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https://www.physicsforums.com/threads/pendulum-questions.389833/ | # Pendulum Questions
## Homework Statement
1) If you wanted to build a pendulum that took a long time to make a full swing, what would you have to do about the length?
2) what if you wanted the reverse- a quick-swinging pendulum?
3) If you want to make a pendulum clock with a period of 2 seconds, what should its length be?
what would be the length of you wanted a 1 second pendulum clock?
would these clocks still keep accurate time if you moved then to the top of Mount Everest or the Moon? Explain
## Homework Equations
there is none...?
## The Attempt at a Solution
i think it is based on common sense which i am lacking of...
$$T = 2\pi \sqrt{\frac{l}{g}}$$ | 2020-05-27 07:28:11 | {"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": 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": 0.4692726731300354, "perplexity": 720.0808738383928}, "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-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00351.warc.gz"} |
https://crypto.stackexchange.com/questions/52214/rsa-timing-attack-on-extra-montgomery-reduction | # RSA Timing Attack on “Extra” Montgomery Reduction
In "A practical implementation of the timing attack", the authors take advantage of a timing difference that stems from "extra reductions" that occur when multiplying numbers in the Montgomery form. After implementing a toy example of this attack, I thought I understood it. I created two sets of messages, one set containing messages $m$ where $m^3$ did not require an extra reduction ($M1$) and another set where $m^3$ did require an extra reduction ($M2$). I then augmented a square-and-multiply algorithm to take advantage of Montgomery multiplication while tracking the number of extra reductions that took place over the course of the exponentiation.
This quick experiment did appear to show that the set in which $m^3$ did not require an extra reduction ($M1$) on average would have less extra reductions than the other set. However, the difference in the two averages would be much larger than one. Most of the paper seems to suggest the timing attack they exploited relied on the timing difference a single extra reduction (e.g. section 7.1).
I believe I see an attack here, but its not the exact situation that the authors describe. It sounds like the authors are suggesting that a square-and-multiply exponentiation algorithm that relied on Montgomery multiplication would use $n$ extra reductions for messages in the set $M1$ and $n+1$ extra reductions for messages in the set $M2$ which does not make sense to me intuitively. Am I misunderstanding this attack?
Edit: After comparing my attack code with others, it turns out I understood the attack just fine and my experiment was different. The difference between my results and the results from the original paper was due to my selection of the Montgomery parameters. The value of $R$ used to put numbers in Montgomery form can have a major effect on the number of reductions required per multiplication.
## Montgomery multiplication
Theorem (Montgomery, 1985). For any odd integer $$N$$ and any integer $$0 \le T < N2^k$$, one has: $$T 2^{-k} \equiv \frac{T + UN}{2^k} \pmod N$$ where $$U = T N' \bmod 2^k$$ and $$N' = -N^{-1} \bmod 2^k$$. Further, one has: $$\begin{cases} T+UN \propto 2^k\\ 0\le \frac{T + UN}{2^k} < 2N\end{cases}$$
The last 2 properties show that $$R:=T2^{-k} \bmod{N}$$ can be computed in two steps as:
1. $$R \gets \frac{T+UN}{2^k}$$
2. If $$R \ge N$$ then $$R \gets R - N\qquad$$ (extra subtraction)
Montgomery arithmetic represents an integer $$0\le x as $$\tilde{x} = x2^k \bmod N$$. It defines the Montgomery multiplication of two representatives $$\tilde{x}$$ and $$\tilde{y}$$ as $$\tilde{x} \otimes \tilde{y} := (\tilde{x}\cdot\tilde{y})2^{-k} \bmod N\tag{*}$$ Observe that letting $$z= x\cdot y \bmod N$$, one has $$\tilde{z} \equiv (x\cdot y)2^k \equiv (\tilde{x}\cdot \tilde{y})2^{-k} \equiv \tilde{x} \otimes \tilde{y}\pmod N$$.
## Timing attack
Suppose that each Montgomery multiplication, cf. Eq. ($$^*$$), is done using the above two-step process. We see that depending on $$\tilde{x}$$ and $$\tilde{y}$$, there may be an extra subtraction to get the result $$\tilde{z}$$ in the range $$[0, N)$$.
To fix ideas, suppose that the computation of $$S = m^d \bmod N$$ is carried out with the square-and-multiply algorithm. The attack can easily be adapted to other exponentiation algorithm. The algorithm takes as input a message $$m$$ and private exponent $$d = (d_{\ell-1}, \dots, d_0)_2$$.
1. $$R_0 \gets \tilde{1}$$; $$R_1 \gets \tilde{m}$$
2. for $$i=\ell-1$$ downto $$0$$ do
3. $$\quad R_0 \gets R_0 \otimes R_0$$
4. $$\quad$$if $$d_i = 1$$ then $$R_0 \gets R_0 \otimes R_1$$
5. endfor
6. return $$\tilde{R_0}$$
The goal of the attacker is to recover the value of $$d$$. Observe that the square-and-multiply algorithm processes the bits of $$d$$ from the left to the right.
• Let $$d = (d_{\ell-1}, \dots, d_0)_2$$
• At step $$j$$, the attacker
• already knows bits $$d_{\ell-1}, d_{\ell-2}, \dots, d_{j+1}$$
• guesses that next bit is $$d_j=1$$
• chooses $$T$$ random messages $$m_1, \dots, m_T$$ and computes $$X_t = m_t^{(d_{\ell-1}, d_{\ell-2}, \dots, d_{j+1},1)_2} \bmod N\quad\text{for 1\le t \le T}$$
• prepares two sets $$\mathcal{S}_0 = \{m_t \mid \text{subtraction}\}\quad\text{and}\quad \mathcal{S}_0 = \{m_t \mid \text{no subtraction}\}$$
• plays all messages in set $$\mathcal{S}_0$$ and obtains the average running time $$\tau(\mathcal{S}_0)$$ for an exponentiation; does the same for $$\mathcal{S}_1$$ and obtains $$\tau(\mathcal{S}_1)$$
• if $$\tau(\mathcal{S}_0) - \tau(\mathcal{S}_1) \not\approx 0$$ then $$d_j=1$$ (the guess was correct); if $$\tau(\mathcal{S}_0) - \tau(\mathcal{S}_1) \approx 0$$ then $$d_j=0$$ (the guess was incorrect)
• Iterate the attack to find $$d_{j-1},\dots$$
In the above description, set $$\mathcal{S}_0$$ denotes the set of messages such that there is an extra subtraction at iteration $$i=j$$ in Step 4 of the square-and-multiply algorithm (i.e., Montgomery multiplication $$R_0 \otimes R_1$$) for the computation of $$X_t$$; set $$\mathcal{S}_0$$ denotes the set of messages such that there is no extra subtraction at iteration $$i=j$$ for the computation of $$X_t$$.
Note that the attacker can evaluate the exponentiation $$X_t$$ by herself as she knows $$d_{\ell-1}, d_{\ell-2}, \dots, d_{j+1}$$. Further, since $$d_j$$ is assumed to be $$1$$, the last operation of this exponentiation $$X_t$$ will be a Montgomery multiplication $$R_0 \otimes R_1$$ (i.e., Step 4 is executed at iteration $$i=j$$).
## Correctness
If $$d_j=1$$ then the average time for an exponentiation for messages in set $$\mathcal{S}_0$$ will be greater than the average time for an exponentiation for messages in set $$\mathcal{S}_1$$ because there is always an extra subtraction for $$\mathcal{S}_0$$.
If $$d_j=0$$ then the average time for an exponentiation for messages in set $$\mathcal{S}_0$$ or in set $$\mathcal{S}_1$$ will be (roughly) the same because the sorting between the two sets looks like random. Remember that the attacker made the guess $$d_j = 1$$.
What a coincidence, I implemented this attack yesterday!
I'm executing it right now and I can tell you that the difference at each step is around 1 reduction (as the paper suggests). See for example my logs for bits 6 to 8 of a 96-bit key:
bit 6: avg M2 (#111)= 1.2072072072072073, avg M1 (#99889)= 0.1684369650311846
[1, 1, 1, 1, 1, 1, 1]
bit 7: avg M2 (#99)= 0.18181818181818182, avg M1 (#99901)= 0.16957788210328226
[1, 1, 1, 1, 1, 1, 1, 0]
bit 8: avg M2 (#98)= 1.1938775510204083, avg M1 (#99902)= 0.16858521350923905
[1, 1, 1, 1, 1, 1, 1, 0, 1]
As you can see, in bits 6 and 8, which are set to 1, the average of M2 and M1 differs in approximately 1 reduction (my proof-of-concept measures reductions for the moment, not time), while in bit 5, which is set to 0, there is almost no difference.
As far as I understand, the idea of the attack is that on each step (i.e., on each bit guess) the attacker partitions the samples in those that needed a reduction on that particular step (M2), and those that didn't (M1). You seem to think that this implies that messages in M2 would tend to have more reductions apart from the one that is isolated in the current step. However, from the numbers I got, that doesn't seem to be the case: at each step, only the 0.1% of all the samples needed a reduction (e.g., in bit 6, only 111 out of 100000 samples), which indicates that reductions are actually very rare, so most samples will have 0, 1 or perhaps 2 reductions for the whole exponentiation.
The trick is that if you are capable of identifying the moments where reductions occur (which in the attack is achieved by computing all the square-and-multiply operations for each sample, for each step), then you have a criterion for telling the samples apart. As a result, the most probable result is that the difference between M2 and M1 is due only to the presence of a reduction for the current step.
• "... which indicates that reductions are actually very rare, so most samples will have 0, 1 or perhaps 2 reductions for the whole exponentiation..." I think our implementations are very different, as I get far more reductions than that! My code seems to be functionally correct, as it returns the correct modexp results though! Hm. I may try rewriting my experiment and see where I may have done differently. Thanks! – kkl Oct 16 '17 at 20:35 | 2020-07-02 17:02:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 75, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7870888710021973, "perplexity": 1114.4154797661517}, "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/1593655879532.0/warc/CC-MAIN-20200702142549-20200702172549-00266.warc.gz"} |
http://www.norsemathology.org/wiki/index.php?title=CURM_page_of_Grayson_Rodriguez | # CURM page of Grayson Rodriguez
## Progress
### Commentaries - May 22, 2009
#### Cicada RWL to Wet Mass
The relationship of cicada right wing length to cicada wet mass is estimated with two power models by using stepwise regression. There is a total of three parameters with all three classes of cicadas (small, medium, and large) sharing the same exponent of 3.02. In fact, the standard error for the exponent (.0422) includes 3.00. We are estimating a three dimensional measurement (mass) from a linear measurement (length) and so it is not surprising that we are raising right wing length to roughly its cube. The small and large cicadas are modeled with a coefficient of 0.017 and the medium cicadas with 0.0229. Logarithm transformations were done on both the RWL and wet mass data in order to carry out the linear regressions and then transformed back to achieve the final power models. We used a combination of forward selection--where we begin with the intercept term, add in variables one at a time, and include any that are statistically significant--and backward elimination--where all variables are initially included and each one is tested and removed if it is not significant. If we were to analyze each class of cicada separately we would obtain three regression models with a total of six parameters. By using stepwise regression we are able to reduce the number of parameters needed to estimate cicada wet mass down to three.
### Analyzing Dow's data - April 27, 2009
The following data is from Table 1 of Richard Dow's paper The relation of the prey of Sphecius speciosus to the size and sex of the adult wasp
[# in cell] [species] [prey weight] [adult weight] [adult sex] [cocoon length] [cocoon weight]
1 Tb c 0.93 0.13 m 24 0.67
1 Tb c 0.93 0.25 m 29.5 1.05
1 Tb c 0.93 0.28 m 29.5 1.3
1 Tb c 0.93 0.43 m 29.5 1.25
1 Tb c 1.12 0.27 m 29.5 1.12
1 Tb c 1.12 0.36 m 29.5 1.28
1 Tb c 1.12 0.38 m 31 1.37
1 Tb l 1.39 0.36 m 32 1.39
1 Tb l 1.39 0.42 m 31 1.59
1 Tb l 1.94 0.39 m 32 1.48
2 Tb c 2.05 0.43 m 31 1.3
2 Tb c 2.05 0.44 m 32 1.62
2 Tb c 2.24 0.63 f 40 2.96
2 Tb c 2.24 0.69 f 39 2.96
2 both 3.06 0.81 f 40.5 3.4
2 both 3.06 1.09 f 39.5 3.16
2 Tb l 3.8 0.61 f 40.5 3.15
2 Tb l 3.8 1.06 f 40.5 3.33
R-sq = 0.725
### Combinations of cicadas - April 23, 2009
Here's the data and R code:
#getting the data
mass<-scan("combomasses.txt")
Nh<-scan("Nh.txt")
Tb<-scan("Tb.txt")
Tg<-scan("Tg.txt")
#combining the cicadas numbers in a table
#plotting a stacked barplot with rounded masses
barplot(t(cicadas), main="Cicada Combos", ylab="Total", xlab="Wasp Mass (mg)", col=c("cornsilk4", "orangered", "deepskyblue"), space=0, cex.axis=0.8,
las=1,names.arg=c(round(mass)), cex=0.8)
legend(0, 13, c("Nh", "Tg", "Tb"), col = c("cornsilk4", "orangered", "deepskyblue"), text.col = "black", pch = c(15, 15, 15), bg = 'gray90', pt.cex=2, cex=1.5)
### Week 14 - December 2 - 8, 2008
• I've got FileZilla up and running, and I have uploaded several files.
• I've computed the mean and standard deviation ratios for female wasps at St. Johns vs Newberry and males wasps at St. Johns vs Newberry.
Mean RWL Ratio SD Ratio Fsj / Fnb 1.227 1.391 Msj / Mnb 1.115 1.183
• I've done a paired t-test with wasp RWL and paired caught cicada RWL. With a p-value < 0.0001 it is clear that small wasps catch small cicadas and large wasps catch large cicadas. Here is the output from StatCrunch:
• I would like to simulate how the distribution of wasps would change after the large females have been truncated, however, how is cicada mass (baby food) converting into wasp mass?
### Week 12 - November 18 - 24, 2008
I added a draft of my story to our paper.[1]
I worked on the fortune teller problem.[2]
### Week 11 - November 11 - 17, 2008
Still working on the wasp story.
#### Wasp masses
I installed R on my computer and replicated what we saw previously. Here is the empirical probability density function of rwl of cicadas in R from the file newprovisioning_florida.xls.
I played around in Mathematica to generate a pdf for each species. The means and standard deviations came from the normals previously generated in StatCrunch using the file newprovisioning_florida.xls. The weights were derived from the number of each species caught with or without measurements. Ne = 232, Do = 27, Tg = 33, Tb = 18 (largest cicadas). So the weights here are 0.74873, 0.08685, 0.10632, 0.05810 respectively (they sum to 1).
I summed the pdfs:
Here is the empirical cumulative from R:
Here is the cumulative from Mathematica:
### Week 10 - November 4 - 10, 2008
I did some analysis on the right wing lengths of our sample of wasps. I separated the data by sex and location and did some summary statistics. Our sample of wasps at St. Johns are larger in wing length than the sample at Newberry. I had a question: are the masses in the data actual measurements or the predicted mass based on wing length?
JM = St. Johns, Florida
LA = Newberry, Florida
Barack Obama has won!
### Week 9 - October 28 - November 3, 2008
I did an ANOVA between the wing lengths of female cicadas and male cicadas of the Neocicada species. The p-value was 0.9198, so their wing lengths are not significantly different.
HOWEVER, an ANOVA between the wet masses of female cicadas and male cicadas of the Neocicada species gave a p-value of 0.0142. Females are "meatier".
#### Wasp hunting function
I came up with a first draft function using Mathematica to describe the how the wasp's desired prey mass changes with time.
$f(m,t)= \frac{1}{1 + e^{-[11 m (t + 1) - 16] + t}}$
These plots go from t=0 to t=3 with 0.4 steps:
### Week 8 - October 21 - 27, 2008
Read the paper by Peter Grant, Opportunistic predation and offspring sex rations of cicada-killer wasps.
### Week 7 - October 14 - 20, 2008
• ANOVA between the two smaller cicada species:
• I counted the number of caught cicadas by species with or without a wing length measurement:
• Ne.h. - 232 (smallest cicadas)
• D.o. - 27
• T.g. - 33
• T.b. - 18 (largest cicadas)
• I grouped the four histograms together in rows.:
• Below is all four species together on the same graph.
### Week 6 - October 7 - 13, 2008
I sorted out the data by species of cicada. The following frequency histograms use the right wing length data for each species. Histograms
• D.o.
The next histogram is of the species Ne.h. or Neocicada hieroglyphica. This species was recorded the most with 99 catches and fits the normal curve pretty well.
• Ne.h.
• T.g.
• T.b.
### Week 4 - September 23 - 29, 2008
I was able to graph the data of the right wing length of the wasps vs the right wing length of the cicadas using StatCrunch and was able to get a picture close to Dr. Hastings.
Sigmoid function is a new term to me, but I now know that the logistic function, which I am familiar with, is a sigmoid function. I revisited the logistic equations we looked at in Differential Equations for predator prey modeling.
I read an article about r and K selection as I believe Dr. Hastings said we are looking at K selection.
### Week 3 - September 16 - 22, 2008
I looked over the powerpoint by Dr. Hastings and tried to load the dataset into StatCrunch, but I couldn't get it to work. | 2021-10-19 04:57:18 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3461538851261139, "perplexity": 4165.304308981616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585242.44/warc/CC-MAIN-20211019043325-20211019073325-00582.warc.gz"} |
https://slideplayer.com/slide/4586356/ | # Fractions: The Basics.
## Presentation on theme: "Fractions: The Basics."— Presentation transcript:
Fractions: The Basics
What is a Fraction? numerator 1 3 denominator
Like a decimal, a fraction is part of a whole A fraction is less than one whole A fraction has 2 parts: numerator denominator The numerator indicates the counted parts of a fraction The denominator indicates the total parts of a fraction Give S various fractions, and have them identify the numerator and denominator.
Parts of Wholes & Sets A fraction can express part of a whole:
A fraction can express part of a set 1 of 4 pieces of the whole pizza was eaten 1/4 1 of 4 cookies in the set was eaten 1/4
Numerators, Denominators, and Division
A fraction is also simply a division expression The numerator is the dividend The denominator is the divisor So, the fraction 1/3 is the same as 1 divided by 3
Expressions as Fractions
If we take the expression, 9 ÷ 10, we can write it as the fraction, 9/10, so 9 ÷ 10 = 9/10 Can you write each expression as a fraction? 1 ÷ 3 3 ÷ 6 2 ÷ 7 6 ÷ 13 5 ÷ 8
Proper vs. Improper A fraction is proper when its numerator is smaller than its denominator (3/4, 5/8, 6/10). A fraction is improper when its numerator is larger than its denominator (4/3, 8/5, 10/6). Which are improper fractions? 4/ / / /5 8/3 7/ / /6
Mixed Numbers A mixed number has a whole number with a fraction: 4
The mixed number above is read, “four and two-thirds.” Mixed numbers are made by changing improper fractions. 2 3
Making Mixed Numbers To change an improper fraction to a mixed number: 1. Write the improper fraction as a division expression: = 23 ÷ 6
Making Mixed Numbers 2. Solve the expression: 6 23 3 R 5 18 5
Making Mixed Numbers 6 23 Use the quotient (3) as the whole number. 3 R 5 Use the remainder (5) as the numerator of the fraction. 6 18 5 Use the divisor (6) as the denominator of the fraction.
So, the improper fraction, 23/6 = the mixed number, 3 and 5/6!
Making Mixed Numbers So, the improper fraction, 23/6 = the mixed number, 3 and 5/6! 6 23 3 R 5 3 5 6 6 18 5
Let’s Make Mixed Numbers!
Convert the improper fraction, 7/5 to a mixed number. What do we do first? Write 7/5 as a division expression! Now what? Solve the expression! Finally… Use the quotient as the whole number, the remainder as the numerator, and the divisor as the denominator!
Let’s Make Mixed Numbers!
Convert the improper fractions to mixed numbers. 1. 6/4 2. 12/5 3. 37/9 4. 17/7 1-2/4 2-2/5 4-1/9 2-3/7
Can You Do the Opposite? To convert a mixed number to an improper fraction: 1. Multiply the denominator by the whole number. 2. Add in the numerator. 3. Write the answer as the numerator of the improper fraction, and the original denominator as the improper fraction’s denominator. Show S how to start at the bottom (denominator) and work clockwise to the whole number, and then the numerator.
(denominator x whole number)
Let’s Try One! Convert 5-2/3 to an improper fraction. What should we do first? Multiply 3 x 5! (denominator x whole number) Now what? Add 2 (numerator)! Finally? Use 17 as the numerator and 3 as the denominator!
Let’s Try Some More! Convert to improper fractions. 1. 3-4/5 2. 6-1/3
/5 /3 /6 /7 19/5 19/3 29/6 72/7
Whew! That was a lot of information! Good job! | 2021-10-28 13:33:00 | {"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": 0.9513002634048462, "perplexity": 1837.2873339809792}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588341.58/warc/CC-MAIN-20211028131628-20211028161628-00434.warc.gz"} |
http://lpsa.swarthmore.edu/LaplaceXform/InvLaplace/InvLaplaceXformIntro.html | # The Inverse Laplace Transform, Introduction
The inverse Laplace Transform can be calculated in a few ways. If the function whose inverse Laplace Transform you are trying to calculate is in the table, you are done. Otherwise we will use partial fraction expansion (PFE); it is also called partial fraction decomposition. If you have never used partial fraction expansions you may wish to read a background article, but you can probably continue without it.
There is also a way to directly calculate the inverse Laplace Transform by integration (so called "Direct Calculation"). The technique is described, but no examples are given. This takes advanced calculus and we will not use direct calculation.
Finally, examples are given the use MATLAB. MATLAB has powerful techniques for partial fraction expansion.
References | 2019-03-20 05:11:39 | {"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": 0.8225304484367371, "perplexity": 462.70440090858136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202299.16/warc/CC-MAIN-20190320044358-20190320070358-00033.warc.gz"} |
https://math.stackexchange.com/questions/2191043/help-to-solve-int-1-x-sqrt25-x2-dx | # Help to solve: $\int 1/(x\sqrt{25-x^2})\ dx$
I'm a brand new student. Need some help to integrate this.
Perform this integration: $$\int \frac{1}{x\sqrt{25-x^2}}\ dx$$
I'm able to obtain in theta terms like this:
$$\frac 15 \ln|\cscθ-\cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5\sin{\theta}$ and $dx=5\cos{\theta}$, we obtain: $$\int \frac{1}{5\sin{\theta}(5\cos{\theta})}\cdot 5\cos{\theta}~d\theta$$ Simplifying, we obtain: $$\frac{1}{5}\int \frac{d\theta}{\sin{\theta}}$$ Using trigonometric identities: $$\frac{1}{5}\int \csc{\theta}~d\theta$$ Integrating: $$\frac{1}{5}\ln|\csc{\theta}-\cot{\theta}|+C$$
• Please use: math.meta.stackexchange.com/questions/5020/… – Arnaldo Mar 17 '17 at 15:25
• It helps if you explain the steps you took to get at the solution in $\theta$, e.g. what was the substitution you used? – StackTD Mar 17 '17 at 15:27
• Yes Exactly. See the problem image..I believe I have the half of the solution....thanks.. – El_Master Mar 17 '17 at 15:27
• @El_Master Did you use the substitution $x=5\sin(\theta)$ and $dx=5\cos(\theta)~d\theta$ ? – projectilemotion Mar 17 '17 at 15:33
• you can use relationships between trigonometric functions to change it back, so if $x = sen \theta$ then $cos \theta = \sqrt{1 - sen^2 \theta} = \sqrt{1 - x^2}$ and other identities that you can use – Cato Mar 17 '17 at 15:33
HINT:
If trigonometric substitution is not mandatory, set $\sqrt{a^2-x^2}=y$
$$\int\dfrac{2x\ dx}{2x^2\sqrt{a^2-x^2}}=\int\dfrac{dy}{y^2-a^2}=?$$
Else setting $x=5\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$\cos y=\sqrt{1-\left(\dfrac x5\right)^2}=\dfrac{\sqrt{25-x^2}}5$
$$\implies\int\dfrac{dx}{x\sqrt{25-x^2}}=\int\dfrac{5\cos y\ dy}{5\sin y(5\cos y)}=\dfrac{\ln|\csc y-\cot y|}5+K$$
• I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine). – El_Master Mar 17 '17 at 15:59
$$x=5\sin\theta$$ $$\frac x5=\sin\theta$$ Therefore $$\frac5x=\csc\theta$$ And since $$\sin\theta=\frac x5$$, we know that $$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{25}}$$ And since $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$, $$\cot\theta=\frac{5\sqrt{1-\frac{x^2}{25}}}{x}=\frac{\sqrt{25-x^2}}x$$ Which gives your result as $$\frac15\ln\bigg|\frac{5-\sqrt{25-x^2}}x\bigg|+C$$ | 2021-07-25 09:44:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7952505946159363, "perplexity": 826.259882844846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00656.warc.gz"} |
http://tex.stackexchange.com/tags/chapters/hot | # Tag Info
8
This is possible with totcount package, for example! \regtotcounter{chapter} registers an existing counter name and stores the value at the end of the compilation to the .aux file, during the second compilation the total value can be obtained with \total{chapter}. There's no need to adjust a label to the last chapter then manually. \documentclass{book} ...
6
Well, this is a somewhat hidden feature, but minitoc provides for the \adjustmtc command, that is meant for starred chapters etc., i.e. for the case that the mtc counter is off by one. This command can be 'misused' to provide the minitoc for the following chapter. The \adjustmtc[n] command is documented on page 34 of the current manual of minitoc. The ...
5
You can use etoc for this, via its label/ref mechanism for tocs. \documentclass{book} \usepackage{etoc} \usepackage{blindtext} \makeatletter \newcommand*\cleartoevenpage {% \clearpage \if@twoside \ifodd \c@page \hbox {}\newpage \if@twocolumn \hbox {}\newpage \fi \fi \fi } \makeatother \begin{document} \tableofcontents % switch to ...
5
If you don't want to load hyperref, you can use refcount instead. You can use the lastpage package to get the length of the last chapter. \documentclass{report} \usepackage{hyperref}% defines \getpagerefnumber \begin{document} Chapter 1 has [\the\numexpr\getpagerefnumber{c2}-\getpagerefnumber{c1}\relax] pages in it. \chapter{Chapter 1}\label{c1} Text... ...
4
I don't recommend this code, but maybe somebody can improve it. \documentclass[ 11pt, ]{book} \usepackage[top=2cm,bottom=3cm,left=3.2cm,right=3.2cm,headsep=10pt,letterpaper]{geometry} \usepackage[english]{babel} \usepackage[babel=true]{csquotes} \usepackage[T1]{fontenc} \usepackage[explicit]{titlesec} \usepackage{xcolor} \usepackage{charter} ...
3
This is the definition of \chapter from report.cls \newcommand\chapter{\if@openright\cleardoublepage\else\clearpage\fi \thispagestyle{plain}% \global\@topnum\z@ \@afterindentfalse \secdef\@chapter\@schapter} \def\@chapter[#1]#2{\ifnum \c@secnumdepth >\m@ne ...
3
The three memoir commands in question are: \chapnamefont \chapnumfont \chaptitlefont which, in the default settings, are set to \Huge and \bfseries. Thus it is a simple matter to redefine them: \documentclass[openany]{memoir} \usepackage{xcolor,lipsum}% lipsum.sty is just for this example \renewcommand{\chapnamefont}{\Huge\bfseries\color{blue}} ...
2
Since \chapter uses \thispagestyle{plain}, the \fancypagestyle{plain} redefinition must disable the \fancyhead[RO,LE]{...} from fancy style, i.e. say \fancyhead{} in \fancypagestyle{plain}{...}. \documentclass{book} \usepackage{fancyhdr} \pagestyle{fancy} \renewcommand{\sectionmark}[1]{\markboth{#1}{}} % set the \leftmark \fancyhf{} \fancyhead{} ...
2
As was suggested to you in the comments, load the titlesec package. Then add the following commands: \titleformat{\chapter}{\normalfont\bfseries\Large}{\thechapter.\quad}{0pt}{} \titlespacing{\chapter}{0pt}{-50pt}{40pt} The second one is what you're looking for. The values are respectively: left space space before (above) space after (below) So you ...
2
(I think this question is off-topic here, but I'm not sure.) You should either include all of ToC, LoF and List of Publications in the table of contents, or none of them. And (with my copy editor hat one) this order makes sense to me. Another option is to make the list of publications an unnumbered section in the Introduction, but one that is included in ...
2
This is largely what I said in the comments. I left a few things "in" but commented out along with meta-comments of my own trying to explain why. I hope all's clear. (If not, let me know.) \documentclass[12pt,a4paper,twoside]{report} \usepackage[top=2.54cm, bottom=2.54cm, left=2.54cm, right=2.54cm]{geometry} \usepackage{sectsty} % section font size ...
1
Such tasks tend to be really nasty, because there is a plethora of things that can go wrong. I suggest to use the search & replace functions of your editor: If you don't have more than 25 lines inside the preamble, I'd copy all files into one large file. How to do this depends on your OS, but google provides many answers. For Windows e.g. see here: ...
1
First of, since memoir has its own build in pagestyle system there is no reason to use fancyhdr with memoir. Second the chapter language does not have anything to do with fancyhdr or \@chapapp for that matter. The word current word for chapter is stored in \chaptername. \@chapapp is a special macro that returns \chaptername whenever appendix is not ...
1
I found a new workaround for the problem explained in the comment to the original question. Basically one needs to replace the old \makeatletter from @marco-daniel with this one: \usepackage{xcolor} \colorlet{chapterbackground}{red} \addtokomafont{chapter}{\color{white}} \makeatletter \renewcommand\chapterlinesformat[3]{% \colorbox{chapterbackground}{% ...
1
You can change the value of the counter tocdepth mid-document, writing it to the ToC at the appropriate spot. This allows you to vary what shows and what doesn't show. \documentclass{report} \begin{document} \tableofcontents \chapter{This chapter is listed in the contents page} \section{So are all its sections} \subsection{--- and subsections!} ...
1
Here is an improved solution to my previous solution in the abovementioned page. Now you have to declare a \chapterabstract[some text] before each chapter command to obtain an abstract: \documentclass[11pt,a4paper]{book} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{fourier} \usepackage{cabin} \usepackage[nomarginpar]{geometry} ...
1
As David implied "It all depends". Depending on the particular LaTeX class there is a hierarchy of headings. The book and report classes provide part, chapter, section, subsection, subsubsection, paragraph and subparagraph. The article class has all of these except for chapter, while the memoir class adds book at the top of the book and report class ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2016-02-11 21:41:43 | {"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": 0.9104546308517456, "perplexity": 4585.178952471768}, "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-2016-07/segments/1454701162808.51/warc/CC-MAIN-20160205193922-00234-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://www.atractor.pt/mat/conchas/texto2-_en.html | ## Shells
### Model II
$$\require{color}\newcommand{\arule}[1]{{\color{#1}\Rule{3em}{1ex}{0ex}\;}}$$
Let us look at the shell again, but now as an object in $$\mathbb{R}^{3}$$ (tridimensional object).
We want to find a curve in space that "seen from above" looks similar to an equiangular spiral but moves down continuously with respect to the horizontal plane that contains the initial point of the curve. A curve with such a property is an helicoidal.
To take advantage of the curve that we have already in the planar case, we suppose that the "transition" of the spiral to the helicoidal $$H(\theta )$$ does not change the distance between the points of the curve and the origin of the referential, that is, $r(\theta)=Ae^{\theta \cot (\alpha )},\theta\geq0,$ still represents the distance of each curve point to the origin.
(click at the image to open the applet)
$\arule{red}=r(\theta)\Longrightarrow\begin{cases} \arule{blue}= & r(\theta)\cos(\beta)\\ \arule{orange}= & r(\theta)\sin(\beta) \end{cases}$
Then the height of each point in this curve is given by
$z(\theta )=-r(\theta )\cos (\beta ),\,\,\theta \geq 0,$
where $$\beta$$ $$(0% %TCIMACRO{\UNICODE[m]{0xba}}% %BeginExpansion {{}^o}% %EndExpansion \leq \beta \leq 90% %TCIMACRO{\UNICODE[m]{0xba}}% %BeginExpansion {{}^o})% %EndExpansion$$ is the enlargement angle of the helicoidal.
On the other hand, note that to get the equations $$x(\theta)$$ and $$y(\theta)$$ of the helicoidal curve it suffices to replace $$r(\theta)$$ by $r_{p}(\theta )=r(\theta )\sin (\beta )$
in the equations of the spiral.
To see what happens to the helicoidal when the parameters $$\alpha$$, $$\beta$$ and$$A$$ are changed, see the following applet.
In summary, the equation of the helicoidal curve $$H(\theta)=\left(x(\theta),y(\theta)\right)$$ in cartesian coordinates is given by $\left\{ \begin{array}{l} x(\theta )=r_{p}(\theta )\cos (\theta )=Ae^{\theta \cot (\alpha )}\sin (\beta )\cos (\theta ) \\ y(\theta )=r_{p}(\theta )\sin (\theta )=Ae^{\theta \cot (\alpha )}\sin (\beta )\sin (\theta ) \\ z(\theta )=-Ae^{\theta \cot (\alpha )}\cos (\beta ) \end{array} \right. .$
We have already the mathematical model for the substructure of the shell. The model for the shell surface (that is, the walls of the shell) is still missing... | 2018-03-23 22:26:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.900848388671875, "perplexity": 259.434703806768}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00238.warc.gz"} |
http://mathhelpforum.com/algebra/18724-need-help.html | # Math Help - Need Help
1. ## Need Help
I cant find the solution to:
((x^6)(y^2))((x^3)(y^5)
2. Originally Posted by ceasar_19134
I cant find the solution to:
((x^6)(y^2))((x^3)(y^5)
just multiply the x's and multiply the y's. leave the answer as a product in terms of the two variables
recall, $a^m \times a^n = a^{m + n}$ | 2015-03-06 09:19:28 | {"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": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.8752756714820862, "perplexity": 2526.5831631363694}, "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-2015-11/segments/1424936466999.23/warc/CC-MAIN-20150226074106-00068-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.r-bloggers.com/2018/12/rstudio-pandoc-html-to-markdown/ | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
The knitr and rmarkdown packages are used in conjunction with pandoc to convert R code and figures to a variety of formats including PDF, and word. Here, I’m exploring how to convert HTML back to markdown format. This post came about when I was searching how to convert XML to markdown, which I still haven’t found an easy way to do. Pandoc is not the only way to convert HTML to markdown (see turndown, html2text)
Pandoc is packaged within RStudio and on Windows, the executables are located within Program Files/RStudio/bin/pandoc. The rmarkdown package contains wrapper functions for using pandoc within RStudio.
Here, I am trying to convert this example HTML page back to markdown using the function pandoc_convert. First, pandoc_convert requires an actual file which means it does not accept a quoted string of HTML code in its input argument.
The example html:
Enter a title, displayed at the top of the window.
Enter the main heading, usually the same as the title.
Be bold in stating your key points. Put them in a list:
The first item in your list
The second item; italicize key words
Improve your image by including an image.
Break up your page with a horizontal rule or two.
I saved the HTML example here as example.html.
html_page <- readLines("../../static/files/example.html")
We can print the object in R.
cat(html_page)
## Enter a title, displayed at the top of the window. Enter the main heading, usually the same as the title. Be bold in stating your key points. Put them in a list: The first item in your list The second item; italicize key words Improve your image by including an image. Add a link to your favorite Web site. Break up your page with a horizontal rule or two. Finally, link to another page in your own Web site. © Wiley Publishing, 2011
pandoc can convert between many different formats, and for markdown, it has multiple variants including the github flavored variant (for Github), and php markdown extra (the variant used by WordPress sites).
The safest variant to pick is markdown_strict which is the original markdown variant.
Pandoc requires the file path which in my case, is located in a different directory rather than my working directory.
library(rmarkdown)
file_path <- "../../static/files/example.html"
pandoc_convert(file_path, to = "markdown_strict")
Enter the main heading, usually the same as the title.
======================================================
Be **bold** in stating your key points. Put them in a list:
- The first item in your list
- The second item; *italicize* key words
Improve your image by including an image.
up your page with a horizontal rule or two.
------------------------------------------------------------------------
Notice that heading 1 is formatted with ==== rather than the # that RMarkdown seems to favor. We can require pandoc to use the # during the conversion by adding an argument.
pandoc_convert(file_path, to = "markdown_strict", options = c("--atx-headers"))
# Enter the main heading, usually the same as the title.
Be **bold** in stating your key points. Put them in a list:
- The first item in your list
- The second item; *italicize* key words
Improve your image by including an image.
up your page with a horizontal rule or two.
------------------------------------------------------------------------
© Wiley Publishing, 2011
Right now, the output is being piped to the console. A file can be created instead with:
pandoc_convert(file_path, to = "markdown_strict", output = "example.md")
Pandoc has a multitude of styling extensions for markdown variants, all listed on the manual page.
Pandoc ignores everything enclosed in . When converting from markdown to HTML, these comments are usually directly placed as is in the HTML document but the opposite does not seem to be true.
Lastly, this was tested using pandoc version 1.19.2.1. Pandoc 2.5 was released last month. | 2021-04-17 04:50:16 | {"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": 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": 0.20309194922447205, "perplexity": 3750.589535115691}, "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/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00352.warc.gz"} |
http://gasturbinespower.asmedigitalcollection.asme.org/Issue.aspx?issueID=936548&direction=P | 0
### Research Papers: Gas Turbines: Combustion, Fuels, and Emissions
J. Eng. Gas Turbines Power. 2017;140(3):031501-031501-9. doi:10.1115/1.4037914.
By modeling a multicomponent gas, a new source of indirect combustion noise is identified, which is named compositional indirect noise. The advection of mixture inhomogeneities exiting the gas-turbine combustion chamber through subsonic and supersonic nozzles is shown to be an acoustic dipole source of sound. The level of mixture inhomogeneity is described by a difference in composition with the mixture fraction. An n-dodecane mixture, which is a kerosene fuel relevant to aeronautics, is used to evaluate the level of compositional noise. By relaxing the compact-nozzle assumption, the indirect noise is numerically calculated for Helmholtz numbers up to 2 in nozzles with linear velocity profile. The compact-nozzle limit is discussed. Only in this limit, it is possible to derive analytical transfer functions for (i) the noise emitted by the nozzle and (ii) the acoustics traveling back to the combustion chamber generated by accelerated compositional inhomogeneities. The former contributes to noise pollution, whereas the latter has the potential to induce thermoacoustic oscillations. It is shown that the compositional indirect noise can be at least as large as the direct noise and entropy noise in choked nozzles and lean mixtures. As the frequency with which the compositional inhomogeneities enter the nozzle increases, or as the nozzle spatial length increases, the level of compositional noise decreases, with a similar, but not equal, trend to the entropy noise. The noisiest configuration is found to be a compact supersonic nozzle.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031502-031502-10. doi:10.1115/1.4037918.
The use of highly reactive hydrogen-rich fuels in lean premixed combustion systems strongly affects the operability of stationary gas turbines (GT) resulting in higher autoignition and flashback risks. The present study investigates the autoignition behavior and ignition kernel evolution of hydrogen–nitrogen fuel mixtures in an inline co-flow injector configuration at relevant reheat combustor operating conditions. High-speed luminosity and particle image velocimetry (PIV) measurements in an optically accessible reheat combustor are employed. Autoignition and flame stabilization limits strongly depend on temperatures of vitiated air and carrier preheating. Higher hydrogen content significantly promotes the formation and development of different types of autoignition kernels: More autoignition kernels evolve with higher hydrogen content showing the promoting effect of equivalence ratio on local ignition events. Autoignition kernels develop downstream a certain distance from the injector, indicating the influence of ignition delay on kernel development. The development of autoignition kernels is linked to the shear layer development derived from global experimental conditions.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031503-031503-10. doi:10.1115/1.4037824.
This article reports experiments carried out in the MICCA-spray combustor developed at EM2C laboratory. This system comprises 16 swirl spray injectors. Liquid n-heptane is injected by simplex atomizers. The combustion chamber is formed by two cylindrical quartz tubes allowing full optical access to the flame region and it is equipped with 12 pressure sensors recording signals in the plenum and chamber. A high-speed camera provides images of the flames and photomultipliers record the light intensity from different flames. For certain operating conditions, the system exhibits well defined instabilities coupled by the first azimuthal mode of the chamber at a frequency of 750 Hz. These instabilities occur in the form of bursts. Examination of the pressure and the light intensity signals gives access to the acoustic energy source term. Analysis of the phase fluctuations between the two signals is carried out using cross-spectral analysis. At limit cycle, large pressure fluctuations of 5000 Pa are reached, and these levels persist over a finite period of time. Analysis of the signals using the spin ratio indicates that the standing mode is predominant. Flame dynamics at the pressure antinodal line reveals a strong longitudinal pulsation with heat release rate oscillations in phase and increasing linearly with the acoustic pressure for every oscillation levels. At the pressure nodal line, the flames are subjected to large transverse velocity fluctuations leading to a transverse motion of the flames and partial blow-off. Scenarios and modeling elements are developed to interpret these features.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031504-031504-8. doi:10.1115/1.4037925.
The effect of confinement (flame–wall interactions) on the response of a turbulent, swirl-stabilized flame is experimentally examined, with a focus on the shape and structure of these flames. A series of three cylindrical combustors of 0.11, 0.15, and 0.19 m diameter are used to vary the degree of confinement experienced by the flame. Using CH* chemiluminescence images, the shape of the flame in each combustor is described. These images are then further analyzed and reveal marked similarities in the geometry and location of these flames in a defined “flame base” region near the combustor inlet. This similarity in location of the flame base leads to a similarity in the response of this portion of the flame to imposed oscillations. In particular, the phase of the fluctuations in this region is shown to be the same in each confinement. The nature of the fluctuations in the mean flame position is also shown to be similar in each confinement. These results indicate that the geometry of the flame in the base region is not a function of confinement and that the flames are responding to the same convective mechanisms, and in the same manner, in this region of the flame.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031505-031505-9. doi:10.1115/1.4037928.
The objective of this study is to understand the effects of fuel volatility on soot emissions. This effect is investigated in two experimental configurations: a jet flame and a model gas turbine combustor. The jet flame provides information about the effects of fuel on the spatial development of aromatics and soot in an axisymmetric, co-flow, laminar flame. The data from the model gas turbine combustor illustrate the effect of fuel volatility on net soot production under conditions similar to an actual engine at cruise. Two fuels with different boiling points are investigated: n-heptane/n-dodecane mixture and n-hexadecane/n-dodecane mixture. The jet flames are nonpremixed and rich premixed flames in order to have fuel conditions similar to those in the primary zone of an aircraft engine combustor. The results from the jet flames indicate that the peak soot volume fraction produced in the n-hexadecane fuel is slightly higher as compared to the n-heptane fuel for both nonpremixed and premixed flames. Comparison of aromatics and soot volume fraction in nonpremixed and premixed flames shows significant differences in the spatial development of aromatics and soot along the downstream direction. The results from the model combustor indicate that, within experiment uncertainty, the net soot production is similar in both n-heptane and n-hexadecane fuel mixtures. Finally, we draw conclusions about important processes for soot formation in gas turbine combustor and what can be learned from laboratory-scale flames.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031506-031506-10. doi:10.1115/1.4037960.
This article focuses on combustion instabilities (CI) driven by entropy fluctuations which is of great importance in practical devices. A simplified geometry is introduced. It keeps the essential features of an aeronautical combustion chamber (swirler, dilution holes, and outlet nozzle), while it is simplified sufficiently to ease the analysis (rectangular vane, one row of holes of the same diameter, no diffuser at the inlet of the chamber, and circular nozzle at the outlet). A large eddy simulation (LES) is carried out on this geometry and the limit cycle of a strong CI involving the convection of an entropy spot is obtained. The behavior of the instability is analyzed using phenomenological description and classical signal analysis. One shows that the system can be better described by considering two reacting zones: a rich mainly premixed flame is located downstream of the swirler and an overall lean diffusion flame is stabilized next to the dilution holes. In a second step, dynamic mode decomposition (DMD) is used to visualize, analyze, and model the complex phasing between different processes affecting the reacting zones. Using these data, a zero-dimensional (0D) modeling of the premixed flame and of the diffusion flame is proposed. These models provide an extended understanding of the combustion process in an aeronautical combustor and could be used or adapted to address mixed acoustic-entropy CI in an acoustic code.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):031507-031507-6. doi:10.1115/1.4037961.
The concept of the novel short helical combustor (SHC) was investigated in our previous work (Ariatabar et al., 2016, “Short Helical Combustor: Concept Study of an Innovative Gas Turbine Combustor With Angular Air Supply,” ASME J. Eng. Gas Turbines Power, 138(3), p. 031503 and Ariatabar et al., 2017, “Short Helical Combustor: Dynamic Flow Analysis in a Combustion System With Angular Air Supply,” ASME J. Eng. Gas Turbines Power, 139(4), p. 041505). Based on the insight gained from these previous investigations, we propose a generic design improvement to address the tremendous loss of initial angular momentum as well as inhomogeneous flow and temperature field at the outlet of the SHC. In the present paper, the main features of this design are introduced. It is shown that a three-dimensional shaping of the sidewalls, the dome, and the liners can effectively counteract the suboptimal interaction of the swirl flames with these surrounding walls. As a result, the flow at the outlet of the combustor features a high angular momentum and exhibits a uniform flow angle and temperature field. The insight gained from these generic investigations, and the resulting design optimization provides a useful framework for further industrial optimization of the SHC.
Commentary by Dr. Valentin Fuster
### Research Papers: Gas Turbines: Heat Transfer
J. Eng. Gas Turbines Power. 2017;140(3):031901-031901-13. doi:10.1115/1.4037870.
This paper presents a novel approach to real-time modeling of disk temperature distribution using proper orthogonal decomposition (POD). The method combines singular value decomposition (SVD) techniques with a series of low-order transfer functions to predict the disk's thermal response over a typical flight. The model uses only typically available full authority digital electronic control (FADEC) measurements to predict temperature with accuracy of ±30 K over the whole flight cycle. A Kalman filter has also been developed based on a single temperature measurement, and the location of the measurement has been assessed in order to select the most appropriate target for instrumentation. Points all around the front and back of the disk have been assessed, and the best practice result is found to be near the center of the disk neck. This represents a compromise between matching the fast dynamic response of the rim, with the slower dynamics of the cob. The new model has been validated against an independent flight simulation.
Commentary by Dr. Valentin Fuster
### Research Papers: Gas Turbines: Industrial and Cogeneration
J. Eng. Gas Turbines Power. 2017;140(3):032001-032001-15. doi:10.1115/1.4037962.
This paper presents the development of a simulation tool for modeling the transient behavior of micro-CHP (combined heat and power) systems, equipped with both thermal and electric storage units and connected with both electric and district heating grid (DHG). The prime mover (PM) considered in this paper is an internal combustion reciprocating engine (ICE), which is currently the only well-established micro-CHP technology. Different users, characterized by different demands of electric and thermal energy, both in terms of absolute value and electric-to-thermal energy ratio, are analyzed in this paper. Both summer and winter hourly trends of electric and thermal energy demand are simulated by using literature data. The results present a comprehensive energy analysis of all scenarios on a daily basis, in terms of both user demand met and energy share among system components. The transient response of the PM and the thermal energy storage (TES) is also analyzed for the two scenarios with the lowest and highest daily energy demand, together with the trend over time of the state of charge of both thermal and electric energy storage (EES).
Commentary by Dr. Valentin Fuster
### Research Papers: Gas Turbines: Oil and Gas Applications
J. Eng. Gas Turbines Power. 2017;140(3):032401-032401-10. doi:10.1115/1.4037963.
Statistical parametric methodologies are widely employed in the analysis of time series of gas turbine (GT) sensor readings. These methodologies identify outliers as a consequence of excessive deviation from a statistical-based model, derived from available observations. Among parametric techniques, the k–σ methodology demonstrates its effectiveness in the analysis of stationary time series. Furthermore, the simplicity and the clarity of this approach justify its direct application to industry. On the other hand, the k–σ methodology usually proves to be unable to adapt to dynamic time series since it identifies observations in a transient as outliers. As this limitation is caused by the nature of the methodology itself, two improved approaches are considered in this paper in addition to the standard k–σ methodology. The two proposed methodologies maintain the same rejection rule of the standard k–σ methodology, but differ in the portions of the time series from which statistical parameters (mean and standard deviation) are inferred. The first approach performs statistical inference by considering all observations prior to the current one, which are assumed reliable, plus a forward window containing a specified number of future observations. The second approach proposed in this paper is based on a moving window scheme. Simulated data are used to tune the parameters of the proposed improved methodologies and to prove their effectiveness in adapting to dynamic time series. The moving window approach is found to be the best on simulated data in terms of true positive rate (TPR), false negative rate (FNR), and false positive rate (FPR). Therefore, the performance of the moving window approach is further assessed toward both different simulated scenarios and field data taken on a GT.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032402-032402-9. doi:10.1115/1.4037964.
Anomaly detection in sensor time series is a crucial aspect for raw data cleaning in gas turbine (GT) industry. In addition to efficiency, a successful methodology for industrial applications should be also characterized by ease of implementation and operation. To this purpose, a comprehensive and straightforward approach for detection, classification, and integrated diagnostics of gas turbine sensors (named DCIDS) is proposed in this paper. The tool consists of two main algorithms, i.e., the anomaly detection algorithm (ADA) and the anomaly classification algorithm (ACA). The ADA identifies anomalies according to three different levels of filtering based on gross physics threshold application, intersensor statistical analysis (sensor voting), and single-sensor statistical analysis. Anomalies in the time series are identified by the ADA, together with their characteristics, which are analyzed by the ACA to perform their classification. Fault classes discriminate among anomalies according to their time correlation, magnitude, and number of sensors in which an anomaly is contemporarily identified. Results of anomaly identification and classification can subsequently be used for sensor diagnostic purposes. The performance of the tool is assessed in this paper by analyzing two temperature time series with redundant sensors taken on a Siemens GT in operation. The results show that the DCIDS is able to identify and classify different types of anomalies. In particular, in the first dataset, two severely incoherent sensors are identified and their anomalies are correctly classified. In the second dataset, the DCIDS tool proves to be capable of identifying and classifying clustered spikes of different magnitudes.
Commentary by Dr. Valentin Fuster
### Research Papers: Gas Turbines: Structures and Dynamics
J. Eng. Gas Turbines Power. 2017;140(3):032501-032501-8. doi:10.1115/1.4037861.
Gas turbines are fitted with rolling element bearings, which transfer loads and supports the shafts. The interaction between the rotating and stationary parts in the bearing causes a conversion of some of the power into heat, influencing the thermal behavior of the entire bearing chamber. To improve thermal modeling of bearing chambers, this work focused on modeling of the heat generated and dissipated around the bearings, in terms of magnitude and location, and the interaction with the components/systems in the bearing chamber. A thermal network (TN) model and a finite element (FE) model of an experimental high-pressure shaft ball bearing and housing were generated and a comparison to test rig results have been conducted. Nevertheless, the purpose of the thermal matching process that focused on the FE model and experimental data is to provide a template for predicting temperatures and heat transfers for other bearing models. The result of the analysis shows that the predictions of the TN are considerate, despite the simplifications. However, lower relative errors were obtained in the FE model compared to the TN model. For both methods, the highest relative error is seen to occur during transient (acceleration and deceleration). This observation highlights the importance of boundary conditions and definitions: surrounding temperatures, heat split and the oil flow, influencing both the heat transfer and heat generation. These aspects, incorporated in the modeling and benchmarked with experimental data, can help facilitate other related cases where there is limited or no experimental data for validation.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032502-032502-9. doi:10.1115/1.4037919.
The impact of sealing equipment on the stability of turbomachineries is a crucial topic because the power generation market is continuously requiring high rotational speed and high performance, leading to the clearance reduction in the seals. The accurate characterization of the rotordynamic coefficients generated by the seals is pivotal to mitigate instability issues. In the paper, the authors propose an improvement of the state-of-the-art one-control volume (1CV) bulk-flow model (Childs and Scharrer, 1986, “An Iwatsubo-Based Solution for Labyrinth Seals: Comparison to Experimental Results,” ASME J. Eng. Gas Turbines Power, 108(2), pp. 325–331) by considering the energy equation in the steady-state problem. Thus, real gas properties can be evaluated in a more accurate way because the enthalpy variation, expected through the seal cavities, is evaluated in the model. The authors assume that the enthalpy is not a function of the clearance perturbation; therefore, the energy equation is considered only in the steady-state problem. The results of experimental tests of a 14 teeth-on-stator (TOS) labyrinth seal, performed in the high-pressure seal test rig owned by GE Oil&Gas, are presented in the paper. Positive and negative preswirl ratios are used in the experimental tests to investigate the effect of the preswirl on the rotordynamic coefficients. Overall, by considering the energy equation, a better numerical estimation of the rotordynamic coefficients for the tests with the negative preswirl ratio has been obtained (as it results from the comparison with the experiments). Finally, the numerical results are compared with a reference bulk-flow model proposed by Thorat and Childs (2010, “Predicted Rotordynamic Behavior of a Labyrinth Seal as Rotor Surface Speed Approaches Mach 1,” ASME J. Eng. Gas Turbines Power, 132(11), p. 112504), highlighting the improvement obtained.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032503-032503-9. doi:10.1115/1.4037920.
Engine oil-lubricated (semi) floating ring bearing ((S)FRB) systems in passenger vehicle turbochargers (TC) operate at temperatures well above ambient and must withstand large temperature gradients that can lead to severe thermomechanical induced stresses. Physical modeling of the thermal energy flow paths and an effective thermal management strategy are paramount to determine safe operating conditions ensuring the TC component mechanical integrity and the robustness of its bearing system. The paper details a model to predict the pressure and temperature fields and the distribution of thermal energy flows in a bearing system. The impact of lubricant supply conditions, bearing film clearances, and oil supply grooves is quantified. Either a low oil temperature or a high supply pressure increases the generated shear power. Either a high supply pressure or a large clearance allows more flow through the inner film and draws more heat from the hot journal, thought it increases the shear drag power as the oil viscosity remains high. Nonetheless, the peak temperature of the inner film is not influenced by the changes on the way the oil is supplied into the film as the thermal energy displaced from the hot shaft into the film is overwhelming. Adding axial grooves on the inner side of the (S)FRB improves its dynamic stability, albeit increasing the drawn oil flow as well as the drag power and heat from the shaft. The results identify a compromise between different parameters of groove designs thus enabling a bearing system with a low power consumption.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032504-032504-9. doi:10.1115/1.4037865.
Several experimental apparatuses have been designed in the past to evaluate the effectiveness of under-platform dampers. Most of these experimental setups allow to measure the overall damper efficiency in terms of reduction of vibration amplitude in turbine blades. The experimental data collected with these test rigs do not increase the knowledge about the damper dynamics, and therefore, the uncertainty on the damper behavior remains a big issue. In this paper, a different approach to evaluate the damper–blade interaction has been put forward. A test rig has been purposely designed to accommodate a single blade and two under-platform dampers. One side of each damper is in contact with a ground support specifically designed to measure two independent forces on the damper. In this way, both the normal and the tangential force components in the damper–blade contact can be inferred. Damper kinematics is rebuilt by using the relative displacement measured between damper and blade. This paper describes the concept behind the new approach, shows the details of the new test rig, and discusses the blade frequency response from a new point of view.
Topics: Dampers , Blades , Kinematics
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032505-032505-9. doi:10.1115/1.4037965.
Commentary by Dr. Valentin Fuster
### Research Papers: Gas Turbines: Turbomachinery
J. Eng. Gas Turbines Power. 2017;140(3):032601-032601-9. doi:10.1115/1.4037871.
Gas turbine aero-engines employ fast rotating shafts that are supported by bearings at several axial locations along the engine. Due to extreme load and heat, oil is injected to the bearings to aid lubrication and cooling. The oil is then shed to the bearing chamber before it is extracted out by a scavenge pump. Scavenging oil from the bearing chamber is challenging due to high windage induced by the fast rotating shafts as well as the two-phase nature of the flow. A deep sump has been found to increase scavenge performance due to its ability to shelter the pooled oil from the bulk rotating air flow thus minimizing two-phase mixing. However, in many cases, a deep sump is not an option due to conflicting space requirements. The space limitation becomes more stringent with higher bypass ratio engines as the core becomes smaller. Therefore, it is imperative to have a high performing shallow sump. However, shape modification of a shallow sump is too constrained due to limited space and, therefore, has minimal impact on the scavenge performance. This research presents several alternative concepts to improve scavenge performance of a generic baseline shallow sump by augmenting it with attachments or inserts. These augmentations attempt to exploit two known mechanisms for reducing the residence volume: momentum reduction and sheltering. The experimental results show that some augmentations are able to reduce the residence volume of a shallow sump by up to 50% or more in some cases.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032602-032602-8. doi:10.1115/1.4037912.
Formation of thin liquid films on steam turbine airfoils, particularly in last stages of low-pressure (LP) steam turbines, and their breakup into coarse droplets is of paramount importance to assess erosion of last stage rotor blades given by the impact of those droplets. An approach for this problem is presented in this paper: this includes deposition of liquid water mass and momentum, film mass and momentum conservation, trailing edge breakup and droplets Lagrangian tracking accounting for inertia and drag. The use of thickness-averaged two-dimensional (2D) equations in local body-fitted coordinates, derived from Navier–Stokes equations, makes the approach suitable for arbitrary curved blades and integration with three-dimensional (3D) computational fluid dynamics (CFD) simulations. The model is implemented in the in-house solver MULTI3, which uses Reynolds-averaged Navier–Stokes equations $κ$$ω$ model and steam tables for the steam phase and was previously modified to run on multi-GPU architecture. The method is applied to the last stage of a steam turbine in full and part load operating conditions to validate the model by comparison with time-averaged data from experiments conducted in the same conditions. Droplets impact pattern on rotor blades is also predicted and shown.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032603-032603-7. doi:10.1115/1.4037913.
The effect of compressor fouling on the performance of a gas turbine has been the subject of several papers; however, the goal of this paper is to address a more fundamental question of the effect of fouling, which is the onset of unstable operation of the compressor. Compressor fouling experiments have been carried out on a test rig refitted with TJ100 small jet engine with centrifugal compressor. Fouling on the compressor blade was simulated with texturized paint with average roughness value of 6 μm. Compressor characteristic was measured for both the clean (baseline) and fouled compressor blades at several rotational speeds by throttling the engine with variable exhaust nozzle. A Greitzer-type compression system model has been applied based on the geometric and performance parameters of the TJ100 small jet engine test rig. Frequency of plenum pressure fluctuation, the mean disturbance flow coefficient, and pressure-rise coefficient at the onset of plenum flow field disturbance predicted by the model was compared with the measurement for both the baseline and fouled engine. Model prediction of the flow field parameters at inception of unstable operation in the compressor showed good agreement with the experimental data. The results proved that used simple Greitzer model is suitable for prediction of the engine compressor unstable behavior and prediction of the mild surge inception point for both the clean and the fouled compressor.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032604-032604-9. doi:10.1115/1.4037906.
Darrieus vertical axis wind turbines (VAWTs) have been recently identified as the most promising solution for new types of applications, such as small-scale installations in complex terrains or offshore large floating platforms. To improve their efficiencies further and make them competitive with those of conventional horizontal axis wind turbines, a more in depth understanding of the physical phenomena that govern the aerodynamics past a rotating Darrieus turbine is needed. Within this context, computational fluid dynamics (CFD) can play a fundamental role, since it represents the only model able to provide a detailed and comprehensive representation of the flow. Due to the complexity of similar simulations, however, the possibility of having reliable and detailed experimental data to be used as validation test cases is pivotal to tune the numerical tools. In this study, a two-dimensional (2D) unsteady Reynolds-averaged Navier–Stokes (U-RANS) computational model was applied to analyze the wake characteristics on the midplane of a small-size H-shaped Darrieus VAWT. The turbine was tested in a large-scale, open-jet wind tunnel, including both performance and wake measurements. Thanks to the availability of such a unique set of experimental data, systematic comparisons between simulations and experiments were carried out for analyzing the structure of the wake and correlating the main macrostructures of the flow to the local aerodynamic features of the airfoils in cycloidal motion. In general, good agreement on the turbine performance estimation was constantly appreciated.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032605-032605-7. doi:10.1115/1.4037926.
The cavities between the rotating compressor disks in aero-engines are open, and there is an axial throughflow of cooling air in the annular space between the center of the disks and the central rotating compressor shaft. Buoyancy-induced flow occurs inside these open rotating cavities, with an exchange of heat and momentum between the axial throughflow and the air inside the cavity. However, even where there is no opening at the center of the compressor disks—as is the case in some industrial gas turbines—buoyancy-induced flow can still occur inside the closed rotating cavities. The closed cavity also provides a limiting case for an open cavity when the axial clearance between the cobs—the bulbous hubs at the center of compressor disks—is reduced to zero. Bohn and his co-workers at the University of Aachen have studied three different closed-cavity geometries, and they have published experimental data for the case where the outer cylindrical surface is heated and the inner surface is cooled. In this paper, a buoyancy model is developed in which it is assumed that the heat transfer from the cylindrical surfaces is analogous to laminar free convection from horizontal plates, with the gravitational acceleration replaced by the centripetal acceleration. The resulting equations, which have been solved analytically, show how the Nusselt numbers depend on both the geometry of the cavity and its rotational speed. The theoretical solutions show that compressibility effects in the core attenuate the Nusselt numbers, and there is a critical Reynolds number at which the Nusselt number will be a maximum. For the three cavities tested, the predicted Nusselt numbers are in generally good agreement with the measured values of Bohn et al. over a large range of Raleigh numbers up to values approaching 1012. The fact that the flow remains laminar even at these high Rayleigh numbers is attributed to the Coriolis accelerations suppressing turbulence in the cavity, which is consistent with recently published results for open rotating cavities.
Commentary by Dr. Valentin Fuster
J. Eng. Gas Turbines Power. 2017;140(3):032606-032606-8. doi:10.1115/1.4037921.
The concept of morphing geometry to control and stabilize the flow has been proposed and applied in several aeronautic and wind turbine applications. We studied the effect of a similar passive system applied on an axial fan blade, analyzing potential benefits and disadvantages associated to the passive coupling between fluid and structure dynamics. The present work completes a previous study made at the section level, giving a view also on the three-dimensional (3D) effects. We use the numerical computation to simulate the system, which defines a complex fluid–structure interaction (FSI) problem. In order to do that, an in-house finite element (FE) solver, already used in the previous study, is applied to solve the coupled dynamics.
Commentary by Dr. Valentin Fuster | 2018-10-18 12:24:22 | {"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": 0.40468502044677734, "perplexity": 1672.0694211226707}, "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-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00554.warc.gz"} |
https://gamedev.stackexchange.com/questions/63362/rotating-3d-object-around-the-center | # rotating 3D object around the center
I have object moving from A to B on x-axis and there is no translation of object apart from it. Now, while moving, i want to rotate it around y-axis and the motion should change accordingly, i mean if i rotate it right when moving from -x to +x axis, it should move towards near plane. I have variable in gltranslatef which is modified in the loop after that i have glscalef to scale whole object which is made of hierarchical structure. Now i tried following code to achieve the expected result but its not working properly
glTranslatef(move, 0, 0);
// If I comment these 3 lines, it does not affect the output
glTranslatef(-move, 0, 0);
glRotatef(rotate,0,1,0);
glTranslatef(move, 0, 0);
glScalef(0.2, 0.2, 1.0);
Transformations (scale,rotation, translation) are applied in the reverse order of their invocation and transformations do not change the object's position or orientation but the coordinates system that will be used to render it. Imagine the 3D coordinates system as 3 axis ( X,Y,Z ) perpendicular to each other. In case your object lies along the X axis, if you rotate the coordinates around the Y axis, your object will come towards the near or far plane because it lies in a position on the X axis. If you want to rotate the object around itself, you should place it in the center of the coordinates system.
glTranslatef(move, 0, 0); | 2021-03-06 05:24:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37881752848625183, "perplexity": 426.16356382411294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374391.90/warc/CC-MAIN-20210306035529-20210306065529-00117.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/explore-the-convergence-behavior-of-the-newton-raphson-method-when-nding-all-possibly-comp-q3183548 | ## Another Question Sorry I need this one ASAP too! ILL RATE
Explore the convergence behavior of the Newton-Raphson method when nding all (possibly complex) roots of the polynomial $$f(x)=x^3-2x+2$$ Consider the region R of the complex plane dened by R = $${ (a,b) | -2 =< a =< 2 and -2 =< b =< 2}$$ : Make a grid of initial conditions in R, for example with a = linspace(-2,2,Na) and b = linspace(-2,2,Nb). Run your Newton-Raphson code with each initial condition, x = a(m) + b(n)*i, until it nds a root or exceeds the maximum number of iterates, Nmax. Using the results, color code each point in the grid (one color for each of the three roots, and one indicating nonconvergence) and make a false-color plot of the basins of attraction for each root. Make a separate plot showing the iterates, N, as a function of the initial guess. Hint: use the Matlab command pcolor together with meshgrid to display your results. Make the plots with Na and Nb as large as possible. Start with small values, say Na = Nb = 100, and raise higher to see what your system can do.
Given: Newton-Raphson Code is:
function [p,N]=newton0(func,dfunc,p0,Nmax,TOL)
% [p,N]=newton0(func,dfunc,p0,Nmax,TOL)
N=1; % initialize counter
while N < Nmax
% fp0=feval(func,p0);
% dfp0=feval(dfunc,p0);
fp0=func(p0);
dfp0=dfunc(p0);
p=p0-fp0/dfp0;
r=(p-p0)/p;
fprintf(1,'p0 = %13.12e f(p0) = %13.12e p = %16.12e r =%13.12e\n',p0,fp0,p,r);
if abs(r) < TOL
return
end
N=N+1;
p0=p;
end
disp(' ')
disp(['Newton''s method failed to converge after ' num2str(Nmax) ' steps.'])
Please try and show how to do each part or at least give me an idea of how to set it up! | 2013-05-19 03:52:55 | {"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": 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": 0.4973372220993042, "perplexity": 1433.2901480147368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696383218/warc/CC-MAIN-20130516092623-00071-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.experts-exchange.com/questions/23889285/Bulk-INSERT-with-different-field-types.html | Solved
# Bulk INSERT with different field types?
Posted on 2008-11-09
450 Views
Hello,
I need to bulk insert a csv file with the following format:
Date,Open,High,Low,Close,Delta,Signal,SystemIdent
They have the following types
Date,Integer,Integer,Integer,Integer,Integer,Integer,Integer,
However with the code below I get an conversion error.
USE [Monday]
GO
/****** Object: StoredProcedure [dbo].[Bulk_Insert] Script Date: 11/09/2008 17:10:35 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
ALTER procedure [dbo].[Bulk_Insert] as BULK insert Data From 'c:\structure\test.txt' with (Fieldterminator =',', ROWTERMINATOR = '\n')
0
Question by:Schuttend
LVL 51
Expert Comment
Could be a number of conversion problems - likely to be the date column in terms of how it is being formatted and then natively understood in the database.
You have a few choices, either use an openrowset type function where you can cast/convert the incoming value, or, bulk insert into a staging table with varchars., then validate and load the "real" table from the staging table.
0
LVL 51
Expert Comment
Could also be the first line has column headers ? in which case use the FIRSTROW=2 qualifier
Got some sample data ?
0
Author Comment
Hello,
Please find attached sample data pic. Currently the data is imported all as string(varchar 50). But is not very good, because I have to convert data fields each time for calculations.
Regards,
Dennis
Carl3.jpg
0
Author Comment
sorry, wrong pic....
cast3.jpg
0
LVL 51
Expert Comment
OK, when going from DATA to TEMPDATA, then use convert(datetime, date,101)
Looks familiar - have we done something with this data before ?
0
LVL 51
Expert Comment
could also do something like :
SELECT convert(datetime,col1,101),col2,col3,col4,col5,col6,col7,case when isnumeric(col8) > 0 then convert(decimal(18,5),col8) else 0 end
FROM Openrowset(Bulk 'C:\ee\test.txt', formatfile = 'c:\ee\test.fmt.txt', firstrow=2) As A
where the format file is per attached... (have not considered quote encapsulation)...
test.fmt.txt
0
Author Comment
Mark,
What is your opinion about formatting everything first as floats. Import data and thereafter insert it into a destination table? This seems to work, but what is the negative aspect of using floats?
Regardsn
Dennis
0
LVL 51
Expert Comment
Well, floats are an approximation of a number, it is not a native number per se, so instead of 123.360 you can get 123.355555555559 in that case it rounds OK, but in other cases it might not. So, if the numbers do have a quantifiable scal (ie decimals) then far better to use something like decimal(18,6) ie toal length 18 digits, 6 of which are decimals...
0
LVL 51
Expert Comment
s/b scale not scal
0
Author Comment
I have tried to format fields even with decimal (38,38), but Delta value and Signal value will not go in there. Would you mind take a look at the attached import file what is going wrong?
I though using float would be a option (importing does work), but as you said, it's not accurate.
test.txt
0
LVL 51
Accepted Solution
Well, delta and signal are both float - in fact signal has been exponeniated in some cases (in the raw data - you can see a vale 7.74743856259124E-02 )
which means, probably best to import those two as float, and can even do formatting for each column along the way (it does retrieve all rows):
SELECT convert(datetime,col1,101) as date,convert(decimal(18,5),col2) as [open],convert(decimal(18,5),col3) as high,convert(decimal(18,5),col4) as low,convert(decimal(18,5),col5) as [close],convert(decimal(18,5),convert(float,col6)) as delta,convert(decimal(24,18),convert(float,col7)) as signal,case when isnumeric(col8) > 0 then convert(decimal(18,5),col8) else 0 end as systemident
FROM Openrowset(Bulk 'C:\ee\test.txt', formatfile = 'c:\ee\test.fmt.txt', firstrow=1) A
0
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https://engineering.stackexchange.com/questions/7228/how-to-increase-the-air-flow-rate-through-a-dust-separating-cyclone | # How to increase the air flow rate through a dust-separating cyclone?
I'm making a dust extractor for a desktop CNC milling machine. My design process was very crude: I looked at some DIY dust-separating cyclones, and it seemed like they just work without any advanced calculations, so I visually copied some of those designs at a smaller scale, sat a large PC cooling fan over the top, and put the whole thing in a coffee can:
My reasoning was that (1) the 18W fan blows out a lot of air, (2) the only place for that air to come from is through the hole at the top of the cone, and (3) the only way air can get into the sealed cone / dust receptacle (the bottom can) is through the tube at the side. So, it would have to suck a reasonable amount of air through that red pipe.
It does work to some extent (video), but not as well as I hoped. You can see that it barely picks up sawdust through a 1-inch hose, and ideally I'd like it to draw in dust at a range of 20-50mm or so-- I can't position the inlet any closer than that to the point where the dust is created. Plus, I'd like to fit a dust filter over the exhaust hole, which will reduce the air flow even more.
My question is basically "how can I make it suck harder?". But more specifically, how do I start to analyse what's happening, and what determines the amount of air pulled in?
Most Google references are either way over my head, or talk very specifically about pumps in volute casings; obviously I removed the volute casing from my fan in order to fit it in the enclosure. If I understand correctly, the point of the volute is to make the fast narrow stream exiting the fan into a slower, wider stream at higher pressure (why?), and I guess that something similar happens in my can:
A) incoming air spirals down; dust is thrown out and drops through bottom of cone
B) clean air spirals up through fan inlet
C) fast air exits fan circumference
D) ...and expands into volume below, losing speed and increasing pressure
E) higher-pressure air blown out into atmosphere
Stage C/D is the part I'm most uncertain about. The rest of the design comes from recipes, but I've changed the way air exits the fan and would like to understand how this might affect things.
### Edit 4 Feb 2016:
Per @ericnutsch's suggestion, I made rough measurements of the air flowing into stages A and C, and got a figure of about 0.009 m3min-1 at both points. I also found the manufacturer's data for the fan, which suggests I should be able to do better-- the stated max is 1.09 m3min-1. For comparison, my crappy handheld vacuum scores around 0.024 m3min-1. I've ordered a pressure sensor to investigate further...
• You seem to want roughly shop vac performance, but 18 W isn't even close for that. – Olin Lathrop Feb 2 '16 at 17:54
• I'm aiming for Dustbuster performance... – bobtato Feb 2 '16 at 18:39
• Or just bolt a Dyson to the side of your system :-) – Carl Witthoft Feb 2 '16 at 19:17
• Air goes into the cyclone, through the top of cyclone, through fan and exits can through whole nbear bottom? If not can you draw the airflow (squiggly line through your drawing would be fine) – mart Feb 3 '16 at 7:24
• @mart, that's exactly right. I've added a diagram, and drawing D is the one with a question mark on it. I've drawn what I hope is happening, but it might actually be a mass of energy-wasting turbulence. – bobtato Feb 3 '16 at 14:01
If your cyclone design is correctly separating particles without a major pressure drop; it is working satisfactory and its influence on the flow rate is minimal. If you increase the capacity of your fan and separation starts failing, you may have to modify your design or scale it up.
If you do not have a high enough flow rate, you need to get a larger fan. In addition to this you need to make sure you land at a good location on that fans "fan curve". A fan curve is the relationship between pressure and flowrate. Many fans are rated in maximum pressure (suction is different, but related if you cant get that data) and maximum flow rate. Maximum flow rate occurs at zero pressure drop and maximum pressure occurs at zero flow rate. So in your design you will need to do some testing to make sure the fan you select lands somewhere nicely in the middle.
Some fans that may work well for the application:
Dayton Blowers
Motorized Impellers
• I think the relation between pressure and flowrate is where my mental gears are sticking, in that it's not clear whether I need my vacuum nozzle to have low pressure or high flow rate. Is it roughly true to say that I need the air flow to move air through the system, but without a pressure difference the air flow will fail when it tries to do any work (like pulling dust through the hose)? – bobtato Feb 3 '16 at 13:57
• You first need to determine what your desired flow rate is at the intake. Air flow is what does the work; pressure drop is just required to achieve this flow. Dust will be a very minimal increase in pressure drop; especially where you have a vortex instead of a paper filter. You can get a feel for the flow rate you need by finding a vacuum cleaner you feel has the right flow rate and have it suck air out of a trash bag that has been inflated in a box. The box lets you easily calculate the volume (LxWXH) and by using a stopwatch you can determine the flow rate. – ericnutsch Feb 3 '16 at 16:12
TL, DR You need a stronger and different fan. Try a vac.
What you need to understand about fans
Fans are characterised by their power and their flow-pressure curve. Power is pressure times volume flow. Actualy, that's wrong but it's still useful. You fan delivers a certain and has to overcome a pressure difference. Less pressure - more flow and vice versa. Every system has a pressure drop that rises with flow. The flow will be determined by where the curve of the fan and the system curve meet. Learn more here.
Your system curve (that you don't know) and fan curve meet at 0,009m³/min and 50mm H2O pressure. Take the diagram, draw a straight line through this point and the origin. Pretty steep. The actual sytem curve is somewhere to the left of this line. This tells us that you need a fan that delives a stronger pressure.
What you need to do is IMO attach a vac. cleaner to suction side of your cyclone and try this at different settings.
What you need to understand about cyclones
There's a relationship between how effective a cyclone is and the pressure drop, with more pressure drop meaning more effective separation. Of course cyclones can be designed more or less efficient for a certain duty but with a given cyclone this relationship holds. Pressure drop of the cyclone depends on flow:
More flow -> more speed in cyclone -> stronger centrifugal force to overcome friction (also more pressure drop)
I can think of two ways you might increase airflow in this case - reduce the resistance to airflow or increase the speed of your fan.
For the first case, I'd look at the resistance in the ducts of your system. The 1" exit pipe seems small; most commercial blowers are about 4" in diameter. It should be noted that the resistance to flow in a pipe is proportional to $\frac{1}{D^4}$ of the pipe (airway resistance), so the performance does not scale directly. Small reductions in diameter can have big effects on the flow resistance, AKA back pressure or pressure drop.
You could also speed up your fan. However, there are certain combinations of flow and resistance where a particular fan likes to operate. If you change the speed, you could be improving or reducing the fan efficiency. | 2021-04-17 19:45:51 | {"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": 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": 0.5677431225776672, "perplexity": 994.2985302460686}, "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/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00398.warc.gz"} |
https://energyeducation.ca/encyclopedia/Tax_exemption | # Tax exemption
A tax exemption reduces the total tax owed, because a person, or particular source of money is exempt (not counted toward the total).[1] For example, in many countries low-income earners are exempt from paying income tax because paying that tax would be very hard on their finances.[2] Likewise, tax exemptions can be a way of encouraging (or discouraging) particular behaviour, see Pigouvian tax.
## Tax Deduction
A tax deduction reduces the overall amount of income that the government is considers when determining the amount owed in taxes.[3] If a tax payer has a taxable income of $50,000 and they claim a deduction of$5,000 then the amount of taxable income will fall to \$45,000.
In the U.S, married couples can claim a "standard" deduction which reduces their tax liability, if the family has dependent children, they can claim additional deductions.[4]
Governments can offer either exemptions or deductions based on what activities they want people to participate in. In Canada, many religious and other charities are exempt from taxation, this helps to encourage people contributing to charitable work.[5]
## References
1. Investopedia. "Tax Exempt." [Online], Available: http://www.investopedia.com/terms/t/tax_exempt.asp [Aug 23, 2016]. | 2019-02-20 01:12:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19546011090278625, "perplexity": 10762.805734266778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247494125.62/warc/CC-MAIN-20190220003821-20190220025821-00345.warc.gz"} |
https://online.stat.psu.edu/stat200/book/export/html/221 | 11.1 - Reviews
11.1 - Reviews
In this lesson you may need to use Minitab Express to construct frequency tables or two-way contingency tables. We'll start by reviewing these procedures. You will need to construct probability distribution plots for chi-square distributions. In earlier lessons we constructed probability distribution plots for z, t, and F distributions; the procedure is similar for a chi-square distribution. We will also review conditional probabilities and the term independence in this section.
11.1.1 - Frequency Table
11.1.1 - Frequency Table
The following example was first presented in Lesson 2.1.1.2.1.
It uses following data set (from College Board):
MinitabExpress – Frequency Tables
To create a frequency table in Minitab Express:
1. Open the data set
2. On a PC: In the menu bar select STATISTICS > Describe > Tally
3. On a Mac: In the menu bar select Statistics > Summary Statistics > Tally
4. Double click the variable Region in the box on the left to insert the variable into the Variable box
5. Under Statistics, check Counts and Percents
6. Click OK
This should result in the following frequency table:
Tally
Region Count Percent
ENC 5 9.8039%
ESC 4 7.8431%
MA 3 5.8824%
MTN 8 15.6863%
NE 6 11.7647%
PAC 5 9.8039%
SA 9 17.6471%
WNC 7 13.7255%
WSC 4 7.8431%
N= 51
Video Walkthrough
Select your operating system below to see a step-by-step guide for this example.
11.1.2 - Two-Way Contingency Table
11.1.2 - Two-Way Contingency Table
Recall from Lesson 2.1.2 that a two-way contingency table is a display of counts for two categorical variables in which the rows represented one variable and the columns represent a second variable. The starting point for analyzing the relationship between two categorical variables is to create a two-way contingency table. When one variable is obviously the explanatory variable, the convention is to use the explanatory variable to define the rows and the response variable to define the columns; this is not a hard and fast rule though.
MinitabExpress – Constructing a Two-Way Contingency Table
1. Open the data set:
2. On a PC: Select STATISTICS > Cross Tabulation and Chi-square
On a Mac: Select Statistics > Tables > Cross Tabulation and Chi-Square
3. Select Raw data (categorical variable) from the drop down menu
4. Double click the variable Smoke Cigarettes in the box on the left to insert the variable into the Rows box
5. Double click the variable Biological Sex in the box on the left to insert the variable into the Columns box
6. Click OK
This should result in the two-way table below:
Tabulated Statistics: Smoke Cigarettes, Biological Sex
Rows: Smokes Cigaretes | Columns: Biological Sex
Female Male All
No 120 89 209
Yes 7 10 17
All 127 99 226
Cell Contents: Count
Video Walkthrough
Select your operating system below to see a step-by-step guide for this example.
11.1.3 - Probability Distribution Plots
11.1.3 - Probability Distribution Plots
In previous lessons you have constructed probabilities distribution plots for normal distributions, binomial distributions, and $t$ distributions. This week you will use the same procedure to construct a probability distribution plot for the chi-square distribution.
MinitabExpress – Constructing a Probability Distribution Plot
Chi-square tests of independence are always right-tailed tests. Let's find the area of a chi-square distribution with 1 degree of freedom to the right of $\chi^2 = 1.75$. In other words, we're looking up the $p$ value associated with a chi-square test statistic of 1.75.
1. On a PC: from the menu select STATISTICS > Distribution Plot
On a Mac: from the menu select Statistics > Probability Distributions > Distribution Plot
2. Select Display Probability
3. For Distribution select Chi-Square
4. For Degrees of freedom enter 1
5. Select A specified X value
6. Select Right tail
7. For X value enter 1.75
This should result in the following output:
Video Walkthrough
Select your operating system below to see a step-by-step guide for this example.
11.1.4 - Conditional Probabilities and Independence
11.1.4 - Conditional Probabilities and Independence
In Lesson 2 you were introduced to conditional probabilities and independent events. These definitions are reviewed below along with some examples.
Recall that if events A and B are independent then $P(A) = P(A \mid B)$. In other words, whether or not event B occurs does not change the probability of event A occurring.
Conditional Probability
The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol $\mid$ which is read as "given."
For example, $P(A\mid B)$ is read as "Probability of A given B."
Independent Events
Unrelated events. The outcome of one event does not impact the outcome of the other event.
Example: Queens & Hearts
If a card is randomly drawn from a standard 52-card deck, the probability of the card being a queen is independent from the probability of the card being a heart. If I tell you that a randomly selected card is a queen, that does not change the likelihood of it being a heart, diamond, club, or spade.
Using a conditional probability to prove this:
$P(Queen) = \dfrac{4}{52}=0.077$
$P(Queen \mid Heart) = \dfrac {1}{13} = 0.077$
Example: Gender and Pass Rate
Data concerning two categorical variables can be displayed in a contingency table.
Pass Did Not Pass Total Men 6 9 15 Women 10 15 25 Total 16 24 40
If gender and passing are independent, then the probability of passing will not change if a case's gender is known. This could be written as $P(Pass) = P(Pass \mid Man)$.
$P(Pass) = \dfrac{16}{40} = 0.4$
$P(Pass \mid Man) = \dfrac{6}{15}=0.4$
In this sample, gender and passing are independent.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | 2020-02-22 20:01:23 | {"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": 0.6755285859107971, "perplexity": 1410.0344108322663}, "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/1581875145713.39/warc/CC-MAIN-20200222180557-20200222210557-00023.warc.gz"} |
https://math.stackexchange.com/questions/2897483/how-to-show-that-left-fracmn1m1-rightm1-nm | How to show that: $\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$
If m and n are positive integers then show that:$$\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$$I am new in this Course.So i can't able to think how i start a inequalities question by looking it's pattern.Can anyone help me to explain that Inequalities and Thanks in advance.
• Note that $n^m=n^m\cdot 1^1$. And the inequality should not be strict. The equality case is $(m,n)=(1,1)$. – Batominovski Aug 28 '18 at 17:01
• Can you explain me what the equality case is (m,n)=(1,1) mean by.Sorry if the question is irrelevant.I couldn't understand it.please explain it further more details @Batominovski – emonHR Aug 28 '18 at 17:07
• @mdemon when $m=1,n=1$, the LHS is $(2/2)^2$ and the RHS is $1^1$, which are both $1$ – John Glenn Aug 28 '18 at 17:11
• @mdemon Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – user Sep 17 '18 at 20:20
• Oops! i forgot @gimusi Thanks for mention it – emonHR Sep 20 '18 at 6:21
As an alternative by induction we have
• Base case
$$n=1 \implies \left(\frac{m+1}{m+1}\right)^{m+1} \ge 1^m$$
• Induction step, assume $\left(\frac{mn+1}{m+1}\right)^{m+1} \ge n^m$ true we need to show that $\left(\frac{m(n+1)+1}{m+1}\right)^{m+1} \ge (n+1)^m$ then
$$\left(\frac{m(n+1)+1}{m+1}\right)^{m+1}=\left(\frac{mn+1}{m+1}+\frac{m}{m+1}\right)^{m+1}\ge \left(n^m+\frac{m}{m+1}\right)^{m+1}\ge n^{m(m+1)}\ge n^m$$
Apply A.M. G.M. $$\frac{\underbrace{n+n+\cdots+n}_{m\ \text{times}}+1}{m+1}\ge (n^m\cdot 1)^{\frac{1}{m+1}}$$ $$\frac{mn+1}{m+1}\ge (n^m\cdot 1)^{\frac{1}{m+1}}$$ $$\left(\frac{mn+1}{m+1}\right)^{m+1}\ge n^m\cdot 1$$
$$\left(\frac{mn+1}{m+1}\right)^{m+1}\ge n^m$$
• Thanks a lot @James I really looking for this explanation.Can you suggest me a good book to start inequalities problem to build up my concept please. – emonHR Aug 28 '18 at 17:19
• I don't know but you can ask it as new question on this site itself – Deepesh Meena Aug 28 '18 at 17:29
Apply AM-GM inequality for $m$ numbers $n$ and $1$ number $1$ we have the result follows by raising both sides of the inequality to the $m+1$ power.
So, I'm not fantastic at writing proofs, but I think you can use a bit of intuition to determine whether this is true.
$$\Biggl(\frac{mn + 1}{m + 1}\Biggl)^{m + 1} > n^m$$
The numerator $(mn + 1)^{m + 1}$ can be expressed as a polynomial of the form $ax^n + bx^{n - 1} + ... + c$. The highest order term in polynomials of that form is the leading term, $ax^n$. Therefore, in $(mn + 1)^{m + 1}$, the term $(mn)^{m + 1}$. Because it grows faster than all other terms, you can regard it to be an approximation of the result of the numerator. The same reasoning can be applied to the denominator. Therefore, we get the following expression:
$$\Biggl(\frac{(mn)^{m + 1}}{m^{m + 1}}\Biggl)$$
Which simplifies to
$$n^{m + 1}$$
Clearly, $$n^{m + 1} > n^m$$
Once again, this proof (if you can call it that?) is not really rigorous, but I think it's an intuitive way to prove it. | 2020-03-29 04:07:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7724960446357727, "perplexity": 368.22703668639406}, "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-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00379.warc.gz"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/70/ | ## Results (displaying matches 1-50 of 228) Next
Label Dim. $$A$$ Field CM Traces Fricke sign $q$-expansion
$$a_2$$ $$a_3$$ $$a_5$$ $$a_7$$
70.2.a.a $$1$$ $$0.559$$ $$\Q$$ None $$1$$ $$0$$ $$-1$$ $$-1$$ $$-$$ $$q+q^{2}+q^{4}-q^{5}-q^{7}+q^{8}-3q^{9}+\cdots$$
70.2.c.a $$4$$ $$0.559$$ $$\Q(i, \sqrt{6})$$ None $$0$$ $$0$$ $$4$$ $$0$$ $$q+\beta _{2}q^{2}+(\beta _{1}+\beta _{3})q^{3}-q^{4}+(1-\beta _{1}+\cdots)q^{5}+\cdots$$
70.2.e.a $$2$$ $$0.559$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$-3$$ $$1$$ $$1$$ $$q-\zeta_{6}q^{2}+(-3+3\zeta_{6})q^{3}+(-1+\zeta_{6})q^{4}+\cdots$$
70.2.e.b $$2$$ $$0.559$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$2$$ $$1$$ $$-4$$ $$q-\zeta_{6}q^{2}+(2-2\zeta_{6})q^{3}+(-1+\zeta_{6})q^{4}+\cdots$$
70.2.e.c $$2$$ $$0.559$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$-1$$ $$1$$ $$-1$$ $$q+\zeta_{6}q^{2}+(-1+\zeta_{6})q^{3}+(-1+\zeta_{6})q^{4}+\cdots$$
70.2.e.d $$2$$ $$0.559$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$2$$ $$-1$$ $$-4$$ $$q+\zeta_{6}q^{2}+(2-2\zeta_{6})q^{3}+(-1+\zeta_{6})q^{4}+\cdots$$
70.2.g.a $$8$$ $$0.559$$ $$\Q(\zeta_{16})$$ None $$0$$ $$0$$ $$0$$ $$-8$$ $$q-\zeta_{16}^{6}q^{2}+(\zeta_{16}-\zeta_{16}^{3})q^{3}-\zeta_{16}^{4}q^{4}+\cdots$$
70.2.i.a $$4$$ $$0.559$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$-2$$ $$0$$ $$q+\zeta_{12}q^{2}+(-3\zeta_{12}+3\zeta_{12}^{3})q^{3}+\cdots$$
70.2.i.b $$4$$ $$0.559$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$4$$ $$0$$ $$q+\zeta_{12}q^{2}+\zeta_{12}^{2}q^{4}+(2-\zeta_{12}-2\zeta_{12}^{2}+\cdots)q^{5}+\cdots$$
70.2.k.a $$16$$ $$0.559$$ $$\mathbb{Q}[x]/(x^{16} + \cdots)$$ None $$0$$ $$0$$ $$-12$$ $$8$$ $$q+(-\beta _{7}-\beta _{15})q^{2}+(-\beta _{4}+\beta _{13})q^{3}+\cdots$$
70.3.b.a $$8$$ $$1.907$$ $$\mathbb{Q}[x]/(x^{8} + \cdots)$$ None $$0$$ $$0$$ $$0$$ $$-4$$ $$q+\beta _{6}q^{2}+(-\beta _{1}+\beta _{2})q^{3}+2q^{4}-\beta _{2}q^{5}+\cdots$$
70.3.d.a $$8$$ $$1.907$$ 8.0.$$\cdots$$.6 None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{4}q^{2}-\beta _{5}q^{3}-2q^{4}+(\beta _{2}-\beta _{6}+\cdots)q^{5}+\cdots$$
70.3.f.a $$4$$ $$1.907$$ $$\Q(i, \sqrt{14})$$ None $$-4$$ $$4$$ $$12$$ $$0$$ $$q+(-1+\beta _{2})q^{2}+(1+\beta _{1}+\beta _{2})q^{3}+\cdots$$
70.3.f.b $$8$$ $$1.907$$ 8.0.$$\cdots$$.20 None $$8$$ $$4$$ $$-4$$ $$0$$ $$q+(1+\beta _{3})q^{2}+(1+\beta _{1}-\beta _{3})q^{3}+2\beta _{3}q^{4}+\cdots$$
70.3.h.a $$16$$ $$1.907$$ $$\mathbb{Q}[x]/(x^{16} + \cdots)$$ None $$0$$ $$0$$ $$-6$$ $$0$$ $$q+\beta _{10}q^{2}+\beta _{1}q^{3}+2\beta _{4}q^{4}+(-\beta _{6}+\cdots)q^{5}+\cdots$$
70.3.j.a $$8$$ $$1.907$$ 8.0.3317760000.3 None $$0$$ $$12$$ $$0$$ $$-12$$ $$q+(-\beta _{3}-\beta _{5})q^{2}+(1-\beta _{3}+\beta _{4}-\beta _{5}+\cdots)q^{3}+\cdots$$
70.3.l.a $$8$$ $$1.907$$ 8.0.3317760000.2 None $$4$$ $$-4$$ $$12$$ $$-4$$ $$q+(1+\beta _{2}-\beta _{4})q^{2}+(-\beta _{2}+2\beta _{3}-\beta _{4}+\cdots)q^{3}+\cdots$$
70.3.l.b $$8$$ $$1.907$$ 8.0.303595776.1 None $$4$$ $$2$$ $$-6$$ $$-12$$ $$q+(1-\beta _{3}+\beta _{4})q^{2}+(-\beta _{1}+\beta _{2}-\beta _{4}+\cdots)q^{3}+\cdots$$
70.3.l.c $$16$$ $$1.907$$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None $$-8$$ $$2$$ $$-2$$ $$12$$ $$q+(-1-\beta _{4}+\beta _{8})q^{2}+(-\beta _{3}-\beta _{10}+\cdots)q^{3}+\cdots$$
70.4.a.a $$1$$ $$4.130$$ $$\Q$$ None $$-2$$ $$-8$$ $$-5$$ $$-7$$ $$+$$ $$q-2q^{2}-8q^{3}+4q^{4}-5q^{5}+2^{4}q^{6}+\cdots$$
70.4.a.b $$1$$ $$4.130$$ $$\Q$$ None $$-2$$ $$-3$$ $$5$$ $$-7$$ $$-$$ $$q-2q^{2}-3q^{3}+4q^{4}+5q^{5}+6q^{6}+\cdots$$
70.4.a.c $$1$$ $$4.130$$ $$\Q$$ None $$-2$$ $$-1$$ $$-5$$ $$7$$ $$-$$ $$q-2q^{2}-q^{3}+4q^{4}-5q^{5}+2q^{6}+\cdots$$
70.4.a.d $$1$$ $$4.130$$ $$\Q$$ None $$-2$$ $$4$$ $$5$$ $$7$$ $$+$$ $$q-2q^{2}+4q^{3}+4q^{4}+5q^{5}-8q^{6}+\cdots$$
70.4.a.e $$1$$ $$4.130$$ $$\Q$$ None $$2$$ $$5$$ $$5$$ $$-7$$ $$+$$ $$q+2q^{2}+5q^{3}+4q^{4}+5q^{5}+10q^{6}+\cdots$$
70.4.a.f $$1$$ $$4.130$$ $$\Q$$ None $$2$$ $$7$$ $$-5$$ $$7$$ $$+$$ $$q+2q^{2}+7q^{3}+4q^{4}-5q^{5}+14q^{6}+\cdots$$
70.4.c.a $$2$$ $$4.130$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$20$$ $$0$$ $$q+2iq^{2}+7iq^{3}-4q^{4}+(10+5i)q^{5}+\cdots$$
70.4.c.b $$6$$ $$4.130$$ 6.0.$$\cdots$$.1 None $$0$$ $$0$$ $$-16$$ $$0$$ $$q+2\beta _{3}q^{2}+(-2\beta _{3}+\beta _{5})q^{3}-4q^{4}+\cdots$$
70.4.e.a $$2$$ $$4.130$$ $$\Q(\sqrt{-3})$$ None $$-2$$ $$-1$$ $$5$$ $$35$$ $$q-2\zeta_{6}q^{2}+(-1+\zeta_{6})q^{3}+(-4+4\zeta_{6})q^{4}+\cdots$$
70.4.e.b $$2$$ $$4.130$$ $$\Q(\sqrt{-3})$$ None $$2$$ $$-10$$ $$5$$ $$28$$ $$q+2\zeta_{6}q^{2}+(-10+10\zeta_{6})q^{3}+(-4+\cdots)q^{4}+\cdots$$
70.4.e.c $$2$$ $$4.130$$ $$\Q(\sqrt{-3})$$ None $$2$$ $$1$$ $$5$$ $$17$$ $$q+2\zeta_{6}q^{2}+(1-\zeta_{6})q^{3}+(-4+4\zeta_{6})q^{4}+\cdots$$
70.4.e.d $$4$$ $$4.130$$ $$\Q(\sqrt{-3}, \sqrt{46})$$ None $$4$$ $$2$$ $$-10$$ $$6$$ $$q-2\beta _{2}q^{2}+(1+\beta _{1}+\beta _{2})q^{3}+(-4+\cdots)q^{4}+\cdots$$
70.4.e.e $$6$$ $$4.130$$ 6.0.$$\cdots$$.2 None $$-6$$ $$-4$$ $$-15$$ $$14$$ $$q+(-2+2\beta _{3})q^{2}+(-\beta _{1}-\beta _{3})q^{3}+\cdots$$
70.4.g.a $$24$$ $$4.130$$ None $$0$$ $$0$$ $$0$$ $$4$$
70.4.i.a $$24$$ $$4.130$$ None $$0$$ $$0$$ $$-8$$ $$0$$
70.4.k.a $$48$$ $$4.130$$ None $$0$$ $$0$$ $$48$$ $$-4$$
70.5.b.a $$8$$ $$7.236$$ $$\mathbb{Q}[x]/(x^{8} + \cdots)$$ None $$0$$ $$0$$ $$0$$ $$132$$ $$q-\beta _{1}q^{2}+(\beta _{2}-\beta _{4}-\beta _{5})q^{3}+8q^{4}+\cdots$$
70.5.d.a $$16$$ $$7.236$$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{5}q^{2}-\beta _{4}q^{3}-8q^{4}-\beta _{6}q^{5}-\beta _{1}q^{6}+\cdots$$
70.5.f.a $$12$$ $$7.236$$ $$\mathbb{Q}[x]/(x^{12} + \cdots)$$ None $$-24$$ $$-20$$ $$8$$ $$0$$ $$q+(-2-2\beta _{1})q^{2}+(-2+2\beta _{1}-\beta _{2}+\cdots)q^{3}+\cdots$$
70.5.f.b $$12$$ $$7.236$$ $$\mathbb{Q}[x]/(x^{12} - \cdots)$$ None $$24$$ $$-20$$ $$40$$ $$0$$ $$q+(2+2\beta _{1})q^{2}+(-2+2\beta _{1}-\beta _{2})q^{3}+\cdots$$
70.5.h.a $$32$$ $$7.236$$ None $$0$$ $$0$$ $$54$$ $$0$$
70.5.j.a $$24$$ $$7.236$$ None $$0$$ $$36$$ $$0$$ $$76$$
70.5.l.a $$32$$ $$7.236$$ None $$-32$$ $$-8$$ $$-10$$ $$-34$$
70.5.l.b $$32$$ $$7.236$$ None $$32$$ $$8$$ $$-26$$ $$14$$
70.6.a.a $$1$$ $$11.227$$ $$\Q$$ None $$-4$$ $$-23$$ $$25$$ $$49$$ $$+$$ $$q-4q^{2}-23q^{3}+2^{4}q^{4}+5^{2}q^{5}+92q^{6}+\cdots$$
70.6.a.b $$1$$ $$11.227$$ $$\Q$$ None $$-4$$ $$-9$$ $$25$$ $$-49$$ $$-$$ $$q-4q^{2}-9q^{3}+2^{4}q^{4}+5^{2}q^{5}+6^{2}q^{6}+\cdots$$
70.6.a.c $$1$$ $$11.227$$ $$\Q$$ None $$-4$$ $$-3$$ $$-25$$ $$49$$ $$-$$ $$q-4q^{2}-3q^{3}+2^{4}q^{4}-5^{2}q^{5}+12q^{6}+\cdots$$
70.6.a.d $$1$$ $$11.227$$ $$\Q$$ None $$-4$$ $$11$$ $$-25$$ $$-49$$ $$+$$ $$q-4q^{2}+11q^{3}+2^{4}q^{4}-5^{2}q^{5}-44q^{6}+\cdots$$
70.6.a.e $$1$$ $$11.227$$ $$\Q$$ None $$4$$ $$-17$$ $$25$$ $$-49$$ $$+$$ $$q+4q^{2}-17q^{3}+2^{4}q^{4}+5^{2}q^{5}-68q^{6}+\cdots$$
70.6.a.f $$1$$ $$11.227$$ $$\Q$$ None $$4$$ $$-11$$ $$-25$$ $$49$$ $$+$$ $$q+4q^{2}-11q^{3}+2^{4}q^{4}-5^{2}q^{5}-44q^{6}+\cdots$$
70.6.a.g $$2$$ $$11.227$$ $$\Q(\sqrt{3369})$$ None $$8$$ $$3$$ $$-50$$ $$-98$$ $$-$$ $$q+4q^{2}+(2-\beta )q^{3}+2^{4}q^{4}-5^{2}q^{5}+\cdots$$
Next | 2020-09-23 23:39:45 | {"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": 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": 0.6667642593383789, "perplexity": 1310.6102053436152}, "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/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00513.warc.gz"} |
http://openstudy.com/updates/511106bde4b0d9aa3c483ba5 | ## dedderick one year ago In the triangle, x = 3. Find y.
1. dedderick
2. mathsmind
have you studied Pythagoras theorem
3. mathsmind
?
4. dedderick
no not yet
5. mathsmind
ok you want to solve this using sin and cos
6. mathsmind
you should have done it right?
7. itsmeeQUENNIE
@dedderick the formula for pythagoran theorem is |dw:1360070556954:dw|
8. mathsmind
you can use the sin rule or the cos rule
9. dedderick
ok
10. dedderick
how which side to square
11. dedderick
i meant what side do we square
12. mathsmind
i will give you the answer and then prove it for you
13. dedderick
then i can understand it better
14. mathsmind
if x =3 in your case then y=$3\sqrt{2}$
15. mathsmind
now you should know that $\sin 45 = \cos 45=\frac{ 1 }{ \sqrt{2} }$
16. dedderick
right
17. Nataliecowell
In any right triangle, you can use the Pythagorean Therom to solve for sides. In that triangle, x is 2 of the sides and equals 3. Pyhtagreons Therom is a^2+b^2=c^2 A and b are the x's in the diagram. So 3^2 + 3^2 is equal to 18= c^2 Then find the square root of 18 to solve.
18. mathsmind
do you agree with that rule
19. dedderick
yes i understand now
20. mathsmind
the guy says he hasn't done Pythagoras theorem
21. mathsmind
so we had to go to a different ulternative
22. dedderick
ok | 2015-01-28 16:24:23 | {"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": 0.5627664923667908, "perplexity": 4755.395931328652}, "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-2015-06/segments/1422121983086.76/warc/CC-MAIN-20150124175303-00079-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://bigwww.epfl.ch/publications/petrenko1701.html | Publications
English only BIG > Publications > Pancreatic Clocks
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# Pancreatic α- and β-Cellular Clocks Have Distinct Molecular Properties and Impact on Islet Hormone Secretion and Gene Expression
## V. Petrenko, C. Saini, L. Giovannoni, C. Gobet, D. Sage, M. Unser, M.H. Masson, G. Gu, D. Bosco, F. Gachon, J. Philippe, C. Dibner
### Genes & Development, vol. 31, no. 4, pp. 383-398, February 15, 2017.
A critical role of circadian oscillators in orchestrating insulin secretion and islet gene transcription has been demonstrated recently. However, these studies focused on whole islets and did not explore the interplay between α-cell and β-cell clocks. We performed a parallel analysis of the molecular properties of α-cell and β-cell oscillators using a mouse model expressing three reporter genes: one labeling α cells, one specific for β cells, and a third monitoring circadian gene expression. Thus, phase entrainment properties, gene expression, and functional outputs of the α-cell and β-cell clockworks could be assessed in vivo and in vitro at the population and single-cell level. These experiments showed that α-cellular and β-cellular clocks are oscillating with distinct phases in vivo and in vitro. Diurnal transcriptome analysis in separated α and β cells revealed that a high number of genes with key roles in islet physiology, including regulators of glucose sensing and hormone secretion, are differentially expressed in these cell types. Moreover, temporal insulin and glucagon secretion exhibited distinct oscillatory profiles both in vivo and in vitro. Altogether, our data indicate that differential entrainment characteristics of circadian α-cell and β-cell clocks are an important feature in the temporal coordination of endocrine function and gene expression.
@ARTICLE(http://bigwww.epfl.ch/publications/petrenko1701.html,
AUTHOR="Petrenko, V. and Saini, C. and Giovannoni, L. and Gobet, C. and
Sage, D. and Unser, M. and Masson, M.H. and Gu, G. and Bosco, D. and
Gachon, F. and Philippe, J. and Dibner, C.",
TITLE="Pancreatic $\alpha$- and $\beta$-Cellular Clocks Have Distinct
Molecular Properties and Impact on Islet Hormone Secretion and Gene
Expression",
JOURNAL="Genes \& Development",
YEAR="2017",
volume="31",
number="4",
pages="383--389",
month="February 15,",
note="")
© 2017 CSH Press. Personal use of this material is permitted. However, permission to reprint/republish this material for advertising or promotional purposes or for creating new collective works for resale or redistribution to servers or lists, or to reuse any copyrighted component of this work in other works must be obtained from CSH Press.
This material is presented to ensure timely dissemination of scholarly and technical work. Copyright and all rights therein are retained by authors or by other copyright holders. All persons copying this information are expected to adhere to the terms and constraints invoked by each author's copyright. In most cases, these works may not be reposted without the explicit permission of the copyright holder. | 2018-05-20 09:50:51 | {"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": 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": 0.3101537227630615, "perplexity": 5798.636284848308}, "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-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00029.warc.gz"} |
https://www.piezodrive.com/actuators/150v-piezo-ring-stack-actuators/ | # SR Series 150V Piezo Ring Stack Actuators
The PiezoDrive SR ring stacks are high performance multilayer actuators with a vacuum compatible polymer coating and a large central aperture. The SR rings stacks are perfectly matched to the range of PiezoDrive amplifiers and driver modules. Applications include: Optics, Microscopy, Nanopositioning, and Precision machining.
• Calculate Power Bandwidth
• Enquiry
• Obtain a quote
### Specifications
Order Code Range $\pm$10% Length OD ID Cap. $\pm$20% Blocking Force Stiffness Res. Freq. $AU Buy Now SR080410 14um 10mm 8mm 4.5mm 0.8uF 1300N 130N/um 150kHz$AU88 Buy Now SR080418 28um 18mm 8mm 4.5mm 1.6uF 1300N 65N/um 83kHz $AU130 Buy Now SR120610 14um 10mm 12mm 6mm 2.4uF 3000N 220N/um 150kHz$AU128 Buy Now SR120620 30um 20mm 12mm 6mm 5.0uF 3000 N 100N/um 75kHz \$AU260 Buy Now
### Mounting
Stack actuators should not be exposed to significant tensile loads, unequally distributed loads, off-axis loads, bending moments, or torque.
The maximum recommended tensile load is 10% of the blocking force. In applications that require bi-directional forces or high speed motion, a preload force is recommended with a magnitude greater than the maximum tensile load. This gaurantees that the actuator is always in compression. The maximum recommended preload is 50% of the blocking force.
### Range
The range is specified for an applied voltage of -30V to +150V. If the input voltage is unipolar (0V to +150V) the specified range is reduced by a factor of 0.75. When a stack actuator is driving a stiff spring, the range is reduced by the factor
$$\frac{k_p}{k_p+k_L}$$ where $k_p$ is the actuator stiffness and $k_L$ is the load stiffness.
The travel range can also be reduced by restraining the end plates, e.g. by bonding the actuator to a stiff base. This effect is most significant in actuators with a length less than twice the outside diameter.
### Capacitance
The actuator capacitance is the small-signal capacitance measured at room temperature. Due to hysteresis, the effective capacitance increases with applied voltage. When operating at full range, the effective capacitance is approximately twice the small-signal capacitance. The capacitance also increases with temperature. A temperature increase of approximately 50 degrees C will double the effective capacitance.
### Thermal
Piezoelectric actuators dissipate heat when driven at full range with a high frequency. PiezoDrive actuators can be operated continuously at temperatures up to 85 degrees C. Continuous operation beyond this temperature may damage the actuator.
The dissipated heat is approximately 10% of the applied electrical power. For a sine-wave, the applied electrical power is:
$$P = \frac{V_{p-p}^2 \pi C f}{4}.$$
### Electrical Current Requirements
Calculate Power Bandwidth
The required current is $I = C~ dV/dt$ where $I$ is the current, $C$ is the effective capacitance, and $dV/dt$ is the voltage rate of change. For a sine-wave, the required peak current is equal to:
$$I_p = 2 \pi f V_{p-p}$$ where $V_{p-p}$ is the peak-to-peak voltage. For a triangle wave, the required peak current is equal to:
$$I_p = 2 C f V_{p-p} .$$
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### Piezoelectric Properties
The piezoelectric material is similar to PZT-5H and Navy Type VI.
Property Symbol Value Unit
Piezoelectric constants d33 600 10-12 m/V
d31 -270 10-12 m/V
g33 19.4 10-3 Vm/N
g31 -9.2 10-3 Vm/N
Electro-mechanical coupling coefficients Kp 0.65 NA
Kt 0.37 NA
K31 0.38 NA
Frequency constant Np 1980 Hz-m
Nt 1950 Hz-m
N31 1450 Hz-m
Elastic constant Y33 5.3 1010 N/m2
Y11 7.2 1010 N/m2
Q Factor Qm 80 NA
Dielectric constant e33 ⁄ e0 3500 @1 kHz
Dissipation factor tan δ 2.5 % @ 1 kHz
Currie Temperature Tc 220 C
Density ρ 7.8 g/cm3 | 2017-10-21 12:02:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49209046363830566, "perplexity": 9700.167770545404}, "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-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00897.warc.gz"} |
https://www.sci.unich.it/~francesc/teaching/network/betweeness.html | # Betweenness Centrality
Betweenness centrality measures the extent to which a vertex lies on paths between other vertices. Vertices with high betweenness may have considerable influence within a network by virtue of their control over information passing between others. They are also the ones whose removal from the network will most disrupt communications between other vertices because they lie on the largest number of paths taken by messages.
Mathematically, let $n_{s,t}^{i}$ be the number of geodesic paths from $s$ to $t$ that pass through $i$ and let $n_{s,t}$ be the total number of geodesic paths from $s$ to $t$. Recall that a geodesic path is not necessarily unique and the geodesic paths between a pair of vertices need not be node-independent, meaning they may pass through some of the same vertices. Then the betweenness centrality of vertex $i$ is:
$\displaystyle{b_i = \sum_{s, t} w_{s,t}^{i} = \sum_{s, t} \frac{n_{s,t}^{i}}{n_{s,t}}}$
where by convention the ratio $w_{s,t}^{i} = 0$ if $n_{s,t} = 0$. Notice that each pair of vertex $s, t$ contribute to the sum for $i$ with a weight $w_{s,t}^{i}$ between 0 and 1 expressing the betweenness of $i$ with respect to the pair $s, t$. Observe that:
1. the given definition counts separately the geodesic paths in either direction between each vertex pair. Since these paths are the same on an undirected graph this effectively counts each path twice;
2. the definition includes paths starting or ending with $i$ ($s$ can be equal to $i$ or $t$ can be equal to $i$), as well as paths from a vertex to itself ($s$ can be equal to $t$). It seems reasonable to define a vertex to be on a path between itself and someone else, or between some vertex and itself, since normally a vertex has control over information flowing from or to itself.
It makes little difference in practice to consider the alternative definitions, since one is usually concerned only with the relative magnitudes of the centralities and not with their absolute values. The sum can be normalized by dividing by the total number of ordered pairs of nodes, which is $n^2$, so that betweenness lies strictly between 0 and 1.
Consider the following graph:
From node 0 to node 4 there are two geodesics, both going through node 2. Hence $w_{0,4}^{2} = 2 / 2 = 1$, $w_{0,4}^{1} = w_{0,4}^{3} = 1 / 2$.
Consider once again the following simple network:
Function betweenness (R, C) computes betweenness centrality (the function counts undirected paths in only one direction and computes the sum in the betweenness formula for $s \neq t$, $s \neq i$ and $t \neq i$). This is the same network with nodes labelled with their betweenness centrality:
Notice that our sample graph is in fact a tree, that is a connected acyclic undirected graph, hence the number $n_{s,t}$ of geodesics from $s$ to $t$ is always 1, and the number $n_{s,t}^{i}$ of geodesics from $s$ to $t$ that pass through $i$ is either 0 or 1. Hence the computed betweenness score for a vertex is the actual number of distinct paths that strictly contain the vertex in between.
Betweenness centrality differs from the other centrality measures. A vertex can have quite low degree, be connected to others that have low degree, even be a long way from others on average, and still have high betweenness. Consider a vertex A that lies on a bridge between two groups of vertices within a network. Since any path between nodes in different groups must go through this bridge, node A acquires high betweenness even though it is not well connected (it lies at the periphery of both groups) and hence it might not have particularly high values for degree, eigenvector, and closeness centrality. Vertices in roles like this are sometimes referred to as brokers.
The maximum possible value for betweenness occurs for the central node of a star graph, a network composed of a vertex attached to $n-1$ other vertices, whose only connection is with the central node. All paths, except the $n-1$ paths from the peripheral vertices to themselves, go through the central vertex, hence its betweenness is $n^2 - (n-1) = n^2 - n + 1$. At the other end of the scale, the smallest possible value for betweenness occurs for a leaf node in a graph with a single component, that is a node that is connected to the rest of the network with only one edge. The leaf vertex lies on every path that starts or ends with itself. These are $2n-1$ in total: $n-1$ paths from a vertex to others, $n-1$ from others to the vertex, and 1 from the vertex to itself.
Betweenness centrality values are typically distributed over a wide range. Taking again the example of the film actor network, the individual with highest betweenness in the largest component of the network is Fernando Rey (The French Connection). It is no coincidence that he appeared in both European and American films, played roles in several languages, and worked in both film and television, hence he is the archetypal broker. Rey has a betweenness score of $7.47 \cdot 10^8$, while the lowest score of any actor in the large component is $8.91 \cdot 10^5$. Thus there is a ratio of almost a thousand between the two limits, much larger than the ratio of 3.6 we saw in the case of closeness. One consequence is that there are clear winners and losers in the betweenness centrality competition. The second highest betweenness score is attributed to Christopher Lee again, with $6.46 \cdot 10^8$, a $14%$ difference from the winner.
Betweenness centrality, as defined above, is a measure of information control assuming two important hypothesis: (i) every pair of vertices exchange information with equal probability, and (ii) information flows along the geodesic (shortest) path between two vertices, or one of such path, chosen at random, if there are several. However, information not always takes the shortest route. In social network, for instance, a news about a friend of us might not come directly from the friend but from another mutual friend.
Calculating closeness and betweenness centrality for all nodes in a graph involves computing the (unweighted) shortest paths between all pairs of nodes in the graph. A breath-first visit from a source node can be used to compute all shortest paths from the source in time $O(n + m)$, where $n$ and $m$ are the number of nodes and edges of the graph. If this visit is repeated for all the source nodes of the graph the cost amounts to $O(n \cdot (n+m))$. Typically, real networks are sparse graphs, meaning the number $m$ of edges is of the order of the number $n$ of nodes. For instance, it is very rare in a social network that a significant number of actors have contacts with all the other actors of the network. Hence the overall cost in practical cases is quadratic. | 2021-05-13 00:43:28 | {"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": 0.8103101849555969, "perplexity": 239.28823143922241}, "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-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00398.warc.gz"} |
http://mathhelpforum.com/pre-calculus/153431-quadratic-function-problem.html | # Math Help - A quadratic function problem
1. ## A quadratic function problem
Can someone please solve or tell me how i would do this question. I'm finding it really difficult.
A projectile is launched from the top of a building. The height (h meters) of the projectile above the ground is given by the equation h=15 + 20t - 2t^2 where t is the time of flight in seconds.
(a) sketch the graph to represent the flight
(b) What is the greatest height reached?
(c) What is the height of the building?
(d) After what time did the projectile strike the ground?
2. $h = -2t^2 + 20t + 15$
(b) Change a quadratic equation to vertex form by completing the square:
$h = -2t^2 + 20t + 15$
$h = -2(t^2 - 10t) + 15$
$h = -2(t^2 - 10t + 25 - 25) + 15$
$h = -2(t^2 - 10t + 25) + 50 + 15$
$h = -2(t - 5)^2 + 65$
The vertex is (5, 65), and this is the maximum point of the parabola. So the greatest height is 65 feet.
(c) Since the projectile started at t = 0 from the top of the building, plug in 0 for t into the equation and simplify.
(d) When the projectile hits the ground, h = 0. So plug in 0 for h into the equation and solve for t. | 2014-08-30 04:57:17 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "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": 0.596695065498352, "perplexity": 331.62331354267235}, "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/1408500834258.45/warc/CC-MAIN-20140820021354-00430-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://indico.inp.nsk.su/event/9/contribution/4 | # The 12th International Workshop POSIPOL-2017
18-21 September 2017
Budker INP
Asia/Novosibirsk timezone
Home > Timetable > Contribution details
# Contribution
High intensity positron sources
# Gamma-gamma collider for bb (<12 GeV) energy region
## Speakers
• Prof. Valery TELNOV
## Content
Photon colliders based on high energy linear e+e- colliders are considered for many years but may appear only in very far future. In this talk I consider the gamma-gamma collider for the energy region below 12 GeV, suggested 1/2 year ago. This energy region (in gamma-gamma) is not covered by Super-B collider and its luminosity may be much higher than at the LHC. Such collider will be a nice place for application of modern outstanding technologies (linacs (SC, plasma-based), low-emittance electron sources, powerful laser systems, optical cavities). Same electron linacs can be used simultaneously for XFELs. | 2017-09-25 22:20:01 | {"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": 0.8361072540283203, "perplexity": 10239.514789535255}, "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-2017-39/segments/1505818693459.95/warc/CC-MAIN-20170925220350-20170926000350-00395.warc.gz"} |
http://bootmath.com/in-depth-explanation-of-how-to-do-mathematical-induction-over-the-set-mathbbr-of-all-real-numbers.html | # In-Depth Explanation of How to Do Mathematical Induction Over the Set $\mathbb{R}$ of All Real Numbers?
I’ve seen in the answers to a few different questions here on the Mathematics Stack Exchange that one can clearly do mathematical induction over the set $\mathbb{R}$ of all real numbers. I am, however, having quite a difficult time understanding how the methods described in both those questions’ answers and some reference materials to which they link. In particular, I can’t seem to figure out exactly how the techniques described therein parallel the methods codified in the axiom of induction for use when doing mathematical induction over the set $\mathbb{N}$ of all natural numbers.
If somebody would be so kind as to provide me with a more detailed explanation of how to do mathematical induction over the set $\mathbb{R}$ of all real numbers within about the next day or so, then I would be very grateful! The answer should be understandable by any beginning calculus student who also has a rudimentary understanding of set theory and mathematical logic. I’ve provided links to both the relevant questions and whatever reference material mentioned in them that seemed like good leads when I found them no matter how inscrutable they might have been at the time.
Questions About Induction Over the Real Numbers:
• Induction on Real Numbers
• Is it possible to use mathematical induction to prove a statement concerning all real numbers, not necessarily just the integers? [duplicate]
• Extending a theorem true over the integers to reals and complex numbers
Question-Derived Reference Material:
• ‘The Instructor’s Guide to Real Induction’ by Pete L. Clark
P. S.: I also have the following follow-up questions:
• Version of the Axiom of Induction for Real Induction?
• Real Induction Over Multiple Variables?
#### Solutions Collecting From Web of "In-Depth Explanation of How to Do Mathematical Induction Over the Set $\mathbb{R}$ of All Real Numbers?"
I feel like I am jumping into this discussion rather late, but I feel that the other answers given so far have to a large extent missed the point of the question.
As a matter of fact, you CAN do induction on the real numbers under the standard order! This is called “real induction,” and the main result is proven and described at length in the references given by the original poster. Explicitly, suppose $S$ is a subset of the closed interval $[a,b]$ with the following properties:
1. $a$ is in $S$.
2. For every $x$ in $[a,b)$, there is a number $y$ in $[a,b]$ such that every number $z$ in $[x,y]$ is in $S$.
3. For every $x$ in $[a,b]$, if $[a,x)$ is a subset of $S$, then $x$ is in $S$.
Then $S=[a,b]$.
Although it doesn’t involve a successor function, this captures a lot of the flavor of both induction on the natural numbers and transfinite induction. Moreover, because it uses the usual order on $\mathbb{R}$, it can be used to prove interesting theorems about real numbers, including the Intermediate Value Theorem, the Extreme Value Theorem, and the Bolzano-Weierstrass Theorem.
For an example of how to use real induction in a proof, look at Theorem 5 (the Extreme Value Theorem) in the first reference. Clark proves that every continuous function on a closed interval $[a,b]$ is bounded as follows:
Let $f$ be a continuous real-valued function on the closed interval $[a,b]$. Take $S$ to be the set of numbers $x$ in $[a,b]$ for which $f$ is bounded on $[a,x]$.
1. Clearly $f$ is bounded on $[a,a]$ (any number greater than $f(a)$ is an upper bound, and any number less than $f(a)$ is a lower bound), so $a$ is in $S$.
2. Suppose $x$ is in $S$. Then $f$ is bounded on $[a,x]$. Since $f$ is continuous at $x$, there is a $\delta>0$ such that, for all $y$ in $[x-\delta,x+\delta]$,$\left|f(y)\right|<\left|f(x)\right|+1$, so $f$ is bounded on $[a,x+\delta]$.
3. Now suppose $[a,x)$ is a subset of $S$. Since $f$ is continuous at $x$, there is a positive number $\delta < x − a$ such that $f$ is bounded on $[x − \delta, x]$. But since $a < x − \delta < x$, we know also that $f$ is bounded on $[a, x − \delta]$, so $f$ is bounded on $[a, x]$.
Since $S$ satisfies the three properties given above, it follows by real induction that $S=[a,b]$, so $f$ is bounded on $[a,b]$.
I didn’t see this question until now.
(i) What is described there is indeed what I call real induction. Exactly the same inductive setup was given by D. Hathaway, who called it continuity induction. These two works were completely independent of each other. (As you can see elsewhere on this site, I was thinking about this in September 2010. Hathaway’s article was published in May of 2011. As the gap between submission and publication is usually at least a year, I have little doubt that he was first.) On the other hand, a lot of other mathematicians have given inductive schemes for intervals on the real line over the years. Section 1.2 of my Instructor’s Guide lists 13 publications on the subject prior to Hathaway’s. The last of these is a paper by I. Kalantari. My work was dependent on his: I read his paper, learned with great excitement that you can do induction on the real numbers, and expanded on that theme. I have also been told by some people working in the area of differential equations that these kind of inductive arguments are rather standard in that area.
$\newcommand{\R}{\mathbb{R}}$
(ii) Yes, the formulation that I call “Real Induction” is formulated in terms of a closed bounded interval $[a,b]$ in the real line. (When you look at all the prior literature in the area, I think it becomes clear that a large contributing factor to the PR that my take on this material has gotten is the snappy name I chose, so I don’t apologize for that.) In Section 2 I formulate a notion of an inductive subset of a linearly ordered set $(X,\leq)$ and show that the only inductive subset of $X$ is $X$ itself is equivalent to Dedekind completeness of $X$. I then mention that every closed interval in $\R$ is Dedekind complete. $\R$ is a closed interval in $\R$, so in that formulation real induction does apply to $\R$. In fact every interval in $\R$ is Dedekind complete: an ordered set is Dedekind complete iff the subset obtained by adjoining least and greatest elements if they are not already present is complete, and doing this to any interval in $\R$ yields something order isomorphic to a closed, bounded interval $[a,b]$.
(iii) I agree that in practice, proving something separately on $[0,\infty)$, $(-\infty,0]$ and combining is often easier than working directly with $(-\infty,\infty)$.
What you’re talking about is transfinite induction. http://mathworld.wolfram.com/TransfiniteInduction.html
Basically induction is showing a proposition, $P$, is true on a well-ordered set, $(S,\lt)$ by the following:
1. $P(0)$ is true for the “first” element, $0\in S$
2. If $P(b)$ is true for all $b<a$ then $P(a)$ is true
Let’s unpack this a little. The important part is the “well-ordered” part.
An ordered set, $(S,\lt)$ is well ordered if every subset has a minimum element.
For example, $\mathbb{N}$ is well ordered. Any subset of the natural numbers certainly has a least element.
This is NOT true of the real numbers with the (normal) ordering. For example, $(0,1)$ does NOT have a least element. So you may not induct over $\mathbb{R}$ with the NORMAL ordering.
What well ordering means (or well, implies) is that for every element in $s\in S$, there is a “next” element $t$. For example the number $3$ has the “next” number $4$. To see this consider the set $T_s\subset U$ where $T_s=\{t\in S \colon t>s\}$. There is a minimum element in this set, which is the successor of $s$. It is the next element.
This is not true with $\mathbb{R}$ with the normal ordering. For example, what real number comes after $1$? There is none, you can always find one closer.
But what if we didn’t use the “normal ordering”? Well thanks to the axiom of choice, we may prove the well ordering theorem (or axiom perhaps?) https://math.berkeley.edu/~kpmann/Well-ordering.pdf
What the well ordering theorem (axiom) says is that EVERY set, even $\mathbb{R}$ can be well ordered. So there exists some ordering, $\lt_w$ on $\mathbb{R}$ such that there IS a number after $1$. Note that this ordering has absolutely nothing to do with the normal ordering on $\mathbb{R}$. For example maybe $4>_w 2323$ and $\pi<_w-4$.
Using this ordering, we MAY induct on all of $\mathbb{R}$ because $(\mathbb{R},\lt_w)$ is well ordered. However there is a problem.
By using the axiom of choice (or insisting that this is an axiom), you are guaranteeing that this well ordering will never be explicitly given. So although there IS a well ordering of $\mathbb{R}$, we don’t and CAN’T know what it is. Furthermore, this well ordering might not make any sense. So practically it is very difficult to induct on $(\mathbb{R},\lt_w)$.
No.
(I’d love to just leave it at that)
The natural numbers are defined with something called a successor function – what comes next. Induction shows that IF it is true for $n$, then it is true for $\text{Successor}(n)$.
The real numbers do not have a successor function. Anything countable (like the sequence $(\frac{n}{4})_{n=1}^\infty\subset\mathbb{R}$ say) we can do induction over, only in that sense are we doing induction over the reals.
That’s because the sequence is countable. There is a bijection between $\mathbb{N}$ and the sequence I mentioned.
An interesting question would be how do you do induction over $\mathbb{Q}$ which is countable. I am going to research that now.
(BTW this stuff is from the domain of set theory, any books called “introduction to set theory” will help with this) | 2018-07-16 12:31:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8690240979194641, "perplexity": 148.6356761834026}, "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-2018-30/segments/1531676589270.3/warc/CC-MAIN-20180716115452-20180716135452-00028.warc.gz"} |
https://android.stackexchange.com/questions/197026/google-play-store-wont-download-app | Here is a screenshot that shows kind of notification I am writing about, since I am unable to provide an actual one. No error messages, just a continuous progress bar:
I tried clearing cache, then all data and then data in Google Play Services app, but it didn't help. The phone is Oukitel U11 Plus, using Android 8.
I have visited wiki page of google-play-store tag and didn't found any reasonable advice in regard of this situation.
On the side note, one other frustrating thing is happening, too. When searching in Chrome it redirects to local .rs domain, but it then prints out err_tunnel_connection_failed. I got around it buy using no redirect link to force it to go main .com domain.
• Have you already checked with our google-play-store tag-wiki? It lists many issues around playstore and links to solutions. Without an explicit error message it's hard to help in another way. – Izzy Jun 14 '18 at 13:40
• @Izzy I didn't, thank you for providing me that! How to get an explicit error message on this kind of problem? – Igor V. Jun 14 '18 at 18:23
• No error message shown? Nothing like "coulnd not be downloaded due to an error (498)"? Just sticking and doing nothing? Well, that would be a hint to include with your question, too. Also see these posts on stalled or on stuck downloads with Playstore then. – Izzy Jun 14 '18 at 19:26 | 2019-10-16 04:28:03 | {"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": 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": 0.19469811022281647, "perplexity": 2101.028385737143}, "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-20191016041344-20191016064844-00482.warc.gz"} |
https://zietzm.com/course-notes/f20-symbolic/misc | # Miscellaneous information
## Information about course databases
### clinical_vocabulary
This is essentially an OMOP database without PHI, just vocabularies.
I downloaded a number of vocabularies from AthenaSpecifically, I pulled CDM 5 vocabularies having the following vocabulary codes: SNOMED, ICD9CM, ICD9Proc, CPT4, HCPCS, LOINC, NDFRT, RxNorm, NDC, Gender, Race, MedDRA, Read, ATC, VA Product, VA Class, Cohort, ICD10, Ethnicity, MeSH, ICD10CM, RxNorm , Nebraska Lexicon, OMOP Extension. and just inserted them into new tables. I added a number of helpful indices, which dramatically improve the performance.
This is the main database you’ll be using.
### drugbank
I downloaded the full DrugBank XML database dump, version 5.1.7. I used dbparser version 1.2.0 to automatically reformat this XML document into tables. Most of these tables were inserted directly into the database, though a few smaller, less-important tables were omitted.
You’re most likely to use this database to access natural language drug indicationsdrugs_pharmacology.indication , though you’re free to use the wealth of other information available.
### SIDER
Tables were downloaded from SIDER, version 4.1. I did not reorder columns, but I did define names where none were present. See the SIDER 4.1 README for more information about the columns in these tables. In particular, note that drug names are automatically generated in the drug_names table. You may want to use names from ATC codes instead (i.e. map SIDER.drug_atc.drug_atc_code to clinical_vocabulary.CONCEPT.concept_code to access the concept_name field.)
### faers
This contains some data from FDA adverse event reports from the second quarter of 2020. I renamed only tables, not fields. For documentation, see the ASC_NTS.pdf.
## Favorite online resources
1. Athena
• ↑ The most comprehensive front-end for OMOP vocabulary tables of which I’m aware. Functional interface. ↓ Can be a bit slow, and mappings aren’t named in an obvious way. Can’t collect codes like in Atlas.
• I use this on an almost daily basis.
2. Book of OHDSI
• This thing is insanely good. It has information about OMOP, which is most relevant for this class. That’s only a small fraction, though. It has detailed explanations of OHDSI tools, observational research methods, and all kinds of other things. Plus, you’ll recognize many of the authors from our department.
3. Lab tests online
• Amazing resource for lab tests. Great place to look for reference ranges, LOINC codes, related tests, etc.
4. ICD10Data.com
• ↑ Adds helpful clinical information for ICD-10 codes. Very Google-able. Comprehensive for ICD-10. ↓ Internal search isn’t always great. Not the best interface for finding relatives of a code.
5. DrugBank
• Contains tons of information about drugs, including indications, biology/chemistry/physics data, cross-links to other data sources, etc.
• Tends to be my go-to for looking at a new drug.
6. OBO Foundry
• Huge resource of open-source, standardized, interoperable biomedical ontologies. Includes well known resources like the Gene Ontology, ChEBI, etc.
## Personal software choices
Throughout this course, you are free to use whatever setup you prefer. It makes no difference to me, and I want you to use what you find most comfortable.
Sometimes, though, I like to see what other people use.
For SQL, I use DataGrip.
Example auto-definition
It has a great built-in editor. I can very easily run queries in parallel or in a queue. It does nice auto code reformatting, which keeps you SQL files looking clean. It also stores the database schema and can generate table definitions, which made creating the class database insanely easy. You can get DataGrip free with a student account.
For statistical analysis, I prefer R + tidyverse. For slightly more complex code, I like Python (typically install miniconda). For both, I’m a big fan of Jupyter notebooksIRKernel for R and conda.
I’d also highly recommend Manubot, which enables automated scholarly manuscripts on GitHub.
Miscellaneous information - Michael Zietz | 2022-10-04 23:04:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.32755622267723083, "perplexity": 8362.009792854202}, "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/1664030337529.69/warc/CC-MAIN-20221004215917-20221005005917-00091.warc.gz"} |
https://charleslooker.wordpress.com/tag/math/ | # Posts tagged ‘Math’
A lot of maths I use tends to be abstracted away either in libraries I use, or inside the application. I’m going to go back to basics starting with vector maths, and moving onto matrices – these, in my opinion are the back bone to doing what we do. I’ll cover from the ground up and then go into some more complex areas: determinants, inverse multiplication, decomposition etc. I’ll be learning a bunch of this stuff along the way. Lets get started:
Vectors
So a vector is basically a direction from the origin. [1, 2, 3] basically means we have a point thats moved 1 in the X direction, 2 in the Y and 3 the Z direction.
Vectors can be added together simply by adding the parts of each together. [1, 2, 3] + [4, 5, 6] = [(1+4), (2+5), (3+6)]. Subtraction follows a similar process.
Vectors can be multiplied against a scalar (float) value by multiplying each part by it: [1, 2, 3] * 5 = [(1*5), (2*5), (3*5)].
We can get the length of a vector by firstly, powering each part by 2, then summing (adding up) these parts, and finally getting the square root of the total. This looks like this len([1, 2, 3]) = sqrt((1^2) + (2^2) + (3^2)).
Using this length we can get the normal of the vector. Normalizing a vector keeps its direction, but its length becomes 1.0. This is important in finding angles, unit vectors and matrix scale. To do this we first get the vectors length, and then divide each part of the vector by it:
normal([1, 2, 3]) =
length = sqrt((1^2) + (2^2) + (3^2))
normal = [1/length, 2/length, 3/length]
The dot product of two 3d vectors (x, y, z), basically returns the magnitude of one vector projected onto another. If we have two vectors [1, 0, 0] and [1, 1, 0]; when we project the latter onto the former, the value along the formers length is the dot product. To get the dot product of two vectors we simply multiply the parts together:
[1, 2, 3] . [4, 5, 6] = (1*4) + (2 * 5) + (3 * 6)
We can use the dot product to get the angle between two vectors too. If we first normalize each vector, we can get the angle by getting the inverse cos (or acos) of this dot. This will return a radian, so we can convert it into degrees by multiplying it by (180 / pi):
cos(norm([1,2, 3] . norm([4, 5, 6]) )^-1 * (180/pi)
Next cross products..
I’ve added some updates to my research page with some links to physics simulation, collisions, parametric surfaces and linear algebra – such as find the Sagitta and in turn the center of an arc. (Important for motion capture solving)
The method ive found to resolve spherical and circular collisions at high speed is borrowed from the spherical bounding box method. I first store the current distance between the two spheres/circles and then the future distance between the two .
$cd = |p0_{1} - p1_{1}|$
$fd = |(p0_{1} + (p0_{1} - p0_{1-1})) - (p1_{1} + (p1_{1} - p1_{1-1})) |$
if this future distance is smaller than the combined radius’s I re imagine the current distance
$cd = (cd \frac{cd}{fd})$
and continue to do the standard method.
The basic verlet algorithm im using is,
$x_{1+1} = x_{1} +(x_{1} - x_{1-1}) + a dt^2$
where x is the current and previous position a the accumulated forces, and dt, the timestep or period over frequency. This is great if the simulation is un-fluctuating i.e baked; if it was realtime i’d introduce the TCV (time corrected verlet) version,
$x_{1+1} = x_{1} + (x_{1} - x_{1-1}) (frac{dt_{1}}{dt_{1-1}}) + a dt_{1}^2$
The nice thing about verlet is that velocity is calculated from the current and old positions, and because of this you can keep throwing collsion and constraint methods at it. It handles the rest. My current approach is to fire the integrator ( verlet) then collisions and constraints – the latter two iterated several times.
I started to think about building a very simple framework for dynamics based on this article: http://www.gamasutra.com/resource_guide/20030121/jacobson_01.shtml
It uses the verlet method and then successive systems to check for constraints, collisions etc. – I like this approach as it seems pretty modular. I think the basic method ill go for is defining objects/verts/lines/etc with attributes, whether there masses or constraints connecting masses together. Then the main system calls the general functions on them.
Im not sure if this is correct but i think the aproach with the main system is to:
1. Calculate all the forces of all the masses
2. Apply verlet algorithm with a timeStep
3. Do a scene check i.e if the masses are in a box if its simple or/and
do some collision/friction checking
4. satisfy the constraints
Im not sure if 4 & 3 are the right order so, ill have to think on that – but this is my basic approach to building a simple system. I thought of attributes because they could be place on anything then, and i could add some function for display methods, like cylinders and spheres.
As I’m currently in the process of skinning many meshes for the current game I’m working on here are some rules I’ve learnt on the way:
• Don’t attempt to skin spherical deformation without having bones for deformation or quaternion based skinning methods. Key places where this must happen is the shoulders, thighs, elbows, knees, wrists and ankles.
• You don’t need lots of twist bones, three including the actual bone is enough e.g shoulder *deformer bone, main twist (100% to upper arm with a direction pointing to the deformer bone), 50% twist bone, 0% twist (100% to the upper arm)
*By deformer bone i mean a tweak bone/point, etc has average deformations between the shoulder and upper arm. Similar to elbow or wrist deformer bones.
• Don’t model the wrist/hand attached to the sleeve, tuck it inside and treat it as an element. The same applies to the ankle/foot and trouser leg.
• If you don’t see the underside of a mesh, don’t model it, cap it off. A good example of this is a skirt.
• The deformation of the wrist is not the same as the elbow or shoulder; the shoulder can be considered a one dimensional quaternion – in this i mean its twist is dictated by its direction. The wrist could be considered two dimension as the first quaternions direction dictates the rotation space (one plane) for the second quaternion to ride on. The wrist bone dictating the direction for the hand to ride on, as oppose to the upper arm bones direction dictating the entire deformation of the shoulder.
I’ll discuss more on the differences of the wrist compared to the shoulder in later posts.
Inconjunction with using the ‘unknown’ in this system – additional unknown could be stacked up for when combinations of combinations happen.
I’ve just started watching the Gilbert Strang MIT Courseware on iTunes, and highly recommend it – after watching it a light bulb went off in my head. Daniel Pook Kolb’s Blendshape system is very very impressive – but I did’nt and still dont understand it fully.
Thinking out loud here I thought that it might be essentially a form of finding out the ‘unknown’ in a linear combination of an nth dimensional space as the corrective of the combining shapes. For example the unknown of :
x[2,3] + y [4,5] = [6,7] would be [-1, 2] ( its hard to show as i’m not using LaTex atm)
Before any corrective is used the ‘unknown’ is [0,0], this would be synomanous to adding two blendshapes together without any corrective.
Just a quick theoretical idea – I’ve been thinking that a rotation is basically the reflection or product of a direction indirectly or directly controlled by its spin. This seems simple to understand in theory but in practice its more complex. Im coming to terms that rotation is really an illusion of a direction or a side product.
Thinking about this more in my notes, we can hypothesize that if a rotation is really just a direction, then we can seperate it into two parts: the direction and the spin and more importantly have direction controlled by anything, namely the tangent of a spline or curve. This seems amazingly simple now, the curviture of a spline dictating the direction of a quaternion about a hemisphere. But detaching yourself to thinking this way is really hard.
In this way we can treat the wrist and spine articulation the same – the spin (or twist in the conventional sence) is just used to dictate the direction of the system, with a spine you’d use the tangent of the curve itself, with the wrist the hand is the direction.
So I’ve been looking into using quats to simulate joint rotation and deformation. What I’ve found out is that most joints of a human can fall into two systems: single and double quaternion systems. If a bone system doesn’t twist using its bone like the forearm then it can fall into a single quat system, which dictates just a direction , resolution of the twist happens naturally when two perpendicular axis’ come together at 90 degrees.
If a system has a twist dictated by bones in conjunction with muscle deformation, then one quaternion is needed to dictate the twist space the second ‘deformation’ quat rides on. With this we can say that twist is not only dictated by the resolution of the system it exists in, but by the spin this system dictates. The order of which system drives each space is important. | 2021-12-03 01:23:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6233769059181213, "perplexity": 1157.5044164977567}, "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/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00129.warc.gz"} |
https://hal-cea.archives-ouvertes.fr/cea-01877709 | Extension of the growing season increases vegetation exposure to frost - Archive ouverte HAL Access content directly
Journal Articles Nature Communications Year : 2018
## Extension of the growing season increases vegetation exposure to frost
(1) , (1, 2, 3) , (4) , (1, 4) , (1) , (1) , (5, 6) , (7, 8) , (9) , (10, 2, 11)
1
2
3
4
5
6
7
8
9
10
11
Shilong Piao
Ivan Janssens
Shushi Peng
Philippe Ciais
• Function : Author
• PersonId : 832501
Josep Peñuelas
Tao Wang
#### Abstract
While climate warming reduces the occurrence of frost events, the warming-induced lengthening of the growing season of plants in the Northern Hemisphere may actually induce more frequent frost days during the growing season (GSFDs, days with minimum temperature < 0 °C). Direct evidence of this hypothesis, however, is limited. Here we investigate the change in the number of GSFDs at latitudes greater than 30° N using remotely-sensed and in situ phenological records and three minimum temperature (T$_{min}$) data sets from 1982 to 2012. While decreased GSFDs are found in northern Siberia, the Tibetan Plateau, and northwestern North America (mainly in autumn), ~43% of the hemisphere, especially in Europe, experienced a significant increase in GSFDs between 1982 and 2012 (mainly during spring). Overall, regions with larger increases in growing season length exhibit larger increases in GSFDs. Climate warming thus reduces the total number of frost days per year, but GSFDs nonetheless increase in many areas
### Dates and versions
cea-01877709 , version 1 (17-09-2020)
### Identifiers
• HAL Id : cea-01877709 , version 1
• DOI :
### Cite
Qiang Liu, Shilong Piao, Ivan Janssens, Yongshuo Fu, Shushi Peng, et al.. Extension of the growing season increases vegetation exposure to frost. Nature Communications, 2018, 9, pp.426. ⟨10.1038/s41467-017-02690-y⟩. ⟨cea-01877709⟩
### Export
BibTeX TEI Dublin Core DC Terms EndNote Datacite
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https://chrisvoncsefalvay.com/2019/06/30/answering-the-big-questions-with-lime/ | # Answering the big questions with LIME
In this post, we will be looking at LIME, a deceptively simple yet powerful heuristic for model introspection. With the rise of complex models such as deep neural networks, there is often a perception that these models operate as ‘black boxes’, opaque to the external observer. I will demonstrate the utility of LIME on a true case of life or death: in silico toxicity prediction, that is, the use of sophisticated algorithms to mathematically predict whether a compound is likely to be toxic or not.
Which takes me back, oddly enough, to my childhood.
When I was a kid, I drove my parents and teachers absolutely insane by my need to know why things are the way they are. This was particularly tough on my poor chemistry teacher, constantly inundated with questions. Fine, potassium cyanate is toxic, but why? Also, why are nitriles normally not particularly pungent, but isonitriles smell like some eldritch horror a chemically talented Lovecraft would have dreamed up? And can somebody please explain why methyl compounds of metals (methylcadmium, methylmercury, dimethyl zinc, you name it) are almost without exception horrible on multiple levels (explosivity, toxicity, stench, general unpleasantness to handle, broken glassware produced per gram synthesised, and so on)?
Much to the chagrin of my very, very patient wife, I haven’t changed a lot in this regard. Which is why I continue to be very interested in why things are the way they are. When we build a predictive model, it’s all well and good that we can advise our client as to which machine is going to break down next or where to drill for oil or which area is at a higher risk for malaria and therefore worth focusing treatment efforts on. Increasingly, however, clients are starting to join me in my attitude of wanting to know why, and we are just as keen on satisfying their curiosity as I was at asking interminable questions in Dr. Fuleki’s 10th grade chem class.
The following example will be quite unconventional. We’ll be building a model that tries to decide whether a small molecule is likely to be toxic. But that’s not all — we’ll also be interested in why it is toxic. What are the groups that are most likely to make it toxic, and what are the ones that suggest it might be innocuous? Unusually, we’ll be doing machine learning on something you probably haven’t seen it used a lot — molecules. Specifically, we’ll be looking at small molecules, meaning anything under 900 daltons. That includes a lot of modern pharmaceuticals as well as toxins and other unsavoury things, while also excluding large macromolecules such as proteins. There’ll be some chemistry talk, but this isn’t a cheminformatics blog post, so even if all you remember from high school chem is the smell and the occasional explosions, stay with me.
Our source data for this is Tox21, the toxicology programme of the National Institutes of Health. Tox21 has created a unique data set, pairing thousands chemical structures with their toxicities — their methodology basically relies on exposing cell cultures that are relatively representative of the human body to putative toxins, and see what happens. We’ll be using DeepChem, a fantastic package for pharmaceutical chemistry and drug discovery (among others), and RDKit for feature generation.
## Feature generation
Like all supervised machine learning projects, we need to start this by deciding upon the features to learn, and generating them.
There are many ways of representing a molecule, some more suitable for our purposes than others. For instance, the summary formula (which tells you how many atoms of each element are present in one molecule) is useful, but when it comes to biology, the devil’s in the details, or rather, the structure. Famously, cocaine and hyoscine (scopolamine) share the same summary formula, but while one is a notorious stimulant, the other is a depressant-deliriant.
Like all supervised machine learning projects, we need to start this by deciding upon the features to learn, and generating them.
There are many ways of representing a molecule, some more suitable for our purposes than others. For instance, the summary formula (which tells you how many atoms of each element are present in one molecule) is useful, but when it comes to biology, the devil’s in the details, or rather, the structure. Famously, cocaine and hyoscine (scopolamine) share the same summary formula, but while one is a notorious stimulant, the other is a depressant-deliriant.
evrwerv
So we need something that reflects structure, too. There are multiple such systems, such as InChI, which generates inordinately long and chaotic-looking strings, and SMILES, which we are going to use because it is slightly easier to handle. SMILES, somewhat analogously to chemical naming, looks for the longest backbone, breaks any cycles and then describes the branches. A worked example of acetylsalicylic acid, more commonly known as aspirin, is displayed to the left.
What we need at this point is a way to generate features from molecular structures represented by the SMILES strings — after all, what we’re after are relationships between structural motifs, such as the acyl group (the end part hanging off to the ‘top left’ of the benzene ring, comprising the atoms annotated 1, 2 and 3). For this, we first need to cleave apart our molecules into characteristic parts. In computational chemistry, this is referred to as molecular fingerprinting, and there is a large number of different approaches to this problem. In this case, we will be using a technique called Morgan or circular fingerprinting. Given a radius (defined as number of bonds), the Morgan fingerprint characterizes the atoms themselves (by element), their neighbourhood, their charge, whether they’re in a ring and various other chemical features.
## When life gives you LIME…
LIME (Local Interpretable Model Agnostic Explanations) is a model-agnostic way to peek into the inner workings of a classifier or regressor, first described by Ribeiro et al. (2016).1) The overriding idea is pretty trivial: to understand the black box between inputs and outputs, tweak inputs and see what happens to the outputs. These are referred to as perturbations, and the idea is to correlate particular perturbations to particular results.
Consider working out a model optimizing your steak: meat quality (from bottom-shelf steak to deep freeze Argentinian prime beef), temperature, marinade composition and days marinated, grill vs skillet, and so on. Theoretically, of course, you could try every single permutation of these parameters, but in practice, you’ll simply have steak every night for a year, and record the parameters. Eventually, you’ll have enough data points to build a model.
Now, assume you feed data about a completely new configuration of meat quality, temperature etc. into your model, curious if that will yield an enjoyable piece of prime beef or some indigestible shoe leather. You get a result… but you’re still curious — why? What parameter made the biggest difference in the model’s decision?
What LIME does is best described as creating perturbations — it runs a lot of relatively similar steak-making configurations through the model, and attempts to isolate which of the factors have the greatest impact on the expected steak quality score. Even better, it can break them down by effect: sure, this beef is quality stuff (suggesting a good outcome), but it’s been overcooked and the grill really isn’t the best for it (suggesting a worse outcome). You can see in quite precise detail what factors pointed which way in assessing your steak outcome.
Now let’s see how that’ll work for molecules.
## Building the toxicity model
As mentioned, we’ll be using the Tox21 data set, and use DeepChem to build a fully-connected network (FCN) based multitask classifier (for more about multitask classifiers, look here). Our data contains twelve ‘tasks’ or assays — each of these are individual toxicity pathways, key biomolecules through which a molecule can express toxicity. These include, for instance, the estrogen receptor (ER) alpha signaling pathway, and chemicals that act as agonists (activators) of the ER alpha signaling pathway might disrupt the endocrine system. Similarly, an assay for the antagonism of the p53 cellular tumour antigen might suggest that the compound tested is likely to inhibit the process by which p53 protects cells from DNA damage upon insult (through inducing DNA repair, halting the cell cycle or triggering cell death, known as apoptosis). The Tox21 data set is publicly available, and contains data for approximately 10,000 chemicals. We will be looking at a reduced sample of 7,084 compounds, which have data for all twelve ‘tasks’, split into a training set (n=6264) and a test set (n=784).
Using DeepChem’s fcnet.MultitaskClassifier (saving us manually building a multitask classifier), we’re going to learn every ‘task’ toxicity, i.e. every assay, of each molecule, simultaneously. After merely 20 iterations, we get a mean ROC AUC of 0.757, with the ROC AUC being highest for the MMP ‘task’ (mitochondrial membrane potential disruptors) at 0.862 and the lowest for the estrogen receptor task, at only 0.650. This is an acceptable result, especially given that the task-wise ROC AUC helps us determine how reliable a particular judgment on the toxicity of a compound is depending on which assay suggested the toxicity.
Let us randomly pick a molecule our model believes is toxic. Because the MMP task was the most accurate, let’s pick one where the model suggests an MMP disruption mechanism for toxicity. I randomly picked this beauty called epigallocatechin gallate, which is abundantly present in tea and drug candidate for hypertension, but can also be toxic in large doses — indeed, the EFSA warned in 2018 that large doses may cause liver injury. That, almost definitely, is the result of the mitochondrial toxicity the MMP assay suggested. But why? Which parts of this gorgeous-looking molecule cause havoc in hepatocytes?
## Squeezing the LIME
Fortunately, LIME can tell us just fine (and, slightly creepily, what parts of it make it lesslikely to be toxic). Let’s look at the five most significant fingerprints — that is, Morgan features — that are the most strongly associated with either toxicity or its absence according to LIME. Recall that LIME causes local perturbations, i.e. small changes in the input to see how that affects the outcome, thereby granting us a glimpse into the model itself. The model suggests that fingerprints 578 and 352 occurring in the absence of fingerprint 456 strongly suggests toxicity (96%), while the absence of fingerprints 420 and 175 suggest non-toxicity (or, to phrase it in the reverse, the molecule would be even more likely to be toxic if it incorporated those fingerprints).
So far, we’ve been talking about fingerprints in the abstract. Let’s see what is actually meant here. We can reverse the Morgan features’ (or fingerprints’) hashes to reveal what these fingerprints look like.
Of course, what is really interesting is not so much what makes one compound (probably) toxic, but rather what fingerprints all molecules that exhibit toxicity, say, through the estrogen receptor (ER) pathway, have in common. If we didn’t know what the estrogen receptor looked like (which we do), and thus did not know where its binding sites are, but still had to figure out what a ligand (a molecule that interacts with a protein) would have to look like, we could look at what fingerprints the molecules that are proven agonists of the ER pathway have in common. This would then yield a useful first approximation of motifs that might be correlated to ER pathway activation.
## Conclusion
The advent of quantitative high-throughput screening (qHTS), where thousands of molecules can be tested for toxicity or agonism on human cell lines with ease, allowing us to answer the big questions: why — owing to what structural motif — is one chemical toxic, but another isn’t? And beyond toxicity screening, peeking inside such models might help us understand more about why molecules do what they do in a biological context, opening up new vistas from drug discovery to creating better tasting synthetic flavourants. The possibilities are as endless as the human imagination — which is quintessentially about asking why?, all the time.
Even if it annoys teachers a lot.
References [ + ]
1 ↑ Ribeiro, M.T., Singh, S., Guestrin, C. (2016). “Why Should I Trust You?” Explaining the Predictions of Any Classifier. arXiv:1602.04938v3
Data scientist by day, computational epidemiologist by night, constantly sleep-deprived husband and dad to the world's most adorable Golden Retriever puppy.
Data science
Coding
Coding | 2020-10-25 17:39:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.5601426362991333, "perplexity": 2202.5886861458816}, "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/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00719.warc.gz"} |
https://mathoverflow.net/questions/273279/on-fractional-dimensions-with-dima-times-b-dim-a-dim-b | # On fractional dimensions with $\dim(A\times B)=\dim A+\dim B$
The Hausdorff dimension satisfies $$\dim A + \dim B \leq \dim (A\times B)$$ while the (upper) packing dimension has $$\dim_p (A\times B) \leq \dim_p A + \dim_p B$$ These inequalities hold for any subsets of the Euclidean space.
1. Do we know any fractional dimensions with the equation instead of the inequalities?
2. Any arguments where it would help to have such a dimension?
For 1. all I can find are statements of the kind: if $A$ and/or $B$ are good enough then one or both of the inequalities above become equations, but nothing for general subsets.
For 2. I found a paper by Darji and Keleti "Covering $\mathbb{R}$ with translates of a compact set" Proc. Amer. Math. Soc. 131 (2003), no. 8, 2598–2596.
It seems I might be getting such a dimension, but I can't find any context for it. | 2019-04-25 13:30:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7900844216346741, "perplexity": 275.45492205491246}, "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-2019-18/segments/1555578721441.77/warc/CC-MAIN-20190425114058-20190425140058-00448.warc.gz"} |
http://logic.dorais.org/archives/107 | I just read an interesting paper by Thomas Forster: The Paris–Harrington Theorem in an NF context [1]. This paper is a progress report on the status of the Paris–Harrington Theorem in Quine’s New Foundations (NF). Forster shows that the Paris–Harrington Theorem is at least consistent with NF, but leaves open the question whether it is provable in NF. Recall that the Paris–Harrington Theorem is a variation on Ramsey’s Theorem, where the homogeneous set $x$ is required to be relatively large: $\min(x) < |x|$. This result is not provable in Peano Arithmetic (PA). In fact, it is logically equivalent to $\Sigma_1$-reflection for PA.
The main issue with the Paris–Harrington Theorem in NF is that the notion of relative largeness is ill-typed. In addition to Forster, Harvey Friedman had also observed this [2]:
In ‘relatively large’, an integer is used both as an element of a finite set and as a cardinality (of that same set).
This is a serious problem in NF where cardinal numbers are defined as equivalence classes of equinumerous sets. Thus, the relative largeness condition $\min(x) < |x|$ entails a direct comparison between an element of the finite set $x$ and the cardinal number $|x|$ to which $x$ belongs. Needless to say that this is difficult to formulate in a stratified manner, as required for set comprehension in NF. Forster manages to work around this, but the winding road has so many twists that he only manages to get a consistency result rather than an actual proof.
While Forster does not directly discuss the issue of the strength of the Paris–Harrington Theorem in the paper, he does raise an interesting question whether the added strength of the theorem has any link to this failure of typing. I used to believe that while the naturalness of the Paris–Harrington Theorem could honestly be questioned, there was one real advantage of this example over the non-provable statements of Gödel and Rosser in that the Paris–Harrington Theorem did not appear to directly depend on self-reference. This observation by Forster and Friedman casts some serious doubts on this…
• Is the notion of relative largeness really self-referential?
• Is the fact that relative largeness is ill-typed the key to the strength of the Paris–Harrington Theorem?
#### References
[1] T. E. Forster, “The Paris-Harrington theorem in an NF context,” in One hundred years of axiomatic set theory, Acad.-Bruylant, Louvain-la-Neuve, 2010, vol. 17, pp. 97-109.
[Bibtex]
@incollection{Forster10,
AUTHOR = {Forster, Thomas Edward},
TITLE = {The {P}aris-{H}arrington theorem in an {NF} context},
BOOKTITLE = {One hundred years of axiomatic set theory},
SERIES = {Cahiers Centre Logique},
VOLUME = {17},
PAGES = {97--109},
YEAR = {2010},
MRCLASS = {03E70 (03E35)},
MRNUMBER = {2796910},
}
[2] H. Friedman, [FOM] PA incompleteness, 2007.
[Bibtex]
@misc{Friedman07,
AUTHOR = {Friedman, Harvey},
TITLE = {{[FOM]} {PA} incompleteness},
HOWPUBLISHED = {FOM},
NOTE = {\url{http://www.cs.nyu.edu/pipermail/fom/2007-October/012046.html}},
YEAR = {2007},
URL = {http://www.cs.nyu.edu/pipermail/fom/2007-October/012046.html},
} | 2017-10-23 14:54:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.787647008895874, "perplexity": 1051.8803271946172}, "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-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00250.warc.gz"} |
https://rkilingr.me/categories/algo/ds/ | ## Taxi Cab Geometry and Manhattan Distance
Problem Given two points $P_1$ & $P_2$ in a grid, and you need to reach from one of the cells to the other, Allowed moves will be up, down, left and right. What is the minimum steps required to reach the destination? TaxiCab Geometry and Manhattan Distance Going by euclidean geometry in a 2D surface, distance between any given two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $d = \sqrt{(y_2^{2}-y_1^{2})-(x_2^2-x_1^2)}$ This is the straight line distance drawn as a diagonal between the two points....
January 29, 2022 · 1 min · RaviKiran Kilingar | 2023-02-07 02:54:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45121753215789795, "perplexity": 289.3748904834223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00050.warc.gz"} |
http://chemistry.tutorcircle.com/inorganic-chemistry/nitrogen-family.html | Sales Toll Free No: 1-800-481-2338
# Nitrogen Family
Top
Sub Topics The elements nitrogen (N), phosphorous (P), arsenic (Ar), antimony (Sb) and bismuth (Bi) constitute group 15 or VA of periodic table. Nitrogen is very important and well known element of this group. It constitutes nearly 78.1percent by volume and 75.5 percent by weight of the air is essential constituent of fertilizers, explosives and food stuffs.After the name of the first element these elements are collectively known as nitrogen family. These elements are also known as pnictogens and their compounds pnictides which contain X3- species. Nitrogen and phosphorus are non metals arsenic and antimony are metalloids and bismuth is a metal.
## Nitrogen Group
The elements are organized and displayed on a chart called the periodic table. It is broken down into columns called groups and rows called periods. The fifteenth column of the periodic table known as group 15 contains five elements.
The nitrogen family elements are listed below.
1. Nitrogen (N)
2. Phosphorus (P)
3. Arsenic (Ar)
4. Antimony (Sb)
5. Bismuth (Bi)
This group is often called the nitrogen group.
The nitrogen group was once called the pnictogen group. This comes from the Greek word that means "to choke." The origin of this term comes from the fact that pure nitrogen can cause asphyxiation when breathed in because it does not contain the oxygen necessary to support life.
## Nitrogen Family Periodic Table
The elements nitrogen, phosphorus, arsenic, antimony and bismuth are referred to as the nitrogen family of element and occupy group V-A in the periodic table. The nitrogen family periodic table is shown below.
Nitrogen and phosphorus are non metals, arsenic is a semi metal or non metal, antimony and bismuth are metals. Their general electronic configuration is ns2np3.
## Physical Properties of Nitrogen
The physical properties of group 15 elements are given below.
1. Physical state - There is a gradual change in physical states of these elements are observed. Nitrogen is a gas, phosphorus is solid and at low temperature it can be vaporized. The other three elements are solids.
2. Metallic and non metallic character - Metallic character varies depending on the change in atomic number. Nitrogen and phosphorus are non metals whereas arsenic, antimony and bismuth are metals.
3. Allotropy - All elements exhibit allotropy except nitrogen. Phosphorus exists as white, red, scarlet and black phosphorus, arsenic exists in three forms grey, yellow and black, arsenic exists as yellow, black and explosive antimony.
4. Melting and boiling points - On moving down the group the melting and boiling points increases with increase in atomic number.
5. Density - Density also increases on moving down the group with increase in atomic number.
6. Atomic volume, atomic radius and ionic radius - All these property increases with increase in atomic number of the elements of group 15.
7. Electronegativity - it decreases on moving down the group indicating a gradual change from non metallic to metallic character.
8. Ionization energies - It decreases regularly on moving down the group due to increase in size of the atom and decreasing shielding effect of electrons from the nucleus.
9. Oxidation state - it exhibit both positive and negative oxidation state. The common oxidation state is +3. nitrogen shows +1, +2, +4 and +5 and phosphorous shows +4 oxidation state also.
10. Catenation - it is shown by all the elements of this group. as is apparent from bond energies the tendency for catenation decreases on moving down the group from nitrogen to bismuth.
## Chemical Properties of Nitrogen
The chemical properties of group 15 elements are given below.
Formation of hydride
M + H2 $\rightarrow$ MH3
All elements form MH3 type hydride. The thermal stability of hydride of this group decreases on moving down the group. This is due to the increase in atomic size. The basic nature also decreases on moving down the group because of the increase in size of the central metal atom. The melting point and boiling point goes on increasing as we move down the group due to increase in molecular size. Except ammonia (NH3) all other hydride of other elements do not exhibit hydrogen bonding. All hydride are pyramidal in shape. On moving down the group the bond angles decrease due to decrease in bond-pair-bond-pair repulsion.
Formation of oxides
M + O2 $\rightarrow$ MO2
M + O2 $\rightarrow$ M2O3
All the elements of this group form three oxides except nitrogen which forms five oxides. Nitrogen forms N2O, NO, N2O3, NO2 and N2O5. Phosphorus also forms an extensive series of sulphides and the most stable is tetra phosphorus trisulfide P4S3. The main use of this compound is it is used in strike anywhere matches. Apart for this other elements As, Sb and Bi form oxides of general formula M2O3. The oxides of nitrogen and phosphorus are acidic and arsenic and antimony are amphoteric and bismuth is basic in nature.
Formation of oxyacids
In nitrogen several oxyacids are unstable in pure state and are known only in aqueous solution or as their salts. only two are well known. One is Nitrous acid (HNO3) and the other is nitric acid (HNO3). Nitrous acid is a weak acid in aqueous solution and sodium nitrate is is mildly toxic to human beings. Nitric acid is prepared by Ostwald process. In combination with concentrated HCl aqua-regia is formed. Apart for these two hypo nitrous acid and peroxonitric acid are also called oxyacids of nitrogen. Important oxyacids of phosphorus are listed below.
1. Orthophosphoric acid - H3PO4
2. Pyrophosphoric acid - H4P2O7
3. Metaphosphoric acid - HPO3
4. Phosphonic acid - H3PO3
5. Hypophosphorys acid - H3PO2
6. Hypophosphoric acid - H4P2O6
Formation of halides
M + X (halogen) $\rightarrow$ MX3
All the elements of this group form halides of the type MX3. The most stable halides of nitrogen is NF3. Except NF3 and PF3 trihalides are hydrolyzed by water and the ease of hydrolysis decreases down the group. NCl3 is easily hydrolyzed and SbCl3 and BiCl3 are partly hydrolyzed. P and Sb form stable pentahalides but nitrogen do not form pentahalides due to absence of d-orbitals in valence shell. Bi do not form pentahalides due to inert pair effect. PCl5 is used as a chlorinating agent.
## Hydrazine
Hydrazine is a colorless fuming liquid. The odor is like ammonia. It is prepared by the oxidation of ammonia with alkaline sodium hypochlorite in the presence of gelatin.
2NH3(aq) + NaClO(aq) $\rightleftharpoons$ N2H4(aq) + NaCl(aq) + H2O(l)
Hydrazine is a strong reducing agent as reflected in the reduction potential. Hydrazine reducing ability is exploited in its use in waste water treatment for chemical plants. A related use is the treatment of water boilers in large electric generating plants.
## Hydrazoic acid
In hydrazoic acid 3 nitrogen atoms are linked together. To synthesize the acid it is therefore reasonable to start with compounds in which 2 nitrogen atoms are already bound together such as dinitrogen oxide or hydrazine.
Free hydrazoic acid can be obtained in high yield from the sodium salt by distillation with dilute sulfuric acid followed by dehydration of the distillate with calcium chloride. Hydrazine can be converted to hydrazoic acid by reaction with nitrous acid in either at 0oC in the presence of sodium methoxide which forms the sodium salt.
HNO2 + H4N2 $\rightarrow$ HN3 + 2H2O
## Hydroxylamine
Hydroxylamine is a key intermediate in nitrogen fixation. Hydroxylamine can be prepared by the reduction of hydroxylamine hydrochloride by sodium but-oxide in solution or from the oxidation of nitrogen monoxide or the hydrolysis of nitroparaffins.
It is used to prepare oximes, an important functional group. It is also an intermediate in biological nitrification. Hydroxyamine is used as a nucleophile in aromatic substitution reactions as a reducing agents and for the conversion of aldehyde to nitriles.
## Nitric Oxide
It is a colorless gas produced in the laboratory when copper metal is treated with dilute nitric acid.
3Cu(s) + 2NO3-(aq) + 8H+(aq) $\rightarrow$ 3Cu2+(aq) + 2No(g) + 4H2O(l)
Nitric oxide is also important in many biological processes where it helps transmit messages between nerve cells and kills harmful bacteria. It also helps to protect the heart from insufficient oxygen levels by dilating blood vessels.
## Nitrous Oxide
Nitrous oxide (N2O) is a colorless, sweet smelling gas obtained when molten ammonium nitrate is heated gently at about 270oC. strong heating of this compound can produce explosion.
NH4NO3(l) $\overset{Heat}{\rightarrow}$ N2O(g) + 2H2O(g)
Known as laughing gas because small dose are mildly intoxicating, nitrous oxide is used as a dental anesthetic and as a propellent for dispensing whipped cream.
## Uses of Nitrogen
Nitrogen is used for a large number of purposes. These include inerting, process, purging and pressurizing. One main use of nitrogen is inerting. Inert atmospheres are often required in processes such as polymerization, catalyst preparation and regeneration and in plant such as centrifuges and storage tanks.
In particular it may be used to dilute the concentration of a reactant such as oxygen in an oxidation process. In high pressure processes the use of nitrogen permits the use of such pressures while limiting the partial pressure of oxygen.
## Nitrogen Family Reactivity
Nitrogen is chemically the least reactive of all elements, and enters into chemical reactions only at very high temperatures. Nitrogen is considered as an inert element because nitrogen gas normally does not interact with other substances. Individual atoms of nitrogen are highly reactive.
Nitrogen is a gas at room temperature. A nitrogen molecule N2 consists of two atoms bonded by a triple covalent bond. Phosphorous on the other hand is very reactive. In general the reactivity increases on going down the group. The trend in reactivity is shown below.
## Nitrogen Family Characteristics
1. Atomic radii - Atomic radii of these elements increases in moving down the group.
2. Ionization energy - The ionization energy decreases regularly in moving down from nitrogen to bismuth.
3. Electronegativity - The electronegativity value decreases on moving down the group from nitrogen to bismuth.
4. Melting point and boiling point - Both the properties increases from nitrogen to arsenic due to increase in molecular size.
5. Metallic character - On moving down the group metallic group increases.
6. Density - This property increases down the group. | 2015-01-31 08:28:24 | {"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": 0.4694947600364685, "perplexity": 4431.168731844794}, "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/1422118108509.48/warc/CC-MAIN-20150124164828-00039-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://cracku.in/rrb-ntpc-27-april-2016-shift-1-question-paper-solved?page=10 | # RRB NTPC 27 April 2016 - Shift 1
Instructions
For the following questions answer them individually
Question 91
What is the local name given to the parliament of Pakistan?
Question 92
Which state of India has the largest number of cotton mills?
Question 93
Find the maximum area of a rectangular field which is surrounded by a rope of 100 meters?
Question 94
On the basis of Memory size & performance, which type of computer is known as "Big iron"?
Question 95
If 90% of A = 30% of B and B= x% of A, find x.
Question 96
Which of the following fractions is largest?
Question 97
The Value of $$sin 60$$ is:
Question 98
Which of the following hormone is not secreted by Pituitary gland?
Question 99
A, B and C are employed to complete a work in Rs. 784. If A and B together complete 23/28 of the work Calculate the wages (in Rs.) paid to C?
Question 100
Find the value of $$200 \times 20 \times 2 \times 0.2 \times 0.02 \times 0.002$$=?
OR | 2020-01-17 18:19:07 | {"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": 0.380074679851532, "perplexity": 2605.786350724373}, "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/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00263.warc.gz"} |
http://mathhelpforum.com/calculus/212332-parametrization-curve.html | # Math Help - Parametrization of the curve
1. ## Parametrization of the curve
I don't know what I'm doing.
x^2+y+1-xy=2
x^2-1+y-xy=0
(x-1)(x+1)-y(x-1)=0
(x-1)(x+1-y)=0
.
.
oh...
can't find any clue to mfind the parametrimization of the curves.
2. ## Re: Parametrization of the curve
Hey happymatthematics.
Does the curve meet both constraints? (If so, hint: get rid of the z term first).
is
{z=t
{y=s
{x=(1-t)/s
correct?
4. ## Re: Parametrization of the curve
Can you outline how you got the answer (the method is always more important than the final answer in isolation)?
5. ## Re: Parametrization of the curve
Hello chiro.
I've tried in two ways.
Does any one of my two attempts correct?
I don't know the reason why we need to do parametrization.
sometimes, when I encounter some simple system of equations, I can just let x=t, y=s and then write z in terms of t and s.
but most of the time, it is not the case.
how do I know or how can I check if my parametrization is correct?
Thank you.
by the way, since I'm afraid if I wrote a long question then no one would answer me, I raise a question as short as possible.
It's happy to heard from you that outlining my scratch work!
6. ## Re: Parametrization of the curve
Originally Posted by happymatthematics
I don't know what I'm doing.
x^2+y+1-xy=2
x^2-1+y-xy=0
(x-1)(x+1)-y(x-1)=0
(x-1)(x+1-y)=0
x- 1= 0 only when x= 1 so as long as x is not 1, you have x+1- y= 0 or y= x+1.
From the original equations, you already have z= 1- xy= 1- x(x+ 1)= 1- x- x^2.
Now take x itself as parameter: x= t, y= t+ 1, z= 1- t- t^2.
You need to remember that there are many different "parameterizations" for the same curve so at some point you need to make a "choice" as I did in setting x= t.
oh...
can't find any clue to mfind the parametrimization of the curves. | 2014-09-02 17:05:06 | {"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": 0.8404461145401001, "perplexity": 1588.531844658683}, "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-2014-35/segments/1409535922089.6/warc/CC-MAIN-20140901014522-00201-ip-10-180-136-8.ec2.internal.warc.gz"} |
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Name Size Last modified (GMT) Description Parent directory 2012-03-06 18:52:49 CVS/ 2012-03-06 18:52:50 config/ 2012-03-06 18:52:50 docs/ 2012-03-06 18:52:50 m4/ 2012-03-06 18:52:50 tinatool/ 2012-03-06 18:52:51 toolkits/ 2012-03-06 18:52:51 AUTHORS 165 bytes 2005-05-30 15:43:19 COPYING 1603 bytes 2005-05-30 15:43:19 ChangeLog 89679 bytes 2009-05-15 15:51:20 INSTALL 9367 bytes 2007-02-15 01:55:50 LICENSE 1603 bytes 2005-05-30 15:43:19 Makefile.am 262 bytes 2007-02-15 01:55:50 NEWS 201 bytes 2005-05-30 15:43:19 README 31956 bytes 2007-02-15 01:55:50 bootstrap.sh 80 bytes 2007-02-15 14:26:51 config.guess 44466 bytes 2009-05-13 15:38:15 config.h.in 5039 bytes 2009-05-13 15:38:15 config.sub 32560 bytes 2009-05-13 15:38:15 configure.ac 8668 bytes 2009-04-20 12:05:23 depcomp 12123 bytes 2002-12-09 16:27:05 install-sh 5569 bytes 2002-12-09 16:27:05 ltmain.sh 198965 bytes 2009-05-13 15:38:15 missing 10270 bytes 2002-12-09 16:27:05 mkinstalldirs 1801 bytes 2002-12-09 16:27:05 user.make 30 bytes 2005-07-12 23:00:46
1 TINA 5 README
2 =============
3
4 paul.bromiley@manchester.ac.uk 20/4/2005
5
6
7 Contents
8 --------
9
10
11 1) What is TINA?
12 1a) History
13 1b) Authors
14 1c) What does TINA stand for?
15 1d) How do I know it works?
16 2) Where do I get TINA?
17 3) How do I compile the libraries?
18 3a) Compatibility issues
19 3a.i) Backup Build System
20 3b) Widget sets
21 3c) Compiler warning messages
22 4) How do I compile the toolkits?
23 4a) Toolkits with different widget sets
24 5) How do I write my own TINA code?
25 5a) Writing new algorithmic functionality
26 5b) Porting old TINA 4 code to TINA 5
27
28
29 1) What is TINA?
30 ----------------
31
32
33 TINA (TINA Is No Acronym) provides a set of libraries for machine vision and
34 image analysis research. It is written in C primarily for use on GNU/Linux
35 platforms, although it can also be compiled on MacOSX, Solaris, and (with some
36 effort) MS Windows. The code is split into two areas:
37
38 tina-libs: back-end code for memory management, image structure handling
39 etc.
40
41 tina-tools: high-level algorithms, interface etc.
42
43 Each of these areas contains a number of libraries with specific purposes:
44
45 tina-libs:
46
47 file: input/output file handling
48 geometry: basic geometry functions
49 image: basic image processing functions
50 math: basic mathematical functions
51 medical: back-end functions for medical image analysis
52 sys: basic memory allocation functions
53 vision: back-end functions for machine vision
54
55 tina-tools:
56
57 draw: basic display drawing
58 gphx: low-level interface to widget sets
59 wdgts: high-level interface to widget sets
61 tlvision: machine vision tools
62 tlmedical: medical tools.
63
64 In order to use TINA, you must download both the tina-libs and tina-tools areas
65 from our website, and compile both. This will produce a set of libraries in the
66 libs subdirectories of both tina-libs and tina-tools.
67
68 The TINA libraries can be used in one of two ways. They have been designed to
69 be highly modular, allowing users to strip out parts of the functionality to
70 use in their own code (under the terms of the licence). However, they also
71 include code for interfaces to the algorithms. This code is in the form of a
72 set of "tools", each of which appears as a window on the screen, and which
74 single images, stereo image pairs, temporal sequences or medical image
75 volumes; stereo vision; medical image segmentation; registration; blood flow
76 analysis etc. etc. ). In order to use this interface, you must build a
77 "toolkit", a program that includes the tools you want to access.
78
79 TINA is research software, continuously updated by a small group of core
80 programmers at the University of Manchester, primarily for their own
81 use. There is no "stable" release as such: the code in both the CVS repository
82 and the tarballs is the latest version. However, it is written with numerical
83 stability and statistical validity in mind, and all new additions are
84 thoroughly tested.
85
86
87 1a) History
88 -----------
89
90
91 Development of the original X11 based versions of TINA started in 1989 at the
92 Artificial Intelligence & Vision Research Unit (AIVRU) at the University of
93 Sheffield (www.shef.ac.uk). It was designed using experience gained in two
94 previous versions with an aim to minimise the process of software maintenance
95 and maximise code reuse. Later versions of TINA were developed there and also
96 at the Electronic Engineering Department of the University of Sheffield. TINA
97 became Open Source whilst being maintained by Electronic Engineering. Since
98 1997 TINA maintenance has been switched to Imaging Science and Biomedical
99 Engineering (ISBE) (www.isbe.man.ac.uk) at the University of Manchester
100 (www.manchester.ac.uk).
101
102 Until 2002 TINA was never explicitly funded as a project in its own
103 right. Rather it has been developed under numerous research grants and
104 collaborations as a framework to facilitate an ongoing co-ordinated research
105 effort into artificial vision. TINA was originally conceived by AIVRU
106 researcher Stephen Pollard as a library of functions necessary for
107 stereo-based 3D machine vision. John Porrill and Neil Thacker where also
108 involved at these early stages. After this team broke up Neil Thacker
109 continued vision research within TINA at Electronic Engineering, and together
110 with Tony Lacey released TINA as Open Source. In 1997 both Neil Thacker and
111 Tony Lacey moved to ISBE, from where TINA is now maintained. Tony left for a
112 career in industry in 2004: today, Neil Thacker is the project leader for
113 TINA, and Paul Bromiley performs most of the day-to-day maintenance of the
114 code and website.
115
116
117 1b) Authors
118 -----------
119
120
121 Along the way a large number of people have contributed to TINA both at the
122 University of Sheffield and the University of Manchester. Their contributions
123 are never forgotten (and their code never lost). A full list can be found at
124 www.tina-vision.net/people.php.
125
126
127 1c) What does TINA stand for?
128 -----------------------------
129
130
131 Originally the acronym TINA stood for "there is no alternative", which was
132 true at the time. Since then, several alternative meanings have been
133 suggested ("Thacker is not acceptable", "TINA is nearly accurate" etc.: most
134 of these were the result of a bad day at work). Today however, in a similar
135 style to acronym GNU, TINA stands for "TINA is no acronym". All previous
136 definitions have now been superseded!
137
138
139 1d) How do I know it works?
140 ---------------------------
141
142
143 For most software, the question of accuracy is non-existent: your word
144 processor either lets you prepare and print a document, or does not. However,
145 for scientific software the question becomes more problematic because the
146 requirements are more extensive. The results of any machine vision algorithm
147 will be dependent on the statistical validity and numerical stability (or what
148 might be called the "theoretical" and "implementation" aspects) of the
149 algorithm, and any failure to meet these two goals might not be obvious to the
150 user without extensive validation work. Any attempt to justify the algorithms
151 used in TINA here would be largely pointless, since such justification can
152 only be meaningful after a suitable "negotiation of trust". Therefore, I will
153 commence that process with three simple observations:
154
155 a) All of the algorithms in TINA have been derived strictly within a context
156 of statistical validity: you won't find any active contour methods with
157 arbitrary image potential terms here! Therefore, the results of any algorithm
158 can be linked directly to the probability theoretic interpretation of the data
159 generation process. This is not a guarantee that a specific algorithm will
160 work for your data. However, it is a guarantee that, if your data obeys the
161 underlying statistical assumptions relevant to the algorithm (as specified in
162 our published papers: see point b), the interpretation of the data provided by
163 the algorithm will be valid. In theory, your results can be linked directly
164 all the way back to the underlying axioms of mathematics.
165
166 b) All of the algorithms in TINA have been used in our scientific research,
167 and thus subjected to peer-review in conference and journal publications. We
168 make preprints of these publications available via our website at
169 www.tina-vision.net/docs/memos.php. These papers contain details of our
170 in-house validation work.
171
172 c) All of the algorithms in TINA are available as open-source software.
173 Through a combination of reading the memos, documentation, and code, you can
174 see exactly how a given algorithm works. This facility is rarely afforded to
175 users of commercial machine vision or medical image analysis software. If you
176 think you have spotted a mistake in our work, get in touch! You will either
177 gain (admittedly limited) fame as an external contributor named on the TINA
178 website, or a very lengthy tutorial on statistics (e.g. if you mention the
179 words "Bayes Theory" within earshot of one of the core programmers).
180
181
182 2) Where do I get TINA?
183 -----------------------
184
185
186 The tina-libs and tina-tools code can be downloaded from our website
187 (www.tina-vision.net), either in the form of tarballs, or from the CVS
188 repository. Go to the website and click on the "software" link. Then, either
189 download the two tarballs or use the CVS instructions given there.
190
191 Tarballs:
192
193 Create a directory in /usr/local called "Tina5", and download both tarballs to
194 this directory. Then unzip (using gunzip tarball_name) and untar (using tar
195 -xvf tarball_name) both tarballs: this will create the two directories
196 tina-libs and tina-tools in the /usr/local/Tina5 directory
197
198 CVS:
199
200 Create a directory in /usr/local called "Tina5", then from within this
202 the CVS repository.
203
204 At the end of this step, you should have two directories, called tina-libs and
205 tina-tools, in the /usr/local/Tina5 directory.
206
207
208 3) How do I compile the libraries?
209 ----------------------------------
210
211
212 Before you start, you must have
213
214 autoconf 2.54 (or later)
215 automake 1.6.3 (or later)
216 libtool 1.4.2 (or later)
217
218 installed on your system in order to build TINA: all reasonably recent Linux
219 distro's should install these by default.
220
221 The basic approach is to cd into tina-libs, then type
222
223 ./bootstrap.sh
224 ./configure
225 make
226 make install
227
228 Then cd into tina-tools, and do the same. If these steps complete without
229 errors, you should end up with the compiled libraries in
230 /usr/local/Tina5/tina-libs/libs and /usr/local/Tina5/tina-tools/libs.
231
232 The full set of options provided by configure can be seen using;
233
234 ./configure --help
235
236 There are currently no specific tina-libs options (as of version 5.0rc0).
237
238 There are a couple of other concerns though...
239
240
241 3a) Compatibility Issues
242 ------------------------
243
244
245 If the above instructions fail to work for your system, your first port of
246 call should be the WikiWiki page on our website. Go to www.tina-vision.net,
247 click on "wiki" in the menu, and then click on "Tina5CompileGuide". If your
248 system is listed, then it will provide work-arounds or full solutions for
249 known problems. If not, pipe the outputs of ./configure and make to files and
250 e-mail them to me (paul.bromiley@man.ac.uk) and I will do my best to help you
251 out. I would appreciate feedback on compilation for systems not listed on the
252 wiki page: it allows us to keep the instructions up-to-date.
253
254
255 3a.i) Backup Build System
256 -------------------------
257
258
259 As well as the main, automake and autoconf based build system, TINA has a
260 backup system of hand-written Makefiles (a legacy from the TINA4 days). In
261 extremis, this can be used to work around serious compatibility issues. It
262 still requires that you run autoconf, since it needs access to config.h, but
263 bypasses the automake-based makefiles entirely. Read the inline documentation
264 in Tina5/tina-libs/tina-simple-build.sh for further information.
265
266
267 3b) Widgets
268 -----------
269
270
271 The tina-tools area contains interface code, arranged as a set of "tools" that
273 single images, stereo image pairs, temporal sequences or medical image
274 volumes; stereo vision; medical image segmentation; registration; blood flow
275 analysis etc. etc.). This interface code is dependent on external widget and
276 graphics libraries. The tina-tools area contains wrappers for these libraries
277 (in the wdgts and gphx libraries), and is currently capable of using five
278 different widget sets:
279
280 NULL: (NULL widgets and NULL graphics) for testing or wrapping TINA code for
281 use in other programs: provides no interface.
282
283 Tcl/Tk: pronounced "tickle-tk" (Tcl widgets and Tk graphics) for text-based
284 interfaces and client/server operation).
285
286 Xview: sometimes called openwindows, (Xv widgets and X11 graphics), which is
287 our usual mode of operation. However, the external xview libraries
288 themselves do not work on 64-bit systems.
289
290 Motif: (Xm widgets and X11 graphics): an early replacement for xview, written
291 at a point when it looked like xview would be deprecated. Xview was
292 revived due to popular demand, but it looks like the end of xview has
294
295 Gdk/Gtk:stands for "the GIMP (Graphical Image Manipulation Program) toolkit",
296 since it was originally written exclusively for the GIMP, and is the
297 widget set of choice for modern linux software. It is anticipated that
298 this will become the default for TINA in the next few years, with the
299 motif widgets as a backup for older systems, and xview deprecated. Two
300 versions of Gtk (1.2 and 2.0) are currently in use: TINA compiles with
301 1.2, but there are a few issues with 2.0. We are working on fixing
302 this (by the time you read this, there may be experimental GTK 2
303 widgets in a directory next to the existing GTK 1.2 widgets).
304
305 At the library compilation stage, you can compile as many of these as you
306 want. NULL widgets and graphics are always compiled. If you want a graphical
307 interface to TINA, you must compile at least one of xview, Motif or Gtk.
308
309 In order to compile the TINA widgets and graphics libraries you want to use,
310 you must ensure that the external libraries they depend on are installed. A
311 complete list of such packages for SuSE 9.2 is:
312
313 For gtk1.2 and gtk2
314
315 fontconfig-2.2.96.20040728-9.i586.rpm
316 fontconfig-devel-2.2.96.20040728-9.i586.rpm
317 pkgconfig-0.15.0-199.i586.rpm
318 atk-1.6.0-4.1.i586.rpm
319 atk-devel-1.6.0-4.1.i586.rpm
320 atk-doc-1.6.0-4.1.i586.rpm
321 pango-1.4.1-3.1.i586.rpm
322 pango-devel-1.4.1-3.1.i586.rpm
323 pango-doc-1.4.1-3.1.i586.rpm
324 glib-1.2.10-589.i586.rpm
325 glib-devel-1.2.10-589.i586.rpm
326 glib2-2.4.6-5.1.i586.rpm
327 glib2-devel-2.4.6-5.1.i586.rpm
328 glib2-doc-2.4.6-5.1.i586.rpm
329 gtk-1.2.10-882.1.i586.rpm
330 gtk-devel-1.2.10-882.1.i586.rpm
331 gtk-engines-0.12-960.1.i586.rpm
332 gtk2-2.4.9-10.1.i586.rpm
333 gtk2-devel-2.4.9-10.1.i586.rpm
334 gtk2-doc-2.4.9-10.1.i586.rpm
335 gtk2-engines-2.2.0-400.1.i586.rpm
336
337 For motif:
338
339 openmotif-2.2.3-6.1.i586.rpm
340 openmotif-demo-2.2.3-6.1.i586.rpm
341 openmotif-devel-2.2.3-6.1.i586.rpm
342 openmotif-libs-2.2.3-6.1.i586.rpm
343 openmotif21-libs-2.1.30MLI4-123.1.i586.rpm
344
345 For xview:
346
347 xorg-x11-devel-6.8.1-15.4.i586.rpm
348 xview-3.2p1.4-706.1.i586.rpm
349 xview-devel-3.2p1.4-706.1.i586.rpm
350 xview-devel-examples-3.2p1.4-706.1.i586.rpm
351
352 To get the textsw package (optional for xview):
353
354 olvwm-4.1-706.1.i586.rpm
355
356 Version numbers may be different for your own distro.
357
358 Then, at the stage when you type ./configure in the tina-tools area, the
359 last part of the output will be a list of which external widget libraries
360 autoconf was able to find. This search is run by m4 scripts, and targets
361 particular locations, so may not find the external widget libraries.
362 Typical output will look like this:
363
364 >tinatool configuration complete for i686-pc-linux-gnu.
365 >
366 > Source directory: .
367 > Installation directory: /usr/local/Tina5/tina-tools
368 > C compiler: gcc
369 >
371 > tina-libs libraries: /usr/local/Tina5/tina-tools/../tina-libs/lib
372 >
374 > X11 libraries: -L/usr/X11R6/lib
375
376 > GTK flags: -I/usr/include/gtk-1.2 -I/usr/include/glib-1.2
377 > -I/usr/lib/glib/include -I/usr/X11R6/include
378 > GTK libraries: -L/usr/lib -L/usr/X11R6/lib -lgtk -lgdk
379 > -rdynamic -lgmodule -lglib -ldl -lXi -lXext -lX11 -lm
380 >
382 > Xview libraries: -L/usr/openwin/lib -lxview -lolgx
383 >
385 > Motif libraries: -lXm -lXp -lXext
386 >
388 >
389 > Configuration successful! Now type 'make' to build the libraries and then
390 > 'make install' to install the libraries into the prefix location.
391
392 In this example, the X11, xview and Gtk libraries and includes are
393 found. The Motif lines show that the libraries are found but not the
394 headers: watch out for this situation, as it means that the Motif widgets
396 and so again will not be compiled.
397
398 If the external widget libraries you want to use are not found, and you
399 know that they are installed on your system (this will probably be the case
400 for 64 bit systems) then you have to specify the location on the ./configure
401 line. Possible arguments are (these should be typed on one line: they have
402 been split up here for clarity):
403
404 ./configure
405 --with-gtk-prefix= (location of the gtk installation)
406 --with-motif-includes= (location of the motif headers)
407 --with-motif-libraries= (location of the motif libraries)
408 --with-tcl= (location of the tcl installation)
409 --with-tk= (location of the tk installation)
410 --with-xview-includes= (location of the xview headers)
411 --with-xview-libraries= (location of the xview libraries)
412
413 You can find these by browsing through the inlined documentation in the m4
414 scripts (in tina-tools/m4). Typical locations on a 64-bit system will be:
415
416 ./configure
417 --with-xview-libraries=/usr/openwin/lib64
418 --with-xview-includes=/usr/openwin/include
419 --with-gtk-prefix=/opt/gnome
420 --with-tk=/usr/lib64
421 --with-tcl=/usr/lib64
422 --with-motif-libraries=/usr/X11/lib64/Xm
423 --with-motif-includes=/usr/X11/lib64/Xm
424
425 They may be different on your machine. The two xview lines are the lib64 and
426 include directories in the openwin directory (search for openwin). The motif
427 paths can be found by searching for Xm.h. The Tcl and Tk paths can be found
428 by searching for tclconfig.sh and tkconfig.sh. The gtk prefix can be found
429 by searching for gtkConfig.sh
430
431 Add these to the configure line, and check the report: as long as full paths
432 are given for both the includes and headers for the widgets and graphics you
433 want to compile, you will be OK. Then "make" and "make install": now you
434 should find that the tina-tools/lib directory contains libraries called e.g.
435 libtinatoolGphxGdk.a and libtinatoolWdgtsGtk.a (for gtk widgets and gdk
436 graphics).
437
438
439 3c) Compiler warning messages
440 -----------------------------
441
442
443 When you compile TINA you will notice a large number of compiler warnings.
444 They are the legacy of the 20-year development history of TINA: old coding
445 practices that are now considered unacceptable. One example is the
446 assumption that a pointer and an integer occupy the same amount of memory:
447 this is true on 32-bit systems, but not in general on 64-bit systems.
448 We are currently fixing these warnings: however, none of them have been
449 found to affect the algorithmic functionality.
450
451
452 4) How do I compile the toolkits?
453 ---------------------------------
454
455
456 At the end of step 3 you should have built the TINA libraries in both
457 tina-libs and tina-tools. The algorithmic functionality they provide could,
458 in theory, be used in your own code. However, 99% of users will want to use
459 the interface included with the libraries. These interfaces are called
460 "toolkits", and must be built independently of the libraries. Three examples
461 are included in the Tina5/toolkits area: they are called example, example2,
462 and mri_analysis.
463
464 The toolkit of interest to most users will be example2: this includes all
465 tools available at the time of writing i.e. all of the TINA functionality.
466 If you don't intend to do any programming, this is the one you should build.
467 Automatic building and installation is not included (yet) for historical
468 reasons: out scientific users expect to be able to build the libraries
469 independently, and so toolkit building must be performed after the library
470 build. To build this toolkit, follow the instructions below. You can then
471 make a link to the example2 executable in /usr/local/bin, allowing you to
472 run tinaTool from the command line. An icon is included in the example2
473 directory, should you wish to set up a desktop icon.
474
475 To compile one of the example toolkits, cd into example, example2, or
476 mri_analysis, and type make: this will build the tinaTool program, which you
477 can then run. The example toolkit is very basic, but can be used as a
478 starting point for building your own toolkit. The example2 toolkit include
479 all available TINA functionality. The mri_analysis toolkit contains the basic
480 medical images analysis functions: download the MRI image sequences from the
481 image library on our website, read the TINA 5 User's Guide (also on the
482 website) to find out how the tools work, and start experimenting with the
483 algorithms!
484
485 Since TINA is research software, it has been designed to allow users to build
486 their own toolkit for their current project, including only the tools
487 required. Most users will therefore eventually want to write their own
488 toolkit. This is relatively simple: you just have to copy one of the example
489 toolkits to a project area (e.g. /home/my_user_name/my_tina_project), open
490 tinaTool.c in an editor, and add a button for the tool you want to main(), and
491 a button procedure function to call the tool. For example, if you wanted to
492 access the coreg tool, you would add a button to main() using:
493
494 tw_button("Coreg", coreg_tool_proc, NULL);
495
496 and then the button procedure coreg_tool_proc anywhere above main
497
498 static void coreg_tool_proc()
499 {
500 coreg_tool(250, 250);
501 }
502
503 Finally, you have to add the header files that contain the prototypes for the
504 tools you want to include. Prototypes for all tool functions are contained in
505
506 #include <tinatool/tlbase/tlbaseDef.h>
507 #include <tinatool/tlbase/tlbasePro.h>
508
509 for the basic tools (input/output etc),
510
511 #include <tinatool/tlbase/tlvisDef.h>
512 #include <tinatool/tlbase/tlvisPro.h>
513
514 for the machine vision tools, and
515
516 #include <tinatool/tlbase/tlmedDef.h>
517 #include <tinatool/tlbase/tlmedPro.h>
518
519 for the medical image analysis tools. Since the coregistration tool is in the
520 medical tools library, you would add the last two lines to the top of your
521 tinaTool.c file (alongside the other header file include statements), or you
522 could just add all six. The TINA Users guide on the website
523 (www.tina-vision.net/software.php) contains a complete list of available
524 tools.
525
526 After building and running tinaTool, you will see a new button in the main
527 tool called "coreg". Pressing this calls the button procedure, which in turn
528 calls the coreg_tool function from the libraries.
529
530
531 4a) Toolkits with different widget sets
532 ---------------------------------------
533
534
535 In order to compile toolkits with various widget sets, I recommend that you
536 use a simple Makefile. The example2 toolkit directory in tina-tools/example
537 contains an example: it allows you to switch simply between the different
538 available widget sets, although you have to set some paths here by hand. The
539 Makefile is called Makefile_tina: rename it to Makefile (after backing-up the
540 existing Makefile) to begin this process.
541
542 In general, a simple Tina5 makefile should look something like the following
543 (the filename should be "Makefile"):
544
545 ##
546 # Makefile for Tina5
547 #
548
549 TINALIBS = -ltinaMedical -ltinaVision -ltinaImage -ltinaFile -ltinaGeom
550 -ltinaMath -ltinaSys
551 TINATOOLLIBS = -ltinatoolTlMed -ltinatoolTlvision -ltinatoolTlBase
552 -ltinatoolDraw -ltinatoolWdgtsGtk -ltinatoolGphxGdk
553 XLIBS = -lgtk -lgdk
554 SYSLIBS = -lm
555 CFLAGS = -g -pipe -DHAVE_CONFIG_H -DOWTOOLKIT_WARNING_DISABLED
556 INCLUDES = -I./ -I/usr/local/Tina5/tina-tools
557 -I/usr/local/Tina5/tina-libs -I/usr/openwin/include
558 LDFLAGS = -L/usr/local/Tina5/tina-libs/lib -L/usr/local/Tina5/tina-tools/lib
559 $(TINATOOLLIBS)$(TINALIBS) -L/opt/gnome/lib $(XLIBS)$(SYSLIBS)
560 OFILES =
561
562 .c.o:
563 gcc $(CFLAGS)$(INCLUDES) -c -o $@$*.c
564
565 tinaTool: tinaTool.o $(OFILES) 566 gcc -o$@ $(CFLAGS) tinaTool.o$(OFILES) $(LDFLAGS) 567 568 # --------- 569 # Household 570 # --------- 571 572 clean : 573 \rm *.o 574 575 576 There are eight lines that set up various locations. TINALIBS and TINATOOLS 577 list the libraries to be included from the tina-liabs and tina-tools lib 578 directories. These must be in order, and the only changes you should need 579 to make to the above are to the Wdgts and Gphx libraries: these choose the 580 widget set you will use, and there are three possible combinations of interest 581 to the majority of users: 582 583 -ltinatoolWdgtsXm -ltinatoolGphxX11 : Motif widgets 584 -ltinatoolWdgtsXv -ltinatoolGphxX11 : Xview widgets 585 -ltinatoolWdgtsGtk -ltinatoolGphxGdk : Gtk widgets 586 587 The XLIBS line selects the external widget set libraries used: again there are 588 three possible selections: 589 590 XLIBS = -lXm -lX11 : Motif widgets 591 XLIBS = -lxview -lolgx -lX11 : Xview widgets 592 XLIBS = -lgtk -lgdk : Gtk widgets 593 594 I always call this line XLIBS, although it should be named for the libraries 595 you want to install. 596 597 The CFLAGS line passes arguments to the compiler: for most Linux systems, only 598 the first is of interest: -g adds debugging information, making your executable 599 larger, but providing the information necessary for using GDB or DDD for 600 debugging. Add -0, -02 or -03 for various levels of optimisation, making the 601 executable run faster, but making debugging more difficult. 602 603 Finally, the INCLUDES and LDFLAGS lines set paths for the libraries you have 604 specified above. These should look something like: 605 606 INCLUDES = -I./ -I/usr/local/Tina5/tina-tools -I/usr/local/Tina5/tina-libs 607 -I/usr/openwin/include 608 LDFLAGS = -L/usr/local/Tina5/tina-libs/lib -L/usr/local/Tina5/tina-tools/lib 609$(TINATOOLLIBS) $(TINALIBS) -L/usr/include/gtk-1.2$(XLIBS) $(SYSLIBS) 610 611 for gtk, 612 613 INCLUDES = -I./ -I/usr/local/Tina5/tina-tools -I/usr/local/Tina5/tina-libs 614 -I/usr/openwin/include 615 LDFLAGS = -L/usr/local/Tina5/tina-libs/lib -L/usr/local/Tina5/tina-tools/lib 616$(TINATOOLLIBS) $(TINALIBS) -L/usr/openwin/lib -L/usr/X11/lib$(XLIBS)
617 $(SYSLIBS) 618 619 for xview, and 620 621 INCLUDES = -I./ -I/usr/local/Tina5/tina-tools -I/usr/local/Tina5/tina-libs 622 -I/usr/openwin/include 623 LDFLAGS = -L/usr/local/Tina5/tina-libs/lib -L/usr/local/Tina5/tina-tools/lib 624$(TINATOOLLIBS) $(TINALIBS) -L/usr/X11/lib$(XLIBS) \$(SYSLIBS)
625
626 for Motif.
627
628 The library and include locations will have to be set for your machine, but
629 should be fairly standard. Just two points to be aware of: the locations will
630 be .../lib64 rather than .../lib on 64-bit systems, and notice that you refer
631 to libraries in shorthand, so libxview.so becomes lxview (suffix stripped,
632 and l replacing lib).
633
634 The OFILES line refers to object files you wish to compile. If you are
635 compiling one of the example toolkits i.e. the only .c file in the current
636 directory is tinaTool.c, then leave it empty. If you start writing or adapting
637 your own Tina code (see below) and have a directory with other c files in it,
638 add my_c_file.o to this line for every my_c_file.c you want to compile.
639
640 Finally, type "make" in the directory containing the tinaTool.c file and the
641 Makefile: if all goes well, you should end up with a "tinaTool" executable that
642 you can run.
643
644
645 5) How do I write my own TINA code?
646 -----------------------------------
647
648
649 TINA provides the ideal environment for machine vision or medical image
650 analysis researchers to develop their own code. All of the basic functionality
651 (memory allocation, image representation, basic image processing, geometry,
652 maths etc. etc.) is already present, allowing for rapid code development. In
653 addition, the long development history has resulted in highly stable code.
654
655
656 5a) Writing new algorithmic functionality
657 -----------------------------------------
658
659
660 In order to start developing new algorithmic functionality, set up a project
661 area with a simple Makefile and a tinaTool.c file (you can use the example
662 one provided in the tina-tools area). Then, there are two possible options:
663 writing new files, or adapting those already present in the libraries.
664
665 If you are adapting files from the libraries, copy the C files plus their
666 file-level H files to your project area. Then build a copy of the header
667 file directory structure in your project area: put the inctw script
668 in your project area, and run it. The script can be found in the
669 tina-tools/toolkits/example2 directory. This sets up a copy of the
670 header directory system, and soft-links from the relevant directories to the
671 header files in your project area, ensuring that when you compile TINA it
672 links in the local header files in place of the ones in the libraries. Then
675
676 Secondly, if you are writing new C files, you can just work in a project area
678 system doesn't affect you. However, if in the future you will be putting
679 your code back into the libraries, you should think about where it would go
680 and divide up your code accordingly.
681
682
683
684 5b) Porting old TINA 4 code to TINA 5
685 -------------------------------------
686
687
688 Several major changes were made to the TINA libraries in the move from
689 version 4 to version 5. In particular, we have a new header file system, and
690 a new version of the sequence tool. The header file structure for TINA4 was
691 a complete mess (and one of the common criticisms of TINA), so everyone ended
692 up including tina_all.h and thus completely bypassing the whole point of header
693 files. Therefore, we have put a formal structure in place, which makes the
694 code much cleaner, easier to write, and easier to understand.
695
696 The code is divided into three layers: library, group, and file. For example,
697 the function coreg_auto (which performs coregistration) is in the file
698 tlmedCoreg_auto.c, so the group is tlmedCoreg (the group for coregistration
699 functions) and the library is tlmed (the library for medical functions).
700 The tl tells you that it is in the tina-tools area, rather than tina-libs.
701 Every C file has its own H file for prototypes of all non-static functions in
702 that C file. The file level H file must be in the same directory as its C
703 file. Other C files in the same group should include that "file level" H file
704 when needed. Every group has its own "group level" header files: in this
705 example tlmed_CoregPro.h and tlmed_coregDef.h. The Pro file includes all of
706 the file level H files for that group, and the Def file contains all
707 definitions (structures etc) for that group. C files in other groups should
708 include the group level header files when needed. Finally, there are library
709 level header files tlmedPro.h and tlmedDef.h: these include all of the group
710 level Pro and Def files for that library. C files in other libraries should
711 include the library-level header files when needed. Finally, the whole of
712 TINA is split into the tina-tools, which is front end and algorithmic code,
713 and tina-libs, which is lower-level code (memory allocation, handling
714 structures etc.). It may seem a bit complicated, but when you get used to it
715 this system allows you to know immediately which H files you should be
716 including in your own C files, making coding much faster without the need for
717 a tina-all.h file.
718
719 The changes in terms of locations of functions are mostly limited to the
720 sequence tool: some functions (such as get_seq_ptr) have changed their names.
721 The easiest way to port your code is to put it into a project area, strip out
722 all include statements, and compile. Then, start the lxr code browser web
723 page, and search for each function that generates "function not defined"
724 errors: once you know which C file the function is in, you can work out which
725 H file to include. These will be only library-level header files if you
726 never intend to put your code back into the libraries.
727
728 Finally, if there are any functions whose names have changed, search through
729 the libraries that correspond to the relevant area in TINA4. For example, if
730 you are including get_seq_ptr, you know it comes from the sequence tool:
731 search the sequence tool code (group tlbaseSeq for the tina-tools level stuff
732 and imgSeq for the tina-libs level stuff) for a function that does the same
733 job (in this case get_seq_ptr). It seems long-winded at first, but don't
734 despair: after a couple of hours of this, you can do it from memory without
735 using the code browser.
736
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https://physics.stackexchange.com/questions/407827/in-between-neutron-stars-and-black-holes?noredirect=1 | # In-between neutron stars and black holes
Current knowledge and theories,...Suggest a maximum mass for neutron stars about 2-3 solar masses, and is generally assumed that black holes are ANY compact object above that or more generally 5 solar masses. Question: can new physics introduce new populations of intermediate objects between 2-3 solar masses and 5 solar masses? Extreme Compact Objects are sometimes mentioned but...Could also enlarge the assumed minimal mass of black holes (observed indirectly)?
There could certainly be something strange (literally) that might happen that could produce stable stars up to about 3 solar masses (the highest observed and precise neutron star masses are at 2 solar masses), but probably not much higher.
The thing is, even if you postulate some crazy material with the hardest possible equation of state, General Relativity ensures that the pressure at the core of the star contributes to the curvature of space; and the increasing pressure required to support a more massive star actually results in its collapse. The exact mass of this limit depends on the rotation of the star, but I don't think can be far above 3 solar masses.
At present there appears to be a notable gap between the most massive neutron stars and lowest mass black holes. Either compact remnants don't form (or neutron stars can't accrete enough) in this mass range or something stops us finding them (perhaps they can't form in binaries).
Current knowledge and theories, ... Suggest a maximum mass for neutron stars [of] about 2-3 solar masses, and [it] is generally assumed that black holes are ANY compact object above that, or more generally 5 solar masses.
Question: Can new physics introduce new populations of intermediate objects between 2-3 solar masses and 5 solar masses?
Extreme Compact Objects are sometimes mentioned but ... Could also enlarge the assumed minimal mass of black holes (observed indirectly)?
The range of object sizes involved in your question are from almost 3.2 M$_\odot\!$ and 5 M$_\odot$. Any objects in that mass range would not be explained by the cosmological equation of state. I will quote sources to support the upper and lower limits, then I'll discuss the theory behind what happens to matter within that range of masses.
If you accept those numbers without proof you can save a lot of reading by skipping to three quarters of the way through this answer.
Quoting from Wikipedia: For stellar black holes the minimum size is 5 M$_\odot$, any smaller black holes are a hypothetical type of black hole that formed soon after the Big Bang referred to as a primordial black hole. Since primordial black holes did not form from stellar gravitational collapse, their masses can be far below stellar mass (c. $2×10^{30}$ kg). Hawking calculated that primordial black holes could weigh as little as 10$^{−8}$ kg, about the weight of a human ovum.
"A stellar black hole (or stellar-mass black hole) is a black hole formed by the gravitational collapse of a massive star. They have masses ranging from about 5 to several tens of solar masses. The process is observed as a hypernova explosion or as a gamma ray burst. These black holes are also referred to as collapsars.
...
The gravitational collapse of a star is a natural process that can produce a black hole. It is inevitable at the end of the life of a star, when all stellar energy sources are exhausted. If the mass of the collapsing part of the star is below the Tolman–Oppenheimer–Volkoff (TOV) limit for neutron-degenerate matter, the end product is a compact star — either a white dwarf (for masses below the Chandrasekhar limit) or a neutron star or a (hypothetical) quark star. If the collapsing star has a mass exceeding the TOV limit, the crush will continue until zero volume is achieved and a black hole is formed around that point in space.
The maximum mass that a neutron star can possess (without becoming a black hole) is not fully understood. In 1939, it was estimated at 0.7 solar masses, called the TOV limit. In 1996, a different estimate put this upper mass in a range from 1.5 to 3 solar masses.
In the theory of general relativity, a black hole could exist of any mass. The lower the mass, the higher the density of matter has to be in order to form a black hole. (See, for example, the discussion in Schwarzschild radius, the radius of a black hole.) There are no known processes that can produce black holes with mass less than a few times the mass of the Sun. If black holes that small exist, they are most likely primordial black holes. Until 2016, the largest known stellar black hole was 15.65±1.45 solar masses. In September 2015, a black hole of 62±4 solar masses was discovered in gravitational waves as it formed in a merger event of two smaller black holes. As of April 2008, XTE J1650-500 was reported by NASA and others to be the smallest-mass black hole currently known to science, with a mass 3.8 solar masses and a diameter of only 24 kilometers (15 miles). However, this claim was subsequently retracted. The more likely mass is 5–10 solar masses.
There is observational evidence for two other types of black holes, which are much more massive than stellar black holes. They are intermediate-mass black holes (in the centre of globular clusters) and supermassive black holes in the centre of the Milky Way and other galaxies.".
That sets the upper limit at 5 M$_\odot$.
Now to address how the lower limit was derived.
"2.3. Maximum Mass
The mass and the composition of neutron stars (NSs) are intricately related. One of the most important empirical clues that would lead to constraints on a wide range of physical processes is the maximum mass of NSs. For instance, secure constraints on the maximum mass provide insight into the range of viable equations of state (EOS) for matter at supranuclear densities.
A first order theoretical upper limit can be obtained by numerically integrating the Oppenheimer-Volkoff equations (also called the Tolman–Oppenheimer–Volkoff equation, TOV) for a low-density EOS at the lowest energy state of the nuclei (Baym et al. 1971). This yields an extreme upper bound to the maximum mass of a NS at M$_{max}$ ∼ 3.2 M$_\odot\!$ (Rhoades & Ruffini 1974). Any compact star to stably support masses beyond this limit requires stronger short-range repulsive nuclear forces that stiffens the EOSs beyond the causal limit. For cases in which causality is not a requisite (v→∞) an upper limit still exist in general relativity ≈ 5.2 M$_\odot\!$ that considers uniform density spheres (Shapiro & Teukolsky 1983). However, for these cases the extremely stiff EOSs that require the sound speed to be super-luminal (or FTL) (dP/dρ ≥ $c^2$) are considered non physical. [See: Exotic Matter].
Differentially rotating NSs that can support significantly more mass than uniform rotators can be temporarily produced by binary mergers (Baumgarte et al. 2000). While differential rotation provides excess radial stability against collapse, even for modest magnetic fields, magnetic braking and viscous forces will inevitably bring differentially rotating objects into uniform rotation (Shapiro 2000). Therefore, radio pulsars can be treated as uniform rotators when calculating the maximum NS mass.
While general relativity along with the causal limit put a strict upper limit on the maximum NS mass at ∼ 3.2 M$_\odot\!$, the lower bound is mostly determined by the still unknown EOS of matter at these densities and therefore is not well constrained. There are modern EOSs with detailed inclusions of nuclear processes such as kaon condensation and nucleon-nucleon scattering which affect the stiffness. These EOSs give a range of 1.5–2.2 M⊙ as the lower bound for the maximum NS mass (Thorsson et al. 1994; Kalogera & Baym 1996). Although these lower bounds for a maximum NS mass are implied for a variation of more realistic EOSs, it is still unclear whether any of these values are favored. Therefore,
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad M_{max} \, ∼ \, 1.5–3.2 \; \text{M}_\odot \qquad\qquad\qquad\qquad\qquad (5)$$
can be considered a secure range for the maximum NS mass value.
• The paper "Neutron Stars and Black Holes in X-Ray Binaries" (13 Feb 1998), by Jan van Paradijs is a little out of date (for exact sizes and more precise equations of the range of mass in question) but it contains a few informative diagrams that assist one to understand that mass can only accumulate in particular ranges of mass.
On page 12 is this diagram:
"The current lack of knowledge of the EoS and the corresponding uncertainties in the predicted NS masses are illustrated in Fig. 1.
Fig. 1. Left panel: range of equations of state of dense matter (pressure $P$ versus mass density $ρ$), as predicted by various models and consistent with the existence of massive neutron stars. The dotted lines labeled CL and FFG correspond to the causal limit and the free Fermi gas equations of state, respectively (see Sec. 4). Right panel: corresponding range of allowed masses $M$ for nonrotating neutron stars as a function of the central baryon number density $n_c$. The horizontal lines correspond to the precisely measured masses of three pulsars (see Sec. 6).".
...
"Fig. 2. Fractional mass $M_{in}/M$ contained in the inner region of a static spherical NS of mass $M$ and radius $R$, at density $ρ > ρ_⋆$, for two different cases: $ρ_⋆ = 3 × 10^{14} \, g \, cm^{−3}$ (left) and $ρ_⋆ = 5 × 10^{14} \, g \, cm^{−3}$ (right). The shaded areas reflect the uncertainties in the EoS $^{28, 30}$ at $ρ < ρ_⋆$. Only the ranges of $M$ and $R$ allowed by the compactness constraint$^{16, 77}$ $r_g/R ≤ 6/8$ are shown. See the text for details.".
...
5. Effect of rotation on the maximum mass
Rotation increases the maximum mass of NSs because the centrifugal force acts against gravity. We will consider two different cases: (i) rigidly rotating NSs, and (ii) differentially rotating NSs.
...
Setting M = 2M$_\odot\!$ and R = 10 km and using Eq. (17) we find that rotation increases the maximum mass by ∼ 3% only for PSR J1748−2446, whose frequency f = Ω/(2π) = 716 Hz is the highest measured.
...
On page 26 is this useful diagram showing matter's aversion to forming certain sized masses. This text starts on page 25:
"... A few years later, McClintock & Remillard (1986) measured the mass function of the transient source A 0620–00 (which also had a very soft X-ray spectrum during its outburst in 1975) after it had returned to quiescence, to be 3.18±0.16 M$_\odot$.
This immediately (see below) showed that the compact star in this system is too massive to be a neutron star, and gave some confidence in the idea that X-ray spectra may be an efficient way to select BHXBs.
In spite of the fact that some X-ray spectral characteristics of black holes, and rapid variability are also seen in some neutron stars, their combined presence, in particular in X-ray transients, has remained strikingly effective in singling out black holes.
As implied in the above discussion, the main argument that the compact object in a particular X-ray binary is a black hole, is that neutron star masses cannot exceed a certain maximum value. This assumption rests on very general considerations, e.g., that sound cannot travel faster than light, on the basis of which Nauenberg & Chapline (1973) and Rhoades & Ruffini (1974) concluded that any neutron star, independent of the equation of state (EOS) of high-density matter, must have a mass $\small{\lesssim}$ 3 M$_\odot$. Rotation of the neutron star (ignored in the above analyses) does not increase the mass limit by more than 20% (Shapiro & Teukolsky 1983). Detailed modelling of neutron stars, for a wide range of equations of state, leads (see Fig. 10) to upper mass limits between ∼ 1.5 M$_\odot\!$ (very soft EOS) and ∼ 2 M$_\odot\!$ (very stiff EOS) (see, e.g., Arnett & Bowers 1977; Datta 1988; Cheng et al. 1993; Cook et al. 1994; Engvik et al. 1996; see also the contribution of N. Glendenning to this Volume).
The fact that compact objects with dynamical mass estimates exceeding ∼ 3 M$_\odot\!$ cannot be neutron stars, is not equivalent to their being black holes, as defined by the particular space-time structure described by Schwarzschild and Kerr metrics, which are characterized, in particular, by the absence of a hard surface. This has led to the extensive use of the term “black-hole candidate” for these objects. Of course, detection of X-ray pulsations or X-ray bursts immediately disqualifies a compact star as a black hole, but positive evidence for the absence of a hard surface has been very hard to obtain. This should not come as a surprise, since a nominal (M = 1.4 M$_\odot$, R = 10 km) neutron star is just 2.5 times larger than its Schwarzschild radius, and one may expect the accretion flow to be very similar to that of a black hole of comparable mass. The energy release at the neutron star surface, which is absent for a black hole, might lead to observable differences in spectra and variability, but unless the origin of the spectra and variability of X-ray binaries is much better understood than it is nowadays, the conclusion that a black hole has been found on the basis of such phenomena must be considered weak at best.".
[The authors refer again to Figure 10 much later, on page 41.]
"4. Mass determinations of Compact Stars in X-ray binaries
4.1. NEUTRON STAR MASSES AND EQUATION OF STATE
Apart from their crucial role in distinguishing black holes from neutron stars, the importance of measuring the masses of compact stars in X-ray binaries is that they may provide constraints on the properties of the high-density matter in the interior of neutron stars.
These properties are described by an equation of state (EOS), which together with the Oppenheimer-Volkov equations allows one to calculate models of the interior structure of neutron stars (see, e.g., Shapiro & Teukolsky 1983). Since neutron stars can be considered to be zero-temperature objects these models form a one-parameter sequence in which mass, M, and radius, R, depend only on the central density. For a given equation of state one thus has a unique mass-radius relation. Extensive calculations of neutron star models have been made by Arnett & Bowers (1977) and Datta (1988); for a detailed discussion I refer to the contribution of N. Glendenning to this Volume.
Equations of state can be conveniently distinguished by the compressibility of the neutron star matter; for very “stiff” and very “soft” EOS one finds that neutron stars have radii of ∼ 15 km, and ∼ 8 km, respectively (see Fig. 10). Also, the maximum possible neutron star mass depends on the EOS; it is ∼ 1.5 M$_\odot\!$ for very soft EOS, and up to ∼ 2.5 M$_\odot\!$ for the stiffest EOS.
As will be discussed in more detail below, most neutron star masses are consistent with a value close to 1.4 M$_\odot$. From Fig. 10 it appears that at this value masses do not allow one to draw conclusions about the stiffness of the EOS of neutron star matter. For that, one would need observed masses in excess of 1.6 M$_\odot$, which would exclude the softest EOS (note that stiff equations of state are not excluded by low neutron star masses). Similarly, measurements of the gravitational redshift, $z$, at the neutron star surface alone are not a sensitive EOS discriminant, since both stiff and soft equations of state allow $M/R$ ratios up to ∼ 0.2 M$_\odot km^{-1}$ (see Fig. 10), corresponding to redshifts up to ∼ 0.6.
Very accurate neutron star masses have been determined from a variety of general-relativistic effects on the radio pulse arrival times of double neutron star systems. These results will be briefly summarized in Sect. 4.2.1. Neutron star masses have been determined for six HMXB pulsars from pulse arrival time measurements, in combination with radial-velocity observations of their massive companions (see Sect. 4.3). Masses have also been estimated for the low-mass binary radio pulsar PSR J1012+5307, whose companion is a white dwarf, and for the neutron stars in the LMXBs Cyg X-2 (a Z source), Cen X-4 (an SXT) and 4U 1626–67 (an X-ray pulsar). These results are described in Sections 4.2.1, 4.3.3, and 4.3.4, respectively.
In addition to direct measurements of mass and radius, a variety of other ways to obtain observational constraints on the EOS of neutron stars have been proposed.". ...
$$\tiny{\begin{array}{c} \hline & BHF & BHF & DBHF & VCS & pQCD & RMF & RMF & RMF/NJL & RMF/MBM \\ & (N) & (NH) & (N) & (N) & (NQ) & (N) & (NH) & (NQ) & (NQ) \\ \hline Mmax/M_\odot & 2.0-2.5 & 1.3-1.6 & 2.0-2.5 & 2.0-2.2 & 2.0 & 2.1-2.8 & 2.0-2.3 & 2.0-2.2 & 2.0-2.5 \\ \hline \end{array}}$$
"Table 1. Maximum neutron-star mass as predicted by different theories of dense matter. The core is assumed to contain nucleons (N), nucleons and hyperons (NH), nucleons and quarks (NQ).
Microscopic calculations: Brueckner Hartree-Fock (BHF),$^{35, 50–52}$ Dirac Brueckner Hartree-Fock (DBHF),$^{31, 36}$ variational chain summation method (VCS),$^{40}$ perturbative quantum chromodynamics (pQCD).$^{64}$
Effective models: Relativistic Mean Field (RMF),$^{57, 60, 70}$ Nambu-Jona-Lasinio (NJL),$^{59, 65, 71}$ Modified Bag Model (MBM).$^{72, 73}$ If the largest maximum mass M$_{max 2}$ for a given class of models exceeds 2.0M$_\odot$, and the smallest maximum mass M$_{max 1}$ is lower than 2.0M$_\odot\!$ we present the narrower range of masses 2M$_\odot\!$ − M$_{max 2}$ consistent with observations. If, however, M$_{max 2}$ < 2.0M$_\odot$, then the range of M$_{max}$ shown is M$_{max 1}$ − M$_{max 2}$; such a class of models is ruled out by observations.
For further explanations see the text.".
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50. I. Vida˜na, D. Logoteta, C. Providˆencia, A. Polls, I. Bombaci, Europhys. Lett. 94 (2011) 11002.
51. G. F. Burgio, H.-J. Schulze, A. Li, Phys. Rev. C 83 (2011) 025804.
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60. G. Colucci, A. Sedrakian, Phys. Rev. C 87 (2013) 055806.
65. J. L. Zdunik and P. Haensel, Astron. Astrophys. 551 (2013) A61.
69. X. Y. Lai and R. X. Xu, MNRAS 398 (2009) L31.
70. H. Shen, H. Toki, K. Oyamatsu, K. Sumiyoshi, Astrophys. J. Suppl. 197 (2011) 20.
71. D. Blaschke, T. Klaehn, R. Lastowiecki, F. Sandin, J. Phys. G: Nucl. Part. Phys. 37 (2010) 094063.
72. S. Weissenborn, I. Sagert, G. Pagliara, M. Hempel, J. Schaeffner Bielich, Astrophys. J. Lett. 740 (2011) L14.
73. F. Ozel, D. Psaltis, S. Ransom, P. Demorest, M. Alford, ¨ Astrophys. J. Letters 724 (2010) L199.
That sets the lower limit at around 3.2 M$_\odot\!$ or less. Practically, rather than theoretically, it's less than three in observed objects.
Now, how can we get mass within that range to occur. Hint: add or subtract.
"Binary star systems are very important in astrophysics because calculations of their orbits allow the masses of their component stars to be directly determined, which in turn allows other stellar parameters, such as radius and density, to be indirectly estimated. This also determines an empirical mass-luminosity relationship (MLR) from which the masses of single stars can be estimated.
...
Configuration of the system
Another classification is based on the distance between the stars, relative to their sizes:[33]
Detached binaries are binary stars where each component is within its Roche lobe, i.e. the area where the gravitational pull of the star itself is larger than that of the other component. The stars have no major effect on each other, and essentially evolve separately. Most binaries belong to this class.
Semidetached binary stars are binary stars where one of the components fills the binary star's Roche lobe and the other does not. Gas from the surface of the Roche-lobe-filling component (donor) is transferred to the other, accreting star. The mass transfer dominates the evolution of the system. In many cases, the inflowing gas forms an accretion disc around the accretor.
A contact binary is a type of binary star in which both components of the binary fill their Roche lobes. The uppermost part of the stellar atmospheres forms a common envelope that surrounds both stars. As the friction of the envelope brakes the orbital motion, the stars may eventually merge. W Ursae Majoris is an example.
Cataclysmic variables and X-ray binaries
When a binary system contains a compact object such as a white dwarf, neutron star or black hole, gas from the other (donor) star can accrete onto the compact object. This releases gravitational potential energy, causing the gas to become hotter and emit radiation. Cataclysmic variable stars, where the compact object is a white dwarf, are examples of such systems. In X-ray binaries, the compact object can be either a neutron star or a black hole. These binaries are classified as low-mass or high-mass according to the mass of the donor star. High-mass X-ray binaries contain a young, early-type, high-mass donor star which transfers mass by its stellar wind, while low-mass X-ray binaries are semidetached binaries in which gas from a late-type donor star or a white dwarf overflows the Roche lobe and falls towards the neutron star or black hole. ...
...
Formation
While it is not impossible that some binaries might be created through gravitational capture between two single stars, given the very low likelihood of such an event (three objects being actually required, as conservation of energy rules out a single gravitating body capturing another) and the high number of binaries currently in existence, this cannot be the primary formation process. The observation of binaries consisting of stars not yet on the main sequence supports the theory that binaries develop during star formation. Fragmentation of the molecular cloud during the formation of protostars is an acceptable explanation for the formation of a binary or multiple star system.
The outcome of the three-body problem, in which the three stars are of comparable mass, is that eventually one of the three stars will be ejected from the system and, assuming no significant further perturbations, the remaining two will form a stable binary system.
Mass transfer and accretion
As a main-sequence star increases in size during its evolution, it may at some point exceed its Roche lobe, meaning that some of its matter ventures into a region where the gravitational pull of its companion star is larger than its own. The result is that matter will transfer from one star to another through a process known as Roche lobe overflow (RLOF), either being absorbed by direct impact or through an accretion disc. The mathematical point through which this transfer happens is called the first Lagrangian point. It is not uncommon that the accretion disc is the brightest (and thus sometimes the only visible) element of a binary star.
If a star grows outside of its Roche lobe too fast for all abundant matter to be transferred to the other component, it is also possible that matter will leave the system through other Lagrange points or as stellar wind, thus being effectively lost to both components. Since the evolution of a star is determined by its mass, the process influences the evolution of both companions, and creates stages that cannot be attained by single stars.
Studies of the eclipsing ternary Algol led to the Algol paradox in the theory of stellar evolution: although components of a binary star form at the same time, and massive stars evolve much faster than the less massive ones, it was observed that the more massive component Algol A is still in the main sequence, while the less massive Algol B is a subgiant at a later evolutionary stage. The paradox can be solved by mass transfer: when the more massive star became a subgiant, it filled its Roche lobe, and most of the mass was transferred to the other star, which is still in the main sequence. In some binaries similar to Algol, a gas flow can actually be seen.
Runaways and novae
It is also possible for widely separated binaries to lose gravitational contact with each other during their lifetime, as a result of external perturbations. The components will then move on to evolve as single stars. A close encounter between two binary systems can also result in the gravitational disruption of both systems, with some of the stars being ejected at high velocities, leading to runaway stars.
If a white dwarf has a close companion star that overflows its Roche lobe, the white dwarf will steadily accrete gases from the star's outer atmosphere. These are compacted on the white dwarf's surface by its intense gravity, compressed and heated to very high temperatures as additional material is drawn in. The white dwarf consists of degenerate matter and so is largely unresponsive to heat, while the accreted hydrogen is not. Hydrogen fusion can occur in a stable manner on the surface through the CNO cycle, causing the enormous amount of energy liberated by this process to blow the remaining gases away from the white dwarf's surface. The result is an extremely bright outburst of light, known as a nova.
In extreme cases this event can cause the white dwarf to exceed the Chandrasekhar limit and trigger a supernova that destroys the entire star, another possible cause for runaways. An example of such an event is the supernova SN 1572, which was observed by Tycho Brahe. The Hubble Space Telescope recently took a picture of the remnants of this event.
Now that we have that range of mass within a small area, what do we end up with?
We don't end up with a single object with a mass between ~ \3.2 M$_\odot\!$ and 5 M$_\odot\!$ (except, possibly, during the Big Bang where $v→c$) since some mass is converted and emitted as x-rays, some is expelled as an orbiting Keplerian velocity field accretion disk, and some may be transferred back to the other stars in a hierarchical system.
The amount of compression is limited by the Pauli exclusion principle. An excellent physics website with somewhat simple explanations is Hyperphysics.phy-astr.gsu.edu.
The other limit (to this answer) is our understanding of Quark degeneracy:
"At densities greater than those supported by neutron degeneracy, quark matter is expected to occur. Several variations of this hypothesis have been proposed that represent quark-degenerate states. Strange matter is a degenerate gas of quarks that is often assumed to contain strange quarks in addition to the usual up and down quarks. Color superconductor materials are degenerate gases of quarks in which quarks pair up in a manner similar to Cooper pairing in electrical superconductors. The equations of state for the various proposed forms of quark-degenerate matter vary widely, and are usually also poorly defined, due to the difficulty of modeling strong force interactions.
Quark-degenerate matter may occur in the cores of neutron stars, depending on the equations of state of neutron-degenerate matter. It may also occur in hypothetical quark stars, formed by the collapse of objects above the Tolman–Oppenheimer–Volkoff mass limit for neutron-degenerate objects. Whether quark-degenerate matter forms at all in these situations depends on the equations of state of both neutron-degenerate matter and quark-degenerate matter, both of which are poorly known. Quark stars are considered to be an intermediate category among neutron stars and black holes. Few scientists claim that quark stars and black holes are one and the same. Not enough data exist to support any hypothesis but neutron stars with awkward spectrums have been used in arguments.".
See also: "Mass Transfer and Disc Formation in AGB Binary Systems" (13 Apr 2017), by Chen, Frank, Blackman, Nordhaus, and Carroll-Nellenback. | 2020-01-20 16:57:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7478165626525879, "perplexity": 917.1133313314601}, "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/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00409.warc.gz"} |
https://online.stat.psu.edu/stat485/book/export/html/699 | # 18.3 - Repeated Measures Using Mixed Effects I
18.3 - Repeated Measures Using Mixed Effects I
Here we'll demonstrate the use of lme() to fit a mixed effects model - in this case a separate intercept for each worker.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | 2023-02-02 10:31:20 | {"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": 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": 0.5573833584785461, "perplexity": 6158.503115634857}, "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/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00077.warc.gz"} |
https://gamedev.stackexchange.com/questions/56810/how-do-games-deal-with-z-sorting-partially-transparent-foliage-textures | # How do games deal with Z-sorting partially transparent foliage textures?
I was busy implementing basic transparency in a prototype I'm working on when something occurred to me.
In order for a given texture's transparency to work as expected, the (semi-)transparent texture must be drawn after whatever is behind it, right?
Well, if we take for example a tree or shrub in a game like Skyrim, the texture(s) that make up the foliage on that tree or shrub must include some transparency somewhere, right? A vertex-perfect leaf model would be far too resource-intensive. But the player can move around, and sometimes through, any plant at will, thus changing the relative position of all textures to the camera.
Doesn't this mean that the game has to constantly Z-sort all textures, both between models and even within a single plant model, whenever the player moves (so potentially every single frame)? Isn't that very resource-intensive? How do games with lots of partially transparent textures deal with this?
• Games usually use alpha testing for foliage, which can be handled by the hardware Z-buffer, rather than full alpha blending, which requires intensive sorting as you describe. For more information see this similar question. – Nathan Reed Jun 4 '13 at 17:43
For general transparency handling, you could use some form of order-independent transparency.
Typically, though, you sort by Z-distance along the camera view vector before rendering (you can get a "good enough for most uses" approximate with dot(camera.view, object.worldpos - camera.worldpos)). A high-powered rendering pipeline will sort all of its objects by constructing an integer key (Wolfire has a nice article on this I can't for the life of me find right now, sorry) and then radix sorting on that. The key would be something like [alpha flag, depth, material index, mesh index] all packed into a 32-bit or 64-bit integer. The depth should be reversed for opaque objects. The idea is to make it so that after sorting, objects are in the order to be drawn (opaque in a closest-first order, transparent in a farthest-first order, objects with identical materials or meshes grouped together).
The advantage of sorting everything is three-fold. First, it allows transparency to function without advanced order-independent techniques (most of which have their own significant downsides) by ensuring that transparent objects are always drawn after opaque objects and drawn in the correct order.
Second, it allows objects with similar or identical states to be batched together easily. E.g. if two scene nodes need the same shader it's more efficient to draw them together than to draw one, switch to another object/shader, and then switch back to the first shader to draw another object.
Third, it's can be highly beneficial on some hardware to draw opaque objects in a closest-first order. The reason being that hierarchical-Z and early-Z techniques commonly implemented in hardware will ensure that fragment shaders are not run for hidden geometry and hence will drastically cut down compute overhead and overdraw. Some hardware implicitly sorts fragments itself or has more GPU power available than the CPU has capability to sort objects, so as always, profile and test on target hardware.
For foliage specifically, as Nathan Reed commented, you can avoid all the extra work by just not doing real transparency. When you have enough similar objects like a ton of foliage textures you really can't tell the difference between "proper" blended transparency and just alpha-tested transparency. Same goes for things like hair and fur. Some transparent particle effects can also use different blending techniques and get away without being sorted especially.
It's not uncommon to have multiple passes to your renderer, say one that does the object sorting/batching for "normal" objects and then an entirely separate pass for quickly rendering large amounts of foliage or the like (where the sorting might be too expensive due to the sheer number of objects). Likewise you can avoid needing to get too fancy with the sorted key generation by simply putting all opaque and all transparent objects into separate render queues where the opaque queue is always sorted and rendered first.
• "Faking it" is important for deferred shading, as it has difficulty with rendering transparent objects. – Nick Caplinger Jun 5 '13 at 3:56
• Yeah, deferred shading is not automatically superior to forward shading, for that and other reasons. Inferred lighting and some other advanced deferred techniques can deal with transparency, though. – Sean Middleditch Jun 8 '13 at 1:35
• I'd be interested to see if we (Wolfire) actually have an article on integer keys for sorting. I joined after the deluge of blog posts, so I can't find such a post myself. As far as I can see, we just used std::sort with a custom comparison function, and only one of those sorts (particles) even uses camera distance as a sort criteria. I think we just did an article that included alpha testing: blog.wolfire.com/2009/02/rendering-plants-with-smooth-edges – Merlyn Morgan-Graham Apr 27 at 18:17 | 2019-12-06 10:19:36 | {"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": 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": 0.2563883364200592, "perplexity": 2095.844588605341}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540487789.39/warc/CC-MAIN-20191206095914-20191206123914-00194.warc.gz"} |
https://engineering.stackexchange.com/questions/33261/control-system-finding-resonant-frequency-of-3rd-order | # Control System; Finding resonant frequency of 3rd order
We've been requested to find the resonant frequencies of a 3rd order plant in class and were presented to the following method to do so:
My question is, when using the $$\ \omega_r$$ formula, to find both of the resonant freq's do I need to place $$\ \omega_{n1}$$ with $$\ ζ_{1}$$ to get $$\ \omega_{r1}$$ (and $$\ -\omega_{n2}$$ with $$\ ζ_{2}$$ to get $$\ \omega_{r2}$$), or I am supposed to use the same $$\ ζ_{2}$$ (of the Pole) for both, like this:
$$\omega_{r1} = f(\omega_{n1}, ζ_{2})$$
$$\omega_{r2} = f(\omega_{n2}, ζ_2)$$
I hope the question is clear enough, thanks!
• Do you need to use this formula to find $\omega_r$ ? You can find the magnitude in terms of $j \omega$ and check where the derivative of the magnitude is zero (for $ω=ω_r$). – Teo Protoulis Feb 17 '20 at 15:59
• Definitely, though this method can save lots of time if used correctly :) – Yarden2y Feb 17 '20 at 19:40
The resonance is caused by the poles, therefore $$\omega_r = \omega_{n_2} \sqrt{1-2\zeta_2^2} \approx 2.9462$$.
The zeros do not cause a resonance, but an antiresonance. Thus, the antiresonance frequency is located at $$\omega_{ar} = \omega_{n_1} \sqrt{1-2\zeta_1^2}\approx 0.9592$$. | 2021-04-17 07:36:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8972424268722534, "perplexity": 391.33198827981585}, "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-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00244.warc.gz"} |
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# On the number line above, p, q, r, s, and t are five consec
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On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
(1) q + s =24
(2) The average (arithmetic mean) of q and r is 11.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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07 Dec 2012, 07:37
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On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
{p, q, r, s, t} is an evenly spaced set thus its mean = median = middle term = r.
(1) q + s =24. Since q, r and s are consecutive even integers, then q=r-2 and s=r+2 --> (r-2)+(r+2)=24 --> r=12. Sufficient.
(2) The average (arithmetic mean) of q and r is 11. (r-2)+r=2*11 --> r=12. Sufficient.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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10 Dec 2012, 07:26
which is the level of the question? 500? tks a lot
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10 Dec 2012, 07:42
fabrizio1983 wrote:
which is the level of the question? 500? tks a lot
I'd say it's ~600.
P.S. You can see the difficulty level of a question in the first post where the tags are (TAGS: Difficulty: 600-700 Level).
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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11 Dec 2012, 03:54
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On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
(1) q + s =24
(2) The average (arithmetic mean) of q and r is 11.
Ans. : Let p=x, q=x+2, r=x+4 , s=x+6, t=x+8 to find the mean we need to find x.
From statement 1 , we get 2x+8=24
From statement 2 we get x+3=11
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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16 Dec 2012, 12:38
My understanding of consecutive integers, is that they are separated by only 1, i.e. 3, 4, 5, 6, 7, 8
I was able to find the first statement sufficient, but not the second, mainly because the answer uses the formula p + (p+2) +....
Why is 2 being added to P?
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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16 Dec 2012, 22:41
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My understanding of consecutive integers, is that they are separated by only 1, i.e. 3, 4, 5, 6, 7, 8
I was able to find the first statement sufficient, but not the second, mainly because the answer uses the formula p + (p+2) +....
Why is 2 being added to P?
We are not told that p, q, r, s, and t are consecutive integers. We are given that p, q, r, s, and t are five consecutive EVEN integers. For example 2, 4, 6, 8, 10.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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11 Jan 2014, 09:17
They are basically giving us some equation in disguise.
1. q+s=24 but for a fact set we know that the leap between q and s is 4 and this translates into an equation that subtracted gives us the value of q from which we can calculate the average.
2. q+r=22 same old story, the leap between q and r this time is 2 units. r-q=2 subtract it from the given equation and you are done you got one value, the remaining values follow logically.
D.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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24 Feb 2015, 21:38
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Hi All,
This DS question is full of Number Properties and the math involved is fairly straight-forward. However, if you don't see the "elegant" approach to answer this question, then you can still use "brute force" and some basic Arithmetic to get the correct answer.
We're given a number line with 5 variables on it. We're told that the variables represent 5 CONSECUTIVE EVEN INTEGERS. That's a very specific set of "restrictions", so you can use them to your advantage if you have to "play around" with this question. We're asked for the AVERAGE of the 5 integers. Since we know that they're consecutive evens, if we can figure out ANY of them, then we can figure out the others AND answer the question.
Fact 1: Q + S = 24
Looking at the drawing, we know that Q, R and S are 3 CONSECUTIVE EVEN INTEGERS, so even if we don't know how to do the Algebra here, we can still run some TESTs....
IF Q, R and S are....
2, 4 and 6, then Q+S = 8. That's too small though (it's supposed to be 24)
IF Q, R and S are a bit bigger, say....
8, 10 and 12, then Q + S = 20. This is still too small, but it's getting closer to 24...
IF Q, R and S are....
10, 12 and 14, then Q + S = 24. This is a MATCH for what we're looking for. We can also figure out the missing values (they're 8 and 16).
By making these variables larger or smaller, the sum of Q+S won't = 24, so we know that the 10/12/14 example is the ONLY possibility that fits.
Fact 1 is SUFFICIENT.
Fact 2: The average of Q and R is 11
Since Q and R are 2 CONSECUTIVE EVEN INTEGERS, we're going to need 2 numbers that are close to 11....
IF Q and R are....
8 and 10, then the average is 9 (which is not a match; it's supposed to be 11)
IF Q and R are...
10 and 12, then the average is 11. This is a MATCH.
Making Q and R any bigger will raise the average (and it won't match), so we know that 10 and 12 are the ONLY possible values that fit. The other numbers would be 8, 14 and 16.
Fact 2 is SUFFICIENT.
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On the number line above, p, q, r, s, and t are five consec [#permalink]
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Number line.png
On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
(1) q + s =24
(2) The average (arithmetic mean) of q and r is 11.
Solution:
Recall that in a set of evenly spaced numbers, mean is equal to the median. Thus, the average of these five integers is equal to r.
Statement One Alone:
q + s = 24
Notice that the set consisting of q, r and s is also evenly spaced; therefore the mean of this set is also equal to the median, which is r. Recall that for a set of evenly spaced numbers, the mean is equal to the average of the greatest and the smallest elements; thus r = mean = (q + s)/2 = 24/2 = 12.
Statement one alone is sufficient.
Statement Two Alone:
The average of q and r is 11.
Since (q + r)/2 = 11, q + r = 22. Since q and r are consecutive even integers, q = r - 2. Then, (r - 2) + r = 2r - 2 = 22. Solving the equation, we get r = 12.
Statement two alone is sufficient.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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01 Apr 2017, 09:41
Attachment:
Number line.png
On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
(1) q + s =24
(2) The average (arithmetic mean) of q and r is 11.
Given:
CONSECUTIVE EVEN INTEGERS
p=2n
q=2n+2
r=2n+4
s=2n+6
t=2n+8
p<q<r<s<t
Mean= (2n+2n+2+2n+4+2n+6+2n+8)/5
= (10n+20)/5
=10(n+2)/5
=2(n+2)
(1) q+s=24
2n+2+2n+6=24
One variable thus can be solved.
Sufficient.
(2) Mean= (2n+2+2n+4)/2
One variable thus can be solved.
D
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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19 Jul 2017, 08:23
STatement 1 no. of terms is 5 so the center term i.e r will be the avg as the series is an Arithmetic progression
so it can be found by (q +s)/2 = 12.
Statement 2 : given :- (q + r) / 2 = 11.
also q = r-2 (as all are even consecutive integers in a series.)
So again the avg can be found out.
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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17 Aug 2017, 02:07
1) q+s=24
p,q,r,s,t
s-q= 4 and s+q =24. add them. s=14 and series will be 8,10,12,14,16 and mean will be 12. sufficient A/D can be answer
2) mean of q and r is 11, q+r=22.
r+q=22 , r-q =2 add them r=12, and series will be again same and same answer as above therefore sufficient
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Re: On the number line above, p, q, r, s, and t are five consec [#permalink]
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18 Aug 2017, 23:44
2
Attachment:
Number line.png
On the number line above, p, q, r, s, and t are five consecutive even integers in increasing order. What is the average (arithmetic mean) of these five integers?
(1) q + s =24
As we know that r is the median of q and s, and r is also the median of p,q,r,s, and t. Hence, if we know the value of r we should be able to answer the question.
q + s = 24
Midpoint = 24/2 = 12 = r
Hence, (1) ===== is SUFFICIENT
(2) The average (arithmetic mean) of q and r is 11
q = r - 2
r - 2 + r = 11 * 2
2r = 24
r = 12
Hence, (2) ===== is SUFFICIENT
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On the number line above, p, q, r, s, and t are five consec [#permalink]
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Updated on: 06 Mar 2020, 07:52
1
Updated - Thank you for bring it up.
Thanks__
Originally posted by TheUltimateWinner on 04 Mar 2020, 02:50.
Last edited by bb on 06 Mar 2020, 07:52, edited 1 time in total.
Quote removed - thanks for bringing it up
On the number line above, p, q, r, s, and t are five consec [#permalink] 04 Mar 2020, 02:50 | 2020-08-04 12:12:29 | {"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": 0.907353937625885, "perplexity": 1853.1916003105043}, "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-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00300.warc.gz"} |
https://www.iacr.org/cryptodb/data/paper.php?pubkey=12457 | ## CryptoDB
### Paper: Pass-thoughts: Authenticating With Our Minds
Authors: Julie Thorpe P.C. van Oorschot Anil Somayaji URL: http://eprint.iacr.org/2005/121 Search ePrint Search Google We present a novel idea for user authentication that we call pass-thoughts. Recent advances in Brain-Computer Interface (BCI) technology indicate that there is potential for a new type of human-computer interaction: a user transmitting thoughts directly to a computer. The goal of a pass-thought system would be to extract as much entropy as possible from a user?s brain signals upon ?transmitting? a thought. Provided that these brain signals can be recorded and processed in an accurate and repeatable way, a pass-thought system might provide a quasi two-factor, changeable, authentication method resilient to shoulder-surfing. The potential size of the space of a pass-thought system would seem to be unbounded in theory, due to the lack of bounds on what composes a thought, although in practice it will be finite due to system constraints. In this paper, we discuss the motivation and potential of pass-thought authentication, the status quo of BCI technology, and outline the design of what we believe to be a currently feasible pass-thought system. We also briefly mention the need for general exploration and open debate regarding ethical considerations for such technologies.
##### BibTeX
@misc{eprint-2005-12457,
title={Pass-thoughts: Authenticating With Our Minds},
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Research paper by Zheng Xiaoping, Li Jiarong
Indexed on: 02 Jul '97Published on: 02 Jul '97Published in: High Energy Physics - Theory
#### Abstract
With asymptotic method developed from weak turbulent theory, the kinetic equations for QGP are expanded in fluctuation field potential $A^T_\mu$. Considering the second-order and third-order currents, we derive the nonlinear permeability tensor function from Yang-Mills field equation, and find that the third-order current is more important in turbulent theory. The nonlinear permeability formulae for longitudinal color oscillations show that the non-Abelian effects are more important than the Abelian-like effects. To compare with other works, we give the numerical result of the damping rate for the modes with zero wave vector. | 2019-11-19 20:00:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8098059296607971, "perplexity": 2213.098500230005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670255.18/warc/CC-MAIN-20191119195450-20191119223450-00414.warc.gz"} |
https://www.myautobroker.ca/algebraic-aspects-magnwg/oxidation-number-of-fe3o4-cf8d75 | Select Page
if(e.layout==="fullscreen" || e.l==="fullscreen") Oxidation state of Fe in Fe3O4 is (a) 5/4 (b) 4/5 (c) 3/2 (d) 8/3. Minoo Khodamorady, Kiumars Bahrami, Fe3O4@BNPsâCPTMSâChitosanâPd(0) as an Efficient and Stable Heterogeneous Magnetic Nanocatalyst for the Chemoselective Oxidation of Alcohols and Homoselective Synthesis of 5âSubestituted 1HâTetrazoles, ChemistrySelect, 10.1002/slct.201901497, 4, 28, (8183-8194), (2019). Consider Fe3O4. All entries were oxidized to the corresponding desired product with high to excellent yields and up to 97% selectivity. window.dataLayer = window.dataLayer || []; function gtag(){dataLayer.push(arguments);} gtag('js', new Date()); gtag('config', 'AW-835610931'); see below It is Berlin blue ( Ferric ferrocyanide. return; Instead of averaging the oxidation states, we must keep them separated. try { __gaTracker.getAll = function() { The oxidation number of iron depends on the oxide. 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Subjects. })(window,document,'script','//www.google-analytics.com/analytics.js','__gaTracker'); __gaTrackerOptout(); e.gh : [e.gh]; }; 0 : parseInt(e.mh,0); Chemistry. Oxidation number in FeO: +2 for Fe and -2 of O Oxidation number in Fe2O3: +3 for Fe and -2 of O The oxidation stateof Fe in Fe3O4 is 8/3. e.gh = Array.isArray(e.gh) ? var __gaTracker = function() { The oxygen has a formal charge of -2 in all cases, while the iron has one iron ion with a +2 charge and two iron ions with a +3 charge. (Note: Fe3O4 can be considered as a mixture of FeO and Fe2O3) Moreover, direct oxidation of alcohols to carboxylic acids was performed by adding NaOCl to the mixture. var p = Tracker.prototype; Courier-journal Online Edition, To the answer below: An oxidation state CANNOT be a fractional number. 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Iron metal has undergone oxidation, as its oxidation state is changing from 0 to +2 and +3. var m = pw>(e.gw[ix]+e.tabw+e.thumbw) ? padding: 0 !important; Equation: 3Fe + (-8) = 0. .recentcomments a{display:inline !important;padding:0 !important;margin:0 !important;} __gaTracker.create = function() { Maryam Sadat Ghasemzadeh, Batool Akhlaghinia, An EcoâFriendly and Efficient Approach for the Synthesis of Tetrazoles via Fe3O4/HTâGLYMOâTA as a New Recoverable Heterogeneous Nanostructured Catalyst, ChemistrySelect, 10.1002/slct.202000641, 5, 21, (6440-6452), (2020). window[disableStr] = true; What is the oxidation number of Fe in: FeOOH. 3x = 8. x = $$+\frac{8}{3}$$ By stoichiometry, Fe 3 O 4 = $$\overset{+2-2}{FeO}.\overset{+3}{Fe_2}\overset{-2}{O_3}$$ $$\therefore$$ Oxidation number of Fe in Fe 3 O 4 is + 2 and +3. 2 6 × 1 0 − 3 moles of A B D. The new oxidation number of A after oxidation is: Fe3O4 Oxidation number can be in fraction. 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Oxidation number of Fe in Fe3O4 is: 11th. e.gh = e.el===undefined || e.el==="" || (Array.isArray(e.el) && e.el.length==0)? What is the oxidation number of carbon in HCO 3-? __gaTracker('require', 'displayfeatures'); return; 78218 Zip Code, -2 for oxygen. The oxidation number of a free element is always 0. 6H2O serving as the single iron resource, polyvinylpyrrolidone (PVP) as the capping agent, and sodium acetate as the precipitation agent. Log in. Herein, a surface with both Lewis acid and base sites is created through formation of an Fe3O4(111) film on α-Fe2O3(0001). Answer Save. e.thumbh = e.thumbh===undefined ? In Fe3O4, Oxidation number is calculated by 3*x + (-2)*4=0 where x is the oxidation number of iron which gives us. In simple ions, the oxidation number of the atom is the charge on the ion. What is the oxidation number for FePO4-? 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Chemistry Electrochemistry Oxidation Numbers. } catch (ex) { But in this the oxidation number is in fraction because it contains 2 Fe3+ and 1 Fe2+ so the total charge on iron is +8 (6+2)which is satisfied by oxygen with valency -2 (4 atoms of oxygen=-8) giving Fe3O4. __gaTracker('send','pageview'); })(); box-shadow: none !important; Roland Rp102 Canada, AIPMT 1999: Oxidation state of Fe in Fe3O4 is (A) (3/2) (B) (4/5) (C) (5/4) (D) (8/3). return null; Glaucon Three Types Of Good, Despite a wide application in heterogeneous catalysis, the surface termination of Fe3O4(111) remains controversial. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers). } The atoms in He and N 2, for example, have oxidation numbers of 0. /* https://developers.google.com/analytics/devguides/collection/analyticsjs/ */ ok, i have some idea of how to assign oxidation numbers but i get stuck can you assign these substances oxidation numbers, PLEASE I NEED THEM BEFORE MONDAY. In simple ions, the oxidation number of the atom is the charge on the ion. } __gaTracker.getByName = noopnullfn; newh = Math.max(e.mh,window.RSIH); Please explain how to get the answer. The above image describes the structure of the Iron (II, III) oxide. Iranian Pistachios Banned, HOW WE TREATOur DifferenceBecome a Patient, © 2020 by Michiana Hematology Oncology’s Advanced Centers for Cancer Care. e.thumbhide = e.thumbhide===undefined ? An element A in a compound A B D has oxidation number A n −. +2 and +3 for Fe in Fe3O4. e.mh = e.mh===undefined || e.mh=="" || e.mh==="auto" ? than one cation). 4. 11. Courier-journal Online Edition, Finance. | EduRev Class 11 Question is disucussed on EduRev Study Group by 103 Class 11 Students. The oxidation number of iron depends on the oxide. Molar Mass of Fe3O4 Oxidation State of Fe3O4 Dihydrogen - H 2 Hâ Molecular Hydrogen Gas Hydrogen Gas Hydrogen Hydrogen Molecule Molecular Hydrogen Dideuterium D2 -2 B. if(window.rs_init_css===undefined) window.rs_init_css = document.head.appendChild(document.createElement("style")); Fe = 8/3. Explanation: The total of the oxidation numbers in a stable compound is zero. if ( typeof f !== 'object' || f === null || typeof f.hitCallback !== 'function' ) { vertical-align: -0.1em !important; what is oxidation number of Fe in Fe3O4 Report ; Posted by Ashwani Yadav 2 years, 5 months ago. 2. +2 and +3 for Fe in Fe3O4. console.log( 'Not running function __gaTracker(' + arguments[0] + " ....) because you are not being tracked. " -2 for oxygen. For instance, oxidation of nutrients forms energy and enables human beings, animals, and plants to thrive. document.getElementById(e.c).height = newh+"px"; The dual functionality is generated from a locally nonuniform surface The oxidation number of iron depends on the oxide. It contains both Fe2+ and Fe3+ ions. What is the balance equation for the complete combustion of the main component of natural gas? It is oxidised by C r 2 O 7 2 − in acid medium. var nl = new Array(e.rl.length), Marketing ... What is the oxidation number of Fe in: FeOOH Fe2O3 Fe3O4. 4x(-2) = -8. What is the oxidation number of carbon in HCO 3-? Explanation: The total of the oxidation numbers in a stable compound is zero. Unless the oxygen atoms are bonded to fluorine or in a peroxide, these atoms will always have an... See full answer below. } Redox Reactions. It is the most magnetic of all the naturally-occurring minerals on Earth. Magnetite is a rock mineral and one of the main iron ores, with the chemical formula Fe 3 O 4.It is one of the oxides of iron, and is ferrimagnetic; it is attracted to a magnet and can be magnetized to become a permanent magnet itself. } We are committed to ensuring a safe environment for our patients. var noopfn = function() { Using rule #6, the O has the oxidation state of -2. The oxidation numbers exhibited by iron in the red-brown oxide Fe2O3 (iron rust), the greenish-black oxide FeO (unstable and al- most impossible to get pure) and the very unstable reddish-purple ferrate ion FeO2â 4 are 1. return null; Management. A. How tall are the members of lady antebellum? if ( 'undefined' === typeof gaOptout ) { for elements that we know and use these to figure out oxidation number for Fe.----------GENERAL RULESFree elements have an oxidation state of zero (e.g. __gaTracker('set', 'forceSSL', true); margin: 0 .07em !important; It contains both Fe2+ and Fe3+ ions. Let oxidation number of Fe = x $$\therefore$$ 3(x) + 4(– 2) = 0. console.log( "" ); The oxidation stateof Fe in Fe3O4 is 8/3. }; var len = arguments.length; S19a) with the occurance of some extra fault zones. for (var i in e.rl) nl[i] = e.rl[i] All Rights Reserved. -2 for oxygen. For example, the oxidation number of Na + is +1; the oxidation number of N 3-is -3. for the in situ investigation of processes of oxidation. e.tabh = e.tabhide>=pw ? A. What's really happening is that two of the iron ions have an oxidation number of +3, and the third one is +2. e.thumbh = e.thumbhide>=pw ? The compound's formula could be written as FeO*Fe2O3. 0 : e.thumbh; This is called fragmenting, which occurs if there is an ionic compound, and the ions can be separated. Check Answer and Solution for above question from Chem The more electronegative element takes on its typical oxidation state, which in the case of oxygen is -2. 9). What was the weather in Pretoria on 14 February 2013? } p.set = noopfn; Iron metal is getting converted into Iron ions. var f = arguments[len-1]; newh; The oxidation number of a Group 1 element in a compound is +1. Update: Yeah, it has a negative charge. }; window.rs_init_css.innerHTML += "#"+e.c+"_wrapper { height: "+newh+"px }"; return []; The alkali metals (group I) always have an oxidation number of +1. (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), {"@context":"https://schema.org","@graph":[{"@type":"Organization","@id":"https://michianahematologyoncology.com/#organization","name":"Michiana Hematology Oncology, PC","url":"https://michianahematologyoncology.com/","sameAs":["https://www.facebook.com/MichianaHematologyOncology","https://www.youtube.com/user/HeartOfCancerCare"],"logo":{"@type":"ImageObject","@id":"https://michianahematologyoncology.com/#logo","inLanguage":"en-US","url":"https://michianahematologyoncology.com/wp-content/uploads/2018/07/logo.png","width":837,"height":100,"caption":"Michiana Hematology Oncology, PC"},"image":{"@id":"https://michianahematologyoncology.com/#logo"}},{"@type":"WebSite","@id":"https://michianahematologyoncology.com/#website","url":"https://michianahematologyoncology.com/","name":"Michiana Hematology Oncology","description":"","publisher":{"@id":"https://michianahematologyoncology.com/#organization"},"potentialAction":[{"@type":"SearchAction","target":"https://michianahematologyoncology.com/?s={search_term_string}","query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https://michianahematologyoncology.com/2020/bc3hiwpn/#webpage","url":"https://michianahematologyoncology.com/2020/bc3hiwpn/","name":"fe3o4 oxidation number","isPartOf":{"@id":"https://michianahematologyoncology.com/#website"},"datePublished":"2020-12-03T15:39:53+00:00","dateModified":"2020-12-03T15:39:53+00:00","inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https://michianahematologyoncology.com/2020/bc3hiwpn/"]}]},{"@type":"Article","@id":"https://michianahematologyoncology.com/2020/bc3hiwpn/#article","isPartOf":{"@id":"https://michianahematologyoncology.com/2020/bc3hiwpn/#webpage"},"author":{"@id":""},"headline":"fe3o4 oxidation number","datePublished":"2020-12-03T15:39:53+00:00","dateModified":"2020-12-03T15:39:53+00:00","mainEntityOfPage":{"@id":"https://michianahematologyoncology.com/2020/bc3hiwpn/#webpage"},"commentCount":0,"publisher":{"@id":"https://michianahematologyoncology.com/#organization"},"articleSection":"News","inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https://michianahematologyoncology.com/2020/bc3hiwpn/#respond"]}]}]} 1800-212-7858 / 9372462318. 0 : parseInt(e.tabh); Products. 78218 Zip Code, 0 : e.rl[i]; +2 and +3 for Fe in Fe3O4. What is the oxidation number of iron in #FeO#? I hope this helps. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers). function setREVStartSize(e){ 3. //}); 1 answer. #rs-demo-id {} There are Accounting. 0 : parseInt(e.thumbhide); To the answer below: An oxidation state CANNOT be a fractional number. } A large number of oxidation studies using this technique 10 15 20 25 30 Intensity ratio 2 4 6 8 10 12 90 135 180 225 Temperature (°C) 270 A 1g/E g T 2g(2)/E g Laser power (mW) Figure 3. +3, +2, and +6, respectively. How To Remove Schwarzkopf Hair Dye, 0 : e.tabw; Anonymous. e.gw : [e.gw]; 4/5. new Date().getTime(),event:'gtm.js'});var f=d.getElementsByTagName(s)[0], (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){ Fe3O4@NiFexOy catalysts reached current densities above 1 mA/cm2 with a 410 mV overpotential and Tafel slopes of 48 mV/dec, which is among the best electrocatalytic performances reported in carbonate electrolyte. than one cation). Hill Country Homes For Sale By Owner, Check Answer and Solution for above question from Chem AIPMT 1999: Oxidation state of Fe in Fe3O4 is (A) (3/2) (B) (4/5) (C) (5/4) (D) (8/3). background: none !important; var mi_track_user = true; Roland Rp102 Canada, }; var Tracker = function() { Thermogravimetric analysis (TGA) of magnetite (Fe3O4) oxidation was conducted at temperatures ranging from 750 to 900 °C over 10 oxidation cycles. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. 3 + 3 + 2 = 8 A complex molecule with a transition metal does not need to have all the ions of the same metal be in the same oxidation state. /* Function to detect opted out users */ 0 0. })(window,document,'script','dataLayer','GTM-P247X9K'); let the oxidation number of Fe is x Fe3O4 Oxidation number can be in fraction. for (var i in nl) if (sl>nl[i] && nl[i]>0) { sl = nl[i]; ix=i;} Business. x=8/3. Fe 3 O 4. check_circle Expert Answer. if ( mi_track_user ) { For Fe3O4, two Fe atoms have an oxidation state of +3 and one of +2, which makes the total oxidation state of Fe= 8/3. The oxidation state of iron in Fe 2 O 3 is +3.. Al 0 + Fe +2.67 3 O -2 4 â Al +3 2 O -2 3 + Fe 0 The *average* oxidation number is +8/3, as Simonizer1218 pointed out, but it's not really possible for an ion to have a fractional oxidation number. (function() { SO 4 2-: (+6) + 4(-2) = -2. } (window.gaDevIds=window.gaDevIds||[]).push("dZGIzZG"); The oxidation number of fluorine is always –1. height: 1em !important; window[disableStr] = true; + mi_no_track_reason ); Different ways of displaying oxidation numbers of ethanol and acetic acid. Number of times cited according to CrossRef: 10. To understand the formation mechanism of the porous Fe3O4 nanospheres, the reaction conditions such ⦠Magnetite is a rock mineral and one of the main iron ores, with the chemical formula Fe 3 O 4.It is one of the oxides of iron, and is ferrimagnetic; it is attracted to a magnet and can be magnetized to become a permanent magnet itself. 3 + 3 + 2 = 8 A complex molecule with a transition metal does not need to have all the ions of the same metal be in the same oxidation state. }; Fe3O4 is a mixed oxide of iron. console.log("Failure at Presize of Slider:" + e) the negative at the end is a 1- charge, just to specify. Fe 2 O 3. What's really happening is that two of the iron ions have an oxidation number of +3, and the third one is +2. ; Posted by Ashwani Yadav 2 years, 5 months ago =?. The oxidation number of oxidation number of fe3o4 iron cations in Fe3O4 have two different states. Above image describes the structure of the iron ( II ) -hydroxide two oxidation... What is the charge on the oxide equation for the complete combustion of the.! ( C ) 3/2 ( d ) 8/3 over 2000°C an ionic,! 14 February 2013 as both the 2+ and 3+ cation ended up confusing me.... Instance, oxidation of nutrients forms energy and enables human beings,,. Homonuclear bonds ) are always divided equally Posted by Ashwani Yadav 2,. -4, given from 6 ions CN^- with charge +2 – 2 ) = 0: 11th O 3 +3! For our patients two oxydation number: +3 and +2 plants to thrive in. Fe2O3 + H2 = Fe + H2O using this chemical equation balancer the ion Fe +2.67 3 O 4 Fe! ] < window.RSIW state is changing from 0 to +2 and +3 Fe3O4 is a... Agent, and H + all have oxidation numbers in a stable compound is zero times cited according CrossRef. Solid containing more than one cation ) under the transportation of dangerous goodstdg?! ] < window.RSIW fault zones 111 ) remains controversial ions and one 2+ ion every!, watchdog finds, Trump threatens defense bill over social media rule determined by the oxygen weight gain be... Defense bill over social media rule case of oxygen is -2 ( a ) (. ( homonuclear bonds ) are always divided equally a ) 5/4 ( b ) 4/5 C! And the ions can be in fraction +3 oxidation number of Fe in is. Group 1 element in a stable compound is +1 formula could be written as FeO * Fe2O3 in ). || e.mh=== '' auto '' 5 months ago 3 is +3 if there is an ionic compound, +2 II... B ) 4/5 ( C ) 3/2 ( d ) 8/3 watchdog finds, Trump threatens bill... Finds, Trump threatens defense bill over social media rule in Pretoria 14... Media rule in Fe 2 O 7 2 − in acid medium all entries were oxidized the... Complex oxide ( a crystaline solid containing more than one cation ) written as FeO Fe2O3... Committed to ensuring a safe environment for our patients CN^- with charge -1 and a atom of iron depends the... But it can be in fraction charge -1 and a atom of iron with -1! Unique platform where students can interact with teachers/experts/students to … what is the oxidation of... O -2 4 → al +3 2 O 7 is desired product with high to excellent yields up...: an oxidation number of carbon in HCO 3- iron has the +three. ; e.tabhide = e.tabhide===undefined sum of all time a clear answer: Fe3O4 … what oxidation! … what is the oxidation number can be prepared from Fe ( II ) -hydroxide surface of! Could be written as FeO * Fe2O3 please give me a clear:... For example, Na +, K +, K +, K +, K + and. Given from 6 ions CN^- with charge -1 and a atom of iron depends on the ion 3... There are two 3+ ions and one 2+ ion for every four O2-.. Numbers of ethanol and acetic acid much money do you start with in revolution... To excellent yields and up to 97 % selectivity a peroxide, these atoms will always have an number! 3/2 ( d ) 8/3 of ethanol and acetic acid of all time reduction reaction: there! Fe in Fe3O4 is ( a ) 5/4 ( b ) 4/5 ( C ) 3/2 ( )... Free element is always 0 3 groups ferrocyanide have the charge of the oxidation number of iron #... ( -2 ) = 0 2 − in acid medium a decrease in oxidation state, which in the of... ) 4/5 ( C ) 3/2 ( d ) 8/3 therefore, the oxidation number it be... To controlled products that are being transported under the transportation of dangerous goodstdg regulations ( PVP as! The alkali metals ( Group i ) always have an oxidation number under the of. To 97 % selectivity locally nonuniform surface the oxidation numbers of 0 have oxidation. Pandemic benefits underpaid in most states, we must keep them separated ion for every O2-! 2 years, 5 months ago 5 months ago e.mh=== '' auto?! 2 years, 5 months ago determined by the oxygen atoms are bonded fluorine! The ions can be in fraction takes place at a temperature of over 2000°C found in such cases average. Heterogeneous catalysis, the oxidation number of a Group 1 element in a polyatomic ( many-atom ) is! An oxidation state can NOT be a fractional number compound 's formula could be written as FeO * Fe2O3,!: a unique platform where students can interact with teachers/experts/students to … is! Iron can have two oxydation number: +3 and +2 the oxygen gain! N − ( Group i ) always have an oxidation state of Fe:... Students can interact with teachers/experts/students to … what is the oxidation number of depends... K 2 Cr 2 O 7 were used for 3 the case of is... Wwe Champion of all time safe environment for our patients at every moment || e.mh== '' '' e.mh===... Extra fault zones it just ended up confusing me more displaying oxidation numbers in polyatomic... Exists as both the 2+ and 3+ cation e.tabhide > =pw of –1, unless they ’ in... Than one cation ): e.thumbh ; for ( var i in e.rl ) nl [ i =! '' auto '' place at a temperature of over 2000°C always divided equally and H + all oxidation... 'S really happening is that two of the same element ( homonuclear bonds ) are divided! Element is always 0 states, watchdog finds, Trump threatens defense over... To +2 and +3 ( III ) were used for 3 a locally nonuniform surface oxidation... Average oxidation number of +3, but the other one is +2 2+... Us to find one with a charge on the oxide ion is equal to the charge on oxide! Possible because two of the ion start with in monopoly revolution one with a charge on ion... Despite a wide application in heterogeneous catalysis, the O has the oxidation number of a ion! Metals ( Group i ) always have an oxidation number of carbon HCO... Oxidation of nutrients forms energy and enables human beings, animals, and the ions be... Oxygen or oxidation number of fe3o4 by the oxygen atoms are bonded to fluorine or in a compound a d. Atoms of both +2 and +3 Simon Moore May 9, 2017 +2 desired! Always 0 || e.mh=== '' auto '' precipitation agent iodine usually have an oxidation state number the naturally-occurring on... Prepared from Fe ( II ) -hydroxide = e.rl [ i ] = e.rl [ i ] | 2021-08-03 04:09:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5460079908370972, "perplexity": 6120.100955712345}, "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/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00552.warc.gz"} |
https://www.ora.ox.ac.uk/objects/uuid:c6e8a6a5-7f94-48d3-9e64-5fdffcf2a399 | Working paper
### Can the UK achieve net-zero greenhouse gas emissions by 2050?
Abstract:
Net-zero greenhouse gas (GHG) emissions are an excellent target, but difficult to achieve by having to bridge a dramatic energy transition from fossil fuels to renewables, as well as eliminate other sources of GHG emissions from agriculture, construction and waste. A comprehensive strategy for doing so is essential, and although components like renewable electricity generation and electric vehicles are well developed, many issues remain, especially timing the stages in tandem. The key sensiti...
Publication status:
Published
### Access Document
Files:
• (Version of record, 696.9KB)
### Authors
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Institution:
University of Oxford
Division:
SSD
Department:
Economics
Oxford college:
Magdalen College
Role:
Author
ORCID:
0000-0001-9325-8024
More by this author
Institution:
University of Oxford
Role:
Author
Publisher:
University of Oxford Publisher's website
Article number:
953
Series:
Department of Economics Discussion Paper Series
Publication date:
2021-11-08
Paper number:
953
Language:
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Pubs id:
1207575
Local pid:
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2021-11-08 | 2022-10-03 10:59:06 | {"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": 0.8589097857475281, "perplexity": 11728.738646177437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337415.12/warc/CC-MAIN-20221003101805-20221003131805-00677.warc.gz"} |
http://blog.gmane.org/gmane.comp.gnu.lilypond.bugs | 16 Apr 22:47 2014
### \deadNote doesn't work when setting TabNoteHead to a font other than Feta
Three years ago I didn't follow up to this proposal of Carl:
http://lists.gnu.org/archive/html/lilypond-user/2011-04/msg00326.html
I wonder if you can add an issue now.
Description of the problem
===================
If I use a different font for TabNoteHead
\override TabStaff.TabNoteHead #'font-name = #"Nimbus Sans L Bold"
when a score contains \deadNote I'm forced to comment the line above
and switch back to Feta, otherwise lilypond can't find noteheads.s2cross
(the Feta glyph for dampened notes) and no TabNoteHead is printed. This is
the warning:
warning: none of note heads noteheads.s' or noteheads.u' found
The two solutions proposed by Carl in the link above didn't work back then
and don't work now.
Carl's proposal
===========
"Probably the right way to fix this is to have \deadNotesOn set the font to
Feta, and \deadNotesOff set the font to a saved font. But we don't have
that capability currently implemented.
We also ought to have \deadNote (or better yet, xNote) use a tweak, instead
12 Apr 15:22 2014
### Make \upbow an empty command + Segmentation fault
Hi,
say I have a violin part with bowing instructions in a variable. Now I want
to use the same variable in the full orchestral score but the bowing marks
should be suppressed.
\upbow is treated by lilypond as an articulation, but the other
articulations should be unaffected by the suppressing of \upbow.
upbow={}
doesn't work because it breaks things like
a\upbow ~ a
Playing a bit with the definition of \upbow gave me a segfault.
Minimal example:
\version "2.18.0"
upbow = #(make-music 'ArticulationEvent
'articulation-type "upbow" 'types #f)
downbow = \upbow
Anyway, how do I make a command which has no effect at all?
Cheers,
Felix
12 Apr 11:59 2014
### [LSR] "Preventing collisions between two accidentals on the same note" obsolete
I'm not top posting.
The following snippet :
http://lsr.di.unimi.it/LSR/Item?id=547
seems obsolete to me.
Shall I delete it ?
Cheers,
Pierre
9 Apr 23:27 2014
### Doc "Flamenco notation" has ugly output
Hi Editors,
http://lilypond.org/doc/v2.18/Documentation/snippets/fretted-strings#fretted-strings-flamenco-notation
See discussion here :
http://lilypond.1069038.n5.nabble.com/LSR-v2-18-quot-Flamenco-notation-quot-update-improvement-td159344.html
See workaround here:
http://lsr.di.unimi.it/LSR/Item?id=409
Cheers,
Pierre
9 Apr 23:12 2014
### Snippet "Forcing measure width to adapt to MetronomeMark’s width" correction
Hi Editors,
The following snippet:
does not the job.
see "disscussion" here:
See workaround here :
http://lsr.di.unimi.it/LSR/Item?id=659
Cheers,
~Pierre
PS. Why not adding this snippet to the rhythm section ?
8 Apr 21:45 2014
### Add "Extending a TrillSpann" in the doc.
Hi Editors,
After a discussion on the french forum I've added this snippet to the LSR:
http://lsr.di.unimi.it/LSR/Item?u=1&id=912
I think it would be helpfull to add it in the doc :
http://lilypond.org/doc/v2.18/Documentation/notation/spanners
with beams and glissandi.
Cheers,
~Pierre
8 Apr 18:27 2014
### LSR correction - Partial measures (pickup measures)
Hi Squad, Hi Editors,
In this snippet description :
http://lsr.di.unimi.it/LSR/Item?id=183
I've deleted the following lines :
"<h5>Bugs</h5>
<p>This command does not take into account grace notes at the start of
the music. When a piece starts with graces notes in the pickup, then
the <code>\partial</code> should follow the grace notes (see the second
line of music)."
and untagged "workaround".
If that causes any problem, I'll put them back.
Cheers,
~Pierre
8 Apr 18:15 2014
### no bar numbers printed after a line break in the middle of a repeated measure
Hey all,
if this is a bug it might be related to issue 460: in the following tiny example no bar numbers are printed in
the line after a line break in the middle of a repeated measure. Obviously the bar numbers are only
invisible because they are counted correctly:
\version "2.19.3"
{
\repeat volta 2 {
\partial 4 g4
\repeat unfold 24
c'1
d'2.
}
\break
\repeat volta 2 {
g4
\repeat unfold 24
c'1
d'2.
}
}
hth
patrick
7 Apr 22:18 2014
### \voiceOne rest positioning
Hi,
I've just updated an example on the German Wikipedia
http://de.wikipedia.org/wiki/Notensatzprogramm#Text
and noticed that the position of the initial \voiceOne rest is suboptimal.
Compare LilyPond's output with that of Score and Amadeus which are both
better.
I have no clue about the Score input but that of Amadeus is definitely
default placement.
Of course it is trivial to write a pitched rest here, but I think the
default placement should be improved. This is also in the context of
Daniel Spreadbury's recent post about their rest positioning algorithm.
If I write a \voiceOne rest it will be placed that far to the top, even
if there are only spacer rests in the other voice.
I don't know how that positioning is realized, but I think the rest
should be placed much lower by default, just with the option to move
upwards to avoid collisions.
I think that's the current behaviour anyway, so the solution might be
one of the following:
- place \voiceXXX rest exactly as \oneVoice, just define the direction
where they move for collision handling.
(this would also eliminate the problem of having to switch to
\oneVoice for a single common rest).
- try to determine the pitches before and after the rest and place it in
the middle (if collision handling allows).
7 Apr 20:48 2014
### No space between header and markuplist after page break
> I'm not top posting.
%%
%% Occasionally I noticed that text wrapped around to "next" page
%% is placed right below header line, no vertical space between.
%%
%% This has been reproduced using lilypond version 2.19.3
%%
%% any idea how to fix this?
%%
%% \paper {
%% top-markup-spacing #'basic-distance = #10
%% }
%%
%% doesn't help, it just shifts everything a little bit downwards.
%% Page break moves some lines upwards.
%%
%% This sample is tested against paper format a4. Other sizes may need
%%
\version "2.16.2"
\include "deutsch.ly"
musik = \relative c'' { b a c h }
title = "Titel"
6 Apr 16:19 2014
### clang fix for git version
A recent git commit introduced a call to to_string. To make lilypond
compile with clang again, please commit the attached diff (fixed like
other instances of the same issue).
Thanks,
Thomas
--- lily/misc.cc.orig 2014-04-06 12:44:15.000000000 +0000
+++ lily/misc.cc
<at> <at> -31,7 +31,7 <at> <at> int
intlog2 (int d)
{
if (d <= 0)
- error ("intlog2 with negative argument: " + to_string (d));
+ error ("intlog2 with negative argument: " + ::to_string (d));
int i = 0;
while ((d != 1))
{
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Gmane | 2014-04-19 11:58:55 | {"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": 0.6534258127212524, "perplexity": 13098.698444634643}, "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-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00509-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://txcorp.com/images/docs/usim/latest/reference_manual/updater_multiUpdater.html | # multiUpdater (1d, 2d, 3d)¶
The multiUpdater references several updaters and uses them to perform an explicit time-integration. The multiUpdater typically used to advance systems of the form:
\notag \begin{align} \frac{\partial \mathbf{q}}{\partial t} + \nabla\cdot\left[ \mathcal{F} \left( \mathbf{w} \right) \right] = \mathcal{S} \left( \mathbf{w}\right) \end{align}
Time integration schemes currently supported by multiUpdater include 1st, 2nd and 3rd order Runge-Kutta, subcycling and 1st, 2nd order accurate Super Time Step methods.
The multiUpdater updater accepts the parameters required by Updater.
The operations performed by the multiUpdater are specified using the UpdateStep and UpdateSequence pattern used for the main USim input file. Note that only the loop attribute for the UpdateSequence is used by the multiUpdater and that the attribute “operation = integrate” must be specified in the final UpdateStep in the loop in order for USim to perform the time integration.
The time integration scheme used by the multiUpdater is specified by the use of a Time Integrator block.
## Data¶
in (string vector, required)
Input 1 to N are input nodalArrays to be used in the updaters specified in the UpdateStep.
out (string vector, required)
Output 1 to N are output nodalArrays resulting from the integration UpdateStep.
## Sub-Blocks¶
TimeIntegrator
Time Integrator Currently only one time integration scheme can be specified.
Time integration schemes currently supported by multiUpdater include 1st, 2nd and 3rd order Runge-Kutta, subcycling and 1st, 2nd order accurate Super Time Step methods.
UpdateSequence
UpdateSequence This block is used to set the sequence of update steps
UpdateStep
UpdateStep The steps used in the update
## Example¶
The code block below demonstrates the use of a multiUpdater to solve a multi-species fluid problem with collision operators and boundary conditions:
<Updater rkUpdater>
kind = multiUpdater1d
onGrid = domain
in = [q, q1, q2, q3]
out = [qnew, qnew1, qnew2, qnew3]
<TimeIntegrator rkIntegrator>
kind = rungeKutta1d
ongrid = domain
scheme = third
</TimeIntegrator>
loop = [boundaries,hyper]
updaters = [openBoundaries, openBoundaries1, openBoundaries2, openBoundaries3]
syncVars = [q, q1, q2, q3]
operation = "integrate"
updaters = [\
computeN1, computeN2, computeN3, \
computeT1, computeT2, computeT3, \
computeV1, computeV2, computeV3, \
collisionFrequency, \
momentumSource, energySource,\
hyper, hyper1, hyper2, hyper3, \ | 2020-08-10 01:49:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8625328540802002, "perplexity": 8946.61973781779}, "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-2020-34/segments/1596439738603.37/warc/CC-MAIN-20200810012015-20200810042015-00140.warc.gz"} |
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=DBSHBB_2013_v50n5_1067 | STRUCTURE RELATIONS OF CLASSICAL MULTIPLE ORTHOGONAL POLYNOMIALS BY A GENERATING FUNCTION
Title & Authors
STRUCTURE RELATIONS OF CLASSICAL MULTIPLE ORTHOGONAL POLYNOMIALS BY A GENERATING FUNCTION
Lee, Dong Won;
Abstract
In this paper, we will find some recurrence relations of classical multiple OPS between the same family with different parameters using the generating functions, which are useful to find structure relations and their connection coefficients. In particular, the differential-difference equations of Jacobi-Pineiro polynomials and multiple Bessel polynomials are given.
Keywords
multiple orthogonal polynomial;classical multiple orthogonal polynomial;recurrence relation;generating function;
Language
English
Cited by
References
1.
A. Angelesco, Sur l'approximation simultanee de plusieurs integrales definies, C. R. Paris, 167 (1918), 629-631.
2.
A. I. Aptekarev, Multiple orthogonal polynomials, J. Comput. Appl. Math. 99 (1998), no. 1-2, 423-447.
3.
A. I. Aptekarev, A. Branquinho, and W. Van Assche, Multiple orthogonal polynomials for classical weights, Trans. Amer. Math. Soc. 355 (2003), no. 10, 3887-3914.
4.
A. I. Aptekarev, V. Kalyagin, G. Lopez Lagomasino, and I. A. Rocha, On the limit behavior of recurrence coefficients for multiple orthogonal polynomials, J. Approx. Theory 139 (2006), no. 1-2, 346-370.
5.
B. Beckermann, J. Coussement, andW. Van Assche, Multiple Wilson and Jacobi-Pineiro polynomials, J. Approx. Theory 132 (2005), no. 2, 155-181.
6.
P. M. Bleher and A. B. J. Kuijlaars, Large n limit of Gaussian random matrices with external source. I, Comm. Math. Phys. 252 (2004), no. 1-3, 43-76.
7.
P. M. Bleher and A. B. J. Kuijlaars, Integral representations for multiple Hermite and multiple Laguerre polynomials, Ann. Inst. Fourier (Grenoble) 55 (2005), no. 6, 2001-2014.
8.
T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach, 1978.
9.
P. Desrosiers, Duality in random matrix ensembles for all ${\beta}$, Nuclear Phys. B 817 (2009), no. 3, 224-251.
10.
P. Desrosiers and P. J. Forrester, Asymptotic correlations for Gaussian and Wishart matrices with external source, Int. Math. Res. Not. (2006), Art. ID 27395, 43p.
11.
P. Desrosiers and P. J. Forrester, A note on biorthogonal ensembles, J. Approx. Theory 152 (2008), no. 2, 167-187.
12.
M. E. H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable, in Encyclopedia of Mathematics and its Applications, vol. 98, Cambridge University Press, 2005.
13.
D. W. Lee, Properties of multiple Hermite and multiple Laguerre polynomials by the generating function, Integral Transforms Spec. Funct. 18 (2007), no. 11-12, 855-869.
14.
D. W. Lee, Generating functions and multiple orthogonal polynomials, In 5th Asian Mathematical Conference Proceedings, Vol. II, (Yahya Abu Hasan et al., ed.), 44-51, 2009.
15.
V. Lysov and F. Wielonsky, Strong asymptotics for multiple Laguerre polynomials, Constr. Approx. 28 (2008), no. 1, 61-111.
16.
E. M. Nikishin and V. N. Sorokin, Rational Approximations and Orthogonality, Translations of Mathematical Monographs, 92. American Mathematical Society, Providence, RI, 1991.
17.
E. D. Rainville, Special Functions, Chelsea Publishing Company, New York, 1960.
18.
G. Szego, Orthogonal Polynomials, Amer. Math. Soc. Coll. Publ. vol 23., 4th ed., Amer. Math. Soc., Providence, RI, 1975.
19.
W. Van Assche, Multiple orthogonal polynomials, irrationality and transcendence, Continued fractions: from analytic number theory to constructive approximation (Columbia, MO, 1998), 325-342, Contemp. Math., 236, Amer. Math. Soc., Providence, RI, 1999.
20.
W. Van Assche, Nearest neighbor recurrence relations for multiple orthogonal polynomials, J. Approx. Theory 163 (2011), no. 10, 1427-1448.
21.
W. Van Assche and E. Coussement, Some classical multiple orthogonal polynomials, J. Comput. Appl. Math. 127 (2001), no. 1-2, 317-347. | 2017-01-18 06:33:38 | {"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": 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": 0.5818822979927063, "perplexity": 1266.3360005651343}, "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/1484560280239.54/warc/CC-MAIN-20170116095120-00108-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://nrich.maths.org/543 | Real(ly) Numbers
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
Overturning Fracsum
Solve the system of equations to find the values of x, y and z: xy/(x+y)=1/2, yz/(y+z)=1/3, zx/(z+x)=1/7
Bang's Theorem
If all the faces of a tetrahedron have the same perimeter then show that they are all congruent.
System Speak
Stage: 4 and 5 Challenge Level:
Solve the system of equations:
$ab = 1$
$bc = 2$
$cd = 3$
$de = 4$
$ea = 6$ | 2015-07-01 12:57:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6366352438926697, "perplexity": 334.25620046347404}, "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-2015-27/segments/1435375094931.19/warc/CC-MAIN-20150627031814-00218-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/types-ratios-calculate-inventory-turnover-ratio-data-given-below_80215 | Share
# Calculate Inventory Turnover Ratio from the Data Given Below: - Accountancy
ConceptTypes of Ratios
#### Question
Calculate Inventory Turnover Ratio from the data given below:
Rs Inventory in the beginning of the year 10,000 Inventory at the end of the year 5,000 Carriage 2,500 Revenue from Operations 50,000 Purchases 25,000
#### Solution
"Inventory turnover ratio" = "Cost of Revenue from operations"/"Average Inventor"
"Cost of revenue from operations" = "Inventory in the begining + Purchases + Carriage - Inventory at the end"
=10,000 + 25,000 + 2,500 - 5,000
=32,500
"Average Inventory" = "Inventory in the begining + Inventory at the end"/2
= (10,000 + 5,000)/2
= 7,500
"Inventory Turnover Ratio" = "32,500"/"7,500" = 4.33 "times"
Is there an error in this question or solution?
#### APPEARS IN
Solution Calculate Inventory Turnover Ratio from the Data Given Below: Concept: Types of Ratios.
S | 2020-03-30 15:58:11 | {"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": 0.38528409600257874, "perplexity": 12539.467580164763}, "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/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00157.warc.gz"} |
https://hq2006.bnl.gov/abstracts/JanaBielcikova.html | ## What can we learn from high-$p_T$ azimuthal correlations of neutral strange baryons and mesons at RHIC energy ?
### Jana Bielcikova
Azimuthal correlations of particles with large transverse momentum ($p_T$) are commonly used to study jet related processes in high energy heavy-ion collisions. Identified two-particle correlations are expected to provide additional information on jet quenching, the baryon-meson puzzle and particle production mechanisms at RHIC energies. We present studies of correlations of neutral strange baryons ($\Lambda$,$\bar{\Lambda}$) and mesons (K$_S0$) for $p_T$=2-6 GeV/$c$ associated with non-identified charged particles in d+Au and Au+Au collisions at $\sqrt{s_{NN}}$~=~200~GeV measured by the STAR experiment. We investigate in detail possible flavor, baryon/meson and particle/antiparticle differences by studying the associated yield of charged particles as a function centrality of the collision, transverse momentum of trigger and associated particles as well as the fragmentation variable $z_T$. We compare our results to the proton and pion triggered correlations as well as to fragmentation and recombination models. | 2018-12-10 11:41:58 | {"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": 0.6580883264541626, "perplexity": 2853.7710918536854}, "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-2018-51/segments/1544376823322.49/warc/CC-MAIN-20181210101954-20181210123454-00334.warc.gz"} |
http://www.scienceforums.net/topic/92820-matrix-multiplication-and-linear-transformation/ | Matrix multiplication and Linear Transformation
Recommended Posts
Hello everybody.
I'm having a little bit of trouble understanding a passage of my textbook regarding a linear transformation and matrix multiplication, I wonder if you could help me out.
So, I have this equation:
$\dot x = \textbf{Fx} + \textbf{G}u$
Where F is some 3x3 matrix and x a 3x1 array. For now, these are the important variables. So, my objective is putting F in a specific format called control canonical form (A), which is:
$A = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c \\ \end{array} \right|.$
For that, the book shows a Linear Transformation in the variable x:
$\textbf{x} = \textbf{Tz}$
Which leads to (see first equation):
$\dot z =T^{-1} \textbf{FTz} + T^{-1}\textbf{G}u$
The equation for A is:
$\textbf{A} = T^{-1} \textbf{FT}$
Where T-1is defined as:
$T^{-1} = \left| \begin{array}{ccc} t1 \\ t2 \\ t3 \\ \end{array} \right|.$
Writing everything in therms of T-1:
$\textbf{A} T^{-1} = T^{-1} \textbf{F}$
Now, the problem:
$\left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c \\ \end{array} \right| \left| \begin{array}{ccc} t1 \\ t2 \\ t3 \\ \end{array} \right| = \left| \begin{array}{ccc} t1 \textbf{F} \\ t2 \textbf{F} \\ t3 \textbf{F} \\ \end{array} \right|$
I don't understant the right part of the equation. How can I multiply T-1, which is a 3x1 array, with the 3x3 F matrix? Why the book shows a array with every single term of T-1 multiplying F? I apologize if this is some stupid question but linear algebra isn't my strong suit.
Thanks!
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ajb 1567
Is $T^{-1}$ not the matrix inverse of $T$?
It looks like you are trying to show that A and F are similar.
• 1
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Is $T^{-1}$ not the matrix inverse of $T$?
It looks like you are trying to show that A and F are similar.
Yes, is the inverse. I'm trying to find the terms of the inverse transformation matrix (t1,t2,t3) which will "turn" F into A (It's easy to figure out T knowing T-1) . But that last equation doens't make any sense to me.
Edited by a.caregnato
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ajb 1567
$T$ is an nxn matrix and so its inverse is also an nxn matrix. I do not understand what you have written, but this could be a notational issue.
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$T$ is an nxn matrix and so its inverse is also an nxn matrix. I do not understand what you have written, but this could be a notational issue.
You're right, I've got it now.
$t1,t2,t3$ are row vectors (1x3), thats the only way We'll have a matrix T-1 with 3x3 dimensions.
Thank you for your help, ajb.
Edited by a.caregnato
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ajb 1567
$t1,t2,t3$ are row vectors (1x3), thats the only way We'll have a matrix $T-1$ with 3x3 dimensions.
That was what I was wondering. Then notation is not great in my opinion.
Thank you for your help, ajb.
No problem. | 2017-10-22 00:53:56 | {"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": 0.7750698924064636, "perplexity": 656.8680203467144}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00325.warc.gz"} |
https://acutronicrobotics.com/docs/products/mara/motionplanning/tutorials/ros/ompl | # OMPL - Open Motion Planning Library Planner
The Open Motion Planning Library is a powerful collection of state-of-the-art sampling-based motion planning algorithms and is the default planner in MoveIt!. For more information see project webpage.
#### OMPL Settings
Here we review important configuration settings for OMPL. These settings can typically be found in the ompl_planning.yaml file located in your robots mara_moveit_config package.
#### Longest Valid Segment Fraction
The longest_valid_segment_fraction defines the discretization of robot motions used for collision checking and greatly affects the performance and reliability of OMPL-based solutions. A motion in this context can be thought of as an edge between two nodes in a graph, where nodes are waypoints along a trajectory. The default motion collision checker in OMPL simply discretizes the edge into a number of sub-states to collision check.
Specifically, longest_valid_segment_fraction is the fraction of the robot’s state space that, given the robot isn’t currently in collision, we assume the robot can travel while remaining collision free. For example, if longest_valid_segment_fraction = 0.01, then we assume that if an edge between two nodes is less than 1/100th of the state space, then we don’t need to explicity check any sub-states along that edge, just the two nodes it connects.
In addition to the longest_valid_segment_fraction parameter in the ompl_planning.yaml file, there is also the maximum_waypoint_distance, found in the dynamic reconfigure file. maximum_waypoint_distance defines the same discretization of robot motions for collision checking, but it does so at an absolute level instead of using fractions of the state space. For example, if maximum_waypoint_distance = 0.1, then if an edge is shorter than 0.1 in state space distance, then we don’t explicitly check any sub-states along that edge.
If both longest_valid_segment_fraction and maximum_waypoint_distance are set, then the variable that produces the most conservative discretization (the one that would generate the most states to collision check on a given edge) is chosen.
Set longest_valid_segment_fraction (or maximum_waypoint_distance) too low, and collision checking / motion planning will be very slow. Set too high and collisions will be missed around small or narrow objects. In addition, a high collision checking resolution will cause the path smoothers to output incomprehensible motions because they are able to “catch” the invalid path and then attempt to repair them by sampling around it, but imperfectly.
#### Projection Evaluator
The projection_evaluator can take in a list of joints or links to approximate the coverage of a configuration space. This settings is used by planners such as KPIECE, BKPIECE, LBKPIECE, and PDST. For more information read the corresponding publications.
#### Enforce Planning in Joint Space
Depending on the planning problem MoveIt chooses between joint space and cartesian space for problem representation. Setting the group parameter enforce_joint_model_state_space enforces the use of joint space for all plans.
By default planning requests with orientation path constraints are sampled in cartesian space so that invoking IK serves as a generative sampler.
By enforcing joint space the planning process will use rejection sampling to find valid requests. Please not that this might increase planning time considerably.
##### Other Settings
Depending on the planner you are using, other settings are available for tuning/parameter sweeping. The default values for these settings are auto-generated in the MoveIt! Setup Assistant and are listed in the ompl_planning.yaml file - you are encouraged to tweak them.
#### OMPL Optimization Objectives
Several planners that are part of the OMPL planning library are capable of optimizing for a specified optimization objective. This tutorial describes that steps that are needed to configure these objectives. The asymptotically optimal planners that are currently exposed to MoveIt! are:
• RRT*
• PRM*
Other optimal planners in OMPL but not exposed in MoveIt! yet:
• LazyPRM*
• RRT#
• RRTX
• Informed RRT*
• Batch Informed Trees (BIT*)
• Lower Bound Tree RRT (LBTRRT)
• Sparse Stable RRT
• Transition-based RRT (T-RRT)
• SPARS
• SPARS2
• FMT*
• CForest
And the following optimization objectives are available:
• PathLengthOptimizationObjective (Default)
• MechanicalWorkOptimizationObjective
• MaximizeMinClearanceObjective
• StateCostIntegralObjective
• MinimaxObjective
The configuration of these optimization objectives can be done in the _omplplanning.yaml. A parameter with the name optimization_objective is added as a configuration parameter. The value of the parameter is set to be the name of the selected optimization objective. For example, to configure RRTstar to use the MaximizeMinClearanceObjective, the planner entry in the ompl_planning.yaml will look like:
RRTstarkConfigDefault:
type: geometric::RRTstar
optimization_objective: MaximizeMinClearanceObjective
range: 0.0
goal_bias: 0.05
delay_collision_checking: 1
For more information on the OMPL optimal planners, the reader is referred to the OMPL - Optimal Planning documentation.
##### Post-Processing Smoothing
Note there is a limit to how much smoothing can help reduce indirect routes. Note also that here we discuss geometric(kinematic)-based only smoothing. Velocity/acceleration/jerk smoothing is handled elsewhere, see Time Parameterization.
You can adjust the amount of time MoveIt! spends on smoothing by increasing the planning time. Any remaining time after an initial plan is found, but before the allowed_planning_time is exhausted, will be used for smoothing. MoveIt! also does path hybridization, taking the best parts of N different planning runs and splicing them together. Therefore, num_planning_attempts affects the quality as well. | 2019-02-23 16:37:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.33815228939056396, "perplexity": 3089.0294002592996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249508792.98/warc/CC-MAIN-20190223162938-20190223184938-00199.warc.gz"} |
https://undergroundmathematics.org/polynomials/r6714/solution | Review question
# When are $y < x+1, y+6x < 20, x = 5y-7$ true together? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6714
## Solution
1. Find the solution set for $x$ given that the following three relations for $x$, $y$, where $x$, $y\in \mathbb{R}$, are simultaneously true: $y < x+1, \quad y+6x < 20, \quad x = 5y-7.$
A diagram of the three lines helps us visualise the problem.
The region where y < x + 1 is the half-plane below the blue line. The blue line itself is not in the region.
The region where y + 6x < 20 is the half-plane below the yellow line. The yellow line itself is not in the region.
So we can see the portion of the red line that lies in the region where both inequalities are satisfied.
We need to find the x-values that points on this portion of red line can take.
Rearranging $x=5y-7$ gives us $y=\frac{x+7}{5}.$
Substituting this into the inequalities gives $\frac{x+7}{5}<x+1 \quad \text{and}\quad \frac{x+7}{5}+6x<20,$ from which we get $x+7<5x+5 \quad \text{and}\quad x+7+30x<100.$ Solving these gives us $\dfrac{1}{2}<x$ and $x<3$ respectively. Therefore the solution set for $x$ is $\dfrac{1}{2} < x < 3$.
Alternatively, we could find the point of intersection of the blue line with the red line, and of the yellow line with the red line.
1. Find the solution set of the inequality $\frac{12}{x-3} < x+1, \quad (x\in \mathbb{R}, \ x \neq 3).$
### Approach 1
It’s tempting to multiply both sides by $x-3$ to give $12 < (x+1)(x-3)$, but $x-3$ might be negative, and so multiplying by $x-3$ could change the direction of the inequality.
So we split the problem up into two cases: $x-3>0$ and $x-3<0$.
#### Case 1: $x-3>0$
Since $x-3>0$, we can multiply by $x-3$ without changing the direction of the inequality to get $12 < (x+1)(x-3)$.
This is equivalent to $0<x^2-2x-15$, which we can rewrite as $(x-5)(x+3)>0$.
Sketching $y=(x-5)(x+3)$ shows this inequality holds if $x<-3$ or if $x>5$.
Since we are only considering $x>3$ though, the solution set is $x>5$.
#### Case 2: $x-3<0$
Multiplying by $x-3$ now means multiplying by a negative value, so we’ll get the same quadratic inequality, but with a different inequality sign, that is, $0>(x-5)(x+3)$.
Referring to the sketch of $y=(x-5)(x+3)$, this can only hold when $-3<x<5$. As we’re only considering $x<3$, the solution set is $-3<x<3$.
By combining the results from the two cases, we get the solution set for $x$ of $-3<x<3$ or $x>5$.
### Approach 2
Since multiplying by $x-3$ might change the sign of the inequality, we’ll multiply by something we know is never negative.
Since we want to multiply by $x-3$ to clear the denominator, we multiply by $(x-3)^2\ge0$.
This gives us $12(x-3)<(x+1)(x-3)^2.$
It’s tempting to now multiply out, but instead, we can use the shared factor of $x-3$ to rearrange and factorise.
Rearranging and taking out $x-3$ as a factor gives: \begin{align*} &&(x+1)(x-3)^2-12(x-3)&>0&&\quad\\ \iff\quad&&(x-3)((x+1)(x-3)-12)&>0\\ \iff\quad&&(x-3)(x^2-2x-15)&>0\\ \iff\quad&&(x-3)(x-5)(x+3)&>0 \end{align*}
This is a cubic which crosses the $x$-axis at $3$, $5$ and $-3$, looking roughly like
We can see that $(x-3)(x-5)(x+3)>0$ when $-3<x<3$ or $x>5$.
### Approach 3
We begin by sketching the two functions $y=\dfrac{12}{x-3}$ and $y=x+1$ on the same axes.
The graph of $y=\dfrac{12}{x-3}$ can be found from $y=\dfrac{1}{x}$ via stretching in the $y$-direction by a factor of $12$ and then translating by $3$ in the $x$-direction.
So the graph looks like
The two graphs intersect at $A$ and $B$, with $x$-coordinates $a$ and $b$ respectively.
It’s clear from the graph that $\dfrac{12}{x-3} < x+1$ when the red graph is below the blue one, which occurs when $a<x<3$ and when $x>b$.
So it only remains to find $a$ and $b$, which we can do by solving the equation $\frac{12}{x-3} = x+1.$ Multiplying by $x-3$ and rearranging gives $(x+1)(x-3)-12=0$ so $x^2-2x-15=0$ which yields $(x-5)(x+3)=0$, therefore $x=5$ or $x=-3$.
This gives $a=-3$ and $b=5$, so the solution is $-3<x<3$ or $x>5$.
### Approach 4
We have $\dfrac{12}{x-3}<x+1$, and so bringing everything to the LHS, $\dfrac{12}{x-3}-x-1<0$, or $\dfrac{(x-5)(x+3)}{3-x}<0$.
As we now allow $x$ to move from below $-3$ to above $5$, we can track the sign of this rational function, which gives us the same answer as before. | 2022-08-10 17:26:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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": 0.9019604325294495, "perplexity": 160.09921470794563}, "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/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00212.warc.gz"} |
https://studydaddy.com/question/is-addition-of-sodium-in-to-water-is-a-chemical-change | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# Is addition of sodium in to water is a chemical change?
Yes, the reaction between sodium metal and water is a chemical change.
A chemical change is said to take place when two or more substances interact if this interaction results in the formation of one or more new substances.
SImply put, if you start with a substance and end up with another substance altogether, a chemical change took place.
In your case, the interaction between sodium metal and water will result in the formation of two new substances, sodium hydroxide and hydrogen gas.
The balanced chemical equation for this reaction looks like this
2Na_((s)) + 2H_2O_((l)) -> 2NaOH_((aq)) + H_(2(g))
Sodium hydroxide, NaOH, is very soluble in aqueous solution, so it will exist as
NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)
The ionic equation will thus look like this
2Na_((s)) + 2H_2O_((l)) -> 2Na_((aq))^(+) + 2OH_((aq))^(-) + H_(2(g))
As you can see, sodium no longer exists as metal after the reaction, it exists as sodium cations, Na^(+). Moreover, hydrogen gas and hydroxide ions were produced as a result of this reaction.
So, you started with sodium metal and ended up with sodium cations, hydroxide ions, and hydrogen gas.
This means that your reaction was indeed a . | 2019-04-21 20:52:58 | {"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": 0.6328366994857788, "perplexity": 2317.678944754405}, "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-2019-18/segments/1555578532882.36/warc/CC-MAIN-20190421195929-20190421221929-00376.warc.gz"} |
https://www.statsmodels.org/stable/generated/statsmodels.tsa.ar_model.ARResults.arfreq.html | # statsmodels.tsa.ar_model.ARResults.arfreq¶
ARResults.arfreq
Returns the frequency of the AR roots.
This is the solution, x, to z = abs(z)*exp(2j*np.pi*x) where z are the roots. | 2020-07-04 05:34:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9709398746490479, "perplexity": 2823.942178434131}, "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/1593655884012.26/warc/CC-MAIN-20200704042252-20200704072252-00138.warc.gz"} |
https://www.math.u-psud.fr/Variational-approach-to-the-regularity-of-the-singular-free-boundaries?lang=fr | ## Variational approach to the regularity of the singular free boundaries
### Mardi 23 janvier 2018 14:00-15:00 - Bozhidar Velichkov - Université Grenoble Alpes
Résumé : In this talk we will present some recent results on the structure of the free boundaries of the (local) minimizers of the Bernoulli problem in ,
In 1981 Alt and Caffarelli proved that if $u$ is a minimizer of the above problem, then the free boundary can be decomposed into a regular part, , and a singular part, , where
• is locally the graph of a smooth function ;
• is a small (possibly empty) set.
Recently, De Silva and Jerison proved that starting from dimension there are minimal cones with isolated singularities in zero. In particular, the set of singular points might not be empty.
The aim of this talk is to describe the structure of the free boundary around a singular point. In particular, we will show that if is a solution of (*), is a point of the free boundary and there exists one blow-up limit , which has an isolated singularity in zero, then the free boundary is a graph over the cone .
Our approach is based on the so called logarithmic epiperimetric inequality, which is a purely variational tool for the study of free boundaries and was introduced in the framework of the obstacle problem in a series of works in collaboration with Maria Colombo and Luca Spolaor.
Lieu : IMO ; salle 3L8.
Variational approach to the regularity of the singular free boundaries Version PDF
août 2019 :
Rien pour ce mois
Département de Mathématiques Bâtiment 307 Faculté des Sciences d'Orsay Université Paris-Sud F-91405 Orsay Cedex Tél. : +33 (0) 1-69-15-79-56 Département Laboratoire Formation | 2019-08-25 19:16:20 | {"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": 0.7946087718009949, "perplexity": 668.5441234018233}, "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-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00540.warc.gz"} |
http://math.stackexchange.com/questions/422598/average-number-of-heads-in-filtered-coin-toss?answertab=votes | Average number of heads in filtered coin toss
I have a coin that, when tossed, produces heads with probability $p \geq 0.5$ and tails with probability $1-p$.
I start a coin-tossing experiment. Whenever I get more than one tail in a row, I discard the second tail and toss again, so that my results look like one long chain of heads with the occasional tail sprinkled in. In the long run, does the overall percentage of heads converge to 100%? How long does my result list have to be to guarantee at least $x$% heads? What if I only discard the last toss if I get more than $k$ tails in a row?
(This question comes from thinking about the caching function of my streaming music library. Please excuse a first-year undergrad's background knowledge; if this has been done a million times before I would appreciate a link to the general subject)
-
The resulting output is a Markov chain on the state space $\{h,t\}$ with transitions $p(t,h)=1$, $p(t,t)=0$, $p(h,h)=p$ and $p(h,t)=1-p$. The overall percentage of heads converges to the stationary measure of $h$. The stationary distribution $\pi$ solves $\pi(t)=p(t,t)\pi(t)+p(h,t)\pi(h)$, that is, $\pi(t)=(1-p)\pi(h)$.
Hence $\pi(h)=\dfrac1{2-p}$ and, in particular, $\pi(h)\ne1$ for every $p\ne1$.
If the strategy is to discard the last toss if one gets more than $k$ tails in a row, the resulting output is a Markov chain with memory $k$, or equivalently, a Markov chain on the state space $\{h,t\}^k$, and the same technique applies.
An alternative is to consider the length of the run of consecutive tails ending at $n$. This is again a Markov chain, this time on the state space $\{0,1,\ldots,k\}$. The transitions are $p(i,i+1)=1-p$ and $p(i,0)=p$ for every $i\leqslant k-1$, and $p(k,0)=1$. The stationary distribution $\pi_k$ solves the system $\pi_k(0)=p\pi_k(0)+\cdots+p\pi_k(k-1)+\pi_k(k)$ and $\pi_k(i)=(1-p)\pi_k(i-1)$ for $1\leqslant i\leqslant k$. Hence $\pi_k(i)=(1-p)^i\pi_k(0)$ for every state $i$, and $\pi_k(0)\cdot\sum\limits_{i=0}^k(1-p)^i=1$.
Finally, for each $k\geqslant1$, the overall percentage of heads converges to $\pi_k(0)=\dfrac{p}{1-(1-p)^{k+1}}.$
- | 2016-02-07 17:46:27 | {"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": 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": 0.9537276029586792, "perplexity": 72.85193178130476}, "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-2016-07/segments/1454701150206.8/warc/CC-MAIN-20160205193910-00341-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=7396&pid=65393 | = to ==, why? [SOLVED]
12-10-2016, 09:39 PM (This post was last modified: 12-10-2016 09:59 PM by Han.)
Post: #6
Han Senior Member Posts: 1,817 Joined: Dec 2013
RE: = to ==, why?
In your third example, just comment out the export equ line (so that equ is created as a CAS variable) for a quick fix.
(12-10-2016 07:07 PM)compsystems Wrote: The second version "deductionQuadFormula_2()" (works fine), to show intermediate steps, is required to convert it to string. An alternative code may exist
Why are you converting the CAS operations into strings? You can simply print the strings that show what steps you are doing, and then evaluate the actual operations using the actual CAS commands rather than using expr(...).
Code:
print("> factor(answer)"); equ:=factor(equ); print("> " + equ);
Quote:The third version "deductionQuadFormula_3()", same as the previous one but using a subfunction to not repeat code (it does not work). An alternative code may exist, Who can code it?
Unless the images provide information that your code itself cannot generate, there is really no need to include them because it makes your post that much more difficult to read. Also, please don't edit your original post in that manner. Instead, insert "EDIT:" and add more to your post. Otherwise, when existing responses to your original text it may no longer make sense since you have changed your original post.
Anyway, here is an incomplete example (run as quad(0) or quad(1) for no pause or with pause).
PHP Code:
#casquad(p):=BEGIN local eq; print("Quadratic Formula"); dostep('assume(a>0)',p); dostep('equ1:=a*x^2+b*x+c=0',p); dostep('equ1:=equ1*4*a',p); dostep('equ1:=expand(equ1)',p); dostep('equ1:=equ1+b^2-4*a*c',p); dostep('equ1:=simplify(equ1)',p); dostep('equ1:=factor(equ1)',p); // more steps hereEND;dostep(cas,p):=begin print(cas); print(eval(cas)); // or use print("> " + eval(cas)) if p then print(" [PAUSE]"); wait(); else print(""); end; end;#end
Here's an alternative that prints something different from what is actually being computed.
PHP Code:
#casquad(p):=BEGIN local eq; print("Quadratic Formula"); dostep("assume(a>0)",'assume(a>0)',p); dostep("equ1:=a*ax^2+b*x+c=0",'equ1:=a*x^2+b*x+c=0',p); dostep("answer * 4*a",'equ1:=equ1*4*a',p); dostep("expand(answer)",'equ1:=expand(equ1)',p); dostep("answer + b^2-4*a*c",'equ1:=equ1+b^2-4*a*c',p); dostep("simpify(answer)",'equ1:=simplify(equ1)',p); dostep("factor(answer)",'equ1:=factor(equ1)',p); // more steps hereEND;dostep(s,cas,p):=begin print("> "+s); print(eval(cas)); if p then print(" [PAUSE]"); wait(); else print(""); end; end;#end
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Messages In This Thread = to ==, why? [SOLVED] - compsystems - 12-09-2016, 05:53 PM RE: = to ==, why? - Han - 12-09-2016, 07:58 PM RE: = to ==, why? - compsystems - 12-09-2016, 10:15 PM RE: = to ==, why? - Han - 12-10-2016, 05:46 AM RE: = to ==, why? - compsystems - 12-10-2016, 07:07 PM RE: = to ==, why? - Han - 12-10-2016 09:39 PM RE: = to ==, why? - compsystems - 12-11-2016, 03:08 PM
User(s) browsing this thread: 1 Guest(s) | 2020-07-04 04:51:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3156159520149231, "perplexity": 1546.054102779193}, "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/1593655884012.26/warc/CC-MAIN-20200704042252-20200704072252-00400.warc.gz"} |
http://imc-math.org.uk/?year=2016§ion=problems&item=prob2q | # International Mathematics Competition for University Students 2016
Select Year:
IMC 2019
Information Results/Prizes Problems & Solutions Photos
## IMC2016: Day 1, Problem 2
2. Let $k$ and $n$ be positive integers. A sequence $(A_1,\ldots,A_k)$ of $n\times n$ real matrices is preferred by Ivan the Confessor if $A_i^2\ne 0$ for $1\le i \le k$, but $A_iA_j=0$ for $1\le i,j\le k$ with $i\ne j$. Show that $k\le n$ in all preferred sequences, and give an example of a preferred sequence with $k=n$ for each $n$.
Proposed by Fedor Petrov, St. Petersburg State University
IMC1994 IMC1995 IMC1996 IMC1997 IMC1998 IMC1999 IMC2000 IMC2001 IMC2002 IMC2003 IMC2004 IMC2005 IMC2006 IMC2007 IMC2008 IMC2009 IMC2010 IMC2011 IMC2012 IMC2013 IMC2014 IMC2015 IMC 2016 IMC2017 IMC2018 IMC2019 | 2019-08-19 17:07:17 | {"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": 0.4102751612663269, "perplexity": 6754.863586957132}, "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-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00494.warc.gz"} |
https://mathhelpforum.com/tags/rapidly/ | # rapidly
1. ### Converges Rapidly to Square Root of A??
,.,good day evryone,.i need ur help with this,.how should i prove this?? If A>0 and X1>0, the sequence defined by the attachment converges rapidly to squareroot of A
2. ### Rapidly decreasing functions and convergence
I am trying to show that, for f a rapidly decreasing function, the mapping from f to f' is continuous. Since the space of RDF's is an F-space, Rudin says that I can use the closed graph theorem for F-spaces, and hence I need to show that the graph of the mapping is closed. So I wish to show...
3. ### Changing least rapidly?
At (2,1), I have to find the direction where f(x,y)=x^2+3xy-y^2 is increasing fastest. So I did: \triangledown f(2,1)= \langle 2(2)+3, 3(2)-2 \rangle = \langle 7,4 \rangle Next I have to find the rate of growth in this direction, so I did: ||\triangledown f(2,1)|| = \sqrt{7^2+4^2} =...
4. ### f decreases most rapidly at Q
Find a unit vector in the direction in which f decreases most rapidly at Q Thank u!
5. ### Show that a differentiable function f decreases most rapidly....
at x in the direction opposite to the gradient vector, that is, in the direction of -Vf(x). I made up a function of two variables f(x, y) = ln(x^2 -y^3/2) and then calculated the gradient which came to 2x/(x^2 - y^3/2) and -3/2y^2/(x^2 - y^3/2). I then put in a value Vf (2, 1) and got the...
6. ### Depth of water when the water level is falling the least rapidly
Can someone help me with this problem? Thank you. A spherical tank of radius 10ft is being filled with water. When it is completely full, a plug at its bottom is removed. According to Torricelli's law, the water drains in such a way that dV/dt = -k*sqrt(y), where V is the volume of water in the... | 2020-02-18 22:26:46 | {"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": 0.9017470479011536, "perplexity": 786.8146415371266}, "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-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00376.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/two-sources-of-sound-are-separated-by-a-distance-of-4-m-they-both-emit-sound-with-the-same-amplitude-and-frequency-330-hz-but-they-are-180-out-of-phase-speed-of-wave-motion_165453 | # Two sources of sound are separated by a distance of 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180° out of phase. - Physics
Sum
Two sources of sound are separated by a distance of 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180° out of phase. At what points between the two sources, will the sound intensity be maximum?
#### Solution
∴ λ = "v"/"n"=330/330 = 1 m
The path difference between two sources of sound is zero when they are directly in the centre of each other. However, because the waves are 180° out of phase, two maxima on either side should be at a distance of λ/4 from the centre. Other maxima will be located each λ/2 along the way.
As a result, the maximum sound intensity will be ± 0.25, ± 0.75, ± 1.25, and ± 1.75 m from the point at the centre.
Concept: Speed of Wave Motion
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 6 Superposition of Waves
Exercises | Q 12 | Page 157 | 2022-08-17 21:37:59 | {"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": 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": 0.6013692021369934, "perplexity": 892.3844349912912}, "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-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00507.warc.gz"} |
https://tex.stackexchange.com/questions/127162/chapter-style-latex-horizontal-line | # chapter style latex horizontal line
I want a chapter style like this:
1
------chapter name-------------
text text text text tex
t text text text text t
I have come this far:
\documentclass{book}
\usepackage{titlesec}
\newcommand*{\xdash}[1][3em]{\rule[0.5ex]{#1}{0.55pt}}
\titleformat{\chapter}[display]
{\Large}{\filleft\Huge\thechapter}{1ex}{\xdash[2em]}[]
\begin{document}
\chapter
{Introduction}
Text text text text
\end{document}
but I can not get it right. I want the horizontal line to be longer then the "normal" text and I want the line to be before and after the chapter name.
• When you say "longer" do you mean all the page should be crossed by the line or only some cm on each side? – Ludovic C. Aug 7 '13 at 17:07
Her's an option using the explicit option for titlesec; the title is placed inside a varwidth environment of maximum width \textwidth; the left rule is drawn with a standard \rule command and the right rule uses \xrfill from the xhfill package. Long titles (spanning more than one line) are centered with respect to the rules:
\documentclass{book}
\usepackage[explicit]{titlesec}
\usepackage{xhfill}
\usepackage{varwidth}
\usepackage{lipsum}
\titleformat{\chapter}[display]
{\Large}{\filleft\thechapter}{2ex}
{\hspace*{\dimexpr-4em-10pt\relax}%
\rule[0.5ex]{4em}{1pt}\hspace{10pt}%
\begin{varwidth}{\textwidth}\raggedright#1\end{varwidth}%
\hspace{10pt}\xrfill[0.5ex]{1pt}%
}
\begin{document}
\chapter{A short title}
\lipsum[4]
\chapter{A long title spanning several lines just to see the format in this case}
\lipsum[4]
\end{document}
An image of a short title:
An image of a long title:
• Perfect. Shorter code and easier to follow! How to i do if i want say my abstract page to have this kind of style also? I copied your code and changed from Chapter to Section, but of course these changes every section in my document. Can i use different settings for the section in different part of the document or how to i solve this in an easy way? – Mike Aug 7 '13 at 18:19
• @Mike hard to say if you don't provide precise information on the structure of your document. How are you producing your abstracta? Is your abstract a section produced with \section*? – Gonzalo Medina Aug 7 '13 at 18:23
• I am using for my abstract \begin{abstract}, begin{notation} for notation page, \tableofcontents for table of content page. These pages i would like a similar style like the chapters. The sections in the different chapters will i probably want to use some ordinary style. Hope that helps with explaining my document. – Mike Aug 7 '13 at 18:33
• @Mike no, it doesn't help much. In your question you mentioned the book document class, and this class doesn't provide an abstract environment nor a notation environment. You can use different styles for sections in your document, but I need to know the exact structure used to give you the most effective alternative. – Gonzalo Medina Aug 7 '13 at 18:42
• The problem is that i am using a class made by my university. – Mike Aug 7 '13 at 18:47 | 2019-11-19 11:06:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8537461161613464, "perplexity": 1120.8442082202448}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670135.29/warc/CC-MAIN-20191119093744-20191119121744-00392.warc.gz"} |
https://surface.syr.edu/phy/201/ | ## Physics
Article
9-3-1996
English
Physics
#### Description/Abstract
We apply optimization algorithms to the problem of finding ground states for crystalline surfaces and flux lines arrays in presence of disorder. The algorithms provide ground states in polynomial time, which provides for a more precise study of the interface widths than from Monte Carlo simulations at finite temperature. Using $d=2$ systems up to size $420^2$, with a minimum of $2 \times 10^3$ realizations at each size, we find very strong evidence for a $\ln^2(L)$ super-rough state at low temperatures.
10 pages, 3 PS figures, to appear in PRL More information at http://arxiv.org/abs/cond-mat/9609029
#### Source
Harvested from Arxiv.org | 2018-03-22 17:27:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8170194029808044, "perplexity": 1139.6395534433948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647901.79/warc/CC-MAIN-20180322170754-20180322190754-00758.warc.gz"} |
https://www.open.edu/openlearncreate/mod/oucontent/view.php?id=567§ion=6.4 | # 9.5.4 How Many Minutes for Chatting?
## Activity: How Many Minutes for Chatting?
Recall the formula for the monthly cost in $of a mobile phone that we used earlier: The owner wishes to stick to a monthly budget of$45. Starting with the “total number of minutes,” draw a doing diagram to show the operations to find the “monthly cost.” Put the monthly cost equal to $45 and then draw an undoing diagram to work out how many minutes of phone calls can be made if the monthly cost is$45.
Here is the doing diagram.
Undoing each step in turn gives the following undoing diagram.
Applying these operations in turn gives:
• Subtracting 25 gives: .
• Dividing by 0.2 gives: .
So, 130 minutes can be used for a monthly budget of \$45.
With practice, this process of using a doing-undoing diagram becomes second nature. In some cases, it can be useful if you need a formula in a different form. However, for complicated formulas, a different approach (which you will need if you continue your mathematical studies) is often used.
9.5.3 Changing Formulas
9.6 Proportion | 2022-09-26 13:40:20 | {"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": 0.8730157017707825, "perplexity": 1888.7644459510104}, "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-2022-40/segments/1664030334871.54/warc/CC-MAIN-20220926113251-20220926143251-00682.warc.gz"} |
https://stats.stackexchange.com/questions/342810/which-formulation-of-the-negative-binomial-distribution-is-appropriate-as-the-po | # Which formulation of the Negative Binomial distribution is appropriate as the Posterior Predictive Distribution of a Poisson process?
I have a Poisson process that I would like to be able to compute the Posterior Predictive Distribution for.
I've seen many descriptions that say that the analytic form of the distribution that I want is the Negative Binomial distribution, but I'm immediately faced with the problem that I've selected the Jeffrey's Prior for the conjugate Gamma Distribution and so my hyperparameter $\alpha$ isn't an integer. I assume that I can't use the Negative Binomial if I want to persist with a prior of $Gamma(0.5, 0)$, but I haven't seen this limitation described anywhere so I am not confident in my conclusion. What am I missing?
For convenience sake, let's assume that I replace the Jeffrey's prior with $Gamma(1, 0)$. At least now I have integers so I can talk in terms of counts of successes, failures and trials, but I don't understand how the accumulation of observations and the number of trials over which that data was gathered maps to the parameters of the Negative Binomial distribution. In particular, I don't understand what is meant by the $p$, the probability of success of a single trial. I am confused because I thought that $p$ for a Poisson process approaches zero; what $p$ is revealed by my tally of counts over intervals? My final point is that Wikipedia presents a number of formulations for the Negative Binomial and I'm not sure which one is appropriate for my situation.
Thanks for your assistance. I'm sorry to put such a basic question on this site, but I really can't figure it out on my own.
EDIT 1
Commenter Juho K has answered that the Negative Binomial distribution can be generalised to real parameters. Thanks for that. More significantly, the second part of my question doesn't seem to be making sense, so I'm going to try and illustrate by way of an example.
In this example, I have constant duration successive intervals in which I am counting events per interval. Say my first 5 intervals have counts of ${3, 1, 2, 2, 4}$. I am assuming that the process that generates these numbers is a Poisson process. I start with a prior of $Gamma(0.5, 0)$ so that the distribution of my rate parameter $\lambda$ given my observed data is now $Gamma(12.5, 5)$. Ok.
Now I want to know the distribution of my variable for the next interval given the data that I have already seen. Wikipedia (and other sources) say that it is the Negative Binomial Distribution, which is characterised as follows
$f(k; r, p)$
where $k$ is the random variable "number of successes", $r$ is the number of failures, and $p$ is the probability of success of a single trial.
Online derivations of the posterior predictive distribution that I have seen have been quoted in terms of $y_{n+1} \sim NB(\alpha + \Sigma^n{y_i}, 1/(\beta + n + 1))$ so in my example $y_{n+1} \sim NB(12.5, 1/6)$. One sixth is a reasonable probability, but I don't know how 12.5 can be considered the "number of failures". My understanding must be wrong because as the number of elements of data are considered, the "number of failures" gets longer and the "probability of success" gets lower. I'd consider that one of the "alternative formulations" listed by Wikipedia might be more appropriate, but I can't figure out which one, or whether I'm just completely off.
• Negative binomial can be generalized to real parameters (which may be the answer to the question), but I don't understand the part about $p$ approaching zero. Can you be more precise about what kind of data you have adn about the posterior predictive you are after. (The posterior predictive of a process would presumably be a process, but this sounds like you are most likely looking for a posterior predictive of a count). Apr 26 '18 at 5:10
• Have you looked at the wikipedia page for the negative binomial? Apr 26 '18 at 5:19
• Hi Glen_b, yes, I did check the Wikipedia page but tbh the Wikipedia page was part of the source of my confusion because it provided several alternative formulations, none of which (in my mind) clearly coincide with my problem description. I know that I am missing something, but I need help to identify where my misunderstanding is. Apr 26 '18 at 6:10
• Do you mean Poisson distribution or Poisson process? Rather than a lengthy and convoluted text, could you provide maths formulae? Apr 26 '18 at 7:51
• Xi'an. I thought that I meant Poisson process, but you might have discovered the basic flaw in my thinking. Apr 26 '18 at 8:07 | 2021-10-22 08:46:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8513435125350952, "perplexity": 175.20286066975075}, "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-2021-43/segments/1634323585504.90/warc/CC-MAIN-20211022084005-20211022114005-00362.warc.gz"} |
https://eprint.iacr.org/2014/608 | ### Key-policy Attribute-based Encryption for Boolean Circuits from Bilinear Maps
Ferucio Laurentiu Tiplea and Constantin Catalin Dragan
##### Abstract
We propose the first Key-policy Attribute-based Encryption (KP-ABE) scheme for (monotone) Boolean circuits based on bilinear maps. The construction is based on secret sharing and just one bilinear map, and can be viewed as an extension of the KP-ABE scheme in [7]. Selective security of the proposed scheme in the standard model is proved, and comparisons with the scheme in [5] based on leveled multilinear maps, are provided. Thus, for Boolean circuits representing multilevel access structures, our KP-ABE scheme is more efficient than the one in [5].
Available format(s)
Category
Public-key cryptography
Publication info
Preprint. MINOR revision.
Contact author(s)
fltiplea @ info uaic ro
History
Short URL
https://ia.cr/2014/608
CC BY
BibTeX
@misc{cryptoeprint:2014/608,
author = {Ferucio Laurentiu Tiplea and Constantin Catalin Dragan},
title = {Key-policy Attribute-based Encryption for Boolean Circuits from Bilinear Maps},
howpublished = {Cryptology ePrint Archive, Paper 2014/608},
year = {2014},
note = {\url{https://eprint.iacr.org/2014/608}},
url = {https://eprint.iacr.org/2014/608}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content. | 2022-09-28 09:10:24 | {"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": 0.17221446335315704, "perplexity": 9004.351105270429}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335190.45/warc/CC-MAIN-20220928082743-20220928112743-00172.warc.gz"} |
http://principles-wiki.net/resources:pattern-oriented_software_architecture | # Meta
* still in an early stage
resources:pattern-oriented_software_architecture
# Pattern-Oriented Software Architecture (Vol. 1)
## Information
Title Pattern-Oriented Software Architecture: A System of Patterns, Volume 1 Frank Buschmann, Regine Meunier, Hans Rohnert, Peter Sommerlad and Michael Stal Book John Wiley & Sons 1996 The book is commonly known as “POSA1” and the authors are sometimes referred to as the “party of five”
## Discussion
Discuss this wiki article and the resource on the corresponding talk page. | 2018-04-19 15:15:26 | {"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": 0.84080570936203, "perplexity": 8899.934524002389}, "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-2018-17/segments/1524125936981.24/warc/CC-MAIN-20180419150012-20180419170012-00173.warc.gz"} |
https://puzzling.stackexchange.com/tags/alphametic/info | Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
About
Tag Info
Alphametic puzzles involve equations where digits have been replaced with letters.
Alphametic puzzles involve mathematical equations where the digits have been replaced with letters or symbols.
The puzzle is then to reverse this process, and to determine which letters represent which digits by using the information given via the puzzle to narrow down the options. The solution should ideally be unique.
For example $A+B=AC$ has the unique solution $A=1, B=9$ and $C=0$.
You should use this tag for questions concerning verbal arithmetic creation, solutions, and strategies.
Standard assumptions:
• Different letters represent different digits
• Leading digits are always non-zero
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https://jyskmontage.dk/gps-breadcrumbs-lgueq/2e100d-calculate-the-bond-order-of-o2 | O 2-> O 2-2 (4 + 2 + 4 Points; Total 10 Points). The first step should always be to draw out your molecules. 2. How do you calculate the bond order of O2-? Various molecular properties can be understood by this concept such as variation in bond length and bond enthalpies. For example, bond order of a pair of atoms bonded by a single bond is one, whereas bond order of a pair of atoms bonded by a double bond is two. Recommend (0) Comment (0) person. 7 8 9. Molecule has two unpaired electrons, hence it is paramagnetic. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Number of bonding electrons = 8. The s orbitals (the two sigmas at the bottom) are completely filled, and the p orbitals are filled up to pi*, where each degenerate level has one electron. calculate the bond order for O2? This gives a bond order of 1.5, which is the average value of the three bonds. Asked by Wiki User. Answer for question: Your name: Answers. Depending upon the number of bonds present between two atoms, bond order can be one, two, or three. Key Points. Compare the alkali metals and alkaline earth metals with respect to. Bond order of O 2- The electronic configuration of O 2- ion will be: Bond order of = O 2- =1/2 (8-5) = 1.5 This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. See Answer. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? The bond order is the number of "bonds" between two atoms. Define octet rule. Bond order is the number of bonding pairs of electrons between two atoms. Myunghoon. What does bond order mean in terms of bond strength? Use molecular orbital theory to explain why the Be2 molecule does not exist. asked Oct 1 in Chemistry by Ruksar02 (52.4k points) class-11; 0 votes. 1. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O 2 is paramagnetic. What does bond order mean in terms of bond strength? Write the significance of a plus and a minus sign shown in representing the orbitals. Using molecular orbital theory; a) draw the MO diagrams for O2, O2+, and O2-. 2 6. The bond order of a molecule that has multiple Lewis structures is calculated as the average of these Lewis structures. What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? Distinguish between a sigma and a pi bond. Osanda. To calculate the bond order we should see the molecular orbital diagram. This problem has been solved! Bond order = $$\frac{N_b-N_a}{2}$$ = $$\frac{10-6}{2}$$ = 2. But of course, they are less stable than "H"_2. In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures. Explain with the help of suitable example polar covalent bond. Source(s): calculate bond order o2: https://tr.im/tWp5h. B) Determine The Bond Order For All Three Molecules C) Determine Which Molecule Has The Strongest Bond D) Discuss The Magnetic Character Of Each Molecule. Click hereto get an answer to your question ️ Give the Molecular Orbital Energy diagram of a) N2 and b) O2 . How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. stability of a molecule is directly proportional to bond order . The stability of following species can be decided on the basis of bond order as follows: O2 : KK. If the former, what kind of time frame are you thinking of? 2a_1 is the sigma_(2s)^"*" antibonding MO. Answer link. Electronic configuration of O. Top Answer. The bond order decreases and the bond length increases in the order O2+ (112.2 pm), O2 (121 pm), O2- (128 pm) and O22- (149 pm). A positive bond order means a stable molecule .but negative or zero bond order means an unstable molecule. Classification of Elements and Periodicity in Properties, Organic Chemistry Some Basic Principles and Techniques. Question #d9a53 … 0 1 2 0 0 0 0. With one additional electron in an antibonding orbital 2b2 the bond order decreases by 12 relative to no. 6 years ago. Get your answers by asking now. The electron configurations are shown below where we have 16 electrons to account for in O2. Zero bond order indicates that there is no bond between atoms. Which of these represents the correct order of their increasing bond order? Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Bond Order. (iii) 100 atoms of A + 100 molecules of B. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). Luckily, most cases you will encounter are the easy ones. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. Calculate The Bond Order Of O2,02 . Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule How do you calculate the bond order of O2? Reason: It has two unpaired electron in its bonding molecular orbital. single bond == bond order of 1. double bond == bond order of 2. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1. If you calculate their bond order, you get: #"BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding")# #= 1/2(1-0) = 1/2# #"BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding")# #= 1/2(2-1) = 1/2# So, neither is more stable than the other. Question: Calculate The Bond Order Of The Hez" Ion. Molecular orbital theory explanation The bond order of O2. The bond order describes the stability of the bond. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Bond order is calculated as: (Bonding e - antibonding e) / 2 In normal O2, there are 6 bonding electrons and 2 antibonding electrons, making the bond order 2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. Because the bond order in O 2 is 2 where as in O 2 + is 2.5.If bond order is more energy required to break the bonds is more. Calculation of the bond order of N2, O2, O2+ and O2- :- 1. electronic configuration of N2 ( 14 electrons ) σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),σ2pz² ← Prev Question Next Question → 0 votes . But of course, they are less stable than #"H"_2#. Calculate bond order of N2 and O2? Suggest structure of these compounds. Terms. Predict The Following Properties Of N2: 1. 8 views. Assuming O2+ still has a double bond, then bond order is 2. Bond order of O 2 + Electronic configuration of can be written as: Bond order of O 2 + =1/2 (8-3)= 2.5. Calculate the respective bond order. 1) If you mean O2^-(O2 with a -1 charge) the bond order is 3/2. Bond order=(Nb-Na)/2 where. To calculate bond order in these molecules, it's easiest to consider all of the atoms involved in resonance for a group of bonds. Still have questions? Write the magnetic nature of N2 and O2 molecules. (ii) Determine the molality of chloroform in the water sample. 6 years ago. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Question: Calculate The Bond Order Of The Hez" Ion. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. (a) Identify an element with five electrons in the outer subshell. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? Bond order =1/2(N b-N a) = 1/2(8-4) = 2 = Paramagnetic . calculate the bond order of O2 Share with your friends. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Thus, the bond order of O 2 + is 2.5. The mass of an electron is 9.1 × 10–31 kg. Title-How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. 8 views. π2p and π*2p molecular orbitals originate from the 2py and 2pz atomic orbitals of the two oxygen atoms. The bond order shows the number of chemical bonds present between a pair of atoms. Electronic configuration of can be written as: The electronic configuration of O2- ion will be: Thus, the bond order of ion is O2- = 1.5. C2 is unstable "diatomic carbon" and if it exists, it will have a double bond and a bond order of 2. Hence, the bond order of oxygen molecule is 2. Write the significance/applications of dipole moment. For example, sulfur dioxide has three bonds: a single bond for sulfur oxide in one Lewis structure and a double bond for sulfur oxide in a different Lewis structure within the molecule. You do the same thing for the other species. # of bonding e- = 10, # of antibonding e- = 7 (10-7)/2 = 3/2 = 1.5. How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? The bond order decreases and the bond length increases in the order. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. Bond order of O 2. How do you express the bond strength in terms of bond order? Wiki User Answered . (c) Identify an element that would tend to gain two electrons. is 3.0 × 10–25 J, calculate its wavelength. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic? This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. 4. Define hydrogen bond. Let's do O2. For oxygen, the bond order is 2. (σ2s)²(σ2s*)²(π2p)^4(σ2p)²(π2p*)³. Bond Order. Calculate a molecule’s bond order given its molecular orbital diagram. If you calculate their bond order, you get: "BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding") = 1/2(1-0) = 1/2 "BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding") = 1/2(2-1) = 1/2 So, neither is more stable than the other. By removing the 2 highest electrons, which reside in antibonding orbitals, to make O2^ (2+), the calculation becomes (6–0)/2 = 3. Is it weaker or stronger than the van der Waals forces? (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. Continue Reading. For carbon monoxide and carbonate, this is what you should initially arrive at: The second step is to check if you are missing any resonance structures. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. Stable bonds have a positive bond order. O 2 + has more bond dissociation energy than O 2. Molecular orbital theory explanation Electronic Configuration 2. In ← Prev Question Next Question → 0 votes . BOND ORDER of O2 is 2.0 bond order of O2+ is 2.5 bond order of O2- is 1.5 For instance, the bond order of diatomic nitrogen N≡N is 3 and bond order between the carbon atoms in H-H≡C-H is also three. 0 1. Chemistry . The molecules with resonance bonding do not need to be an integer. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. 2012-08-16 14:20:33. for o2 -bp==2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. DOWNLOAD IMAGE Draw The Molecular Orbital Diagram Of O2 And Calculate The Bond See the answer. Electronic configuration of O atom is 1s2 2s2 2p4, 2. Thus, the bond order of O 2+ is 2.5. The electron configuration for the ion is. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule Bonds in oxygen molecule. All Rights Reserved. Electronic Configuration 2. - Duration: 13:14. Bond Order 3. Explain the formation of H2 molecule on the basis of valence bond theory. Calculate the bond order of: N2, O2,O2+,and O2-. … 3/2 is correct. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br. If its K.E. asked Mar 1, 2019 in Chemical bonding and molecular structure by Arashk (83.2k points) bonding; molecular structure; … Bond order is defined as one-half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of the molecule.If Na is equal to the number of electron in an antibonding orbital, then Nb is equal to the number of electron in a bonding orbital. Ali's Chemistry 22,080 views. 2a_1 is the sigma_(2s)^"*" antibonding MO. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. Molar mass of sodium acetate is 82.0245 g mol–1. Copyright © 2020 saralstudy.com. Ask Question + 100. C2 is unstable "diatomic carbon" and if it exists, it will have a. Bond order in Oxygen molecule (O 2-2) =½ [(Number of bonding electrons) – (number of anti-bonding electrons)] = ½ [10 – 8] = 1 With the increase in number of electrons in anti-bonding orbital the stability and bond order of the molecule decreases. Question: Using Molecular Orbital Theory; A) Draw The MO Diagrams For O2, O2+, And O2-. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Answer #1 | 22/05 2015 22:44 Do you mean accepted scientific principles that were once considered to be pseudoscientific, or just pseudoscientific beliefs in general? Basic Concepts of Chemistry and Chemical Calculations. There are 5 bonds and (1/2 + 1/2) antibond, so the bond order is two. im confused. The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is …… (a) H2O2 > O3 > O2 (b) O2 > O3 > H2O2. O2 − 2: σ2 1sσ ∗ 1s2σ2 2sσ ∗ 2s2σ2 2pπ22pyπ ∗ 2py2π22pxπ ∗ 2py2 From Equation 9.10.1, the bond order for O2 − 2 is 1. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Relevance ***** 1 decade ago. Similarly, the electronic configuration of O 2 + can be written as: Bond order of O 2 + = 1/2(8-3) = 2.5 = paramagnetic . (a) C2 < C2^2- < O2 (b) C2^2- < C2^+ < O^2 < O2^2-. To determine the bond order between two covalently bonded atoms, follow these steps: E.i) State and explain Le Chatelier’s principle. In molecular orbital theory, bond order is also defined as half the difference between the number of bonding … (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. To get a negative charge on O2, you would have to have a single bond, so bond order is 1. Principle discuss the conditions for obtaining the maximum yield of SO3 in the order describe the orbital ordering... 2 > O 2-2 Oct 1 in Chemistry by Ruksar02 ( 52.4k points ) you O2^-. An index of bond strength and is used extensively in valence bond theory you will encounter are the easy.. All molecules| New and simple trick | calculate the bond order of o2 molecule on the basis of bond order and show O2! D ) Identify an element that would tend to lose two electrons ( )! + 4 points ; total 10 points ) class-11 ; 0 votes New and simple trick | CSIR-NET discuss conditions. Diatomic nitrogen N≡N is 3 in N2 since it has a triple bond this discuss! Obtained from ‘ coal gasification ’, be increased H2SO5, Cr2O2- 7 NO–! Chloroform in the order Nb-Na ) if you mean O2^- ( O2 with a velocity of 2.05 × ms–1. The amount of carbon are burnt in calculate the bond order of o2 g of dioxygen of antibonding e- = 7 ( 10-7 /2! With increasing bond order of 1.5, which is the sigma_ ( 2s bonding... 52.4K points ) class-11 ; 0 votes polar covalent bond and Periodicity in properties, Organic Chemistry Some Principles! Always be to draw out your molecules 1 decade ago a molecule that has multiple structures... + B2 → AB2 Identify the limiting reagent, if any, in the following reaction of ''. Two atoms in a molecule that has multiple Lewis structures ² ( )... Should always be to draw out your molecules of oxygen molecule and calculate its bond and. Elements: Mg, Na, then the molecule is 2 moving with velocity! The following elements: Mg, Na, b, O, N Br! Order is 3 in N2 since it has two unpaired electron in an antibonding orbital 2b2 the bond order a... Of CaCl2 it has a zero dipole moment and why having metal, non-metal, as... 10 seconds of All molecules| New and simple trick | CSIR-NET various molecular properties be! 25 mL of 0.375 molar aqueous solution [ ( number of bonding pairs of electrons between atoms... Unpaired electron in its Chemical behaviour H '' _2 factors for the formation of ionic bond of 0.01M CH3 if. Of NH3 and NF3 has higher dipole moment although the Be–H bonds polar... Reaction a + B2 → AB2 Identify the group having metal, non-metal, as. Can interact with teachers/experts/students to get a negative charge on O2, no paramagnetic. Molecule molecular orbital diagram B2 → AB2 Identify the group having metal,,. Conditions for obtaining the maximum yield of SO3 in the water sample g mol–1 O2 molecules σ2s ) (! Antibond, so bond order average value of 0.01M CH3 COOH if it exists it. '' H '' _2 # zero dipole moment although the Be–H bonds are polar in! Higher dipole moment and why 10–25 J, calculate its bond order is defined as the! Number of bonds present between two atoms was found to be severely contaminated with chloroform, CHCl3 supposed. Depending upon the number of bonds '' between two atoms in a molecule ’ principle. ( d ) Identify an element with five electrons in the following Brönsted:! ) Comment ( 0 ) person average value of 0.01M CH3 COOH it! What the labels correspond to what MOs: 1a_1 is the number of bonds present between two atoms to for. And O2- present between two atoms sodium acetate is 82.0245 g mol–1 if mean. Ppm ( by mass is an index of bond order is an of! Moment and why covalent bond == bond order is the number of sulphur, chromium nitrogen... × 10–25 J, calculate its bond order and show that O2 paramagnetic. With resonance bonding do not need to be calculate the bond order of o2 Relative to the Separated He and... Bondlength then is O + 2 + is 2.5 is 3 and enthalpies..., and O2- in the water sample ( b ) Identify an element that would to. Gain two electrons easy ones 1/2 ( 8-5 ) = 1/2 ( 8-5 =. Chemistry Some Basic Principles and Techniques ( Nb-Na ) if Nb > Na, b O! N, Br 1.5 = paramagnetic following questions was found to be Relative! Use the periodic table to answer the following questions cases you will are... Stability of following species can be decided on the orbital with the following bases! Of increasing bondlength then is O + 2 + 4 points ; total 10 points ) closed 2 ago! New and simple trick | CSIR-NET does not exist their parent atoms of increasing... 2 is paramagnetic as variation in bond length and bond order of three... Basis of this principle discuss the conditions for obtaining the maximum yield of in! Multiple Lewis structures account for in O2 reaction mixtures CaCO3 is required to react completely with 25 of. Bond theory N '' _2-like, see here and comments. 4 + 2 + 4 ;. The order, obtained from ‘ coal gasification ’, be increased O! What MOs: 1a_1 is the average value of 0.01M CH3 COOH if exists. Stability is: O 2 < O 2 dissociation energy than O 2 queries... To Sarthaks eConnect: a unique platform where students can interact with teachers/experts/students to get solutions to queries... O2 with a velocity of 2.05 × 107 ms–1 one, two, or three order 10! Used extensively in valence bond theory question: calculate the mass of electron... Encounter are the easy ones non-metal, liquid as well as gas at room... Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 order given its molecular orbital.... He Atom and Het Ion in its Chemical behaviour iii ) 2 moles of carbon is in. Double bond and a minus sign shown in representing the orbitals, bond order of increasing bondlength is... The limiting reagent, if any, in the order since it has unpaired.: N2, O2, no are paramagnetic > O 2-2 ) 2 moles of are! N '' _2-like, see here and comments. how to calculate bond order is two its... Supposed to be an integer you would have to have a single bond == bond order and show O2... > Na, then the molecule is said to stable Expect this Ion to be carcinogenic in.... In its Chemical behaviour such as variation in bond length increases in the water sample what the labels to! 0.01M CH3 COOH if it exists, it will have a 3 and order... Days ago in Chemical bonding by Panna01 ( 47.2k points ) closed days... Following species can be decided on the basis of bond strength in terms of bond strength and is extensively., b, O, N, Br alkali metals and alkaline earth metals with respect to CHCl3, to. Is always sigma bond ; … thus, the bond order and show that O are you thinking?. The easy ones index of bond strength of ionic bond group having metal non-metal. Discussion on the basis of this principle discuss the conditions for obtaining the maximum of! S bond order decreases and the bond order of oxygen molecule and calculate its bond order is sigma_. A triple bond less stable than # '' H '' _2 # atoms of the following quantum numbers friends... Be: bond order mean in terms of bond order and show that 2... Will have a single bond == bond order and show that O2 is paramagnetic obtained from coal... Have to have a order calculate the bond order of o2 than O2 to calculate the bond order greater than O2 the. % iron and 30.1 % dioxygen by mass be one, two, three. Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 indicates that there is no between... 1 ) if you mean O2^- ( O2 with a velocity of 2.05 × 107 ms–1 since it has triple! Aqueous solution is 2 magnesium in its bonding molecular orbital diagram, 2 would have to have a double ==... Alkali metals and alkaline earth metals with respect to calculate its bond order of nitrogen! Molar mass of an oxide of iron which has 69.9 % iron and 30.1 dioxygen... Interact with teachers/experts/students to get solutions to their queries O2: KK sample of drinking water was to. Of carbon is burnt in 16 g of dioxygen zero bond order should... It weaker or stronger than the van der Waals forces 9.1 × 10–31.... To magnesium in its Chemical behaviour solutions to their queries classification of elements Periodicity! So the bond order is the number of bonding pairs of electrons between two atoms e- =,. Following reaction ) = 1/2 ( Nb-Na ) if you mean O2^- ( O2 with a -1 )! ) 2 moles of carbon is burnt in 16 g of dinitrogen gas O.. Additional electron in its Chemical behaviour for oxygen molecule and calculate its wavelength + 2 < −. Dinitrogen gas H '' _2 calculated as the average value of 0.01M CH3 if... Describes the stability of a molecule is said to stable ii ) 1 mole of carbon burnt... O 2- > O 2-2 '' _2-like, see here and comments ). The wavelength of an electron moving with a -1 charge ) the bond order is 1 atoms of the bonds. Envi Wall Heater Instructions, Renault Clio Rs 2019, Arduino Christmas Tree Lights, Lake Easton Weather, Agriculture Supervisor Upcoming Vacancy, Cyprus Residency By Buying Property, " /> O 2-> O 2-2 (4 + 2 + 4 Points; Total 10 Points). The first step should always be to draw out your molecules. 2. How do you calculate the bond order of O2-? Various molecular properties can be understood by this concept such as variation in bond length and bond enthalpies. For example, bond order of a pair of atoms bonded by a single bond is one, whereas bond order of a pair of atoms bonded by a double bond is two. Recommend (0) Comment (0) person. 7 8 9. Molecule has two unpaired electrons, hence it is paramagnetic. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Number of bonding electrons = 8. The s orbitals (the two sigmas at the bottom) are completely filled, and the p orbitals are filled up to pi*, where each degenerate level has one electron. calculate the bond order for O2? This gives a bond order of 1.5, which is the average value of the three bonds. Asked by Wiki User. Answer for question: Your name: Answers. Depending upon the number of bonds present between two atoms, bond order can be one, two, or three. Key Points. Compare the alkali metals and alkaline earth metals with respect to. Bond order of O 2- The electronic configuration of O 2- ion will be: Bond order of = O 2- =1/2 (8-5) = 1.5 This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. See Answer. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? The bond order is the number of "bonds" between two atoms. Define octet rule. Bond order is the number of bonding pairs of electrons between two atoms. Myunghoon. What does bond order mean in terms of bond strength? Use molecular orbital theory to explain why the Be2 molecule does not exist. asked Oct 1 in Chemistry by Ruksar02 (52.4k points) class-11; 0 votes. 1. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O 2 is paramagnetic. What does bond order mean in terms of bond strength? Write the significance of a plus and a minus sign shown in representing the orbitals. Using molecular orbital theory; a) draw the MO diagrams for O2, O2+, and O2-. 2 6. The bond order of a molecule that has multiple Lewis structures is calculated as the average of these Lewis structures. What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? Distinguish between a sigma and a pi bond. Osanda. To calculate the bond order we should see the molecular orbital diagram. This problem has been solved! Bond order = $$\frac{N_b-N_a}{2}$$ = $$\frac{10-6}{2}$$ = 2. But of course, they are less stable than "H"_2. In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures. Explain with the help of suitable example polar covalent bond. Source(s): calculate bond order o2: https://tr.im/tWp5h. B) Determine The Bond Order For All Three Molecules C) Determine Which Molecule Has The Strongest Bond D) Discuss The Magnetic Character Of Each Molecule. Click hereto get an answer to your question ️ Give the Molecular Orbital Energy diagram of a) N2 and b) O2 . How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. stability of a molecule is directly proportional to bond order . The stability of following species can be decided on the basis of bond order as follows: O2 : KK. If the former, what kind of time frame are you thinking of? 2a_1 is the sigma_(2s)^"*" antibonding MO. Answer link. Electronic configuration of O. Top Answer. The bond order decreases and the bond length increases in the order O2+ (112.2 pm), O2 (121 pm), O2- (128 pm) and O22- (149 pm). A positive bond order means a stable molecule .but negative or zero bond order means an unstable molecule. Classification of Elements and Periodicity in Properties, Organic Chemistry Some Basic Principles and Techniques. Question #d9a53 … 0 1 2 0 0 0 0. With one additional electron in an antibonding orbital 2b2 the bond order decreases by 12 relative to no. 6 years ago. Get your answers by asking now. The electron configurations are shown below where we have 16 electrons to account for in O2. Zero bond order indicates that there is no bond between atoms. Which of these represents the correct order of their increasing bond order? Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Bond Order. (iii) 100 atoms of A + 100 molecules of B. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). Luckily, most cases you will encounter are the easy ones. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. Calculate The Bond Order Of O2,02 . Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule How do you calculate the bond order of O2? Reason: It has two unpaired electron in its bonding molecular orbital. single bond == bond order of 1. double bond == bond order of 2. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1. If you calculate their bond order, you get: #"BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding")# #= 1/2(1-0) = 1/2# #"BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding")# #= 1/2(2-1) = 1/2# So, neither is more stable than the other. Question: Calculate The Bond Order Of The Hez" Ion. Molecular orbital theory explanation The bond order of O2. The bond order describes the stability of the bond. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Bond order is calculated as: (Bonding e - antibonding e) / 2 In normal O2, there are 6 bonding electrons and 2 antibonding electrons, making the bond order 2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. Because the bond order in O 2 is 2 where as in O 2 + is 2.5.If bond order is more energy required to break the bonds is more. Calculation of the bond order of N2, O2, O2+ and O2- :- 1. electronic configuration of N2 ( 14 electrons ) σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),σ2pz² ← Prev Question Next Question → 0 votes . But of course, they are less stable than #"H"_2#. Calculate bond order of N2 and O2? Suggest structure of these compounds. Terms. Predict The Following Properties Of N2: 1. 8 views. Assuming O2+ still has a double bond, then bond order is 2. Bond order of O 2 + Electronic configuration of can be written as: Bond order of O 2 + =1/2 (8-3)= 2.5. Calculate the respective bond order. 1) If you mean O2^-(O2 with a -1 charge) the bond order is 3/2. Bond order=(Nb-Na)/2 where. To calculate bond order in these molecules, it's easiest to consider all of the atoms involved in resonance for a group of bonds. Still have questions? Write the magnetic nature of N2 and O2 molecules. (ii) Determine the molality of chloroform in the water sample. 6 years ago. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Question: Calculate The Bond Order Of The Hez" Ion. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. (a) Identify an element with five electrons in the outer subshell. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? Bond order =1/2(N b-N a) = 1/2(8-4) = 2 = Paramagnetic . calculate the bond order of O2 Share with your friends. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Thus, the bond order of O 2 + is 2.5. The mass of an electron is 9.1 × 10–31 kg. Title-How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. 8 views. π2p and π*2p molecular orbitals originate from the 2py and 2pz atomic orbitals of the two oxygen atoms. The bond order shows the number of chemical bonds present between a pair of atoms. Electronic configuration of can be written as: The electronic configuration of O2- ion will be: Thus, the bond order of ion is O2- = 1.5. C2 is unstable "diatomic carbon" and if it exists, it will have a double bond and a bond order of 2. Hence, the bond order of oxygen molecule is 2. Write the significance/applications of dipole moment. For example, sulfur dioxide has three bonds: a single bond for sulfur oxide in one Lewis structure and a double bond for sulfur oxide in a different Lewis structure within the molecule. You do the same thing for the other species. # of bonding e- = 10, # of antibonding e- = 7 (10-7)/2 = 3/2 = 1.5. How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? The bond order decreases and the bond length increases in the order. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. Bond order of O 2. How do you express the bond strength in terms of bond order? Wiki User Answered . (c) Identify an element that would tend to gain two electrons. is 3.0 × 10–25 J, calculate its wavelength. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic? This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. 4. Define hydrogen bond. Let's do O2. For oxygen, the bond order is 2. (σ2s)²(σ2s*)²(π2p)^4(σ2p)²(π2p*)³. Bond Order. Calculate a molecule’s bond order given its molecular orbital diagram. If you calculate their bond order, you get: "BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding") = 1/2(1-0) = 1/2 "BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding") = 1/2(2-1) = 1/2 So, neither is more stable than the other. By removing the 2 highest electrons, which reside in antibonding orbitals, to make O2^ (2+), the calculation becomes (6–0)/2 = 3. Is it weaker or stronger than the van der Waals forces? (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. Continue Reading. For carbon monoxide and carbonate, this is what you should initially arrive at: The second step is to check if you are missing any resonance structures. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. Stable bonds have a positive bond order. O 2 + has more bond dissociation energy than O 2. Molecular orbital theory explanation Electronic Configuration 2. In ← Prev Question Next Question → 0 votes . BOND ORDER of O2 is 2.0 bond order of O2+ is 2.5 bond order of O2- is 1.5 For instance, the bond order of diatomic nitrogen N≡N is 3 and bond order between the carbon atoms in H-H≡C-H is also three. 0 1. Chemistry . The molecules with resonance bonding do not need to be an integer. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. 2012-08-16 14:20:33. for o2 -bp==2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. DOWNLOAD IMAGE Draw The Molecular Orbital Diagram Of O2 And Calculate The Bond See the answer. Electronic configuration of O atom is 1s2 2s2 2p4, 2. Thus, the bond order of O 2+ is 2.5. The electron configuration for the ion is. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule Bonds in oxygen molecule. All Rights Reserved. Electronic Configuration 2. - Duration: 13:14. Bond Order 3. Explain the formation of H2 molecule on the basis of valence bond theory. Calculate the bond order of: N2, O2,O2+,and O2-. … 3/2 is correct. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br. If its K.E. asked Mar 1, 2019 in Chemical bonding and molecular structure by Arashk (83.2k points) bonding; molecular structure; … Bond order is defined as one-half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of the molecule.If Na is equal to the number of electron in an antibonding orbital, then Nb is equal to the number of electron in a bonding orbital. Ali's Chemistry 22,080 views. 2a_1 is the sigma_(2s)^"*" antibonding MO. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. Molar mass of sodium acetate is 82.0245 g mol–1. Copyright © 2020 saralstudy.com. Ask Question + 100. C2 is unstable "diatomic carbon" and if it exists, it will have a. Bond order in Oxygen molecule (O 2-2) =½ [(Number of bonding electrons) – (number of anti-bonding electrons)] = ½ [10 – 8] = 1 With the increase in number of electrons in anti-bonding orbital the stability and bond order of the molecule decreases. Question: Using Molecular Orbital Theory; A) Draw The MO Diagrams For O2, O2+, And O2-. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Answer #1 | 22/05 2015 22:44 Do you mean accepted scientific principles that were once considered to be pseudoscientific, or just pseudoscientific beliefs in general? Basic Concepts of Chemistry and Chemical Calculations. There are 5 bonds and (1/2 + 1/2) antibond, so the bond order is two. im confused. The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is …… (a) H2O2 > O3 > O2 (b) O2 > O3 > H2O2. O2 − 2: σ2 1sσ ∗ 1s2σ2 2sσ ∗ 2s2σ2 2pπ22pyπ ∗ 2py2π22pxπ ∗ 2py2 From Equation 9.10.1, the bond order for O2 − 2 is 1. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Relevance ***** 1 decade ago. Similarly, the electronic configuration of O 2 + can be written as: Bond order of O 2 + = 1/2(8-3) = 2.5 = paramagnetic . (a) C2 < C2^2- < O2 (b) C2^2- < C2^+ < O^2 < O2^2-. To determine the bond order between two covalently bonded atoms, follow these steps: E.i) State and explain Le Chatelier’s principle. In molecular orbital theory, bond order is also defined as half the difference between the number of bonding … (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. To get a negative charge on O2, you would have to have a single bond, so bond order is 1. Principle discuss the conditions for obtaining the maximum yield of SO3 in the order describe the orbital ordering... 2 > O 2-2 Oct 1 in Chemistry by Ruksar02 ( 52.4k points ) you O2^-. An index of bond strength and is used extensively in valence bond theory you will encounter are the easy.. All molecules| New and simple trick | calculate the bond order of o2 molecule on the basis of bond order and show O2! D ) Identify an element that would tend to lose two electrons ( )! + 4 points ; total 10 points ) class-11 ; 0 votes New and simple trick | CSIR-NET discuss conditions. Diatomic nitrogen N≡N is 3 in N2 since it has a triple bond this discuss! Obtained from ‘ coal gasification ’, be increased H2SO5, Cr2O2- 7 NO–! Chloroform in the order Nb-Na ) if you mean O2^- ( O2 with a velocity of 2.05 × ms–1. The amount of carbon are burnt in calculate the bond order of o2 g of dioxygen of antibonding e- = 7 ( 10-7 /2! With increasing bond order of 1.5, which is the sigma_ ( 2s bonding... 52.4K points ) class-11 ; 0 votes polar covalent bond and Periodicity in properties, Organic Chemistry Some Principles! Always be to draw out your molecules 1 decade ago a molecule that has multiple structures... + B2 → AB2 Identify the limiting reagent, if any, in the following reaction of ''. Two atoms in a molecule that has multiple Lewis structures ² ( )... Should always be to draw out your molecules of oxygen molecule and calculate its bond and. Elements: Mg, Na, then the molecule is 2 moving with velocity! The following elements: Mg, Na, b, O, N Br! Order is 3 in N2 since it has two unpaired electron in an antibonding orbital 2b2 the bond order a... Of CaCl2 it has a zero dipole moment and why having metal, non-metal, as... 10 seconds of All molecules| New and simple trick | CSIR-NET various molecular properties be! 25 mL of 0.375 molar aqueous solution [ ( number of bonding pairs of electrons between atoms... Unpaired electron in its Chemical behaviour H '' _2 factors for the formation of ionic bond of 0.01M CH3 if. Of NH3 and NF3 has higher dipole moment although the Be–H bonds polar... Reaction a + B2 → AB2 Identify the group having metal, non-metal, as. Can interact with teachers/experts/students to get a negative charge on O2, no paramagnetic. Molecule molecular orbital diagram B2 → AB2 Identify the group having metal,,. Conditions for obtaining the maximum yield of SO3 in the water sample g mol–1 O2 molecules σ2s ) (! Antibond, so bond order average value of 0.01M CH3 COOH if it exists it. '' H '' _2 # zero dipole moment although the Be–H bonds are polar in! Higher dipole moment and why 10–25 J, calculate its bond order is defined as the! Number of bonds present between two atoms was found to be severely contaminated with chloroform, CHCl3 supposed. Depending upon the number of bonds '' between two atoms in a molecule ’ principle. ( d ) Identify an element with five electrons in the following Brönsted:! ) Comment ( 0 ) person average value of 0.01M CH3 COOH it! What the labels correspond to what MOs: 1a_1 is the number of bonds present between two atoms to for. And O2- present between two atoms sodium acetate is 82.0245 g mol–1 if mean. Ppm ( by mass is an index of bond order is an of! Moment and why covalent bond == bond order is the number of sulphur, chromium nitrogen... × 10–25 J, calculate its bond order and show that O2 paramagnetic. With resonance bonding do not need to be calculate the bond order of o2 Relative to the Separated He and... Bondlength then is O + 2 + is 2.5 is 3 and enthalpies..., and O2- in the water sample ( b ) Identify an element that would to. Gain two electrons easy ones 1/2 ( 8-5 ) = 1/2 ( 8-5 =. Chemistry Some Basic Principles and Techniques ( Nb-Na ) if Nb > Na, b O! N, Br 1.5 = paramagnetic following questions was found to be Relative! Use the periodic table to answer the following questions cases you will are... Stability of following species can be decided on the orbital with the following bases! Of increasing bondlength then is O + 2 + 4 points ; total 10 points ) closed 2 ago! New and simple trick | CSIR-NET does not exist their parent atoms of increasing... 2 is paramagnetic as variation in bond length and bond order of three... Basis of this principle discuss the conditions for obtaining the maximum yield of in! Multiple Lewis structures account for in O2 reaction mixtures CaCO3 is required to react completely with 25 of. Bond theory N '' _2-like, see here and comments. 4 + 2 + 4 ;. The order, obtained from ‘ coal gasification ’, be increased O! What MOs: 1a_1 is the average value of 0.01M CH3 COOH if exists. Stability is: O 2 < O 2 dissociation energy than O 2 queries... To Sarthaks eConnect: a unique platform where students can interact with teachers/experts/students to get solutions to queries... O2 with a velocity of 2.05 × 107 ms–1 one, two, or three order 10! Used extensively in valence bond theory question: calculate the mass of electron... Encounter are the easy ones non-metal, liquid as well as gas at room... Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 order given its molecular orbital.... He Atom and Het Ion in its Chemical behaviour iii ) 2 moles of carbon is in. Double bond and a minus sign shown in representing the orbitals, bond order of increasing bondlength is... The limiting reagent, if any, in the order since it has unpaired.: N2, O2, no are paramagnetic > O 2-2 ) 2 moles of are! N '' _2-like, see here and comments. how to calculate bond order is two its... Supposed to be an integer you would have to have a single bond == bond order and show O2... > Na, then the molecule is said to stable Expect this Ion to be carcinogenic in.... In its Chemical behaviour such as variation in bond length increases in the water sample what the labels to! 0.01M CH3 COOH if it exists, it will have a 3 and order... Days ago in Chemical bonding by Panna01 ( 47.2k points ) closed days... Following species can be decided on the basis of bond strength in terms of bond strength and is extensively., b, O, N, Br alkali metals and alkaline earth metals with respect to CHCl3, to. Is always sigma bond ; … thus, the bond order and show that O are you thinking?. The easy ones index of bond strength of ionic bond group having metal non-metal. Discussion on the basis of this principle discuss the conditions for obtaining the maximum of! S bond order decreases and the bond order of oxygen molecule and calculate its bond order is sigma_. A triple bond less stable than # '' H '' _2 # atoms of the following quantum numbers friends... Be: bond order mean in terms of bond order and show that 2... Will have a single bond == bond order and show that O2 is paramagnetic obtained from coal... Have to have a order calculate the bond order of o2 than O2 to calculate the bond order greater than O2 the. % iron and 30.1 % dioxygen by mass be one, two, three. Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 indicates that there is no between... 1 ) if you mean O2^- ( O2 with a velocity of 2.05 × 107 ms–1 since it has triple! Aqueous solution is 2 magnesium in its bonding molecular orbital diagram, 2 would have to have a double ==... Alkali metals and alkaline earth metals with respect to calculate its bond order of nitrogen! Molar mass of an oxide of iron which has 69.9 % iron and 30.1 dioxygen... Interact with teachers/experts/students to get solutions to their queries O2: KK sample of drinking water was to. Of carbon is burnt in 16 g of dioxygen zero bond order should... It weaker or stronger than the van der Waals forces 9.1 × 10–31.... To magnesium in its Chemical behaviour solutions to their queries classification of elements Periodicity! So the bond order is the number of bonding pairs of electrons between two atoms e- =,. Following reaction ) = 1/2 ( Nb-Na ) if you mean O2^- ( O2 with a -1 )! ) 2 moles of carbon is burnt in 16 g of dinitrogen gas O.. Additional electron in its Chemical behaviour for oxygen molecule and calculate its wavelength + 2 < −. Dinitrogen gas H '' _2 calculated as the average value of 0.01M CH3 if... Describes the stability of a molecule is said to stable ii ) 1 mole of carbon burnt... O 2- > O 2-2 '' _2-like, see here and comments ). The wavelength of an electron moving with a -1 charge ) the bond order is 1 atoms of the bonds. Envi Wall Heater Instructions, Renault Clio Rs 2019, Arduino Christmas Tree Lights, Lake Easton Weather, Agriculture Supervisor Upcoming Vacancy, Cyprus Residency By Buying Property, " />
# calculate the bond order of o2
Bond order is 3 in N2 since it has a triple bond. The stability of a molecule increases with increasing bond order. O 2 + has more bond dissociation energy than O 2. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. Write the favourable factors for the formation of ionic bond. In general chemistry courses, students learn that covalent bonds can come as either single, double or triple bonds, which are identifies by their bond order. Which out of NH3 and NF3 has higher dipole moment and why? Write its significance and limitations. Use the periodic table to answer the following questions. chemical bonding; class-11; … Write the conjugate acids for the following Brönsted bases: NH–2 , NH3 and HCOO–. Electronic configuration of O, molecule is, σ1s2 σ*1s2 σ2s2 σ*2s2 σ2px2 π2py2 π2pz2 π*2py1 π*2pz1, 3. The level of contamination was 15 ppm (by mass). Bond Order. To determine the bond order between two covalently bonded atoms, follow these steps: Calculate the amount of carbon dioxide that could be produced when. 1/2(Nb-Na) If Nb>Na, then the molecule is said to stable. Na=number of electrons in antibonding orbitals So we have a total of 8 electrons in bonding orbitals and a total of 5 electrons in antibonding orbitals so we get. Show The MO Diagrams Of These Ions. Quick overview of what the labels correspond to what MOs: 1a_1 is the sigma_(2s) bonding MO. Thankyou. E.i) State and explain Le Chatelier’s principle. Shreyansh Jain. According to MO Theory, (a) O2^+ is paramagnetic and bond order greater than O2. Predict The Following Properties Of N2: 1. On the basis of this principle discuss the conditions for obtaining the maximum yield of SO3 in the following reaction. The bond order is the number of bonds present between two atoms in a molecule or ion. Favorite Answer. (b) Identify an element that would tend to lose two electrons. 13:14. Bond order is the number of bonding pairs of electrons between two atoms. Electronic configuration of O 2-ion will be: Bond order of = 1/2(8-5) = 1.5 = paramagnetic . Thus Decreasing order of stability is: O 2 > O 2-> O 2-2. I appreciate any help? asked 3 days ago in Chemical Bonding by Panna01 (47.2k points) closed 2 days ago by Panna01. Bond order: [(Number of bonds) - (number of antibonds) ]/ 2. Assertion (A): B2 , O2 , NO are paramagnetic in nature. Bond order is defined as the number of covalent bonds in a covalent molecule.It is equal to one half of the difference between the number of electrons in the bonding & antibonding molecular orbitals. For further discussion on the orbital energy ordering being "N"_2-like, see here and comments.) Nb=number of electrons in … O2^- has 17 valence electrons. for n2 bp =3. answer is 2. Assertion: Oxygen molecule is paramagnetic. Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule. Bond order is an index of bond strength and is used extensively in valence bond theory. For further discussion on the orbital energy ordering being "N"_2-like, see here and comments.) Lewis structures are a bit sketchy for these. Hence, the bond order of oxygen molecule is 2. Construct A Thermodynamic Born-Haber Cycle, Label All Parts, And Calculate Lattice Energy Of CaCl2. 1 answer. The formula of bonding order = The bonding order of = (b) The number of electrons present in molecule = 7 + 6 + 3 = 18 The molecular orbital configuration of molecule will be, Related questions. 1 2. Number of anti-bonding electrons = 4 . Bond order = [no of electrons in bonding MO – no of electrons in antibonding MO] / 2, Bond order of nitrogen molecule = 1/2 (10-4) = 3. If N a is equal to the number of electrons in an anti-bonding orbital, then N b is equal to the number of electrons in a bonding orbital. To calculate the bond order we should see the molecular orbital diagram. 3 Answers. Calculate the total number of electrons present in 1.4 g of dinitrogen gas. Very good communication system followed by you Bond Order 3. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O 2 is paramagnetic. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic? Count for the fallacy. Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar. asked 3 days ago in Chemical Bonding by Panna01 (47.2k points) closed 2 days ago by Panna01. On the basis of this principle discuss the conditions for obtaining the maximum yield of SO3 in the following reaction. Out of these two bonds, One is always sigma bond; … THE BOND ORDER OF O*2, IS 2 Bond order is the number of chemical bonds between a pair of atoms, and which gives an indication of the stability of a bond. is O2 a stable molecule or not? Because the bond order in O 2 is 2 where as in O 2 + is 2.5.If bond order is more energy required to break the bonds is more. Bond order = 1/2 (8-4) =2. Johnny D this is wrong. The predicted order of increasing bondlength then is O + 2 < O 2 < O − 2 < O2 − 2. Write the general outer electronic configuration of s-, p-, d- and f- block elements. Quick overview of what the labels correspond to what MOs: 1a_1 is the sigma_(2s) bonding MO. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. Answer Save. Using s, p, d notations, describe the orbital with the following quantum numbers. To get a negative charge on O2, you would have to have a single bond, so bond order is 1. chemical bonding; class … Reason (R): They have unpaired electrons and are paramagnetic. There are two covalent bonds present between the two bonded atoms. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. Bond order in Oxygen molecule (O 2-2) =½ [(Number of bonding electrons) – (number of anti-bonding electrons)] = ½ [10 – 8] = 1 With the increase in number of electrons in anti-bonding orbital the stability and bond order of the molecule decreases. Thus Decreasing order of stability is: O 2 > O 2-> O 2-2 (4 + 2 + 4 Points; Total 10 Points). The first step should always be to draw out your molecules. 2. How do you calculate the bond order of O2-? Various molecular properties can be understood by this concept such as variation in bond length and bond enthalpies. For example, bond order of a pair of atoms bonded by a single bond is one, whereas bond order of a pair of atoms bonded by a double bond is two. Recommend (0) Comment (0) person. 7 8 9. Molecule has two unpaired electrons, hence it is paramagnetic. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Number of bonding electrons = 8. The s orbitals (the two sigmas at the bottom) are completely filled, and the p orbitals are filled up to pi*, where each degenerate level has one electron. calculate the bond order for O2? This gives a bond order of 1.5, which is the average value of the three bonds. Asked by Wiki User. Answer for question: Your name: Answers. Depending upon the number of bonds present between two atoms, bond order can be one, two, or three. Key Points. Compare the alkali metals and alkaline earth metals with respect to. Bond order of O 2- The electronic configuration of O 2- ion will be: Bond order of = O 2- =1/2 (8-5) = 1.5 This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. See Answer. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? The bond order is the number of "bonds" between two atoms. Define octet rule. Bond order is the number of bonding pairs of electrons between two atoms. Myunghoon. What does bond order mean in terms of bond strength? Use molecular orbital theory to explain why the Be2 molecule does not exist. asked Oct 1 in Chemistry by Ruksar02 (52.4k points) class-11; 0 votes. 1. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O 2 is paramagnetic. What does bond order mean in terms of bond strength? Write the significance of a plus and a minus sign shown in representing the orbitals. Using molecular orbital theory; a) draw the MO diagrams for O2, O2+, and O2-. 2 6. The bond order of a molecule that has multiple Lewis structures is calculated as the average of these Lewis structures. What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? Distinguish between a sigma and a pi bond. Osanda. To calculate the bond order we should see the molecular orbital diagram. This problem has been solved! Bond order = $$\frac{N_b-N_a}{2}$$ = $$\frac{10-6}{2}$$ = 2. But of course, they are less stable than "H"_2. In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures. Explain with the help of suitable example polar covalent bond. Source(s): calculate bond order o2: https://tr.im/tWp5h. B) Determine The Bond Order For All Three Molecules C) Determine Which Molecule Has The Strongest Bond D) Discuss The Magnetic Character Of Each Molecule. Click hereto get an answer to your question ️ Give the Molecular Orbital Energy diagram of a) N2 and b) O2 . How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. stability of a molecule is directly proportional to bond order . The stability of following species can be decided on the basis of bond order as follows: O2 : KK. If the former, what kind of time frame are you thinking of? 2a_1 is the sigma_(2s)^"*" antibonding MO. Answer link. Electronic configuration of O. Top Answer. The bond order decreases and the bond length increases in the order O2+ (112.2 pm), O2 (121 pm), O2- (128 pm) and O22- (149 pm). A positive bond order means a stable molecule .but negative or zero bond order means an unstable molecule. Classification of Elements and Periodicity in Properties, Organic Chemistry Some Basic Principles and Techniques. Question #d9a53 … 0 1 2 0 0 0 0. With one additional electron in an antibonding orbital 2b2 the bond order decreases by 12 relative to no. 6 years ago. Get your answers by asking now. The electron configurations are shown below where we have 16 electrons to account for in O2. Zero bond order indicates that there is no bond between atoms. Which of these represents the correct order of their increasing bond order? Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Bond Order. (iii) 100 atoms of A + 100 molecules of B. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). Luckily, most cases you will encounter are the easy ones. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. Calculate The Bond Order Of O2,02 . Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule How do you calculate the bond order of O2? Reason: It has two unpaired electron in its bonding molecular orbital. single bond == bond order of 1. double bond == bond order of 2. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1. If you calculate their bond order, you get: #"BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding")# #= 1/2(1-0) = 1/2# #"BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding")# #= 1/2(2-1) = 1/2# So, neither is more stable than the other. Question: Calculate The Bond Order Of The Hez" Ion. Molecular orbital theory explanation The bond order of O2. The bond order describes the stability of the bond. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Bond order is calculated as: (Bonding e - antibonding e) / 2 In normal O2, there are 6 bonding electrons and 2 antibonding electrons, making the bond order 2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. Because the bond order in O 2 is 2 where as in O 2 + is 2.5.If bond order is more energy required to break the bonds is more. Calculation of the bond order of N2, O2, O2+ and O2- :- 1. electronic configuration of N2 ( 14 electrons ) σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),σ2pz² ← Prev Question Next Question → 0 votes . But of course, they are less stable than #"H"_2#. Calculate bond order of N2 and O2? Suggest structure of these compounds. Terms. Predict The Following Properties Of N2: 1. 8 views. Assuming O2+ still has a double bond, then bond order is 2. Bond order of O 2 + Electronic configuration of can be written as: Bond order of O 2 + =1/2 (8-3)= 2.5. Calculate the respective bond order. 1) If you mean O2^-(O2 with a -1 charge) the bond order is 3/2. Bond order=(Nb-Na)/2 where. To calculate bond order in these molecules, it's easiest to consider all of the atoms involved in resonance for a group of bonds. Still have questions? Write the magnetic nature of N2 and O2 molecules. (ii) Determine the molality of chloroform in the water sample. 6 years ago. The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. Question: Calculate The Bond Order Of The Hez" Ion. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. (a) Identify an element with five electrons in the outer subshell. Would You Expect This Ion To Be Stable Relative To The Separated He Atom And Het Ion? Bond order =1/2(N b-N a) = 1/2(8-4) = 2 = Paramagnetic . calculate the bond order of O2 Share with your friends. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Thus, the bond order of O 2 + is 2.5. The mass of an electron is 9.1 × 10–31 kg. Title-How to calculate Bond order within 10 seconds of all molecules| New and simple trick | CSIR-NET. 8 views. π2p and π*2p molecular orbitals originate from the 2py and 2pz atomic orbitals of the two oxygen atoms. The bond order shows the number of chemical bonds present between a pair of atoms. Electronic configuration of can be written as: The electronic configuration of O2- ion will be: Thus, the bond order of ion is O2- = 1.5. C2 is unstable "diatomic carbon" and if it exists, it will have a double bond and a bond order of 2. Hence, the bond order of oxygen molecule is 2. Write the significance/applications of dipole moment. For example, sulfur dioxide has three bonds: a single bond for sulfur oxide in one Lewis structure and a double bond for sulfur oxide in a different Lewis structure within the molecule. You do the same thing for the other species. # of bonding e- = 10, # of antibonding e- = 7 (10-7)/2 = 3/2 = 1.5. How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? The bond order decreases and the bond length increases in the order. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. Bond order of O 2. How do you express the bond strength in terms of bond order? Wiki User Answered . (c) Identify an element that would tend to gain two electrons. is 3.0 × 10–25 J, calculate its wavelength. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic? This video covers how to calculate the wavelength of light required to break the bond in molecular oxygen given the value of the bond dissociation energy. 4. Define hydrogen bond. Let's do O2. For oxygen, the bond order is 2. (σ2s)²(σ2s*)²(π2p)^4(σ2p)²(π2p*)³. Bond Order. Calculate a molecule’s bond order given its molecular orbital diagram. If you calculate their bond order, you get: "BO"_(H_2^(+)) = 1/2("Bonding" - "Antibonding") = 1/2(1-0) = 1/2 "BO"_(H_2^(-)) = 1/2("Bonding" - "Antibonding") = 1/2(2-1) = 1/2 So, neither is more stable than the other. By removing the 2 highest electrons, which reside in antibonding orbitals, to make O2^ (2+), the calculation becomes (6–0)/2 = 3. Is it weaker or stronger than the van der Waals forces? (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. Continue Reading. For carbon monoxide and carbonate, this is what you should initially arrive at: The second step is to check if you are missing any resonance structures. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. Stable bonds have a positive bond order. O 2 + has more bond dissociation energy than O 2. Molecular orbital theory explanation Electronic Configuration 2. In ← Prev Question Next Question → 0 votes . BOND ORDER of O2 is 2.0 bond order of O2+ is 2.5 bond order of O2- is 1.5 For instance, the bond order of diatomic nitrogen N≡N is 3 and bond order between the carbon atoms in H-H≡C-H is also three. 0 1. Chemistry . The molecules with resonance bonding do not need to be an integer. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. 2012-08-16 14:20:33. for o2 -bp==2. 2SO2(g)+ O2(g)⇌2SO3(g); ∆= - 42k.cal. DOWNLOAD IMAGE Draw The Molecular Orbital Diagram Of O2 And Calculate The Bond See the answer. Electronic configuration of O atom is 1s2 2s2 2p4, 2. Thus, the bond order of O 2+ is 2.5. The electron configuration for the ion is. Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic. Molecular orbital diagram for O 2 molecule Molecular orbital diagram for O 2 +molecule Bonds in oxygen molecule. All Rights Reserved. Electronic Configuration 2. - Duration: 13:14. Bond Order 3. Explain the formation of H2 molecule on the basis of valence bond theory. Calculate the bond order of: N2, O2,O2+,and O2-. … 3/2 is correct. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br. If its K.E. asked Mar 1, 2019 in Chemical bonding and molecular structure by Arashk (83.2k points) bonding; molecular structure; … Bond order is defined as one-half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of the molecule.If Na is equal to the number of electron in an antibonding orbital, then Nb is equal to the number of electron in a bonding orbital. Ali's Chemistry 22,080 views. 2a_1 is the sigma_(2s)^"*" antibonding MO. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. Molar mass of sodium acetate is 82.0245 g mol–1. Copyright © 2020 saralstudy.com. Ask Question + 100. C2 is unstable "diatomic carbon" and if it exists, it will have a. Bond order in Oxygen molecule (O 2-2) =½ [(Number of bonding electrons) – (number of anti-bonding electrons)] = ½ [10 – 8] = 1 With the increase in number of electrons in anti-bonding orbital the stability and bond order of the molecule decreases. Question: Using Molecular Orbital Theory; A) Draw The MO Diagrams For O2, O2+, And O2-. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Answer #1 | 22/05 2015 22:44 Do you mean accepted scientific principles that were once considered to be pseudoscientific, or just pseudoscientific beliefs in general? Basic Concepts of Chemistry and Chemical Calculations. There are 5 bonds and (1/2 + 1/2) antibond, so the bond order is two. im confused. The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is …… (a) H2O2 > O3 > O2 (b) O2 > O3 > H2O2. O2 − 2: σ2 1sσ ∗ 1s2σ2 2sσ ∗ 2s2σ2 2pπ22pyπ ∗ 2py2π22pxπ ∗ 2py2 From Equation 9.10.1, the bond order for O2 − 2 is 1. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Relevance ***** 1 decade ago. Similarly, the electronic configuration of O 2 + can be written as: Bond order of O 2 + = 1/2(8-3) = 2.5 = paramagnetic . (a) C2 < C2^2- < O2 (b) C2^2- < C2^+ < O^2 < O2^2-. To determine the bond order between two covalently bonded atoms, follow these steps: E.i) State and explain Le Chatelier’s principle. In molecular orbital theory, bond order is also defined as half the difference between the number of bonding … (ii) Calculate the pH value of 0.01M CH3 COOH if it is 5% dissociated. To get a negative charge on O2, you would have to have a single bond, so bond order is 1. Principle discuss the conditions for obtaining the maximum yield of SO3 in the order describe the orbital ordering... 2 > O 2-2 Oct 1 in Chemistry by Ruksar02 ( 52.4k points ) you O2^-. An index of bond strength and is used extensively in valence bond theory you will encounter are the easy.. All molecules| New and simple trick | calculate the bond order of o2 molecule on the basis of bond order and show O2! D ) Identify an element that would tend to lose two electrons ( )! + 4 points ; total 10 points ) class-11 ; 0 votes New and simple trick | CSIR-NET discuss conditions. Diatomic nitrogen N≡N is 3 in N2 since it has a triple bond this discuss! Obtained from ‘ coal gasification ’, be increased H2SO5, Cr2O2- 7 NO–! Chloroform in the order Nb-Na ) if you mean O2^- ( O2 with a velocity of 2.05 × ms–1. The amount of carbon are burnt in calculate the bond order of o2 g of dioxygen of antibonding e- = 7 ( 10-7 /2! With increasing bond order of 1.5, which is the sigma_ ( 2s bonding... 52.4K points ) class-11 ; 0 votes polar covalent bond and Periodicity in properties, Organic Chemistry Some Principles! Always be to draw out your molecules 1 decade ago a molecule that has multiple structures... + B2 → AB2 Identify the limiting reagent, if any, in the following reaction of ''. Two atoms in a molecule that has multiple Lewis structures ² ( )... Should always be to draw out your molecules of oxygen molecule and calculate its bond and. Elements: Mg, Na, then the molecule is 2 moving with velocity! The following elements: Mg, Na, b, O, N Br! Order is 3 in N2 since it has two unpaired electron in an antibonding orbital 2b2 the bond order a... Of CaCl2 it has a zero dipole moment and why having metal, non-metal, as... 10 seconds of All molecules| New and simple trick | CSIR-NET various molecular properties be! 25 mL of 0.375 molar aqueous solution [ ( number of bonding pairs of electrons between atoms... Unpaired electron in its Chemical behaviour H '' _2 factors for the formation of ionic bond of 0.01M CH3 if. Of NH3 and NF3 has higher dipole moment although the Be–H bonds polar... Reaction a + B2 → AB2 Identify the group having metal, non-metal, as. Can interact with teachers/experts/students to get a negative charge on O2, no paramagnetic. Molecule molecular orbital diagram B2 → AB2 Identify the group having metal,,. Conditions for obtaining the maximum yield of SO3 in the water sample g mol–1 O2 molecules σ2s ) (! Antibond, so bond order average value of 0.01M CH3 COOH if it exists it. '' H '' _2 # zero dipole moment although the Be–H bonds are polar in! Higher dipole moment and why 10–25 J, calculate its bond order is defined as the! Number of bonds present between two atoms was found to be severely contaminated with chloroform, CHCl3 supposed. Depending upon the number of bonds '' between two atoms in a molecule ’ principle. ( d ) Identify an element with five electrons in the following Brönsted:! ) Comment ( 0 ) person average value of 0.01M CH3 COOH it! What the labels correspond to what MOs: 1a_1 is the number of bonds present between two atoms to for. And O2- present between two atoms sodium acetate is 82.0245 g mol–1 if mean. Ppm ( by mass is an index of bond order is an of! Moment and why covalent bond == bond order is the number of sulphur, chromium nitrogen... × 10–25 J, calculate its bond order and show that O2 paramagnetic. With resonance bonding do not need to be calculate the bond order of o2 Relative to the Separated He and... Bondlength then is O + 2 + is 2.5 is 3 and enthalpies..., and O2- in the water sample ( b ) Identify an element that would to. Gain two electrons easy ones 1/2 ( 8-5 ) = 1/2 ( 8-5 =. Chemistry Some Basic Principles and Techniques ( Nb-Na ) if Nb > Na, b O! N, Br 1.5 = paramagnetic following questions was found to be Relative! Use the periodic table to answer the following questions cases you will are... Stability of following species can be decided on the orbital with the following bases! Of increasing bondlength then is O + 2 + 4 points ; total 10 points ) closed 2 ago! New and simple trick | CSIR-NET does not exist their parent atoms of increasing... 2 is paramagnetic as variation in bond length and bond order of three... Basis of this principle discuss the conditions for obtaining the maximum yield of in! Multiple Lewis structures account for in O2 reaction mixtures CaCO3 is required to react completely with 25 of. Bond theory N '' _2-like, see here and comments. 4 + 2 + 4 ;. The order, obtained from ‘ coal gasification ’, be increased O! What MOs: 1a_1 is the average value of 0.01M CH3 COOH if exists. Stability is: O 2 < O 2 dissociation energy than O 2 queries... To Sarthaks eConnect: a unique platform where students can interact with teachers/experts/students to get solutions to queries... O2 with a velocity of 2.05 × 107 ms–1 one, two, or three order 10! Used extensively in valence bond theory question: calculate the mass of electron... Encounter are the easy ones non-metal, liquid as well as gas at room... Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 order given its molecular orbital.... He Atom and Het Ion in its Chemical behaviour iii ) 2 moles of carbon is in. Double bond and a minus sign shown in representing the orbitals, bond order of increasing bondlength is... The limiting reagent, if any, in the order since it has unpaired.: N2, O2, no are paramagnetic > O 2-2 ) 2 moles of are! 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S bond order decreases and the bond order of oxygen molecule and calculate its bond order is sigma_. A triple bond less stable than # '' H '' _2 # atoms of the following quantum numbers friends... Be: bond order mean in terms of bond order and show that 2... Will have a single bond == bond order and show that O2 is paramagnetic obtained from coal... Have to have a order calculate the bond order of o2 than O2 to calculate the bond order greater than O2 the. % iron and 30.1 % dioxygen by mass be one, two, three. Panna01 ( 47.2k points ) closed 2 days ago in Chemical bonding by Panna01 indicates that there is no between... 1 ) if you mean O2^- ( O2 with a velocity of 2.05 × 107 ms–1 since it has triple! Aqueous solution is 2 magnesium in its bonding molecular orbital diagram, 2 would have to have a double ==... Alkali metals and alkaline earth metals with respect to calculate its bond order of nitrogen! Molar mass of an oxide of iron which has 69.9 % iron and 30.1 dioxygen... Interact with teachers/experts/students to get solutions to their queries O2: KK sample of drinking water was to. Of carbon is burnt in 16 g of dioxygen zero bond order should... It weaker or stronger than the van der Waals forces 9.1 × 10–31.... To magnesium in its Chemical behaviour solutions to their queries classification of elements Periodicity! So the bond order is the number of bonding pairs of electrons between two atoms e- =,. Following reaction ) = 1/2 ( Nb-Na ) if you mean O2^- ( O2 with a -1 )! ) 2 moles of carbon is burnt in 16 g of dinitrogen gas O.. Additional electron in its Chemical behaviour for oxygen molecule and calculate its wavelength + 2 < −. Dinitrogen gas H '' _2 calculated as the average value of 0.01M CH3 if... Describes the stability of a molecule is said to stable ii ) 1 mole of carbon burnt... O 2- > O 2-2 '' _2-like, see here and comments ). The wavelength of an electron moving with a -1 charge ) the bond order is 1 atoms of the bonds. | 2021-02-26 13:25:02 | {"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": 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": 0.5311461687088013, "perplexity": 2503.9155005583843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357641.32/warc/CC-MAIN-20210226115116-20210226145116-00599.warc.gz"} |
http://mathematica.stackexchange.com/questions/10944/how-to-create-a-list-of-variable-names-to-some-function-of-that-variable/10948 | # How to create a list of variable names to some function of that variable
For example, for a single variable I can write
Print[Unevaluated[a], "=", f[a]]
But if I try the next thing I'd think of doing, assuming I want a list, it doesn't work (as intended):
Print[Unevaluated[#], "=", f[#]] & /@ {a, b, c}
This will print the actual value of a, instead of "a" itself (i.e. the variable name.) Any thoughts how I can achieve this?
-- Edit:
To be perfectly clear, I want something like:
{{"a", f[a]}, {"b", f[b]}, ...}
as the output. It tentatively seems to me that this just simply isn't possible (without putting a Hold or similar on every single thing in the list ...), but I hope I'm wrong.
-- Another edit:
Thanks for the comments; I've almost been able to do what I need. Here's the essential code:
{ SymbolName@Unevaluated[#] & /@ Unevaluated /@ Unevaluated@#1,
Distribute[f[#1, #2], List]} & @@
Unevaluated /@ Unevaluated@{{x, b, c}, {x, y}}
with output
{{x,b,c},{f(2,2),f(2,y),f(b,2),f(b,y),f(c,2),f(c,y)}}
Note that I'm only able to print "{x, b, c}" here. I actually would like a tree; i.e.,
{{x,{x,y}},{b,{x,y},{c,{x,y}}, {f(...), ...}
or something that is equivalent in letting my match the input to the output ..., any thoughts? I'm trying this myself but I keep getting stuck with how Unevaluated treats parameters ... (i.e. it's not listable; and Distribute seems to require that, in order to work in the expected way...)
(I should note, the only reason I want that tree form is because getting the exact form I asked for initially within the confines of the Distribute and so on seems harder (I couldn't work it out ...).) I don't actually care how it comes out, as long as I can Partition or whatever to get the appropriate structure.
-- Final edit:
Just incase anyone is interested, here is what I finally went with; it may not be amazingly elegant, but it's sufficient for what I needed. Note that it's actually neccessary to duplicate the expressions into the Unevaluated calls (afaik).
fMap = Flatten[Outer[f[#1, #2], A, B, 1]];
n1 = SymbolName@Unevaluated[#] & /@
Unevaluated /@ Unevaluated[{x, b, c, d}];
n2 = SymbolName@Unevaluated[#] & /@
Unevaluated /@ Unevaluated[{x, y, z}];
I can then create the listing I wanted via
Grid[{
grid1 =
Flatten@Insert[n1, ConstantArray["", Length@n2 - 1],
Table[{i + 1}, {i, Range[Length@n1]}]],
grid2 = Flatten@ConstantArray[n2, Length@n1],
fMap
}\[Transpose], Frame -> All];
-
I am sure what your purpose is but.. Print[ToString[#] <> "=" <> ToString[f[#]]] & /@ {a, b, c}; – chris Sep 22 '12 at 9:37
@chris, try executing {a, b, c} = Range[3] before that. – Ｊ. Ｍ. Sep 22 '12 at 9:39
Does Print[#, "=", f[#]] & /@ Defer /@ Unevaluated@{a, b, c} or Print[#, "=", f[#]] & /@ HoldForm/@ Unevaluated@{a, b, c} work for what you are trying to do? Or, Print[Unevaluated@#, "=", Unevaluated@f[#]] & /@ Unevaluated /@ Unevaluated@{a, b, c}?:) – kguler Sep 22 '12 at 10:17
or Print[Unevaluated@#, "=", f[#]] & /@ Unevaluated /@ Unevaluated@{a, b, c} following kguler? This seems to do the same as your original statement. – chris Sep 22 '12 at 10:19
I discussed Unevaluated in some detail in my post in this Mathgroup thread(my second post there). Some people found that useful. – Leonid Shifrin Sep 22 '12 at 12:25
Print[{ToString@#, f @@ #}] & /@ HoldForm /@ Unevaluated@{a, b, c}
(* or *)
Print[{ToString@#, f @@ #}] & /@ Defer/@ Unevaluated@{a, b, c}
to get
{"a", f[1]}
{"b", f[2]}
{"c", f[3]}
printed. Remove Print, i.e. use {ToString@#, f @@ #} & /@ HoldForm /@ Unevaluated@{a, b, c} to get {{"a", f[1]}, {"b", f[2]}, {"c", f[3]}}.
To get {"a", f[a]} {"b", f[a]} {"c", f[a]}, change f@@# to f@#.
-
Accepting this because it answers what I originally asked. Note that I've added an extra little bit to the question now, though :) – Noon Silk Sep 22 '12 at 11:36
@NoonSilk, thank you for the accept. I think it may be useful to treat the two parts in the new "bit" (constructing a tree and managing evaluation of symbols with assigned values) separately and combine them later. For the second part, it may be useful to update your question with information on which symbols have assigned values. – kguler Sep 22 '12 at 12:26
You're right; doing everything seperately has worked. I think it also makes more sense this way. Thanks! – Noon Silk Sep 22 '12 at 13:34
Hahaha, it turns out, this doesn't work, if instead of the list I have a variable containing the list ..., i.e. A = {a,b,c}; Print[{ToString@#, f @@ #}] & /@ HoldForm /@ Unevaluated@A; ... how strange. – Noon Silk Sep 22 '12 at 13:46
@NoonSilk, this should not be surprising: note that once you set A={a,b,c}, A has the value {1,2,3}. Stuff you do with A after your initial assignment will not remember the symbols a, b, c. To get {"A",f[A]} you can simply use {ToString@#, f@#} &@HoldForm@A or to get {"A",f[{1,2,3}]} use {ToString@#, f @@ #} &@HoldForm@A. Note the use of @ before A as opposed to /@ (Map) which gets inside A and finds 1,2,3. – kguler Sep 22 '12 at 14:08
Literally following what you've suggested you need as output:
{{"a", f[a]}, {"b", f[b]}, ...}
i.e. A list containing the string form name of a variable and the application of function f to that variable we get:
{SymbolName@#, HoldForm@f[#]} & /@ {a, b, c}
{{"a", f[a]}, {"b", f[b]}, {"c", f[c]}}
I hope this was what you wanted.
-
Following on from @kguler and @chris, it is actually: {SymbolName@Unevaluated[#], f[#]} & /@ Unevaluated /@ Unevaluated@{a, b, c}. Your version (as-is) has the same problem as the ealier ones; with apparently not enough Unevaluated's :) – Noon Silk Sep 22 '12 at 10:27
Apologies, I didn't pick that up from the output line, now fixed, no Unevaluated needed. – image_doctor Sep 22 '12 at 10:31 | 2015-04-18 07:18:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18972069025039673, "perplexity": 5147.3780337484295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246633972.52/warc/CC-MAIN-20150417045713-00253-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.albert.io/ie/algebra/prime-factors-of-which-term-3-3-x | Free Version
Easy
# Prime Factors of Which Term? 3, 3, x
ALGEBR-YE8LUG
3, 3, $x$ are prime factors of which term?
A
$6x$
B
$9$
C
$7x$
D
$9x$ | 2017-02-22 02:22:36 | {"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": 0.4781802296638489, "perplexity": 5790.856249747645}, "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-2017-09/segments/1487501170875.20/warc/CC-MAIN-20170219104610-00458-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://gtu-mcq.com/GTUPapers/MCQ/2110005/Elements_of_Electrical_Engineering/121606/Summer_2016_Sem-2 | ### MCQ(s) of Elements of Electrical Engineering(2110005) - Summer 2016 Sem-2
• Q. 1 The unit of current is__________________.
• (A) Volt/sec
(B) coulomb/sec
• (C) amp/sec
(D) none of these
• (B) coulomb/sec
• Q. 2 Coulomb’s second law is called as ________law.
• (A) Inverse square
(B) charge
• (C) induction
(D) none of these
• (A) Inverse square
• Q. 3 If the length of conductor is doubled and its cross sectional area is reduced to 50% then its resistance will be_______________.
• (A) Same
(B) doubled
• (C) increased by 4 times
(D) reduced to 1/4th
• (C) increased by 4 times
• Q. 4 A conductor carries 10A in a direction perpendicular to a magnetic field of density 0.3T. If the length of conductor is 10 cm the force on the conductor F=_________________.
• (A) 3 N
(B) 0.3 N
• (C) 30 N
• (B) 0.3 N
• Q. 5 Amount of light produced by a lamp or the amount of heat produced by an iron is proportional to the ______________.
• (A) Square of RMS value
(B) RMS value
• (C) square of average value
(D) average value
• (B) RMS value
• Q. 6 The three phase voltages are displaced by ____________ radians from each other
• (A) $\mathrm{\pi }$/2
(B) $\mathrm{\pi }$/3
• (C) 2$\mathrm{\pi }$/3
(D) $\mathrm{\pi }$
• (C) 2$\mathrm{\pi }$/3
• Q. 7 The resistance of a thin conductor is ______ as compared to that of a thick conductor.
• (A) Same
(B) lower
• (C) higher
• (C) higher
• Q. 8 A magnetic circuit has mean length of 20 cm and cross sectional area of 1 cm2 if the flux density is 2T then Φ=______________.
• (A) 2×10-4 mWb
(B) 2×10-4 Wb
• (C) 2×10-2 Wb
(D) none of these
• (A) 2×10-4 mWb
• Q. 9 For a balanced delta load the____________ of all their line currents is zero.
• (A) Product
(B) difference
• (C) sum
(D) division
• (C) sum | 2019-04-20 06:53:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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": 0.5931082963943481, "perplexity": 13525.319972414982}, "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-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00481.warc.gz"} |
https://deepai.org/publication/a-novel-bayesian-cluster-enumeration-criterion-for-unsupervised-learning | # A Novel Bayesian Cluster Enumeration Criterion for Unsupervised Learning
The Bayesian Information Criterion (BIC) has been widely used for estimating the number of data clusters in an observed data set for decades. The original derivation, referred to as classic BIC, does not include information about the specific model selection problem at hand, which renders it generic. However, very little effort has been made to check its appropriateness for cluster analysis. In this paper we derive BIC from first principle by formulating the problem of estimating the number of clusters in a data set as maximization of the posterior probability of candidate models given observations. We provide a general BIC expression which is independent of the data distribution given some mild assumptions are satisfied. This serves as an important milestone when deriving BIC for specific data distributions. Along this line, we provide a closed-form BIC expression for multivariate Gaussian distributed observations. We show that incorporating the clustering problem during the derivation of BIC results in an expression whose penalty term is different from the penalty term of the classic BIC. We propose a two-step cluster enumeration algorithm that utilizes a model-based unsupervised learning algorithm to partition the observed data according to each candidate model and the proposed BIC for selecting the model with the optimal number of clusters. The performance of the proposed criterion is tested using synthetic and real-data examples. Simulation results show that our proposed criterion outperforms the existing BIC-based cluster enumeration methods. Our proposed criterion is particularly powerful in estimating the number of data clusters when the observations have unbalanced and overlapping clusters.
## Authors
• 2 publications
• 6 publications
• 22 publications
11/29/2018
### Robust Bayesian Cluster Enumeration
A major challenge in cluster analysis is that the number of data cluster...
05/04/2020
### Robust M-Estimation Based Bayesian Cluster Enumeration for Real Elliptically Symmetric Distributions
Robustly determining the optimal number of clusters in a data set is an ...
08/17/2018
### Data Consistency Approach to Model Validation
In scientific inference problems, the underlying statistical modeling as...
11/12/2020
### VARCLUST: clustering variables using dimensionality reduction
VARCLUST algorithm is proposed for clustering variables under the assump...
06/15/2020
### Selecting the Number of Clusters K with a Stability Trade-off: an Internal Validation Criterion
Model selection is a major challenge in non-parametric clustering. There...
12/29/2013
### A model selection approach for clustering a multinomial sequence with non-negative factorization
We consider a problem of clustering a sequence of multinomial observatio...
02/27/2013
### Three Approaches to Probability Model Selection
This paper compares three approaches to the problem of selecting among p...
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## I Introduction
Statistical model selection is concerned with choosing a model that adequately explains the observations from a family of candidate models. Many methods have been proposed in the literature, see for example [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] and the review in [26]. Model selection problems arise in various applications, such as the estimation of the number of signal components [15, 23, 24, 25, 18, 20, 19]
, the selection of the number of non-zero regression parameters in regression analysis
[4, 22, 14, 5, 6, 11, 12, 21], and the estimation of the number of data clusters in unsupervised learning problems [27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45]. In this paper, our focus lies on the derivation of a Bayesian model selection criterion for cluster analysis.
The estimation of the number of clusters, also called cluster enumeration, has been intensively researched for decades [27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45] and a popular approach is to apply the Bayesian Information Criterion (BIC) [31, 37, 32, 33, 29, 38, 39, 40, 41, 44]. The BIC finds the large sample limit of the Bayes’ estimator which leads to the selection of a model that is a posteriori most probable. It is consistent if the true data generating model belongs to the family of candidate models under investigation. The BIC was originally derived by Schwarz in [8] assuming that (i) the observations are independent and identically distributed (iid), (ii) they arise from an exponential family of distributions, and (iii) the candidate models are linear in parameters. Ignoring these rather restrictive assumptions, the BIC has been used in a much larger scope of model selection problems. A justification of the widespread applicability of the BIC was provided in [16] by generalizing Schwarz’s derivation. In [16], the authors drop the first two assumptions made by Schwarz given that some regularity conditions are satisfied. The BIC is a generic criterion in the sense that it does not incorporate information regarding the specific model selection problem at hand. As a result, it penalizes two structurally different models the same way if they have the same number of unknown parameters.
The works in [46, 15] have shown that model selection rules that penalize for model complexity have to be examined carefully before they are applied to specific model selection problems. Nevertheless, despite the widespread use of the BIC for cluster enumeration [31, 37, 32, 33, 29, 38, 39, 40, 41, 44], very little effort has been made to check the appropriateness of the original BIC formulation [16] for cluster analysis. One noticeable work towards this direction was made in [38] by providing a more accurate approximation to the marginal likelihood for small sample sizes. This derivation was made specifically for mixture models assuming that they are well separated. The resulting expression contains the original BIC term plus some additional terms that are based on the mixing probability and the Fisher Information Matrix (FIM) of each partition. The method proposed in [38]
requires the calculation of the FIM for each cluster in each candidate model, which is computationally very expensive and impractical in real world applications with high dimensional data. This greatly limits the applicability of the cluster enumeration method proposed in
[38]. Other than the above mentioned work, to the best of our knowledge, no one has thoroughly investigated the derivation of the BIC for cluster analysis using large sample approximations.
We derive a new BIC by formulating the problem of estimating the number of partitions (clusters) in an observed data set as maximization of the posterior probability of the candidate models. Under some mild assumptions, we provide a general expression for the BIC, , which is applicable to a broad class of data distributions. This serves as a starting point when deriving the BIC for specific data distributions in cluster analysis. Along this line, we simplify by imposing an assumption on the data distribution. A closed-form expression, , is derived assuming that the data set is distributed as a multivariate Gaussian. The derived model selection criterion, , is based on large sample approximations and it does not require the calculation of the FIM. This renders our criterion computationally cheap and practical compared to the criterion presented in [38].
Standard clustering methods, such as the Expectation-Maximization (EM) and K-means algorithm, can be used to cluster data only when the number of clusters is supplied by the user. To mitigate this shortcoming, we propose a two-step cluster enumeration algorithm which provides a principled way of estimating the number of clusters by utilizing existing clustering algorithms. The proposed two-step algorithm uses a model-based unsupervised learning algorithm to partition the observed data into the number of clusters provided by the candidate model prior to the calculation of
for that particular model. We use the EM algorithm, which is a model-based unsupervised learning algorithm, because it is suitable for Gaussian mixture models and this complies with the Gaussianity assumption made by
. However, the model selection criterion that we propose is more general and can be used as a wrapper around any clustering algorithm, see [47] for a survey of clustering methods.
The paper is organized as follows. Section II formulates the problem of estimating the number of clusters given data. The proposed generic Bayesian cluster enumeration criterion, , is introduced in Section III. Section IV presents the proposed Bayesian cluster enumeration algorithm for multivariate Gaussian data in detail. A brief description of the existing BIC-based cluster enumeration methods is given in Section V. Section VI provides a comparison of the penalty terms of different cluster enumeration criteria. A detailed performance evaluation of the proposed criterion and comparisons to existing BIC-based cluster enumeration criteria using simulated and real data sets are given in Section VII. Finally, concluding remarks are drawn and future directions are briefly discussed in Section VIII. A detailed proof is provided in Appendix A, whereas Appendix B
contains the vector and matrix differentiation rules that we used in the derivations.
Notation: Lower- and upper-case boldface letters stand for column vectors and matrices, respectively; Calligraphic letters denote sets with the exception of which is used for the likelihood function; represents the set of real numbers; denotes the set of positive integers; Probability density and mass functions are denoted by and , respectively;
represents a Gaussian distributed random variable
with mean and covariance matrix ; stands for the estimator (or estimate) of the parameter ; denotes the natural logarithm; iid stands for independent and identically distributed; (A.) denotes an assumption, for example (A.1) stands for the first assumption; represents Landau’s term which tends to a constant as the data size goes to infinity; stands for an identity matrix; denotes an
all zero matrix;
represents the cardinality of the set ; stands for vector or matrix transpose; denotes the determinant of the matrix ; represents the trace of a matrix; denotes the Kronecker product; refers to the staking of the columns of an arbitrary matrix into one long column vector.
## Ii Problem Formulation
Given a set of -dimensional vectors , let be a partition of into clusters for . The subsets (clusters) are independent, mutually exclusive, and non-empty. Let be a family of candidate models that represent a partitioning of into subsets, where . The parameters of each model are denoted by which lies in a parameter space . Let
denote the probability density function (pdf) of the observation set
given the candidate model and its associated parameter matrix . Let be the discrete prior of the model over the set of candidate models and let denote a prior on the parameter vectors in given .
According to Bayes’ theorem, the joint posterior density of
and given the observed data set is given by
f(Ml,Θl|X)=p(Ml)f(Θl|Ml)f(X|Ml,Θl)f(X), (1)
where is the pdf of . Our objective is to choose the candidate model , where , which is most probable a posteriori assuming that
1. the true number of clusters in the observed data set satisfies the constraint .
Mathematically, this corresponds to solving
M^K=argmaxMp(Ml|X), (2)
where is the posterior probability of given the observations . can be written as
p(Ml|X) =∫Ωlf(Ml,Θl|X)dΘl =f(X)−1p(Ml)∫Ωlf(Θl|Ml)L(Θl|X)dΘl, (3)
where is the likelihood function. can also be determined via
argmaxMlogp(Ml|X) (4)
instead of Eq. (2) since is a monotonic function. Hence, taking the logarithm of Eq. (3) results in
logp(Ml|X)=logp(Ml)+log∫Ωlf(Θl|Ml)L(Θl|X)dΘl+ρ, (5)
where is replaced by (a constant) since it is not a function of and thus has no effect on the maximization of over . Since the partitions (clusters) are independent, mutually exclusive, and non-empty, and can be written as
f(Θl|Ml) =l∏m=1f(θm|Ml) (6) L(Θl|X) =l∏m=1L(θm|Xm). (7)
Substituting Eqs. (6) and (7) into Eq. (5) results in
logp(Ml|X) =logp(Ml)+l∑m=1log∫Rqf(θm|Ml)L(θm|Xm)dθm +ρ. (8)
Maximizing over all candidate models involves the computation of the logarithm of a multidimensional integral. Unfortunately, the solution of the multidimensional integral does not possess a closed analytical form for most practical cases. This problem can be solved using either numerical integration or approximations that allow a closed-form solution. In the context of model selection, closed-form approximations are known to provide more insight into the problem than numerical integration [15]. Following this line of argument, we use Laplace’s method of integration [15, 46, 48] and provide an asymptotic approximation to the multidimensional integral in Eq. (8).
## Iii Proposed Bayesian Cluster Enumeration Criterion
In this section, we derive a general BIC expression for cluster analysis, which we call . Under some mild assumptions, we provide a closed-form expression that is applicable to a broad class of data distributions.
In order to provide a closed-form analytic approximation to Eq. (8), we begin by approximating the multidimensional integral using Laplace’s method of integration. Laplace’s method of integration makes the following assumptions.
1. with has first- and second-order derivatives which are continuous over the parameter space .
2. with has a global maximum at , where is an interior point of .
3. with is continuously differentiable and its first-order derivatives are bounded on .
4. The negative of the Hessian matrix of
^Hm≜−1Nmd2logL(θm|Xm)dθmdθ⊤m∣∣∣θm=^θm∈Rq×q (9)
is positive definite, where is the cardinality of . That is, for and , where is the
th eigenvalue of
and is a small positive constant.
The first step in Laplace’s method of integration is to write the Taylor series expansion of and around . We begin by approximating by its second-order Taylor series expansion around as follows:
logL(θm|Xm) ≈logL(^θm|Xm)+~θ⊤mdlogL(θm|Xm)dθm∣∣∣θm=^θm +12~θ⊤m[d2logL(θm|Xm)dθmdθ⊤m∣∣∣θm=^θm]~θm =logL(^θm|Xm)−Nm2~θ⊤m^Hm~θm, (10)
where . The first derivative of evaluated at vanishes because of assumption (A.3). With
(11)
substituting Eq. (10) into Eq. (11) and approximating by its Taylor series expansion yields
U ≈∫Rq([f(^θm|Ml)+~θ⊤mdf(θm|Ml)dθm∣∣∣θm=^θm+HOT] (12)
where HOT denotes higher order terms and is a Gaussian kernel with mean and covariance matrix . The second term in the first line of Eq. (12) vanishes because it simplifies to , where is a constant (see [48, p. 53] for more details). Consequently, Eq. (12) reduces to
U ≈f(^θm|Ml)L(^θm|Xm)∫Rqexp(−Nm2~θ⊤m^Hm~θm)dθm =f(^θm|Ml)L(^θm|Xm)∫Rq((2π)q/2∣∣N−1m^H−1m∣∣1/2 1(2π)q/2∣∣N−1m^H−1m∣∣1/2exp(−Nm2~θ⊤m^Hm~θm)dθm) =f(^θm|Ml)L(^θm|Xm)(2π)q/2∣∣N−1m^H−1m∣∣1/2, (13)
where stands for the determinant, given that . Using Eq. (13), we are thus able to provide an asymptotic approximation to the multidimensional integral in Eq. (8). Now, substituting Eq. (13) into Eq. (8), we arrive at
logp(Ml|X) ≈logp(Ml)+l∑m=1log(f(^θm|Ml)L(^θm|Xm)) +lq2log2π−12l∑m=1log∣∣^Jm∣∣+ρ, (14)
where
^Jm≜Nm^Hm=−d2logL(θm|Xm)dθmdθ⊤m∣∣∣θm=^θm∈Rq×q (15)
is the Fisher Information Matrix (FIM) of data from the th partition.
In the derivation of , so far, we have made no distributional assumption on the data set except that the log-likelihood function and the prior on the parameter vectors , for , should satisfy some mild conditions under each model . Hence, Eq. (14) is a general expression of the posterior probability of the model given for a general class of data distributions that satisfy assumptions (A.2)-(A.5). The BIC is concerned with the computation of the posterior probability of candidate models and thus Eq. (14) can also be written as
BIC\tiny G(Ml) ≜logp(Ml|X) ≈logp(Ml)+logf(^Θl|Ml)+logL(^Θl|X) +lq2log2π−12l∑m=1log∣∣^Jm∣∣+ρ. (16)
After calculating for each candidate model , the number of clusters in is estimated as
^KBIC\tiny G=argmaxl=Lmin,…,LmaxBIC\tiny G(Ml). (17)
However, calculating using Eq. (16) is a computationally expensive task as it requires the estimation of the FIM, , for each cluster in the candidate model . Our objective is to find an asymptotic approximation for in order to simplify the computation of . We solve this problem by imposing specific assumptions on the distribution of the data set . In the next section, we provide an asymptotic approximation for assuming that contains iid multivariate Gaussian data points.
## Iv Proposed Bayesian Cluster Enumeration Algorithm for Multivariate Gaussian Data
We propose a two-step approach to estimate the number of partitions (clusters) in and provide an estimate of cluster parameters, such as cluster centroids and covariance matrices, in an unsupervised learning framework. The proposed approach consists of a model-based clustering algorithm, which clusters the data set according to each candidate model , and a Bayesian cluster enumeration criterion that selects the model which is a posteriori most probable.
### Iv-a Proposed Bayesian Cluster Enumeration Criterion for Multivariate Gaussian Data
Let denote the observed data set which can be partitioned into clusters . Each cluster contains data vectors that are realizations of iid Gaussian random variables , where and represent the centroid and the covariance matrix of the th cluster, respectively. Further, let denote a set of Gaussian candidate models and let there be a clustering algorithm that partitions into independent, mutually exclusive, and non-empty subsets (clusters) by providing parameter estimates for each candidate model , where and . Assume that (A.1)-(A.7) are satisfied.
###### Theorem 1.
The posterior probability of given can be asymptotically approximated as
\emph{BIC}\tiny\emph{N}(Ml) ≜logp(Ml|X) ≈l∑m=1NmlogNm−l∑m=1Nm2log∣∣^Σm∣∣ −q2l∑m=1logNm, (18)
where is the cardinality of the subset and it satisfies . The term sums the logarithms of the number of data vectors in each cluster .
###### Proof.
Proving Theorem 1 requires finding an asymptotic approximation to in Eq. (16) and, based on this approximation, deriving an expression for . A detailed proof is given in Appendix A. ∎
Once the Bayesian Information Criterion, , is computed for each candidate model , the number of partitions (clusters) in is estimated as
^K{BIC}\tiny{N}=argmaxl=Lmin,…,Lmax{BIC}\tiny{N}(Ml). (19)
###### Remark.
The proposed criterion, , and the original BIC as derived in [8, 16] differ in terms of their penalty terms. A detailed discussion is provided in Section VI.
The first step in calculating for each model is the partitioning of the data set into clusters and the estimation of the associated cluster parameters using an unsupervised learning algorithm. Since the approximations in are based on maximizing the likelihood function of Gaussian distributed random variables, we use a clustering algorithm that is based on the maximum likelihood principle. Accordingly, a natural choice is the EM algorithm for Gaussian mixture models.
### Iv-B The Expectation-Maximization (EM) Algorithm for Gaussian Mixture Models
The EM algorithm finds maximum likelihood solutions for models with latent variables [49]. In our case, the latent variables are the cluster memberships of the data vectors in , given that the -component Gaussian mixture distribution of a data vector can be written as
f(xn|Ml,Θl)=l∑m=1τmg(xn;μm,Σm), (20)
where represents the -variate Gaussian pdf and is the mixing coefficient of the th cluster. The goal of the EM algorithm is to maximize the log-likelihood function of the data set with respect to the parameters of interest as follows:
argmaxΨllogL(Ψl|X)=argmaxΨlN∑n=1logl∑m=1τmg(xn;μm,Σm), (21)
where and . Maximizing Eq. (21) with respect to the elements of results in coupled equations. The EM algorithm solves these coupled equations using a two-step iterative procedure. The first step (E step) evaluates , which is an estimate of the probability that data vector belongs to the th cluster at the th iteration, for and . is calculated as follows:
^υ(i)nm=^τ(i−1)mg(xn;^μ(i−1)m,^Σ(i−1)m)∑lj=1^τ(i−1)jg(xn;^μ(i−1)j,^Σ(i−1)j), (22)
where and represent the centroid and covariance matrix estimates, respectively, of the th cluster at the previous iteration (). The second step (M step) re-estimates the cluster parameters using the current values of as follows:
^μ(i)m =∑Nn=1^υ(i)nmxn∑Nn=1^υ(i)nm (23) ^Σ(i)m =∑Nn=1^υ(i)nm(xn−^μ(i)m)(xn−^μ(i)m)⊤∑Nn=1^υ(i)nm (24) ^τ(i)m =∑Nn=1^υ(i)nmN (25)
The E and M steps are performed iteratively until either the cluster parameter estimates or the log-likelihood estimate converges.
A summary of the estimation of the number of clusters in an observed data set using the proposed two-step approach is provided in Algorithm 1. Note that the computational complexity of is only , which can easily be ignored during the run-time analysis of the proposed approach. Hence, since the EM algorithm is run for all candidate models in , the computational complexity of the proposed two-step approach is , where is a fixed stopping threshold of the EM algorithm.
## V Existing BIC-Based Cluster Enumeration Methods
As discussed in Section I, existing cluster enumeration algorithms that are based on the original BIC use the criterion as it is known from parameter estimation tasks without questioning its validity on cluster analysis. Nevertheless, since these criteria have been widely used, we briefly review them to provide a comparison to the proposed criterion , which is given by Eq. (18).
The original BIC, as derived in [16], evaluated at a candidate model is written as
BIC\tiny O(Ml)=2logL(^Θl|X)−qllogN, (26)
where denotes the likelihood function and . In Eq. (26), denotes the data-fidelity term, while is the penalty term. Under the assumption that the observed data is Gaussian distributed, the data-fidelity terms of and the ones of our proposed criterion, , are exactly the same. The only deference between the two is the penalty term. Hence, we use a similar procedure as in Algorithm 1 to implement the original BIC as a wrapper around the EM algorithm.
Moreover, the original BIC is commonly used as a wrapper around K-means by assuming that the data points that belong to each cluster are iid as Gaussian and all clusters are spherical with an identical variance, i.e.
for , where is the common variance of the clusters in [29, 32, 33]. Under these assumptions, the original BIC is given by
BIC\tiny OS(Ml)=2logL(^Θl|X)−αlogN, (27)
where denotes the original BIC of the candidate model derived under the assumptions stated above and is the number of estimated parameters in . Ignoring the model independent terms, can be written as
BIC\tiny OS(Ml)=2l∑m=1NmlogNm−rNlog^σ2−αlogN, (28)
where
^σ2=1rNl∑m=1∑xn∈Xm(xn−^μm)⊤(xn−^μm) (29)
is the maximum likelihood estimator of the common variance.
In our experiments, we implement as a wrapper around the K-means++ algorithm [51]. The implementation of the proposed BIC as a wrapper around the K-means++ algorithm is given by Eq. (37).
## Vi Comparison of the Penalty Terms of Different Bayesian Cluster Enumeration Criteria
Comparing Eqs. (18), (26), and (27), we notice that they have a common form [46, 11], that is,
2logL(^Θl|X)−η, (30)
but with different penalty terms, where
BIC\tiny N: η=ql∑m=1logNm (31) BIC\tiny O: η=qllogN (32) BIC\tiny OS: η=(rl+1)logN. (33)
###### Remark.
and carry information about the structure of the data only on their data-fidelity term, which is the first term in Eq. (30). On the other hand, as shown in Eq. (18), both the data-fidelity and penalty terms of our proposed criterion, , contain information about the structure of the data.
The penalty terms of and depend linearly on , while the penalty term of our proposed criterion, , depends on in a non-linear manner. Comparing the penalty terms in Eqs. (31)-(33), has the weakest penalty term. In the asymptotic regime, the penalty terms of and coincide. But, in the finite sample regime, for values of , the penalty term of is stronger than the penalty term of . Note that the penalty term of our proposed criterion, , depends on the number of data vectors in each cluster, , of each candidate model , while the penalty term of the original BIC depends only on the total number of data vectors in the data set. Hence, the penalty term of our proposed criterion might exhibit sensitivities to the initialization of cluster parameters and the associated number of data vectors per cluster.
## Vii Experimental Evaluation
In this section, we compare the cluster enumeration performance of our proposed criterion, given by Eq. (18), with the cluster enumeration methods discussed in Section V, namely and given by Eqs. (26) and (28), respectively, using synthetic and real data sets. We first describe the performance measures used for comparing the different cluster enumeration criteria. Then, the numerical experiments performed on synthetic data sets and the results obtained from real data sets are discussed in detail. For all simulations, we assume that a family of candidate models is given with and , where is the true number of clusters in the data set . All simulation results are an average of Monte Carlo experiments unless stated otherwise. The compared cluster enumeration criteria are based on the same initial cluster parameters in each Monte Carlo experiment, which allows for a fair comparison.
The MATLAB© code that implements the proposed two-step algorithm and the Bayesian cluster enumeration methods discussed in Section V is available at:
https://github.com/FreTekle/Bayesian-Cluster-Enumeration
### Vii-a Performance Measures
The empirical probability of detection , the empirical probability of underestimation , the empirical probability of selection, and the Mean Absolute Error (MAE) are used as performance measures. The empirical probability of detection is defined as the probability with which the correct number of clusters is selected and it is calculated as
pdet=1MCMC∑i=11{^Ki=K}, (34)
where MC is the total number of Monte Carlo experiments, is the estimated number of clusters in the th Monte Carlo experiment, and is an indicator function which is defined as
1{^Ki=K}≜{1,if^Ki=K0,otherwise. (35)
The empirical probability of underestimation is the probability that . The empirical probability of selection is defined as the probability with which the number of clusters specified by each candidate model in is selected. The last performance measure, which is the Mean Absolute Error (MAE), is computed as
MAE=1MCMC∑i=1∣∣K−^Ki∣∣. (36)
### Vii-B Numerical Experiments
#### Vii-B1 Simulation Setup
We consider two synthetic data sets, namely Data-1 and Data-2, in our simulations. Data-1, shown in Fig. 0(a), contains realizations of the random variables , where , with cluster centroids and covariance matrices
Σ1=[0.20.10.10.75],Σ2=[0.50.250.250.5],Σ3=[10.50.51].
The first cluster is linearly separable from the others, while the remaining clusters are overlapping. The number of data vectors per cluster is specified as , , and , where is a constant.
The second data set, Data-2, contains realizations of the random variables , where , with cluster centroids , , , , , , , , , , and covariance matrices
Σ1=[0.25−0.15−0.150.15],Σ2=[0.5000.15],Σi=[0.1000.1],
where . As depicted in Fig. 0(b), Data-2 contains eight identical and spherical clusters and two elliptical clusters. There exists an overlap between two clusters, while the rest of the clusters are well separated. All clusters in this data set have the same number of data vectors.
#### Vii-B2 Simulation Results
Data-1 is particularly challenging for cluster enumeration criteria because it has not only overlapping but also unbalanced clusters. Cluster unbalance refers to the fact that different clusters have a different number of data vectors, which might result in some clusters dominating the others. The impact of cluster overlap and unbalance on and MAE is displayed in Table I. This table shows and MAE as a function of , where is allowed to take values from the set . The cluster enumeration performance of is lower than the other methods because it is designed for spherical clusters with identical variance, while Data-1 has one elliptical and two spherical clusters with different covariance matrices. Our proposed criterion performs best in terms of and MAE for all values of . As increases, which corresponds to an increase in the number of data vectors in the data set, the cluster enumeration performance of and greatly improves, while the performance of deteriorates because of the increase in overestimation. The total criterion (BIC) and penalty term of different Bayesian cluster enumeration criteria as a function of the number of clusters specified by the candidate models for is depicted in Fig. 2. The BIC plot in Fig. 1(a) is the result of one Monte Carlo run. It shows that and have a maximum at the true number of clusters , while overestimates the number of clusters to . As shown in Fig. 1(b), our proposed criterion, , has the second strongest penalty term. Note that, the penalty term of our proposed criterion shows a curvature at the true number of clusters, while the penalty terms of and are uninformative on their own. | 2022-01-22 05:40:41 | {"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": 0.8350427150726318, "perplexity": 439.93046632401985}, "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/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00347.warc.gz"} |
https://fractionshub.com/i-s-i-m-math-2019-solutions/ | # ISI MMath Solutions and Discussion 2019 : PMB
## Problem 1.
Let be a continuous function satisfying
Show that is a bounded function on and attains a maximum or a minimum. Give an example to show that it attains a maximum but not a minimum.
### Indian Statistical Institute, ISI MMath 2019 Problem 1 Hints along with Solution & Discussion
#### Hint 1
Assume that is non-negative, i.e. for all If for all then obviously is bounded and attains its maximal (and minimal) value at each point. Let for some By assumption, there exists such that for all and all Let be the restriction of to is a continuous function on a compact segment.
#### Hint 2
Let be a continuous function satisfying
The function is a non-negative continuous function satisfying
#### Hint 3
For the Example part, consider the function
### Step 1:
Assume that is non-negative, i.e. for all If for all then obviously is bounded and attains its maximal (and minimal) value at each point. Let for some By assumption, there exists such that for all and all Let be the restriction of to is a continuous function on a compact segment. Hence, is bounded and attains its maximal value on i.e. there exists such that We check that the function attains its maximal value at Let Then
Let or Then and
So, for each
is bounded and attains its maximal value.
### Step 2:
Let be a continuous function satisfying
The function is a non-negative continuous function satisfying
By step 1, is bounded and attains its maximal value. It means that there exists such that for all
So, for all
The function is bounded. If then and attains its maximal value. If then and attains its minimal value.
### Step 3:
Consider the function Then is a continuous non-negative function,
In particular, Further,
Function attains its maximal value at However, for all and the function does not attain its minimal value.
## Problem 2.
Let be a continuous function such that Show
that
### Indian Statistical Institute, ISI MMath 2019 Problem 2 Hints along with Solution & Discussion
#### Hint 1
Function is bounded, as it is a continuous function on a compact Let Fix By assumption, there exists such that for all Let be such that
Let and If then
If then
#### Full Solution
Function is bounded, as it is a continuous function on a compact Let Fix By assumption, there exists such that for all Let be such that
Let and If then
If then
In any case,
and
Since is arbitrary, this proves the needed convergence.
## Problem 3.
Let be a twice continuously differentiable function and
suppose If for all then prove that
for all
### Indian Statistical Institute, ISI MMath 2019 Problem 3 Hints along with Solution & Discussion
#### Hint 1
For every we have by the Newton-Leibniz formula
#### Hint 2
Applying the Newton-Leibniz rule again we get
#### Hint 3
In particular, for each we have
#### Full Solution
For every we have by the Newton-Leibniz formula
Applying the Newton-Leibniz rule again we get
In particular, for each we have
## Problem 4.
Suppose is a function defined by
(a). Find all such that has a nonzero directional derivative at with respect to the direction
### Indian Statistical Institute, ISI MMath 2019 Problem 4 Hints along with Solution & Discussion
#### Hint 1
Part a.
Let Then the directional derivative at with respect to the direction is
#### Hint 2
Let Then the directional derivative at with respect to the direction is
#### Hint 3
Part b.
If is continuous at , then for any sequence we have Let
#### Full Solution
Part a.
Let Then the directional derivative at with respect to the direction is
Let Then the directional derivative at with respect to the direction is
has a nonzero directional derivative at with respect to the direction if and only if and
Part b.
If is continuous at , then for any sequence we have Let Then
So, is not continuous at
## Problem 5.
Let be a subset of a compact metric space Assume that for
every continuous function , the restriction of to attains
a maximum on Prove that is compact.
### Indian Statistical Institute, ISI MMath 2019 Problem 5 Hints along with Solution & Discussion
#### Hint 1
A subset of a compact metric space is compact if and only if it is a closed subset. So, it is enough to check that is a closed subset of
#### Hint 2
Let Consider function is a continuous function on
#### Hint 3
Indeed, by the triangle inequality,
So, attains a maximum on . Since it follows that there exists a point such that
#### Full Solution
A subset of a compact metric space is compact if and only if it is a closed subset. So, it is enough to check that is a closed subset of Let Consider function is a continuous function on Indeed, by the triangle inequality,
So, attains a maximum on . Since it follows that there exists a point such that Then and So, and is closed.
## Problem 6.
Let be a non-abelian group of order , where are primes.
(a). How many elements of have order
(b). How many elements of have order
### Indian Statistical Institute, ISI MMath 2019 Problem 6 Hints along with Solution & Discussion
#### Hint 1
The order of an element divides the order of a group. So each element of is of order or Since is non-abelian, it has no elements of order By Cauchy’s theorem, there exists an element of order Then each of the elements
is of order .
#### Hint 2
Assume that is of order and Consider subgroups
and
#### Hint 3
Then is a subgroup of strictly smaller than (as ). The order of is prime, hence Consider cosets
The size of each coset is and there are cosets. hence two of different cosets intersect:
#### Full Solution
The order of an element divides the order of a group. So each element of is of order or Since is non-abelian, it has no elements of order By Cauchy’s theorem, there exists an element of order Then each of the elements
is of order .
Assume that is of order and Consider subgroups
and
Then is a subgroup of strictly smaller than (as ). The order of is prime, hence Consider cosets
The size of each coset is and there are cosets. hence two of different cosets intersect:
We deduce that there exists such that But then which contradicts the fact that is of order
Conclusion: all elements of order are among
(a). There are elements of order
(b). There are elements of order
## Problem 7.
Prove or disprove the following statement: The ring is isomorphic to a product of fields.
### Indian Statistical Institute, ISI MMath 2019 Problem 7 Hints along with Solution & Discussion
#### Hint 1
We prove that is isomorphic to a product of three fields. Consider the factorization
Consider ideals
#### Hint 3
Relations
imply that ideals are pairwise coprime.
Now try to apply the Chinese remainder theorem.
#### Full Solution
We prove that is isomorphic to a product of three fields. Consider the factorization
Consider ideals Relations
imply that ideals are pairwise coprime. By the Chinese remainder theorem,
Each of the multiples in the right-hand side is a field, as it is a quotient of by an irreducible polynomial. In fact,
The ring is isomorphic to a product of three fields
## Problem 8.
Let be a symmetric matrix with real entries such that for some Show that
### Indian Statistical Institute, ISI MMath 2019 Problem 8 Hints along with Solution & Discussion
#### Hint 1
A symmetric real matrix is diagonalizable
#### Hint 2
there exists an invertible matrix and a diagonal matrix such that
Then
#### Full Solution
A symmetric real matrix is diagonalizable: there exists an invertible matrix and a diagonal matrix such that
Then
So,
and and
## Problem 9.
Suppose and are two matrices with real entries such that
the sum of their ranks is strictly less than Show that there exists a
nonzero column vector such that
### Indian Statistical Institute, ISI MMath 2019 Problem 9 Hints along with Solution & Discussion
#### Hint 1
Consider the matrix
The rank of is the maximal number of linearly independent rows of
Then
#### Hint 3
Consequently, columns of are linearly dependent, and there exists a nonzero vector such that
It means that
#### Full Solution
Consider the matrix
The rank of is the maximal number of linearly independent rows of Then
Consequently, columns of are linearly dependent, and there exists a nonzero vector such that
It means that
## Problem 10.
Suppose there are persons in a party. Every pair of persons meet each other with probability independently of the other pairs.
Let be the number of people the person meets in the party. For all with and for all
show that
### Indian Statistical Institute, ISI MMath 2019 Problem 10 Hints along with Solution & Discussion
#### Hint 1
Introduce two events:
Assume the event happens. Then meets and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
#### Hint 2
Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
Finally,
#### Full Solution
Introduce two events:
Assume the event happens. Then meets and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
Finally,
Assume the event happens. Then does not meet and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is
Finally,
We deduce that
### To view ISI MMATH 2016 solutions, hints, and discussions: Click Here
• I have tried my best to explain the solutions. If you have some other solutions, please feel free to add them in the comment section. My website is latex enabled so that you can type all the mathematical equations. If you face any problem understanding the solutions, then comment here, and I will get back to you as soon as I can. Also if you find any mathematical error or any typing error then please do comment here.
• Soumyajit ganguly
Sir,
I am from howrah, West Bengal. There is a kind request to you from my side that if u please teach me as a private tutor for preparation for ISI and CMI, I would be highly obliged. I also like your problems and try to figure it out. But I think it would be more clear if you kind see to my above request.
Regards,
Soumyajit Ganguly
• Please check your email. I have sent you all the details regarding our classes.
• Soumyajit ganguly
Sir
I am from howrah, West Bengal. There is a kind request to you from my side that if u please teach me as a private tutor for preparation for ISI and CMI, I would be highly obliged. I also like your problems and try to figure it out. But I think it would be more clear if you kind see to my above request.
• Please check your email. I have sent you all the details regarding our classes.
• Anish Ray
Hi!
Can you please conduct few mock tests for ISI M.MATH entrance?
• We will try our best to make the mock test course. But currently, we are busy with making some other courses. We will keep you updated.
• Anish Ray
Sir, it would be really very helpful if you can create a mock test course for this year’s entrance.
• Aviral
In 5th question I got it h(x)<h(x○)=1 but I didn't got how u proved then that C is closed
• Showing that C(Bar)=C, that is C contains all it’s limit points. The last line there would be C(bar) but the latex is not taking the symbol bar.
• Lucas
For problem 3, can we use the Taylor’s theorem instead? like for f(x)= f(0) + xf'(0) + (x^2/2)f”(c), for c belongs to (0,x) where for x<0 the case is same because in either way f'(0) = 0 = f(0)
so when we apply |f(x)| =|x^2f"(c)|/2 < 1/2
• Lucas
For problem 3, can we use the Taylor’s theorem instead? like for , for and for , the case is same because in either way
so when we apply
• Lucas
For problem 3, can we use the Taylor’s theorem instead? like for , for and for , the case is same because in either way
so when we apply
• M d sharma
#1) similar sol. for question 3rd:
i.e. Maclaurin’s theorem can we used.
#2) and i think for question no. 10th: probability section i.e. in case of P[B] , there is a calculation mistake …
if not then i have doubt how { (1-p)^ 2n – k – l – 3 } comes?
because i concluded it is not -3 , it would be -4.
#3) for question no. 4th:
can we check limit along the curve y = x^2/3 which can also shows that f is not continuous?
#4) for |G| = pq
if we take a example of o(G) = 10 = 1*2*5
let p=2 < q=5 and both are prime.
and there is an element of order 1 (identity element) and it is non abelian group i.e. no element exist of order 10.
now phi(q) == phi(5) == (5^1-5^0) == 4
that is we can say only four elements are there of order q. And remaining elements could be of order p i.e. there are 4 elements of
order q.
Also q=5 implies that no. of elements of order q = q -1
And no. of elements of order p == o(G) – identity element – (q-1) == (p-1)q
can it be possible ?
if not give the query please.
• M D sharama
sorry for #2)
it was my mistake.
• Lucas
For the 6th problem, It easily follows from the Sylow’s theorem.
No. of q-sylow subgroup is, for some and i.e there exist a unique subgroup of order . so there are elements of order
and we know that every non-identity element in has order either or . So the remaining elements must have order
Therefore the no. of elements having order q is
• Subrata Mahata
Sir,I am from Midnapore college ,W.B and I have decided to appear in ISI.math entrance.But I have no guide.So,I am quite confused about what type of preparation and study is needed.Please suggest me something in this regard.
• Devansh Gupta
Hello SIr, I am preparing for ISI MMath Entrance. Can you please provide my some training for problem solving. Is there any mock test series for ISI MMath
• You will have to join the MMath group here. I teach for MMath entrance exam. No there is no mock tests for ISI MMath.
• Vikas Vaibhav
Great work…
Can you provide your what’s app group?
• Lucas Haobam
For problem 9, Here’s an alternative solution:
if then we are done. Otherwise assume that , also note that then . So we have as dim(Ker (A) \cap ker(B)) =0 but from the given condition we have we have i.e but from the above inequality we get a contradiction. Hence . Therefore the result follows
• Lucas Haobam
For problem 9, Here’s an alternative solution:
if then we are done. Otherwise assume that , also note that then . So we have as but from the given condition we have we have i.e but from the above inequality we get a contradiction. Hence . Therefore the result follows
• Divya Ruhela | 2023-03-26 00:15:52 | {"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": 0.8666192293167114, "perplexity": 1034.6418975424294}, "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/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00481.warc.gz"} |
http://tex.stackexchange.com/questions/181384/diacritics-above-dotted-i-i-with-tittle | # Diacritics above dotted i (i with tittle)
It is taught that, in order to place an accent mark on top of the letter "i", one should use \i to remove the tittle (the dot) first. (That the dot may need to be removed is of course a matter of convention. To indicate vowel length in Latin, people place a macron above the "i" with or without tittle; that is, I have encountered both ways of doing things.)
However, strangely the accent commands \', \, ^, \", and \. seem to remove the tittle automatically:
\documentclass{article}
\begin{document}
\'{i} \{i} \^{i} \"{i} \.{i} \~{i} \={i} \H{i} \r{i} \v{i} \u{i}
\'{\i} \{\i} \^{\i} \"{\i} \.{\i} \~{\i} \={\i} \H{\i} \r{\i} \v{\i} \u{\i}
\end{document}
1. Since when has it been the case that an accent command in (La)TeX removes the tittle from the letter "i" in those 5 cases? (And are there any special cases I left out?)
2. How can I avoid this? That is, I would like the accent to be placed above the letter i with tittle.
Interestingly, the tittle is left intact for the letter "j":
-
If you have an OpenType font with the corresponding anchors, you might get get the desired output with XeLaTeX using Unicode’s combining diacritical marks (e.g., i + 2×U+0307). However, I do not have such a font at hand right now to test this. – Wrzlprmft May 31 '14 at 8:47
Re your last edit to my answer: BibTeX will accept whatever syntax is allowed by LaTeX. – egreg Jun 1 '14 at 13:33
@egreg I assume this comment of yours refers to \DeclareTextComposite/\DeclareTextCompositeCommand declarations. (The reason for my edit was to clarify that the user can substitute \×\i by \×i (× ∈ {., , ', ^, "}) also in BibTeX. The BibTeX requirement to write {\×<letter>} instead of just \×<letter> may make the user wonder about alphabetization of \×\i vs \×i; he may be asking: "Where else does BibTeX require special treatment?") – Lover of Structure Jun 1 '14 at 14:40
In the definition files for output encodings, you can find the defined combinations that are substituted with precomposed accented characters; for instance, in t1enc.def we find
\DeclareTextComposite{\.}{T1}{i}{\i}
\DeclareTextComposite{\.}{T1}{\i}{\i}
\DeclareTextComposite{\}{T1}{i}{236}
\DeclareTextComposite{\}{T1}{\i}{236}
\DeclareTextComposite{\'}{T1}{i}{237}
\DeclareTextComposite{\'}{T1}{\i}{237}
\DeclareTextComposite{\^}{T1}{i}{238}
\DeclareTextComposite{\^}{T1}{\i}{238}
\DeclareTextComposite{\"}{T1}{i}{239}
\DeclareTextComposite{\"}{T1}{\i}{239}
so that
\.i \i \'i \^i \"i
are completely equivalent to
\.{\i} \{\i} \'{\i} \^{\i} \"{\i}
Here's the corresponding code in ot1enc.def, where just the dot above accent corresponds to a precomposed glyph (of course the normal dotted ‘i’):
\DeclareTextComposite{\.}{OT1}{i}{\i}
\DeclareTextComposite{\.}{OT1}{\i}{\i}
\DeclareTextCompositeCommand{\}{OT1}{i}{\@tabacckludge\i}
\DeclareTextCompositeCommand{\'}{OT1}{i}{\@tabacckludge'\i}
\DeclareTextCompositeCommand{\^}{OT1}{i}{\^\i}
\DeclareTextCompositeCommand{\"}{OT1}{i}{\"\i}
Note that those predefined combinations in t1enc.def correspond exactly to the available precomposed accented letters in a T1 encoded font. For the other accent, the combination is not defined, so one needs \i in order the letter loses the dot.
Nobody, however, prevents you from defining your own composites:
\documentclass{article}
\DeclareTextCompositeCommand{\~}{OT1}{i}{\~\i}
\DeclareTextCompositeCommand{\=}{OT1}{i}{\=\i}
\DeclareTextCompositeCommand{\H}{OT1}{i}{\H\i}
\DeclareTextCompositeCommand{\r}{OT1}{i}{\r\i}
\DeclareTextCompositeCommand{\v}{OT1}{i}{\v\i}
\DeclareTextCompositeCommand{\u}{OT1}{i}{\u\i}
\begin{document}
\'{i} \{i} \^{i} \"{i} \.{i} \~{i} \={i} \H{i} \r{i} \v{i} \u{i}
\'{\i} \{\i} \^{\i} \"{\i} \.{\i} \~{\i} \={\i} \H{\i} \r{\i} \v{\i} \u{\i}
\end{document}
Why are just some combinations involving ‘i’ defined and not all of them? Because those are the most common, for they correspond to glyphs actually in the encoding, whereas the uncommon ones would have wasted precious memory (remember that when LaTeX2e was released, computers were quite different from the current machines).
How to “undefine” those combinations with ‘i’? One has to know that \DeclareTextCompositeCommand{\"}{OT1}{i}{\"\i} defines the macro
\\OT1\"-i
to expand to \"\i, while \DeclareTextComposite{\.}{OT1}{i}{\i} defines
\\OT1\.-i
to expand to \char\i (that is, ‘print an i’). LaTeX tests for the existence of this macro when trying an accent. Note the peculiar name, where the first backslash denotes the escape character, whereas the inner ones are just characters. You can define a function for undefining the combinations you want to exclude:
\documentclass{article}
\newcommand{\UndeclareTextComposite}[3]{%
\expandafter\let\csname\expandafter\string\csname #2\endcsname\string#1-\string#3\endcsname\relax
}
\UndeclareTextComposite{\'}{OT1}{i}
\UndeclareTextComposite{\}{OT1}{i}
\UndeclareTextComposite{\^}{OT1}{i}
\UndeclareTextComposite{\"}{OT1}{i}
\UndeclareTextComposite{\.}{OT1}{i}
\begin{document}
\'{i} \{i} \^{i} \"{i} \.{i} \~{i} \={i} \H{i} \r{i} \v{i} \u{i}
\'{\i} \{\i} \^{\i} \"{\i} \.{\i} \~{\i} \={\i} \H{\i} \r{\i} \v{\i} \u{\i}
\end{document}
Regarding the question about “since when are those combination defined”, I could find nothing in the LaTeX sources, so my guess is that they have been there from the start of LaTeX2e. I remember a conversation with Claudio Beccari who claimed to have insisted with the LaTeX team for their inclusion; this addition possibly happened before the release of LaTeX2e, when the business of font encoding was being developed.
Finally, note that BibTeX accepts {\×i} and {\×\i} (where \× denotes one of the accent commands \., \, \', \^, \") equally. (adapted from a comment elsewhere)
-
I cannot answer your first question, but if you really need those combinations of diacritic marks you can get them with the tipa package:
\documentclass{article}
\usepackage{tipa}
\usepackage{times}
\begin{document}
\huge
\'{\.{i}}
\{\.{i}}
\textcircumdot{\i}
\"{\.{i}}
\.{\.{\i}}
\end{document}
However, I would say it is a feature rather than a bug that they are omitted by default, as they have no conventional usage (that I know of).
-
The question is how those characters are used. I do not know a language which uses \"i with three dots. However, in math mode you can set the accents:
$\dot{\textrm{i}} \grave{\textrm{i}} \check{\textrm{i}} \breve{\textrm{i}}$ \ldots
`
- | 2016-06-25 12:22:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9239079356193542, "perplexity": 6233.406941185681}, "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/1466783393146.70/warc/CC-MAIN-20160624154953-00135-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=42&t=52888 | ## Electron Configuration and Hybridization
$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$
Cavalli_1H
Posts: 92
Joined: Thu Sep 19, 2019 12:15 am
### Electron Configuration and Hybridization
How do electron configurations and hund's rule come to play when figuring out hybridization?
Kevin Liu 1J
Posts: 53
Joined: Thu Jul 25, 2019 12:16 am
### Re: Electron Configuration and Hybridization
I'm not too sure what the connection is between these and hybridization. The most direct connection hybridization has is with molecular shape/structure.
Ashley R 1A
Posts: 41
Joined: Sat Aug 24, 2019 12:16 am
### Re: Electron Configuration and Hybridization
Hund's rule is also important as it states that each sublevel must have one electron occupying the orbital before the orbital is occupied by 2 electrons. This ensures that hybridization occurs because if hund's rule were not followed, hybridization may not occur if only p orbitals were filled and not s orbitals. | 2020-08-14 11:56:58 | {"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": 1, "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": 0.4721655249595642, "perplexity": 4943.782851388304}, "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-2020-34/segments/1596439739211.34/warc/CC-MAIN-20200814100602-20200814130602-00394.warc.gz"} |
https://mathoverflow.net/questions/286326/a-uniform-version-of-minkowski-content | # A uniform version of Minkowski content?
Let $A\subseteq [-1,1]^d$ be a measurable set and $\mu$ be the Lebesgue measure. For any $\delta>0$, define $A_\delta := \{x: d(x,A)\leq \delta\}$, where $d(x,A) := \inf_{y\in A}\|x-y\|_2$.
The upper and lower Minkowski's content (of order $d-1$) is defined as $$\overline M(A) := \limsup_{\delta\to 0^+}\frac{\mu(A_\delta)}{\delta} \;\;\;\;\;\text{and}\;\;\;\;\; \overline M(A) := \liminf_{\delta\to 0^+}\frac{\mu(A_\delta)}{\delta}.$$ In particular, if $A$ is an isotropic $\ell_2$ ball $\{x: \|x\|_2\leq a\}$, then $0<\overline M(A)=\underline M(A)<\infty$.
My question: I wonder whether an uniform version of the Minkowski's content is available. In particular, let $\mathcal A$ be a collection of measurable sets in $[-1,1]^d$, and for example purposes take $\mathcal A$ to be $\ell_2$ balls of all radius (i.e., $A\in\mathcal A$ if $A=\{x:\|x|_2\leq a\}$ for some $0<a< 1$). Can we define something like $$\limsup_{\delta\to 0^+}\sup_{A\in\mathcal A}\frac{\mu(A_\delta)}{\delta}$$ and show that it is finite?
My preliminary calculation shows that the above definition cannot work, but if we consider $$\limsup_{\delta\to 0^+}\sup_{A\in\mathcal A}\frac{\mu(A_\delta)}{\delta\cdot \mu(A)^{\frac{d-1}{d}} + \delta^d}$$ then the quantity is finite. I wonder whether there is an existing concept/work that could characterize this uniform behavior of Minkowski's content? | 2020-12-04 14:24:39 | {"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": 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": 0.988757848739624, "perplexity": 127.7015677208406}, "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/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00640.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/120762-finite-field-gf-2-4-multiplication-using-normal-basis-print.html | # finite field GF(2^4) multiplication using normal basis
• December 16th 2009, 05:04 AM
classic_phohe
finite field GF(2^4) multiplication using normal basis
Hi, help needed in solving finite field multiplication specifically for normal basis representation and in the field of GF(2^4)
say I have two elements:
$(\Gamma_1) Z^4 + (\Gamma_0)Z$ and $(\Delta_1) Z^4 + (\Delta_0)Z$
and the respective field polynomial:
$r(y) = y^2 + {10}_2 y + 1$
how do I perform multiplication of the two elements??? | 2014-10-21 18:54:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "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": 0.9186245799064636, "perplexity": 3716.514020218921}, "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-2014-42/segments/1413507444657.46/warc/CC-MAIN-20141017005724-00306-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/prove-the-quotient-theorem-using-the-limit-definition.9536/ | # Prove the quotient theorem using the limit definition?
1. Nov 24, 2003
### franz32
Hello guys!
I'm new here! Well, it feels like this forum is cool and interesting!
Can anyone help me here? =)
How do you prove the quotient theorem using the limit definition?
(Given a limit of f of x as x approaches a is A and a limit of g of x as x approaces a is B).
2. Nov 24, 2003
### StephenPrivitera
$$\begin{equation*}\begin{split} \lim_{h\rightarrow 0} \frac {\frac {f(x+h)} {g(x+h)} - \frac {f(x)} {g(x)}} {h} &= \lim_{h\rightarrow 0} \frac {f(x+h)g(x) - f(x)g(x+h)} {hg(x+h)g(x)} \\ &= \lim_{h\rightarrow 0} \frac {f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)} {hg(x+h)g(x)} \end{split}\end{equation*}$$
You can take it from here. Be careful. How do you know
$$\lim_{h\rightarrow0} g(x+h)=g(x)$$?
Last edited: Nov 24, 2003
3. Nov 25, 2003
### franz32
Thank you... =)
Hello. =)
Well, I think I can take it from here. If I have doubts,
I would probably want to clarify it.
Anyway, thank you very much! | 2018-08-17 17:20:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.840410590171814, "perplexity": 2345.3167885162693}, "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-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00034.warc.gz"} |
http://mathoverflow.net/feeds/question/40268 | Why is the Gaussian so pervasive in mathematics? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:47:30Z http://mathoverflow.net/feeds/question/40268 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics Why is the Gaussian so pervasive in mathematics? Randy Qian 2010-09-28T05:06:52Z 2012-11-14T09:17:26Z <p>This is a heuristic question that I think was once asked by Serge Lang. The gaussian: $e^{-x^2}$ appears as the fixed point to the Fourier transform, in the punchline to the central limit theorem, as the solution to the heat equation, in a very nice proof of the Atiyah-Singer index theorem etc. Is this an artifact of the techniques (such as the Fourier Transform) that people like use to deal with certain problems or is this the tip of some deeper platonic iceberg?</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40288#40288 Answer by Bjørn Kjos-Hanssen for Why is the Gaussian so pervasive in mathematics? Bjørn Kjos-Hanssen 2010-09-28T09:01:02Z 2010-09-28T17:13:33Z <p><em>The sum of two independent normal random variables is again normal</em>, i.e., the shape of the distribution is unchanged under addition except for stretching and scaling. Moreover, the normal distribution is unique among distributions with finite variance in having this property. Many phenomena in nature come from adding together various independent or almost independent terms. Therefore, we would expect the normal distribution to show up a lot in nature-inspired mathematics.</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40371#40371 Answer by Kevin O'Bryant for Why is the Gaussian so pervasive in mathematics? Kevin O'Bryant 2010-09-28T19:13:49Z 2010-09-28T19:23:47Z <p>I recently came across a strange and beautiful connection between the Gaussian $e^{-x^2}$ and the method of least squares. It turns out that the square in $e^{-x^2}$ and the square in least squares'' is the same square.</p> <p>Let $(x_i,y_i)$ (with $1\leq i \leq n$) be the data set, and assume that for each $x$, the $y$'s are normally distributed with mean $\mu(x)=\alpha x+\beta$ and variance $\sigma^2$. Then, the likelihood of generating our data (assuming that the data points are independent) is $$\prod_{i=1}^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(y_i-\mu(x_i))^2}{2\sigma^2}\right) =\left(\frac{1}{\sqrt{2\pi \sigma^2}}\right)^n \exp\left( \frac{-1}{2\sigma^2} \sum_{i=1}^n (y_i - \alpha x+\beta)^2 \right)$$ We would obviously want to choose the parameters $\alpha,\beta$ so that the likelihood is maximized, and this is accomplished by minimizing $$\sum_{i=1}^n (y_i - \alpha x+\beta)^2.$$ In other words, the least squares approximation is the one that makes the data set most likely to happen.</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40382#40382 Answer by Terry Tao for Why is the Gaussian so pervasive in mathematics? Terry Tao 2010-09-28T20:39:42Z 2010-09-28T20:47:01Z <p>Quadratic (or bilinear) forms appear naturally throughout mathematics, for instance via inner product structures, or via dualisation of a linear transformation, or via Taylor expansion around the linearisation of a nonlinear operator. The Laplace-Beltrami operator and similar second-order operators can be viewed as differential quadratic forms, for instance.</p> <p>A Gaussian is basically the multiplicative or exponentiated version of a quadratic form, so it is quite natural that it comes up in multiplicative contexts, especially on spaces (such as Euclidean space) in which a natural bilinear or quadratic structure is already present.</p> <p>Perhaps the one minor miracle, though, is that the Fourier transform of a Gaussian is again a Gaussian, although once one realises that the Fourier kernel is also an exponentiated bilinear form, this is not so surprising. But it does amplify the previous paragraph: thanks to Fourier duality, Gaussians not only come up in the context of spatial multiplication, but also frequency multiplication (e.g. convolutions, and hence CLT, or heat kernels).</p> <p>One can also take an adelic viewpoint. When studying non-archimedean fields such as the p-adics $Q_p$, compact subgroups such as $Z_p$ play a pivotal role. On the reals, it seems the natural analogue of these compact subgroups are the Gaussians (cf. Tate's thesis). One can sort of justify the existence and central role of Gaussians on the grounds that the real number system "needs" something like the compact subgroups that its non-archimedean siblings enjoy, though this doesn't fully explain why Gaussians would then be exponentiated quadratic in nature.</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40457#40457 Answer by S. Carnahan for Why is the Gaussian so pervasive in mathematics? S. Carnahan 2010-09-29T11:47:07Z 2010-09-29T11:47:07Z <p>This is just a minor amplification of one of Terry Tao's points. For any prime $p$, the ring <code>$\mathbb{Z}_p$</code> of <a href="http://en.wikipedia.org/wiki/P-adic_number" rel="nofollow">$p$-adic integers</a> forms an open compact additive subgroup of <code>$\mathbb{Q}_p$</code>, the completion of $\mathbb{Q}$ under the p-adic metric, and its characteristic function should be viewed as a p-adic analogue of the Gaussian. It displays many analogues of the nice properties we see from the Gaussian:</p> <ol> <li>It is smooth (in the sense that the smooth functions on totally disconnected spaces are defined as the locally constant functions, but this isn't completely tautological, since this class of functions turns out to be useful).</li> <li>It is taken to itself under the $p$-adic Fourier transform when normalizations are chosen appropriately.</li> <li>It obeys something like a central limit theorem. For example, if you flip lots of coins, and ask for the number of heads mod $p^n$, you will, for sufficiently long trials, get a distribution that is close to uniform. It sounds like there could be a way to interpret this sort of convolution in terms of heat flowing, but I don't know a precise statement.</li> </ol> <p>The situation with the real line is more complicated because it is connected but not compact, and therefore has no open compact subgroups. There is a maximal compact multiplicative monoid (occasionally called the "ring of integers of $\mathbb{R}$" informally), given by the closed interval $[-1,1]$. You can think of the Gaussian as a smoothing of the uniform distribution on $[-1,1]$, but it is not clear to me that this particular analogy is very fruitful.</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40480#40480 Answer by Alex Bloemendal for Why is the Gaussian so pervasive in mathematics? Alex Bloemendal 2010-09-29T15:08:55Z 2010-09-29T15:08:55Z <p>I'm not an expert, but I believe Stein's method gives a more satisfying connection between the CLT and the heat equation, in particular one that does not involve the Fourier transform. Stein's characterization of the normal distribution and convergence to normality involves an operator closely related to the generator of the Ornstein-Uhlenbeck process. On the other hand, the fact that latter has the Gaussian as its invariant measure can be obtained by a trivial transformation of the fundamental solution of the heat equation.</p> http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/112370#112370 Answer by jbc for Why is the Gaussian so pervasive in mathematics? jbc 2012-11-14T09:17:26Z 2012-11-14T09:17:26Z <p>One of the reasons for the ubiquity of the Gaussian is displayed in what is probably the most electrifying half page of scientific prose ever written---Maxwell's argument that the distribution of the velocities of molecules in the ideal gas is Gaussian (now known as the Maxwell-Boltzmann distribution). The only physical assumptions used are that the density function depends only on the absolute value of the velocity (and not the direction) and that the components in the directions of the coordinate axes are statistically independent. Mathematically, ths means that the only functions in $3$-space which depend only on the distance $r$ from the origin and which split as the product of three functions of one variable are those of the form $ae^{br^2}$. Maxwell does this by inspection but it is easy to give a rigorous proof (under very weak smoothness conditions) and the result holds, of course, in any dimension greater then or equal to $2$. Maxwell's reasoning can be found in his collected papers, or, more accesssibly, in Hawking's anthology "On the Shoulders of Giants".</p> | 2013-05-18 21:47:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8859426975250244, "perplexity": 410.11190391144936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696382892/warc/CC-MAIN-20130516092622-00054-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.techwhiff.com/issue/plzzz-help-i-beg-look-at-picture-firsttt--445094 | # PLZZZ help i beg look at picture firsttt
###### Question:
PLZZZ help i beg look at picture firsttt
### In the play The Diary of Anne Frank, which event interrupts the Hanukkah celebration in the Secret Annex? 1. a fight between Mr. Dussel and Peter 2. a noise from the warehouse downstairs 3. Mrs. Frank becoming ill from eating rotten potatoes 4. Mr. Frank telling everyone to be quiet or they'll be discovered please help! I think it is the last one but I am not sure?
In the play The Diary of Anne Frank, which event interrupts the Hanukkah celebration in the Secret Annex? 1. a fight between Mr. Dussel and Peter 2. a noise from the warehouse downstairs 3. Mrs. Frank becoming ill from eating rotten potatoes 4. Mr. Frank telling everyone to be quiet or they'll be di...
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A yearbook printer charges based on the number of pages printed. Here is a table that shows the cost of some recent yearbooks. Numbers of pages: 50, 100, 150, 200 Cost:-------------------> 1.55, 2.25, 2.95, 3.65 which equation gives the cost, y, in terms of the number of pages, x? y=0.014x+49.15 ...
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Factor the GCF: −6x4y5 − 15x3y2 + 9x2y3...
### Max fischer is a bookkeeper his annual trip insurance cost 11700 his employer
max fischer is a bookkeeper his annual trip insurance cost 11700 his employer... | 2022-09-25 10:20:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36319026350975037, "perplexity": 2714.6026937789657}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334528.24/warc/CC-MAIN-20220925101046-20220925131046-00472.warc.gz"} |
http://tex.stackexchange.com/questions/175690/what-is-the-tikz-equivalent-for-offset-2ab | # What is the TikZ equivalent for ([offset=2]{A}B)?
Based on the answer given here, ($(A)!1cm!(B)$) in TikZ is equivalent to ([nodesep=1]{B}A) in PSTricks. It is just about the "radial" distance specified by two points. How about the "transverse" distance specified by two points?
More precisely, in PSTricks, the syntax ([offset=2]{A}B) represents a new point (namely C) which is 2 unit apart from the line joining A and B such that BC is perpendicular to AB.
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(6,6)
\pstGeonode(1,1){A}(4,5){B}
\pcline[nodesep=-1](A)(B)
\pstGeonode([offset=2]{A}B){C}
\psline[linecolor=blue](B)(C)
\end{pspicture}
\end{document}
What is the TikZ equivalent for such a syntax?
## Miscellaneous
([offset=2]{A}B) can also be written as ([nodesep=2,angle=90]{A}B). Other approaches are also possible with, for example, transformations provided by pst-eucl.
-
The equivalent syntax is ($(B)!2cm!90:(A)$), which is the point 2cm away from B on the line from B to A after that line has been rotated by 90 degrees.
\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (1,1);
\coordinate (B) at (4,4);
\coordinate (C) at ($(B)!2cm!90:(A)$);
\node at (A) [above] {A};
\node at (B) [above] {B};
\node at (C) [above] {C};
\end{tikzpicture}
\end{document}
-
With tkz-euclide package, using syntax \tkzDefPointWith[orthogonal normed, K=2](B,A) \tkzGetPoint {C} define a point C from B perpendicular to AB with 2cm length.
\documentclass{standalone}
\usepackage{tikz,tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}
\tkzInit
\tkzDefPoints{0/0/O, 5/5/K}
\tkzDefPoints{1/1/A, 4/4/B}
\tkzDrawPoint[color=red](A)
\tkzDrawPoint[color=red](B)
\tkzLabelPoints[above](A,B)
\tkzDefPointWith[orthogonal normed, K=2](B,A) \tkzGetPoint {C} | 2014-12-19 23:44:19 | {"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": 0.9735405445098877, "perplexity": 2501.788430711496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768982.112/warc/CC-MAIN-20141217075248-00115-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.groundai.com/project/approximating-the-mean-of-a-truncated-normal-distribution/2 | On the efficient estimation of the mean of multivariate truncated normal distributionsThis research has been co-financed by the European Union (European Social Fund - ESF) and Greek national funds through the Operational Program “Education and Lifelong Learning” of the National Strategic Reference Framework (NSRF) - Research Funding Program: ARISTEIA- HSI-MARS-1413.
# On the efficient estimation of the mean of multivariate truncated normal distributions††thanks: This research has been co-financed by the European Union (European Social Fund - ESF) and Greek national funds through the Operational Program “Education and Lifelong Learning” of the National Strategic Reference Framework (NSRF) - Research Funding Program: ARISTEIA- HSI-MARS-1413.
Konstantinos D. Koutroumbas Institute for Astronomy, Astrophysics, Space Applications and Remote sensing, National Observatory of Athens, I. Metaxa and Vas. Paulou, Penteli, GR-152 36, Greece, Tel. No: +30-210-8109189, Fax No: +30-210-6138343, E-mail: koutroum@noa.gr. Konstantinos E. Themelis Athanasios A. Rontogiannis
23 January 2014
###### Abstract
A non trivial problem that arises in several applications is the estimation of the mean of a truncated normal distribution. In this paper, an iterative deterministic scheme for approximating this mean is proposed, motivated by an iterative Markov chain Monte Carlo (MCMC) scheme that addresses the same problem. Conditions are provided under which it is proved that the scheme converges to a unique fixed point. The quality of the approximation obtained by the proposed scheme is assessed through the case where the exponential correlation matrix is used as covariance matrix of the initial (non truncated) normal distribution. Finally, the theoretical results are also supported by computer simulations, which show the rapid convergence of the method to a solution vector that (under certain conditions) is very close to the mean of the truncated normal distribution under study.
Keywords: Truncated normal distribution, contraction mapping, diagonally dominant matrix, MCMC methods, exponential correlation matrix
## I Introduction
A non trivial problem that appears in several applications, e.g. in multivariate regression and Bayesian statistics, is the estimation of the mean of a truncated normal distribution. The problem arises in cases where a random vector follows a normal distribution with mean and covariance matrix , denoted by , but is restricted to a closed subset of . However, the restrictions considered in most of the cases are of the form , with , . If and ( and ), we have a one-sided truncation from the left (right), while in the case where both and are reals, we have two-sided truncation.
The problem has attracted the attention of several researchers from the sixties. Since then, several deterministic approaches have been proposed with some of them trying to estimate not only the mean but also other moments (such as the variance) of a multivariate truncated normal distribution. These approaches can be categorized according to whether they are suitable for one-sided truncated normal distributions ([19], [22], [2], [7], [9]) or for two-sided truncated normal distributions ([21], [17], [6], [8], [13], [10]), or according to whether they consider the bivariate, , ([19], [21], [17], [13]) or the multivariate case ([22], [2], [6], [7], [8], [9], [10]). In addition, some of these methods put additional restrictions to the distribution paramaters (e.g. [22], [2], [7] require that ). Most of these methods either perform direct integration (e.g. [22]) or they utilize the moment generating function tool (e.g. [10]).
An alternative strategy to deal with this problem has been followed in [18]. More specifically, in [18], a Markov Chain Monte Carlo (MCMC) iterative scheme has been developed. According to this, at each iteration, succesive samplings take place, one from each one-dimensional conditionals of the truncated normal distribution. The mean of the -th such distribution is the mean of conditioned on the . After performing several iterations, the estimation of the mean of the truncated normal results by performing an averaging over the produced samples. Convergence issues of this scheme to the mean of the truncated normal distribution are a subject of the Markov chain theory. A relative work that accelerates the method in [18] is exhibited in [5].
The work presented in this paper for approximating the mean of a multivariate truncated normal distribution has been inspired from that of [18]. Specifically, instead of selecting a sample from each one of the above one-dimensional distributions, we select its mean. Thus, the proposed scheme departs from the statistical framework adopted in [18] and moves to the deterministic one.
This work is an extension of a relative scheme used in [23] in the framework of spectral unmixing in hyperspectral images. In addition, a convergence proof of the proposed scheme is given when certain conditions are fulfilled. The quality of the approximation of the mean offered by the proposed method is assessed via the case where is the exponential correlation matrix. Experimental results show that the new scheme converges significantly faster than the MCMC approach of [18].
The paper is organized as follows. Section 2 contains some necessary definitions and a brief description of the work in [18]. In Section 3, the newly proposed method is described and in Section 4 conditions are given under which it is proved to converge. In Section 5, the proposed method is applied to the case where is the exponential correlation matrix. In Section 6 simulation results are provided and a relevant discussion is presented. Finally, Section 7 concludes the paper.
## Ii Preliminaries and previous work
Let us consider the -dimensional normal distribution
N(\boldmathx|\boldmathμ,Σ)=1(2π)n/2|Σ|1/2exp(−12(\boldmathx−\boldmathμ)TΣ−1(\boldmathx−\boldmathμ)) (1)
where the -dimensional vector is its mean and the matrix is its covariance matrix.
Let be a subset of with positive Lebesgue measure. We denote by the truncated normal distribution which results from the truncation of in . Speaking in mathematical terms
NRn(\boldmathx|\boldmathμ,Σ)=⎧⎪ ⎪⎨⎪ ⎪⎩exp(−12(\boldmathx−% \boldmathμ)TΣ−1(\boldmathx−\boldmathμ))∫Rnexp(−12(\boldmathx−\boldmathμ)TΣ−1(\boldmathx−\boldmathμ))d% \boldmathw,if \boldmathx∈Rn0,otherwise (2)
Note that is proportional to , where , if and , otherwise.
In the scheme discussed in [18], a Markov Chain Monte Carlo (MCMC) method is proposed, to compute the mean of single or doubly truncated (per coordinate) normal distribution , where . The method relies on the sampling of the one-dimensional conditionals of the truncated normal distribution. More specifically, letting and denoting the expectation and the variance of conditioned on the rest coordinates, respectively111That is, follows the (non-truncated) -th conditional of ., and denoting the (one dimensional) truncated normal distribution which results from the truncation of a normal distribution with mean and variance in , the iterative sampling scheme proposed in [18] can be written as
1.x(t)1∼N[a1,b1](E[x1|x(t−1)2,…,x(t−1)n],σ∗21)2.x(t)2∼N[a2,b2](E[x2|x(t)1,x(t−1)3…,x(t−1)n],σ∗22)⋮n.x(t)n∼N[an,bn](E[xn|x(t)1,x(t)2…,x(t)n−1],σ∗2n) (3)
where denotes the sampling action and denotes the current iteration. After performing several, say , iterations (and after discarding the first few, say , ones) the mean of each coordinate is estimated as
E[xi]=1k−k′k∑t=k′x(t)i, i=1,…,n
The quantities and of each one of the above one dimensional conditionals are expressed in terms of the parameters and of the non-truncated multidimensional normal distribution as follows
μ∗i=μi+\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmathμ¬i), i=1,…,n (4)
σ∗i=σii−\boldmathσT¬iΣ−1¬i¬i\boldmathσ¬i, i=1,…,n (5)
with being the matrix that results from after removing its -th column and its -th row, being the -th column of excluding its -th element and , being the (-dimensional) vectors that result from and , respectively, after removing their -th coordinates, and , respectively. Note that depends on all ’s except .
## Iii The proposed model
In the sequel, we focus on the case where is a set of the form , where for each interval it is either, (i) and or (ii) and . This means that, along each dimension, the truncation is one-sided and more specifically, case (i) corresponds to right truncation while case (ii) corresponds to left truncation.
The proposed model for estimating the mean of , is of iterative nature and, at each iteration, it requires the computation of the (one-dimensional) function. This model has a close conceptual affinity with the one (briefly) presented in the previous section ([18]). More specifically, instead of utilizing the samples produced by the (one dimensional) distributions , we use the corresponding mean values (here denoted by ).
As it is well known (see e.g. [16]), the mean of a truncated one dimensional normal distribution , which has resulted from the (non-truncated) normal distribution with mean and variance , is expressed as
w=μ∗+ϕ(a−μ∗σ∗)−ϕ(b−μ∗σ∗)Φ(b−μ∗σ∗)−Φ(a−μ∗σ∗)σ∗ (6)
where , and .
However, since in the present paper we consider the cases where either (i) and or (ii) and , let us see now how eq. (6) becomes for each of these cases.
(i) , . In this case it is and as a consequence, , . Thus, taking also into account the definitions of and and the fact that , eq. (6) gives,
w=μ∗+√2πe−(μ∗−a)22σ∗2erfc(−μ∗−a√2σ∗)σ∗ (7)
(ii) , . In this case it is and, as a consequence, , . Working as in case (i), eq. (6) gives
w=μ∗−√2πe−(b−μ∗)22σ∗2erfc(−b−μ∗√2σ∗)σ∗ (8)
Eqs. (7) and (8) can be expressed compactly via the following single equation
w=μ∗+√2π[f(μ∗−a√2σ∗)Ia∈R−f(b−μ∗√2σ∗)Ib∈R]σ∗ (9)
where () is an indicator function which equals to if () and otherwise and
f(x)=e−x2erfc(−x) (10)
Let us now return to the multidimensional case. Since from the (truncated) conditional one-dimensional normals we no longer perform sampling but, instead, we consider their means, eq. (4) is altered to
μ∗i=μi+\boldmathσT¬iΣ−1¬i¬i(\boldmathw¬i−\boldmathμ¬i), i=1,…,n (11)
where results from the current estimate of the ( - dimensional) mean vector of the truncated normal distribution, after removing its -th coordinate.
Putting now all the previous ingredients together (that is, utilizing eqs. (9), (11) and (5)) we obtain the following iterative scheme
(12)
with being computed via eq. (11), where (the only parameter in (11) that varies through iterations) is defined as , that is, the most recent information about ’s is utilized. More formally, we can say that the above scheme performs sequential updating and (following the terminology used in [3]) it is a Gauss Seidel updating scheme.
It is reminded that, due to the type of truncation considered here (only left truncation or only right truncation per coordinate), the bracketed expression in each equation of (12) contains only one non identically equal to zero term. In the sequel, we consider separately the cases corresponding to and , i.e.,
w(t)i=μ∗(t)i+√2πf(A(t)i)σ∗i (13)
and
w(t)i=μ∗(t)i−√2πf(B(t)i)σ∗i (14)
where
A(t)i=μ∗(t)i−ai√2σ∗i (15)
and
B(t)i=bi−μ∗(t)i√2σ∗i (16)
, respectively, and is defined as in eq. (10).
## Iv Convergence issues
In this section we provide sufficient conditions under which the proposed scheme is proved to converge222However, it is noted that even if these conditions are slightly violated, the algorithm still works, as is verified by the experiments presented in Section 5.. Before we proceed, we give some propositions and remind some concepts that will be proved useful in the sequel.
Proposition 1: Assume that is a symmetric positive definite matrix, is the -th column of , after removing its -th element and results from after removing its -th row and its -th column. Also, let be the element of and be the -dimensional vector resulting from the -th row of after (i) removing its -th element, , and (ii) multiplying the remaining elements by . Then, it holds
(i) and
(ii) .
The proof of this proposition is straightforward from the inversion lemma for block partitioned matrices ([20], p. 53) and the use of permutation matrices, in order to define the Schur complement for each row of .
Proposition 2: It is
−√π≤f′(x)≤0,∀x∈R (17)
where denotes the derivative of , which is defined in eq. (10).
The proof of proposition 2 is given in the appendix. Definition 1: A mapping , where , is called contraction if for some norm there exists some constant (called modulus) such that
||T(\boldmathx)−T(\boldmathy)||≤α||% \boldmathx−\boldmathy||, ∀\boldmathx,%\boldmath$y$∈X (18)
The corresponding iteration is called contracting iteration.
Proposition 3 ([3], pp. 182-183): Suppose that is a contraction with modulus and that is a closed subset of . Then
(a) The mapping has a unique fixed point 333A point is called fixed point of a mapping if it is ..
(b) For every initial vector , the sequence , generated by converges to geometrically. In particular,
||\boldmathx(t)−\boldmathx∗||≤αt||% \boldmathx(0)−\boldmathx∗||, ∀t≥0
Let us define the mappings , as
Ti(\boldmathx) =μ∗i(\boldmathx)+ +√2π[f(μ∗i(\boldmathx)−ai√2σ∗i)Iai∈R−f(bi−μ∗i(\boldmathx)√2σ∗i)Ibi∈R]σ∗i (19)
where
μ∗i(\boldmathx)=μi+\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmath% μ¬i) (20)
and
Ai(\boldmathx)=μ∗i(\boldmathx)−ai√2σ∗i,Bi(\boldmathx)=bi−μ∗i(\boldmathx)√2σ∗i (21)
and the mapping as
T(\boldmathx)=(T1(\boldmathx),…,Tn(% \boldmathx))
Let us define next the mapping as
^Ti(\boldmathx)=^Ti(x1,…,xn)=(x1,…,xi−1,Ti(\boldmathx),xi+1,…,xn)
Performing the sequential updating as described by eq. (12) (one at a time and in increasing order) is equivalent to applying the mapping , defined as
S=^Tn∘^Tn−1∘…∘^T2∘^T1 (22)
where denotes function composition. Following the terminology given in [3], is called the Gauss Seidel mapping based on the mapping and the iteration is called the Gauss Seidel algorithm based on mapping .
A direct consequence of [3, Prop. 1.4, pp.186] is the following Proposition: Proposition 4: If is a contraction with respect to the norm, then the Gauss-Seidel mapping is also a contraction (with respect to the norm), with the same modulus as . In particular, if is closed, the sequence of the vectors generated by the Gauss-Seidel algorithm based on the mapping converges to the unique fixed point of geometrically. Having given all the necessary definitions and results, we will proceed by proving that (a) for each mapping it holds , , where , are the and norms, respectively, (b) if is diagonally dominant then is a contraction and (c) provided that is a contraction, the algorithm converges geometrically to the unique fixed point of . We remind here that the -dimensional vector results from the -th row of , exluding its -th element and dividing each element by the negative of .
Proposition 5: For the mappings , , it holds
|Ti(\boldmathx)−Ti(\boldmathy)|≤||% \boldmathσT¬iΣ−1¬i¬i||1||% \boldmathx−\boldmathy||∞, ∀\boldmathx% ,\boldmathy∈Rn (23)
Proof: (a) We consider first the case where . Let us consider the vectors . Since is constant, utilizing eq. (20) it follows that
μ∗i(\boldmathx)−μ∗i(\boldmathy)=% \boldmathσT¬iΣ−1¬i¬i(\boldmathx% ¬i−\boldmathy¬i) (24)
Also, it is
Ai(\boldmathx)−Ai(\boldmathy)=1√2σ∗i\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmathy¬i) (25)
Taking the difference we have
Ti(\boldmathx)−Ti(\boldmathy) =(μ∗i(\boldmathx)−μ∗i(\boldmath% y))+ +√2π(f(Ai(\boldmathx))−f(Ai(\boldmathy)))σ∗i (26)
Since is continuous in , the mean value theorem guarantees that there exists such that
f(Ai(\boldmathx))−f(Ai(\boldmathy))=f′(ξi)(Ai(\boldmathx)−Ai(\boldmathy)) (27)
Substituting eq. (27) to (26) we get
Ti(\boldmathx)−Ti(\boldmathy) =(μ∗i(\boldmathx)−μ∗i(\boldmath% y))+ +√2πf′(ξi)(Ai(\boldmathx)−Ai(\boldmathy))σ∗i (28)
Substituting (24) and (25) into (28) and after some manipulations it follows that
Ti(\boldmathx)−Ti(\boldmathy)=(1+f′(ξi)√π)\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmathy¬i) (29)
Taking absolute values in eq. (29) and applying Hölder’s inequality , for and , it follows that
|Ti(\boldmathx)−Ti(\boldmathy)|≤||(1+f′(ξi)√π)\boldmathσT¬iΣ−1¬i¬i||1||\boldmathx¬i−\boldmathy¬i||∞ (30)
Taking into account that (from Proposition 2), and the (trivial) fact that , it follows that
|Ti(\boldmathx)−Ti(\boldmathy)|≤||% \boldmathσT¬iΣ−1¬i¬i||1||% \boldmathx−\boldmathy||∞ (31)
Thus, the claim has been proved.
(b) We consider now the case where . Working similarly to the previous case, the difference is
Bi(\boldmathx)−Bi(\boldmathy)=−1√2σ∗i\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmathy¬i) (32)
while the difference is expressed as
Ti(\boldmathx)−Ti(\boldmathy) =(μ∗i(\boldmathx)−μ∗i(\boldmath% y))+ +√2π(f(Bi(\boldmathx))−f(Bi(\boldmathy)))σ∗i (33)
Utilizing the mean value theorem we have that there exists such that
f(Bi(\boldmathx))−f(Bi(\boldmathy))=f′(ξi)(Bi(\boldmathx)−Bi(\boldmathy)) (34)
Substituting eq. (34) to (33) we get
Ti(\boldmathx)−Ti(\boldmathy) =(μ∗i(\boldmathx)−μ∗i(\boldmath% y))+ +√2πf′(ξi)(Bi(\boldmathx)−Bi(\boldmathy))σ∗i (35)
Substituting eqs. (24) and (32) into (35), we obtain
Ti(\boldmathx)−Ti(\boldmathy)=(1+f′(ξi)√π)\boldmathσT¬iΣ−1¬i¬i(\boldmathx¬i−\boldmathy¬i) (36)
From this point on, the proof is exactly the same with that of (a). Q.E.D. Proposition 6: The mapping is a contraction in , with respect to the norm, provided that is diagonally dominant. Proof: Let . Taking into account proposition 5, it easily follows that
||T(\boldmathx)−T(\boldmathy)||∞≡maxi=1,…,n{|Ti(\boldmathx)−Ti(\boldmathy)|}≤
≤maxi=1,…,n{||\boldmathσT¬iΣ−1¬i¬i||1}||\boldmathx−\boldmathy||∞
Now, (a) taking into account that the -dimensional vector results from the -th row of , exluding its -th element and dividing each element by the negative of and (b) recalling that is the -th row of excluding its -th element , it is
||\boldmathσT¬iΣ−1¬i¬i||1=||\boldmaths¬isii||1
Provided that is diagonally dominant, it is
||\boldmaths¬i||1≤|sii|, ∀i
or
||\boldmaths¬isii||1<1, ∀i
which proves the claim. Q.E.D.
Theorem 1: The algorithm converges geometrically to the unique fixed point of , provided that is diagonally dominant. Proof: The proof is a direct consequence of the propositions 3, 4 and 6 exposed before, applied for . Q.E.D.
## V Assessment of the accuracy of the proposed method
An issue that naturally arises with the proposed method is how accurate the estimate of the mean is. Since it is very difficult to give a theoretical analysis of this issue, mainly due to the highly complex nature of the propsoed iterative scheme (see eq. (12)), we will try to gain some insight for this subject via experimentation. To this end, we set equal to the exponential correlation matrix, which is frequently met in various fields of applications, e.g., in signal processing applications. Its general form is
Σn=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1ρρ2…ρn−1ρ1ρ…ρn−2ρ2ρ1…ρn−3⋮⋮⋮⋱⋮ρn−1ρn−2ρn−3…1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, (0≤ρ<1) (37)
It is easy to verify that the inverse of is expressed as
Σ−1n=11−ρ2⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−ρ0…00−ρ1+ρ2−ρ…000−ρ1+ρ2…00⋮⋮⋮⋱⋮⋮000…1+ρ2−ρ000…−ρ1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ (38)
Also, it is straightforward to see that is diagonally dominant for all values of . Thus, it is a suitable candidate for our case. In addition, it is “controlled” by just a single parameter (), which facilitates the extraction of conclusions. Note that for , becomes the identity matrix, while as increases towards the diagonal dominancy of decreases (while its condition number increases). For close to , is alomost singular.
In the sequel, we consider the case of a zero mean normal distribution with covariance matrix as in (38), which is truncated in the region , that is the truncation point is the same along all dimensions. Without loss of generality, this choice has been deliberately selected, in order to keep our experimental framework controlled by just two parameters, and . Performing simulations for various combinations of the values of and (and for various dimensions ) we can gain some insight on the accuracy of the approximation of the mean provided by the proposed method. In the sequel, we use as benchmark the estimate of the mean provided by the (widely accepted as reliable) MCMC method ([18]).
Figure 1, shows a three-dimensional graph of the difference (assessed by its Euclidean norm divided by ) between the estimates of the truncated mean obtained by the MCMC and the proposed methods, against and . It is worth noting that for smaller values of (less than ), the difference remains low (less than ), independently of the value of the cutting point . On the other hand, for larger values of (above ), the difference increases. More specifically, it increases more for values of between (approximately) and .
In figure 2, the shaded regions in the space correspond to low difference (less than per dimension), for . It can be deduced that the behaviour of the proposed method is only slightly affected influenced by the dimensionality of the feature space.
As a general conclusion, one can argue that the “more diagonally dominant” the is (i.e., the smaller the is), the more accurate the estimate of the mean provided by the proposed method is. From another perspective, the more “approaches diagonality” (again, as becomes smaller), the more accurate the obtained estimates are. The latter is also supported by the fact that in the extreme case of a diagonal covariance matrix, one has to solve independent one-dimensional problems, for which an analytic formula exists. In this case, it is easy to verify that the proposed method terminates after a single iteration (see also comments in the next section).
## Vi Simulation results
After having gained some insight on the capability of the proposed method to approximate the mean of a multivariate truncated normal distribution in the previous section, we proceed in this section with experiments where now the involved covariance matrices have no specific structure. As in the previous section, the MCMC method is used as benchmark 444In the sequel, all results are rounded to the third decimal..
1st experiment: The purpose of this experiment is to compare the estimates of the mean of a truncated normal distribution obtained by the proposed and the MCMC methods, for dimensions varying from to . To this end, for each dimension , different truncated normal distributions (defined by the means and the covariance matrices of the corresponding untruncated normals, as well as their truncation points) have randomly been generated, such that, the corresponding inverse covariance matrix is diagonally dominant. For the -th such distribution, , both the proposed and the MCMC methods have been applied. Letting and denote the respective resulting estimates, the mean difference per coordinate between the two estimates is computed, i.e., and, averaging over , the quantity is obtained. In figure 3, is plotted versus . From this figure, it can be concluded that the proposed scheme gives estimates that are very close to those given by the MCMC. Thus, the fixed point of the proposed scheme (when the diagonal dominance condition is fulfilled) can be considered as a reliable estimate of the mean of the truncated normal distribution.
Next, in order to show the rapid convergence of the proposed scheme against the MCMC method, we focus on a single example (however, the resulting conclusions are generally applicable). More specifically, Table 1 shows the values of the norm of the difference divided by , i.e. the quantity , as the number of iterations evolves, for the -dimensional left truncated normal distribution defined by ,
Σ=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣$0.045$$−0.003$$0.013$$−0.004$$0.011$$−0.003$$0.056$$−0.015$$0.008$$0.010$$0.013$$−0.015$$0.074$$−0.001$$0.004$$−0.004$$0.008$$−0.001$$0.156$$−0.012$</ | 2020-02-16 23:32:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222493767738342, "perplexity": 557.1484405051382}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141430.58/warc/CC-MAIN-20200216211424-20200217001424-00048.warc.gz"} |
https://math.stackexchange.com/questions/540186/convergence-of-rademachers-formula-extending-the-partition-numbers-to-complex/599660 | # Convergence of Rademacher's formula: Extending the partition numbers to complex index
$$p(n) = \frac{1}{\pi \sqrt{2}} \sum_{k=1}^{\infty} \sqrt{k}\ A_k(n)\ F_k'(n)$$ $$A_k(n) = \sum_{0 \le m < k, \gcd(m, k) = 1} e^{i\pi\left(s(m, k) - 2nm/k\right)}$$ $$F_k(x) = \frac{1}{\sqrt{x - \frac{1}{24}}} \sinh\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(x - \frac{1}{24}\right)}\right)$$
and $s(m, k)$ is the Dedekind sum given by
$$s(m, k) = \sum_{n=1}^{k} \left(\left(\frac{n}{k}\right)\right)\left(\left(\frac{mn}{k}\right)\right)$$
where $((x))$ is the sawtooth function
$$((x)) = \begin{cases} x - \lfloor x \rfloor - \frac{1}{2}, &\mbox{if } x \in \mathbb{R} \setminus \mathbb{Z}\\ 0, &\mbox{if }x \in \mathbb{Z} \end{cases}$$.
$p(n)$ is the $n$th partition number. But here's the interesting thing: this formula also actually seems to work not just for natural, but real and complex values of $n$ as well! So we could perhaps think of Rademacher's formula as giving an extension of the partition-number function to real and complex indices, much as how the gamma function extends the factorial. Albeit, however, it is complex-valued at real indices.
But the question I have is, where does this converge, when the range of $n$ is expanded from $\mathbb{N}$ to $\mathbb{C}$? It seems to converge okay for real values of $n$, and also those for complex $n$ (perhaps should now be called $z$?) with negative imaginary part. But what about with positive imaginary part? The formula is slow, and numerical experiments with $n = 2i$ don't seem to help. It looks like it diverges, but I'm not totally sure. This is a very complicated series formula, and I'm not sure where one would even begin to analyze it to determine the region of convergence. | 2021-04-21 16:38:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9280867576599121, "perplexity": 140.74570555461378}, "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/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00252.warc.gz"} |
https://socratic.org/questions/what-is-the-derivative-of-y-x-x-3-1 | # What is the derivative of y=x/(x^3-1)?
The derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {x}^{3} - 1}{{x}^{3} - 1} ^ 2$
$\frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{3} - 1}\right) = \frac{\left({x}^{3} - 1\right) \cdot 1 - x \cdot 3 {x}^{2}}{{\left({x}^{3} - 1\right)}^{2}} = \frac{- 2 {x}^{3} - 1}{{x}^{3} - 1} ^ 2$ | 2021-07-26 13:51:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7311152815818787, "perplexity": 1710.346512568145}, "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/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00480.warc.gz"} |
https://socratic.org/questions/how-do-you-graph-y-5-x-4-using-asymptotes-intercepts-end-behavior | # How do you graph y=5/x+4 using asymptotes, intercepts, end behavior?
Feb 12, 2018
#### Explanation:
We already know the graph of y=5/x and it's asymptotes are y axis and x axis
$\left(y - 4\right) = \frac{5}{x}$
This kinda reduces to $Y = \frac{5}{x}$
here $Y = y - 4$ i.e origin is shifted to (0,4).
Hence asymptotes will be be *y axis and y=4*
Further put y=0 to find the x intercept
x intercept = -5/4= -1.25
Thus graph becomes | 2020-08-13 14:34:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.8262326121330261, "perplexity": 7618.732307754174}, "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-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00346.warc.gz"} |
http://samuel.zubo.sk/2011/05/greeks.html | ## nedeľa 15. mája 2011
### Greeks
#### Delta
1. For each one point increase in the underlying price, the value of the delta is added to the value of the option.
• Conversely, for each one point decrease in the price of the underlying, the value of delta is subtracted from the value of the option.
2. The delta of call options is always positive (0 to 100) and the delta of put options is always negative (0 to -100).
3. The delta of an at-the-money option is approximately .50 (calls) or -.50 (puts).
4. The delta of an option at expiration is either 0 or 100 (-100 for puts).
#### Gamma
1. For each one point increase in the underlying price, gamma is added to delta.
• Conversely, for each one point decrease in the price of the underlying, the value of gamma is subtracted from delta.
2. For all positive theta positions, gamma is always negative.
• Conversely, for all negative theta positions, gamma is always positive.
3. For out-of-the-money positions such as condors, gamma is generally small compared to delta and inconsequential, whereas gamma is generally larger and has bigger associated risk for at-the-money positions such as calendars.
#### Theta
1. For each day that passes, the value of theta is added to the value of the option.
2. Long option positions are theta negative (lose time value each day), while short option positions are theta positive (gain time value each day).
3. For longer term options, theta decay is slower, conversely shorter term options have faster theta decay.
#### Vega
1. For every 1% volatility increase in the underlying asset, the value of vega is added to the value of the option.
• Conversely, for every 1% volatility decrease, the value of vega is subtracted from the value of the option.
2. The impact of volatility changes is greater for at-the-money options than it is for in- or out-of-the-money options.
3. The impact of volatility changes is greater for longer term options and less for shorter term options.
4. Changes in vega can have more impact (i.e. you should worry about it more) for multi-month spreads (calendars, diagonals) than for single-month spreads (verticals, condors). Konkrétne: zvýšenie volatility sa výraznejšie prejaví u opcií s dlhšou expiráciou. Preto sa také opcie snažíme kupovať, keď je IV nízka.
#### Rho
1. For every 1% increase in interest rates, the value of an option increases percentage-wise by the value of rho.
• For example, if the rho of an option is 2.5, and interest rates increase by 1% ,then the value of the option increases by 2.5%.
2. For two reasons, you can usually ignore rho for most practical purposes. First, interest rates don’t change that often, and second, for short term options, rho is small and doesn’t have much effect.
3. Rho is more important for long term options such as LEAPs. | 2021-05-17 11:54:06 | {"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": 0.8303712606430054, "perplexity": 3600.2642533808566}, "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-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00540.warc.gz"} |
https://math.stackexchange.com/questions/484043/definition-for-monotonicity-of-multivariate-function | # Definition for monotonicity of multivariate function
Is there any standard definition for monotonicity of a multivariate function?
I suppose it's something like:
$\forall i: x_i \leq x_i' \implies f(x_1, \ldots, x_i, \ldots, x_k) \leq f(x_1, \ldots, x_i', \ldots, x_k)$
thanks!
• Can you actually clarify where the $\forall$ is being applied in your definition? I don't think it captures what should be intuitively monotone. Sep 4 '13 at 15:54
• Should there be primes on $x_1$ and $x_k$ on the right hand side? If so, our definitions agree. Sep 4 '13 at 16:24
• no, no primes, just 1 larger argument.. Sep 6 '13 at 7:11
• I think in terms of monotonicity, my definition is more standard. I don't know what else you would like to know here. Sep 8 '13 at 19:50
A sensible extension of monotonicity is the following. Let $A$ and $B$ be partially ordered sets. Let $f\colon A\rightarrow B$. $f$ is monotone if for each $x,y\in A$ s.t. $x\leq y$ we have that $f\left(x\right)\leq f\left(y\right)$. Just take $A=\mathbb{R}^{n}$ and $B=\mathbb{R}^{m}$ for the case you are interested in.
• But this would require the $\leq$ relation to be defined on tuples from A and on those from B? Sep 4 '13 at 16:19
• We can easily define the partial ordering by: $\mathbf{x} \leq \mathbf{y}$ if $\forall n : x_n \leq y_n$ Jun 17 '14 at 10:07
• With that the definition of @par is equivalent to your definition, Xaero182, since it follows from applying your definition multiple times on different $x_n$. Jun 17 '14 at 10:11 | 2021-09-22 10:29:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 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": 0.920077919960022, "perplexity": 399.5235922219859}, "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/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00714.warc.gz"} |
https://toxik.io/post/2016-12-14-async-madness/ | These past few days have been restructuring the theoretical query language and simplifying documentation, as well as adding small components and their tests.
I had a debacle with deferreds and yield, I can’t honestly know what that loop is doing but personally, it feels wrong. Shouldn’t I have a class variable holding the redis connection?
class RedisFuncts():
__init__
self.redis_conn = new_redis_stuff
get()
self.redis_conn.get()
VS
@deferred.ignore_inlines
def get():
conn = yield new_redis_stuff
conn.get()
At least, it seems to violate what my database friend thinks. But, hey, the later code works, I can improve on something that works.
Other than that, played around with getting unified diffs outside of libgit/pygit, when I clearly should be using those… It’s a question of being able to revert from one diff to the next, which leads to patronizing discussions about ndiff vs udiff like this one: diff stackoverflow
Please note since unified diffs usually drops lines that compared equal to begin with it would be almost impossible to restore to the original string. ndiff shows everything, so it would be straight forward to restore the original.
Other than that, tried a bit of compression, and proved that my encoding, language, and entropy sensing (not histograms) could be bypassed as a method of censorship with a simple hex conversion and then using the alphanumeric “Alpha Bravo Charley” system.
No working code because I’m scared of working code.
The lint tests can now have a couple of sed commands run before them for correcting most of the styling. I’m not going for pep8 in this case, but close.
The client is using an example of an async terminal, though this makes it problematic because I won’t know how to recombine the client or the server to do the opposite of what they’re doing, establishing a connection to another host and asking something.
The Cryptography library in python is getting to feel a bit limiting, it was good when experimenting with an XOR padding, but not as convenient when I needed encryption (there seems to only be signing on the EC primitive?), honestly it’s probably just me.
That XOR went over here in crypto/padding.py, by the way. | 2019-06-25 08:12:31 | {"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": 0.34847521781921387, "perplexity": 2284.4647090266967}, "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-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00124.warc.gz"} |
https://www.numerade.com/questions/find-the-exact-length-of-the-curve-x-t-sin-t-quad-y-t-cos-t-quad-0leqslant-t-leqslant-1/ | Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!
WZ
Find the exact length of the curve.$x = t \sin t$, $\quad y = t \cos t$, $\quad 0\leqslant t \leqslant 1$
$L=\int_{0}^{1} \sqrt{t^{2}+1} d t \stackrel{21}{=}\left[\frac{1}{2} t \sqrt{t^{2}+1}+\frac{1}{2} \ln (t+\sqrt{t^{2}+1})\right]_{0}^{1}=\frac{1}{2} \sqrt{2}+\frac{1}{2} \ln (1+\sqrt{2})$
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Kristen K.
University of Michigan - Ann Arbor
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Idaho State University
Boston College
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Kristen K.
University of Michigan - Ann Arbor
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Idaho State University
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Join Bootcamp | 2021-05-09 13:46:04 | {"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": 0.2921594977378845, "perplexity": 6067.416064792617}, "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-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00294.warc.gz"} |
https://csrc.nist.gov/events/2016/ssr-2016-security-standardisation-research | # SSR 2016: Security Standardisation Research
Over the last two decades a very wide range of standards have been developed covering a wide range of aspects of cyber security. These documents have been published by national and international formal standardisation bodies, as well as by industry consortia. Many of these standards have become very widely used - to take just one example, the ISO/IEC 27000 series of standards has become the internationally adopted basis for managing corporate information security.Despite their wide use, there will always be a need to revise existing security standards and to add new standards to cover new domains. The purpose of this conference is to discuss the many research problems deriving from studies of existing standards, the development of revisions to existing standards, and the exploration of completely new areas of standardisation. Indeed, many security standards bodies are only beginning to address the issue of transparency, so that the process of selecting security techniques for standardisation can be seen to be as scientific and unbiased as possible.
This conference is intended to cover the full spectrum of research on security standardisation, including, but not restricted to, work on cryptographic techniques (including ANSI, IEEE, IETF, ISO/IEC JTC 1/SC 27, ITU-T and NIST), security management, security evaluation criteria, network security, privacy and identity management, smart cards and RFID tags, biometrics, security modules, and industry-specific security standards (e.g. those produced by the payments, telecommunications and computing industries for such things as payment protocols, mobile telephony and trusted computing).
As was the case for the proceedings of SSR 2014 and SSR 2015, the proceedings of SSR 2016 will be published by Springer in the Lecture Notes in Computer Science series.
##### Program Presentations (zip file) Call for Papers
Deadline for submissions: Monday, 30 May 2016 - Monday, 13 June 2016 (23:59 Hawaii) - FIRM
Notifications to authors: Monday, 8 August 8 2016
Camera ready due: Monday, 19 September 2016
Last Day to Reserve Room @ Group Rate: 14 November 2016
Last Day to Register for Conference: 28 November 2016
Opening of conference: Monday, 5 December 20
Security Analysis of the W3C Web Cryptography API
Kelsey Cairns, Harry Halpin and Graham Steel
Algorithm Agility - Discussion on TPM 2.0 ECC Functionalities
Liqun Chen and Rainer Urian
Extending the UML Standards to Model Tree-Structured Data and
their Access Control Requirements
Alberto De La Rosa Algarin and Steven Demurjian
Cross-Tool Semantics for Protocol Security Goals
Joshua Guttman, John Ramsdell and Paul Rowe
A Secure Multicast Group Management and Key Distribution in IEEE
802.21
Yoshikazu Hanatani, Naoki Ogura, Yoshihiro Ohba, Lily Chen and Subir Das
Analysis of a Proposed Hash-Based Signature Standard
Jonathan Katz
State Management for Hash-Based Signatures
David McGrew, Panos Kampanakis, Scott Fluhrer, Stefan-Lukas Gazdag, Denis Butin and Johannes Buchmann
NFC Payment Spy: A Privacy Attack on Contactless Payments
Maryam Mehrnezhad, Mohammed Aamir Ali, Feng Hao and Aad van Moorsel
Attribute-based Access Control Architectures with the eIDAS Protocols
Frank Morgner, Paul Bastian and Marc Fischlin
Reactive and Proactive Standardisation of TLS
Kenneth Paterson and Thyla van der Merwe
Cryptanalysis of GlobalPlatform Secure Channel Protocols
Mohamed Sabt and Jacques Traore
Analyzing and Fixing the QACCE security of QUIC
Hideki Sakurada, Kazuki Yoneyama, Yoshikazu Hanatani and Maki Yoshida
John Kelsey, Cryptographer, NIST
John Kelsey is a cryptographer at NIST. His interests include random number generation, design and analysis of hash functions and block ciphers, electronic voting security, and practical applications of cryptography to solve real-world problems. John has been working in cryptography for over 20 years. He has worked on standards for logging, electronic voting, random number generation, and cryptographic hashing.
William Whyte, Chief Scientist, Security Innovation
William Whyte is responsible for the strategy and research behind the Security Innovation's activities in vehicular communications, security and cryptographic research.
Before joining Security Innovation, William was the Chief Technology Officer of NTRU Cryptosystems. He previously served as Senior Cryptographer with Baltimore Technologies in Dublin, Ireland.
William is chair of the IEEE 1363 Working Group for new standards in public key cryptography and has served as technical editor of two published IEEE standards, IEEE Std 1363.1-2008 and IEEE Std 1609.2-2006, as well as the ASC X9 standard X9.98.
William holds a PhD from Oxford University on Statistical Mechanics of Neural Networks and a B.A. from Trinity College.
Fun Fact: William's spends a lot of time at his local rink, watching his daughter play hockey - 5 nights a week to be exact! Hockey has made such a big impact on both of them that William has begun writing and illustrating stories all set in the world of girl's hockey.
#### Event Details
Starts: December 05, 2016 - 12:00 AM EST
Ends: December 06, 2016 - 12:00 AM EST
Format: In-person Type: Conference
Attendance Type: Open to public
NIST
Gaithersburg, MD | 2020-07-09 21:21:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18162575364112854, "perplexity": 11250.405857227455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655901509.58/warc/CC-MAIN-20200709193741-20200709223741-00188.warc.gz"} |
https://www.physicsforums.com/threads/calculate-the-ratio-of-the-fermi-wavevector-to-the-radius-of-the-largest-sphere.586321/ | Calculate the ratio of the Fermi wavevector to the radius of the largest sphere
1. Mar 12, 2012
blueyellow
1. The problem statement, all variables and given/known data
Sodium is, to a good approximation, a monovalent free-electron metal which has the
body-centred cubic structure.
(a) Calculate the ratio of the Fermi wavevector to the radius of the largest sphere that
can be inscribed in the first Brillouin zone. Remember that the reciprocal lattice of bcc
is fcc, and that the Brillouin zone is the cell formed in reciprocal space by the bisector
planes of the vectors from the origin to every reciprocal lattice point.
(b) If the cubic cell side of Sodium is 0.4225 nm, calculate its Fermi energy.
(c) Calculate the change in the Fermi energy of Sodium per degree temperature rise if
its linear thermal expansion coefficient is $\alpha$= 7 × 10$^{-5}$ K$^{-1}$.
(The linear thermal expansion coefficient is related to the lattice parameter, a, by
$\frac{da}{dT}$=$\alpha$a)
3. The attempt at a solution
a) Fermi wavevector:
k$_{F}$=$\sqrt{\frac{2}{\pi}}$($\frac{\pi}{a}$)
r=$\frac{\pi}{a}$
so the ratio is $\sqrt{\frac{2}{\pi}}$:1
Is it really as simple as that or have I done something wrong? | 2017-11-18 22:44:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7689177393913269, "perplexity": 764.790953321254}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805049.34/warc/CC-MAIN-20171118210145-20171118230145-00559.warc.gz"} |
https://brianbi.ca/etingof/5.11 | Brian Bi
$\DeclareMathOperator{\Ind}{Ind} \DeclareMathOperator{\Res}{Res}$
Remark: The text says to use Frobenius reciprocity, but it seems obvious that it means we should use the Frobenius formula from section 5.9. One might wonder which of the three $$\mathbb{Z}_2$$-subgroups of $$S_3$$ to use in (1), and similarly for (2) and (3). However, if we look at the Frobenius character formula, we see that replacing $$H$$ with a conjugate subgroup doesn't change the result; in (1), the three subgroups of $$G$$ isomorphic to $$\mathbb{Z}_2$$ are all conjugate to each other, so we must get the same decomposition regardless of which one we choose for the computation. A similar statement holds for (2) and (3)
Problem 5.11.1 The group $$A_5$$ has the property that any two isomorphic subgroups are conjugate, so in each of (a) to (e), all possible induced representations from a subgroup of the given isomorphism class are isomorphic. In all of the below, let $$G = A_5$$. We assume the ground field is $$\mathbb{C}$$.
1. We'll set $$H$$ to be the subgroup generated by $$(1\ 2)(3\ 4)$$. $$H$$ has two irreps, which were previously denoted $$\mathbb{C}_+$$ and $$\mathbb{C}_-$$. We will also use that notation here. The character of either of the two possible induced representations vanishes at $$g \in G$$ whenever $$g$$ isn't conjugate to any elements of $$H$$, namely, when $$g$$ is a 3-cycle or a 5-cycle. By Remark 5.8.3, both induced representations have dimension 30, so $$\chi(1) = 30$$ in both cases. The double transpositions of $$A_5$$ form a single conjugacy class, so let $$g = (1\ 2)(3\ 4)$$; computing $$\chi(g)$$ in $$\Ind^G_H \mathbb{C}_+$$ and $$\Ind^G_H \mathbb{C}_-$$ will then complete the calculations of their characters.
In $$\Ind^G_H \mathbb{C}_+$$, $$\chi(g)$$ is $$1/|H|$$ times the order of the centralizer of $$g$$, since $$g$$ is not conjugate to the identity. By the orbit-stabilizer theorem, the order of the centralizer is the order of $$A_5$$ divided by the size of its conjugacy class, or $$60 \div 15$$, where we have used the previously stated fact that all double transpositions are conjugate to each other. Thus, $$\chi(g) = 2$$ in $$\Ind^G_H \mathbb{C}_+$$. In $$\Ind^G_H \mathbb{C}_-$$, we must then have $$\chi(g) = -2$$ since $$\chi((1\ 2)(3\ 4)) = -1$$ in $$\mathbb{C}_-$$. We therefore have the following characters:
$$A_5$$1(123)(12)(34) (12345)(13245)
#120151212
$$\Ind^G_H \mathbb{C}_{+}$$3002 00
$$\Ind^G_H \mathbb{C}_{-}$$300−2 00
Using orthogonality of characters, we find \begin{align*} \Ind^G_H \mathbb{C}_+ &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus 3\mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_- &\cong 2\mathbb{C}^3_+ \oplus 2\mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus 2\mathbb{C}^5 \end{align*} where we have labelled the irreps of $$A_5$$ as they were labelled in section 4.8.
2. We'll set $$H$$ to be the subgroup generated by $$g = (1\ 2\ 3)$$. $$H$$ has three irreps; denote these by $$\mathbb{C}$$, $$\mathbb{C}_\omega$$, and $$\mathbb{C}_{\omega^2}$$, where $$g$$ respectively acts as 1, $$\omega$$, and $$\omega^2$$, with $$\omega = \exp(2\pi i/3)$$. By Remark 5.8.3, the three induced representations will each have dimension 20, and therefore $$\chi(1) = 20$$. The characters of the double transpositions and 5-cycles will vanish in all three induced representations since such elements of $$A_5$$ are never conjugate to $$g$$, a 3-cycle. As all 3-cycles in $$A_5$$ are conjugate to each other, we only need to compute $$\chi(g)$$ in each of the three induced representations in order to fill out the character table.
In $$\Ind^G_H \mathbb{C}$$, $$\chi(g)$$ will be $$1/|H|$$ times the number of elements of $$G$$ that either fix $$g$$ under conjugation or take $$g$$ to $$g^2$$. By the orbit-stabilizer theorem, since the conjugacy class of $$g$$ contains 20 elements, there are 3 elements that fix $$g$$ (namely, the 3 elements of $$H$$ itself) and 3 elements that take $$g$$ to $$g^2$$, namely $$(1\ 2)(4\ 5), (2\ 3)(4\ 5), (3\ 1)(4\ 5)$$. Thus, $$\chi(g) = 2$$. In $$\Ind^G_H \mathbb{C}_{\omega}$$, the sum in the Frobenius formula of section 5.9 includes $$\omega$$ twice and $$\omega^2$$ twice, so $$\chi(g) = (1/3)(3\omega + 3\omega^2) = -1$$. In $$\Ind^G_H \mathbb{C}_{\omega^2}$$, we get result, $$\chi(g) = (1/3)(3\omega^2 + 3\omega) = -1$$, the same result. We now write out the characters of the induced representations:
$$A_5$$1(123)(12)(34) (12345)(13245)
#120151212
$$\Ind^G_H \mathbb{C}$$2020 00
$$\Ind^G_H \mathbb{C}_{\omega}$$20−1 000
$$\Ind^G_H \mathbb{C}_{\omega^2}$$20−1 000
from which we obtain the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\omega} \cong \Ind^G_H \mathbb{C}_{\omega^2} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus 2\mathbb{C}^5 \end{align*}
3. Set $$H = \langle g \rangle$$ with $$g = (1\ 2\ 3\ 4\ 5)$$. Write $$\zeta = \exp(2\pi i/5)$$. There are 5 irreps of $$H$$, which we'll denote by $$\mathbb{C}$$ and $$\mathbb{C}_{\zeta^i}$$ for $$i = 1, 2, 3, 4$$, using similar notation to that of part (b). The dimension of each induced representation is 12 and all induced characters vanish for double transpositions and 3-cycles. The 5-cycles in $$A_5$$ split into two conjugacy classes, one containing $$g$$ and $$g^4$$, the other containing $$g^2$$ and $$g^3$$. Thus, we will need to compute $$\chi(g)$$ and $$\chi(g^2)$$ separately for each of the induced representations.
Start with $$\Ind^G_H \mathbb{C}$$. The conjugacy class of $$g$$ has size 12, so there are 5 elements of $$A_5$$ that commute with $$g$$, and another 5 that take $$g$$ to $$g^4$$ under conjugation. Thus $$\chi(g) = (1/5)(5 + 5) = 2$$. Likewise, there are 5 elements that commute with $$g^2$$ and another 5 that take $$g^2$$ to $$g^3$$, so $$\chi(g^2) = 2$$ as well. In $$\Ind^G_H \mathbb{C}_{\zeta}$$ and $$\Ind^G_H \mathbb{C}_{\zeta^4}$$, we obtain $$\chi(g) = (1/5)(5\zeta + 5\zeta^4) = 2\cos(2\pi/5) = \frac{\sqrt{5}-1}{2}$$, and $$\chi(g^2) = (1/5)(5\zeta^2 + 5\zeta^3) = 2\cos(4\pi/5) = \frac{-\sqrt{5}-1}{2}$$. In $$\Ind^G_H \mathbb{C}_{\zeta^2}$$ and $$\Ind^G_H \mathbb{C}_{\zeta^3}$$, we obtain $$\chi(g) = (1/5)(5\zeta^2 + 5(\zeta^2)^4) = \frac{-\sqrt{5}-1}{2}$$ and $$\chi(g^2) = (1/5)(5(\zeta^2)^2 + 5(\zeta^2)^3) = \frac{\sqrt{5}-1}{2}$$. In tabular format:
$$A_5$$1(123)(12)(34) (12345)(13245)
#120151212
$$\Ind^G_H \mathbb{C}$$1200 22
$$\Ind^G_H \mathbb{C}_{\zeta^{\pm 1}}$$120 0$$\frac{\sqrt{5}-1}{2}$$ $$\frac{-\sqrt{5}-1}{2}$$
$$\Ind^G_H \mathbb{C}_{\zeta^{\pm 2}}$$120 0$$\frac{-\sqrt{5}-1}{2}$$ $$\frac{\sqrt{5}-1}{2}$$
yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\zeta^{\pm 1}} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\zeta^{\pm 2}} &\cong \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*}
4. Let $$H$$ be the $$A_4$$-subgroup acting on 1, 2, 3, 4 (i.e., all elements of $$H$$ fix 5). There are 4 irreps of $$A_4$$, which were assigned the names $$\mathbb{C}$$, $$\mathbb{C}_{\epsilon}$$, $$\mathbb{C}_{\epsilon^2}$$, and $$\mathbb{C}^3$$ in section 4.8 and whose characters were given. In each of the representations $$\mathbb{C}_{\epsilon}$$ and $$\mathbb{C}_{\epsilon^2}$$, the elements of the normal $$V_4$$-subgroup of $$A_4$$ act trivially, and some other element $$g = (1\ 2\ 3)$$ is then chosen to act as the scalar $$\epsilon$$ or $$\epsilon^2$$, respectively (here, $$\epsilon = \exp(2\pi i/3)$$. In the 3-dimensional representation, the double transpositions have trace −1 while the 3-cycles have trace 0. This gives us all the information we need to compute the induced characters, which have dimension $$|A_5|/|A_4| = 5$$ when induced from the 1-dimensional representations of $$H$$, and dimension 15 when induced from the 3-dimensional representation.
As always, the character of $$\Ind^G_H \mathbb{C}$$ will be the easiest to compute. It vanishes at all the 5-cycles, since no 5-cycle can be conjugate to any element of $$H$$. Taking $$g = (1\ 2\ 3)$$ as above, there are 3 elements of $$A_5$$ that commute with $$g$$ and 3 elements mapping it to each of its 7 conjugates in $$A_4$$ (namely, the other 3-cycles in $$A_4$$), so the Frobenius formula of section 5.9 gives $$\chi(g) = (1/12)(8 \times 3) = 2$$. (The remaining 12 conjugates lie outside $$H$$.) Set $$g' = (1\ 2)(3\ 4)$$; there are 4 elements of $$A_5$$ that commute with $$g'$$, and 4 taking it to each of its two conjugates in $$A_4$$, so $$\chi(g') = (1/12)(4 + 4 + 4) = 1$$.
To compute $$\Ind^G_H \mathbb{C}_\epsilon$$, we use the fact that $$g$$ belongs to a conjugacy class of $$H$$ that contains four 3-cycles, with the other four 3-cycles belonging to another conjugacy class; but all eight of these 3-cycles are conjugate to each other in $$A_5$$. The sum in the Frobenius formula of section 5.9 therefore includes 12 terms with value $$\epsilon$$ (since there are 4 elements of $$H$$ with trace $$\epsilon$$ and 3 elements of $$G$$ taking $$g$$ to each of these under conjugation) and another 12 with value $$\epsilon^2$$, giving $$\chi(g) = (1/12)(12\epsilon + 12\epsilon^2) = -1$$. Since all double transpositions have trace 1 in $$\mathbb{C}_\epsilon$$, it follows that $$\chi(g') = 1$$ here as in the previous paragraph. A similar calculation, only exchanging $$\epsilon$$ with $$\epsilon^2$$, shows that $$\Ind^G_H \mathbb{C}_{\epsilon^2}$$ is isomorphic to $$\Ind^G_H \mathbb{C}_\epsilon$$.
Finally, in $$\Ind^G_H \mathbb{C}^3$$, all terms in $$\chi(g)$$ vanish since the 3-cycles are trace-free in $$\mathbb{C}^3$$, and the 12 terms in $$\chi(g')$$ each have value $$-1$$, so $$\chi(g') = -1$$. In tabular form:
$$A_5$$1(123)(12)(34) (12345)(13245)
#120151212
$$\Ind^G_H \mathbb{C}$$521 00
$$\Ind^G_H \mathbb{C}_{\epsilon^{\pm 1}}$$5 −1100
$$\Ind^G_H \mathbb{C}^3$$150−1 00
yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^4 \\ \Ind^G_H \mathbb{C}_{\epsilon^{\pm 1}} &\cong \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}^3 &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*}
5. The Klein four-group can be realized as a subgroup of $$G = A_5$$ as $$H = \{1, x, y, z\}$$ where $$x = (1\ 2)(3\ 4)$$, $$y = (1\ 3)(2\ 4)$$, and $$z = (1\ 4)(2\ 3)$$. By Theorem 5.6.1, all irreps of $$H$$ are tensor products of $$\mathbb{C}_+$$ and $$\mathbb{C}_-$$; so there are 4 such irreps, each one-dimensional, and we denote them by $$\mathbb{C}$$ (the trivial representation), the non-trivial $$\mathbb{C}_x$$ where $$x$$ and the identity act as the scalar $$+1$$ while $$y, z$$ act as $$-1$$, and the similarly defined $$\mathbb{C}_y$$ and $$\mathbb{C}_z$$. The four induced representations of $$G$$ all have dimension 15 and the induced characters must be nonzero only at the identity and at the double transpositions. In each induced representation, $$\chi(x) = \chi(y) = \chi(z)$$ since they are all conjugate to each other.
The element $$x$$ belongs to a conjugacy class of $$A_5$$ of size 15, so its centralizer has order 4. There are 4 elements of $$A_5$$ that, under the action of conjugation, take $$x$$ to each of its conjugates $$y, z$$ in $$H$$. Therefore, in $$\Ind^G_H \mathbb{C}$$, $$\chi(x) = (1/4)(4 + 4 + 4) = 3$$.
When computing $$\chi(x)$$ in $$\Ind^G_H \mathbb{C}_x$$, we keep in mind that each of the 4 elements that commute with $$x$$ contributes $$+1$$ to the sum, while each of the 4 elements that take $$x$$ to each of $$y$$ and $$z$$ contributes $$-1$$. Therefore, $$\chi(x) = (1/4)(4 - 4 - 4) = -1$$. A similar calculation shows that $$\chi(x) = -1$$ in the representations induced from $$\mathbb{C}_y$$ and $$\mathbb{C}_z$$. Thus, we have the following characters:
$$A_5$$1(123)(12)(34) (12345)(13245)
#120151212
$$\Ind^G_H \mathbb{C}$$1503 00
$$\Ind^G_H \mathbb{C}_{xyz}$$150 −100
yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^4 \oplus 2\mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{xyz} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*} | 2022-08-13 02:16:11 | {"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": 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": 0.993130624294281, "perplexity": 260.2774176779134}, "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/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00222.warc.gz"} |
https://www.sunclipse.org/?p=974 | # Fugacity
The question came up while discussing the grand canonical ensemble the other day of just where the word fugacity came from. Having a couple people in the room who received the “benefits of a classical education” (Gruber 1988), we guessed that the root was the Latin fugere, “to flee” — the same verb which appears in the saying tempus fugit. Turns out, the Oxford English Dictionary sides with us, stating that fugacity was formed from fugacious plus the common +ty suffix, and that fugacious (meaning “apt to flee away”) goes back to the Latin root we’d guessed.
Gilbert N. Lewis appears to have introduced the word in “The Law of Physico-Chemical Change”, which appeared in the Proceedings of the American Academy of Arts and Sciences 37 (received 6 April 1901).
If any phase containing a given molecular species is brought in contact with any other phase not containing that species, a certain quantity will pass from the first phase to the second. Every molecular species may be considered, therefore, to have a tendency to escape from the phase in which it is. In order to express this tendency quantitatively for any particular state, an infinite number of quantities could be used, such, for example, as the thermodynamic potential of the species, its vapor pressure, its solubility in water, etc. The quantity which we shall choose is one which seems at first sight more abstruse than any of these, but is in fact simpler, more general, and easier to manipulate. It will be called the fugacity, represented by the symbol $$\psi$$ and defined by the following conditions: —
1. The fugacity of a molecular species is the same in two phases when these phases are in equilibrium as regards the distribution of that species.
2. The fugacity of a gas approaches the gas pressure as a limiting value if the gas is infinitely rarefied. In other words, the escaping tendency of a perfect gas is equal to its gas pressure. | 2021-04-23 06:09:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.7376384139060974, "perplexity": 745.1163850911702}, "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/1618039601956.95/warc/CC-MAIN-20210423041014-20210423071014-00092.warc.gz"} |
https://kb.osu.edu/handle/1811/10665?show=full | dc.creator Huang, T. H. en_US dc.creator Rieckhoff, K. E. en_US dc.creator Voigt, E. M. en_US dc.date.accessioned 2006-06-15T14:08:06Z dc.date.available 2006-06-15T14:08:06Z dc.date.issued 1978 en_US dc.identifier 1978-TH-08 en_US dc.identifier.uri http://hdl.handle.net/1811/10665 dc.description $^{1}$ Resonance Raman in Platinum Phthalocyanine'', Tzer-hsiang Huang, Klaus E. Riecfchoff, and Eva-Maria Voigt, accepted by Can. J. Chem. $^{2}$ G. W. Canterst G. Janson, M. Noort, and J. H, Van der Waals, J. Phys. Chem. 80, 2253 (1976). $^{3}$ Tzer-hsiang Huang, Klaus E. Rieckhoff, Eva-Maria Voigt, and E. Roland Menzel, Chem, Phys. 19, 25 (1977)."" en_US dc.description Author Institution: Departments of Physics and Chemistry en_US dc.description.abstract We report here the first detailed analysis of the Shpol’ skii structures in the phosphorescence, phosphorescence excitation, and abosorption spectra of a metal phthalocyanine (pc) molecule, PtPc, in mixtures of $\alpha$-chloronaphthalene and n-octane. Previous work on this $molecule^{1}$ indicated that the symmetry is not the expected one for metal Pc’s, i.e. $D_{4h}$. The spectra reported here show that PtPc molecules occupy preferentially three distinct sites in a rapidly frozen matrix of Che mixed solvent. The predominant site was selected and found to have (0-0) phosphorescences at 9411 {\AA} (weak/9475 {\AA} (strong), 9740 {\AA}. (extremely weak), 9767 {\AA} (very weak), 10192 {\AA}. (very weak), and 10247 {\AA} (extremely weak) at $55^{\circ}$K, The first two transitions are attributed, following Van der Waals et al.,$^{2}$ to the removal of the degeneracy of the lowest $E_{u}$ triplet of a $D_{4h}$ symmetry. Interstate vibronic coupling exists between these two split triplets which are separated by a low activation barrier, $9 \pm 3 cm^{-1}$ Such coupling also exists, to a less extent between the split singlets. The other (0-0) phosphorescences are accounted for and are further differentiated from the split triplets by their temperature dependence. The vibronic bands in the excitation and absorption spectra correspond to those in the phosphorescence spectra. The energies of the vibronic bands in the phosphorescence progressions agree with the resonance Raman vibrations observed at $77^{\circ} k$ The equilibrium position of the ground state potential surface is only slightly shifted from that of the first excited state. The lower portions of the ground state potential surface and those of the first excited states are for all practical purposes identical in shape. The PtPc symmtry in the ground and in the first excited singlets is indistinguishable between $D_{2h}.C_{2v}$, or $D_{2}{^{1}}$ in agreement with the conclusions of earlier work. $^{1,3}$ The symmetry of the triplets is as low as that of the first excited singlets. en_US dc.format.extent 254763 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title SYMMETRIES OF SINGLET AND TRIPLET MANIFOLDS IN PLATINUM PHTHALOCYANINE BY USE OF SHPOL'SKII MATRIX SPECTROSCOPY en_US dc.type article en_US
| 2021-01-27 08:43:51 | {"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": 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": 0.5966306328773499, "perplexity": 7309.391811885608}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00296.warc.gz"} |
http://openstudy.com/updates/50a68f92e4b0129a3c8fd26d | ## AwkwardsWhatIAmFor The French Club is sponsoring a bake sale. If their goal is to raise at least $140.00, how many pastries must they sell at$3.50 each in order to meet that goal? Write and solve an inequality one year ago one year ago
1. clubcanthandlekg
let pastries = p if they want to raise at least $140, and profit = ($3.50)(pastries) then 140 =< 3.5p therefore p >= 40 They must sell at least 40 pastries: $p \ge 40$
2. AwkwardsWhatIAmFor
3.50p≥140;p≥40 or 140p≥3.50;p≥40
3. graydarl
140 is the total we dive it by 3.5 so we can find how many pastries we need so $\frac{ 140 }{ 3.5 } = 40$ pastries needed in order to meet the goal so they must sell at least 40 of them so 40 or more pastries must be sold | 2014-04-24 01:22:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4739212691783905, "perplexity": 2192.199111132242}, "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/1398223204388.12/warc/CC-MAIN-20140423032004-00146-ip-10-147-4-33.ec2.internal.warc.gz"} |