all together.
Wulff and Morgenthaler’s WuMo for the 24th shows a parent struggling with mathematics, billed as part of “the terrible result of homeschooling your kids”. It’s a cameo appearance. It’d be the same if Mom were struggling with history or English. This is just quick for the comic strip reader to understand.
Andrés J. Colmenares’s Wawawiwa for the 25th sets several plants in a classroom. They’re doing arithmetic. This, too, could be any course; it just happens to be mathematics.
Sam Hurt’s Eyebeam for the 25th is built on cosmology. The subject is a blend of mathematics, observation, and metaphysics. The blackboard full of mathematical symbols gets used as shorthand for describing the whole field, not unfairly. The symbols as expressed don’t come together to mean anything. I don’t feel confident saying they don’t mean anything, though.
This is enough for today. I keep all my Reading the Comics posts at this link, and should have another one later this week. And I am trying to get my All 2020 Mathematics A-to-Z ready, with nominations open for the first several letters of the alphabet already. Thank you for reading.
## Reading the Comics, May 23, 2020: Parents Can’t Do Math Edition
This was a week of few mathematically-themed comic strips. I don’t mind. If there was a recurring motif, it was about parents not doing mathematics well, or maybe at all. That’s not a very deep observation, though. Let’s look at what is here.
Liniers’s Macanudo for the 18th puts forth 2020 as “the year most kids realized their parents can’t do math”. Which may be so; if you haven’t had cause to do (say) long division in a while then remembering just how to do it is a chore. This trouble is not unique to mathematics, though. Several decades out of regular practice they likely also have trouble remembering what the 11th Amendment to the US Constitution is for, or what the rule is about using “lie” versus “lay”. Some regular practice would correct that, though. In most cases anyway; my experience suggests I cannot possibly learn the rule about “lie” versus “lay”. I’m also shaky on “set” as a verb.
Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 18th shows a mathematician talking, in the jargon of first and second derivatives, to support the claim there’ll never be a mathematician president. Yes, Weinersmith is aware that James Garfield, 20th President of the United States, is famous in trivia circles for having an original proof of the Pythagorean theorem. It would be a stretch to declare Garfield a mathematician, though, except in the way that anyone capable of reason can be a mathematician. Raymond Poincaré, President of France for most of the 1910s and prime minister before and after that, was not a mathematician. He was cousin to Henri Poincaré, who founded so much of our understanding of dynamical systems and of modern geometry. I do not offhand know what presidents (or prime ministers) of other countries have been like.
Weinersmith’s mathematician uses the jargon of the profession. Specifically that of calculus. It’s unlikely to communicate well with the population. The message is an ordinary one, though. The first derivative of something with respect to time means the rate at which things are changing. The first derivative of a thing, with respect to time being positive means that the quantity of the thing is growing. So, that first half means “things are getting more bad”.
The second derivative of a thing with respect to time, though … this is interesting. The second derivative is the same thing as the first derivative with respect to time of “the first derivative with respect to time”. It’s what the change is in the rate-of-change. If that second derivative is negative, then the first derivative will, in time, change from being positive to being negative. So the rate of increase of the original thing will, in time, go from a positive to a negative number. And so the quantity will eventually decline.
So the mathematician is making a this-is-the-end-of-the-beginning speech. The point at which the the second derivative of a quantity changes sign is known as the “inflection point”. Reaching that is often seen as the first important step in, for example, disease epidemics. It is usually the first good news, the promise that there will be a limit to the badness. It’s also sometimes mentioned in economic crises or sometimes demographic trends. “Inflection point” is likely as technical a term as one can expect the general public to tolerate, though. Even that may be pushing things.
Gary Wise and Lance Aldrich’s Real Life Adventures for the 19th has a father who can’t help his son do mathematics. In this case, finding square roots. There are many ways to find square roots by hand. Some are iterative, in which you start with an estimate and do a calculation that (typically) gets you a better estimate of the square root you want. And then repeat the calculation, starting from that improved estimate. Some use tables of things one can expect to have calculated, such as exponentials and logarithms. Or trigonometric tables, if you know someone who’s worked out lots of cosines and sines already.
Henry Scarpelli and Craig Boldman’s Archie rerun for the 20th mentions romantic triangles. And Moose’s relief that there’s only two people in his love triangle. So that’s our geometry wordplay for the week.
Bill Watterson’s Calvin and Hobbes repeat for the 20th has Calvin escaping mathematics class.
Julie Larson’s The Dinette Set rerun for the 21st fusses around words. Along the way Burl mentions his having learned that two negatives can make a positive, in mathematics. Here it’s (most likely) the way that multiplying or dividing two negative numbers will produce a positive number.
This covers the week. My next Reading the Comics post should appear at this tag, when it’s written. Thanks for reading.
## Reading the Comics, May 15, 2020: Squared Away Edition
The end of last week offered just a few more comic strips, and some pretty casual mathematics content. Let me wrap that up.
Daniel Beyer’s Long Story Short for the 13th has the “math department lavatory” represented as a door labelled $1 \pm 2$. It’s an interesting joke in that it reads successfully, but doesn’t make sense. To match the references to the commonly excreted substances they’d want $\frac32 \pm \frac12$.
On funny labels, though, I did once visit a mathematics building in which the dry riser had the label N Bourbaki. Nicholas Bourbaki was not a member of that college’s mathematics department, of course. This is why the joke was correctly formed and therefore funny.
Keith Tutt and Daniel Saunders’s Lard’s World Peace Tips for the 13th features the rounding-up-sheep joke.
Gary Larson’s The Far Side strips for the 14th includes the famous one of Albert Einstein coming so close to working out $E = mc^2$. The usual derivations for $E = mc^2$ don’t start with that and then explore whether it makes sense, which is what Einstein seems to be doing here. Instead they start from some uncontroversial premises and find that they imply this $E = mc^2$ business. Dimensional analysis would also let you know that, if c is involved, it’s probably to the second power rather than anything else.
But that doesn’t mean we can’t imagine Einstein assuming there must be a relationship between energy and mass, finding one that makes sense, and then finding a reason it’s that rather than something else. That’s a common enough pattern of mathematical discovery. Also, a detail I hadn’t noticed before, is that Einstein tried out $E = mc^3$, rejected it, and then tried it again. This is also a common pattern of discovery.
Mark Litzler’s Joe Vanilla for the 14th has a vague recollection of the Pythagorean Theorem be all that someone says he remembers of mathematics.
Niklas Eriksson’s Carpe Diem for the 15th depicts a couple ancient Greek deep-thinkers. A bit of mathematics, specifically geometry, is used as representative of that deep thinking.
This wraps up the past week’s mathematically-themed comics. Read this and next week’s comic strips at this link. Thank you.
## Reading the Comics, May 12, 2020: Little Oop Counts For More Edition
The past week had a fair number of comic strips mentioning some aspect of mathematics. One of them is, really, fairly slight. But it extends a thread in the comic strip that I like and so that I will feature here.
Jonathan Lemon and Joey Alison Sayers’s Little Oop for the 10th continues the thread of young Alley Oop’s time discovering numbers. (This in a storyline that’s seen him brought to the modern day.) The Moo researchers of the time have found numbers larger than three. As I’d mentioned when this joke was first done, that Oop might not have had a word for “seven” until recently doesn’t mean he wouldn’t have understood that seven of a thing was more than five of a thing, or less than twelve of a thing. At least if he could compare them.
Sam Hurt’s Eyebeam for the 11th uses heaps of mathematical expressions, graphs, charts, and Venn diagrams to represent the concept of “data”. It’s spilled all over to represent “sloppy data”. Usually by the term we mean data that we feel is unreliable. Measurements that are imprecise, or that are unlikely to be reliable. Precision is, roughly, how many significant digits your measurement has. Reliability is, roughly, if you repeated the measurement would you get about the same number?
Nate Fakes’s Break of Day for the 12th is the anthropomorphic numerals joke for the week.
Ryan North’s Dinosaur Comics for the 12th talks about immortality. And what the probability of events means when there are infinitely many opportunities for a thing to happen.
We’re accustomed in probability to thinking of the expectation value. This is the chance that something will happen, given some number N opportunities to happen, if at each opportunity it has the probability p of happening. Let me assume the probability is always the same number. If it’s not, our work gets harder, although it’s basically the same kind of work. But, then, the expectation value, the number of times we’d expect to see the thing happen, is N times p. Which, as Utahraptor points out, we can expect has to be at least 1 for any event, however unlikely, given enough chances. So it should be.
But, then, to take Utahraptor’s example: what is the probability that an immortal being never trips down the stairs? At least not badly enough to do harm? Why should we think that’s zero? It’s not as if there’s a physical law that compels someone to go to stairs and then to fall down them to their death. And, if there’s any nonzero chance of someone not dying this way? Then, if there are enough immortals, there’s someone who will go forever without falling down stairs.
That covers just the one way to die, of course. But the same reasoning holds for every possible way to die. If there’s enough immortals, there’s someone who would not die from falling down stairs and from never being struck by a meteor. And someone who’d never fall down stairs and never be struck by a meteor and never fall off a cliff trying to drop an anvil on a roadrunner. And so on. If there are infinitely many people, there’s at least one who’d avoid all possible accidental causes of death.
More. If there’s infinitely many immortals, then there are going to be a second and a third — indeed, an infinite number — of people who happen to be lucky enough to never die from anything. Infinitely many immortals die of accidents, sure, but somehow not all of them. We can’t even say that more immortals die of accidents than don’t.
My point is that probability gets really weird when you try putting infinities into it. Proceed with extreme caution. But the results of basic, incautious, thinking can be quite heady.
Bill Amend’s FoxTrot Classics for the 12th has Paige cramming for a geometry exam. Don’t cram for exams; it really doesn’t work. It’s regular steady relaxed studying that you need. That and rest. There is nothing you do that you do better for being sleep-deprived.
Bob Weber Jr and Jay Stephens’s Oh Brother for the 12th has Lily tease her brother with a story problem. I believe the strip’s a rerun, but it had been gone altogether for more than a year. It’s nice to see it returned anyway.
And while I don’t regularly cover web-only comics here, Norm Feuti has carried on his Gil as a Sunday-only web comic. The strip for the 10th of May has Gil using a calculator for mathematics homework, with a teacher who didn’t say he couldn’t. I’m surprised she hadn’t set a guideline.
This carries me through half a week. I’ll have more mathematically-themed comic strips at this link soon. Thanks for reading.
## Reading the Comics, May 9, 2020: Knowing the Angles Edition
There were a couple more comic strips in the block of time I want to write about. Only one’s got some deeper content and, I admit, I had to work to find it.
Bob Scott’s Bear With me for the 7th has Bear offering the answer from mathematics class, late.
Jerry Bittle’s Shirley and Sons Classic rerun for the 7th has Louis struggling on an arithmetic test.
Olivia Jaimes’s Nancy for the 8th has Nancy and Sluggo avoiding mathematics homework. Or, “practice”, anyway. There’s more, though; Nancy and Sluggo are doing some analysis of viewing angles. That’s actual mathematics, certainly. Computer-generated imagery depends on it, just like you’d imagine. There are even fun abstract questions that can give surprising insights into numbers. For example: imagine that space were studded, at a regular square grid spacing, with perfectly reflective marbles of uniform size. Is there, then, a line of sight between any two points outside any marbles? Even if it requires tens of millions of reflections; we’re interested in what perfect reflections would give us.
Using playing cards as a makeshift protractor is a creative bit of making do with what you have. The cards spread in a fanfold easily enough and there’s marks on the cards that you can use to keep your measurements reasonably uniform. Creating ad hoc measurement tools like this isn’t mathematics per se. But making a rough tool is a first step to making a precise tool. And you can use reason to improve your estimates.
It’s not on-point, but I did want to share the most wondrous ad hoc tool I know of: You can use an analog clock hand, and the sun, as a compass. You don’t even need a real clock; you can draw the time on a sheet of paper and use that. It’s not a precise measure, of course. But if you need some help, here you go. You’ve got it.
Tony Rubino and Gary Markstein’s Daddy’s Home for the 9th has Elliot avoiding doing his mathematics homework.
And that’s got the last week covered. Some more comic strips should follow at a link here, soon. And I hope to have some other stuff to announce here, soon.
## Reading the Comics, May 7, 2020: Getting to Golf Edition
Last week saw a modest number of mathematically-themed comic strips. Then it threw in a bunch of them all on Thursday. I’m splitting the week partway through that, since it gives me some theme to this collection.
Tim Rickard’s Brewster Rockit for the 3rd of May is a dictionary joke, with Brewster naming each kind of chart and making a quick joke about it. The comic may help people who’ve had trouble remembering the names of different kinds of graphs. I doubt people are likely to confuse a pie chart with a bar chart, admittedly. But I could imagine thinking a ‘line graph’ is what we call a bar chart, especially if the bars are laid out horizontally as in the second panel here.
The point of all these graphs is to understand data geometrically. We have fair intuitions about relatives lengths and areas. Bar charts represent relative magnitudes in lengths. Pie charts and bubble charts represent magnitudes in area. We have okay skills in noticing structures in complex shapes. Line graphs and scatter plots use that skill. So these pictures can help us understand some abstraction or something we can’t sense using a sense we do have. It’s not necessarily great; note that I said our intuitions were ‘fair’ and ‘okay’. But we hope to use reason helped by intuition to better understand what we are doing.
Jef Mallett’s Frazz for the 3rd is a resisting-the-story-problem joke. It’s built not just on wondering the point of story problems at all, but of these story problems during the pandemic. (Which Mallett on the 27th of April, would be taking “some liberties” with the real world. It’s a respectable decision.)
And, yes, in the greater scheme of things, any homework or classwork problem is trivial. It’s meant to teach how to calculate things we would like to know. The framing of the story is meant to give us a reason to want to know a thing. But they are practice, and meant to be practice. One practices on something of no consequence, where errors in one’s technique can be corrected without breaking anything.
It happens a round of story problems broke out among my family. My sister’s house has some very large trees. There turns out to be a poorly-organized process for estimating the age of these trees from their circumference. This past week saw a lot of chatter and disagreement about what the ages of these trees might be.
Jason Poland’s Robbie and Bobby for the 4th riffs on the difference between rectangles and trapezoids. It’s also a repeat, featured here just five years ago. Amazing how time slips on like that.
Samson’s Dark Side of the Horse for the 4th is another counting-sheep joke. It features one of those shorthands for large numbers which often makes them more manageable.
Michael Fry’s Committed rerun for the 7th finally gets us to golf. The Lazy Parent tries to pass off watching golf as educational, with working out the distance to the pin as a story problem. Structurally this is just fine, though: a golfer would be interested to know how far the ball has yet to go. All the information needed is given. It’s the question of whether anyone but syndicated cartoonists cares about golf that’s a mystery.
Bill Amend’s FoxTrot Classics for the 7th is another golf and mathematics joke. Jason has taken the homonym of ‘fore’ for ‘four’, and then represented ‘four’ in a needlessly complicated way. Amend does understand how nerd minds work. The strip originally ran the 21st of May, 1998.
That’s enough comics for me for today. I should have the rest of last week’s in a post at this link soon. Thank you.
## Reading the Comics, May 2, 2020: What Is The Cosine Of Six Edition
The past week was a light one for mathematically-themed comic strips. So let’s see if I can’t review what’s interesting about them before the end of this genially dumb movie (1940’s Hullabaloo, starring Frank Morgan and featuring Billie Burke in a small part). It’ll be tough; they’re reaching a point where the characters start acting like they care about the plot either, which is usually the sign they’re in the last reel.
Patrick Roberts’s Todd the Dinosaur for the 26th of April presents mathematics homework as the most dreadful kind of homework.
Jenny Campbell’s Flo and Friends for the 26th is a joke about fumbling a bit of practical mathematics, in this case, cutting a recipe down. When I look into arguments about the metric system, I will sometimes see the claim that English traditional units are advantageous for cutting down a recipe: it’s quite easy to say that half of “one cup” is a half cup, for example. I doubt that this is much more difficult than working out what half of 500 ml is, and my casual inquiries suggest that nobody has the faintest idea what half of a pint would be. And anyway none of this would help Ruthie’s problem, which is taking two-fifths of a recipe meant for 15 people. … Honestly, I would have just cut it in half and wonder who’s publishing recipes that serve 15.
Ed Bickford and Aaron Walther’s American Chop Suey for the 28th uses a panel of (gibberish) equations to represent deep thinking. It’s in part of a story about an origami competition. This interests me because there is serious mathematics to be done in origami. Most of these are geometry problems, as you might expect. The kinds of things you can understand about distance and angles from folding a square may surprise. For example, it’s easy to trisect an arbitrary angle using folded squares. The problem is, famously, impossible for compass-and-straightedge geometry.
Origami offers useful mathematical problems too, though. (In practice, if we need to trisect an angle, we use a protractor.) It’s good to know how to take a flat, or nearly flat, thing and unfold it into a more interesting shape. It’s useful whenever you have something that needs to be transported in as few pieces as possible, but that on site needs to not be flat. And this connects to questions with pleasant and ordinary-seeming names like the map-folding problem: can you fold a large sheet into a small package that’s still easy to open? Often you can. So, the mathematics of origami is a growing field, and one that’s about an accessible subject.
Nate Fakes’s Break of Day for the 29th is the anthropomorphic-symbols joke for the week, with an x talking about its day job in equations and its free time in games like tic-tac-toe.
Bill Holbrook’s On The Fastrack for the 2nd of May also talks about the use of x as a symbol. Curt takes eagerly to the notion that a symbol can represent any number, whether we know what it is or not. And, also, that the choice of symbol is arbitrary; we could use whatever symbol communicates. I remember getting problems to work in which, say, 3 plus a box equals 8 and working out what number in the box would make the equation true. This is exactly the same work as solving 3 + x = 8. Using an empty box made the problem less intimidating, somehow.
Dave Whamond’s Reality Check for the 2nd is, really, a bit baffling. It has a student asking Siri for the cosine of 174 degrees. But it’s not like anyone knows the cosine of 174 degrees off the top of their heads. If the cosine of 174 degrees wasn’t provided in a table for the students, then they’d have to look it up. Well, more likely they’d be provided the cosine of 6 degrees; the cosine of an angle is equal to minus one times the cosine of 180 degrees minus that same angle. This allows table-makers to reduce how much stuff they have to print. Still, it’s not really a joke that a student would look up something that students would be expected to look up.
… That said …
If you know anything about trigonometry, you know the sine and cosine of a 30-degree angle. If you know a bit about trigonometry, and are willing to put in a bit of work, you can start from a regular pentagon and work out the sine and cosine of a 36-degree angle. And, again if you know anything about trigonometry, you know that there are angle-addition and angle-subtraction formulas. That is, if you know the cosine of two angles, you can work out the cosine of the difference between them.
So, in principle, you could start from scratch and work out the cosine of 6 degrees without using a calculator. And the cosine of 174 degrees is minus one times the cosine of 6 degrees. So it could be a legitimate question to work out the cosine of 174 degrees without using a calculator. I can believe in a mathematics class which has that as a problem. But that requires such an ornate setup that I can’t believe Whamond intended that. Who in the readership would think the cosine of 174 something to work out by hand? If I hadn’t read a book about spherical trigonometry last month I wouldn’t have thought the cosine of 6 a thing someone could reasonably work out by hand.
I didn’t finish writing before the end of the movie, even though it took about eighteen hours to wrap up ten minutes of story. My love came home from a walk and we were talking. Anyway, this is plenty of comic strips for the week. When there are more to write about, I’ll try to have them in an essay at this link. Thanks for reading.
## Reading the Comics, April 25, 2020: Off Brand Edition
Comic Strip Master Command decided I should have a week to catch up on things, and maybe force me to write something original. Of all the things I read there were only four strips that had some mathematics content. And three of them are such glancing mentions that I don’t feel it proper to include the strip. So let me take care of this.
Mark Anderson’s Andertoons for the 20th is the Mark Anderson’s Andertoons for the week. Wavehead apparently wants to know whether $\frac{3}{4}$ or $\frac{6}{8}$ is the better of these equivalent forms. I understand the impulse. Rarely in real life do we see two things that are truly equivalent; there’s usually some way in which one is better than the other. There may be two ways to get home for example, both taking about the same time to travel. One might have better scenery, though, or involve fewer difficult turns or less traffic this time of day. This is different, though: $\frac{3}{4}$ or $\frac{6}{8}$ are two ways to describe the same number. Which one is “better”?
The only answer is, better for what? What do you figure to do with this number afterwards? I admit, and suppose most people have, a preference for $\frac{3}{4}$. But that’s trained into us, in large part, by homework set to reduce fractions to “lowest terms”. There’s honest enough reasons behind that. It seems wasteful to have a factor in the numerator that’s immediately divided out by the denominator.
If this were 25 years ago, I could ask how many of you have written out a check for twenty-two and 3/4 dollars, then, rather than twenty-two and 75/100 dollars? The example is dated but the reason to prefer an equivalent form is not. If I know that I need the number represented by $\frac{3}{4}$, and will soon be multiplying it by eight, then $\frac{6}{8}$ may save me the trouble of thinking what three times two is. Or if I’ll be adding it to $\frac{5}{8}$, or something like that. If I’m measuring this for a recipe I need to cut in three, because the original will make three dozen cookies and I could certainly eat three dozen cookies, then $\frac{3}{4}$ may be more convenient than $\frac{6}{8}$. What is the better depends on what will clarify the thing I want to do.
A significant running thread throughout all mathematics, not just arithmetic, is finding equivalent forms. Ways to write the same concept, but in a way that makes some other work easier. Or more likely to be done correctly. Or, if the equivalent form is more attractive, more likely to be learned or communicated. It’s of value.
Jan Eliot’s Stone Soup Classics rerun for the 20th is a joke about how one can calculate what one is interested in. In this case, going from the number of days left in school to the number of hours and minutes and even seconds left. Personally, I have never had trouble remembering there are 24 hours in the day, nor that there are 86,400 seconds in the day. That there are 1,440 minutes in the day refuses to stick in my mind. Your experiences may vary.
Thaves’s Frank and Ernest for the 22nd is the Roman Numerals joke for the week, shifting the number ten to the representation “X” to the prefix “ex”.
Harry Bliss’s Bliss for the 23rd speaks of “a truck driver with a PhD in mathematical logic”. It’s an example of signifying intelligence through mathematics credentials. (It’s also a bit classicist, treating an intelligent truck driver as an unlikely thing.)
I’m caught up! This coming Sunday I hope to start discussingthis week’s comics in a post at this link. And for this week? I don’t know; maybe I’ll figure something to write. We’ll see. Thanks for reading.
## Reading the Comics, April 17, 2020: Creating Models Edition
And now let me close out a week ago, in the comics. It was a slow week and it finished on a bunch of casual mentions of mathematical topics.
Gary Larson’s The Far Side compilation “Hands Off My Bunsen Burner” features this panel creating a model of how to get rights out of wrongs. The material is a joke, but trying to find a transformation from one mathematical object to another is a reasonable enough occupation.
Ted Shearer’s Quincy rerun for the 15th is one in the lineage of strips about never using mathematics in later life. Quincy challenges us to think of a time a reporter asks the President how much is 34 times 587.
That’s an unpleasant multiplication to do. But I can figure some angles on it. 34 is just a bit over one-third of 100. 587 is just a bit under 600. So, 34 times 587 has to be tolerably near one-third of 100 times 600. So it should be something around 20,000. To get it more exact: 587 is 13 less than 600. So, 587 times one-third of a hundred will be 600 times one-third of a hundred minus 13 times one-third of a hundred. That’s one-third of 130, which is about 40. So the product has to be something close to 19,960. And the product has be some number which ends in an 8, what with 4 times 7 being 28. So the answer has to be one of 19,948, 19,958, or 19,968. And, indeed, it’s 19,958. I doubt I could do that so well during a press conference, I’ll admit. (If I wanted to be sure about that second digit, I’d have worked out: the tens unit in 34 times the ones in 587 is three times seven which is 21; the ones unit in 34 times the tens unit in 587 is four times eight which is 32; and the 4 times 7 being 28 gives me a 2 in the tens unit. So, 1 plus 2 plus 2 is 5, and there we go.)
Brian Anderson’s Dog Eat Doug for the 15th uses blackboards full of equations to represent deep thinking. I can’t make out what the symbols say. They look quite good, though, and seem to have the form of legitimate expressions.
Terri Liebenson’s The Pajama Diaries for the 17th imagines creating a model for the volume of a laundry pile. The problem may seem trivial, but it reflects an important kind of work. Many processes are about how something that’s always accumulating will be handled. There’s usually a hard limit to the rate at which whatever it is gets handled. And there’s usually very little reserve, in either capacity or time. This will cause, for example, a small increase in traffic in a neighborhood to produce great jams, or how a modest rain can overflow the whole city’s sewer systems. Or how a day of missing the laundry causes there to be a week’s backlog of dirty clothes.
And a little final extra comic strip. I don’t generally mention web comics here, except for those that have fallen in with a syndicator like GoComics.com. (This is not a value judgement against web comics. It’s that I have to stop reading sometime.) But Kat Swenski’s KatRaccoon Comics recently posted this nice sequence with a cat facing her worst fear: a calculus date.
And that’s my comics for a week ago. Later this week I’ll cover the past week’s handful of comics, in an essay at this link. Thanks for reading. | 2022-12-05 10:40:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5050693154335022, "perplexity": 1206.6641918338619}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00590.warc.gz"} |
https://www.pharmanotes.org/2022/04/conductometry.html | # Conductometry - Pharmaceutical Analysis 1 B. Pharma 1st semester
Conductometry
Contents
• Conductometry
• Principle involved
• Measurement of conductivity
• Pros and cons of conductometric titrations
• Precautions to be taken
• Procedure
• Comparison of potentiometry vs conductometry
• Applications
Objectives
By the end of this session, students will be able to:
• Define conductometry
• Define and explain the principle involved in conductometric titrations
• Discuss the pros and cons Conductometry
• Explain precautions to be taken for conductometric titrations
• Brief the applications of conductometric titrations
Conductometry
• Measurement of conductivity of a solution
• Due to mobility of cations and anions towards respective electrodes
• Conductivity (C) is inversely proportional to resistance (R) of a solution
C = 1/R
• Unit of conductivity is mhos or ohms-1
• Conductivity of a solution depends upon-
o Number of ions (concentration)
o Charge of ions
o Size of ions
o Temperature
• Resistance of a solution is given by
R = E/I
Where E = potential difference
I = current which flows through
Unit of resistance (R) is ohms
Potential difference (E) is volts
Current (I) is amperes
• Resistance of a solution depends upon length (l) and cross resistance (a) of the conductor through which conductivity takes place
R = ρl/a
• ρ is specific resistance
Specific resistance (ρ) is the resistance offered by a substance of 1cm length and 1 sq.cm surface area, Unit of measurement is ohm cm
Specific conductivity (kv) is the conductivity offered by a substance of 1cm length and 1 sq.cm surface area, Unit of measurement is mhos cm-1
Equivalent conductivity (λv) is the conductivity of a solution containing equivalent weight of the solute between electrodes 1 cm and 1 sq.cm, Unit of measurement is mhos cm-1
• Molar conductivity (μv) is the conductivity of a solution containing molecular weight of the solute between electrodes 1 cm apart and 1 sq.cm surface area
• Molar conductivity = specific conductivity x volume of solution containing one molecular weight of the electrolyte
Measurement of Conductivity
• Conductivity may be measured by applying an alternating electrical current (I) to two electrodes immersed in a solution and measuring the resulting voltage (V)
• Cations migrate to the negative electrode, the anions to the positive electrode and the solution acts as an electrical conductor
• Conductivity is typically measured in aqueous solutions of electrolytes/ ions
• For the actual determination of conductivity, we need
• Wheatstone bridge circuit and Conductivity cell
• Conductivity cells are of different types
• Made up of platinum and coated with platinum black
• If the electrodes are old, platinisation can be done be done by using 3% solution of chloroplatinic acid and 0.02-0.03% of lead acetate to get uniform coating
• Different electrodes used depends upon the conductivity of the solution is high or low
• Commonly used are platinum electrodes
• Wheatstone bridge circuit consists of
• Standard resistance in one of its arms
• Other arm contains a conductivity cell (platinum electrode) dipped into the solution whose conductivity is to be determined
• Galvanometer shows the deflection of standard resistance with that of resistance of unknown solution
• R2/R1 = Resistance of BC/ Resistance of BA
• R2 is resistance of unknown solution
• R1 is standard resistance
• R2 = BC/BA x R1
• Conductivity of unknown solution = BA/BC x R1
• Observed conductivity is not always the specific conductivity
• Dimensions of the platinum electrode of various manufacturers are not same
• Distance between the electrodes and surface area of electrodes varies
• Value of cell constant to be calculated
• Cell constant (x) = l/a
• Where l = distance between electrodes
• a = area of electrode
• Relation between specific conductivity and observed conductivity can be derived as R = ρl/a
• R/ρ = l/a
• x = R/ρ = 1/observed conductivity / 1/specific conductivity
• x = R/ρ = specific conductivity / observed conductivity
• Specific conductivity = x * observed conductivity
• Specific conductivity = cell constant x observed conductivity
• Determination of cell constant
• Cell constant of a conductivity cell is determined by measuring the conductivity of a known strength of potassium chloride at specific temperature
• Conductivity of 0.02 KCl at 25 0C, cell constant is
• 2765/ observed conductivity of 0.02 KCl at 25 0C in µmhos
• Conductivity of 0.01 KCl at 25 0C, cell constant is
• 1221/ observed conductivity of 0.02 KCl at 25 0C in µmhos
Conductometric Titrations
• End point determination by conductivity measurements
• Conductivity solution depends on
• Change in number of ions
• Mobility of ions
• Graph of conductivity vs volume of titrant added
Pros
• Determination of specific conductivity is not required
• Not necessary to use conductivity water
• No indicator is required
• Titrations can be done with colored or dilute or turbid solutions
• Incompletion at end point doesn’t affect results as measurements before and after end point are sufficient
• End point is determined graphically, errors are minimized and can get accurate end point
• Cell constant need not be determined provided the same electrode is used throughout the experiment
• Temperature need not be known provided it is maintained constant throughout the titration
Apparatus required
• Titration vessel (beaker)
• Stirrer for mixing
• Automatic or manual burette to deliver titrant
• Conductivity meter with a conductivity cell (platinum electrode)
Procedure
• Conductivity is measured in millimhos or micromhos
• Titrant is added in small increments like 0.5 ml – 1.0 ml
• Solution is mixed properly and conductivity readings are taken
• Readings were taken before and after end point
• Graph is plotted- conductivity vs volume of titrant added
• Point of intersection is found
• Corresponds to end point or volume of titrant required to neutralize the reactants or sample present in titration vessel
Precautions to be taken
• Initial volume of titrating substance and final volume after titration are not same
• Conductivity measurements made during titration are subject to error
• Correction factor is included to know actual conductivity
• Actual conductivity = observed conductivity X (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 + 𝑣𝑜𝑙.𝑜𝑓 𝑡𝑖𝑡𝑟𝑎𝑛𝑡 𝑎𝑑𝑑𝑒𝑑 /𝑖𝑛𝑖𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒)
• Temperature should be maintained constant
• Heat of neutralization may affect the temperature and it effects the conductivity of solution
Acid base Titrations
Strong acid vs Strong base
• HCl + NaOH à NaCl + H2O
• HCl in beaker as titrate- high conductivity
• Strong acid- dissociation is complete
• After the end point, when all the H+ has reacted- addition of NaOH increases the concentration of OH- ions – conductivity starts increasing
• First part of curve shows steep fall in conductivity because of decrease in H+ ions
• Second part of curve shows gradual increase because of increase in OH- ions Strong Acid vs Weak Base
Strong Acid vs Weak Base
• HCl + NH4OH à NH4Cl + H2O
• HCl in beaker as titrate- high conductivity
• Strong acid- dissociation is complete
• After the end point, when all the H+ has reacted- addition of NH4OH doesn’t cause increase in the concentration of OH- ions
• Poor dissociation- conductivity remains constant
• First part of curve shows steep fall in conductivity because of decrease in H+ ions
• Second part of curve shows plateau
Weak Acid vs Strong Base
• CH3COOH + NaOH à CH3COONa + H2O
• CH3COOH in beaker as titrate- initial conductivity is low
• Weak acid- doesn’t dissociate into H+ ions
• NaOH is added as titrant – slight increase in conductivity till end point
• After the end point, addition of NaOH causes increase in the concentration of OH- ions
• Conductivity starts to increase steeply
• First part of curve shows gradual increase
• Second part of curve shows steep increase because of increase in OH- ions
Weak Acid vs Weak Base
• CH3COOH + NH4OH à CH3COONH4 + H2O
• CH3COOH in beaker as titrate- initial conductivity is low
• Weak acid- doesn’t dissociate into H+ ions
• NH4OH is added as titrant – ammonium acetate salt has better conductivity gradually increases after every addition
• After the end point, when all the CH3COOH has reacted addition of NH4OH causes no increase in the conductivity
• Plateau is obtained
• First part of curve shows gradual increase in conductivity because of ammonium acetate salt
• Second part of curve shows plateau because or poor dissociation of NH4OH
Comparison
Potentiometric titration Conductometric titration Parameter measured Potential in mv Conductivity in mhos Parameter not necessary Potential of reference electrode Cell constant At end point Rate of change of potential is maximum Sharp change in conductivity occurs End point determination Normal curve, first derived curve, second derivative curve End point shown by intersection of two lines Strength of titrant Same as that of titrate 5 or 10 times stronger than titrate Dependency Temperature dependent Temperature dependent
Applications
• Solubility of sparingly soluble salts- silver chloride, barium sulfate, lead sulfate
• Ionic product of water
• Basicity of organic acids- number of carboxylic groups present in the molecule- tartaric acid, oxalic acid
• Purity of water- specific conductivity of pure water is 5 x 10-8 ohm-1 cm-1
• Quantitative analysis
• Salinity of sea water
• Equilibrium in ionic reactions- progress of ionic reactions can be determined
Summary
• Measurement of conductivity of a solution- Due to mobility of cations and anions towards respective electrodes
• Resistance of a solution depends upon length (l) and cross resistance (a) of the conductor through which conductivity takes place
• Conductivity may be measured by applying an alternating electrical current (I) to two electrodes immersed in a solution and measuring the resulting voltage (V)
• Cations migrate to the negative electrode, the anions to the positive electrode and the solution acts as an electrical conductor
• Observed conductivity is not always the specific conductivity
Conductivity solution depends on
Change in number of ions
Mobility of ions
Graph of conductivity vs volume of titrant added
• Conductivity is measured in millimhos or micromhos | 2022-09-25 15:31:09 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8132460713386536, "perplexity": 8660.067391612742}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334579.46/warc/CC-MAIN-20220925132046-20220925162046-00407.warc.gz"} |
https://copingwithredundancy.blogspot.com/2009/ | ## Sunday, 14 June 2009
### Making introductions
And so you start your first day at the new job.
There are two sorts of starts.
1. The informal induction
2. The formal induction.
The informal induction.
It goes like this.
'Here's your desk, here's your PC, there is some paperwork for you to complete, see you in November'.
IT haven't set your log in, you therefore do not officially exist, you can't access your email and your default printer is set to print in another office on another floor but no one tells you for a week. As you do not exist you cannot book holidays and the IT help line is always on answerphone. HR do not seem to have heard about you joining and ask for your bank details five times.
You now spend six months trying to find out what your job is, what it is you do and what you do does. Forget it, you'll never find out, just fill in forms A/11/C1957 and fax them to 5543 as soon as you have done so and don't forget the pinks go to ACCAT7 and the blues in that tray there.
You never find out what or who is ACCAT7 or whether there is a 1 - 6 version either.
You spend the first four months walking half a mile to the nearest loo and then find out that there is one around the corner.
You are startled by the fire alarm once a week, looking around to see if any other members of staff are making any attempt to leave only to find out it is the regular test. Then, when the fire alarm goes off on a different day and time you look around to see if any of the other members of staff are making any attempt to leave and can't hep feeling concerned that they aren't when you can hear approaching sirens and notice there does seem to be a lot of smoke coming from the stationery cupboard.
For the first year you can only navigate your way from the entrance to your desk and back and do not realise that there are 1500 other members of staff on the same site, on different floors only nobody has told you anything about them.
You are told that there is a probationary period before you become a full member of staff and that performance objectives will be set. You never hear anything more about either.
You hear stories about a staff canteen but never find it. You eat your sandwiches at your work station whilst all around you your colleagues disappear for two hours, for lunch, but you don't know where they go.
You have never seen the MD of the organisation but see his/her car park space immediately outside the main door. You however have a quarter mile trek from the staff car park along a muddy path. You believe the MD is the one whose presence causes everyone to quit looking at the BBC news site and eBay on their PCs and look intensely busy as he/her strides through the office looking neither left nor right
You finally make contact with the two key people in the whole organisation - the keeper of the stationery cupboard and the one person in IT who can actually make your PC work without first asking you to reboot it.
The formal introduction.
It goes like this.
You have four days of tightly scheduled presentations from members of staff who you never see again in your whole time with the organisation.
The presenters never start or finish on time and about 33% mysteriously never turn up causing the over jolly person from HR to go into meltdown and end each day session early.
You are given an introduction to the aims and goals of the organisation by the highest member of the senior management team HR have convinced to turn up. This could be the cleaner, though their introduction is usually an improvement on the one given by the senior manager who clearly thinks that achieving his/her performance bonus is the major aim of the business. They tell you about the structure of the business. It looks like someone has upturned a bowl of spaghetti but presume it makes sense to someone somewhere. Actually it never does.
85% of the presenters start off by apologising for the boring nature of their subject. They do not lie. 95% then go on to overrun their alloted slot causing you to think longingly of blunt objects with which to strike them and thinking maybe signing on once every two weeks wasn't that bad after all.
90% of presenters believe that a good presentation depends on them standing in front of the new staff with their back to them reading directly from 173 densely written Powerpoint slides which they use as their script. They do not notice they lose their audience from slide 2 and neither do they understand why, after two long deadly dull hours of talking in a monotone no one has a question. Everyone is now comatose. The new staff only have one question - how long before I can go home and tell myself this is all a horrible dream.
IT haven't set your log in, you therefore do not officially exist, you can't access your email and your default printer is set to print in an office in France but no one tells you for a month as you frantically try to find the scurrilous emails your friend has sent you and you have sent to print. As you do not exist electronically you cannot book holidays and the IT help line is always on answerphone.
You finally make contact with the two key people in the whole organisation - the keeper of the stationery cupboard and Kevin in IT who can actually make your PC work without first asking have you rebooted it. You have, 26 times that day for a start.
You are required to sign an additional 23 forms telling you about data protection, eating at your workstations, staff socials, joining a Trades Union ('We welcome it.' They don't.), DSE, GSE, ABH, DDT, ABAGH and other abbreviations and acronyms you can't understand.
Then, on the first day at real work, you get a 'local' induction. 'Here's your desk, here's your PC, there is some paperwork for you to complete, see you in November'.
You are told about PDRs or PDPs leading to NVQs or possibly RACs and NFIs. You are so full of information after the first five days you can no longer absorb any more. You stagger back to the staff car park and drive home, exhausted.
Welcome back to work.
## Tuesday, 26 May 2009
### In three words I can sum up everything I've learned about life: it goes on*
So where did it go right? Why, after 2 years, 82 applications and six interviews did I finally get a job?
I have no idea. It's a mystery.
That is not very helpful is it?
But it's true.
Why my carefully constructed, honed and polished CVs made no impact when applying for jobs that you'd have thought I absolutely must be the best candidate for the job got nowhere, yet I managed to get a job in a totally different business sector will, for ever, be a mystery to me.
However I have learned somethings along the way (covering all 5 redundancies, we must learn from history) that I feel impelled to inflict on you. It may help, it may stop you making the same mistakes or you could always print them out and make a draught excluder from them.
Prepare for rejection (and more rejection).
Did I mention rejection?
Expect to be out of work for some time. Expect rejection, many rejections. Expect not to hear anything from most applications. Move on immediately - remember the HOE curve (yes rejection is very hard you don't have to tell me about it, now a fully qualified Master of Being Rejected) and keep looking. Keep a file of jobs applied for - you may need it as proof that you are actively job seeking.
Stay positive.
There will be good days, there will be bad days and there will be mind bendingly awful days. However there is a lot of silliness in the world. Sometimes it is very hard to see it but it’s there. And it can make you laugh. Well it made me laugh. Whatever happens try and bounce back. And there is life after redundancy. It may not be the life you had but it might very well be better.
It might have more kittens and yawning puppies.
Use every way you can to find a new job.
Use every channel you can think of to find a job and keep scouring them.
Do not stop using them even if you think they may not be working. You will not be able to predict easily where a job might be found. There are many channels: on-line jobs (Monster, Total Jobs, Fish4jobs), national newspapers (Times on line, Daily Telegraph, Guardian), local papers, local library, notices in shops, referrals, information from your friends and acquaintances and so. Get creative and think of any others that may work for you. Use them. Don’t stop.
And if you keep doing the same thing you will keep getting the same result. Evolve and adapt.
Stop spending.
Now.
You may have a reasonable redundancy payout, statutory redundancy pay or even nothing. What ever you have you need to stop spending now because you do not know how long this is going to last. Cancel all non-essential spending and start budgeting and get real. The kids may hate it, your partner may hate it, it may put you in a difficult position with your employed friends but that’s their problem. If they are that insensitive then have nothing to do with them because they will only vex you more - or ‘get in your grill’ as my kids say. You can do without many things - stay solvent and, if you have any money left when you get the new job, then is the time to spend.
Keep fit. Learn something new.
This time may be gruelling, will sap your motivation and test your sense of humour. Don’t sit in the house all day telling yourself life is crap, get fit, walk, go running, do something, get creative, learn something new. You’ll feel a whole lot better and able to face the job hunt. And it is a way of demonstrating a positive response to this difficulty to a potential employer.
Believe me chewing endlessly over and over the subject of 'no one replies to my job applications' tends to empty the room you are currently sitting in quite effectively.
Your partner might be at home or still doing whatever they were doing before your job loss. Respect their position and their space. You are going to need them for lots of support and it won’t help trailing after them all day around the house like a demented toddler following their mother. They will not want to hear your ills and moans all the time - try and remember that and be supportive to them.
Don't pay good money to snake oil merchants.
There are many organisations that are waiting to take your severance pay, savings or JSA. They are very seductive and promise much - some are very expensive. But none of them will find you a job, that's always down to you. So you might as well save your money.
Sign on. Don’t be proud.
This is, admittedly, not the most rewarding experience you will ever have. Be prepared for quite a demeaning process which may include giving all your private financial details to a complete stranger in an open office. However you are entitled to State Support (subject to a means test) and your National Insurance will be paid. There is some help in retraining available and there may be jobs available that the Job Centre Team can put you in touch with though usually in Fife I found as a sous chef or CNC operator.
Sell yourself properly with your CV.
Treat finding a new job as a job in itself.
Set aside time every day to look for a job or do something positive in finding a new position. There is always something that you can do
.
Don’t apply for jobs you’re not qualified for.
This is difficult but unless you want more rejection then do not apply for jobs you are patently not qualified for. Remember there are many more applicants who really will be better qualified so why beat yourself up?
Three steps backwards to go forwards.
There comes a time when you might have to accept the lesser paid job, take a considerable drop in salary and perks. That was then, this is now. Ask yourself do you want the money and see it as a way of fighting back up the ladder. Or do you continue to wait for the 'right job'. Ask yourself 'Do you feel lucky?' Well do you?
Voluntary contributions
Almost the subject of a blog in its own right, local voluntary organisations are looking for volunteers to muddle through the many layers of impenetrable bureaucracy as we speak. If you'd like to spend time helping others, just jog on down to your local volunteer centre, making sure it's not an Army recruiting office and ending up in Helmand Province.
With a little help from your friends...
Keep in contact, don't become isolated - contact by email, Skype, iChat, smoke signals, two tin cans with a piece of taut string. What ever it takes don't become isolated and, whilst I'm at it, thanks Neil, Dennis and Dick for your invaluable support.
And that's it, down to the irreducible level - forget the books, forget the seminars, this is what it comes down to.
And don't give up.
*Robert Frost
## Monday, 25 May 2009
### Life, but not as we know it. Hmmm
Sometimes you need to count the number of buses at the bus stop.
I'd walked into town last week to photostat all the documents required to prove that I exist for the new job. We'll ignore the rather obvious one of actually being in the room at the time of the interview which suggested to me that, unless I was some phantasm, my corporeal existence could be taken for granted.
No we wanted, or rather they wanted: birth certificate, driving license, passport, evidence of NI number, copies of educational certificates (I just knew someone would want to see my 'O' level in Agriculture), photos of me as a baby, saliva sample, DNA, palm prints, iris shots, and my Cub Scout badge for using a phone box with a 'Push button A' and 'B' (I cheated, I asked a passerby to phone Akela for me). Then, just as I was about to post the signed T&Cs I thought 'Sod this, I'm vastly overqualified for this job, I'm going to ask for more money' and walked home. There, on the door mat, was an invitation for an interview for a BETTER job with MORE money but two days after I potentially start the new job.
OMG, two years, over 75 applications, five interviews the last one being six months ago and now, in the space of just three weeks two interviews and a firm job offer. Now what do I do? It's true what they say about buses you don't see one for ages and then you find out they are big, run on diesel and carry upwards of 60 passengers to somewhere where you don't really want to go at a time that is massively inconvenient.
Well the first thing to do was to dither. Then procrastinate, then consider the HOE curve and finally cry hot tears of frustration. OK not cry, but is that frustrating or what?
I felt the only course of action was to first swear loudly, and for some time, in the fortunately empty house then call them to ask for more money...and I got it. So I decided to take the job where there was a firm job offer (and therefore money, even more since I asked) and sadly turn down the interview for the other where I might not get it - wouldn't I feel foolish then? After two years of determined searching that hurt I can tell you.
And now work. Next week.
But this feels odd.
I've had to check I've got enough shirts to wear during the week. I haven't.
I had to check if my ties were still in the proximity of fasionable. They are not, they are not even in the same neighbourhood.
I had to check if my suits still fitted. They don't, I've lost so much weight during the last two years.
It is going to cost a small fortune going to work just in buying clothes.
Then, as Mrs EoTP also has a job we will, for the first time in our married lives, be out at work at the same time. So in our ever changing world we are having to evolve and adapt once again to deal with this. We've managed to fit our respective requirements for Mrs EoTP's little blue car (lbc) around our lives over the last two years but it now seems that I will also have to have a lbc of my own. Good grief that's a quarter of my yet to come salary already gone and I haven't actually started yet.
That can't be correct, can it? My WIGAJ list involves lots of expensive electrical gadgets for me and not necessary transport.
And yet it is. For the cruel fact is that it costs to go to work.
And yet I am not complaining. Not yet anyway, that will come after five months when the halo effect of having a job has worn off, when I discover that most of my colleuagues are paid at least as twice as much as me but collectively have half the qualifications and that I could do all their jobs without even breaking into a sweat. At least I think that is what happens at work, it is all in the distant past.
In the end I don't think there was a choice. Yep, I'd have loved to have a job with a salary near to one I enjoyed two years ago and all the rest of the 'package' but it's a very tough world out there right now as we all know and this might very well be the career change that gets me out of the industry that keeps making me redundant. I'd quite like a spell away from the Job Centre where they will be buying me my own seat soon, I've been such a regular visitor over the last few years.
So the JSA back-to-work form has been completed and posted to Fife - well that's where all the jobs ever seemed to be on the Job Centre Plus site, the contract of employment has been posted to the new employer, I've bought some shirts and ties (and Mrs EoTP has returned them and bought something more suitable) and we are good to go.
Once I've negotiated the use of the lbc before I buy my own.
## Monday, 18 May 2009
### Because I'm worth it?
I've been offered a job.
No really, after all this time, after all these applications, after all this searching on every job site in the known universe an organisation has offered me a job.
During the interview I batted every question out off the pitch (that's a cricketing metaphor I believe but as I loathe sport of all descriptions it might well be a reference to tennis for all I know) and kept thinking 'Is that the best you can do, come on make me really think.' Anyway the organisation rang later that day and said the job offer would be in the shredder. Post. Sorry that's just a habit after two years searching.
The thing is, and I say this with all due modesty, I am vastly over-qualified for the job and therefore, naturally, will be vastly underpaid to do it. Think Nurses. However as it represents 100% more than I am being paid now I think that's not a bad deal really and I'm very happy to be employed again. It's also a totally different business to one I've been in most of my working life i.e it doesn't lose money and most people currently want what it offers, unlike the car industry where I've come from where the reverse seems to be true. The last time I made a bid to leave the automotive industry in, oh let me see, 1985, this proved to be such an unmitigated disaster that I rejoined it two years later only to see my career path prove to be an unmitigated disaster for the next 20 years. There's consistency for you. Still I did get to travel around the world selling who-ha's to anyone who wanted them.
A couple of light years ago, in one blog, I explained my concept of intelligent capitulation - you can even Google the phrase now (you must have read it, 'Brilliant concept' The Times) - sometimes you just cannot cross the bottomless chasm with two planks, some string, a candle and an oil drum, and have to walk away. Thats walk away from the edge of the bottomless chasm of course, because if you went in the other direction you'd fall down it. That's how I felt with this job offer. I posed some questions to myself:
1. Have you got a job?
2. Are you anywhere near getting another job?
3. Is the economy in free-fall, think it's on a bungy jump but has forgotten to tie the harness securely around its waist?
4. Can you find any other jobs in the desired salary range?
5. Are you already fed up with signing on?
6. Has some one just offered a position you can make a demonstrable difference to and is willing to pay you to do so?
7. Are you desperate?
Well it was 'No' to number 7, but the others had fairly self evident answers and anyway I believe I can make a difference and, in a couple of year's time, the economic outlook will have changed and we can see what happens then.
And finally...I signed on last Thursday, once again.
'Do you use the JobCentre Plus web site to search for jobs' the jolly Job Centre person enquired?
'No' I said 'I think the site is quite poor and just keeps offering me jobs as a sous chef in Fife, I prefer to use Monster or Reed Jobs.'
Patronising smile. 'Shall I search for you right now?'
And the result of the search?
An admin job for £10,000 p.a in Perth. 'Oh' she said 'I see what you mean. You carry on doing it your way.'
Point proved I think.
I've always done it my way.
## Thursday, 7 May 2009
### Routine matters
You see when you have a job you have a porpoise. You have a purpose as well (Blogger spell check not working too well today).
You can waste another hour as you try and get the photocopier to work. Photocopiers have three states: warming up, on and jammed. 'Warming up' takes about a day, 'on' lasts for exactly the amount of time it takes you to get to the machine needing 45 collated, stapled copies for a stroky beard meeting in a hurry when it goes immediately into the 'jammed' state. Only one person in the entire company knows how to resolve the 'jammed' state and she is on holiday in Florida. Very occasionally the photocopier enters a new state, 'Add toner'. This is guaranteed to happen when you are wearing your newest, brightest, whitest shirt or blouse and you end up looking like a Friesian cow.
And so the week continues.
Tuesday is still an opportunity to have many 'coffee machine' meetings ostensibly for informal internal lobbying but really to moan about the management whilst drinking scalding, bland, liquid from plastic cups that are marginally thinner than the average condom.
Wednesday is a difficult day being half way between the two weekend states and generally, and unhappily, this is when the work often has to be done.
Thursday is a chance to catch up on the paperwork, emails and office gossip and CC in everyone else on the email network slowing the server down to the pace of an asthmatic snail.
Friday is, of course, only half a day long and that is mostly spent talking about the upcoming weekend and wishing everyone a 'good one'. For many companies, Fridays are enlivened by a dress down policy (which is interpreted to mean dress up) and where half of admin wear clothes that would be more suited to a club environment and are therefore deemed 'inappropriate' by HR but somehow 95% of the male members of staff find a compelling reason to visit admin on that day. 60% of the management team wear patterned jumpers and Rupert Bear type trousers that are more typical of a particularly brash Florida golf course. 10% of management wear clothes that would be more suited to a club environment and somehow several members of admin find compelling reasons to visit their offices or desks to discuss urgent admin problems concerning stapler supplies. 10% of management always forget about dress down days and dress in suits and try and pretend they knew all along but have client meetings and the balance is HR and no one knows where their offices are so can't recall what they wear anyway. And Windows takes 60 minutes to shut down and 'save your settings'. Where is it saving them, in Nepal?
And then the next week starts all over again.
But if you don't have a job...
You don't have a routine. There is nothing you actually have to do - well, apart from searching for and applying for jobs and that is quite important really, but that can be done at any time of the day or night. You can clean, shop and watch the 18 episodes of 'The Wire' that you have recorded at any time. Dress down day is everyday and I can't find my watch anywhere as I no longer need to wear it all the time. But I think it is important to find some rhythm to life even if that rhythm is a longer beat than it used to be. It takes work to deal with the lack of routine but that in itself can be quite liberating. Don't get me wrong I want to get back to work as soon as I can but just how often in life can you be largely free of the routine of work? Especially when I can send Mrs EoTP out to earn money.
Hedy Lamarr said 'Some men like a dull life-they like the routine of eating breakfast, going to work, coming home, petting the dog, watching TV, kissing the kids, and going to bed. Stay clear of it-it's often catching.'
There's something in that.
Now what shall I do next?
## Thursday, 23 April 2009
### This blog contains Graphic language.
I believe that each week starts, as I've said in earlier blogs, full of hope and promise. I like to think that I'm a fairly positive sort of guy (or a complete prat, take your pick) in the circumstances. But sometimes the pheasant of promise is shot down by the gamekeeper of despair.
I think most of us travel in hope. You know the sort of stuff: your teenager might spontaneously clean their room without being threatened by the withholding of their pocket money, they might change their underwear more than once a week, they might have a conversation with you that includes words with more than one syllable and lasts longer than 20 seconds, someone in a call centre is actually able to sort out your problem, that sort of thing. However hope is usually powered by experience and often our travel plans end up in Welshpool bus station late at on a Saturday night after the last one has left and there's not another bus until Monday. Next month.
So it is with applying for jobs. Unless you are applying for a job so ludicrous and so far beyond your abilities and qualifications (and I still don't understand why the White House won't let me run for President they are so narrow minded) then you cannot but help but hope some teeny weeny goes against all probability speck of hope that you'll get the job. And, because you have this teeny weeny goes against all probability speck of hope then, like a grain of sand in an oyster, a little pearl of optimism starts to grow and glow faintly - go on, don't deny it, it does doesn't it, and you start to visualise yourself in that very job.
Well it's no good, this has to stop for your own good.
Therefore I have produced a scientifically based series of graphs to demonstrate this tendency in a variety of job seeking circumstances and to help you all (well all three of you readers) quit hoping unnecessarily - a bit (but only a bit) like the NHS stop smoking campaign except with a lot less money and no pile of fag stubs outside the door where we all go outside with our coffees to have a drink, smoke and serious slagging off of the organisation and the boss and have you seen what they've done to our budgets, slashed them how can I run a department on 35p a year? I call this the EoTP HOE curve.
HOE = Hope Over Experience.
Let's start with the 'applying for a job on a on-line job site'. Here you can see that once you have submitted your CV you may as well go and feed the hamster, wash the car, disconnect from the broadband and go and live in remotest Peru because you are never going to hear from them. Ever. Again. Notice how one doesn't even start off with any hope at all as we all know that a giant electronic points system is directing all CVs into space as part of the CETI project and, even now, aliens on the planet Thorg are involved in the universe's largest ever job paper sift preventing them from launching their Earth invasion fleet until 2506 at the earliest.
Next we have the job application where your skills and experience exactly match the job specification, so much so that your Mum must have written it. Note how you start off with such high hopes and then, as time passes, those hopes decay a little and then you start hoping again, then fading steeply and rising so that the graph looks like a little range of mountains such as Hobbits might have to climb with the Ring. Perhaps tomorrow the call to an interview will come, they've all simultaneously gone down with the vomiting bug that's why they haven't called. You fool you.
Now here we have the graph that shows the HOE curve for those jobs that we think we might have a bit of a chance with. You know dark horse, got to be in to win. Hmmm.
Note here that despite all the evidence and knowing that there are 603 applicants for every job we still can't stop ourselves having just a glimmer of hope and that we'll hear. Something.
And then finally, in this current series, we have the 'Job Centre insists you apply for three jobs a week' HOE curve. Even though there are zero vacancies in your sector and yet 35627 job seekers we know, they (the Job Centre) know, the recruiter knows, even the aliens on Thorg know that this is just plain silly. But then you never know.
So there we have it. Scientifically graphed evidence that demonstrates that you might as well forget about every job application the moment it leaves your hand/PC/Mac/quill and indeed you might as well shred some of them yourself straight away as it saves time later in the process - if you are gong to be contacted then you will be, so no point worrying unnecessarily. The Gamekeeper of Despair has just reloaded both barrels. You're not going to make his day are you?
But...
Mrs EoTP applied for a job last year after not working (in paid employment, I know, I know child care and looking after the house is a 26 hour a day, 8 days a week job) for 16 years; we must not forget her HOE curve because it shows that, sometimes, the pheasant of opportunity gets away and leaves a very large message on the head of the gamekeeper of despair. And that message says 'never give up'.
## Wednesday, 15 April 2009
### Applying one's self
Is there any more fiendish way to get one's blood boiling than trying to complete a job application sent to you by a prospective employer in a template created in Word? I ask, as one who has made many applications for jobs over the last few years and whose heart sinks and soul dies a little more each time every time one of these doozies turns up.
There are, as we all know, many ways to apply for a job.
There's the click, fire and forget method of the on-line job sites where you can apply for (and be subsequently ignored by) many jobs with the practiced ease of a professional job seeker. 'Look at me apply for this job with my my back to the computer and using just a mirror to use the mouse whilst juggling two apples'.
We have the carefully tuned, honed and polished CV where you spend hours skillfully attempting to match the skills and competencies outlined in the job description and that you email off and then are subsequently ignored.
There's the 'Oh just send a standard CV as the company or agency doing the recruiting gives just palimpsest details of what the job entails' and then are subsequently ignored.
We have the proactive approach CV where you have forensically targeted a potential employer and sent in 'Let's meet up and talk about how I can turn your company around in 48 hours CV' and then are subsequently ignored.
However with all the above approaches there is one positive aspect to the whole process and that it is you, with whatever word processing package you use, who are in control of the layout, formatting and aesthetics of the CV. If the final result looks like a dog's dinner and hamster's nest remember that you that prepared it.
But then, but then, we have the templates, the very output of Beelzebub.
In Word.
Or created in Word but turned into a pdf.
These seem to turn up mainly from public sector organisations who have clearly had a giant big strokey beard conference some years ago to decide that no single public sector organisation will ever ask for the same information in the same way or same format.
Now this is all OK'ish if you make one job application every, say, 10 years. Any higher frequency than that and the whole thing makes you want to tear cushions apart with your bare teeth which goes down badly in the waiting area of the doctor's surgery I've found. And, furthermore, the application forms are created by people who never have to fill them in and have only been on the Word Perfect basic training course in 1992. If they did they would realise that some boxes are too small for people with addresses in Wales. Try fitting Llanfihangel-yng-Nqwynfa into a fixed text box.
Try entering Llanfihangel-yng-Nqwynfa into your SatNav after a good night out come to that.
There are those that want to know every job you have ever had, not just the employer but every title, every start date and finish date, all starting salaries and ending salaries and details of the job responsibilities. I was sorely tempted to put 'take bosses dog for walk daily and clean car weekly' for my first job with a major car manufacturer to see if any one read it. I can't remember what I did last week let alone 25 years ago.
A particular favourite of mine are those that want to know the year/month/date/time of exam/number of questions/name of invigilator/grade of each and every 'O' level/'A' level you have taken and the make of pen you used to complete the exam paper. Never, in all my working life, have I ever been asked to prove that I have the qualifications I say I have, so to find the original certificates would be a small miracle after all these years. I just ignore these boxes and say I have stuff, lots of stuff.
Let me give you some other examples after first taking a strong sedative.
I particularly like the forms that invite you to add additional details to support your application. These invariably permit you to type free text in small box that, when you exceed the length of the page, cause the text to mysteriously disappear on the next page because Word can't handle the page break. You can then spend hours of what remains of the rest of your life trying to find a way to make the text reappear on the following page without messing up the formatting of the rest of the document and swearing like a Marine as you fight a losing battle with Word.
And, because the forms all differ, you cannot cut and paste the information from one form to another as you can with a CV that you have created. Oh no no no, that would be too easy, all the boxes are different lengths, sizes and widths. And can anyone tell me why Word does not line up text exactly in columns? Why not? I can see the formatting marks, I can see they are exactly the same and I can also see that freakin' Word does not line up. Exasperation.
Sometimes the potential employer gets cunning, or more stupid, I can't decide which, by sending the document to you as a pdf. Presumably this is to encourage you to complete the form in your bestest hand writing. As I don't have bestest hand writing, or even averagest hand writing I convert these forms into Word and complete them as normal - see above for comments, and send them back in and let them work out how I did it. And then get subsequently ignored.
My favourite example, in the recent past is one I completed last week. There was no electronic version of the application form, just the old fashioned 'complete in ink' paper form except, except that it said 'You may complete the form by pasting in appropriate answers to the sections'. So, let me understand this clearly, there is no electronic version but I can type my responses in Word, print out the result once I have worked out the size and shape of the boxes and then cut them out and glue them into the various sections. Yes. A masterclass in wasting time if I ever saw one and presumably that won an award at the public sector conference for imagination and innovation in application form design
I suppose it wouldn't matter if you only did this occasionally but, as a dedicated job seeker now fully signed on, it's a regular thing. However one of the requirements of having a Job Seekers Allowance is that you apply for three jobs a week. I pointed out that there weren't three jobs a week that I could apply for that were in any way suitable for my skills and competencies. So we agreed I would just apply for any three - but you can bet they won't be for public sector organisations.
## Monday, 30 March 2009
### Tales of the unexpected
Looking for a job is a lot like buying a house really.
You lie back in your sofa in your current home, just after switching off 'Grand Designs' where Kevin McCloud has once again waffled on about the architectural narrative remaining coherent ('I think it does'), finish your second bottle of wine and decide to move to the house of your dreams.
But you have to remember that for most of us buying a house follows this inevitable sequence:
What you want.
What you will accept.
What you end up with.
So it is with a job, especially when there are at least 10 applicants for every job, according to the Trades Union Council a few weeks ago. Ten, I should be so lucky, there seem to be the entire population of the Isle of Wight applying for every post that I think looks interesting.
When I last lost my job I once again went through the 'no need to panic, plenty of time, important to find a position that fits my skills, knowledge and interest' rationalisation, then panicked. Nah, I didn't I'm a job seeker vet, man don't panic, man gets on the street does a little hussling, y'know whad I'm sayin?
Well 70 plus unsuccessful applications later I'm beginning to doubt my strategy a little. Having spent all this time carefully targeting selected organisations, finessing my CV to meet the job spec and writing covering letters that Mr. W Shakespeare himself would have been pleased with and to have achieved nothing at all sort of tells me something and this is what it tells me...
If you keep doing the same thing you keep getting the same results.
So is it time to abandon the strategy, shout everyman for himself, push the women folk and children aside, grab a lifebelt and throw myself into any job that looks like it might pay something and the 'fit to skills, knowledge and interest bit' can go to hell in a hand cart? Is it time to respond to those advertisements tied to lampposts at road junctions that ask do you want to earn £2000 A WEEK? Just ring this premium number in Columbia and ask for Serge, own AK47 an advantage. Or should I consider a paper round?
I was in my school's CCF (Combined Cadet Force), RAF section. They probably have a MI6 section these days. Yes, that sort of school. Anyway we went on mandatory annual camps to military establishments all over Europe where we got to do all sorts of spiffing military stuff like fire big, noisy weapons (must tell you about the Bloodhound missile I fired at a Russian jet sometime) and get cruelly bullied by the regular squaddies who had very innovative uses for Kiwi boot polish. At one camp we had the task, as a teams of five, to cross the mythical bottomless ravine using only two planks, a teddy bear, tube of toothpaste, four sticking plasters, 10 foot length of rope, three oil cans, a woman's bra, (empty, pity) and a copy of yesterdays Daily Mail. I learned two valuable lessons here, apart from never volunteer and just how hard boot polish is to remove from the body:
1. Sometimes it is possible, with imagination, teamwork, innovation, planning and not quite enough time, to figure out how to get the team safely across the bottomless ravine.
2. Sometimes, no matter how imaginative you are, how well you work as a team, how superb your planning and despite having more than enough time it is actually not possible, with the equipment to hand, to cross the bottomless ravine.
What the military seem to be looking for in 2 is the concept of what I'm going to call Intelligent Capitulation. That is to say you realise that what is being asked for is just not possible so you stop wasting your time and give up and go and do something else. Or wait to be captured. Or helicoptered out. Or blow something up. I don't know I didn't join the army, stop asking.
So now we have two concepts to consider. I just like to get you thinking.
• If you keep doing the same thing you keep getting the same results.
• Intelligent Capitulation.
The problem comes back to process of buying a house. Do I now abandon my dream of having a replacement job with a salary that was similar to one I had? Do I dilute my expectations to that of looking for a job that is somewhat less than the package I had? Do I just accept anything at just about any salary as long as there is some income coming in? What do I tell the Job Centre when I sign on? Shall I tell them about my theory of intelligent capitulation and see if they still want to pay the Job Seekers allowance?
Or do I just accept that with my background , skills and qualifications then hell will freeze over before any one offers me a job with a significantly lower salary than I used to have and I might as well stop bothering applying for circular meat patty high temperature rotating operative jobs.
Now then, I still have the bear, bra and one plank left - just place it on the Teddy Bear's head and tie it tightly. I'm sure we can cross the ravine that way.
## Thursday, 19 March 2009
### Let us prey
Do you know what gets on my toot and really, really annoys me? Many things actually, so many that I need two bouncers with a red rope rail only allowing selected annoying things through, so that there is often quite a queue to get on my toot at any one time of the day or night. There is health and safety everywhere these days you know, the bouncers even have to carry those clicker counters to ensure that the maximum capacity of my toot is not exceeded.
What annoys me are those organisations that prey on your deep felt sense of insecurity as you search for a job.
First up are those that clearly come from such on line job sites such as Monster - I'm not picking on Monster - no actually I am, they all seem to come from that source now I think of it. The emails start off innocently enough as the mark is identified; 'I've just reviewed your CV which is of considerable interest to me' (heart begins to speed up slightly) 'and I think it may be worth us meeting to consider your CV and possible career moves.' (Blood pressure rising, adrenalin starts kicking in, it might be an interview, it might be, OMG, a job offer). And then, so so seductively, the letter continues its tantalising message of hope until the sign off, 'Let's meet' it oozes, 'at our nearest office to you for a chat.' Then you notice, but only in the mouse type or by checking out its web site, that actually it is not a potential employer but an agency trying to flog you a course in how to find a job, for a big fat fee of course. It doesn't promise you will find a job. Just that it will take your money from you. And it doesn't tell you how much money either, presumably not until it has ensnared you in its web of promise and seduction.
Next on my list of let's profit from other people's misery shall we is the 'Is your CV being targeted at the right headhunters? Just send us £500 and we'll make sure that we distribute your CV to selected headhunters who will then whoop with unabashed joy, and jump in the air pumping their arms with clenched fists at seeing exactly the right candidate for the job they are seeking to fill, fall into their hands with no effort on their part just as they were about to despair. The job is practically yours right now'. This sentence is, of course, correct up to bit in italics. OMG people fall for this? Well of course we do, we are desperate, we need a job; any thing, including handing over large amounts of our fast disappearing cash has to be worth it doesn't it? Er no, in my experience. Handing over large amounts of cash to Snake Oil merchants is the last thing you want to be doing. There are plenty of organisations that help you for free - just look at the wonderful web site www.copingwithredundancy.org.uk and you'll see what I mean. For £5. Kidding, it's free.
We have the pernicious CV writing companies. 'Just send us your CV for a free review' they gush and we'll tell you how to write a CV that targets your desired job with the accuracy of an American missile.' Oh yes? How so I ask? If CVs were that easy to manipulate that potential employers would immediately jump into their company car, drive to your house, go down on their knees and beg you to work for them the instant they glanced at the carefully crafted document don't you think that might, just somehow, have leaked out or potential employers become inured to them?
Now don't misunderstand me. I'm a pretty positive sort of guy that sees the start of each week as full of hope and promise (or a complete prat, you can take your pick really). But these sort of things just get me going. So much so I feel that I have to go to Tescos and see if I can buy some hope and promise there as these organisations do so well in destroying exactly that by about Tuesday morning.
I see the Government believes that there is an opportunity to fast track refugees from the business world into teaching. As some one still intimately involved in the world of education (Mrs EoTP and her job and with children still at school) you can only gasp open mouthed at the sheer audacity of the scheme. Now I accept that many of us ended up in the wrong industry, and some of us are still wondering how we can escape. I ended up in the car industry because Big Thirsty Cars of America offered me a job first during the milk round at Uni and hinted that a company car would follow in short order - and it did, a 2.0 saloon no less, with a brown vinyl roof covering, velour seat covers and optional push button radio (LW and MW, eat your heart out). One of the students I knew stayed on to to complete teacher training after Uni because he fancied two blond babes on our course. The rest of us could see that they would rather run around naked on the campus in mid Winter than have anything to do with him, though he and I might have voted for that if we'd had a choice and Mrs EoTP (to be, though I didn't know it then) was away for the weekend. However that decision cost him dear as he is still on the fringes of teaching and can't escape either.
I went to one of my child's parent teacher evenings last week. I get annoyed before I even get to them, and I've been to many now, and therefore been annoyed many times You'd think it would be the easiest thing in the world to organise slots for clots (i.e, us parents) and rotate the meetings in a controlled and organised manner thus enabling us all to talk to about our little darlings in the time allowed. You'd think. What actually happens is anarchy, every time, as parents/children/teachers fail to get a grasp of what is going on and mill around ineffectually. We never did get to see, literally, one teacher, who was surrounded by layers of parents even though it is meant to be a one-on-one meeting. The Government clearly believes that bringing in experienced people from industry will give a much needed boost to these apparently ineffectual academic types and sort things out. Well it won't because, in industry, the well paid executives are far more incompetent than in academia - it's just that they can hide their giant, enormous mistakes much easier than you can in public life and in the public eye - I present the evidence of the finance industry M'lord but, if you want back up, then the automotive industry is quite instructive. For example GM. And Ford. And, well, possibly many other car manufacturers. 30% over capacity in a booming world economy? No wonder they are in trouble now. I wouldn't want to be a teacher. I think that they do a great job with such variable inputs (i.e. kids) but one does question their ability to manage an organisation of more than three people. At times. Such as parents' evenings.
Anyway back to retraining. I have been extensivley retrained I can reveal. It was suggested to me (and when Mrs EoTP suggests you'd better listen) that we needed a grout cleaning brush, as my bathroom cleaning came under some detailed scrutiny a week or so back. Apparently the grouting wasn't clean enough and needed a good scrubbing and therefore I concluded I needed a good scrubber. The old ones are the best I feel. A grout scrubber was bought (men - Lakeland - sells everything for the house that you never knew you needed and then some) My grout now gleams with the intensity of three white hot suns as do my toilet bowls and basins.
See, we men can multi-skill and multi-task and if the Job Centre ever finds out I'll be cleaning the municipal toilets as part of my next career step.
## Friday, 13 March 2009
### Cold comfort
Mrs EoTP has a job (yippee).
Mrs EoTP is home at the moment, and has been for two days, with a streaming cold, which she has very kindly shared with me.
Thank you, Mrs EoTP.
As someone without much work to do, (but don't get me started on the toilets and bathroom which 'need cleaning'. Mrs EoTP doesn't take much notice when she's out at work but she has the eyes of an eagle at home) when you don't feel well you just stay in bed as everyone leaves for school/work. Or you wave them off and then dive back into bed and feel very sorry for oneself all day, such a luxury.
Of course, at work, there are all these procedures to follow when you've been off sick.
• Notify someone who is the slightest bit interested. That's not always easy as all your colleagues are immediately ensconced in their own work on arrival at work or at least arguing about who is going to make the coffee. Of course you have to tell them this in a voice that suggests that you could come into work if it was absolutely necessary and would be sure that your slumped, fever wracked body in the corridor wouldn't get in the way too much but if it is OK with them you'll just hang on at home with your tissue box clutched to your side.
• Find out if you have to call in sick on subsequent days. Nobody ever seems to know this.
• On returning to work try and figure out how on earth you report your sickness and decide how graphically to describe the symptoms which are generally asked for in a big box on the sickness reporting form. Too little and there may be just the hint that you've been swinging the lead; too much and you've been Googling for just the right words to elicit full sympathy and plausibility and noted down an illness that only three people in the history of medicine have ever had.
Many smaller organisations seem not to be too bothered about all this reporting stuff. They wave an airy corporate hand at you suggesting 'oh don't worry about all that malarky' but actually means 'We'll make sure we make it up from you in the next year as you work twice the hours you are actually paid for.' And of course we do make it up - or you do if you are in work. I don't have to at the moment but just find that the ironing mountain is three times higher than when I last looked at it.
Then there is this fiendish system called the Bradford factor: using a cunning formulae and algorithms devised by the Devils's own little imps it measures the disruption of short, unplanned absences thus;
$B = S^2 \times D$ (and you thought you'd left algebra behind for ever didn't you?). Don't say that this blog doesn't teach you anything. This identifies malingerers with the accuracy of an American missile. Sort of like a speed camera for illness at work really.
Of course for most illnesses the most sensible thing to do is take to your bed and ram Ibrprofen down your throat until everything seems a whole lot better. In just about all the illnesses I've presented to my doctor in the last 10 years that has been his stock response : 'Oh I've had that just take some Nurofen and stay at home.' No more signing off from work for a week. Though I did ask for a little more help than that when I fell headlong down the stairs and dislocated my shoulder.
If you are of a certain age then you may remember Mother's remedys. Mrs EoTP and I do.
• Colds, No tissues just one hankie which would be soaked after an hour as would the pockets of your school trousers, skirt (or both if it was a public school). Your nose would be the colour of the plastic noses for Comic Relief. I tried to dry mine (hankie not nose, there was health and safety even in the 1960's) on the fire guard of the coal fires in the classroom. Yes, in the classroom, Big cast iron stoves. With coal scuttles.
• A warm onion in a bag for earache.
• Warm olive oil poured into your ear to cure earache.
• Syrup of figs for the going problems
• Milk of Magnesia for the stopping problems
• Calamine lotion - for making you look stupid with pink residue drying on your skin
The other thing that always helps is a Brandy. Surely it's not too early for one of those? Well I'm working on Doctor EoTP's orders so mustn't ignore them, must I?
## Monday, 9 March 2009
### I'll be back
The one thing you discover about searching for a job is that...it's just a job.
The last month has been a very difficult one with a death in the family. Naturally that takes precedence over all other things - like looking for a job.
Still, the show must go on, and I still have things to ramble on about. And will be doing so later this week.
I will be back.
EoTP
## Tuesday, 27 January 2009
### People who need people are the luckiest people
There are a number of phrases that set my in-built danger bells ringing loudly.
• 'Dad can you help me with my maths homework.'
• 'Dad I'm going out I don't know where I am going or what time I'll be back.'
• 'Dad I'm just going to use the phone I won't be on it long.'
• 'You've won a weeks free accommodation in Florida, call this premium rate phone number to Croatia to find out more.'
• Can you cook the meal tonight I'm going out?'
• 'I'm calling from BT to tell you about our new calling rates.'
and
• 'People are our greatest asset'
This last statement is generally uttered with the implication that the CEO/MD is also saying 'I love our people so much, so very much, that I want to take them all home with me every weekend for a party and give then give them all picture of a kitten and a puppy each because I love them so much did I tell you that?'
The last phrase sends shivers down my spine as it often associated with not investing in any other physical asset to enable staff to do their job with any degree of efficiency. There is also the proudly uttered 'We are investors in people' - great, but can you get a new biro from the fearsome Keepers Of The Stationery or more paper for the photocopier, can you hell as like.
You see, from my perspective, telling your employees that they are the company's greatest asset and then actively preventing them from undertaking the job they are paid to do by not giving them the correct tools is, well, unfortunately far too common. And if organisations are not actively preventing the staff doing their job then, by not fixing the problems, they clearly are condoning the whole can of wiggly things. I know that W Edwards Deeming covered all this back in the 50's so I won't go on about it (well not more than I am already doing) - but in my experience most people come to work with the intention of doing a good job, or at least the one they are paid to do, and then spend the day fighting the system which seems intent in preventing them doing exactly that. This is also true of volunteer work as well where the great British spirit of muddling through is alive and well and being honed to new levels of muddling not hitherto considered possible.
And then the recession.
'Get rid of people and get rid of them now'
'But I thought people were our greatest asset?
'Who said that, I never said that, I hate people, they are a liability, there are too many people around me I feel crowded get rid of them, preferably lots of them and they are drinking too much coffee and eating my biscuits. And shred those pictures of kittens and puppies, they make me feel nauseous.'
'But without people we will not be able to provide the high level of service to our customers that we, er, like to pretend we are delivering.'
'You still here, just lose the people and quickly. And you can go too.'
'Oh...'
Well that is Capitalism for you I suppose, in all its ruthlessness, but when you hear of all the people now without jobs, and know there is more to come, it sort of reminds me of the first letter I ever had making me redundant - 'You are surplus to requirements'.
That's not how you deal with your greatest asset, that's how you treat a filing cabinet.
## Friday, 23 January 2009
### Door stops
It has been so cold in the house during the day over the last month I've had to use an ice scraper on the computer monitor bef
ore I could use it.
EoTP develops a new law of finance
When you have a much reduced income then the sensible thing to do is to reduce your spending at the same time. However there is an immutable law of the universe that states 'a drastic fall in your income and a sensible and matched reduction in spending is immediately followed by a corresponding substantial rise in your utility bills and an expensive failure of several vital domestic appliances.' Thus our gas and electricity monthly bill has just risen by 74%. Now I have appealed, in vain, to the accounts department of Horrendous Gas Bills and Laughingly Large Electricity Costs Ltd and thought I'd proved that, actually, our fuel consumption had fallen over the year - cunningly I used their own graphs and data but they remain unmoved.
So back to the cold house. You see, when you work you are so pampered, well at least many of us are/were, with warm buildings and, sometimes, even aircon during the summer. Mrs EoTP would like to point out here that she works in an Elizabethan building where the only way to warm it up was throw another Catholic on the fire. That was in the 16th Century of course. As a result, at home, the heating does not get switched on during the day and I have to wear so many layers to keep warm that the Michelin Man looks anorexic compared to me. I can't bend my arms to use the keyboard and have to rock from side to side to press the keys. thiz hus teken mi al dai to tipe.
Being at home also involves talking to a lot more people who knock at the door during the day, people I didn't know existed before unemployment. There is of course the Postie. Consistently, through snow and sun, rain and fog, she never fails to drop yet another substantial bill through the letterbox but never, ever any job offers. Don't get me wrong, after being home by myself all day with just my head goblins to chat with, the opportunity of talking to another human is one I can eagerly grasp.
But there are the others. We have the proselytisers, naturally, a steady stream of people most weeks, with the convictions of their faith, wishing to share it with you. I have no problems with anyone trying to convince others of their faith and will listen politely at my door for up to several seconds before declining their various tracts. Live and let live as long as it doesn't scare the horses I say. They seem to travel in twos: good proselytiser, bad proselytiser? Sometimes it is necessary to be a little more assertive in convincing them to leave, they having mistaken my smile as I greet them for someone desperate to be converted to Pantheism, become a Jedi Warrior, Man from Uncle or whatever. It is the ones who presume on my innate politeness by coming back for a second or third go over the next few weeks that, frankly, start to get on my er, er, er, wick (that'll do). I do point out that I don't come to their door and attempt to convert them to paganism or atheism on multiple occasions even though they have made it clear that they are not interested but we could make a pact and, if they don't stop calling, I will dress in a long white sheet and pointy hat and dance round in circles, singing and clapping on their drive whilst waving an inflated pigs bladder on a stick at each turn of the Equinox and other divers ancient ceremonial dates and is that OK with them?
We also have many itinerant peddlers call. They launch into their sales pitch for Latvian Gonks, Peruvian nose flute music CDs or super cloths (super cloths??) the instant the door opens, presenting you with an 'Identity card' that they have clearly made on a 1950's John Bull childs' printing set with a wonky, blurry picture of what looks like a Womble stuck on. I generally laugh at this identity card and tell them my kids could make a better one. Last week one presented me with a 'Peddlers' licence', allegedly signed by Sergeant in a Nottingham police station. I scoffed and he huffed and said 'Well, call them then it's pukka.' 'Right', I said, 'I will.' and made for the phone. He departed so fast he left scorch marks on the gravel on the drive.
Outside the house the Council are repairing the path. They are changing it from the 4 x4 testing ground it has resembled for many years to a blacktop surface with the smoothness of baize on a snooker table. Anyway they have this sign up - see the top of this blog. I wondered then, what sort of assistance they might offer me?
'Can you help me with my job seeking?'
'Yes go and talk to Albert there on the concrete mixer, he's a renowned expert and talks regularly on the Radio 4 'Today' programme to John Humphries amongst others. Albert is so motivational and passionate on the subject he'll soon put you right and get you back to work.'
'I'm confused about Marxism-Leninism could you explain the difference?'
'Of course, Brian driving the dumper truck completes the Wikipedia entry on that subject and lectures in many countries during his holidays - good friend of Castro actually.'
'I've never had cause to use the horse's hoof stone extracting tool on my Swiss Army knife, why do they still include one in the tool set?'
'Ah, now Jenny can answer that, she's an expert on medieval armour as well as our JCB driver.'
'I've been writing this blog on people who call at the door now I am at home a lot and don't know how to finish it.'
'That's an easy one, I can answer that for you, all you have to do is...'
## Wednesday, 7 January 2009
### Testing, testing
OK so we've had a letter saying they might, just, not promising anything though, it's only a thought, look I still have my fingers crossed whilst I'm saying this, be interested in you.
For a job.
But first, before they even start actually speaking to you, before they even deign to actually even start to pay you any attention whatsoever, you have to be tested.
Tested.
What with? Needles, samples and involving latex gloves and bend over Mr EoTP, that sort of testing?
No, scarier than that.
Psychometric testing.
You see in the old days you'd get a tap on the shoulder and a 'You're alright, just turn up for the interview more or less on time, try and find the right office, don't dribble too much, pretend you like the boss's wife and the jobs yours'. Then all those HR managers were targeted by recruitment professionals selling their snake oil and, suddenly, an interview with the pre-meeting shoulder tap was no longer good enough. HR had been convinced that some of us could blag our way through interviews without knowing anything about the job, managing people, team building and yet still be a complete and utter power crazed sociopath who would ruin most of the business before moving on. Many of those types found their way to the top I found. The HR team wanted to know more, much more about your psyche, team playing abilities and what your head goblins were saying to you when all was quiet and still.
So they came up with testing.
Now I am the first to admit that relying on an interview alone is a poor way of deciding who should get a job particularly if it is me that hasn't got the job. If I have got the job then it is a first class way of recruiting exactly the right person and don't tell me otherwise.
My first test was handwriting analysis. They asked me to write all the letters of the alphabet as neatly as I could between the lines on a page in a notebook using a pen with a nib and an inkwell. No, false memory there, that was at primary school. Shame I accidentally spilt all that ink over the dress of the little girl sitting next to me. No my first test came, strangely, at the end of an interview process. I was reasonably confident that I'd got the job and 100% confident I wouldn't accept it. However the HR manager remarked, seemingly out of no where, that I should have filled out the initial application form in ink, in my very bestest handwriting, and not have printed it out on a computer. Did he not know how many bleedin' hours I'd spent making sure the print came out in exactly the right places on the application form after lining it up in Word? Well no, he didn't so he made me write it out in longhand like some form of after-school detention. Now I'm left handed and wondered whether all the resultant Rorschach inkblots would count against me and they, too, would be analysed - 'Be careful of this one he thinks he has special yellow friends living on the ceiling.' No a week later he rang up and said, simply, 'You're OK.'
OK?
OK?
Is that all the handwriting analyst could come up with on their professional scale? What was the scale then?
1. The ruthlessness of Genghis Khan
2. The sheer terror of Margret Thatcher
3. The team building skills of Stalin
4. OK
6. The personality of Eeyore.
7. The intelligence of George W Bush
I turned the job down.
Next job application came with the computerised version. Left alone at a screen and keyboard I was given an hour to complete the test. Twenty minutes later I'd finished and, eventually, the HR person returned and was suitably impressed at my speed. 'That was fast' she said 'Some of those maths questions are very tricky, you did well to do them in your head.' 'Head' I thought 'I used a calculator I had in my pocket.' Well as no one had said not to I decided not to say anything - and took the offered job.
Of course HR are not so easily diverted anymore and the tests have become longer and more involved. I still have the results of several to hand. Let's share:
Actual quotes here.
• 'EoTP is unlikely to be reticent about coming forward.' - er, aggressive?
• 'He doesn't seem to have problems with stage fright in large groups.' Show off?
• 'Naturally curious'. Stop pushing your fingers in that electric socket - that was my Mother.
• 'Strongly creative.' Yes I see that of course I am, but 'don't let him work in finance'.
• Rules - likely to exercise a degree of flexibility.' Yes, I treat them as guidelines.
• 'You should employ him now.' Are you listening?
Here's another one, only 25 pages long this one. My own Mum wouldn't recognise me.
• 'Affable, socially confident.' Lock the drinks cabinet and hide the party food.
• 'Very independent.' Well I think that means very independent what do you mean you disagree, I don't care I'm going to do it anyway.
Should I take any notice of it, should anyone? In some ways they are like horoscopes as they all talk in very positive and flattering terms about someone very important to me i.e. me. Whether they truly make a difference in recruitment when the potential employer finally groks my age and goes 'No we wanted young and cheap, I don't mind if they have the personality and moral scruples of Pol Pot tell the older guy we actually wanted uncreative and totally biddable.'
So I leave you with just one question?
Why?
## Tuesday, 6 January 2009
### Consultation period
I could see the gleam of hope kindle and then blaze in her eyes with the heat of a white hot sun. She'd read my CV, discussed my career aspirations, and now she pounced with the zeal of someone revealing the entire secrets of the universe to the uninitiated.
'You could be,' she paused dramatically, 'a consultant' and then leaned back in her chair with the triumphant look of someone who has just bought the last bag of brussel sprouts in Tescos before Christmas whilst all around her the unlucky ones wailed and gnashed their teeth at their misfortune (well they do where I live).
And that was it, 50% of the result of the career counsellor's output after exhaustive psychometric testing, CV evaluation and interview skills. Oh that and be an interim manager which proved to be the other 50% and we've already had a little look at those roles. Having developed a poker face, as a result of so many people telling me 'Oh you'll soon get a job with your qualifications', and then not having done so, I asked, simply, 'Consult who with what?'. Had I asked how to fix the Large Hadron Collider for just £5.30 I feel I might have got a more fulfilling answer. 'Ah' she said, 'I've just forgotten that I had another client, er right now in Mumbai, must go'.
So let's put the beefburger of consultancy over the flames of possibility and see what catches fire. Well I've been an independent 'consultant' and it falls, in my experience, somewhere between the fun you have hitting your face with a house brick repeatedly and running over your foot with the car. Both interesting experiences in their own ways but you are very glad when they stop.
Firstly, consult what? Well, when you've had many years in the wacky and fun filled world of the sandwich-in-a-balloon industry you might feel uniquely qualified to be able to consult with the said purveyors of sandwiches-in-balloons. Of course, you first have to get over the hurdle that the the employer that has just ejected you might not feel like having you back on the premises, in the same town, county or even country and the fact that their competitors probably loathe them and their staff with a passion bordering on an obsession. The moment you left you were erased completely from corporate memory unless you figure in the same sentence as 'Ah that was the fault of XYZ, he/she didn't complete it before they left.' Your departure is good for at least six months of backside saving.
Then you discover that, strange though it seems, there are 65 other people who also claim to be to be consultants on the sandwiches-in-a-balloon business. You've never heard of them, you don't know where they've come from but they all seem to be pitching for the same teeny weeny bit of work. And all have strangely sounding business pedigrees that somehow make yours just that little bit less lovely and cuddly.
How much do charge? Oh this is so hard, so very hard. You are desperate for the work but don't want to sell yourself short because, once you name a day rate, hell will freeze over before you can raise it. £500 a day sounds a lot. But how many days will you work for and how any days do you have to spend not being paid looking for the next job. And you don't get paid when you are on holiday or have a sickie - pulling a sickie now costs you money (and that's a bad thing). Remember consultants often have the same, or slightly worse, survival rate of a soldier in the trenches during the First World War. I once turned down a week's work at £250 per day on the basis that it was half my usual day rate and then spent that week at home not earning anything. Still I had my pride. And an empty wallet. But I had my pride and that's what counts Tesco, that's what counts why are you escorting me out of the store?
The car, what car do you buy? Right, the big shiny company car has gone leaving only a small oily stain and few weeds on the drive. Now you have to buy your own. Prudence says a 10 year old Fiesta will do - but what will the client think? Too big a motor and they will think you are charging them too much, a pile of rust and they will think nobody else will employ you and have they made a major error. But how can you be a consultant without a sharp suit and a big shiny car? Everyone else has a big shiny car, why can't I Mummy?
And then there is the actual work.
'So what value exactly will you add to our company Mr EoTP?'
'Well I know stuff.'
'What stuff do you know and will that stuff enable us to sell more of our stuff for more money?'
'Well I know stuff about the industry and stuff about selling and stuff about marketing. And I have a kitten.'
'OK we like the kitten but not you - the kitten can stay.'
Fundamentally you will only get work as a consultant if you can actually show that, by the end of the paid period, that things are actually better in some way. Compare that to most people's jobs where they are satisfied if things are not actually materially worse than when they started out that morning or, at the very least, somebody else has been blamed for it.
And don't think that a groovy web site that lists your qualifications and Alma Mater will make the blindest bit of difference: it won't. Neither will funky headed paper - I had really very funky and colourful headed paper and I couldn't find any work after 18 months and therefore no one to send it to. Anyone like to buy 15 reams whilst we're here? The people who pay you money want the knowledge in your head extracted and put into a bottle and then they can throw the husk of your body out onto the cold, cold streets.
There are the accounts and the Inland Revenue to deal with and I still have nightmares about the day the VAT man sent in the
bailiffs. Yes I was VAT registered. Pay the VAT before the children eat, before the mortgage is paid, above all else do not fail to pay the VAT man as they have super powers that transcend all us mortals. Pay them and all will be well, they said. Well I did and the one time that the VAT system paid my cheque into the wrong account, through their error, I had the bailiffs knocking on the door three days later demanding anything that wasn't too large to get through the front door but not including Mrs EoTP fortunately. Oh it was all sorted out, how I laughed, though I'm still waiting for an apology 10 years later.
In the end I gave up the consultation work. To be fair it gave me up first. All my contacts dried up or moved on and in the end there wasn't really a market for what I was offering. Marketing support for small to medium enterprises since you ask. Though I was prepared to do anything by the end of the period. That work as an ice trucker paid handsomely thank you. As if.
So before you decide that consulting is the very thing for you, in the worst recession the country has seen since the dinosaurs were wiped out by a giant meteorite or some one said 'there's a big black rat just got off that boat do you think it might be carrying a funny foreign disease', just be sure that the stuff you are selling is the stuff organisations are buying.
Or you'll be stuffed. | 2020-09-18 13:37:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24271772801876068, "perplexity": 1957.7862381940915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00429.warc.gz"} |
https://physics.meta.stackexchange.com/questions/7569/what-is-our-position-on-answers-that-consist-almost-entirely-of-copied-content/9592 | # What is our position on answers that consist almost entirely of copied content?
We know that plagiarism is bad, and unacknowledged citations will be dealt with.
However, what about answers that properly acknowledge their sources, but consist of nothing but quotes?
I believe such answers do not consititute sufficient effort to be actually counted as answers, and should be deleted. If a user cannot be bothered to phrase the answer to a question in their own words, or to explain and give context to their quotes, they do not deserve to derive reputation from such a post.
• But the flip side of it -- if something is so complete and well-written to begin with, why "make it worse" by requiring somebody to paraphrase it? Shouldn't people get reputation for the depth of knowledge to know of an absolutely perfect source in the field? I think, like so many other things, there is a wide enough range here that we cannot rule in or out anything and it is up to the community to weed out the bad from the good. – tpg2114 Mar 2 '16 at 18:53
• @tpg2114 In the case that prompted this question the user was quoting large swaths of the wikipedia that only almost addressed the question. And while most wikipedia articles are pretty good, there aren't many that I consider to be excellent. So I think there is a case to be made that the perfect quote can be the perfect answer, I don't think that it represent the most common case. – dmckee --- ex-moderator kitten Mar 2 '16 at 19:19
• @dmckee Yeah, I usually see the bad ones too. But my only point was that there is a spectrum from horrendous to brilliant and I'm not sure we can decide on a policy that clearly and objectively works. I think it's best left up to the down votes and the closing if it doesn't answer the question or is unclear – tpg2114 Mar 2 '16 at 21:23
• @tpg2114 How do you close an answer? – Emilio Pisanty Mar 3 '16 at 11:25
• @EmilioPisanty Flag it as not an answer, or low quality, depending on the case. I was writing the comment on my phone and got tired of fighting it, so I didn't really want to go back and expand out what I meant at the time. – tpg2114 Mar 3 '16 at 11:30
• @tpg2114 Just saying - language matters. On the flags - using low quality for posts which do attempt an answer has been a bone of contention for as long as I can remember, and not-an-answer only applies if it is actually not an answer. The way to deal with bad answers which do attempt to address the question is votes and comments, unless it's really egregiously bad. – Emilio Pisanty Mar 3 '16 at 11:32
• @EmilioPisanty Agreed on all points. The point I was making is the same point you made in your answer -- I feel we already have the tools we need to address the different ways these types of answers show up. Through voting, commenting, and flagging appropriately we can do what needs to be done when needed and I don't think there is any "policy" we can write down to specifically target this issue. – tpg2114 Mar 3 '16 at 11:35
• @ACuriousMind: it is not clear to me what the purpose of this site is - is it to provide answers? If so, a link and a brief response should be sufficient; this is all that I would do for a student. Or are we writing our own Wikipedia articles? It seems to me that an answer which answers the OP should be sufficient, though a well-written answer (as voted on by the participants) is worth more "credits". – Peter Diehr Mar 3 '16 at 22:07
• Just re-read your last sentence: "...they do not deserve to derive reputation...". I would argue that if someone finds a relevant link, they could comment with that link ("see also []()"), but that doesn't provide a permanent record; they can put that link in an answer, to be flagged "link only", or they can go the extra mile and provide the relevant section from the link. I think that's already three steps up from doing nothing. As a community we can choose whether to reward that; but I don't think we should discourage it. – Floris Mar 3 '16 at 22:16
• @Floris: Your comment (2 up from this one) should be an answer here ;) – Kyle Kanos Mar 4 '16 at 14:45
• This is a really bad problem on Astronomy SE. Worst case I've seen - somebody posts a fairly lame question that could easily have been answered by looking at wikipedia; they then get upvoted several times; then the person that asked the question posts an answer that simply quotes the wikipedia page and that then gets upvoted! Go figure. See astronomy.stackexchange.com/q/89 or astronomy.stackexchange.com/q/4 – ProfRob Mar 4 '16 at 14:53
• @RobJeffries: I think you have the timeline wrong there: In both cases the user who asked the question immediately self-answered it. That is, they prepared both question and answer ahead of time, and purposefully decided to ask a question without research effort so that both that and their copy-pasted answer would get upvoted. Not that that's any better. – ACuriousMind Mar 4 '16 at 14:58
• @ACuriousMind No, I hadn't noticed that - it's even worse. – ProfRob Mar 4 '16 at 15:04
• @RobJeffries: Well, self-answering questions is fine (I've done that, too). Self-answering effortless questions with Wikipedia quotes seems less fine to me. I think the self-answer feature is there so that you can write something up which you haven't seen in that way anywhere else, but which you spent some time figuring out. – ACuriousMind Mar 4 '16 at 15:26
Unacknowledged or unmarked quotes are plagiarism, plain and simple, and they deserve the full weight of community displeasure to be heaved on them: comments, downvotes, and eventual deletion.
On the other hand, if an answer consists completely of well-acknowledged quotations then it deserves to stay on the site - it's not doing anything against the rules, so deletion is not appropriate. In general, though, such answers are of very low quality, and do very little to actually address the problem, so they deserve what all bad answers do: to be downvoted into grayness, with comments expressing this displeasure.
There are indeed cases where "something is so complete and well-written to begin with" that it would "make it worse" to require the poster to paraphrase it, and if that's the case - the quotation is completely on the mark that it fully answers the question in a self-contained but well-referenced fashion - then kudos to the poster and an upvote for providing a useful answer.
Otherwise, though, I say let votes show that we don't like this.
• If you want to be clear on language, may I suggest changing the sentence: "In general, though, such answers are of very low quality, and do very little to actually address the problem, so they deserve what all bad answers do: to be downvoted into grayness, with comments expressing this displeasure."? It could imply that you want people to use the VLQ flag, but that seems to be inconsistent with your reply to my comment above. Unless that is what you intended to say here, in which case carry on! – tpg2114 Mar 3 '16 at 11:44
• A late thought: plagiarism is bad, yes, but fixed plagiarism is not. So if a post contains an unacknowledged or unmarked quote, I think editing it to properly mark and attribute the quote (if there is enough information to do so) is better than downvoting and deleting, and once it's edited, everything is fine. Of course commenting to help the OP understand why they shouldn't plagiarize in the future is still a good idea. – David Z Mar 14 '16 at 14:57
As Floris remarks | 2021-05-17 19:55:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39410242438316345, "perplexity": 913.1925349012915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00546.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&oldid=88463 | # 2017 AMC 8 Problems/Problem 5
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 5
What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$?
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$
## Solution
We evaluate both the top and bottom: $\frac{40320}{36}$. This simplifies to $1120$, or $B$. | 2021-01-22 07:26:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3666064739227295, "perplexity": 5102.731880871106}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00770.warc.gz"} |
https://www.wikihow.com/Identify-Dependent-and-Independent-Variables | # How to Identify Dependent and Independent Variables
Whether you’re conducting an experiment or learning algebra, understanding the relationship between independent and dependent variables is a valuable skill. Learning the difference between them can be tricky at first, but you’ll get the hang of it in no time. A dependent variable is an outcome that depends on other factors, like the effects of a medicine depend on the dose. On the other hand, an independent variable is the cause of an outcome, and it isn’t affected by any other factors.
### Method 1 of 3: Understanding Independent and Dependent Variables
1. 1
Think of an independent variable as a cause that produces an effect. A variable is a category or characteristic that’s measured in an equation or experiment. An independent variable stands alone and isn’t affected by other variables. In a scientific experiment, a researcher changes an independent variable to see how it affects other variables.[1]
• For example, if a researcher wants to see how well different doses of a medication work, the dose is the independent variable.
• Suppose you want to see if studying more improves your test scores. The amount of time you spend studying is the independent variable.
2. 2
Treat the dependent variable as an outcome. A dependent variable is an effect or result, and it always depends on another factor. The goal of an experiment or study is to explain or predict the dependent variables caused by the independent variable.[2]
• Say a researcher is testing an allergy medication. Allergy relief after taking the dose is the dependent variable, or the outcome caused by taking the medicine.
3. 3
Remember that a dependent variable can’t change an independent variable. When distinguishing between variables, ask yourself if it makes sense to say one leads to the other. Since a dependent variable is an outcome, it can’t cause or change the independent variable. For instance, “Studying longer leads to a higher test score” makes sense, but “A higher test score leads to studying longer” is nonsense.[3]
Tip: When you encounter variables, plug them into this sentence: “Independent variable causes Dependent Variable, but it isn't possible that Dependent Variable could cause Independent Variable.
For example: “A 5 mg dose of medication causes allergy relief, but it isn’t possible that allergy relief could cause a 5 mg dose of medication.”
### Method 2 of 3: Identifying Variables in Equations
1. 1
Use letters to represent variables in word problems. Turning statements with variables into math equations makes it easy to see which variable is which. For example, suppose your parents give you \$3 for every chore you complete. You want to figure out how much you'll earn if you do a certain number of chores.[4]
• The \$3 per chore is a constant. Your parents set that in stone, and that number isn't going to change. On the other hand, the number of chores you do and the total amount of money you earn aren't constant. They're variables that you want to measure.
• To set up an equation, use letters to represent the chores you do and the money you'll earn. Let t represent the total amount of money you earn and n stand for the number of chores you do.
2. 2
Set up an equation with the variables. If you get \$3 for every chore you complete, say out loud, “The total amount of money I'll earn (or t) equals \$3 times the number of chores I do (or n).” That gives you the equation ${\displaystyle t=3n}$.[5]
• Notice that the amount of money you'll earn depends on the number of chores to do. Since it depends on other variables, it's the dependent variable.
3. 3
Practice solving equations to see how variables are connected. If, in the chores example, ${\displaystyle t=3n}$, and you do 5 chores, then ${\displaystyle t=(3)(5)=15}$. Doing 5 chores causes t to equal \$15, so t depends on n.[6]
• Say an episode of your favorite TV show is 30 minutes. The total time in minutes (m) you'll spend watching TV equals 30 times the number of episodes (e) you watch. That gives you the equation ${\displaystyle m=30e}$. If you watch 3 episodes, ${\displaystyle m=(30)(3)=90}$.
4. 4
Plug different values into the independent variable. Remember that, in an experiment, a researcher changes the independent variable to see how it affects other variables. Equations work the same way! Try solving your practice equations using different numbers for the independent variables.[7]
• Say you want to know how much you'll earn if you do 8 chores instead of 5. Plug 8 into n: ${\displaystyle t=(3)(8)=24}$. It's the same principle as a researcher changing the dose of a medication from 2 mg to 4 mg to test its effects.
### Method 3 of 3: Graphing Independent and Dependent Variables
1. 1
Create a graph with x and y-axes. Draw a vertical line, which is the y-axis. Then make the x-axis, or a horizontal line that goes from the bottom of the y-axis to the right. The y-axis represents a dependent variable, while the x-axis represents an independent variable.
• Say you sell apples and want to see how advertising affects your sales. The amount of money you spent in a month on advertising is the independent variable, or the factor that causes the effect you’re trying to understand. The number of apples you sold that month is the dependent variable.
2. 2
Label the x-axis with units to measure your independent variable. Next, make dashes in even increments along the horizontal line. The line should now look a bit like a ruler. These dashes will stand for units, which you’ll use to measure your independent variables.
• Suppose you’re trying to see if advertising more increases the number of apples you sold. Divide the x-axis into units to measure your monthly advertising budget.
• If you’ve spent between \$0 and \$500 a month in the last year on advertising, draw 10 dashes along the x-axis. Label the left end of the line “\$0.” Then label each dash with a dollar amount in \$50 increments (\$50, \$100, \$150, and so on) until you’ve reached the last dash, or “\$500.”
3. 3
Draw dashes along the y-axis to measure the dependent variable. As with the x-axis, make dashes along the y-axis to divide it into units. If you're studying the effects of advertising on your apple sales, the y-axis measures how many apples you sold per month.
• Suppose your monthly apple sales have ranged between 60 and 250 over the last year. Draw 10 dashes across the y-axis, label the first “50,” and label the rest of the dashes in increments of 25 (50, 75, 100, and so on), until you’ve written 275 next to the last dash.
4. 4
Enter your variables' coordinates onto the graph. Use your variables’ number values as coordinates, and place a dot on the corresponding point on your graph. The coordinate is where invisible lines running from the x and y-axes cross each other.
• For instance, if you spent \$350 on advertising last month, find the dash labeled “350” on the x-axis. If last month’s apple sales totaled 225, find the dash labeled “225” on the y-axis. Draw a dot at the point at the graph coordinate (\$350, 225), then continue graphing points for the rest of your monthly numbers.
5. 5
Look for patterns in the points you’ve graphed. If the dots form a recognizable pattern, such as a roughly organized line, there’s a relationship between the independent and dependent variables. The independent variable probably doesn’t affect the dependent variable if the dots are randomly scattered across the graph without any recognizable order.[8]
• For example, say you’ve graphed your advertising expenses and monthly apple sales, and the dots are arranged in an upward sloped line. This means that your monthly sales were higher when you spent more on advertising.
## Community Q&A
Search
• Question
What is the independent and dependent variable for y+5 = x^2 / 3+1?
The variable that is expressed in the first degree (having an exponent of 1) is the dependent variable. In this case, it's y.
200 characters left
## Tips
Submit a Tip
All tip submissions are carefully reviewed before being published
Thanks for submitting a tip for review!
Co-authored by:
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 26,054 times.
Co-authors: 4
Updated: April 19, 2020
Views: 26,054
Categories: Mathematics
Thanks to all authors for creating a page that has been read 26,054 times. | 2020-12-02 18:58:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4571830928325653, "perplexity": 945.9530441535761}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141715252.96/warc/CC-MAIN-20201202175113-20201202205113-00418.warc.gz"} |
https://ec.gateoverflow.in/1855/gate-ece-1991-question-1-4 | 10 views
Two two-port networks are connected in cascade. The combination is to be represented as a single two-port network. The parameters of the network are obtained by multiplying the individual
1. $z$-parameter matrix
2. $h$-parameter matrix
3. $y$-parameter matrix
4. $\textsf{ABCD}$ parameter matrix | 2022-10-04 04:41:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7613361477851868, "perplexity": 2838.4955807264287}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337473.26/warc/CC-MAIN-20221004023206-20221004053206-00309.warc.gz"} |
https://toph.co/p/xor-gun | # Practice on Toph
Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.
# XOR-Gun
By YouKnowWho · Limits 2s, 256 MB
YouKn0wWho has an integer sequence $a_1, a_2, \ldots, a_n$. He will perform the following operation until the sequence becomes empty: Select a non-empty subsequence of $a$ such that the bitwise XOR of the elements of the subsequence is $x$, and then erase the elements of that subsequence.
He is wondering if it is possible to erase the whole sequence by performing the operation as many times as he wants. If possible, then he wants to find the minimum number of operations that he has to perform to erase the whole sequence. Help YouKn0wWho solve the problem as he is busy preparing a contest.
You will be given $q$ queries each containing an integer $x$. Solve the problem independently for each $x$.
## Input
The first line contains two space-separated integers $n$ and $q$ ($1 \le n, q \le 3 \cdot 10^5$).
The second line contains $n$ space-separated integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \lt 2^{30}$).
Each of the following $q$ lines contains an integer $x$ ($0 \le x \lt 2^{30}$).
## Output
For each query -
• Print -1, if it is not possible to erase the whole sequence by performing the aforementioned operation one or more times.
• Otherwise, output the minimum number of operations that YouKn0wWho has to perform to erase the whole sequence.
## Sample
InputOutput
5 5
1 3 2 5 5
7
10
3
0
8
2
-1
2
1
-1
In the first query, in the first operation erase the subsequence $[1, 3, 5]$ as $1 \oplus 3 \oplus 5 = 7$. So the modified sequence will be $a=[2, 5]$. Now erase the whole sequence as $2 \oplus 5 = 7$. It can be shown that it is not possible erase the whole sequence using less than $2$ operations.
In the second query, it is not possible to erase the whole sequence.
### Statistics
75% Solution Ratio
NirjhorEarliest, 1M ago
LUMBERJACK_MANFastest, 0.1s
prodip_bsmrstuLightest, 1.0 MB
serotoninShortest, 571B | 2022-01-23 22:31:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3942441940307617, "perplexity": 904.781276065633}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00519.warc.gz"} |
http://www.aimsciences.org/article/doi/10.3934/cpaa.2010.9.1161 | # American Institute of Mathematical Sciences
• Previous Article
Homogenization limit and asymptotic decay for electrical conduction in biological tissues in the high radiofrequency range
• CPAA Home
• This Issue
• Next Article
Imperfect bifurcations in nonlinear elliptic equations on spherical caps
2010, 9(5): 1161-1188. doi: 10.3934/cpaa.2010.9.1161
## Kirchhoff systems with nonlinear source and boundary damping terms
1 Dipartimento di Matematica e Informatica, Università degli Studi di Perugia, Via Vanvitelli 1, I–06123 Perugia, Italy, Italy
Received August 2009 Revised November 2009 Published May 2010
In this paper we treat the question of the non--existence of global solutions, or their long time behavior, of nonlinear hyperbolic Kirchhoff systems. The main $p$--Kirchhoff operator may be affected by a perturbation which behaves like $|u|^{p-2} u$ and the systems also involve an external force $f$ and a nonlinear boundary damping $Q$. When $p=2$, we consider some problems involving a higher order dissipation term, under dynamic boundary conditions. For them we give criteria in order that $|| u(t,\cdot) ||_q\to\infty$ as $t \to\infty$ along any global solution $u=u(t,x)$, where $q$ is a parameter related to the growth of $f$ in $u$. Special subcases of $f$ and $Q$, interesting in applications, are presented in Sections 4, 5 and 6.
Citation: Giuseppina Autuori, Patrizia Pucci. Kirchhoff systems with nonlinear source and boundary damping terms. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1161-1188. doi: 10.3934/cpaa.2010.9.1161
2017 Impact Factor: 0.884 | 2018-08-20 02:57:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5335360169410706, "perplexity": 1088.3052481019563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215523.56/warc/CC-MAIN-20180820023141-20180820043141-00400.warc.gz"} |
https://socratic.org/questions/what-is-the-the-vertex-of-y-x-2-6x-6 | # What is the the vertex of y = x^2-6x+6?
Jun 11, 2016
vertex: $\left(3 , - 3\right)$
#### Explanation:
The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
Given
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x + 6$
$\Rightarrow$
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - \textcolor{c y a n}{6} x \textcolor{\mathmr{and} a n \ge}{+} {\left(\frac{\textcolor{c y a n}{6}}{2}\right)}^{2} + 6 \textcolor{\mathmr{and} a n \ge}{-} {\left(\frac{\textcolor{c y a n}{6}}{2}\right)}^{2}$
$\textcolor{w h i t e}{\text{XXX")y=(x-color(red)(3))^2+color(blue)(} \left(- 3\right)}$
which is the vertex form with vertex at $\left(\textcolor{red}{3} , \textcolor{b l u e}{- 3}\right)$
For verification purposes, here is a graph of the original equation:
graph{x^2-6x+6 [-3.506, 7.595, -3.773, 1.774]} | 2021-09-18 16:29:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4989253878593445, "perplexity": 3119.199132783938}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00401.warc.gz"} |
https://math.stackexchange.com/questions/2130781/p-mid-qr1-prove-that-either-2r-mid-p-1-or-p-mid-q2-1 | # $p \mid q^r+1$, Prove that either $2r\mid p-1$ or $p \mid q^2-1$
Let $p$ be an odd prime. $q,r$ are primes such that $p \mid q^r+1$. Prove that either $2r\mid p-1$ or $p \mid q^2-1$ $\def \m {\; \text{mod}\;}$
My Work
$$q^r\equiv -1 \m p\\ q^{2r} \equiv 1 \m p$$
From Fermat's Little theorem we have - $$q^{p-1} \equiv 1 \m p$$
From this two equation -
$$q^{(2r,p-1)} \equiv 1 \m p$$
As $p$ is odd prime so $p-1$ is even. Then $(2r,p-1) = 2$. So - $$q^2 \equiv 1 \m p \implies p \mid q^2 -1$$
So, my solution is saying always $p \mid q^2-1$ but I was supposed to prove that only one of them is right. Where I am mistaken?
I'd also like to know different approaches to solve this problem.
$(2r,p-1)$ is not always $2$, since $r$ is a prime and $p-1$ is even, if it is not $2$, it means $r| p-1$, so $2r|p-1$.
• $\def \m {\; \text{mod}\; }$ $q^{2r} \equiv 1 \m p \\ (q^2)^r \equiv 1 \m p \\ p \mid q^2-1 \\ \text{Then both are true .. :|}$ – Rezwan Arefin Feb 5 '17 at 20:16
• If gcd is not $2$ then $r \mid p-1$... I think it is not obvious. – Rezwan Arefin Feb 5 '17 at 20:18
I got another approach. $\def \m {\; \text{mod}\;}$
Let $d$ be minimum positive integer such that $q^d \equiv 1 \m p$. Then $$q^r \equiv -1 \not \equiv 1 \m p\; \text{and}\\ q^{2r} \equiv -1 \m p$$
From above 3 equations- $d \mid 2r$ but $d \not \mid r$. Since $r$ is prime, then $d=2\; or\; 2r$.
If $d = 2$ then $q^2 \equiv 1 \m p \implies p \mid q^2-1$
If $d=2r$ then from Fermat's $q^{p-1} \equiv 1 \m p$. So from 1st congruence we have $2r \mid p-1$
Actually this is same solution indeed. But from this it is easily understandable why $(2r,p-1)$ can be $2r$. | 2021-08-03 17:21:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9851857423782349, "perplexity": 188.6810752324009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00684.warc.gz"} |
https://intelligencemission.com/free-energy-gravity-water-pump-electricity-free-dishwasher.html | Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
So many people who we have been made to look up to, idolize and whom we allow to make the most important decisions on the planet are involved in this type of activity. Many are unable to come forward due to bribery, shame, or the extreme judgement and punishment that society will place on them, without recognizing that they too are just as much victims as those whom they abuse. Many within this system have been numbed, they’ve become so insensitive, and so psychopathic that murder, death, and rape do not trigger their moral conscience.
We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
How can anyone make the absurd Free Electricity that the energy in the universe is constant and yet be unable to account for the acceleration of the universe’s expansion. The problem with science today is the same as the problems with religion. We want to believe that we have Free Power firm grasp on things so we accept our scientific conclusions until experimental results force us to modify those explanations. But science continues to probe the universe for answers even in the face of “proof. ” That is science. Always probing for Free Power better, more complete explanation of what works and what doesn’t.
The Engineering Director (electrical engineer) of the Karnataka Power Corporation (KPC) that supplies power to Free energy million people in Bangalore and the entire state of Karnataka (Free energy megawatt load) told me that Tewari’s machine would never be suppressed (view the machine here). Tewari’s work is known from the highest levels of government on down. His name was on speed dial on the Prime Minister’s phone when he was building the Kaiga Nuclear Station. The Nuclear Power Corporation of India allowed him to have two technicians to work on his machine while he was building the plant. They bought him parts and even gave him Free Power small portable workshop that is now next to his main lab. ”
Vacuums generally are thought to be voids, but Hendrik Casimir believed these pockets of nothing do indeed contain fluctuations of electromagnetic waves. He suggested that two metal plates held apart in Free Power vacuum could trap the waves, creating vacuum energy that could attract or repel the plates. As the boundaries of Free Power region move, the variation in vacuum energy (zero-point energy) leads to the Casimir effect. Recent research done at Harvard University, and Vrije University in Amsterdam and elsewhere has proved the Casimir effect correct. (source)
Try two on one disc and one on the other and you will see for yourself The number of magnets doesn’t matter. If you can do it width three magnets you can do it with thousands. Free Energy luck! @Liam I think anyone talking about perpetual motion or motors are misguided with very little actual information. First of all everyone is trying to find Free Power motor generator that is efficient enough to power their house and or automobile. Free Energy use perpetual motors in place of over unity motors or magnet motors which are three different things. and that is Free Power misnomer. Three entirely different entities. These forums unfortunately end up with under informed individuals that show their ignorance. Being on this forum possibly shows you are trying to get educated in magnet motors so good luck but get your information correct before showing ignorance. @Liam You are missing the point. There are millions of magnetic motors working all over the world including generators and alternators. They are all magnetic motors. Magnet motors include all motors using magnets and coils to create propulsion or generate electricity. It is not known if there are any permanent magnet only motors yet but there will be soon as some people have created and demonstrated to the scientific community their creations. Get your semantics right because it only shows ignorance. kimseymd1 No, kimseymd1, YOU are missing the point. Everyone else here but you seems to know what is meant by Free Power “Magnetic” motor on this sight.
Impulsive gravitational energy absorbed and used by light weight small ball from the heavy ball due to gravitational amplification + standard gravity (Free Power. Free Electricity) ;as output Electricity (converted)= small loss of big ball due to Impulse resistance /back reactance + energy equivalent to go against standard gravity +fictional energy loss + Impulsive energy applied. ” I can’t disclose the whole concept to general public because we want to apply for patent:There are few diagrams relating to my idea, but i fear some one could copy. Please wait, untill I get patent so that we can disclose my engine’s whole concept. Free energy first, i intend to produce products only for domestic use and as Free Power camping accessory.
No “boing, boing” … What I am finding is that the abrupt stopping and restarting requires more energy than the magnets can provide. They cannot overcome this. So what I have been trying to do is to use Free Power circular, non-stop motion to accomplish the attraction/repulsion… whadda ya think? If anyone wants to know how to make one, contact me. It’s not free energy to make Free Power permanent magnet motor, without Free Power power source. The magnets only have to be arranged at an imbalanced state. They will always try to seek equilibrium, but won’t be able to. The magnets don’t produce the energy , they only direct it. Think, repeating decimal…..
Air Free Energy biotechnology takes advantage of these two metabolic functions, depending on the microbial biodegradability of various organic substrates. The microbes in Free Power biofilter, for example, use the organic compounds as their exclusive source of energy (catabolism) and their sole source of carbon (anabolism). These life processes degrade the pollutants (Figure Free Power. Free energy). Microbes, e. g. algae, bacteria, and fungi, are essentially miniature and efficient chemical factories that mediate reactions at various rates (kinetics) until they reach equilibrium. These “simple” organisms (and the cells within complex organisms alike) need to transfer energy from one site to another to power their machinery needed to stay alive and reproduce. Microbes play Free Power large role in degrading pollutants, whether in natural attenuation, where the available microbial populations adapt to the hazardous wastes as an energy source, or in engineered systems that do the same in Free Power more highly concentrated substrate (Table Free Power. Free Electricity). Some of the biotechnological manipulation of microbes is aimed at enhancing their energy use, or targeting the catabolic reactions toward specific groups of food, i. e. organic compounds. Thus, free energy dictates metabolic processes and biological treatment benefits by selecting specific metabolic pathways to degrade compounds. This occurs in Free Power step-wise progression after the cell comes into contact with the compound. The initial compound, i. e. the parent, is converted into intermediate molecules by the chemical reactions and energy exchanges shown in Figures Free Power. Free Power and Free Power. Free Power. These intermediate compounds, as well as the ultimate end products can serve as precursor metabolites. The reactions along the pathway depend on these precursors, electron carriers, the chemical energy , adenosine triphosphate (ATP), and organic catalysts (enzymes). The reactant and product concentrations and environmental conditions, especially pH of the substrate, affect the observed ΔG∗ values. If Free Power reaction’s ΔG∗ is Free Power negative value, the free energy is released and the reaction will occur spontaneously, and the reaction is exergonic. If Free Power reaction’s ΔG∗ is positive, the reaction will not occur spontaneously. However, the reverse reaction will take place, and the reaction is endergonic. Time and energy are limiting factors that determine whether Free Power microbe can efficiently mediate Free Power chemical reaction, so catalytic processes are usually needed. Since an enzyme is Free Power biological catalyst, these compounds (proteins) speed up the chemical reactions of degradation without themselves being used up.
Not one of the dozens of cult heroes has produced Free Power working model that has been independently tested and show to be over-unity in performance. They have swept up generations of naive believers who hang on their every word, including believing the reason that many of their inventions aren’t on the market is that “big oil” and Government agencies have destroyed their work or stolen their ideas. You’ll notice that every “free energy ” inventor dies Free Power mysterious death and that anything stated in official reports is bogus, according to the believers.
### The solution to infinite energy is explained in the bible. But i will not reveal it since it could change our civilization forever. Transportation and space travel all together. My company will reveal it to thw public when its ready. My only hint to you is the basic element that was missing. Its what we experience in Free Power everyday matter. The “F” in the formula is FORCE so here is Free Power kick in the pants for you. “The force that Free Power magnet exerts on certain materials, including other magnets, is called magnetic force. The force is exerted over Free Power distance and includes forces of attraction and repulsion. Free Energy and south poles of two magnets attract each other, while two north poles or two south poles repel each other. ” What say to that? No, you don’t get more out of it than you put in. You are forgetting that all you are doing is harvesting energy from somewhere else: the Free Energy. You cannot create energy. Impossible. All you can do is convert energy. Solar panels convert energy from the Free Energy into electricity. Every second of every day, the Free Energy slowly is running out of fuel.
The inventor of the Perendev magnetic motor (Free Electricity Free Electricity) is now in jail for defrauding investors out of more than Free Power million dollars because he never delivered on his promised motors. Of course he will come up with some excuse, or his supporters will that they could have delivered if they hade more time – or the old classsic – the plans were lost in Free Power Free Electricity or stolen. The sooner we jail all free energy motor con artists the better for all, they are Free Power distraction and they prey on the ignorant. To create Free Power water molecule X energy was released. Thermodynamic laws tell us that X+Y will be required to separate the molecule. Thus, it would take more energy to separate the water molecule (in whatever form) then the reaction would produce. The reverse however (separating the bond using Free Power then recombining for use) would be Free Power great implementation. But that is the bases on the hydrogen fuel cell. Someone already has that one. Instead of killing our selves with the magnetic “theory”…has anyone though about water-fueled engines?.. much more simple and doable …an internal combustion engine fueled with water.. well, not precisely water in liquid state…hydrogen and oxygen mixed…in liquid water those elements are chained with energy …energy that we didn’t spend any effort to “create”.. (nature did the job for us).. and its contained in the molecular union.. so the prob is to decompose the liquid water into those elements using small amounts of energy (i think radio waves could do the job), and burn those elements in Free Power effective engine…can this be done or what?…any guru can help?… Magnets are not the source of the energy.
## Remember the Free Power Free Power ? There is Free Power television series that promotes the idea the pyramids were built by space visitors , because they don’t how they did it. The atomic bomb was once thought impossible. The word “can’t” is the biggest impediment to progress. I’m not on either side of this issue. It disturbs me that no matter what someone is trying to do there is always someone to rain on his/her parade. Maybe that’s Free Power law of physics as well. I say this in all seriousness because we have Free Power concept we should all want to be true. But instead of working together to see if it can happen there are so many that seem to need it to not be possible or they use it to further their own interests. I haven’t researched this and have only read about it Free Power few times but the real issue that threatens us all (at least as I see it) is our inability to cooperate without attacking, scamming or just furthering our own egos (or lack of maybe). It reminds me of young children squabbling about nonsense. Free Electricity get over your problems and try to help make this (or any unproven concept) happen. Thank you for the stimulating conversations. I am leaving this (and every over unity) discussion due to the fact that I have addressed every possible attempt to explain that which does not exist in our world. Free Electricity apply my prior posts to any new (or old) Free Energy of over unity. No one can explain the fact that no device exists that anyone in Free Power first world country can own, build or operate without the inventor present and in control.
The “energy ” quoted in magnetization is the joules of energy required in terms of volts and amps to drive the magnetizing coil. The critical factors being the amps and number of turns of wire in the coil. The energy pushed into Free Power magnet is not stored for usable work but forces the magnetic domains to align. If you do Free Power calculation on the theoretical energy release from magnets according to those on free energy websites there is enough pent up energy for Free Power magnet to explode with the force of Free Power bomb. And that is never going to happen. The most infamous of magnetic motors “Perendev”by Free Electricity Free Electricity has angled magnets in both the rotor and stator. It doesn’t work. Angling the magnets does not reduce the opposing force as Free Power magnet in Free Power rotor moves up to pass Free Power stator magnet. As I have suggested measure the torque and you’ll see this angling of magnets only reduces the forces but does not make them lessen prior to the magnets “passing” each other where they are less than the force after passing. Free Energy’t take my word for it, measure it. Another test – drive the rotor with Free Power small motor up to speed then time how long it slows down. Then do the same test in reverse. It will take the same time to slow down. Any differences will be due to experimental error. Free Electricity, i forgot about the mags loseing their power.
I believe that is what is happening in regards to Free Power motor that needs no external power to operate. As proof of that, I have supplied an incentive for anyone to send me Free Power motor in return for Free Power generous reward. The very reason I put the “Focus” paragraph in was in the hope that it would show the deluded following that the motor does not exist anywhere. Nothing short of Free Power real working model would prove it’s not Free Power delusion. Stay focused on that and you will see the truth of what I am saying. Harvey1A magical magnetic motor? Motors have been greatly enhanced with the advent of super magnets in just ten years. Smaller and more powerful to say the least. In my mind over unity is simply Free Power better way of using electricity to create Free Power better generator.
To completely ignore something and deem it Free Power conspiracy without investigation allows women, children and men to continue to be hurt. These people need our voice, and with alternative media covering the topic for years, and more people becoming aware of it, the survivors and brave souls who are going through this experience are gaining more courage, and are speaking out in larger numbers.
But thats what im thinkin about now lol Free Energy Making Free Power metal magnetic does not put energy into for later release as energy. That is one of the classic “magnetic motor” myths. Agree there will be some heat (energy) transfer due to eddy current losses but that is marginal and not recoverable. I takes Free Power split second to magnetise material. Free Energy it. Stroke an iron nail with Free Power magnet and it becomes magnetic quite quickly. Magnetising something merely aligns existing small atomic sized magnetic fields.
The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
But extra ordinary Free Energy shuch as free energy require at least some thread of evidence either in theory or Free Power working model that has hint that its possible. Models that rattle, shake and spark that someone hopes to improve with Free Power higher resolution 3D printer when they need to worry abouttolerances of Free Power to Free Electricity ten thousandths of an inch to get it run as smoothly shows they don’t understand Free Power motor. The entire discussion shows Free Power real lack of under standing. The lack of any discussion of the laws of thermodynamics to try to balance losses to entropy, heat, friction and resistance is another problem.
This expression has commonly been interpreted to mean that work is extracted from the internal energy U while TS represents energy not available to perform work. However, this is incorrect. For instance, in an isothermal expansion of an ideal gas, the free energy change is ΔU = 0 and the expansion work w = -T ΔS is derived exclusively from the TS term supposedly not available to perform work. | 2019-02-16 02:23:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5495619177818298, "perplexity": 1550.9342517939672}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479729.27/warc/CC-MAIN-20190216004609-20190216030609-00603.warc.gz"} |
https://www.ctcms.nist.gov/fipy/documentation/tutorial/index.html | # How to Read the Modules Documentation¶
Each chapter describes one of the main sub-packages of the fipy package. The sub-package fipy.package can be found in the directory fipy/package/. In a few cases, there will be packages within packages, e.g. fipy.package.subpackage located in fipy/package/subpackage/. These sub-sub-packages will not be given their own chapters; rather, their contents will be described in the chapter for their containing package.
Last updated on Jun 15, 2022. Created using Sphinx 5.0.1. | 2022-09-29 01:39:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3207387924194336, "perplexity": 3845.367426262802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335303.67/warc/CC-MAIN-20220929003121-20220929033121-00406.warc.gz"} |
https://socratic.org/questions/how-do-you-factor-25x-50y | # How do you factor 25x + 50y?
$25 \left(x + 2 y\right)$
Taking 25 common we get, $25 \left(x + 2 y\right)$. This cannot be factored further. | 2019-12-12 13:37:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.529390275478363, "perplexity": 939.6834087031554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540543850.90/warc/CC-MAIN-20191212130009-20191212154009-00312.warc.gz"} |
http://math.stackexchange.com/questions/615901/intersection-of-trig-functions | # Intersection of Trig Functions
The questions asks to find the intersections of
$$f(x) = 2 \sin(x-7) + 6$$ and $$g(x) = \cos(2x-10) + 8$$
within the interval $[6,14]$.
So my general strategy was, 1) equate the functions, 2) get all the $X$s on one side and 3) convert to the same trig function.
So
$$2 \sin(x-7) + 6 = \cos(2x-10) + 8$$
I recognized the double angle in the cosine function, so
$$2 \sin(x-7) + 6 = \cos[ 2 (x-5) ] + 8$$
then
$$2 \sin(x-7) + 6 = \cos^2(x-5) - \sin^2(x-5) + 8$$
$\cos^2$ can be replaced with an identity, so
$$2 \sin(x-7) + 6 = 1 - \sin^2(x-5) - \sin^2(x-5) + 8$$
Group like terms and move then around,
$$2 \sin(x-7) + 2 \sin^2(x-5) = 3$$
Extracting the $2$ from the left side.
$$\sin(x-7) + \sin^2(x-5) = \frac 3 2$$
So here is where I hit a mental wall.
I could use the sine addition formula, but that would reintroduce cosine.
I can't simplify the terms any further since the angles are different.
Where would I go from here? Or is my approach off completely?
-
Is it $x-7$ or $x+7$? – Amzoti Dec 22 '13 at 22:09
Maybe consider plotting, including a parametric plot, see both using WA. Look at the point $6$ in the parametric plot, which is a hint. – Amzoti Dec 22 '13 at 22:22
Excuse my typos. I corrected it to x-7. – Marty B. Dec 22 '13 at 22:23
I graphed the functions on Desmos and I can see that there are two solutions in the given interval. However, I still don't know how to derive the answer. – Marty B. Dec 22 '13 at 22:31
## 1 Answer
$$\sin(x-7)+\sin^2(x-5)=\frac32\iff\sin(t-2)+\sin^2t=\frac32$$
$$\sin t\cdot\cos(-2)+\cos t\cdot\sin(-2)+\sin^2t=\frac32$$
$$\sin^2t+\cos2\cdot\sin t-\sin2\cdot\cos t=\frac32\iff y^2+\cos2\cdot y-\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)=\frac32$$
$$y^2+\cos2\cdot y-\frac32=\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)\iff\bigg(y^2+\cos2\cdot y-\frac32\bigg)^2=\sin^22\cdot(1-y^2)$$ Now you're left with solving a quartic equation in $y=\sin t=\sin(x-5)\iff x=5+\arcsin y$
-
Thank you @Lucian. Trying it out but it's not working. Substituting on the right, I get cos [ 2 (t+2) ] + 8. Then, cos^2 (t+2) - sin^2 (t+2) + 8, which becomes 9 - 2sin^2 (t+2), followed by 9 - 2 ( sin t cos 2 + cos t sin 2 )^2, and 9 - 2 ( sin^2 t cos^2 2 + 2 sin t cos t sin 2 cos 2 + cos^2 t sin^2 2 ). Replacing cos^2 with 1-sin^2 I get, 9 - 2 ( sin^2 t + sin^2 2 - 2sin^2 t sin^2 2 + 2 sin t cost t sin 2 cos 2 ). It just gets messier. It appears that I am misunderstanding your solution. Please advise. – Marty B. Dec 25 '13 at 12:52
@MartyB.: Done! – Lucian Dec 25 '13 at 16:37
Thank you @Lucian. This was an involved problem. – Marty B. Dec 28 '13 at 20:45 | 2016-05-30 05:25:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9001012444496155, "perplexity": 1091.3829677602598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049288709.66/warc/CC-MAIN-20160524002128-00065-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.zsquad.pl/data1/Dec-363-Tue-313/ | ## our products
D&Q Mining is a high-tech company integrating R&D, production and sales. It provides mature products and solutions such as crushers, sand making, milling equipment, mobile crushing stations, etc., for aggregate, mining and waste recycling.
### Extracting iron Iron and aluminium GCSE Chemistry
Iron is extracted from iron ore. in a huge container called a blast furnace. Iron ores such as haematite contain iron(III) oxide, Fe 2 O 3 . The oxygen must be removed from the iron(III) oxide in
More
### EXTRACTION OF IRON- EXTRACTION OF IRON FROM HAEMATITE
Iron is extracted from its oxide: ore called HAEMATITE (Fe 2 O 3). PRINCIPLE OF EXTRACTION : Extraction of iron is based on the reduction of HAEMATITE (Fe 2 O 3) with carbon.. DETAILS OF EXTRACTION : The process of the extraction of iron is carried out by the following steps: Concentration of ore Calcination or Roasting of ore
More
### Extraction of Iron from Haematite Grade 12 Science Notes
The slag is lighter than molten iron and to floats on the surface of the iron. The formation of prevents the oxidation of iron. d. Zone of Reduction:-This is the most important zone and has temperature of 600-700 0 c. In this zone Fe 2 O 3 is reduced to iron by co in three steps. 3Fe 2 O 3 + CO → 2Fe 3 O 4 + CO 2
More
### Extraction of Iron Class 12, General Principles and
May 15, 2020· Extraction of Iron. The cast iron is usually extracted from its oxide ore (haematite). This process involves the following steps: 1) Concentration. The ore is crushed in jaw crushers and is broken to small pieces of about 1 inch in size. The crushed ore is concentrated by gravity separation process in which it is washed with water to remove
More
### GCSE CHEMISTRY Extraction of Iron Haematite
Extraction of Metals. Extraction of Iron.. Iron is extracted from its ore in the blast furnace.. The main iron ore is called haematite. Haematite is iron(III) oxide Fe 2 O 3. The iron ore contains impurities, mainly silica (silicon dioxide). Limestone (calcium carbonate) is added to the iron ore which reacts with the silica to form molten calcium silicate in the blast furnace.
More
### How is iron extracted from haematite?
Iron is extracted from its ore, haematite in a blast furnace. The ore is led into the top of the furnace along with coke and limestone. The limestone decomposes in the hot furnace, forming calcium oxide. This reacts with the sandy impurities (silicon dioxide) to form a slag.
More
### Extracting iron Redox, extraction of iron and transition
Iron is extracted from iron ore. in a huge container called a blast furnace. Iron ores such as haematite contain iron(III) oxide, Fe 2 O 3 . The oxygen must be removed from the iron(III) oxide in
More
### Extraction of Iron Metallurgy Blast Furnace and Reactions
The extraction of iron from its ore is a long and subdued process, that helps in separating the useful components from the waste materials such as slag. The Haematite reacts with Carbon in the cast iron to give pure iron and carbon monoxide gas which escapes. $$Fe_{2}O_{3}+ 3 C \rightarrow 2Fe + 3 CO$$ Limestone is then added as flux, and
More
### Extraction of Iron (haematite) by blast furnace (in hindi
Aug 22, 2018· Extraction of Iron (haematite) by blast furnace (in hindi): Metallurgy. This video explain how to extrac pure iron from iron ore. the blast furnace is used t...
More
### EXTRACTION OF IRON IN A BLAST FURNACE reduction
The most common ore of iron is called haematite (iron(iii) oxide). Its formula is Fe 2 O 3. Haematite is added to the top of the furnace along with coke (i.e. carbon) and limestone. Three reactions take place during this extraction. Firstly, the carbon in the blast furnace burns with the hot air to form carbon dioxide. This reaction
More
### Extraction of Iron.docx Extraction of Iron Hematite
Extraction of Iron Hematite (Fe 2 O 3) is a form of raw material that is used to obtain iron.It is a reddish-black mineral consisting of ferric oxide, it is an important ore of iron. This ore contains iron(III) oxide which is then converted into iron in a blast furnace. The process used to extract iron from its’ ore is called smelting. (A 30m tall chimney like shaped tower is called a blast
More
### Extraction Of Iron From Its Ore Haematite
Extraction Of Iron From Its Ore Haematite Different Types of Iron Ore. 17 Mar 2017 . The iron minerals that are at present used as ores are hematite, magnetite, limonite, and siderite; also, occasionally ankerite, goethite, and. More Details Assaying for Iron Determination Methods. 5 Feb 2018 .
More
### The extraction of iron for iron ore (haematite), using
The extraction of iron for iron ore (haematite), using coke, limestone and air in a blast furnace. Iron is displaced from its ore (haematite) by carbon (from coke). 2Fe2O3 + 3C 4Fe + 3CO2 It is also displaced by carbon monoxide. Fe203 + 3CO 2Fe + 3CO2 Limestone reacts with
More
### Extraction Of Iron From Haematite Ore
Extraction Of Iron In Haematite Flow Chart. Extraction Of Iron From Its Ore Process. Extraction of iron in haematite flow chart. The early extraction of iron from its ore has been done in the direct process The line for the reaction of carbon dioxide formation runs nearly horiontally in the diagram oxide and reducing element the faster is the reaction of iron ore reduction avoiding either an
More
### details on the extraction of iron
EXTRACTION OF IRON- EXTRACTION OF IRON FROM HAEMATITEDETAILS OF EXTRACTION : The process of the extraction of iron is carried out by the following steps: Concen
More
### Iron Extraction Crushing Of Hematite
Iron Extraction Crushing Of Hematite. Iron ores are rocks and minerals from which metallic iron can be economically extracted.The ores are usually rich in iron oxides and vary in color from dark grey, bright yellow, or deep purple to rusty red.The iron itself is usually found in the form of magnetite fe.Ores containing very high quantities of hematite or magnetite greater than.
More
### extraction of iron from hematite
iron extraction process separation of hematite. iron extraction crushing of hematite Iron processing, use of a smelting process to turn the ore into a form from which used as a source of iron before 3000 bc, but extraction of the metal from ores . minerals are oxides, and iron ores consist mainly of hematite Fe2O3, which is.
More
### extraction of iro from hematite
3 Iron is extracted from its ore hematite in the blast furnace. Which gas is produced as a waste product? A carbon dioxide. B hydrogen. C nitrogen. D oxygen. Extracting and using metals ores native WordPress. Extraction and uses of Iron. Reduction is used to extract iron from its ore haematite which is mainly iron III oxide.
More
### extraction of iron from its ore outline
extraction of iron from it is ores igogreenfoundation. solution extraction iron from its ore salafi. solution extraction iron from its ore_CHEMISTRY The extraction of Iron and its impact on Sep 04, 201332Impacts on the environment due to the extraction of iron from haematite does not only start during the extraction of iron fr. Get Price.
More
### flow chart of extraction of iron from an ore
flow chart for iron ore extraction Grinding Mill China. flow chart involes in mining gold ore.. Flow chart extraction of iron crusher iron ore extraction <p>iron ore processing flow ..flow chart of extraction fe from its one Ore Beneficiation Plants » Learn More. flow Flowchart Of Extraction Of Metal From Haematite. Iron Extraction Process ROGESA.
More
### Assertion : For the extraction of iron, haematite ore is
Assertion : For the extraction of iron, haematite ore is used. Reason : Haematite is a carbonate ore fo iron. (1) Both the assertion and reson are correct, but the reason is not hte correct explanation for the assertion. (2) Both the assertion and reason are correct and the reason is
More
### The Extraction of Iron Chemistry LibreTexts
Aug 15, 2020· This page looks at the use of the Blast Furnace in the extraction of iron from iron ore, and the conversion of the raw iron from the furnace into various kinds of steel. Contributors and Attributions. Jim Clark (Chemguide.co.uk) Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
More
### Chemistry Project On Extraction Of Iron From Haematite
Chemistry Project On Extraction Of Iron From Haematite. Process In Extraction Of Iron From Iron Ore Extraction Of Metalsiron Wikiversity 2020425Iron is extracted from its ore haematite by reduction with carbon This takes place in a blast furnace named after blasts of air 20 oxygen heated to 1000 C which are blasted into the bottom of the furnace A mixture of crushed iron ore coke an almost
More
### Extraction of metals Metals Quiz Quizizz
Q. Iron is made by burning its ore, hematite. Hematite contains silica. What reacts with this impurity to remove it? answer choices . Carbon. Carbon Dioxide. Which statement is correct about extraction of iron from its ore? answer choices . Iron is more difficult to extract than copper.
More | 2021-06-21 16:57:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4789504408836365, "perplexity": 8396.58882575015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00415.warc.gz"} |
https://www.physicsforums.com/threads/charge-on-the-surface-of-a-conducting-sphere.376683/ | # Charge on the surface of a conducting sphere.
I need to find the charge and charge density on the surface of a conducting sphere of radius r=0.15m and of a potential of V=200V.
I know the surface charge density, which is Q/A. The surface area is 0.28274, but I can't find the charge. | 2020-11-30 15:19:19 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8812218308448792, "perplexity": 223.89666517118647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00397.warc.gz"} |
http://openstudy.com/updates/4f2c38f1e4b0571e9cba03e7 | ## hosein 2 years ago the wave function is "A exp[ikx]" , what's the possibility of reflection this, when face to V(x)=V0 delta(x) "delta function "
1. hosein
$\psi(x)=A \exp[ikx]$ $V(x)=V _{0}δ(x)$
2. hosein
wants R for that function but i can't solve it $R=(J _{reflected})/J _{incident})$ J is flux of wave func
3. Carl_Pham
4. Not the answer you are looking for?
Search for more explanations.
Search OpenStudy | 2014-12-19 08:52:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6543291211128235, "perplexity": 6173.685138264487}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768352.71/warc/CC-MAIN-20141217075248-00055-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://crypto.stackexchange.com/questions?page=12&sort=unanswered | All Questions
0answers
124 views
Existential Unforgeability of signature scheme against Adaptive Chosen Message Attack
While reading a literature on signature schemes, I came across the concept of Existential Unforgeability of signature scheme against Adaptive Chosen Message Attack. Can anyone point me to the paper ...
0answers
72 views
Need help solving a shift cipher problem?
A three letter alphabet {a, b, c} has probabilities p(a) = .5 and p(b) = p(c) = .25. A shift cipher is applied to pairs of letters. More precisely: plaintexts consist of two-letter sequences of ...
0answers
50 views
Forward Secrecy with pseudorandom functions
Let $H_1$, $H_2$ be keyed hash functions (e.g. $H_i(x) = SHA_{256}(s_i||x)$ for pseudorandom $s_1$, $s_2$). Let $s_n = H_1^k(s_0)$, $k_n = H_2(s_n)$, where $s_0$ is a secret (pseudorandomly chosen ...
0answers
53 views
Generating a public key certificate with ECDSA params
I have an ECDSA key generated in a HSM and I am able to retrieve it's public key components via the PKCS11 library. I would like to create a public x509 certificate with the public params for the ...
0answers
68 views
Bit level permutation
Could anyone explain how secure is bit level permutation? What is the most serious threat against the security of this kind of cipher? Thank you
0answers
131 views
how to find key matrix in hill cipher
I want to solve this problem but there are 3 known plaintext-ciphertext pairs. The key of Hill cipher is a 3*3 matrix as k=[k1,k2,3; k4,k5,k6; k7,k8,k9] where the unknown ...
0answers
70 views
side channel attacks on AES
Say you have a web application that's performing AES encryption. What sorts of side channel attacks should one keep an eye out for? Timing attacks affect RSA more than symmetric ciphers in-so-far as ...
0answers
69 views
Need help solving message encrypted on an Enigma machine?
So basically I'm in a class an one problem we were given a while ago was one involving an enigma machine and a few messages. I do not have any idea of how to approach it, and I really need help. The ...
0answers
71 views
advantages of hashing over elliptic curve signatures for a proof of work protocol
I'm trying to create a proof-of-work protocol for a proof-of-concept software, and it's basically something like this: ...
0answers
50 views
How are key length and size of group elements related?
I am reading a paper on ring signature where the author has mentioned a correlation between key length requirement and size of group elements. Can anyone hint me how he arrived at this conclusion out ...
0answers
122 views
Possible Digital Signature Hack?
I am starting to learn more about cryptography and I just read more about how asymmetric keys can make a digital certificate, and I would like if someone could explain me why the following case can't ...
0answers
62 views
Don't Understand these Parameters
I don't understand section 3 of the paper Fully Homomorphic Encryption over the Integers. How can one calculate these paramters and make these equations in pictures into code?
0answers
34 views
impersonating attack against a protocol
recently I'm reading Efficient eCK-secure Authenticated Key Exchange Protocols in the Standard Model, and I sense that the scheme in the following(the reader can refer to the paper for details) is ...
0answers
20 views
Ability to locate free sectors on an encrypted volume
I am trying to understand, generally how could this be a security issue ? Assuming the volume's backing media has this weakness (lets say an NTFS sparse file based TrueCrypt container) what could the ...
0answers
60 views
Problem understanding the construction of block cipher distinguishers
I know a block cipher distinguisher is an algorithm which can predict with high probability whether a system is a random permutation or a block cipher by asking a few queries to the system. My ...
0answers
77 views
Montgomery Multiplication in FPGA explanation
I'm trying to implement an RSA module in a FPGA. I choose to use Montgomery multiplication. I've found a document where the Montgomery Multiplication is pretty clear. I don't understand only a step. ...
0answers
43 views
Need: Fast bulk signature verification followed by fast non-interactive multisignature aggregation
Q: Is there an efficient way to batch-verify signatures (e.g. some may be incorrect) and then non-interactively aggregate the correct ones into a multisignature (they are of the same message) such ...
0answers
69 views
Protocol for establishing physical ownership of a connected/IoT device?
If I'm connected to a device e.g. using BLE are there rules/protocols for establishing that I'm also in physical possession of that device (at least for the first time I connect to it)? Crypto is in ...
0answers
32 views
Perform general computation using a crypto algorithm as a building block
I have a somewhat special question. In our software we use a existing dongle which got cracked by our customers. The protection offered by the dongle isn't very sophisticated. For the moment I can't ...
0answers
29 views
Plain cascade and Xor cascade
I have seen in few places (mostly papers) talking about plain cascade and XOR cascade. I know what is cascade cipher is but I am not clear about what these are. Can anyone help to understand using ...
0answers
118 views
fuzzy commitment implemenation to make secure biometric template
I want implement some ideas from an article that is about making secure biometric templates using fuzzy commitment and fuzzy vault. I need to implement just the fuzzy commitment (fuzzy commitment is a ...
0answers
62 views
Scalar multiplication of elliptic curve point by a fraction
I'm implementing an algorithm that works on a generic finite cyclic group written in the classic multiplicative notation: (G,*) = < g > , n = |g| At a ...
0answers
25 views
Calculating cost of variables and equation in Groth-Sahai (DLIN) proof system
In Groth-Sahai proof system under DLIN assumption [Page 29], they have provided a few number denoting the cost of each variable and equation. Can anyone please explain how did they calculated the ...
0answers
47 views
Partial Homomorphic Schemes with padding
As mentioned in wikipedia there are many Partial Homomorphic Encryption(PHE) scheme like RSA, Elgamal, Pailler etc. But out of them only unpadded RSA scheme seems to be partial(multiplicative) ...
0answers
66 views
Self-Convergent Hashing
What is the name of an algorithm that hashes its result as an embed of larger set of data that results back to the hash? For instance where f(x) is the hashed result and x is the dataset, and ...
0answers
44 views
Is there a practical succinct interactive argument for fortress draws in chess endgames?
(In theory, it is known that collision-resistant hashing suffices for the type of protocol that I'm asking about: page 3.) Is there a known practical candidate protocol for succinctly (and ...
0answers
33 views
Hard problems in composite order group even when factorization is known
Composite discrete log problem has been proved to be reducible to hardness of factorization and discrete log on the prime factor groups. Are there any problems apart from that in composite order ...
0answers
37 views
Size of Messages Exchanged by PRV and VER for Schnorr Protocol
In this file Elliptic Curve Based Zero Knowledge Proofs and Their Applicability on Resource Constrained Devices I don't understand the Table 6 (Table 6: Size of Messages Exchanged by the Prover(PRV) ...
0answers
51 views
Advantages of using intermediate hash over full hash in digital signature application
Description of intermediate hashes Intermediate (or partial) hashes are canonical forms of digest state that can be transferred from one hash implementation to another, so that the other, limited ...
0answers
129 views
RSA and ECDSA Certificate Sizes
Is there a table (or a whitepaper from official sources) that compares the size of X509 certificates generated with RSA (starting from 1024 bits) and ECDSA (starting from 160 bits) ? Thanks for the ...
0answers
65 views
0answers
82 views
Example of CL PKC
I am trying to solve the example of algorithms of Certificateless Signature by manually and solving the mathematics of Algorithms but not succeed while doing mathematical calculations. Please ...
0answers
76 views
hybrid PKE scheme CCA2 insecurity
I'm reading a paper here say: As we know, in hybrid PKE schemes XOR alone cannot perfectly hide challenge bit to the CCA2 adversary The author does not define hybrid PKE schemes. What is ...
15 30 50 per page | 2015-04-01 01:18:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6867011189460754, "perplexity": 2333.394190254664}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131302428.75/warc/CC-MAIN-20150323172142-00251-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://www.electro-tech-online.com/threads/hate-register.104516/ | # Hate register!
#### Hero999
##### Banned
Hero, the world now is not the same as it was 20 years ago, or the same as it was 20 before that.
According to random net statistics, world population as of 2008 was approximatly 6.9 billion people. In 1988 there were only approximately 5 billion people, in 1968 it was 3.5 billion.
You can't just double the world population in 40 years and expect government and social systems to stay the same, your opinion that it's worse is only that, your opinion. Things will HAVE to change to account for this dramatic change in population, especially considering it's typically centered in cities or other population centers. The more dense the population is the harder it is for people to be governed after a point the government tries to 'crack down' on common themes, it's the nature of government.
If you're afraid of 'big brother', get over it, it's here now and it's going get 'worse' it's not necessarily bad either.
Last edited:
#### HarveyH42
##### Banned
Come on now, 5 years old? Normal kid stuff, also important for healthy development. Pretty scary stuff, training kids to be sheep, probably teach 'Climate Change Hypothesis' to them as well. Giving kids direction is good, but they also need a chance to explore on their own, handle a few problems of their own choosing. A chance to be individuals, not members of a mind-numb cult.
#### Torben
##### Well-Known Member
Change is inevitable. Many changes are good. It does not follow that any given change is good.
On this one, I am very solidly with Hero. If you want to just lay down and let the government strip you of right after right that's your account.
Judging by the number of people living in terror of everything (right down to schoolyard taunts) I'd have to say that the terrorists have won.
This is an appalling step by the government responsible.
Torben
##### Banned
HarveyH42, what does your post have to do with what is under discussion here? Normal kid stuff should not be taken for granted, that 'normal kid stuff' is not always normal.
Last edited:
#### Pommie
##### Well-Known Member
It's funny that the law recognizes that a child under the age of 10 is not responsible for their own actions but the nanny state is going to hold them responsible from the age of 5. On the other hand, it may enable individuals like Venables and Thomson to be identified before they commit some heinous crime.
So, I have mixed thoughts on the matter. Also, it is The Daily Mail which is known for sensationalizing things.
Mike.
#### tcmtech
##### Banned
WTF? Some of the feelings Nazis would pee themselves if they heard what my 8 year old niece and I say at times to each other.
Why? Because I and her father are not going to have a little girl who goes crying to teacher or anyone else if someone calls her poopie head. Her dad an I both agree that she will have the skin and confidence to take what ever life throws at her for the worst.
More people on this planet does not mean be softer and give up easy. It means you will need to stand your ground and make sure no one takes advantage of you that much more.
Words are words and nothing else unless you make them into something. And she will have more than enough verbal guns to tear down anyone who provokes her justly enough to need to use them.
I can hardly wait until she is old enough to be on the debate teams. I bet she will make a few boys cry too some day!
In a few more years when here typing skills mature and her tactfulness levels out I may even bring her here and let her have a go at a few chat subjects for a while.
Last edited:
##### Well-Known Member
HarveyH42, what does your post have to do with what is under discussion here? Normal kid stuff should not be taken for granted, that 'normal kid stuff' is not always normal.
I think that what HarveyH42 was getting at was that children in a 5 to 7 year old age group will frequently do things somewhat out of the ordinary and among those behaviors some can just be considered kid stuff and normal development behavior.
I agree that what some consider "normal kid stuff" may not be normal in the eyes or thinking of another. Generally this in itself is not a problem however, it can be a problem. If a 7 year old does something a teacher views as not normal should that child's name be placed in a national data-base soley on what a single teacher perceived to be abnormal behavior. Should a single person like a teacher have the power to make that decision?
Questions:
Little Johnny is 7 years old and little Susie is also 7 years old. A teacher finds Lil' Johnny and Lil' Susie behind a large oak tree during recess playing I'll show you mine if you show me yours. Now should both children simply be disciplined and their parents notified or should this teacher have the power to place the names of both children in a national data-base register as potential someday sexual predators? Or, would this be normal child behavior?
Little Johnny and Little Susie are playing peacefully in the school yard when little Susie takes little Johnny's pencils and won't return them. She taunts him with the pencils a little. Little Johnny finally gets angry and pushes little Susie as he screams "I hate you". This last part is witnessed by a teacher. Should little Johnny be placed in a national data-base register for potentially violent people? Or, would this be considered normal child behavior?
I guess it all depends on who decides what is or is not "normal" child behavior.
Ron
Last edited:
#### Hero999
##### Banned
I can't believe that some people here actually seem to be in support of this.
I also get the feeling that many of you who have commented have not read the article.
For those who haven't got the time to read the entire thing I'll post a couple of quotes:
Heads will be forced to list children as young as five on school 'hate registers' over everyday playground insults.
Even minor incidents must be recorded as examples of serious bullying and details kept on a database until the pupil leaves secondary school.
Somerset, Peter Drury, were told that his name would be put on a register and his behaviour monitored while he remained at school.
The boy was reported after he called a friend 'gay boy'. His parents fear the record of homophobic bullying will count against him throughout his school career and even into adulthood.
In another incident last year a six-year-old girl, Sharona Gower, was reported for 'racist bullying' at her school near Tunbridge Wells in Kent.
Sharona was chased by two 11-year-old girls, one of whom taunted her that she had chocolate on her face.
The six-year-old responded to one of the girls, who was black: 'Well, you've got chocolate on yours.'
The government are recording every incident of playgound banter as bullying. This covers insults ranging from racist, to homophobic to disability related, even if they're blatantly light-hearted.
I don't think there's a single person at the school I went to who hasn't called someone a retard, batty boy or some other insult which would go on the register.
Using certain words in a derogatory sense is also an offence, i.e. you can't say "I hate English, it's seriously retarded", "Homework's gay" or "That's dark" without it being recorded on the stupid register. All of the aforementioned words have been used by British schoolchildren in recent years to describe something inferior or bad and have caused minimal offence at the time.
Even people who are gay or black can often enter into some banter about their differences without being offended, it depends on the person and the nature of the banter. My brother's gay and some of the banter he enters into at work regarding his sexuality would put nearly all his colleagues on the list. It's true that there's often a fine line between banter and bullying but I think people should try to assume the best of others as much as possible.
#### Nigel Goodwin
##### Super Moderator
Our government are going worse.
Children as young as 7 now get their names recorded on file if they're witnessed insulting their fellow pupils in a politically incorrect manner. This is worse than some of the nannying on this forum.
Pupils aged five on hate register: Teachers must log playground taunts for Government database | Mail Online
I've never heard any suggestion of such a thing, and my wife works in a Primary school - anyway, any decent school would flatly refuse to participate in such a ludicrous idea.
Sounds more like a Daily Mail made up story than anything else?.
#### Hero999
##### Banned
I haven't been able to find another source of information but if this is made up the Daily Mail are going to be subject to pretty big libel suit.
#### Nigel Goodwin
##### Super Moderator
I haven't been able to find another source of information but if this is made up the Daily Mail are going to be subject to pretty big libel suit.
Considering the papers seem to make up most of what they print it never seems much of a concern?.
#### HarveyH42
##### Banned
HarveyH42, what does your post have to do with what is under discussion here? Normal kid stuff should not be taken for granted, that 'normal kid stuff' is not always normal.
First, kids mimic things they hear and see, doesn't mean they actually understand what they are doing. Name calling isn't always hateful, and usually better than violence. Children who are teased and taunted, usually rise to the challenge, and strive to overcome and improve themselves.
Kids need to develop emotionally, which is something that has to be done by the individual, you have to earn it through experience.
There have always been defective kids, those that do some pretty horrific things, and grow up to do even worse. Doubt it had much to do with school yard taunting, probably something at home, they can either choose to change what sets them apart, try to fit in, or continue with their strange ways.
Basically, having the school step in at every incident, is going to turn out some emotional weak individuals, and another bunch afraid to express themselves. Pretty much just sheep in a flock, depend on a Shepard to lead them through life, and protect them from harm, lost, without such a provider. Guess that would be ideal, for some organizations, but not great for society as a whole.
Everybody has unique qualities, and usually excel in a few things, that others struggle at. Some might be really good at physical activities, like sports. Others do well in some academic subjects, or music, arts... Those with common strengths usually form groups, which usually also share the same weaknesses. The 'jock' are noted to excel in studies, the 'nerds' not some much for sports. Some do okay at most anything.
Now when one kid says something hurtful to another, out of anger, things change, and both kids know they did something bad to each other, and will eventually make peace. If the school steps in, the verbal kid that was originally wronged, is punished, and the other kid is rewarded. It's okay to provoke other kids, because they are going to get into trouble if they get caught reacting. The time-bomb of rage, just hold it in, until you explode...
#### shortbus=
##### Well-Known Member
I lay the blame to the way society is today on Dr. Spock (no not that Spock, Klingon) Benjamin Spock - Wikipedia, the free encyclopedia I was born in that era ,but, my Mom never read the book!
From that time forward it seems that no one is held responsible for their own actions. It's always somebody else thats wrong. Letting kids learn to figure out how to work things out on their own is much better in the long run than parents or government doing it for them. Someday they will need to do it for themselves, but don't know how because they never learned.
The same people behind this went after Criminal Rights. And look where we're at today, criminals have rights, but lawful people need to watch out every where they go.
##### Banned
Harvey, and how do you expect to watch out for these types of behavior if the kids aren't monitored? An official register makes sense with fewer teachers expected to take care of more kids. There was absolutely nothing whatsoever said in the article about consequence, just that the behavior was monitored. If you think that this is big brother monitoring what we can and can not say and trying to force youngsters to conform to some absurd ideal then you've gone off on your own tanget as there is nothing in that article that would suggest that is what's occurring.
If the teachers log this kind of behavior and are able to view other log entries over time this will build up an image of the childs behavioral makeup to at least a lesser degree, you'll at least have something to compare to other students, it would empower the teachers that might not be able to mentally correlate all of this information to have a simple record of behavior. If a trend emerges then parents should be notified so that they are made aware of it.
#### Hero999
##### Banned
Fair enough, I would accept it, if only serious incidents were recorded but surely not harmless banter?
Come on the remark about a girl having chocolate on her face should have never been recorded and neither should the gay boy remark.
I would not have a problem if yellow and red card type of system were introduced, i.e. if a kid says something to make another child cry, not silly banter, they get a yellow card, if it happens again it's a red card, then if it happens a third time within a year they get badly punished and the black mark stays there for a year or so.
But a UK wide scheme run by the government?
What a complete waste of money? I'd rather they spend in on tackling the real causes of bullying: counselling and confidence building sessions for victims so they can stand up for themselves and a decent discipline and behaviour reform scheme for the perpetrators.
I think the bully register is is very dangerous because it can be easily abused. Suppose a child learns that someone saying a certain word gets them in trouble? The might make a malicious false allegation about another child calling them a bad name. The trouble is that the rules state that every silly incident needs to be recorded so it can't be ignored so the kid making the allegation wins every time.
#### Nigel Goodwin
##### Super Moderator
Get a grip - you're getting excited over one unconfirmed article in a newspaper renowned for their imagination. There are no such rules, and heads would refuse to do it if anyone was daft enough to try and introduce such a silly scheme.
#### HarveyH42
##### Banned
Harvey, and how do you expect to watch out for these types of behavior if the kids aren't monitored? An official register makes sense with fewer teachers expected to take care of more kids. There was absolutely nothing whatsoever said in the article about consequence, just that the behavior was monitored. If you think that this is big brother monitoring what we can and can not say and trying to force youngsters to conform to some absurd ideal then you've gone off on your own tanget as there is nothing in that article that would suggest that is what's occurring.
If the teachers log this kind of behavior and are able to view other log entries over time this will build up an image of the childs behavioral makeup to at least a lesser degree, you'll at least have something to compare to other students, it would empower the teachers that might not be able to mentally correlate all of this information to have a simple record of behavior. If a trend emerges then parents should be notified so that they are made aware of it.
Schools are run like a business, they have to control risk and liability. If there is a serious problem, and they new about it before hand, but only wrote it down. They would risk being sued, which happens often enough already. Even if somebodies kid comes home with a bump or bruise, from a little school yard brawl, the school could face consequences. A black-eye doesn't always tell who initiated the fight either, nor is it related to anything said in the past. But when the kid's name gets checked on the list...
The point is, if the school has information, that potentially puts them at risk, they will have to act on it. The less you know, the better off you are.
Now, I will admit to not having been around big city, high society, type mentality, and probably a few decades behind the times, perhaps a little barbaric by some standards. I've always lived in small towns, grew up on the side of a huge mountain, so the big city issues were never part of my life. Will probably never understand much of big-city lifestyle. I just grew up to be independent and self sufficient. There isn't always someone else around to help you through the rough spots, and just sitting around waiting could, cost you your life. Rivers rise quickly when it rains, the temperature drops quickly when the sun sets, not to mention the wild animals roaming for food. We had to develop inner strength, learn how to function through unpleasant circumstances, stuff you learn through experience, you have to live it, and survive. Some things can be taught, but there are many things we must learn for ourselves.
#### Thunderchild
##### New Member
it's the childs patrents names that need to go on the register ! for some reason goverments miss it everytime, a child can be taught in school whatever the gov wants but home life and the example parents give is the determining factor in a childs development (and no I don't need a pediatric degree to know that !). With people having children long before they should and for the wrong reasons (ie the gov benefits they get and yes they DO do it for that) its small wonder that some parents are little more than tramps that do have a house and a car but their attitude to society is appaling. I've seen young parents in aour local shopping centre who have a hard time controlling their kids and don't seem to understand where that fine line between discipline and nasty/bullying attitude lies. unfortunately the goverment can't control people directly and only blows things out of proportion when it tries the direct method like this one. what I'd do is stop people having children in the first place if they are no good at bringing them up but then who is to be the judge of that. of course removing child benefit for unmarried or onviously unstable couples may bring down the birth of children into undesirable circumstances and reduce the "redundant" population. Unfortunatelt the goverment seems to think that the way to tackle things is to ever more treat children like adults and let adults behave like children. our recent credit crunch is a good example, ok it was wrong of banks to "give away" so much money to people that clearly could not afford it but hey any normal adult should know what they can affort, I recently submitted an application for a mortage and yes I weighed up very carefully how much I can afford, others in the past have been happy to take on loans they could not afford and say "what will be will be" or be that incompetent that they could not see what they were doing, and by the way I'm talking about the same people that are supposed to be bringing up children and showing them how to live as adults - great example !
what I would like to ask the gov is: why did nothing happen to the children aged 14 and 16 and their parents when they had sex (illegally) and produced a child who's future I have no faith in at all. Why were the parents of the kids that killed a man not loose custody of all of their other children and the same for other children/their parents that have commited MURDER. when will this goverment WAKE UP ?????????????? bloody paper work loving idiots and incapable of commn sense,
Last edited:
##### Well-Known Member
HarveyH42 says in part:
Schools are run like a business, they have to control risk and liability.
You got that right. Currently going on in Detroit, Michigan US:
In what experts say could be a landmark decision, a Michigan school district has been ordered to pay $800,000 this week to a student who claimed the school did not do enough to protect him from years of bullying, some sexually tinged. This week's jury verdict against Hudson Area Schools puts districts on notice that it's not enough to stop a student from bullying another. There needs to be a concerted effort to stop systemic bullying, too. Essentially, the federal court ruling says schools can be held responsible for what students do, if there is a pattern of harassment or if they don't do enough to provide a safe environment. "This is going to have implications across the nation," said Glenn Stutzky, a Michigan State University instructor and an expert on bullying. The district's attorney, however, says the verdict puts schools in the tricky position of being held liable for student behavior. The district plans to appeal. "You're never going to completely stop kids from being mean to kids," said Timothy Mullins of Giamarco, Mullins and Horton of Troy. The case It started with name-calling in middle school and escalated as Dane Patterson entered high school. Some of the harassment was bullying, such as being shoved into lockers. Other harassment was decidedly sexual in nature. He was called sexual insults, his locker and notebook were defaced with similar names, and worse. He and his parents say they reported the abuse, and yet it continued. Finally, in 10th grade, he was taunted in a locker room by a naked student rubbing against him. That was the last straw for the Patterson family. In 2005, they sued Hudson Area Schools under Title IX, the Equal Opportunity in Education Act, using the sexually tinged bullying as the basis for a sexual harassment lawsuit. This week a jury in U.S. District Court told the school district to pay$800,000 in damages to Patterson, now 19. Anti-bullying proponents say the case will send a message to all school districts that they are responsible for sexual harassment and, by extension, bullying.
For the Pattersons, however, the verdict is much simpler. It's vindication.
"I can't even put into words the pain and suffering that I went through for years," Dane Patterson said. "It's something that I would not want anyone else to go through."
While Patterson said he feels vindicated and is trying to move forward, his mother can't help but look back on their ordeal.
"I don't know how you get back eight years," Dena Patterson said. She said her son is so emotionally damaged by his experiences, he can't even go away to college and live in a dorm with other students. "We said it was worth standing up. We don't want another student, another parent to endure what we have seen."
Hudson schools, like most school districts, has an anti-bullying policy, and it took action against individual students when the bully could be identified. What the district failed to do is stop the pattern of abuse, said Terry Heiss, attorney for Patterson. For example, the school could have done more anti-bullying education, instituted more monitors or other measures to stop the pattern.
But this case makes it clear that having a policy, or even punishing individual bullies is not enough to stop a school from being liable, said Stutzky.
School officials will now have to show they were not indifferent, and that they made sure there wasn't a broader pattern of harassment beyond the individual case that went unchecked.
"If you only deal with things on an isolated case, that doesn't meet the standard for an effective response," Stutzky said.
But Mullins said the verdict leaves schools in a difficult situation.
"It sounds simple, but when you've got 500 kids and you're supposed to predict what any two or three or one are going to do in advance, well good luck," Mullins said. "If somebody writes dirty names on a boy's locker and you can't identify who it is, you can't punish the whole school."
The Patterson case initially was dismissed by the lower court in Detroit. But in October, the United States Court of Appeals for the Sixth Circuit reversed that decision, saying the family had demonstrated that there was enough of a question of whether the district's response was adequate to go forward with the trial.
Patterson was called pig or sexual insults and pushed into lockers, among other types of harassment, "on a daily basis" in sixth grade. When they complained, the family was sometimes told, "Kids will be kids; it's middle school," according to the court papers.
The harassment escalated in seventh grade, according to the lawsuit. Dane Patterson wanted to quit school, and his grades slipped.
Eighth grade was better. He began going to a resource room, a kind of study hall, to be counseled, and the lawsuit says the resource room teacher was helping him cope with the problems.
But ninth grade meant a change to high school. The Pattersons wanted their child to continue working with the resource room teacher, who had been successful the previous year. But the school district said no. The bullying continued in 10th grade, culminating in the locker room incident.
"It's a terrible thing, and I'm hoping with this verdict that schools will have to enforce stricter sexual harrassment and bullying policies," Patterson said.
Relatively large award for being bullied. I wonder what else the schools should be held accountable for. Gee, I remember when a school was only responsible for teaching. I guess the reach can be much further and all at a cost to who? Oh wait, that would be me.
Ron | 2019-03-21 16:26:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1959209144115448, "perplexity": 2187.4947159166522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00418.warc.gz"} |
https://tex.stackexchange.com/questions/331918/tikz-cones-with-a-wide-base?noredirect=1 | # Tikz: cones with a wide base
In every example of cone which I found so far there is an unpleasant mistake at the base since the sides should be tangent to the bottom ellipse but they are never. Drawing a larger base the problem gets bigger.
Here are examples of wrong cones: How to draw a simple cone with height and radius with TikZ?
Is there a fast way to make a good cone? This is my attempt:
\documentclass[border=.5cm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\begin{scope}
\clip (-3,0) -- (3,0) -- (0,4) -- cycle ;
\draw[dashed] (0,0) circle(3cm and 0.6cm);
\end{scope}
\begin{scope}
\clip (-3,0) rectangle (3,-1cm);
\draw (0,0) circle(3cm and 0.6cm);
\end{scope}
\begin{scope}
\draw (-3,0) -- (3,8);
\draw (3,0) -- (-3,8);
\end{scope}
\begin{scope}
\clip (-3,8) rectangle (3,9cm);
\draw[dashed] (0,8) circle(3cm and 0.6cm);
\end{scope}
\begin{scope}
\clip (-3,8) rectangle (3,7cm);
\draw (0,8) circle(3cm and 0.6cm);
\end{scope}
\end{tikzpicture}
\end{document}
This is a perspective issue. If you would look at a cone up front, without perpesctive you'd see a triangle, so to make it look like a cone we draw an ellipse in the bottom part, the problem is that the more perspective we want the thicker gets the ellipse, then these 2D issues appear.
There are two ways to deal with it: either decrease the ellipse thickness to loose a little perspective or make the ellipse start and end in an "inclined" position (I don't know how to put this to words if someone can improve the language please do). I made an MWE to show what I mean.
\documentclass{article}
\usepackage{tikz}
\begin{document}
\newcommand*{\coneheight}{6}
\newcommand*{\perspective}{0.5}
\pgfmathsetmacro{\startangle}{10*\perspective}
\pgfmathsetmacro{\endangle}{180-\startangle}
\begin{tikzpicture}
\coordinate (r) at (\startangle:\coneradius cm and \perspective cm);
\coordinate (-r) at (\endangle:\coneradius cm and \perspective cm);
\coordinate (h) at (0,\coneheight cm);
% Base circle
\fill[
top color=gray!50,
bottom color=gray!10,
opacity=0.25
] (0,0) circle (\coneradius cm and \perspective cm);
%Hat filling
\fill[
left color=gray!50!black,
right color=gray!50!black,
middle color=gray!50,
opacity=0.25
] (r) -- (h) -- (-r) arc (\endangle:360+\startangle:\coneradius cm and \perspective cm);
%Surrounding lines
\draw[dashed] (r) arc (\startangle:\endangle:\coneradius cm and \perspective cm);
\draw (-r) arc (\endangle:360+\startangle:\coneradius cm and \perspective cm);
\draw (-r) -- (h) --(r);
\draw[dashed] (r) -- node[below] {$r$} (0,0) -- node[left] {h} (h) ;
\draw (0,8pt) -- ++(8pt,0) -- (8pt,0);
\end{tikzpicture}
\end{document}
You define the normal cone parameters (height and radius) plus a perspective parameter, which is the b value for the ellipse. Then a starting angle will make the line look tangent to the ellipse in the bottom part, I used 10 times the perspective value out of try and error but that can be changed.You can play around with the perspective and the multiplier to see how the results come up best to you.
Result:
• I consider it pure sorcery but I appreciate very much! – ThePunisher Sep 29 '16 at 21:06
• @lanzariel Glad you appreciate it. There's actually mathematics behind it to get perfect tangency but I was a little bit lazy and it wasn't that easy math. ;D – Guilherme Zanotelli Sep 30 '16 at 5:34
• @GuilhermeZanotelli I tried \newcommand*{\coneheight}{2}, and I got bad cone. – minhthien_2016 Feb 5 '19 at 7:33
• @minhthien_2016 yeah, this implementation requires that the tip of the cone lie above the circumference of the base. A combination of perspective and cone radius will require a minimum cone height otherwise there will be non intersecting lines which will give you a bad cone. – Guilherme Zanotelli Feb 10 '19 at 12:59
I think this just about does it. Note the annoying use of \rx+0 to get around the required space in angle and arc radius specifications.
\documentclass[border=5]{standalone}
\usepackage{tikz}
\begin{document}
\def\b{2}
\def\h{2}
\begin{tikzpicture}
\foreach \p [count=\i from 0,
evaluate={\rx=\b/2; \ry=\rx*\p; \ta=90-atan2(\h,\ry);}]
in {0.1,0.2,...,0.6}{
\begin{scope}[shift={({mod(\i,3)*\b*1.25},{-floor(\i/3)*\h*1.25})}]
\fill [gray!50]
(0, \h) -- (\ta:\rx+0 and \ry) arc (\ta:180-\ta:\rx+0 and \ry) -- cycle;
\draw [dashed] (\ta:\rx+0 and \ry) arc (\ta:180-\ta:\rx+0 and \ry);
\draw (0, \h) -- (\ta:\rx+0 and \ry) arc (\ta:-180-\ta:\rx+0 and \ry) -- cycle;
\draw [dotted] (\rx,0) -| (0, \h);
\end{scope}
}
\end{tikzpicture}
\end{document}
Which in close up looks like: | 2020-04-04 22:46:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8421292304992676, "perplexity": 3469.232698187015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370525223.55/warc/CC-MAIN-20200404200523-20200404230523-00152.warc.gz"} |
http://megasoft-rapid.com/Iowa/error-series-fourier.html | Address 22 S Frederick Ave, Oelwein, IA 50662 (319) 283-1188 http://www.cdwwireless.com
# error series fourier Oelwein, Iowa
When Buffy comes to rescue Dawn, why do the vampires attack Buffy? Generated Fri, 14 Oct 2016 17:34:44 GMT by s_wx1094 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection This expression is needed to compute the Fourier coefficients (by writing the cosines in terms of complex exponentials). The derivatives of $\cos x/2$ are essentially itself and $\sin x/2$, so direct computation shows that the Fourier coefficients of any derivative of $\cos x/2$ decay like $1/n$, so that the
Which option did Harry Potter pick for the knight bus? The Fourier coeffcients are calculated in the normal way using $$a_r=\frac{2}{L}\int_{-L/2}^{L/2}f(x)cos(\frac{2\pi rx}{L})dx$$ Since the f(x) isn't periodic, it in order to compute these $a_r$ then the approximation Please try the request again. How would you help a snapping turtle cross the road?
When plotting the calculated Fourier series against the actual value, the difference in their values increases a lot as x reaches $-L/2$ or $L/2$. Your cache administrator is webmaster. Rankings of the historic universities in Europe How to handle a senior developer diva who seems unaware that his skills are obsolete? Your cache administrator is webmaster.
Suppose we have two functions, f(t) and g(t), defined over t=[0,T]. Which option did Harry Potter pick for the knight bus? Continuous FT) in terms of Sobolev order-3A question about pointwise convergence of Fourier transform in $N$-dimensions2What is the Fourier transform of this function?4Eliminating Gibbs phenomenon, and approximating with jumping functions in up vote 0 down vote favorite I've been given the task of computing the first 30 coefficients for the Fourier series of a Gaussian wavepacket given by: $$f(x)=exp(\frac{-x^2}{2\sigma^2})cos(kx)$$ for
Your cache administrator is webmaster. It can be seen from Figure 1 that the finite Fourier Series converges fairly quickly to f(t). This can be done via the use of the integral: [Equation 2] Note that the double brackets ||f-g|| means "the norm of f-g" (a norm, or a metric, is a distance question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science Other Stack Overflow
Proof: Define $$g(t)=\frac{f(x-t)-f(x)}{\sin (t/2)}$$ so that $$s_N(f;x)-f(x)= \frac{1}{2\pi}\int_{-\pi}^\pi \left[g(t)\cos \frac{t}{2}\right]\sin Nt\,dt + \frac{1}{2\pi}\int_{-\pi}^\pi \left[g(t)\sin \frac{t}{2}\right]\cos Nt\,dt$$ For Rudin, an application of the Riemann-Lebesgue lemma ends the proof here. Please try the request again. I initially thought it was due to Gibbs phenomena, however, the function is even and hence should not have a discontinuity if repeated. Generated Fri, 14 Oct 2016 17:34:44 GMT by s_wx1094 (squid/3.5.20)
Browse other questions tagged reference-request fourier-analysis fourier-transform or ask your own question. Security Patch SUPEE-8788 - Possible Problems? Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the The system returned: (22) Invalid argument The remote host or network may be down.
more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science Since you know $g$, you can find $N$ that will achieve the required precision. Are there any rules or guidelines about designing a flag? How do we know how close x1 is to x2?
asked 3 years ago viewed 1828 times active 3 years ago Related 0Deriving fourier series using complex numbers - introduction2Upper bound on truncation error of a fourier series approximation of a How to convert a set of sequential integers into a set of unique random numbers? Generation of Dictionary in Python Does chilli get milder with cooking? Not the answer you're looking for?
I would like an upper bound for $|f(x_0)-\sum_{n=-N}^N a_n e^{inx_0}|$. How do I explain that this is a terrible idea What is that the specific meaning of "Everyone, but everyone, will be there."? Thank you! more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed
The books that I consulted usually explain the Gibbs phenomenon for discontinuous functions, but $L^2$ error estimates are provided only when $u$ is regular. To give an idea of the convergence, let's look again at the square function from the complex coefficients page. Not the answer you're looking for? In a sense, we want to take the squared difference of each component, add them up and take the square root.
How do we do this for functions? UPDATE heap table -> Deadlocks on RID Deutsche Bahn - Quer-durchs-Land-Ticket and ICE Sum of neighbours Which fonts support Esperanto diacritics? Your cache administrator is webmaster. Ideally, I would like an answer in the spirit of estimating the error term for Taylor series.
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the share|cite|improve this answer answered Oct 2 '13 at 23:01 user98326 Thanks, this is great! –Steven Spallone Oct 6 '13 at 3:20 add a comment| Your Answer draft saved But that doesn't give an error estimate. Are "ŝati" and "plaĉi al" interchangeable?
Validity of "stati Schengen" visa for entering Vienna What is that the specific meaning of "Everyone, but everyone, will be there."? Browse other questions tagged fourier-series estimation or ask your own question. For example, if $f$ is a periodic function so that $f(x)=x$ on $(-\pi,\pi)$, and I require a partial Fourier series which is within $\frac{1}{2}$ of $f(x_0)$ at, say, $x_0=1$, I want What advantages does Monero offer that are not provided by other cryptocurrencies?
more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science We are interested in the distance (MSE) between gN(t) and f(t). Specifically, we are interested in knowing about the convergence of the Fourier Series Sum, g(t) (equation [3]), with the original periodic function f(t): [Equation 3] To get an idea of the The system returned: (22) Invalid argument The remote host or network may be down.
asked 2 years ago viewed 78 times active 2 years ago Related 1How does this error depend on h?2Why does this relative error work?3Derive error term by using Taylor series expansions.0Fourier Likely due to the approximation $$\int_{-L/2}^{L/2}\exp(-a(x+id)^2)\,dx\approx \sqrt{\frac{\pi}{a}}$$ which is fine when $L\gg d$, but it breaks down badly when $d$ is sizable compared to $L$. How would you help a snapping turtle cross the road? Detect if runtime is device or desktop (ARM or x86/x64) How is the Heartbleed exploit even possible?
I can't figure why this is. Katznelson's An introduction to harmonic analysis.) Do you need an estimate for the rate of convergence when the function happens to be piecewise continuous? –Joonas Ilmavirta Oct 6 '14 at 9:47 Please try the request again. Can Communism become a stable economic strategy?
And what about "double-click"? I've seen the wiki page, is there a particular section that would answer this particular question? –Steven Spallone Oct 1 '13 at 8:52 add a comment| 1 Answer 1 active oldest | 2018-11-19 17:35:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2298644483089447, "perplexity": 1713.5192092439427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746061.83/warc/CC-MAIN-20181119171420-20181119193420-00107.warc.gz"} |
https://socratic.org/questions/what-is-the-integral-of-ln-7x | # What is the integral of ln(7x)?
##### 1 Answer
Dec 9, 2014
Integration by Parts
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
Let $u = \ln \left(7 x\right) \text{ }$ $\text{ } \mathrm{dv} = \mathrm{dx}$
$\implies \mathrm{du} = \frac{\mathrm{dx}}{x} \text{ }$ $\text{ } \implies v = x$
By Integration by Parts,
$\int \ln \left(7 x\right) \mathrm{dx} = \ln \left(7 x\right) \cdot x - \int x \cdot \frac{\mathrm{dx}}{x}$
$= x \ln \left(7 x\right) - \int \mathrm{dx} + C$
$= x \ln \left(7 x\right) - x + C$
I hope that this was helpful. | 2019-03-26 18:33:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9593613147735596, "perplexity": 3378.466244970755}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912205600.75/warc/CC-MAIN-20190326180238-20190326202238-00302.warc.gz"} |
http://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-1-section-1-5-equations-1-5-exercises-page-56/21 | ## Precalculus: Mathematics for Calculus, 7th Edition
Published by Brooks Cole
# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 21
#### Answer
$2(1-x)=3(1+2x)+5$ Solution: $x=-\frac{3}{4}$
#### Work Step by Step
1. $2(1-x)=3(1+2x)+5$ 2. Distribute the 2 and the 3:$2-2x=3+6x+5$ 3. Combine like terms:$2-2x=8+6x$ 4. Add $2x$ to each side: $2=8+8x$ 5. Subtract 8 from each side: $-6=8x$ 6. Divide each side by 8:$x=-\frac{6}{8}$ 7. Reduce fraction: $x=-\frac{3}{4}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-02-21 01:36:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7387398481369019, "perplexity": 2762.0520841968464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813187.88/warc/CC-MAIN-20180221004620-20180221024620-00653.warc.gz"} |