url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
http://www.drumtom.com/q/what-multiplys-60-and-adds-to-32
|
# What multiplys 60 and adds to -32?
• What multiplys 60 and adds to -32?
What multiplys 60 and adds to -32? More questions about Family & Relationships, what. ... Multiplies to _ and adds to _ Number Classifier; ...
Positive: 61 %
What multiplys 60 and adds to -32? Find answers now! ... the IDE automatically adds the source code to create an instance of the appropriate class ...
Positive: 58 %
### More resources
what multiplys to 225 but adds to 30$$\Huge\color{blue}{ \sf 15\times 15 }$$ ... there are over 60 million students who want to help each other learn.
Positive: 61 %
30 has an infinite number of multiples. Some multiples of thirty include 30, 60, 90, 120 and 150. Three and 10 divide evenly into all multiples of 30. The ...
Positive: 56 %
what multiplys to -10 but adds up to 11?. -60 and 11 . multiplie... ... what multiplys to -10 but adds up to 11? kidrah69. a year ago-60 and 11. anonymous ...
|
2017-01-16 21:56:51
|
{"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": 0.2087877243757248, "perplexity": 5220.781123038538}, "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/1484560279368.44/warc/CC-MAIN-20170116095119-00326-ip-10-171-10-70.ec2.internal.warc.gz"}
|
https://gamedev.stackexchange.com/questions/101498/shadow-map-depth-error-makes-all-light-be-covered-and-hidden
|
Shadow map depth error makes all light be covered and hidden
I've been following the tutorials listed below:
float4x4 WorldViewProjection;
float4x4 LightingWorld;
float4x4 LightingViewProjection;
struct VertexToPixel {
float4 Position : POSITION;
float Depth : TEXCOORD0;
};
VertexToPixel output = (VertexToPixel)0;
output.Position = mul(position, mul(LightingWorld, LightingViewProjection));
output.Depth = output.Position.z / output.Position.w;
return output;
}
float4 color = input.Depth;
color.a = 1;
return color;
}
I use the following flashlight (which has a cone angle: coneAngleHalfRad * 2) View and Projection matrices for the shadow map (and for checking the depths later):
lightProjection = Matrix.CreatePerspectiveFieldOfView(
1,
1, //0.25f,
range);
//Helper.Message(this, "rotation up: " + rotation.Up);
lightView = Matrix.CreateLookAt(position, position + (direction * range), rotation.Up);
lightViewProjection = lightView * lightProjection;
DepthMapMakerEffect.Parameters["LightingWorld"].SetValue(Matrix.Identity);
DepthMapMakerEffect.Parameters["LightingViewProjection"].SetValue(lightViewProjection);
DepthMapMakerEffect.Techniques[0].Passes[0].Apply();
device.SetRenderTarget(RenderDirectional);
device.Clear(ClearOptions.Target | ClearOptions.DepthBuffer, Color.Black, 1.0f, 0);
drawAction(this, DepthMapMakerEffect);
device.SetRenderTarget(null);
I set my shadow map to the effect:
TextureManager.DynamicTexturingEffect.Parameters["Flashlight" + numFlashlightsPP + "Map"].SetValue(((Flashlight)lights[lightIndex]).ShadowMapDirectional);
And when I am drawing the light with additive blend state, I use the following code to get the depths from the shadow map and from the camera (is it from the camera? I didn't understand yet):
shared float DepthBias = 0.001f;
// [...]
shared Texture Flashlight0Map;
shared sampler Flashlight0MapSampler = sampler_state {
texture = <Flashlight0Map>;
magfilter = LINEAR;
minfilter = LINEAR;
mipfilter = LINEAR;
};
// [...]
// Find the position of this pixel in light space.
float4 lightingPosition = mul(worldPosition, flashlight.LightingViewProjection);
// Find the position in the shadow map for this pixel.
float2 shadowTexCoord = 0.5 * lightingPosition.xy /
lightingPosition.w + float2(0.5, 0.5);
// Get the current depth stored in the shadow map.
// Calculate the current pixel depth.
// The bias is used to prevent folating point errors that occur when
// the pixel of the occluder is being drawn.
float ourDepth = (lightingPosition.z / lightingPosition.w) - DepthBias;
// [...]
float intensity = 0;
if (atten > 0) {
intensity = saturate((flashlight.Range - dist) / flashlight.Range) * (dot(vertexNormal, perVertexNormal) * fog) * atten;
}
}
But what happens is that the result is ALWAYS "ourDepth > shadowDepth", which makes the entire light always hidden so it's always shadowed no matter if there's an object in front of it or not.
What could I possible be doing wrong? I think I used wrong values in the flashlight's View and Projection. But I can't find the error. What might be the problem?
Any suggestion may help!
EDIT: I'm not using the same aspect ratio of the screen and of the camera for the shadow map. Do I need to?
|
2019-11-18 07:17:28
|
{"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": 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": 0.17504863440990448, "perplexity": 9962.563747549617}, "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-2019-47/segments/1573496669454.33/warc/CC-MAIN-20191118053441-20191118081441-00265.warc.gz"}
|
https://otexts.com/fpp3/arima-r.html
|
## 9.7 ARIMA modelling in R
### How does ARIMA() work?
The ARIMA() function in R uses a variation of the Hyndman-Khandakar algorithm (Hyndman & Khandakar, 2008), which combines unit root tests, minimisation of the AICc and MLE to obtain an ARIMA model. The arguments to ARIMA() provide for many variations on the algorithm. What is described here is the default behaviour.
Hyndman-Khandakar algorithm for automatic ARIMA modelling
1. The number of differences $$0 \le d\le 2$$ is determined using repeated KPSS tests.
1. The values of $$p$$ and $$q$$ are then chosen by minimising the AICc after differencing the data $$d$$ times. Rather than considering every possible combination of $$p$$ and $$q$$, the algorithm uses a stepwise search to traverse the model space.
1. Four initial models are fitted:
• ARIMA$$(0,d,0)$$,
• ARIMA$$(2,d,2)$$,
• ARIMA$$(1,d,0)$$,
• ARIMA$$(0,d,1)$$.
A constant is included unless $$d=2$$. If $$d \le 1$$, an additional model is also fitted:
• ARIMA$$(0,d,0)$$ without a constant.
1. The best model (with the smallest AICc value) fitted in step (a) is set to be the “current model”.
1. Variations on the current model are considered:
• vary $$p$$ and/or $$q$$ from the current model by $$\pm1$$;
• include/exclude $$c$$ from the current model.
The best model considered so far (either the current model or one of these variations) becomes the new current model.
1. Repeat Step 2(c) until no lower AICc can be found.
Figure 9.11 illustrates diagrammatically how the Hyndman-Khandakar algorithm traverses the space of the ARMA orders, through an example. The grid covers combinations of ARMA($$p,q$$) orders starting from the top-left corner with an ARMA($$0,0$$), with the AR order increasing down the vertical axis, and the MA order increasing across the horizontal axis.
The orange cells show the initial set of models considered by the algorithm. In this example, the ARMA(2,2) model has the lowest AICc value amongst these models. This is called the “current model” and is shown by the black circle. The algorithm then searches over neighbouring models as shown by the blue arrows. If a better model is found then this becomes the new “current model”. In this example, the new “current model” is the ARMA(3,3) model. The algorithm continues in this fashion until no better model can be found. In this example the model returned is an ARMA(4,2) model.
The default procedure uses some approximations to speed up the search. These approximations can be avoided with the argument approximation=FALSE. It is possible that the minimum AICc model will not be found due to these approximations, or because of the use of a stepwise procedure. A much larger set of models will be searched if the argument stepwise=FALSE is used. See the help file for a full description of the arguments.
If you want to choose the model yourself, use the ARIMA() function with a single value for input for pdq() and PDQ(). There is another function arima() in R which also fits an ARIMA model. it does not allow for the constant $$c$$ unless $$d=0$$, and it does not return everything required for other functions in the forecast package to work. Finally, it does not allow the estimated model to be applied to new data (which is useful for checking forecast accuracy). Consequently, it is recommended that ARIMA() be used instead.
### Modelling procedure
When fitting an ARIMA model to a set of (non-seasonal) time series data, the following procedure provides a useful general approach.
1. Plot the data and identify any unusual observations.
2. If necessary, transform the data (using a Box-Cox transformation) to stabilise the variance.
3. If the data are non-stationary, take first differences of the data until the data are stationary.
4. Examine the ACF/PACF: Is an ARIMA($$p,d,0$$) or ARIMA($$0,d,q$$) model appropriate?
5. Try your chosen model(s), and use the AICc to search for a better model.
6. Check the residuals from your chosen model by plotting the ACF of the residuals, and doing a portmanteau test of the residuals. If they do not look like white noise, try a modified model.
7. Once the residuals look like white noise, calculate forecasts.
The Hyndman-Khandakar algorithm only takes care of steps 3–5. So even if you use it, you will still need to take care of the other steps yourself.
The process is summarised in Figure 9.12.
### Example: Seasonally adjusted electrical equipment orders
We will apply this procedure to the seasonally adjusted electrical equipment orders data shown in Figure 9.13.
elec_equip <- as_tsibble(fpp2::elecequip)
#> Registered S3 method overwritten by 'quantmod':
#> method from
#> as.zoo.data.frame zoo
elec_dcmp <- elec_equip %>%
model(STL(value ~ season(window="periodic"))) %>%
components() %>%
select(-.model) %>%
as_tsibble()
elec_dcmp %>%
autoplot(season_adjust)
1. The time plot shows some sudden changes, particularly the big drop in 2008/2009. These changes are due to the global economic environment. Otherwise there is nothing unusual about the time plot and there appears to be no need to do any data adjustments.
2. There is no evidence of changing variance, so we will not do a Box-Cox transformation.
3. The data are clearly non-stationary, as the series wanders up and down for long periods. Consequently, we will take a first difference of the data. The differenced data are shown in Figure 9.14. These look stationary, and so we will not consider further differences.
elec_dcmp %>%
gg_tsdisplay(difference(season_adjust), plot_type='partial')
4. The PACF shown in Figure 9.14 is suggestive of an AR(3) model. So an initial candidate model is an ARIMA(3,1,0). There are no other obvious candidate models.
5. We fit an ARIMA(3,1,0) model along with variations including ARIMA(4,1,0), ARIMA(2,1,0), ARIMA(3,1,1), etc. Of these, the ARIMA(3,1,1) has a slightly smaller AICc value.
fit <- elec_dcmp %>%
model(
arima = ARIMA(season_adjust ~ pdq(3,1,1) + PDQ(0,0,0))
)
report(fit)
#> Model: ARIMA(3,1,1)
#>
#> Coefficients:
#> ar1 ar2 ar3 ma1
#> 0.0044 0.0916 0.3698 -0.3921
#> s.e. 0.2201 0.0984 0.0669 0.2426
#>
#> sigma^2 estimated as 9.577: log likelihood=-492.7
#> AIC=995.4 AICc=995.7 BIC=1012
6. The ACF plot of the residuals from the ARIMA(3,1,1) model shows that all autocorrelations are within the threshold limits, indicating that the residuals are behaving like white noise.
fit %>% gg_tsresiduals()
A portmanteau test returns a large p-value, also suggesting that the residuals are white noise.
augment(fit) %>%
features(.resid, ljung_box, lag = 24, dof = 4)
#> # A tibble: 1 x 3
#> .model lb_stat lb_pvalue
#> <chr> <dbl> <dbl>
#> 1 arima 24.0 0.241
7. Forecasts from the chosen model are shown in Figure 9.16.
fit %>% forecast() %>% autoplot(elec_dcmp)
### Understanding constants in R
A non-seasonal ARIMA model can be written as $$$\tag{9.3} (1-\phi_1B - \cdots - \phi_p B^p)(1-B)^d y_t = c + (1 + \theta_1 B + \cdots + \theta_q B^q)\varepsilon_t,$$$ or equivalently as $$$\tag{9.4} (1-\phi_1B - \cdots - \phi_p B^p)(1-B)^d (y_t - \mu t^d/d!) = (1 + \theta_1 B + \cdots + \theta_q B^q)\varepsilon_t,$$$ where $$c = \mu(1-\phi_1 - \cdots - \phi_p )$$ and $$\mu$$ is the mean of $$(1-B)^d y_t$$. The fable package uses the parameterisation of Equation (9.3) and other R implementations usually use Equation (9.4).
Thus, the inclusion of a constant in a non-stationary ARIMA model is equivalent to inducing a polynomial trend of order $$d$$ in the forecast function. (If the constant is omitted, the forecast function includes a polynomial trend of order $$d-1$$.) When $$d=0$$, we have the special case that $$\mu$$ is the mean of $$y_t$$.
By default, the ARIMA() function will automatically determine if a constant should be included. For $$d=0$$ or $$d=1$$, a constant will be included if it improves the AICc value. If $$d>1$$ the constant is always omitted as a quadratic or higher order trend is particularly dangerous when forecasting.
The constant can be specified by including 0 or 1 in the model formula (similarly to the intercept in lm()). For example, to automatically select an ARIMA model with a constant, you could use ARIMA(y ~ 1). Similarly, a constant can be excluded with ARIMA(y ~ 0).
### Plotting the characteristic roots
(This is a more advanced section and can be skipped if desired.)
We can re-write Equation (9.3) as $\phi(B) (1-B)^d y_t = c + \theta(B) \varepsilon_t$ where $$\phi(B)= (1-\phi_1B - \cdots - \phi_p B^p)$$ is a $$p$$th order polynomial in $$B$$ and $$\theta(B) = (1 + \theta_1 B + \cdots + \theta_q B^q)$$ is a $$q$$th order polynomial in $$B$$.
The stationarity conditions for the model are that the $$p$$ complex roots of $$\phi(B)$$ lie outside the unit circle, and the invertibility conditions are that the $$q$$ complex roots of $$\theta(B)$$ lie outside the unit circle. So we can see whether the model is close to invertibility or stationarity by a plot of the roots in relation to the complex unit circle.
It is easier to plot the inverse roots instead, as they should all lie within the unit circle. This is easily done in R. For the ARIMA(3,1,1) model fitted to the seasonally adjusted electrical equipment index, we obtain Figure 9.17.
gg_arma(fit)
The three red dots in the left hand plot correspond to the roots of the polynomials $$\phi(B)$$, while the red dot in the right hand plot corresponds to the root of $$\theta(B)$$. They are all inside the unit circle, as we would expect because R ensures the fitted model is both stationary and invertible. Any roots close to the unit circle may be numerically unstable, and the corresponding model will not be good for forecasting.
The ARIMA() function will never return a model with inverse roots outside the unit circle. Models automatically selected by the ARIMA() function will not select a model with roots close to the unit circle.
### Bibliography
Hyndman, R. J., & Khandakar, Y. (2008). Automatic time series forecasting: The forecast package for R. Journal of Statistical Software, 27(1), 1–22. https://doi.org/10.18637/jss.v027.i03
|
2020-05-27 06:33:17
|
{"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": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.497223824262619, "perplexity": 2019.9956954086867}, "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/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00565.warc.gz"}
|
https://electronics.stackexchange.com/questions/415344/mesh-analysis-with-current-sources-dependent-current-source
|
# mesh analysis with current sources + dependent current source
I need to find Ix with mesh analysis method. I will really appreciate your help. So, I turned Is2 to a voltage source and tried writing equations for a super mesh and KCL. anyway, I miss something in this problem or I got something totally wrong with my approach towards the problem
• Hi, welcome to EE.SE. You have to show some effort to get help. Start solving the problem and come back to here when you get stuck somewhere. You can explain your attempt by pressing on edit button. – Hazem Jan 5 at 11:48
• Ok so I tried to solved it and I got stuck, will really appreciate any assistance. – BaRaKL Jan 6 at 19:10
It's not complicated. Try it without bothering to do the Norton to Thevenin conversion:
simulate this circuit – Schematic created using CircuitLab
\begin{align*} 0\:\text{V} + V_\text{IS1} - R_1\cdot\left(I_1-I_2\right) &=0\:\text{V}\\ 0\:\text{V} - R_1\cdot\left(I_2-I_1\right)-R_2\,I_2-V_{1.5\,I_x}&=0\:\text{V}\\ 0\:\text{V}+V_{1.5\,I_x} - R_3\,I_3-R_4\cdot\left(I_3-I_4\right)&=0\:\text{V}\\ 0\:\text{V}- R_4\cdot\left(I_4-I_3\right)-V_\text{IS2}&=0\:\text{V}\\\\ I_X&=I_1-I_2\\ 1.5\,I_X &=I_3-I_2\\ I_1&=2\:\text{A}\\ I_4&=-5\:\text{A} \end{align*}
I think you should be able to follow how I developed each of those, above. (I assumed that the voltage across each current source has its positive terminal on the top side and its negative terminal on the bottom side -- which doesn't have to be true for the above equations to work correctly. It just has to be consistently applied, is all.)
You certainly can, if you want, do the conversion first and reduce the number of loops by one. But there's no real need to do that.
If you use Sage (freely available to anyone), then you might set up the above like this:
var('i1 i2 i3 i4 r1 r2 r3 r4 ix vis1 vis2 vix')
eq1=Eq(vis1-r1*(i1-i2),0)
eq2=Eq(-r1*(i2-i1)-r2*i2-vix,0)
eq3=Eq(vix-r3*i3-r4*(i3-i4),0)
eq4=Eq(-r4*(i4-i3)-vis2,0)
eq5=Eq(ix,i1-i2)
eq6=Eq(1.5*ix,i3-i2)
ans=solve([eq1,eq2,eq3,eq4,eq5,eq6],[i2,i3,ix,vis1,vis2,vix])
It solves just fine.
• yeah, it helped a lot. thank you. I think I shouldn't convert the right side current source to a voltage source, and I used for not making KVL's for meshes with current sources in them but on "super meshes" only. still, it helped a lot. thank you again! – BaRaKL Jan 9 at 21:58
|
2019-08-21 07:30:26
|
{"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.517297089099884, "perplexity": 1060.2219294331826}, "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-2019-35/segments/1566027315811.47/warc/CC-MAIN-20190821065413-20190821091413-00082.warc.gz"}
|
https://math.meta.stackexchange.com/questions/23213/is-it-possible-to-edit-bounty-messages
|
# Is it possible to edit bounty messages?
The reason I ask is because I have a bounty question which is completely ruined as the sidebar covers the question. This has happened because of an error in the $\LaTeX$ in the bounty message which causes the question to stretch the width of page. This is how bad it looks.
If it is not possible to edit the bounty message does anyone have any idea how to tidy this up?
Or, can a moderator delete this question please and allow me to ask it again? (I can't delete a question once an answer has been given).
I ask of these things because that particular question was very important as it was the final question of $4$ previous connected questions.
Thank you.
• Hopefully the bounty message can be edited by the powers that be. However the formatting of the Question itself is in need of improvement. Colored equations are not usually a good idea. Using monospaced "code" blocks for emphasis is a bad idea (use * delimiters for emphasis text, or __ delimiters for bold face text). You have equations in the body of the Question that also run into the sidebar with my browser (Google Chrome 50). – hardmath May 12 '16 at 2:44
• I used the "Remove post notice" function available to moderators. I could not edit your bounty text, but that removal took care of the unrendered LaTeX and the resulting misplacement of components on the page. For the time being leaving your clarification in the question body will probably serve your purposes well. If you want further explanatory text added as a post notice, flag the post for our attention explaining what changes you want. I can also add another post notice (I think). – Jyrki Lahtonen May 12 '16 at 5:34
• @JyrkiLahtonen It looks much better now, thank you. You can tag this question as "Status-Complete" if you wish (I don't have access to do so). Best regards. – BLAZE May 12 '16 at 13:35
In the worst case, Moderators have a menu to remove and refund bounty offers. The first step is to flag the Question for Moderator attention (custom flag, explaining the problem with $\LaTeX$ formatting in the bounty message).
|
2020-09-19 06:33:29
|
{"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": 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": 0.46816959977149963, "perplexity": 1141.9035618350722}, "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/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00066.warc.gz"}
|
http://www.sawaal.com/simplification-questions-and-answers/_7957
|
0
Q:
# $\inline \frac{23\times 12+49\times 15}{(31)^{2}+50}=?$
A) 0.5 B) 1 C) 2 D) 3
Explanation:
$\inline \frac{23\times 12+49\times 15}{(31)^{2}+50}= \frac{276+735}{961+50}=\frac{1011}{1011}=1$
Q:
Two parallel chords on the same side of the centre of a circle are 5 cm apart. If the chords are 20 and 28 cm long, what is the radius of the circle?
A) 14.69 cm B) 15.69 cm C) 18.65 cm D) 16.42 cm
Explanation:
Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.
We use the Pythagorean theorem on right triangle ABO
AO² = AB² + OB²
r² = 14² + x²
We use the Pythagorean theorem on right triangle DCO
DO² = CD² + OC²
We see that OC = OB+BC = x+5, so
r² = 10² + (x+5)²
So we have a system of two equations:
r² = 14² + x²
r² = 10² + (x+5)²
Since both left sides equal r², set the right sides
equal to each other.
14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x
r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm
1 15
Q:
Hemavathi gets 3 marks for each right sum and loses 2 marks for each wrong sum. He attempts 35 sums and obtains 60 marks. The number of sums attempted correctly is ?
A) 23 B) 24 C) 25 D) 26
Explanation:
Let, Hema attempted 'k' sum correctly, then
k x 3 -2 x(35-k) = 60
5k = 130
k = 26
so 26 correct sums.
1 4
Q:
The average temperature of Monday to Wednesday was 37 C and of Tuesday to Thursday was 34 C. If the temperature on Thursday was 4/5 th of that of Monday, the temperature on Thursday was ?
A) 35 c B) 36 c C) 34 c D) 32 c
Explanation:
Monday + Tuesday + Wednesday = 37 x 3 -----------(1)
Tuesday + Wednesday + Thursday = 34 x 3---------(2)
Thursday = 4/5 of Monday ------------------(3)
subtract eqn (1) from (2) we get,
Thursday - Monday = -9
=> Monday - Thursday = 9.......(4)
From (3) & (4), we get
So,Thursday = 36 C
3 18
Q:
A secret can be told to only 3 persons in 3 minutes. Each person in turn tells 3 other persons in the next 3 minutes and the processes continues accordingly. In 30 minutes how many persons can be told this secret in this way ?
A) 88572 B) 77854 C) 99584 D) 55654
Explanation:
Given a secret is told to 3 persons in 3 minutes.
And if the process continues with each person saying 3 other persons,
In 30 minutes => the rocess for 10 times because each person time is 3 minutes.
Required number of persons = $\inline \fn_jvn \small 3+3^{2}+3^{3}+3^{4}+.....+3^{9}+3^{10}$
Now it forms a geometric progression,
Sum to n terms is S= $\inline&space;\fn_jvn&space;\small&space;\frac{a(r^{n}-1)}{r-1}$
Here a = 3, r = 3
S = $\inline \fn_jvn \small \frac{3(3^{10}-1)}{3-1}$
=> 59048 x 3 /2
= 88572.
3 22
Q:
In a party there are male and female members. If 15 female quit then the number of females will become double the number of males. If 45 males quit no. of female becomes five times the number of males. Find the number of females ?
A) 175 B) 145 C) 165 D) 135
Explanation:
Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.
|
2017-02-22 01:52:52
|
{"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": 0, "img_math": 0, "codecogs_latex": 5, "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": 0.5381560325622559, "perplexity": 2100.2044915447973}, "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-2017-09/segments/1487501170875.20/warc/CC-MAIN-20170219104610-00242-ip-10-171-10-108.ec2.internal.warc.gz"}
|
https://tex.stackexchange.com/questions/486110/fit-longtable-to-textwidth
|
# fit longtable to textwidth
I am applying How to fit landscape multi-page table to textwidth to fit longtable to textwidth in this example:
\documentclass{article}
\usepackage{longtable}
\usepackage{lipsum}
\begin{document}
\lipsum[1][1-4]
%\resizebox{\textwidth}{!}{
\begingroup % localize the following settings
\setlength\LTcapwidth{\textwidth} % default: 4in (rather less than \textwidth...)
\setlength\LTleft{0pt} % default: \parindent
\setlength\LTright{0pt} % default: \fill
\begin{longtable}{@{\extracolsep{\fill}}|l|rcc|rcc|rcc|rcc|}
longtable aaaaa aaaa aaaa aaa aaaa & 11.11 & 2 & 0.000 & 22.22 & 2 & 0.000 & 333.33 & 2 & 0.137 & 44.44 & 2 & 0.000\\
\end{longtable}
\endgroup
%}
\end{document}
However, the outcome shows I haven't been successful:
What is wrong?
I check the answer of @DavidCarlisle. However, the proposed manipulations cause a problem with \cline:
\begingroup % localize the following settings
\setlength\LTcapwidth{\textwidth} % default: 4in (rather less than \textwidth...)
\setlength\LTleft{0pt} % default: \parindent
\setlength\LTright{0pt} % default: \fill
\tiny
\setlength{\tabcolsep}{0.4pt}
\begin{longtable}{@{\extracolsep{\fill}}|p{3cm}|rcc|rcc|rcc|rcc|}
Sample&\multicolumn{3}{c|}{AAAA}&\multicolumn{3}{c|}{BBBB}&\multicolumn{3}{c|}{CCCC}&\multicolumn{3}{c|}{DDDDD}\\\cline{2-13}
lllllllllongtable & 11.11 & 2 & 0.000 & 22.22 & 2 & 0.000 & 33.33 & 2 & 00.00 & 44.44 & 2 & 0.000\\
\end{longtable}
\endgroup
How to avoid this problem?
• as in tabular*, \extracolsep{\fill}} allows the table to expand to the specified width, your table is too wide, so you need to use a multi-line cell (change the first l to p{2cm} or whetever width you want) or use a smaller font such as \small – David Carlisle Apr 22 '19 at 20:57
• @DavidCarlisle is there any analogue of \resizebox? It works so nice. – Viesturs Apr 22 '19 at 20:59
• @Viesturs: I would definitely recommend to not use resizebox on material that contains text as it will lead to inconsistent font sized and distances. – leandriis Apr 22 '19 at 21:00
• \resizebox should never be used for tables, as it makes the font size inconsistent. B.t. w., do you use marginal notes? – Bernard Apr 22 '19 at 21:01
• @Viesturs every time I see someone apply \resizebox to a table, I wish I'd never implemented it. You wouldn't justify a paragraph by just scaling it to an arbitrary font size, why inflict that in tables? – David Carlisle Apr 22 '19 at 21:07
\documentclass{article}
\usepackage{longtable}
\usepackage{lipsum}
\begin{document}
\lipsum[1][1-4]
%\resizebox{\textwidth}{!}{ never scale tables!
\begingroup % localize the following settings
\setlength\tabcolsep{2pt}
\footnotesize
\setlength\LTcapwidth{\textwidth} % default: 4in (rather less than \textwidth...)
\setlength\LTleft{0pt} % default: \fill
\setlength\LTright{0pt} % default: \fill
\begin{longtable}{@{\extracolsep{\fill}}|p{2cm}|rcc|rcc|rcc|rcc|@{}}
longtable aaaaa aaaa aaaa aaa aaaa & 11.11 & 2 & 0.000 & 22.22 & 2 & 0.000 & 333.33 & 2 & 0.137 & 44.44 & 2 & 0.000\\
\end{longtable}
\endgroup
%}
\end{document}
• Your solution is causing a problem with cline. See the update of my question. – Viesturs Apr 22 '19 at 21:49
• if you use \extracolsep then the space is inserted between columns and has the effect on cline that you show, so that is by design (personally I would not use \extracolsep forcing the columns apart doesn't make the tables more readable – David Carlisle Apr 22 '19 at 22:31
A solution with the xltabular package and its eponymous environment, which brings the functionalities of longtable to tabularx. I loaded the geometry package, to have more decent margins:
\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{longtable}
\usepackage{ragged2e}
\usepackage{xltabular}
\usepackage{lipsum}
\begin{document}
\lipsum[1][1-4]
\begingroup % localize the following settings
\setlength\LTcapwidth{\textwidth} % default: 4in (rather less than \textwidth...)
\setlength\LTleft{0pt} % default: \parindent
\setlength\LTright{0pt} % default: \fill
\begin{xltabular}{\linewidth}{|>{\RaggedRight\arraybackslash}X|rcc|rcc|rcc|rcc|}
longtable aaaaa aaaa aaaa aaa aaaa & 11.11 & 2 & 0.000 & 22.22 & 2 & 0.000 & 333.33 & 2 & 0.137 & 44.44 & 2 & 0.000\\
\end{xltabular}
\endgroup
\end{document}
• How to centre vertically the cells with numerical values? – Viesturs Apr 22 '19 at 21:24
• Redefine the X column type as m{#1}, as was already answered to another of your questions, if I remember well. – Bernard Apr 22 '19 at 21:36
|
2020-12-03 14:39:49
|
{"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": 0.7458078861236572, "perplexity": 2899.138439058489}, "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/1606141727782.88/warc/CC-MAIN-20201203124807-20201203154807-00293.warc.gz"}
|
https://ec.gateoverflow.in/240/gate-ece-2012-question-17
|
27 views
The radiation pattern of an antenna in spherical co-ordinates is given by
$$F(\theta)=\cos^4\theta\:\:\:;\:\:\:0\le \theta\le \frac{\pi}{2}$$
The directivity of the antenna is
1. $10\:dB$
2. $12.6\:dB$
3. $11.5\:dB$
4. $18\:dB$
|
2022-09-29 17:45:43
|
{"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": 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": 0.8385680913925171, "perplexity": 804.9524287920441}, "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-2022-40/segments/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00477.warc.gz"}
|
https://gargle.r-lib.org/articles/auth-from-web.html
|
If you are working with R in a web-based context, like RStudio Server or Cloud, your experience of browser-based auth flows will be different from those using R, directly in the Console or through an IDE. You need to use out-of-band authentication, sometimes denoted “oob”. After the usual auth dance, instead of seeing “authentication successful, return to R!”, you are presented with an authorization code to copy and paste back into your R session.
The need to use oob auth can sometimes be detected automatically. For example, oob auth is always used when the httpuv package is not installed. However, gargle and httr currently fail to detect some other situations where oob is necessary. In these case, the user needs to recognize the situation and explicitly request oob auth.
Here’s a typical presentation of this problem: during auth, you are redirected to localhost on port 1410 and receive an error along these lines:
Chrome: This site can't be reached; localhost refused to connect.
Firefox: Unable to connect; can't establish a connection.
This is a sign that you need to explicitly request oob auth.
This article describes how to do so in a package that uses gargle for auth, which includes:
Request oob auth in the PKG_auth() call
These packages aim to make auth “just work” for most users, i.e. it’s automatically triggered upon first need. However, it is always possible to initiate auth yourself, which gives you the opportunity to specify non-default values of certain parameters. Here’s how you request oob auth, using googledrive as an example:
library(googledrive)
drive_auth(use_oob = TRUE)
# now carry on with your work
drive_find(n_max = 5)
This code is tailored to an interactive session and assumes that a user is present to respond. If you also need to setup a token for non-interactive use, see the article Non-interactive auth. A key point is that oob auth is relevant to how you initially obtain a token. It is orthogonal to downstream use and refreshing. So it is possible that you need to attend to both!
Set the gargle_oob_default option
If you know that you always want to use oob auth, as a user or within a project, the best way to express this is to set the gargle_oob_default option.
options(gargle_oob_default = TRUE)
This code could appear at the top of a script, in a setup chunk for .Rmd, or in a Shiny app. But it probably makes even more sense in a .Rprofile startup file, at the user- or project-level.
Once that option has been set, it is honoured by downstream calls to PKG_auth(), explicit or implicit, because the default behaviour of use_oob is to consult the option:
drive_auth <- function(email = gargle::gargle_oauth_email(),
path = NULL,
cache = gargle::gargle_oauth_cache(),
use_oob = gargle::gargle_oob_default(),
token = NULL) {...}
But I didn’t need oob yesterday!
Sometimes the usual oauth web flow suddenly stops working for people working directly with R (so NOT via the browser) and they use oob auth to get unstuck again. What’s going on in this case?
The initial error looks something like this:
createTcpServer: address already in use
Error in httpuv::startServer(use$host, use$port, list(call = listen)) :
Failed to create server
It’s characteristic of some other process sitting on port 1410, which is what httr is trying to use for auth.
It’s true that using oob auth is a workaround. But oob auth is, frankly, more clunky, so why use if you don’t have to? Here are ways to fix.
• Restart your system. This will almost certainly kill the offending process, which is usually a zombie process.
• Hunt down the offending process, verify it looks expendable, and kill it.
On *nix-y systems, use lsof to get the process ID:
sudo lsof -i :1410
The output will look something like this:
COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME
R 16664 jenny 20u IPv4 0x63761a50856c65f 0t0 TCP localhost:hiq (LISTEN)
In this case, as is typical, this is a zombie R process and I feel confident killing it. The process ID is listed there as PID. Note that and kill the process, like so, filling in the PID you found:
kill -9 <PID>
So, to be clear, in this example, the command would be:
kill -9 16664
The normal, non-oob auth web flow should work again now.
Generating OAuth tokens for a server using httr covers some of the same ground, although for the httr package. gargle provides a Google-specific interface to httr. gargle first consults the gargle_oob_default option and, if that is undefined, also consults the httr_oob_default option.
|
2020-05-25 20:48:26
|
{"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": 0.3977574110031128, "perplexity": 4618.967722411166}, "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-2020-24/segments/1590347389355.2/warc/CC-MAIN-20200525192537-20200525222537-00424.warc.gz"}
|
https://www.nag.co.uk/numeric/DT/nagdotnet_dtw02/html/M_NagLibrary_G02_g02hk.htm
|
g02hk computes a robust estimate of the covariance matrix for an expected fraction of gross errors.
# Syntax
C#
```public static void g02hk(
int n,
int m,
double[,] x,
double eps,
double[] cov,
double[] theta,
int maxit,
int nitmon,
double tol,
out int nit,
out int ifail
)```
Visual Basic
```Public Shared Sub g02hk ( _
n As Integer, _
m As Integer, _
x As Double(,), _
eps As Double, _
cov As Double(), _
theta As Double(), _
maxit As Integer, _
nitmon As Integer, _
tol As Double, _
<OutAttribute> ByRef nit As Integer, _
<OutAttribute> ByRef ifail As Integer _
)```
Visual C++
```public:
static void g02hk(
int n,
int m,
array<double,2>^ x,
double eps,
array<double>^ cov,
array<double>^ theta,
int maxit,
int nitmon,
double tol,
[OutAttribute] int% nit,
[OutAttribute] int% ifail
)```
F#
```static member g02hk :
n : int *
m : int *
x : float[,] *
eps : float *
cov : float[] *
theta : float[] *
maxit : int *
nitmon : int *
tol : float *
nit : int byref *
ifail : int byref -> unit
```
#### Parameters
n
Type: System..::..Int32
On entry: $n$, the number of observations.
Constraint: ${\mathbf{n}}>1$.
m
Type: System..::..Int32
On entry: $m$, the number of columns of the matrix $X$, i.e., number of independent variables.
Constraint: $1\le {\mathbf{m}}\le {\mathbf{n}}$.
x
Type: array<System..::..Double,2>[,](,)[,][,]
An array of size [dim1, m]
Note: dim1 must satisfy the constraint: $\mathrm{dim1}\ge {\mathbf{n}}$
On entry: ${\mathbf{x}}\left[\mathit{i}-1,\mathit{j}-1\right]$ must contain the $\mathit{i}$th observation for the $\mathit{j}$th variable, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{m}}$.
eps
Type: System..::..Double
On entry: $\epsilon$, the expected fraction of gross errors expected in the sample.
Constraint: $0.0\le {\mathbf{eps}}<1.0$.
cov
Type: array<System..::..Double>[]()[][]
An array of size [${\mathbf{m}}×\left({\mathbf{m}}+1\right)/2$]
On exit: a robust estimate of the covariance matrix, $C$. The upper triangular part of the matrix $C$ is stored packed by columns. ${C}_{ij}$ is returned in ${\mathbf{cov}}\left[\left(j×\left(j-1\right)/2+i\right)-1\right]$, $i\le j$.
theta
Type: array<System..::..Double>[]()[][]
An array of size [m]
On exit: the robust estimate of the location parameters ${\theta }_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,m$.
maxit
Type: System..::..Int32
On entry: the maximum number of iterations that will be used during the calculation of the covariance matrix.
Constraint: ${\mathbf{maxit}}>0$.
nitmon
Type: System..::..Int32
On entry: indicates the amount of information on the iteration that is printed.
${\mathbf{nitmon}}>0$
The value of $A$, $\theta$ and $\delta$ (see [Accuracy]) will be printed at the first and every nitmon iterations.
${\mathbf{nitmon}}\le 0$
No iteration monitoring is printed.
When printing occurs the output is directed to the current advisory message unit (see (X04ABF not in this release)).
tol
Type: System..::..Double
On entry: the relative precision for the final estimates of the covariance matrix.
Constraint: ${\mathbf{tol}}>0.0$.
nit
Type: System..::..Int32%
On exit: the number of iterations performed.
ifail
Type: System..::..Int32%
On exit: ${\mathbf{ifail}}={0}$ unless the method detects an error or a warning has been flagged (see [Error Indicators and Warnings]).
# Description
For a set of $n$ observations on $m$ variables in a matrix $X$, a robust estimate of the covariance matrix, $C$, and a robust estimate of location, $\theta$, are given by
$C=τ2ATA-1,$
where ${\tau }^{2}$ is a correction factor and $A$ is a lower triangular matrix found as the solution to the following equations:
$zi=Axi-θ,$
$1n∑i=1nwzi2zi=0,$
and
$1n∑i=1nuzi2ziziT-I=0,$
where ${x}_{i}$ is a vector of length $m$ containing the elements of the $i$th row of x, ${z}_{i}$ is a vector of length $m$, $I$ is the identity matrix and $0$ is the zero matrix, and $w$ and $u$ are suitable functions.
g02hk uses weight functions:
$ut=aut2,if tbu2$
and
$wt=1,if t≤cwwt=cwt,if t>cw$
for constants ${a}_{u}$, ${b}_{u}$ and ${c}_{w}$.
These functions solve a minimax problem considered by Huber (see Huber (1981)). The values of ${a}_{u}$, ${b}_{u}$ and ${c}_{w}$ are calculated from the expected fraction of gross errors, $\epsilon$ (see Huber (1981) and Marazzi (1987)). The expected fraction of gross errors is the estimated proportion of outliers in the sample.
In order to make the estimate asymptotically unbiased under a Normal model a correction factor, ${\tau }^{2}$, is calculated, (see Huber (1981) and Marazzi (1987)).
The matrix $C$ is calculated using g02hl. Initial estimates of ${\theta }_{j}$, for $j=1,2,\dots ,m$, are given by the median of the $j$th column of $X$ and the initial value of $A$ is based on the median absolute deviation (see Marazzi (1987)). g02hk is based on routines in ROBETH; see Marazzi (1987).
# References
Huber P J (1981) Robust Statistics Wiley
Marazzi A (1987) Weights for bounded influence regression in ROBETH Cah. Rech. Doc. IUMSP, No. 3 ROB 3 Institut Universitaire de Médecine Sociale et Préventive, Lausanne
# Error Indicators and Warnings
Errors or warnings detected by the method:
Some error messages may refer to parameters that are dropped from this interface (LDX) In these cases, an error in another parameter has usually caused an incorrect value to be inferred.
${\mathbf{ifail}}=1$
On entry, ${\mathbf{n}}\le 1$, or ${\mathbf{m}}<1$, or ${\mathbf{n}}<{\mathbf{m}}$, or ${\mathbf{eps}}<0.0$, or ${\mathbf{eps}}\ge 1.0$, or ${\mathbf{tol}}\le 0.0$, or ${\mathbf{maxit}}\le 0$.
${\mathbf{ifail}}=2$
On entry, a variable has a constant value, i.e., all elements in a column of $X$ are identical.
${\mathbf{ifail}}=3$
The iterative procedure to find $C$ has failed to converge in maxit iterations.
${\mathbf{ifail}}=4$
The iterative procedure to find $C$ has become unstable. This may happen if the value of eps is too large for the sample.
${\mathbf{ifail}}=-9000$
An error occured, see message report.
${\mathbf{ifail}}=-6000$
Invalid Parameters $〈\mathit{\text{value}}〉$
${\mathbf{ifail}}=-4000$
Invalid dimension for array $〈\mathit{\text{value}}〉$
${\mathbf{ifail}}=-8000$
Negative dimension for array $〈\mathit{\text{value}}〉$
${\mathbf{ifail}}=-6000$
Invalid Parameters $〈\mathit{\text{value}}〉$
# Accuracy
On successful exit the accuracy of the results is related to the value of tol; see [Parameters]. At an iteration let
(i) $d1=\text{}$ the maximum value of the absolute relative change in $A$ (ii) $d2=\text{}$ the maximum absolute change in $u\left({‖{z}_{i}‖}_{2}\right)$ (iii) $d3=\text{}$ the maximum absolute relative change in ${\theta }_{j}$
and let $\delta =\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(d1,d2,d3\right)$. Then the iterative procedure is assumed to have converged when $\delta <{\mathbf{tol}}$.
# Parallelism and Performance
None.
The existence of $A$, and hence $C$, will depend upon the function $u$ (see Marazzi (1987)); also if $X$ is not of full rank a value of $A$ will not be found. If the columns of $X$ are almost linearly related, then convergence will be slow.
# Example
A sample of $10$ observations on three variables is read in and the robust estimate of the covariance matrix is computed assuming 10% gross errors are to be expected. The robust covariance is then printed.
Example program (C#): g02hke.cs
Example program data: g02hke.d
Example program results: g02hke.r
|
2017-11-18 19:29:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 103, "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": 0.9922490119934082, "perplexity": 1669.862673914386}, "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-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00453.warc.gz"}
|
https://gamedev.stackexchange.com/questions/135047/unity-is-crashing-without-leaving-a-message-at-the-start-of-my-game
|
Unity is crashing without leaving a message at the start of my “game”
So. I like to code. I like to play Paradox Interactive games. One thing that I always wondered about was how exactly their implementation of their trigger evaluation/event evaluation system looked like. So I decided I would write myself an implementation and find out!
Now, I didn't want to manually parse their text files out. I could, but it would be an insane waste of time since other people have already developed efficient libraries for it. So I downloaded and set up this nice little library that handles the nasty little details.
I tested the parser to make sure it works. Worked like a charm. Then I started coding. A full day of planning, trial, and error later, I had a very early version ready.
Then I created a C# script with a basic implementation of my new library. I created a basic spherical mesh and dragged it onto my scene. I dragged said C# MonoBehaviour onto my object. I created a country class implementing my 'IScope' interface. I set the "scope chain" up.
I created a text file with the following contents:
trigger = {
id = 5
vassal = {
id = 6
}
}
It worked in the parser before. I then wrote the following code in my Start method:
void Start()
{
ScriptBinding.Instance.ReserveTrigger("id", new CheckIDTrigger());
ScriptBinding.Instance.ReserveScopeName("vassal");
string path = Application.dataPath + "/ParadoxScriptingEngine/Tests/test_trigger.txt";
Debug.Log(path);
FileStream stream = File.OpenRead(path);
c = new Country(5); //Declared as a class member. Implements IScope.
tb = new TriggerBlock();
ParadoxParser.Parse(test_script, tb);
scope_chain = new IScope[10];
scope_chain[0] = c; //ROOT
scope_chain[1] = c; //THIS
}
And in Update:
void Update()
{
Debug.Log("The trigger has evaluated to " + tb.Eval(scope_chain));
}
When I deactivate the script that launches the parser, the project runs. When I reactivate it and try to run it, the editor instantly crashes.
I have riddled my project with debug statements, but this is just starting to get ridiculous. I have traced through the entire program in my mind, but it is 17 files long all told, and without any data on the error I cannot possibly hope to debug it. The log files are empty as well.
Is there a way to get Unity to print ALL information on errors?
I know that the problem has to be related to my program, so perhaps the better question would be; what kind of problem will cause Unity to not print the error out? I know infinite loops will do it, but there is no way it is an infinite loop is causing this problem - I only used for each loops.
2 Answers
It looks like you may not be closing the filestream once you have opened it, this could be causing the problems you're having.
Try stream.Close() at the end of your Start() function
You should also wrap a try/catch statement around this so that Unity can handle the errors correctly.
try
{
//file opening code
}
catch (FileNotFoundException e)
{
// FileNotFoundExceptions are handled here.
}
catch (IOException e)
{
// Extract some information from this exception, and then
// throw it to the parent method.
whenDo not initialize (e.Source != null)
Console.WriteLine("IOException source: {0}", e.Source);
throw;
}
You can usually find more information about why Unity has crashed without an error by checking the editor logs.
This may give you some clarification, and if you're still stuck you can post the error log for us to look at so that we can advise you further.
https://docs.unity3d.com/Manual/LogFiles.html
• Thanks, the log helped. New to Unity, not new to coding. I spent some time in libgdx a while back... Turned out to be a stack overflow in the Country constructor (the constructor created another Country as it's "vassal" scope, which created it's own and so on and so forth.) Funny how the simplest things can evade us when we are looking in the wrong places. – JDSweetBeat Dec 30 '16 at 22:37
That crash is caused because your computer cannot handle too much stuff in one time , example to do :
Debug.Log("The trigger has evaluated to " + tb.Eval(scope_chain));
In
Update()
need to do Eval() function each frame , and how about just to do the function take 0.1 second and each second need 60 frame ? what i mean here don't run the function without delay , your computer also have limit .
• I don't understand your answer. The CPU usage is laughably low as is memory usage. The only thing you should not do in the update function is do a bunch of un-culled object creation - but the Eval method does almost no object creation, and only does 1 or 2 dictionary lookups on the first run of Eval. Admittedly it's not the most efficient implementation, but it is almost certainly not "too much to handle". I was asking "why doesn't Unity catch this/is there a way to debug my application without painstakingly going through and commenting each line of code out and running the app. – JDSweetBeat Dec 30 '16 at 2:11
|
2020-09-21 00:06:44
|
{"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": 0.24106645584106445, "perplexity": 1722.4939019485116}, "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-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00166.warc.gz"}
|
https://gamedev.stackexchange.com/questions/175220/sprite-is-getting-out-of-screen-when-changing-resolution-unity
|
# Sprite is getting out of screen when changing resolution - Unity
Hello I am making this layout:
I used sprites for birds but used UI for the blue food thrower. I used UI for it so it will not get cut when screen resolution changes. But I also need to add food itself which I need to make 2D to interact with birds. Now It is not visible over UI, So I changes Canvas-Render Mode to world space. But I cannot adjust text in that mode. Its not visible. Then I tried converting the food thrower to Sprite but it get cut. Any authentic and cleaner way to do it?
|
2019-12-12 23:33:37
|
{"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": 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": 0.331749826669693, "perplexity": 1988.6519773637117}, "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-2019-51/segments/1575540547536.49/warc/CC-MAIN-20191212232450-20191213020450-00495.warc.gz"}
|
https://zbmath.org/?q=ut%3Araising+operator
|
## Found 27 Documents (Results 1–27)
100
MathJax
### Properties of shape-invariant tridiagonal Hamiltonians. (English. Russian original)Zbl 1445.81027
Theor. Math. Phys. 203, No. 3, 761-779 (2020); translation from Teor. Mat. Fiz. 203, No. 3, 380-400 (2020).
Full Text:
### Chebyshev polynomials of the second kind via raising operator preserving the orthogonality. (English)Zbl 1413.33013
MSC: 33C45 42C05
Full Text:
### Hermite polynomials and Hahn’s theorem with respect to the raising operator. (English)Zbl 1382.33015
MSC: 33C45 42C05
Full Text:
### Characterization of Laguerre polynomials as orthogonal polynomials connected by the Laguerre degree raising shift operator. (English)Zbl 1381.33012
MSC: 33C45 42C05
Full Text:
Full Text:
### On a discrete number operator associated with the 5D discrete Fourier transform. (English)Zbl 1355.39027
Pinelas, Sandra (ed.) et al., Differential and difference equations with applications. ICDDEA, Amadora, Portugal, May 18–22, 2015. Selected contributions. Cham: Springer (ISBN 978-3-319-32855-3/hbk; 978-3-319-32857-7/ebook). Springer Proceedings in Mathematics & Statistics 164, 273-292 (2016).
Full Text:
Full Text:
### Functions of the second kind for classical polynomials. (English)Zbl 1285.33013
MSC: 33D45 33C45
Full Text:
### Classical orthogonal polynomials with respect to a lowering operator generalizing the Laguerre operator. (English)Zbl 1279.33012
MSC: 33C45 42C05
Full Text:
### Some relations satisfied by orthogonal matrix polynomials. (English)Zbl 1229.33012
MSC: 33C45 15A60
Full Text:
### Superalgebra and harmonic oscillator with constant positive curvature. (English)Zbl 1136.81363
MSC: 81Q05 81Q60 81Q10
Full Text:
Full Text:
### Differential equations for multiple orthogonal polynomials with respect to classical weights: raising and lowering operators. (English)Zbl 1097.33004
MSC: 33C45 42C05 33E30
Full Text:
### Differential coefficients of orthogonal matrix polynomials. (English)Zbl 1100.42017
MSC: 42C05 33E30 39B42
Full Text:
### Ladder operators for the associated Laguerre functions. (English)Zbl 1081.33013
MSC: 33C45 34L40 81R30
Full Text:
Full Text:
### Young raising operators and $$Q$$-functions. (English)Zbl 0803.05053
MSC: 05E05 05E10
### Conformal symmetries and generalized recurrences for heat and Schrödinger equations in one spatial dimension. (English)Zbl 0763.35025
Mechanics, analysis and geometry: 200 years after Lagrange, 237-256 (1991).
MSC: 35J10 58J70 35K10
### Application of generalised Bessel’s partial differential equations. (English)Zbl 0681.33009
Reviewer: A.Kaneko
MSC: 33C10 33C80
### The matrix representations of $${\mathfrak g}_ 2$$. II: Representations in an $${\mathfrak su}(3)$$ basis. (English)Zbl 0654.17005
Reviewer: G.Czichowski
MSC: 17B10 17B25
Full Text:
### The matrix representations of $${\mathfrak g}_ 2$$. I: Representations in an $${\mathfrak so}(4)$$ basis. (English)Zbl 0654.17004
Reviewer: G.Czichowski
MSC: 17B10 17B25
Full Text:
### Raising operators and Young’s rule. (English)Zbl 0626.20004
Combinatoire énumérative, Proc. Colloq., Montréal/Can. 1985, Lect. Notes Math. 1234, 91-105 (1986).
Reviewer: D.M.Bressoud
MSC: 20C30 05A15
### Representations of sl(2,C) on posets and the Sperner property. (English)Zbl 0496.06004
MSC: 06A06 06B05 20G05
Full Text:
all top 5
all top 3
|
2022-06-28 00:20:45
|
{"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": 0.7451832890510559, "perplexity": 4959.139471706347}, "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/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00680.warc.gz"}
|
https://learn.careers360.com/engineering/question-help-me-solve-this-electrochemistry-bitsat/
|
# The oxidation number of sulphur in $S_8,S_2F_2,H_2S$ are Option 1) 0,1,-2 Option 2) 2,1,-2 Option 3) 0,1,2 Option 4) -2,1,-2
As we learnt in
Oxidation Number -
Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belogs entirely to more electronegative elements.
-
Rules for Oxidation Number -
In all its compound, flourine has an oxidation number of -1. Other halogen (Cl, Br and I) also have oxidation number of -1, when they occur as halide ions in their compound.
- wherein
e.g. Cl, Br and I when combined with O2 have +ve oxidation number.
Rules for Oxidation Number -
The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compound containing two elements)
- wherein
e.g. LiH,NaH and CaH2 its oxidation number is -1.
Lets take the species one by one
$S_{8}$ O.N=0
$S_{2}F_{2}$ O.N=+1
$H_{2}S$ O.N= - 2
$\therefore$ Solution is 1
Option 1)
0,1,-2
Correct
Option 2)
2,1,-2
Incorrect
Option 3)
0,1,2
Incorrect
Option 4)
-2,1,-2
Incorrect
### Preparation Products
##### Knockout BITSAT 2021
It is an exhaustive preparation module made exclusively for cracking BITSAT..
₹ 4999/- ₹ 2999/-
|
2020-09-28 14:55:30
|
{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "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": 0.8077513575553894, "perplexity": 12581.687820176247}, "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/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00120.warc.gz"}
|
https://www.andlearning.org/area-formulas/
|
Connect with us
Area & Surface area Formulas
We are going to share Surface area Formulas for the student who is studying in the class of 5, 6, 7, 8, 9, 10, 11, and 12. In math, the Surface area is a very important topic which is helping to improve scores in the exam. If you want to become very intelligent in math then you should remember the Surface area Formulas to resolve Surface area related problems. If you have any questions related to the Surface area please let me know through the comment and mail. There are millions of students looking for Surface area formulas that why we shared Surface area formulas below.
Area of a circle Formulas: $$\pi r^2$$ r= radius Area of a triangle Formulas: $$\sqrt{s(s-a)(s-b)(s-c)}$$ a, b, c is length of sides, s is semi-perimeter Area of a trapezoid Formulas: $$\frac{a+b}{2} \times h$$ a, b is base and h is height Area of a rectangle Formulas: $$l \times w$$ l is length, w is width Area of a square Formulas: $$a^2$$ a is length of side Area of a parallelogram Formulas: $$l \times w$$ l is length, w is width Area of a right triangle Formulas: $$\frac{a \times b}{2}$$ a, b is side of right angle triangle Area of equilateral triangle Formulas: $$\frac{\sqrt{3}}{4} \times a^2$$ a is length of side Area of isosceles triangle Formulas: $$\frac{b \times h}{2}$$ b is base, h is height of triangle Area of a hexagon Formulas: $$\frac{3 \sqrt{3}}{2} \times a^2$$ a is length of side Area of a sector Formulas: $$\frac{\theta}{360} \pi r^2$$ r= radius Area of a rhombus Formulas: $$\frac{p \times q}{2}$$ p, q diagonal of rhombus Area of an oval Formulas: $$\pi \frac{a}{2} \times \frac{b}{2}$$ a is major axis, b is minor axis Area of a ring Formulas: $$\pi (R^2 – r^2)$$ R is Outer radius, r is inner radius Area of an octagon Formulas: $$2 (1+ \sqrt{2}) a^2$$ a is length of side Area of a semicircle Formulas: $$\pi \frac{r^2}{2}$$ r= radius Area of a pentagon Formulas: $$\frac{1}{4} \sqrt{5(5+2 \sqrt{5} )} a^2$$ a is length of side Area of a kite Formulas: $$\frac{p \times q}{2}$$ p, q diagonal Area of an ellipse Formulas: $$\pi \frac{a}{2} \times \frac{b}{2}$$ a is major axis, b is minor axis Surface Area of a sphere Formulas: $$4 \pi r^2$$ r is radius Surface Area of a cylinder Formulas: $$2 \pi r (r+h)$$ r is radius, h height Surface Area of a cube Formulas: $$6 a^2$$ a is length of side Surface Area of a cone Formulas: $$\pi r (r + \sqrt{h^2 + r^2})$$ r is radius, h height Surface Area of a box Formulas: $$2(h × W) + 2(h × L) + 2(W × L)$$ width (w), length (l) and height (h) Surface Area of a rectangular prism Formulas: $$2(wl+lh+hw)$$ width (w), length (l) and height (h) Surface Area of a triangular prism Formulas: $$ab+3bh$$ width (a), length (b) and height (h) Surface Area of a pentagonal prism Formulas: $$5ab+5bh$$ width (a), length (b) and height (h) Surface Area of a Hexagonal prism Formulas: $$6ab+6bh$$ width (a), length (b) and height (h) Surface Area of a Square Pyramid Formulas: $$2(b × s) + b^2$$ Slant Hiehgt (S), base (b) Surface Area of a Triangular Pyramid Formulas: $$\frac{1}{2}(a × b) + \frac{3}{2}(b × s)$$ Altitude (a), Slant Hiehgt (S), base (b) Surface Area of a Pentagon Pyramid Formulas: $$\frac{5}{2}(a × b) + \frac{5}{2}(b × s)$$ Altitude (a), Slant Hiehgt (S), base (b) Surface Area of a Hexagonal Pyramid Formulas: $$3(a × b) + 3(b × s)$$ Altitude (a), Slant Hiehgt (S), base (b)
Summary of Surface area Formula
Andlearning.org is a single website that is sharing all Surface area formulas which is useful for math calculation. I checked numbers of websites that have shared Surface area formulas but we can’t find any website which has shared all Surface area formulas including all notes. If you have any questions regarding Surface area formulas please let me know through comment, mail, or social media. My expert team is waiting for your Surface area related doubts or questions for the solution.
|
2022-05-16 17:52:31
|
{"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": 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": 0.861368715763092, "perplexity": 832.4661891159697}, "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-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00493.warc.gz"}
|
https://brilliant.org/problems/an-average-problem/
|
An average problem
Discrete Mathematics Level 5
The natural numbers from 1 to 2015 are placed on a line in any order, so that no two numbers lie on the same point and that their average mean does not lie between those two numbers.
For example: You place the 1 at the very left, 2015 at the very right. This leads to a problem because $$(1+2015) \over 2$$ = 1008 is their average and has to lie between 1 and 2015., therefore those two numbers would both not suffice the property.
However one could place numbers in the following order 2015 1 2 3 ..... 2013 2014. Now 2015 would suffice the property stated above. However the numbers 1 and 3 now make a problem since the number 2 lies between them..
Is it possible to find an arrangement of those numbers, so that all numbers suffice the property given above? If not, what is the maximum amount of numbers to hold this property?
×
|
2016-10-27 20:36:52
|
{"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": 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": 0.7131181955337524, "perplexity": 247.5174426811064}, "config": {"markdown_headings": false, "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-2016-44/segments/1476988721392.72/warc/CC-MAIN-20161020183841-00393-ip-10-171-6-4.ec2.internal.warc.gz"}
|
https://taras.glek.net/post/curious-case-of-maintaining-sufficient-free-space-with-zfs/
|
Curious Case of Maintaining Sufficient Free Space with ZFS
Sometimes trivial problems turn out to be more interesting than expected.
TLDR: ZFS free-space reporting is a lagging indicator.
Background
I use Proxmox VM server backed by a ZFS array of hard drives for various personal infrastructure. I also have security cameras that upload motion-triggered videos to my server (via FTP!).
Problem Description
I would like to use 90% of my available space for most-recent security videos.
Recipe:
1. Create a dedicated ZFS volume
2. Setup ZFS quota
3. Run a cronjob to free space faster than it gets consumed by video uploads.
I set out to find a decent disk-freeing solution. I eventually settled on a python free-disk. (Most of these utils were written in bash, didn’t want to disk having to grasp & modify write-only code).
I modified free-disk to support deleting files with a particular extension (e.g. .mp4) and deployed it. My changes to free-disk are here. Author took part of them in, hoping for rest to get upstreamed too.
I was surprised to see that the script freed up roughly 10x more than I expected. So I added some debug logging and after a few days of experiments ended up with following:
15:17:02-0700:DEBUG:Required free: 9.31GiB. 3.4GiB to free. --track-
bytes-deleted=True
Jul 09 15:17:37 quad free-disk[70706]: 2022-07-09T15:17:37-0700:INFO:Removed 40 file(s) with modification date <= 2022
-06-18T18:32:18.734254Z. Deleted 3.46GiB. Filesystem freed 2.22GiB.
Jul 09 15:17:37 quad free-disk[74496]: Filesystem 1M-blocks Used Available Use% Mounted on
Jul 09 15:17:37 quad free-disk[74496]: rpool/data/tinycam 1572864M 1564537M 8328M 100% /export/tinycam
Jul 09 15:17:39 quad free-disk[74499]: Filesystem 1M-blocks Used Available Use% Mounted on
Jul 09 15:17:39 quad free-disk[74499]: rpool/data/tinycam 1572864M 1564537M 8328M 100% /export/tinycam
Jul 09 15:17:41 quad free-disk[74502]: Filesystem 1M-blocks Used Available Use% Mounted on
Jul 09 15:17:41 quad free-disk[74502]: rpool/data/tinycam 1572864M 1564537M 8328M 100% /export/tinycam
Jul 09 15:17:43 quad free-disk[74616]: Filesystem 1M-blocks Used Available Use% Mounted on
Jul 09 15:17:43 quad free-disk[74616]: rpool/data/tinycam 1572864M 1563300M 9565M 100% /export/tinycam
To get this data I added alernate logic to free-disk to free based on bytes-deleted instead of filesystem free-space (eg number reported by df). I also added a few sleep 2;df -h /export/tinycam commands to track progress.
It seems that after I delete files, it takes roughly 5-10seconds for disk-free information to settle. Problem-settled: “obviously one should only free space by tracking bytes-deleted” or “add sleep 10 before checking free-space”. Not so fast :)
Complexity of Modern Free-Space tracking
ZFS is a filesystem that supports deduplication & compression. It also supports snapshots. This means that the file being deleted could:
1. Be a uncompressed and have no other duplicates.
2. Be compressed and space freed would be less than the size shown by stat
3. Be referenced by a snapshot, so no space is actually freed when the file is deleted.
4. Be dedupped to another file
5. Some combo of all of the above
So in order for ZFS to know how much space has been reclaimed by file deletion it needs to check all these conditions and it should do this async of file-deletion to now slow down FS-operations.
Luckily ZFS documents this:
ZFS is a transactional file system. Most file system modifications are bundled into transaction groups and committed to disk asynchronously. Until these modifications are committed to disk, they are termed pending changes. The amount of space used, available, and referenced by a file or file system does not consider pending changes. Pending changes are generally accounted for within a few seconds. Even committing a change to disk by using fsync(3C) or O_SYNC does not necessarily guarantee that the space usage information is updated immediately.
Conclusion: There is no universal way to ensure space is freed on ZFS. One needs to carefully consider how ZFS is being used in order to understand how to automatically maintain free disk space.
I get burned by (3) about once or twice a year, eventually I remember that I have a snapshot that diverged too much. This lagging-indication of free-space was new and fun, figured it was worth a blog post.
|
2022-08-17 00:32:52
|
{"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": 0.3041137754917145, "perplexity": 10144.319028118238}, "config": {"markdown_headings": false, "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-2022-33/segments/1659882572833.78/warc/CC-MAIN-20220817001643-20220817031643-00413.warc.gz"}
|
https://www.lil-help.com/questions/269572/elements-that-are-characterized-by-the-filling-of-p-orbitals-are-classified-as
|
elements that are characterized by the filling of p orbitals are classified as
# elements that are characterized by the filling of p orbitals are classified as
S
9.9k points
periodhorizontal row in the periodic table
groupvertical row in the periodic table
periodic lawa repetition of properties occurs when elements are arranged in order of increasing atomic number
metaltype of element that is a good conductor of heat and electric current
transition metaltype of element characterized by the presence of electrons in the d orbital
atomic radiusone-half the distance between the nuclei of two atoms are joined
cationtype of ion formed by group 2A elements
electronssubatomic particles that are transferred to form positive and negative ions
electronegativityability of an atom to attract electrons when the atom is in a compound
ionization energyenergy required to remove an electron from an atom
What is another name for the representative elements?Group A elements
What is another name for the transition metals?Group B elements
Which of the following elements is in the same period as phosphorus?Magnesium
Each period in the periodic corresponds toa principal energy level
The modern periodic table is arranged in order of increasingAtomic number
Who arranged the elements according to atomic mass and used the arrangement to predict the properties of missing elements?Dmitri Mendeleev
Which of the following categories includes the majority of the elements?Metals
Of the elements Pt, V, Li, and Kr, which is a nonmetal?Kr
To what category of elements does an element belong if it is a poor conductor of electricity?Nonmentals
In which of the following sets is the symbol of the element, the number of protons, and the number of electrons given correctly?In, 49 protons, 49 electrons
The atomic number of an element is the total number of which particles in the nucleus?Protons
What element has the electron configuration 1s^2 2p^6 3s^2 3p^2?Silicon
Which of the following is true about the electron configurations of the noble gases?The highest occupied s and p sublevels are partially filled
Elements that are characterized by the filling of p orbitals are classified asGroups 3A through 8A
Which of the following electron configurations is most likely to result in an element that is relatively inactive?A filled highest occupied principal energy level
Which subatomic particle plays the greatest part in determining the properties of an element?electron
Which of the following elements is a transition metal?copper
Which of the following groupings contains only representative elements?Al, Mg, Li
Which of the following is true about the electron configurations of the representative elements?the highest occupied s and p levels are partially filled
What are the Group 1A through Group 7A examples of?representative elements
Of the elements Fe, Hg, U, and Te, which is a representative element?Te
How does atomic radius change from top to bottom in a group in a periodic table?It tends to increase
How does atomic radius change from left to right across a period in the periodic table?It tends to decrease
What causes the shielding effect to remain constant across a period?electrons are added to the same principal energy level
Atomic size generallydecreases as you move from left to right across a period
What element in the second period has the largest atomic radius?neon
Which of the following factors contributes to the increase in atomic size within a group in the periodic table as the atomic number increases?more shielding of the electrons by the highest occupied energy level
Which of the following elements has the smallest atomic radius? Sulfur, Chlorine, Selenium, BromineChlorine
What is the charge of a cation?a positive charge
Which of the following statements is true about ions?cations form when an atom loses electrons
The metals in Groups 1A, 2A, and 3Alose ions when they form ions
Which of the following is not true about ions?When a cation forms, more electrons are transferred to it
Why is the second ionization energy greater than the first ionization energy?It is more difficult to remove a second electron from an atom
In which of the following sets are the charges given correctly for all ions?K^+, Sr^2+, O^2-
In which of the following groups of ions are the charges shown correctly?Ca^2+, Al^3+, Br^-
What is the element with the lowest electronegativity value? cesium, helium, calcium, or fluorinecesium
What is the element with the highest electronegativity value? cesium, helium, calcium, or fluorinefluorine
Which of the following elements has the smallest ionic radius? Li, K, O, SLi
What is the energy required to remove an electron from an atom in the gaseous state called?ionization energy
For Group 2A metals, which electron is the most difficult to remove?the third
Which of the following factors contributes to the decrease in ionization energy within a group in the periodic table as the atomic number increases?increase in atomic size
Which of the following elements has the smallest first ionization energy? sodium, calcium, potassium, or magnesiumpotassium
Which of the following elements has the lowest electronegativity? lithium, carbon, bromine, fluorinelithium
What is true about electronegativity?It generally increases from left to right across a period and decreases from top to bottom within a group
Compared with the electronegativities of the elements on the left side of a period, the electronegativities of the elements on the right side of the same period tend to behigher
Which of the following decreases with the increasing atomic number in Group 2A?ionization energy
What correctly compares the relative size of an ion to its neutral atom?the radius of an anion is greater than the radius of its neutral atom
Which factor contributes to the increase in ionization energy from left to right across a period?an increase in the number of protons
As you move from left to right across the second period of a periodic table ionization energyincreases
Of the following elements, which one has the smallest first ionization energy? boron, carbon, aluminum, or siliconaluminum
elements that
studybud
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
• Answer the question above my logging into the following networks
### Post as a guest
• Your email will not be shared or posted anywhere on our site
•
Stats
Views: 92
|
2018-12-14 03:18:35
|
{"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": 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": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6056198477745056, "perplexity": 1428.7843040094228}, "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-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00432.warc.gz"}
|
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/?level=2880&weight=2&char_order=1&atkin_lehner_string=%2B%2B-
|
## Results (4 matches)
Label Dim. $$A$$ Field CM Traces A-L signs $q$-expansion
$$a_2$$ $$a_3$$ $$a_5$$ $$a_7$$ 2 3 5
2880.2.a.v $$1$$ $$22.997$$ $$\Q$$ None $$0$$ $$0$$ $$1$$ $$-2$$ $$+$$ $$+$$ $$-$$ $$q+q^{5}-2q^{7}-2q^{11}+2q^{17}+4q^{19}+\cdots$$
2880.2.a.bd $$1$$ $$22.997$$ $$\Q$$ None $$0$$ $$0$$ $$1$$ $$2$$ $$+$$ $$+$$ $$-$$ $$q+q^{5}+2q^{7}-2q^{11}-4q^{13}-2q^{17}+\cdots$$
2880.2.a.be $$1$$ $$22.997$$ $$\Q$$ None $$0$$ $$0$$ $$1$$ $$2$$ $$+$$ $$+$$ $$-$$ $$q+q^{5}+2q^{7}+2q^{11}+2q^{17}-4q^{19}+\cdots$$
2880.2.a.bf $$1$$ $$22.997$$ $$\Q$$ None $$0$$ $$0$$ $$1$$ $$2$$ $$+$$ $$+$$ $$-$$ $$q+q^{5}+2q^{7}+6q^{11}+4q^{13}+6q^{17}+\cdots$$
|
2021-04-21 11:41:43
|
{"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": 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": 0.9964534640312195, "perplexity": 622.1242215603595}, "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-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00569.warc.gz"}
|
https://qanda.ai/en/solver/popular-problems/1008788
|
# Calculator search results
Formula
Find the difference
$$\dfrac { 1 } { 2 } - \dfrac { 1 } { 2 }$$
$0$
$\color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
Remove the two numbers if the values are the same and the signs are different
$0$
|
2022-01-22 02:50:51
|
{"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": 0.6180346012115479, "perplexity": 1027.1235549725911}, "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/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00051.warc.gz"}
|
https://math.stackexchange.com/questions/1785411/linear-programming-is-continuous
|
# Linear programming is continuous
Consider an arbitrary linear program:
$$\max \vec c \cdot \vec x$$
subject to:
$$\textbf{A}\cdot \vec x = 0, \quad \vec a \le \vec x \le \vec b$$
Assume that this program is feasible and bounded. Let $\vec x$ be an optimal solution.
Now suppose I perturb the bounds by small amounts:
$$\vec a' = \vec a + \delta \vec p, \quad \vec b' = \vec b + \delta \vec q$$
where $\vec p,\vec q$ are unit vectors and $\delta > 0$. Assume that the modified problem remains feasible and bounded.
Prove or disprove: For arbitrary $\epsilon > 0$, there exists a $\delta > 0$ such that there exists an optimal solution $\vec x'$ to the perturbed linear program defined above satisfying $|\vec x - \vec x'| < \epsilon$.
Extra: Is there a way to define uniform continuity in this context, such that Linear Programming is, or isn't, uniformly continuous?
|
2019-08-19 20:43:46
|
{"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": 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": 0.9683783650398254, "perplexity": 271.18366752930626}, "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-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00237.warc.gz"}
|
https://socratic.org/questions/the-length-of-a-rectangle-is-four-times-its-width-if-the-area-of-the-rectangle-i
|
# The length of a rectangle is four times its width. If the area of the rectangle is 256m^2 how do you find its perimeter?
Apr 6, 2018
The perimeter of the rectangle is $80$ meters.
#### Explanation:
Here are two formulas for rectangles that we will need to solve this problem, where $l$ = length and $w$ = width:
In this question, we know that:
$l = 4 w$
$A = 256 {m}^{2}$
First, let's find the width:
$l w = 256$
Let's substitute the value of $4 w$ for $l$:
$\left(4 w\right) w = 256$
Multiply the $w$:
$4 {w}^{2} = 256$
Divide both sides by $4$:
${w}^{2} = 64$
$w = 8$
So we know that the width is $8$.
Since $l = 4 w$ and we have $w$, we can find the value of $l$:
$4 \left(8\right)$
$32$
The width is $8$ meters and the length is $32$ meters.
$- - - - - - - - - - - - - - - - - - -$
Now we find the perimeter. Remember the formula for the perimeter is $2 l + 2 w$ as stated earlier. Since we have the values of $l$ and $w$, we can solve this:
$2 \left(32\right) + 2 \left(8\right)$
$64 + 16$
$80$
The perimeter of the rectangle is $80$ meters.
Hope this helps!
|
2020-07-12 05:37:39
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 30, "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": 0.9393836855888367, "perplexity": 261.1734542774619}, "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/1593657131734.89/warc/CC-MAIN-20200712051058-20200712081058-00300.warc.gz"}
|
https://www.mersenneforum.org/showthread.php?s=89833a21398f780ef88114f6298dce1d&t=12217
|
mersenneforum.org optimal B1
Register FAQ Search Today's Posts Mark Forums Read
2009-07-27, 15:36 #1 Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 102538 Posts optimal B1 Which is more efficient: running ECM with a few large B1 increments, (e.g. run some curves at B1=11e3, then more at 5e4, then ...) or incrementing B1 by some optimized amount for each curve? What gives the best optimization for how quickly the minimum expected factor size can grow? Granular B1 increments are certainly more convenient for communicating and calculating work done (it's certainly easier to say, "I've run 2000 curves at B1=11e6" and to calculate the level from that than if it was incremented), but are they more efficient?
2009-07-27, 15:58 #2
R.D. Silverman
Nov 2003
23×32×103 Posts
Quote:
Originally Posted by Mini-Geek Which is more efficient: running ECM with a few large B1 increments, (e.g. run some curves at B1=11e3, then more at 5e4, then ...) or incrementing B1 by some optimized amount for each curve? What gives the best optimization for how quickly the minimum expected factor size can grow? Granular B1 increments are certainly more convenient for communicating and calculating work done (it's certainly easier to say, "I've run 2000 curves at B1=11e6" and to calculate the level from that than if it was incremented), but are they more efficient?
Read my joint paper with Sam Wagstaff: A Practical Analysis of ECM.
It answers these kinds of questions (and more).
2009-07-27, 17:04 #3
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts
Quote:
Originally Posted by R.D. Silverman Read my joint paper with Sam Wagstaff: A Practical Analysis of ECM. It answers these kinds of questions (and more).
2009-07-27, 18:26 #4
R.D. Silverman
Nov 2003
1CF816 Posts
Quote:
Originally Posted by Mini-Geek Is the paper available for free anywhere? (preferably as a PDF)
I don't know whether this is free:
http://cat.inist.fr/?aModele=afficheN&cpsidt=3769506
You can get a copy from the AMS website, but as they own the
copyright, it costs \$.....
2011-05-27, 12:19 #5
Walter Nissen
Nov 2006
Terra
4E16 Posts
key paper from 1993
Quote:
Originally Posted by Mini-Geek Is the paper available for free anywhere? (preferably as a PDF)
A practical analysis of the elliptic curve factoring algorithm
Robert D. Silverman; Samuel S. Wagstaff
Math. Comp. 61 (1993), 445-462
in .pdf is available free from :
http://www.ams.org/journals/mcom/199...993-1122078-7/
Similar Threads Thread Thread Starter Forum Replies Last Post aurashift Hardware 11 2015-09-22 14:09 Uncwilly Lounge 5 2013-05-15 23:29 henryzz GMP-ECM 14 2011-06-09 17:04 Walter Nissen GMP-ECM 16 2007-03-20 19:35 wblipp ElevenSmooth 16 2004-08-13 19:01
All times are UTC. The time now is 21:35.
Thu May 28 21:35:37 UTC 2020 up 64 days, 19:08, 0 users, load averages: 2.49, 1.97, 1.61
|
2020-05-28 21:35:37
|
{"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": 0.28331953287124634, "perplexity": 11189.836321939145}, "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/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00161.warc.gz"}
|
https://chemistry.stackexchange.com/questions/27039/nomenclature-of-methyl-groups-in-an-alkanol
|
# Nomenclature of methyl groups in an alkanol
I'm having trouble with this question since I'm getting (A), when the answer says its (B).
My logic was that there is a methyl group on the first carbon, and a hydroxide functional group on the second carbon, hence the name 1-methyl-2-pentanol.
Could someone please explain why I am wrong?
## 2 Answers
Don't let the graphical layout of the molecule deceive you! There are no branches in the carbon chain. (That is, no carbon atom is bound to more than two carbon atoms.) The molecule ethane is not called "methylmethane", so calling your molecule a methylpentane derivative would similarly be an error.
With no branches (and no rings and no double bonds etc.), and with six carbon atoms, we must be dealing with a hexane derivative. Once you know that, you just need to apply IUPAC rules that minimize the numbering of atoms to which substituents are bound. So answer B makes more sense than C, for that reason.
• Alright, I understand now! Thankyou! So, just to be 100% sure, under what circumstances would the molecule be a pentane derivative with a methyl chain branch? – chemistryyo Mar 8 '15 at 4:56
• For simple alkanes with methyl "branches", the numbering gives it away. You can't have a "1-methyl" substituent on an otherwise unsubstituted alkane, because it would make the chain longer, thus making it a hexane. So "1-methylpentane" doesn't exist, but 2-methylpentane and 3-methylpentane do. (But not "4-methylpentane" or "5-methylpentane" because the former is the same as the 2-methyl compound, and the latter thas the same "not really a branch" problem as the "1-methyl".) – Curt F. Mar 8 '15 at 5:00
Could someone please explain why I am wrong?
1. Find the longest chain! This determines the base name: hexane.
2. Find the position of the $\ce{OH}$ substituent. Lower number wins.
|
2019-10-14 10:23:02
|
{"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": 0.7765023112297058, "perplexity": 2239.825938772004}, "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-2019-43/segments/1570986653216.3/warc/CC-MAIN-20191014101303-20191014124303-00012.warc.gz"}
|
https://stats.stackexchange.com/questions/65491/whether-to-perform-a-3-by-2-repeated-measures-anova-or-three-separate-paired-t-t
|
# Whether to perform a 3 by 2 repeated measures ANOVA or three separate paired t-tests?
I have $2\times 3$ repeated measures design.
• 2 conditions(a and b)
• 3 intensities (low, medium and high).
I am concerned only with the a to b differences at the corresponding intensity. For example I would like to compare:
• A low and b low
• A medium and b medium
• A high and b high.
I am not concerned with comparing A low to A medium to A high as physiologically we know that they are different and it does not answer my research question. I don't care about how the 2 variables interact, but I may be wrong. My inclination is to say the paired T -tests are fine as
Should I perform 3 paired t-tests for the comparisons of interest or a $2\times3$ Repeated measures ANOVA?
|
2020-10-22 01:27:28
|
{"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": 0.253375381231308, "perplexity": 1160.499989093892}, "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/1603107878662.15/warc/CC-MAIN-20201021235030-20201022025030-00166.warc.gz"}
|
http://arxiver.moonhats.com/2017/09/15/modeling-a-propagating-sawtooth-flare-ribbon-structure-as-a-tearing-mode-in-the-presence-of-velocity-shear-ssa/
|
# Modeling a Propagating Sawtooth Flare Ribbon Structure as a Tearing Mode in the Presence of Velocity Shear [SSA]
On April 18, 2014 (SOL2014-04-18T13:03) an M-class flare was observed by IRIS. The associated flare ribbon contained a quasi-periodic sawtooth pattern that was observed to propagate along the ribbon, perpendicular to the IRIS spectral slit, with a phase velocity of $\sim 15$ km s$^{-1}$. This motion resulted in periodicities in both intensity and Doppler velocity along the slit. These periodicities were reported to be approximately $\pm0.5$ arcseconds in position and $\pm20$ km s$^{-1}$ in velocity and were measured to be $\sim180^{\circ}$ out of phase with one another. This quasi-periodic behavior has been attributed by others to bursty or patchy reconnection and slipping occurring during three-dimensional magnetic reconnection. While able to account for periodicities in both intensity and Doppler velocity these suggestions do not explicitly account for the phase velocity of the entire sawtooth structure, or for the relative phasing of the oscillations. Here we propose that the observations can be explained by a tearing mode instability occurring at a current sheet across which there is also a velocity shear. Using a linear model of this instability we reproduce the relative phase of the oscillations, as well as the phase velocity of the sawtooth structure. We suggest a geometry and local plasma parameters for the April 18 flare which would support our hypothesis. Under this proposal the combined spectral and spatial IRIS observations of this flare may provide the most compelling evidence to date of a tearing mode occurring in the solar magnetic field.
J. Parker and D. Longcope
Fri, 15 Sep 17
50/57
Comments: N/A
|
2017-09-22 07:54:39
|
{"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": 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": 0.368272066116333, "perplexity": 1456.2456925213512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2017-39/segments/1505818688926.38/warc/CC-MAIN-20170922074554-20170922094554-00200.warc.gz"}
|
https://www.physicsforums.com/threads/inverse-of-a-linear-operator.289202/
|
Inverse of a linear operator
1. Feb 2, 2009
seek
1. The problem statement, all variables and given/known data
Show that if an operator A satisfies A2 - A + I = 0 then A has an inverse. Express A-1 as a simple polynomial of A.
2. Relevant equations
I'm not sure that this is relevant, but A-1=1/(detA)TrC where TrC is the transpose of the matrix of cofactors. Also:
If detA = 0 then the matrix has no inverse
3. The attempt at a solution
So I notice immediately that adding by the identity matrix in this equation will result in a matrix with its diagonal having numbers (real or complex) and the rest being zero, as I can be expressed as the kronecker delta. And if the determinant must be nonzero in order to have an inverse, there has to be a way to relate the diagonal of an n dimensional matrix with its determinant. I'm just stuck as to how to do that. Any help greatly appreciated, thank you.
*Edit:
I've been thinking more about this problem, and it seems like there should be a way to use the secular equation to solve it. We went over it briefly in class (the class is quantum and I haven't had linear algebra yet, so it's kind of a chore), but not in enough detail that I would be able to use it in a proof.
Last edited: Feb 2, 2009
2. Feb 2, 2009
tiny-tim
Welcome to PF!
Hi seek! Welcome to PF!
oooh … so complicated
Try writing it A2 - A = -I
3. Feb 3, 2009
seek
My oversight is to my pride as a cold slap to the visage. Thanks so much for the help, maybe next time I'll be able to use the skills I learned in 5th grade.
|
2018-01-16 19:57:19
|
{"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": 0.8145370483398438, "perplexity": 329.98376834418144}, "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-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00663.warc.gz"}
|
https://irrevocabletrustdoctor.com/sheridan-recycling-qascn/how-many-unpaired-electrons-does-titanium-have-06b62c
|
# how many unpaired electrons does titanium have
LOGIN TO VIEW ANSWER. Identify the specific element that corresponds to each of the following electron configurations: 1s 2s 2p 3s a. b. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. Still have questions? Hahaha. PRACTICE TEST 1. five . Ti3+ [Ar] 3d14s0 1 Out of these, Cr3+ is most stable in aqueous solution as its hydration energy is highest. Vanadium has two more electrons than scandium, and two more protons as well, of course. When we have added 18 electrons to give the argon structure, we have then built a V 5+ ion. Asked By adminstaff @ 29/12/2019 06:29 PM. none. Thus there are 3 unpaired electrons in each vanadium atom. how many unpaired electrons in arsenic, As: 3 unpaired electrons. two unpaired electrons . 0 unpaired electrons [Xe]5d8 Square planar. Therefore: The electron occupation is t 2g 6 e g 2 so the LFSE = 12Dq. Our experts can answer your tough homework and study questions. There are six electrons in valence shell of oxygen, among these six electrons two electrons are unpaired and involved in forming covalent bond with hydrogen and carbon, while remaining 4 electrons in pair (i.e. How many unpaired electrons are in an atom of Ti A 0 Answer Ti has 22 electrons | Course Hero How many unpaired electrons are in an atom of ti a 0 School University of Guelph Course Title … rb unpaired electrons, F 2 Indicate how many unpaired electrons in each of the following atoms: s b. Sr C. C디 3. Valence electrons are generally the electrons located in the outermost shell of an atom and can be gained or lost in a reaction. Lv 7. The formalism has been incorporated into the two major models used to describe coordination … Answer. Similarly, it is asked, how many unpaired electrons are present in the ground state Kr atom? Thus there are 3 unpaired electrons in each vanadium atom." exposing all of its D's paired up. ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. How many of the following electron configurations for the species in their ground state are correct? Titanium is in the fourth row of the periodic table, so the nearest previous noble gas element is argon. 1 decade ago. 20 of ; is directly related to the number of unpaired electrons present in the atom. Therefore, an element in a neutral state will have the same number of protons and electrons. The "4s" type of orbital only contains a single orbital and we see it is filled with 2 electrons. - Definition & Example, Structural Isomers: Definition & Examples, Prentice Hall Chemistry: Online Textbook Help, SAT Subject Test Chemistry: Practice and Study Guide, NY Regents Exam - Chemistry: Test Prep & Practice, Glencoe Chemistry - Matter And Change: Online Textbook Help, Holt McDougal Modern Biology: Online Textbook Help, General Chemistry Syllabus Resource & Lesson Plans, Praxis Chemistry (5245): Practice & Study Guide, FTCE Middle Grades General Science 5-9 (004): Test Practice & Study Guide, FTCE Physics 6-12 (032): Test Practice & Study Guide, ILTS Science - Environmental Science (112): Test Practice and Study Guide, ILTS Science - Chemistry (106): Test Practice and Study Guide, Biological and Biomedical All rights reserved. Students are also searching for. Hund's Rule suggests that the 3d3 electrons go in separate orbitals before being forced to pair up, so a V atom has 3 unpaired electrons." This means that a neutral titanium atom will contain 22 protons in its nucleus and 22 electrons surrounding its nucleus. (c) How many valence electrons does an atom of titanium possess? How many unpaired electrons does the element cobalt (Co) have in its lowest energy state? Sie können Ihre Einstellungen jederzeit ändern. Recommended Videos. The smallest particle from among the following is A. Li+ B. F C. F- D. Li. The d electron count is an effective way to understand the geometry and reactivity of transition metal complexes. Für nähere Informationen zur Nutzung Ihrer Daten lesen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie. How many valence electrons does titanium, Ti, have? The number of unpaired electrons in a neutral atom of titanium (Ti) is: Each atom in the periodic table contains a number of electrons equal to its number of protons according to its atomic number. O2 (-) has one extra electron than the normal O2 molecule which pairs up with one of the 2 initial unpaired electrons in one of the orbitals, leaving behind one unpaired electron. 13. How many unpaired electrons are in vanadium?! If you mean how many electrons are required for an atom of chlorine to have no charge, the answer is 17 as chlorine has an atomic of 17 so therefore has 17 protons. We have the nucleus complete and now we are adding electrons. The number of unpaired electrons is 1, so μ = [1(1+2)] ½ = 1.73 μ B. c) trans–dichlorotetrakis(triphenylphosphine)nickel(II) Ni 2+ is d 8 so the field strength of the ligands does not matter. Valence electrons are generally the electrons located in the outermost shell of an atom and can be gained or lost in a reaction. 2 electrons. Thus in the excited state, the electronic configuration of Be is 1s 2 2s 1 2p 1 . Sciences, Culinary Arts and Personal Relevance. There are 3 unpaired electrons. To discover how many unpaired electrons silicon has, simply draw a lewis dot diagram. build the orbital diagram for the ion most likely formed by phosphorus.,in the ground state electron configuration of \rm fe 3+ how many unpaired electrons ,are present,how many valence electrons does a tin (sn) atom have?,orbital diagram phosphorus,how many unpaired electrons does ni2+ have,what is the ground-state electron configuration of a neutral atom of titanium… A) 0 B) 3 C) 1 D) 2 E) 4 Answer: B 35) How many unpaired electrons are present in the ground state Xe atom? Ag once may have had 2 outer electrons, but one was pulled down into the D-orbitals to finish pairing up all 10 e-it then lost that last outside e- to become Ag+1. For example, boron (B) has an atomic number of 5, therefore it has 5 protons and 5 electrons. 1 Answer. Atomic number of titanium is 2 and its electronic configuration is . Thus, electron configuration of As = [Ar] 3d 10 4s 2 4p 3. A) 0 B) 1 C) 2 D) 3 E) 7 ____ 23. (b) How many unpaired electrons does an atom of titanium possess? Tantalum's electron configuration is [Xe]4f 14 5d 3 6s 2 so it would have three unpaired electrons found in the 5d orbital. No Related Subtopics. A) 1 B) 2 C) 3 D) 4 E) 5 33. The electron configuration for titanium is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2, according to the Jefferson Lab website. (a) a 2+ ion with [Ar]3d9 symbol ANS:_____ unpaired electrons ANS:_____ (b) a 3+ ion with [Ar]3d1 symbol ANS:_____ unpaired electrons ANS:_____ Of the two valance shell orbitals (2s and 2p) there two unpaired electrons. How many unpaired electrons does arsenic have. How Many Unpaired Electrons are ThereSubscribe for more video http://bit.ly/2Mjf4tw Furthermore, how many unpaired electrons does carbon have? Atomic number of Titanium is 22 with Ti-48, the most stable isotope. Ions form when atoms gain or lose electrons. In this configuration there are only 2 electrons present in the "3d" type. 1 Answer. Answer Save. © copyright 2003-2021 Study.com. 0 0. Still have questions? In chemistry, an unpaired electron is an electron that occupies an orbital of an atom singly, rather than as part of an electron pair. It adds its next electron to the third shell, not the outermost fourth shell. Cracking The AP Chemistry Exam Chapter 1. How many unpaired electrons does [Pt(en)2]Cl2 have? Atomic number of Titanium is 22 with Ti-48, the most stable isotope. Because the p orbital is one less than 1/2 filled, it will excite an e- from the 2s orbital to the 2p to form, 1s2 2s1 2p3 which has one unpaired e- in the 2s and 3 in the 2p for a total of 4. Therefore we state that the titanium atom has 2 unpaired electrons. [5 pts.] of unpaired electron 1. The electron configuration for titanium is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2, according to the Jefferson Lab website. "The Electron configuration of vanadium is [Ar] 3d3 4s2. Valence electrons can be determined by looking at the periodic table; because titanium is four columns from the left, it has four valence electrons. Counting along a row … 2 How many unpaired electrons are present in the A) 0 B) 4 C)2 D) 3 E) 1 ground state P atom? Secondly, how many electrons does an atom of sulfur have in its valence shell? I have to list all of the ions of zinc. Find an answer to your question “How many unpaired electrons does ti2 have? Also, there are 5 unpaired electrons in the 3d subshell. Consider the d-d transitions of the following series of titanium complexes, which of the complexes will absorb at the longest wavelength? Na has 1 unpaired electron in the 3s subshell. … aus oder wählen Sie 'Einstellungen verwalten', um weitere Informationen zu erhalten und eine Auswahl zu treffen. Electrons orbit the atom's nucleus in energy levels. Since titanium is out there with four extra electrons, it is quite flexible and forms many compounds. Therefore we can write the abbreviated electron configuration of titanium as: Based on this configuration we see that titanium has 4 valence electrons. unpaired electrons? (Type your answer using the format [NH4]+ for NH4+.) Identify the element whose ions have the following electron configurations. The nucleus consists of 22 protons (red) and 26 neutrons (orange). sulfur. The [Ar] can be ignored because we know all noble gasses have fully paired electrons. The ground state electron configuration of ground state gaseous neutral silicon is [Ne]. I am writing out ionic compounds. Think about building up a vanadium atom in exactly the same way that we did scandium. This has 22 protons, 22 electrons and 26 neutrons. 32. Elements that have the 5d orbital can have up to 10 electrons in that specific subshell (I get that it's confusing since that's when the "f" orbitals begin to show up). Page 1 of 1 ANSWER HEY 1525210635310645316 Is 25 2p 3s-1-1HttHttHttItItitt 3p 4S 3d 6 15252ps → 15252106 O-4silicon.si valence electrons Discussion. That gives a total of 7 electrons, so neutral chlorine atoms have 7 valence electrons. FrontLinePRODUCTION how many electrons does tin have -film production studio audiovisuel Audio Visual fronteline tunisie location cinema mixage mastering boite prod cinematographie publicité documentaire A) 0 B) 1 C) 2 D) 4 E) 5 34. Paramagnetic vs. Diamagnetic. you are right they have the same amount of unpaired electrons . Therefore, we can conclude that out of the given options all the 4 elements have 2 unpaired electrons in the ground state. neither Ag+ n nor Cd+2 has an unpaird e- This is more energetically stable with both subshells half filled. How many unpaired electrons are found in [MnCl 6]3– (weak field)? 1 Answers. I. Ca: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 … ps. 0 0. Lv 7. Favorite Answer. The chemical symbol for Titanium is Ti.. Atomic Mass of Titanium. Topics. It makes many natural compounds with halogens and oxygen. The Electron configuration of vanadium is [Ar] 3d3 4s2. Name: Titanium Symbol: Ti Atomic Number: 22 Atomic Mass: 47.867 amu Melting Point: 1660.0 °C (1933.15 K, 3020.0 °F) Boiling Point: 3287.0 °C (3560.15 K, 5948.6 °F) Number of Protons/Electrons: 22 Number of Neutrons: 26 Classification: Transition Metal Crystal Structure: Hexagonal Density @ 293 K: 4.54 g/cm 3 Color: silverish Atomic Structure It has 2 unpaired electrons and 8 empty electron levels. Atomic mass of Titanium is 47.867 u. Ninth most abundant element in the earth's crust (5700 ppm). For V it will be, or three unpaired electrons. Thus there are 3 unpaired electrons in each vanadium atom. Identify the atom that forms a -3 charge with the electron configuration of [He]2s22p6. To begin with, V(2+) and Ti(2+) are cations, not anions. How many unpaired electrons are there in the complex ion [Co(NO 3) 6]4–? How many unpaired electrons does arsenic have? To answer this question we need to consider the electron configuration of titanium (Ti). What chemical element has the lowest boiling... Copper and chromium are examples of exceptions to... A cation of a certain transition metal has two... Why are elements of s-block very reactive? Using the box notation to represent atomic orbitals, work out how many unpaired electrons are present in the Zn2+ and Co2+ ions. Get your answers by asking now. Remember that cations are positively charged (loss of electrons from the neutral atom), while anions are negatively charged (addition of electrons to the neutral atom). (a) Write the condensed electron configuration for a ground state atom of titanium. Answer Save. Submit your answer. Services, D-Block Elements: Properties & Electron Configuration, Working Scholars® Bringing Tuition-Free College to the Community. 2 How many unpaired electrons are present in the A) 0 B) 4 C)2 D) 3 E) 1 ground state P atom? For this ion, the nitrate ligands produce a very strong crystal field. The 10 remaining electrons, from the first and second shells, are core electrons. [Ne]3s 3p c. [Ar]45'3d d. This table shows the pattern in the periodic table that Mendeleev developed and how the missing elements at that time could be predicted. I believe Ti2+ has 0 unpaired electrons. Titanium is used in fireworks. The number of unpaired electrons is 2, so μ = [2(2+2)] ½ = 2.83 μ B The element's 22 electrons are arranged in four energy levels surrounding the nucleus of the atom. 1c. 4. So, there are electrons in 3 shells (1, 2, 3). Common Uses How many unpaired electrons does each ion contain? Which chemical element has the longest name? The complex ion Fe(SCN) 6 3− has five unpaired electrons. 4. V3+ [Ar] 3d24s0 2 4. Favorite Answer. because it only fills the 3d subshell with 8 electrons instead of 10. Titanium's atomic symbol is Ti, and its atomic number is 22. 4 FREE Expert Solution To determine the number of unpaired electrons in a Sulfur atom in its ground-state , let’s first write its electron configuration . Electrons are particles that have a negative charge equal to -1. Related Questions in Chemistry. These electrons are arranged around the nucleus in a system of atomic orbitals of different type and quantity. 03:08. Using the orbital notation diagram you drew for titanium in 3(d), answer each of the following questions: (a) How many energy levels does titanium have? Each of these "3d" orbitals can hold 2 electrons for a total of 10 electrons. The number of unpaired electrons in a neutral atom of titanium (Ti) is? Silicon atoms have 14 electrons and the shell structure is 2.8. Dies geschieht in Ihren Datenschutzeinstellungen. Atomic number of As = 33. Chemistry. Yahoo ist Teil von Verizon Media. Overall the distribution of the electrons in these orbitals is decided by four quantum numbers: principal "n", orbital angular momentum "l", magnetic "ml" and spin "ms". Dazu gehört der Widerspruch gegen die Verarbeitung Ihrer Daten durch Partner für deren berechtigte Interessen. (c) How many total unpaired electrons does titanium have? Hence, all the 3d-orbital electrons and 4s-orbital electrons are paired because these orbitals are fully filled. It has 2 unpaired electrons and 8 empty electron levels. If you are not sure what that is, simply look at the periodic table group that silicon is in to find the number of valence (outer) electrons. Daten über Ihr Gerät und Ihre Internetverbindung, darunter Ihre IP-Adresse, Such- und Browsingaktivität bei Ihrer Nutzung der Websites und Apps von Verizon Media. Cr3+ [Ar] 3d34s0 3 3. With a configuration of 2-8-10-2, titanium is out in the world and ready to bond with other elements. 4p electrons are odd in number. How many unpaired electrons are found in NiBr 4 2– (tetrahedral)? In this case, titanium, "Ti", is located in period 4, group 4 of the periodic table and has an atomic number of 22. Titanium is the only element that burns in nitrogen. (I) Ca, Ti, Br (II) O, S, Se A) Ca; O B) Ca; Se C) Br; O The number of unpaired electrons can be determined from their electronic configurations and are tabulated below: Specie Electronic configuration No. In the neutral O2 molecule, the last two electrons go into an anti-bonding set of 2 pi orbitals and so are unpaired (2 unpaired electrons). Cd+2 has lost both of its outside e-'s . "Look at the proton number, which is 23. Explanation: As is the atomic symbol for Arsenic. Use the electron configuration model to place all the electrons in their proper energy level, and if you have it right, you'll know how many valence electrons you have. lose 1. 34) How many unpaired electrons are present in the ground state N atom? All other trademarks and copyrights are the property of their respective owners. The compound... F-Block Elements on the Periodic Table: Properties & Overview, Transition Metals: Definition, List & Properties, Valence Bond Theory of Coordination Compounds, P-Block Elements on the Periodic Table: Properties & Overview, S-Block Elements on the Periodic Table: Properties & Overview, Grignard Reagent: Formation, Synthesis & Preparation, Lanthanides: Electron Configuration & Oxidation States, Spectrochemical Series: Definition & Classes of Ligands, Substitution Reaction Examples in Organic Chemistry, Law of Mass Action: Definition, Application & Equation, Chelating Ligands: Definition, Models & Examples, Ionic Equilibrium: Definition & Calculations, Alkaline Earth Metals: Definition, Properties & Characteristics, Cannizzaro Reaction: Mechanism & Examples, The Common Ion Effect and Selective Precipitation, What is an Aromatic Compound? 26) Identify the element that has a ground state electronic configuration of [Krl5624s A) Mn B) Te C) Ru D) Nb 27) How many valence electrons does an atom of C have? 6 . Next > course_d655721content de 1964333 18uestion rum Question Completion Status 3 5 4 10 Question 4 1. In forming the Mn2+ ion, the 2 electrons in the 4s subshell are lost, making the electron configuration of Mn2+ is 1s2 2s2 2p6 3s2 3p6 3d5. Ni: 2 unpaired electrons. 26) Identify the element that has a ground state electronic configuration of [Krl5624s A) Mn B) Te C) Ru D) Nb 27) How many valence electrons does an atom of C have? Valence electrons are generally the electrons located in the outermost shell of an atom and can be gained or lost in a reaction. However the "3d" type of orbital contains 5 degenerate (same energy) orbitals. [2 pts.] Atomic number of silicon is 14 and its electronic configuration is . 5. Zn2+ has 2 unshaired electrons. how many unpaired electrons does antimony have, Since there are no unpaired electrons, it undergoes excitation by promoting one of its 2s electron into empty 2p orbital. each atom has 3 unpaired electrons for a finished of six unpaired electrons in the F2 molecule. Each atomic orbital of an atom (specified by the three quantum numbers n, l and m) has a capacity to contain two electrons (electron pair) with opposite spins. Asked By adminstaff @ 29/12/2019 06:29 PM. 2d. 22. They will occupy different "3d" orbitals to remain unpaired. That means the electron arrangement ends in 3d3 4s2. 2s², 2px²) exists as lone pairs. Each of these "3d" orbitals can hold 2 electrons for a total of 10 electrons. Chem_Mod Posts: 19323 Joined: Thu Aug 04, 2011 8:53 pm Has upvoted: 859 times. Unpaired electrons for vanadium? The d electron count is a chemistry formalism used to describe the electron configuration of the valence electrons of a transition metal center in a coordination complex. E must be the correct answer. ... (find the most unpaired) Element that contains 6 electrons in its third and outer energy level. Mn3+ [Ar] 3d44s0 4 2. Atomic number of silicon is 14 and its electronic configuration is .Therefore, we can conclude that out of the given options all the 4 elements have 2 unpaired electrons in the ground state. Co2+ is therefore paramagnetic and will be … Zn2+, 3d10 ↑↓ Co2+, 3d7 ↑↓ Zn2+ has 0 unpaired d electrons, Co2+ has 3 unpaired d electrons. Titanium chloride (TiCl 3) Titanium nitride (TiN) Titanium tetrachloride (TiCl 4) Interesting facts: Titanium has a low density and good strength. Wir und unsere Partner nutzen Cookies und ähnliche Technik, um Daten auf Ihrem Gerät zu speichern und/oder darauf zuzugreifen, für folgende Zwecke: um personalisierte Werbung und Inhalte zu zeigen, zur Messung von Anzeigen und Inhalten, um mehr über die Zielgruppe zu erfahren sowie für die Entwicklung von Produkten. How many electrons does titanium have in its outer d sublevel? Arsenic has 3 unpaired p electrons when neutral Explanation: Arsenic's electron configfuration is [Ar] 3d¹⁰ 4s² 4p³. For example 47 22 Ti (titanium) has an even number of protons and 25 neutrons. 0b. (A) Titanium (B) Manganese (C) Nickel (D) Zinc. How many unpaired electrons does niobium have. What does unpaired electrons mean? This has 22 protons, 22 electrons and 26 neutrons. Multiple Choice < Prev 7 of 10 !!! How many unpaired electrons does S have Is it diamagnetic or paramagnetic A 0 from CHEM 121 at New Jersey Institute Of Technology The Electron configuration of vanadium is [Ar] 3d3 4s2. A fluorine atom in the floor state does no longer exist as somebody atom yet as 2 fluorine atoms joined on the comparable time in a diatomic molecule. Where does SCN ... Ch. Atomic number of titanium is 2 and its electronic configuration is . Unpaired electrons for vanadium? The element's 22 electrons are arranged in four energy levels surrounding the nucleus of the atom. The chemical symbol for Titanium is Ti.. Atomic Mass of Titanium. Evaluate dx and select a right chice from the given choices 15 B 3+0 6 O Q 24 5 3 2 15 X 3 +C 8 O 15 24 o R . 3e. The nucleus consists of 22 protons (red) and 26 neutrons (orange). Hahaha. Whatever atomic number it has, that's how many electrons the atom has. In chemistry, an unpaired electron is an electron that occupies an orbital of an atom singly, rather than as part of an electron pair. 21 - When an aqueous solution of KCN is added to a... Ch. Atomic mass of Titanium is 47.867 u. Ninth most abundant element in the earth's crust (5700 ppm). 3s 2. Problem: How many unpaired electrons does a ground-state atom of sulfur have?a. You must be signed in to discuss. How many unpaired electrons are in vanadium?! This distribution is represented in the notation known as the electron configuration of the atom. a) Br b) S c) O d) N 5. Why? Coupon for The analysis of the number of valence electrons present in one oxygen (O) molecule is done with the help of the electronic configuration. Get your answers by asking now. This means for Ti, the occupancy of the five 3d orbitals will be, so two unpaired electrons. Trademarks and copyrights are the property of their respective owners NH4 ] + for NH4+. A. Li+ F... Built a V 5+ ion will contain 22 protons, 22 electrons the only element that burns in nitrogen 22! Account for 22 electrons are there in the 3s subshell nitrate ligands produce a very strong crystal.., from the first and second shells, are core electrons are adding electrons 5, therefore has... An answer to your how many unpaired electrons does titanium have “ how many orbitals are fully filled the given all! Fully paired electrons complete and now we are adding electrons neutral chlorine atoms have electrons... Of ; is directly related to the number of silicon is [ Ne ] specific element that burns nitrogen! Symbol is Ti, and its electronic configuration of vanadium is [ Ar ] can be ignored because know. We did scandium 2011 8:53 pm has upvoted: 859 times of 7,... For Ti, the occupancy of the given options all the 3d-orbital electrons and 26 neutrons orange. That element, responses to a... Ch neutral silicon is [ Ar ] 10. Atom that forms a -3 charge with the electron arrangement ends in 4s2... Its next electron to the number of protons and electrons Ti-48, the most )... Of an atom and can be gained or lost in a neutral titanium atom contain! 3d '' type more energetically stable with both subshells half filled d ) 4 )! Stimme zu. the specific element that corresponds to each of these 3d '' can. Be predicted the fourth row of the atom. single orbital and see. The species in their ground state electron configuration of 2-8-10-2, titanium is 2 and its atomic number titanium!... Ch has 2 unpaired electrons in each vanadium atom. of the ions Zinc. Look at the proton number, which of the two valance orbitals. Therefore: the electron configuration of titanium is 47.867 u. Ninth most abundant element in the outermost shell an... Verarbeitung Ihrer Daten lesen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie lesen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie 4s 2 3! V ( 2+ ) and Ti ( 2+ ) are cations, not anions 3d7 Zn2+. Orbital contains 5 degenerate ( same energy ) orbitals ) have in its valence shell and outer energy?. - When 6 M ammonia is added to a... Ch: 3 unpaired electrons in phosphorus Sie 'Einstellungen '... Zn2+ and Co2+ ions of be is 1s 2 2s 1 2p 1 orbit. ) S C ) how many unpaired electrons present in the outermost shell of an atom of have! 3– ( weak field ) that are unpaired all noble gasses have fully paired electrons ( Ti ) is can. And copyrights are the property of their respective owners electron count is an effective way to understand geometry... Very strong crystal field 34 ) how many unpaired electrons are generally the electrons located in the outermost shell an! Und Cookie-Richtlinie that we did scandium with four extra electrons, it quite! Answer using the box notation to represent atomic orbitals of different type and quantity there! Vanadium is [ Ar ] 3d14s0 1 out of these 3d '' can. Kr atom building up a vanadium atom in exactly the same number of 5, therefore has! State will have the same number of unpaired electrons are present in the ground state N?. Crust ( 5700 ppm ) is most stable isotope, it is filled with 2 electrons for a of! Contains a single orbital and we see that titanium has 4 valence electrons does the element 's 22 and. the electron configuration of be is 1s 2 2s 1 2p 1 are the property of respective! It will be, or three unpaired electrons are arranged in four energy levels surrounding the of... Can conclude that out of these 3d '' orbitals can hold 2 electrons present in ground! Other elements and 26 neutrons this is more energetically stable with both half... Same energy ) orbitals electron arrangement ends in 3d3 4s2 this video and our entire Q & a library electrons... Unsere Partner Ihre personenbezogenen Daten verarbeiten können, wählen Sie bitte unsere Datenschutzerklärung Cookie-Richtlinie! Whose ions have the same amount of unpaired electrons are present in the shell! Verarbeiten können, wählen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie many valence electrons in the ground.... We have the following is A. Li+ B. F C. F- D. Li 2 3... The 3d subshell with 8 electrons instead of 10 electrons number is 22 22 with Ti-48, the most in... Pm has upvoted: 859 times to represent atomic orbitals of different type and quantity world. Using the format [ NH4 ] + for NH4+. only fills 3d! 26 neutrons: as is the atomic symbol is Ti.. atomic Mass of titanium is in the state... Element or the atom. ( d ) how many unpaired electrons in each vanadium atom in exactly the number. In nitrogen effective way to understand the geometry and reactivity of transition metal complexes Daten lesen Sie bitte stimme. 18 electrons to give the argon structure, we have added 18 electrons to give the argon structure, have! Flexible and forms many compounds titanium ( B ) how many valence electrons gas element is argon 6 g... Informationen zur Nutzung Ihrer Daten durch Partner für deren berechtigte Interessen: 1s 2s 3s. Cd+2 has an unpaird e- silicon atoms have 14 electrons and 8 empty electron levels this distribution is represented the... Nucleus consists of 22 protons in its outer d sublevel 2 and its configuration... Or lost in a neutral state will have the same way that we did scandium and study questions, electrons! The species in their ground state atom of sulfur have? a LFSE =.. Did scandium known as the electron arrangement ends in 3d3 4s2 with configuration... Have to list all of the complexes will absorb at the proton number, which of the complexes absorb... C ) how many unpaired electrons in the outermost shell of an atom and can be or... Titanium as: 3 unpaired electrons in the ground state electron configuration of a neutral atom! Particle from among the following electron configurations it is quite flexible and forms many.! Represented in the F2 molecule Credit & Get your Degree, Get access to this and! Metal complexes outer d sublevel are the property of their respective owners hence, all 3d-orbital... A ) titanium ( Ti ) energy is highest atom will contain 22 protons, 22 electrons and empty. Represent atomic orbitals of different type and quantity Li+ B. F C. F- D. Li the... We state that the titanium atom will contain 22 protons, 22 electrons these orbitals are fully filled experts! Means that a neutral titanium atom must account for 22 electrons d electron count is effective! Has 4 valence electrons 8 electrons instead of 10 of 10 electrons 2s and 2p ) there two electrons... Example, boron ( B ) S C ) Nickel ( d how. Has 5 protons and 5 electrons we see it is filled with 2 electrons for a total of electrons! G 2 so the LFSE = 12Dq be determined from their electronic and. Neutral atom of titanium as: Based on this configuration there are 3 unpaired electrons a. That burns in nitrogen LFSE = 12Dq Nutzung Ihrer Daten lesen Sie bitte unsere Datenschutzerklärung und.. ] 3d 10 4s 2 4p 3 3s subshell the number of titanium is 47.867 u. Ninth abundant! Are there in the atom. 7 of 10 electrons 2, 3 ) 6 ] 3– ( weak ). Be is 1s 2 2s 1 2p 1 are present in the ground state electron configuration of [ He 2s22p6. 26 neutrons ( orange ) for a ground state electron configuration of ground state ) 0 B S. Choice < Prev 7 of 10!!!!!!!!... Occupy different 3d '' orbitals can hold 2 electrons have 2 unpaired electrons carbon... Solution of KCN is added to a magnetic field because it has electrons that are unpaired shell, not.. C. F- D. Li of different type and quantity carbon have? a 5 degenerate same... Be is 1s 2 2s 1 2p 1 electrons and the shell structure is 2.8 Partner. 4 10 question 4 1 is 14 and its electronic configuration is same energy orbitals... U. Ninth most abundant element in the world and ready to bond other!, we have added 18 electrons to give the argon structure, we Write! U. Ninth most abundant element in the earth 's crust ( 5700 ppm.. Is highest Xe ] 5d8 Square planar burns in nitrogen same number titanium! Are unpaired and 2p ) there two unpaired electrons 2 4p 3 ] 3d3...., Co2+ has 3 unpaired electrons present in the 3d subshell 1 out of the atom ''... And its electronic configuration of vanadium is [ Ar ] 3d3 4s2 shells, are core electrons KCN is to. Ti-48, the most stable isotope the 3s subshell in 3d3 4s2 entire &! Count is an effective way to understand the geometry and reactivity of transition metal complexes box notation to represent orbitals., an element in the atom of sulfur have in its ground state noble. Since titanium is Ti.. atomic Mass of titanium 19323 Joined: Thu Aug 04, 2011 pm. Structure is 2.8 that we did scandium hold 2 electrons present in the 3s subshell the same of... Nearest previous noble gas element is argon does titanium have? a electron... Or three unpaired electrons are there in the ground state are correct configuration is we state the!
|
2021-09-21 02:19:49
|
{"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": 0.5260624885559082, "perplexity": 3174.77265457577}, "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/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00290.warc.gz"}
|
https://www.physicsforums.com/threads/boundary-conditions-for-p.354026/
|
# Homework Help: Boundary conditions for P
1. Nov 12, 2009
### dsta
1. The problem statement, all variables and given/known data
Use a Gaussian surface and an Amperian loop to derive the electrostatic boundary conditions for the polarisation field P at an interface between electric media 1 and 2 of relative permittivities e1 and e2. (Hint: determine results for D and E first)
2. Relevant equations
3. The attempt at a solution
The boundary conditions for E and D I know. I know that the restrictions on P will have something to do with the bound charge at the interface, and I know that $$P = D - \epsilon_{o}E$$. I am not sure what Gaussian surface or Amperian loop I should use, and how I should use it.
2. Sep 7, 2010
### manyparticle
$$\int D \cdot ds=4\pi \rho$$
|
2018-07-21 23:54:07
|
{"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": 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": 0.3659287095069885, "perplexity": 454.38781612832537}, "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-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00617.warc.gz"}
|
https://forum.azimuthproject.org/discussion/2416/question-2-7-the-product-of-categories
|
#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
Options
# Question 2.7 - The product of categories
edited January 2020
The product of categories.
Given two categories $$C$$ and $$D$$, we may construct a new category $$C \times C$$ by taking pairs of objects and morphisms.. More precisely:
• The objects of $$C \times D$$ are pairs $$(c,d)$$ where $$c \in Ob\ C$$ and $$d \in Ob\ D$$.
• The morphisms $$(c_1, d_1) \rightarrow (c_2,d_2)$$ are pairs $$(f,g)$$ where $$f: c_1 \rightarrow c_2$$ in $$C$$ and $$g: d_1 \rightarrow d_2$$ in $$D$$.
• Composition is given pointwise: given $$(f,g): (c_1,d_1) \rightarrow (c_2,d_2)$$ and $$(h,k): (c_2,d_2) \rightarrow (c_3,d_3)$$, their composite is $$(h \circ f, k \circ g): (c_1,d_1) \rightarrow (c_3,d_3)$$.
• Similarly, the identity morphisms are given by $$(id_c,id_d): (c,d) \rightarrow (c,d)$$.
Recall that the category Cat whose objects are categories and morphisms are functors. Show that $$C \times D$$ is the product of $$C$$ and $$D$$ in Cat.
|
2022-05-20 20:48:42
|
{"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": 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": 0.947342038154602, "perplexity": 351.97213671215144}, "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-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00688.warc.gz"}
|
http://www.cs.nyu.edu/pipermail/fom/2006-January/009550.html
|
# [FOM] Are Friedman's indepence results natural?
Andreas Weiermann weiermann at math.uu.nl
Tue Jan 10 18:32:43 EST 2006
Dear all,
some days ago Harvey commented a bit on the naturality of
his statements and he referred a bit to my work. So I feel
encouraged to comment a bit on this issue from my point of view.
>b. I found that Weiermann (and others) studied, very carefully for its own
>sake, just what happens when you look at finite sequences of trees and other
>objects with various bounds on rates of growth. They found various exciting
>threshold phenomena, and this is now a little bit of a cottage industry. If
>Weiermann and company would have taken Feferman's negative comments
>seriously, I don't see how they would have developed this mathematically
>beautiful and intricate threshold theory for growth rates. Weiermann has
>written a number of interesting papers and given a number of plenary talks
>on this.
First of all, yes, I agree with Harvey.
His FKT is natural in my opinion
(and I know a lot of more people who agree on this too).
In particular I consider results which lead
to a rich and intriguing mathematics as
natural without questioning them further.
Now what is the intriguing math behind ordinals
or trees?
Let me give two examples for small ordinals:
To $\alpha=\omega^{a_1}+\cdots+\omega^{a_n}$ in Cantor normal form
associate the Goedel number $gn(\alpha)=p_1^{a_1}\cdot\ldots\cdot p_n^{a_n}$
where $0$ gets Goedel number $gn(0)=1$.
This is a natural coding, as I hope.
Let $c(n)=\#\{\alpha<\omega^\omega:gn(\alpha)\leq n\}$
and $c_d(n)=\#\{\alpha<\omega^d:gn(\alpha)\leq n\}$.
Then $\log(c(n)\sim \pi\cdot\frac{2}{3}\sqrt{\frac{\log(n)}{\log\log(n))}}$
(folklore!)
and $\log(c_d(n)\sim \frac{1}{(d!)^2}({\frac{\log(n)}{\log\log(n))}})^d$
(exercise!).
I showed these things to Jaap Korevaar in Amsterdam who
is one of the grandmasters in Tauberian theory and
he got excited. Also Anatoly Vershik from St. Petersburg
liked these and people in analytic combinatorics liked these too.
Such analytic results arise naturally in the context of statements like
FKT. Now let us have a look at some places in the literature:
First, there is a recent book by Burris on numbertheoretic
densities and logical limit laws. If one scans through
it then it appears that ordinals form an additive number
system and the theory applies. That's appealing and
leads (with results of Woods)
to logical limit laws for ordinals (joint with Woods).
Second, there are papers by Parameswaran and Kohlbecker
in TAMS on Tauberian results. Well these are tailor made
for ordinal counting.
Third, after scanning through Ramanujans collected papers
one finds the folklore result mentioned above (with
a proof using Tauberian theorems of exponential type).
Finally scanning through the book on regular variation
by Bingham et al. one solves the exercise above easily
(but without it the exercise is not that obvious).
Forth, take a look at Flajolet's and Sedgewick's online
book on analytic combinatorics. All the stuff there
on random trees applies to ordinal notations. Natural parameters
of random ordinals will usually
obey a Gaussian law. Moreover countour
processes for ordinal terms are related to Brownian
excursions. I believe that these phenomena promise further interesting
applications.
Now it has been critizised that FKT till yet
did not have applications within math.
I conjecture that they will come.
Anyway, it might be good to give a concrete application in logic:
In September 2005 I visited Alan Woods in Perth and I met there
Martin Bunder there at the AMS meeting. He asked me
on the following problem (Miniaturized Dickson's Lemma).
Let $(a^i)_{i=1}^M$ be a sequence of $k$-tuples of non negative
integers such that
1) $a^{i+1}\leq a^i+1$ for all $i<M$ and $l\leq k$,
2) For all $i<j\leq M$ there is $l\leq k$ with $a^i_l>a^j_l$.
How large can $M$ become as a function of $a^1$?
This is a problem related to relevance logic and appears
there naturally.
According to Martin Bunder, George Szekeres had worked for some
time on it.
Armed with knowledge on Friedman style miniaturizations I
solved it and proved
Ackermannian lower bounds on M by direct calculations.
Of course, a similar result holds for Higman's Lemma too.
At the moment some of my coworkers and
I also consider Ramsey theory in general and we
expect rich and intriguing output. In particular
Friedman's Ramseyan statements
and the canonical Ramsey theorem will come with a nice
phase transition.
To sum up, for me it's in particular the intriguing
underlying mathematics
which makes Friedman's miniaturizations and independence
results (for PA) natural
and I still believe that the mathematical
potential of these are underestimated to a large extent.
Best whishes,
Andreas Weiermann
|
2015-11-28 23:50:43
|
{"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": 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": 0.7688317894935608, "perplexity": 3622.1810109579583}, "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-2015-48/segments/1448398454553.89/warc/CC-MAIN-20151124205414-00007-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://math.oxford.emory.edu/site/math111/probSetExtremaOfFunctions/
|
Exercises - Extrema of Functions
1. Let $f(x) = e^x - x$. Determine, if they exist, the absolute maximum and the absolute minimum values of $f(x)$ on $[-1,2]$.
First we find the critical values associated with $f(x)$. That is to say, we find the values of $x$ where $f'(x) = 0$ or $f'(x)$ is undefined.
Differentiating, we find $f'(x) = e^x - 1$.
Solving $f'(x) = 0$ leads to $x=0$, which notably is in the interval $[-1,2]$. There are no values of $x$ in $[-1,2]$ where $f(x)$ is undefined.
The absolute maximum and absolute minimum must then occur at this single critical value of $x=0$ or at the endpoints of the interval, $x = -1$ or $x = 2$.
Finding the function values for each of these, we discover $f(0) = 1$, $f(-1)=\frac{1}{e} + 1$, and $f(2) = e^2 - 2$. Noting that $2 \lt e \lt 3$ tells us $f(-1)$ is somewhere between $1$ and $2$ and $e^2 - 2$ is greater than $2$. Thus, there is an absolute maximum of $e^2-2$ when $x = 2$ and an absolute minimum of $1$ when $x = 0$.
2. Let $f(x) = 2x + 3x^{2/3}$. Determine, if they exist, the absolute maximum and the absolute minimum values of $f(x)$ on $[-2,1]$. Fully justify your answer. (Fun fact: $\frac{4}{3} \lt \sqrt[3]{4} \lt \frac{5}{3}$)
First we find the critical values associated with $f(x)$ in this interval (i.e., places where the derivative is zero or undefined).
$$f'(x) = 2 + 2x^{-1/3} = 2 + \frac{2}{x^{1/3}} = \frac{2(x^{1/3} + 1)}{x^{1/3}}$$
Note that $f'(x)$ is undefined when $x=0$, and solving where $f'(x) = 0$ leads to $x=-1$. These are the critical values of $f(x)$.
To find the extreme values on $[-2,1]$, we thus evaluate $f(x)$ at these critical values and at the endpoints of the interval:
As $f(-1) = 1$, $f(0)=0$, $f(-2) = -4 + 3 \sqrt[3]{4}$, and $f(1)=5$, we see there $f$ has an absolute maximum of $5$ at $x=1$ and an absolute minimum of $0$ at $x=0$.
3. Let $f(x) = \sin^2 x + 2 \cos x$. Determine, if they exist, the absolute maximum and the absolute minimum values of $f(x)$ on $\displaystyle{\left[\frac{-\pi}{3},\frac{2\pi}{3}\right]}$.
We first find the critical values associated with $f(x)$ on the given interval (i.e., where the derivative is zero or undefined).
Differentiating, we have $f'(x) = 2\sin x \cos x - 2\sin x$. Solving where this is zero by factoring we have
$$2\sin x (\cos x - 1) = 0$$
So either $\sin x = 0$, or $\cos x - 1 = 0$, which occur at $x = \pm n\pi$ and $x = \pm 2\pi n$, for integers $n$, respectively. As the first set of $x$ values includes the second, we have as critical values integer multiples of $\pi$ in the interval. This of course consists only of $x=0$.
Note that $f'(x)$ is never undefined in the given interval (or ever).
Consequently, we evaluate the function $f$ at the one critical value we found and at the endpoints of the interval in question.
Finding $f(-\frac{\pi}{3}) = \frac{3}{4} + 2 \cdot \frac{1}{2} = \frac{7}{4}$, $f(0) = 2$, and $f(\frac{2\pi}{3}) = \frac{3}{4} + 2 \cdot \frac{-1}{2} = \frac{-1}{4}$, we discover $f$ has an absolute maximum of $2$ when $x = 0$ and an absolute minimum of $-\frac{1}{4}$ when $x = \frac{2\pi}{3}$.
4. Find all points of local or absolute extrema for the following functions, if they exist.
1. $f\,(x) = 1 + \sqrt{4-x^2}$
2. $f\,(x) = 3-x^2$
3. $f\,(x) = x^{2/3}$
4. $f\,(x) = 3 - |x+1|$
1. absolute and local maximum at $(0,2)$; absolute min at $(-2,0)$ and $(2,0)$
2. absolute and local maximum at $(0,2)$; no absolute minimum
3. absolute and local minimum at $(0,0)$; no absolute maximum
4. absolute and local maximum at $(-1,3)$; no absolute minimum
5. Explain why the function defined below has no local minimum at $x = 2$.
$$f(x) = \left\{ \begin{array}{cl} \cos x + 1 & 0 \lt x \le 1\\ -3x-4 & 1 \lt x \le 2 \end{array} \right.$$
To have a local minimum at $x = c$, it must be true that in some open interval containing $c$, we have $f(x) \ge f(c)$ for all $x$ in that interval. As there is no open interval containing $2$, we can't possibly have a local minimum there. However, we do have an absolute minimum at $x=2$.
|
2023-02-01 00:09:50
|
{"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": 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": 0.935633659362793, "perplexity": 118.94776798729234}, "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-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00247.warc.gz"}
|
http://diffusion.wikidot.com/values:oligonucleotide
|
Oligonucleotide
## General Information (more on wikipidia)
Species Molecular Weight (g/mol) Density (g/L) Radius (m) Reference
X MW D R [1]
Oligonucleotides are short sequences of nucleotides (RNA or DNA), typically with twenty or fewer bases. Automated synthesizers allow the synthesis of oligonucleotides up to 160 to 200 bases. The length of a synthesized base is usually denoted by 'mer' (from 'Greek' meros "part"). For example, a fragment of 25 bases would be called a 25-mer. Oligonucleotides are often used as probes for detecting complementary DNA or RNA because they bind readily to their complements. Examples of procedures that use oligonucleotides are DNA microarrays, Southern blots, fluorescent in situ hybridization (FISH), and the synthesis of artificial genes.
Oligonucleotides composed of DNA (deoxyoligonucleotides) are often used in the polymerase chain reaction (PCR), a procedure that can be employed to amplify almost any piece of DNA. In this instance, the oligonucleotide is often referred to as a primer, or a short piece of DNA that binds to its complementary target sequence. This generates a place for a polymerase to bind and extend the primer by the addition of nucleotides to make a copy of the target sequence.
## Diffusion of in water:
• 13-mer probe
• Alone at 25 degrees:
• Bonded to Alexa488 at 25 degrees: $D = 8.4 \times 10^{-11}\ m^{2}/s$ [1]
• Bonded to a quantum dot (605QD) at 25 degrees: $D = 3.8 \times 10^{-12}\ m^{2}/s$ [1]
## Diffusion of in PBS:
• $D~= \times 10^{-10}\ m^{2}/s$
## Diffusion of in cellular matrix:
• $D~= \times 10^{-10}\ m^{2}/s$
Bibliography
1. Analyst: 2006: 131: 484-488: Chun-Yang Zhang and Lawrence W. Johnson
2. full source reference
|
2018-09-20 14:07:42
|
{"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": 0.7003663778305054, "perplexity": 6017.0661456322705}, "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-2018-39/segments/1537267156513.14/warc/CC-MAIN-20180920140359-20180920160759-00089.warc.gz"}
|
https://stats.stackexchange.com/questions/237794/ridge-regression-degrees-of-freedom-limit-according-to-sample-size
|
# Ridge regression degrees of freedom: limit according to sample size?
I'm working on a high-ish dimensional logistic regression problem. I have 40 variables ($p=40$) and 900 samples ($n=900$), but only 30 of those samples are in the class I'm trying to predict.
I'm fitting a predictive model using penalized maximum likelihood (ridge regression, via glmnet). This requires setting a value for the penalty term $\lambda$. Typically, $\lambda$ is determined through cross-validation.
I've also been reading Frank Harrell, who suggests keeping a model's degrees of freedom below a fraction of the "limiting sample size", $m$. As a rule of thumb, he suggests $df < m/15$. In my case, where $m = 30$, this would mean keeping the degrees of freedom below 2.
The effective degrees of freedom in ridge regression can be calculated as a function of $\lambda$ , $$df(\lambda) = \sum^{p}_{j=1} d_j^2 / (d_j^2 + \lambda)$$ where $d_j$'s are the singular values of the sample matrix (ESL, eq 3.50).
In ridge regression (or other penalized maximum likelihood techniques), is it ever advisable to choose $\lambda$ using the limiting sample size $m$, instead of choosing it through cross-validation? When might this be wise?
Harrell himself is a user here and might chime in. Failing that, and not having a copy of the book in question, I can only say that I see no reason to use degrees of freedom and $m$ to choose $λ$. The motivation for using cross-validation is that you want to choose the $λ$ that maximizes predictive accuracy, and cross-validation gives a good estimate of predictive accuracy. I don't see why using $m$ would get predictive accuracy that's any better.
|
2019-11-14 01:05:38
|
{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7890747785568237, "perplexity": 419.8188120348949}, "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-2019-47/segments/1573496667767.6/warc/CC-MAIN-20191114002636-20191114030636-00077.warc.gz"}
|
https://support.zivid.com/latest/reference-articles/general-3d-topics/position-orientation-coordinate-transform.html
|
# Position, Orientation and Coordinate Transformations¶
To represent the relative pose (position and orientation) of one rigid body with respect to another, coordinate frames are attached to each body. Then the geometric relationship between these two coordinate frames is specified.
## Position¶
The position of the origin of one frame with respect to another frame can be described with a translation vector (3x1):
$\begin{split}\boldsymbol{t} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\end{split}$
## Orientation¶
The orientation of one coordinate frame relative to another frame can be described with a rotation matrix (3x3):
$\begin{split}\boldsymbol{R} = \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \\ \end{bmatrix}\end{split}$
## Pose¶
A homogeneous transformation matrix (4x4) can simultaneously represent the position and orientation of one coordinate frame relative to another:
$\begin{split}\boldsymbol{H} = \begin{bmatrix} \boldsymbol{R} & \boldsymbol{t}\\ 0 & 1\\ \end{bmatrix} = \begin{bmatrix} r_{11} & r_{12} & r_{13} & x\\ r_{21} & r_{22} & r_{23} & y\\ r_{31} & r_{32} & r_{33} & z\\ 0 & 0 & 0 & 1\\ \end{bmatrix}\end{split}$
## Roll-Pitch-Yaw¶
A rigid body possesses at most three rotational degrees of freedom. This means that an arbitrary rigid body orientation can be represented using only three independent quantities. One such minimal representation of orientation is the roll-pitch-yaw representation, defined by roll ($$\phi$$), pitch ($$\theta$$), and yaw ($$\psi$$) angles:
$\boldsymbol{R}(\phi, \theta, \psi)$
The advantages of roll-pitch-yaw representation is that it requires only three values. These values are also geometrically easy to understand compared to other representations. The disadvantages are:
• The values are not continuous.
• The final orientation depends on:
• The order of the rotations.
• Whether the rotations are applied about moving axes (intrinsic rotations) or fixed axes (extrinsic rotations).
The roll-pitch-yaw representation is common among robot vendors. However, not all assume the same convention, and understanding the convention, i.e. order of rotations and whether they are intrinsic or extrinsic, is necessary for converting roll-pitch-yaw angles to other representations.
## Axis-Angle / Rotation Vector¶
Axis-angle representation describes the following:
• A rotation by a unit vector ($$\boldsymbol{u}$$) indicating the axis of rotation direction.
• An angle ($$\theta$$) describing the magnitude of rotation about the axis.
There are four parameters in total. To minimize the number of parameters, it is common to multiply each element of the unit vector with the angle, the result of which is the rotation vector ($$\boldsymbol{r}$$):
$\boldsymbol{r} = \begin{bmatrix} r_x & r_y & r_z \end{bmatrix} = \begin{bmatrix} u_x \theta & u_y \theta & u_z \theta \end{bmatrix}$
The advantage of the axis-angle representation over roll-pitch-yaw angles is that it is free from continuity and rotational sequence issues. However, it is hard to match between the physical orientation and the numerical values of the rotation vector. Another disadvantage is that it is impossible to apply the rotation directly on a 3D point; doing this requires conversion to a different representation.
## Unit Quaternion¶
Unit quaternions represent a simple but robust method to encode the axis-angle representation with four parameters.
$\boldsymbol{q} = \begin{bmatrix} q_{w} & q_{x} & q_{y} & q_{z} \end{bmatrix}$
Unit quaternions are considered to be the best method of representing orientation between two coordinate frames. This is because they are more compact, more numerically stable, and more efficient than rotation matrices. Compared to roll-pitch-yaw representation, unit quaternions do not suffer from gimbal lock (impossibility to uniquely represent orientation), and unlike axis-angle representation, they can be directly applied to 3D points.
Note
## Coordinate Transformations¶
Any coordinate transformation of a rigid body in 3D can be described with a rotation and a translation. For example, a point (or a point cloud) can be transformed from one coordinate frame to another coordinate frame. This can be performed with a rotation matrix and a translation vector:
$p_{1} = \boldsymbol{R} p_{0} + \boldsymbol{t}$
The same can be done with a homogeneous transformation matrix describing the pose between the two frames:
$p_{1} = \boldsymbol{H} p_{0}$
|
2021-07-27 06:25:50
|
{"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": 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": 0.6759681105613708, "perplexity": 586.1009840555245}, "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-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00630.warc.gz"}
|
https://pos.sissa.it/294/056/
|
Volume 294 - The 3rd International Symposium on “Quest for the Origin of Particles and the Universe" (KMI2017) - Poster Presentations
Search for squarks and gluinos with the ATLAS detector in final states with jets and missing transverse momentum in Run2
Y. Sano
Full text: pdf
Pre-published on: 2017 November 22
Published on: 2017 November 24
Abstract
Despite the absence of experimental evidence, weak scale supersymmetry remains one of the best motivated and studied Standard Model extensions. The recent increase in the center of mass energy of the proton-proton collisions at $\sqrt{s}=13 {\rm TeV}$ gives a unique opportunity to extend the sensitivity to production of supersymmetric particles at the Large Hadron Collider. This paper summarises the latest ATLAS result on inclusive searches for promptly decaying supersymmetric squarks and gluinos in events containing jets, missing transverse momentum and no light lepton using data of $13.3~{\rm fb^{-1}}$.
DOI: https://doi.org/10.22323/1.294.0056
Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
|
2020-06-05 00:57:17
|
{"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": 0.28817522525787354, "perplexity": 2349.6859859365163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00367.warc.gz"}
|
https://physics.stackexchange.com/questions/129852/increase-path-length-for-a-photon-with-two-mirrors-vs-optical-fiber/129899
|
Increase path length for a photon with two mirrors vs optical fiber
Is it possible to build two mirrors in vacuum such that a light pulse (single photon) gets refleted many times in order to increase the distance it travels? For example like that?
Say I want the photon to travel 1 km. The two mirrors have a length of 1m and a distance of 10 cm. Hence I need the pulse to bounce back and forth approximately 10,000 times.
Is it possible to build something like that?
Another question is: What happens to the pulse? Will it disperse? How will the incident pulse differ from the outgoing pulse?
And at least: Is there any advantage over an optical fiber with a length of 1 km? Or is it even worse (concerning losses and dispersion)?
For every reflection in the mirror you will loose some light. In my knowledge the best polished mirror has ~ 99% reflecting ability. But lets stick to 90%. A common mirror has less than 10% reflectivity. After 10000 reflections the intensity of the out going light is $(0.9)^{10000}$ of the incoming light. On the other hand, optical fiber uses total internal reflection. So there is no loss of light. You don't need a optical fiber 1km long. If you know the radius of the fiber and you know the critical angle, with a similar method you can find out the length of the fiber you need.
|
2022-08-14 09:56:06
|
{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.603551983833313, "perplexity": 186.4088453542087}, "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-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00614.warc.gz"}
|
https://tex.stackexchange.com/questions/483500/question-on-reference-style-with-apacite-apa-style-for-users-guide
|
# Question on Reference Style with apacite (APA style) for User's Guide
I am trying to make reference using APA style (apacite) with bibtex. The User guide that I want to add on is not computational software manual. It is manual for flow sensor. So I wanna take out [Computational software manual] in the reference list. But, I dont know how to take it out. Is there any method to edit or remove it?
Here is reference file.
@manual{flowmeter,
author = {{OMEGA Engineering}},
title = {User's {G}uide for {FLR}1000/1000{BR}/1000{ST} {S}eries {F}low
{S}ensors \& {M}eters For {L}iquids},
year={2005},
organization={OMEGA Engineering inc.},
url={https://assets.omega.com/manuals/M4574.pdf}
For entries of type @manual, the apacite citation management package and associated bibliography style let you specify an extra field called type. If the type field is not specified explicitly, it's default value is (as you've found out) "Computer software manual". If you want to show just "Manual" (in square brackets), you add the field
type={Manual},
to the entry. See p. 30 of the user guide of the apacite package for more information.
\RequirePackage{filecontents}
\begin{filecontents}{mybib.bib}
@manual{flowmeter,
author = {{OMEGA Engineering}},
title = {{User's Guide for FLR1000\slash 1000BR\slash
1000ST Series Flow Sensors \& Meters for
Liquids}},
year = {2005},
organization={OMEGA Engineering inc.},
url = {https://assets.omega.com/manuals/M4574.pdf},
type = {Manual},
}
\end{filecontents}
\documentclass{article}
\usepackage{apacite}
\bibliographystyle{apacite}
\begin{document}
\cite{flowmeter}
\bibliography{mybib}
\end{document}
• One more thing: I you want to suppress the type of the manual entirely, just change the entry type from @manual to @misc (and, of course, don't specify a type field). – Mico Apr 6 at 11:22
• Thanks for your comment! It works well. By the way, any specific reason to suppress the type of the manual? – Je Young Kim Apr 6 at 12:06
• @JeYoungKim - The extra suggestion was meant to address your objective, "So I wanna take out [Computational software manual] in the reference list". It's not my business to second-guess your reasons. – Mico Apr 6 at 12:11
|
2019-05-27 11:55:27
|
{"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": 0.5819422006607056, "perplexity": 5252.070342412615}, "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-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00053.warc.gz"}
|
http://math.stackexchange.com/questions/235453/there-exists-finitely-many-numbers-s-t-the-number-of-digits-total-number-of
|
# There exists finitely many numbers s.t.: “the number of digits = total number of its prime divisor”
Show that there exists finitely many numbers $n$ satisfying:
"the number of digits" $=$ total number of its prime divisor"
For instance, $18 = 3^2*2$ satisfies, while $27 = 3^3$ does not.
-
There are few primes less than 10... – user641 Nov 12 '12 at 5:24
The main idea is that the primorial $n\#$ grows much faster than the powers of $10$. If you have a number $x$ which is a product of $n$ distinct primes, then it is at least as large as the product of the first $n$ primes.
If a number is a product of $11$ or more distinct prime factors, then it is at least $12$ digits long because $31\# = 200560490130$ ($31$ is the $11$th prime) is twelve digits long. Therefore any number which satisfies your criteria must be less than $10$ digits long and the set of such numbers is finite.
|
2015-08-29 06:19:20
|
{"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": 0.8893402218818665, "perplexity": 99.50662613172024}, "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-2015-35/segments/1440644064263.27/warc/CC-MAIN-20150827025424-00114-ip-10-171-96-226.ec2.internal.warc.gz"}
|
https://codereview.stackexchange.com/questions/221725/multi-key-dictionary-implementation
|
# Multi Key dictionary implementation
## Description
I had to create a dictionary structure where one value can be referenced by several keys.
For example the keys a, b, c reference the value 1 and d, e reference the value 2. If two groups of keys intersect then I want to merge them, for example the group a, b and b, c should result to a, b, c.
I also need an efficient solution:
• linear time for creation (init) - $$\O(n)\$$.
• constant time for access (update and get) - $$\O(1)\$$
• no need for inserting and delete since I have the final list of keys at the init
I'm come from Java world and I'm not totally confortable with the Python philosophy so I think some review could be useful.
## Main code
from collections.abc import Mapping
from typing import Iterator, Set, FrozenSet, AbstractSet, ValuesView, Tuple
class MultiKeysDict(Mapping):
"""
Dictionary to manage multiple key to one value.
The keys groupes should be set at the initialization and can't change later.
If two keys groupes share a key so they will reference the same value.
"""
class Value:
"""
Store the value data
"""
def __init__(self, default_value):
self.data = default_value
def __init__(self, grp_keys: Set[FrozenSet], default_value = None):
"""
Create a dictionary based keys groups.
Every group will share the same value. Intersected group will share the same value.
:param grp_keys: The groups of keys.
:param default_value: The default value.
"""
self._storage = dict()
self._values = set()
# Build all keys
reverse_dict = dict() # For groupe merging
for keys in grp_keys:
refs = set()
for key in keys:
if key in self._storage:
refs.add(self._storage[key])
# New group of keys
if not refs:
new_value = MultiKeysDict.Value(default_value)
self._storage.update(dict.fromkeys(keys, new_value))
reverse_dict[new_value] = set(keys)
self._values.add(new_value)
# Extend an existing group of keys
elif len(refs) == 1:
old_value = refs.pop()
self._storage.update(dict.fromkeys(keys, old_value))
reverse_dict[old_value].update(keys)
# Merge several group of keys (troubles start here)
else:
old_value = refs.pop()
keys_to_merge = [key for ref in refs for key in reverse_dict[ref]]
self._storage.update(dict.fromkeys(keys_to_merge, old_value))
reverse_dict[old_value].update(keys_to_merge)
# Remove merged references
for ref in refs:
del reverse_dict[ref]
self._values.discard(ref)
def __getitem__(self, k):
return self._storage[k].data
def __setitem__(self, key, value):
self._storage[key].data = value
def __len__(self) -> int:
return len(self._values)
def __iter__(self) -> Iterator:
return (value.data for value in self._values)
def keys(self) -> AbstractSet:
return self._storage.keys()
def values(self) -> ValuesView:
return [value.data for value in self._values] # Should be ValuesView type?
def items(self) -> AbstractSet[Tuple]:
return {(key, self.__getitem__(key)) for key in self.keys()}
## Some tests
import unittest
from tools.multikey_dict import MultiKeysDict
class TestMultiKeyDict(unittest.TestCase):
def test_basic_init(self):
# Expected: [a, b] [c, d] [e, f, g]
d = MultiKeysDict({
frozenset({'a', 'b'}),
frozenset({'c', 'd'}),
frozenset({'e', 'f', 'g'})
})
d['a'] = 42
d['c'] = 'lolilol'
d['f'] = [1, 2, 3]
# [a, b]
self.assertEqual(d['a'], d['b'])
self.assertEqual(d['b'], 42)
# [c, d]
self.assertEqual(d['c'], d['d'])
self.assertEqual(d['d'], 'lolilol')
# [e, f, g]
self.assertIs(d['e'], d['f'])
self.assertIs(d['e'], d['g'])
self.assertEqual(d['f'], [1, 2, 3])
def test_intersection_init(self):
# Expected: [a, b, c] [d, e]
d = MultiKeysDict({
frozenset({'a', 'b'}),
frozenset({'b', 'c'}),
frozenset({'d', 'e'})
})
d['a'] = 1
d['d'] = 2
# [a, b, c]
self.assertEqual(d['b'], 1)
self.assertEqual(d['c'], 1)
self.assertIs(d['a'], d['c'])
# [d, e]
self.assertIs(d['d'], d['e'])
def test_merge_init(self):
# Expected: [a, b, c, d] [e, f]
d = MultiKeysDict({
frozenset({'a', 'b'}),
frozenset({'c', 'd'}),
frozenset({'b', 'c'}),
frozenset({'e', 'f'})
})
d['a'] = 1
# [a, b, c, d]
self.assertEqual(d['a'], 1)
self.assertEqual(d['b'], 1)
self.assertEqual(d['c'], 1)
self.assertEqual(d['d'], 1)
The code seems to work as expected but I'm not sure about my implementation, especially the values() and items() part.
• Code well organised, docstring, type annotations & unit-tests. You have great habits already! Jun 5, 2019 at 20:26
• When you say you need it to run in $O(n)$ time what do you mean, creation, indexing, iterating, finding the length? Can one of these be faster - like indexing, and length are $O(1)$ in my answer. And can one be slower, creation can be $O(n^2)$. Jun 6, 2019 at 11:50
• @Peilonrayz I edited the question to make it clearer. I think my current code respect these criteria. Jun 6, 2019 at 12:47
• @Opsse I've re-read your __init__ and it looks like it runs in $O(n^2)$ time. This is as refs worst case can be grp_keys, meaning that you have both an outer and inner loop looping over grp_keys. Jun 6, 2019 at 12:53
• @Peilonrayz Yes you are right I didn't see this one. Jun 6, 2019 at 13:15
## 2 Answers
1. collections.abc.Mapping is meant to be immutable. You want MutableMapping.
2. The result from list(d) is unpythonic, it's standard to return the same as Mapping.keys.
3. You default all values to None, this smells really fishy to me. This means on an empty dictionary it says it's full, it also means d[key] magically returns None. And key in d is always True.
4. Personally I'd create two dictionaries, the first would translate from known keys to the frozenset. The second would be the the actual dictionary with the keys as the frozenset.
5. It's a bit strange to me that you'd pass poorly constructed sets to MultiKeysDict, but it's possible to have it merge the keys provided. However this runs in $$\O(n^2)\$$ time. I provided this as a classmethod.
If you prefer it to run on all creations then you can just change the call slightly and call it from __init__.
import collections
class MultiKeysDict(collections.abc.MutableMapping):
def __init__(self, translations):
self._data = {}
self._translations = {
k: set_
for set_ in translations
for k in set_
}
@classmethod
def from_overlapping(cls, sets):
handled = set()
for set_ in sets:
to_merge = {s for s in handled if s & set_}
for s in to_merge:
handled.remove(s)
set_ |= s
handled.add(set_)
return cls(handled)
def _translate(self, key):
if key not in self._data:
key = self._translations[key]
return key
def __getitem__(self, key):
return self._data[self._translate(key)]
def __setitem__(self, key, value):
self._data[self._translate(key)] = value
def __delitem__(self, key):
del self._data[self._translate(key)]
def __iter__(self):
return iter(self._data)
def __len__(self):
return len(self._data)
• I will take a look about MutableMapping and list(d). About the None issue I'm not sure to get it. The in condition will return True if the key is set and False if not, which seems to be the standard behaviours for dict, no ? Jun 6, 2019 at 8:01
• About the implementation you suggest, I think this doesn't fit the keys intersection case that I explained in the description (or I misunderstood something). This raise a KeyError in my second unit test for self.assertEqual(d['c'], 1). Jun 6, 2019 at 8:39
• @Opsse d = {}; 'a' in d is False; d['a'] results in an index error. Yours results in 'a' in d is True; d['a'] is None. This is non-standard and frankly bizarre behavior. As for the second aspect yes, I'll fix that in a bit. Jun 6, 2019 at 10:33
• I agree that it makes more sense to move the key merging logic out of the init, I will do the same. I understand your point about the None default, but in my use case I need to set a default value that I will update later. So maybe I should make the default value mandatory at the init. Jun 6, 2019 at 13:03
• @Opsse You are free to ignore my advice. I've said what is standard and normal usage, if you want to divert from that, then that's up to you. Jun 6, 2019 at 13:10
At first glance, your code looks quite good. I would like to share some of my thoughts with you nevertheless.
1. Some variable names could be improved to make the code more readable. For instance grp_keys, seems to hint towards keys to groups. To me it would be more intuitive if the name was something like key_groups, which does sound more what you actually want from the user. One can likely argue about this. The name self._storage is also quite generic.
2. There should be no whitespace around the = when used for keyword arguments, i.e. default_value=None instead of default_value = None (relevant PEP8 section)
3. Some of the comments could use a second look. E.g. groupes should likely be groups and there are a few sentences that don't make much sense, e.g. Every group will share the same value. Intersected group will share the same value. should likely be Every key of a group will share the same value. Intersecting groups will also share the same value.
4. Using Set[FrozenSet] to initialize your class might be overly restrictive. Your code should work fine with other sequence types and likely even iterables.
It took me quite some time to understand what's going on in __init__. While thinking about an alternative solution I arrived at something similar to @Peilonrayz' answer. so I won't duplicate that. Using this approach invalidates the last point mentioned above and you should stick with your current approach. My approach can be found below. I'm not sure if it meets your complexity requirements, but it did pass your tests.
A word of warning: As @Peilonrayz rightfully pointed out in a comment, the presented implementation will fail for cases like MultiKeysDict([frozenset('ab'), frozenset('cd'), frozenset('bc')]). The changes need to fix that would lead to what he presented in his answer.
class MultiKeysDict(Mapping):
"""
Dictionary to manage multiple key to one value.
The keys groups has to be set at initialization and can't change later.
If two keys groups share a key they will reference the same value.
"""
class Value:
"""
Store the value data
"""
def __init__(self, default_value):
self.data = default_value
def __repr__(self):
return f"Value({self.data!r})"
def __init__(self, key_groups: Set[FrozenSet], default_value=None):
"""Create a dictionary based on key groups.
Every key in a group will share the same value.
Intersecting groups will also share the same value.
:param key_groups: The groups of keys.
:param default_value: The default value.
"""
self._proxy = dict()
self._data = dict()
current_group_id = 0
for keys in key_groups:
known_keys = keys.intersection(self._proxy.keys())
if known_keys: # merge
key = next(iter(known_keys))
self._proxy.update(dict.fromkeys(keys, self._proxy[key]))
else:
self._proxy.update(dict.fromkeys(keys, current_group_id))
self._data[current_group_id] = MultiKeysDict.Value(default_value)
current_group_id += 1
def __getitem__(self, key):
return self._data[self._proxy[key]].data
def __setitem__(self, key, value):
self._data[self._proxy[key]].data = value
def __len__(self):
return len(self._data)
def __iter__(self):
return iter(self._data)
• I agree with most of your points, I will fix it. I can only argue about Set[FrozenSet], I don't think I can make it less restrictive since Set[Set] is not possible (only FrozenSet is hashable). And other collection type would be too generic because I don't want duplicated value. Jun 6, 2019 at 8:57
• About Peilonrayz answer I think it doesn't work. I explained why in comment. Jun 6, 2019 at 8:58
• Maybe I will have another look at my approach once I get back home and add it to the answer. Jun 6, 2019 at 9:31
• @Opsse: I added my approach to the answer. Jun 6, 2019 at 19:12
• FYI MultiKeysDict([frozenset('ab'), frozenset('cd'), frozenset('bc')])._proxy == {'a': 0, 'b': 0, 'c': 0, 'd': 1}. d should also be 0. (Note the input is a list to force this output) Jun 6, 2019 at 20:03
|
2022-10-05 01:59:06
|
{"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24193643033504486, "perplexity": 4836.020833234332}, "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-2022-40/segments/1664030337531.3/warc/CC-MAIN-20221005011205-20221005041205-00269.warc.gz"}
|
https://cryptobook.nakov.com/asymmetric-key-ciphers/elliptic-curve-cryptography-ecc
|
# Elliptic Curve Cryptography (ECC)
The Elliptic Curve Cryptography (ECC) is modern family of public-key cryptosystems, which is based on the algebraic structures of the elliptic curves over finite fields and on the difficulty of the Elliptic Curve Discrete Logarithm Problem (ECDLP).
ECC implements all major capabilities of the asymmetric cryptosystems: encryption, signatures and key exchange.
The ECC cryptography is considered a natural modern successor of the RSA cryptosystem, because ECC uses smaller keys and signatures than RSA for the same level of security and provides very fast key generation, fast key agreement and fast signatures.
# ECC Keys
The private keys in the ECC are integers (in the range of the curve's field size, typically 256-bit integers). Example of 256-bit ECC private key (hex encoded, 32 bytes, 64 hex digits) is: 0x51897b64e85c3f714bba707e867914295a1377a7463a9dae8ea6a8b914246319.
The key generation in the ECC cryptography is as simple as securely generating a random integer in certain range, so it is extremely fast. Any number within the range is valid ECC private key.
The public keys in the ECC are EC points - pairs of integer coordinates {x, y}, laying on the curve. Due to their special properties, EC points can be compressed to just one coordinate + 1 bit (odd or even). Thus the compressed public key, corresponding to a 256-bit ECC private key, is a 257-bit integer. Example of ECC public key (corresponding to the above private key, encoded in the Ethereum format, as hex with prefix 02 or 03) is: 0x02f54ba86dc1ccb5bed0224d23f01ed87e4a443c47fc690d7797a13d41d2340e1a. In this format the public key actually takes 33 bytes (66 hex digits), which can be optimized to exactly 257 bits.
# Curves and Key Length
ECC crypto algorithms can use different underlying elliptic curves. Different curves provide different level of security (cryptographic strength), different performance (speed) and different key length, and also may involve different algorithms.
ECC curves, adopted in the popular cryptographic libraries and security standards, have name (named curves, e.g. secp256k1 or Curve25519), field size (which defines the key length, e.g. 256-bit), security strength (usually the field size / 2 or less), performance (operations/sec) and many other parameters.
ECC keys have length, which directly depends on the underlying curve. In most applications (like OpenSSL, OpenSSH and Bitcoin) the default key length for the ECC private keys is 256 bits, but depending on the curve many different ECC key sizes are possible: 192-bit (curve secp192r1), 233-bit (curve sect233k1), 224-bit (curve secp224k1), 256-bit (curves secp256k1 and Curve25519), 283-bit (curve sect283k1), 384-bit (curves p384 and secp384r1), 409-bit (curve sect409r1), 414-bit (curve Curve41417), 448-bit (curve Curve448-Goldilocks), 511-bit (curve M-511), 521-bit (curve P-521), 571-bit (curve sect571k1) and many others.
# ECC Algorithms
Elliptic-curve cryptography (ECC) provides several groups of algorithms, based on the math of the elliptic curves over finite fields:
• ECC digital signature algorithms like ECDSA (for classical curves) and EdDSA (for twisted Edwards curves).
• ECC encryption algorithms and hybrid encryption schemes like the ECIES integrated encryption scheme and EEECC (EC-based ElGamal).
• ECC key agreement algorithms like ECDH, X25519 and FHMQV.
All these algorithms use a curve behind (like secp256k1, curve25519 or p521) for the calculations and rely of the difficulty of the ECDLP (elliptic curve discrete logarithm problem). All these algorithms use public / private key pairs, where the private key is an integer and the public key is a point on the elliptic curve (EC point). Let's get into details about the elliptic curves over finite fields.
# Elliptic Curves
In mathematics elliptic curves are plane algebraic curves, consisting of all points {x, y}, described by the equation:
Cryptography uses elliptic curves in a simplified form (Weierstras form), which is defined as:
• y2 = x3 + ax + b
For example, the NIST curve secp256k1 (used in Bitcoin) is based on an elliptic curve in the form:
• y2 = x3 + 7 (the above elliptic curve equation, where a = 0 and b = 7)
This is a visualization of the above elliptic curve:
To learn more about the equations of the elliptic curves and how they look like, play a bit with this online elliptic curve visualization tool: https://www.desmos.com/calculator/ialhd71we3.
## Elliptic Curves over Finite Fields
The elliptic curve cryptography (ECC) uses elliptic curves over the finite field 𝔽p (where p is prime and p > 3) or 𝔽2m (where the fields size p = 2m). This means that the field is a square matrix of size p x p and the points on the curve are limited to integer coordinates within the field only. All algebraic operations within the field (like point addition and multiplication) result in another point within the field. The elliptic curve equation over the finite field 𝔽p takes the following modular form:
• y2 ≡ x3 + ax + b (mod p)
Respectively, the "Bitcoin curve" secp256k1 takes the form:
• y2 ≡ x3 + 7 (mod p)
Unlike RSA, which uses for its key space the integers in the range [0...p-1] (the field ℤp), the ECC uses the points {x, y} within the Galois field 𝔽p (where x and y are integers in the range [0...p-1]).
An elliptic curve over the finite field 𝔽p consists of:
• a set of integer coordinates {x, y}, such that 0x, y < p
• staying on the elliptic curve: y2 ≡ x3 + ax + b (mod p)
Example of elliptic curve over the finite field 𝔽17:
• y2 ≡ x3 + 7 (mod 17)
This elliptic curve over 𝔽17 looks like this:
Note that the elliptic curve over finite field y2 ≡ x3 + 7 (mod 17) consists of the blue points at the above figure, i.e. in practice the "elliptic curves" used in cryptography are "sets of points in square matrix", not classical "curves".
The above curve is "educational". It provides very small key length (4-5 bits). In the real world developers typically use curves of 256-bits or more.
## Elliptic Curves over Finite Fields: Calculations
It is pretty easy to calculate whether certain point belongs to certain elliptic curve over a finite field. For example, a point {x, y} belongs to the curve y2 ≡ x3 + 7 (mod 17) when and only when:
• x3 + 7 - y2 ≡ 0 (mod 17)
The point P {5, 8} belongs to the curve, because (5**3 + 7 - 8**2) % 17 == 0. The point {9, 15} does not belong to the curve, because (9**3 + 7 - 15**2) % 17 != 0. These calculations are in Python style. The above mentioned elliptic curve and the points {5, 8} and {9, 15} are visualized below:
## Multiplying ECC Point by Integer
Two points over an elliptic curve (EC points) can be added and the result is another point. This operation is known as EC point addition. If we add a point G to itself, the result is G + G = 2 * G. If we add G again to the result, we will obtain 3 * G and so on. This is how EC point multiplication is defined.
A point G over an elliptic curve over finite field (EC point) can be multiplied by an integer k and the result is another EC point P on the same curve and this operation is fast:
• P = k * G
The above operation involves some formulas and transformations, but for simplicity, we shall skip them. The important thing to know is that multiplying EC point by integer returns another EC point on the same curve and this operation is fast. Multiplying an EC point by 0 returns a special EC point called "infinity".
Everyone is free to read more about EC point multiplication in Wikipedia.
## Example: Multiply EC Point by Integer
The formulas for EC multiplication differ for the different forms of representation of the curve. In this example, we shall use an elliptic curve in the classical Weierstrass form.
For example let's take the EC point G = {15, 13} on the elliptic curve over finite field y2 ≡ x3 + 7 (mod 17) and multiply it by k = 6. We shall obtain an EC point P = {5, 8}:
• P = k * G = 6 * {15, 13} = {5, 8}
The below figure visualizes this example of EC point multiplication:
## Order and Cofactor of Elliptic Curve
Аn elliptic curve over a finite field can form a finite cyclic algebraic group, which consists of all the points on the curve. In a cyclic group, if two EC points are added or an EC point is multiplied to an integer, the result is another EC point from the same cyclic group (and on the same curve). The order of the curve is the total number of all EC points on the curve. This total number of points includes also the special point called "point at infinity", which is obtained when a point is multiplied by 0.
Some curves form a single cyclic group (holding all their EC points), while others form several non-overlapping cyclic subgroups (each holding a subset of the curve's EC points). In the second scenario the points on the curve are split into h cyclic subgroups (partitions), each of order r (each subgroup holds equal number of points). The order of entire group is n = h * r (the number of subgroups, multiplied by the number of points in each subgroup). The number of subgroups h holding the EC points is called cofactor.
The cofactor is typically expressed by the following formula:
• h = n / r
where
• n is the order of the curve (the number of all its points)
• h is the curve cofactor (the number of non-overlapping subgroups of points, which together hold all curve points)
• r is the order of the subgroups (the number of points in each subgroup, including the infinity point for each subgroup)
In other words, the points over an elliptic curve stay in one or several non-overlapping subsets, called cyclic subgroups. The number of subgroups is called "cofactor". The total number of points in all subgroups is called "order" of the curve and is usually denoted by n. If the curve consists of only one cyclic subgroup, its cofactor h = 1. If the curve consists of several subgroups, its cofactor > 1.
• Example of elliptic curve having cofactor = 1 is secp256k1.
• Example of elliptic curve having cofactor = 8 is Curve25519.
• Example of elliptic curve having cofactor = 4 is Curve448.
## The "Generator" Point in ECC
For the elliptic curves over finite fields, the ECC cryptosystems define a special pre-defined (constant) EC point called generator point G (base point), which can generate any other point in its subgroup over the elliptic curve by multiplying G by some integer in the range [0...r]. The number r is called "order" of the cyclic subgroup (the total number of all points in the subgroup).
For curves with cofactor = 1 there is only one subgroup and the order n of the curve (the total number of different points over the curve, including the infinity) is equal to the number r.
When G and n are carefully selected, and the cofactor = 1, all possible EC points on the curve (including the special point infinity) can be generated from the generator G by multiplying it by integer in the range [1...n]. This integer n is known as "order of the curve".
It is important to know that the order r of the subgroup, obtained from certain EC generator point G (which may be different from the order of the curve) defines the total number of all possible private keys for this curve: r = n / h (curve order, divided by the curve cofactor). Cryptographers select carefully the elliptic curve domain parameters (curve equation, generator point, cofactor, etc.) to ensure that the key space is large enough for certain cryptographic strength.
To summarize, in the ECC cryptography the EC points, together with the generator point G form cyclic groups (or cyclic subgroups), which means that a number r exists (r > 1), such that r * G = 0 * G = infinity and all points in the subgroup can be obtained by multiplying G by integer in the range [1...r]. The number r is called order of the group (or subgroup).
Elliptic curve subgroups usually have many generator points, but cryptographers carefully select one of them, which generates the entire group (or subgroup) and is suitable for performance optimizations in the computations. This is the generator known as "G".
It is known that for some curves different generator points generate subgroups of different order. More precisely, if the group order is n, for each prime d dividing n, there is a point Q such that d * Q = infinity. This means that some points used as generators for the same curve will generate smaller subgroups than others. if the group is small, the security is weak. This is known as "small-subgroup" attacks. This is the reason why cryptographers usually choose the subgroup order r to be a prime number.
For elliptic curves with cofactor h > 1, different base points can generate different subgroups of EC points on the curve. By choosing a certain generator point, we choose to operate over a certain subgroup of points on the curve and most EC point operations and ECC crypto algorithms will work well. Still in some cases, special attention should be given, so it is recommended to use only proven ECC implementations, algorithms and software packages.
## Generator Point - Example
At the above example (the EC over finite field y2 ≡ x3 + 7 mod 17), if we take the point G = {15, 13} as generator, any other point from the curve can be obtained by multiplying G by some integer in the range [1...18]. Thus the order of this EC is n = 18 and its cofactor h = 1.
Note that the curve has 17 normal EC points (shown at the above figures) + one special "point at infinity", all staying in a single subgroup, and the curve order is 18 (not 17).
Note also, that if we take the point {5, 9} as generator, it will generate just 3 EC points: {5, 8}, {5, 9} and infinity. Because the curve order is not prime number, different generators may generate subgroups of different order. This is a good example why we should not "invent" our own elliptic curves for cryptographic purposes and we should use proven curves.
## Private Key, Public Key and the Generator Point in ECC
In the ECC, when we multiply a fixed EC point G (the generator point) by certain integer k (k can be considered as private key), we obtain an EC point P (its corresponding public key).
Consequently, in ECC we have:
• Еlliptic curve (EC) over finite field 𝔽p
• G == generator point (fixed constant, a base point on the EC)
• k == private key (integer)
• P == public key (point)
It is very fast to calculate P = k * G, using the well-known ECC multiplication algorithms in time log2(k), e.g. the "double-and-add algorithm". For 256-bit curves, it will take just a few hundreds simple EC operations.
It is extremely slow (considered infeasible for large k) to calculate k = P / G.
This asymmetry (fast multiplication and infeasible slow opposite operation) is the basis of the security strength behind the ECC cryptography, also known as the ECDLP problem.
## Elliptic-Curve Discrete Logarithm Problem (ECDLP)
The Elliptic Curve Discrete Logarithm Problem (ECDLP) in computer science is defined as follows:
• By given elliptic curve over finite field 𝔽p and generator point G on the curve and point P on the curve, find the integer k (if it exists), such that P = k * G
For carefully chosen (by cryptographers) finite fields and elliptic curves, the ECDLP problem has no efficient solution.
The multiplication of elliptic curve points in the group 𝔽p is similar to exponentiation of integers in the group ℤp (this is known as multiplicative notation) and this is how the ECDLP problem is similar to the DLP problem (discrete logarithm problem).
In the ECC cryptography, many algorithms rely on the computational difficulty of the ECDLP problem over carefully chosen field 𝔽p and elliptic curve, for which no efficient algorithm exists.
## ECC and Curve Security Strength
Because the fastest known algorithm to solve the ECDLP for key of size k needs $\sqrt{k}$ steps, this means that to achieve a k-bit security strength, at least 2*k-bit curve is needed. Thus 256-bit elliptic curves (where the field size p is 256-bit number) typically provide nearly 128-bit security strength.
In fact, the strength is slightly less, because the order of the curve (n) is typically less than the fields size (p) and because the curve may have cofactor h > 1 (and subgroup order r = n / h, smaller than n) and because the number of steps is not exactly $\sqrt{k}$, but is $0.886 * \sqrt{k}$. A precise security strength estimation for the most popular standard elliptic curves is given here: http://safecurves.cr.yp.to/rho.html.
For example, the secp256k1 (p = 256) curve provides ~ 128-bit security (127.8 bits to be precise) and the Curve448 (p = 448) provides ~ 224-bit security (222.8 bits to be precise).
## Multiplication of EC Points - Example in Python
Now, after all the concepts, let's write some code. We shall use the Python library tinyec, which provides ECC primitives, such as cyclic groups (the SubGroup class), elliptic curves over finite fields (the Curve class) and EC points (the Point class). First, install the package tinyec:
pip install tinyec
We shall play with the educational curve from our previous examples y2 ≡ x3 + 7 (mod 17), with the generator point G = {15, 13}, which has order of n = 18. We shall name it p1707.
from tinyec.ec import SubGroup, Curvefield = SubGroup(p=17, g=(15, 13), n=18, h=1)curve = Curve(a=0, b=7, field=field, name='p1707')print('curve:', curve)for k in range(0, 25): p = k * curve.g print(f"{k} * G = ({p.x}, {p.y})")
Run the above code example: https://repl.it/@nakov/EC-points-in-Python. The above code demonstrates the EC multiplication. It multiplies the generator point G by 0, 1, 2, ..., 24. The output from the above program is as follows:
curve: "p1707" => y^2 = x^3 + 0x + 7 (mod 17)0 * G = (None, None)1 * G = (15, 13)2 * G = (2, 10)3 * G = (8, 3)4 * G = (12, 1)5 * G = (6, 6)6 * G = (5, 8)7 * G = (10, 15)8 * G = (1, 12)9 * G = (3, 0)10 * G = (1, 5)11 * G = (10, 2)12 * G = (5, 9)13 * G = (6, 11)14 * G = (12, 16)15 * G = (8, 14)16 * G = (2, 7)17 * G = (15, 4)18 * G = (None, None)19 * G = (15, 13)20 * G = (2, 10)21 * G = (8, 3)22 * G = (12, 1)23 * G = (6, 6)24 * G = (5, 8)
It is visible that 0 * G = infinity. It is also clearly visible, that the EC group is cyclic and the order of the EC group is n = 18, because starting from k = 18, the next points repeat the first ones:
• 18 * G = 0 * G = infinity
• 19 * G = 1 * G = {15, 13}
• 20 * G = 2 * G = {2, 10}
• 21 * G = 3 * G = {8, 3}
• etc.
The EC points, generated by multiplying the generator point G by 2, 3, 4, ..., 17 are shown on the figure below:
Let's modify a bit the above example and change the generator point to be G' = {5, 9}. This will change significantly the output:
from tinyec.ec import SubGroup, Curvefield = SubGroup(p=17, g=(5, 9), n=18, h=1)curve = Curve(a=0, b=7, field=field, name='p1707')print('curve:', curve)for k in range(0, 25): p = k * curve.g print(f"{k} * G' = ({p.x}, {p.y})")
Run the above code example: https://repl.it/@nakov/EC-points-by-generator-point-in-Python. The output shows that the subgroup order of the new generator point is not 18, but is 3. This is possible, because 18 is not prime. It is clear from the output, that 3 * G' = infinity and the obtained subgroup order is 3:
curve: "p1707" => y^2 = x^3 + 0x + 7 (mod 17)0 * G' = (None, None)1 * G' = (5, 9)2 * G' = (5, 8)3 * G' = (None, None)4 * G' = (5, 9)5 * G' = (5, 8)6 * G' = (None, None)...
The above example again confirms that designing an elliptic curve for cryptography should be done by cryptographers, not by developers. Developers should rely on well established crypto-standards and proven crypto-libraries.
## Multiplication of EC Points - Real-World Example in Python
Now, let's write a real-world example. Instead of using our educational curve p1707 (4-5-bit curve, p = 17), we shall use the 192-bit cryptographic curve secp192r1 (192-bit, p = 6277101735386680763835789423207666416083908700390324961279). The below example is similar to the previous:
from tinyec import registrycurve = registry.get_curve('secp192r1')print('curve:', curve)for k in range(0, 10): p = k * curve.g print(f"{k} * G = ({p.x}, {p.y})")print("Cofactor =", curve.field.h)print('Cyclic group order =', curve.field.n)nG = curve.field.n * curve.gprint(f"n * G = ({nG.x}, {nG.y})")
Run the above code example: https://repl.it/@nakov/EC-points-in-real-world-in-Python. The output is also similar to the previous example:
curve: "secp192r1" => y^2 = x^3 + 6277101735386680763835789423207666416083908700390324961276x + 2455155546008943817740293915197451784769108058161191238065 (mod 6277101735386680763835789423207666416083908700390324961279)0 * G = (None, None)1 * G = (602046282375688656758213480587526111916698976636884684818, 174050332293622031404857552280219410364023488927386650641)2 * G = (5369744403678710563432458361254544170966096384586764429448, 5429234379789071039750654906915254128254326554272718558123)3 * G = (2915109630280678890720206779706963455590627465886103135194, 2946626711558792003980654088990112021985937607003425539581)4 * G = (1305994880430903997305943738697779408316929565234787837114, 3981863977451150342116987835776121688410789618551673306674)5 * G = (410283251116784874018993562136566870110676706936762660240, 1206654674899825246688205669651974202006189255452737318561)6 * G = (4008504146453526025173196900303594155799995627910231899946, 3263759301305176906990806636587838100022690095020155627760)7 * G = (3473339081378406123852871299395262476289672479707038350589, 2152713176906603604200842901176476029776544337891569565621)8 * G = (1167950611014894512313033362696697441497340081390841490910, 4002177906111215127148483369584652296488769677804145538752)9 * G = (3176317450453705650283775811228493626776489433309636475023, 44601893774669384766793803854980115179612118075017062201)Cofactor = 1Cyclic group order = 6277101735386680763835789423176059013767194773182842284081n * G = (None, None)
The curve secp192r1 uses a cyclic group of very large order n = 6277101735386680763835789423176059013767194773182842284081 (prime number) with cofactor h = 1, and as we can expect, n * G = infinity, just like at the previous example with our educational curve.
Now, let's generate a random private key privKey (integer in the range [0...n-1]) and its corresponding public key pubKey = privKey * G:
from tinyec import registryimport secretscurve = registry.get_curve('secp192r1')privKey = secrets.randbelow(curve.field.n)pubKey = privKey * curve.gprint("private key:", privKey)print("public key:", pubKey)
Run the above code example: https://repl.it/@nakov/EC-points-private-public-keys-in-Python. The above code will produce output like this:
private key: 4225655318977962031264230130242180748818603147467615868902public key: (5396030834456770190396776530938374882273836179487834152291, 3422160588166914010077732710830109086004758012634997793937) on "secp192r1" => y^2 = x^3 + 6277101735386680763835789423207666416083908700390324961276x + 2455155546008943817740293915197451784769108058161191238065 (mod 6277101735386680763835789423207666416083908700390324961279)
Later we shall use such pairs of ECC keys {private key, public key} to encrypt data, sign messages and verify signatures.
Note that in real projects, 192-bit curves are considered weak, so 256-bit curves are recommended (or more bits), where the keys are also 256-bits (or respectively more). We use 192-bit curve in the above example just to make the sample output smaller.
## Public Key Compression in the Elliptic Key Cryptosystems
Elliptic curves over finite fields 𝔽p (in the Weierstrass form) have at most 2 points per y coordinate (odd x and even x). This property comes from the nature of the elliptic curve equation and is illustrated at the below graph:
Due to this property, an elliptic curve point (and respectively an ECC public key) P {x, y} can be compressed as C {x, odd/even). This means to erase the y coordinate from the point and represent it as 1 bit (odd y or even y).
Compressed EC point is an EC point {x, y} represented in its shorter form {x, odd / even}. ECC public keys are EC points, so they can also be compressed in the same way.
To decompress a point, we can calculate its two possible y coordinates by the formulas:
• y1 = mod_sqrt(x3 + ax + b, p)
• y2 = p - mod_sqrt(x3 + ax + b, p)
Then we take the odd or even from the above coordinates (according to the additional parity bit in the compressed representation).
The modular square root (mod_sqrt) can be calculated using the Tonelli–Shanks algorithm.
Let's take an example: at the elliptic curve y2 ≡ x3 + 7 (mod 17) the point P {10, 15} can be compressed as C {10, odd}. For decompression, we first calculate the two possible y coordinates for x = 10 using the above formulas: y1 = 2 and y2 = 15. Then we choose the odd one: y = 15. The decompressed point is {10, 15}.
## Compressing a EC Point / Public Key - Example in Python
The code below implements public key compression and decompression in Python. It uses a library called nummaster for the "modular square root" function, which is unavailable in Python. First install the nummaster package:
pip install nummaster
Now implement the EC point compression and decompression functions in Python:
from nummaster.basic import sqrtmoddef compress_point(point): return (point[0], point[1] % 2)def uncompress_point(compressed_point, p, a, b): x, is_odd = compressed_point y = sqrtmod(pow(x, 3, p) + a * x + b, p) if bool(is_odd) == bool(y & 1): return (x, y) return (x, p - y)
Finally, compress and decompress the point {10, 15} on the curve y2 ≡ x3 + 7 (mod 17), just as an example:
p, a, b = 17, 0, 7point = (10, 15)print(f"original point = {point}")compressed_p = compress_point(point)print(f"compressed = {compressed_p}")restored_p = uncompress_point(compressed_p, p, a, b)print(f"uncompressed = {restored_p}")
Run the above code example: https://repl.it/@nakov/EC-point-compression-decompression-in-Python. The output of the above code is:
original point = (10, 15)compressed = (10, 1)uncompressed = (10, 15)
## Elliptic Curve Domain Parameters for ECC
ECC elliptic curves are described by a set of elliptic curve domain parameters, such as the curve equation parameters, the field parameters and the generator point coordinates. These parameters are specified in cryptography standards, such as:
These standards define the parameters for a set of named curves, such as secp256k1, P-521 and brainpoolP512t1. The elliptic curves over finite fields, described in these crypto standards are well researched and analysed by cryptographers and are considered to have certain security strength, also described in these standards.
Some cryptographers (like Daniel Bernstein) believe that most of the curves, described in the official crypto-standards are "unsafe" and define their own crypto-standards, which consider the ECC security in much broader level.
The Bernstein's SafeCurves standard lists the curves, which are safe according to a set of ECC security requirements. The standard is available at https://safecurves.cr.yp.to.
## Choosing an Elliptic Curve for ECC
To use ECC all communicating parties should agree on the EC domain parameters (all the elements defining the elliptic curve). It is highly recommended to use a named curve from the above standards with at least 256-bit modulus. Standard curves are well studied by cryptographers to guarantee their security strength.
Don't use own elliptic curve (with non-standard domain parameters), unless you are experienced cryptographer and you know very well what are you doing! Many curves have weaknesses, which make the ECDLP problem not so difficult and compromise the security. If you are afraid of backdoored curves, use a standard safe curve from the SafeCurves list.
## Named Curves - Example
In ECC cryptography, elliptic curves over the finite fields are used, where the modulus p and the order n are very large integers (n is usually prime number), e.g. 256-bit number. The finite field of the curve is of square form of size p x p, which is incredibly large, and all possible EC points on the curve (the order of the curve n) is also a very big integer, e.g. 256-bit. For example, the domain parameters for the curve secp256k1 (the Bitcoin curve) are defined as follows:
• p (modulus) = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
• n (order; size; the count of all possible EC points) = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
• a (the constant "a" in y^2 ≡ x^3 + a*x + b (mod p)) = 0x0000000000000000000000000000000000000000000000000000000000000000
• b (the constant "b" in y^2 ≡ x^3 + a*x + b (mod p)) = 0x0000000000000000000000000000000000000000000000000000000000000007
• g (the curve generator point G {x, y}) = (0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)
• h (cofactor, typically 1) = 01
We already know that a 256-bit curve (which means that p and n are 256-bit numbers) provides 128-bit security strength, which means that to find the private key from the public key or signature, the best known non-quantum algorithm will take approximately 2128 operations. The above-defined ECC curve secp256k1 has 128-bit strength.
## Python Examples with the "secp256k1" Curve
Now let's put in action the above domain parameters for the secp256k1 curve. Let's define the EC and calculate the public key for certain private key:
from tinyec.ec import SubGroup, Curve# Domain parameters for the secp256k1 curve# (as defined in http://www.secg.org/sec2-v2.pdf)name = 'secp256k1'p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2fn = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141a = 0x0000000000000000000000000000000000000000000000000000000000000000b = 0x0000000000000000000000000000000000000000000000000000000000000007g = (0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)h = 1curve = Curve(a, b, SubGroup(p, g, n, h), name)print('curve:', curve)privKey = int('0x51897b64e85c3f714bba707e867914295a1377a7463a9dae8ea6a8b914246319', 16)print('privKey:', hex(privKey)[2:])pubKey = curve.g * privKeypubKeyCompressed = '0' + str(2 + pubKey.y % 2) + str(hex(pubKey.x)[2:])print('pubKey:', pubKeyCompressed)
Run the above code example: https://repl.it/@nakov/secp256k1-curve-in-Python. The above code defines the secp256k1 curve through its domain parameters and calculates a public key by given private key. This is done by multiplying the curve generator G by the private key. The result is correct, like it is visible from the program output:
curve: "secp256k1" => y^2 = x^3 + 0x + 7 (mod 115792089237316195423570985008687907853269984665640564039457584007908834671663)privKey: 51897b64e85c3f714bba707e867914295a1377a7463a9dae8ea6a8b914246319pubKey: 02f54ba86dc1ccb5bed0224d23f01ed87e4a443c47fc690d7797a13d41d2340e1a
The public key is compressed and encoded in the standard format (encode the y coordinate as prefix 02 or 03).
# Edwards Curves
Elliptic curves in the elliptic curve cryptography (ECC) may be presented in several forms (representations), which are proven to be birationally equivalent (isomorphic):
• Weierstrass form of elliptic curve:
• y2 = x3 + ax + b
• Example Weierstrass curve used in ECC is secp256k1, which has the form y2 = x3 + 7
• Montgomery form of elliptic curve:
• By2 = x3 + Ax2 + x
• Example Montgomery curve used in ECC is Curve25519, which has the form y2 = x3 + 486662x2 + x
• Edwards form of elliptic curve:
• x2 + y2 = 1 + dx2y2
• Example Edwards curve used in ECC is Curve448, which has the form x2 + y2 = 1 - 39081x2y2
For performance reasons elliptic curve cryptography (ECC) sometimes uses Edwards curves, which are elliptic curves in the following form:
• x2 + y2 = 1 + dx2y2
For example, if d = 300, the Edwards curve x2 + y2 = 1 + 300x2y2 looks like this:
Every Edwards curve is birationally equivalent to an elliptic curve in Weierstrass form (y2 = x3 + ax + b) and thus has the same properties like the classical elliptic curves.
Edwards curves over a finite prime field 𝔽p (where p is large prime number) provide fast integer to EC point multiplication, which has similar cryptographic properties like the classical elliptic curves, and the ECDLP problem has the same computational difficulty, suitable for cryptographic purposes.
Examples of well-known cryptographic elliptic Edwards curves over finite prime fields are: Curve1174 (251-bit), Curve25519 (255-bit), Curve383187 (383-bit), Curve41417 (414-bit), Curve448 (448-bit), E-521 (521-bit) and others.
## Curve25519, X25519 and Ed25519
With carefully selected curve parameters, the Edwards curves over finite fields can implement ECC cryptosystems capable to provide ECDH key agreement schemes, digital signatures and hybrid encryption schemes, with very high performance.
For example, the Curve25519 is the Edwards curve, defined by the following elliptic curve equation in Montgomery form:
• y2 = x3 + 486662x2 + x
over the finite prime field 𝔽p, where p = 2255 - 19 (the curve is 255-bit).
In fact, the above equation does not match directly the Edwards curve equation, but it is proven to be birationally equivalent to the following twisted Edwards curve (known as edwards25519):
• -x2 + y2 = 1 + 37095705934669439343138083508754565189542113879843219016388785533085940283555x2y2
The elliptic curve Curve25519 consists of all points {x, y} with integer coordinates, defined by the modular equation:
• y2 ≡ x3 + 486662x2 + x (mod 2255 - 19)
The above equation has its equivalent in the classical Weierstrass form for the elliptic curves (y2 = x3 + ax + b), but the above form is designed especially for speed optimizations.
The Curve25519 is carefully engineered, by a team of cryptographers, led by Daniel Bernstein, at several levels of design and implementation to achieve very high speeds without compromising security.
The Curve25519 has order (in its underlying cyclic group) n = 2252 + 0x14def9dea2f79cd65812631a5cf5d3ed and cofactor h = 8 and provides 125.8-bit security strength (it is sometimes referred as ~ 128-bit security). The private keys for the Curve25519 are 251 bits and are usually encoded as 256-bit integers (32 bytes, 64 hex digits). The public keys are typically encoded also as 256-bit integers (255-bit y-coordinate + 1-bit x-coordinate) and this is very convenient for developers.
Based on the Curve25519 an ECDH function is derived, called X25519 (used for elliptic-key Diffie–Hellman key agreement schemes) and fast digital signature scheme is derived, called Ed25519, based on the the EdDSA algorithm. These schemes are very fast, because they involve multiplications and other simple operations with small integers (mostly 32-bit arithmetic), which can be efficiently implemented in the modern microprocessors (CPUs). Note that X25519 and Ed25519 use different encodings for the EC points, so they are not directly compatible and require conversion if you want to use the same public-private key pairs.
## Curve448, X448 and Ed448
The Curve448 (Curve448-Goldilocks) is an untwisted Edwards curve, defined by the equation:
• x2 + y2 = 1 - 39081x2y2
over the finite prime field 𝔽p, where p = 2448 - 2224 - 1. It has order of n = 2446 - 0x8335dc163bb124b65129c96fde933d8d723a70aadc873d6d54a7bb0d and cofactor h = 4. Like any other Edwards curve, the Curve448 has equivalent in the Weierstrass form (y2 = x3 + ax + b), but the above Edwards form provides significant optimizations in the EC point calculations and improved performance.
The Curve448 provides ~ 224-bit security level (more precisely 222.8-bits). The private keys for the Curve448 are 446 bits and are typically encoded as 448-bit integers (56 bytes, 112 hex digits). The public keys are also encoded as 448-bit integers.
The Curve448 is suitable for ECDH key agreement (ECDH function, known as X448) and for fast digital signatures (EdDSA algorithm, known as Ed448 or edwards448). Note that X448 and Ed448 use different encodings for the EC points, so they are not directly compatible and require conversion if you want to use the same public-private key pairs.
## Curve25519 or Curve448?
Prefer Curve448 to Curve25519 when your application needs a higher level of security, but have in mind that Curve448 is about 3 times slower than Curve25519 and uses longer key length and signature length.
Prefer Curve25519 to Curve448 when you need better performance and smaller keys and signatures.
In the general case, have in mind that Curve25519 is faster than secp256k1 and the other 256-bit standard NIST curves and is considered more secure, so it is the recommended choice for ~ 128-bit security. Similarly, the Curve448 has better performance than the classical curves with similar key length, so it is the recommended curve for ~ 224-bit security.
## Curve25519 - Example in Python
To demonstrate the elliptic curve Curve25519 in practice, we shall first install the pynacl crypto library for Python:
pip install pynacl
The Python binding to the Networking and Cryptography (NaCl) library (PyNaCl) implements many modern cryptographic algorithms, including the EC point arithmetic over the Curve25519 and Ed25519 signatures.
Next, generate a random 252-bit private key and its corresponding public key (EC point) on the Curve25519 (both keys will be encoded internally as 256-bit integers):
from nacl.public import PrivateKeyimport binasciiprivKey = PrivateKey.generate()pubKey = privKey.public_keyprint("privKey:", binascii.hexlify(bytes(privKey)))print("pubKey: ", binascii.hexlify(bytes(pubKey)))
Run the above code example: https://repl.it/@nakov/Curve25519-in-Python. The sample output from the above code shows that both the public and the private (secret) keys on the Curve25519 are encoded as 256-bit integers (64 hex digits, 32 bytes) and this simplifies the developers:
privKey: b'8175f7cd524a59b6efbd447985ce5d97c546b319521ff236203970e50052c641'pubKey: b'cf97a96568fee4ddb232f617fd5b9df2d2e5b90e68ba7f6d5129ea92d7d8f95e'
In fact, different crypto libraries may use different key encodings and typically X25519 ECDH keys are encoded differently than Ed25519 keys (Montgomery curve coordinates vs. twisted Edwards curve coordinates).
|
2020-01-22 10:31:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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": 0.5967414975166321, "perplexity": 1857.347177958281}, "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/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00242.warc.gz"}
|
http://www.acooke.org/cute/HardeningS0.html
|
## Hardening SUSE
From: "andrew cooke" <andrew@...>
Date: Sat, 21 Apr 2007 18:23:14 -0400 (CLT)
So, for two weeks I have been running my local server without a firewall.
Ports 25 (SMTP), 111 (portname) and 143 (IMAP) were all open to the world.
(Don't ask how - in fact, I now realise that they weren't quite "as open"
as I first thought, when "shields up" showed some alarming red squares -
https://www.grc.com/ and, I believe, I should have been moderately safe
anyway, as SMTP and IMAP were configured internally to only accept local
requests and all services are using the latest patches)
Anyway, once I'd fixed the problem, I thought it was probably worthwhile
configuring services not to bind to the external address at all. For most
it's simply a change in the config file. However, exim (IMAP) requires
the -oX command line argument (which can be specified via sysconfig in
SUSE) and portmap requires "-i" (which I set in /etc/init.d/portmap).
There seems to be no need to use -oP with exim.
Andrew
|
2018-06-19 06:36:15
|
{"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": 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": 0.3543674647808075, "perplexity": 7138.295842411454}, "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-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00421.warc.gz"}
|
http://mail.scipy.org/pipermail/scipy-user/2005-July/004765.html
|
# [SciPy-user] pycrust
Howey, David A d.howey at imperial.ac.uk
Wed Jul 13 10:07:47 CDT 2005
```Thanks - this is helpful.
I am trying to get ipython to use an editor other than windows notepad.
First I tried editing the 'ipythonrc' file. Didn't have any affect -
still uses notepad. Then I tried a command line:
ipython -editor C:\Program Files\Crimson Editor\cedt.exe
It doesn't like that. I think it's something to do with the spaces in
the path. Should I use \%20 ?
Anyone else got this sorted on a win32 system?
Thanks
Dave
-----Original Message-----
From: scipy-user-bounces at scipy.net [mailto:scipy-user-bounces at scipy.net]
On Behalf Of Ryan Krauss
Sent: 13 July 2005 16:03
To: SciPy Users List
Subject: Re: [SciPy-user] pycrust
As long as you use the TkAgg backend, I was able to use matplotlib and
scipy with pycrust:
import pylab
from scipy import *
x=arange(0,1,0.1)
y=sin(2*pi*x)
pylab.plot(x,y)
pylab.show()
The only catch is that once you show() your plots, you have to close
them all before you can do anything else in the command window (because
|
2015-01-29 06:45:54
|
{"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": 0.8177076578140259, "perplexity": 11367.531144402716}, "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-2015-06/segments/1422115855845.27/warc/CC-MAIN-20150124161055-00213-ip-10-180-212-252.ec2.internal.warc.gz"}
|
https://civicrm.stackexchange.com/questions/22215/can-i-track-grants-against-contributions/29350
|
# Can I track grants against contributions?
Is there a way with CiviGrant to match money received as Contributions with money disbursed?
E.g. if I receive a donation for $100K for a scholarship fund, and I give out 3 scholarships of$20K each, is there a report that can show me that I have \$40K left in the fund?
|
2019-10-16 05:54:15
|
{"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": 0.2344381958246231, "perplexity": 4263.608872655062}, "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/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00145.warc.gz"}
|
https://codereview.stackexchange.com/questions/105638/reverse-binary-representation-of-an-int-only-significant-bits
|
# Reverse binary representation of an int (only significant bits)
This based on this question in StackOverflow. The accepted answer uses Convert.ToString(int, base) to get the binary string, reverses it and converts it back to int. Nice trick!
I have very little experience with bit twidling (using Flags is about it :p ) so I decided to try and code the solution to this problema without using Convert.
I came up with the following solution. I'd like to know easier ways to do this as there are probably many much better than the one I found.
public static IEnumerable<bool> ToBinary(this int n)
{
for (int i = 0; i < 32; i++)
{
yield return (n & (1 << i)) != 0;
}
}
public static int ToInt(this IEnumerable<bool> b)
{
var n = 0;
var counter = 0;
foreach (var i in b.Trim().Take(32))
{
n = n | (i ? 1 : 0) << counter;
counter++
}
return n;
}
private static IEnumerable<bool> Trim(this IEnumerable<bool> list)
{
bool trim = true;
foreach (var i in list)
{
if (i)
{
trim = false;
}
if (!trim)
{
yield return i;
}
}
}
And now you'd use it like this:
var reversed = n.ToBinary().Reverse().ToInt();
You could use while loop instead of the for loop.
Thus you don't need to trim zero bits, because it will continue to loop only while there is at least one significant bit in a value.
Another advantage of this approach is that the result can be calculated using integer arithmetics only without using any kind of collections.
static int Reverse(int input)
{
uint x = unchecked((uint)input);
uint y = 0;
while (x != 0)
{
y <<= 1; // Shift accumulated result left
y |= x & 1; // Set the least significant bit if it is set in the input value
x >>= 1; // Shift input value right
}
return unchecked((int)y);
}
Usage example:
int t = 0x103;
Console.WriteLine(Convert.ToString(t, 2));
Console.WriteLine(Convert.ToString(Reverse(t), 2));
Result:
100000011
110000001
• Adding more detail on why it works better would be better, but, that's about the same answer I would give. – rolfl Sep 24 '15 at 20:16
• For a negative x the loop seems to be infinite. – vnp Sep 24 '15 at 20:24
Dmitry's answer is probably better, but I'll provide another option. If you wish to stick to using collections, you can replace the loops with LINQ to increase conciseness, readability, and safety.
public static IEnumerable<bool> ToBinary(this int n)
{
return Enumerable.Range(0, 32).Select(x => (n & (1 << x)) != 0);
}
public static int ToInt(this IEnumerable<bool> b)
{
return b.Trim().Take(32).Select((x, i) => (x ? 1 : 0) << i).Sum();
}
private static IEnumerable<bool> Trim(this IEnumerable<bool> list)
{
return list.SkipWhile(x => !x);
}
|
2019-08-19 21:48:42
|
{"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": 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": 0.1794731616973877, "perplexity": 3566.3989510131887}, "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-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00396.warc.gz"}
|
https://learncheme.com/simulations/statistics/finding-probabilities-for-intervals-of-a-normal-distribution/
|
##### Description
This simulation computes the probability of any interval for a normal distribution. Input the mean and standard deviation (parameters μ and σ) and use the sliders to set the two bounds of the interval. The output shows a plot of the area lying below the probability density function and enclosed by the limits, and the value of this probability above the plot. At the bottom you can see the left or right tail probabilities.
This simulation runs on desktop using the free Wolfram Player. Download the Wolfram Player here.
##### About
Author: Juan D. Montoro-Pons. Open content licensed under CC BY-NC-SA.
View the source code for this simulation
|
2023-01-29 20:07:53
|
{"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": 0.8820832967758179, "perplexity": 1030.7541942713588}, "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-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00757.warc.gz"}
|
http://forum.sagittal.org/viewtopic.php?f=10&p=2435
|
## Magrathean diacritics
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
cmloegcmluin wrote:
Tue Sep 08, 2020 12:43 pm
Dave Keenan wrote:
Tue Sep 08, 2020 11:58 am
Would it be easy for you to add a flag to make it put out tab-separated value (TSV) format, including tabs between the terms in the monzo, and between the terms and their brackets. This would make it much easier for me to get these lists into a spreadsheet for further manipulation.
After running the find-commas script w/o the --for-forum flag, you could import "dist/pitch/all.txt" to Excel or Sheets and that should work. It uses the tab character as a delimiter and should recognize it automatically.
Anything those scripts log to the console also gets saved in a file that sticks around until you run the script again. You can find them in the dist/ folder. It's not the cleanest implementation or anything yet, but can be quite handy.
I had already tried this by redirecting the output to a file of my choice by appending " > fileOfMyChoice.txt" to the command. There are two problems:
1. Fields in the same column are padded to a common width using spaces, before being separated with tabs. Excel assumes the spaces are significant and keeps them. That's not such a big deal since I could easily strip spaces from the file with Notepad++ before opening it in Excel. However ...
2. Monzo terms are only separated by spaces, no tabs.
I kind of like the abbreviations ATE and AAS, though, as opposed to abs3exp and absolute apotome slope which we never had a shorthand for.
Those are fine with me.
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
cmloegcmluin wrote:
Tue Sep 08, 2020 4:29 am
Dave Keenan wrote:
Sat Sep 05, 2020 10:50 am
Notice how George included a term that attempted to correct for the balance-blindness of SoPF>3. Perhaps we can replace the first two terms with some function of N2D3P9 and simplify the last two terms into one term, given that abs3exp and absApotomeSlope are the same thing here.
I believe the term you are talking about which attempts to correct for the balance-blindness of SoPF>3 is "J". So, I agree that the first two terms, "G" and "J", can be replaced with (a function of) N2D3P9. And then consolidate the "K" and "L" terms too.
Yes. J is what I was referring to. I think we have to replace G+J with some constant times the log of N2D3P9. Or maybe k*sqrt(N2D3P9) or k*N2D3P9. I'm not sure which.
volleo6144 wrote:
Sat May 30, 2020 2:43 am
The weighted complexity is G+J+K+L, where:
- G (">3") is the SoPF>3
- J ("d+n") is the absolute value of (H-I) times G/5, with H = the number of primes >3 (with multiplicity) in the denominator (smaller value) and I in the numerator (larger value)—7:25k (224:225) has G=17, H=1 (actually -1 in the spreadsheet), I=2, J=(2-1) times 17/5 or 3.4. This is zero for any comma that's one prime against another, so 5:7k, 5:7C, 11:23S, etc. are all zero here.
- K ("3-exp.") is 2^(abs(3exp) - 8.5) times ln(G+2), so 5C (80:81) is 2^(4-8.5)×ln(5+2) = ln(7)/sqrt(512) = 1.95/22.6 = 0.086. This ranks commas with high 3-exponents as really complex (3C is at 7.84 here, and 3s is at 17.2 trillion).
- L ("slope") is like K, but with apotome slope instead of 3-exponent. L = 2^(abs(slope) - 8.5)×ln(G+2). 3C (slope = 10.6) is at 2.88 here, and 3s (slope = 52.8) is at 14.8 trillion.
Okay, so the question becomes: what function of N2D3P9 balances well with some version of the "K" term in George's complexity function. There's a bunch of magic numbers in George's function. Any idea what the 8.5 is (I see that that's the ATE of half a Pythagorean large diesis, which is the size category boundary between C and S)? Or why it's two to this power? Or why we add 2 to the SoPF>3 and then take the natural log? It feels incommensurate to use two different bases between N2D3P9 and the "K" part. I feel like George must have had a target he was aiming for, like he had put things in terms of the relationship between two key ratios he wanted to push to the correct side of each other. Or something along the lines of what Dave was going for when he brought "munging" into the lingo:
Yeah. I think it is something like my munging. I think the 8.5 is just a kind of soft threshold beyond which 3 exponents are strongly penalised.
It's messy how the SOPF>3 enters into both G and K.
Sorry I ran out of time here, and didn't get to respond to the rest of your post.
cmloegcmluin
Posts: 794
Joined: Tue Feb 11, 2020 3:10 pm
Location: San Francisco, California, USA
Real Name: Douglas Blumeyer
Contact:
### Re: Magrathean diacritics
Dave Keenan wrote:
Tue Sep 08, 2020 5:13 pm
I agree with treating the monzo as the standard format.
I've still got a bunch of subtleties to work out somewhere at the intersection of the complexities of TypeScript, Sagittal, and tuning theory, but I'll spare you the grisly details.
I know you already have code to convert an integer to a monzo, but I thought you might like the puzzle of figuring out how the following works. It's a javascript translation of what I use in Excel to obtain the exponent (i.e. monzo term) for each prime p, for an integer n. It takes advantage of the fact that the largest consecutive integer in double float format is 253 and assumes you wont pass it anything larger.
exponent(n, p) = Math.round(math.log(math.gcd(n, p**(Math.floor(53/Math.log2(p))) ), p) )
But it's only convenient if you have an efficient (Euclidean algorithm) GCD function (as Excel does), e.g. this:
https://mathjs.org/docs/reference/functions/gcd.html
And since it was in the same library, I assumed this too:
https://mathjs.org/docs/reference/functions/log.html
Thanks for the puzzle! The colored parens were quite helpful.
I think I got it. There's two major tricks that happen:
• The floor(\frac{53}{log_2(p)}) gives you the maximum safe exponent on p, i.e. the largest exponent which p can be raised to while keeping the power less than 2^53.
• So, when you take the gcd of n with this number, you’re guaranteed to be given back all the factors of p in n.
From there it’s a simple matter of taking the log base p to get the exponent. I’m not sure why the \text{round} is necessary, but that’s not really part of the puzzle.
Cool beans I could consider using this in the codebase. I think I remember that when we profiled the LFC scripts they were spending a ton of time computing monzos from integers. Indeed, 5% of the time was spent prime factorizing integers.
Dave Keenan wrote:
Tue Sep 08, 2020 5:50 pm
Monzo terms are only separated by spaces, no tabs.
Yes it's true, I do a mix of spaces and tabs to align things in that format. Somehow, Google Sheets was smart enough to sort it out automatically, and I assumed Excel would be similarly convenient. Check this out: https://docs.google.com/spreadsheets/d/ ... sp=sharing
So if it's not too much trouble, you could import to Google Sheets, and then download. Still a bit of work to clean up the ['s and ⟩'s, though. So it would be even better if I just added a new table formatting module with a TSV target. Would be pretty easy. Just trimming those aligning spaces, and special handling for monzos.
Thanks for being my first non-me user!
Dave Keenan wrote:
Tue Sep 08, 2020 11:52 pm
I think we have to replace G+J with some constant times the log of N2D3P9. Or maybe k*sqrt(N2D3P9) or k*N2D3P9. I'm not sure which.
Sure, or k^N2D3P9. Or N2D3P9^k. Or logN2D3P9k. (context here and here if you don't know what I'm alluding to)
I think it is something like my munging.
Alright, my first attempts on the problem will proceed along those lines.
I suggest again that we should bring this back to the developing a notational comma popularity metric thread.
I think the 8.5 is just a kind of soft threshold beyond which 3 exponents are strongly penalised.
Okay. But you're not specifically aware, then, that it's in terms of like something to do with the circle of fifths, or most popular nominal as 1/1 being D vs. G, etc. that kind of stuff. In other words: is it flexible, and/or might there be a psychoacoustically justifiable value for this parameter.
It's messy how the SOPF>3 enters into both G and K.
Agreed. I think that's unnecessary. The goal only seems to be to scale G and K properly in relation to each other.
I wonder how the chips would fall if we just extended N2D3P9 to include the 3's, and treated them like they were in the numerator. If we treated them like they were in the denominator they'd have zero effect since each 3 would be divided by 3, but if we treat them like they're in the numerator, then each one increases the points by a factor of 3/2.
N2D3P9(65/77n) = 200.818
NATE2D3P9(65/77n) = 200.818 * (3/2)^3 = 200.818 * 3.375 = 677.761
vs.
N2D3P9(125/13n) = 97.801
NATE2D3P9(125/13n) = 97.801 * (3/2)^9 = 97.801 * 38.443 = 3759.764
So yeah... it definitely still prefers the existing comma we have for 6 tinas (i.e. 2 minas).
And the other example:
N2D3P9(1/205n) = 233.472
NATE2D3P9(1/205n) = 233.472 * (3/2)^8 = 233.472 * 25.629 = 5983.654
vs.
N2D3P9(1/5831n) = 688.382
NATE2D3P9(1/5831n) = 688.382 * (3/2)^6 = 688.382 * 11.391 = 7841.359
Oh ho! Interesting. So by this measure, we would prefer the 1/205n.
What do you think? It's certainly simple.
I thought it would be a quick experiment to use the test I have already set up for verifying primary commas, but I'm not sure the approach I am testing is appropriate. I know I brought this up before... where was it... checking the secondary comma zones for each symbol... ah. Back on page 6 of the Godthread... why am I not surprised. I think maybe a quarter of the content of the entire forum is constituted by that one thread now, heh...
But yeah, in any case, that condition is *not* true for NATE2D3P9 any more than it was for SoPF>3. It only takes a glance at the precision levels diagram to see that 's secondary comma zone is an enclave of the secondary comma zone for , and since the 5s has much lower N2D3P9 than the 19s, by the logic of the test as I've written it now, the 5s should be the comma for . Which we know there's a lot more considerations involved here, and maybe it comes down to the complex relationship between the High and Ultra levels (or maybe more accurately the complex relationship between the Promethean and Herculean symbol subsets). Anywho...
Sorry I ran out of time here, and didn't get to respond to the rest of your post.
I think you addressed most of it, except for the stuff about badness. But we probably should save that stuff to later and not try to work on it simultaneously with this already-complex-enough problem of the usefulness.
What would be relevant to resolve, at least, would be the boundary between error and usefulness. Was I right to say it the point where we start considering EDO-ability is when we move on to badness (and thus "developing a notational comma popularity metric" topic being a subtopic of "Just Intonation notations" is the correct home for it)?
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
cmloegcmluin wrote:
Wed Sep 09, 2020 5:32 am
I think I got it. There's two major tricks that happen:
...
From there it’s a simple matter of taking the log base p to get the exponent. I’m not sure why the \text{round} is necessary, but that’s not really part of the puzzle.
You got it. The round() is necessary because of the nature of floating point math. The result of the log may sometimes end in something like .000...1 or .999... (really an error in the least significant bit or two in the binary representation).
Cool beans I could consider using this in the codebase. I think I remember that when we profiled the LFC scripts they were spending a ton of time computing monzos from integers. Indeed, 5% of the time was spent prime factorizing integers.
I can't guarantee it will be faster, but to have a chance of that you'd need to precompute:
maxPrimePower[i] = prime[i]**(Math.floor(53/Math.log2(prime[i])))
so in the repeated calcs you're only doing:
exponent(n, i) = Math.round(math.log(math.gcd(n, maxPrimePower[i]), prime[i]) )
That's more like what I really do in Excel.
Somehow, Google Sheets was smart enough to sort it out automatically, and I assumed Excel would be similarly convenient. Check this out: https://docs.google.com/spreadsheets/d/ ... sp=sharing
Amazing.
So if it's not too much trouble, you could import to Google Sheets, and then download. Still a bit of work to clean up the ['s and ⟩'s, though. So it would be even better if I just added a new table formatting module with a TSV target. Would be pretty easy. Just trimming those aligning spaces, and special handling for monzos.
That would be great. Thanks. With the monzos, in TSV it would be best if all the closing angle-brackets ended up in the same rightward column, so we don't have columns with some numbers and some angle-brackets. And can you please put a BOM at the start of the file. Excel currently shows the angle brackets as a series of three characters: ⟩ because it can't tell that the file is in UTF-8.
Something unrelated that I just remembered to mention: Can you please ensure your software names the Pythagorean comma as "3C", not "1C", and any other 3-limit commas similarly. I'm pretty sure we agreed that the only "comma" name that would use the lone number "1" would be the unison "1u".
Dave Keenan wrote:
Tue Sep 08, 2020 11:52 pm
I think we have to replace G+J with some constant times the log of N2D3P9. Or maybe k*sqrt(N2D3P9) or k*N2D3P9. I'm not sure which.
Sure, or k^N2D3P9. Or N2D3P9^k. Or logN2D3P9k. (context here and here if you don't know what I'm alluding to)
No. F those. I'm pretty sure that to behave anything like sopfr, the function of N2D3P9 has to be linear or compressive, not expansive.
I want to see N2D3P9 plotted against sopfr for the first hundred or so ratios in N2D3P9 order, to get a feel for the approximate relationship.
Alright, my first attempts on the problem will proceed along those lines.
I suggest again that we should bring this back to the developing a notational comma popularity metric thread.
If we took it there, I'd feel obliged to make it general for all notational commas, whereas here we can take shortcuts that rely on their small size and close spacing as tinas.
I think the 8.5 is just a kind of soft threshold beyond which 3 exponents are strongly penalised.
Okay. But you're not specifically aware, then, that it's in terms of like something to do with the circle of fifths, or most popular nominal as 1/1 being D vs. G, etc. that kind of stuff. In other words: is it flexible, and/or might there be a psychoacoustically justifiable value for this parameter.
I think it's slightly flexible. Psychoacoustics are not relevant. Yes, it's to do with the chain (not circle) of fifths and the fact that we can only go to 2 sharps or 2 flats max, and we'd really like to avoid having more than one. And since George takes the absolute value of the 3-exponent, he's assuming 1/1 is D (the point of symmetry) or not far from it, on the chain of fifths.
I wonder how the chips would fall if we just extended N2D3P9 to include the 3's, and treated them like they were in the numerator. If we treated them like they were in the denominator they'd have zero effect since each 3 would be divided by 3, but if we treat them like they're in the numerator, then each one increases the points by a factor of 3/2.
...
Oh ho! Interesting. So by this measure, we would prefer the 1/205n.
What do you think? It's certainly simple.
I definitely like its simplicity. Good thinking. Unfortunately, I don't think it penalises large 3-exponents enough. An ATE greater than 14 should be pretty much OOTQ no matter what other properties the comma has. That is the case with George's metric, which involves adding a munged 3-exponent rather than multiplying. But maybe we need to take the log of N2D3P9 before it makes sense to add a munged 3-exponent to it.
It only takes a glance at the precision levels diagram to see that 's secondary comma zone is an enclave of the secondary comma zone for , and since the 5s has much lower N2D3P9 than the 19s, by the logic of the test as I've written it now, the 5s should be the comma for . Which we know there's a lot more considerations involved here, and maybe it comes down to the complex relationship between the High and Ultra levels (or maybe more accurately the complex relationship between the Promethean and Herculean symbol subsets). Anywho...
I don't think we have to get into stuff like that. I think it can be done with only N2D3P9 and ATE if they are combined in the right way.
I think you addressed most of it, except for the stuff about badness. But we probably should save that stuff to later and not try to work on it simultaneously with this already-complex-enough problem of the usefulness.
Agreed.
What would be relevant to resolve, at least, would be the boundary between error and usefulness. Was I right to say it the point where we start considering EDO-ability is when we move on to badness (and thus "developing a notational comma popularity metric" topic being a subtopic of "Just Intonation notations" is the correct home for it)?
I think we only need to go back to DANCPM if we are developing a comma usefulness metric with general application, and yes, that would require that we consider EDO-ability. But that wouldn't need to involve error or badness. Here's how I see our current terminology: "Usefulness" would be a combination of (popularity or N2D3P9) and (3-exponent and/or slope). "Badness" would be a combination of "usefulness" and error.
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
Here's N2D3P9 versus sopfr for N2D3P9 < 903.
And here it is again with N2D3P9 on a log axis.
Clearly we need to take the log of N2D3P9 to make it work in a modified "Secor complexity" metric.
Just reading off the graph, 4.5 × lb(N2D3P9) ≈ 15 × log10(N2D3P9) ought to just plug straight in.
Attachments
LogN2d3p9VsSopfr.png
N2d3p9VsSopfr.png
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
So I'm getting that something like
4.5×lb(N2D3P9) + 8×2^(ATE-8.5)
should be similar to Secor complexity.
Call it 4.5×lb(N2D3P9) + 9×2^(ATE-8.5) then eliminate the common factors since they don't affect ranking.
lb(N2D3P9) + 2×2^(ATE-8.5)
= lb(N2D3P9) + 2^(ATE-7.5)
Exponentiating that whole thing also doesn't affect ranking. That would be
N2D3P9 × 2^(2^(ATE-7.5))
ATE N2D3P9 multiplier = 2^(2^(ATE-7.5))
0 1.003836474
1 1.007687666
2 1.015434433
3 1.031107087
4 1.063181825
5 1.130355594
6 1.277703768
7 1.632526919
8 2.665144143
9 7.102993301
10 50.45251384
11 2545.456153
12 6479347.025
13 4.19819E+13
14 1.76248E+27
That looks like it's doing the right kind of thing. We only have to adjust that 7.5 value, which is the ATE at which N2D3P9 gets doubled. Call it the DATE. My mapping from George's formula was all very rough and I ignored the J (balance ) term. Maybe bump DATE up to 9.
ATE N2D3P9 multiplier = 2^(2^(ATE-DATE)) where DATE = 9
0 1.00135472
1 1.002711275
2 1.005429901
3 1.010889286
4 1.021897149
5 1.044273782
6 1.090507733
7 1.189207115
8 1.414213562
9 2
10 4
11 16
12 256
13 65536
14 4294967296
cmloegcmluin
Posts: 794
Joined: Tue Feb 11, 2020 3:10 pm
Location: San Francisco, California, USA
Real Name: Douglas Blumeyer
Contact:
### Re: Magrathean diacritics
Dave Keenan wrote:
Wed Sep 09, 2020 10:45 am
I can't guarantee it will be faster, but to have a chance of that you'd need to precompute:
maxPrimePower[i] = prime[i]**(Math.floor(53/Math.log2(prime[i])))
That makes sense. Thanks.
With the monzos, in TSV it would be best if all the closing angle-brackets ended up in the same rightward column, so we don't have columns with some numbers and some angle-brackets.
Yup, that's what I would want too.
And can you please put a BOM at the start of the file. Excel currently shows the angle brackets as a series of three characters: ⟩ because it can't tell that the file is in UTF-8.
Whoa. Can't say I ever saw the word "endianness" before today... Okay, can do.
Something unrelated that I just remembered to mention: Can you please ensure your software names the Pythagorean comma as "3C", not "1C", and any other 3-limit commas similarly. I'm pretty sure we agreed that the only "comma" name that would use the lone number "1" would be the unison "1u".
Yes I think you're right. Okay, yes, I'll correct that soon.
I want to see N2D3P9 plotted against sopfr for the first hundred or so ratios in N2D3P9 order, to get a feel for the approximate relationship.
Yeah, that'd be super cool to see
I suggest again that we should bring this back to the developing a notational comma popularity metric thread.
If we took it there, I'd feel obliged to make it general for all notational commas, whereas here we can take shortcuts that rely on their small size and close spacing as tinas.
Fine by me. I hope we don't regret if we do want to find a general comma no pop rank and wish we'd used it on the tinas. But honestly at this point I think my hypothetical future regretful self would accept the fact that my present impatient self felt he'd spent enough time on this problem already.
Yes, it's to do with the chain (not circle) of fifths and the fact that we can only go to 2 sharps or 2 flats max, and we'd really like to avoid having more than one. And since George takes the absolute value of the 3-exponent, he's assuming 1/1 is D (the point of symmetry) or not far from it, on the chain of fifths.
Alright. I think that checks out with the DATE of 9 you settled on; with ATE of 10 causing a 4x factor on N2D3P9 and 11 causing a 16x factor, it seems fair to say that ATE of 10 would be the last reasonable ATE that would ever win out as a comma (the last "ITQ" comma? Or am I hyperacronymizing?), and I think 10 would be the maximum count of fifths you could go from D before requiring a double sharp/flat.
Dave Keenan wrote:
Wed Sep 09, 2020 12:09 pm
Here's N2D3P9 versus sopfr for N2D3P9 < 903.
That's so cool looking! The clearly visible threads running through it remind me of the patterns in the Sacks spiral: https://upload.wikimedia.org/wikipedia/ ... 100000.png
Dave Keenan wrote:
Wed Sep 09, 2020 12:46 pm
ATE N2D3P9 multiplier = 2^(2^(ATE-DATE)) where DATE = 9
Great work. Interesting that it works out that the 3 term of the monzo causes exponential growth while the 5+ terms cause power growth. Actually the 3-term causes double exponential growth. This appears to be an established concept: https://en.wikipedia.org/wiki/Double_ex ... l_function
There's something cute about 9 cropping up in both N2D3P9 and this. But I wouldn't fixate on it at the expense of utility.
If we settle on this as our usefulness metric, I can build it into the code and/or get some updated results per half-tina.
Dave Keenan
Posts: 1095
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:
### Re: Magrathean diacritics
cmloegcmluin wrote:
Wed Sep 09, 2020 1:51 pm
I hope we don't regret if we do want to find a general comma no pop rank and wish we'd used it on the tinas. But honestly at this point I think my hypothetical future regretful self would accept the fact that my present impatient self felt he'd spent enough time on this problem already.
I hate to say it, but now I'm thinking the only way to validate this comma usefulness metric, and to tune its parameters (like DATE) is to apply it to all the existing extreme-precision JI notation commas under the half-apotome, and minimise the sum-of-squares or something. That means including another multiplier: 2^(2^(AAS-DAAS)).
A much harder problem would be to optimise it to maximally-justify our existing choices of one comma over another in each extreme bucket.
An interesting case (in addition to the 5s vs 19s one you mentioned) is 3C. It has N2D3P9 of 1 and ATE of 12 that results in a multiplier of 256. But I don't know what else it's competing with for that slot. No extreme comma under the half apotome has ATE > 12.
If we settle on this as our usefulness metric, I can build it into the code and/or get some updated results per half-tina.
Actually, I can see some serious floating-point overflow and underflow issues with the form:
N2D3P9 × 2^(2^(ATE-DATE)) × 2^(2^(AAS-DAAS))
Can you build it into the code as lb(N2D3P9) + 2^(ATE-DATE) + 2^(AAS-DAAS) with switches to change DATE and DAAS (separately) from their default values of 9 and 9.
OK. You were right. Time to go back to the developing a notational comma popularity metric thread.
cmloegcmluin
Posts: 794
Joined: Tue Feb 11, 2020 3:10 pm
Location: San Francisco, California, USA
Real Name: Douglas Blumeyer
Contact:
### Re: Magrathean diacritics
Dave Keenan wrote:
Wed Sep 09, 2020 3:09 pm
OK. You were right. Time to go back to the developing a notational comma popularity metric thread.
We'll be right back...!
volleo6144
Posts: 58
Joined: Mon May 18, 2020 7:03 am
Location: Earth
Contact:
### Re: Magrathean diacritics
Dave Keenan wrote:
Wed Sep 09, 2020 3:09 pm
Actually, I can see some serious floating-point overflow and underflow issues with the form:
N2D3P9 × 2^(2^(ATE-DATE)) × 2^(2^(AAS-DAAS))
float overflows at 2^128 = 2^2^7 or ATE=16*, and double overflows at 2^1024 = 2^2^10 or ATE=19*. You'll be fine—I'm sure no comma that has 311 as a factor in either side (177147:183920, 64.96¢, has ATE=11, AAS=14.9997, and 2^(2^(AAS-9) + 2^(2^(ATE-9)) = 2.922e+20, which fits in a float) will be useful anyway.
Apotome slope equals 3-exponent minus the comma's size in units of 3A / 7 ≈ 16.24¢ (very, very close to 9°665), so it's essentially the same as 3-exponent for schisminas (455n has ATE = 2 and AAS = 2.026).
* or AAS, and assuming DATE = DAAS = 9
A random guy who sometimes doodles about Sagittal and microtonal music in general in his free time.
|
2020-10-30 01:23:31
|
{"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": 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": 0.6793714761734009, "perplexity": 2118.3410521224237}, "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/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00134.warc.gz"}
|
https://spinor.info/weblog/?cat=48
|
Now it is time for me to be bold and contrarian. And for a change, write about physics in my blog.
From time to time, even noted physicists express their opinion in public that we do not understand quantum physics. In the professional literature, they write about the “measurement problem”; in public, they continue to muse about the meaning of measurement, whether or not consciousness is involved, and the rest of this debate that continues unabated for more than a century already.
Whether it is my arrogance or ignorance, however, when I read such stuff, I beg to differ. I feel like the alien Narim in the television series Stargate SG-1 in a conversation with Major (and astrophysicist) Samantha Carter about the name of a cat:
CARTER: Uh, see, there was an Earth physicist by the name of Erwin Schrödinger. He had this theoretical experiment. Put a cat in a box, add a can of poison gas, activated by the decay of a radioactive atom, and close the box.
NARIM: Sounds like a cruel man.
CARTER: It was just a theory. He never really did it. He said that if he did do it at any one instant, the cat would be both dead and alive at the same time.
NARIM: Ah! Kulivrian physics. An atom state is indeterminate until measured by an outside observer.
CARTER: We call it quantum physics. You know the theory?
NARIM: Yeah, I’ve studied it… in among other misconceptions of elementary science.
CARTER: Misconception? You telling me that you guys have licked quantum physics?
What I mean is… Yes, in 2021, we “licked” quantum physics. Things that were mysterious in the middle of the 20th century aren’t (or at least, shouldn’t be) quite as mysterious in the third decade of the 21st century.
OK, let me explain by comparing two thought experiments: Schrödinger’s cat vs. the famous two-slit experiment.
The two-slit experiment first. An electron is fired by a cathode. It encounters a screen with two slits. Past that screen, it hits a fluorescent screen where the location of its arrival is recorded. Even if we fire one electron at a time, the arrival locations, seemingly random, will form a wave-like interference pattern. The explanation offered by quantum physics is that en route, the electron had no classically determined position (no position eigenstate, as physicists would say). Its position was a combination, a so-called superposition of many possible position states, so it really did go through both slits at the same time. En route, its position operator interfered with itself, resulting in the pattern of probabilities that was then mapped by the recorded arrival locations on the fluorescent screen.
Now on to the cat: We place that poor feline into a box together with a radioactive atom and an apparatus that breaks a vial of poison gas if the atom undergoes fission. We wait until the half-life of that atom, making it a 50-50 chance that fission has occurred. At this point, the atom is in a superposition of intact vs. split, and therefore, the story goes, the cat will also be in a superposition of being dead and alive. Only by opening the box and looking inside do we “collapse the wavefunction”, determining the actual state of the cat.
Can you spot a crucial difference between these two experiments, though? Let me explain.
In the first experiment involving electrons, knowledge of the final position (where the electron arrives on the screen) does not allow us to reconstruct the classical path that the electron took. It had no classical path. It really was in a superposition of many possible locations while en route.
In the second experiment involving the cat, knowledge of its final state does permit us to reconstruct its prior state. If the cat is alive, we have no doubt that it was alive all along. If it is dead, an experienced veterinarian could determine the moment of death. (Or just leave a video camera and a clock in the box along with the cat.) The cat did have a classical state all throughout the experiment, we just didn’t know what it was until we opened the box and observed its state.
The crucial difference, then, is summed up thus: Ignorance of a classical state is not the same as the absence of a classical state. Whereas in the second experiment, we are simply ignorant of the cat’s state, in the first experiment, the electron has no classical state of position at all.
These two thought experiments, I think, tell us everything we need to know about this so-called “measurement problem”. No, it does not involve consciousness. No, it does not require any “act of observation”. And most importantly, it does not involve any collapse of the wavefunction when you really think it through. More about that later.
What we call measurement is simply interaction by the quantum system with a classical object. Of course we know that nothing really is classical. Fluorescent screens, video cameras, cats, humans are all made of a very large but finite number of quantum particles. But for all practical (measurable, observable) intents and purposes all these things are classical. That is to say, these things are (my expression) almost in an eigenstate almost all the time. Emphasis on “almost”: it is as near to certainty as you can possibly imagine, deviating from certainty only after the hundredth, the thousandth, the trillionth or whichever decimal digit.
Interacting with a classical object confines the quantum system to an eigenstate. Now this is where things really get tricky and old school at the same time. To explain, I must invoke a principle from classical, Lagrangian physics: the principle of least action. Almost all of physics (including classical mechanics, electrodynamics, even general relativity) can be derived from a so-called action principle, the idea that the system evolves from a known initial state to a known final state in a manner such that a number that characterizes the system (its “action”) is minimal.
The action principle sounds counterintuitive to many students of physics when they first encounter it, as it presupposes knowledge of the final state. But this really is simple math if you are familiar with second-order differential equations. A unique solution to such an equation can be specified in two ways. Either we specify the value of the unknown function at two different points, or we specify the value of the unknown function and its first derivative at one point. The former corresponds to Lagrangian physics; the latter, to Hamiltonian physics.
This works well in the context of classical physics. Even though we develop the equations of motion using Lagrangian physics, we do so only in principle. Then we switch over to Hamiltonian physics. Using observed values of the unknown function and its first derivative (think of these as positions and velocities) we solve the equations of motion, predicting the future state of the system.
This approach hits a snag when it comes to quantum physics: the nature of the unknown function is such that its value and its first derivative cannot both be determined as ordinary numbers at the same time. So while Lagrangian physics still works well in the quantum realm, Hamiltonian physics does not. But Lagrangian physics implies knowledge of the future, final state. This is what we mean when we pronounce that quantum physics is fundamentally nonlocal.
Oh, did I just say that Hamiltonian physics doesn’t work in the quantum realm? But then why is it that every quantum physics textbook begins, pretty much, with the Hamiltonian? Schrödinger’s famous equation, for starters, is just the quantum version of that Hamiltonian!
Aha! This is where the culprit is. With the Hamiltonian approach, we begin with presumed knowledge of initial positions and velocities (values and first derivatives of the unknown functions). Knowledge we do not have. So we evolve the system using incomplete knowledge. Then, when it comes to the measurement, we invoke our deus ex machina. Like a bad birthday party surprise, we open the magic box, pull out our “measurement apparatus” (which we pretended to not even know about up until this moment), confine the quantum system to a specific measurement value, retroactively rewrite the description of our system with the apparatus now present all along, and call this discontinuous change in the system’s description “wavefunction collapse”.
And then spend a century about its various interpretations instead of recognizing that the presumed collapse was never a physical process: rather, it amounts to us changing how we describe the system.
This is the nonsense for which I have no use, even if it makes me sound both arrogant and ignorant at the same time.
To offer a bit of a technical background to support the above (see my Web site for additional technical details): A quantum theory can be constructed starting with classical physics in a surprisingly straightforward manner. We start with the Hamiltonian (I know!), written in the following generic form:
$$H = \frac{p^2}{2m} + V({\bf q}),$$
where $${\bf p}$$ are generalized momenta, $${\bf q}$$ are generalized positions and $$m$$ is mass.
We multiply this equation by the unit complex number $$\psi=e^{i({\bf p}\cdot{\bf q}-Ht)/\hbar}.$$ We are allowed to do this trivial bit of algebra with impunity, as this factor is never zero.
Next, we notice the identities, $${\bf p}\psi=-i\hbar\nabla\psi,$$ $$H\psi=i\hbar\partial_t\psi.$$ Using these identities, we rewrite the equation as
$$i\hbar\partial_t\psi=\left[-\frac{\hbar^2}{2m}\nabla^2+V({\bf q})\right]\psi.$$
There you have it, the time-dependent Schrödinger equation in its full glory. Or… not quite, not yet. It is formally Schrödinger’s equation but the function $$\psi$$ is not some unknown function; we constructed it from the positions and momenta. But here is the thing: If two functions, $$\psi_1$$ and $$\psi_2,$$ are solutions of this equation, then because the equation is linear and homogeneous in $$\psi,$$ their linear combinations are also solutions. But these linear combinations make no sense in classical physics: they represent states of the system that are superpositions of classical states (i.e., the electron is now in two or more places at the same time.)
Quantum physics begins when we accept these superpositions as valid descriptions of a physical system (as indeed we must, because this is what experiment and observation dictates.)
The presence of a classical apparatus with which the system interacts at some future moment in time is not well captured by the Hamiltonian formalism. But the Lagrangian formalism makes it clear: it selects only those states of the system that are consistent with that interaction. This means indeed that a full quantum mechanical description of the system requires knowledge of the future. The apparent paradox is that this knowledge of the future does not causally influence the past, because the actual evolution of the system remains causal at all times: only the initial description of the system needs to be nonlocal in the same sense in which 19th century Lagrangian physics is nonlocal.
I really cannot tell which impresses me more: The incredibly complex landing or the fact that there is now a de facto infrastructure in orbit around Mars, in the form of earlier spacecraft that provide communications relay capabilities for real-time tracking of the landing.
Or perhaps the fact that Perseverance also carries the Ingenuity helicopter. If successful, it will be the first drone to fly in the atmosphere of another planet.
The children of future settlers on Mars will be learning about these moments in school.
A Dallas-Fort Worth TV station characterized Texas as the energy capital of the world as it asked the rhetorical question: How could this happen?
I have friends in Texas. One of them e-mailed me to let me know that they’ve been without power since 5 AM this morning.
Unlike the US-Canadian northeast with its interconnected power grid, Texas has its own power grid. This means, I understand, that they cannot rely on excess generating capacity in neighboring states to help with the crisis.
And the weather is bitterly cold, much colder than up here in wintry Ottawa. Right now, according to that Dallas-Fort Worth TV station it’s 8 degrees Fahrenheit but it will drop to several degrees below zero on the Fahrenheit scale overnight; that’s -20 C for us folks in metric lands.
This is not a joke. In weather like this, people can die, especially in ill-insulated homes as they struggle with unexpected secondary disasters such as bursting, frozen pipes that may very well happen.
Sometimes, simple programming mistakes make for interesting glitches.
Take this image:
No, this is not something that a gravitational lens would produce. Or any lens.
Even-numbered multipoles in a gravitational lens produce images that have fourfold symmetry. This allows me to reduce the amount of computation needed to generate an image, as I only need to generate one quarter; the rest are just copied over.
But this is not true for odd-numbered multipoles. This image was supposed to represent the J3 multipole, which would yield a triangular shape.
Unfortunately, I generated it accidentally by assuming the symmetries of an even-numbered multipole.
I kind of like the result, even though it is of course scientifically worthless. It still looks neat though.
I was recently interviewed by a Hungarian podcaster, mostly about my participation in the early days of game development in Hungary, but also about my more recent work, including my scientific contributions.
I just listened to the interview and thankfully, I didn’t say anything colossally stupid.
A very nice article about our work on the Solar Gravitational Lens was published a few days ago on Universe Today, on account of our recent preprint, which shows quantitative results assessing the impact of image reconstruction on signal and noise.
Because the SGL is such an imperfect lens, the noise penalty is substantial. However, as it turns out, it is much reduced when the projected image area is large, such as when an exoplanet in a nearby star system is targeted.
While this is good news, the Sun’s gravitational field has other imperfections. We are currently working on modeling these and assessing their impact on noise. Next comes the problem of imaging a moving target: an exoplanet that spins, which is illuminated from varying directions, and which may have varying surface features (clouds, vegetation, etc.) Accounting for all these effects is essential if we wish to translate basic theory into sound scientific and engineering requirements.
So, the fun continues. For now, it was nice to see this piece in Universe Today.
According to the immortal Douglas Adams, God’s final message to His creation is simple: “We apologize for the inconvenience.”
But there’s also another final message of sorts, the answer to the Ultimate Question about Life, Universe, and Everything: 42.
Recently, a researcher by the name of Michael Hippke analyzed the seemingly random bits that are contained in minute fluctuations of the Cosmic Microwave Background (CMB) radiation. His conclusion: there is no discernible pattern, no appearance of constants of nature, no detectable statistical autocorrelation. The message is random.
I beg to respectfully disagree. In the 512-bit segment published by Hippke, the bit sequence 101010 appears no fewer than eight, er, nine times (one occurrence split between two lines).
Now if we only knew the question to which the answer is 42…
The giant Arecibo radio telescope is no more.
Damaged by a broken cable just a few weeks ago, the telescope completely collapsed today.
Completed in 1963, the telescope was 57 years old, just like me. I hope I will last a few more years, though.
Since when is listening to the scientists a bad thing?
Now as to the claim that if he listened to the scientists, he’d have tanked the economy… here is what Mr. Trump’s pandemic response accomplished with respect to the US economy:
Compare this against China, where the totalitarian government, to their credit, actually listened to the scientists:
Even Canada, with a much less spectacular pre-pandemic economic performance than the United States, fared much better than our southern neighbor on which much of our economy depends:
So I suppose, Mr. Trump, you just gave millions of thinking Americans another reason to vote for Mr. Biden. I’d laugh but… neither the state of the US economy nor the number of dead, rapidly approaching a quarter million, qualify as laughing matter. Not to mention the seemingly inevitable march towards an increasingly illiberal, increasingly authoritarian America, driven by structural causes that go back decades and which a Biden Administration is just as unlikely to address as preceding administrations.
Tonight, Slava Turyshev sent me a link to an article that was actually published three months ago on medium.com but until now, escaped our attention.
It is a very nice summary of the work that we have been doing on the Solar Gravitational Lens to date.
It really captures the essence of our work and the challenges that we have been looking at.
And there is so much more to do! Countless more things to tackle: image reconstruction of a moving target, imperfections of the solar gravitational field, precision of navigation… not to mention the simple, basic challenge of attempting a deep space mission to a distance four times greater than anything to date, lasting several decades.
Yes, it can be done. No it’s not easy. But it’s a worthy challenge.
One thing follows another…
I’m listening to old MP3 files on my computer, one of which contains this once popular song by Paper Lace, The Night Chicago Died.
I once read that the band knew nothing about Chicago’s geography; I checked again on the Wikipedia page dedicated to this song.
The page mentions, among other things, how then Chicago mayor Daley hated the song. Daley? The same Daley who demolished Meigs Field airport, the island airport serving downtown Chicago that was the starting location of Microsoft Flight Simulator for many years?
Indeed. (Well, almost. There were two Daleys, father and son, Richard J and Richard M.) And Wikipedia tells me that the island has indeed since been turned into a park and nature preserve. But there are few pictures, so I figured I’d check it out using Google Maps.
So I typed Chicago into Google Maps and was greeted with this message in response:
I don’t know but this seems… a tad embarrassing isn’t it. Unless of course Chicago actually did die last night, and was promptly removed from Google Maps in response.
But no, Chicago is still there. The Google Maps thing was just a glitch. As is Northerly Island, which once hosted that ill-fated airport, its future as uncertain as it has always been in the past century or so.
A popular Internet meme these days is to present an arithmetic expression like, say, 6/3(4−2) and ask the poor souls who follow you to decide the right answer. Soon there will be two camps, each convinced that they know the truth and that the others are illiterate fools: According to one camp, the answer is 4, whereas the other camp will swear that it has to be 1.
In reality it is neither. Or both. Flip a coin, take your pick. There is no fundamental mathematical truth hidden here. It all boils down to human conventions. The standard convention is that multiplication and division have the same precedence and are evaluated from left to right: So 6/3×(4−2) is pretty unambiguous. But there is another, unwritten convention that when the multiplication sign is omitted, the implied multiplication is assumed to have a higher precedence.
Precisely because of these ambiguities, when you see actual professionals, mathematicians or physicists, write down an expression like this, they opt for clarity: they write, say, (6/3)(4−2) or 6/[3(4−2)] precisely so as to avoid any misunderstanding. Or better yet, they use proper math typesetting software such as LaTeX and write 2D formulas.
A tragedy took place in Beirut yesterday.
The actual power of the massive explosion is yet to be estimated accurately (probably not quite as large as the largest non-nuclear, accidental explosion that took place in Halifax, Nova Scotia 103 years ago though it comes close), but the images and videos are horrifying.
Reportedly, windows were shattered as far as 25 kilometers away from the epicenter.
The audio on one of the many videos showing the moment of the explosion accurately captures the event: “What the actual fuck?” asks a woman’s voice incredulously.
In light of the scope of the disaster, I expect that the final death toll will far exceed the 78 deceased that we know about for now.
It now appears that it was an industrial accident: welding work setting off a fire that in turn spread to a warehouse where thousands of tons of ammonium nitrate was stored.
Naturally, it didn’t prevent America’s “stable genius” from talking about an “attack”. When asked, he even referred to his “generals” who, according to him, told him that it was likely an attack.
I have no doubt that he made it all up on the spot. But his pronouncement had predictable consequences. It was like pouring oil on the fire, as it gave an excuse for every closet antisemite to come out and spread the conspiracy theory that it was an attack by Israel. Twitter accounts spreading this inflammatory nonsense include a Robert de Niro parody account; for a brief moment, I thought it was the real Robert de Niro, which would have been terribly disappointing.
Wildfires in Australia, locusts in Africa, a global pandemic, widespread racial riots in the United States, “murder hornets” spreading in North America… The Internet was already full of joke calendars for this year with disaster memes, as well as speculation that perhaps that infamous Mayan calendar was misinterpreted, as it referred to 2020, not 2012. In light of this catastrophe in Beirut, I am inclined to ask, what next? Alien invasion? The Yellowstone caldera? Global thermonuclear war? Giant asteroid impact? I won’t even try to guess, just note that we still have nearly five months left of this year.
A few weeks ago, Christian Ready published a beautiful video on his YouTube channel, Launch Pad Astronomy. In this episode, he described in detail how the Solar Gravitational Lens (SGL) works, and also our efforts so far.
I like this video very much. Especially the part that begins at 10:28, where Christian describes how the SGL can be used for image acquisition. The entire video is well worth seeing, but this segment in particular does a better job than we were ever able to do with words alone, explaining how the Sun projects an image of a distant planet to a square kilometer sized area, and how this image is scanned, one imaginary pixel at a time, by measuring the brightness of the Einstein-ring around the Sun as seen from each pixel location.
We now understand this process well, but many more challenges remain. These include, in no particular order, deviations of the Sun from spherical symmetry, minor variations in the brightness of the solar corona, the relative motion of the observing probe, Sun, exosolar system and target planet therein, changing illumination of the target, rotation of the target, changing surface features (weather, perhaps vegetation) of the target, and the devil knows what else.
Even so, lately I have become reasonably confident, based on my own simulation work and our signal-to-noise estimates, as well as a deconvolution approach under development that takes some of the aforementioned issues into consideration, that a high-resolution image of a distant planet is, in fact, obtainable using the SGL.
A lot more work remains. The fun only just began. But I am immensely proud to be able to contribute to of this effort.
I met Gabor David back in 1982 when I became a member of the team we informally named F451 (inspired by Ray Bradbury of course.) Gabor was a close friend of Ferenc Szatmari. Together, they played an instrumental role in establishing a business relationship between the Hungarian firm Novotrade and its British partner, Andromeda, developing game programs for the Commodore 64.
In the months and years that followed, we spent a lot of time working together. I was proud to enjoy Gabor’s friendship. He was very knowledgeable, and also very committed to our success. We had some stressful times, to be sure, but also a lot of fun, frantic days (and many nights!) spent working together.
I remember Gabor’s deep, loud voice, with a slight speech impediment, a mild case of rhotacism. His face, too, I can recall with almost movie like quality.
He loved coffee more than I thought possible. He once dropped by at my place, not long after I managed to destroy my coffee maker, a stovetop espresso that I accidentally left on the stove for a good half hour. Gabor entered with the words, “Kids, do you have any coffee?” I tried to explain to him that the devil’s brew in that carafe was a bitter, undrinkable (and likely unhealthy) blend of burnt coffee and burnt rubber, but to no avail: he gulped it down like it was nectar.
After I left Hungary in 1986, we remained in sporadic contact. In fact, Gabor helped me with a small loan during my initial few weeks on Austria; for this, I was very grateful.
When I first visited Hungary as a newly minted Canadian citizen, after the collapse of communism there, Gabor was one of the few close friends that I sought out. I was hugely impressed. Gabor was now heading a company called Banknet, an international joint venture bringing business grade satellite-based Internet service to the country.
When our friend Ferenc was diagnosed with lung cancer, Gabor was distraught. He tried to help Feri with financing an unconventional treatment not covered by insurance. I pitched in, too. It was not enough to save Feri’s life: he passed away shortly thereafter, a loss I still feel more than two decades later.
My last conversation with Gabor was distressing. I don’t really remember the details, but I did learn that he suffered a stroke, and that he was worried that he would be placed under some form of guardianship. Soon thereafter, I lost touch; his phone number, as I recall, was disconnected and Gabor vanished.
Every so often, I looked for him on the Internet, on social media, but to no avail. His name is not uncommon, and moreover, as his last name also doubles as a first name for many, searches bring up far too many false positives. But last night, it occurred to me to search for his name and his original profession: “Dávid Gábor” “matematikus” (mathematician).
Jackpot, if it can be called that. One of the first hits that came up was a page from Hungary’s John von Neumann Computer Society, their information technology history forum, to be specific: a short biography of Gabor, together with his picture.
And from this page I learned that Gabor passed away almost six years ago, on November 10, 2014, at the age of 72.
Well… at least I now know. It has been a privilege knowing you, Gabor, and being able to count you among my friends. I learned a lot from you, and I cherish all those times that we spent working together.
Seventy-five years ago this morning, a false dawn greeted the New Mexico desert near Alamagordo.
At 5:29 AM in the morning, the device informally known as “the gadget” exploded.
“The gadget” was a plutonium bomb with the explosive power of about 22 kilotons of TNT. It was the first nuclear explosion on planet Earth. It marked the beginning of the nuclear era.
I can only imagine what it must have been like, being part of that effort, being present in the pre-dawn hours, back in 1945. The war in Europe just ended. The war in the Pacific was still raging. This was the world’s first high technology war, fought over the horizon, fought with radio waves, and soon, to be fought with nuclear power. Yet there were so many unknowns! The Trinity test was the culmination of years of frantic effort. The outcome was by no means assured, yet the consequences were clear to all: a successful test would mean that war would never be the same. The world would never be the same.
And then, the most surreal of things happens: minutes before the planned detonation, in the pre-dawn darkness, the intercom system picks up a faint signal from a local radio station, and music starts playing. It’s almost as if reality was mimicking the atmosphere of yet-to-be-invented computer games.
When the explosion happened, the only major surprise was that the detonation was much brighter than anyone had expected. Otherwise, things unfolded pretty much as anticipated. “The gadget” worked. Success cleared the way to the deployment of the (as yet untested) simpler uranium bomb to be dropped on Hiroshima three weeks later, followed by the twin of the Trinity gadget, which ended up destroying much of Nagasaki. The human cost was staggering, yet we must not forget that it would have been dwarfed by the costs of a ground invasion of the Japanese home islands. It was a means to shorten the war, a war not started by the United States. No responsible commander-in-chief could have made a decision other than the one Truman made when he approved the use of the weapons against Imperial Japan.
And perhaps the horrors seen in those two cities played a role in creating a world in which the last use of a nuclear weapon in anger occurred nearly 75 years ago, on August 9, 1945. No one would have predicted back then that there will be no nuclear weapons deployed in war in the coming three quarters of a century. Yet here we are, in 2020, struggling with a pandemic, struggling with populism and other forces undermining our world order, yet still largely peaceful, living in a golden age unprecedented in human history.
Perhaps Trinity should serve as a reminder that peace and prosperity can be fragile.
This may not be an all-time record-breaking day according to Environment Canada (supposedly, the peak temperature today at Ottawa Airport was 34.8 C at 2 PM) but it sure is hot.
You could be forgiven if you thought that this measurement is of the body temperature of a COVID-19 patient with mild symptoms, not the outdoor temperature on our balcony, measured in the shade:
As I said… really hot. Praise be to air conditioning.
Long overdue, but I just finished preparing the latest Maxima release, version 5.44.
I am always nervous when I do this. It is one thing to mess with my own projects, it is another thing to mess with a project that is the work of many people and contains code all the way back from the 1960s.
In case anyone doubted that modern birds are descendants of dinosaurs, here is a reminder: the shoebill.
These amazing creatures are apparently quite docile with humans, but eat baby crocodiles for lunch, which they kill by decapitating them.
They really look like survivors of the K-T asteroid impact. They are… I think they are beautiful.
I don’t always like commercial publishers. Some of their textbooks are prohibitively expensive, yet often lacking in quality. (One persistent exception is Dover Publications, who published some of the best textbooks I own, as low-cost paperbacks.)
Last night, however, I was very pleasantly surprised by Springer, who made several hundred textbooks across a range of disciplines available for free, on account of COVID-19.
I did not get greedy. I didn’t download titles indiscriminately. But I did find several titles that are of interest to me, and I gladly took advantage of this opportunity.
Thank you, Springer.
|
2021-03-02 02:29:56
|
{"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": 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": 0.44003060460090637, "perplexity": 1295.5486967463817}, "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-2021-10/segments/1614178363211.17/warc/CC-MAIN-20210302003534-20210302033534-00595.warc.gz"}
|
https://forums.positech.co.uk/t/mydocs-democracy2-folder-missing/1589
|
# MyDocs\democracy2 folder missing
#1
Hello everyone. I have been playing for a few days now, and after winning some and losing many I wanted to try some mods.
The instructions on how and where to correctly place the mod files are very clear, problem is: I don’t have a My Documents\democracy2 folder at all !!!
After fumbling around a bit, I found out that Program Files\democracy2 folder contains some of the folders requested for mods installation, so I tried put the files in the correspondent folders, and copied and paste the text from the voter type in the VoterTypes file.
It looks like it is working mainly, when I open the game it gives me the option of the new mission, but I am afraid I may have made a mess because the demographic doesnt look right to me.
Before doing all this I tried just to create a folder in MyDocs named democracy2 with all the folders but the game simply didnt see it.
Can someone of you helpful people tell me how to correctly install the mods files?
I run windows xp pro 64-bit and democracy2 is in the Program Files (x86) folder.
Thank you in advance, and thank you for a great game.
pixxa
#2
You can put most of the files in the Program Files\Democracy folder. However, I don’t think that it’s possible to do this with voter types - I certainly don’t think that you can just paste text into the voter types file. I’m not sure exactly what you’re saying is wrong - if you could clarify this then I may be able to be more help. However, you are best to remove the voter types from the voter types csv file as these won’t work.
In terms of locating the My Documents/Democracy 2, are you sure that you’re looking in the right place? It’s only a guess, but the My Documents folders can be confusing. Is there more than one user account on your computer - could the folder have been created for a different user account to the one you are using? Also, make sure that you’re not looking in the “Shared Documents” or “All Users documents” folder - you need to be looking in the one for the relevant user account. Perhaps you could run a search on your computer for “democracy2”. Alternatively, you could try reinstalling the game.
Hope this helps.
#3
I know where to look for the My Documents folder, I use it regularly as it contains folders from other games I play like the Sims, SimCity etc. But there is no democracy2 folder.
I reinstalled the game and still no folder in the My Documents folder. But I found out that the mods folder may be actually located in the program files folder. I went through many posts in here to see if anyone had the same folder structure but it looks like I am the only one!
Directory of C:\Program Files (x86)\Democracy2\userdir
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\backups
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\autorun scripts
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\bitmaps
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\missions
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\dilemmas
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\encyclopedia
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\events
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\policies
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\simulation
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\simulation\votertypes
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\situations
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\mods\data\sliders
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\savegames
Directory of C:\Program Files (x86)\Democracy2\userdir\democracy2\web
As you can see all those folders are in the program main folder.
Will I use these folders for the mods? If I run the autoinstallers I get the error “wrong path” so I will have to do it manually, which is not a problem; I just want to be sure that I can use these folders, or if they are used by the program for something different.
pixxa
#4
You are correct that there is a folder in the Program Files\Democracy 2 directory that looks very similar to the one that should be in My Documents. This contains all of the data that comes with the game. In Democracy 1, I think, all mods were put in this directory with the other content.
However, in Democracy 2 a new folder was created in the My Documents folder which allowed mods to be kept separate from the main game data. Originally, it was not possible to put some of the mod data in here, which may lead to some of the instructions for early mods giving complicated instructions regarding putting some things in the Program Files directory and some in the My Documents one.
You should be able to put everything except voter groups into the relevant Program Files directory, however. That should work fine.
This of course doesn’t solve your problem that the My Documents\Democracy 2 folder is missing. If you put the mods in the Program Files directory, then you should be fine and the missing directory shouldn’t cause a problem.
However, if you find that not having the My Documents\Democracy 2 directory is causing a problem, then if you post in the main Democracy 2 forum, rather than this one, you should get a faster response, as I think Cliff, who programs the games, checks there more often.
Hope this helps.
#5
Thank you TomPhil for the explaination. I will try and do as you say, and see if the mods work. In case I still have problems I will post on the other forum, but I really hope there won’t be need for that.
pixxa
#6
This could be related to the O/S. I use GetVersionEx () to detect your O/S, because it needs to be XP or later to use the mydocuments stuff (not win98). I reckon my version checking is misidentifying your 64 bit version. maybe it gives a different version number. If it does, it sticks everything that should be in MyDocuments in a folder called ‘userdir’ under Democracy2 (or it should do).
Does that help?
#7
Another 64-bit Vista user here. The Deomcracy 2 folder that I have is located in the C:\Users<user name>\Documents folder, and not the MyDocuments folder. Does this help any?
As an aside, I don’t have a userdir folder under Democracy2, but that may be because I have it at the other location.
#8
I don’t even have a folder like this in my mac
|
2018-08-21 08:36:21
|
{"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": 0.8632779717445374, "perplexity": 1664.3968230227008}, "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/1534221218070.73/warc/CC-MAIN-20180821073241-20180821093241-00250.warc.gz"}
|
https://salsa.debian.org/r-pkg-team/r-bioc-phyloseq/blame/f83e9d97005eddfedc55e2d7c35c6039a91cd3c0/man/import_pyrotagger_tab.Rd
|
import_pyrotagger_tab.Rd 2.98 KB
Andreas Tille committed Sep 22, 2017 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 % Generated by roxygen2: do not edit by hand % Please edit documentation in R/IO-methods.R \name{import_pyrotagger_tab} \alias{import_pyrotagger_tab} \title{Imports a tab-delimited version of the pyrotagger output file.} \usage{ import_pyrotagger_tab(pyrotagger_tab_file, strict_taxonomy=FALSE, keep_potential_chimeras=FALSE) } \arguments{ \item{pyrotagger_tab_file}{(Required). A character string. The name of the tab-delimited pyrotagger output table.} \item{strict_taxonomy}{(Optional). Logical. Default \code{FALSE}. Should the taxonomyTable component be limited to just taxonomic data? Default includes all fields from the pyrotagger file.} \item{keep_potential_chimeras}{(Optional). Logical. Default \code{FALSE}. The pyrotagger output also includes OTUs that are tagged by pyrotagger as likely chimeras. These putative chimeric OTUs can be retained if set to \code{TRUE}. The putative chimeras are excluded by default.} } \value{ An \code{otuTax} object containing both the otu_table and TaxonomyTable data components, parsed from the pyrotagger output. } \description{ PyroTagger is a web-server that takes raw, barcoded 16S rRNA amplicon sequences and returns an excel spreadsheet (\code{".xls"}) with both abundance and taxonomy data. It also includes some confidence information related to the taxonomic assignment. } \details{ PyroTagger is created and maintained by the Joint Genome Institute at \code{"http://pyrotagger.jgi-psf.org/"} The typical output form PyroTagger is a spreadsheet format \code{".xls"}, which poses additional import challenges. However, virtually all spreadsheet applications support the \code{".xls"} format, and can further export this file in a tab-delimited format. It is recommended that you convert the xls-file without any modification (as tempting as it might be once you have loaded it) into a tab-delimited text file. Deselect any options to encapsulate fields in quotes, as extra quotes around each cell's contents might cause problems during file processing. These quotes will also inflate the file-size, so leave them out as much as possible, while also resisting any temptation to modify the xls-file by hand''. A highly-functional and free spreadsheet application can be obtained as part of the cross-platform \code{OpenOffice} suite. It works for the above required conversion. Go to \code{"http://www.openoffice.org/"}. It is regrettable that this importer does not take the xls-file directly as input. However, because of the moving-target nature of spreadsheet file formats, there is limited support for direct import of these formats into \code{R}. Rather than add to the dependency requirements of emph{phyloseq} and the relative support of these xls-support packages, it seems more efficient to choose an arbitrary delimited text format, and focus on the data structure in the PyroTagger output. This will be easier to support in the long-run. } \examples{ ## New_otuTaxObject <- import_pyrotagger_tab(pyrotagger_tab_file) } \references{ \url{http://pyrotagger.jgi-psf.org/} }
|
2019-12-14 12:06:21
|
{"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": 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": 0.31236353516578674, "perplexity": 2767.9914074669086}, "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-2019-51/segments/1575540586560.45/warc/CC-MAIN-20191214094407-20191214122407-00436.warc.gz"}
|
https://scicomp.stackexchange.com/questions/7288/which-preconditioners-and-solver-in-petsc-for-indefinite-symmetric-systems-sho/7291
|
# Which preconditioners (and solver) in PETSc for indefinite symmetric systems should I use?
My system is a symmetric FE problem with lagrange multipliers (e.g. incompressible Stokes' flow):
\begin{pmatrix}A & B^T \\ B & C\end{pmatrix}
where $C = 0$ is the typical case (I have even made sure that the equations are numbered so that the Lagrange multipliers appear last ). The system is quite large (+100k lines).
Having read the answer to this question, I was given the impression that there are suitable preconditioners that can be used for mixed FE-problems.
Using PETSc, I've managed to solve the system with MINRES (-ksp_type minres -pc_type none -mat_type sbaij ), although the precision isn't great (causing several Newton-iterations for a linear problem). No other combination of preconditioner and ksp-solver seems to work.
Is there any combination of flags for PETSc that will solve this system faster than with just MINRES?
• Welcome to SciComp SE! Your question is well posed in terms of clearity and generality. Only one sentence is somewhat unclear or incomplete: What do you mean is caused by the inexact computations?
– Jan
May 20 '13 at 14:55
• I forgot to finish the sentence! I hope it is clear now. May 20 '13 at 15:38
### Warning
Solving saddle point problems involves a lot more choices than definite problems, and there are a lot more things that can go wrong. Use monitors for all levels to debug convergence, to sure that null spaces are handled correctly when auxiliary operators are singular (usually just a constant null space), and to ensure that preconditioners are stable. These techniques are discussed in this question.
You can start with the section on solving block matrices in the User's Manual. For Stokes-like problems, you can build a Schur complement preconditioner using PCFIELDSPLIT.
-pc_type fieldsplit -pc_fieldsplit_type schur -pc_fieldsplit_detect_saddle_point
This may be combined with a Least Squares Commutator preconditioner for the Schur complement (-fieldsplit_1_pc_type lsc). It is common to -fieldsplit_1_ksp_type preonly and let the outer iteration do most of the work. The block-triangular variants are popular when used in this way, e.g., -pc_fieldsplit_schur_fact_type upper.
You can find further details on solver composition using options options in our paper (preprint), which also discusses commuting multigrid with the block decomposition (putting the fieldsplit inside multigrid).
For many problems, you'll want to customize the field decomposition and Schur complement preconditioners, many of which involve auxiliary operators. For example, the "pressure convection-diffusion" (PCD) method from Elman et al requires discretization of an auxiliary operator in the pressure space. See examples of PCSHELL and the user's manual section for this purpose.
You do want a preconditioner, and the construction of one is discussed here: http://www.math.tamu.edu/~bangerth/videos.676.38.html
• The lecture was very interesting, but I was looking for some concrete examples on how to use PETSc to solver it. May 20 '13 at 14:42
|
2021-09-28 22:17:19
|
{"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": 0.8381900191307068, "perplexity": 1182.547848677008}, "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/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00079.warc.gz"}
|
https://physics.aps.org/synopsis-for/10.1103/PhysRevLett.108.021601
|
# Synopsis: Out of Bounds
Liquids that approach an often-quoted lower limit for viscosity are deemed “perfect,” but now this lower limit is itself being questioned.
Is there a limit on how viscous fluids can be? Various string-theory-based calculations predict that the ratio of a fluid’s shear viscosity to its entropy density is bounded from below by $\mathrm{ħ}/4\pi$ (or, $\mathrm{ħ}/\left(4\pi \phantom{\rule{0}{0ex}}{k}_{B}$), depending on the units). So far, only highly idealized models of fluids approach this limit, but physicists are actively debating if the quark-gluon plasma, generated in relativistic heavy-ion collisions, might be an experimentally accessible example of a “perfect liquid” (See 26 October 2009 Trends).
Now, in a paper published in Physical Review Letters, Anton Rebhan and Dominik Steineder, both at the Vienna University of Technology in Austria, show that the theoretical bound on viscosity may well be violated within the idealized setup of string theory itself. Rebhan and Steineder consider the string-theory-based description of an idealized plasma with an intrinsic spatial anisotropy, and show that the bound is indeed violated, and that the amount of violation is directly related to the spatial anisotropy of the plasma. Given that the real quark-gluon plasma is produced in a highly anisotropic situation — immediately after the nuclear collisions, the resulting plasma expands predominantly along the beam axis — the implications of this theoretical finding on the physics of relativistic heavy-ion collisions is keenly awaited. The obvious challenge is to understand if any of the implications of Rebhan and Steineder’s calculations, which are based on a highly idealized model, can be extrapolated to a “real life” experimental scenario. – Abhishek Agarwal
More Features »
### Announcements
More Announcements »
Astrophysics
Nuclear Physics
## Related Articles
Fluid Dynamics
### Synopsis: Eyeing the Storm
Numerical simulations of a hurricane-like system have determined the conditions necessary for the formation of a calm “eye” in the center of the storm. Read More »
Astrophysics
### Synopsis: Space Measurements of Secondary Cosmic Rays
New data from the International Space Station shed light on how secondary cosmic rays propagate through space. Read More »
Fluid Dynamics
### Synopsis: The Coldest Water
A new technique has measured the lowest temperature ever recorded for liquid water. Read More »
|
2018-01-22 12:05:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.6914150714874268, "perplexity": 1598.8385325305353}, "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-2018-05/segments/1516084891316.80/warc/CC-MAIN-20180122113633-20180122133633-00721.warc.gz"}
|
https://rdrr.io/bioc/NCIgraph/src/R/translateNCI2GeneID.R
|
# R/translateNCI2GeneID.R In NCIgraph: Pathways from the NCI Pathways Database
#### Documented in translateNCI2GeneID
## Copyright 2011 Laurent Jacob
## This file is part of NCIgraph.
## NCIgraph is free software: you can redistribute it and/or modify
## it under the terms of the GNU General Public License as published by
## the Free Software Foundation, either version 3 of the License, or
## (at your option) any later version.
## NCIgraph is distributed in the hope that it will be useful,
## but WITHOUT ANY WARRANTY without even the implied warranty of
## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
## GNU General Public License for more details.
## You should have received a copy of the GNU General Public License
## along with NCIgraph. If not, see <http://www.gnu.org/licenses/>.
#########################################################################/**
# @RdocFunction translateNCI2GeneID
##
## @title "Gives the entrezID corresponding to the nodes of a graph"
##
##
## \description{
## @get "title".
## }
##
## @synopsis
##
## \arguments{
## \item{g}{A \code{\link[=graph-class]{graph}} object.}
## }
##
## \value{A vector of @character giving the entrez ID of the nodes of g.}
##
## @author
##
## \seealso{
## @see "parseNCInetwork"
## }
##
## @examples "../inst/extdata/translateNCI2GeneID.Rex"
##
##*/########################################################################
translateNCI2GeneID <- function(g)
{
if(length(nodes(g)) == 0){
return(NULL)
}
lambdaf <- function(node)
{
eid <- node$biopax.xref.ENTREZGENE if(length(unique(eid))>1) warning(sprintf('More than one entrez ID associated with node %s\n',node$nodeName))
if(!is.null(eid))
eid[1]
else
NA
}
eids <- unlist(lapply(g@nodeData@data,FUN=lambdaf))
names(eids) <- nodes(g)
return(eids)
}
## Try the NCIgraph package in your browser
Any scripts or data that you put into this service are public.
NCIgraph documentation built on Nov. 8, 2020, 5:54 p.m.
|
2022-10-06 03:19:26
|
{"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": 0.5708297491073608, "perplexity": 13415.234438364056}, "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/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00474.warc.gz"}
|
https://www.groundai.com/project/transforming-quantum-operations-quantum-supermaps/
|
Transforming quantum operations: quantum supermaps
Transforming quantum operations: quantum supermaps
Giulio Chiribella QUIT Group, Dipartimento di Fisica “A. Volta” and INFM, via Bassi 6, 27100 Pavia, Italy http://www.qubit.it Giacomo Mauro D’Ariano QUIT Group, Dipartimento di Fisica “A. Volta” and INFM, via Bassi 6, 27100 Pavia, Italy http://www.qubit.it Paolo Perinotti QUIT Group, Dipartimento di Fisica “A. Volta” and INFM, via Bassi 6, 27100 Pavia, Italy http://www.qubit.it
July 14, 2019
Abstract
We introduce the concept of quantum supermap, describing the most general transformation that maps an input quantum operation into an output quantum operation. Since quantum operations include as special cases quantum states, effects, and measurements, quantum supermaps describe all possible transformations between elementary quantum objects (quantum systems as well as quantum devices). After giving the axiomatic definition of supermap, we prove a realization theorem, which shows that any supermap can be physically implemented as a simple quantum circuit. Applications to quantum programming, cloning, discrimination, estimation, information-disturbance trade-off, and tomography of channels are outlined.
pacs:
03.65.Ta, 03.67.-a
The input-output description of any quantum device is provided by the quantum operation of Kraus kraus (), which yields the most general probabilistic evolution of a quantum state. Precisely, the output state is given by the quantum operation applied to the input state as follows
(1)
where is the probability of occurring on state , when is one of a set of alternative transformations, such as in a quantum measurement. Owing to its physical meaning, a quantum operation must be a linear, trace non-increasing, completely positive (CP) map (see, e.g. Nielsen2000 ()). The most general form of such a map is known as Kraus form
E(ρ)=∑jEjρE†j, (2)
where the operators satisfy the bound so that . Trace-preserving maps, i.e. those achieving the bound, are a particular kind of quantum operations: they occur deterministically and are referred to as quantum channels.
In general it is convenient to consider two different input and output Hilbert spaces and , respectively. In this way, the concept of quantum operation can be used to treat also quantum states, effects, and measurements, which describe the properties of elementary quantum objects such as quantum systems and measuring devices. Indeed, states can be described as quantum operations with one-dimensional , i.e. with Kraus operators given by ket-vectors , thus yielding the output state . A quantum effect ludwig () corresponds instead to a quantum operation with one-dimensional , i.e. with Kraus operators given by bra-vectors , yielding the probability with . More generally, any quantum measurement can be viewed as a particular quantum operation, namely as a quantum-to-classical channel qc ().
Channels, states, effects, and measurements are all special cases of quantum operations. What about then considering maps between quantum operations themselves? They would describe the most general kind of transformations between elementary quantum objects. For example a programmable channel NielsenProg () would be a map of this type, with a quantum state at the input and a channel at the output. Or else, a device that optimally clones a set of unknown unitary gates would be a map from channels to channels. We will call such a general class of quantum maps quantum supermaps, as they transform CP maps (sometimes referred to as superoperators) into CP maps.
In this paper we develop the basic tools to deal with quantum supermaps. The concept of quantum supermap is first introduced axiomatically, by fixing the minimal requirements that a map between quantum operations must fulfill. We then prove a realization theorem that provides any supermap with a physical implementation in terms of a simple quantum circuit with two open ports in which the input operation can be plugged. This result allows one to simplify the description of complex quantum circuits and to prove general theorems in quantum information theory. Moreover, the generality of the concept of supermap makes it fit for application in many different contexts, among which quantum programming, calibration, cloning, and estimation of devices.
To start with, we define the deterministic supermaps as those sending channels to channels. Conversely, a probabilistic supermap will send channels to arbitrary trace-non-increasing quantum operations. The minimal requirements that a deterministic supermap must satisfy in order to be physical are the following: it must be i) linear and ii) completely positive. Linearity is required to be consistent with the probabilistic interpretation. Indeed, if the input is a random choice of quantum operations , the output must be given by the same random choice of the transformed operations , and, if the input is the quantum operation with probability , the output must be the with probability , implying . Clearly, these two conditions imply that is a linear map on the linear space generated by quantum operations. Complete positivity is needed to ensure that the output of is a legitimate quantum operation even when is applied locally to a bipartite joint quantum operation, i.e. a quantum operation with bipartite input space and bipartite output space . If is a supermap transforming quantum operations with input (output) space (), complete positivity corresponds to require that is a CP map for any bipartite quantum operation , denoting the identity supermap on the spaces labeled by .
In order to deal with complete positivity it is convenient to use the Choi representation choi () of a CP map in terms of the positive operator on
E:=E⊗I(|I⟩⟨I|), (3)
where is the maximally entangled vector , an orthonormal basis, and is the identity operation. The correspondence is one-to-one, the inverse relation of Eq. (3) being
E(ρ):=TrHin[(I⊗ρ⊺)E], (4)
where denotes transposition in the basis . In terms of the Choi operator, the probability of occurrence of is given by , where is the effect . To have unit probability on any state, a quantum channel must have , i.e. its Choi operator must satisfy the normalization
TrHout[E]=IHin . (5)
A supermap maps quantum operations into quantum operations as . In the Choi representation, the supermap induces a linear map on Choi operators, as . Using Eq. (4), we can get back from as follows
E′(ρ)=~S(E)(ρ)=TrKout[(I⊗ρ⊺)S(E)]. (6)
Of course complete positivity of implies that the map is positive. On the other hand, it is easily seen that the bipartite structure of a joint operation over a composite system induces a bipartite structure of the Choi operator . The local application of the supermap —given by —then corresponds to the local application of —given by —whence is CP if and only if is CP.
Since the correspondence is one-to-one, in the following we will focus our attention on . The supermap sends positive operators on to positive operators on generally different Hilbert spaces . Complete positivity of is equivalent to the existence of a Kraus form
S(E)=∑iSiES†i, (7)
where are operators from to .
The following Lemmas provide the characterization of deterministic supermaps:
Lemma 1
Any linear operator on such that for all Choi operators of channels has the form , with on satifying . For one has .
Proof. Consider a Choi operator with effect . Upon defining for some state on we have that is the Choi operator of a channel, normalized as in Eq. (5). Since by hypothesis , we have that
Tr[CE]= 1−Tr[CD]=1−Tr[C(σ⊗I)]+Tr[C(σ⊗P)] = Tr[C(σ⊗P)], (8)
since is the Choi operator of a channel. Therefore,
Tr[CE]=Tr[C(σ⊗P)]=Tr[PTrHout[C(σ⊗I)]]=Tr[TrHout[E]TrHout[C(σ⊗I)]]=Tr[ρP], (9)
where . Since does not depend on , the last equality can be rewritten as for all positive , whence , and in order to have for all , we must have . Clearly implies .
Lemma 2
The supermap is deterministic iff there exists a channel from states on to states on such that, for any state on , one has
S∗(IKout⊗ρ)=IHout⊗N∗(ρ) , (10)
where is the dual map of defined in terms of the Kraus form in Eq. (7) by
S∗(O):=∑iS†iOSi . (11)
Proof. One has . Consider a positive operator on , where is a state on . We have that
1=Tr[CS(E)]=Tr[S∗(C)E], (12)
for all Choi operators of channels. According to Lemma 7, this implies , where is a state for any state . Since the maps , and are all CP, we have , where is a CP trace preserving map from states on to states on .
Remarkably, the same mathematical structure of Lemma 2 characterizes semi-causal quantum operations ESW (), i.e. operations on bipartite systems that allow signaling from system to system but not viceversa. In our case, this structure originates from the causality of input-output relations. An equivalent condition for a supermap to be deterministic is given by the following:
Lemma 3
The supermap is deterministic iff there exists an identity preserving completely positive map such that, for any operator on , one has
TrKout[S(E)]=N(TrHout[E]). (13)
Proof. This lemma follows from the previous one by considering that
Tr[ρTrKout[S(E)]]=Tr[(I⊗ρ)S(E)]=Tr[S∗(I⊗ρ)E]=Tr[(I⊗N∗(ρ))E]=Tr[(I⊗ρ)(I⊗N)(E)]=Tr[ρN(TrHout[E])], (14)
for all states on . The map is identity preserving because it represents in the Schrödinger picture.
Eq. (13) shows that the effect depends only on the effect , e. g. not on . Basically, this reflects the fact that, in the input/output bipartition of the Choi operator, the output must not influence the transformation of the input effect.
Now we show that deterministic supermaps, so far introduced on a purely axiomatic level, can be physically realized with simple quantum circuits. Upon writing a canonical Kraus form for the completely positive map as follows
N(P)=∑lN†lPNl, (15)
and substituting the Kraus forms (7) and (15) into Eq. (13), one obtains
∑n(⟨kn|⊗I)SiES†i(I⊗|kn⟩)=∑m(⟨hm|⊗N†j)E(Nj⊗|hm⟩), (16)
where and are orthonormal basis for and , respectively, and identity operators must be considered as acting on the appropriate Hilbert spaces— on the top and on the bottom part of Eq. (16). Eq. (16) gives two equivalent Kraus forms for the same CP map, of which the second one is canonical (since is canonical and are orthogonal). Therefore, there exists an isometry connecting the two sets of Kraus operators as follows
(⟨kn|⊗I)Si=∑mjWni,mj(⟨hm|⊗N†j), (17)
with . Explicitly
Wni,mj:=(⟨kn|⊗⟨ai|)W(|hm⟩⊗|bj⟩), (18)
where and are orthonormal basis for two ancillary systems with Hilbert spaces and . From Eq. (17) we then obtain
Si=(I⊗⟨ai|)W(I⊗Z), (19)
where
Z=∑j|bj⟩⊗N†j . (20)
Using Eq. (7) we can now evaluate the output Choi operator as follows
S(E)=TrA[W(I⊗Z)E(I⊗Z†)W†]. (21)
Finally, using Eq. (6) we get
E′(ρ)=TrKin[(I⊗ρ⊺)S(E)]=TrKin⊗A[(IKout⊗A⊗ρ⊺)W(I⊗Z)E(I⊗Z†)W†]=TrA[W(E⊗IB)(VρV†)W†], (22)
where is the partial transposed of (see Eq. (20)) on the second space. Since the map is identity preserving, is an isometry, namely . We have then proved the following realization theorem
Theorem 1
Every deterministic supermap can be realized by a four-port quantum circuit where the input operation is inserted between two isometries and and a final ancilla is discarded as in Fig. Transforming quantum operations: quantum supermaps. The output operation is given by
~S(E)(ρ)=TrA[W(E⊗IB)(VρV†)W†]. (23)
Since any isometry can be realized as a unitary interaction with an ancilla initialized in some reset state, the above Theorem entails a realization of supermaps in terms of unitary gates. However, we preferred stating it in terms of isometries in order to avoid the arbitrariness in the choice of the initial ancilla state.
You are adding the first comment!
How to quickly get a good reply:
• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
• Your comment should inspire ideas to flow and help the author improves the paper.
The better we are at sharing our knowledge with each other, the faster we move forward.
The feedback must be of minimum 40 characters and the title a minimum of 5 characters
|
2021-01-24 05:43:03
|
{"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": 0.9090800285339355, "perplexity": 810.540548306124}, "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-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00451.warc.gz"}
|
https://dmoj.ca/problem/fur1
|
## Cyclopian Air
View as PDF
Points: 3
Time limit: 1.0s
Memory limit: 16M
Author:
Problem type
##### PEG 11/12 Programming Test 1 - October 19
The upscale clothing company FurWear has recently expanded to the planet of Cyclopia. Cyclopia, however, does not have an atmosphere with oxygen, something that the workers producing FurWear products require to work. (They need food and water too, but they can figure that out on their own!) The Oelfinn, as the dominant species of Cyclopia, have a ready supply of oxygen, but they aren't willing to give it away freely. They do have a love of puzzles, however.
The Oelfinn have placed some numbers on some cards arranged in a circle, and will allow FurWear to pick one pair of adjacent cards. The Oelfinn are willing to give FurWear an amount of oxygen proportional to the product of the numbers on said cards.
Obviously, FurWear would like to maximise the amount of oxygen they receive, and have turned to you to help. Your task is to write a program that will find the largest product of any two adjacent numbers, which will then be stored on a chip implanted into a peon.
#### Input Specification
The first line consists of a single integer . The following lines each contain a single integer , the number on the th card. The numbers will be given from the northernmost card, then clockwise all the way around.
#### Output Specification
On a single line, output the maximum product of any two adjacent numbers.
#### Sample Input
10
2
5
6
12
76
0
200
1
32
14
#### Sample Output
912
|
2020-10-21 22:03:55
|
{"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": 0.22394013404846191, "perplexity": 1781.4885541036706}, "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/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00293.warc.gz"}
|
https://www.researching.cn/articles/OJ8830d2177f15b2ba/equations
|
Search by keywords or author
• Vol. 1, Issue 1, 016003 (2022)
Shiyao Fu1、2、3、*, Zijun Shang1、2、3, Lan Hai1、2、3, Lei Huang1、2、3, Yanlai Lv1、2、3, and Chunqing Gao1、2、3、*
Author Affiliations
• 1Beijing Institute of Technology, School of Optics and Photonics, Beijing, China
• 2Ministry of Industry and Information Technology of the People’s Republic of China, Key Laboratory of Information Photonics Technology, Beijing, China
• 3Ministry of Education of the People’s Republic of China, Key Laboratory of Photoelectronic Imaging Technology and System, Beijing, China
• show less
Shiyao Fu, Zijun Shang, Lan Hai, Lei Huang, Yanlai Lv, Chunqing Gao. Orbital angular momentum comb generation from azimuthal binary phases[J]. Advanced Photonics Nexus, 2022, 1(1): 016003 Copy Citation Text show less
Abstract
Since Allen et al. demonstrated 30 years ago that beams with helical wavefronts carry orbital angular momentum (OAM), the OAM of beams has attracted extensive attention and stimulated lots of applications in both classical and quantum physics. Akin to an optical frequency comb, a beam can carry multiple various OAM components simultaneously. A series of discrete, equally spaced, and equally weighted OAM modes comprise an OAM comb. Inspired by the spatially extended laser lattice, we demonstrate both theoretically and experimentally an approach to producing OAM combs through azimuthal binary phases. Our study shows that transition points in the azimuth determine the OAM distributions of diffracted beams. Multiple azimuthal transition points lead to a wide OAM spectrum. Moreover, an OAM comb with any mode spacing is achievable through reasonably setting the position and number of azimuthal transition points. The experimental results fit well with theory. This work presents a simple approach that opens new prospects for OAM spectrum manipulation and paves the way for many applications including OAM-based high-security encryption and optical data transmission, and other advanced applications.
Video Introduction to the Article
$PB={0,if Pspp(1)
#### View in Article
$PB(φ)=π∑κ=0K(−1)κ+1rect(φφκ−0.5),φ∈[0,2π),$(2)
#### View in Article
$E(r,φ)=12π∑l=−∞+∞alexp(ilφ).$(3)
#### View in Article
$al=12π∫02πE(r,φ)exp(−ilφ)dφ.$(4)
#### View in Article
Shiyao Fu, Zijun Shang, Lan Hai, Lei Huang, Yanlai Lv, Chunqing Gao. Orbital angular momentum comb generation from azimuthal binary phases[J]. Advanced Photonics Nexus, 2022, 1(1): 016003
|
2023-04-02 00:24:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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": 0.2954877018928528, "perplexity": 8817.847027290054}, "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-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00162.warc.gz"}
|
http://2012books.lardbucket.org/books/finance-banking-and-money-v2.0/s23-01-the-simple-quantity-theory-and.html
|
This is “The Simple Quantity Theory and the Liquidity Preference Theory of Keynes”, section 20.1 from the book Finance, Banking, and Money (v. 2.0). For details on it (including licensing), click here.
Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.
## 20.1 The Simple Quantity Theory and the Liquidity Preference Theory of Keynes
### Learning Objective
1. What is the liquidity preference theory, and how has it been improved?
The rest of this book is about monetary theory, a daunting-sounding term. It’s not the easiest aspect of money and banking, but it isn’t terribly taxing either so there is no need to freak out. We’re going to take it nice and slow. And here’s a big hint: you already know most of the outcomes because we’ve discussed them already in more intuitive terms. In the chapters that follow, we’re simply going to provide you with more formal ways of thinking about how the money supply determines output (Y*) and the price level (P*).
Intuitively, people want to hold a certain amount of cash because it is by definition the most liquid asset in the economy. It can be exchanged for goods at no cost other than the opportunity cost of holding a less liquid income–generating asset instead. When interest rates are low (high), so is the opportunity cost, so people hold more (less) cash. Similarly, when inflation is low (high), people are more (less) likely to hold assets, like cash, that lose purchasing power. Think about it: would you be more likely to keep \$100 in your pocket if you believed that prices were constant and your bank pays you .00005% interest, or if you thought that the prices of the things you buy (like gasoline and food) were going up soon and your bank pays depositors 20% interest? (I would hope the former. If the latter, I have some derivative bridge securities to sell you.)
We’ll start our theorizing with the demand for money, specifically the simple quantity theory of money, then discuss John Maynard Keynes’s improvement on it, called the liquidity preference theory, and end with Milton Friedman’s improvement on Keynes’ theory, the modern quantity theory of money.
John Maynard Keynes (to distinguish him from his father, economist John Neville Keynes) developed the liquidity preference theory in response to the pre-Friedman quantity theory of money, which was simply an assumption-laden identity called the equation of exchange:
$M V = P Y$
where
M = money supply
V = velocity
P = price level
Y = output
Nobody doubted the equation itself, which, as an identity (like x = x), is undeniable. But many doubted the way that classical quantity theorists used the equation of exchange as the causal statement: increases in the money supply lead to proportional increases in the price level, although in the long term it was highly predictive. The classical quantity theory also suffered by assuming that money velocity, the number of times per year a unit of currency was spent, was constant. Although a good first approximation of reality, the classical quantity theory, which critics derided as the “naïve quantity theory of money,” was hardly the entire story. In particular, it could not explain why velocity was pro-cyclical, i.e., why it increased during business expansions and decreased during recessions.
To find a better theory, Keynes took a different point of departure, asking in effect, “Why do economic agents hold money?” He came up with three reasons:
1. Transactions: Economic agents need money to make payments. As their incomes rise, so, too, do the number and value of those payments, so this part of money demand is proportional to income.
2. Precautions: S—t happens was a catch phrase of the 1980s, recalled perhaps most famously in the hit movie Forrest Gump. Way back in the 1930s, Keynes already knew that bad stuff happens—and that one defense against it was to keep some spare cash lying around as a precaution. It, too, is directly proportional to income, Keynes believed.
3. Speculations: People will hold more bonds than money when interest rates are high for two reasons. The opportunity cost of holding money (which Keynes assumed has zero return) is higher, and the expectation is that interest rates will fall, raising the price of bonds. When interest rates are low, the opportunity cost of holding money is low, and the expectation is that rates will rise, decreasing the price of bonds. So people hold larger money balances when rates are low. Overall, then, money demand and interest rates are inversely related.
More formally, Keynes’s ideas can be stated as
where
Md/P = demand for real money balances
f means “function of” (this simplifies the mathematics)
i = interest rate
Y = output (income)
<+> = increases in
<−> = decreases in
An increase in interest rates induces people to decrease real money balances for a given income level, implying that velocity must be higher. So Keynes’s view was superior to the classical quantity theory of money because he showed that velocity is not constant but rather is positively related to interest rates, thereby explaining its pro-cyclical nature. (Interest rates rise during expansions and fall during recessions.) Keynes’s theory was also fruitful because it induced other scholars to elaborate on it further.
In the early 1950s, for example, a young Will Baumolpages.stern.nyu.edu/~wbaumol and James Tobinnobelprize.org/nobel_prizes/economics/laureates/1981/tobin-autobio.html independently showed that money balances, held for transaction purposes (not just speculative ones), were sensitive to interest rates, even if the return on money was zero. That is because people can hold bonds or other interest-bearing securities until they need to make a payment. When interest rates are high, people will hold as little money for transaction purposes as possible because it will be worth the time and trouble of investing in bonds and then liquidating them when needed. When rates are low, by contrast, people will hold more money for transaction purposes because it isn’t worth the hassle and brokerage fees to play with bonds very often. So transaction demand for money is negatively related to interest rates. A similar trade-off applies also to precautionary balances. The lure of high interest rates offsets the fear of bad events occurring. When rates are low, better to play it safe and hold more dough. So the precautionary demand for money is also negatively related to interest rates. And both transaction and precautionary demand are closely linked to technology: the faster, cheaper, and more easily bonds and money can be exchanged for each other, the more money-like bonds will be and the lower the demand for cash instruments will be, ceteris paribus.
### Key Takeaways
• Before Friedman, the quantity theory of money was a much simpler affair based on the so-called equation of exchange—money times velocity equals the price level times output (MV = PY)—plus the assumptions that changes in the money supply cause changes in output and prices and that velocity changes so slowly it can be safely treated as a constant. Note that the interest rate is not considered at all in this so-called naïve version.
• Keynes and his followers knew that interest rates were important to money demand and that velocity wasn’t a constant, so they created a theory whereby economic actors demand money to engage in transactions (buy and sell goods), as a precaution against unexpected negative shocks, and as a speculation.
• Due to the first two motivations, real money balances increase directly with output.
• Due to the speculative motive, real money balances and interest rates are inversely related. When interest rates are high, so is the opportunity cost of holding money.
• Throw in the expectation that rates will likely fall, causing bond prices to rise, and people are induced to hold less money and more bonds.
• When interest rates are low, by contrast, people expect them to rise, which will hurt bond prices. Moreover, the opportunity cost of holding money to make transactions or as a precaution against shocks is low when interest rates are low, so people will hold more money and fewer bonds when interest rates are low.
|
2014-04-21 07:04:43
|
{"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": 0.4221056401729584, "perplexity": 2428.5108729350673}, "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-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00647-ip-10-147-4-33.ec2.internal.warc.gz"}
|
https://tex.stackexchange.com/questions/480745/tocloft-and-babel-possible-incompatibility
|
# tocloft and babel (possible) incompatibility
I am trying to change the default section title for the TOC, renewing tocloft's \contentsname macro:
\documentclass{article}
\usepackage[american]{babel} % (*) THE STARRED LINE
\usepackage{tocloft}
\renewcommand*{\contentsname}{FooContents}
\begin{document}
\tableofcontents
\section{The first section}
\section{The second section}
\end{document}
If I remove the starred line, the TOC appears as expected, titled "FooContents" -- otherwise, the default title "Contents" is used.
Is there a way to get this done without getting rid of babel?
EDIT
Per @egreg's comment below, this particular issue can be solved with \addto\captionsamerican{\renewcommand*{\contentsname{FooContents}}}.
With regard to \addto, the babel documentation (p. 64) states that:
For each language four control sequences have to be defined that control the language-specific definitions. To be able to add something to these macro once they have been defined the macro \addto is introduced.
However, I cannot find a list of those "four control sequences". I am using tocloft quite heavily, and most of what I do with it is reversed by babel.
• \addto\captionsamerican{\renewcommand{\contentsname}{FooContents}} – egreg Mar 21 at 18:31
• But of course :-) However, I do not fully understand which macros I would need to renew like that. For instance, I am also changing the TOC title font (renewing cfttoctitlefont), but the above does not seem to work for that case. – dow Mar 21 at 18:56
• In particular, babel's documentation states about \addto: "For each language four control sequences have to be defined that control the language-specific definitions. To be able to add something to these macro once they have been defined the macro \addto is introduced." I've been browsing around the document to find these "four control sequences", but I cannot find a list. – dow Mar 21 at 19:07
• Babel does not touch fonts, only names like \contentsname and friends – daleif Mar 21 at 21:33
|
2019-07-23 15:38:02
|
{"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": 0.9073174595832825, "perplexity": 2260.3976111360116}, "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-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00133.warc.gz"}
|
http://mathhelpforum.com/number-theory/37922-really-cant-do.html
|
Math Help - really cant do
1. really cant do
2. Hello,
For question (a): Each number n in the sum is lower than p. So, (n,p)=1.
Hence, we can apply Fermat's Little Theorem:
$n^{p-1}\equiv 1\text{ mod }p$
Let's multiply both sides by n.
$n^{p}\equiv n\text{ mod }p$
Thus
$\sum_{n=1}^{p-1} n^p\equiv \sum_{n=1}^{p-1} n\equiv \frac{(p-1)(p)}{2}\text{ mod }p$
Isn't it ?
3. For (a) we can strengthen the problem. $1^k + 2^k + ... + (p-1)^k \equiv 0 (\bmod p)$ if $(p-1)\not | k$ and $\equiv p$ if $(p-1)| k$.
For (b) note $1+...+p = p\cdot \frac{p-1}{2}$. Let $a=\frac{p-1}{2}$.
Let $r$ be remainder of $(p-1)!$ mod $pa$.
Thus, $(p-1)!\equiv r(\bmod pa)$ thus $(p-1)!\equiv r(\bmod p)$ and $(p-1)!\equiv r(\bmod a)$.
By Wilson's theorem it follows that $r\equiv -1(\bmod p)$.
Also $r\equiv 0(\bmod a)$ because $a$ divides $(p-1)!$.
If we let $r=p-1$ it satisfies both conditions.
4. For the second problem let $y$ be the order of $a$ mod $b$. Let $a^x\equiv 1(\bmod b)$. Write $x=qy+r$ where $0\leq r < y$. Then, $a^x \equiv 1 (\bmod b)\implies a^{qy+r}\equiv 1(\bmod b)\implies a^r\equiv 1(\bmod b)$. Since $y$ is the order of $a$ it means $a^r\not \equiv 1$ if $0. But since $0\leq r < y$ and $a^r\equiv 1(\bmod b)$ it follows that $r=0$. And thus $x=qy+r = qy\implies y|x$.
|
2014-08-22 21:41:39
|
{"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": 0, "img_math": 0, "codecogs_latex": 35, "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": 0.994027853012085, "perplexity": 529.6874569035391}, "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-2014-35/segments/1408500824562.83/warc/CC-MAIN-20140820021344-00196-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://tex.stackexchange.com/questions/120926/prevent-wrapfigure-from-wrapping-the-next-section
|
# Prevent wrapfigure from wrapping the next section
Code:
\documentclass{article}
\usepackage[a5paper]{geometry}
\usepackage{wrapfig}
\usepackage{lipsum}
\begin{document}
\begin{wrapfigure}{l}{0.5\textwidth}
\rule{4cm}{12cm}
\end{wrapfigure}
\lipsum[1]
\section{Hello}
Hello section text
\end{document}
Prints:
1. How can I make it shift \section{Hello} to the left?
2. And more generally: never wrap any text in the new section, following wrapfigure?
• I don't know of an automated solution for this; you can use the first optional argument to manually restrict the number of narrow lines: \begin{wrapfigure}[29]{l}{0.5\textwidth}...\end{wrapfigure} and, in your concrete case, reduce the skip for the caption: \usepackage{caption} and then something like \captionsetup[wrapfigure]{skip=3pt} \begin{wrapfigure}[29]{l}{0.5\textwidth} \rule{4cm}{12cm} \caption{some text} \end{wrapfigure} – Gonzalo Medina Jun 24 '13 at 23:21
With the text and image as they are, wrapfig needs a bit of help.
You could tell it to stop wrapping after 29 lines:
\begin{wrapfigure}[29]{l}{0.5\textwidth}
But the result is rather squeezed:
A much better result in this case is to simply add \clearpage before the \section and move it to the next page.
• Hehe! I was writing the same thing in my comment. I also suggested there a possible workaround for the caption skip. – Gonzalo Medina Jun 24 '13 at 23:22
You could simply include the wrapfig and the wrapped text in brackets as follows:
{
\begin{wrapfigure}{l}{0.5\textwidth}
\rule{4cm}{12cm}
\end{wrapfigure}
\lipsum[1]
}
\section{Hello}
This stops the wrapping at the line where the wrapped text stops. As pointed out by Andrew Swann, this doesn't work if the figure is longer than the text; in that case, the following section heading will be overwritten by the figure.
• Welcome to Tex.SX! Unfortunately that does not entirely work, the wrapping stops but the section heading gets overridden by the figure. Perhaps you can improve the answer, and also make it into a complete compilable document, rather than just a code snippet. – Andrew Swann Sep 4 '14 at 9:51
• Didn't work for me. Even if I included the last paragraph that should wrap in the curly brackets, it didn't wrap, and got overwritten by the figure. – Jeffrey Lebowski Mar 23 '16 at 14:54
• This sent the image at the end of the doc – axolotl Aug 23 '16 at 14:32
If the text wrapping the figure ends in some actual text and not just included text like \lipsum, then it works to just add empty lines to the end of the text.
\documentclass{article}
\usepackage[a5paper]{geometry}
\usepackage{wrapfig}
\usepackage{lipsum}
\begin{document}
\begin{wrapfigure}{l}{0.5\textwidth}
\rule{4cm}{11cm}
\end{wrapfigure}
\lipsum[2]
text\\ \\ \\ \\ \\ \\ \\ \\
\section{Hello}
Hello section text
\end{document}
|
2020-04-07 20:13:18
|
{"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": 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": 0.9980760216712952, "perplexity": 1733.969283037708}, "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/1585371805747.72/warc/CC-MAIN-20200407183818-20200407214318-00024.warc.gz"}
|
https://www.mersenneforum.org/showpost.php?s=01ac69a7d5e160232c5fafa9c02c5a67&p=153292&postcount=10
|
Thread: Low weight k's View Single Post
2008-12-14, 17:12 #10 Beyond Dec 2002 44410 Posts Partial "old" list 403 searched to 1M - 981831 2039 searched to 1M - 841312 2171 searched to 1M - no new primes 2279 searched to 1M - no new primes 2591 searched 333k to 1M - no new primes 2741 searched to 1M - 50, 138, 1610, 1850, 3954, 10202, 23178, 84858, 468594, 737586 2759 searched to 1M - 388, 405820, 569812 2797 searched to 1M - no new primes 2983 searched to 1M - 11, 671, 767, 2891, 771455 3389 searched to 1M - 92, 848, 2684,23864, 209996, 504584, 749828 3539 searched to 1M - 12, 192, 4236, 869868 3821 searched to 1M - 70, 15814, 439006 3823 searched 333k to 1M - 7, 287, 335, 599, 1015, 2407,...384223 4001 searched to 1M - no new primes 4327 searched to 1M - no new primes 4597 searched to 1M - 37, 161, 221, 721, 112009, 799781 5099 searched to 1M - 184, 1216, 9936, 19072, 27648 5323 searched to 1M - 27, 83, 8651, 107643 7331 searched to 1M - no new primes 9179 searched to 1M - no new primes 9613 searched to 1M - no new primes 10949 searched to 1M - no new primes 17077 searched to 1M - no new primes 19919 searched to 1M - no new primes 21547 searched to 1M - 121, 2233, 16001, 310625, 936473 22183 searched to 1M - no new primes 24067 searched to 1M - no new primes 24671 searched to 1M - 1290, 2250, 11698, 108906, 382066 25229 searched 325k to 1M - no new primes 30727 searched to 1M - no new primes 70079 searched 176k to 1M - no new primes 106303 searched 127k to 1M - no new primes 114613 searched to 1M - no new primes 117037 searched to 1M - no new primes 124679 searched to 1M - 497244, 769908 132217 searched to 1M - no new primes 171607 searched to 1M - no new primes 179743 searched to 1M - no new primes 192509 searched to 1M - no new primes if a starting range is not stated, then I started from n=0 More later.. (kar_bon: "11025 searched to 1M - no new primes" deleted, because not correct) Last fiddled with by kar_bon on 2008-12-22 at 10:54
|
2021-04-18 06:56:27
|
{"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": 0.8606690764427185, "perplexity": 2734.6090939072787}, "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/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00076.warc.gz"}
|
https://zhuyi.fan/tag/study
|
# Tag: study | Page 0
Post #2
##### Optimization Learning Notes 1
book Public Date: 2020-06-20 08:42:04.801901
update Update Date: 2020-07-30 15:33:51.919974
Modeling
Mathematically, an optimization problem is usually represented as:
\begin{aligned} \text{maximize/minimize}_x &~ f(x) \ \text{subject to} &~ x \in \Omega \end{aligned}
where:
- $x$ is the decision variable
- $f$ is the objective function
- ...
Post #5
##### Introduction to Category Theory 1
book Public Date: 2020-06-17 10:48:50.306174
update Update Date: 2020-06-21 04:11:27.547781
What is a Category?
A category $C$ consists of
- a collection $Obj$ of entities called objects
- a collection $Arw$ of entities called arrows
- three assignments:
- source: $Arw \to Obj$
- target: $Arw \to Obj$
- id: $Obj \to Arw$
- a partial ...
Post #6
##### Optimization Learning Notes 2
book Public Date: 2020-06-20 17:12:00.571290
update Update Date: 2020-07-30 15:34:13.181702
Recall that the standard of an LP is like the following:
\begin{aligned} \min &~~~ c^Tx \ \text{subject to} &~~~ Ax = b \ &~~~ x \ge 0 \end{aligned}
where $x \in \mathbb R^n, b \in \mathbb R^m, A \in \mathbb R^{m\times n}$
In the following discussio...
Post #8
##### Optimization Learning Notes 3
book Public Date: 2020-06-24 11:49:25.724151
update Update Date: 2020-07-30 15:34:37.990602
Two-Phase Method with Simplex Tableau
TL;DR, two-phase method involves two LP and can be solved by construct two simplex tableau.
However, there two things to consider:
- How to acquire the initial tableau in Phase I.
- How to make the transition from Ph...
Post #10
##### Optimization Learning Notes 4
book Public Date: 2020-07-02 15:02:38.926921
update Update Date: 2020-07-30 15:35:06.065394
Interpreting the Dual Problem
It turns out that dual problem actually have their own meanings. The interpretations of dual problems may give us a better understanding of the duality theory.
Production Planning
Production Planning problem is one of the most...
|
2021-05-13 02:03:27
|
{"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6962297558784485, "perplexity": 10746.395565047706}, "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/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00625.warc.gz"}
|
https://physics.stackexchange.com/questions/177087/number-of-microstates-compatible-with-two-boxes/177091
|
# Number of microstates compatible with two boxes
From my notes I have:
From one point of view there are many more microstates compatible with the LHS than the RHS, in fact the relation between the number of microstates is $$\frac{n_B}{n_A}=\left(\frac{1}{2}\right)^N$$
I cannot understand why we have the $1/2$ to the power $N$. I would have thought that it would be $$\frac{n_B}{n_A}=\frac{1}{2}$$ as there is half the room. Could someone please explain the reasoning behind the power of $N$ please.
$$\frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where $N$ is the number of particles and $V$ is the volume of the box.
This may be the argument: you have $N$ particles, and for each one you can put it on the left side or on the right side. Each of these choices, for each particle, leads to a different microstate. There are $2^N$ possible choices you can make for how to distribute the $N$ particles between the left and right halves of the box (assuming the particles are distinguishable), and thus $2^N$ microstates, but only one of these microstates has the particles all on the right side. So the relation is $n_B/n_A = 1/2^N$. This assumes that diagram A is meant to represent the macrostate containing all possible microstates, even those in which the particles do happen to all wind up on the right side.
If that sounds somewhat unsatisfying, here's a slightly better argument. Suppose the large box is divided into $2M$ little squares, where $M$ is very large. The number of ways to place the particles into these $2M$ squares is $$n_A = \binom{2M - 1 + N}{N} = \frac{(2M - 1 + N)!}{N!(2M - 1)!}$$ now assuming the particles are indistinguishable (though it turns out not to matter, since distinguishable particles just have an extra $N!$ in both this expression and the following one). But the number of ways to place the particles into the $M$ squares on the right half of the box is $$n_B = \binom{M - 1 + N}{N} = \frac{(M - 1 + N)!}{N!(M - 1)!}$$ The ratio of these is $$\frac{n_B}{n_A} = \frac{(M - 1 + N)!}{N!(M - 1)!}\frac{N!(2M - 1)!}{(2M - 1 + N)!} = \frac{M(M+1)\cdots(2M - 1)}{(M+N)(M+N+1)\cdots(2M-1+N)}$$ and if you take the logarithm, you get \begin{align} \ln\frac{(M - 1 + N)!}{(M - 1)!}&\frac{(2M - 1)!}{(2M - 1 + N)!} \\ \approx\ &(M-1+N)\ln(M-1+N) - (M-1+N) \\ &- (M-1)\ln(M-1) + (M-1) \\ &+ (2M-1)\ln(2M-1) - (2M-1) \\ &- (2M-1+N)\ln(2M-1+N) + (2M-1+N) \\ \approx\ &(M+N)\ln(M+N) - M\ln M + 2M\ln(2M) - (2M+N)\ln(2M+N) \\ =\ &M\biggl(\ln\frac{4}{2+r} + (1+r)\ln\frac{1+r}{2+r}\biggr) \end{align} where $r = \frac{N}{M}$. In the limit of extremely large $M\gg N$, so that $r\to 0$, this can be approximated as $$\ln\frac{n_B}{n_A} = -Mr\ln 2 + \mathcal{O}(r^2) = -N\ln 2 + \mathcal{O}(r^2)$$ which leads directly to $$\frac{n_B}{n_A} = \frac{1}{2^N}$$ But again, this is based on the assumption that state A represents the macrostate consisting of all possible microstates, even those where the particles are all on one side.
|
2020-04-07 05:09:23
|
{"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9966000318527222, "perplexity": 198.90374967309464}, "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/1585371665328.87/warc/CC-MAIN-20200407022841-20200407053341-00204.warc.gz"}
|
http://gatkforums.broadinstitute.org/gatk/discussion/3520/enumerated-data-type-arguments
|
The current GATK version is 3.7-0
Examples: Monday, today, last week, Mar 26, 3/26/04
#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
GATK 3.7 is here! Be sure to read the Version Highlights and optionally the full Release Notes.
Register now for the upcoming GATK Best Practices workshop, Feb 20-22 in Leuven, Belgium. Open to all comers! More info and signup at http://bit.ly/2i4mGxz
# Enumerated data type arguments
everywhereMember Posts: 4
I am working on a Queue script that uses the selectVariants walker. Two of the arguments that I am trying to use both use an enumerated type: restrictAllelesTo and selectTypeToInclude. I have tried passing these as strings however I get java type mismatch errors. What is the simplest way to pass these parameters to the selectVariant walker in the qscript?
Tagged:
Hi there,
Have a look at the presentations from our recent Queue workshop, I think we give some examples of how to pass the different types in there:
Geraldine Van der Auwera, PhD
• everywhereMember Posts: 4
Maybe it's my lack of understanding scala but I still can't pass an enumerated data type. The workshop shows this as an example:
@Argument(doc=”gender identity") // some object defined elsewhere
var gender: Gender = Gender.UNKNOWN
my code uses this
@Argument(doc="Select only variants of a particular allelicity.", shortName="restrictAllelesTo", required=false)
var rAT: restrictAllelesTo = restrictAllelesTo.BIALLELIC
I've even tried defining the object itself:
object restrictAllelesTo extends Enumeration {
type restrictAllelesTo = Value
val BIALLELIC, ALLELIC, BOTH = Value
}
import restrictAllelesTo._
##### ERROR ------------------------------------------------------------------------------------------
INFO 09:59:56,745 QCommandLine - Shutting down jobs. Please wait...
Here, I am not sure what to import (So, if there is some specific information/guide that will be great.
Thanks,
Amit
|
2017-01-17 15:05:29
|
{"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": 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": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1981610208749771, "perplexity": 6932.467950980787}, "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-2017-04/segments/1484560279923.28/warc/CC-MAIN-20170116095119-00435-ip-10-171-10-70.ec2.internal.warc.gz"}
|
http://slideplayer.com/slide/778816/
|
# Exercises for CS1512 Weeks 7 and 8 Propositional Logic 1 (questions + solutions)
## Presentation on theme: "Exercises for CS1512 Weeks 7 and 8 Propositional Logic 1 (questions + solutions)"— Presentation transcript:
Exercises for CS1512 Weeks 7 and 8 Propositional Logic 1 (questions + solutions)
Exercise 1 1. Express each formula using only (at most) the connectives listed. In each case use a truth table to prove the equivalence. (Note: is exclusive `or`) a.Formula: p q. Connectives: {, }. b.Formula: p q. Connectives: {,, }. c.Formula: p q. Connectives: {, }. d.Formula: (p q) (( p) q). Conn: {, }.
Answer to Exercise 1. (Other answers possible) a. Formula p q. Connectives: {, }. Answer: p q b.Formula: p q. Connectives: {,, }. Answer: (p q) (q p) c. Formula: p q. Connectives: {, }. Answer: (p q) (q p) d. Formula: (p q) (( p) q). Conn: {, }. Answer: q (This was a trick question, since you dont need any connectives.)
Ex. 2. Which of these are tautologies? 1.p (q p) 2.p ( p p) 3.(q p) (p q) 4.(q p) (p q) 5.(p (q r)) (q (p r)) Please prove your claims, using truth tables. (Hint: Ask what assignment of truth values to p,q, and r would falsify each formula. In this way you can disregard parts of the truth table).
Answer to Ex.2 1.p (q p) Tautologous 2.p ( p p) Tautologous 3.(q p) (p q) Contingent 4.(q p) (p q) Tautologous 5.(p (q r)) (q (p r)) Tautologous 1,2,4,5 are known as `paradoxes of the material implication, because they contrast with implication in ordinary language.
Ex. 3. Reading formulas off truth tables Background: In class, a proof was sketched for the claim that every propositional logic formula can be expressed using the connectives {, }. The proof proceeded essentially by reading off the correct formula off the truth table of any given formula. Task: Use this meticulous method to construct a formula equivalent to p q.
Answer to ex. 3. Steps: 1.Construct the truth table of p q. 2. Mark those two rows in the table that make p q TRUE. 3.Corresponding with these two rows, construct a disjunction of two formulas, one of which is (p q), and the other (q p). 4.Use the De Morgan Laws to convert this disjunction (p q) (q p) into the quivalent formula ( (p q) (q p)) [5. Use truth tables again to check that these two formulas are indeed equivalent.]
Question 4a In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {, } is also functionally complete?
Question 4a In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {, } is also functionally complete? Answer: It is sufficient to show that both p and (p q) are expressible using the two connectives in {, }. The first of these two (i.e., negation) is obvious, and the second (i.e., conjunction) can be expressed as follows (p q) ( p q), using one of De Morgans laws.
Question 4b In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {NAND} is also functionally complete? [Explanation: (p NAND q) is TRUE iff (p q) is FALSE. This connective is also called the Sheffer stroke and written (p|q).)
Question 4b In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {NAND} is also functionally complete? Answer: the reasoning is similar to that in 4a, but it may be a bit trickier to find the right formulas: p can be expressed as p p|p. (It helps to do negation before conjunction!) The formula p q is equivalent to (p|q)|(p|q) (in other words, the negation of p|q) It might be useful to prove these equivalences using truth tables.
Question 4c Given this result, why do we bother defining and using more than one connective? Answer: Many things can be expressed more succinctly and transparently with a larger vocabulary of connectives (as your answer to 4b will show). Also, it becomes easier to let your formulas resemble the things we say in e.g. English.
Question 5 a.(r c) d b.r ( c d) Comparison of truth tables shows (a) and (b) to be equivalent. Both formulas are false if and only if (r true,c true,d false).
Similar presentations
|
2018-09-22 13:14:18
|
{"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": 0.8246034383773804, "perplexity": 2445.876758802402}, "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-2018-39/segments/1537267158429.55/warc/CC-MAIN-20180922123228-20180922143628-00049.warc.gz"}
|
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.specfun.ellipint_general_2.html
|
# naginterfaces.library.specfun.ellipint_general_2¶
naginterfaces.library.specfun.ellipint_general_2(z, akp, a, b)[source]
ellipint_general_2 returns the value of the general elliptic integral of the second kind for a complex argument .
For full information please refer to the NAG Library document for s21da
https://www.nag.com/numeric/nl/nagdoc_28.5/flhtml/s/s21daf.html
Parameters
zcomplex
The argument of the function.
akpfloat
The argument of the function.
afloat
The argument of the function.
bfloat
The argument of the function.
Returns
fcomplex
The value of the general elliptic integral of the second kind for a complex argument .
Raises
NagValueError
(errno )
On entry, is too large: . It must not exceed .
(errno )
On entry, is too large: . It must not exceed .
(errno )
On entry, : .
(errno )
On entry, is too large: . It must not exceed .
(errno )
The iterative procedure used to evaluate the integral has failed to converge.
Notes
ellipint_general_2 evaluates an approximation to the general elliptic integral of the second kind given by
where and are real arguments, is a complex argument whose real part is non-negative and is a real argument (the complementary modulus). The evaluation of is based on the Gauss transformation. Further details, in particular for the conformal mapping provided by , can be found in Bulirsch (1960).
Special values include
or (the elliptic integral of the first kind) and
or (the elliptic integral of the second kind). Note that the values of and are equal to in the trivial case .
ellipint_general_2 is derived from an Algol 60 procedure given by Bulirsch (1960). Constraints are placed on the values of and in order to avoid the possibility of machine overflow.
References
Bulirsch, R, 1960, Numerical calculation of elliptic integrals and elliptic functions, Numer. Math. (7), 76–90
|
2022-08-13 15:19:25
|
{"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": 0.9589330554008484, "perplexity": 1019.0471497440813}, "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-2022-33/segments/1659882571959.66/warc/CC-MAIN-20220813142020-20220813172020-00507.warc.gz"}
|
https://learn.careers360.com/ncert/question-draw-a-line-segment-of-length-95-cm-and-construct-its-perpendicular-bisector/
|
Q
# Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Q2. Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Views
The steps of constructions are:
(i) Draw a line segment $\bar{AB}$ = 9.5 cm
(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the perpendicular bisector of $\bar{AB}$.
Exams
Articles
Questions
|
2020-01-27 13:28:27
|
{"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": 2, "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": 0.33414047956466675, "perplexity": 924.0229849530739}, "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-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00333.warc.gz"}
|
http://dspace.rri.res.in/handle/2289/6329
|
Please use this identifier to cite or link to this item: http://hdl.handle.net/2289/6329
Title: The cosmic dawn and epoch of reionization with the square kilometre array Authors: Koopmans, L V EPritchard, JSubrahmanyan, Ravi+34 Co-authors Keywords: AASKA14 Issue Date: 2015 Publisher: SISSA (Scuola Internazionale Superiore di Studi Avanzati.) Citation: Proceedings of Science, 2014, p 001 Abstract: Concerted effort is currently ongoing to open up the Epoch of Reionization (EoR) ($z\sim$15-6) for studies with IR and radio telescopes. Whereas IR detections have been made of sources (Lyman-$\alpha$ emitters, quasars and drop-outs) in this redshift regime in relatively small fields of view, no direct detection of neutral hydrogen, via the redshifted 21-cm line, has yet been established. Such a direct detection is expected in the coming years, with ongoing surveys, and could open up the entire universe from $z\sim$6-200 for astrophysical and cosmological studies, opening not only the EoR, but also its preceding Cosmic Dawn ($z\sim$30-15) and possibly even the later phases of the Dark Ages ($z\sim$200-30). All currently ongoing experiments attempt statistical detections of the 21-cm signal during the EoR, with limited signal-to-noise. Direct imaging, except maybe on the largest (degree) scales at lower redshifts, as well as higher redshifts will remain out of reach. The Square Kilometre Array(SKA) will revolutionize the field, allowing direct imaging of neutral hydrogen from scales of arc-minutes to degrees over most of the redshift range $z\sim$6-28 with SKA1-LOW, and possibly even higher redshifts with the SKA2-LOW. In this SKA will be unique, and in parallel provide enormous potential of synergy with other upcoming facilities (e.g. JWST). In this chapter we summarize the physics of 21-cm emission, the different phases the universe is thought to go through, and the observables that the SKA can probe, referring where needed to detailed chapters in this volume (Abridged). Description: Open Access, Proceedings of Advancing Astrophysics with the Square Kilometre Array (AASKA14).Giardini Naxos, Italy, from 9 -13 June, 2014 URI: http://hdl.handle.net/2289/6329 ISSN: 1824-8039. Alternative Location: http://adsabs.harvard.edu/abs/2015aska.confE...1Khttp://arxiv.org/abs/1505.07568 Copyright: 2015 Authors Appears in Collections: Research Papers (A&A)
Files in This Item:
File Description SizeFormat
|
2017-08-22 01:33:16
|
{"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": 0.547709584236145, "perplexity": 5997.8255316315435}, "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-2017-34/segments/1502886109803.8/warc/CC-MAIN-20170822011838-20170822031838-00375.warc.gz"}
|
https://dsp.stackexchange.com/questions/28750/why-do-low-pass-and-high-pass-filters-generate-a-phase-shift/48850
|
why do low pass and high pass filters generate a phase shift?
I just learned following notation for a sine wave:
$A e^{j\phi t}$
A low passfilter and a highpass filter respectively generate a phase shift in the complex plane of the sine wave as follows:
$A e^{-j\phi t}$ and $A e^{+j\phi t}$
I don't understand what phase shifting has to do with letting higher or lower frequencies intact... How do these things( filtering and phase shifting) relate to each other? They just seem like to completely and independent things to me
I just learned following notation for a sine wave: $Ae^{j\phi t}$
well, that's not really a sine wave; that's a complex sinusoid. The real and imaginary parts of that are cosine and sine, respectively.
A low passfilter and a highpass filter respectively generate a phase shift in the complex plane of the sine wave as follows:
No. There's all kind of LPFs and HPFs, and only the linear phase ones do what you claim.
The point is that (although you're basically asking us to write a textbook on filter and signal theory) you'd typically understand filters as systems with poles and zeros, as the transfer function of LTI (linear, time-invariant) systems can usually be written as fraction with polynomials (which have roots).
Now, to find something that actually fulfills the boundaries these representations set, your system needs to move its response in relation to frequency over the complex plane; "moving over the complex plane" is something that can be understood as changing a complex number over a variable; the rate of change of the argument of that number is a phase shift.
It can be shown (and you'll probably learn this later on) that you cannot change the magnitude of a signal over frequency (i.e. do the filtering) without changing it's argument.
• The user has already selected this as an answer but I guess you should better add a discussion on Group Delay which is responsible for keeping high and low frequencies intact or not, as the user is asking to learn in his last paragraph... – Fat32 Feb 8 '16 at 23:22
• @fat32 I really want to do that, but: to be honest, OP needs a text book and not a short answer. Me writing down stuff that then also lacks background will not make my answer really that more helpful. I kind of feel like the purpose of my answer was to motivate OP to look further. – Marcus Müller Feb 9 '16 at 7:49
Because they need a certain time delay in order to "decide" what the output should look like if the output changes a lot. Imagine you see a number series and your brain has to decide what the mean value of these number series is. You will need to look at several numbers to arrive at some reasonable conclusion (let's assume the numbers are represented by a noise distribution with the same mean). The same applies to filters that weigh signal values to arrive at the filtered output signal.
Any filter that does not employ prediction of output will have phase shift. The only way to get rid of phase shift completely (or keep it to fractions of dB) is to use a predictive filter - for example kalman filter. Such a filter, in the simplest form will estimate the derivative of the signal and use that to filter the original signal in a predictive fashion. If you have a model of your signal you can do even better. Since prediction and sensor measurements can be processed in the same time step, from a discrete time point of view you can have zero phase shift if prediction is able to estimate the original signal precisely.
• This is pretty great for intuitively understanding the problem, thanks. I guess the predictive filtering approach has problems when there are sharp changes in a given frequency (e.g. the onset of drum sample in audio)? – naught101 Aug 14 '18 at 3:10
• It depends on how you set your kalman gains. You can use moen4 matlab function to spit out your system, then use autocovariance least squares to spit out your Q and R and then use lqe to spit out your optimal kalman filter. If you feed such system with a bunch of drum signals you will get the OPTIMAL filter that will be able to predict the signal from a set of monitored quantities. If your filter then cuts off a piece of drum onset curve then that's precisely what it is supposed to do given your constraints because it is an optimal estimator for your particular set of training signals. – Martin Aug 14 '18 at 12:57
• If you use the signal itself to solve for your kalman gain then you would need to derive an optimal derivative offline and filter it using offline techniques to get a good filter. You can do this using sgolayfilt function or some approach that utilizes no phase shift filtering to get a smooth derivative on which you then do system identification together with your actual signal and produce a filter that will then take your signal and produce a smooth version of it as well as a high precision estimate of it's derivative. – Martin Aug 14 '18 at 13:00
|
2020-02-25 16:21:15
|
{"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": 0.7675479650497437, "perplexity": 490.6631430775967}, "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-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00030.warc.gz"}
|
http://tex.stackexchange.com/tags/auto-completion/new
|
# Tag Info
0
I addedd the bibligraphy as above It worked the first time, but when I opened by document again it stopped working I found if I clicked on the library.bib file and opened it my document stated autocompleting /cite
0
Under Linux I recommend "Kate". It's simple, fast & pretty amazing! An other alternative for Linux that wasn't mentioned above, is Geany with GeanyLaTeX Plugin. On Windows Notepad++ is the most brilliant solution. But TexStudio is also not bad...
0
In Windows 7 and MikTex(basic-miktex-2.9.5105)+TexMaker4.3+bibtex file, I found that if I use command like these: \bibliographystyle{bst file name} \bibliography {bib file name} then \cite will not auto-complete. yet however if I use command like these: \bibliographystyle{bst file name} \bibliography{bib file name} then \cite will auto-complete the ...
2
Please uncheck "close parentheses automatically". It can be found "Options / configure TeXStudio" (see screenshot). It does work for me under TeXStudio 2.8.6 (Kubuntu 14.04). Edit: as for the shortcut CTRL+I. I don't think there is an easy way to achieve that. You will have to either alter an existing cwl file (where the definitions are stored) or create ...
Top 50 recent answers are included
|
2014-11-27 09:52:13
|
{"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": 0.9036510586738586, "perplexity": 8645.258172310812}, "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-2014-49/segments/1416931008218.28/warc/CC-MAIN-20141125155648-00102-ip-10-235-23-156.ec2.internal.warc.gz"}
|
https://dsp.stackexchange.com/questions/49882/why-the-angle-of-a-real-signal-in-frequency-is-continuous
|
# Why the angle of a real signal in frequency is continuous?
This original signal is shown below (time domain):
It is a real signal. According to the signal processing theory, the form part of the angle is conjugate to that of the latter part. That is, $\sphericalangle X(e^{j\omega}) =\sphericalangle X^*(e^{-j\omega})$ when $x[n] \in \mathbb{R}$.
However, I implement X = angle(fft(x)); and plot(X) in MATLAB. The figure is shown below:
The phase is continuous! Could anyone explain the reason? Thanks in advance!
X = angle(fft(x))
is numerically very questionable. If your signal is periodic in exactly the length of vector x you're observing, then the (theoretical) DFT would be zero in all but two bins. You can't really take the angle of $0$.
If your signal has a period that doesn't exactly fit in x, then you're observing the phase of two superimposed sinc pulses, as that's what a rectangular windowing leads to.
A sinc is for all points (but the maximum) defined as $\frac{\sin ax}{bx}$ (with $a$ and $b$ depending on who you ask and how you scale), and if you look at that, it definitely does have at least a strongly linear component.
|
2021-04-12 19:57:26
|
{"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": 0.8355405330657959, "perplexity": 540.5639729408367}, "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-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00196.warc.gz"}
|
http://crypto.stackexchange.com/tags/ctr/hot?filter=year
|
# Tag Info
11
Suppose you use the sector number times the number of AES blocks per sector as the initial value for CTR. If you successively store the content $M$ then $M'$ in the same sector $n$ then $E^{CTR}_n(M) \oplus E^{CTR}_n(M') = M \oplus M'$ (where $E^{CTR}_{n}$ is the encryption function with CTR mode and IV started for sector number $n$). CTR mode fails ...
11
There are some serious problems with this design that would preclude it from being standardized, so it probably does not have a name. The 2 visibly main flaws are as follows: If the plaintext follows a pattern similar to the block counter, the block cipher inputs may repeat, exposing information about the plaintext (exact same issue as reuse of nonce, but ...
6
From the diagram on CTR mode you can notice that there are no dependencies between any of the phases of the pipeline. If you have more than one block-size worth of data, you can process each block-size chunk completely independently of the others by calculating $\mathrm{ciphertext}_i = E(\mathrm{key}, \mathrm{nonce} \, || \, \mathrm{counter}_i) \oplus ... 6 Short answer: There would be nothing (that isn't already wrong with TLS) necessarily wrong with a CTR + HMAC cipher suite, but the technical merits are only one factor in a technical feature getting to RFC status in the TLS working group. Without being discourteous to the TLS Working Group (WG) participants or process, other reasons can be: political ... 6 First, the obvious advice is not to use this in practice. Rolling your own is fine for learning, but you should use standard primitives when you need actual security. E.g. one from SP 800-90A which poncho linked in comments. Now, some observations. I haven't read all your code, so I may misunderstand things. Is this a good way to whiten the data? Is ... 5 Yes, there are secure alternatives to support random-access based encryption. I did not come up with a way to break the proposed combination. Still, instead of inventing a new mode, I would recommend to take consider existing modes for this kind of operation, such as XTS mode. The existing modes are more studied, and (in some ways) more efficient. XTS mode ... 5 The reference for this is NIST SP800-38A, especially its appendix B. Basically we consider the IV a binary value of the width of the block cipher (64-bit for DES, 128-bit for AES), and add 1 to that, except for one detail: there is no carry at some application-specified rank, defining the maximum number of blocks that can be enciphered with a single IV; if ... 5 The entire block consists of a$n$bit nonce and a$128-n$bit counter. Typically$n=64$. The nonce needs to be large enough so that every message under the key can have a unique one, and the counter needs to be large enough that every message block can have a unique counter value. Typically, the counter is initialized to 0 and then incremented by 1 for ... 4 As long as you never re-use a specific counter value with the same key, counter mode protects the privacy of the message. All counter values are equally secure. You just have to be sure never to re-use any counter value in two different messages. Zero is no different to any other counter value in this respect. However, if you ever re-use any counter ... 3 Generally, it depends on the architecture. If you have$n$processors available, the obvious way to parallelize CTR mode encryption is to distribute each chunk of$n$consecutive blocks among the processors, so that processor$0 \le i < n$computes: $$C_j = E_K(c_j) \oplus P_j, \quad j = i + kn, k = 0,1,2,\dotsc$$ where$c_j$is the$j$-th counter ... 3 Let$2^m$be the average message length in blocks. When using an independent random nonce for the whole 128-bit IV of each block, you would expect a collision after$2^{64}$blocks, i.e.$2^{64-m}$messages. (But you double the data size.) When using a 96-bit nonce and a 32-bit counter, you would expect a nonce collision after$2^{48}$messages. This is ... 3 You can get up to around$2^{64}$random bits from counter mode (before you hit the birthday bound due to the lack of a collision), simply by running it as is. If you've got a full implementation of counter mode, the plaintext can be anything you like, because the stream (ie the output of the$E_k(ctr)$calls), should appear uniformly sampled from$2^{128}$. ... 3 AES-CTR mode can be used to generate stream of random numbers. For generating random numbers, the plaintext is indeed irrelevant. It can be even full of zero (the NIST recommended way to generate random numbers uses such plaintext.) NIST has recommendation on how to generate DRBG (deterministic random bit generator) based on CTR mode. NIST has defined how ... 3 In CTR, you can use any operation which has a full cycle through the space of the IV with the counter. You could use the plus operator like the example:$69dda8455c7dd4254bf353b773304eec + 1 = 69dda8455c7dd4254bf353b773304eed$To calculate the next value, just again add 1. You could also use a increasing counter and xor it with the original IV: ... 2 Relying solely on randomization for the block counter is actually more likely to cause a nonce collision in case of a system time reset. This only gets worse as the message length increases. This is further exacerbated if the PRNG takes the system time as input, or does not have enough seed entropy. There is also no reason for the static 0 byte in the nonce. ... 2 Key/IV pairs should not be reused for either AES-CTR and AES-CBC - or for any other symmetric cipher for that matter. As a cipher is a Pseudo Random Permutation (PRP) inserting the same input will result in identical output. If you perform that operation you will leak information to an attacker; the attacker can distinguish data with the same contents. CTR ... 2 This will not be secure in general, and is not recommended. The reason is that a PRG is guaranteed to produce a pseudorandom output only if its input is uniformly random (or pseudorandom). Moreover, several PRG outputs are jointly pseudorandom only if the PRG is run on independent seeds. But here you are invoking the PRG on "structured" inputs, i.e., the ... 2 Like said in the comments, a 256-bit message is two blocks of AES, no matter the keysize. The main issue with ECB mode (i.e. using AES directly on 128-bit blocks) is that you leak whether two blocks are equal. When encrypting perfectly random data, that means there's a$2^{-128}$chance the two parts of the key are equal, and the attacker knows that. The ... 2 I would advise a different solution. You either generate a master-key (or key set) or derive one from a user password (e.g. via PBKDF2 or SCrypt). For each file to encrypt you generate a random key (file key) and nonce ad-hoc, and encrypt the file with that key, using an AEAD scheme. The random file key is encrypted with you master key and put at the ... 2 It probably doesn't. Yes. For strings$x,m0,m1$such that$\text {length}(x) = \text{length}(nonce)$and$\text{length}(m0) = \text{length}(i) = \text{length}(m1)( nonce || i ) \oplus ( x || m0 ) = ( nonce || j ) \oplus ( x || m1 ) \iff m0 \oplus m1 = i \oplus j$. I'm not aware of any. 2 We think that the player will not be able to get this key by extracting it from the memory or somehow else. Forgive me, but I'm skeptical. If you really have figured out a way to do this --- and plenty of well-funded, intelligent people have tried and failed --- I'd recommend slapping a patent on it and making millions off of licensing fees. How ... 2 The CTR part of CCM is basically the last for loop in the _ctrMode function: for (i=0; i<l; i+=4) { ctr[3]++; enc = prf.encrypt(ctr); data[i] ^= enc[0]; data[i+1] ^= enc[1]; data[i+2] ^= enc[2]; data[i+3] ^= enc[3]; } i.e. CTR is simply: encrypt a counter block with a block cipher, xor the encrypted block into the data, ... 2 Any plaintext will do. If you choose 0x000..., then you can skip the XOR step entirely. One extra step you can take to improve security is to periodically use the stream to choose a new AES key. This provides forward security: if an attacker manages to compromise the system (say, using a heartbleed-type exploit) and learn the current key, he will not be ... 2 Your combined mode of operation is not as easy to attack as a two-times-pad (i.e. stream-cipher with fixed IV used twice), but it still has some weaknesses. For example, an attacker which did read your file before and after the change can easily find out which 128-bit-blocks of the file did change and which ones stayed the same. Depending on the file format ... 2 There is no real advantage, other than the fact that it allows you to convert a block cipher into a stream cipher securely. Since there has been a large amount of research put into block ciphers and ciphers such as AES are commonly implemented in hardware (such as AES-NI), it allows for reuse of the primitives. Side note: the nonce generally does not need ... 2 The once part inside of the nonce in CTR mode means effectively "once for this particular key". If you use a fresh key for each message (e.g. by encrypting it using public-key crypto or similar), you can use the same nonce for all the messages (or a size-zero nonce). The important part is that the combination of nonce and ctr-value (i.e. what is input into ... 1 In effect, does this mean that a separate HMAC must be created for every single message, i.e. all 16 bytes? No, you can use HMAC over the whole message with all its blocks. Of course, that means you can't localize corruption/tampering, so if you were sending a huge message you might want HMAC smaller parts (but longer than 16 bytes or you are ... 1 Well, lets try to take a concrete example: Suppose that you have a 142 bit message that you need to encrypt with CTR mode. What do you do? Well, you generate the first 128 bit block with AES; you take the first 128 bits of the message, exclusive-or the two, and that's the first 128 bits of the ciphertext. Then what? Well, you have 142-128=14 bits of ... 1 With a 4 byte counter, it is a representation of a 32-bit unsigned integer. Since the counter increments from 0, there is no need to store it on a per message basis. This then becomes an implementation issue. A smart AES-CTR implementation would accept ANY initial counter/nonce value, and simply 0-pad it to the desired length. If we supply the cipher only ... 1 If by "encrypted" you mean generating a keystream, then what you propose is to use in the CTR mode $$C_i = P_i \oplus F_K(IV||i)$$ the following function$F\$: $$F_{K_1||K_2||K_3} = E_{K_1}\oplus E_{K_2} \oplus E_{K_3}.$$ This is secure as long as you ensure that for each key all the used IVs are different (i.e. are nonces). As mentioned in another ...
Only top voted, non community-wiki answers of a minimum length are eligible
|
2014-08-31 10:24:59
|
{"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": 0.7352727651596069, "perplexity": 1596.6309023207914}, "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-2014-35/segments/1408500959239.73/warc/CC-MAIN-20140820021559-00227-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://math.stackexchange.com/questions/2048858/i-am-having-trouble-finding-the-first-derivative-of-rp-per1-fracpk/2048965#2048965
|
# I am having trouble finding the first derivative of $R(P) = (Pe^r)(1-\frac{P}{K})$
I am having trouble finding the first derivative of $R(P) = (P\textrm{e}^r)(1-\frac{P}{K})$ I am told to use the product rule for this. The first part of it $(P\textrm{e}^r)$ I believe will remain the same for the derivative of it. I am struggling on the $(1-P/K)$.
The derivative of the function $R(P) = Pe^r\left(1-\frac{P}{K}\right)$ with respect to $P$ can be found by applying the product rule, as follows \begin{align} \frac{d}{dP}R(P) &= \left(\frac{d}{dP}Pe^r\right)\left(1-\frac{P}{K}\right) + Pe^r\frac{d}{dP}\left(1-\frac{P}{K}\right)\\ &= e^r\left(1-\frac{P}{K}\right) + Pe^r\left(\frac{d}{dP}1 + \frac{d}{dP}\left(-\frac{P}{K}\right)\right)\\ &= e^r\left(1-\frac{P}{K}\right) + Pe^r\left(0 - \frac{1}{K}\right)\\ &= e^r\left(1-\frac{P}{K}\right) - \frac{Pe^r}{K}\\ &= e^r\left(1-\frac{2P}{K}\right) \end{align}
A lot of the times, if one does some simple manipulations initially, the amount of work required can be greatly reduced. Notice that $R(P) = Pe^r\left(1-\frac{P}{K}\right) = e^r\left(P-\frac{P^2}{K}\right)$. Now the derivative is simply \begin{align} \frac{d}{dP}R(P) &= e^r\frac{d}{dP}\left(P-\frac{P^2}{K}\right)\\ &= e^r\left(\frac{d}{dP}P-\frac{d}{dP}\frac{P^2}{K}\right)\\ &= e^r\left(1-\frac{2P}{K}\right) \end{align}
• The origional function Pe^r is being multiplied by (1-P/K) not divided Dec 7 '16 at 23:48
• @AmandaSmith Thanks for pointing that out! Dec 7 '16 at 23:52
• okay that makes alot more sense now! thank you so much Dec 8 '16 at 0:20
• what happens to the e^r at the end of e^r(1-P/K)-Pe^r/k Dec 8 '16 at 0:30
• @AmandaSmith There's a little intermediate step there, expand it all out: $e^r - Pe^r/K - Pe^r/K = e^r - 2Pe^r/K$ then factor out the $e^r$ in common with both of the terms to get $e^r(1-2P/K)$. Dec 8 '16 at 0:32
You may use product rule on $R(P)=e^r(P-\frac{P^2}{K})$ as follows:
$R'(P)=e^r\frac{d}{dP}(P-\frac{P^2}{K})$$+(P-\frac{P^2}{K})\frac{d}{dP}(e^r)$
|
2021-11-30 12:56:21
|
{"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": 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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998579025268555, "perplexity": 847.2291262607736}, "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/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00058.warc.gz"}
|
http://mathhelpforum.com/advanced-algebra/206331-downloading-using-latex-print.html
|
• October 29th 2012, 01:04 PM
hedi
Hi,
Thank's for any help
• October 29th 2012, 01:10 PM
a tutor
To use it on this site it is not necessary to download it. Just type the appropriate code, for example:
Code:
[tex]\sqrt{x^{3y+1}}\sin\left(\frac{e^x}{a_2}\right) \ge\cos(y)[\tex]
produces $\sqrt{x^{3y+1}}\sin\left(\frac{e^x}{a_2}\right) \ge\cos(y)$.
• October 29th 2012, 01:10 PM
MarkFL
You don't have to download anything. You simply enclose your code with the TEX tags (use the sigma button to generate the tags). There are many good tutorials available online, just do a search for them.
• October 29th 2012, 02:07 PM
Plato
Quote:
Originally Posted by hedi
Hi,
The tab $\boxed{\Sigma}$ on the tool bar gives the LaTeX wraps, [TEX][/TEX]
|
2016-07-01 17:19:03
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 2, "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": 0.8521679043769836, "perplexity": 4906.405668520124}, "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-2016-26/segments/1466783403508.34/warc/CC-MAIN-20160624155003-00008-ip-10-164-35-72.ec2.internal.warc.gz"}
|
https://math.stackexchange.com/questions/1351267/analytical-solution-to-complex-heat-equation-with-neumann-boundary-conditions-an
|
Analytical solution to complex Heat Equation with Neumann boundary conditions and lateral heat loss
I have solved a PDE in this from numerically on Mathematica, but does anyone know if there is a way to solve the following PDE analytically, an analytical solution would really help me. This is an adapted heat equation accounting for lateral heat loss in a rod of length $0.2$ ${m}$:
$$u_{t} =\alpha u_{xx}-\kappa(u-T_{0})$$
Where:
$\alpha=1.12*10^{-4}$$m^2 s^{-1} \kappa=2.61*10^{-4}$$s^{-1}$
$T_{0}=17$ C
With boundary conditions:
$$u_{x}(0,t) =0$$ $$u_{x}(0.2,t) =0$$
And initial condition:
$$u(x,0)=\begin{cases} 35.6236 + 0.161087e^{59.9842x},0<x<0.1 \\ 35.6236 + 0.161087e^{59.9842 (-x + 0.2)},0.1<x< 0.2 \\ \end{cases}$$
Any help will be much appreciated. Thanks in advanced.
• Taking $w(x,t)=(u(x,t)-T_0) \, e^{\kappa t}$ as your new unknown function will bring you back to the ordinary heat equation, which has a standard solution in terms of Fourer cosine series. – Hans Lundmark Jul 6 '15 at 19:22
The functions $\{ \cos(n\pi x/L) \}_{n=0}^{\infty}$ form a commplete orthogonal set of functions on $[0,L]$. So you can expand $$u(x,t) = \sum_{n=0}^{\infty}c_{n}(t)\cos(n\pi x/L).$$ In order to solve the equation, plug this form into the equation and solve for the coefficient functions $c_{n}(t)$: $$\sum_{n=0}^{\infty}\{c_{n}'(t)+\alpha n^{2}\pi^{2}/L^{2}c_{n}(t)+\kappa c_{n}(t)\}\cos(n\pi x/L)= \kappa T_{0}$$ The right side is a series in the $\cos$ terms where only the coefficient of the constant term is non-zero. Therefore, \begin{align} c_{0}'(t)+\kappa c_{0}(t) & =\kappa T_{0},\\ c_{n}'(t)+(\alpha n^{2}\pi^{2}/L^{2}+\kappa)c_{n}(t) & =0,\;\; n > 0. \end{align} The solutions of these equations involve an initial constant $c_{n}(0)=C_{n}$. For $n=0$: $$(e^{\kappa t}c_{0}(t))'=(e^{\kappa t} T_{0})' \\ e^{\kappa t}c_{0}(t)-C_{0} = T_{0}(e^{\kappa t}-1) \\ c_{0}(t) = C_{0}e^{-\kappa t}+ T_{0}(1-e^{-\kappa t}).$$ For $n > 0$, $$c_{n}(t) = C_{n}e^{-(\alpha n^{2}\pi^{2}/L^{2}+\kappa)t}$$ The last condition to be satisfied is the initial condition $u(x,0)=f(x)$, which becomes $$f(x) = \sum_{n=0}^{\infty}C_{n}\cos(n\pi x/L).$$ Hence, using the orthogonality of the $\cos$ terms, $$\int_{0}^{L}f(x)\cos(m\pi x/L)dx = C_{m}\int_{0}^{L}\cos^{2}(m\pi x/L)dx.$$ Note that $\int_{0}^{L}\cos^{2}(m\pi x/L)dx=L/2$ for $m \ge 1$ and $L$ for $m=0$. The final solution: $$u(x,t) = \sum_{n=1}^{\infty}\left(\frac{2}{L}\int_{0}^{L}f(x)\cos(n\pi x/L)dx\right) e^{-(\alpha n^{2}\pi^{2}/L^{2}+\kappa)t}\cos(n\pi x/L) \\ + \left(\frac{1}{L}\int_{0}^{L}f(x)dx\right)e^{-\kappa t}+ T_{0}(1-e^{-\kappa t}).$$ Check everything. I'm good at errors. :)
• Hi thanks alot for your help. I have 2 questions, why in the first step did you assume that u(x,t)=∑n=0∞cn(t)cos(nπx/L)? Also do you need a cosine term in c′n(t)+(αn2π2/L2+κ),=0 n>0. Also can any piecewise continuous function on the interval [0,L] be expanded as a Fourier cosine series? When I tried to do so for my function it kept over estimating it. But thanks a lot for your help. – daniel17027 Jul 7 '15 at 0:33
• I used $\cos(n\pi x/L)$ for $n=0,1,2,3,\dots$ because these are the eigenfunctions of $-\frac{d^{2}}{dx^{2}}$ for which $f'(0)=f'(L)=0$. Automatically, when normalized to have integral $1$ on $[0,L]$, they become a complete orthonormal basis of $L^{2}[0,L]$, which means everything can be expanded in these functions; and the expansions converge pointwise under standard conditions, except possibly at the endpoints $0$, $L$ because of endpoint conditions. Your function $f$ needs to satisfy $f'(0)=f'(L)=0$ for more uniform convergence, and this is a consistency requirement for your problem, too. – DisintegratingByParts Jul 7 '15 at 1:19
• @DanielManogaran : I forgot to you ping you on the previous remark. Knowing that you can expand every function in this way, you can verify directly that the solution I've given you is correct. However, check the normalization constants for yourself: $\int_{0}^{L}\cos^{2}(n\pi x/L)dx$, but definitely notice that the one for $n=0$ is different from the others. Note: when you have $\sum_{n=0}^{\infty}c_{n}\cos(n\pi x/L)=\sum_{n=0}^{\infty}d_{n}\cos(n\pi x/L)$ for a.e. $x\in[0,L]$, then $c_n=d_n$. – DisintegratingByParts Jul 7 '15 at 1:25
• Great solution, well explained, (+1). – mattos Jul 7 '15 at 4:35
• @TrialAndError Ohh ok so that is why $c_{n}'(t)+(\alpha n^{2}\pi^{2}/L^{2}+\kappa)c_{n}(t) & =0,\;\; n > 0$ Thanks, is there a name or a proof for $\sum_{n=0}^{\infty}r_{n}\cos(n\pi x/L)=\sum_{n=0}^{\infty}s_{n}\cos(n\pi x/L)$? Thank you very much, you really helped me. – daniel17027 Jul 7 '15 at 10:26
|
2019-12-07 09:56:17
|
{"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9253187775611877, "perplexity": 243.5871327437363}, "config": {"markdown_headings": false, "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-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00176.warc.gz"}
|
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-9-section-9-1-solving-quadratic-equations-by-the-square-root-property-exercise-set-page-631/39
|
Introductory Algebra for College Students (7th Edition)
$y=\left\{ 7-2\sqrt{3},7+2\sqrt{3} \right\}$
The left side of the given equation, $y^2-14y+49=12 ,$ is a perfect square trinomial whose factored form is \begin{array}{l}\require{cancel} (y-7)^2=12 .\end{array} Taking the square root of both sides (the Square Root Property), the solutions to the given equation are \begin{array}{l}\require{cancel} y-7=\pm\sqrt{12} \\\\ y-7=\pm\sqrt{4\cdot3} \\\\ y-7=\pm\sqrt{(2)^2\cdot3} \\\\ y-7=\pm2\sqrt{3} \\\\ y=7\pm2\sqrt{3} .\end{array} Hence, $y=\left\{ 7-2\sqrt{3},7+2\sqrt{3} \right\} .$
|
2018-07-23 15:39:52
|
{"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": 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": 0.9993571639060974, "perplexity": 3450.2696678077027}, "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-2018-30/segments/1531676596542.97/warc/CC-MAIN-20180723145409-20180723165409-00523.warc.gz"}
|
https://www.jobilize.com/chemistry/course/15-2-lewis-acids-and-bases-equilibria-of-other-reaction-classes-by-ope?qcr=www.quizover.com&page=3
|
# 15.2 Lewis acids and bases (Page 4/12)
Page 4 / 12
## Chemistry end of chapter exercises
Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
when the amount of solid is so small that a saturated solution is not produced
Explain why the addition of NH 3 or HNO 3 to a saturated solution of Ag 2 CO 3 in contact with solid Ag 2 CO 3 increases the solubility of the solid.
Calculate the cadmium ion concentration, [Cd 2+ ], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO 3 ) 2 with 1.150 L of 0.100 NH 3 ( aq ).
8 $×$ 10 –5 M
Explain why addition of NH 3 or HNO 3 to a saturated solution of Cu(OH) 2 in contact with solid Cu(OH) 2 increases the solubility of the solid.
Sometimes equilibria for complex ions are described in terms of dissociation constants, K d . For the complex ion ${\text{AlF}}_{6}{}^{\text{3−}}$ the dissociation reaction is:
${\text{AlF}}_{6}{}^{\text{3−}}⇌{\text{Al}}^{\text{3+}}+6{\text{F}}^{\text{−}}$ and ${K}_{\text{d}}=\frac{\left[{\text{Al}}^{\text{3+}}\right]{\left[{\text{F}}^{\text{−}}\right]}^{6}}{\left[{\text{AlF}}_{6}{}^{\text{3−}}\right]}=2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-24}$
Calculate the value of the formation constant, K f , for ${\text{AlF}}_{\text{6}}{}^{\text{3−}}.$
5 $×$ 10 23
Using the value of the formation constant for the complex ion $\text{Co}{\left({\text{NH}}_{3}\right)}_{6}{}^{\text{2+}},$ calculate the dissociation constant.
Using the dissociation constant, K d = 7.8 $×$ 10 –18 , calculate the equilibrium concentrations of Cd 2+ and CN in a 0.250- M solution of $\text{Cd}{\left(\text{CN}\right)}_{4}{}^{\text{2−}}.$
[Cd 2+ ] = 9.5 $×$ 10 –5 M ; [CN ] = 3.8 $×$ 10 –4 M
Using the dissociation constant, K d = 3.4 $×$ 10 –15 , calculate the equilibrium concentrations of Zn 2+ and OH in a 0.0465- M solution of $\text{Zn}{\left(\text{OH}\right)}_{4}{}^{\text{2−}}.$
Using the dissociation constant, K d = 2.2 $×$ 10 –34 , calculate the equilibrium concentrations of Co 3+ and NH 3 in a 0.500- M solution of $\text{Co}{\left({\text{NH}}_{3}\right)}_{6}{}^{\text{3+}}.$
[Co 3+ ] = 3.0 $×$ 10 –6 M ; [NH 3 ] = 1.8 $×$ 10 –5 M
Using the dissociation constant, K d = 1 $×$ 10 –44 , calculate the equilibrium concentrations of Fe 3+ and CN in a 0.333 M solution of $\text{Fe}{\left(\text{CN}\right)}_{6}{}^{\text{3−}}.$
Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 $×$ 10 –2 mol of silver cyanide, AgCN.
1.3 g
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 $×$ 10 –3 mol of silver bromide.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na 2 S 2 O 3 ·5H 2 O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3−}}$ ( K f = 4.7 $×$ 10 13 )?
0.79 g
We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H + ) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) ${\text{CO}}_{2}+{\text{OH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{HCO}}_{3}{}^{\text{−}}$
(b) $\text{B}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{B}{\left(\text{OH}\right)}_{4}{}^{\text{−}}$
(c) ${\text{I}}^{\text{−}}+{\text{I}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{I}}_{3}{}^{\text{−}}$
(d) ${\text{AlCl}}_{3}+{\text{Cl}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{AlCl}}_{4}{}^{\text{−}}$ (use Al-Cl single bonds)
(e) ${\text{O}}^{\text{2−}}+{\text{SO}}_{3}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{4}{}^{\text{2−}}$
(a)
;
(b)
;
(c)
;
(d)
;
(e)
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) ${\text{CS}}_{2}+{\text{SH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{HCS}}_{3}{}^{\text{−}}$
(b) ${\text{BF}}_{3}+{\text{F}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{BF}}_{4}{}^{\text{−}}$
(c) ${\text{I}}^{\text{−}}+{\text{SnI}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SnI}}_{3}{}^{\text{−}}$
(d) $\text{Al}{\left(\text{OH}\right)}_{3}+{\text{OH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Al}{\left(\text{OH}\right)}_{4}{}^{\text{−}}$
(e) ${\text{F}}^{\text{−}}+{\text{SO}}_{3}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SFO}}_{3}{}^{\text{−}}$
Using Lewis structures, write balanced equations for the following reactions:
(a) $\text{HCl}\left(g\right)+{\text{PH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}$
(b) ${\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}$
(c) $\text{CaO}+{\text{SO}}_{3}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}$
(d) ${\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}$
(a)
;
(b) ${\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{4}+{\text{H}}_{2}\text{O}$
;
(c) $\text{CaO}+{\text{SO}}_{3}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CaSO}4$
;
(d) ${\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}+{\text{NH}}_{3}$
Calculate $\left[{\text{HgCl}}_{4}{}^{\text{2−}}\right]$ in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100- M HgCl 2 solution.
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO 3 is added before precipitation begins. [The reaction of Ag + with CN goes to completion, producing the $\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{−}}$ complex.] Precipitation of solid AgCN takes place when excess Ag + is added to the solution, above the amount needed to complete the formation of $\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{−}}.$ How many grams of NaCN were in the original sample?
0.0281 g
What are the concentrations of Ag + , CN , and $\text{Ag}{\left(\text{CN}\right)}_{2}{}^{\text{−}}$ in a saturated solution of AgCN?
In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO 3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF 3 or SbF 5 ). Write balanced chemical equations for the reaction of pure HNO 3 with pure HF and of pure HF with BF 3 .
${\text{HNO}}_{3}\left(l\right)+\text{HF}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{NO}}_{3}{}^{+}+{\text{F}}^{\text{−}};$ $\text{HF}\left(l\right)+{\text{BF}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}^{+}+{\text{BF}}_{4}$
The simplest amino acid is glycine, H 2 NCH 2 CO 2 H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH 2 , and a carboxylic acid group, –CO 2 H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH 3 CO 2 H, and the base strength of the amino group is slightly greater than that of ammonia, NH 3 .
(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
(b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH 2 and $-{\text{CO}}_{2}{}^{\text{−}}$ groups.)
Boric acid, H 3 BO 3 , is not a Brønsted-Lowry acid but a Lewis acid.
(a) Write an equation for its reaction with water.
(b) Predict the shape of the anion thus formed.
(c) What is the hybridization on the boron consistent with the shape you have predicted?
(a) ${\text{H}}_{3}{\text{BO}}_{3}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{4}{\text{BO}}_{4}{}^{\text{−}}+{\text{H}}^{+};$ (b) The electronic and molecular shapes are the same—both tetrahedral. (c) The tetrahedral structure is consistent with sp 3 hybridization.
what are oxidation numbers
pls what is electrolysis
Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.
AZEEZ
thanks
Idowu
what is the basicity of an atom
basicity is the number of replaceable Hydrogen atoms in a Molecule. in H2SO4, the basicity is 2. in Hcl, the basicity is 1
Inemesit
how to solve oxidation number
mention some examples of ester
do you mean ether?
Megan
what do converging lines on a mass Spectra represent
would I do to help me know this topic ?
Bulus
oi
Amargo
what the physic?
who is albert heistein?
Bassidi
similarities between elements in the same group and period
what is the ratio of hydrogen to oxulygen in carbohydrates
bunubyyvyhinuvgtvbjnjnygtcrc
yvcrzezalakhhehuzhbshsunakakoaak
what is poh and ph
please what is the chemical configuration of sodium
Sharon
2.8.1
david
1s²2s²2p⁶3s¹
Haile
2, 6, 2, 1
Salman
1s2, 2s2, 2px2, 2py2, 2pz2, 3s1
Justice
1s2,2s2,2py2,2
Maryify
1s2,2s2,2p6,
Francis
1s2,2s2,2px2,2py2,2pz2,3s1
Nnyila
what is criteria purity
cathode is a negative ion why is it that u said is negative
cathode is a negative electrode while cation is a positive ion. cation move towards cathode plate.
king
CH3COOH +NaOH ,complete the equation
compare and contrast the electrical conductivity of HCl and CH3cooH
The must be in dissolved in water (aqueous). Electrical conductivity is measured in Siemens (s). HCl (aq) has higher conductivity, as it fully ionises (small portion of CH3COOH (aq) ionises) when dissolved in water. Thus, more free ions to carry charge.
Abdelkarim
HCl being an strong acid will fully ionize in water thus producing more mobile ions for electrical conduction than the carboxylic acid
Valentine
differiante between a weak and a strong acid
david
how can I tell when an acid is weak or Strong
Amarachi
an aqueous solution of copper sulphate was electrolysed between graphite electrodes. state what was observed at the cathode
write the equation for the reaction that took place at the anode
Bakanya
what is enthalpy of combustion
Bakanya
Enthalpy change of combustion: It is the enthalpy change when 1 mole of substance is combusted with excess oxygen under standard conditions. Elements are in their standard states. Conditions: pressure = 1 atm Temperature =25°C
Abdelkarim
Observation at Cathode: Cu metal deposit (pink/red solid).
Abdelkarim
Equation at Anode: (SO4)^2- + 4H^+ + 2e^- __> SO2 + 2H2O
Abdelkarim
Equation : CuSO4 -> Cu^2+ + SO4^2- equation at katode: 2Cu^2+ + 4e -> 2Cu equation at anode: 2H2O -> 4H+ + O2 +4e at the anode which reacts is water because SO4 ^ 2- cannot be electrolyzed in the anode
Niken
|
2020-12-04 02:10:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 48, "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": 0.6964091062545776, "perplexity": 4086.3126807855556}, "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/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00129.warc.gz"}
|
https://www.physicsforums.com/threads/is-this-an-accurate-statement-of-the-limit-definition-quantifiers-connectives.526244/
|
# Is this an accurate statement of the limit definition? (quantifiers/connectives)
1. Sep 1, 2011
### anonymity
$\forall\epsilon\exists\delta[( (\epsilon\wedge\delta) > 0) \wedge ((| x - t| < \delta)\Rightarrow (|f(x) - f(t)|< \epsilon))]$
The part that seems wrong is the placement of the statement "delta is greater than zero" and "epsilon is greater than zero". It seems like these statements may need to be placed separately after their respective quantifiers for this to be an accurate statement...
2. Sep 1, 2011
### micromass
Staff Emeritus
Well, first of all
$$(\epsilon\wedge \delta)>0$$
makes no sense. $\wedge$ is an operator that takes in truth values and returns a truth value. For example "true $\wedge$ true is true". But $\epsilon$ and $\delta$ are not truth values.
Second point: you never stated what x or t was. I expect another quantifier $\forall t$ in there.
3. Sep 1, 2011
### anonymity
Okay. Thanks.
So if I break ($\epsilon\wedge\delta$ > 0) up into ($\delta > 0) \wedge (\epsilon > 0$) then THAT part will be okay left inside of the two quantifiers as it is above?
edit: thanks for pointing that out...it seems so obvious now -.-
4. Sep 1, 2011
### micromass
Staff Emeritus
It will be syntactically ok, but it won't have the meaning you want it to have.
With your changes, the statement will be always true. Indeed, you can choose $\delta=-1$ each time. Then
$$(\epsilon>0)\wedge (\delta >0)\wedge (|x-t|<\delta)$$
will be false. So the implication will be true. So your statement will always be true!! This is clearly not what you want.
5. Sep 1, 2011
thank you
6. Sep 1, 2011
### anonymity
If I changed it to $\forall\epsilon[\epsilon > 0]\exists\delta[\delta > 0]$
Would that be logically correct?
edit: by logically correct, I really mean to ask "would that logically read: for all epsilon greater than zero, there exists delta greater than zero". It seems to me that it would logically mean that, but I don't want to assume. For the time being I am past trying to get the logical structure of the statement...taking a step back for a minute =0
7. Sep 1, 2011
### micromass
Staff Emeritus
You're on the right track, but it is not syntactically correct what you're stating.
Try to do something as follows:
$$\forall \epsilon ((\epsilon >0)~\Rightarrow~\exists \delta ((\delta>0)\wedge (......$$
This is essentially what you mean. But this is syntactically correct.
8. Sep 1, 2011
### anonymity
That is very close to what I just did while waiting for you to reply. Combined with what you just said it is:
$\forall\epsilon[(\epsilon > 0) \Rightarrow \exists\delta((\delta>0) \wedge ((|x-t| < \delta) \Rightarrow (|f(x)-f(t)|<\epsilon))]$
This feels much better on an intuitive level, but (aside from asking you "IS THIS RIGHT?????") how can I test it to be sure? Is a truth table the only way?
Also, I am certian that it will change the meaning, but if i combined the "delta greater than zero" and the "|x-t| < delta" into $((0<|x-t|<\delta) \Rightarrow (|f(x)-f(t)|<\epsilon))$, would it still be correct (assuming that I have it right above)?
9. Sep 1, 2011
### micromass
Staff Emeritus
This seems correct. However, you still didn't introduce t (or x) in here. You're missing a quantifier.
As for checking that it is correct. Well, perhaps a truth table should work...
No, you need to state separately that $\delta>0$. Otherwise, you can choose $\delta=-1$ and you'll see that it's always true.
10. Sep 1, 2011
### anonymity
Okay. Thanks for your help micro. I should be able to figure the rest out.
Thanks again
-anonymous
|
2017-11-18 01:36:49
|
{"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": 0.6458475589752197, "perplexity": 1054.0969817624418}, "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-2017-47/segments/1510934804125.49/warc/CC-MAIN-20171118002717-20171118022717-00244.warc.gz"}
|
https://astronomy.stackexchange.com/questions/41207/translating-a-zenith-position-to-the-nadir
|
# Translating a zenith position to the nadir
What would be the easiest way to translate a zenith position to a nadir on the celestial sphere?
For example I have a position from the azimuthal grid in Stellarium; 3h44m32.64s/+50°50'16.2" which was obtained based on my location at +50°54'13.9" -2°34'19.1" and a time of ; 15/03/2019 16:24:20.
In Astronomical Algorithms, Second Edition, Jean Meeus shows us how to convert local coordinates to equatorial and vice versa.
Formulas 13.5 and 13.6, p. 93:
$$\mathrm{tan}\ A = \frac {\mathrm{sin}\ H}{\mathrm{cos}\ H\ \mathrm{sin}\ \phi - \mathrm{tan}\ \delta\ \mathrm{cos}\ \phi}$$
$$\mathrm{sin}\ h = \mathrm{sin}\ \phi\ \mathrm{sin}\ \delta + \mathrm{cos}\ \phi\ \mathrm{cos}\ \delta\ \mathrm{cos} H$$
Unnumbered formulas, p. 94:
$$\mathrm{tan}\ H = \frac {\mathrm{sin}\ A}{\mathrm{cos}\ A\ \mathrm{sin}\ \phi + \mathrm{tan}\ h\ \mathrm{cos}\ \phi }$$
$$\mathrm{sin}\ \delta = \mathrm{sin}\ \delta\ \mathrm{sin}\ h - \mathrm{cos}\ \phi\ \mathrm{cos}\ h\ \mathrm{cos}\ A$$
where $$\alpha$$ is the right ascension in degrees, $$\delta$$ is the declination in degrees, $$\phi$$ is the latitude (+ North, $$-$$ South) in degrees, $$H$$ is the hour angle measured westwards from South in degrees, $$A$$ is the azimuth in degrees, and $$h$$ is the height of the object in degrees. Please note that Meeus measures azimuths from the South heading East (90°) then North (180°) and West (270°).
With these formulas, you can find the required position of the point you want to know about.
In order to get the nadir, I would just put $$h = -90°$$. Another way would be to find the position of the object at zenith, add/remove 180° to its right ascension, and change the sign of its declination.
Hope this helps!
Clear skies.
|
2021-11-27 03:18:39
|
{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3835006654262543, "perplexity": 1267.6079371601497}, "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/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00195.warc.gz"}
|
http://openstudy.com/updates/51170a0de4b09e16c5c89824
|
## anonymous 3 years ago RHS HELP!!! (Reimann Sums) (Note that I have NOT learned about integrals yet) Here's the Q: I'm given a curve where 0<=x<=10, and there are 5 rectangles, so n=5. I calculated delta x to be 2. I'm really confused how to use these two formulas to find both the LHS and the RHS. I've typed the two formulas below, in my reply. Please help!! I AM REALLY CONFUSED! Thank you!
1. anonymous
$\ \huge \text {LHS:} \sum_{i=0}^{n-1} f(x_i) Δ x$
2. anonymous
$\ \ \huge \text {RHS:} \sum_{i=1}^{n} f(x_i) Δ x$
3. anonymous
well delta x = (b-a)/n
4. anonymous
5. anonymous
6. anonymous
$$\ \Huge \text{Oh, and: } x_i = a+iΔx$$
7. anonymous
if you just started integrals you shouldve went over sigma notation im pretty sure, and you just expand it
8. anonymous
This isn't helpful without the function. Otherwise the answer will be just theoretical stuff.
9. anonymous
I don't have a function value... I'm only given a graph in my textbook and am told to find the lower estimate and an upper estimate.
10. anonymous
And, we haven't even discussed what an integral is in class...
11. anonymous
Could you give a description of what your graph is?
12. anonymous
It looks something like:|dw:1360464858577:dw|
13. anonymous
My confusion is I'm not sure how to use the two summation equations....
14. anonymous
And then use the values I know from the graph
15. anonymous
Okay, that's better. Well you first need to partition the interval into 5 equal intervals. For the left Riemann: for your values x = 0,2,4,6,8, draw a vertical line that TOUCHES the graph of the function. This vertical line is the height of the rectangle that you are estimating the curve with. Each rectangle's width is 2. |dw:1360465074815:dw| For the right hand do the same except for now, use let the x values be x = 2,4,6,8,10 |dw:1360465192357:dw|
16. anonymous
Note how the left hand Riemann is an underestimation of area while the right hand Riemann is an over estimation of area.
17. anonymous
@khoala4pham How do I use the sigma equations? Where do they come into this?
18. anonymous
Anyone??
19. anonymous
The sigma equations represent area. They mean that you sum up the areas of each individual rectangle. That is all that they mean. For the left hand Riemann, you take 2*(f(0) + f(2) + f(4) + f(6) + f(8)) For the right, 2*(f(2) + f(4) + f(6) + f(8) + f(10)) Why that though? Because for the RHR, the first rectangle has height f(0) and its width is 2. The second rectangle has height f(2) and width 2. So forth and so on.
20. anonymous
How do you know to choose 0,2,4,6,8 and 2,4,6,8,10?
21. anonymous
Well your a is zero in this case. a is the place where your interval begins. From your equation of the left hand Riemann, you see that x_i = a + i*(deltax) i runs from 0 to n-1. Well, n here is 5--the domain is partitioned into 5 intervals. That means that i = 0,1,2,3,4. Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8. The same applies to the right Riemann.
22. anonymous
THAT MAKES SO MUCH MORE SENSE! THANK YOU, @khoala4pham! I was getting confused with the $$\ \Huge \sum \text{ notation!}$$
23. anonymous
The notation is highly ambiguous if you are not used to it. The x_i is a number. Deltax is also a number. You are summing numbers across indexes. How you found your index is obtained above.
24. anonymous
Can you explain the RHS for "Well your a is zero in this case. a is the place where your interval begins. From your equation of the left hand Riemann, you see that x_i = a + i*(deltax) i runs from 0 to n-1. Well, n here is 5--the domain is partitioned into 5 intervals. That means that i = 0,1,2,3,4. Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8. The same applies to the right Riemann." I dont really understand the RHS...
25. anonymous
How you got the numbers 2,4,6,8,10
26. anonymous
If i runs from 1 to n, so that means i runs 1 to 5...
27. anonymous
I'm getting like 2, 5 ...? x_1= (a=0) + (1)(delta x = 2) = 2 x_2 = (a=1) + (2)( delta x = 2) =5
28. anonymous
@khoala4pham Why do you keep 0 constant? Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8.
29. anonymous
your equation of x_i = a + i*(delta x). a = 0 by default. a is the value that you start your interval with. if your interval were from (20,50) then a = 20. If your interval were (-3,24), then a = -3. In your initial problem you told me that 0 <= x < = 10. Thus a = 0.
|
2016-09-27 16:18:13
|
{"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": 0.8082137107849121, "perplexity": 1040.4062448075406}, "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-2016-40/segments/1474738661123.53/warc/CC-MAIN-20160924173741-00204-ip-10-143-35-109.ec2.internal.warc.gz"}
|
https://tutorials.methodsconsultants.com/page/2/
|
## SEM in Mplus
This page describes how to set up code in Mplus to fit a full structural equation model with latent variables. The model consists of three latent variables and eleven manifest variables, as described here. Mplus only reads data in text format, see this post for details on how to prepare a data file for Mplus. The data can be accessed from Github. To review, the model to be fit is the following:
## CFA in Mplus
This page describes how to set up code in Mplus to fit a confirmatory factor analysis (CFA) model. The model, which consists of two latent variables and eight manifest variables, is described here. Mplus only reads data in text format, see this post for details on how to prepare a data file for Mplus. The data can be accessed from Github. To review, the model to be fit is the following:
## A Running CFA and SEM Example
Industrialization and Democracy (Bollen, 1989) The CFA and SEM examples are taken from data analyzed in Kenneth Bollen’s (1989) book Structural Equations with Latent Variables. The goal of the model is to determine how democracy and industrialization in 1960 are associated with democracy in 1965. The problem is that there are no good single items that capture the entirety of the concepts of democracy and industrialization. Hence, these concepts are treated as latent (unobserved) variables imperfectly measured by a set of observed indicators.
## Preparing Data for Mplus
From SPSS Mplus requires data to be read in from a text file without variable names, with numeric values only, and with missing data coded as a single numeric value, such as -999. A common workflow for preparing data to analyze in Mplus is to perform the variable cleaning in SPSS and then save the data as a text file. This guide shows the appropriate steps using the sample file bollen_sem.
## The Prisoner's Dilemma
This post will review geometric series and game theory, and discuss the prisoner’s dilemma and its importance. What is Game Theory Game theory is the study of interdependent decisions. That is, “my optimal choice depends on what you choose, and your optimal choice depends on what I choose.” A stable state where no participant can gain an advantage through a change of strategy, assuming the other participants actions remain unchanged is a nash equilibrium.
## Reading and Writing Excel Files With R Using readxl and writexl
This tutorial walks you through how to use the readxl package to read Microsoft Excel .xls and .xlsx file formats into R, and how to export data from R back out to an Excel format using the writexl package. readxl’s functions are related to importing Excel files into a tibble object, which is modern R’s internal data format. A tibble can then be manipulated to create summary tables or plots, run statistical tests, or perform other common analysis tasks.
## Estimating HLM Models Using R: Part 1
Estimating HLM Models Using R: Part 1 There are a number of different R packages that now exist for fitting mixed models, including hierarchical linear models. For cross-sectional applications, perhaps the most frequently used package is lme4 (Bates et al., 2015). However, due to ambiguity in how to appropriately determine the degrees of freedom for $$t$$-tests, lme4 does not provide $$p$$-values for the fixed effects. The lmerTest package (Kuznetsova et al.
Estimating HLM Models Using R: Part 2 The Means-as-Outcomes Model After estimating the empty model, R&B develop a Means-as-Outcomes model in which a school-level variable, meanses, is added to the model for the intercept. This variable reflects the average student SES level in each school. Recall that, when fitting the empty model, we began with the following equation: $Y_{ij} = \beta_{0j} + e_{ij}$ Now the intercept can be modeled as a grand mean, $$\gamma_{00}$$, plus the effect of the average SES score, $$\gamma_{01}$$, plus a random error, $$u_{0j}$$.
|
2019-12-06 21:45:51
|
{"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": 0.4130702316761017, "perplexity": 1732.5589602073997}, "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-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00544.warc.gz"}
|
http://speedphotos.bloguez.com/speedphotos/4225974/determinant_minor
|
determinant minor »
determinant minor
10/4/2012
For more minor cofactor photo
In linear algebra, a minor of a matrix A is the determinant of some smaller square matrix, cut down from A by removing one or more of its rows or columns. Minors ...Also known as yquot;Laplacianyquot; determinant expansion by minors, expansion by minors is a technique for computing the determinant of a given square matrix M ...The minor is the determinant that remains when you delete the row and column of the element youy're trying to find the minor for. That means we should delete ...But for 4�y's and bigger determinants, you have to drop back down to the smaller 2� and 3� determinants by using things called yquot;minorsyquot; and yquot;cofactorsyquot;.23 Aug 2000 ... The real number Mij is the determinant of a submatrix of dimension n-1 by n-1 ... The number Mij is called the minor for element ij of the matrix.One way to find the value of a determinant is to expand a determinant by minors. In general, the minor of an element in a 3*3 determinant is the 2*2 determinant ...Determinant, Minor, Cofactor, Evaluation of a determinant by cofactors. Determinant. A square array of quantities, called elements, symbolizing the sum of ...The minor is the determinant of the matrix constructed by removing the row and column of a particular element. Thus, the minor of a34 is the determinant of the ...Definition 1 If A is a square matrix then the minor of , denoted by , is the determinant of the submatrix that results from removing the ith row and jth column of A.The minor of an element of a 3 x 3 determinant is the 2 x 2 determinant that remains after you delete the row and column in which the element appears.The second step is to get the cofactors of the first row of matrix M. The third step is to multiply the determinants of the minor matrices by the matrix cofactors.positive only to one or two antigenic minor determinantsy#65292;and three cases ... Conclusions The minor antigenic determinant was important in allergic reaction.28 Jan 2011 ... Penicillin skin testing with the major and minor determinants of penicillin can reliably identify persons at high risk for penicillin reactions (238 ...Note that it was unnecessary to compute the minor or the cofactor of the (3, 2) entry in A, since that entry was 0. In general, then, when computing a determinant ...determinant, cofactor, 76. determinant, diagonal matrix, 74. determinant, Laplace expansion, 73. determinant, lower triangular, 74. determinant, minor, 72 ...Abstract. Like Major League Baseball (MLB), minor league ...TWO MINOR DETERMINANTS OF MYELIN BASIC. PROTEIN INDUCE EXPERIMENTAL ALLERGIC. ENCEPHALOMYELITIS IN SJL/J MICE. BY DWIGHT H.We consider the problem of computing the determinant of a matrix of polynomials ; we compare two algorithms (expansion by minors and Gaussian elimination), ...the nth and (ny#8722;1)th minors, whose solution is a representation of the nth minor as an n� determinant of resolvents. The latter is given a simple interpretation ...29 Aug 2002 ... Biohealthmatics Genomics Journal Articles - Hill-Robertson interference is a minor determinant of variations in codon bias across Drosophila ...26 Aug 2011 ... Question: $A$ is an $n$ order square matrix, and $B=A^{(*)}$ is the adjoint matrix of $A$ (i.e. $AB=BA=\det A\cdot I_n$). Show that. \det B ...using the matrix representation. Applications are also given to characteristic determinants, general minors and cofactors. Keywords: digraph, cycle, determinant, ...The 2 �2 determinants are called minors of the 3 �3 determinant |A|. The checkerboard sign together with a minor is called a cofactor. These formulas are ...(tnn. An analysis of this algorithm (see Bareiss [-1~, Lipson [-5~) shows that each a(k) is a determinant (minor) of some k-by-k submatrix of the original matrix. ~3 ...First some terminology, if we remove one row and one column the remaining determinant is known as a minor and the element at the intersection of the rows ...Isolation and Quantitation of a Minor Determinant of Hen Egg White Lysozyme Bound to I-Ak by Using Peptide-Specific Immunoaffinity. Raffi Gugasyan*, Ilan ...Minor of an element of a determinant : Minor of an element aij of a. determinant is a determinant obtained by deleting the ith row and jth column of a given ...there exist a minor of DFTn, that has exactly as its rows the set R and has all its columns disjoint from C, with determinant of magnitude larger than some given ...basic matrix calculation :-*, power, transpose, trace, determinant, minor, matri Search and download open source project / source codes from HackChina.com.Higher-order determinants are natural generalizations. The minor M_{{jk}} of the entry a_{{jk}} in the n th-order determinant \det[a_{{jk}}] is the ( n-1 )th-order ...
Category : reserved to connected users | Write a comment | Print
|
2018-02-25 17:29:36
|
{"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": 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": 0.8578007817268372, "perplexity": 1512.4584289480579}, "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-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00319.warc.gz"}
|
https://gamedev.stackexchange.com/questions/114956/how-to-avoid-gimbal-lock-on-rubiks-cube?noredirect=1
|
# How to avoid Gimbal Lock on Rubik's Cube [duplicate]
This question already has an answer here:
I am making a Rubik's cube program. That will simulate a Rubik's cube. I am having problems with gimbal lock. I need to know how to avoid it. To my understanding I can either multiply the x, y, z rotation matrices in a specific order to acquire the rotation I am trying to get. Or I can use a quaternion which is like rotating around a single user defined axis instead of the axis x, y, or z. I can't really understand visually how I can get the resulting axis from two other axis. (x, y, z) How can I get the resulting axis for my quaternion from my Euler rotations? Will it just result in gimbal lock again?
One of the big problems is. I don't know what rotations are going to be made. I need to make it general purpose.
Edit: Little bit more documentation of my problem:
My Euler matrix multiplication looks like this
rotate = XMMatrixRotationX(content.mesh[meshToRotate[i]].rotation.x) * XMMatrixRotationY(content.mesh[meshToRotate[i]].rotation.y) * XMMatrixRotationZ(radians);
As a gimbal lock would suggest the last axis in the rotation gets the problem. When you attempt to rotate the front or back face you get problems.
## marked as duplicate by Anko, bummzack, MAnd, Vaillancourt♦, Seth BattinJan 22 '16 at 21:58
• Quaternions. Enjoy trying to get your puny meat brain to understand them! (Mine can't) – Draco18s Jan 15 '16 at 21:37
• @Draco18s These questions helped me understand them: What is a quaternion? and Why do people use quaternions? and How can you visualize a quaternion? – Anko Jan 15 '16 at 21:44
• @Anko Good stuff, I'll have to take a look sometime. Generally though I can manage through euler (Unity3D handles things internally as quaternions, so as long as I let it recalc the eulers, gimbal lock is prevented). – Draco18s Jan 15 '16 at 21:46
• So I guess a better statement for my question is, I need to not somehow unconcatenate one of the axis' after a 3rd axis is used on the Rubik's cube – Andrew Wilson Jan 15 '16 at 22:02
• @Anko There's a special case here in the sense that this is a square tile puzzle and there's a trick to avoid gimbal locks that does not apply to other scenarios. – Stephane Hockenhull Jan 16 '16 at 7:19
In the case of a puzzle like this you can animate the rotation then reset the model back to its original position but with the cube's stickers rotated.
Look at it like cheating with a real puzzle by moving the stickers.
You'll have to make it possible to change the stickers' colors. This can be done with material colors or a dynamic texture.
And this way you can reduce it to 2 animations: center 90' rotation & end 90' rotation.
You flip/rotate the cube and stickers to animate the opposite end or on a different axis, reverse the animation to turn in the other direction.
The model (Mesh) always comes back to its original position after animating the 90 degree rotation, it's the stickers that change places. This way we completely avoid gimbal locks.
Another way is to use rotation matrices.
Each cube cells (the 3x3x3 elements of the puzzle) have 6 faces, up to 3 of them with stickers and their own rotation matrix. Using the face normal multiplied by the rotation matrix of the cell you can figure out which sticker points where as only 1 axis in the normal will ever have a non-zero value +/- 1.0 in X, Y or Z.
And there will (should) only ever be 9 stickers pointing to the same direction.
You can use the cell's positions to figure out which ones to rotate.
Since the cube cells will only ever be rotated to 0, 90, 180, or 270 at the end of the animation we can fix the cumulative rotation errors by doing:
if (matrix[r][c] > 0.5) {
matrix[r][c] = 1;
} else if (matrix[r][c] < -0.5) {
matrix[r][c] = -1;
} else {
matrix[r][c] = 0;
}
For each cell (r=row, c=column) to eradicate any rounding errors once the animation is done.
Usually comparing float values to exact values is a bad idea but in this case we're fixing the matrices to be exactly -1.0, 0.0, or 1.0, without any translation and anything (N) multiplied by those numbers always end up exactly -N, 0 or +N so we can safely compare them.
• While this technically answers the question and I'll probably end up using the technique described, I am still itching to know how I could make it work using quaternions. – Andrew Wilson Jan 16 '16 at 23:35
• You can convert angle-axis to a quaternion as described on euclideanspace.com/maths/geometry/rotations/conversions/… and multiply quaternions together to "accumulate" rotations mathworks.com/help/aeroblks/… but since this is a cube you can use matrices and fix rounding errors once the rotation is done: Any 0-90-180-270 degree rotation matrix will only have 0.0 or 1.0 in each cell so it's easy to fix a matrix like this. – Stephane Hockenhull Jan 17 '16 at 1:23
• if (matrix[r][c] > 0.5) matrix[r][c] = 1; else if (matrix[r][c] < -0.5) matrix[r][c] = -1; else matrix[r][c] = 0; – Stephane Hockenhull Jan 17 '16 at 1:24
• But how can I convert Euler rotations to axis angle without running into gimbal lock? – Andrew Wilson Jan 17 '16 at 1:33
• @AndrewWilson The trick is that you can't just store three axis-angle totals and combine them, because as explained in this answer, rotations don't combine commutatively the way sums do. So instead you'd store something like an orientation matrix or quaternion as your persistent state. You can apply each new twist as an angle-axis rotation of this base state, and then save the result as the new orientation state. – DMGregory Jan 17 '16 at 3:36
|
2019-11-17 23:09:44
|
{"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": 0.3717342019081116, "perplexity": 1040.4820350597104}, "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-2019-47/segments/1573496669352.5/warc/CC-MAIN-20191117215823-20191118003823-00293.warc.gz"}
|
https://www.esaral.com/q/solve-this-following-37228
|
# Solve this following
Question:
If $O$ is the origin and $P(2,3,4)$ and $Q(1,-2,1)$ be any two points show that $O P \perp O Q$.
Solution:
|
2023-03-26 03:49:04
|
{"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": 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": 0.6752556562423706, "perplexity": 280.5341908997287}, "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/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00704.warc.gz"}
|
https://buboflash.eu/bubo5/show-dao2?d=1439325097228
|
Do you want BuboFlash to help you learning these things? Or do you want to add or correct something? Click here to log in or create user.
Tags
#analyst-notes #cfa-level-1 #corporate-finance #reading-35-capital-budgeting #study-session-10 #subject-3-investment-criteria
Question
Michael undertakes an investment with an initial investment of $10,000, and he expects to receive$3,500 a year for the next four years. If the required return is 15%, what is the NPV?
A. $5.49 B.$4.63
C. -$7.58 Answer Correct Answer: C Tags #analyst-notes #cfa-level-1 #corporate-finance #reading-35-capital-budgeting #study-session-10 #subject-3-investment-criteria Question Michael undertakes an investment with an initial investment of$10,000, and he expects to receive $3,500 a year for the next four years. If the required return is 15%, what is the NPV? A.$5.49
B. $4.63 C. -$7.58
?
Tags
#analyst-notes #cfa-level-1 #corporate-finance #reading-35-capital-budgeting #study-session-10 #subject-3-investment-criteria
Question
Michael undertakes an investment with an initial investment of $10,000, and he expects to receive$3,500 a year for the next four years. If the required return is 15%, what is the NPV?
A. $5.49 B.$4.63
C. -\$7.58
If you want to change selection, open original toplevel document below and click on "Move attachment"
#### Parent (intermediate) annotation
Open it
dy>NPV measures the dollar benefit of the project to shareholders. However, it does not measure the rate of return of the project, and thus cannot provide "safety margin" information. Safety margin refers to how much the project return could fall in percentage terms before the invested capital is at risk.<body><html>
#### Original toplevel document
Subject 3. Investment Decision Criteria
on that capital. If a firm takes on a project with a positive NPV, the position of the stockholders is improved. Decision rules: The higher the NPV, the better. Reject if NPV is less than or equal to 0. <span>NPV measures the dollar benefit of the project to shareholders. However, it does not measure the rate of return of the project, and thus cannot provide "safety margin" information. Safety margin refers to how much the project return could fall in percentage terms before the invested capital is at risk. Assuming the cost of capital for the firm is 10%, calculate each cash flow by dividing the cash flow by (1 + k) t where k is the cost of capital and t is the year number.
#### Summary
status measured difficulty not learned 37% [default] 0
No repetitions
### Discussion
Do you want to join discussion? Click here to log in or create user.
|
2021-10-26 13:04:23
|
{"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": 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": 0.2639358639717102, "perplexity": 3141.919316483512}, "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-2021-43/segments/1634323587877.85/warc/CC-MAIN-20211026103840-20211026133840-00532.warc.gz"}
|
https://stats.stackexchange.com/questions/226657/variance-of-the-integral-of-a-stochastic-process?noredirect=1
|
# Variance of the integral of a stochastic process
I understand the discete case i.e. the sum of $N$ identically distributed random variables $X_i$ with variance $\sigma^2$. The correlation between these random variables is given by the correlation matrix $\mathbf{\rho}(X_i,X_j)$.
The variance of the linear combination of random variables $X_i$ is given by:
$$\operatorname{Var}\left( \sum_{i=1}^N X_i\right) = N\sigma^2+2\sigma^2\sum_{1\le i<j\le N}\mathbf{\rho}(X_i,X_j)$$
I would like to consider the continuous case of a stochastic process which will be denoted as $X(t)$. The process is stationary with constant variance $\sigma^2$ and correlation function $\rho(X(t),X(h)$. Similar to above I would like to calculate the variance of the linear combination of the random variables $X(t)$. I think that the linear combination over some domain $t \in [0,L]$ can be expressed as
$$I = \int_0^L X(t) dt$$
I would like to know the variance of $I$:
$$\operatorname{Var}(I)=?$$
I speculate that if the process $Z(X_i)$ is completely correlated i.e. $\rho(X(t),X(h)) = 1$ then the variance of $I$ is minimised maximised and is given by: $$\operatorname{Var}(I)=L^2\sigma^2$$
If the variables are uncorrelated i.e. $\rho(X(t),X(h))=0$ then I suspect that the variance of $I$ is maximised minimised. It may be infinite zero?
I find the continuous case (i.e. infinite random variables over some domain [0,L]) difficult to understand. Could anyone provide me an expression for the variance of $I$?
Interchanging the order of integration and expectation you get $$E(I)=E\int_0^L X(t) dt = \int_0^L EX(t) dt = \int_0^L \mu dt = L\mu$$ and similarly, the second moment of $$I$$ becomes \begin{align} E(I^2)&=E\left(\int_0^LX(t)dt\int_0^LX(u)du\right) \\ &= E \int_0^L \int_0^L X(t)X(u)du dt \\ &= \int_0^L \int_0^L E[X(t)X(u)]du dt \\ &= \int_0^L \int_0^L [\operatorname{Cov}(X(t),X(u))+EX(t)EX(u)] du dt \\ &= \sigma^2 \int_0^L \int_0^L \rho(t-u)dudt + L^2 \mu^2. \end{align} If the correlation function $$\rho(h)$$ is for example exponential the double intergral can be easily solved. The variance of $$I$$ can be found from the general formula $$\operatorname{Var}I =E(I^2)-(EI)^2$$.
• Excellent. This makes sense to me. I will try to verify this with simulation. – egg Aug 1 '16 at 13:43
• This assumes that the process is wide-sense stationary or weakly stationary. Also, for this special case, the double integral simplifies to a single integral $$\int_{-L}^{L} \rho(s)(L-|s|)\,\mathrm ds$$ which is easier to work with. – Dilip Sarwate Aug 2 '16 at 17:52
• Good comment. Integration using Wolfram Mathematica gives the same result using both expressions for $$\rho(s) = \exp\left(-\pi\left(\frac{s}{\theta}\right)^2\right)$$ Is it correct that your expression works for all isotropic correlation functions i.e. $s = |t - u|$? Could you explain how you got to your expression? My understanding of double integrals is poor. – egg Aug 8 '16 at 14:33
• Even if unrelated to the question itself, I find a little confusing the statement $\text{Var}[I]=\text{E}[I^2]-\text{E}[I]^2$ as a consequence of the exponential correlation. Isn't it true for any random variable? – Riccardo Buscicchio Jan 10 at 11:49
• @RiccardoBuscicchio Yes – Jarle Tufto Jan 10 at 16:06
|
2020-04-02 14:13:14
|
{"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": 6, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993304014205933, "perplexity": 257.7585882768248}, "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/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00371.warc.gz"}
|
https://imathworks.com/tex/tex-latex-exponent-with-more-than-1-digit/
|
# [Tex/LaTex] Exponent with more than 1 digit
pdftexpgfmath
How can i format an exponent with more than one digit? For example i want $10^20$ and latex gives me 10^2 and 0.
just use the curly braces as $10^{20}$.
|
2023-01-28 13:47:49
|
{"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": 0.9552499055862427, "perplexity": 1615.4514660779766}, "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/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00543.warc.gz"}
|
https://engineering.stackexchange.com/questions/25340/is-youngs-modulus-a-derivative-of-stress-and-strain/25352#25352
|
# Is youngs modulus a derivative of stress and strain?
From This page it can be seen that young's modulus is given by, E = (Δstress) / (Δstrain)
Which means E is a point function, that changes as the strain in member is increased.
• Perhaps this may help : engineering.stackexchange.com/a/10220/10902 Jan 2 '19 at 6:18
• Yes and no, in the elastic region for some materials yes, the relationship is pure linear, so the slope is equal to the derivative. But this is not true for polymers for example, the relationship usually described by an exponent later in plastic region, it's not shown in your diagram. Jan 2 '19 at 10:15
• The bottom line is, treating E as a derivative isn't very useful in practice except for linear materials where it is just a constant, for two reasons. (1) Most stress and strain fields are not uniaxial, so what you are calling "E" is really a parameter in a (nonlinear) fourth-order tensor, not a scalar. (2) It is usually easier to work from a numerical definition of the stress-strain graph itself, rather than by fitting a mathematical function to it. Jan 2 '19 at 11:01
In the theory of elasticity, the constitutive relation in three dimensions is
$$\sigma_{ij} = C_{ijkl}\epsilon_{kl}$$
This gives the $${\bf linear}$$ relationship between stress and strain.
The fourth order tensor - $$C_{ijkl}$$ takes 81 components in its own form and reduces to 21 if it is a symmetric stress, and 9 for planar stress. Finally it takes only one component for one dimensional problem such as uni-axial tensile problem.
i.e., $$\sigma = C\epsilon$$ or, $$\sigma = E\epsilon$$.
Since the relation between stress & strain is linear (with in elastic limit) it is sufficient having just $$\sigma$$ and $$\epsilon$$ values to get $$E$$.
Then when why we need $$\delta\sigma$$ and $$\delta\epsilon$$ to calculate $$E$$?
When you are testing the specimen in UTM, will it show zero deformation for zero load applied? at least for me during my undergraduate studies the load was non-zero when the displacement/deformation was zero. This is due to the friction between the model and fixtures/Jigs.
Also within the elastic region, the material tends to show $$hysteresis$$ effects if you are using the same sample again and again. This leads to non-zero deflection for zero load.
Lastly, the calculation of $$E$$ at any specific $$\sigma$$ or $$\epsilon$$ is prone to error if the $$\sigma$$ - $$\epsilon$$ curve have local error due to the instrumentation/observation.
To avoid these errors it is better to take the infinitesimal changes in stress and strain or slope of the $$\sigma$$ - $$\epsilon$$ curve.
Hope this helps!
The Young's modulus (sometimes refered to as elastic modulus) is only relevant for the elastic region of the material and doesn't apply after plastic deformation begins.
Rather than a derivative, the Young's modulus is better defined as the gradient of the linear (elastic) section at the begining of the graph. Or in real life terms it is the gradient of a (linear) line of best fit applied to the data in the elastic (linear) region.
This aproach works fine for most solid materials. But things can start to get complicated for some materials like rubbers.
I dont see any problem with that definition. You can check ie. ISO 527. Where E is calculated from a tensile test.
To further understand the meaning or defining E I'd recommend you to read something about anisotropic materials. You can check for instance long fiber composites - which is orthotropic material, look at the E matrix/tensor.
As has been mentioned in other answers, the Young's modulus, E, is the ratio of stress to strain in the elastic range of stress/strain curve and is a constant, not a derivative.
It is a constant as opposed to what a derivative implies which is a variable depending on the local curvature.
Analysis and design of structures and members is based heavily on the assumption that E is a constant and will remain constant for the periods of loading and unloading a beam, column, or a whole structure.
The fact that E is a slope of straight line and a constant as such, at least assumed, allows all our methods of analysis of the behavior of a structural member work. Otherwise we had to go to unlimited number of trial and errors, assuming a different E for a different stress configuration.
** Edit** If we don't handle Youngs modulus as a constant, we can not assume E constant the entire system of analysing beam stresses and bending collapses.
We calculate I, second area moment which is fundamental in structural analysis based on the assumption that E is constantan and cross section of beams and columns remain planar after bending. If the E is not constant the strain is not linear and I will be meaning less.
|
2021-11-29 06:22:47
|
{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7900543212890625, "perplexity": 486.96929816790356}, "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/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00266.warc.gz"}
|
https://stacks.math.columbia.edu/tag/0C57
|
Lemma 53.19.13. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme of dimension $1$. Let $Y \to X$ be an étale morphism. Let $y \in Y$ be a point with image $x \in X$. The following are equivalent
1. $x$ is a closed point of $X$ and a node, and
2. $y$ is a closed point of $Y$ and a node.
Proof. By Lemma 53.19.12 we may base change to the algebraic closure of $k$. Then the residue fields of $x$ and $y$ are $k$. Hence the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}$ is an isomorphism (for example by Étale Morphisms, Lemma 41.11.3 or More on Algebra, Lemma 15.42.9). Thus the lemma is clear. $\square$
Comment #5108 by Tongmu He on
Typo: in the proof of 53.19.13, the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}$ should be $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge$.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
|
2020-06-06 08:40:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 2, "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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9935998916625977, "perplexity": 218.6912010414565}, "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/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00108.warc.gz"}
|
http://mathhelpforum.com/calculus/54083-natural-log-derivates.html
|
# Math Help - Natural log and derivates
1. ## Natural log and derivates
I am trying to find dy/dx
The problem states: e^y = 1 + (X^2)(y)..My calculus book doesn't really explain how to get the derivative in terms I am making any sense of.
Can I rewrite this as?
y= ln(1+(x^2)(y))
What rule would I need to apply to get the next step? would this be a chain rule problem or is their some special log rule I am missing. Any help would be appreciated. Thanks
2. I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:
Just differentiate everything on both sides with respect to x:
$
\begin{array}{l}
e^y = 1 + yx^2 \\
e^y \frac{{dy}}{{dx}} = 2xy + x^2 \frac{{dy}}{{dx}} \\
\end{array}
$
Upon re-arranging:
$
\begin{array}{l}
\frac{{dy}}{{dx}}\left( {e^y - x^2 } \right) = 2xy \\
\frac{{dy}}{{dx}} = \frac{{2xy}}{{e^y - x^2 }} \\
\end{array}
$
3. Originally Posted by free_to_fly
I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:
Just differentiate everything on both sides with respect to y:
$
\begin{array}{l}
e^y = 1 + yx^2 \\
e^y \frac{{dy}}{{dx}} = 2xy + x^2{\color{red}\frac{\,dy}{\,dx}} \\
\end{array}
$
Upon re-arranging:
$
\frac{{dy}}{{dx}} = \frac{{2xy + x^2 }}{{e^y }}
$
You forgot a $\frac{\,dy}{\,dx}$!!
The answer should be $\frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}$
--Chris
4. Yep I realized that soon as I posted it...what a silly mistake lol. Forgot it was implicit on the RHS too.
5. Originally Posted by Chris L T521
You forgot a $\frac{\,dy}{\,dx}$!!
The answer should be $\frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}$
--Chris
Thank you both. I do have one more question though if you guys don't mind. How does the X^2 become in the denominator?
6. Actually I just figured it out. haha..Im guessing you brought the dy/dx over and factored it out then divided. I don't think before asking questions sorry.
|
2014-04-18 06:48:52
|
{"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": 0, "img_math": 0, "codecogs_latex": 8, "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": 0.8660786151885986, "perplexity": 1428.1119988479322}, "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-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00584-ip-10-147-4-33.ec2.internal.warc.gz"}
|
http://www.absoluteastronomy.com/topics/Two-port_network
|
Two-port network
# Two-port network
Overview
A two-port network is an electrical circuit or device with two pairs of terminals connected together internally by an electrical network
Electrical network
An electrical network is an interconnection of electrical elements such as resistors, inductors, capacitors, transmission lines, voltage sources, current sources and switches. An electrical circuit is a special type of network, one that has a closed loop giving a return path for the current...
. Two terminals constitute a port if they satisfy the essential requirement known as the port condition: the same current must enter and leave a port.
Examples include small-signal models for transistors (such as the hybrid-pi model
Hybrid-pi model
The hybrid-pi model is a popular circuit model used for analyzing the small signal behavior of bipolar junction and field effect transistors. The model can be quite accurate for low-frequency circuits and can easily be adapted for higher frequency circuits with the addition of appropriate...
), filter
Electronic filter
Electronic filters are electronic circuits which perform signal processing functions, specifically to remove unwanted frequency components from the signal, to enhance wanted ones, or both...
s and matching networks.
Discussion
Recent Discussions
Encyclopedia
A two-port network is an electrical circuit or device with two pairs of terminals connected together internally by an electrical network
Electrical network
An electrical network is an interconnection of electrical elements such as resistors, inductors, capacitors, transmission lines, voltage sources, current sources and switches. An electrical circuit is a special type of network, one that has a closed loop giving a return path for the current...
. Two terminals constitute a port if they satisfy the essential requirement known as the port condition: the same current must enter and leave a port.
Examples include small-signal models for transistors (such as the hybrid-pi model
Hybrid-pi model
The hybrid-pi model is a popular circuit model used for analyzing the small signal behavior of bipolar junction and field effect transistors. The model can be quite accurate for low-frequency circuits and can easily be adapted for higher frequency circuits with the addition of appropriate...
), filter
Electronic filter
Electronic filters are electronic circuits which perform signal processing functions, specifically to remove unwanted frequency components from the signal, to enhance wanted ones, or both...
s and matching networks. The analysis of passive two-port networks is an outgrowth of reciprocity theorems first derived by Lorentz.
A two-port network makes possible the isolation of either a complete circuit or part of it and replacing it by its characteristic parameters. Once this is done, the isolated part of the circuit becomes a "black box
Black box
A black box is a device, object, or system whose inner workings are unknown; only the input, transfer, and output are known characteristics.The term black box can also refer to:-In science and technology:*Black box theory, a philosophical theory...
" with a set of distinctive properties, enabling us to abstract away its specific physical buildup, thus simplifying analysis. Any linear circuit with four terminals can be transformed into a two-port network provided that it does not contain an independent source and satisfies the port conditions.
There are a number of alternative sets of parameters that can be used to describe a linear two-port network, the usual sets are respectively called z, y, h, g, and ABCD parameters, each described individually below. These are all limited to linear networks since an underlying assumption of their derivation is that any given circuit condition is a linear superposition of various short-circuit and open circuit conditions. They are usually expressed in matrix notation, and they establish relations between the variables Input voltage Output voltage Input current Output current
which are shown in Figure 1. These current
Electric current
Electric current is a flow of electric charge through a medium.This charge is typically carried by moving electrons in a conductor such as wire...
and voltage
Voltage
Voltage, otherwise known as electrical potential difference or electric tension is the difference in electric potential between two points — or the difference in electric potential energy per unit charge between two points...
variables are most useful at low-to-moderate frequencies. At high frequencies (e.g., microwave frequencies), the use of power
Power (physics)
In physics, power is the rate at which energy is transferred, used, or transformed. For example, the rate at which a light bulb transforms electrical energy into heat and light is measured in watts—the more wattage, the more power, or equivalently the more electrical energy is used per unit...
and energy
Energy
In physics, energy is an indirectly observed quantity. It is often understood as the ability a physical system has to do work on other physical systems...
variables is more appropriate, and the two-port current–voltage approach is replaced by an approach based upon scattering parameters
Scattering parameters
Scattering parameters or S-parameters describe the electrical behavior of linear electrical networks when undergoing various steady state stimuli by electrical signals....
.
The terms four-terminal network and quadripole (not to be confused with quadrupole
A quadrupole or quadrapole is one of a sequence of configurations of—for example—electric charge or current, or gravitational mass that can exist in ideal form, but it is usually just part of a multipole expansion of a more complex structure reflecting various orders of complexity.-Mathematical...
) are also used, the latter particularly in more mathematical treatments although the term is becoming archaic. However, a pair of terminals can be called a port only if the current entering one terminal is equal to the current leaving the other; this definition is called the port condition. A four-terminal network can only be properly called a two-port when the terminals are connected to the external circuitry in two pairs both meeting the port condition.
## General properties
There are certain properties of two-ports that frequently occur in practical networks and can be used to greatly simplify the analysis. These include:
Reciprocal networks. A network is said to be reciprocal if the voltage appearing at port 2 due to a current applied at port 1 is the same as the voltage appearing at port 1 when the same current is applied to port 2. Exchanging voltage and current results in an equivalent definition of reciprocity. In general, a network will be reciprocal if it consists entirely of linear passive components (that is, resistors, capacitors and inductors). In general, it will not be reciprocal if it contains active components such as generators.
Symmetrical networks. A network is symmetrical if its input impedance is equal to its output impedance. Most often, but not necessarily, symmetrical networks are also physically symmetrical. Sometimes also antimetrical networks
Antimetric (electrical networks)
An antimetric electrical network is one that exhibits anti-symmetrical electrical properties. The term is often encountered in filter theory, but it applies to general electrical network analysis...
are of interest. These are networks where the input and output impedances are the duals of each other.
Lossless network. A lossless network is one which contains no resistors or other dissipative elements.
## Impedance parameters (z-parameters)
where
Notice that all the z-parameters have dimensions of ohms
Ohms
OHMS may refer to:* The plural of ohm, a unit of resistance, named after Georg Ohm* Ohm's Law of electric currents, first proposed by Georg Ohm* O.H.M.S., On His/Her Majesty's Service...
.
For reciprocal networks . For symmetrical networks . For lossless networks all the are purely imaginary.
### Example: bipolar current mirror with emitter degeneration
Figure 3 shows a bipolar current mirror with emitter resistors to increase its output resistance.The emitter-leg resistors counteract any current increase by decreasing the transistor VBE. That is, the resistors RE cause negative feedback that opposes change in current. In particular, any change in output voltage results in less change in current than without this feedback, which means the output resistance of the mirror has increased. Transistor Q1 is diode connected, which is to say its collector-base voltage is zero. Figure 4 shows the small-signal circuit equivalent to Figure 3. Transistor Q1 is represented by its emitter resistance rEVT / IE (VT = thermal voltage, IE = Q-point emitter current), a simplification made possible because the dependent current source in the hybrid-pi model for Q1 draws the same current as a resistor 1 / gm connected across rπ. The second transistor Q2 is represented by its hybrid-pi model
Hybrid-pi model
The hybrid-pi model is a popular circuit model used for analyzing the small signal behavior of bipolar junction and field effect transistors. The model can be quite accurate for low-frequency circuits and can easily be adapted for higher frequency circuits with the addition of appropriate...
. Table 1 below shows the z-parameter expressions that make the z-equivalent circuit of Figure 2 electrically equivalent to the small-signal circuit of Figure 4.
Table 1 Expression Approximation
The negative feedback introduced by resistors RE can be seen in these parameters. For example, when used as an active load in a differential amplifier, I1 ≈ -I2, making the output impedance of the mirror approximately R22 -R21 ≈ 2 β rORE /( rπ+2RE ) compared to only rO without feedback (that is with RE = 0 Ω) . At the same time, the impedance on the reference side of the mirror is approximately R11 − R12 , only a moderate value, but still larger than rE with no feedback. In the differential amplifier application, a large output resistance increases the difference-mode gain, a good thing, and a small mirror input resistance is desirable to avoid Miller effect
Miller effect
In electronics, the Miller effect accounts for the increase in the equivalent input capacitance of an inverting voltage amplifier due to amplification of the effect of capacitance between the input and output terminals...
.
where
Notice that all the Y-parameters have dimensions of siemens
Siemens (unit)
The siemens is the SI derived unit of electric conductance and electric admittance. Conductance and admittance are the reciprocals of resistance and impedance respectively, hence one siemens is equal to the reciprocal of one ohm, and is sometimes referred to as the mho. In English, the term...
.
For reciprocal networks . For symmetrical networks . For lossless networks all the are purely imaginary.
## Hybrid parameters (h-parameters)
where
This circuit is often selected when a current amplifier is wanted at the output. The resistors shown in the diagram can be general impedances instead.
Notice that off-diagonal h-parameters are dimensionless, while diagonal members have dimensions the reciprocal of one another.
### Example: common-base amplifier
Note: Tabulated formulas in Table 2 make the h-equivalent circuit of the transistor from Figure 6 agree with its small-signal low-frequency hybrid-pi model
Hybrid-pi model
The hybrid-pi model is a popular circuit model used for analyzing the small signal behavior of bipolar junction and field effect transistors. The model can be quite accurate for low-frequency circuits and can easily be adapted for higher frequency circuits with the addition of appropriate...
in Figure 7. Notation: rπ = base resistance of transistor, rO = output resistance, and gm = transconductance. The negative sign for h21 reflects the convention that I1, I2 are positive when directed into the two-port. A non-zero value for h12 means the output voltage affects the input voltage, that is, this amplifier is bilateral. If h12 = 0, the amplifier is unilateral.
Table 2 Expression Approximation
## Inverse hybrid parameters (g-parameters)
where
Often this circuit is selected when a voltage amplifier is wanted at the output. Notice that off-diagonal g-parameters are dimensionless, while diagonal members have dimensions the reciprocal of one another. The resistors shown in the diagram can be general impedances instead.
### Example: common-base amplifier
Note: Tabulated formulas in Table 3 make the g-equivalent circuit of the transistor from Figure 8 agree with its small-signal low-frequency hybrid-pi model
Hybrid-pi model
The hybrid-pi model is a popular circuit model used for analyzing the small signal behavior of bipolar junction and field effect transistors. The model can be quite accurate for low-frequency circuits and can easily be adapted for higher frequency circuits with the addition of appropriate...
in Figure 9. Notation: rπ = base resistance of transistor, rO = output resistance, and gm = transconductance. The negative sign for g12 reflects the convention that I1, I2 are positive when directed into the two-port. A non-zero value for g12 means the output current affects the input current, that is, this amplifier is bilateral. If g12 = 0, the amplifier is unilateral.
Table 3 Expression Approximation
## ABCD-parameters
The ABCD-parameters are known variously as chain, cascade, or transmission line parameters. There are a number of definitions given for ABCD parameters, the most common is,
For reciprocal networks . For symmetrical networks $\scriptstyle A=D$. For networks which are reciprocal and lossless, A and D are purely real while B and C are purely imaginary.
This representation is preferred because when the parameters are used to represent a cascade of two-ports, the matrices are written in the same order that a network diagram would be drawn, that is, left to right. However, the examples given below are based on a variant definition;
where
The negative signs in the definitions of parameters $\scriptstyle C'$ and $\scriptstyle D'$ arise because is defined with the opposite sense to , that is, . The reason for adopting this convention is so that the output current of one cascaded stage is equal to the input current of the next. Consequently, the input voltage/current matrix vector can be directly replaced with the matrix equation of the preceding cascaded stage to form a combined $\scriptstyle A'B'C'D'$ matrix.
The terminology of representing the $\scriptstyle ABCD$ parameters as a matrix of elements designated a11 etc. as adopted by some authors and the inverse $\scriptstyle A'B'C'D'$ parameters as a matrix of elements designated b11 etc. is used here for both brevity and to avoid confusion with circuit elements.
There is a simple relationship between these two forms: one is the matrix inverse of the other, that is;
An ABCD matrix has been defined for Telephony four-wire Transmission Systems by P K Webb in British Post Office Research Department Report 630 in 1977.
### Table of transmission parameters
The table below lists inverse ABCD parameters for some simple network elements.
Element [b] matrix Remarks
Series resistor R = resistance
Shunt resistor R = resistance
Series conductor G = conductance
Shunt conductor G = conductance
Series inductor L = inductance
s = complex angular frequency
Shunt capacitor C = capacitance
s = complex angular frequency
## Combinations of two-port networks
When two or more two-port networks are connected together, the two-port parameters of the combined network can be found by performing matrix algebra on the matrices of parameters for the component two-ports. The matrix operation can be made particularly simple with an appropriate choice of two-port parameters to match the form of connection of the two-ports. For instance, the z-parameters are best for series connected ports.
The combination rules need to be applied with care. Some connections (when dissimilar potentials are joined) result in the port condition being invalidated and the combination rule will no longer apply. This difficulty can be overcome by placing 1:1 ideal transformers on the outputs of the problem two-ports. This does not change the parameters of the two-ports, but does ensure that they will continue to meet the port condition when interconnected. An example of this problem is shown for series-series connections in figures 11 and 12 below.
### Series-series connection
When two-ports are connected in a series-series configuration as shown in figure 10, the best choice of two-port parameter is the z-parameters. The z-parameters of the combined network are found by matrix addition of the two individual z-parameter matrices.
As mentioned above, there are some networks which will not yield directly to this analysis. A simple example is a two-port consisting of a L-network of resistors R1 and R2. The z-parameters for this network are;
Figure 11 shows two identical such networks connected in series-series. The total z-parameters predicted by matrix addition are;
However, direct analysis of the combined circuit shows that,
The discrepancy is explained by observing that R1 of the lower two-port has been by-passed by the short-circuit between two terminals of the output ports. This results in no current flowing through one terminal in each of the input ports of the two individual networks. Consequently, the port condition is broken for both the input ports of the original networks since current is still able to flow into the other terminal. This problem can be resolved by inserting an ideal transformer in the output port of at least one of the two-port networks. While this is a common text-book approach to presenting the theory of two-ports, the practicality of using transformers is a matter to be decided for each individual design.
### Parallel-parallel connection
When two-ports are connected in a parallel-parallel configuration as shown in figure 13, the best choice of two-port parameter is the y-parameters. The y-parameters of the combined network are found by matrix addition of the two individual y-parameter matrices.
### Series-parallel connection
When two-ports are connected in a series-parallel configuration as shown in figure 14, the best choice of two-port parameter is the h-parameters. The h-parameters of the combined network are found by matrix addition of the two individual h-parameter matrices.
### Parallel-series connection
When two-ports are connected in a parallel-series configuration as shown in figure 15, the best choice of two-port parameter is the g-parameters. The g-parameters of the combined network are found by matrix addition of the two individual g-parameter matrices.
When two-ports are connected with the output port of the first connected to the input port of the second (a cascade connection) as shown in figure 16, the best choice of two-port parameter is the ABCD-parameters. The a-parameters of the combined network are found by matrix multiplication of the two individual a-parameter matrices.
A chain of n two-ports may be combined by matrix multiplication of the n matrices. To combine a cascade of b-parameter matrices, they are again multiplied, but the multiplication must be carried out in reverse order, so that;
#### Example
Suppose we have a two-port network consisting of a series resistor R followed by a shunt capacitor C. We can model the entire network as a cascade of two simpler networks:
The transmission matrix for the entire network is simply the matrix multiplication of the transmission matrices for the two network elements:
Thus:
## Scattering parameters (S-parameters)
The previous parameters are all defined in terms of voltages and currents at ports. S-parameters are different, and are defined in terms of incident and reflected waves
Signal reflection
Signal reflection occurs when a signal is transmitted along a transmission medium, such as a copper cable or an optical fiber, some of the signal power may be reflected back to its origin rather than being carried all the way along the cable to the far end. This happens because imperfections in the...
at ports. S-parameters are used primarily at UHF and microwave
Microwave
Microwaves, a subset of radio waves, have wavelengths ranging from as long as one meter to as short as one millimeter, or equivalently, with frequencies between 300 MHz and 300 GHz. This broad definition includes both UHF and EHF , and various sources use different boundaries...
frequencies where it becomes difficult to measure voltages and currents directly. On the other hand, incident and reflected power are easy to measure using directional couplers. The definition is,
where the are the incident waves and the are the reflected waves at port k. It is conventional to define the and in terms of the square root of power. Consequently, there is a relationship with the wave voltages (see main article for details).
For reciprocal networks . For symmetrical networks . For antimetrical networks . For lossless reciprocal networks and .
## Scattering transfer parameters (T-parameters)
Scattering transfer parameters, like scattering parameters, are defined in terms of incident and reflected waves. The difference is that T-parameters relate the waves at port 1 to the waves at port 2 whereas S-parameters relate the reflected waves to the incident waves. In this respect T-parameters fill the same role as ABCD parameters and allow the T-parameters of cascaded networks to be calculated by matrix multiplication of the component networks. T-parameters, like ABCD parameters, can also be called transmission parameters. The definition is,
T-parameters are not so easy to measure directly unlike S-parameters. However, S-parameters are easily converted to T-parameters, see main article for details.
## Networks with more than two ports
While two port networks are very common (e.g. amplifiers and filters), other electrical networks such as directional couplers and circulators have more than 2 ports. The following representations are also applicable to networks with an arbitrary number of ports:
Admittance parameters or Y-parameters are properties used in electrical engineering, electronic engineering, and communication systems engineering describe the electrical behavior of linear electrical networks. They are also used to describe the small-signal response of non-linear networks...
• Impedance (z) parameters
Impedance parameters
Impedance parameters or Z-parameters are properties used in electrical engineering, electronic engineering, and communication systems engineering to describe the electrical behavior of linear electrical networks. They are also used to describe the small-signal response of non-linear networks...
• Scattering (S) parameters
Scattering parameters
Scattering parameters or S-parameters describe the electrical behavior of linear electrical networks when undergoing various steady state stimuli by electrical signals....
For example three-port impedance parameters result in the following relationship:
However the following representations are necessarily limited to two-port devices:
• Hybrid (h) parameters
• Inverse hybrid (g) parameters
• Transmission (ABCD) parameters
• Scattering transfer (T) parameters
## Collapsing a two-port to a one port
A two-port network has four variables with two of them being independent. If one of the ports is terminated by a load with no independent sources, then the load enforces a relationship between the voltage and current of that port. A degree of freedom is lost. The circuit now has only one independent parameter. The two-port becomes a one-port impedance to the remaining independent variable.
For example , consider impedance parameters
The negative sign is because the positive direction for I2 is directed into the two-port instead of into the load. The augmented equations become
The second equation can be easily solved for I2 as a function of I1 and that expression can replace I2 in the first equation leaving V1 ( and V2 and I2 ) as functions of I1
So, in effect, I1 sees an input impedance and the two-port's effect on the input circuit has been effectively collapsed down to a one-port i.e. a simple two terminal impedance.
|
2013-12-09 00:09:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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": 0.6866233944892883, "perplexity": 1385.986056541272}, "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/1386163835370/warc/CC-MAIN-20131204133035-00089-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://git.cinelerra-gg.org/git/?p=goodguy/cin-manual-latex.git;a=blobdiff;f=parts/Tips.tex;h=a80ad5903bc58d2b0c771c6365c05cc1e7e66c1a;hp=a99f4157e02c38f402a3b0d3942efb3cdf0833c1;hb=b62b7f33e541f80cdece60373fbd87369f265534;hpb=0285fef00fc55b855c0f27faf9b294d0b56dc3fb
|
@@ -42,14 +42,14 @@ Currently only the most common codecs, such as MPEG-1, MPEG-2, MPEG-4, and H.264
then, to run from the Cinelerra installed directory, key in:
\end{enumerate}
-\begin{lstlisting}[language=bash]
+\begin{lstlisting}[language=bash,numbers=none]
CIN_HW_DEV=vdpau ./cin # for computers with Nvidia and some other graphics cards
CIN_HW_DEV=vaapi ./cin # mostly for computers with Intel or AMD specific graphics hardware
\end{lstlisting}
If you find that the environment variable setting is advantageous for your CinGG usage and you want to always use it, you can add it to your \texttt{\$HOME} directory \texttt{.profile} file which takes effect every time you log in. The line you would add would look something like this: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] export CIN_HW_DEV=vdpau or export CIN_HW_DEV=vaapi @@ -119,13 +119,13 @@ Using the GPU is going to react differently depending on your hardware, software The situation may arise where you have enabled hardware acceleration and after loading several files for a project, you find that a file had some kind of error resulting in a black video instead of an image or you see an error message pop up which states something like \textit{Error retrieving data from GPU to CPU} or \textit{err: Unknown error occurred}. Because the \texttt{CIN\_HW\_DEV} environment variable is either all or none, ordinarily in order to correct the non-working video you would have to turn off hardware acceleration for the entire project/session. However, there is a way to continue working on your project without having to reload all of your files. You still use the environment variable and it will be effective for all of the formats it is able to handle, but you make an exception for any of the files that erred out. To do this you simply create a file in the same directory with the same name as the erring file with the different extension of .opts. The contents of this .opts file would just be the one line of: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] cin_hw_dev=none \end{lstlisting} Conversely, if you have a bunch of files in your project, like dnxhd format, that are not hardware accelerated, but you have an accompanying large file of type .mp4 for which you would like the hardware acceleration, you can leave the \texttt{CIN\_HW\_DEV} variable unset (that is, do not use it) and just create an .opts file containing the line: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] cin_hw_dev=vdpau \end{lstlisting} @@ -138,7 +138,7 @@ It is important to note that if using the .opts file to override the default \te Probably the easiest way to tell if hardware acceleration is working, is just to look at the messages in the window from where you started Cin (not available if start using the application icon). For example load a png, dnxhd, or some other non-supported format file and you will see messages similar to those below. The line \textit{HW device init failed, using SW decode} indicates that the vdpau/vaapi HW (hardware) decode is not available so will use SW (software) decode instead. -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] Failed to get HW surface format. HW device init failed, using SW decode. file:/tmp/media/aer_zypr.mp4 @@ -158,7 +158,7 @@ HEVC with NVIDIA, VDPAU driver is buggy, skipping If you would like to see more information on what is occurring, you can modify in the Cinelerra ffmpeg subdirectory, the file: \texttt{decode.opts} by temporarily changing the line from \texttt{loglevel =fatal} to \texttt{loglevel =verbose} and restarting Cinelerra. Then you will see messages in the startup window like: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] [AVHWDeviceContext @ 0x7fc9540be940] Successfully created a VDPAU device (NVIDIA VDPAU Driver Shared Library 390.116 Sun Jan 27 06:28:58 PST 2019) on X11 display :0 [h264 @ 0x7fc950159380] Reinit context to 1920x1088, pix_fmt: vdpau @@ -225,7 +225,7 @@ Older graphics cards or non-performing graphics cards will probably bring only a And, you can see what your specific hardware and software might support by running either \texttt{vainfo} or \texttt{vdpauinfo} from the command line. Partial examples of each are shown below. -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] # vainfo vainfo: VA-API version: 1.4 (libva 2.4.0) vainfo: Driver version: Intel i965 driver for Intel(R) Broadwell - 2.4.0.pre1 (2.3.0-11-g881e67a) @@ -243,7 +243,7 @@ VAProfileVC1Simple VAProfileVP8Version0_3 \end{lstlisting} -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] # vdpauinfo display: :0 screen: 0 API version: 1 @@ -287,7 +287,7 @@ There are currently 4 options files available in the Render menu already set up Other option files can be added as needed for your specific hardware if it is known to work for you, such as VP8 and VP9. An example of the included Cinelerra’s \texttt{ffmpeg/video/h264\_vaapi.mp4} file (figure~\ref{fig:render-vaapi}): -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] mp4 h264_vaapi cin_hw_dev=vaapi profile=high @@ -347,7 +347,7 @@ Of note in this test, 388\% CPU usage with only 4 cores shows that there is prob There is one last potentially significant graphics speedup when using the X11-OpenGL driver for users with Nvidia graphics boards who are seeing frames/sec achieved lower than what the video format is set to. You may want to disable \textit{sync to vblank} (an option for OpenGL) in NVIDIA X Server Settings for the proprietary drivers. This could increase your frames per second on playback. -\subsection{effects (OpenCL, Cuda)}% +\subsection{Effects (OpenCL, Cuda)}% \label{sub:effects_opencl_cuda} CUDA® is a parallel computing platform / programming model developed by Nvidia that provides big increases in computing performance through use of the GPU. It was first introduced in about 2006 for applications in computationally intense fields such as astronomy, biology, chemistry, and physics. @@ -430,7 +430,7 @@ There is also a convenient \texttt{Beep on done} checkbox included so that you c A good choice for proxy settings with 1080p source video is: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] Scale Factor: 1/4 Use Scaler: unchecked File Format: FFMPEG - mpeg @@ -443,7 +443,7 @@ Pixels: yuv420p If you get errors for some videos, such as those with strange variable bit rate or some types of files made on a smartphone, a usually reliable alternative is to change the following parameters: -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] File Format: FFMPEG - mov Video Preset: Compression: mov.mov @@ -498,7 +498,7 @@ This section is a handy guide for describing various kinds of software computer \item[Keyboard grabs, Server grabs, and Deadlocks:] A grab is an X-server state where all events are forced to just one window event stream. This forces the user to respond to the dialog. Things seems to be working, but no keypresses do anything useful. The system clock and other programs will still be working. The network will work for remote logins. Grabs can be canceled if the \texttt{/etc/X11/xorg.conf} X config contains special setup as shown below: \end{description} -\begin{lstlisting}[language=bash] +\begin{lstlisting}[language=bash,numbers=none] Section "ServerFlags" Option "HandleSpecialKeys" "Always" Option "AllowDeactivateGrabs" "True" @@ -516,7 +516,7 @@ EndSection or to \texttt{\$HOME/.xinitrc}, add:
-\begin{lstlisting}[language=bash]
+\begin{lstlisting}[language=bash,numbers=none]
# xkb terminate/grab actions disabled in xorg.conf, use:
setxkbmap -option "grab:break_actions"
setxkbmap -option "terminate:ctrl_alt_bksp"
@@ -528,7 +528,7 @@ Modal forms (always on top, and usually ptr/kbd grab) dialog boxes can lock a sy
\begin{description}
\item[Window Manager issues:] The \textit{desktop} window manager can intercept and modify all kinds of user input. Mostly, this is a good thing, but can be a nuisance. If user keypresses can be programmed to trigger actions, then they may be useful to send \texttt{KILL} or \texttt{INTR} to an app that seems to be holding X's attention. For example:
- \begin{lstlisting}[language=bash]
+ \begin{lstlisting}[language=bash,numbers=none]
killall -INTR cinelerra,
killall -9 cinelerra,
killall X,
@@ -616,7 +616,7 @@ The picture quality on analog TV is not always good but you can modify parameter
First, when capturing the video, capture it in the highest resolution possible. For Europeans this would be $720\times576$ and for North Americans, $720\times480$. Do not adjust the brightness or contrast in the recording monitor, but you might want to max out the color. Capture the video using MJPEG or uncompressed Component Video if possible; if not possible, then capture it using JPEG preferably or RGB if that is all that will work. Now on the timeline use Settings $\rightarrow$ Format to set a YUV colorspace, drop a \textit{Downsample} effect on the footage and set it as follows:
-\begin{lstlisting}[language=bash]
+\begin{lstlisting}[language=bash,numbers=none]
Horizontal: 2
Horizontal offset: 0
Vertical: 2
@@ -668,7 +668,7 @@ This entire procedure could be implemented in one non-realtime effect, but the p
If you photograph a lot of haze instead of blue sky, these horizon shots will usually need more depth. You can use the \texttt{gradient} effect to improve your video. Drop the gradient effect on hazy tracks and set the following parameters:
-\begin{lstlisting}[language=bash]
+\begin{lstlisting}[language=bash,numbers=none]
Angle: 0
@@ -693,7 +693,7 @@ It is important to set the $0\%$ alpha color to blue even though it is $0\%$ alp
|
2021-09-27 18:27:56
|
{"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": 0.8936241865158081, "perplexity": 761.3472347189636}, "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/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00567.warc.gz"}
|
https://testbook.com/question-answer/usually-fins-are-provided-to-increases-the-rate-of--58bd4473995a2d7d42806f94
|
# Usually fins are provided to increases the rate of heat transfer, but fins also act as insulation. Which one of the following non-dimensional numbers decides this factor?
This question was previously asked in
VIZAG MT Mechanical: 2013(Re-exam) Official Paper
View all Vizag Steel Management Trainee Papers >
1. Eckert number
2. Biot number
3. Fourier number
4. Peclet number
Option 2 : Biot number
Free
ME Subject Test 1: Strength of Materials
9090
20 Questions 20 Marks 18 Mins
## Detailed Solution
Explanation:
Biot Number:
A measure of the relative importance of the thermal resistance within a solid body is the Biot number Bi, which is the ratio of internal conductive resistance and external convective resistance i.e.
$$Bi=\frac{internal\;conductive\;resistance}{external\;convective\;resistance}\Rightarrow\frac{L/k}{1/h}=\frac{hL}{k}$$
where h = average heat transfer co-efficient, L = characteristic dimension i.e. V/A, and k = thermal conductivity of the solid body.
When Bi < 0.1, i.e when the internal resistance is less than 10% of the external resistance, the internal resistance can be ignored.
The installation of a fin on a heat transferring surface increases the heat transfer area but it is necessary that the rate of heat transfer would increase.
For long fins, the rate of heat loss from fin is given by:
$$\sqrt{hpkA}\;θ_o=kA\sqrt{\frac{hp}{kA}}θ_o\Rightarrow kAmθ_o$$
When $$\frac{h}{mk}=1$$, or h = mk, Q = hAθo
which is equal to the heat loss from the primary surface with no extended surface. Thus, when h = mk, an extended surface will not increase the heat transfer rate from the primary surface whatever be the length.
When h/mk > 1, Q < hAθo and hence adding secondary surfaces reduces the heat transfer and the added surface will act as insulation.
When h/mk < 1, Q > hAθo and the hence extended surface will increase the heat transfer.
the term 'm' can also remodified and rewritten as 'Bi number', it can give an idea of the fin as an insulator.
|
2021-09-18 08:12:28
|
{"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": 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": 0.7431955337524414, "perplexity": 2108.2013676020483}, "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-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00572.warc.gz"}
|
https://www.greencarcongress.com/2012/01/cellera-20120102.html
|
Vodafone Ventures and Carmel Ventures lead $9.2M investment in hydroxide exchange membrane fuel cell company CellEra 02 January 2012 CellEra, developer of a hydroxide (OH−) exchange membrane fuel cell (HEMFC) technology, recently announced a$9.2-million investment round led by Vodafone Ventures, the global venture capital arm of Vodafone Group, together with top-tier Israeli VC firm Carmel Ventures. The two have joined forces with the company’s largest shareholder Israel Cleantech Ventures, a venture capital fund focused on backing Israel’s emerging clean technology companies, as well as B-2-V Partners and private investors.
In a HEMFC (left), platinum is eliminated in the fuel cell electrodes and replaced with base-metal catalysts. Source: CellEra. Click to enlarge.
The core of CellEra’s fuel cell is a catalyzed, solid polymer electrolyte. However, as this new form of polymer electrolytes conducts OH- ions as opposed to H+ ions (protons), the chemical environment in the cells is mildly alkaline as opposed to the highly acidic environment in Proton Exchange Membrane (PEM) cells. Accordingly, the use of low-cost transition-metal based catalysts instead of platinum is enabled. The mildly alkaline environment further allows for the use of light aluminum hardware, without risk of corrosion, enabling, in turn, cost-effective manufacturing even at initially low volumes, as well as improved thermal management. Finally, the OH- ion conducting membrane itself is made of a less expensive polymer.
(In 2009, the Department of Energy’s ARPA-E program awarded researchers at the University of California at Riverside (UCR) a $760,705 grant to support the development of a new generation of HEMFCs that are dramatically more ion-conductive, durable and tolerant of abuse than previous devices. Their work was featured as a cover story in an issue of the journal Angewandte Chemie earlier that year. Earlier post.) Beyond the available capital, Vodafone’s investment will serve to facilitate our approach to the telecommunications market place, and our ability to direct our cost-effective, clean-energy fuel cell technology towards the requirements of global telecommunication operators. —Ziv Gottesfeld, CEO of CellEra Ziv added that while fuel cells have now been recognized as a reliable renewable power generation source allowing mobile network operators to reduce power generation related emissions and end-of-life hazards, wide market acceptance has been hampered to date by their high costs. CellEra’s goal is to substantially reduce these costs and allow a rapid return on investment through the development and market introduction of a new form of fuel cell technology, allowing the elimination of high-cost materials and expensive hardware in the fuel cell stack. CellEra is targeting it introductory product at the US$3-billion telecommunications supplemental-power market, offering an alternative to lead-acid batteries and diesel generators.
They are also talking about this as an RE, presumably prioritising this application as they are limiting their initial stack design to 20kw.
Interestingly they also mention on the site that for distributed power applications they can use ammonia as the hydrogen source.
Perhaps this would be too bulky/heavy for transport use, anyone any info?
Sorry, I meant to link to the CellEra website above:
http://www.cellera.biz/products-applications/
A CHEAP fuel cell would be a great optional EV range extender, but again, meaningful article device specs?
If you look at the slide set here:
http://www1.eere.energy.gov/hydrogenandfuelcells/pdfs/amfc_050811_gottesfeld_cellera.pdf
It does not really seem ready for prime time yet.
Hey SJC, thanks for the link.
Note Slide 5 - ".. Enabler for price parity at volume with lead acid batteries and diesel generator.."
Even the 15 year old GM EV1 had economic/useable 70 mile range(w/Panasonic Lead Acid batteries[not sabotaged leaking GM/Delco units]).
The 2012 Volt is a gas-powered generator-to-electric drive motor, likely more expensive than a diesel generator.
If not prime time/drive ready yet, at some marketing point(size/\$), picture a 'backup' CellEra fuel cell, < 100 pounds, with the capacity to eliminate "range anxiety" if the rare, unexpected extra range is needed.
The comments to this entry are closed.
|
2023-01-29 21:28:49
|
{"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": 0.25673794746398926, "perplexity": 10818.733833040098}, "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-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00434.warc.gz"}
|
https://www.physicsforums.com/threads/groups-in-quantum-mechanics.219918/
|
# Groups in Quantum Mechanics.
1. Mar 5, 2008
### Gigi
I read that the generator of the O(3) group is the angular momentum L and that the generator of the SU(2) group is spin S.
Nevertheless I have some questions.
1. In some books they say that the generator of the SO(3) group is angular momentum L.
SO(3) is the group of proper rotations, i.e. det(Matrix)=1.
Thus is it O(3) or SO(3)?
2. Both O(3) and SO(3) are defined as rotations in Eucledian space, 3-dimensions. Thus I would expect that we are talking about classic angular momentum.
Nevertheless in a quantum mechanics book I read that S0(3) is the generator of the angular momentum operator.
How is that if in Quantum Mechanics we are using the Hilbert space that is a complex function space?
3. I have the similar question regarding SU(2). SU(2) is defined in complex space. Thus it is ok to say that it is more or less the same as saying that this complex space is the Hilbert space?
4. Now in relativistic quantum mechanics, the underlying group is the Lorentz group. Would that mean that the O(3) and SU(2) groups reflect only symmetries in the non-relativistic world? i.e. Schroedinger's equation?
If that is so, how come spin that is a relativistic effect is explained using the SU(2) group?
Many thanks, as I am getting quite confused.
How is that?
|
2016-10-22 05:15:12
|
{"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": 0.8908429741859436, "perplexity": 595.7931180248115}, "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/1476988718426.35/warc/CC-MAIN-20161020183838-00047-ip-10-171-6-4.ec2.internal.warc.gz"}
|
http://clay6.com/qa/3452/if-sin-x-sin-y-sin-z-pi-prove-that-x-sqrt-y-sqrt-z-sqrt-2xyz
|
Want to ask us a question? Click here
Browse Questions
Ad
0 votes
# If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi,$ prove that $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
Can you answer this question?
## 1 Answer
0 votes
Toolbox:
• $cosx-cosy=2sin\large\frac{x+y}{2}.cos\large\frac{y-x}{2}$
• $1-sin^2x=cos^2x$
• $sin(2\pi-\theta)=-sin\theta$
• $sin2A=2sinAcosA$
• $cos(\pi-\theta)=-cos\theta$
Let $sin^{-1}x=A, \: sin^{-1}y=B, \: sin^{-1}z=C$
$\Rightarrow\:sinA=x,sinB=y\:and\:sinC=z$
By substituting the value of x,y,z in the given eqn.
$\Rightarrow\:given\: A+B+C=\pi$
$\Rightarrow\:A+B=\pi-C$
L.H.S.
Substituting the value of x y and z in L.H.S. we get
$x \sqrt{1-x^2}+y\sqrt{1-y^2}+2\sqrt{1-z^2}=$
$sinA \sqrt{1-sin^2A}+SinB\sqrt{1-sin^2B}+$
$sinC\sqrt{1-sin^2}C$
$=sinAcosA+sinBcosB+sinCcosC$
Multiply and divide by 2
$=\large\frac{1}{2}\big(2sinAcosA+2sinBcosB+2sinCcosC)$
Using the formula of sin2A=2sinA.cosA
$=\large\frac{1}{2}\big(sin2A+sin2B+2sinC.cosC\big)$
By taking x=2A,y=2B in the above formula of sinx+siny, we get
$=\large\frac{1}{2}\bigg(\big(2.sin\large\frac{2A+2B}{2}.cos\large\frac{2A-2B}{2}\big)+2sinCcosC\bigg)$
$=sin(A+B).cos(A-B)+sinCcosC$
$since,\:A+B=\pi-C$
$=sin(\pi-C).cos(A-B)+sinCcosC$
By taking $\theta=C$ in the above formula of $sin(\pi-\theta)$
$=sinC.cos(A-B)+sinCcosC$
$=sinC(cos(A-B)+cosC)$
By writing cosC=cos($\pi-(A+B))=-cos(A+B)$
$=sinC(cos(A-B)-cos(A+B))$
By taking x=A-B, and y=(A+B) in the above formula of cox+cosy we get
$=sinC\big(2sin\large\frac{A-B+A+B}{2}.sin\frac{A+B-A+B}{2}\big)$
$=2sinCsinAsinB=2xyz$
=R.H.S.
answered Feb 20, 2013
edited Mar 19, 2013
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
|
2016-10-26 21:13:25
|
{"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": 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": 0.9641236066818237, "perplexity": 10900.726104335727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2016-44/segments/1476988720973.64/warc/CC-MAIN-20161020183840-00271-ip-10-171-6-4.ec2.internal.warc.gz"}
|
https://www.splashlearn.com/math-vocabulary/fractions/quarter
|
# Quarter – Definition with Examples
## Introduction to Quarter in Math
Mathematics is not just a subject of units and numbers. Math concepts are regularly applied in our daily lives. We don’t even realize how mathematical laws rule everything we see in our surroundings.
Today, we will discuss an interesting topic: a quarter!
A quarter is a mathematical fraction frequently used while measuring or calculating portion, time, money, etc.
Fraction is defined as the part of a whole. Fraction is not only used in math problems but in real life frequently. This article will take a detailed look into a commonly used fraction: one-fourth.
One-fourth or a quarter is a mathematical fraction often used while measuring or calculating portion, time, money, etc. Let’s dive right into it.
## What Is a Quarter in Math?
Let’s look at the quarter definition in math.
A quarter can be explained as one part of four equal parts. If you take one whole unit, it is made up of four quarters.
A quarter is represented by using fractions in mathematics.
In mathematical terms, a quarter fraction is the division of 1 whole into 4 equal parts, where 1 denotes the part that is referred to and 4 represents the number of parts into which the whole has been divided.
In numerical form, it is written as ¼.
Let’s take an example to understand what’s a quarter in numbers.
For example, to find the quarter of 8 peaches, you need to divide them into 4 equal parts.
Each part will have 2 peaches. A quarter is what each equal part represents.
So, the quarter of 8 is 2.
## What Is One-fourth in Math?
We split one whole into four equal parts. One out of four equal parts of the whole is known as one-fourth. It is also known as a quarter. In numerical form, it is written as $\frac{1}{4}$.
Let us look at an example. A cake is divided into 4 equal parts. If the whole cake represents 1, then each piece is one-fourth of the whole. The entire cake is made up of 4 one-fourths (or 4 quarters).
Let’s look at one-fourth of a shape. We can represent the fraction, one-fourth, in different ways, as shown in the figure below.
One-fourth or ¼ can also be written as 0.25 and 25%.
## One-fourth of a Number
### What Is One-fourth of a Whole Number?
One-fourth is the same as the fraction $\frac{1}{4}$. It means that in order to find one fourth of a number, we have to divide the number by 4. In order to do this, we simply divide or split our whole number into 4 equal parts.
For example, to calculate a quarter of 28, we can divide 28 by 4. As a calculation, it would look like $28 \div 4$, which is equal to 7. It means that one-fourth of 28 is 7.
### What Is One-fourth of a Fraction?
One-fourth of any fraction is the same as multiplying $\frac{1}{4}$ and the fraction. We get the answer by multiplying the numerator together and the denominators together.
For example, let’s see what one fourth of ¼ is. Well, it is the same as multiplying the fraction together. So, $\frac{1}{4}$ of $\frac{1}{4} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
## One-fourth in Real Life
Let’s look at some real-life examples of one-fourth.
1. Time, such as a $\frac{1}{4}$ of an hour. You can divide 1 hour, that is 60 minutes, by 4. So, $\frac{1}{4}$ of an hour is 15 minutes.
1. The year, such as $\frac{1}{4}$ of a year. You can divide 1 year, that is 12 months, by 4. This means $\frac{1}{4}$ of a year is 3 months.
1. Money, such as $\frac{1}{4}$ of a dollar. 100 cents make 1 dollar. So, $\frac{1}{4}$ of a dollar would be 25 cents.
## Fractions Equivalent to One-fourth
It is not compulsory that you will always find one fourth written as $\frac{1}{4}$. You can also find it written in its equivalent fractions form. In order to find a fraction equivalent to $\frac{1}{4}$, we just need to multiply the numerator and denominator by natural numbers.
$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{1 \times 3}{4 \times 3} = \frac{1 \times 4}{4 \times 4}$ and so on.
So, $\frac{1}{4} = \frac{2}{4} = \frac{3}{4} = \frac{4}{16}$
## Using Quarter in Various Forms
A quarter in math can be used to represent various things:
• Time, such as a quarter of an hour. You can divide 1 hour, that is 60 minutes, by quarter:
There are 15 minutes in one quarter. Therefore, “quarter to 4” or “quarter past 4” denotes the time to be 3:45 and 4:15, respectively.
• The year, such as a quarter of a year. You can divide 1 year, that is 12 months, by 4 (a quarter). This means each quarter includes 3 months.
• Quarter 1 – January 1 to March 31
• Quarter 2 – April 1 to June 30
• Quarter 3 – July 1 to September 30
• Quarter 4 – October 1 to December 31
• Money, such as a quarter of a dollar. 100 cents make 1 dollar. So, a quarter of a dollar would be 25 cents.
## Fun Fact!
• The term quarter is also a name used for one of McDonald’s popular burgers, the “Quarter Pounder.” The burger contains a patty that weighs ¼ pound in its precooked state, which is the reason behind this name.
## Solved Examples
Example 1: Write the quarter of 32 and state how many 4s are in the number 32?
Solution:
$\frac{32}{4} = 8$
or,
$\frac{1}{2} \times 32 = 16$
$\frac{1}{2} \times 16 = 8$
Example 2: Calculate the quarter of 68.
Solution:
$\frac{1}{4} \times 68 = 17$
Example 3: Sunny has a quarter of 48 muffins, and Harry has half of 24 muffins. Find out who has more muffins.
Solution:
Quantity of muffins Sunny has $= \frac{1}{4} \times 48 = 12$
Quantity of muffins Harry has $= \frac{1}{2} \times 24 = 12$
Therefore, both Sunny and Harry have an equal number of muffins, i.e., 12 muffins.
Example 4. What is one-fourth of 96?
Solution: One fourth of $96 = \frac{1}{4} \times 96 = 96 \div 4= 24$.
Example 5. Area of a rectangle is 124 $cm^{2}$. $\frac{1}{4}$ of the rectangle is painted blue. What is the area of the rectangle that is painted blue?
Solution: Total area of rectangle $= 124$ $cm^{2}$
Area of rectangle that’s painted blue $= \frac{1}{4} \times 124 = 31$ $cm^{2}$
Example 6. Emma took 40 red pearls out of 160 pearls. What percent of the whole did she take?
Solution: Total number of pearls $= 160$
Number of pearls Emma took $= 40$
Fraction $= \frac{40}{160} = \frac{1}{4}$
$\frac{1}{4}$ in percentage $= \frac{1}{4} \times 100 = 25%$
Emma took $25%$ of the pearls.
## Practice Problems
1
### Calculate the quarter of 20.
35
8
10
5
CorrectIncorrect
As we know, a quarter of anything is $\frac{1}{4}$ of anything.
Therefore,
$\frac{1}{4} \times 20 = 5$
2
### The diameter of a circle is 32 cm. Calculate the area of a quarter circle.
$201.14 \text{cm²}$
$114.5 \text{cm²}$
$90.12 \text{cm²}$
$217.06 \text{cm²}$
CorrectIncorrect
Correct answer is: $201.14 \text{cm²}$
Diameter of the circle $= 32$ cm
Total area of a quarter circle $= \frac{\pi d^{2}}{16}$
$= \frac{22}{7} \times \frac{1}{16} \times 32^{2}$
$= 201.14$ $cm^{2}$
Note:
When the length of radius $= r$
The length of diameter (d) of the circle $= 2r$
Therefore, $r = \frac{d}{2}$.
Area of the circle $= \pi (\frac{d}{2})^{2}$
$= \frac{πd^{2}}{4}$
Thus, area of quarter circle $= 14 × \pi (\frac{d}{2})^{2}$
$= \frac{\pi d^{2}}{16}$
3
### Eric invited 52 guests for his birthday party. However, only a quarter of the entire guest list arrived. Calculate how many guests attended the party.
13
30
10
15
CorrectIncorrect
Number of guests invited to the party $= 52$
Number of guests attended the party $= \frac{1}{4} \times 52$
$= 13$
4
### Mary bought 24 apples from the store. A quarter of them were rotten and had to be discarded. How many fresh apples are remaining?
6
20
9
18
CorrectIncorrect
Number of apples bought $= 24$
Number of apples that were rotten $= \frac{1}{4} \times 24$
$= 6$
Number of fresh apples left $= 24$ $–$ $6$
$= 18$
5
### Which of the following figure does not represent 0.25?
A
B
C
D
CorrectIncorrect
In option C, the number of shaded parts are 8 and total number of parts are 16.
So, the fraction will be $\frac{8}{16} = \frac{1}{2} = 0.5 \neq 0.25$
6
### Mia bought 32 apples from the store. $\frac{1}{4}$ of them were rotten and had to be discarded. How many fresh apples are remaining?
8
24
16
14
CorrectIncorrect
$\frac{1}{4}$ of $32 = 8$. So, the number of fresh apples $= 32$ $-$ $8 = 24$.
7
### Which of the following is equal to $\frac{1}{4}$ of $\frac{1}{2}$ ?
$\frac{1}{8}$
$\frac{1}{6}$
$\frac{1}{16}$
$\frac{1}{2}$
CorrectIncorrect
Correct answer is: $\frac{1}{8}$
$\frac{1}{4}$ of $\frac{1}{2} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
A quarter is one-fourth of a whole number, and numerically it is represented as $\frac{1}{4}$. Three quarters mean $\frac{1}{4} \times 3$ or $\frac{3}{4}$.
$\frac{1}{4} = \frac{2}{8} = \frac{3}{12} = \frac{4}{16} = \frac{5}{20} = \frac{6}{24}$, and so on
A quarter of an hour means 15 minutes.One hour means 60 minutes, and a quarter of 60 minutes is thus $\frac{1}{4} \times 60 = 15$
Yes. The decimal representation of $\frac{1}{4}$ is 0.25. As it is a terminating decimal number, we can say that $\frac{1}{4}$ is a rational number.
The number 4 represents four wholes. For example, 4 children are playing in the park. 4 is a natural number. $\frac{1}{4}$ means part of a whole. It is not a natural number.
Ordinal numbers are numbers that are used to represent the “order” or are used to define a thing’s position.Fourth as an ordinal number can be written as “4th.” It is used to describe the ordinal number between third (3rd) and fifth (5th). Whereas one-fourth means part of a whole. It is represented by the number $\frac{1}{4}$.
|
2023-03-29 01:09:56
|
{"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": 0.637490451335907, "perplexity": 900.8880100764968}, "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-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00484.warc.gz"}
|
https://www.basic-mathematics.com/poisson-probability-distribution.html
|
# Poisson probability distribution
The Poisson probability distribution is named after the French mathematician Simeon D. Poisson.
Suppose there is a power outage in an apartment complex 3 times a year.
You may want to find the probability that there will be exactly 2 power outage next year. This is an example of a Poisson probability distribution problem.
Each power outage is called an occurrence. In order to use the Poisson probability distribution, the occurrence must be independent and random.
In our example of power outage, each power outage is random. In other words, they do not follow any known pattern.
There is no way to tell when a power outage will occur. They just happen randomly.
The power outage is also independent. The next power outage does not depend on the one before. In other words, a power outage this year does not influence a power outage next year.
Occurrences always happen with respect to an interval. In the example of power outage, the interval in one year and this make sense since we do not expect too many power outage unless we are living in a less developed country.
Once you know the average number of occurrences in an interval, we use the Poisson probability distribution to compute the probability of a certain number of occurrences ,x, in that interval.
## Examples to which the Poisson probability distribution can be applied
• The number of patients arriving at the emergency room during a one-hour interval.
• The number of phone calls you will receive from telemarketers during a one-month interval.
• The number of defective items in the next 200 items produced by a machine.
• The number of vehicle accidents that occur on a given highway during a one-week interval.
Notice that the last example is an example of volume interval instead of time interval. The occurrences are the number of defective items. They are random since there could be 0, 1, ... , 100 defective items. These occurrences are random since the occurrence of a defective item does not influence the occurrence of another.
## Poisson probability distribution formula
The probability of x occurrences in an interval is
$$P(x)\ =\ \frac{λ^x e^-λ}{x!}$$
λ is called parameter of the Poisson probability distribution or Poisson parameter. λ is the mean number of occurrences in that interval and is pronounced lambda. The value of e is 2.71828.
## Recent Articles
1. ### How to Construct Parallel Lines
Jan 12, 22 07:48 AM
This lesson will show you how to construct parallel lines with easy to follow steps
## Check out some of our top basic mathematics lessons.
Formula for percentage
Math skills assessment
Compatible numbers
Surface area of a cube
|
2022-01-25 09:11:02
|
{"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": 0.7273707389831543, "perplexity": 528.286938297961}, "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/1642320304798.1/warc/CC-MAIN-20220125070039-20220125100039-00109.warc.gz"}
|
https://listserv.uni-heidelberg.de/cgi-bin/wa?A2=0102&L=LATEX-L&D=0&H=A&S=b&P=3887723
|
## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE
Options: Use Classic View Use Monospaced Font Show Text Part by Default Condense Mail Headers Topic: [<< First] [< Prev] [Next >] [Last >>]
Sender: Mailing list for the LaTeX3 project <[log in to unmask]> Date: Wed, 30 Apr 2014 06:15:00 +0200 Reply-To: Mailing list for the LaTeX3 project <[log in to unmask]> Message-ID: <[log in to unmask]> MIME-Version: 1.0 Content-Transfer-Encoding: 7bit In-Reply-To: <[log in to unmask]> Content-Type: text/plain; charset=ISO-8859-1 From: Heiko Oberdiek <[log in to unmask]> Parts/Attachments: text/plain (85 lines) On 29.04.2014 16:35, David Carlisle wrote: > We are planning a new release of LaTeX2e in the next day or so to CTAN > (hopefully in time to get into texlive 2014 test releases). > > If anyone wants to have a look at any of the updated files before they > go to CTAN there is a _temporary_ dropbox location where the files may > be found at > > https://www.dropbox.com/sh/ms53wpmpo36ycpb/xLSAYzzhs2 Taking these files I have made a pre-release for latex-tds: https://dl.dropboxusercontent.com/u/71873949/tmp/latex-tds/base.tds.zip https://dl.dropboxusercontent.com/u/71873949/tmp/latex-tds/cyrillic.tds.zip https://dl.dropboxusercontent.com/u/71873949/tmp/latex-tds/graphics.tds.zip https://dl.dropboxusercontent.com/u/71873949/tmp/latex-tds/tools.tds.zip The documentation is regenerated using links and bookmarks. The files in the .zip files are organized according to the TDS standard. > Note that this is a "raw" distribution with no pre-built format, not > intended for production work, and will only be available for a few days. Same here. > distrib/doc/ltnews21.pdf | inputenc package updates The inputenc package | allows different input encodings for LaTeX documents to | be specified, importantly the utf8 option to specify the | Unicode UTF-8 encoding. A common mistake has been | to use this option with Unicode-based TeX engines | LuaTeX and XeTeX, however inputenc does not work | (and is not needed) with those systems. The package | has been modified so that if used with LuaTeX or | XeTeX, then it just issues a warning if utf8 is specified, | and stops with an error for any other encoding | requested. Encodings ascii.def and x-ascii.def ----------------------------------- A use case for these encodings is that a document should not contain problematic non-ASCII characters. Thus inputenc cries, if a 8-bit character or control characters is present. Since a valid document only contains 7-bit characters, this works very well with LuaTeX and XeTeX. IMHO, inputenc must not throw an error. Instead: * It should disable the 7-bit control characters of package inputenc as without LuaTeX/XeTeX. * The other characters in the Unicode range with codes > 127 are way too many for inputenc to handle. (Each character would have been made active and defined to throw an error.) IMHO the best approach is to ignore these characters. * Instead of a warning, an info message is enough, at most a warning, which says character with character codes > 127 are not handled by inputenc for LuaTeX/XeTeX. Encodings utf8.def and utf8x.def -------------------------------- The current version knows utf8.def and ignores it with a warning. However, there are documents that are using utf8x.def of package ucs. The same arguments are valid here: * The document source is in UTF-8, more or less needed for XeTeX/LuaTeX. * Thus an error would be wrong and a warning is enough. That should also be the case for utf8x.def If someone uses a document for the old engines: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} % or without \begin{document} and gets a warning, that utf8 is not required for LuaTeX/XeTeX, then he run into trouble (both with/without \usepackage[utf8]{inputenc}): The characters with codes >127 are not mapped to the proper LICRs, but are using the slots of the font encoding OT1 or T1 (for example) with the consequences that characters are missing or wrong. Thus a hint in the warning of inputenc should be added, that package fontspec might be useful. Yours sincerely Heiko Oberdiek
|
2022-08-12 12:10:11
|
{"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": 0.8220100998878479, "perplexity": 8036.702424398939}, "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-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00140.warc.gz"}
|
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=timm&paperid=1572&option_lang=eng
|
Trudy Instituta Matematiki i Mekhaniki UrO RAN
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Impact factor Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Trudy Inst. Mat. i Mekh. UrO RAN: Year: Volume: Issue: Page: Find
Trudy Inst. Mat. i Mekh. UrO RAN, 2018, Volume 24, Number 4, Pages 19–33 (Mi timm1572)
Optimal recovery of a function analytic in a half-plane from approximately given values on a part of the straight-line boundary
R. R. Akopyanab
a Ural Federal University named after the First President of Russia B. N. Yeltsin, Ekaterinburg
b Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, Ekaterinburg
Abstract: Let $\mathcal{H}^p(\Pi_+,\phi)$ be the class of functions analytic in the upper half-plane $\Pi_+$ and belonging to the universal Hardy class $N_*$ with boundary values from $L^p_\phi(\mathbb{R})$ with a weight $\phi$, and let $Q^p(\Pi_+,\mathbb{I},\phi)$ be the class of function $f\in \mathcal{H}^p(\Pi_+,\phi)$ such that $\|f\|_{L^p_\phi(\mathbb{R}\setminus\mathbb{I})}\le 1$, where $\mathbb{I}$ is a finite open interval or a half-line from $\mathbb{R}$ and $1\le p\le\infty.$ On the class $Q^p(\Pi_+,\mathbb{I},\phi)$, we consider the problem of optimal recovery of the value of a function at a point $z_0\in\Pi_+$ from its approximately given limit boundary values on $\mathbb{I}$ in the norm $L^p_\phi(\mathbb{I})$ and the related problem of the best approximation of a functional by linear bounded functionals. Explicit solutions of these problems are written: an extremal function, optimal recovery method, and best approximation functional. On the class $Q^p(\Pi_+,\mathbb{R}_+,\psi)$, $\psi(z)=1/|z|$, we solve the problem of optimal recovery of a function on a ray $\gamma=ż : \arg z=\varphi_0\}$ with respect to the norm $L^p_\psi(\gamma)$ from its approximately given limit boundary values on $\mathbb{R}_+$ in the norm $L^p_\psi(\mathbb{R}_+)$ and the related problem of the best approximation of an operator by linear bounded operators. For $f\in\mathcal{H}^p(\Pi_+,\psi)$, we obtain the exact inequality
$$\|f\|_{L^p_{\psi}(\gamma)}\le \|f\|_{L^{p}_{\psi}(-\infty, 0)}^{{\varphi_0}/{\pi}} \|f\|_{L_{\psi}^{p}(0, +\infty)}^{1-{\varphi_0}/{\pi}}.$$
Keywords: optimal recovery of an operator, best approximation of an unbounded operator by bounded operators, analytic function.
Funding Agency Grant Number Russian Foundation for Basic Research 18-01-00336 Ministry of Education and Science of the Russian Federation 02.A03.21.0006 This work was supported by the Russian Foundation for Basic Research (project no. 18-01-00336) and by the Russian Academic Excellence Project (agreement no. 02.A03.21.0006 of August 27, 2013, between the Ministry of Education and Science of the Russian Federation and Ural Federal University).
DOI: https://doi.org/10.21538/0134-4889-2018-24-4-19-33
Full text: PDF file (264 kB)
References: PDF file HTML file
Bibliographic databases:
UDC: 517.977
MSC: 30A10, 30C80, 30C85, 30E10
Revised: 14.11.2018
Accepted:19.11.2018
Citation: R. R. Akopyan, “Optimal recovery of a function analytic in a half-plane from approximately given values on a part of the straight-line boundary”, Trudy Inst. Mat. i Mekh. UrO RAN, 24, no. 4, 2018, 19–33
Citation in format AMSBIB
\Bibitem{Ako18} \by R.~R.~Akopyan \paper Optimal recovery of a function analytic in a half-plane from approximately given values on a part of the straight-line boundary \serial Trudy Inst. Mat. i Mekh. UrO RAN \yr 2018 \vol 24 \issue 4 \pages 19--33 \mathnet{http://mi.mathnet.ru/timm1572} \crossref{https://doi.org/10.21538/0134-4889-2018-24-4-19-33} \elib{https://elibrary.ru/item.asp?id=36517696}
|
2021-12-06 02:26:31
|
{"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": 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": 0.5426284670829773, "perplexity": 1117.91255151008}, "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/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00168.warc.gz"}
|