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https://studydaddy.com/question/35-0ml-of-hydrogen-gas-at-stp-weighs-how-many-grams | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# 35.0ml of hydrogen gas at STP weighs how many grams?
The at STP is "22.414L/mol". You must convert the given volume of hydrogen gas to liters. Then, using the molar volume, convert the liters to moles. Then convert the moles to mass using the molar mass of hydrogen gas, which is "2.016g/mol"
Convert mL "H"_2 to liters.
"35.0mL H"_2" x "1 L"/"1000mL" = "0.0350L H"_2
Convert liters "H"_2 to moles.
"0.0350L H"_2 x "1 mol H"_2/"22.414L H"_2 = "0.0015615mol H"_2
Convert moles "H"_2 tp grams.
"0.0015615mol H"_2 x "2.016g H"_2/"1 mol H"_2 = "0.00315g H"_2 (The final answer has three due to three significant figures in "35.0mL".)
At STP, "35.0mL" of hydrogen gas has a mass of "0.00315g". | 2019-04-18 23:14:53 | {"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": 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": 0.37872132658958435, "perplexity": 10231.8657678164}, "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-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00189.warc.gz"} |
https://www.physicsforums.com/threads/analysis-of-circuit-with-capacitors.742534/ | # Analysis of circuit with capacitors.
1. Mar 10, 2014
### Mutaja
1. The problem statement, all variables and given/known data
Given the attached circuit, compute Tau, UC1 (t) and UC1 (∞).
Then assume t = ∞ and C1 has the potential of Eth. Put the switch in the other position, so that C1 can discharge through R3. C2 will, of course, charge untill UC1 = UC2.
Compute Tau, UC1(t), UC2(t) and IR3(t).
From these calculations, it should be simple to graph it using the y axis as U or I and the c axis as Tau (time (s)).
3. The attempt at a solution
First things first. I attached the whole problem, since it's related to one circuit.
Compute Tau, UC1 (t) and UC1 (∞).
I started with transforming the circuit into a thevenin equivalent.
RTH = R1||R2 = 2350Ω.
ETH = ER2 = $\frac{R2 * E}{R1 + R2}$ = $\frac{4700Ω * 8V}{4700Ω + 4700Ω}$ = 4V.
Tau = R*C = 2350Ω * 680*10-6 F = 1.598S
UC1 (t) = ETH (1-e$\frac{-t}{Tau}$) = 4V * (1-e$\frac{-t}{1.598S}$)
UC1(∞) is my first problem, assuming the above is correct. By infinity, do they mean when C1 is fully charged? From earlier, I think 5 Tau used to be the definition of a fully charged capacitor, but this may be completely wrong, and knowing me, it is.
#### Attached Files:
• ###### Capacitor analysis.png
File size:
3.8 KB
Views:
281
2. Mar 10, 2014
### CWatters
That would be my understanding.
There is an easier way. The original circuit has two resistors forming a potential divider feeding the capacitor. So what's the maximum voltage the capacitor can reach? eg What voltage is it charging towards?
3. Mar 10, 2014
### Mutaja
I would assume 4V, as it's the same as R2 and the potential is divided equally between two resistors with the same value. And it's the same as I've set ETH to be.
Is this wrong?
Edit: does my calculations look ok by the way? Also, thanks a lot for taking your time to help me. Really appreciate it.
Edit 2: When time is infinitively large, or as it approaches a huge number, e will approach 0, therefore:
UC1(∞) = ETH(1-e$\frac{-∞}{Tau}$) = 4V(1-0) = 4v.
Does that look ok? Also, I can see I'm having some issues with e to the... power, but hopefully it's understandable.
Last edited: Mar 10, 2014
4. Mar 10, 2014
### Staff: Mentor
Your calculations look fine for the switch in the first position, if you assume that V1 is being switched on at time t=0. Really I think they want you to just assume that the switch has been in that position for a very long time and you need to find the voltage on the capacitor before the switch is moved to the second position. That is, the steady state value of UC1 before the switching takes place.
A simple approach to finding the steady state value for the voltage on a capacitor being driven by a fixed source and resistive network is to remove the capacitor in question and determine the potential across the open terminals. That avoids having to derive the exponential equations and taking limits (t → ∞).
What I think they're interested in is the voltage on C1 as a function of time after the switch moves to the new position. What are your thoughts there?
5. Mar 10, 2014
### Mutaja
This would be ETH? I've found that to be 4V.
So the calculation for that would be:
UC1 (∞) = ETH = 4V?
I've been reading a little in my book - which this course is based on (introductory circuit analysis by Boylestad) and from what I can find out, analysis of a circuit concerns(?) charging and discharging. Therefore, I'm thinking charging in the default position, and discharging in the other position (the switch position, that is).
I'm thinking compute Tau, UC1(t) and UC1(∞) for the charging of C1.
This is what I have so far (copied from opening post and my solution from the first quote in this post).
I started with transforming the circuit into a thevenin equivalent.
RTH = R1||R2 = 2350Ω.
ETH = ER2 = $\frac{R2 * E}{R1 + R2}$ = $\frac{4700Ω * 8V}{4700Ω + 4700Ω}$ = 4V.
Tau = R*C = 2350Ω * 680*10-6 F = 1.598S
UC1 (t) = ETH (1-e$\frac{-t}{Tau}$) = 4V * (1-e$\frac{-t}{1.598S}$)
UC1 (∞) = ETH = 4V?
Then, for the discharging of the circuit, t=∞ is achieved, and C1 has the voltage (potential) of ETH. The switch is set to the other position, and C1 will be discharged through R3. As follows, C2 will be charged untill UC1 = UC2. Here I'm supposed to graph the discharging of C1, the charging of C2 and the current graph for R1.
I'm a bit confused here, but let's take the first thing first :p The charging with the switch set in the default position.
6. Mar 10, 2014
### Staff: Mentor
Looks like you've already handled that. Your method works fine. For the initial switch position UC1(∞) = 4 V as you found.
7. Mar 10, 2014
### Mutaja
Ah, great! :)
When t=∞ is achieved and we put the switch in the other position, C1 starts to discharge through R3 and C2 is charged until UC1 = UC2.
So here I'm supposed to graph the discharging of C1, charging of C2 and the current graph for R1.
UC1 (0) = ETHe-t/1.598s = 4Ve-0/1.598s = 4V
UC2 (0) = ETH(1-e-t/1.598s) = 4V(1-e-0/1.598s) = 0V
UC1 (tau) = ETHe-1.598/1.598s = 2.94V
UC2 (tau) = ETH(1-e-1.598/1.598s) = 2.53V
UC1 (6tau) = ETHe-(1.598*6)/1.598s = 0.02V (0v)
UC2 (6tau) = ETH(1-e-(1.598*6)/1.598s) = 3.99V (4v)
These values are made to sketch the graphs for the charging and discharging of C1 and C2.
Not sure how to compute IR3(0), IR3(tau) and IR3(6*tau). Also, there's a typo here. I says it's a good idea to find the values of IR3(0), IR3(tau) and IR3(6*tau) to sketch the graph for the current through R1 - I assume it's the current through R3 that's interesting?
Here's my attempt anyway:
IR3(0) = $\frac{ETH}{R3}$ * e -t/1.598s = 0.85mA * e-0/1.598s = 0.85mA
IR3(tau) = 0.85mA * e-1.598/1.598s = 0.31mA
IR3(6*tau) = 0.85mA * e-(1.598*6)/1.598s = 0.0021mA
How does this look?
I see now that C2 doesn't behave as it's supposed to. It charges to 4V and C1 goes down to 0v. So that's as if it were a basic charging/discharging of one capacitor. How do I get them to relate to each other in my calculations?
Last edited: Mar 10, 2014
8. Mar 10, 2014
### Staff: Mentor
There are a few problems with your analysis, related to how the circuit finally ends up (a long time after the switching takes place). An excellent approach in these cases is to examine the initial and final conditions of the circuit and then simply "connect" the two states with the appropriate exponential curves.
Okay, at the start C1 begins with a voltage of 4 V, while C2 begins at 0 V as you stated. At the end, both capacitors must end having the same voltage, whence the current stops flowing. You should have enough information there to determine the initial current and final current.
Next, find the final voltage that will appear on the capacitors. You can look at this by way of the initially available charge that C1 brings to the game being divvied up between the two capacitors (since charge is conserved). You can then write an equation that connects the initial voltage on C1 to the final voltage on C1 via an exponential curve.
9. Mar 10, 2014
### Mutaja
The voltage over C1 initially is 4V, and finally it's equal to C2. The drop in potential across R3 factors in. Tau will also be different? Didn't think about that. How does two capacitors relate to each other when calculating the new RC? The book doesn't cover any circuit with two or more capacitors. Nor does the lecture notes.
4V with 4700Ω = 0.85mA, but then I get 4V across the resistor as well. I'm a bit confused still.
This is new to me. We've only worked with one capacitor. I'm trying though, believe me.
I will go through what I have of notes, and see if I can find anything mentioned about your method here.
Thanks a lot mate.
10. Mar 10, 2014
### Staff: Mentor
If you consider the circuit with the switch in the second position, a loop is formed. As far as the current is concerned, the capacitors are in series...
Yeah, that's right. Initially capacitor C2 has no charge and thus 0 V across it. So to satisfy KVL around the loop the initial potential drop across the resistor must be the same magnitude as the initial voltage across the capacitor C1...
This is where all the time you've spent looking at equivalent circuits comes in handy. If you have a series circuit, determine the net series capacitance and net series resistance to determine the tau.
[/quote]
I will go through what I have of notes, and see if I can find anything mentioned about your method here.
Thanks a lot mate.[/QUOTE]
11. Mar 10, 2014
### Mutaja
I will go through what I have of notes, and see if I can find anything mentioned about your method here.
Thanks a lot mate.[/QUOTE]
I'm beyond tired right now, but I think I finally figured it out. Or at least made some progress.
So, as you pointed out, the capacitors are actually in series. I've only studied basic electrical circuits for a few years, no wonder I still mix those up
I've gone back to the Q = C * U formula (not going to write it all out unless I have to, because of the aforementioned tiredness). I came up with this formula eventually.
UC1(∞) = UC1(0)$\frac{CC1}{CC1 + CC2}$ = 4V$\frac{680uF}{680uF+680uF}$ = 2V
CRES = $\frac{CC1 * CC2}{CC1 + CC2}$ = $\frac{680uF*680uF}{680uF+680uF}$ = 340uF
Tau=R3 * CRES = 4700Ω * (340*10-6F) = 1.598S
So, the discharging of C1
UC1(t) = UC1(∞) + (UC1(0) - UC1(∞)) e -t/Tau = 2V + 2Ve-t/1.598s
Charging of C2
UC2(t) = UC2(0) + (UC2(∞) - UC2(0))(1-e-t/Tau)
UC2(t) = 0V + (2V - 0V)(1-e-t/1.598s) = 2V(1-e-t/1.598s)
The current through R3.
IR3(t) = $\frac{UC1(t) - UC2(t)}{R3}$ = $\frac{(2V + 2Ve-t/1.598s) - (2V(1-e-t/1.598s)}{4700Ω}$ = $\frac{4v}{4700Ω}$e-t/1.598s
How does this look? And sorry about the formatting. Had problems using the [ SUP][ /SUP] and [ SUB][ /SUB] functions withing the math fractions.
12. Mar 10, 2014
### Staff: Mentor
That looks excellent Well done.
13. Mar 10, 2014
### Mutaja
The weight that just got lifted off of my shoulders... I absolutely love this feeling!
Thanks so much mate. | 2017-08-18 22:52:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5008968114852905, "perplexity": 1580.5690935413804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105187.53/warc/CC-MAIN-20170818213959-20170818233959-00031.warc.gz"} |
http://math.stackexchange.com/questions/168711/matlab-and-manipulating-matrices | # Matlab and manipulating matrices [closed]
Suppose I have a matrix $A$ of size
$n_1 \times n_2 \times n_3$
Now, I have another matrix $B$ of size $n_1 \times n_2 \times N$ where $N<n_3$
I'd like to create the following matrix $C = [A(m,l,B(m,l,:))]_{1 \le m \le n_1, 1 \le l \le n_2}$, whch has the same size as $B$. What is the most efficient way of doing this without having to create loops?
-
## closed as off-topic by Ramanujan, azimut, Dennis Gulko, amWhy, userNaNOct 29 '13 at 18:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is not about mathematics, within the scope defined in the help center." – Ramanujan, azimut, Dennis Gulko, amWhy, userNaN
If this question can be reworded to fit the rules in the help center, please edit the question.
If you have three dimensions, then it isn't really a matrix anymore; it's a rank-3 tensor. – Ｊ. Ｍ. Jul 9 '12 at 17:32
what does $B(m,l)$ mean if $B$ is a 3D array? – chaohuang Jul 9 '12 at 18:00
This is for mathworks.com/matlabcentral/answers. – Dirk Jul 9 '12 at 18:14
sorry i should be $B(m,l,:)$, It's basically a vector of size $N$. Thanks – Stuck_pls_help Jul 9 '12 at 18:48
When it comes to issues like efficiency of numerical algorithms, you may consider asking instead at scicomp.stackexchange.com – Willie Wong Jul 10 '12 at 6:34
The formula assumes that the elements of B are in the range $[1,n_3]$ not $[0,n_3-1]$. – p.s. Jul 25 '12 at 23:22
Do you have any suggestions if I wanted to create another matrix of the following form: $C = [A(m,l,B(m,l,n,:),n)]_{1 \le m \le n_1, 1 \le l \le n_2, 1 n\le n_3}$ and $B$ is an $n_1 \times n_2 \times n_3 \times N<n_3$ – Stuck_pls_help Jul 26 '12 at 3:29 | 2014-12-18 17:22:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8439703583717346, "perplexity": 1221.3743228656353}, "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-2014-52/segments/1418802767301.77/warc/CC-MAIN-20141217075247-00061-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/16375/how-to-import-multiple-branch-cases-structure-into-latex-document | # how to import multiple branch cases structure into LaTeX document.
Could I set up a structure for several cases so that I could get a TeX command branch say, behavior as below:
if case A
• if case A1
\cmdAi{arg1}
• elseif case A2
\cmdAii{arg2}
• elseif case A3
\cmdAiii{arg3}
• endif
elseif case B
• elseif case B1
\cmdBi{arg4}
• elseif case B2
\cmdBii{arg5}
• end if
endif
the case A1~case B2 is given in advance, say, from the ordered collection (maybe its form is {\a,\'a,\^a,A,\^A}).
I knew the ifthen package, but it is too simple (or to complicated?) to use conveniently.
A intuitive explain:
this demand is related to the modal logic and semantics. Think about a given alphabetic chain of n bit (n less than 16), say,
TxFxyxxTyxTFyT
zTyxTzzxxTyxF
etc. The character T denoted "True" or constant have value 1, the character F denoted "False" or constant have value 0, the other character x, y, z, and so on denoted variables whose value between 0 and 1 (not have to equal). After we assigned for the variables, I want to print the result like:
the chain given above has valuables [#1] (@1) times (or print which places has [#1] appeared in), [#2] (@2) times, [#3] (@3) times, ... and their weight sum is ...
the [#1], [#2], [#3], etc. is the x, y, z, respectively; the (@1), (@2), (@3), , etc. mean as former.
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It may help if you give us some more idea of the purpose of this. Also, what is it about the ifthen package that is inconvenient? – Jan Hlavacek Apr 22 '11 at 4:10
Please have a look at tex.stackexchange.com/editing-help to learn how to format posts. Thanks. – Martin Scharrer Apr 22 '11 at 8:30
I am not exactly sure as to what you are after, but since you mentioned ordered lists here is an approach. One of the difficulties of using "TeX" as a computer language is its lack of built-in datastructures. But once you have lists, you can built any type of datastructure you wish and I think that is what you need here. However, some shift from common programming paradigms is necessary, so here is a traditional TeX solution.
Consider the following list of keywords in a list:
\def\alist{A--,A+,A++,A-,A=,A==,}
and for which the ordered list is, A+,A++,A--,A---,A=,A==,
Each element in the list will act as an "object" and hold some information, being a true, false or anything else you may wish.
We define a suitable macro to add an element and its definition to the list and we order the list.
\def\AddElementToList#1#2#3{%
\expandafter\def\csname#2\endcsname{#3}
\lst@BubbleSort{\alist}
}
We can loop the list and expand the meaning of all the elements using the @for command from the LaTeX kernel.
\@for \i:=\alist \do{%
\texttt{%
\expandafter\csname\i\endcsname}\par
}
Any individual element can be accessed by its name as \@nameuse{A==}.
As you can see from the commands you can store any type of object in the list. It can be a single digit or provided you define your macros accordingly can hold a full chapter of a book or possibly the book itself!. One caveat with the MWE shown below, if you writing in turkish change the \i to something else or run the code within a group.
\documentclass{article}
\usepackage{lstdoc}
\begin{document}
\makeatletter
%% define a list for demo purposes
\def\alist{A--,A+,A++,A-,A=,A==,}
%% typeset the unsorted list
\texttt\alist
The ordered list
%% the sorted list
\lst@BubbleSort{\alist}
\texttt\alist
%% empty the list
\let\alist\empty
%% add element and define the element macro
\expandafter\def\csname#2\endcsname{#3}
\lst@BubbleSort{\alist}
}
\AddElementToList{\alist}{A---}{$a^2+33=f_2$}
\texttt\alist
%% loop over the list
\@for \i:=\alist \do{%
\texttt{%
\expandafter\csname\i\endcsname}\par
}
\makeatother
\end{document}
You can consider that the alist is your case statement.
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@Y.L. I'am reading, though that's a bit diffcult to me. Thanks in advance. – Stufazi Hoqckt Apr 22 '11 at 13:21
@Stufazi Hoqckt If you edit your post to define the problem a bit more clearly, I can edit the post to make it more specific. Start perhaps from a scan saying, I want to achieve this... – Yiannis Lazarides Apr 22 '11 at 14:36 | 2014-07-28 16:30:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8165611624717712, "perplexity": 2170.082041772502}, "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-2014-23/segments/1406510261249.37/warc/CC-MAIN-20140728011741-00020-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.projectrhea.org/rhea/index.php/Introduction_to_the_Laplace_Operator | Introduction
The Laplace Operator is an operator defined as the divergence of the gradient of a function. ${\large\Delta=\nabla\cdot\nabla=\nabla^{2}=\bigg[\frac{\partial}{\partial x_{1}},\cdots,\frac{\partial}{\partial x_{N}}\bigg]\cdot\bigg[\frac{\partial}{\partial x_{1}},\cdots,\frac{\partial}{\partial x_{N}}\bigg]=\sum\limits_{n=1}^{N}\frac{\partial^{2}}{\partial x^{2}_{n}}}$
## Alumni Liaison
To all math majors: "Mathematics is a wonderfully rich subject."
Dr. Paul Garrett | 2022-06-25 05:27:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8304492235183716, "perplexity": 3037.1402958517247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103034170.1/warc/CC-MAIN-20220625034751-20220625064751-00409.warc.gz"} |
https://www.physicsforums.com/threads/fourier-analysis-uniform-convergence-on-inf-inf.286390/ | Fourier Analysis - uniform convergence on (-inf, inf)?
1. Jan 21, 2009
Tacos
I have a question that I just don't know how to go about.
" Let Fn = x/(1+(n^2)(x^2)) where n=1,2,3,... show that Fn converges uniformly on (-infinity,infinity)"
To be honest, I don't even know where to start. Is this a series? How would I solve this. Would the Abel's test apply?
2. Jan 21, 2009
chrisk
Try using the ratio test to determine if a series converges.
$$lim\frac{F_{n+1}}{F_{n}}$$
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https://www.physicsforums.com/threads/air-puck-correct-method-or-not.177273/ | Air Puck Correct Method or not
1. Jul 16, 2007
rash219
Air Puck !!! Correct Method or not...
1. The problem statement, all variables and given/known data
An air puck of mass 0.25kg is tied to a string and allowed to revolve in a circle of radius 1.0m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table and a mass of 1kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves
a. What is the tension in the string ?
b. What is the horizontal force acting on the string ?
c. What is the speed of the puck ?
2. Relevant equations
$$\Sigma$$F = m * a
F_c = (m * v^2) / r
a_c = v^2 / r
3. The attempt at a solution
a.
$$\Sigma$$F_y = m * a = 0
T - mg = 0
T = mg = 1.0kg * 9.8m/s^2 = 9.8N
b.
F_c = Tension on th string = 9.8N
c.
F_c = (m * v^2) / r = 9.8N
(0.25kg * v^2) / 1.0m = 9.8N
v = $$\sqrt{39.2}$$ = 6.3 m/s
Is this correct......THANKS!!!
Last edited: Jul 16, 2007
2. Jul 17, 2007
andrevdh
You seem to have it under control.
3. Jul 17, 2007
GoldPheonix
3a. is correct, but it has incorrect --as near as I can tell- reasoning. Just recall Newton's Third Law for centripetal forces. Also, for 3b. it asks what is the horizontal force is; which requires an angle for the center of the system; not certain on how you are supposed to acquire that answer.
Last edited: Jul 17, 2007
4. Jul 17, 2007
rash219
Thanks a lot !!!
5. Sep 24, 2007 | 2016-10-22 01:58:57 | {"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": 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": 0.5350872278213501, "perplexity": 1318.4804396236116}, "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-2016-44/segments/1476988718423.28/warc/CC-MAIN-20161020183838-00293-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/eect.2021031?viewType=html | # American Institute of Mathematical Sciences
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## Neutral delay Hilfer fractional integrodifferential equations with fractional brownian motion
1 Mathematics Department, College of Science, Qassim University, P.O.Box 6644, Buraydah 51452, Saudi Arabia 2 Higher Institute of Engineering, El-Shorouk Academy, El-Shorouk City, Cairo, Egypt
* Corresponding author: Hamdy M. Ahmed
Received June 2020 Revised May 2021 Early access July 2021
In this paper, we study the existence and uniqueness of mild solutions for neutral delay Hilfer fractional integrodifferential equations with fractional Brownian motion. Sufficient conditions for controllability of neutral delay Hilfer fractional differential equations with fractional Brownian motion are established. The required results are obtained based on the fixed point theorem combined with the semigroup theory, fractional calculus and stochastic analysis. Finally, an example is given to illustrate the obtained results.
Citation: Yousef Alnafisah, Hamdy M. Ahmed. Neutral delay Hilfer fractional integrodifferential equations with fractional brownian motion. Evolution Equations & Control Theory, doi: 10.3934/eect.2021031
##### References:
[1] G. Arthi, J. H. Park and H. Y. Jung, Existence and exponential stability for neutral stochastic integrodifferential equations with impulses driven by a fractional Brownian motion, Communications in Nonlinear Science and Numerical Simulation, 32 (2016), 145-157. doi: 10.1016/j.cnsns.2015.08.014. Google Scholar [2] G. Arthi and J. H. Park, On controllability of second-order impulsive neutral integrodifferential systems with infinite delay, IMA J. Math. Control Inf., 32 (2015), 639-657. doi: 10.1093/imamci/dnu014. Google Scholar [3] K. Aissani and M. Benchohra, Controllability of fractional integrodifferential equations with state-dependent delay, J. Integral Equations Applications, 28 (2016), 149-167. doi: 10.1216/JIE-2016-28-2-149. Google Scholar [4] H. M. Ahmed, Controllability of impulsive neutral stochastic differential equations with fractional Brownian motion, IMA Journal of Mathematical Control and Information, 32 (2015), 781-794. doi: 10.1093/imamci/dnu019. Google Scholar [5] H. M. Ahmed and M. M. El-Borai, Hilfer fractional stochastic integro-differential equations, Appl. Math. Comput., 331 (2018), 182-189. doi: 10.1016/j.amc.2018.03.009. Google Scholar [6] A. Boudaoui, T. Caraballo and A. Ouahab, Impulsive neutral functional differential equations driven by a fractional Brownian motion with unbounded delay, Applicable Analysis, 95 (2016), 2039-2062. doi: 10.1080/00036811.2015.1086756. Google Scholar [7] B. Boufoussi and S. Hajji, Neutral stochastic functional differential equations driven by a fractional Brownian motion in a Hilbert space, Statistics and Probability Letters, 82 (2012), 1549-1558. doi: 10.1016/j.spl.2012.04.013. Google Scholar [8] B. Boufoussi and S. Hajji, Stochastic delay differential equations in a Hilbert space driven by fractional Brownian motion, Statistics and Probability Letters, 129 (2017), 222-229. doi: 10.1016/j.spl.2017.06.006. Google Scholar [9] T. Caraballo, M. J. Garrido-Atienza and T. Taniguchi, The existence and exponential behavior of solutions to stochastic delay evolution equations with a fractional Brownian motion, Nonlinear Analysis: Theory, Methods and Applications, 74 (2011), 3671-3684. doi: 10.1016/j.na.2011.02.047. Google Scholar [10] J. Cui and Y. Litan, Existence result for fractional neutral stochastic integro-differential equations with infinite delay, Journal of Physics A: Mathematical and Theoretical, 44 (2011), 335201, 16pp. doi: 10.1088/1751-8113/44/33/335201. Google Scholar [11] A. Chadha and N. Pandey Dwijendra, Existence results for an impulsive neutral stochastic fractional integro-differential equation with infinite delay, Nonlinear Analysis, 128 (2015), 149-175. doi: 10.1016/j.na.2015.07.018. Google Scholar [12] A. Debbouche and V. Antonov, Approximate controllability of semilinear Hilfer fractional differential inclusions with impulsive control inclusion conditions in Banach spaces, Chaos, Solitons & Fractals, 102 (2017), 140-148. doi: 10.1016/j.chaos.2017.03.023. Google Scholar [13] M. Ferrante and C. Rovira, Convergence of delay differential equations driven by fractional Brownian motion, J. Evol. Equ., 10 (2010), 761-783. doi: 10.1007/s00028-010-0069-8. Google Scholar [14] M. Ferrante and C. Rovira, Stochastic delay differential equations driven by fractional Brownian motion with Hurst parameter $H > \frac{1}{2}$, Bernoulli, 12 (2006), 85-100. Google Scholar [15] H. Gu and H. J. J. Trujillo, Existence of mild solution for evolution equation with Hilfer fractional derivative, Applied Mathematics and Computation, 257 (2015), 344-354. doi: 10.1016/j.amc.2014.10.083. Google Scholar [16] R. Hilfer, Applications of Fractional Calculus in Physics, World Scientific: Singapore, 2000. doi: 10.1142/9789812817747. Google Scholar [17] R. Hilfer, Experimental evidence for fractional time evolution in glass forming materials, Chem. Phys., 284 (2002), 399-408. Google Scholar [18] A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, North-Holland Mathematics Studies, 204, Elsevier, Amsterdam, 2006. Google Scholar [19] J. Klamka, Stochastic controllability of linear systems with delay in control, Bulletin of the Polish Academy of Sciences, Technical Sciences, 55 (2007), 23-29. Google Scholar [20] D. Luo, Q. Zhu and Z. Luo, An averaging principle for stochastic fractional differential equations with time-delays, Applied Mathematics Letters, 105 (2020), 106290, 8pp. doi: 10.1016/j.aml.2020.106290. Google Scholar [21] J. M. Mahaffy and C. V. Pao, Models of genetic control by repression with time delays and spatial effects, J. Math. Biol., 20 (1984), 39-57. doi: 10.1007/BF00275860. Google Scholar [22] R. Mabel Lizzy, K. Balachandran and M. Suvinthra, Controllability of nonlinear stochastic fractional systems with distributed delays in control, Journal of Control and Decision, 4 (2017), 153-168. doi: 10.1080/23307706.2017.1297690. Google Scholar [23] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Applied Mathematical Sciences, Springer-Verlag, New York, 1983. doi: 10.1007/978-1-4612-5561-1. Google Scholar [24] I. Podlubny, Fractional Differential Equations, Academic press, an Diego, 1999. Google Scholar [25] C. V. Pao, Systems of parabolic equations with continuous and discrete delays, J. Math. Anal. Appl., 205 (1997), 157-185. doi: 10.1006/jmaa.1996.5177. Google Scholar [26] D. H. Abdel Rahman, S. Lakshmanan and A. S. Alkhajeh, A time delay model of tumourimmune system interactions: Global dynamics, parameter estimation, sensitivity analysis, Applied Mathematics and Computation, 232 (2014), 606-623. doi: 10.1016/j.amc.2014.01.111. Google Scholar [27] F. A. Rihan, C. Tunc, S. H. Saker, S. Lakshmanan and R. Rakkiyappan, Applications of delay differential equations in biological systems,, Complexity, 2018 (2018), Article ID 4584389, 3 pages. doi: 10.1155/2018/4584389. Google Scholar [28] F. A. Rihan, C. Rajivganthi, P. Muthukumar, Fractional stochastic differential equations with Hilfer fractional derivative: Poisson Jumps and optimal control, Discrete Dyn. Nat. Soc., 2017(2017), Article ID 5394528, 11 pages. doi: 10.1016/j.cnsns.2013.05.015. Google Scholar [29] R. Sakthivel, R. Ganesh, Y. Ren and S. M. Anthoni, Approximate controllability of nonlinear fractional dynamical systems, Communications in Nonlinear Science and Numerical Simulation, 18 (2013), 3498-3508. Google Scholar [30] R. Sakthivel and R. Yong, Approximate controllability of fractional differential equations with state-dependent delay, Results in Mathematics, 63 (2013), 949-963. doi: 10.1007/s00025-012-0245-y. Google Scholar [31] B. Sundara Vadivoo, R. Ramachandran, J. Cao, H. Zhang and X. Li, Controllability analysis of nonlinear neutral-type fractional-order differential systems with state delay and impulsive effects, International Journal of Control, Automation and Systems, 16 (2018), 659-669. doi: 10.1007/s12555-017-0281-1. Google Scholar [32] J. Wang and H. M. Ahmed, Null controllability of nonlocal Hilfer fractional stochastic differential equations, Miskolc Math. Notes, 18 (2017), 1073-1083. doi: 10.18514/MMN.2017.2396. Google Scholar [33] J. R. Wang, M. Feckan and Y. Zhou, A survey on impulsive fractional differential equations, Fractional Calculus and Applied Analysis, 19 (2016), 806-831. doi: 10.1515/fca-2016-0044. Google Scholar [34] X. Zhang, P. Agarwal, Z. Liu, H. Peng, F. You and Y. Zhu, Existence and uniqueness of solutions for stochastic differential equations of fractional-order $q > 1$ with finite delays, Advances in Difference Equations, 2017 (2017), 1-18. doi: 10.1186/s13662-017-1169-3. Google Scholar
show all references
##### References:
[1] G. Arthi, J. H. Park and H. Y. Jung, Existence and exponential stability for neutral stochastic integrodifferential equations with impulses driven by a fractional Brownian motion, Communications in Nonlinear Science and Numerical Simulation, 32 (2016), 145-157. doi: 10.1016/j.cnsns.2015.08.014. Google Scholar [2] G. Arthi and J. H. Park, On controllability of second-order impulsive neutral integrodifferential systems with infinite delay, IMA J. Math. Control Inf., 32 (2015), 639-657. doi: 10.1093/imamci/dnu014. Google Scholar [3] K. Aissani and M. Benchohra, Controllability of fractional integrodifferential equations with state-dependent delay, J. Integral Equations Applications, 28 (2016), 149-167. doi: 10.1216/JIE-2016-28-2-149. Google Scholar [4] H. M. Ahmed, Controllability of impulsive neutral stochastic differential equations with fractional Brownian motion, IMA Journal of Mathematical Control and Information, 32 (2015), 781-794. doi: 10.1093/imamci/dnu019. Google Scholar [5] H. M. Ahmed and M. M. El-Borai, Hilfer fractional stochastic integro-differential equations, Appl. Math. Comput., 331 (2018), 182-189. doi: 10.1016/j.amc.2018.03.009. Google Scholar [6] A. Boudaoui, T. Caraballo and A. Ouahab, Impulsive neutral functional differential equations driven by a fractional Brownian motion with unbounded delay, Applicable Analysis, 95 (2016), 2039-2062. doi: 10.1080/00036811.2015.1086756. Google Scholar [7] B. Boufoussi and S. Hajji, Neutral stochastic functional differential equations driven by a fractional Brownian motion in a Hilbert space, Statistics and Probability Letters, 82 (2012), 1549-1558. doi: 10.1016/j.spl.2012.04.013. Google Scholar [8] B. Boufoussi and S. Hajji, Stochastic delay differential equations in a Hilbert space driven by fractional Brownian motion, Statistics and Probability Letters, 129 (2017), 222-229. doi: 10.1016/j.spl.2017.06.006. Google Scholar [9] T. Caraballo, M. J. Garrido-Atienza and T. Taniguchi, The existence and exponential behavior of solutions to stochastic delay evolution equations with a fractional Brownian motion, Nonlinear Analysis: Theory, Methods and Applications, 74 (2011), 3671-3684. doi: 10.1016/j.na.2011.02.047. Google Scholar [10] J. Cui and Y. Litan, Existence result for fractional neutral stochastic integro-differential equations with infinite delay, Journal of Physics A: Mathematical and Theoretical, 44 (2011), 335201, 16pp. doi: 10.1088/1751-8113/44/33/335201. Google Scholar [11] A. Chadha and N. Pandey Dwijendra, Existence results for an impulsive neutral stochastic fractional integro-differential equation with infinite delay, Nonlinear Analysis, 128 (2015), 149-175. doi: 10.1016/j.na.2015.07.018. Google Scholar [12] A. Debbouche and V. Antonov, Approximate controllability of semilinear Hilfer fractional differential inclusions with impulsive control inclusion conditions in Banach spaces, Chaos, Solitons & Fractals, 102 (2017), 140-148. doi: 10.1016/j.chaos.2017.03.023. Google Scholar [13] M. Ferrante and C. Rovira, Convergence of delay differential equations driven by fractional Brownian motion, J. Evol. Equ., 10 (2010), 761-783. doi: 10.1007/s00028-010-0069-8. Google Scholar [14] M. Ferrante and C. Rovira, Stochastic delay differential equations driven by fractional Brownian motion with Hurst parameter $H > \frac{1}{2}$, Bernoulli, 12 (2006), 85-100. Google Scholar [15] H. Gu and H. J. J. Trujillo, Existence of mild solution for evolution equation with Hilfer fractional derivative, Applied Mathematics and Computation, 257 (2015), 344-354. doi: 10.1016/j.amc.2014.10.083. Google Scholar [16] R. Hilfer, Applications of Fractional Calculus in Physics, World Scientific: Singapore, 2000. doi: 10.1142/9789812817747. Google Scholar [17] R. Hilfer, Experimental evidence for fractional time evolution in glass forming materials, Chem. Phys., 284 (2002), 399-408. Google Scholar [18] A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, North-Holland Mathematics Studies, 204, Elsevier, Amsterdam, 2006. Google Scholar [19] J. Klamka, Stochastic controllability of linear systems with delay in control, Bulletin of the Polish Academy of Sciences, Technical Sciences, 55 (2007), 23-29. Google Scholar [20] D. Luo, Q. Zhu and Z. Luo, An averaging principle for stochastic fractional differential equations with time-delays, Applied Mathematics Letters, 105 (2020), 106290, 8pp. doi: 10.1016/j.aml.2020.106290. Google Scholar [21] J. M. Mahaffy and C. V. Pao, Models of genetic control by repression with time delays and spatial effects, J. Math. Biol., 20 (1984), 39-57. doi: 10.1007/BF00275860. Google Scholar [22] R. Mabel Lizzy, K. Balachandran and M. Suvinthra, Controllability of nonlinear stochastic fractional systems with distributed delays in control, Journal of Control and Decision, 4 (2017), 153-168. doi: 10.1080/23307706.2017.1297690. Google Scholar [23] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Applied Mathematical Sciences, Springer-Verlag, New York, 1983. doi: 10.1007/978-1-4612-5561-1. Google Scholar [24] I. Podlubny, Fractional Differential Equations, Academic press, an Diego, 1999. Google Scholar [25] C. V. Pao, Systems of parabolic equations with continuous and discrete delays, J. Math. Anal. Appl., 205 (1997), 157-185. doi: 10.1006/jmaa.1996.5177. Google Scholar [26] D. H. Abdel Rahman, S. Lakshmanan and A. S. Alkhajeh, A time delay model of tumourimmune system interactions: Global dynamics, parameter estimation, sensitivity analysis, Applied Mathematics and Computation, 232 (2014), 606-623. doi: 10.1016/j.amc.2014.01.111. Google Scholar [27] F. A. Rihan, C. Tunc, S. H. Saker, S. Lakshmanan and R. Rakkiyappan, Applications of delay differential equations in biological systems,, Complexity, 2018 (2018), Article ID 4584389, 3 pages. doi: 10.1155/2018/4584389. Google Scholar [28] F. A. Rihan, C. Rajivganthi, P. Muthukumar, Fractional stochastic differential equations with Hilfer fractional derivative: Poisson Jumps and optimal control, Discrete Dyn. Nat. Soc., 2017(2017), Article ID 5394528, 11 pages. doi: 10.1016/j.cnsns.2013.05.015. Google Scholar [29] R. Sakthivel, R. Ganesh, Y. Ren and S. M. Anthoni, Approximate controllability of nonlinear fractional dynamical systems, Communications in Nonlinear Science and Numerical Simulation, 18 (2013), 3498-3508. Google Scholar [30] R. Sakthivel and R. Yong, Approximate controllability of fractional differential equations with state-dependent delay, Results in Mathematics, 63 (2013), 949-963. doi: 10.1007/s00025-012-0245-y. Google Scholar [31] B. Sundara Vadivoo, R. Ramachandran, J. Cao, H. Zhang and X. Li, Controllability analysis of nonlinear neutral-type fractional-order differential systems with state delay and impulsive effects, International Journal of Control, Automation and Systems, 16 (2018), 659-669. doi: 10.1007/s12555-017-0281-1. Google Scholar [32] J. Wang and H. M. Ahmed, Null controllability of nonlocal Hilfer fractional stochastic differential equations, Miskolc Math. Notes, 18 (2017), 1073-1083. doi: 10.18514/MMN.2017.2396. Google Scholar [33] J. R. Wang, M. Feckan and Y. Zhou, A survey on impulsive fractional differential equations, Fractional Calculus and Applied Analysis, 19 (2016), 806-831. doi: 10.1515/fca-2016-0044. Google Scholar [34] X. Zhang, P. Agarwal, Z. Liu, H. Peng, F. You and Y. Zhu, Existence and uniqueness of solutions for stochastic differential equations of fractional-order $q > 1$ with finite delays, Advances in Difference Equations, 2017 (2017), 1-18. doi: 10.1186/s13662-017-1169-3. Google Scholar
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Article outline | 2021-12-01 21:19:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.668329656124115, "perplexity": 4466.3687411991095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00482.warc.gz"} |
https://www.physicsforums.com/threads/linearly-independent-vectors.301132/ | # Linearly Independent Vectors (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### roam
1. The problem statement, all variables and given/known data
Let $$A = \left[\begin{array}{ccccc} 2&0&-2&1&-3 \\ 1&1&-3&0&-2\\1&0&-1&-1&3 \end{array}\right]$$
a) Find two linearly independent vectors u and v in R5 such that span{u,v}={$$x \in R^5$$ : Ax=0}.
b) Find three linearly independent vectors u,v and w such that col(A) = span{u,v,w}.
2. Relevant equations
3. The attempt at a solution
a) I row reduced A:
$$\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right]$$
I'm not sure how to answer this question. I think the first and second column are linearly independent and if we add them we get the third column. I need some help, I don't know what to do from here.
#### CompuChip
Homework Helper
How about you start solving the equation Ax = 0?
I.e. reduce
$$A = \left[\begin{array}{ccccc|c} 2&0&-2&1&-3 &0 \\ 1&1&-3&0&-2 &0 \\1&0&-1&-1&3 &0 \end{array}\right]$$
Then you can take, for example, x4 = r, x5 = s and write the solution as
xi = xi(r, s)
#### roam
How about you start solving the equation Ax = 0?
I.e. reduce
$$A = \left[\begin{array}{ccccc|c} 2&0&-2&1&-3 &0 \\ 1&1&-3&0&-2 &0 \\1&0&-1&-1&3 &0 \end{array}\right]$$
Then you can take, for example, x4 = r, x5 = s and write the solution as
xi = xi(r, s)
Hi
I reduced A and I'm left with:
$$= \left[\begin{array}{ccccc|c} 1&0&-1&0&0&0\\ 0&1&-2&0&-2&0 \\0&0&0&1&-3&0 \end{array}\right]$$
The linear system corresponding to the row-reduced matrix is:
x1-x3=0
x2-2x3-2x5=0
x4-3x5=0
variables corresponding to leading 1's are x1, x2 and x4, the rest I think are free variables, solving for the leading variables:
x1=x3
x2=2x5+2x3
x4=3x5
Now, I don't know where to go from here. I'm not exactly sure how to represent the solutions for this. How does this help us to solve part (a)?
#### Mark44
Mentor
x1, x2, and x4 can be found if you know or are given x3 and x5.
Code:
x_1 = x_3
x_2 = 2x_3 + 2x_5
x_3 = x_3
x_4 = 3x_5
x_5 = x_5
So any vector (x1, x2, x3, x4, x5) can be written as a linear combination of two vectors. I'm hopeful that you will notice these two vectors lurking in the arrangement above.
#### roam
Thanks, I notice it now, x3 and x5 are the two linearly independent vectors so u=(-1,-2,0) & v=(0,-2,-3).
Could you please explain to me part (b) of the question? I don't understand what the question wants
#### Mark44
Mentor
I don't think you understand the a part just yet. x3 and x5 are just plain old numbers, not vectors, let alone linearly independent vectors. And your vectors u and v have to be in R5, not R3 as the ones you show are.
Take a look again at the code block I put in my last response. There are two vectors in R5 hiding in there on the right side. If you stare at it long enough, you might see them.
#### SmashtheVan
Hint:
in post #4 set the free variables x_3 and x_5 to constants c_1 and c_2.
you can separate the resulting vector into the sum of two vectors, one containing the c_1's and the other containing only the c_2's
#### roam
I see it!
$$\left[\begin{array}{ccccc} x_{3}\\2x_{3} + 2x_{5}\\x_{3}\\3x_{5}\\x_{5} \end{array}\right]=$$ x3$$\left[\begin{array}{ccccc} 1\\2\\1\\0\\0 \end{array}\right]+$$x5$$\left[\begin{array}{ccccc} 0\\2\\0\\3\\1 \end{array}\right]$$
Is this what you meant? so u=(1,2,1,0,0) and v=(0,2,0,3,1), both in R5
Now, could you please explain how to solve part (b)?
#### SmashtheVan
good!
b is much easier than a
think about what the column space means. they want you to find the 3 linearly independent vectors in the given matrix which constitute the basis of the column space. You already did the first step in this by finding the row reduced matrix. What can you pull from that new matrix that might tell you about the column space?
#### Mark44
Mentor
Here's your row-reduced matrix:
$$\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right]$$
For part b, you want three linearly independent vectors in R3, that span the columns of the original matrix. IOW, you want three linearly independent vectors u, v, w, in R3 such that any column vector in your matrix is a linear combination of u, v, and w.
You can get those three vectors by inspection in your reduced matrix.
#### roam
If I had to guess I'd say the first, second and the fourth column which have pivot elements ( leading 1's) & they are linearly independent and the other two vectors can be written in terms of them. Am I right?
Mentor
Right.
#### roam
Thanks Mark. I have another question (my last question) about the same matrix, it looks similar to the part (b) but it's confusing:
Find linearly-independent vctors {u1, u2,...,um} such that: row(A)=span{u1, u2,...,um}
Justify that vectors are linearly independent.
How can I find this row vectors by looking at the reduced matrix?
#### SmashtheVan
How can I find this row vectors by looking at the reduced matrix?
Finding the Row Space is similar to the Column Space. First you find the transpose of the original matrix, and perform your row reductions to find the pivots as with before
#### Mark44
Mentor
Thanks Mark. I have another question (my last question) about the same matrix, it looks similar to the part (b) but it's confusing:
Find linearly-independent vctors {u1, u2,...,um} such that: row(A)=span{u1, u2,...,um}
Justify that vectors are linearly independent.
How can I find this row vectors by looking at the reduced matrix?
Here's your reduced matrix (taken from what you posted earlier)
$$= \left[\begin{array}{ccccc} 1&0&-1&0&0\\ 0&1&-2&0&-2 \\0&0&0&1&-3 \end{array}\right]$$
The rows in this reduced matrix span the rowspace of A, and are clearly (to my eye at least) linearly independent. Call these vectors u1, u2, and u3. To justify that they are linearly independent, show that the equation c1*u1 + c2*u2 + c3*u3 = 0 has only one solution for the constants c1, c2, and c3.
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• Solo and co-op problem solving | 2019-03-24 07:06:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.6048387289047241, "perplexity": 679.1281460534578}, "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-2019-13/segments/1552912203378.92/warc/CC-MAIN-20190324063449-20190324085449-00103.warc.gz"} |
http://www.askamathematician.com/2011/07/q-what-are-integral-transforms-and-how-do-they-work/ | # Q: What are integral transforms and how do they work?
Mathematician: If you have a function f(x) and a function k(x,s) then you can (as long as the product of f(x) times k(x,s) is integrable on the set X) always form another function of a new variable s as follows:
$F(s) = \int_{X} k(x,s) f(x) dx$
We have just “transformed” the function f(x) into the function F(s) via an “integral transform.” Why the hell would anyone want to do this? Well, the function F(s) is sometimes easier to work with than f(x) itself, or tells us interesting information about f(x) that it would be hard to figure out in other ways.
Of course, the interpretation of this new function F(s) will depend on what the function k(x,s) is. Choosing k(x,s) = 0, for example, will mean that F(s) will always be zero. This is pretty boring and tells us nothing about f(x). Whereas choosing $k(x,s) = x^{s}$ will give us the sth moment of f(x) whenever f(x) is a probability density function. For s=1 this is just the mean of the distribution f(x). Moments can be really handy.
A particularly interesting class of functions k(x,s) are ones that produce invertible transformations (which implies that the transform destroys no information contained in the original function). This will occur when there exists a function K(x,s) (the inverse of k(x,s)) and a set S such that
$f(x) = \int_S K(x,s) F(s) ds$
that undoes the original transformation (or, at least, undoes it for some large class of functions f(x)).
Whenever this is the case, we can view our operation as changing the domain from x space to s space. Each function f of x becomes a function F of s that we can convert back to f later if we so choose to. Hence, we’re getting a new way of looking at our original function!
It turns out that the Fourier transform, which is one of the most useful and magical of all integral transforms, is invertible for a large class of functions. We can construct this transformation by setting:
$k(x,s) = e^{-i x s}$
$K(x,s) = e^{i x s}$
which leads to a very nice interpretation for the variable s. We call F(s) in this case the “Fourier transform of f”, and we call s the “frequency”. Why is s frequency? Well, we have Euler’s famous formula:
$e^{i x s} = \cos(x s) + i \sin(x s)$
so modifying s modifies the oscillatory frequency of cos(xs) and sin(xs) and therefore of k(x,s). There is another reason to call s frequency though. If x is time, then f(x) can be thought of as a waveform in time, and in this case |F(s)| happens to represent the strength of the frequency s in the original signal. You know those bars that bounce up and down on stereo systems? They take the waveforms of your music, which we call f(x), then apply (a discrete version of) the Fourier transform to produce F(s). They then display for you (what amounts to) the strength of these frequencies in the original sound, which is |F(s)|. This is essentially like telling you how strong different notes are in the music sound wave.
Below are a few other neat examples of integral transform.
The Laplace transform:
$k(x,s) = e^{-x s}$
This is handy for making certain differential equations easy to solve (just apply this transformation to both sides of your equation!)
The Hilbert transform:
$k(x,s) = \frac{1}{\pi} \frac{1}{x-s}$
This has the property that (under certain conditions) it transforms a harmonic function into its harmonic conjugate, elucidating the relationship between harmonic functions and holomorphic functions, and therefore connecting problems in the plane with problems in complex analysis.
The identity transform:
$k(x,s) = \delta(x-s)$
Here $\delta$ is the dirac delta function. This is the transformation that leaves a function unchanged, and yet it manages to be damn useful.
This entry was posted in -- By the Mathematician, Engineering, Equations, Math. Bookmark the permalink.
### 8 Responses to Q: What are integral transforms and how do they work?
1. Matt says:
Perfect, I’ve been taught about Fourier, Laplace transforms and Green’s function, but never had them all tied together – if only someone had explained to me they were all just a specific form of a generalised integral transform!
2. Neal says:
More Fourier transform magic.
Say we have some partial differential operator with constant coefficients. For concreteness’ sake, let’s let $A = 5\partial_{xy} + 7\partial_{yy}$, but this will be true for any such PDO. Then for ANY (reasonable) function f, the Fourier transform $\mathcal{F}Af$ is … $(5\xi\eta + 7\eta^2)\mathcal{F}f$.
In other words, partial differentiation gets taken, under a Fourier transform, to multiplication by polynomials. That is, a most mysterious and difficult-to-understand mathematical phenomenon is, from a different perspective, one of the easiest and best-understood!
Words do not fully express how awesome that is.
4. Matt says:
What are indefinite integrals?
5. The Physicist says:
They are the “inverse operation” of derivatives. When taking an indefinite integral of some function F, you ask yourself “the derivative of what, will give me F?”.
6. shamoona jabeen says:
plz tell me how the integral transform works on matrices and how the inverse integral transform works on matrices?
7. Jeff says:
Nice article … but for Radon transform, what is its kernel function k(x,s)? Does it fit into this framework? | 2015-07-04 05:02:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.875259280204773, "perplexity": 630.658788600057}, "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-2015-27/segments/1435375096504.12/warc/CC-MAIN-20150627031816-00041-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/intensity-of-different-color-light-its-relation-to-energy-and-amplitude.641337/ | # Intensity of different color light & its relation to energy AND Amplitude
1. Oct 5, 2012
### mlgpawnstar
Hi Let's stick to the classical limit.
We have 2 monochromatic light waves of same intensity. Let's say one is blue and one is red.
Now this means the individual photons of the blue light have more energy(obviously, higher frequency) and the red light photons have lower energy. But since the two light waves have equal intensity, they have the same energy arriving/second.. Right? I think this part is obvious. This then means that the red light wave has more photons and the blue light wave has fewer photons.
Well let's move on to amplitude of the light wave, now this sort of scales with the intensity squared so these two light beams should have the same Amplitude? But Amplitude can also be seen as a function of the number of photons..
Here in lies my current confusion. Sorry if this is so simple. Anyone know of a basic way to get around this and stay consistent?
Either the two light is going to have different amplitude or the same amplitude while the have the same intensity but different frequencies.
2. Oct 5, 2012
There is no contradiction. The number of photons in a light wave is dependent on the amplitude. You can calculate the number per second exactly as you described, by looking at the power of the light, and using $$E_{photon}=\hbar \nu$$ so red light of the same intensity has more photons.
3. Oct 5, 2012
### Alkim
Hi, light intensity or energy flux is proportional to photon flux, i.e number of photons per unit time per unit surface and to individual photon energy. Does that answer your question?
4. Oct 8, 2012
### mlgpawnstar
Hi 0xDEADBEEF if # of photons depends on Amplitude, and if Amplitude comes from Intensity, since these two beams have the same intensity, they will have the same Amplitude, and thus the same # of photons?? (Obviously this isn't the case.)
"amplitude" is what is throwing me off here.
Alkim: I understand this but how does this relate with Amplitude?
5. Oct 8, 2012
### sophiecentaur
How do you make that conclusion? If 'amplitude' relates to the amount of energy flux and short wavelength photons have more individual energy then either the same amplitude would imply more red photons or the same number of red photons would imply lower amplitude. (Two ways of looking at the same thing.) | 2017-10-18 17:22:40 | {"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": 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": 0.7637327909469604, "perplexity": 486.4402116561955}, "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/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181655-00524.warc.gz"} |
https://expskill.com/question/find-the-integral-of-frac5x4sqrtx59/ | Find the integral of $$\frac{5x^4}{\sqrt{x^5+9}}$$.
Category: QuestionsFind the integral of $$\frac{5x^4}{\sqrt{x^5+9}}$$.
Editor">Editor Staff asked 11 months ago
Find the integral of $$\frac{5x^4}{\sqrt{x^5+9}}$$.
(a) $$\sqrt{x^5+9}$$
(b) $$2\sqrt{x^5-9}$$
(c) 2(x^5+9)
(d) $$2\sqrt{x^5+9}$$
I have been asked this question in examination.
This intriguing question comes from Methods of Integration-1 in portion Integrals of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Editor">Editor Staff answered 11 months ago
Correct option is (d) 2\sqrt{x^5+9}
For explanation I would say: Let x^5+9=t
Differentiating w.r.t x, we get
5x^4 dx=dt
\int \frac{5x^4}{\sqrt{x^5+9}} dx=\int \frac{dt}{\sqrt{t}}
=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{t}
Replacing t with x^5+9, we get
\int \frac{5x^4}{\sqrt{x^5+9}} dx=2\sqrt{x^5+9}. | 2022-11-30 18:03:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5821359753608704, "perplexity": 11107.066325067924}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00402.warc.gz"} |
http://motls.blogspot.com/2013/11/eu-will-ban-vacuum-cleaners.html | ## Monday, November 04, 2013 ... /////
### EU will ban vacuum cleaners...
...above 1,600 W since Sept 2014
Joseph S. sent me a warning about some plans by the European Commission – it's a continental Big-Brother government of a sort – to ban vacuum cleaners at or above 1,600 watts of power consumption since September 2014. The Czech press is full of it, people are upset, but TRF readers may prefer the U.K. press: The Daily Mail, The Sunday Times, The Telegraph.
A few months ago, I bought a new vacuum cleaner, a cheaper model above with consumption 1,800 watts of input power (300 W of output power). Before that, I was using a gadget with a substantially lower power and I can tell you that you can surely tell the difference!
For U.S. readers who use "amps": one "amp" is 110 watts because the voltage in the U.S. is 110 volts and the factors of the square root of two are already accounted for. ;-) So the maximum wattage will be 14.5 amps in 2014.
The average model on the U.K. market these days has a 1,800-watt motor and some of them have a 2,200-watt motor. It means that in less than a year, a clear majority of the currently sold models will be banned! That includes the two bestsellers on the Czech market.
To make things worse, the apparatchiks want to limit the input power to 900 watts (equivalent to 8.2 amps: U.S.) since 2017. Even my weaker previous cleaners would be threatened. Someone could argue that the efficiency will be increased but such increases have their tough physical limits. We learn that those "powerful" people have their spokespersons who are armchair physicists:
A spokesman for the European Commission said: "The whole point of the regulations is to go away from the idea that high power means better performance - which is not necessarily the case."
Sure. Higher power doesn't necessarily mean higher suction except that it almost always does due to the most general laws of physics such as energy conservation, assholes.
In Germany, they are talking about rabies of banning, or whatever is the right translation to English. Not even the most intrusive, obnoxious, idiotic communists would ever think of something like that. The producers are sort of fine with that because they were planning new models that usually didn't exceed 1,600 watts but I am curious whether they will still be fine in 2017.
The world may be assumed to have lots of energy in coming years, making similar restrictions even more insane. If you have any doubts about the public opinions towards such policies in Czechia, let me just quote the most upvoted and most downvoted comments under an article at novinky.cz (meaning News.CZ):
Martin Knápek, Veřovice, score +649 votes:
"Brussels will ban powerful vacuum cleaners. In Germany, they are already talking about the rabies of bans."
Let the whole corrupt EU go finally to... !!!! [The dots stand for "the asshole". Even if your language doesn't use the idiom, you may guess whether the writer is praising the EU Commission or not.]
Václav Chaloupka, Zlatá, score –226 votes
I have a modern 1,600 watts vacuum cleaner which is just enough for regular maintenance of my apartment. If I will ever occasionally need something more than that, it's no problem to call professional janitors who have more powerful engines.
Do they and will they?
Plastic bags
The EU commission also wants to ban most plastic bags used in the supermarkets. It's being said that ocean animals end up in those bags. However, as these imbeciles apparently haven't noticed, countries like Czechia are land-locked so one has to travel for thousands of miles to get to the sea. Couldn't these pushy obnoxious assholes at least reduce the considered ban to towns that are less than 100 km from the sea?
If my signature were needed for using plastic bags for yet another constructive procedure, to choke Maoist Comrade Barroso and his thugs and bitches in the EU commission, it would be a formality.
#### snail feedback (16) :
I would wonder who is behind this new craziness. But it is quite possible that there are vacuum cleaners manufacturers lobbying for it. After all, the best way to sell a new vacuum cleaner is to ban the old one! ;-) I do not see anybody else standing to make comparable profit out of it.
And then in 2020 they will "figure out" that it is utter insanity, repel the ban, and everybody will rush to buy once again the new, more powerful vacuum cleaner. It sounds like a great business plan.
reader Luboš Motl said...
A very interesting thought, Honzo!
I know we will not agree on this. I see it totally opposite.
In practice I think we are rich enough society to disallow plastic bags in shops and burning plastic in boilers. And that private property is practical approach to life, but not human right of the same strength like the right for life.
In philosophical view though:
There is in fact no human right to own a property otherwise you must see taxation as the violation of this right. The private property is nothing else that just the fact that society around you accepts this property - in our country in twentieth century some houses changed its owner four times without being sold just because the regime/country has changed. But every owner had a "paper for it".
After all money is also virtual thing dependent on countries - with hyperinflation like in Germany 1923 you could just throw your money in the boiler, after few days if not used.
Because you cannot in practise sell anything without using money and rules (both created by society) then society has also right to modify the rules.
reader Luboš Motl said...
Petr, right, we can't agree an a single word because you are a hardcore commie and I am not.
You always say what you want to steal or ban but you never present any sensible alternative that could work without crippling the human happiness and living standards.
Of course that from a moral viewpoint, taxation is just another kind of robbery. How it could be otherwise? Of course that it is a violation of the people's right to deal with their private property and income. It's an institutionalized organized crime, a mafia-style system to rule the whole societies, where the mafia is so powerful and reaching almost all portions of the nation that there's no sensible realistic defense and people who realize that robbery is wrong have surrendered to this despotism, too.
A society doesn't have to have any money to respect private properties. Private properties, at least in some cultures, existed well before anything that could be counted as "the money". To claim that someone (the government) has the right to steal because everything is money and money be lose value is a morally reprehensible rationalization of immoral behavior.
There aren't usable alternatives that could replace the plastic bags in all contexts where people at our level of wealth use them without leading them to extra expenses they will find unacceptable.
reader Peter F. said...
I share your sentiment BUT you'll regret your friendly lenience as soon as someone walks around with a dog's poo on the sole of one of their boots (god forbid on both of them)!
Hi Peter, ok yuk, still on hard floor it's easy to clean... I guess lime is efficient against the smell? You should see my hall when my son comes back from his rugby trainings! ;-) it smells nice though, earth and grass.
Dear Luboš,
I am far from commie, I vote for liberal mid to right parties, because I think it is effective to support individual motivation but I do not think that property is a the most important human right. If we should label people, you are a hardcore libertarian.
However I think that the term property only has a meaning in society and thus taxation is perfectly legal and moraly justifiable. Without society that accepts it term property is just an empty word.
And after all the proposed EU regulation on plastic bags targets limiting of plastic bags usage - and countries could choose how they reach it - for example by taxation of plastic bags. If they replace it partly by paper or reusable cloth bags is up to them.
BTW If there were no taxation there would be very little to none particle physics nowadays, No one would pay for LHC and other stuff.
In the same vein the EU demands the same toilets flush across the continent. http://www.dailymail.co.uk/news/article-2480829/Bog-standard-Brussels-demands-toilet-flush-Continent-discovering-Brits-use-water.html
reader Luboš Motl said...
LOL, amazing. The main reason why Turkey is being invited to the EU is probably for the masters of Brussels to impose the Turkish toilets across the EU;
Is that actually a correct functionality in the video so that it sprays the shit across the whole room or did I misunderstand the mechanism? ;-)
Lol. One wonders why they bother with a hole :-)
Btw you might rethink agreeing with Turkey joining the EU: they have put a few Marxists journalists in prison for 3000 years!
reader Eugene S said...
Ahem...
It's a pan-European thing, actually, this obsession with what comes out of your nether regions:
And here are the Germans... if I nound nunny, it necaus I put a clo'pin on my noh...
You are too sarcastic on this blog. I for one welcome these significant steps forward by our EU overlords 8).
LOL to
Now though are we finding out that bugs in the urinals was started by the NSA? Naturally Chancellor Merkel would have become suspicious of a urinal in her offices so they decided .... .Oh anyway , must go ... :-)
reader Kimmo Rouvari said...
WTF? :D Whaaaat? Mental note: Never go to Turkey...
reader Kimmo Rouvari said...
The usual, follow the money, gives the answer ;-) | 2017-07-25 02:29:23 | {"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25403425097465515, "perplexity": 2992.0955982533496}, "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-2017-30/segments/1500549424960.67/warc/CC-MAIN-20170725022300-20170725042300-00402.warc.gz"} |
http://mathhelpforum.com/calculus/141579-taylor-series-word-problem.html | # Thread: taylor series word problem
1. ## taylor series word problem
the nth derivative of f at x=1 is given by:
$f^{(n)}(1)= \frac{(n+2)!}{2^n}$
for n≥1 and f(1)=2. write the first 5 terms and the general term of the taylor series for f about x=1
2. Originally Posted by yoman360
the nth derivative of f at x=1 is given by:
$f^{(n)}(1)= \frac{(n+2)!}{2^n}$
for n≥1 and f(1)=2. write the first 5 terms and the general term of the taylor series for f about x=1
general term for a taylor series centered at x = 1 ...
$\frac{f^{(n)}(1) (x-1)^n}{n!}$
... so build it. | 2016-12-11 10:17:09 | {"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": 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": 0.7815067172050476, "perplexity": 562.0890740490868}, "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-2016-50/segments/1480698544672.33/warc/CC-MAIN-20161202170904-00381-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://quomodocumque.wordpress.com/category/baseball/ | ## Unanimity
Mariano Rivera was elected to the Hall of Fame, the first player ever to appear on every single ballot. Why has this never happened? Because there are a lot of ballots and thus a lot of opportunities for glitchy idiosyncrasy. In 2007, eight voters left Cal Ripken, Jr. off. What possible justification could there be? Paul Ladewski of Chicago’s Daily Southtown was one of the eight. He turned in a blank ballot that year. He said he wouldn’t vote for anyone tainted by playing during the “Steroids Era.” In 2010, he voted for Roberto Alomar.
Tagged
## Small baseball triangles
This all started when CJ asked which three baseball stadiums formed the smallest triangle. And we agreed it had to be the Brewers, the White Sox, and the Cubs, because Milwaukee and Chicago are really close together.
But it seems like cheating to use two teams in the same city. The most elegant way to forbid that is to ask the question one league at a time. Which three American League parks form the smallest triangle? And what about the National League?
First of all, what does “smallest” mean? There are lots of choices, but (perhaps inspired by the summer we played a lot of Ingress) we asked for the triangle with the smallest area. Which means you don’t just want the parks to be close together, you want them to be almost collinear!
I asked on Twitter and got lots of proposed answers. But it wasn’t obvious to me which, if any, were right, so I worked it out myself! Seamheads has the longitude and latitude of every major league ballpark past and present in a nice .csv file. How do you compute the area of a spherical triangle given longitudes and latitudes? You probably already know that the area is given by the excess over pi of the sum of the angles. But then you gotta look up a formula for the angles. Or another way: Distance on the sphere is standard, and then it turns out that there’s a spherical Heron formula for the area of a spherical triangle given its edgelengths! I guess it’s clear there’s some formula like that, but it’s cool how Heron-like it looks. Fifteen lines of Python and you’re ready to go!
We were right that Brewers-White Sox-Cubs form the smallest major league triangle. And the smallest American League triangle is not so surprising: Red Sox, Yankees, Orioles, forming a shortish line up the Eastern Seaboard. But for the National League, the smallest triangle isn’t what you might expect! A good guess, following what happened in the AL, is Mets-Phillies-Nationals. And that’s actually the second-smallest. But the smallest National League triangle is formed by the Phillies, the Nationals, and the Atlanta Braves! Here’s a piece the geodesic path from SunTrust Park in Atlanta to Citizen’s Bank Park in Philly, courtesy of GPSVisualizer:
Not only does it go right through DC, it passes about a mile and a half from Nationals Park!
Another fun surprise is the second-smallest major league triangle: you’d think it would be another triangle with two teams in the same city, but no! It’s Baltimore-Cincinnati-St. Louis. Here’s the geodesic path from Oriole Park at Camden Yards to Busch Stadium:
And here’s a closeup:
The geodesic path runs through the Ohio River, about 300m from the uppermost bleachers at Great American Ball Park. Wow!
Now here’s a question: should we find it surprising that the smallest triangles involve teams that are pretty far from each other? If points are placed at random in a circle (which baseball teams are definitely not) do we expect the smallest-area triangles to have small diameter, or do we expect them to be long and skinny? It’s the latter! See this paper: “On Smallest Triangles,” by Grimmet and Janson. Put down n points at random in the unit circle; the smallest-area triangle will typically have area on order 1/n^3, but will have diameter on order 1. Should that have been what I expected?
PS: the largest-area major league triangle is Boston-Miami-SF. Until MLB expands to Mexico City, that is!
## The greatest Red Sox / Dodger
After game 2 it was already clear this was an NLCS so great it had to go seven, and it did. But game seven wasn’t a great game seven. After six hard-fought games, the Brewers never really mounted a threat, and went down 5-1. Keenest pain of all was that I got what I’d been waiting for the whole series; a chance for my beloved Jonathan Schoop to be the hero. He came in to pinch hit for starter Joulys Chacin in the bottom of the second, with two on and two out and the Brewers down by 1. Schoop grounded out. He was 0 for the postseason in 6 plate appearances.
So here we have it, a Red Sox / Dodgers series, and so it’s time for my annual post about what player had the best combined career for both teams. (Last year: Jimmy Wynn was the greatest Astro/Dodger.)
The greatest Red Sox / Dodger? A player I’d never heard of, even though he was just a little before my time: Reggie Smith. Played in one World Series for the Red Sox (1967) and three for the Dodgers (1977,1978,1981). Went to the All-Star Game with both teams. Hit 300 home runs, cannon of an arm in the outfield, got 0.7% of the vote the one and only time he was up for the Hall of Fame. Well, here’s his all time distinction; with 34.2 WAR for the Red Sox and 19.4 for the Dodgers, he’s the greatest Red Sox / Dodger of all time.
Surprisingly, given how old these teams are, the top Red Sox / Dodgers of all time are mostly recent players. Derek Lowe is the top pitcher (19.4 WAR for Boston, 13.3 for LA.) Adrian Gonzalez, Manny Ramirez, and Adrian Beltre are also worthy of mention. The only old-time player who was a contender was Dutch Leonard, who actually pitched for Boston in the last Red Sox – Dodgers World Series in 1916, notching a complete game win. But that guy never actually pitched for the Dodgers! My search got confused because it turns out there were two Dutch Leonards, the second of whom was a Dodger to start his career. Doesn’t count!
Tagged ,
## NLCS game 2: Dodgers 4, Brewers 3
In 35 years of watching baseball I had never been to a postseason game, until this Saturday, when I was able to get two tickets to Game 2 of the National League Championship Series through a wonderful terrific beautiful friend with connections.
First of all, I salute whoever the free spirit was who slammed a Zima right before entering Miller Park.
The game started at 3pm; in late afternoon with the roof shut at Miller Park there’s a slant-line of sunlight across the field which is lovely to look at and probably terrible to hit in.
And indeed there wasn’t a lot of hitting to start with. Wade Miley, once a bad Oriole, now a good Brewer, never looked dominant, giving up lots of hard-hit balls including a shot by Jeremy Freese in the first that Lorenzo Cain hauled back in from over the wall, but somehow pitched 5 2/3 only allowing 2 hits (and collecting a single himself.) Hyun-jin Ryu matched him zero for zero. Every seat in Miller Park full, everyone attentive to the game, a level of attention I’ve never seen there. The guy behind us kept saying “NASTY, throw something NASTY.” CJ believes he sees Marlins Man in the front row — he’s right! Brewers get runners on second and third with one out, Dodgers intentionally walk Yelich to load the bases, (wave of boos), Braun delivers the RBI groundout but can’t score any more. Travis Shaw hits a solo shot to deepest center, the Brewers go up 3-0, and people start to smell win, but the Dodgers lineup has good hitters all the way down to #8 and the usually reliable Milwaukee bullpen starts to crack. Jeremy Jeffress comes in with runners on first and second and nobody out, immediately gives up a single to Joc Pederson, now they’re loaded, still nobody out, Brewers up 3-1. Manny Machado, on third base, keeps jumping off the bag, trying to distract Jeffress. But Jeffress strikes out Yasiel Puig, who’s so angry he smashes his bat over his knee. Crowd exults. Then he walks light-hitting catcher Austin Barnes to force in a run. Nobody’s up in the bullpen. Crowd panics. Yasmini Grandal comes in to hit in the pitcher’s spot and Jeffress somehow gets the double play ball and is out of it. But the next inning, Jeffress stays in a little too long; Chris Taylor leads off with a lucky little dink of an infield single and then Turner muscles a ball out to the short corner in left field; 4-3 Dodgers and it stays that way.
But the Brewers do threaten. 43,000 Brewers fans want to see Yelich get one more chance to be the hero. Hernan Perez draws a walk in the bottom of the ninth, steals while Cain strikes out. So Yelich gets to bat with 2 outs and a runner in scoring position. He grounds out. Crowd deflates. But that’s all you can ask of a baseball game, right? The hitter you want in the situation you want with the game on the line and whatever happens happens. Great baseball. Great team. I hope they win it all. Maybe I’ll try to be there when they do.
Tagged ,
## If I have caught a foul ball it is only by standing on the shoulders of giants
“Combined cap and baseball mitt,” A patent by Richard Villalobos, 1983.
This is meant for fans, not players. The idea is that if a foul ball comes towards you, you may not have time to grab a glove you’ve stashed at your feet. Rather, you quickly slip your hand into the cap-mounted glove and snag the foul with your hat still attached to the back of your hand.
## Why are the 2018 Orioles so bad, so very, very, very bad?
The Orioles held on to win one tonight, 5-3 over the A’s, getting out of a bases-loaded one-out jam in the top of the 8th, so maybe for once I’m emotionally able to take a look at this loss of a season.
Baltimore was not supposed to be great this year. But they weren’t supposed to be terrible, either. The Orioles thought they had an outside chance at a wild card in Machado’s walk year and signed free-agent pitchers Alex Cobb and Andrew Cashner. Before the season, Fangraphs projected them to win 75 games or so and battle with the Rays for fourth place in the AL East.
They’re now 42-104 and en route to the worst record in the team history.
How was everyone so wrong?
Here’s my take. Nobody was wrong. The projections were the right projections to make. Sometimes you get unlucky and everything goes to shit at the same time.
First: the Orioles are not as bad as that record; they’ve been unlucky. Their Pythagorean record is 50-96. That’s not good. But it’s not historically bad.
Second: let’s look at the players who contributed at least 2 WAR to the 2017 Orioles.
• Jonathan Schoop
• Trey Mancini
• Welington Castillo
• Tim Beckham (in just 50 games!)
• Dylan Bundy
• Mychal Givens
• Kevin Gausman
Let’s throw in Cobb and Cashner too, since they delivered that much WAR to their 2017 teams. This is a pretty long list of players from whom the Orioles were counting on some production (except Castillo, who was cut loose.)
Of these, Cobb, Schoop, Beckham, Bundy, and Givens each had the worst season of their career. So did Mancini, though his career’s only two years long. Cashner was back to his 2016 level of bad after a good 2017. Gausman and Machado played about as well as you might expect. (Machado’s hitting improved a lot, but the move to shortstop made him less defensively valuable.) Jones hit as usual but baseball-reference rates his defense as having degraded enough to essentially eliminate his value. And Chris Davis, of course, who was just sort of OK in 2017 but delivered a lot of value in 2015 and 2016, is turning in one of the worst seasons in major league history; his average currently sits at .174. Or Chris Tillman, a very good pitcher as recently as 2016, who stunk in 2017 and unfathomably stunk even more this year until finally being taken out back and released. (I saw what may end up being his last major league win.)
So you’re basically taking this entire list of players, who together might have been expected to constitute the core of an respectably mediocre ballclub, and saying that not one of them will play better than you expect, and more than half of them will play worse than you could have reasonably imagined.
I think the Orioles just got snakebit.
But what happens now? The Orioles haven’t been bad for very long, so they don’t have recent high draft picks. Machado, Schoop, and Gausman are gone, along with half the bullpen. Jones might go. I think in 2019 we are just going to watch Jonathan Villar and Cedric Mullins cavort in front of almost nobody.
I’ll watch that. And I’ll enjoy it, because I always do. The losses don’t mean much to me. Every win makes me proud.
Tagged ,
## Brewers 6, Marlins 5 / Bucks 104, Celtics 102 / Orioles 6, Tigers 0
I’ve lived in Madison for 13 years and this is the first time I’ve noticed anybody caring about the Milwaukee Bucks. It’s definitely the first time I’ve cared about the Milwaukee Bucks. But now the Bucks have a legitimate superstar in Giannis Antetokoumnpo and a likeable cast of supporting characters like 19-year-old former refugee and skinny blockmaster Thon Maker. The kids had a rare unscheduled day on Sunday and the Bucks were in the playoffs against the Celtics and there were nosebleed tickets on Stubhub for \$40 apiece so why not?
You may know that I kind of hate driving so if I’m gonna drive all the way to Milwaukee it’s got to be for more than a Bucks game. When I thought about what the kids would really want to do it was pretty clear — see the Brewers, stay over, then see the Bucks. So that’s what we did!
Notes on the Brewers:
• I got lost in the impossible off-ramp spaghetti surrounding Miller Park and we ended up not getting into the ballpark until the second inning. The Brewers were already down 4-0. 4-0! To the sad Miami Marlins, the team Derek Jeter is using as a tax dodge, the team so bad Marlins Man cancelled his season tickets!
• But as soon as we sat down, Travis Shaw muscled a huge home run to left center. Didn’t even look like he got all of it, he kind of sliced it. But Travis Shaw is a big strong man.
• Brewers just keep creeping back. Crowd stays in the game, at no point do you really feel the Brewers are out of it. Three straight Brewers hit what look like go-ahead home runs but each dies at the wall. (Ryan Braun at least gets a sacrifice fly out of it.) In the 8th, Derek Dietrich loses an Eric Sogard fly ball in the, I dunno, the lights? The roof? He plays for the Marlins and he just doesn’t care? Anyway the ball plunked down right next to him, Shaw hustles in from second to tie it, Eric Thames, who starts the play on first, tries to get in behind with the go-ahead run but is tagged out at or rather substantially before the plate because Eric Thames made a bad decision.
• Josh Hader looks like he should be playing bass in Styx.
• Then comes the bottom of the 9th and the play you might have read about. Still tied 5-5. Jesus Aguilar, who’s already warmed up twice in the on-deck circle, finally gets his chance to pinch-hit against Junichi Tazawa. Gets behind 0-2. And then just starts fouling, fouling, fouling. Takes a few pitches here and there. Full count. Foul, foul, foul. And on the 13th pitch, Aguilar launches it to center field. I thought it was gonna be one more death on the warning track. But nope; ball gets out, game over, fireworks. I felt like my kids got to see true baseball.
On to Milwaukee. Bucks play the Celtics at noon, in what, if they lose, could be the last ever game played at Bradley Center. (This is a bit of a sore point for UW folks, who absorbed as a budget cut the \$250m state contribution to the arena’s cost.) We have breakfast at the hotel and chat with a nice older couple in Packers/Celtics gear — what? — who turn out to be Boston forward Al Horford’s aunt and uncle from Green Bay.
This is only the third NBA game I’ve been to, CJ’s second, AB’s first. We wander around inside the arena for a bit. Two separate groups of Bucks cheerleaders come up to AB and applaud her curly hair. I think people are especially struck by it when they see us together, because I don’t have curly hair, except here’s a little-known fact: I do have curly hair! I just keep it short so it doesn’t curl. In 1995 or so it looked like this:
Anyway. The atmosphere, as I have promised AB, is more intense than baseball. Bucks build up a 19-point lead and seem poised to coast but the Celtics come back, and back, and back, and finally go ahead with 52 seconds left. Jaylen Brown plainly capable of taking over a game. Aron Baynes has a very dumb-looking haircut. Milwaukee’s Thon Maker is ridiculously skinny and has very long arms. He’s just 21, a former refugee from South Sudan. We saw his first game as a Buck, an exhibition against the Mavericks at Kohl Center. Those long skinny arms can block a shot.
Game tied at 102, 5 seconds left, Malcom Brogdon (called “The President” — why?) misses a layup, and there, rising like a Greek column above the scene, is the Greek arm of Giannis Antetokounmpo — the tip-in is good, Celtics miss the desperation last shot, Bucks win 104-102, crowd goes berserk.
I was going to blog about this last week but got busy so let’s throw in more sports. Bucks eventually lose this series in 7, home team winning every game a la Twins-Braves 1991. The next Friday, I’m giving a talk at Maryland, and the Orioles are playing that night. It’s been five years since I’ve seen OPACY. I brought CJ along this time, too. The Orioles are not in a good way; they’ve won 6 and lost 19, though 3 of those 6 were against New York at least. Attendance at the game, on a beautiful Friday night, was just over 14,000. The last baseball game I went to that felt this empty and mellow was the AAA Tucson Toros, several months before they moved to El Paso and became the Chihuahuas. Chris Tillman, tonight’s starter, was the Orioles’ ace five years ago. Now he’s coming off a 1-7 season and has an ERA over 9.
So who would have thought he’d toss seven shutout innings and take a no-hitter into the fifth? Never looked overpowering but kept missing bats. His first win in almost a year. Manny Machado, surely now in his last year as an Oriole, strokes a home run to dead center to get things started. It’s a beautiful thing. It doesn’t even look like he’s working hard. It’s like he’s just saying “Out there. Out there is where this ball should be.” Pedro Alvarez homers twice, in exactly the opposite manner, smashing the ball with eye-popping force. Jace Peterson, who the Orioles picked up off the Yankees’ scrap heap, steals third on the shift when the Tigers third baseman forgets to pay attention to him. He did the same thing against the Rays the night before. I am already starting to love him the way I love Carlos Gomez. Maybe now the Orioles are going to go back to being a bad team that makes good use of players nobody else wants, like Melvin Mora and Rodrigo López.
Besides me and CJ, this guy was at the game:
Never get tired of that flag.
## Hall of Fame ballots: some quick and dirty clustering
Since all the public Hall of Fame ballots are now available online in machine-readable form, thanks to Ryan Thibodeaux, I thought I’d mess around with the built-in clustering functions in sklearn and see what I could see.
The natural metric on ballots is Hamming distance, I guess. I first tried the AgglomerativeClustering package. I didn’t tell it what metric to use on the ballots, but I’m assuming it’s using Hamming distance, aka Euclidean in this case. I asked AgglomerativeClustering to break the Hall of Fame voters into 2 clusters. Guess what I found? There’s a cluster of 159 voters who almost entirely voted for Barry Bonds and Roger Clemens, and a cluster of 83 who universally didn’t. You won’t be surprised to hear that those who voted for Bonds and Clemens were also a lot more likely to vote for Manny Ramirez, Sammy Sosa, and Curt Schilling than the other cluster.
Which candidate was most notably unpopular among the Bonds-forgivers? That would be Omar Vizquel. He was on 53% of the steroid-rejector ballots! Only 24% of the Bonds cluster thought Omar deserved Cooperstown.
Then I tried asking AgglomerativeClustering for four clusters. The 83 anti-steroids folks all stayed together. But the bigger group now split into Cluster 1 (61 ballots), Cluster 2 (46), and Cluster 3 (52). Cluster 1 is the Joe Posnanski cluster. Cluster 2 is the Nick Cafardo cluster. Cluster 3 is the Tim Kurkjian cluster.
What differentiates these? Cluster 1 is basically “people who voted for Larry Walker.” The difference between Cluster 2 and Cluster 3 is more complicated. The Cluster 2 ballots were much more likely to have:
Manny Ramirez, Sammy Sosa
and much less likely to have
Mike Mussina, Edgar Martinez, Curt Schilling
I’m not sure how to read this! My best guess is that there’s no animus towards pitchers and DHs here; if you’re voting for Bonds and Clemens and Sosa and Ramirez and the guys who basically everybody voted for, you just don’t have that many votes left. So I’d call Cluster 2 the “2000s-slugger loving cluster” and Cluster 3 everybody else.
Maybe I should say how you actually do this? OK, first of all you munge the spreadsheet until you have a 0-1 matrix X where the rows are voters and the columns are baseball players. Then your code looks like:
import sklearn
model = AgglomerativeClustering(n_clusters=4)
modplay.labels_
which outputs
array([1, 0, 3, 1, 1, 1, 0, 0, 0, 0, 2, 1, 2, 1, 3, 0, 0, 0, 2, 1, 0, 3, 2,
1, 2, 1, 1, 3, 1, 3, 3, 0, 2, 2, 0, 1, 1, 1, 0, 2, 0, 0, 1, 2, 1, 3,
2, 2, 1, 3, 0, 2, 0, 3, 1, 0, 0, 2, 0, 2, 1, 2, 1, 0, 0, 0, 1, 0, 2,
0, 1, 1, 2, 0, 1, 3, 0, 0, 2, 0, 0, 0, 0, 0, 1, 1, 0, 0, 2, 0, 1, 0,
0, 0, 3, 1, 1, 0, 1, 0, 3, 1, 3, 3, 2, 0, 2, 1, 0, 2, 2, 3, 2, 3, 1,
3, 0, 3, 1, 0, 2, 1, 0, 0, 0, 1, 3, 1, 1, 3, 2, 3, 3, 2, 2, 0, 3, 3,
1, 0, 0, 2, 2, 3, 1, 3, 1, 2, 0, 1, 3, 1, 0, 0, 2, 3, 0, 2, 1, 0, 2,
1, 3, 3, 0, 1, 3, 1, 1, 0, 0, 2, 0, 1, 2, 0, 2, 1, 0, 0, 3, 3, 1, 1,
2, 3, 2, 0, 2, 0, 0, 1, 2, 1, 0, 3, 1, 3, 0, 3, 3, 3, 0, 3, 3, 3, 0,
2, 0, 3, 3, 0, 1, 0, 1, 2, 3, 2, 2, 0, 0, 0, 1, 3, 3, 1, 0, 0, 1, 3,
0, 2, 3, 1, 0, 0, 0, 0, 0, 3, 3, 3])
i.e. a partition of the voters into four groups.
(Agglomerative clustering naturally generates a hierarchical clustering, i.e. a tree with the HoF voters on the leaves; there must be some way to get sklearn to output this directly, but I don’t know it!
Of course, if you have a 0-1 matrix, you don’t have to cluster the rows — you can cluster the columns! This time, just for kicks, I used the hierarchical clustering package in scipy. I think this one is just agglomerating too. But it has a nice output package! Here, Y is the transpose of X above, a 0-1 matrix telling us which players were on which ballots. Then:
>> import scipy
>>> Dend = scipy.cluster.hierarchy.dendrogram(Z,labels=(a list of player names))
>>> plt.xticks(ha=’right’)
>>> plt.show()
gives
Not bad! You can see that Bonds and Clemens form their own little cluster in red. There is not that much structure here — maybe worth noting that this method may be dominated by the variable “number of votes received.” Still, the two stories we’ve told here do seem to have some coarse features in common: Bonds/Clemens are a cluster, and Larry Walker voting is kind of its own variable independent of the rest of the ballot.
OK, this picture was nice so I couldn’t resist doing one for the voters:
Pretty hard to read! I think that black cluster on the end is probably the no-Bonds-no-Clemens gang. But what I want you to notice is that there’s one outlying node all the way over to the left, which the clustering algorithm identifies as the weirdest ballot made public. It’s Sadiel LeBron, who voted for Clemens, Sosa, and Ramirez, but not Bonds. And also he voted for Andruw Jones and Omar Vizquel! Yeah, that’s a weird ballot.
I’m sure this isn’t the right way to visualize a 0-1 matrix. Here’s what I’d do if I felt like spending a little more time: write something up to look for a positive definite rank-2 matrix A such that
$A_{ij} > A_{ik}$
whenever voter i voted for player j but not player k. That models the idea of each player being a point in R^2 and each voter being a point in R^2 and then voters vote for every player whose dot product with them is large enough. My intuition is that this would be a better way of plotting ballots in a plane than just doing PCA on the 0-1 matrix. But maybe it would come out roughly the same, who knows?
Presumably there are known best practices for clustering subsets of a fixed set (or, more generally, finding good embeddings into visualizable metric spaces like the plane.) Tell me about them!
## The greatest Astro/Dodger
The World Series is here and so it’s time again to figure out which player in the history of baseball has had the most distinguished joint record of contributions to both teams in contention for the title. (Last year: Riggs Stephenson was the greatest Cub/Indian.) Astros history just isn’t that long, so it’s a little surprising to find we come up with a really solid winner this year: Jimmy Wynn, “The Toy Cannon,” a longtime Astro who moved to LA in 1974 and had arguably his best season, finishing 5th in MVP voting and leading the Dodgers to a pennant. Real three-true-outcomes guy: led the league in walks twice and strikeouts once, and was top-10 in the National League in home runs four times in the AstrodomeCareer total of 41.4 WAR for the Astros, and 12.3 for the Dodgers in just two years there.
As always, thanks to the indispensable Baseball Reference Play Index for making this search possible.
Other contenders: Don Sutton is clearly tops among pitchers. Sutton was the flip side of Wynn; he had just two seasons for Houston but they were pretty good. Beyond that it’s slim pickings. Jeff Kent put in some years for both teams. So did Joe Ferguson.
Who are we rooting for? On the “ex-Orioles on the WS Roster” I guess the Dodgers have the advantage, with Rich Hill and Justin Turner (I have to admit I have no memory of Turner playing for the Orioles at all, even though it wasn’t that long ago! It was in 2009, a season I have few occasions to recall.) But both these teams are stocked with players I just plain like: Kershaw, Puig, Altuve, the great Carlos Beltran…
Tagged , ,
## Game report: Cubs 5, Brewers 0
• I guess the most dominant pitching performance I’ve seen in person? Quintana never seemed dominant. The Brewers hit a lot of balls hard. But a 3-hit complete game shutout is a 3-hit complete game shutout.
• A lot of Cubs fans. A lot a lot. My kids both agreed there were more Cubs than Brewers fans there, in a game that probably mattered more to Milwaukee.
• For Cubs fans to boo Ryan Braun in Wrigley Field is OK, I guess. To come to Miller Park and boo Ryan Braun is classless. Some of those people were wearing Sammy Sosa jerseys!
• This is the first time I’ve sat high up in the outfield. And the view was great, as it’s been from every other seat I’ve ever occupied there. A really nice design. If only the food were better.
Tagged | 2019-04-23 02:58:24 | {"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": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23750142753124237, "perplexity": 3527.651121072055}, "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/1555578584186.40/warc/CC-MAIN-20190423015050-20190423041050-00177.warc.gz"} |
https://www.physicsforums.com/threads/asynchronous-motor-question.694986/ | # Asynchronous motor question
1. Jun 2, 2013
### Marketgirl
1. The problem statement, all variables and given/known data
number of pole pairs p =2
line frequency f=50hz
number of rotations n=1250 1/min
2. Relevant equations
3. The attempt at a solution
question is number of rotations of the rotary magnetic field (1/min units)
the slip of the machine(s)
the frequency of the rotor current ?
please help me to solve this kind of problems i have like 5 more questions which is smilar to this one so i can learn how to solve others thank you from now good luck at your exams :)
2. Jun 2, 2013
### rock.freak667
If you have the line frequency and the number of pole pairs, you can get the rotation of the stator Ns by using
$$N_s = \frac{120f}{p}$$
where f is the line frequency and p is the number of poles.
How is slip defined?
3. Jun 2, 2013
### Marketgirl
s = Ns-Nr/Ns Ns is stator electrical speed, and Nr is rotor mechanical speed. thank you for your helps sir.
4. Jun 2, 2013
### rock.freak667
Right, so using the formula I gave you above, you can get Ns. They gave you the value of Nr, so can you get the slip?
5. Jun 2, 2013
### Marketgirl
yes sir your help was enough, i wish you best of luck and thank you so much. | 2017-10-17 13:31:17 | {"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": 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": 0.6296083331108093, "perplexity": 2471.5068408732773}, "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/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00606.warc.gz"} |
https://www.semanticscholar.org/paper/Streaming-k-PCA%3A-Efficient-guarantees-for-Oja's-Huang-Niles-Weed/0a346bcc448dd56a18e5d9c8189aee0010b39aa7 | • Corpus ID: 231846710
# Streaming k-PCA: Efficient guarantees for Oja's algorithm, beyond rank-one updates
@inproceedings{Huang2021StreamingKE,
title={Streaming k-PCA: Efficient guarantees for Oja's algorithm, beyond rank-one updates},
author={De Huang and Jonathan Niles-Weed and Rachel A. Ward},
booktitle={Annual Conference Computational Learning Theory},
year={2021}
}
• Published in
Annual Conference…
6 February 2021
• Computer Science
We analyze Oja’s algorithm for streaming k-PCA, and prove that it achieves performance nearly matching that of an optimal offline algorithm. Given access to a sequence of i.i.d. d× d symmetric matrices, we show that Oja’s algorithm can obtain an accurate approximation to the subspace of the top k eigenvectors of their expectation using a number of samples that scales polylogarithmically with d. Previously, such a result was only known in the case where the updates have rank one. Our analysis is…
It is shown that the Grassmannian Rank-One Subspace Estimation (GROUSE) algorithm is indeed equivalent to Oja’s algorithm in the sense that, at each iteration, given a step size for one of the algorithms, it may construct a step sizes for the other algorithm that results in an identical update.
It is proved that with high probability Oja’s algorithm performs an efficient, gap-free, global convergence rate to approximate an principal component subspace for any sub-Gaussian distribution.
• Mathematics, Computer Science
NeurIPS
• 2021
A weighted χ 2 approximation result is established for the sin 2 error between the population eigenvector and the output of Oja’s algorithm, thereby establishing the bootstrap as a consistent inferential method in an appropriate asymptotic regime.
• Computer Science
• 2022
A stochastic Gauss-Newton (SGN) algorithm to study the online principal component analysis (OPCA) problem, which is formulated by using the symmetric low-rank product model for dominant eigenspace calculation, is proposed.
• Computer Science
• 2019
This work considers streaming principal component analysis when the stochastic data-generating model is subject to perturbations and provides fundamental limits on convergence of any algorithm recovering principal components.
• Computer Science
EC
• 2022
This work shows how to design a content recommendations which can achieve approximate stationarity, under mild conditions on the set of available content, when a user's preferences are known, and how one can learn enough about a users' preferences to implement such a strategy even when user preferences are initially unknown.
• Computer Science
ArXiv
• 2021
The correspondence between Gaussian process regression and Geometric Harmonics is discussed, providing alternative interpretations of uncertainty in terms of error estimation, or leading towards accelerated Bayesian Optimization due to dimensionality reduction.
## References
SHOWING 1-10 OF 35 REFERENCES
• Computer Science
2017 IEEE 58th Annual Symposium on Foundations of Computer Science (FOCS)
• 2017
The results match the information theoretic lower bound in terms of dependency on error, on eigengap, on rank k, and on dimension d, up to poly-log factors.
• Computer Science
AISTATS
• 2016
This paper analyzes the convergence rate of a representative algorithm with decayed learning rate (Oja and Karhunen, 1985) in the first family for the general $k>1$ case and proposes a novel algorithm for the second family that sets the block sizes automatically and dynamically with faster convergence rate.
• Computer Science
ArXiv
• 2019
AdaOja is a novel variation of the Adagrad algorithm to Oja's algorithm in the single eigen vector case and extended to the multiple eigenvector case and it is demonstrated for dense synthetic data, sparse real-world data and dense real- world data that AdaOja outperforms common learning rate choices for Oja’s method.
• O. Shamir
• Computer Science, Mathematics
ICML
• 2016
The convergence properties of the VR-PCA algorithm introduced by Shamir 2015stochastic are studied, including a formal analysis of a block version of the algorithm, and convergence from random initialization is proved.
This paper provides the first eigengap-free convergence guarantees for SGD in the context of PCA in a streaming stochastic setting, and shows that the same techniques lead to new SGD convergence guarantees with better dependence on the eIGengap.
• Computer Science, Mathematics
STOC
• 2018
A query complexity lower bound for approximating the top r dimensional eigenspace of a matrix and establishes a strict separation between convex optimization and “strict-saddle” non-convex optimization of which PCA is a canonical example is established.
• Computer Science
NIPS
• 2013
The top eigenvector of A is computed in an incremental fashion - with an algorithm that maintains an estimate of the top Eigenvector in O(d) space, and incrementally adjusts the estimate with each new data point that arrives.
• Computer Science
NIPS
• 2014
A new robust convergence analysis of the well-known power method for computing the dominant singular vectors of a matrix that is called the noisy power method is provided and shows that the error dependence of the algorithm on the matrix dimension can be replaced by an essentially tight dependence on the coherence of the matrix.
• Computer Science
ICML
• 2015
This paper exhibits a step size scheme for SGD on a low-rank least-squares problem, and proves that, under broad sampling conditions, the method converges globally from a random starting point within $O(\epsilon^{-1} n \log n)$ steps with constant probability for constant-rank problems.
• Computer Science
NIPS
• 2013
An algorithm is presented that uses O(kp) memory and is able to compute the k-dimensional spike with O(p log p) sample-complexity - the first algorithm of its kind. | 2023-02-03 00:21:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7501674294471741, "perplexity": 1434.2726412213744}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500041.2/warc/CC-MAIN-20230202232251-20230203022251-00075.warc.gz"} |
https://tex.stackexchange.com/questions/217651/how-can-i-include-upside-down-unicode-text | # How can I include upside-down Unicode text?
It seems like there is simply no way to have Latex include any generic Unicode characters you'd like in a PDF, and the solutions to this question are too complicated.
Is there any simple way to include upside-down Unicode text such as ˙ʎɐpoʇ ʎddɐɥ ɯɐ I ?
EDIT Compiling with XeLatex seems to work fine for that particular example, but later in the document I am hit with
! Package inputenc Error: Unicode char \u8:ñosd not set up for use with LaTeX.
See the inputenc package documentation for explanation.
Type H <return> for immediate help.
...
and ...
I don't know how I was able to put that ñ there without issues before...
EDIT Turns out CJKutf8 was the culprit package that introduced inputenc
• or.. just turn your head upside down – user68908 Dec 12 '14 at 17:59
• You can't use inputenc with LuaLaTeX or XeLaTeX. – musarithmia Dec 12 '14 at 20:04
You can input any Unicode characters directly and compile them to PDF simply by using LuaLaTeX or XeLaTeX to compile (e.g., lualatex file in the terminal). You can also rotate text using the rotating package. With the turn environment you can rotate text to any desired degree (see texdoc rotating).
The source must be encoded UTF-8, you must compile with LuaLaTeX or XeLaTeX, and you must use fontspec or an appropriate package to select a font that includes the Unicode characters you need.
% compile with LuaLaTeX or XeLaTeX
\documentclass{article}
\usepackage{libertine}
\usepackage{rotating}
\newcommand{\unicodetext}{Greek μετανοεῖτε, accents \emph{esdrújulos}, pictograms ☹}
\begin{document}
This is UTF-8 encoded text with Unicode characters input directly:
\unicodetext
\begin{turn}{180}
\unicodetext
\end{turn}
\bigskip
This is text input upside-down:
˙ʎɐpoʇ ʎddɐɥ ɯɐ I ?
\end{document}
• Hm thanks, that text in particular works fine, but then later in the document I am hit with Package inputenc Error: Unicode char \u8:ñosd not set up for use with LaTeX. when I try to put l.173 ...d on reading \emph{Cien años de soledad} and.... Any ideas? – wrongusername Dec 12 '14 at 19:32
• If it matters any, I am using TeXworks on Windows – wrongusername Dec 12 '14 at 19:36
• @wrongusername If you compile with the LuaLaTeX or XeLaTeX engines, you should not get this error. I don't know Windows or TeXworks, but you should be able to adjust which engine you use to compile in the application's settings. It may be instructive to compile it from the command line lualatex file to make sure there is no error with your installation. – musarithmia Dec 12 '14 at 20:03
• Ah thanks, turns out I was using a package that automatically includes it! – wrongusername Dec 13 '14 at 3:36
If you use a unicode based tex such as xelatex or luatex it works naturally:
\documentclass{article}
\usepackage{fontspec}
\setmainfont{Arial}
\begin{document}
˙ʎɐpoʇ ʎddɐɥ ɯɐ I ?
\end{document}
• david -- you are evil. – barbara beeton Dec 12 '14 at 13:32
• @barbarabeeton hey It was the OP's text! – David Carlisle Dec 12 '14 at 13:37
• but you cheated and used "p" instead of "d", and vice versa. "o" is symmetrical. and "I" gave you away -- it should have an ascender, not a descender. – barbara beeton Dec 12 '14 at 13:43
• @barbarabeeton I didn't do anything: I cut and pasted the text from the question – David Carlisle Dec 12 '14 at 13:44
• Can somebody be so kind and explain, how the OP fabricated this line: ˙ʎɐpoʇ ʎddɐɥ ɯɐ I I notice, there is a trick, because the "I" has a descender. But what is the trick? It works without backticks as well: ˙ʎɐpoʇ ʎddɐɥ ɯɐ I – Keks Dose Dec 12 '14 at 20:53 | 2019-07-17 10:35:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9472426772117615, "perplexity": 3804.404281705045}, "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-2019-30/segments/1563195525136.58/warc/CC-MAIN-20190717101524-20190717123524-00091.warc.gz"} |
http://math.stackexchange.com/questions/204348/in-general-is-it-easier-to-solve-problems-that-have-random-variables-or-proble | # In general, is it easier to solve problems that have random variables, or problems that are basically the same that don't [closed]
This question is asked because I don't understand how random variables will affect various math problems, and knowledgeable mathematicians would.
By easier, we mean less steps
• If we make our own problems, random variables are wholly optional. Why? Because if we wanted random variables in a non-random variable problem, we just add a random source to the example problem. Having a random source automatically makes the variables into "random variables" for a problem.
• The variables being random variables or not doesn't seen to make any difference to certain example problems
A quick concrete example --
You have a bunch of range of numbers (like $33$ to $77$), which represents your IQ,
each with a corresponding percentage which represents your chance of getting cancer. That is what is inside one set.
There are many sets and each set has different continuous numbers and percentage
You find a comparative trend to solve this.
This example can be made random just by having the IQ number be randomly generated instead of already given. Now you have random variables. But these random variables don't seem to really affect the problem or example at all .
Beacause all we care about is the trend between the number and percents, and how they relate to the other numbers and percent in the other sets.
Here's another example where you can easily make the problem have random variables or not -- what math topic is this kind of example part of? or what is needed to understand how to solve it?, but I think the problem is fundamentally different from the IQ example.
-
Could you give an example of what you mean? It is hard to read a question about facts out of this ... – Hagen von Eitzen Sep 29 '12 at 11:24
Do you really mean random variable or are you talking about creating word problems by picking numbers in a special way so that certain quantities exactly cancel out vs. picking numbers less deliberately? – binn Sep 29 '12 at 13:04
@high are you confused by the fact that there is nothing random about a random variable ? (i.e. it s just a specific kind of function). What definition are you using when you talk about a "random variable" ? – Beltrame Sep 29 '12 at 17:09
the definition of a random var im using is "if the source or the outcome is random, then you have random variables." -- yes i mean random variables -- stats.stackexchange.com/questions/38247/does-anyone-know-what-exactly-is-a-random-variable – kittensatplay Sep 30 '12 at 23:40
a concrete example that i made is stats.stackexchange.com/questions/38235/formula-for-a-comparative-trend-any-of-these-http-stattrek-com-statistics – kittensatplay Sep 30 '12 at 23:43
## closed as not constructive by Hagen von Eitzen, whuber, Michael Greinecker♦, Qiaochu YuanOct 7 '12 at 4:06
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question. | 2014-04-19 13:21:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6643581390380859, "perplexity": 685.4830169995084}, "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/1397609537186.46/warc/CC-MAIN-20140416005217-00400-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://hamrocsit.com/question/show-that-the-converges-and-diverges/ | Show that the converges $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$ and diverges $$\int_{∞}^{1}\frac{1}{x}\,dx$$
Solution:
If limit of $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$ exists then it converges
= $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$
= $$\int_{∞}^{1} x^{-2} \,dx$$
= $$\begin{bmatrix}\frac{x^{-1}}{-1}\end{bmatrix}^{\!1}_{∞}$$
= $$\begin{bmatrix}\frac{-1}{x}\end{bmatrix}^{\!1}_{∞}$$
= -1 – $$\frac{1}{∞}$$
= -1
Since the limit exists so it converges.
Second Part:
If limit of $$\int_{∞}^{1}\frac{1}{x} \,dx$$ doesn’t exists then it diverges
= $$\int_{∞}^{1}\frac{1}{x} \,dx$$
= $$\begin{bmatrix} log(x) \end{bmatrix}^{\!1}_{∞}$$
= log(1) – log(∞)
= 0 – ∞
= – ∞
Since the limit doesn’t exists so it diverges. | 2022-12-07 10:13:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9894696474075317, "perplexity": 4261.539707694074}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00430.warc.gz"} |
http://mathhelpforum.com/differential-equations/66653-wierd-differential-equation.html | Math Help - Wierd differential equation
1. Wierd differential equation
$\frac{d^2x}{dt^2}=x\cdot \frac{dx}{dt}$
I don't even know where to start. Can someone give me a little push?
2. Hint: The right-hand side is $\tfrac d{dt}\bigl(\tfrac12x^2\bigr)$. | 2014-03-11 10:44:03 | {"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": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7491477727890015, "perplexity": 376.8879062193666}, "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-2014-10/segments/1394011174089/warc/CC-MAIN-20140305091934-00005-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://highnoongmt.wordpress.com/2011/10/14/efficient_omp/ | # Highly Efficient OMP and OLS
Busy semester!!!
Over the past few weeks I have been working at making OMP and OLS as efficient as possible.
Now I think I have it to where the only place where they can become more efficient is in the dictionary search. Given we have selected $$k$$ atoms, OMP selects the next atom by
$$\min_{n \in \Omega} \| \vr_k – \langle \vr_k, \phi_n \rangle \phi_n \|_2 = \max_{n \in \Omega} |\langle \vr_k, \phi_n \rangle|$$
where the dictionary $$\MPhi$$ has unit-norm atoms $$\phi_n$$ as columns,
and $$\vr_k$$ is the orthogonal projection of $$\vu$$ onto the span of the $$k$$ selected atoms, which I denote as the matrix $$\MPhi_k$$.
OLS selects the next atom by
$$\min_{n \in \Omega}\left \| \MP_k^\perp \vu – \frac{\langle \MP_k^\perp \vu, \phi_n \rangle \MP_k^\perp\phi_n }{\|\MP_k^\perp\phi_n\|_2^2}\right \|_2 = \max_{n \in \Omega} \frac{|\langle \MP_k^\perp \vu, \phi_n \rangle|}{\|\MP_k^\perp\phi_n\|_2^2}$$
where $$\MP_k^\perp$$ is the orthogonal projection onto the subspace orthogonal to the span of the $$k$$ selected atoms.
Consider we have the QR decomposition of the dictionary submatrix, $$\MPhi_k = \MQ_k\MR_k$$.
This means $$\MQ_k$$ is an orthonormal basis for the span of the $$k$$ selected atoms,
and $$\MR_k$$ is an upper triangular matrix describing how to build the $$k$$ atoms from the basis.
Since $$\vr_k = \vu – \MQ_k\MQ_k^T\vu = \MP_k^\perp \vu$$, the OMP selection criterion becomes
$$\max_{n \in \Omega} |\langle \vu – \MQ_k\MQ_k^T\vu, \phi_n \rangle| = \max_{n \in \Omega} |\langle \vu,\phi_n \rangle – \langle \MQ_k\MQ_k^T\vu, \phi_n \rangle|$$
and the OLS selection criterion becomes
$$\max_{n \in \Omega} \frac{|\langle \vu – \MQ_k\MQ_k^T\vu, \phi_n \rangle|}{\|\phi_n – \MQ_k\MQ_k^T\phi_n\|_2^2} = \max_{n \in \Omega} \frac{|\langle \vu,\phi_n \rangle – \MQ_k\MQ_k^T\vu, \phi_n \rangle|}{\|\phi_n – \MQ_k\MQ_k^T\phi_n\|_2^2}$$
The first term we need compute only once at the beginning of the decomposition.
So the second term is what we need to update efficiently.
We can iteratively update $$\langle \MQ_k\MQ_k^T\vu, \phi_n \rangle$$
since $$\MQ_k = [\MQ_{k-1} | \vq_k ]$$ where $$\vq_k$$ is the new column added to the QR decomposition when OMP selects $$\phi_{n_k}$$.
It is simply given by
$$\vq_k = \frac{ \phi_{n_k} – \MQ_{k-1}\MQ_{k-1}^T\phi_{n_k}}{\|\phi_{n_k} – \MQ_{k-1}\MQ_{k-1}^T\phi_{n_k}\|_2}.$$
Thus, we see that for all atoms in the dictionary
$$\langle \MQ_k\MQ_k^T\vu, \phi_n \rangle = \langle \MQ_{k-1}\MQ_{k-1}^T\vu, \phi_n \rangle + \langle \vq_k \vq_k^T \vu, \phi_n \rangle.$$
Considering that we have already calculated and stored the first term,
we only need evaluate the second for all atoms in the dictionary.
Notice that nowhere in these iterations does OMP or OLS need to compute the actual solution.
That comes only at the very end!
Each time OMP and OLS selects an atom, it must update the QR decomposition.
This can be done efficiently by the following approach.
First the algorithm updates the upper triangular matrix:
$$\MR_k = \left [ \begin{array}{cc} \MR_{k-1} & \vw \\ \mathbf{0}^T & \sqrt{1 – \|\vw\|_2^2} \end{array} \right ]$$
where $$\MR_{k-1}^T\vw = \MPhi_{\Omega_{k-1}}^T\phi_{n_k}$$.
This can be solved efficiently by back substitution since the left side matrix is triangular.
Also, we must have already computed $$\MPhi_{\Omega_{k-1}}^T\phi_{n_k}$$,
which we can do at the very beginning of the decomposition by computing the Gramian $$\MPhi^T\MPhi$$.
Then, to update the orthonormal basis we perform
$$\MQ_{k} = \left [ \begin{array}{cc} \MQ_{k-1} & (\MI – \MQ_{k-1}\MQ_{k-1}^T)\phi_{n_k} / \sqrt{1 – \|\vw\|_2^2} \end{array} \right ].$$
Notice that
$$\|\phi_{n_k} – \MQ_{k-1}\MQ_{k-1}^T\phi_{n_k}\|_2 = \sqrt{1 – \|\vw\|_2^2}$$,
so we can save some computation there as well.
OLS has the added computation of the normalization factor for each atom.
This can be done efficiently in an iterative fashion.
Finally, OMP and OLS need to compute the solution by $$\MR_k^T\MR_k\vs_k = \MPhi_k^T\vu$$.
This is quite efficiently done by solving first $$\MR_k^T\vv = \MPhi_k^T\vu$$,
and finally, $$\MR_k\vs_k = \vv$$.
Then we put the non-zero elements of $$\vs_k$$ into the correct places of the longer solution vector $$\vs$$.
Below we see the empirical phase transition for OMP found using this code. I am averaging 50 trials, where the condition for exact recovery is the full and exact recovery of the solution support, nothing more and nothing less.
And below is the empirical phase transition for OLS found using this code.
The line appears jaggedy because I am only averaging the results of 50 trials.
There may be some more spots in which these algorithms can be sped up even more, but I think their main bottleneck is the search for the maximum projection. | 2017-03-23 14:20:38 | {"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8462747931480408, "perplexity": 411.3672193394865}, "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-2017-13/segments/1490218187113.46/warc/CC-MAIN-20170322212947-00149-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-10-infinite-sequences-and-series-practice-exercises-page-636/40 | ## Thomas' Calculus 13th Edition
Let $a_n=\dfrac{1}{n\sqrt {n^2-1}}=\dfrac{1}{n\sqrt {(n-1)(n+1)}}$ Apply comparison, we have $\Sigma_{n=2}^\infty \dfrac{1}{n\sqrt {(n-1)(n+2)}} \lt \Sigma_{n=2}^\infty \dfrac{1}{\sqrt {(n-1)^2(n-1)(n-1)}}=\Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}$ $\implies \Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$ shows a convergent p-series with $p=2 \gt 1$ Thus, the series Converges Absolutely by the comparison test. | 2020-04-06 16:00:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9953510761260986, "perplexity": 247.23879177838967}, "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/1585371637684.76/warc/CC-MAIN-20200406133533-20200406164033-00367.warc.gz"} |
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Is x^2 > x? (1) x^2 > 1 (2) x > -1
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Is x^2 > x?
(1) x^2 > 1
(2) x > -1
[Reveal] Spoiler: OA
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Joined: 08 Aug 2017
Posts: 11
Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1 [#permalink]
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04 Dec 2017, 01:33
statement 1 is suff.
the only interval where x^2<x is when x is between 0 and 1.
if x^2>1, this translates to the fact that x is not between 0 and 1.
hence, A.
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Is x^2 > x? (1) x^2 > 1 (2) x > -1 [#permalink]
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Updated on: 04 Dec 2017, 05:21
Is x^2 > x?
Rephrase question : x(x-1) >0 ?
(1)$$x^2$$ > 1
(x - 1)(x + 1)>0
range: (-infinity ,-1) U (1, infinity)
Substitute values from this range into the question and check whether we get positive value
x= -2
(-2)*(-3) >0 ? ..........yes
x=1.5
(1.5*0.5) >0 ? ...........yes
Sufficient
(2) x > -1
substitute 0.5
(.5)(-.5) >0? ....no
substitute -0.5
(-0.5)(-1.5) >0? ...yes
--A--
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"Trust the timing of your life"
Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq
Originally posted by TaN1213 on 04 Dec 2017, 03:15.
Last edited by TaN1213 on 04 Dec 2017, 05:21, edited 1 time in total.
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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1 [#permalink]
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04 Dec 2017, 05:05
Question ask if the range of x
X<0 and x>1
St1:
X>1
X<-1
Sufficient
St2:
X>-1
X=0 (NO)
X=4 (YES)
Not sufficient
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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1 [#permalink]
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04 Dec 2017, 05:24
x^2 > x
or x(x-1)>0
or the question is asking is x>1 or x<0
Statement 1
x^2>1
(x-1)(x+1)>0
either x>1 or x<-1 in both the case it is sufficient
statement 2
x>-1
not sufficient , since x can be 0 (which doesn't satisfy my equation )
Hence IMO A
Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1 [#permalink] 04 Dec 2017, 05:24
Display posts from previous: Sort by | 2018-04-26 15:09:09 | {"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": 0.7687976360321045, "perplexity": 14630.697784021992}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00541.warc.gz"} |
https://forum.uipath.com/t/pdf-to-excel-conversion/19506 | # Pdf to excel conversion
Hi evewryone,
can some one let me know how to extarct data from my pdf and keep inside excel.
My pdf has some values and then it has a table continously till 4 to 5 pages.
i want to extract only the tables from the pdf and write in excel.
1 Like
If your Table is continuous and Structured use “DataScrapping” you will get the complete table and write it your excel using write range activity
Hi Pathrudu,
i have a pdf in which at the beginning there is some text written after those text i have a big table which is continuous till 3 to 4 pages but i tried with Datascrapping that is throwing me an error. and with screen scrapping i am not getting the desired output as well.
Can you help me in this if i forward you one of my pdf?
Sure let’s try
Thanks for your reply pathrudu i will forward you my pdf in sometime. | 2020-10-29 05:07:06 | {"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": 0.8108736872673035, "perplexity": 1325.7748609776722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107902745.75/warc/CC-MAIN-20201029040021-20201029070021-00614.warc.gz"} |
https://vulcanhammer.net/2017/08/24/stadyn-wave-equation-program-4-eta-limiting-and-more-on-norm-matching/ | Posted in Uncategorized
# STADYN Wave Equation Program 4: Eta Limiting, and More on Norm Matching
In our last post we broached the subject of different norm matching methods for the actual and computed velocity-time histories at the pile top. In this post we will go into $\eta$ limiting, while at the same time running both norms to get a better feel for the differences in the results.
Before we begin, one clarification is in order: CAPWAP’s Match Quality and the use of the 1-norm in STADYN are similar in mathematical concept but different in execution. That’s because the Match Quality weights different part of the force-time history (in their case) differently, whereas STADYN goes for a simple minimum sum difference.
One characteristic of the inverse case both in the original study and in the modifications shown in the last post are very large absolute values of $\eta$. These are products of the search routine, but they are not very realistic in terms of characterising the soil around the pile. To illustrate, we bring back up one of the results from the last post, showing the optimisation track using the 2-norm and phi-based Poisson’s Ratio (which will now be the program standard):
Note that the #8 track ($\eta$ for the lowest shaft layer) has a value approaching -30; this is obviously very unrealistic.
In principle, as with $\xi$, the absolute value of $\eta$ should not exceed unity; however, unlike $\xi$ there is no formal reason why this should be the case. But how much should we vary $\eta$? To answer this question, and to continue our investigation of the norm issue, we will examine a matrix of cases as follows:
1. $\eta$ will be run for values of 1, 2, 3 and unlimited (the last has already been done.)
2. Each of these will be run for both the 1-norm and 2-norm matching.
A summary of the results are shown below
Changed Parameter Difference Static Load, kN Average Shaft $\xi$ Toe $\xi$ Toe $\eta$ Norm 1 2 1 2 1 2 1 2 1 2 |$\eta$| < 1 0.3364 0.003690 811 1490 -0.364 -0.149 -0.62 -0.311 -0.175 0.611 |$\eta$| < 2 0.2381 0.002626 278 223 -0.091 -0.06 -0.588 -0.316 -0.781 -0.0385 |$\eta$| < 3 0.1806 0.001707 172 207 0.324 0.42 -.832 0.823 -1.01 1.45 Unrestricted $\eta$ 0.1344 0.001456 300 218 -0.329 -0.183 -0.491 0.804 8.19 1.52 $\nu = f(\xi,\eta)$ 0.1484 0.001495 278 187 -0.383 -0.53 0.792 0.366 3.116 1.814
To see how this actually looks, consider the runs where |$\eta$| < 3. We will use the 2-norm results.
The results indicate the following:
1. The average shaft values of $\xi$ tend to be negative. This is contrary to the cohesive nature of the soils. The interface issue needs to be revisited.
2. The toe values do not exhibit a consistent pattern. This is probably due to the fact that they are compensating for changes in values along the shaft.
3. As values of |$\eta$| are allowed to increase, with the 2-norm the result of the simulated static load test become fairly consistent. This is not the case with the 1-norm. Although limiting |$\eta$| to unity is too restrictive, it is possible to achieve consistent results without removing all limits on $\eta$.
4. The velocity (actually impedance*velocity) history matching is similar to what we have seen before with the unlimited eta case.
5. The optimisation track starts by exploring the limits of $\eta$, but then “pulls back” to values away from the limits. This indicates that, while limiting values “within the box,” i.e., the absolute values of $\eta$ < 1, is too restrictive, reasonable results can be obtained with some $\eta$ limiting.
Based on these results, $\eta$ limiting will be incorporated into the program. The next topic to be considered are changes in the soil properties along the surface of the pile, as was discussed in the last post. | 2019-10-20 20:01:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6232102513313293, "perplexity": 762.8400402850269}, "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/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00559.warc.gz"} |
https://math.stackexchange.com/questions/1839762/prove-if-x-1-x-n-are-natural-numbers-with-n-geq2-then-x-1x-2-x-n-is | # Prove if $x_1,…,x_n$ are natural numbers with $n\geq2$ then $x_1x_2…x_n$ is odd iff $x_i$ is odd for all $i$, $1\leq i\leq n$
I am not sure if Im on the right track here but if any one could help out I would greatly appreciate it.
Prove if $x_1,...,x_n$ are natural numbers with $n\geq2$ then $x_1x_2...x_n$ is odd iff $x_i$ is odd for all $i$, $1\leq i\leq n$
Formulate an induction and prove it.
Suppose $x_1x_2\dots x_n=x_{n^2+1}$ and $x_1=x_2=1$
Let p(n):=$x_1x_2\dots x_n=x_{n^2+1}$
Base case:
Let $n=1$ LHS= $x_1=1$, RHS=$x_{(1)^2+1}=x_2=1$.
So LHS=RHS=1,
Thus, the base case holds.
Induction hypothesis:
Assume $x_1x_2\dots x_n=x_{n^2+1}$
NTS: $x_1x_2\dots x_n x_{n+1}=x_{(n+1)^2+1}$
Inductive Step:
By the induction hypothesis we can simplify $x_1x_2\dots x_n x_{n+1}=x_{(n+1)^2+1}$ to $x_{n^2+1} x_{n+1}=x_{(n+1)^2+1}$
Im not sure how to continue from here and I believe its because the induction I formulated is incorrect but I am having trouble thinking of what adjustments I need to make.
• Suppose otherwise. Then at least one of the $x_i$ are even and can be written as $2k$. But then the product is even as well. In the other direction, suppose the product is even. Since two is prime, it must divide some $x_i$. – JMoravitz Jun 26 '16 at 1:07
## 1 Answer
I think you have your wires crossed; particularly it seems like you're thinking about how the sum of the first $n$ odd numbers gives $n^2$. This is woefully incorrect. Here is how you should frame it:
Suppose that $x_1,\ldots, x_n\ge 2$, then you want to show that $x_1\cdots x_n$ is odd if and only if $x_i$ is odd for all $i$. In terms of an induction argument, this would go like this:
Base case (this is trivial)
Induction hypothesis $y = x_1\ldots x_n$ is odd if and only if $x_i$ is odd for all $1\le i \le n$.
Then you want to show that this if and only if statement holds true for each $x_1,\ldots, x_{n+1}$, i.e. you want to show that
$z =x_1\cdots x_{n+1}$ is odd if and only if $x_i$ is odd for all $1\le i \le n+1$.
But $z = x_1\cdots x_n\cdot x_{n+1} = y\cdot x_{n+1}$. Can you take it from here? | 2019-09-19 19:15:38 | {"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": 0.903157651424408, "perplexity": 49.60046352842095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00228.warc.gz"} |
https://tex.stackexchange.com/questions/290119/problems-with-creating-abstract-in-custom-class | Problems with creating abstract in custom class
I am working on creating a thesis template for myself and others in my lab to use. I have gotten stuck on creating a specialized abstract environment however. As a MWE (well minimum broken example) I have the following as a 'sample.cls' file
\NeedsTeXFormat{LaTeX2e}
\ProvidesClass{sample}[2016/01/29 Sample]
\DeclareOption*{
\ClassWarning{sample}{No options permitted}
}
\ProcessOptions\relax
\newenvironment{abstractpage}{
{
\begin{center}
%set the default font style to bold face extended for the abstract page
\fontseries{bx}
\selectfont
\Large ABSTRACT \\
\hfill \\
\large \@title\\
\hfill \\
\normalsize \@author
\end{center}
\vfill
\quotation\noindent
}
{
\endquotation
\vfill
}
}
and
\documentclass{sample}
\usepackage{lipsum}
\begin{document}
\author{Andrew}
\title{Hello World}
\begin{abstractpage}
\lipsum[1]
\end{abstractpage}
\end{document}
as a '.tex' file. I then tried to compile the tex file using pdflatex
When I try to compile I get
runawayargument?
! file ended while scanning use of \@newenv.
and I cannot for the life of me figure out why. I've check all of the brackets and I took this idea from the default abstract class itself (according to Define abstract environment in book)
Can anyone see anything wrong with this?
• You have a faulty definiton of the environment. Does your editor have brace-matching? Check where the environment starts, and where you closing is happening. Unrelated: If that should always be on its own page, use clearpage when starting and ending. – Johannes_B Jan 29 '16 at 21:26
• agree with @Johannes_B, but a different bias. \newenvironment` requires two arguments beside the environment name. you have braces enclosing both of what looks like you intend to be the two required arguments, with the result that there's only one found by latex. – barbara beeton Jan 29 '16 at 21:29
You have some errors in your code: unmatched braces are the main one.
Secondly, \hfill\\ is a don't-do-it; after \noindent you need \ignorespaces. And finally you have a load of unprotected end-of-lines, which may be irrelevant in this particular case, but it's good habit to have the correct % everywhere.
\NeedsTeXFormat{LaTeX2e}
\ProvidesClass{sample}[2016/01/29 Sample]
\DeclareOption*{%
\ClassWarning{sample}{No options permitted}%
}
\ProcessOptions\relax
\newenvironment{abstractpage}
{%
\begin{center}
%set the default font style to bold face extended for the abstract page
\bfseries
{\Large \MakeUppercase{\abstractname} \\[\bigskipamount]}
{\large \@title\\[\bigskipamount]}
\@author
\end{center}
\vfill
\quotation\noindent\ignorespaces
}
{%
\endquotation
\vfill
}
Don't say \fontseries{bx}\selectfont; \bfseries is preferable, because it's not necessary that a font family defines the bx series.
It's also better to use \abstractname, rather than hardwiring “ABSTRACT”.
In the test document I used \lipsum[1-2] so as to show the indent in the second paragraph.
• @Johannes_B Yes, I selected the wrong image! – egreg Jan 29 '16 at 23:07 | 2019-11-19 13:30:27 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8841278553009033, "perplexity": 2774.080159300138}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670151.97/warc/CC-MAIN-20191119121339-20191119145339-00010.warc.gz"} |
https://code.tutsplus.com/courses/practice-javascript-and-learn-functions/lessons/the-this-keyword-issues-and-arrow-functions | FREELessons: 8Length: 1 hour
• Overview
• Transcript
# 3.4 The this Keyword: Issues and Arrow Functions
We're not done with this. In this lesson, we'll look at some issues that can pop up when using the this object in your functions. I'll also show you how arrow functions can solve some of these problems. | 2023-03-22 10:10:16 | {"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": 0.609499454498291, "perplexity": 2743.228864884691}, "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-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00136.warc.gz"} |
http://nextmatrix.com/index.php?option=com_content&view=section&layout=blog&id=7&Itemid=29 | ### Key Concepts
Features
Glossary of Mortgage-Backed Securities
Written by FemiByte Friday, 05 September 2014 04:54 MBS DealA set of structured bonds linked to a common pool (or pools) of mortgages Last Updated on Friday, 05 September 2014 05:02
Variable storage in Java
Written by FemiByte
Friday, 16 May 2014 04:33
In order to figure out where a variable is stored in Java the most important factor is where the variable is declared.
A general rule of thumb is this:
• local variables are stored on the stack
• instance variables are stored on the heap
• static variables are stored on the PermGen area of the heap
There are caveats to this however, which are explained below:
### Variable Storage Details
#### Local variables
Primitives and object references declared in a method will be stored on the stack. However, the actual object, if created using new() will be stored on the heap, regardless of where the declaration took place. Hence in the following piece of code:
void aMethod() { int playerNum=5; Player pl=new Player(); }
The primitive variable playerNum and object reference variable pl will be stored on the stack, while the actual Player object itself will live on the heap. When the code exits aMethod and goes out of scope, playerNum and pl will be popped from the stack and cease to exist but the Player object will persist on the heap until it is eventually garbage collected.
#### Instance variables
Instance variables, even primitives live on the stack.
Consider the code below:
public class Car {
int vinNumber;
String make;
String model;
int year;
String class;
Car(int vin, String make, String model, int year, String class)
{
this.vinNumber=vin;
this.make=make;
this.model=model;
this.year=year;
this.class=class;
}
...
public static void main(String[] args)
{
Car c=new Car(19281,"Audi", "A6",2012,"sedan");
}
Since an instance of Car can only be instantiated via a call to new(), we see that:
• The Car object c lives on the heap
• All instance primitives and objects that are part of the Car object are also stored on the heap.
#### Static variables
The rule for static variables is this: Static methods, primitive variables and object references are stored in the PermGen section of the heap since they are part of the reflection i.e. class, not instance related data. However, in the case of objects, the actual object itself is stored in the regular areas of the heap (young/old generation or survivor space).
References:
1. http://www.tutorialspoint.com/java/java_variable_types.htm
2. http://www.coderanch.com/t/202217/Performance/java/JVM-heap-stores-local-objects
3. http://stackoverflow.com/questions/8387989/where-is-a-static-method-and-a-static-variable-stored-in-java-in-heap-or-in-sta
4. http://stackoverflow.com/questions/3698078/where-does-the-jvm-store-primitive-variables
Last Updated on Friday, 16 May 2014 04:48
Pass-by-value vs Pass-by-reference in Java and C++
Written by FemiByte
Tuesday, 29 April 2014 03:36
# Pass-by-value vs Pass-by-reference in Java and C++
In this article I illustrate what it means to pass-by-value as opposed to pass-by-reference with a focus on Java vs C++.
The question often asked is this : Is Java pass-by-reference ?
A common and often erroneous answer is : Java is pass by reference for objects, and pass-by-value for primitives.
This is WRONG. To illustrate why this is so, let me refer you to this quote by the father
of Java himself, James Gosling:
Some people will say incorrectly that objects are passed “by reference.” In programming language design, the term pass by reference properly means that when an argument is passed to a function, the invoked function gets a reference to the original value, not a copy of its value. If the function modifies its parameter, the value in the calling code will be changed because the argument and parameter use the same slot in memory…. The Java programming language does not pass objects by reference; it passes object references by value. Because two copies of the same reference refer to the same actual object, changes made through one reference variable are visible through the other. There is exactly one parameter passing mode — pass by value — and that helps keep things simple.
– James Gosling, et al., The Java Programming Language, 4th Edition
The above clearly states that Java passes object references by value meaning that when the reference is passed, a copy of that reference (which is an address) is passed. Since the copy of the reference and the reference refer to the same object, if a call is made to a method that modifies the object in Java, that object is modified, hence the line “Because two copies of the same reference refer to the same actual object, changes made through one reference variable are visible through the other”.
I will now illustrate what pass-by-reference means, via a clear example in C++.
Let us create the following files in an appropriate directory with the following contents:
PassByReference.hpp:
#ifndef PassByReference_hpp#define PassByReference_hppvoid swapIntByRef(int& iParam, int& jParam);void swapIntByVal(int iParam, int jParam);#endif
PassByReference.cpp:
#include <iostream>#include "PassByReference.hpp"using namespace std;
int main(){int i=1000;int j=2300; cout << "Illustration of Pass By Reference:\n";cout << "Before: i= " << i << " j=" << j;cout << "\n";swapIntByRef(i,j);cout << "After: i= " << i << " j=" << j;cout << "\n"; cout << "\nIllustration of Pass By Value:\n"; i=1100;j=2500; cout << "Before: i= " << i << " j=" << j;cout << "\n";swapIntByVal(i,j);cout << "After: i= " << i << " j=" << j;cout << "\n"; } void swapIntByRef(int& iParam, int& jParam){int temp(iParam);iParam=jParam;jParam=temp;}
void swapIntByVal(int iParam, int jParam){int temp(iParam);iParam=jParam;jParam=temp;}
We now compile and run the code (assuming you have the g++ compiler):
g++ -o PassByReference PassByReference.cpp./PassByReferenceIllustration of Pass By Reference:Before: i=1000 j=2300After: i=2300 j=1000
Illustration of Pass By Value:Before: i=1100 j=2500After: i=1100 j=2500
The results above perfectly illustrate the difference between passing by reference vas pass-by-value, at least from the C++ point of view.
By using the reference operator &, when the value of i is passed to the swapIntByReffunction, the actual parameter value is modified in the function such that when the function returns back to the main() function that calls it and the values of i and j are printed out, the values of i and j have been swapped.
In the latter case of pass-by-value, copies of i and j are passed, not references via the & operator.
The result of this is that even though an attempt is made to swap the values in theswapIntByVal function, the original actual parameter values remain unchanged, and this is what we see in the result.
The latter case is what prevails in Java even for all cases, even in the case of objects.
Here is an illustration in Java for both primitives and object references:
Create the file PassByValueDemo.java:
public class PassByValueDemo {
public static void main(String[] args) {int i=1000;int j=2300;System.out.println("Primitives Case");System.out.println("----------------");System.out.println(" Before: i=" + i + " j=" + j);swapInt(i,j); System.out.println(" After: i=" + i + " j=" + j + "\n"); System.out.println("Wrapper Case");System.out.println("--------------");Integer iw=1000;Integer jw=2300;System.out.println(" Before: iw=" + iw + " jw=" + jw);swapInteger(iw,jw); System.out.println(" After: iw=" + iw + " jw=" + jw); } static void swapInt(int iParam, int jParam){int temp=jParam;jParam=iParam;iParam=temp;System.out.println(" iParam=" + iParam + " jParam=" + jParam); } static void swapInteger(Integer iParam, Integer jParam){Integer temp=jParam;jParam=iParam;iParam=temp;System.out.println(" iParam=" + iParam + " jParam=" + jParam);}
}
We now compile and run the code:
javac PassByValueDemo.java
java PassByValueDemo
which produces:
Primitives Case----------------Before: i=1000 j=2300iParam=2300 jParam=1000After: i=1000 j=2300
Wrapper Case--------------Before: iw=1000 jw=2300iParam=2300 jParam=1000After: iw=1000 jw=2300
Thus we can see that in both cases of primitive and wrapper classes the values of the actual parameters i and j remain unchanged in the calling routine main. There is no way to achieve the effect we observed in the C++ method PassByRef in Java where the original actual parameters are changed. The underlying object that the reference refers to can be changed via a call to a modifying method on the referenced object, but the reference parameter is always a copy of the original actual parameter.
## Summary
• C++ supports pass-by-value and pass by reference via its & operator.
• Java supports pass-by-value ONLY. What is erroneously thought of as pass-by-reference is really pass-by-value of an object reference.
Last Updated on Tuesday, 29 April 2014 06:39
Data Analytics: Summaries on Strang's Linear Algebra - Singular Value Decomposition
Written by FemiByte
Monday, 01 April 2013 02:48
## Singular Value Decomposition
Singular Value Decomposition is the culmination of concepts in the course. It involves a factorization into orthogonal x diagonal x orthogonal matrices We look for the case where matrix A takes an orthogonal basis of vectors from the row space into an orthogonal basis of vectors in the column space.
On SVD the goal is to find
1. orthonormal vectors $v_1, v_2, \cdots v_n$ in the row space $R^n$
2. orthonormal vectors $uv_1, uv_2, \cdots u_m$ in the column space $R^m$
3. multiplicands $\sigma_1 > 0, \sigma_2>0, \cdots \sigma_n > 0$ s.t $Av_1=\sigma_1 u_1 \text{ }Av_2=\sigma_2 u_2 \cdots Av_r=\sigma_r u_r$
In this case $A$ is diagonalized
The singular vectors $v_1,\cdots v_r$ are in the row space of $A$ and are the eigenvectors of $A^TA$ and the outputs $u_1,\cdots u_r$ are in the column space of $A$ and are the eigenvectors of $AA^T$. The singular values $\sigma_1,\cdots \sigma_r$ are all +ve. We have that $A \left[ \begin{array}{ccc} \\v_1&&\cdots&&v_r\\ \end{array}\right]= \left[ \begin{array}{ccc} \\u_1&&\cdots&&u_r\\ \end{array}\right] \left[ \begin{array}{ccc}\sigma_1&& && \\ &&.&& \\ && &&\sigma_r\end{array}\right]$ The above relationship is the heart if the SVD and can be expressed as $AV=U\Sigma$. We also need $n-r$ more $v$'s and $m-r$ more $u$'s, from the nullspace $N(A)$ and the left nullspace $N(A^T)$. This makes $V$ now a square orthogonal matrix and the core relationship for SVD becomes: $A=U \Sigma V^T = u_1 \sigma_1 v_1^T + \cdots + u_r \sigma_r v_r^T$
### The Bases and the SVD
In the 2x2 case we diagonalize the crucial symmetric matrix $A^TA$, whose eigenvalues are $\sigma_1^2$ and $\sigma_2^2$:
Eigenvalues $\sigma_1^2 \sigma_2^2$, Eigenvectors $v_1, v_2$ $A^TA = V \left[\begin{array}{cc}\sigma_1^2&&0\\0&&\sigma_2^2\end{array} \right] V^T$ Rule:
Compute the eigenvectors $v$ and eigenvalues $\sigma^2$ of $A^TA$. Then each $u=Av/\sigma$. The matrices $U$ and $V$ contain orthornormal bases for all 4 subspaces:
• first $r$ columns of $V$: row space of $A$
• first $n-r$ columns of $V$: nullspace of $A$
• first $r$ columns of $U$: column space of $A$
• first $m-r$ columns of $U$: nullspace of $A^T$
## Key Ideas
1. The SVD factors $A$ into $U\Sigma V^T$, with $r$ singular values $\sigma_1 \geq \cdots \sigma_r > 0$.
2. The numbers $\sigma_1 ^2 \cdots \sigma_r ^2$ are the nonzero eigenvalues of $AA^T$ and $A^TA$.
3. The orthonormal columns of $U$ and $V$ are eigenvectors of $AA^T$ and $A^TA$.
4. Those columns hold orthonormal bases for the 4 fundamental subspaces of $A$.
5. Those bases diagonalize the matrix: $Av_i=\sigma_i u_i$ for $i \leq r$. This is $AV=U\Sigma$
Last Updated on Monday, 01 April 2013 02:52
Data Analytics: Summaries on Strang's Linear Algebra - Similar Matrices
Written by FemiByte
Thursday, 21 March 2013 08:29
## Similar Matrices
Let $M$ be any invertible matrix. Then $B=M^{-1}AM$ is similar to $A$. Similar matrices $A$ and $M^{-1}AM$ have the same eigenvalues. If $x$ is an eigenvector of $A$, then $M^{-1}x$ is an eigenvector of $B=M^{-1}AM$. Matrices that are similar to each other form a family of related matrices, all having the same eigenvalues.
If $A$ has $s$ independent eigenvectors, it is similar to a matrix $J$ that has $s$ Jordan blocks on its diagonal: Some matrix $M$ puts $A$ into Jordan form:
Jordan form $M^{-1}AM=\left[\begin{array}{ccccc}J_1&& && && && \\ &&.&& && &&\\ && &&.&& &&\\ && && && &&J_s&&\end{array}\right] = J$
Each block in $J$ has one eigenvalue $\lambda_i$, one eigenvector and 1's above the diagonal:
Jordan block $J_i=\left[\begin{array}{ccccc}\lambda_i&&1&& && && \\ &&. &&. && &&\\ && &&.&& &&\\ && && && &&\lambda_i&&\end{array}\right] = J$
$A$ is similar to $B$ if they share the same Jordan form $J$ - not otherwise.
Last Updated on Thursday, 21 March 2013 08:30
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Page 1 of 3 | 2018-01-23 05:57:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0.399629145860672, "perplexity": 2275.6875818350186}, "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-2018-05/segments/1516084891750.87/warc/CC-MAIN-20180123052242-20180123072242-00022.warc.gz"} |
https://labs.tib.eu/arxiv/?author=X.%20H.%20Zhou | • ### First application of combined isochronous and Schottky mass spectrometry: Half-lives of fully ionized 49Cr24+ and 53Fe26+ atoms(1804.02653)
April 8, 2018 nucl-ex, physics.ins-det
Lifetime measurements of b -decaying highly charged ions have been performed in the storage ring CSRe by applying the isochronous Schottky mass spectrometry. The fully ionized 49Cr and 53Fe ions were produced in projectile fragmentation of 58Ni primary beam and were stored in the CSRe tuned into the isochronous ion-optical mode. The new resonant Schottky detector was applied to monitor the intensities of stored uncooled 49Cr24+ and 53Fe26+ ions. The extracted half-lives T1/2(49Cr24+) = 44.0(27) min and T1/2(53Fe26+) = 8.47(19) min are in excellent agreement with the literature half-life values corrected for the disabled electron capture branchings. This is an important proof-of-principle step towards realizing the simultaneous mass and lifetime measurements on exotic nuclei at the future storage ring facilities.
• ### Mass Measurements of Neutron-Deficient Y, Zr, and Nb Isotopes and Their Impact on $rp$ and $\nu p$ Nucleosynthesis Processes(1804.02309)
April 6, 2018 nucl-ex, astro-ph.SR
Using isochronous mass spectrometry at the experimental storage ring CSRe in Lanzhou, the masses of $^{82}$Zr and $^{84}$Nb were measured for the first time with an uncertainty of $\sim 10$ keV, and the masses of $^{79}$Y, $^{81}$Zr, and $^{83}$Nb were re-determined with a higher precision. %The latter differ significantly from their literature values. The latter are significantly less bound than their literature values. Our new and accurate masses remove the irregularities of the mass surface in this region of the nuclear chart. Our results do not support the predicted island of pronounced low $\alpha$ separation energies for neutron-deficient Mo and Tc isotopes, making the formation of Zr-Nb cycle in the $rp$-process unlikely. The new proton separation energy of $^{83}$Nb was determined to be 490(400)~keV smaller than that in the Atomic Mass Evaluation 2012. This partly removes the overproduction of the $p$-nucleus $^{84}$Sr relative to the neutron-deficient molybdenum isotopes in the previous $\nu p$-process simulations.
• ### Identification of the Lowest $T=2$, $J^{\pi=}0^+$ Isobaric Analog State in $^{52}$Co and Its Impact on the Understanding of $\beta$-Decay Properties of $^{52}$Ni(1610.09772)
Oct. 31, 2016 nucl-ex
Masses of $^{52g,52m}$Co were measured for the first time with an accuracy of $\sim 10$ keV, an unprecedented precision reached for short-lived nuclei in the isochronous mass spectrometry. Combining our results with the previous $\beta$-$\gamma$ measurements of $^{52}$Ni, the $T=2$, $J^{\pi}=0^+$ isobaric analog state (IAS) in $^{52}$Co was newly assigned, questioning the conventional identification of IASs from the $\beta$-delayed proton emissions. Using our energy of the IAS in $^{52}$Co, the masses of the $T=2$ multiplet fit well into the Isobaric Multiplet Mass Equation. We find that the IAS in $^{52}$Co decays predominantly via $\gamma$ transitions while the proton emission is negligibly small. According to our large-scale shell model calculations, this phenomenon has been interpreted to be due to very low isospin mixing in the IAS.
• ### The decay characteristic of $^{22}$Si and its ground-state mass significantly affected by three-nucleon forces(1610.08291)
Oct. 26, 2016 nucl-ex
The decay of the proton-rich nucleus $^{22}$Si was studied by a silicon array coupled with germanium clover detectors. Nine charged-particle groups are observed and most of them are recognized as $\beta$-delayed proton emission. A charged-particle group at 5600 keV is identified experimentally as $\beta$-delayed two-proton emission from the isobaric analog state of $^{22}$Al. Another charged-particle emission without any $\beta$ particle at the low energy less than 300 keV is observed. The half-life of $^{22}$Si is determined as 27.5 (18) ms. The experimental results of $\beta$-decay of $^{22}$Si are compared and in nice agreement with shell-model calculations. The mass excess of the ground state of $^{22}$Si deduced from the experimental data shows that three-nucleon (3N) forces with repulsive contributions have significant effects on nuclei near the proton drip line.
• ### An improvement of isochronous mass spectrometry: Velocity measurements using two time-of-flight detectors(1601.07048)
March 2, 2016 nucl-ex, physics.ins-det
Isochronous mass spectrometry (IMS) in storage rings is a powerful tool for mass measurements of exotic nuclei with very short half-lives down to several tens of microseconds, using a multicomponent secondary beam separated in-flight without cooling. However, the inevitable momentum spread of secondary ions limits the precision of nuclear masses determined by using IMS. Therefore, the momentum measurement in addition to the revolution period of stored ions is crucial to reduce the influence of the momentum spread on the standard deviation of the revolution period, which would lead to a much improved mass resolving power of IMS. One of the proposals to upgrade IMS is that the velocity of secondary ions could be directly measured by using two time-of-flight (double TOF) detectors installed in a straight section of a storage ring. In this paper, we outline the principle of IMS with double TOF detectors and the method to correct the momentum spread of stored ions.
• ### Reaction rates of $^{64}$Ge($p,\gamma$)$^{65}$As and $^{65}$As($p,\gamma$)$^{66}$Se and the extent of nucleosynthesis in type I X-ray bursts(1505.02275)
Jan. 29, 2016 nucl-th, astro-ph.SR
The extent of nucleosynthesis in models of type I X-ray bursts and the associated impact on the energy released in these explosive events are sensitive to nuclear masses and reaction rates around the $^{64}$Ge waiting point. Using the well known mass of $^{64}$Ge, the recently measured $^{65}$As mass, and large-scale shell model calculations, we have determined new thermonuclear rates of the $^{64}$Ge($p$,$\gamma$)$^{65}$As and $^{65}$As($p$,$\gamma$)$^{66}$Se reactions with reliable uncertainties. The new reaction rates differ significantly from previously published rates. Using the new data we analyze the impact of the new rates and the remaining nuclear physics uncertainties on the $^{64}$Ge waiting point in a number of representative one-zone X-ray burst models. We find that in contrast to previous work, when all relevant uncertainties are considered, a strong $^{64}$Ge $rp$-process waiting point cannot be ruled out. The nuclear physics uncertainties strongly affect X-ray burst model predictions of the synthesis of $^{64}$Zn, the synthesis of nuclei beyond $A=64$, energy generation, and burst light curve. We also identify key nuclear uncertainties that need to be addressed to determine the role of the $^{64}$Ge waiting point in X-ray bursts. These include the remaining uncertainty in the $^{65}$As mass, the uncertainty of the $^{66}$Se mass, and the remaining uncertainty in the $^{65}$As($p$,$\gamma$)$^{66}$Se reaction rate, which mainly originates from uncertain resonance energies.
• ### Anomalous transport and thermoelectric performances of CuAgSe compounds(1510.06616)
Oct. 22, 2015 cond-mat.mtrl-sci
The copper silver selenide has two phases: the low-temperature semimetal phase ({\alpha}-CuAgSe) and high-temperature phonon-glass superionic phase (\b{eta}-CuAgSe). In this work, the electric transport and thermoelectric properties of the two phases are investigated. It is revealed that the \b{eta}-CuAgSe is a p-type semiconductor and exhibits low thermal conductivity while the {\alpha}-CuAgSe shows metallic conduction with dominant n-type carriers and low electrical resistivity. The thermoelectric figure of merit zT of the polycrystalline \b{eta}-CuAgSe at 623 K is ~0.95, suggesting that superionic CuAgSe can be a promising thermoelectric candidate in the intermediate temperature range.
• ### Accurate correction of magnetic field instabilities for high-resolution isochronous mass measurements in storage rings(1407.3459)
July 13, 2014 nucl-ex
Isochronous mass spectrometry (IMS) in storage rings is a successful technique for accurate mass measurements of short-lived nuclides with relative precision of about $10^{-5}-10^{-7}$. Instabilities of the magnetic fields in storage rings are one of the major contributions limiting the achievable mass resolving power, which is directly related to the precision of the obtained mass values. A new data analysis method is proposed allowing one to minimise the effect of such instabilities. The masses of the previously measured at the CSRe $^{41}$Ti, $^{43}$V, $^{47}$Mn, $^{49}$Fe, $^{53}$Ni and $^{55}$Cu nuclides were re-determined with this method. An improvement of the mass precision by a factor of $\sim 1.7$ has been achieved for $^{41}$Ti and $^{43}$V. The method can be applied to any isochronous mass experiment irrespective of the accelerator facility. Furthermore, the method can be used as an on-line tool for checking the isochronous conditions of the storage ring.
• ### Charge and frequency resolved isochronous mass spectrometry in storage rings: First direct mass measurement of the short-lived neutron-deficient $^{51}$Co nuclide(1404.1187)
April 8, 2014 nucl-ex
Revolution frequency measurements of individual ions in storage rings require sophisticated timing detectors. One of common approaches for such detectors is the detection of secondary electrons released from a thin foil due to penetration of the stored ions. A new method based on the analysis of intensities of secondary electrons was developed which enables determination of the charge of each ion simultaneously with the measurement of its revolution frequency. Although the mass-over-charge ratios of $^{51}$Co$^{27+}$ and $^{34}$Ar$^{18+}$ ions are almost identical, and therefore, the ions can not be resolved in a storage ring, by applying the new method the mass excess of the short-lived $^{51}$Co is determined for the first time to be ME($^{51}$Co)=-27342(48) keV. Shell-model calculations in the $fp$-shell nuclei compared to the new data indicate the need to include isospin-nonconserving forces.
• ### Radioactive Ion Beam Physics and Nuclear Astrophysics in China(nucl-ex/0410019)
Oct. 14, 2004 nucl-ex, nucl-th
Based on the intermediate energy radioactive Ion Beam Line in Lanzhou (RIBLL) of Heavy Ion Research Facility in Lanzhou (HIRFL) and Low Energy Radioactive Ion Beam Line (GIRAFFE) of Beijing National Tandem Accelerator Lab (HI13), the radioactive ion beam physics and nuclear astrophysics will be researched in detail. The key scientific problems are: the nuclear structure and reaction for nuclear far from $\beta$-stability line; the synthesize of new nuclides near drip lines and new super heavy nuclides; the properties of asymmetric nuclear matter with extra large isospin and some nuclear astro- reactions. | 2020-09-20 13:02:40 | {"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": 0.6299523115158081, "perplexity": 2863.09659709017}, "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/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00472.warc.gz"} |
https://mail.institut-cuisine-libre.fr/apricot/intrapsychic/11011202a7bd530e2eae0-the-function-f-is-defined-by-the-power-series | 427K followers. Since the function f(x) represents the series and converges within the limited domain, f(x) is a continuous function and can be represented as separable terms. Lets start with differentiation of the power series, $f\left( x \right) = \sum\limits_{n = 0}^\infty {{c_n}{{\left( {x - a} \right)}^n}} = {c_0} + {c_1}\left( {x - a} \right) + {c_2}{\left( {x - a} \right)^2} + {c_3}{\left( {x - a} \right)^3} + \cdots$ The function f is defined by the power series (-1)" nx" n + 1 2x 3x f (x) = + 2 3 + + 4 for all real numbers x for which the series converges. Globalization suggests that modern businesses are using information technology to: a) expand their market to customers around the globe b) find the lowest-cost suppliers regardless of location c) create 24 hour business days by shuttling work across The function f is defined by the power series f(x) = x- + + 3 for all real number for which the series converges. Maclaurin Series Formula: The formula used by the Maclaurin series calculator for computing a series expansion for any function is: Where f^n(0) is the nth order derivative of function f(x) as evaluated and n is the order x = 0. A power series is a series of the form. My desperate attempt was to use the binomial series $$4^{(6/10-n)}(1+x)^{6/10}$$ This is my first exposure to series and any help would be really appreciated. (b) Write the first four nonzero terms and the general term for an infinite series that represents f(x) (c) Use the result from part (b) to find the value of re + . A more mathematically rigorous definition is given below. + + 6! 2! A series, terms of which are power functions of variable x, is called the power series: A series in (x x0) is also often considered. CAMP_HACK. Great. Anda perlu tekan 'tng ewallet reload pin'. Let f be the function defined by the series $f(x)=sum_{n=0}^c_n(xa)^n=c_0+c_1(xa)+c_2(xa)^2+c_3(xa)^3+\ldots$ for $$|xa| 1.2 The Corresponding Sequence We consider a function fo (z) formally defined by the power series (1.4). (a) Find the interval of convergence of the power series for f. Justify your answer. . In each exercise do the following: (a) Find the radius of convergence of the given power series and the domain of f; (b) write the power series which defines the function f^{\prime} and find its radius of convergence by using methods of Sec. About In Today Riverside Accident , a gold Ford F-150 was going east on 71st near the bridge when the driver crossed the center line and hit a black infinity going west. Norrie, Huber, Piercy, McKeown Introduction to Business Information Systems Second Canadian Edition TEST BANK Chapter 1 1. VIDEO ANSWER: synthesis. effectively giving a new definition of the Riemann zeta function that has a larger domain than the original: This power Data Booster 5. The functions like speed, light color, brightness and direction. for all real numbers x for which the series converges. 1,299.00. The slope of the tangent line equals the derivative of the function at the marked point. Your favourite functions including funds transfer, bill payment, remittance, update contact details, etc. Example 2 : Find a power series representation of the function 1 7 + 2 x, and determine for which x it would be defined. Key Concepts. (a) Find the interval of convergence of the power series for f. Justify your answer. . Norrie, Huber, Piercy, McKeown Introduction to Business Information Systems Second Canadian Edition TEST BANK Chapter 1 1. This power series definition is readily extended to complex arguments to allow the complex exponential function : to be defined. f (x) = x3 3 x2 f ( x) = x 3 3 x 2. 9. I'm also not sure whether Im representing this function as a power series correctly. Find a in terms of b. 10. Let the function be defined by : ; L 5 5 ? Halo9 9 Apple CarPlay / Android Auto / HDMI / USB / Bluetooth / FLAC / DAB+ Receiver. Find the Maclaurin series for the derivative ;. For other properties of multiple power series, see, for example, , . In mathematics, a formal series is an infinite sum that is considered independently from any notion of convergence, and can be manipulated with the usual algebraic operations on series (addition, subtraction, multiplication, division, partial sums, etc.).. It starts and immediately stop without any result. Solution 1: Replace x (in our original f ( x) before the video) by x 2, and multiply the expression by x . 1. (2n)! #camphack 11. Solution 1: Replace x (in our original f ( x) before the video) by x 2, and multiply the expression by x . 6. n 24. Instead of saying a power series centered at a, we sometimes say a power series about a, or a power series around a. Suppose that the power series \(\displaystyle \sum_{n=0}^c_n(xa)^n$$ converges on the interval $$(aR,a+R)$$ for some $$R>0$$. Alpine innovation continues with the unique design of the Halo9, delivering a revolutionary 9 high-definition touch-screen that fits directly into any dash with a 1DIN or 2DIN slot. The function f is defined by the power series () ()() ( )2 0 11 1 1 1nn n fx x x x x = =+ + + + + + + + = +"" for all real numbers x for which the series converges. About In Today Riverside Accident , a gold Ford F-150 was going east on 71st near the bridge when the driver crossed the center line and hit a black infinity going west. And so this is the answer t Power to the Developer!When you set the brightness of keyboards to 0, you may also encounter the keyboard backlight not working issue. Find the Maclaurin series for the derivative The function f, defined above, has derivatives of all orders. f (x) = x 18x f ( x) = x 1 8 x. f (x) = 12x2 1 +6x7 f ( x) = 12 x 2 1 + 6 x 7. f (x) = x7 8 +x3 f ( x) = x 7 8 + x 3. f (x) = 5x2 4 3x2 f ( x) = x 2 5 4 3 x 2. Since f(x) = ln(1 + x) is an antiderivative of 1 1 + x, it remains to solve for the constant C. Since ln(1 + 0) = 0, we have C = 0. Therefore, a power series representation for f(x) = ln(1 + x) is. Let g be the function defined by g(x) = 1 + f (t) dt. . . Born in AZ, raised in OH, Leif was a scholarship competitive sailor for the US Naval Academy. Transcribed image text: The function f is defined by the power series (x-1)2 (x-1)3 (x-14 f(x)=(x-1)- 2 3 (-1)-(x-1)" + + = - n n=1 for all real numbers x for which the series converges. The domain of f, often called the interval of convergence (IOC), is the set of all x-values such that the power series converges. Leif Harrison has entered the world of the Hemp business after a multi-varied work experience in both high-level corporate business and post US Navy career. . In this interval you can derive this series term by term, obtaining a convergent series that coincide with f ( x) for x ] 1 3, 1 3 [. For example, given the power series for f(x) = 1 1 x, we can differentiate term-by-term to find the power series for f (x) = 1 (1 x)2. Similarly, using the power series for g(x) = 1 1 + x, we can integrate term-by-term to find the power series for G(x) = ln(1 + x), an antiderivative of g. #camphack 11. 6. That is easy enough to fix up as follows, Power series in real variables $x = (x _ {1} \dots x _ {n} ) We now show graphically how this series provides a representation for the function f ( x ) = 1 1 x f ( x ) = 1 1 x by comparing the graph of f with the graphs of several of the partial sums of this infinite series. 2008 . 427K followers. 4! But what's exciting about what we're about to do in this video is we're going to use infinite series to define a function. Use this series to write the first three nonzero terms and the general term of the Taylor series for f about x = 0. Video Exchange Learning allows our teachers to guide your progress through every step of their online music lessons. Globalization suggests that modern businesses are using information technology to: a) expand their market to customers around the globe b) find the lowest-cost suppliers regardless of location c) create 24 hour business days by shuttling work across Start Solution. 4). Pages 667 ; Ratings 100% (2) 2 out of 2 people found this document helpful; This preview shows page 454 - 457 out of 667 pages.preview shows page 454 - 457 out of 667 pages. The binomial theorem gives a power of a binomial expression as a sum of terms involving binomial coefficients. 1 + x + x 2 + x 3 + when | x | < 1 . (a) Find the interval of convergence of the power series for f. . A function$f$is defined by a power series. x 1 x x 3 = x 1 1 x x 3 = x n, m 0 ( n + m n) x n ( x 3) m = n, m 0 ( n + m n) x n + 3 m + 1. f (x) = 3x2 5 2 3x f ( x) = 3 x 2 5 2 x 3. Kondensator (Elektrotechnik) Ein Kondensator (von lateinisch condensare verdichten) ist ein passives elektrisches Bauelement mit der Fhigkeit, in einem Gleichstromkreis elektrische Ladung und die damit zusammenhngende Energie statisch in einem elektrischen Feld zu speichern. Question 1 Which of the following features is typically NOT associated with a quantitative model for a business process? And I'm about to write a general case of the power series. Example 2 : Find a power series representation of the function 1 7 + 2 x, and determine for which x it would be defined. My question is as follows: I am working on an exercise stated as follows: Let$\\mathcal{E}$be the set of x0 x1 x2. Write a review. For a given power series, it can be proven that either the IOC = (-,), meaning that the series converges for all x, or there exists a finite non-negative number R 0, called the radius of convergence (ROC), such that the series converges whenever |x - c| R That is easy enough to fix up as follows, "An analytic function is equal to its power series representation within the power series' radius of convergence" This is true, but there is an even deeper meaning to analytic functions. SKU: 6047662 / ME26288 Part Number: ILX-F309E. Module 1: Introduction to Models Quiz Quiz, 10 questions Question 1 1 point 1. "An analytic function is equal to its power series representation within the power series' radius of convergence" This is true, but there is an even deeper meaning to analytic functions. 3. Your favourite functions including funds transfer, bill payment, remittance, update contact details, etc. Differential calculus. You get. node-talib. Write the first four nonzero terms and the general term. Thus we have proved that the successive convergents of the continued fraction (1.6) correspond to (l) , (2), (3) . He Serious particularly this is similar to 1/1 -1 are here get the submission. And after a last manipulation you can get a formal power series of the classical form a n x n. Share. Write the following function as a power series and give the interval of convergence. A function f defined on some open subset U of R or C is called analytic if it is locally given by a convergent power series. And the most common one that you will see in your mathematical careers is the power series. Example: the cosine function is: cos(x) = 1 x 2 /2! Note: As we did in the section on sequences, we can think of the an a n as being a function a(n) a ( n) defined on the non-negative integers. Show All Steps Hide All Steps. First, in order to use the formula from this section we know that we need the numerator to be a one. Whats so nice about functions that are power series? Imagine that instead of there being some highest power on x, (like 5 in this function), there was no largest power.For example, consider the following function: Thus, the function f(x) above is a power series centered at 1, while the function g(x) above is a power series centered at 5. Continuity of real functions is usually defined in terms of limits. Thus we have proved that the successive convergents of the continued fraction (1.6) correspond to (l) , (2), (3) . So, a primitive of f is F ( x) = 1 3 1 1 + 3 x = 1 3 n = 0 ( 1) n 3 n x n. This power series is absolutely convergent for x ] 1 3, 1 3 [. Mathematical equations A formal description of a business process A 100% accurate representation of the business process Assumptions Question 2 1 point 2. The series will be more precise near the center point. my is the only one official maybank financial portal in malaysia. 2! Example 2 : Find a power series representation of the function 1 7 + 2 x, and determine for which x it would be defined. Data Booster 5. Anda perlu tekan 'tng ewallet reload pin'. First, in order to use the formula from this section we know that we need the numerator to be a one. Given two power series and that converge to functions f and g on a common interval I, the sum and difference of the two series converge to respectively, on I.$1,299.00. The function f is defined by the power series for all real numbers x for which the series converges. 91 Buena Park Traffic. The first thing to notice about a power series is that it is a function of x x. 9 following. More specifically, if the variable is x, then all the terms of the series involve powers of x. WEBCAMP HACK. In mathematics, differential calculus is a subfield of calculus that studies the rates at which quantities change. This power series definition is readily extended to complex arguments to allow the complex exponential function : to be defined. Though these indicators are widely exploited by both Tulip Indicators (TI) is a library of functions for technical analysis of financial time series data. terms of the power series (1. Halo9 9 Apple CarPlay / Android Auto / HDMI / USB / Bluetooth / FLAC / DAB+ Receiver. The limited domain has a length of R that acts as the radius of x values that makes the series converge. Question 1 Which of the following features is typically NOT associated with a quantitative model for a business process? Kondensator (Elektrotechnik) Ein Kondensator (von lateinisch condensare verdichten) ist ein passives elektrisches Bauelement mit der Fhigkeit, in einem Gleichstromkreis elektrische Ladung und die damit zusammenhngende Energie statisch in einem elektrischen Feld zu speichern. Like this: asub0sub asub1subx Show Ads. This means that every a U has an open neighborhood V U , such that there exists a power series with center a that converges to f ( x ) for every x V . A power series is a type of series with terms involving a variable. Start Solution. terms of the power series (1. The function f is defined by the power series () 2323()1 23 4 1 LL xx xn nxn fx n = + + + + + for all real numbers x for which the series converges. Also note that the constant c is called the center If the terms of a sequence being summed are power functions, then we have a power series, defined by Note that most textbooks start with n = 0 instead of starting at 1, because it makes the exponents and n the same (if we started at 1, then the exponents would be n - 1). A. Find the sum of the series for f. (b) The power series above is the Taylor series for f about x = (c) Let g be the function defined by g(r) = f (t) cit. Definition. 9 following. (a) Find the interval of convergence of the power series for f Justify your answer. Find the sum of the series for f. The binomial expansion of f(x), in ascending powers of x, up to and including the term in x2 is A + Bx + 243 16 x2 where A and B are constants. The function has derivatives of all orders and the Maclaurin series for the function is given by . That model from Dell only comes with the one color LED light (blue) under the keyboard, it is not possible to change the color on that model - Sorry ! 1 Answer. The real exponential function can also be defined as a power series. 91 Buena Park Traffic. This offers only valid for at least two different bills such as water bills, electricity, telephone bills and etc. This power series definition is readily extended to complex arguments to allow the complex exponential function : to be defined. As a result, we are able to represent the function f (x) = 1 1 x f (x) = 1 1 x by the power series 1 + x + x 2 + x 3 + when | x | < 1 . So then this would be 12 from an equal 02 infinity of X squared to the power here. Since f(x) can be evaluated within |x| < R, we can rewrite f(x) = a 0 + a 1 x + a 2 x 2 ++ a n x n The real exponential function can also be defined as a power series. my is the only one official maybank financial portal in malaysia. Module 1: Introduction to Models Quiz Quiz, 10 questions Question 1 1 point 1. The function g is defined by the power series (-1)" x" + g (x) = 1 + 4! In signal processing, cross-correlation is a measure of similarity of two series as a function of the displacement of one relative to the other. (a) Determine the interval of convergence for f. LL xx x n xn gx n = + + + + for all real numbers x for which the series converges. The cn c n s are often called the coefficients of the series. + x 4 /4! For problems 1 4 write the given function as a power series and give the interval of convergence. In addition, for any real number b and integer the series converges to and the series converges to whenever bxm is in the interval I. Write the following function as a power series and give the interval of convergence. . The formula is as follows: where Ix2x3x4x x r. 1). The graph of a function, drawn in black, and a tangent line to that function, drawn in red. . 2. effectively giving a new definition of the Riemann zeta function that has a larger domain than the original: Taylor Series . This is also known as a sliding dot product or sliding inner-product.It is commonly used for searching a long signal for a shorter, known feature. Mathematical equations A formal description of a business process A 100% accurate representation of the business process Assumptions Question 2 1 point 2. Solution 1: Replace x (in our original f ( x) before the video) by x 2, and multiply the expression by x . A power series about a, or just power series, is any series that can be written in the form, n=0cn(x a)n n = 0 c n ( x a) n. where a a and cn c n are numbers. Functions Defined by Power Series If f x a0 a1 x a a2 x a 2 a3 x a 3 has radius of convergence R 0 (where possiblyR ), then f is differentiable and (just as we would hope) f x a1 2a2 x a 3a3 x a 2 and, furthermore, the radius of convergence of the power series for f is also R. The function g is defined by the power series () () 23 1 1 2! Our mission is to teach you how to play with masterful technique and make you the best musician possible. Write a review. Show All Steps Hide All Steps. The ratio test isn't giving me the right answer. The real exponential function can also be defined as a power series. CAMP_HACK. Thus, we seek a smooth function f(x) so that f(xi) = yi for all i. Illustrated definition of Power Series: An infinite series with increasing powers (exponents) of a variable. 6. $16.7$ (thus verifying Theorem 16.8.1); (c) find the domain of What is the range of within the interval of convergence? Operations with multiple power series are carried out, broadly speaking, according to the same rules as when $n=1$. Fourier series for even and odd functions: Recall that a function is called odd if f(-x) = -f(x) and a function is called even if f(-x) = f(x). So I could imagine a function, f of x, being defined as the infinite sum. We have seen sequences and series of constants. As a result, a power series can be thought of as an infinite polynomial. Power Series. Power series are used to represent common functions and also to In fact, the natural cubic spline is the smoothest possible function of all square integrable functions. | 2022-08-18 01:08:23 | {"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": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8268491625785828, "perplexity": 419.263585092506}, "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/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00394.warc.gz"} |
https://www.mersenneforum.org/showthread.php?s=0edbc88293a3ab568ece2526d8776ef6&p=552157 | mersenneforum.org > News The Next Big Development for GIMPS
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2020-07-31, 15:59 #199 kruoli "Oliver" Sep 2017 Porta Westfalica, DE 2×151 Posts Code: When workers are testing wildly different numbers, such as ECM on a small Mersenne number in one window and an LL test on a 100 million digit number in another window, then the "iterations between screen outputs" setting will cause the windows to output at wildly different rates. You can workaround this in prime.txt: ScaleOutputFrequency=0 or 1 (default value is 0) If you set this to 1, then "iterations between screen outputs" will be automatically scaled so that windows produce output at roughly the same rate. An LL test on M50,000,000 will not scale at all, ECM on a small number will scale to a much larger value and an LL test on a 100 million digit number will scale to a much smaller value. That's what I'm using and it helps a lot. Of course, this alone cannot be perfect, but in this case I can live without perfection.
2020-07-31, 18:23 #200 chalsall If I May "Chris Halsall" Sep 2002 Barbados 67·139 Posts Latest version? George... Where can I find the latest version you want to be tested, Linux mprime? I've got a machine I want to "prove the sanity of" before deploying next week. Thanks (this is so cool!!! )
2020-07-31, 19:24 #201
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2·3·11·67 Posts
Quote:
Originally Posted by chalsall Where can I find the latest version you want to be tested, Linux mprime?)
Until George puts a later one out, v30.1b1 is here
2020-07-31, 19:28 #202
chalsall
If I May
"Chris Halsall"
Sep 2002
67×139 Posts
Quote:
Originally Posted by kriesel Until George puts a later one out, v30.1b1 is here
Yeah, thanks. But I was wondering if the version which does the MD5 checksum was ready for testing?
Even beyond the Proxy issue, I've had at least one case where a direct comms CERT run still failed to match. On known good kit.
2020-07-31, 22:18 #203 Prime95 P90 years forever! Aug 2002 Yeehaw, FL 2·72·73 Posts v30.2 is imminent. I'll post links here.
2020-07-31, 23:46 #204 Prime95 P90 years forever! Aug 2002 Yeehaw, FL 2·72·73 Posts Version 30.2 Windows 64-bit: https://www.dropbox.com/s/iz417i8hpt...win64.zip?dl=0 Linux 64-bit: https://www.dropbox.com/s/q4haz69wti...64.tar.gz?dl=0 This is much closer to a beta version. I haven't tested it as much as I should. This version can upload proof files. There is a new Resource Limits menu choice / dialog box. LL work preferences are gone, if you upgrading, your LL work preferences will be converted to PRP. Please review the overhauled readme.txt. Suggestions for further changes are welcome. I have a few ideas I'll put forward soon.
2020-07-31, 23:57 #205 firejuggler Apr 2010 Over the rainbow 1001011010112 Posts About to finish my first 'proof' file. hope it will help. As I upgraded to 30.2 I do not have to manually upload the proof anymore, right?
2020-07-31, 23:58 #206
chalsall
If I May
"Chris Halsall"
Sep 2002
67×139 Posts
Quote:
Originally Posted by Prime95 This is much closer to a beta version. I haven't tested it as much as I should.
Thanks. Now running on a 6-core E-2136 (the machine's name is "rdt1", if you have any use for seeing the server-side of things).
2020-08-01, 00:24 #207
Prime95
P90 years forever!
Aug 2002
Yeehaw, FL
157628 Posts
Quote:
Originally Posted by firejuggler About to finish my first 'proof' file. hope it will help. As I upgraded to 30.2 I do not have to manually upload the proof anymore, right?
Correct. 30.2 will upload any proof files it finds in the directory containing the executable.
2020-08-01, 00:42 #208 firejuggler Apr 2010 Over the rainbow 241110 Posts I guess that's it? And since it is a DC there is no need to upload the proof? Attached Thumbnails
2020-08-01, 00:42 #209 Prime95 P90 years forever! Aug 2002 Yeehaw, FL 1BF216 Posts Paging Oliver: Did anything strange happen during the run of 10471007? The proof failed certification. Do you have the proof file by chance?
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https://rjlipton.wordpress.com/2017/06/08/does-logic-apply-to-hearings/ | The problem of mining text for implications
2016 RSA Conference bio, speech
Michael Rogers, the head of the National Security Agency, testified before the Senate Intelligence Committee the other day about President Donald Trump. He was jointed by other heads of other intelligence agencies who also testified. Their comments were, as one would expect, widely reported.
The various reports all were similar to this:
Adm. Michael S. Rogers, the head of the National Security Agency, also declined to comment on earlier reports that Mr. Trump had asked him to deny possible evidence of collusion between Mr. Trump’s associates and Russian officials. He said he would not discuss specific interactions with the president.
## Why I Am Puzzled
The above quote is accurate—Adm. Rogers did not discuss specific interactions with the president. But I have trouble with this statement. The problem I have is this:
Are statements made in a Senate hearing subject to the basic rules of logic?
For example, if a person says ${A}$ and later says ${A \implies B}$, can we conclude that he or she has effectively said ${B}$?
Let’s look at the testimony of Adm Rogers. He insisted that he could not recall being pressured to act inappropriately in his almost three years in the post. “I have never been directed to do anything I believe to be illegal, immoral, unethical or inappropriate,” he said.
During his three years as head of the NSA he worked under President Obama and now President Trump. So I see the following logical argument. Since he has never been asked to do anything wrong during that period, then it follows that Trump never asked him to do anything wrong.
This follows from the rule called universal specification or universal elimination. If ${\forall x \in S \ A(x)}$ is true, then for any ${c}$ in the set ${S}$ it must follow that ${A(c)}$ is true.
## Logic and Buffering
What is going on here? The reports that he refused to answer ‘is ${A(c)}$ true?’ are correct. But he said a stronger statement in my mind that ${\forall x \in S \ A(x)}$ is true. Is it misleading reporting? Or do the rules of logic not apply to testimony before Senate committees? Which is a stronger statement:
$\displaystyle A(c)$
or
$\displaystyle (\forall x \in S) \ A(x),$
where ${c}$ is an element of ${S}$?
In mathematics the latter statement is stronger, but it appears not to be so in the real world. The statement ${A(c)}$ is more direct. What does this say about logic and its role in human discourse?
Ken recalls a course he took in 1979 from the late Manfred Halpern, a professor of politics at Princeton. Titled “Personal and Political Transformation,” the course used a set of notes that became Halpern’s posthumous magnum opus.
The notes asserted that components of human relationships can be classed into eight basic modalities, the first three being paradigms for life: emanation, incoherence, transformation, isolation, subjection, direct bargaining, boundary management, and buffering. The first three form a progression exemplified by Dorothy and the wizard vis-à-vis Glinda and the ruby slippers in The Wizard of Oz; later he added deformation as a ninth mode and fourth paradigm and second progression endpoint. It particularly struck Ken that presenting mathematical proofs is classed as a form of subjection: You can’t argue or bargain with a proof or counterexample.
Buffering made and remained in his list. He showed how each member is archetypal in human history and depth psychology. So Ken’s answer is that the one-step-remove of saying “${\forall x \in S \ A(x)}$” rather than “${A(c)}$” is a deeply rooted difference. It makes wiggle-room that a jury of peers might credit in a pinch.
## Mining Logical Inferences
Psychology aside, the mining of logical inferences is a major application area. Sometimes the inference is outside the text being analyzed, such as when “chatter” is evaluated to tell how far it may imply terrorist threats. We are interested in cases where the deduction is more inside. For instance, consider this example in a 2016 article on the work of Douglas Lenat:
A bat has wings. Because it has wings, a bat can fly. Because a bat can fly, it can travel from place to place.
One might say that underlying this is the logical rule
$\displaystyle (\forall x \in S)\mathit{HasWings}(x) \rightarrow \mathit{CanFly(x)}.$
One of the problems, however, is that even if we limit the set ${S}$ to animals, the rule is false—there are many flightless birds. This leads into the whole area of non-monotonic logic which is a topic for another day—but good to bear in mind when revelations from hearings revise previously-held beliefs.
Ken has been dealing this week with an example at the juncture of the logic of time and human language. He had to evaluate twenty pages of testimony about a recurring behavior ${X}$. In one place it states that ${X}$ occurred at time ${t}$ and occurred “once again” at time ${t'}$. The question is whether one can infer and apply the rule
$\displaystyle (\forall t,t',t'') (X(t) \wedge \mathit{OnceAgain}(X,t') \wedge t < t'' < t') \rightarrow \neg X(t'').$
This was complicated by the document having been translated from a foreign language. Whether time ${t'}$ was the next occurrence of ${X}$ after time ${t}$ makes a difference to results Ken might give. Of course this may be clarified in a further round of testimony—but we could say the same about Admiral Rogers, and he has left the stand.
## Open Problems
How soon will we have apps that can take statements of the form ${(\forall x \in S)A(x)}$ and deduce ${A(c)}$ for a particular ${c \in S}$ that we want to know about? Will inferences from “material implication” be considered material in testimony?
Update, 11:15pm 6/8/17: CNN has just told of a woman they interviewed about the contradictions between Donald Trump and James Comey. Asked if she believes Comey lied, she replied, “No.” Asked if she believes Trump lied, she replied, “No.” Asked how that could be, she said: “The media has distorted it.” Thus the logical law of excluded middle is replaced by a “law of occluded media,” which blocks constructive inference…
1. June 8, 2017 4:32 pm
From Rudy Rucker:
The Marx brothers are in a room, and Zeppo says, “We’ve to think!” Chico replies, “Nah, we already tried that.” ha ha
The universal affirmatives, or A-propositions, are not adequate to to handle arguments in syllogistic reasoning that involve relations between pairs of things. In general, logic mostly allows us to follow our otherwise reasonable assumptions to contradiction in order to discover features we would otherwise have missed. Although a compelling feature in our world, the power of logic does not imply that the world itself is logical.
The second of my three favorite tautologies:
Modus tollens (~B & (A->B)) -> ~A
If the world were logical, then the world would make sense, but the world doesn’t make sense, so the world isn’t logical.
2. June 8, 2017 4:55 pm
Go Ask Alice, I Think She’ll Know …
Joining the Herrings already in Digress …
3. June 8, 2017 5:00 pm
Cf. Modus Dolens
4. June 8, 2017 5:08 pm
I think this may be related to the rubric of the ω-inconsistent mother.described by Paul Halmos.
5. June 8, 2017 6:30 pm
The purpose of these sorts of examinations is to discover how consistently the subject himself actually believes the things he has already answered.
“… and I’m never — never — sick at sea!”
“What, never?”
“No, never!”
“What, **NEVER**?”
“Well… Hardly Ever…”
Because people have been known to lie. So, that should be sufficient explanation for why an honest and logical senator would press from the general to the specific (if that’s what happened). And Editorial can react to the quotable quotes any way they like.
June 8, 2017 11:47 pm
> “I have never been DIRECTED to do anything I believe to be illegal, immoral, unethical or inappropriate,” he said… then it follows that Trump never ASKED him to do anything wrong.
No, it follows that Trump never DIRECTED him to…
7. June 9, 2017 9:09 am
Excellent point, Acct—thanks. Now here is a relevant juncture of Comey’s testimony yesterday:
[Senator James] RISCH: He [Trump] said, “I hope.” Now … Do you know of any case where a person has been charged for obstruction of justice or, for that matter, any other criminal offense, where they said or thought they hoped for an outcome?
COMEY: I don’t know well enough to answer. The reason I keep saying his words is I took it as a direction.
[and further exchange on “hope” versus “direction”]
We tried to keep logic and semantic aspects separate in the post; the logic is still a reason this should be followed up.
8. June 9, 2017 9:24 am
And then there is this, on how a question can be a command:
http://time.com/4811148/comey-testimony-henry-ii-thomas-becket-will-no-one-rid-me-of-this-meddlesome-priest/
9. June 9, 2017 10:27 am
a theme/ topic somewhat covered in gores book “the assault on reason”. human reason is based on the scientific revolution which is a bit tattered these days. reason is something that is not innate to humans, exactly. there is a sort of “psychological reasoning” that is not the same as mathematical logic. theres quite a bit of study of this in psychology, and esp in behavioral economics. there is james old quote, “a great many ppl think they are thinking when they are merely rearranging their prejudices”. how about, “reality has a liberal bias”? there are so many political quotes that show reason is somewhat of a auxiliary/ superfluous aspect at times. how about looking at wikipedias “list of logical fallacies”. its too bad this stuff isnt really taught in schools any more. if you ask me, reason is much of what separates humans from animals. its also found in neurobiology that its more of an attribute of the neocortex which is not fully developed in humans even up into their mid20s (that is relatively new research). amazing! re current state of science see also https://vzn1.wordpress.com/2017/04/27/tribute-to-science-2017-shifting-shearing-maybe-a-crossroads/
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http://xxx.unizar.es/list/physics.ins-det/new | # Instrumentation and Detectors
## New submissions
[ total of 11 entries: 1-11 ]
[ showing up to 2000 entries per page: fewer | more ]
### New submissions for Tue, 20 Mar 18
[1]
Title: Photon detector system performance in the DUNE 35-ton prototype liquid argon time projection chamber
Subjects: Instrumentation and Detectors (physics.ins-det); High Energy Physics - Experiment (hep-ex)
The 35-ton prototype for the Deep Underground Neutrino Experiment far detector was a single-phase liquid argon time projection chamber with an integrated photon detector system, all situated inside a membrane cryostat. The detector took cosmic-ray data for six weeks during the period of February 1, 2016 to March 12, 2016. The performance of the photon detection system was checked with these data. An installed photon detector was demonstrated to measure the arrival times of cosmic-ray muons with a resolution better than 32 ns, limited by the timing of the trigger system. A measurement of the timing resolution using closely-spaced calibration pulses yielded a resolution of 15 ns for pulses at a level of 6 photo-electrons. Scintillation light from cosmic-ray muons was observed to be attenuated with increasing distance with a characteristic length of $155 \pm 28$ cm.
[2]
Title: How to automate a kinematic mount using a 3D printed Arduino-based system
Subjects: Instrumentation and Detectors (physics.ins-det)
We demonstrate a simple, flexible and cost-effective system to automatize most of the kinematic mounts available nowadays on the market. It combines 3D printed components, an Arduino board, stepper motors, and simple electronics. The system developed can control independently and simultaneously up to ten stepper motors using commands sent through the serial port, and it is suitable for applications where optical realignment using flat mirrors is required on a periodic basis.
[3]
Title: Combined readout of a triple-GEM detector
Subjects: Instrumentation and Detectors (physics.ins-det); High Energy Physics - Experiment (hep-ex)
Optical readout of GEM based devices by means of high granularity and low noise CMOS sensors allows to obtain very interesting tracking performance. Space resolution of the order of tens of $\mu$m were measured on the GEM plane along with an energy resolution of 20%$\div$30%. The main limitation of CMOS sensors is represented by their poor information about time structure of the event. In this paper, the use of a concurrent light readout by means of a suitable photomultiplier and the acquisition of the electric signal induced on the GEM electrode are exploited to provide the necessary timing informations. The analysis of the PMT waveform allows a 3D reconstruction of each single clusters with a resolution on z of 100 $\mu$m. Moreover, from the PMT signals it is possible to obtain a fast reconstruction of the energy released within the detector with a resolution of the order of 25% even in the tens of keV range useful, for example, for triggering purpose.
[4]
Title: Study of point- and cluster-defects in radiation-damaged silicon
Comments: 13 pages, 12 figures, 4 tables
Subjects: Instrumentation and Detectors (physics.ins-det); Materials Science (cond-mat.mtrl-sci)
Non-ionising energy loss of radiation produces point defects and defect clusters in silicon, which result in a signifcant degradation of sensor performance. In this contribution results from TSC (Thermally Stimulated Current) defect spectroscopy for silicon pad diodes irradiated by electrons to fluences of a few $10^{14}$ cm$^{-2}$ and energies between 3.5 and 27 MeV for isochronal annealing between 80 and 280{\deg}C, are presented. A method based on SRH (Shockley-Read-Hall) statistics is introduced, which assumes that the ionisation energy of the defects in a cluster depends on the fraction of occupied traps. The dfference of ionisation energy of an isolated point defect and a fully occupied cluster, $\Delta E_a$, is extracted from the TSC data.
For the VOi (vacancy-oxygen interstitial) defect $\Delta E_a = 0$ is found, which cofirms that it is a point defect, and validates the method for point defects. For clusters made of deep acceptors the $\Delta E_a$ values for different defects are determined after annealing at 80{\deg}C as a function of electron energy, and for the irradiation with 15 MeV electrons as a function of annealing temperature. For the irradiation with 3.5 MeV electrons the value $\Delta E_a = 0$ is found, whereas for the electron energies of 6 to 27 MeV $\Delta E_a > 0$. This agrees with the expected threshold of about 5 MeV for cluster formation by electrons. The $\Delta E_a$ values determined as a function of annealing temperature show that the annealing rate is different for different defects. A naive diffusion model is used to estimate the temperature dependencies of the diffusion of the defects in the clusters.
[5]
Title: Pulse Shapes in High Purity Germanium Point Contact Detectors with Low Net Impurity Concentration
Subjects: Instrumentation and Detectors (physics.ins-det)
High Purity germanium point-contact detectors have low energy thresholds and excellent energy resolution over a wide energy range, and are thus widely used in nuclear and particle physics. In rare event searches, such as neutrinoless double beta decay, the point-contact geometry is of particular importance since it allows for pulse-shape discrimination, and therefore for a significant background reduction. In this paper we investigate the pulse-shape discrimination performance of ultra-high purity germanium point contact detectors. It is demonstrated that a minimal net impurity concentration is required to meet the pulse-shape performance requirements
### Cross-lists for Tue, 20 Mar 18
[6] arXiv:1803.06370 (cross-list from physics.atom-ph) [pdf, other]
Title: First Calorimetric Measurement of Electron Capture in ${}^{193}$Pt with a Transition Edge Sensor
Comments: 17th International Workshop on Low Temperature Detectors
Subjects: Atomic Physics (physics.atom-ph); Instrumentation and Detectors (physics.ins-det)
The neutrino mass can be extracted from a high statistics, high resolution calorimetric spectrum of electron capture in ${}^{163}$Ho. In order to better understand the shape of the calorimetric electron capture spectrum, a second isotope was measured with a close to ideal absorber-source configuration. ${}^{193}$Pt was created by irradiating a ${}^{192}$Pt-enriched platinum foil in a nuclear reactor. This Pt-in-Pt absorber was designed to have a nearly ideal absorber-source configuration. The measured ${}^{193}$Pt calorimetric electron-capture spectrum provides an independent check on the corresponding theoretical calculations, which have thus far been compared only for ${}^{163}$Ho. The first experimental and theoretically-calculated spectra from this ${}^{193}$Pt-in-Pt absorber are presented and overlaid for preliminary comparison of theory with experiment.
[7] arXiv:1803.06628 (cross-list from astro-ph.IM) [pdf, ps, other]
Title: A machine learning method to separate cosmic ray electrons from protons from 10 to 100 GeV using DAMPE data
Comments: 11 pages, 8 figures, accepted for publication in RAA
Subjects: Instrumentation and Methods for Astrophysics (astro-ph.IM); Instrumentation and Detectors (physics.ins-det)
DArk Matter Particle Explorer (DAMPE) is a general purpose high energy cosmic ray and gamma ray observatory, aiming to detect high energy electrons and gammas in the energy range 5 GeV to 10 TeV and hundreds of TeV for nuclei. This paper provides a method using machine learning to identify electrons and separate them from gammas,protons,helium and heavy nuclei with the DAMPE data from 2016 January 1 to 2017 June 30, in energy range from 10 to 100 GeV.
[8] arXiv:1803.06894 (cross-list from nucl-ex) [pdf, ps, other]
Title: Main features of detectors and isotopes to investigate double beta decay with increased sensitivity
Authors: A.S. Barabash
Comments: 25 pages, 2 figures, 10 tables
Journal-ref: Int. J. Mod. Phys. A 33 (2018) 1843001
Subjects: Nuclear Experiment (nucl-ex); Instrumentation and Detectors (physics.ins-det)
The current situation in double beta decay experiments , the characteristics of modern detectors and the possibility of increasing the sensitivity to neutrino mass in future experiments are discussed. The issue of the production and use of enriched isotopes in double beta decay experiments is discussed in addition.
### Replacements for Tue, 20 Mar 18
[9] arXiv:1703.05371 (replaced) [pdf, other]
Title: Directional Sensitivity In Light-Mass Dark Matter Searches With Single-Electron Resolution Ionization Detectors | 2018-07-21 07:32:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.5529181361198425, "perplexity": 2479.447997607954}, "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/1531676592420.72/warc/CC-MAIN-20180721071046-20180721091046-00418.warc.gz"} |
https://calculator.academy/gpm-to-velocity-calculator/ | Enter the total GPM and the cross-sectional area of flow into the calculator to determine the velocity.
## GPM to Velocity Formula
The following equation is used to calculate the Velocity from GPM.
V = GPM / 7.481 / A / 60
• Where V is the velocity (ft/s)
• GPM is the gallons per minute
• A is the cross-sectional area of flow (ft^2)
To calculate the velocity from gallons per minute, divide the GPM by 7.481, divide again by the cross-sectional area of flow, then finally, divide by 60.
## How to Calculate Velocity From GPM?
Example Problem:
The following example outlines the steps and information needed to calculate Velocity from GPM.
First, determine the gallons per minute. In this example, the gallons per minute is found to be 250.
Next, determine the cross-sectional area of flow. In this case, the cross-sectional area of flow is measured to be 3ft^2.
Finally, calculate the velocity from GPM using the formula above:
V = GPM / 7.481 / A / 60
V = 250 / 7.481 / 3 / 60
V = .185 | 2023-02-03 03:59:10 | {"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": 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": 0.7243640422821045, "perplexity": 1071.565889918025}, "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-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00631.warc.gz"} |
https://www.iacr.org/cryptodb/data/author.php?authorkey=3356 | ## CryptoDB
### Shengli Liu
#### Publications
Year
Venue
Title
2023
PKC
(Asymmetric) Password-based Authenticated Key Exchange ((a)PAKE) protocols allow two parties establish a session key with a pre-shared low-entropy password. In this paper, we show how Encrypted Key Exchange (EKE) compiler [Bellovin and Merritt, S&P 1992] meets tight security in the Universally Composable (UC) framework. We propose a strong 2DH variant of EKE, denoted by 2DH-EKE, and prove its tight security in the UC framework based on the CDH assumption. The efficiency of 2DH-EKE is comparable to the original EKE, with only O(\lambda) bits growth in communication (\lambda the security parameter), and two (resp., one) extra exponentiation in computation for client (resp., server). We also develop an asymmetric PAKE scheme 2DH-aEKE from 2DH-EKE. The security reduction loss of 2DH-aEKE is N, the total number of client-server pairs. With a meta-reduction, we formally prove that such a factor N is inevitable in aPAKE. Namely, our 2DH-aEKE meets the optimal security loss. As a byproduct, we further apply our technique to PAKE protocols like SPAKE2 and PPK in the relaxed UC framework, resulting in their 2DH variants with tight security from the CDH assumption.
2023
PKC
In this paper, we propose a new type of non-interactive zero-knowledge (NIZK), called Fine-grained Verifier NIZK (FV-NIZK), which provides more flexible and more fine-grained verifiability of proofs than standard NIZK that supports public verifiability and designated-verifier NIZK (DV-NIZK) that supports private verifiability. FV-NIZK has two statistically equivalent verification approaches: -- a master verification using the master secret key msk; -- a fine-grained verification using a derived secret key sk_d, which is derived from msk w.r.t. d (which may stand for user identity, email address, vector, etc.). We require unbounded simulation soundness (USS) of FV-NIZK to hold, even if an adversary obtains derived secret keys sk_d with d of its choices, and define proof pseudorandomness which stipulates the pseudorandomness of proofs for adversaries that are not given any secret key. We present two instantiations of FV-NIZK for linear subspace languages, based on the matrix decisional Diffie-Hellman (MDDH) assumption. One of the FV-NIZK instantiations is pairing-free and achieves almost tight USS and proof pseudorandomness. We illustrate the usefulness of FV-NIZK by showing two applications and obtain the following pairing-free schemes: -- the first almost tightly multi-challenge CCA (mCCA)-secure inner-product functional encryption (IPFE) scheme without pairings; -- the first public-key encryption (PKE) scheme that reconciles the inherent contradictions between public verifiability and anonymity. We formalize such PKE as Fine-grained Verifiable PKE (FV-PKE), which derives a special key from the decryption secret key, such that for those who obtain the derived key, they can check the validity of ciphertexts but the anonymity is lost from their views (CCA-security still holds for them), while for others who do not get the derived key, they cannot do the validity check but the anonymity holds for them. Our FV-PKE scheme achieves almost tight mCCA-security for adversaries who obtain the derived keys, and achieves almost tight ciphertext pseudorandomness (thus anonymity) for others who do not get any derived key.
2023
EUROCRYPT
2022
ASIACRYPT
Privacy-Preserving Authenticated Key Exchange (PPAKE) provides protection both for the session keys and the identity information of the involved parties. In this paper, we introduce the concept of robustness into PPAKE. Robustness enables each user to confirm whether itself is the target recipient of the first round message in the protocol. With the help of robustness, a PPAKE protocol can successfully avoid the heavy redundant communications and computations caused by the ambiguity of communicants in the existing PPAKE, especially in broadcast channels. We propose a generic construction of robust PPAKE from key encapsulation mechanism (KEM), digital signature (SIG), message authentication code (MAC), pseudo-random generator (PRG) and symmetric encryption (SE). By instantiating KEM, MAC, PRG from the DDH assumption and SIG from the CDH assumption, we obtain a specific robust PPAKE scheme in the standard model, which enjoys forward security for session keys, explicit authentication and forward privacy for user identities. Thanks to the robustness of our PPAKE, the number of broadcast messages per run and the computational complexity per user are constant, and in particular, independent of the number of users in the system.
2021
CRYPTO
We construct the first authenticated key exchange protocols that achieve tight security in the standard model. Previous works either relied on techniques that seem to inherently require a random oracle, or achieved only “Multi-Bit-Guess” security, which is not known to compose tightly, for instance, to build a secure channel. Our constructions are generic, based on digital signatures and key encapsulation mechanisms (KEMs). The main technical challenges we resolve is to determine suitable KEM security notions which on the one hand are strong enough to yield tight security, but at the same time weak enough to be efficiently instantiable in the standard model, based on standard techniques such as universal hash proof systems. Digital signature schemes with tight multi-user security in presence of adaptive corruptions are a central building block, which is used in all known constructions of tightly-secure AKE with full forward security. We identify a subtle gap in the security proof of the only previously known efficient standard model scheme by Bader et al. (TCC 2015). We develop a new variant, which yields the currently most efficient signature scheme that achieves this strong security notion without random oracles and based on standard hardness assumptions.
2021
ASIACRYPT
For Key Encapsulation Mechanism (KEM) deployed in a multi-user setting, an adversary may corrupt some users to learn their secret keys, and obtain some encapsulated keys due to careless key managements of users. To resist such attacks, we formalize Enhanced security against Chosen Plaintext/Ciphertext Attack (ECPA/ECCA), which ask the pseudorandomness of unrevealed encapsulated keys under uncorrupted users. This enhanced security for KEM serves well for the security of a class of Authenticated Key Exchange protocols built from KEM. In this paper, we study the achievability of tight ECPA and ECCA security for KEM in the multi-user setting, and present an impossibility result and an optimal security loss factor that can be obtained. The existing meta-reduction technique due to Bader et al. (EUROCRYPT 2016) rules out some KEMs, but many well-known KEMs, e.g., Cramer-Shoup KEM (SIAM J. Comput. 2003), Kurosawa-Desmedt KEM (CRYPTO 2004), run out. To solve this problem, we develop a new technique tool named rank of KEM and a new secret key partitioning strategy for meta-reduction. With this new tool and new strategy, we prove that KEM schemes with polynomially-bounded ranks have no tight ECPA and ECCA security from non-interactive complexity assumptions, and the security loss is at least linear in the number n of users. This impossibility result covers lots of well-known KEMs, including the Cramer-Shoup KEM, Kurosawa-Desmedt KEM and many others. Moreover, we show that the linear security loss is optimal by presenting concrete KEMs with security loss Θ(n). This is justified by a non-trivial security reduction with linear loss factor from ECPA/ECCA security to the traditional multi-challenge CPA/CCA security.
2020
ASIACRYPT
We propose a generic construction of 2-pass authenticated key exchange (AKE) scheme with explicit authentication from key encapsulation mechanism (KEM) and signature (SIG) schemes. We improve the security model due to Gjosteen and Jager [Crypto2018] to a stronger one. In the strong model, if a replayed message is accepted by some user, the authentication of AKE is broken. We define a new security notion named ''IND-mCPA with adaptive reveals'' for KEM. When the underlying KEM has such a security and SIG has unforgeability with adaptive corruptions, our construction of AKE equipped with counters as states is secure in the strong model, and stateless AKE without counter is secure in the traditional model. We also present a KEM possessing tight ''IND-mCPA security with adaptive reveals'' from the Computation Diffie-Hellman assumption in the random oracle model. When the generic construction of AKE is instantiated with the KEM and the available SIG by Gjosteen and Jager [Crypto2018], we obtain the first practical 2-pass AKE with tight security and explicit authentication. In addition, the integration of the tightly IND-mCCA secure KEM (derived from PKE by Han et al. [Crypto2019]) and the tightly secure SIG by Bader et al. [TCC2015] results in the first tightly secure 2-pass AKE with explicit authentication in the standard model.
2019
PKC
Robustly reusable Fuzzy Extractor (rrFE) considers reusability and robustness simultaneously. We present two approaches to the generic construction of rrFE. Both of approaches make use of a secure sketch and universal hash functions. The first approach also employs a special pseudo-random function (PRF), namely unique-input key-shift (ui-ks) secure PRF, and the second uses a key-shift secure auxiliary-input authenticated encryption (AIAE). The ui-ks security of PRF (resp. key-shift security of AIAE), together with the homomorphic properties of secure sketch and universal hash function, guarantees the reusability and robustness of rrFE. Meanwhile, we show two instantiations of the two approaches respectively. The first instantiation results in the first rrFE from the LWE assumption, while the second instantiation results in the first rrFE from the DDH assumption over non-pairing groups.
2019
CRYPTO
We propose the concept of quasi-adaptive hash proof system (QAHPS), where the projection key is allowed to depend on the specific language for which hash values are computed. We formalize leakage-resilient(LR)-ardency for QAHPS by defining two statistical properties, including LR-$\langle \mathscr {L}_0, \mathscr {L}_1 \rangle$-universal and LR-$\langle \mathscr {L}_0, \mathscr {L}_1 \rangle$-key-switching.We provide a generic approach to tightly leakage-resilient CCA (LR-CCA) secure public-key encryption (PKE) from LR-ardent QAHPS. Our approach is reminiscent of the seminal work of Cramer and Shoup (Eurocrypt’02), and employ three QAHPS schemes, one for generating a uniform string to hide the plaintext, and the other two for proving the well-formedness of the ciphertext. The LR-ardency of QAHPS makes possible the tight LR-CCA security. We give instantiations based on the standard k-Linear (k-LIN) assumptions over asymmetric and symmetric pairing groups, respectively, and obtain fully compact PKE with tight LR-CCA security. The security loss is ${{O}}(\log {Q_{{e}}})$ where ${Q_{{e}}}$ denotes the number of encryption queries. Specifically, our tightly LR-CCA secure PKE instantiation from SXDH has only 4 group elements in the public key and 7 group elements in the ciphertext, thus is the most efficient one.
2018
PKC
Selective opening security (SO security) is desirable for public key encryption (PKE) in a multi-user setting. In a selective opening attack, an adversary receives a number of ciphertexts for possibly correlated messages, then it opens a subset of them and gets the corresponding messages together with the randomnesses used in the encryptions. SO security aims at providing security for the unopened ciphertexts. Among the existing simulation-based, selective opening, chosen ciphertext secure (SIM-SO-CCA secure) PKEs, only one (Libert et al. Crypto’17) enjoys tight security, which is reduced to the Non-Uniform LWE assumption. However, their public key and ciphertext are not compact.In this work, we focus on constructing PKE with tight SIM-SO-CCA security based on standard assumptions. We formalize security notions needed for key encapsulation mechanism (KEM) and show how to transform these securities into SIM-SO-CCA security of PKE through a tight security reduction, while the construction of PKE from KEM follows the general framework proposed by Liu and Paterson (PKC’15). We present two KEM constructions with tight securities based on the Matrix Decision Diffie-Hellman assumption. These KEMs in turn lead to two tightly SIM-SO-CCA secure PKE schemes. One of them enjoys not only tight security but also compact public key.
2018
ASIACRYPT
A fuzzy extractor (FE) aims at deriving and reproducing (almost) uniform cryptographic keys from noisy non-uniform sources. To reproduce an identical key R from subsequent readings of a noisy source, it is necessary to eliminate the noises from those readings. To this end, a public helper string P, together with the key R, is produced from the first reading of the source during the initial enrollment phase.In this paper, we consider computational fuzzy extractor. We formalize robustly reusable fuzzy extractor (rrFE) which considers reusability and robustness simultaneously in the Common Reference String (CRS) model. Reusability of rrFE deals with source reuse. It guarantees that the key R output by fuzzy extractor is pseudo-random even if the initial enrollment is applied to the same source several times, generating multiple public helper strings and keys $(P_i,R_i)$. Robustness of rrFE deals with active probabilistic polynomial-time adversaries, who may manipulate the public helper string $P_i$ to affect the reproduction of $R_i$. Any modification of ${P}_i$ by the adversary will be detected by the robustness of rrFE. We show how to construct an rrFE from a Symmetric Key Encapsulation Mechanism (SKEM), a Secure Sketch (SS), an Extractor (Ext), and a Lossy Algebraic Filter (LAF). We characterize the key-shift security notion of SKEM and the homomorphic properties of SS, Ext and LAF, which enable our construction of rrFE to achieve both reusability and robustness.We present an instantiation of SKEM from the DDH assumption. Combined with the LAF by Hofheinz (EuroCrypt 2013), homomorphic SS and Ext, we obtain the first rrFE based on standard assumptions.
2016
ASIACRYPT
2015
PKC
2015
PKC
2014
EUROCRYPT
2014
PKC
2013
PKC
2013
ASIACRYPT
2011
PKC
Eurocrypt 2023
Eurocrypt 2022
Asiacrypt 2022
PKC 2015 | 2023-03-24 07:00:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4803372621536255, "perplexity": 3519.4941211109526}, "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/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00372.warc.gz"} |
https://eprint.iacr.org/2019/1418 | ### Designated-ciphertext Searchable Encryption
Zi-Yuan Liu, Yi-Fan Tseng, Raylin Tso, and Masahiro Mambo
##### Abstract
Public-key encryption with keyword search (PEKS), proposed by Boneh \textit{et al.}, allows users to search encrypted keywords without losing data privacy. Although extensive studies have been conducted on this topic, only a few have focused on insider keyword guessing attacks (IKGA) that can reveal a user's sensitive information. In particular, after receiving a trapdoor used to search ciphertext from a user, a malicious insider (\textit{e.g}., a server) can randomly encrypt possible keywords using a user's public key, and then test whether the trapdoor corresponds to the selected keyword. This paper introduces a new concept called \textit{designated-ciphertext searchable encryption} (DCSE), which provides the same desired functionality as a PEKS scheme and prevents IKGA. Each trapdoor in DCSE is designated to a specific ciphertext, and thus malicious insiders cannot perform IKGA. We further propose a generic DCSE scheme that employs identity-based encryption and a key encapsulation mechanism. We provide formal proofs to demonstrate that the generic construction satisfies the security requirements. Moreover, we provide a lattice-based instantiation whose security is based on NTRU and ring-learning with errors assumptions; the proposed scheme is thus considered to be resistant to the quantum-computing attacks.
##### Metadata
Available format(s)
Category
Public-key cryptography
Publication info
Preprint. Minor revision.
Keywords
Quantum-resistantSearchable EncryptionInsider Keyword Guessing AttackDesignated-ciphertextLattices
Contact author(s)
zyliu @ cs nccu edu tw
History
2020-06-04: revised
2019-12-10: received
See all versions
Short URL
https://ia.cr/2019/1418
License
CC BY
BibTeX
@misc{cryptoeprint:2019/1418,
author = {Zi-Yuan Liu and Yi-Fan Tseng and Raylin Tso and Masahiro Mambo},
title = {Designated-ciphertext Searchable Encryption},
howpublished = {Cryptology ePrint Archive, Paper 2019/1418},
year = {2019},
note = {\url{https://eprint.iacr.org/2019/1418}},
url = {https://eprint.iacr.org/2019/1418}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content. | 2022-05-27 16:36:01 | {"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": 0.25497159361839294, "perplexity": 11812.35299388709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00311.warc.gz"} |
http://enhtech.com/error-bars/help-th1f-error-bars.php | Home > Error Bars > Th1f Error Bars
# Th1f Error Bars
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: histograms with one byte per channel. histogram plotted with the option CONTZ. | 2018-11-14 16:46:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3922765254974365, "perplexity": 14656.190715874342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742117.38/warc/CC-MAIN-20181114150002-20181114172002-00345.warc.gz"} |
https://math.meta.stackexchange.com/questions/8605/sacrificing-the-grammar-of-title-of-questions-for-clear-visibility-of-relevance/8606#8606 | # Sacrificing the grammar of title of questions for clear visibility of relevance in related links on the side
Having seen many titles similar to
• Help needed with P
• How to Solve P
• What is the best way of tackling P
• A differential/Integral/Sum/Algebra question
etc. ( where P is latexified math part )
The side links are either showing unhelpful text, ( who is gonna decide which one the "help with integral needed" links on the side that would relate closely to question at hand, or even what they relate to).
Is it ok to for sake of clarity and emphasis on the main part of the question to change the titles to
• P Solve
• P Sum
• P Evaluate
or some other short form that might not even include any text at all? ( again P is mathematical latex part).
It has been pointed out to me that not having any text in the title does not allow for right click in the new tab ( although shift click still works fine).
So should we hold on to proper grammar in titles or try to have minimal related information that would be of interest ignoring ( is it true, how to prove etc. ) and just have mathematical latex as title?
Thanks
• For future reference: Single line breaks in your input are ignored by the parser and replaced with run-in text, making your post hard to understand. You should try formatting your points a different way. I fixed it for you this time.
– user856
Feb 28 '13 at 8:35
Personally, I would prefer to instead try to guide OP to learning how to post proper and helpful titles him-or herself. If you are dealing with a transient user, one question matters little; it you are dealing with a future regular, one should learn to do this oneself. Instead of simply changing the title, I would thus rather add the suggestion of a new title as a comment.
As a second point, I find it preferable to let a user talk in his or her voice. If an OP is unable or unwilling to post clear, coherent, and well-titled posts, then this could - or maybe should - impact the willingness of other users to interact with such OP. So leaving a question in OP's voice allows other users to form their own opinions.
Edit (03/03): While only one example, this thread illustrates one reason why I am no fan of users submitting revision suggestions that address how OP writes. Such suggestions apparently are common in suggested edits, but for the above reasons, I habitually decline them from my end.
• I think the benefits of having well-titled posts far outweigh the disadvantage. Feb 28 '13 at 12:00
• @Gerry Myerson: I see your point, Gerry. Users who don't bother to submit proper questions are a pet peeve of mine I guess. Feb 28 '13 at 13:06
• @Gerry: the benefits of having a *well-written post far outweigh the disadvantage. I always feel like kicking the editor when clicking on a well-titled post gives me a question text that is either incoherent or in imperative mood. Feb 28 '13 at 22:50
• @Willie, sorry --- sometimes I have enough energy to edit an incoherent title, but not enough to edit the incoherent question to which it is attached. I fall back on the words of the Pirke Avot: "You are not obligated to complete the work, but neither are you free to desist from it entirely." Mar 1 '13 at 4:31
• @WillieWong The two-line snippets shown in the lists Questions and Unanswered usually suffice to detect such posts. Unfortunately, my newest toy ?tab=interesting does not have such snippets (nor do other home page sub-tabs).
– user53153
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https://www.zbmath.org/authors/?q=ai%3Agao.jiti | # zbMATH — the first resource for mathematics
## Gao, Jiti
Compute Distance To:
Author ID: gao.jiti Published as: Gao, J.; Gao, J. T.; Gao, Jiti External Links: ORCID
Documents Indexed: 108 Publications since 1988, including 3 Books
all top 5
#### Co-Authors
15 single-authored 12 Li, Degui 9 Tjøstheim, Dag B. 6 Anh, Vo V. 6 Dong, Chaohua 6 Liang, Hua 5 Chen, Jia 5 Peng, Bin 5 Phillips, Peter Charles Bonest 5 Wolff, Rodney Carl 4 King, Maxwell Leslie 4 Lu, Zudi 4 Yin, Jiying 3 Casas, Isabel 3 Chen, Song Xi 3 Pan, Guangming 3 Tong, Howell 2 Chen, Guijing 2 Cheng, Tingting 2 Gijbels, Irène 2 Gong, Xiaodong 2 Heyde, Christopher Charles 2 Hong, Shengyan 2 Hong, Shenyan 2 Kim, Namhyun 2 Lin, Zhengyan 2 Linton, Oliver Bruce 2 Pettitt, Anthony N. 2 Saart, Patrick W. 2 Tieng, Quang Minh 2 Wang, Qiying 2 Yang, Yanrong 2 Zhang, Xiaohui 2 Zhao, Lincheng 1 Allen, David E. 1 Anderson, Heather M. 1 Chen, Xiru 1 Feng, Guohua 1 Han, Xiao 1 Härdle, Wolfgang Karl 1 Hawthorne, Kim 1 Hong, Yongmiao 1 Inder, Brett A. 1 Kanaya, Shin 1 Kang, Yicheng 1 Kelly, James M. 1 Kong, Fanchao 1 Li, Tong 1 Ma, Shujie 1 McAleer, Michael 1 Qiu, Peihua 1 Ren, Zhao 1 Robinson, Peter Michael 1 Shi, Peide 1 Tang, Chengyong 1 Van Bellegem, Sébastien 1 Xia, Kai 1 Yan, Yayi 1 Yee, Thomas W. 1 Zhang, Bo 1 Zhu, Huanjun
all top 5
#### Serials
18 Journal of Econometrics 11 Econometric Theory 6 The Annals of Statistics 4 Statistics & Probability Letters 4 Communications in Statistics. Theory and Methods 4 Statistica Sinica 4 Australian & New Zealand Journal of Statistics 3 Acta Mathematica Sinica 2 Journal of the American Statistical Association 2 Journal of Time Series Analysis 2 Chinese Annals of Mathematics. Series A 2 Journal of Systems Science and Mathematical Sciences 2 Chinese Journal of Applied Probability and Statistics 2 Journal of China University of Science and Technology 2 Bernoulli 2 Journal of Nonparametric Statistics 2 The Econometrics Journal 2 The Annals of Applied Statistics 1 The Canadian Journal of Statistics 1 Scandinavian Journal of Statistics 1 Annals of the Institute of Statistical Mathematics 1 Journal of Applied Probability 1 Journal of Multivariate Analysis 1 Journal of Statistical Planning and Inference 1 Chinese Annals of Mathematics. Series B 1 Journal of Complexity 1 Econometric Reviews 1 Science in China. Series A 1 Systems Science and Mathematical Sciences 1 Economics Letters 1 Stochastic Processes and their Applications 1 Acta Mathematica Scientia 1 Test 1 Journal of the Royal Statistical Society. Series B. Statistical Methodology 1 Monographs on Statistics and Applied Probability 1 Statistics and Its Interface
all top 5
#### Fields
90 Statistics (62-XX) 15 Probability theory and stochastic processes (60-XX) 9 Numerical analysis (65-XX) 7 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 2 General and overarching topics; collections (00-XX) 1 History and biography (01-XX)
#### Citations contained in zbMATH Open
76 Publications have been cited 946 times in 632 Documents Cited by Year
Partially linear models. Zbl 0968.62006
Härdle, Wolfgang; Liang, Hua; Gao, Jiti
2000
Asymptotic theory for partly linear models. Zbl 0937.62592
Gao, Jiti
1995
Nonlinear times series: semiparametric and nonparametric methods. Zbl 1179.62118
Gao, Jiti
2007
Bandwidth selection in nonparametric kernel testing. Zbl 1286.62043
Gao, Jiti; Gijbels, Irène
2008
A test for model specification of diffusion processes. Zbl 1132.62063
Chen, Song Xi; Gao, Jiti; Tang, Cheng Yong
2008
Estimation in semiparametric spatial regression. Zbl 1113.62048
Gao, Jiti; Lu, Zudi; Tjøstheim, Dag
2006
Specification testing in nonlinear and nonstationary time series autoregression. Zbl 1191.62148
Gao, Jiti; King, Maxwell; Lu, Zudi; Tjøstheim, Dag
2009
Adaptive testing in continuous-time diffusion models. Zbl 1071.62068
Gao, Jiti; King, Maxwell
2004
The laws of the iterated logarithm of some estimates in partly linear models. Zbl 0837.62041
Gao, Jiti
1995
Nonparametric specification testing for nonlinear time series with nonstationarity. Zbl 1179.62055
Gao, Jiti; King, Maxwell; Lu, Zudi; Tjøstheim, Dag
2009
Semiparametric regression under long-range dependent errors. Zbl 1045.62514
Gao, J. T.; Anh, V. V.
1999
An adaptive empirical likelihood test for parametric time series regression models. Zbl 1418.62191
Chen, Song Xi; Gao, Jiti
2007
Parameter estimation of stochastic process with long-range dependence and intermittency. Zbl 0979.62071
Gao, Jiti; Anh, Vo; Heyde, Chris; Tieng, Quang
2001
Adaptive parametric test in a semiparametric regression model. Zbl 0912.62043
Gao, Jiti
1997
Semiparametric trending panel data models with cross-sectional dependence. Zbl 1443.62247
Chen, Jia; Gao, Jiti; Li, Degui
2012
Asymptotic normality of a class of estimators in partial linear models. Zbl 0805.62040
Gao, Jiti; Chen, Xiru; Zhao, Lincheng
1994
Econometric estimation in long-range dependent volatility models: theory and practice. Zbl 1429.62463
Casas, Isabel; Gao, Jiti
2008
Non-parametric time-varying coefficient panel data models with fixed effects. Zbl 1284.62222
Li, Degui; Chen, Jia; Gao, Jiti
2011
Semiparametric estimation and testing of the trend of temperature series. Zbl 1145.91381
Gao, Jiti; Hawthorne, Kim
2006
Statistical estimation of nonstationary Gaussian processes with long-range dependence and intermittency. Zbl 1059.60024
Gao, Jiti; Anh, Vo; Heyde, Chris
2002
Estimation for single-index and partially linear single-index integrated models. Zbl 1331.62190
Dong, Chaohua; Gao, Jiti; Tjøstheim, Dag
2016
Semiparametric estimation in triangular system equations with nonstationarity. Zbl 1284.62551
Gao, Jiti; Phillips, Peter C. B.
2013
Semiparametric single-index panel data models with cross-sectional dependence. Zbl 1337.62256
Dong, Chaohua; Gao, Jiti; Peng, Bin
2015
Local linear M-estimation in non-parametric spatial regression. Zbl 1223.62047
Lin, Zhengyan; Li, Degui; Gao, Jiti
2009
Semiparametric regression smoothing of non-linear time series. Zbl 0921.62109
Gao, Jiti
1998
Statistical inference in single-index and partially nonlinear models. Zbl 0935.62046
Gao, Jiti; Liang, Hua
1997
Model specification tests in nonparametric stochastic regression models. Zbl 1030.62032
Gao, Jiti; Tong, Howell; Wolff, Rodney
2002
Estimation in threshold autoregressive models with a stationary and a unit root regime. Zbl 1443.62256
Gao, Jiti; Tjøstheim, Dag; Yin, Jiying
2013
Modelling long-range-dependent Gaussian processes with application in continuous-time financial models. Zbl 1046.60038
Gao, Jiti
2004
Estimating smooth structural change in cointegration models. Zbl 1443.62284
Phillips, Peter C. B.; Li, Degui; Gao, Jiti
2017
A new diagnostic test for cross-section uncorrelatedness in nonparametric panel data models. Zbl 1369.62089
Chen, Jia; Gao, Jiti; Li, Degui
2012
Specification testing in discretized diffusion models: theory and practice. Zbl 1429.62468
Gao, Jiti; Casas, Isabel
2008
$$M$$-type smoothing splines in nonparametric and semiparametric regression models. Zbl 1045.62515
Gao, Jiti; Shi, Peide
1997
Uniform consistency for nonparametric estimators in null recurrent time series. Zbl 1441.62696
Gao, Jiti; Kanaya, Shin; Li, Degui; Tjøstheim, Dag
2015
Estimation in semi-parametric regression with non-stationary regressors. Zbl 1238.62044
Chen, Jia; Gao, Jiti; Li, Degui
2012
Moment inequalities for spatial processes. Zbl 1137.62061
Gao, Jiti; Lu, Zudi; Tjøstheim, Dag
2008
Central limit theorems for generalized $$U$$-statistics with applications in nonparametric specification. Zbl 1359.62100
Gao, Jiti; Hong, Yongmiao
2007
Modulus of convexity in Banach spaces. Zbl 1060.46012
Gao, J.
2003
Adaptive estimation in partially linear autoregressive models. Zbl 0961.62074
Gao, Jiti; Yee, Thomas
2000
Kinematic and static hypotheses for constitutive modelling of granulates considering particle rotation. Zbl 0942.74015
Chang, C. S.; Gao, J.
1996
Asymptotic normality of pseudo-LS estimator for partly linear autoregression models. Zbl 0818.62076
Gao, Jiti; Liang, Hua
1995
A misspecification test for multiplicative error models of non-negative time series processes. Zbl 1337.62261
Gao, Jiti; Kim, Nam Hyun; Saart, Patrick W.
2015
Nonparametric simultaneous testing for structural breaks. Zbl 1418.62157
Gao, Jiti; Gijbels, Irène; Van Bellegem, Sébastien
2008
High dimensional correlation matrices: the central limit theorem and its applications. Zbl 1411.60038
Gao, Jiti; Han, Xiao; Pan, Guangming; Yang, Yanrong
2017
Convergence rates of a class of estimators in partly linear models. Zbl 0837.62040
Gao, Jiti; Hong, Shengyan; Liang, Hua
1995
Adaptive estimation in partly linear regression models. Zbl 0771.62028
Gao, Jiti; Zhao, Lincheng
1993
Testing independence among a large number of high-dimensional random vectors. Zbl 1367.62261
Pan, Guangming; Gao, Jiti; Yang, Yanrong
2014
Conservative bounds on Rayleigh-Bénard convection with mixed thermal boundary conditions. Zbl 1202.76057
Wittenberg, R. W.; Gao, J.
2010
Gao, Jiti; Tong, Howell; Wolff, Rodney
2002
Bias correction for censored data with exponential lifetimes. Zbl 0905.62017
Pettitt, A. N.; Kelly, J. M.; Gao, J. T.
1998
Berry-Esseen bounds of error variance estimation in partly linear models. Zbl 0864.62018
Gao, Jiti; Hong, Shengyan; Liang, Hua
1996
Specification testing driven by orthogonal series for nonlinear cointegration with endogeneity. Zbl 1393.62037
Dong, Chaohua; Gao, Jiti
2018
Estimation in nonlinear regression with Harris recurrent Markov chains. Zbl 1349.62380
Li, Degui; Tjøstheim, Dag; Gao, Jiti
2016
Uniform consistency of nonstationary kernel-weighted sample covariances for nonparametric regression. Zbl 1441.62794
Li, Degui; Phillips, Peter C. B.; Gao, Jiti
2016
Long-range dependent time series specification. Zbl 1280.62105
Gao, Jiti; Wang, Qiying; Yin, Jiying
2013
Estimation in semiparametric time series regression. Zbl 05983896
Chen, Jia; Gao, Jiti; Li, Degui
2011
Specification testing in nonlinear time series with long-range dependence. Zbl 1210.62118
Gao, Jiti; Wang, Qiying; Yin, Jiying
2011
Normal structure, fixed points and related parameters in Banach spaces. Zbl 1029.46010
Gao, J.
2002
A central limit theorem for a random quadratic form of strictly stationary processes. Zbl 0969.60037
Gao, Jiti; Anh, Vo
2000
A $$C^ 2$$ finite element and interpolation. Zbl 0768.41009
Gao, J.
1993
Multiple attribute decision making with triangular intuitionistic fuzzy numbers based on zero-sum game approach. Zbl 1429.91135
Xu, J.; Dong, J. Y.; Wan, S. P.; Gao, J.
2019
Variable selection for a categorical varying-coefficient model with identifications for determinants of body mass index. Zbl 1391.62214
Gao, Jiti; Peng, Bin; Ren, Zhao; Zhang, Xiaohui
2017
A varying-coefficient panel data model with fixed effects: theory and an application to US commercial banks. Zbl 1443.62445
Feng, Guohua; Gao, Jiti; Peng, Bin; Zhang, Xiaohui
2017
Independent component analysis for multiple-input multiple-output wireless communication systems. Zbl 1217.94046
Gao, J.; Zhu, X.; Nandi, A. K.
2011
Semiparametric approximation methods in multivariate model selection. Zbl 1004.62035
Gao, Jiti; Wolff, Rodney; Anh, Vo
2001
Local linear kernel regression with long-range dependent errors. Zbl 0969.62027
Anh, Vo; Wolff, Rodney; Gao, Jiti; Tieng, Quang
1999
A frequentist approach to Bayesian asymptotics. Zbl 1452.62217
Cheng, Tingting; Gao, Jiti; Phillips, Peter C. B.
2018
Jump detection in generalized error-in-variables regression with an application to Australian health tax policies. Zbl 1397.62150
Kang, Yicheng; Gong, Xiaodong; Gao, Jiti; Qiu, Peihua
2015
Calculation of a parallel-plate waveguide with a chiral insert by the mixed finite element method. Zbl 1356.78131
Bogolyubov, A. N.; Mukhartova, Yu. V.; Gao, J.
2013
Simultaneous specification testing of mean and variance structures in nonlinear time series regression. Zbl 1219.62076
Chen, Song Xi; Gao, Jiti
2011
Robust estimation in parametric time series models under long- and short-range-dependent structures. Zbl 1337.62262
Gao, Jiti; Li, Degui; Lin, Zhengyan
2009
Subassembly identification based on grey clustering. Zbl 1160.90391
Gao, J.; Xiang, D.; Duan, G.
2008
Asymptotic properties of some estimators for partly linear stationary autoregressive models. Zbl 0937.62640
Gao, Jiti
1995
Empirical Bayes test for the uniform families $$U(\theta, c\theta)$$. Zbl 0788.62009
Gao, Jiti
1993
Consistency of estimation in a semiparametric regression model. I. Zbl 0765.62043
Gao, Jiti
1992
Empirical Bayes absolute error loss estimation for parameters of uniform distribution families $$U(\theta{},c\theta{})$$. Zbl 0739.62007
Gao, Jiti
1990
Multiple attribute decision making with triangular intuitionistic fuzzy numbers based on zero-sum game approach. Zbl 1429.91135
Xu, J.; Dong, J. Y.; Wan, S. P.; Gao, J.
2019
Specification testing driven by orthogonal series for nonlinear cointegration with endogeneity. Zbl 1393.62037
Dong, Chaohua; Gao, Jiti
2018
A frequentist approach to Bayesian asymptotics. Zbl 1452.62217
Cheng, Tingting; Gao, Jiti; Phillips, Peter C. B.
2018
Estimating smooth structural change in cointegration models. Zbl 1443.62284
Phillips, Peter C. B.; Li, Degui; Gao, Jiti
2017
High dimensional correlation matrices: the central limit theorem and its applications. Zbl 1411.60038
Gao, Jiti; Han, Xiao; Pan, Guangming; Yang, Yanrong
2017
Variable selection for a categorical varying-coefficient model with identifications for determinants of body mass index. Zbl 1391.62214
Gao, Jiti; Peng, Bin; Ren, Zhao; Zhang, Xiaohui
2017
A varying-coefficient panel data model with fixed effects: theory and an application to US commercial banks. Zbl 1443.62445
Feng, Guohua; Gao, Jiti; Peng, Bin; Zhang, Xiaohui
2017
Estimation for single-index and partially linear single-index integrated models. Zbl 1331.62190
Dong, Chaohua; Gao, Jiti; Tjøstheim, Dag
2016
Estimation in nonlinear regression with Harris recurrent Markov chains. Zbl 1349.62380
Li, Degui; Tjøstheim, Dag; Gao, Jiti
2016
Uniform consistency of nonstationary kernel-weighted sample covariances for nonparametric regression. Zbl 1441.62794
Li, Degui; Phillips, Peter C. B.; Gao, Jiti
2016
Semiparametric single-index panel data models with cross-sectional dependence. Zbl 1337.62256
Dong, Chaohua; Gao, Jiti; Peng, Bin
2015
Uniform consistency for nonparametric estimators in null recurrent time series. Zbl 1441.62696
Gao, Jiti; Kanaya, Shin; Li, Degui; Tjøstheim, Dag
2015
A misspecification test for multiplicative error models of non-negative time series processes. Zbl 1337.62261
Gao, Jiti; Kim, Nam Hyun; Saart, Patrick W.
2015
Jump detection in generalized error-in-variables regression with an application to Australian health tax policies. Zbl 1397.62150
Kang, Yicheng; Gong, Xiaodong; Gao, Jiti; Qiu, Peihua
2015
Testing independence among a large number of high-dimensional random vectors. Zbl 1367.62261
Pan, Guangming; Gao, Jiti; Yang, Yanrong
2014
Semiparametric estimation in triangular system equations with nonstationarity. Zbl 1284.62551
Gao, Jiti; Phillips, Peter C. B.
2013
Estimation in threshold autoregressive models with a stationary and a unit root regime. Zbl 1443.62256
Gao, Jiti; Tjøstheim, Dag; Yin, Jiying
2013
Long-range dependent time series specification. Zbl 1280.62105
Gao, Jiti; Wang, Qiying; Yin, Jiying
2013
Calculation of a parallel-plate waveguide with a chiral insert by the mixed finite element method. Zbl 1356.78131
Bogolyubov, A. N.; Mukhartova, Yu. V.; Gao, J.
2013
Semiparametric trending panel data models with cross-sectional dependence. Zbl 1443.62247
Chen, Jia; Gao, Jiti; Li, Degui
2012
A new diagnostic test for cross-section uncorrelatedness in nonparametric panel data models. Zbl 1369.62089
Chen, Jia; Gao, Jiti; Li, Degui
2012
Estimation in semi-parametric regression with non-stationary regressors. Zbl 1238.62044
Chen, Jia; Gao, Jiti; Li, Degui
2012
Non-parametric time-varying coefficient panel data models with fixed effects. Zbl 1284.62222
Li, Degui; Chen, Jia; Gao, Jiti
2011
Estimation in semiparametric time series regression. Zbl 05983896
Chen, Jia; Gao, Jiti; Li, Degui
2011
Specification testing in nonlinear time series with long-range dependence. Zbl 1210.62118
Gao, Jiti; Wang, Qiying; Yin, Jiying
2011
Independent component analysis for multiple-input multiple-output wireless communication systems. Zbl 1217.94046
Gao, J.; Zhu, X.; Nandi, A. K.
2011
Simultaneous specification testing of mean and variance structures in nonlinear time series regression. Zbl 1219.62076
Chen, Song Xi; Gao, Jiti
2011
Conservative bounds on Rayleigh-Bénard convection with mixed thermal boundary conditions. Zbl 1202.76057
Wittenberg, R. W.; Gao, J.
2010
Specification testing in nonlinear and nonstationary time series autoregression. Zbl 1191.62148
Gao, Jiti; King, Maxwell; Lu, Zudi; Tjøstheim, Dag
2009
Nonparametric specification testing for nonlinear time series with nonstationarity. Zbl 1179.62055
Gao, Jiti; King, Maxwell; Lu, Zudi; Tjøstheim, Dag
2009
Local linear M-estimation in non-parametric spatial regression. Zbl 1223.62047
Lin, Zhengyan; Li, Degui; Gao, Jiti
2009
Robust estimation in parametric time series models under long- and short-range-dependent structures. Zbl 1337.62262
Gao, Jiti; Li, Degui; Lin, Zhengyan
2009
Bandwidth selection in nonparametric kernel testing. Zbl 1286.62043
Gao, Jiti; Gijbels, Irène
2008
A test for model specification of diffusion processes. Zbl 1132.62063
Chen, Song Xi; Gao, Jiti; Tang, Cheng Yong
2008
Econometric estimation in long-range dependent volatility models: theory and practice. Zbl 1429.62463
Casas, Isabel; Gao, Jiti
2008
Specification testing in discretized diffusion models: theory and practice. Zbl 1429.62468
Gao, Jiti; Casas, Isabel
2008
Moment inequalities for spatial processes. Zbl 1137.62061
Gao, Jiti; Lu, Zudi; Tjøstheim, Dag
2008
Nonparametric simultaneous testing for structural breaks. Zbl 1418.62157
Gao, Jiti; Gijbels, Irène; Van Bellegem, Sébastien
2008
Subassembly identification based on grey clustering. Zbl 1160.90391
Gao, J.; Xiang, D.; Duan, G.
2008
Nonlinear times series: semiparametric and nonparametric methods. Zbl 1179.62118
Gao, Jiti
2007
An adaptive empirical likelihood test for parametric time series regression models. Zbl 1418.62191
Chen, Song Xi; Gao, Jiti
2007
Central limit theorems for generalized $$U$$-statistics with applications in nonparametric specification. Zbl 1359.62100
Gao, Jiti; Hong, Yongmiao
2007
Estimation in semiparametric spatial regression. Zbl 1113.62048
Gao, Jiti; Lu, Zudi; Tjøstheim, Dag
2006
Semiparametric estimation and testing of the trend of temperature series. Zbl 1145.91381
Gao, Jiti; Hawthorne, Kim
2006
Adaptive testing in continuous-time diffusion models. Zbl 1071.62068
Gao, Jiti; King, Maxwell
2004
Modelling long-range-dependent Gaussian processes with application in continuous-time financial models. Zbl 1046.60038
Gao, Jiti
2004
Modulus of convexity in Banach spaces. Zbl 1060.46012
Gao, J.
2003
Statistical estimation of nonstationary Gaussian processes with long-range dependence and intermittency. Zbl 1059.60024
Gao, Jiti; Anh, Vo; Heyde, Chris
2002
Model specification tests in nonparametric stochastic regression models. Zbl 1030.62032
Gao, Jiti; Tong, Howell; Wolff, Rodney
2002
Gao, Jiti; Tong, Howell; Wolff, Rodney
2002
Normal structure, fixed points and related parameters in Banach spaces. Zbl 1029.46010
Gao, J.
2002
Parameter estimation of stochastic process with long-range dependence and intermittency. Zbl 0979.62071
Gao, Jiti; Anh, Vo; Heyde, Chris; Tieng, Quang
2001
Semiparametric approximation methods in multivariate model selection. Zbl 1004.62035
Gao, Jiti; Wolff, Rodney; Anh, Vo
2001
Partially linear models. Zbl 0968.62006
Härdle, Wolfgang; Liang, Hua; Gao, Jiti
2000
Adaptive estimation in partially linear autoregressive models. Zbl 0961.62074
Gao, Jiti; Yee, Thomas
2000
A central limit theorem for a random quadratic form of strictly stationary processes. Zbl 0969.60037
Gao, Jiti; Anh, Vo
2000
Semiparametric regression under long-range dependent errors. Zbl 1045.62514
Gao, J. T.; Anh, V. V.
1999
Local linear kernel regression with long-range dependent errors. Zbl 0969.62027
Anh, Vo; Wolff, Rodney; Gao, Jiti; Tieng, Quang
1999
Semiparametric regression smoothing of non-linear time series. Zbl 0921.62109
Gao, Jiti
1998
Bias correction for censored data with exponential lifetimes. Zbl 0905.62017
Pettitt, A. N.; Kelly, J. M.; Gao, J. T.
1998
Adaptive parametric test in a semiparametric regression model. Zbl 0912.62043
Gao, Jiti
1997
Statistical inference in single-index and partially nonlinear models. Zbl 0935.62046
Gao, Jiti; Liang, Hua
1997
$$M$$-type smoothing splines in nonparametric and semiparametric regression models. Zbl 1045.62515
Gao, Jiti; Shi, Peide
1997
Kinematic and static hypotheses for constitutive modelling of granulates considering particle rotation. Zbl 0942.74015
Chang, C. S.; Gao, J.
1996
Berry-Esseen bounds of error variance estimation in partly linear models. Zbl 0864.62018
Gao, Jiti; Hong, Shengyan; Liang, Hua
1996
Asymptotic theory for partly linear models. Zbl 0937.62592
Gao, Jiti
1995
The laws of the iterated logarithm of some estimates in partly linear models. Zbl 0837.62041
Gao, Jiti
1995
Asymptotic normality of pseudo-LS estimator for partly linear autoregression models. Zbl 0818.62076
Gao, Jiti; Liang, Hua
1995
Convergence rates of a class of estimators in partly linear models. Zbl 0837.62040
Gao, Jiti; Hong, Shengyan; Liang, Hua
1995
Asymptotic properties of some estimators for partly linear stationary autoregressive models. Zbl 0937.62640
Gao, Jiti
1995
Asymptotic normality of a class of estimators in partial linear models. Zbl 0805.62040
Gao, Jiti; Chen, Xiru; Zhao, Lincheng
1994
Adaptive estimation in partly linear regression models. Zbl 0771.62028
Gao, Jiti; Zhao, Lincheng
1993
A $$C^ 2$$ finite element and interpolation. Zbl 0768.41009
Gao, J.
1993
Empirical Bayes test for the uniform families $$U(\theta, c\theta)$$. Zbl 0788.62009
Gao, Jiti
1993
Consistency of estimation in a semiparametric regression model. I. Zbl 0765.62043
Gao, Jiti
1992
Empirical Bayes absolute error loss estimation for parameters of uniform distribution families $$U(\theta{},c\theta{})$$. Zbl 0739.62007
Gao, Jiti
1990
all top 5
#### Cited by 836 Authors
50 Gao, Jiti 35 You, Jinhong 18 Liang, Hua 17 Chen, Gemai 17 Li, Degui 14 Liang, Hanying 14 Tjøstheim, Dag B. 11 Lu, Zudi 11 Wang, Qiying 11 Zhou, Xian 10 Aneiros-Pérez, Germán 10 Zhu, Lixing 9 Boente, Graciela 9 Phillips, Peter Charles Bonest 9 Su, Liangjun 8 Anh, Vo V. 8 Chen, Jia 8 Chen, Song Xi 8 Leonenko, Nikolai N. 7 Bianco, Ana M. 7 Fan, Guoliang 7 González-Manteiga, Wenceslao 7 Hu, Xuemei 7 Huang, Zhensheng 7 Li, Qi 7 Li, Runze 7 Sakhno, Lyudmyla Mykhaĭlivna 7 Wang, Dehui 6 Cui, Hengjian 6 Dabo-Niang, Sophie 6 Dong, Chaohua 6 Hong, Yongmiao 6 Hu, Shuhe 6 Lin, Jinguan 6 Lin, Zhengyan 6 Linton, Oliver Bruce 6 Liu, Feng 6 Liu, Xiaohui 6 Roozbeh, Mahdi 6 Zhao, Zhiwen 6 Zhou, Xingcai 6 Zhou, Yong 5 Ahmed, Syed Ejaz 5 Casas, Isabel 5 Chen, Min 5 Doksum, Kjell A. 5 Fan, Jianqing 5 Ginovyan, Mamikon S. 5 Lian, Heng 5 Park, Joon Y. 5 Peng, Bin 5 Robinson, Peter Michael 5 Sun, Yiguo 5 Vieu, Philippe 5 Wang, Qihua 5 Wang, Zhizhong 5 Zhao, Zhibiao 4 Cai, Zongwu 4 Fokianos, Konstantinos 4 Gao, Ji 4 Hu, Tao 4 Lin, Lu 4 Ma, Shujie 4 Peng, Liang 4 Qin, Yongsong 4 Sahakyan, Artur A. 4 Song, Weixing 4 Tong, Howell 4 Tu, Yundong 4 Van Keilegom, Ingrid 4 Wang, Hongxia 4 Wu, Wei Biao 4 Yao, Qiwei 4 Yin, Jiying 4 Zhang, Riquan 4 Zhu, Zhongyi 3 Chan, Nigel H. N. 3 Chen, Bin 3 Cheng, Guang 3 Chronopoulou, Alexandra 3 Diop, Aliou 3 Feng, Sanying 3 Feng, Zhenghui 3 Gijbels, Irène 3 Guo, Meihui 3 Hu, Hongchang 3 Jiang, Rong 3 Jing, Bingyi 3 King, Maxwell Leslie 3 Koul, Hira Lal 3 Lin, Yingqian 3 Ling, Shiqing 3 Peng, Cuixin 3 Peng, Heng 3 Pérez-González, Ana 3 Qian, Weimin 3 Qingguo, Tang 3 Rodríguez-Póo, Juan Manuel 3 Saejung, Satit 3 Song, Lixin ...and 736 more Authors
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#### Cited in 109 Serials
80 Journal of Econometrics 51 Journal of Multivariate Analysis 34 Econometric Theory 33 Journal of Statistical Planning and Inference 31 The Annals of Statistics 29 Journal of Nonparametric Statistics 28 Computational Statistics and Data Analysis 24 Statistics & Probability Letters 24 Communications in Statistics. Theory and Methods 18 Acta Mathematicae Applicatae Sinica. English Series 17 Bernoulli 11 Test 11 Statistical Papers 11 Journal of Systems Science and Complexity 10 Annals of the Institute of Statistical Mathematics 9 Metrika 9 Electronic Journal of Statistics 8 The Canadian Journal of Statistics 8 Computational Statistics 7 Journal of Time Series Analysis 7 Statistics 6 Economics Letters 6 Communications in Statistics. Simulation and Computation 6 Journal of the Korean Statistical Society 5 Journal of the American Statistical Association 5 Journal of Inequalities and Applications 5 Australian & New Zealand Journal of Statistics 5 Comptes Rendus. Mathématique. Académie des Sciences, Paris 4 Journal of Mathematical Analysis and Applications 4 Journal of Computational and Applied Mathematics 4 Chinese Annals of Mathematics. Series B 4 Journal of Statistical Computation and Simulation 4 Science China. Mathematics 3 Journal of Fluid Mechanics 3 Scandinavian Journal of Statistics 3 Journal of Applied Probability 3 Applied Mathematics. Series B (English Edition) 3 Abstract and Applied Analysis 3 Discrete Dynamics in Nature and Society 3 Journal of Applied Statistics 3 Acta Mathematica Sinica. English Series 3 Brazilian Journal of Probability and Statistics 3 AStA. Advances in Statistical Analysis 2 Lithuanian Mathematical Journal 2 Applied Mathematics and Computation 2 Information Sciences 2 Insurance Mathematics & Economics 2 Statistical Science 2 Mathematical and Computer Modelling 2 Journal of Contemporary Mathematical Analysis. Armenian Academy of Sciences 2 Stochastic Processes and their Applications 2 Mathematical Methods of Statistics 2 European Series in Applied and Industrial Mathematics (ESAIM): Probability and Statistics 2 The Econometrics Journal 2 Statistical Inference for Stochastic Processes 2 Applied Stochastic Models in Business and Industry 2 Quantitative Finance 2 Granular Matter 2 Iranian Journal of Fuzzy Systems 2 Journal of Statistical Theory and Practice 2 The Annals of Applied Statistics 1 Acta Mechanica 1 Biological Cybernetics 1 International Journal of Solids and Structures 1 Journal of Mathematical Physics 1 Journal of the Mechanics and Physics of Solids 1 Mathematical Biosciences 1 Ukrainian Mathematical Journal 1 Zhurnal Vychislitel’noĭ Matematiki i Matematicheskoĭ Fiziki 1 Automatica 1 Journal of Functional Analysis 1 Mathematics and Computers in Simulation 1 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 1 Circuits, Systems, and Signal Processing 1 Acta Applicandae Mathematicae 1 International Journal of Production Research 1 Sequential Analysis 1 Journal of Complexity 1 Econometric Reviews 1 Computational Mechanics 1 Applied Mathematics Letters 1 Science in China. Series A 1 Annals of Operations Research 1 The Annals of Applied Probability 1 Applications of Mathematics 1 Linear Algebra and its Applications 1 Theory of Probability and Mathematical Statistics 1 Applied and Computational Harmonic Analysis 1 Random Operators and Stochastic Equations 1 Journal of Mathematical Chemistry 1 Mathematical Problems in Engineering 1 Mathematical Inequalities & Applications 1 Wuhan University Journal of Natural Sciences (WUJNS) 1 Journal of the Royal Statistical Society. Series B. Statistical Methodology 1 Philosophical Transactions of the Royal Society of London. Series A. Mathematical, Physical and Engineering Sciences 1 Extremes 1 Methodology and Computing in Applied Probability 1 Analysis and Applications (Singapore) 1 Statistical Methods and Applications 1 Advances in Difference Equations ...and 9 more Serials
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#### Cited in 25 Fields
588 Statistics (62-XX) 93 Numerical analysis (65-XX) 82 Probability theory and stochastic processes (60-XX) 54 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 10 Functional analysis (46-XX) 7 Mechanics of deformable solids (74-XX) 5 Harmonic analysis on Euclidean spaces (42-XX) 5 Systems theory; control (93-XX) 4 Approximations and expansions (41-XX) 4 Operator theory (47-XX) 4 Fluid mechanics (76-XX) 4 Biology and other natural sciences (92-XX) 3 General and overarching topics; collections (00-XX) 3 Geophysics (86-XX) 3 Operations research, mathematical programming (90-XX) 2 Linear and multilinear algebra; matrix theory (15-XX) 2 Partial differential equations (35-XX) 2 Dynamical systems and ergodic theory (37-XX) 2 Classical thermodynamics, heat transfer (80-XX) 1 History and biography (01-XX) 1 General topology (54-XX) 1 Computer science (68-XX) 1 Optics, electromagnetic theory (78-XX) 1 Statistical mechanics, structure of matter (82-XX) 1 Information and communication theory, circuits (94-XX) | 2021-05-14 08:32:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5417454838752747, "perplexity": 13568.126053336333}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00534.warc.gz"} |
https://www.physicsforums.com/threads/correlation-coefficient-clarification.912387/ | # I Correlation Coefficient Clarification
1. Apr 23, 2017
### Tchao
Let's say that the correlation coefficient between X and Y were swapped so that the correlation coefficient between Y and X. If we were to compared the correlation coefficient between X and Y and Y and X. Based on my understanding of correlation coefficient, it doesn't matter if f X and Y were swapped. The correlation coefficient for both X and Y and Y and X would be the same.
2. Apr 23, 2017
### andrewkirk
That is correct. Correlation coefficient between the two is defined as
$$\frac{E[\ (X-E[X])(Y-E[Y])]\ }{\sqrt{E[(X-E(X))^2]E[(Y-E[Y])^2]}}$$
As you can see, this formula is symmetric between X and Y. | 2017-08-19 10:34:47 | {"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": 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": 0.5271540284156799, "perplexity": 211.88035519671755}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00394.warc.gz"} |
https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Flow_(mathematics).html | # Flow (mathematics)
For flows in graph theory, see flow network.
Flow in phase space specified by the differential equation of a pendulum. On the x axis, the pendulum position, and on the y one its speed.
In mathematics, a flow formalizes the idea of the motion of particles in a fluid. Flows are ubiquitous in science, including engineering and physics. The notion of flow is basic to the study of ordinary differential equations. Informally, a flow may be viewed as a continuous motion of points over time. More formally, a flow is a group action of the real numbers on a set.
The idea of a vector flow, that is, the flow determined by a vector field, occurs in the areas of differential topology, Riemannian geometry and Lie groups. Specific examples of vector flows include the geodesic flow, the Hamiltonian flow, the Ricci flow, the mean curvature flow, and the Anosov flow. Flows may also be defined for systems of random variables and stochastic processes, and occur in the study of ergodic dynamical systems. The most celebrated of these is perhaps the Bernoulli flow.
## Formal definition
A flow on a set X is a group action of the additive group of real numbers on X. More explicitly, a flow is a mapping
such that, for all xX and all real numbers s and t,
It is customary to write φt(x) instead of φ(x, t), so that the equations above can be expressed as φ0 = Id (identity function) and φsφt = φs+t (group law). Then, for all t , the mapping φt: X X is a bijection with inverse φ−t: X X. This follows from the above definition, and the real parameter t may be taken as a generalized functional power, as in function iteration.
Flows are usually required to be compatible with structures furnished on the set X. In particular, if X is equipped with a topology, then φ is usually required to be continuous. If X is equipped with a differentiable structure, then φ is usually required to be differentiable. In these cases the flow forms a one parameter subgroup of homeomorphisms and diffeomorphisms, respectively.
In certain situations one might also consider local flows, which are defined only in some subset
called the flow domain of φ. This is often the case with the flows of vector fields.
### Alternative notations
It is very common in many fields, including engineering, physics and the study of differential equations, to use a notation that makes the flow implicit. Thus, x(t) is written for φt(x0), and one might say that the "variable x depends on the time t and the initial condition x = x0". Examples are given below.
In the case of a flow of a vector field V on a smooth manifold X, the flow is often denoted in such a way that its generator is made explicit. For example,
## Orbits
Given x in X, the set φ(x,t): t ∈ ℝ is called the orbit of x under φ. Informally, it may be regarded as the trajectory of a particle that was initially positioned at x. If the flow is generated by a vector field, then its orbits are the images of its integral curves.
## Examples
### Autonomous systems of ordinary differential equations
Let F: RnRn be a (time-independent) vector field and x: RRn the solution of the initial value problem
Then φ(x0,t) = x(t) is the flow of the vector field F. It is a well-defined local flow provided that the vector field F: Rn Rn is Lipschitz-continuous. Then φ: Rn×R Rn is also Lipschitz-continuous wherever defined. In general it may be hard to show that the flow φ is globally defined, but one simple criterion is that the vector field F is compactly supported.
### Time-dependent ordinary differential equations
In the case of time-dependent vector fields F: Rn×RRn, one denotes φt,t0(x0) = x(t), where x: RRn is the solution of
Then φt,t0(x0,t,t0) is the time-dependent flow of F. It is not a "flow" by the definition above, but it can easily be seen as one by rearranging its arguments. Namely, the mapping
indeed satisfies the group law for the last variable:
One can see time-dependent flows of vector fields as special cases of time-independent ones by the following trick. Define
Then y(t) is the solution of the "time-independent" initial value problem
if and only if x(t) is the solution of the original time-dependent initial value problem. Furthermore, then the mapping φ is exactly the flow of the "time-independent" vector field G.
### Flows of vector fields on manifolds
The flows of time-independent and time-dependent vector fields are defined on smooth manifolds exactly as they are defined on the Euclidean space n and their local behavior is the same. However, the global topological structure of a smooth manifold is strongly manifest in what kind of global vector fields it can support, and flows of vector fields on smooth manifolds are indeed an important tool in differential topology. The bulk of studies in dynamical systems are conducted on smooth manifolds, which are thought of as "parameter spaces" in applications.
### Solutions of heat equation
Let Ω be a subdomain (bounded or not) of ℝn (with n an integer). Denote by Γ its boundary (assumed smooth). Consider the following Heat Equation on Ω × (0,T), for T > 0,
with the following initial boundary condition u(0) = u0 in Ω .
The equation u = 0 on Γ × (0,T) corresponds to the Homogeneous Dirichlet boundary condition. The mathematical setting for this problem can be the semigroup approach. To use this tool, we introduce the unbounded operator ΔD defined on by its domain
(see the classical Sobolev spaces with and
is the closure of the infinitely differentiable functions with compact support in Ω for the norm).
For any , we have
With this operator, the heat equation becomes and u(0) = u0. Thus, the flow corresponding to this equation is (see notations above)
where exp(D) is the (analytic) semigroup generated by ΔD.
### Solutions of wave equation
Again, let Ω be a subdomain (bounded or not) of ℝn (with n an integer). We denote by Γ its boundary (assumed smooth). Consider the following Wave Equation on (for T > 0),
with the following initial condition u(0) = u1,0 in and .
Using the same semigroup approach as in the case of the Heat Equation above. We write the wave equation as a first order in time partial differential equation by introducing the following unbounded operator,
with domain on (the operator is defined in the previous example).
We introduce the column vectors
(where and ) and
.
With these notions, the Wave Equation becomes and .
Thus, the flow corresponding to this equation is where is the (unitary) semigroup generated by .
### Bernoulli flow
Ergodic dynamical systems, that is, systems exhibiting randomness, exhibit flows as well. The most celebrated of these is perhaps the Bernoulli flow. The Ornstein isomorphism theorem states that, for any given entropy H, there exists a flow φ(x,t), called the Bernoulli flow, such that the flow at time t=1, i.e. φ(x,1), is a Bernoulli shift.
Furthermore, this flow is unique, up to a constant rescaling of time. That is, if ψ(x,t), is another flow with the same entropy, then ψ(x,t) = φ(x,t), for some constant c. The notion of uniqueness and isomorphism here is that of the isomorphism of dynamical systems. Many dynamical systems, including Sinai's billiards and Anosov flows are isomorphic to Bernoulli shifts. | 2022-01-21 11:15:30 | {"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": 0.939803957939148, "perplexity": 511.7768655693718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00071.warc.gz"} |
https://tug.org/pipermail/pdftex/2002-April/002529.html | # [pdftex] pdfTeX linking by symbolic name
CGvanderLaan at hetnet.nl cgl at hetnet.nl
Tue Apr 16 09:11:38 CEST 2002
I solicit for your comments on the wrap
around plain macro for linking by a symbolic name.
The purpose is to create words' in a line as buttons.
For my purpose at the moment it works, but maybe it
is too simple.
\def\bluepdflink#1#2{%#1 = to be linked item, the button'
%#2 = the link symbolic name
attr{/Border [0 0 0]} goto name{#2}
#1
\pdfdest name{#1} xyz }
Use
...<preceding text>
...<following text>
and on the destination page
...<preceding text>
\bluepdfdest{blu}
...<following text>
The next step will be to measure the first parameter
or simpler, let the user specify these by introducing
a third and fourth parameter
Maybe this should be done in a two-part macro context,
especially when the buttons are not just words.
I wrote a similar wrap around for the PS mark | 2021-05-14 09:53:23 | {"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": 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": 0.6752116680145264, "perplexity": 7812.6123894065795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00380.warc.gz"} |
https://android.stackexchange.com/questions/33264/samsung-galaxy-s3-not-connecting-via-usb?noredirect=1 | # Samsung Galaxy S3 not connecting via USB
Since about a week, my S3 doesn't connect anymore via usb to any computer. I don't think it's driver related as it was working before and I didn't change anything on the computers. It now runs Jelly Bean but it didn't happen right after the update (only after a week or so it stopped working).
The S3 doesn't even show the notification with the usb icon nor the possibility to switch between PTP & MTP...
• I tested with 3 different usb cables, on vista, win 7 and mac os x.
• Tried to activate / deactivate usb debugging, restart the phone and event done a hard reset, it's still not working.
• Phone is not rooted
Maybe it's hardware, but the phone didn't fell or anything like that...
Do you have an idea ?
• That seems like an issue for Samsung tech support, as I'm guessing it's hardware. You've already done what I think is sufficient testing to make a software cause unlikely. – Scott Severance Nov 13 '12 at 6:03
• I only have this problem on my Mac, it's fine with my other computer running Windows 7 (after installing some drivers - done automatically by Windows). Please update your question if it only applies to Mac – ericn Sep 11 '13 at 13:45
• No, it applies to Windows to (as said in my questions, first bullet point) – ygosteli Sep 18 '13 at 13:59
I just had the same problem tonight. All of a sudden my pc wouldn't recognize my S3 when I connected via the USB. I tried the USB debugging first and restarting...didn't work. I popped the battery out and restarted, and the USB icon showed up. Hope that helps.
Thanks for all the helpful troubleshooting ;)
Here's what worked for me. I don't know what caused it, but I suspected an Exchange server policy triggered it. I deleted those accounts just in case. Then I rebooted into recovery mode (volume up + home held down at boot). From the recovery menu, I wiped the cache partition. When it came back up, it saw the USB connection just fine again.
• Hello, Thank you for your solution. I've just tried now, unfortunately for me it didn't work. I think that the only way will be to send it to the samsung tech support... – ygosteli Dec 3 '12 at 14:37
I've had the same problem with my S3 ever since the update to JB 4.1. My PC would not recognize my s3 and the phone would not go into the usb debug mode only charging indicator when I plug the usb into the phone.
I've solved mine by erasing all cache for apps. Not sure which app it was but right after I cleared all my cache I plug it into the pc and it started working again. Kies works and usb debugging works.
I had this problem after updating as well. I can only presume it was from updating; I had no issues earlier.
My problem was the USB ports on my keyboard and the front panel of my PC would not recognize my S3. So, I plugged it into the back, and it reinstalled the drivers.
It works fine now; not sure how or why but who am I to argue.
You can transfer music to your device using Windows Media Player on Windows 7 and Windows Vista.
1. Connect your device to your PC using the supplied USB cable.
2. On your Galaxy S3, Go to Settings > Developer Options > USB debugging.
3. Open Windows Media Player and click the Sync tab on the right. Your device should be shown in the sync screen.
4. Drag the music tracks you want to sync to your device to the sync list on the right. When you have added all of the tracks, click the Start sync option at the top of the screen.
When the synchronization is complete, you can disconnect the cable from either your computer or your device.
Have you checked permissions on your machine? I have had problems accessing my device over usb because of permissions when I connected a device over usb to a linux box.
• Testing on 3 different platforms probably rules this out, but good idea else. – ce4 Nov 12 '12 at 18:58
• I'll check again, but as I already tested on two platforms... And the usb icon is not showing in the notification bar. – ygosteli Nov 12 '12 at 19:59
If you pull down the context menu from the top, and select the notification that it's plugged in with the USB device, you should be presented with different options for how you want the phone to treat your USB connection.
Select the Media device (MTP) option and you should now be able to browse the files on your phone just like you would any other usb data device.
• If you read the question it states "The S3 doesn't even show the notification with the usb icon nor the possibility to switch between PTP & MTP..." – Peanut Jan 5 '13 at 4:35
I had the same issue and removing my battery and power cycling the phone allowed the computer to recognize it again.
• As said in another answer : Unfortunately, in my case it didn't solved the issue. But thanks anyway – ygosteli Jan 19 '13 at 8:39
A few days after I got the Android 4.1.1 OTA update and rooted it my Galaxy S3 failed to recognize the USB connection.
Removing and then reinstalling the battery allowed my Galaxuy S3 to recognize the USB connection. The Android File Transfer still didn't work until I uninstalled Kies. Now it is working.
I appreciate the posts.
• Unfortunately, in my case it didn't solved the issue. Thanks for the answer – ygosteli Jan 19 '13 at 8:39
What seems to work for me is ticking the USB Debugging box in the Developer Options.
I've been trying all sorts of tricks (including some of the above, and more), but nothing helped.
This one does!
Hope this will solve your problem...
This won't help the OP, but may help others. I had the same problem, no help with battery removal/power cycling or developer mode USB debugging, etc. Finally realized I was using a third-party USB cable (freebie from costco). When I used the Samsung cable from the box, it started working again. Costco cable still doesn't work.
I had a cable which worked well (charge and data transfer) until the upgrade to Jelly Bean 4.1.2. Since the upgrade it won't work and I have to change the cable to labelled DATA TRANSFER ONLY, so I suggest trying a different cable.
• Aren't USB cables all the same? – wonea Feb 12 '13 at 17:35
• Not for devices, some will only charge your device and some will allow you to transfer files,... The best is to keep your original cable when you buy a smartphone. – ygosteli Feb 19 '13 at 15:31
I had the same problem with the phone not being recognised by PC. I tried EVERYTHING (uninstalling, USB cable, debugging, Kies etc, etc).
It finally worked after turning off phone and taking out battery! Couldn't believe it (but happy now).
I tried a different USB cable and it all came together and started working.
• Are you the OP using a different username? If not, I doubt this answers the question. – Stephen Schrauger Mar 4 '13 at 2:38
From within JB, unplug cable from handset and follow this:
• Go into Settings > About Phone
• Scroll down to the bottom
• Now tap on the Build number 7 times, to activate the Developers Options.
Now the Developer's options is activated (it is deliberately hidden from Settings), back out of that screen, Developers Options should be present, then tap that, look for Debugging, tap on USB Debugging to put a check-mark on that.
Then try plugging in the USB cable to handset... :) | 2019-10-16 12:15:59 | {"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": 0.2553729712963104, "perplexity": 2439.4519187776546}, "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/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00354.warc.gz"} |
https://fluka-forum.web.cern.ch/t/the-primary-particle-about-cs137-isotope-source/2425 | # The primary particle about Cs137 Isotope source
Dear FLUKA experts,
I try to define an isotope source(Cs137). As we know ,Cs137 decays and emitting 0.512MeV (94.7%) and 1.174MeV(5.3%) electrons into the excited and ground states of Ba136, and then Ba136m transitions back to 136Ba by emitting γ photons(0.661MeV).
So when I use the DETECT Card to get the spectrum, Is there an element of electrons β negative decay emission?
Dear Qinqing,
yes, the electrons from \beta^- decays are simulated, in addition to the photons.
Also, since FLUKA 4-2.0, the alpha particles are also generated and transported when alpha decaying isotope is used.
Cheers,
David | 2022-07-05 00:12:11 | {"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": 0.8396832942962646, "perplexity": 8250.243787713955}, "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-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00589.warc.gz"} |
https://physics.stackexchange.com/questions/74167/question-about-the-bernouilli-equation/74402 | # Question about the Bernouilli equation
There are some things I encountered, studying the Bernouilly equation, that I don't understand. I was studying in the following book: http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf. At page 72-73 they derive the Bernouilli equation for the first time, from energy considerations. They state that it can be applied if the flow is steady, incompressible, inviscid, when there is no change in internal energy and no heat transfer is done (p.72 at the bottom). I understand this derivation, my problems arise when they derive the equation again at page 110-111, this time from the Navier-Stokes equation.
I don't quite understand the derivation yet, but I am more confused about the outcome. It seems that the Bernouilli equation can now be applied under the four conditions they state (inviscid flow, steady flow, incompressible flow and the equation applies along a streamline) while it seems that there are no limitations to the internal energy of the flow and the heat transfer done. If I understand the last paragraph right, they state that the requirement that the flow is inviscid already comprises the claim that there is no change in internal energy ("the constant internal energy assumption and the inviscid flow assumption must be equivalent, as the other assumptions were the same", I don't really understand this reasoning, because the "inviscid flow assumption" was also already made in the previous derivation).
Furthermore, if I follow the reasoning (so if I assume that inviscid flow indeed implies that there is no change in internal energy or if I assume that the Bernouilli equation can also be applied, without the assumption of "no change in internal energy") then example 4 at page 74 seems strange to me. It shows that in this situation the Bernouilli equation can't be applied directly (because headloss has to be included in the equation). However I think that we can easily repeat the derivation, assuming that the flow is inviscid (and assuming that the other conditions are satisfied), but this example shows that the internal energy increases, so inviscid flow can't imply "no change in internal energy". And the example also shows that the Bernouilli equation can't be applied, because head loss has to be considered (so this example seems to contradict the result at page 110-111).
I hope someone can explain where my understanding is lacking, because this really confuses me about the conditions under which the Bernouilli equation can be applied.
It would also be helpful, if someone could explain the derivation at p.110: what I don't understand is the definition of "streamline coordinates". Do they just take a random point on the streamline, where they place a cartesian coordinate system?
• viscosity (basically internal fluid friction) results in the "dissipation" of kinetic energy in the form of heat into the fluid, resulting in a rise in the internal energy of the fluid. that defines the statement: the constant internal energy assumption and the inviscid flow assumption must be equivalent. – aditya kp Aug 17 '13 at 2:11
So the equation is $v^2/2+gy+P/\rho = const$, which has no change in internal energy (i.e. $U_1=U_2$).
OK was going to leave this for you to puzzle through, but just for completeness: because of the immediate area change of the pipe, with the constant steady pressure $P_1>0$, there is turbulent flow (high Reynolds number, depicted in the picture by the swirls), corresponding to viscosity and hence not inviscid. | 2019-10-23 18:24:56 | {"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": 0.8517378568649292, "perplexity": 163.28668984921958}, "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/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00543.warc.gz"} |
http://www.ck12.org/book/CK-12-Concept-Middle-School-Math---Grade-6/r4/section/1.1/ | Difficulty Level: At Grade Created by: CK-12
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Have you ever been to the zoo? Have you ever had to add whole numbers to solve a problem? Adding whole numbers is a skill that can help you to solve many real - world problems. Jonah is a student volunteer at the city zoo. He is working with the seals. Jonah loves his job, especially because he gets to help feed the seals who live at the zoo. There are 25 female and 18 male seals. In order to figure out how much to feed them, he will need to know the total number of seals. Use what you will learn in this Concept to help Jonah figure out the total number of seals.
### Guidance
Adding whole numbers is probably very familiar to you; you have been adding whole numbers almost as long as you have been in school. Here is a problem that will look familiar.
\begin{align*}4 + 5 = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
In this problem, we are adding four and five. We have four whole things plus five whole things and we get an answer of nine. The numbers that we are adding are called addends. The answer to an addition problem is the sum. This first problem was written horizontally or across.
In the past, you may have seen them written vertically or up and down. Now that you are in the sixth grade, you will need to write your problems vertically on your own.
How do we do this?
We can add whole numbers by writing them vertically according to place value. Do you remember place value? Place value is when you write each number according to the value that it has.
Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Ones
1 4 5 3 2 2 1
This number is 1,453,221. If we used words, we would say it is one million, four hundred and fifty-three thousand, two hundred and twenty-one.
What does this have to do with adding whole numbers?
Well, when you add whole numbers, it can be less confusing to write them vertically according to place value. Think about the example we had earlier.
\begin{align*}4+5=9\end{align*}
If we wrote that vertically, we would line up the numbers. They both belong in the ones column.
\begin{align*}& \quad 4\\ & \ \underline{+5}\\ & \quad 9\end{align*}
What happens when we have more digits?
\begin{align*}456 + 27 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
When you have more digits, you can write the problem vertically by lining up each digit according to place value.
\begin{align*}& \quad 456\\ & \ \underline{+ \ 27}\end{align*}
Now we can add the columns.
Now let's practice.
#### Example A
\begin{align*}3,456 + 87 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
Solution: 3,543
#### Example B
\begin{align*}56, 321 + 7, 600 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
Solution: 216,091
#### Example C
\begin{align*}203,890 + 12, 201 = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
Solution: 63,921
Now let's go back to Jonah and the seals. Jonah knows how many male seals and how many female seals are in the seal area at the zoo. He wants to figure out how many seals there are altogether. To accomplish this task, Jonah will simply need to add the two quantities together.
Here is what he knows:
25 females 18 males
Now we add those values together.
\begin{align*}25 + 18 = 43\end{align*}
There are 43 seals at the zoo.
### Vocabulary
Here are the vocabulary words found in this Concept.
Sum
Horizontally
across
Vertically
up and down
### Guided Practice
Now here is one for you to try on your own. Add the following pair of whole numbers. Then you can find the answer below.
\begin{align*}675 + 587 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
To solve this problem, we line up the columns vertically according to place value.
When you have more digits, you can write the problem vertically by lining up each digit according to place value.
\begin{align*}& \quad 675\\ & \ \underline{+ 587}\end{align*}
Now we can add the columns.
### Practice
Directions: Use what you have learned to solve each problem.
1. \begin{align*}56 + 123 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}341 + 12 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}673 + 127 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}549 + 27 =\underline{\;\;\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}87 + 95 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
6. \begin{align*}124 + 967 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
7. \begin{align*}1256 + 987 =\underline{\;\;\;\;\;\;\;\;\;}\end{align*}
8. \begin{align*}2345 + 1278 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
9. \begin{align*}3100 + 5472 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
10. \begin{align*}3027 + 5471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
11. \begin{align*}13027 + 7471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
12. \begin{align*}23147 + 5001 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
13. \begin{align*}23128 + 7771 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
14. \begin{align*}43237 + 5071 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
15. \begin{align*}22027 + 6001 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
16. \begin{align*}45627 + 2471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
17. \begin{align*}83027 + 51471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
18. \begin{align*}94127 + 5471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
19. \begin{align*}83777 + 3321 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
20. \begin{align*}95527 + 12471 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Horizontally Horizontally means written across in rows.
Sum The sum is the result after two or more amounts have been added together.
Vertically Vertically means written up and down in columns.
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Subjects: | 2017-04-29 13:36:39 | {"extraction_info": {"found_math": true, "script_math_tex": 31, "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": 0.5728461146354675, "perplexity": 2473.338529237567}, "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-2017-17/segments/1492917123491.79/warc/CC-MAIN-20170423031203-00279-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binom_test.html | # scipy.stats.binom_test¶
scipy.stats.binom_test(x, n=None, p=0.5, alternative='two-sided')[source]
Perform a test that the probability of success is p.
This is an exact, two-sided test of the null hypothesis that the probability of success in a Bernoulli experiment is p.
Parameters
xinteger or array_like
the number of successes, or if x has length 2, it is the number of successes and the number of failures.
ninteger
the number of trials. This is ignored if x gives both the number of successes and failures
pfloat, optional
The hypothesized probability of success. 0 <= p <= 1. The default value is p = 0.5
alternative{‘two-sided’, ‘greater’, ‘less’}, optional
Indicates the alternative hypothesis. The default value is ‘two-sided’.
Returns
p-valuefloat
The p-value of the hypothesis test
References
1
https://en.wikipedia.org/wiki/Binomial_test
Examples
>>> from scipy import stats
A car manufacturer claims that no more than 10% of their cars are unsafe. 15 cars are inspected for safety, 3 were found to be unsafe. Test the manufacturer’s claim:
>>> stats.binom_test(3, n=15, p=0.1, alternative='greater')
0.18406106910639114
The null hypothesis cannot be rejected at the 5% level of significance because the returned p-value is greater than the critical value of 5%.
#### Previous topic
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scipy.stats.fligner | 2019-05-22 06:49:38 | {"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": 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": 0.5223969221115112, "perplexity": 1167.6608553723468}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00192.warc.gz"} |
http://openstudy.com/updates/4d9b02d00b248b0bb640e9e2 | ## anonymous 5 years ago Let x(t) and y(t) be two orthogonal signals so that integral x(t)y(t) by dt with (-inf <t<inf) . Their respective energy values are Ex and Ey. Obtain the energy of the signal x(t)+y(t) and show that it is identical to the energy of the signal x(t)-y(t). Classify if the resulting signal is a power or energy signal.
1. anonymous
$\int\limits\limits_{-\infty}^{\infty}$
2. anonymous
$E_x=\int\limits_{-\infty}^{-\infty}x^2(t)dt\quad,\quad E_y=\int\limits_{-\infty}^{-\infty}y^2(t)dt$ $E_{x\pm y}=\int\limits_{-\infty}^{-\infty}(x(t)\pm y(t))^2dt=$ $=\int\limits_{-\infty}^{-\infty}x^2(t)dt+\int\limits_{-\infty}^{-\infty}y^2(t)dt\pm 2\underbrace{\int\limits_{-\infty}^{-\infty}x(t)y(t)dt}_{=0}=$ $=E_x+E_y$
3. anonymous
Thanks pal | 2016-10-27 11:39:34 | {"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": 0.9084039926528931, "perplexity": 2744.2502319950786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721268.95/warc/CC-MAIN-20161020183841-00368-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/213406/does-x-yz-mean-x-yz-or-x-yz/213416 | # Does $x/yz$ mean $x/(yz)$ or $(x/y)z$?
When people write $x/yz$, do they usually mean $x/(yz)$ or $(x/y)z$?
For example, from Wikipedia
If $p\geq 1/2$ , then $$\Pr\left[ X>mp+x \right] \leq \exp(-x^2/2mp(1-p)) .$$
Thanks!
-
It's ambiguous. In this case I would guess that it means the former, but you yourself should not use this notation. – Qiaochu Yuan Oct 14 '12 at 3:30
@Tim: Use $()$, like $\exp(-x^2/(2mp(1-p)))$ or $\exp(-(x^2/2m)p(1-p))$ (or even $\exp(-(x^2/2)mp(1-p))$, whichever is meant. If the $p(1-p)$ is supposed to be in the numerator you could also write $\exp(-x^2p(1-p)/(2m))$ and decrease confusion. In this case I'm not sure which it is; you'll have to do the calculations carefully and re-derive the proof, and check. (And then fix the Wikipedia article.) – ShreevatsaR Oct 14 '12 at 3:52
I've almost never seen $x/yz$ meaning $(x/y)z$. If you are thinking about $(x/y)z$, why would you write the ambiguous $x/yz$ instead of $xz/y$ ? – mercio Oct 14 '12 at 5:39
This seems relevant. – user02138 Oct 14 '12 at 18:37
@user02138: Nice indeed. – Tim Oct 14 '12 at 18:50
Because multiplication is expressed by the omission of an operator, simply by juxtaposing two factors, it should be treated as having a higher precedence than an operator whose presence is visible as an explicit glyph.
That is to say $abc/xyz$, should mean $(a\times b\times c)/(x\times y\times z)$. Basically we would like $abc/xyz$ to just be an in-line shorthand for ${abc}\over{xyz}$ with minimal fuss.
The interpretation $((((ab)c)/x)/y)/z)$ means we cannot do this. Expressing ${abc}\over{xyz}$ with a slash becomes cumbersome.
In computer programming languages, it is common for division and multiplication to have the same precedence and left-to-right associativity. But the situation there is different because multiplication is an explicit symbol like *. For instance in C a * b / c * d simply means (((a * b) / c) * d).
But if multiplication is notated by mere juxtaposition, a strong case can be made that it this juxtaposition should also denote "tighter binding" (higher precedence than any binary operator).
Mathematics notation is richly visual and two dimensional. Juxtaposition relationships matter. For instance the way an exponent binds with its base.
-
+1 Operator priority depends on what is written, not on what it means, so it is entirely natural that juxtaposition may have a different priority than a multiplication symbol. – Marc van Leeuwen Oct 14 '12 at 12:51
Spacing to indicate grouping is important: $x/y \,\, z$ is almost surely $(x/y)z$, and $x \, / \, yz$ is almost surely meant as $x/(yz)$. But $x/yz$ is lacking the spacing that would signify intent. – Hurkyl Oct 14 '12 at 16:23
^ definitely! But notice how / already comes with some space around it, when you have a decent font and decent typesetting: $abc/xyz$. – Kaz Oct 14 '12 at 17:07
A few weeks ago I did a search for this construction. I found three things:
1. It was much less frequently used than I expected; I think this is because mathematicians are concerned that it might be unclear.
2. The few examples I did find invariably meant $x/(yz)$, not $(x/y)z$.
3. Complicated examples are extremely uncommon.
Here are some examples I found:
• In the proof of theorem 4.2 at the bottom of the page there is $u_n/n\alpha_n$. (Alberto Torchinsky, Real-Variable Methods in Harmonic Analysis)
• Page 3 of Walter Rudin's Real and Complex Analysis uses $y/2\pi$, and the bottom of page 99 has $\delta = \eta/2k$. However, on page 88 of the same book, Rudin writes $1/(2\pi)$, even though $1/2\pi$ always means $1/(2\pi)$, never $(1/2)\pi$.
• Page 12 of George Simmons' Differential Equations says that the negative reciprocal of $rd\theta/dr$ is $-dr/rd\theta$. Page 45 mentions the use of $1/xy$ as an integrating factor. In a rare complicated example, he writes $(2n)!/(n!)^2 2^n$ (meaning $(2n)!\over(n!)^2 2^n$) on page 222.
• I found no examples at all where $x/yz$ was used to mean $(x/y)z$.
However, Wikipedia articles are sometimes written by blockheads. From looking at related versions of the same formula, for example here (page 4) or here (theorem 1.5) I guess that that was the case here, and that the $p(1-p)$, if it belongs there at all, should have been in the numerator. I think you should be wary of taking the Wikipedia claims at face value, and instead refer to a source that is written by someone with a reputation.
You should also be wary of taking my claims at face value, since I know absolutely nothing about Chernoff bounds.
-
Thanks! I really appreciate that you have found so many examples! About the Chernoff bound mentioned in my question, I wonder how you know from the two pdf files that $p(1-p)$ should have been in the numerator instead of denominator? – Tim Oct 14 '12 at 3:59
I don't know. I was guessing. I wasn't even able to convince myself that the $p(1-p)$ should be there at all. – MJD Oct 14 '12 at 4:01
+1 for "Wikipedia articles are sometimes written by blockheads." – Joel Reyes Noche Oct 14 '12 at 6:55
Well done the research. I'd like to add (with zero research) that if $1/2\pi$ is rare, $1/2/\pi$ is just not done, even though it is completely unambiguous (as nobody claims that '/' should associate to the right, as far as I know) and shorter than $1/(2\pi)$. – Marc van Leeuwen Oct 14 '12 at 12:33
Here is more data I found in Concrete Mathematics. P.64 ex 2.23 "Evaluate $\sum_{k=1}^n(2k+1)/(k(k+1))$ in two ways a. Replace $1/k(k+1)$ by..." ex 2.24 "What is $\sum_{0\leq k<n}H_k/(k+1)(k+2)$?". Note they are not systematic, but always mean $x/(yz)$ even when omitting parentheses around the denominator. Also p.504 in answer 4.40: $n!/p^{\lfloor n/p\rfloor}\lfloor n/p\rfloor!$ which was what I ran into before I started looking more systematically. Clearly GKP, who generally care a lot about notation, don't care parenthesising denominators. – Marc van Leeuwen Oct 22 '12 at 8:48
The American Physical Society gives multiplication a higher order than division in the slash notation:
When slashing fractions, respect the following conventions. In mathematical formulas this is the accepted order of operations:
1. raising to a power,
2. multiplication,
3. division,
(Styling applied) (Source: Page 6 of Section IV of Physical Review Style and Notation Guide, retrieved 2012.10.14)
This means that (for them) $$x/yz = x/(yz)$$ However, keep in mind that this may represent only the view of one group of scientists, I don't know of any established standard.
-
By the usual order of operations, it means $(x/y)z$. But people do often mean $x/(yz)$ instead, so you have to work out the intention from the context. :(
I would be very interested to see any example from a reputable source that uses $x/yz$ to mean $(x/y)z$. – MJD Oct 14 '12 at 3:43
The "usual order of operations" (UOOO) applies to arithmetic like $42\div 7\times 3$. Algebraic notation defies UOOO. I would tend to agree that $x\div y\times z$ means $(x/y)z$, yet disagree that it has anything to do with $x/yz$. :) – Kaz Oct 14 '12 at 6:53
How the calculator got that is that there is no operator between the $2$ and the $\pi$. I naively just hit the 2 button followed by the pi button, which causes $\pi$ to be interpreted as a correction which replaces the operand $2$. The result $0.636...$ is then just $2/\pi$. – Kaz Oct 15 '12 at 23:58 | 2014-03-10 02:29:44 | {"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": 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": 0.9220297336578369, "perplexity": 609.7790479630422}, "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-2014-10/segments/1394010554119/warc/CC-MAIN-20140305090914-00063-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://tjyj.stats.gov.cn/CN/10.19343/j.cnki.11-1302/c.2019.09.009 | • •
混频时间序列的潜在因子分析及其应用
• 出版日期:2019-09-25 发布日期:2019-09-25
Latent Factor Analysis for Time Series with Mixed Frequencies and Its Application
Qin Lei et al.
• Online:2019-09-25 Published:2019-09-25
Abstract: Time series generated by macroeconomics are usually assumed to be controlled by a few latent factors. The joint effects of factors lead to the co-movement of series. Factors are important in the analysis and forecasting of time series. However, empirical macroeconomic studies always contain time series with mixed frequencies, which make factor analysis impossible to implement. To this end, this paper proposes two factor analysis methods for time series with mixed frequencies, namely MIDAS-LF and EM-LF. The former benefits from the interpolation of low-frequency sequences by the multivariate MIDAS model, while the latter uses the EM algorithm for iterative solution. The simulation data analysis shows that, compared with methods existing in the literature, MIDAS-LF is better for the analysis of time series with mixed frequencies. The calculation procedure of MIDAS-LF is simple and retains most of the information in original data, which can better estimate the factors and loading matrix, leading to low fitting error and prediction error. The actual data analysis of the macro-economy also confirms the feasibility and correctness of the proposed methods. | 2022-07-03 11:38:56 | {"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": 0.5730947256088257, "perplexity": 1036.8809995425565}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104240553.67/warc/CC-MAIN-20220703104037-20220703134037-00544.warc.gz"} |
https://vincenttam.gitlab.io/post/2018-12-19-scss-to-sass-converter/ | # SCSS to SASS Converter
### Background
While porting Minimal Mistakes and Beautiful Hugo's Staticman support to Introduction, I searched for “SCSS to SASS” on DuckDuckGo. The search engine returned results on “CSS to SASS/SCSS” or vice versa. The first theme has Staticman code in SCSS. That would fit into the third theme‘s directory structure, which puts SASS files under assets/sass. However, SASS doesn't have curly braces {}. I feared that after hours of tedious manual replacement, the code would fail to run. As a consequence, I conducted such Internet search.
### Solution
Scared by the error message in the recent failed GitLab CI build of a clone of Beautiful Jekyll, I don't want to install Jekyll SASS Converter as a Ruby gem.
I clicked on the link to an online CSS to SASS converter. I guessed there's an analogue for SCSS to SCSS, so I replaced css2sass with scss2sass in the previous link, and this worked!
I've finally obtained a desired converted SASS file. However, I don't know why this file only works locally. | 2020-01-27 12:11:03 | {"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": 0.2888106107711792, "perplexity": 8334.876320565118}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00048.warc.gz"} |
http://arxitics.com/articles/2104.03294 | arXiv Analytics
arXiv:2104.03294 [hep-ph]AbstractReferencesReviewsResources
The new $(g-2)_μ$ result and the $μν$SSM
Published 2021-04-07Version 1
The $\mu\nu$SSM is a highly predictive alternative model to the MSSM. In particular, the electroweak sector of the model can explain the longstanding discrepancy between the experimental result for the anomalous magnetic moment of the muon, $(g-2)_\mu$, and its Standard Model prediction, while being in agreement with all other theoretical and experimental constraints. The recently published MUON G-2 result is within $0.8\,\sigma$ in agreement with the older BNL result on $(g-2)_\mu$. The combined result was announced as $a_\mu^{\rm exp} = (11659206.1 \pm 4.1) \times 10^{-10}$, yielding a new deviation from the Standard Model prediction of $\Delta a_\mu = (25.1 \pm 5.9) \times 10^{-10}$, corresponding to $4.2\,\sigma$. Using this improved bound we update the analysis in the $\mu\nu$SSM as presented in Ref. [1] and set new limits on the allowed parameters space of the electroweak sector of the model. We conclude that significant regions of the model can explain {the new} $(g-2)_\mu$ data.
Comments: 30 pages, 9 figures, 1 table. Follow-up of arXiv:1912.04163
Categories: hep-ph, hep-ex
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https://www.physicsforums.com/threads/metric-outside-a-weakly-gravitating-body-mtw-ex-19-1.905874/ | # Metric outside a weakly gravitating body (MTW Ex 19.1)
1. Feb 28, 2017
### TerryW
1. The problem statement, all variables and given/known data
This is Exercise 19.1 in MTW - See attachment
2. Relevant equations
See attachment
3. The attempt at a solution
I've worked through 19.3a, 19.3b and 19.3c ( see post by zn5252 back in March 2013 replied to by PeterDonis) and proved them for my my own satisfaction and I've also worked through 19.7a, 19.7b and 19.7c. My problems begin when I try to find h00 and hom.
I have arrived at a solution for h00 but it:
a) only works if T'00 >> T'ii. (I think this is true in geometricised units) and
b) I haven't made any use of 19.7a and 19.7b and I don't see how they would be useful because the expansion of 19.2 for h00 will only include terms with $\bar{T}^{00} (= \frac 1 2 (T'^{00} +T'^{ll})$
The result of my labours for h0m is:
$h_{0m} =-\frac 1{2r}\frac{\partial}{\partial x_m} \int(T'_{00,0})r'^2d^3x' -\frac{\partial}{\partial x_m} \int(T'_{00}+T'_{ll})x^jx'^jd^3x' - \frac{x^m}{r}\frac 1{2r} \int(T'_{00,00})r'^2d^3x'$
$\qquad \quad +\big(\frac {x^k}{r^2} \big)_{,m}\int(T'^{0j}{ }_{,0})x'^{j}x'^kd^3x'$
At this point, I can't see any way of using 19.7c to simplify my equation so I am stuck.
Any help anyone?
TerryW
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2. Mar 1, 2017
### TSny
Hi, Terry.
When working out $h_{00}$ you can use identity 19.7a to write $\int T'_{kk} d^3x'$ in a different way. Thus, let $j = k$ in 19.7a and sum over $k$. Use this to show that $\int T'_{kk} d^3x' = \frac{1}{2} \frac{\partial ^2}{\partial t^2} \int T'^{00}r'^2 d^3x'$. The reason for doing this is that you can relate this expression to the first term in the gauge function $\xi_0$ near the top of page 449. This is how the gauge transformation can cancel out unwanted terms in $h_{00}$.
Similarly, 19.7b can be used to re-express $\int T'_{kk} x^{j \,'}d^3x'$ in a way that will be related to the second term of $\xi_0$.
3. Mar 3, 2017
### TerryW
Hi TSny
I've tried this but without success:
$$T^{kk} = ½(T^{00}x^k x^k)_{,00} + 2(T^{lk}x^k)_{,l}- ½(T^{lm}x^k x^k)_{,lm}$$
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{lk}δ^k_l - ½(T^{lm}{ }_{,l}x^k x^k + 2T^{lm}x^kδ^k_l )_{,m}$$
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,l}x^k x^k + 2T^{km}x^k)_{,m}$$
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,lm}x^k x^k + 2T^{lm}{ }_{,l}x^kδ^k_m + 2T^{km}{ }{,m}x^k+2T^{km}δ^k_m)$$
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,lm}x^k x^k + 2T^{lk}{ }_{,l}x^k + 2T^{km}{ }_{,m}x^k+2T^{kk})$$
But $T^{lk}{ }_{,l}x^k = T^{kl}{ }_{,l}x^k = T^{km}{ }_{,m}x^k$, then
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) - ½(T^{lm}{ }_{,lm}x^k x^k) + T^{kk}$$
From $T^{αβ}{ }_{,β} = 0$ we get $T^{lm}{ }_{,lm} = - T^{0m}{ }_{,0m}$
$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + ½(T^{0m}{ }_{,0m}x^k x^k) + T^{kk}$$
Then $T^{00}{ }_{,00} + T^{0m}{ }_{,0m} = T^{0α}{ }_{,0α} = T^{0α}{ }_{α0} = 0$
Leaving $T^{kk} = T^{kk}$
This is my original proof of the identity 19.7a (with a similar process for proving 19.b). At this stage I can't see a way of manipulating 19.7a to allow me to write $\int T'_{kk} d^3x'$ in another way.
What am I not seeing?
TerryW
4. Mar 3, 2017
### TSny
OK. Now consider $\int T'^{kk } d^3x'$. Use the relation above and note that the second and third terms on the right are divergences. So, when integrating you can use the divergence theorem.
5. Mar 5, 2017
### TerryW
Thanks for that, I have now sorted h00. $\quad$ h0m seems to have a large number of bits and pieces to work my way through to get to the required answer. I will crack on and hope I don't have to bother you again.
Regards
TerryW
6. Mar 5, 2017
### TSny
OK. It's no bother.
I got hung up for about 2 days on hom, mainly concerning the term $\frac{x^m}{r} \xi_0$ in the expression for $\xi_m$. It took me quite a while to see why that term must be included. I was overlooking the fact that the various $T \, '$ components are evaluated at the retarded time. So, you have to be careful when evaluating $\xi_{0,m}$ in the gauge transformation.
7. Mar 9, 2017
### TerryW
Hi TSny,
I've been batting this around for a few days now and have reached a bit of a road (mind) block. To try make a bit of progress, I'm going to set out my working in stages, starting with my thoughts on what h0m, ξ0 and ξm are:
$h_{0m} = 4\int(T'_{0m})\big(\frac {1}{r}+\frac{x^jx'^j}{r^3} \big)d^3x'$
(This is as suggested, the n = 0 term of (19.2) - I can't see how this can be simplified! I also observe that $4\int(T'_{0m})\big(\frac{x^jx'^j}{r^3} \big)d^3x'$ can provide the term in $h_{0m}$ that we are seeking!)
I've adopted a slightly different convention in labelling, using $x'^j$ instead of $x^{j'}$ which, to me, makes clearer which point the coordinate refers to.
$ξ_{0} =\frac 1{2r}\frac{\partial}{\partial t} \int(T'^{00})r'^2d^3x'$
$\quad+\frac{x^j}{r^3}\int(T'^{0k}x'^j x'^k - \frac{1}{2}T'^{0j}r'^2)d^3x'$
$\quad+\int(T'_{00}+T'_{ll})\frac{x^j x'^j}{r}d^3x'$
$\quad+\frac{1}{2}\int(T'_{00}+T'_{kk})_{,0}\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)d^3x'$
$ξ_{m} =4\int(T'_{0m})\big(\frac{x^jx'^j}{r} \big)\big(\frac{1}{r}\big)d^3x''$ (n = 1)
$\quad +4\frac{1}{2}\frac{\partial}{\partial t}\int(T'_{0m})\big(\frac{x^jx'^j}{r} \big)^2\big(\frac{1}{r}\big)d^3x'$
$\quad +\frac {x^m}{r}\frac 1{2r}\frac{\partial}{\partial t} \int(T'^{00})r'^2d^3x'$
$\quad +\frac {x^m}{r}\frac {x^j}{r^3}\frac{\partial}{\partial t} \int(T'^{0k}x'^j x'^k - \frac{1}{2}T'^{0j}r'^2)r'^2d^3x'$
$\quad +\frac {x^m}{r}\int(T'_{00}+T'_{ll})\frac{x^j x'^j}{r^2}d^3x'$
$\quad +\frac {x^m}{r}\frac{1}{2}\int(T'_{00}+T'_{kk})_{,0}\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)d^3x'$
$\quad -\frac {1}{2}\big(\frac 1{r}\big)_{,m}\int(T'_{00})r'^2d^3x'$
$\quad -\big(\frac{x^k}{r^2}\big)_{,m} \int(T'^{0j}x'^j x'^k - \frac{1}{2}T'^{0k}r'^2)d^3x'$
$\quad -\frac{1}{2}\int(T'^{00}+T'^{kk})\big[\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)\big]_{,m}d^3x'$
Do these look OK to you?
Regards
TerryW
8. Mar 9, 2017
### TSny
OK. The first term on the right in your expression for $h_{0m}$ can be shown to equal zero via MTW (19.4a). Yes, the second term can be rewritten in terms of the angular momentum $S^k$ and an extra term that will get canceled in the gauge transformation. Also, $h_{0m}$ will include the summation over n for n ≥ 1. But this will get canceled in the gauge transformation.
I have not checked all your terms in $ξ_{0}$ and $ξ_{m}$. I did not modify any of the terms in these expressions as given in MTW. I've attached a figure that shows the role of the various terms of the gauge transformation for $h_{0m}$ as I see it.
#### Attached Files:
• ###### MTW 19.1.png
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9. Mar 14, 2017
### TerryW
Hi TSny,
I've worked my way through your rationale for getting rid of the unwanted terms in $h^{new}_{0m} = h^{old}_{0m} - (ξ_{0,m}+ξ_{m,0})$ but I think that there are still quite a few bits hanging around!
1. The expressions in the orange boxes cancel.... agree
2. ....and the expressions in the purple boxes cancel - I think that the sum of the purple boxes is -
$$\sum_{n=0}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_kk)_{,m}\frac {(r-|x-x'|)^n}{|x-x'|} d^3 x' \quad (A)$$
3. The contribution from the dark blue box is zero due to choosing the origin at the centre of mass....agreed.
4. The expression of the brown boxes can be shown to cancel using identity (19.7b)....agreed.
5. The expression in the green box will cancel expressions in $h_{0m}$ .... $h_{0m}$ contains only terms with $T'{0m}$ so $h^{new}_{0m}$ will still end up with a term -
$$\frac{2x^j}{r^3}\int T'_{00,0}x^j x^{m'} d^3 x' \quad (B)$$
6. The expressions in the light blue boxes will be of order 1/r3 and can be ignored. I don't think this is correct. In 19.5, the term $\frac{x^l}{r^3}$ isn't ignored. The unprimed co-ordinates refer to the location of the observer and as such, some of the values of xl will be of the same order as r. You could maybe argue that all the terms in $ξ_{0,m}$ disappear by taking the partial derivative under the integral sign and then use the divergence theorem but then you would bring back terms from $ξ_{m,0}$ that were eliminated above!
7. The expression in the red box will cancel terms in $ξ_{0,m}$ due to the dependence on the T's on retarded time. If you had made this your final statement, it would have become invalid because there are no parts of $ξ_{0,m}$ left to cancel. It just appears to add four more terms to my collections of terms not yet eliminated.
To summarise, I still have the following terms hanging around:
From $ξ_{0,m}$ :
$$\frac {\partial}{\partial x^m}'\big( \frac {x^j}{r^3} \int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x' \big)\quad(1)$$
$$\frac {\partial}{\partial x^m}\big( \int (T'_{00} + T'_{ll})(\frac {3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4} d^3x' \big) \quad(2)$$
$$\sum_{n=0}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_kk)_{,m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (3)$$
From $ξ_{m,0}$ :
$$\frac{2x^j}{r^3}\int T'_{00,0}x^j x^{m'} d^3 x' \quad(4)$$
$$\frac {x^m}{r} ξ_{0,0} \quad (5)$$
If it is indeed OK to take the partial derivative behind the integral sign and then use the divergence theorem, then we can eliminate (2), but that still leaves four terms to argue away, with $\frac {x^m}{r} ξ_{0,0}$ looking particularly challenging.
I'll continue to explore this to see if 19.7a,b and c can provide further assistance.
Regards
TerryW
10. Mar 14, 2017
### TSny
Every term in $\xi_0$ produces a "retarded time" term when evaluating $\xi_{0, m}$. All of these retarded time terms are canceled by terms generated by the red box term in $\xi_m$ when evaluating $\xi_{m, 0}$. So, the orange and purple boxes cancel after taking this retarded time cancellation into account. Thus, your extra term that you are getting from the purple boxes is canceled by a particular term generated by the red box when evaluating $\xi_{m, 0}$. [Edited to correct subscripts.] (The summation in your left-over purple term should start at n = 2.)
More explicitly, in your left-over term you have $(T'_{00}+T'_{kk})_{,m}$. Show that due to the retarded time dependence, this is equal to $\left( -\frac{x^m}{r} \right) (T'_{00}+T'_{kk})_{,0}$. Then show that the red box will generate a term that will cancel your left-over term.
Similarly, all the terms in $\xi_{0,m}$ that are due to the dependence on the retarded-time will be canceled by terms in $\xi_{m, 0}$ that come from the red box expression in $\xi_m$. [Edited for clarity.]
Show that this equals zero due to choosing the center-of-momentum frame.
Right, $\frac{x^l}{r^3}$ is of order 1/r2 for a least one value of $l$. However, note that $\left( \frac{x^l}{r^3} \right)_{,m} = \frac{\delta_{ml}}{r^3} - \frac{3x^lx^m}{r^5}$, which is of order 1/r3. [Edited to correct a typo.] The expressions in the light blue boxes are of order 1/r2. But when taking the spatial derivative with respect to $x^m$, you either get terms of order 1/r3, or you get "retarded-time" terms that are canceled by terms generated by the red box.
In taking the spatial derivatives with respect to $x^m$ in (1) and (2), you will either make the terms of order 1/r3 or you will produce "retarded-time" terms that will be canceled by terms in (5). Likewise, (3) gets canceled by (5). (4) is zero in the center-of-momentum frame.
Last edited: Mar 14, 2017
11. Mar 14, 2017
### TSny
I have updated my figure to make the comments more accurate.
#### Attached Files:
• ###### MTW 19.1b.png
File size:
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31
12. Mar 17, 2017
### TerryW
Hi TSny,
Your previous post and updated figure has clarified the process of elimination and I can now see that the terms:
$$\frac {\partial}{\partial x^m}'\big( \frac {x^j}{r^3} \int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x' \big)\quad(1)$$
$$\frac {\partial}{\partial x^m}\big( \int (T'_{00} + T'_{ll})(\frac {3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4} d^3x' \big) \quad(2)$$
disappear.
But I still think that there are two terms unaccounted for.
1. In one of my earlier posts, I had incorrectly written the first term of the green box as
$$-\frac {2x^j}{r^3} \int T_{00} x^{j}x^{m'} d^3 x'$$
which you commented on and said that it is zero in the centre of momentum frame. A close look at MTW shows this first term to be
$$-\frac {2x^j}{r^3} \int T_{00} x^{j'}x^{m'} d^3 x'$$
which I don't believe is zero in the centre of momentum frame.
2. I still don't agree that the expressions in the purple boxes cancel. The retarded time term from $ξ_{0,m}$ has already been cancelled by a term from the red box, but it will still produce two terms when the differentiation wrt xm is carried into the integral. One of these terms is cancelled by the purple box on the right hand side, but we are still left with:
$$\sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_{kk})_{,m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x'$$
I've tried to find a way to use 19.7 a,b and c to get these two terms to cancel without success.
If you can help me to resolve this, I promise not to do a post if I can't sort $h_{mn}$!
Regards
TerryW
13. Mar 17, 2017
### TSny
OK. But I don't understand the prime on the derivative operator in (1).
Right, this term is not zero. But when forming $\xi_{m,0}$, this term will cancel an unwanted term in $h_{0m}$ when doing the gauge transformation.
I think this is the retarded time term that gets canceled by a term from the red box.
14. Mar 17, 2017
### TerryW
Sorry, that's a typo!
It can't. All terms in $h_{0m}$ contain $T_{0m}$ and the second term in your green box covers all the terms in $h_{0m}$ for n>0.
I don't think it is. The retarded time term is the term in the left hand purple box differentiated wrt x0 (t) and it includes the -$\frac {x^m}{r}$ term, like the terms in the red box. My term arises from the differentiation of the product $(T'_{00}+T’_{kk})\frac {(r-|x-x'|)^n}{|x-x'|}$ wrt xm which produces two parts, only one of which is cancelled your right hand purple box.
15. Mar 17, 2017
### TSny
After some manipulations and the use of (19.7c) I get (to order 1/r^3) the following:
$h_{om} = -2 \epsilon_{mkl}S^k\frac{x^l}{r^3} - 2 \frac{x^j}{r^3}\int \left( T^{00 '}x^{m'}x^{j'} \right)_{,0} d^3x'$ + a summation term over n.
The middle term is canceled by the first term of the green box and the summation term is canceled by the second term of the green box.
All of the terms that I call "retarded time terms" are terms that come from differentiation of the retarded time with respect to $x^m$ when forming $\xi_{0,m}$. Thus,
$\frac{\partial}{\partial x^m} T^{00'} = -\frac{x^m}{r} T^{00'}\, _{,0}$.
Similarly for differentiating any other $T '$ component with respect to $x^m$.
So, when forming $\xi_{0, m}$, the purple box will yield a contribution containing $(T'_{00}+T’_{kk})_{,m} =- \frac{x^m}{r} (T'_{00}+T’_{kk})_{,0}$. This well get canceled by a term generated by the red box when calculating $\xi_{m,0}$.
Last edited: Mar 17, 2017
16. Mar 19, 2017
### TerryW
Hi TSny,
I probably won't be able to get back to you for a few days - partly other commitments and partly trying to work a few things out, particularly the 'retarded time' effect.
regards
TerryW
17. Mar 22, 2017
### TerryW
Hi TSny
I've spent a bit of time working out that:
If ξm = T0k
then $ξ_{0,m} = \frac{\partial}{\partial x^m} T^{0k} - \frac{x^m}{r}\frac{\partial}{\partial t} T^{0k}$
From this, using the fact that $\int \frac{\partial}{\partial x^m} T^{0k} d^3x' = 0,$ I get:
$ξ_{0,m}$ :
$$\big( \frac{1}{2r}\big)_{,m}\frac{\partial}{\partial t}\int T'^{00}r'^2 d^3x'\quad (1)$$
$$- \frac{x^m}{r}\frac{1}{2r}\frac{\partial}{\partial t}\int T'^{00}\ _{,0} r'^2 d^3x'\quad(2)$$
$$+\big(\frac{x^j}{r^3}\big)_{,m}\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x'\quad(3)$$
$$-\frac{x^m}{r}\big(\frac{x^j}{r^3}\big)\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )_{,0}d^3x'\quad(4)$$
$$-\frac{x^m}{r} \int (T'_{00} + T'_{ll})_{,0}\big[\frac {x^{j}x'^{j}}{r^2}+\frac {(3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4}\big] d^3x'\quad(5)$$
$$-\frac{x^m}{r} \sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})_{,0}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (6)$$
$ξ_{m,0}$ :
$$-\frac{2x^j}{r^3}\int T'_{00,0}x^{j'} x^{m'} d^3 x' \quad(7)$$
$$+4 \sum_{n=1}^\infty \frac{1}{n!}\frac{\partial^{n}}{\partial t^{n}}\int T'_{0m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (8)$$
$$-\big( \frac{1}{2r}\big)_{,m}\frac{\partial}{\partial t}\int T'^{00}r'^2 d^3x'\quad (9)$$
$$-\big(\frac{x^j}{r^2}\big)_{,m}\int (T'^{0j} x'^j x'^k - \frac {1}{2} T'^{0k} r'^2 )_{,0} d^3x'\quad(10)$$
$$-\sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})\big[\frac {(r-|x-x'|)^n}{|x-x'|}\big]_{,m}d^3 x' \quad (11)$$
$\frac {x^m}{r} ξ_{0,0}:$
$$\frac{x^m}{r}\frac{1}{2r}\frac{\partial}{\partial t}\int T'^{00}\ _{,0} r'^2 d^3x'\quad(12)$$
$$+\frac{x^m}{r}\big(\frac{x^j}{r^3}\big)\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )_{,0}d^3x'\quad(13)$$
$$+\frac{x^m}{r} \int (T'_{00} + T'_{ll})_{,0}\big[\frac {x^{j}x'^{j}}{r^2}+\frac {(3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4}\big] d^3x'\quad(14)$$
$$+\frac{x^m}{r} \sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})_{,0}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (15)$$
My expression for $h_{0m}$ is:
$h_{0m} = 4\int T'_{0m} \big( \frac{1}{r} +\frac{x^j x'^j}{r^3} +O\big( \frac{1}{r^3} \big) \big) +4 \sum_{n=1}^\infty \frac{1}{n!}\frac{\partial^{n}}{\partial t^{n}}\int T'_{0m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x'\quad(16)$
I am now working on these expressions to see if I can get the desired result.
For the moment, do you agree that a) (1) to (16) are correct and that b) there are no other terms?
Regards
TerryW
18. Mar 22, 2017
### TSny
I'm not sure I'm following your way of writing it.
In the expressions $ξ_0$ and $ξ_m$, all of the $T ^ \prime$ components are evaluated at the primed spatial coordinates $x^{j \, \prime}$ and at the retarded time $t - r$. For example, the functional dependence of $T^{0k \, \prime}$ is $T^{0k \, \prime} \left(t-r, x^{1'}, x^{2'}, x^{3'} \right)$. So, when taking the derivative of $T^{0k \, \prime}$ with respect to the unprimed spatial coordinate $x^m$, the only contribution is from the dependence of the retarded time on $x^m$ (via $r$). So, $\frac{\partial}{\partial x^m} T^{0k \prime}= - \frac{x^m}{r}\frac{\partial}{\partial t} T^{0k}$. Maybe that's the same thing as you are saying.
All your expressions (1) through (16) look correct to me. So, hopefully, things will now simplify to what you need.
19. Mar 24, 2017
### TerryW
I don't think there are two sets of co-ordinates here. I think MTW is using the prime to indicate the volume round the origin containing the matter, and the non-primed co-ordinates belong to the stationary observer. So when we are taking derivatives $\frac{\partial x'^i}{ \partial x^m} = δ^i _m \quad$ but $\frac{\partial x^i}{ \partial x^m} = 0$ because the observer is stationary.
I derived my formula for $ξ_{0,m}$ as follows:
If $ξ_{m}= T'_{mk} x'^k$ (I've changed this to get the indices in balance)
then
$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^m} dx^m + \frac {\partial {(T'_{mk} x'^k)}}{ \partial x^0} \frac {\partial |x-x'|}{ \partial x^m} dx^m$
i.e. the change in $ξ_{m}$ has a component arising from the change of position and a component arising from the need to account for the retarded time effect at the observer.
So
$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^m} dx^m + \frac {\partial {(T'_{mk} x'^k)}} { \partial x^0} \frac {\partial [(|x-x'|)^2]^½} { \partial x^m} dx^m$
$\quad \frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {1}{2} \frac {1}{(x^i - x'^i)^½} \frac {\partial }{ \partial x^m}[x^i x^i - 2x^i x'^i +x'^i x'^i]$
$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {1}{2} \frac {1}{|x^i - x'^i|} [- 2x^iδ^i _m+x'^i δ^i _m]$ (Observer is stationary)
$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {x^m - x'^m}{|x^i -x'^i|}$
$\quad$ and as x >> x'
$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = - \frac {x^m}{r}$
Therefore:
$\frac {\partial {ξ_{m}}}{\partial x^m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x'^m} - \frac {\partial {(T'_{mk} x'^k)}}{ \partial x^0} \frac {x^m}{ r}$
I'm pleased that you agree I've now correctly identified the necessary components for the problem, but so far, I haven't had any success boiling them down to the required result.
Regards
TerryW
20. Mar 24, 2017
### TSny
We agree that there is only one coordinate system. The primed coordinates and the unprimed coordinates are coordinates in this one coordinate system. Nevertheless, the primed coordinates are independent of the unprimed coordinates, unless there is some sort of constraint that imposes a functional dependence of the primed and unprimed coordinates. In the expressions in $\xi_m$ and $\xi_0$, there is one such constraint. The primed time in the $T’$ components is constrained to be the retarded time. That is, we have the functional constraint $t’ = t – r$. This is the only constraint. There is no constraint that imposes any functional dependence of the primed spatial coordinates and the unprimed spatial coordinates. So, $\frac{\partial x \, ' ^i}{\partial x^m} = 0$ for all $i, m = 1, 2, 3$.
Also, $\frac{\partial x^i}{\partial x^j} = \delta^i_j$. This is true whether or not the observer is stationary. It is just an expression of the independence of the three spatial coordinates of the observer.
Note that in the equation above, the left-hand side is $Δξ_{m}$, which is not summed over $m$. But each term of the right-hand side has $m$ occurring three times. So, it is not clear if there is supposed to be a summation over $m$.
To avoid confusion, suppose we want the change in $ξ_{m}$ due to a small change in the particular unprimed coordinate $x^3$, say. I would write this as
$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^3} dx^3$
Then, we note that the only dependence of $T'_{mk} x'^k$ on $x^3$ is in the $r$ in the retarded time $t' = t - r$. So, we can write
$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \frac {\partial t'}{ \partial x^3} dx^3 = \frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \left( - \frac {\partial r}{ \partial x^3}\right)dx^3 =\frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \left( - \frac{x^3}{r} \right) dx^3$
Finally, we can replace $\partial t'$ by $\partial t$ since $t' = t - r$. So, we finally arrive at
$Δξ_{m} = - \frac{x^3}{r} ξ_{m,0}dx^3$
This is equivalent to $\frac{\partial{ξ_{m}}}{\partial x^3} = - \frac{x^3}{r} ξ_{m,0}$
Last edited: Mar 24, 2017 | 2017-08-20 07:25:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.7530540227890015, "perplexity": 436.7057381391366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00440.warc.gz"} |
https://scioly.org/wiki/index.php?title=Boca_Raton_Community_High_School&oldid=71658 | # Boca Raton Community High School
Boca Raton Community High School
Information
Member of SO since 2008
Results History
Best regional 1st (2009-2017)
Best state 1st (2011-2017, 2019)
Best nationals 12th (2017)
Social Media
Website Team Website
Boca Raton Community High School, commonly shortened to Boca Raton High School or Boca High, is a Division C school located in Boca Raton, Florida. They placed first at their regional tournament and the Florida state tournament for many years in a row, until the 2018 season, where they received a second place at both competitions. Nevertheless, they have attended the National competition since 2010 and have been increasing their scores at the invitational over time. The school's website can be found here.
## History
### 2008-2010: Early Beginnings
The team was started in the late 2000s, meaning that the team is around ten years old. The official season that Boca High started was likely 2008. In 2008, the team was much smaller and had little experience in Science Olympiad. However, team members put in adequate amounts of work when preparing for their events and Boca High ended up getting 6th place at their regional tournament, the Southeast Florida Regional hosted at the nearby Florida Atlantic University in Boca Raton, FL. Considering that Boca High was a new team who turned a 6th place at one of their first regional tournaments showed that the team was ambitious and ready to overcome challenges. Despite their success as a first team, the 6th place was not sufficient to qualify the team for the state competition that year. Nevertheless, Boca continued to study into the 2009 season.
The 2009 season showed that Boca High was preparing better and more efficiently in terms of their events. They continued to study for their events, motivated by their 6th place score at the previous regional. While Boca's improvement overall was not noticeable in the early part of the year, it showed in their placing at their regional tournament in 2009 - a 1st place. Since they placed first at the regional competition, they were able to move onto the state tournament in that year, marking the team's first state tournament. At the state tournament, Boca High scored a 6th place. While this did not qualify the team for the national-level tournament, it was an incredible score for the team's first state-level competition. Similar to 2008, Boca High's scores would motivate them to continue in Science Olympiad into the 2010 season.
In 2010, the team began its year like previous years, by continuously studying until the regional tournament. Like 2009, it was able to score 1st at the regional-level tournament at Florida Atlantic University. As the 2010 Florida state tournament rolled around, the team had been vigorously studying material in preparation. Due to this, Boca High was able to score a 2nd place at the 2010 Florida state tournament. They had fallen behind Gulf Breeze High School, which scored 146, while Boca High scored a 234 - a difference of almost 100 points. While there was a significant difference between the scores of Boca High and Gulf Breeze, this would not stop them from pursuing the state-level tournament in the future.
Additionally, since they received a 2nd place at the state-level tournament, the team was allowed to head to the 2010 National tournament - the first Nationals tournament Boca High had ever attended. At the tournament, the team scored a 51st place, with over 1,000 points. This result was not too good of a score, considering that there were 60 teams at the Nationals tournament that year. Since it was their first Nationals competition, the team's score was not looked upon so poorly throughout Boca High. It offered Boca High to improve, which they would do for a great deal of time to come.
### 2011-2014: Improving Scores
In 2011, Boca Raton Community High School won both the regional-level tournament and state-level tournament for the first time in its history. This of course allowed the team to continue to the 2011 Nationals tournament, where they would score 38th with 829 points - a large improvement from the previous year.
Boca High was able to continue the pattern of scoring 1st at both the regional-level and state-level tournaments for many years, until 2018. It is during this period that Archimedean Upper Conservatory began to emerge as a competitor at both the regional and state tournament. The school from Miami was slowly improving, but they never won the regional-level or state-level tournament until 2018 because of Boca High's competitive nature, which caused them to consistently win the 1st place at both tournaments.
Boca High also had a pair of team members place in an event at Nationals for the first time at the 2012 National Tournament at the University of Central Florida, coincidentally enough. While it was a landmark achievement for Boca High to have won their first individual placing, it was also an achievement because it was Boca High's first 1st place individual placing. The pair placed in Water Quality with a 1st place. Boca High would also place in Water Quality in 2013 and 2014, with 2nd and 5th places, respectively.
### 2015 - Present: Invitationals and Nationals Placements
In 2015, Boca High attended its first invitational, the MIT Invitational, with both Team 1 and Team 2. At this invitational, Boca High Team 1 earned a 13th place (with 158 points) and Team 2 earned a 29th place (with 304). Later that year, they also attended the Cypress Falls Invitational. Here, they earned their first placement at an invitational tournament, earning a 4th place and a score of 252. These were the only 2 invitationals that Boca High attended in 2015. Through 2017, the team did not attend more than two invitationals per year. However, in 2018, the team attended 4 invitationals. In 2019, they repeated this practice.
Also, since 2015, members of Boca High have received countless medals at the Nationals tournaments over the year. In 2015, members of the team received individual awards in 3 events. In 2016 and 2017, members of the team received individual placements in 5 events. These years showed a great increase in the number of awards given to individual competitors, as at the 2014 National tournament, the team only received an individual placement in one event.
In 2018, the team did not do as well as they had done in the years before. The team's invitational scores were not all that bad, but the team lost some streaks it had in the regional and state tournament. This included the team getting a 2nd place at the regional competition and the state tournament, breaking two long-lasting streaks. Nevertheless, the team still managed to make a 27th place at Nationals.
In the 2019 season, the team did spectacular. They placed at multiple invitationals (including the MIT, Penn, and Northview invitationals) and won 1st place at the state-level tournament. Despite this, the team lost to Archimedean Upper Conservatory, another competitive school, at the regional tournament by a margin of 2 points.
### Invitationals
Boca Raton began to attend invitationals in 2015, starting with the MIT Invitational, where they brought both Team 1 and Team 2. They also decided to also attend the Cypress Falls Invitational the same year with Team 1 only. They found relatively good success at these invitationals, including placing 4th overall at Cypress Falls. These were the only 2 invitational competitions that the team attended in 2015. Since 2015, the team has consistently attended the MIT Invitational, however, the team has not attended Cypress Falls since their visit in 2015.
In 2016, the team chose to attend the Cornell Invitational, along with the MIT Invitational. They brought both Team 1 and Team 2 to both invitationals. While neither team was able to place at the MIT Invitational, Team 1 placed 3rd overall at the Cornell Invitational. It appears that the team did not think too poorly of the Cornell Invitational, attending again in 2018.
In 2017, Boca Raton tried the Golden Gate Invitational with Team 1, along with taking both teams to the MIT Invitational as usual. Team 1 placed 6th overall at the Golden Gate Invitational, but neither team placed at the MIT Invitational again. Like Cornell, the team appeared to enjoy the Golden Gate Invitational, since it returned in 2019. However, getting to the Golden Gate Invitational can be quite a hassle for the team, since it is located in California, almost all the way across the country.
In 2018, the team attended 4 invitationals for the first time. This included bringing Team 1 to the Northview Invitational and the Princeton Invitational for the first time, as well. Both teams were taken to MIT and Cornell. Even though this was not the team's best year, Team 1 did manage to place 6th overall at the Cornell Invitational.
In 2019, the team attended 4 invitationals again. This year, they tried the Penn Invitational, taking both teams. Boca Raton was very good at the invitationals it attended, with Team 1 placing at all invitationals. The placements earned by Team 1 included three 2nd places (Northview, MIT, and Golden Gate) and one 6th place (Penn).
## Team Setup
The team is coach-run with high stress on equality among members. The majority of preparation is done during after-school gatherings in the coach's classroom on weekdays. Over the weekends, students may meet up together to review their events, including some student-led gatherings at a nearby university.
The entire program consists of 5 teams:
• Team 1 is the team that attends every tournament, including all invitationals, the regional tournament, state tournament, and national tournament.
• Team 2 is considered a good introductory team for those new to Science Olympiad and hope to advance in the future. This teams attends some invitationals, as well as the regional tournament and state tournament.
• Teams 3, 4, and 5 are the teams that often only attend regionals, but if a team does well at the regional tournament then they may attend the state-level competition.
The placement of students on teams is determined mainly by their scores on placement exams that occur usually around the middle to end of September. While these are the main factor, other factors play into the decision as well, such as the events the student is interested in and how often they come after school. It is frequent for students to move up and down teams over the years based on their performance, however, most students do not move during the year unless something severe has happened on a team.
In previous seasons, Teams 3, 4, and 5 were named after colors instead of numbers in order to make the teams appear more equal.
## Streaks & Records
### Southeast Florida Regional
Throughout its history, Boca High has won a multitude of awards at the Southeast Florida Regional, including many 9 1st place awards and 2 2nd place awards. Additionally, they held a 9-year streak of getting 1st place from 2009 to 2017.
### Florida State Tournament
At the state-level tournament, Boca High has earned many high placements, similar to their performance at the Southeast Florida Regional. Over its history, it has received 8 1st place medals, making Boca High the team in Florida with the second-highest number of 1st place medals, behind Gulf Breeze High School.
Additionally, in 2015, Team 1 and Team 2 broke a record at the Florida state tournament for being the first school to attain both 1st and 2nd place rankings at the 2015 States competition. However, since both teams were from the same school, only Team 1 was permitted to the National tournament. Archimedean Upper Conservatory, the school that scored third place, won the second bid by default, and went to the 2015 National competition in place of Boca High's Team 2. In 2019, Team 1 and Team 2 won first and second place again and Archimedean Upper Conservatory was allowed to Nationals.
## Results History
Year National Points State Points Regional Points
2008 - - - - 6th 227
2009 - - 6th 227 1st
2010 51st 1019 2nd 234 1st 65
2011 38th 829 1st 181 1st 74
2012 28th 644 1st 97 1st 80
2013 24th 583 1st 91 1st 64
2014 21st 526 1st 85 1st
2015 19th 545 1st 124 1st 73
2016 20th 534 1st 103 1st 102
2017 12th 426 1st 126 1st 125
2018 27th 626 2nd 145 2nd 119
2019 1st 82 2nd 83
### Trophies
#### Regional Tournament
1st Boca Raton Community High School received a 1st place trophy at the 2009 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2010 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2011 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2012 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2013 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2014 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2015 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2016 Southeast Florida Regional.
1st Boca Raton Community High School received a 1st place trophy at the 2017 Southeast Florida Regional.
2nd Boca Raton Community High School received a 2nd place trophy at the 2018 Southeast Florida Regional.
2nd Boca Raton Community High School received a 2nd place trophy at the 2019 Southeast Florida Regional.
#### State Tournament
6th Boca Raton High School received a 6th place trophy at the 2009 Florida State Tournament.
2nd Boca Raton High School received a 2nd place trophy at the 2010 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2011 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2012 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2013 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2014 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2015 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2016 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2017 Florida State Tournament.
2nd Boca Raton High School received a 2nd place trophy at the 2018 Florida State Tournament.
1st Boca Raton High School received a 1st place trophy at the 2019 Florida State Tournament.
## Invitational Results History
Invitational Tournament Results
Year Northview MIT Cypress Falls Golden Gate Cornell Princeton UPenn
Team 1 Team 2 Team 1 Team 2 Team 1 Team 2
2019 2nd (167) 2nd (241) 18th (556) - 2nd (205) - - 6th (255) 16th (418)
2018 9th (308) 21st (570) 30th (735) - - 6th (165) 8th (240) 9th (386) -
2017 - 11th (405) 26th (693) - 6th (255) - - -
2016 - 11th (313) 24th (608) - Didn't exist 3rd (158) 9th (304) Didn't exist Didn't exist
2015 - 13th (405) 29th (648) 4th (252) -
### Trophies
4th Boca Raton High School Team 1 received a 4th place trophy at the 2015 Cypress Falls Invitational.
3rd Boca Raton High School Team 1 received a 3rd place trophy at the 2016 Cornell Invitational.
6th Boca Raton High School Team 1 received a 6th place trophy at the 2017 Golden Gate Invitational.
6th Boca Raton High School Team 1 received a 6th place trophy at the 2018 Cornell Invitational.
2nd Boca Raton High School Team 1 received a 2nd place trophy at the 2019 Northview Invitational.
2nd Boca Raton High School Team 1 received a 2nd place trophy at the 2019 MIT Invitational.
2nd Boca Raton High School Team 1 received a 2nd place trophy at the 2019 Golden Gate Invitational.
6th Boca Raton High School Team 1 received a 6th place trophy at the 2019 UPenn Invitational.
## National Tournament Medals
Year Medals
2018
2017 Electric Vehicle (2nd), Microbe Mission (2nd), Invasive Species (5th), Forensics (6th), Hydrogeology (6th)
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... Now we have to equate the coefficients of x and y terms and find the values of constants. If the equation balances, then the answers are correct: $$2x + y = 8$$ when $$x = 3$$ and $$y = 2$$. Accelerating the pace of engineering and science. In this example the equations will need to be subtracted from each other as, If the equations were added together, then, into either equation to find the value of. Thank You. 2x+2 =−x+5 3x=3. But I need a matlab code to evaluate this. 5 Answers. Keep the answer exact or give decimal approximations. So...we have Equation A and Equation B. The full sets of solutions to the system of equations are the two sets of points formed by all possible combinations of the values in solx and soly. Substitute into the second equation: 6x + 3(1 - 2x) = 3 6x + 3 - 6x = 3 3 = 3. Hi Star Strider, the coeffs function was what I was looking for! , which has a coefficient of 1 in each equation. This fact can be used to help solve the two simultaneous equations at the same time and find the values of $$x$$ and $$y$$. Commented: Philip Hoskinson on 19 Jan 2017 I need to find the possible solutions for (x) using the following equations: (Where A,B,C are some numbers) 0 = A + B*cos(x) + C*sin(x) 1 = cos^2(x) + sin^2(x) I looked through the solve functions to no avail 1 Comment. I have two equations, one with known parameters and one with unknown. 2) Get the logarithms of both sides of the equation. I would like to ask for a simple way to solve the system that equates the parameters for the two equations, eq. The solution of an equation with one variable is a number. The other line has an equation of y = 3x – 1 which also has a slope of 3. eq. If you're seeing this message, it means we're having trouble loading external resources on our website. Now for reasons we will discuss later in the coordinate geometry module, any single equation with just x and y (neither variable raised to a power or in a fraction), can be represented by a line in the x-y plane. The unknowns of $$x$$ and $$y$$ have the same value in both equations. Type in any equation to get the solution, steps and graph Use the isolation strategy, with just a few new tricks, to solve complicated radical equations. Note that solx and soly are the two sets of solutions to x and to y. Active 3 years ago. EQUATION A: Y = (1750/X)(308/417) EQUATION B: Y = 1750/(X+90) i need to obtain the value of "x", i can't apply elimination method :(please help me...thanks :)) Answer Save. I would like to ask for a simple way to solve the system that equates the parameters for the two equations e.g. Solve two equations for unknown. Based on your location, we recommend that you select: . Others have mixtures of multiple angles and single angles with the same variable. Answered: Star Strider on 4 Feb 2019 My doubt is how to solve a system in which the number of equations is lower than the number of unknowns, using Matlab. You want to get the variables by themselves, remove the radicals one at a time, solve the leftover equation, and check all known solutions. Sal solves the equation -16 = x/4 + 2. Steps to Solve Exponential Equations using Logarithms. In this example the equations will need to be subtracted from each other as $$y - y = 0$$ . Vote. When you write a new equation by equating the two, you get the following new equation: 2 x + 3 y − 5 = x − y ≡ x + 4 y − 5 = 0 Note that this does not mean every (x, y) that satisfies the first also satisfies the second. Unable to complete the action because of changes made to the page. Choose a web site to get translated content where available and see local events and offers. The inverse of adding 9 is subtracting 9, so subtract 9 from each side: Check the answers by substituting both values into the other original equation. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Learn more about symbolic, equation Lv 7. Sign in, choose your GCSE subjects and see content that's tailored for you. To reduce the expression, combine like terms, for example everything with the same variable. Please help. Read about our approach to external linking. I have two expressions: Exp1: $\text{erfc}\left(\sqrt{\phi }\right)$ Exp2:\frac{\Gamma (k+1) \left(\frac{\sqrt{2 \pi } \Gamma \left(k-\frac{3}{2}\right)}{\left(\frac{1}{2 k-3}\right)^{3/2} ... Stack Exchange Network. It relies on the fact that two expressions are identical precisely when corresponding coefficients are equal for … This can be done if the coefficient of one of the letters is the same, regardless of sign. For example, To be able to solve an equation like this, another equation needs to be used alongside it. Ask Question Asked 3 years ago. I just made this video for a class. Learn more about it and how to test solutions to 2-variable equations. How does the solution of a 2-variable equation look like? Simply equate the two. $\begin{array}{ccccc} 3x & + & y & = & 11 \\ - && - && - \\ 2x & + & y & = & 8 \\ = && = && = \\ x &&& = & 3 \end{array}$. See answer buentea6ira is waiting for your help. 1 = (a2 - a1 - a3 + 1)*x1 + (2*a1 - a2 - 3)*x2 + (3 - a1)*x3 1 decade ago. How to solve a system of two equations with 3 unknowns?? The remaining unknown can then be calculated. Equate these two equations for acceleration and express, for constant mass, FΔt. Swamy. The basics of solving radical equations are still the same. This algebra video tutorial explains how to solve linear equations with fractions. . In this example this is the letter $$y$$, which has a coefficient of 1 in each equation. 3) Solve for the variable. You don’t need to worry about how will you find th… An exponential equation is an equation in which a variable occurs as an exponent. That way it is possible to find the only pair of values that solve both equations at the same time. In this example the equations will need to be subtracted from each other as $$y - y = 0$$. It means that every (x, y) that satisfy the first and the second also satisfy the third. Either add or subtract the two equations from each other to eliminate the letter $$y$$. Relevance. Hussain514 Hussain514 Acceleration is the rate at which velocity changes. Simultaneous equations require algebraic skills to find the values of letters within two or more equations. Vote. They are called simultaneous equations because the equations are solved at the same time. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. Reload the page to see its updated state. 0. In mathematics, the method of equating the coefficients is a way of solving a functional equation of two expressions such as polynomials for a number of unknown parameters. There are many ways to solve linear system equations. If you want to solve an algebraic expression, first understand that expressions, unlike equations, are mathematical phrase that can contain numbers and/or variables but cannot be solved. 0. Substitute $$x = 3$$ into either $$3x + y = 11$$ or $$2x + y = 8$$. by factoring we may get two linear factors and that will be the separate equations for the straight line. It takes two steps because he first has to subtract 2 from both sides and then multiply both sides by 4. Note that if these equations had the same y-intercept, they would be the same line instead of parallel. Solve the following simultaneous equations: First, identify which unknown has the same coefficient. However this method solves the solution into a structure p.a1, p.a2 and p.a3 which is not convient (I would like to have it a vector) And second I have to manually create the sym vectors v1 and v2 which again is not convenient when I have to repeat the same thing for different equations of different orders. You may receive emails, depending on your. 1) Keep the exponential expression by itself on one side of the equation. Other MathWorks country sites are not optimized for visits from your location. Overlay them on the plot of the equations. The remaining unknown can then be calculated. Some trigonometry equations contain more than one trig function. In this example this is the letter. I'm trying to equate two equations to solve for two unknowns in excell. Find the value of $$y$$ using inverse operations to solve equations. Add your answer and earn points. Since both equations are in the formy=f(x) we can equate the right hand sides of the equations and solve forx. So the seven solutions that we mentioned, those are the seven dots on this diagram. of one of the letters is the same, regardless of sign. There is a simplest way to perform this. For example, $$2x + y = 10$$ could be solved by: To be able to solve an equation like this, another equation needs to be used alongside it. The value of $$x$$ can now be substituted into either equation to find the value of $$y$$. The inverse of adding 9 is subtracting 9, so subtract 9 from each side: Transformation of curves - Higher - Edexcel, Home Economics: Food and Nutrition (CCEA). The end goal is to … Of course in this simple case it is easy to see that a1=2.043, a2=1.339 and a3=0.177. I have these equations, e1=S^2-k2*s-k1 and e2=s^2+0.7*s+0.12 and e1=e2 now, by the naked eye you can see, k1=-0.12 and k2=-0.7. equate the parameters of two equations in symbolic. Equations that have more than one unknown can have an infinite number of solutions. These are known as simultaneous equations. It takes two steps because he first has to subtract 2 from both sides and then multiply both sides by 4. Some examples of such equations include 3cos 2 x = sin 2 x, 2sec x = tan x + cot x, cos 2x + cos x + 1 = 0, and sin x cos x = 1/2. Opportunities for recent engineering grads. Well, it is an ordered pair. Viewed 67 times -1. Favorite Answer. Learn how to solve exponential equations. You can use any bases for logs. Z2 = (Zt)*dT + (Zp)*dP In this case I want to separate terms so I only have dT terms in one equation, and only dP terms in the second equation. You can easily create the vector yourself in the format you choose from these results: eq1 = (a2 - a1 - a3 + 1)*x1 + (2*a1 - a2 - 3)*x2 + (3 - a1)*x3; [a1, a2, a3] = solve(C1 == C2, [a1, a2, a3]). Either add or subtract the two equations from each other to eliminate the letter, . And they all lie on a straight line. Let us see an example problem for better understanding. If you've got two radicals in your equation, don't panic. The most common method for solving simultaneous equations is the elimination method which means one of the unknowns will be removed from each equation. In the example the java code solve for two variables USING Matrix method but you can modify to perform 3 variables calculations. , 4x + 2 is an equation that says that two ratios are equivalent hussain514 acceleration is same! Solves the equation that way it is possible to find the treasures in matlab Central and discover how the can. ) how to equate two equations Hoskinson on 1 Sep 2016 and find the treasures in matlab Central discover... Exponential equation is an equation like this, another equation needs to be subtracted from equation... And the second also satisfy the third be substituted into either equation to findy: (... Variable x '', a2=1.339 and a3=0.177 find the only pair of values that solve equations... Find the treasures in matlab Central and discover how the community can help you Get the logarithms of both of! Equate two equations from each other to eliminate the letter \ ( y\ ), has... The third of the unknowns will be removed from each other as \ ( y - y = +. To findy: y=2 ( 1 ) Keep the exponential expression by itself on one of... The rate at which velocity changes the page also satisfy the first and the background music if you 've two! And see local events and offers to test solutions to x and y and... Variables calculations, it means that every ( x, y ) satisfy. … learn how to test solutions to x and y terms and find the pair. Simultaneous equations is the leading developer of mathematical computing software for engineers and scientists in excell first has subtract. At the same variable ( y\ ), which has a coefficient of 1 each. Pardon the crappy microphone and the background music if you dislike it strategy. ( 1 ) Keep the exponential expression by itself on one side of the letters is leading... Of mathematical computing software for engineers and scientists changes made to the page:... Problem for better understanding answer_176288, https: //www.mathworks.com/matlabcentral/answers/213028-equate-the-parameters-of-two-equations-in-symbolic # answer_176288, https: //www.mathworks.com/matlabcentral/answers/213028-equate-the-parameters-of-two-equations-in-symbolic answer_176264. Solves the equation -16 = x/4 + 2 2 is an expression with known parameters and with... Solving simultaneous equations is the elimination method which means one of the unknowns of \ ( y\ USING. Not optimized for visits from your location 3x + 5, therefore it s!, 4x + 2 that a1=2.043, a2=1.339 and a3=0.177 equations will need to be used alongside.! On your location mixtures of multiple angles and single angles with the.! System of two equations from each other as \ ( x\ ) can now substitutex= 1 into either equation findy! Recommend that you select: of parallel equate these two lines are.. Got two radicals in your equation, do n't panic equations that have more than one trig function have same. The second also satisfy the third it ’ s slope is 3 simple! Computing software for engineers and scientists 2-variable equation look like country sites are not optimized for visits from your,! Satisfy the third Keep the exponential expression by itself on one side of equation! I need a matlab code to evaluate this engineers and scientists equate two equations to solve equations x/4 2! Simple case it is possible to find the treasures in matlab Central and discover how the community can you! Is 3 Philip Hoskinson on 1 Sep 2016 learn more about it and how to solve an equation says! This diagram equates the parameters for the two equations, eq... now we have equate! And discover how the community can help you through be removed from each other \! Survivors will help you and express, for example, to solve an equation in which a variable as! Then multiply both sides of the unknowns will be removed from each other to eliminate letter... Variable occurs as an exponent equations to solve complicated radical equations are still the same coefficient we mentioned those... Web site to Get translated content where available and see content that 's tailored for.... The treasures in matlab Central and discover how the community can help you through example everything with the same.... Has a coefficient of one of the unknowns will be removed from each other as \ ( )! Then multiply both sides and then multiply both sides of the equation radicals in equation! Angles with the same coefficient it takes two steps because he first has to subtract 2 from sides. Are identical, these two equations e.g resources on our website on the other hand is an equation y! A matlab code to evaluate this be the same, regardless of sign, they would be the same in! Music if you dislike it 2 from both sides of the letters is the same variable acceleration express. Called simultaneous equations require algebraic skills to find the only pair of values that solve both equations see a1=2.043! Y ) that satisfy the third -16 = x/4 + 2 to complete action! The solution of a 2-variable equation look like exponential equation is an equation of y = 3x + 5 therefore. //Www.Mathworks.Com/Matlabcentral/Answers/213028-Equate-The-Parameters-Of-Two-Equations-In-Symbolic # answer_176288, https: //www.mathworks.com/matlabcentral/answers/213028-equate-the-parameters-of-two-equations-in-symbolic # comment_280301 ) that satisfy the third solve complicated radical equations two. Exponential and logarithmic equations with all the steps we 're having trouble loading external resources on our.. Would be the same time local events and offers eliminate the letter \ ( y - y = ). And \ ( y\ ), which has a coefficient of 1 in each equation within. For acceleration and express, for constant mass, FΔt is the letter \ y! I need a matlab code to evaluate this Javier Fernández Alonso on 4 Feb 2019 this! A variable occurs as an exponent it is possible to find the values of constants x... Will need to worry about graphing those to see that a1=2.043, a2=1.339 and a3=0.177 translated... A2=1.339 and a3=0.177, one with known parameters and one with known parameters and one with parameters... Two lines are parallel Hoskinson on 1 Sep 2016 as \ ( y\ ), has! The following simultaneous equations require algebraic skills to find the only pair values! Is the same coefficient system that equates the parameters for the two equations to solve equations:... Sides of the letters is the same, regardless of sign logarithmic with. 'M trying to equate the coefficients of x and to y is to … learn how to solve equations and... And then multiply both sides how to equate two equations then multiply both sides and then multiply both sides and then multiply sides... This is the same, regardless of sign and single angles with the same, regardless sign. Two or more equations for you -16 = x/4 + 2 is a number radical, exponential and equations. Can now be substituted into either equation to find the value of \ ( y\ ) equations because the will... As, have the same variable a simple way to solve linear system equations equation of y 3x! Example everything with the same coefficient was what i was looking for to perform 3 variables.. Equation in which a variable occurs as an exponent the first and the background if... Was what i was looking for have mixtures of multiple angles and single angles with the same variable equations. Content that 's tailored for you for you: how do i equate this equations. Of constants with just a few new tricks, to be used alongside it number..., have the same, regardless of sign strategy, with just a few new,. We 're having trouble loading external resources on our website, identify which unknown the... One side of the letters is the same time will help you a2=1.339 and.... Itself on one side of the letters is the leading developer of mathematical computing software for engineers and scientists the... In excell hi Star Strider, the first and the second also the! These are known as, have the same coefficient by 4 and single angles with same! Equation B same y-intercept, they would be the same variable have than... A proportion on the other hand is an equation like this, another equation needs to be able solve! To solve a system of two equations e.g method which means one the. Have to equate the coefficients of x and to y first has subtract. Tailored for you, combine like terms, for example everything with the same, regardless of.. Equations from each other to eliminate the letter \ ( y\ ) angles with the same time have a! Trying to equate two equations, one with unknown y ) that satisfy the first and the second satisfy... Have the same value in both equations subtract the two sets of solutions equations that have more than one function... Angles and single how to equate two equations with the same y-intercept, they would be the same line instead parallel! Solve the system that equates the parameters for the two equations, eq the., 4x + 2 means that every ( x, y ) that satisfy the third common.: first, identify which unknown has the same variable both equations the. Hi Star Strider, the first and the background music if you 've got two radicals in your,! For you Star Strider, the coeffs function was what i was looking for, therefore it ’ slope. Mass, FΔt about it and how to test solutions to x and y terms and the! To see that a1=2.043, a2=1.339 and a3=0.177 y ) that satisfy the third 3x +,... Have to equate the coefficients of x and to y your GCSE subjects and see that!, which has a coefficient of 1 in each equation are not optimized for visits your. Value in both equations the seven solutions that we mentioned, those are two... Means we 're having trouble loading external resources on our website 1 into either equation to findy: y=2 1. | 2021-01-20 22:41:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0.6739543080329895, "perplexity": 528.3923178238995}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00674.warc.gz"} |
http://mymathforum.com/number-theory/346308-intersection-two-sets-question-2.html | My Math Forum Intersection of Two Sets Question
Number Theory Number Theory Math Forum
May 2nd, 2019, 07:35 PM #11
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Originally Posted by Maschke Didn't you say it was 1?
I said the intersection between a Vitali set and the rationals would have a cardinality of 1. You said it had a measure of 0 and I'm fine with that too, but that's wholly irrelevant. Here I thought maybe you were just alluding to the fact that you thought the measure of the relevant intersection would be 0 since the measure of $\mathcal{C}$ is zero, but now I realize you really didn't have a clue what this thread was about. double
May 2nd, 2019, 07:39 PM #12 Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40 Let's pose a more generalized version of the question for you then. Are there sets A and B where A $\cap$ B = A, the measure of A is undefined, and the measure of B is 0?
May 2nd, 2019, 08:33 PM #13
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Originally Posted by AplanisTophet now I realize you really didn't have a clue what this thread was about.
I'm truly sorry I wasted my time on you tonight.
May 2nd, 2019, 08:38 PM #14
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Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by AplanisTophet I get it now, you were neglecting the possibility that the measure of the intersection between $V$ and $\mathcal{C}$ could be undefined, especially if the intersection were to equal $V$. You can't just assume that the intersection would have a measure of $0$ simply because $\mathcal{C}$ does as this says nothing of whether or not the measure of the intersection may be undefined.
He is neglecting that possibility because it isn't a possibility. It's a fundamental fact that if $A$ has (Lebesgue) measure zero, then every subset of $A$ is measurable and also has measure 0. In fact, this is true of any complete measure which follows as an easy consequence of the Carathéodory (sp?) completion procedure.
Last edited by skipjack; May 2nd, 2019 at 08:58 PM.
May 2nd, 2019, 11:09 PM #15
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Quote:
Originally Posted by AplanisTophet Let's pose a more generalized version of the question for you then. Are there sets A and B where A $\cap$ B = A, the measure of A is undefined, and the measure of B is 0?
Hey man I'm sorry I overreacted. That silly goose remark got to me for some reason. You're right, you said cardinality and I misread it. All the best.
May 8th, 2019, 06:11 PM #16
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Quote:
Originally Posted by Maschke Hey man I'm sorry I overreacted. That silly goose remark got to me for some reason. You're right, you said cardinality and I misread it. All the best.
No worries. You told me what a Vitali set was in the first place.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2019-05-21 15:02:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8492825627326965, "perplexity": 1742.4326720159972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256426.13/warc/CC-MAIN-20190521142548-20190521164548-00342.warc.gz"} |
https://brainiak.in/443/an-electromagnetic-wave-can-produced-charge-amoving-constant-velocity | # An electromagnetic wave can be produced, when charge is Amoving with a constant velocity
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An electromagnetic wave can be produced, when charge is
(A) moving with a constant velocity
(B) moving in a circular orbit
(C) falling in an electric field
(D) both B and C
## 1 Answer
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Correct option is D) both B and C
An accelerated charge is the source of electromagnetic waves (EMWs). When the charge is in a circular motion, the direction of its velocity continuously changes and thus it is in accelerated motion and produces EMWs.
A charge falling in an electric field is accelerated by the electric force and thus produces EMWs
.
1 answer
1 answer
1 answer
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1 answer
1 answer | 2022-10-03 17:13:14 | {"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": 0.8323099613189697, "perplexity": 1727.528993476976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337428.0/warc/CC-MAIN-20221003164901-20221003194901-00127.warc.gz"} |
http://hal.in2p3.fr/in2p3-00023436 | # An angular distribution analysis of $\Lambda_b$ decays
Abstract : A study of the angular distributions of the processes $\Lambda_b \to {\Lambda} V(1^-)$ with $V ({J/\Psi,\rho}^0)$ is performed. Emphasis is put on the initial $\Lambda_b$ polarization and the polarization density-matrices are derived to perform tests of both Time-Reversal (TR) and CP violation
Document type :
Conference papers
http://hal.in2p3.fr/in2p3-00023436
Contributor : Jeanine Pellet Connect in order to contact the contributor
Submitted on : Tuesday, December 14, 2004 - 11:56:20 AM
Last modification on : Tuesday, April 20, 2021 - 12:00:07 PM
### Citation
O. Leitner, Ziad Ajaltouni, E. Conte. An angular distribution analysis of $\Lambda_b$ decays. International Conference on Structure of Baryons 10 (Baryons 2004), Oct 2004, Palaiseau, France. pp.435-438, ⟨10.1016/j.nuclphysa.2005.03.051⟩. ⟨in2p3-00023436⟩
Record views | 2021-10-24 14:37:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 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": 0.27716830372810364, "perplexity": 6710.355820151401}, "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/1634323586043.75/warc/CC-MAIN-20211024142824-20211024172824-00348.warc.gz"} |
https://mathoverflow.net/questions/409633/embedding-quadric-bundles | Let $$\pi:X\rightarrow W$$ be a morphism of smooth projective varieties over a field $$k$$ whose generic fiber is a smooth quadric, and let $$r$$ be the dimension of the fibers of $$\pi$$.
Does there always exists a rank $$r+2$$ vector bundle $$f:\mathcal{E}\rightarrow W$$ on $$W$$ such that $$X$$ can be embedded in $$\mathbb{P}(\mathcal{E})$$ as a divisor and $$X\cap f^{-1}(w) = \pi^{-1}(w)$$ for all $$w\in W$$?
When $$r=1$$ this is true. Indeed, $$\omega_{X}^{-1}$$ is relatively very ample and we can take $$\mathcal{E} = \pi_{*}\omega_{X}^{-1}$$. Does there exist a similar construction for $$r\geq 2$$?
In this paper
ARNAUD BEAUVILLE, Variétés de Prym et jacobiennes intermédiaires, Annales scientifiques de l’É.N.S. 4e série, tome 10, no 3 (1977), p. 309-391
A quadric bundle over $$\mathbb{P}^2$$ is defined as a smooth projective variety $$X$$ with a morphism $$\pi:X\rightarrow\mathbb{P}^2$$ whose fibers are isomorphic to $$r$$-dimensional quadrics.
According to Proposition $$1.2$$ there exists a vector bundle $$\mathcal{E}\rightarrow\mathbb{P}^2$$ of rank $$r+2$$ and a form $$q\in H^0(\mathbb{P}^2,Sym^2\mathcal{E}(h))$$ such that $$X$$ identifies with the zero locus of $$q$$ in the projective bundle $$\mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$$.
No. For instance, take your favorite $$\mathbb{P}^1$$-bundle $$Y \to W$$ which is not a projectivization of a rank 2 vector bundle and set $$X := Y \times \mathbb{P}^1.$$ This is a quadric surface bundle over $$W$$, but if there is an embedding $$X \hookrightarrow \mathbb{P}(\mathcal{E})$$, then its restriction to any fiber of $$X$$ over $$\mathbb{P}^1$$ would give a rational section of $$Y \to W$$, which is impossible.
• I just fund a paper by Beauville saying that the answer is positive for quadric bundles over $\mathbb{P}^2$, and his proof does not seem to be related to the fact that the base is $\mathbb{P}^2$. I added some explanation in my question. However, I do not understand why your example shouldn't fit Beauville's definition of quadric bundle. For $W = \mathbb{P}^2$ the quadric bundle $X\rightarrow W$ you considered is flat and seems to fit Beauville's definition.
– Arty
Nov 29 '21 at 20:06
• There are two possible interpretations of the word "quadric". One, is "an algebraic variety, isomorphic to a hypersurface of degree 2 in a projective space". The other is "an algebraic variety which after base change to an algebraic closure of the base field is isomorphic to a hypersurface of degree 2 in a projective space". Beauville is using the former interpretation, and I was using the latter. Nov 29 '21 at 20:41
• If I understand correctly you are saying that the restriction of the relative hyperplane section of $\mathbb{P}(\mathcal{E})$ determines a point on each fiber of $Y\rightarrow W$ and hence gives a section or $W\rightarrow Y$. The existence of such section would imply that $Y\rightarrow W$ is locally trivial. Did I get it right?
– Arty
Nov 30 '21 at 11:29
• It seems that you construction is related to the fact that you can distinguish in family which one of the two $\mathbb{P}^1$'s giving the quadrics is the one coming from a fiber of $Y\rightarrow W$.
– Arty
Nov 30 '21 at 11:32
• No, I am not saying this. There are examples of quadric surface bundles with a relative hyperplane but without points. Nov 30 '21 at 18:36 | 2022-01-28 06:35:23 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.94703209400177, "perplexity": 78.18608578660138}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00341.warc.gz"} |
https://helptest.net/mathematics/59614/ | What best describes 15x+8
This expression can not be factored with rational numbers. The polynomial is not factorable with rational numbers.
It could be $15 meal for x people plus$8 tip
Only authorized users can leave an answer! | 2021-12-08 12:52:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.2757224440574646, "perplexity": 1890.153790355924}, "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-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00080.warc.gz"} |
https://www.groundai.com/project/large-deviations-performance-of-consensusinnovations-distributed-detection-with-non-gaussian-observations/ | Large Deviations Performance of Consensus+Innovations Distributed Detection with Non-Gaussian Observations
# Large Deviations Performance of Consensus+Innovations Distributed Detection with Non-Gaussian Observations
Dragana Bajović, Duan Jakovetić, José M. F. Moura, João Xavier, and Bruno Sinopoli The work of the first, second, fourth, and fifth authors is partially supported by grants CMU-PT/SIA/0026/2009, SFRH/BD/33517/2008 (through the Carnegie Mellon/Portugal Program managed by ICTI) and by grant PTDC/EEA-CRO/104243/2008 from Fundação para a Ciência e Tecnologia and also by ISR/IST plurianual funding (POSC program, FEDER). The work of the third author is partially supported by NSF under grants CCF-1011903 and CCF-1018509, and by AFOSR grant FA95501010291. Dragana Bajović and Duan Jakovetić hold fellowships from the Carnegie Mellon/Portugal Program.Dragana Bajović and Duan Jakovetić are with the Institute for Systems and Robotics (ISR), Instituto Superior Técnico (IST), Technical University of Lisbon, Lisbon, Portugal, and with the Department of Electrical and Computer Engineering, Carnegie Mellon University, Pittsburgh, PA, USA, dbajovic@andrew.cmu.edu, djakovet@andrew.cmu.eduJosé M. F. Moura and Bruno Sinopoli are with the Department of Electrical and Computer Engineering, Carnegie Mellon University, Pittsburgh, 15213 PA, USA, moura@ece.cmu.edu, brunos@ece.cmu.eduJoão Xavier is with the Institute for Systems and Robotics (ISR), Instituto Superior Técnico (IST), Technical University of Lisbon, Lisbon, Portugal, jxavier@isr.ist.utl.pt
###### Abstract
We establish the large deviations asymptotic performance (error exponent) of consensus+innovations distributed detection over random networks with generic (non-Gaussian) sensor observations. At each time instant, sensors 1) combine theirs with the decision variables of their neighbors (consensus) and 2) assimilate their new observations (innovations). This paper shows for general non-Gaussian distributions that consensus+innovations distributed detection exhibits a phase transition behavior with respect to the network degree of connectivity. Above a threshold, distributed is as good as centralized, with the same optimal asymptotic detection performance, but, below the threshold, distributed detection is suboptimal with respect to centralized detection. We determine this threshold and quantify the performance loss below threshold. Finally, we show the dependence of the threshold and performance on the distribution of the observations: distributed detectors over the same random network, but with different observations’ distributions, for example, Gaussian, Laplace, or quantized, may have different asymptotic performance, even when the corresponding centralized detectors have the same asymptotic performance.
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Keywords: Consensus+innovations, performance analysis, Chernoff information, non-Gaussian distributions, distributed detection, random network, information flow, large deviations.
## I Introduction
Consider a distributed detection scenario where sensors are connected by a generic network with intermittently failing links. The sensors perform consensus+innovations distributed detection; in other words, at each time , each sensor updates its local decision variable by: 1) sensing and processing a new measurement to create an intermediate variable; and 2) weight averaging it with its neighbors’ intermediate decision variables. We showed in [1] that, when the sensor observations are Gaussian, the consensus+innovations distributed detector exhibits a phase transition. When the network connectivity is above a threshold, then the distributed detector is asymptotically optimal, i.e., asymptotically equivalent to the optimal centralized detector that collects the observations of all sensors.
This paper establishes the asymptotic performance of distributed detection over random networks for generic, non-Gaussian sensor observations. We adopt as asymptotic performance measure the exponential decay rate of the Bayes error probability (error exponent). We show that phase transition behavior emerges with non-Gaussian observations and demonstrate how the optimality threshold is a function of the log-moment generating function of the sensors’ observations and of the number of sensors . This reveals a very interesting interplay between the distribution of the sensor observations (e.g., Gaussian or Laplace) and the rate of diffusion (or connectivity) of the network (measured by a parameter defined in Section II): for a network with the same connectivity, a distributed detector with say, Laplace observations distributions, may match the optimal asymptotic performance of the centralized detector, while the distributed detector for Gaussian observations may be suboptimal, even though the centralized detectors for the two distributions, Laplace and Gaussian, have the same optimal asymptotic performance.
For distributed detection, we determine the range on the detection threshold for which each sensor achieves exponentially fast decay of the error probability (strictly positive error exponent), and we find the optimal that maximizes the error exponent. Interestingly, above the critical (phase transition) value for the network connectivity , the optimal detector threshold is , mimicking the (asymptotically) optimal threshold for the centralized detector. However, below the critical connectivity, we show by a numerical example that the optimal distributed detector threshold might be non zero.
Brief review of the literature. Distributed detection has been extensively studied, in the context of parallel fusion architectures, e.g., [2, 3, 4, 5, 6, 7, 8], consensus-based detection [9, 10, 11, 12], and, more recently, consensus+innovations distributed inference, see, e.g., [13, 14, 15, 16, 17] for distributed estimation, and [18, 19, 20, 21, 22, 23, 24] for distributed detection. Different variants of consensus+innovations distributed detection algorithms have been proposed; we analyze here running consensus, the variant in [20].
Reference [20] considers asymptotic optimality of running consensus, but in a framework that is very different from ours. Reference [20] studies the asymptotic performance of the distributed detector where the means of the sensor observations under the two hypothesis become closer and closer (vanishing signal to noise ratio (SNR)), at the rate of , where is the number of observations. For this problem, there is an asymptotic, non-zero, probability of miss and an asymptotic, non-zero, probability of false alarm. Under these conditions, running consensus is as efficient as the optimal centralized detector, [25], as long as the network is connected on average. Here, we assume that the means of the distributions stay fixed as grows. We establish, through large deviations, the rate (error exponent) at which the error probability decays to zero as goes to infinity. We show that connectedness on average is not sufficient for running consensus to achieve the optimality of centralized detection; rather, phase change occurs, with distributed becoming as good as centralized, when the network connectivity, measured by , exceeds a certain threshold.
We distinguish this paper from our prior work on the performance analysis of running consensus. In [26], we studied deterministically time varying networks and Gaussian observations, and in [27], we considered a different consensus+innovations detector with Gaussian observations and additive communication noise. Here, we consider random networks, non-Gaussian observations, and noiseless communications. Reference [1] considers random networks and Gaussian, spatially correlated observations. In contrast, here the observations are non-Gaussian spatially independent. We proved our results in [1] by using the quadratic nature of the Gaussian log-moment generating function. For general non-Gaussian observations, the log-moment generating function is no longer quadratic, and the arguments in [1] no longer apply; we develop a more general methodology that establishes the optimality threshold in terms of the log-moment generating function of the log-likelihood ratio. We derive our results from generic properties of the log-moment generating function like convexity and zero value at the origin. Finally, while reference [1] and our other prior work considered zero detection threshold , here we extend the results for generic detection thresholds . Our analysis reveals that, when is above its critical value, the zero detector threshold is (asymptotically) optimal. When is below the critical value, we compute the best detector threshold , which may be non-zero in general.
Our analysis shows the impact of the distribution of the sensor observations on the performance of distributed detection: distributed detectors (with different distributions of the sensors observations) can have different asymptotic performance, even though the corresponding centralized detectors are equivalent, as we will illustrate in detail in Section IV.
Paper outline. Section II introduces the network and sensor observations models and presents the consensus+innovations distributed detector. Section III presents and proves our main results on the asymptotic performance of the distributed detector. For a cleaner exposition, this section proves the results for (spatially) identically distributed sensor observations. Section IV illustrates our results on several types of sensor observation distributions, namely, Gaussian, Laplace, and discrete valued distributions, discussing the impact of these distributions on distributed detection performance. Section V extends our main results to non-identically distributed sensors’ observations. Finally, Section VI concludes the paper.
Notation. We denote by: the -th entry of a matrix ; the -th entry of a vector ; , , and , respectively, the identity matrix, the column vector with unit entries, and the -th column of ; the ideal consensus matrix ; the vector (respectively, matrix) -norm of its vector (respectively, matrix) argument; the Euclidean (respectively, spectral) norm of its vector (respectively, matrix) argument; the -th largest eigenvalue; and the expected value and probability operators, respectively; the indicator function of the event ; the product measure of i.i.d. observations drawn from the distribution with measure ; and the first and the second derivatives of the function at point .
## Ii Problem formulation
This section introduces the sensor observations model, reviews the optimal centralized detector, and presents the consensus+innovations distributed detector. The section also reviews relevant properties of the log-moment generating function of a sensor’s log-likelihood ratio that are needed in the sequel.
### Ii-a Sensor observations model
We study the binary hypothesis testing problem versus . We consider a network of nodes where is the observation of sensor at time , where ,
###### Assumption 1
The sensors’ observations are independent and identically distributed (i.i.d.) both in time and in space, with distribution under hypothesis and under :
Yi(t)∼{ν1,H1ν0,H0,i=1,…,N,t=1,2,… (1)
The distributions and are mutually absolutely continuous, distinguishable measures. The prior probabilities and are in .
By spatial independence, the joint distribution of the observations of all sensors
Y(t):=(Y1(t),…,YN(t))⊤ (2)
at any time is under and under . Our main results in Section III are derived under Assumption 1. Section V extends them to non-identical (but still independent) sensors’ observations.
### Ii-B Centralized detection, log-moment generating function (LMGF), and optimal error exponent
The log-likelihood ratio of sensor at time is and given by
Li(t)=logf1(Yi(t))f0(Yi(t)),
where, , is 1) the probability density function corresponding to , when is an absolutely continuous random variable; or 2) the probability mass function corresponding to , when is discrete valued.
Under Assumption 1, the log-likelihood ratio test for time observations from all sensors, for a threshold is: 111In (3), we re-scale the spatio-temporal sum of the log-likelihood ratios by dividing the sum by . Note that we can do so without loss of generality, as the alternative test without re-scaling is: with
D(k):=1Nkk∑t=1N∑i=1Li(t)[H0]H1≷γ. (3)
Log-moment generating function (LMGF). We introduce the LMGF of and its properties that play a major role in assessing the performance of distributed detection.
Let () denote the LMGF for the log-likelihood ratio under hypothesis :
Λl:R⟶(−∞,+∞],Λl(λ)=logE[eλL1(1)|Hl]. (4)
In (4), replaces , for arbitrary , and , due to the spatial and temporal identically distributed observations, see Assumption 1.
###### Lemma 1
Consider Assumption 1. For and in (4) the following holds:
1. is convex;
2. , for , , and , ;
3. satisfies:
Λ1(λ)=Λ0(λ+1),forλ∈R. (5)
###### Proof.
For a proof of (a) and (b), see [28]. Part (c) follows from the definitions of and , which we show here for the case when the distributions and are absolutely continuous (the proof for discrete distributions is similar):
Λ1(λ)=logE[eλL1(1)|H1] =log∫y∈R(f1(y)f0(y))λf1(y)dy =log∫y∈R(f1(y)f0(y))1+λf0(y)dy=Λ0(1+λ).
We further assume that the LMGF of a sensor’s observation is finite.
###### Assumption 2
, .
In the next two remarks, we give two classes of problems when Assumption 2 holds.
Remark I. We consider the signal+noise model:
Yi(t)={m+ni(k),H1ni(k),H0. (6)
Here is a constant signal and is a zero-mean additive noise with density function supported on ; we rewrite , without loss of generality, as , where is a constant. Then, the Appendix shows that Assumption 2 holds under the following mild technical condition: either one of (7) or (8) and one of (9) or (10) hold:
limy→+∞g(y)|y|τ+ = ρ+,forsomeρ+,τ+∈(0,+∞) (7) limy→+∞g(y)(log(|y|))μ+ = ρ+,forsomeρ+∈(0,+∞),μ+∈(1,+∞) (8) limy→−∞g(y)|y|τ− = ρ−, forsomeρ−,τ−∈(−∞,0) (9) limy→−∞g(y)(log(|y|))μ− = ρ−,forsomeρ−∈(0,−∞),μ−∈(1,+∞). (10)
In (8) and (10), we can also allow either (or both) to equal 1, but then the corresponding is in . Note that need not be symmetric, i.e., need not be equal to . Intuitively, the tail of the density behaves regularly, and grows either like a polynomial of arbitrary finite order in , or slower, like a power , , or like a logarithm . The class of admissible densities includes, e.g., power laws , , or the exponential families , , with: 1) the Lebesgue base measure ; 2) the polynomial, power, or logarithmic potentials ; and 3) the canonical set of parameters [29].
Remark II. Assumption 2 is satisfied if has arbitrary (different) distributions under and with the same, compact support; a special case is when is discrete, supported on a finite alphabet.
Centralized detection: Asymptotic performance. We consider briefly the performance of the centralized detector that will benchmark the performance of the distributed detector. Denote by , It can be shown [30] that and . Now, consider the centralized detector in (3) with constant thresholds , for all , and denote by:
α(k,γ)=P(D(k)≥γ|H0),β(k,γ)=P(D(k)<γ|H1),:Pe(k,γ)=α(k,γ)π0+β(k,γ)π1, (11)
respectively, the probability of false alarm, probability of miss, and Bayes (average) error probability. In this paper, we adopt the minimum Bayes error probability criterion, both for the centralized and later for our distributed detector, and, from now on, we refer to it simply as the error probability. A standard Theorem (Theorem 3.4.3., [30]) says that, for any choice of , the error probability decays exponentially fast to zero in . For , the error probability does not converge to zero at all. To see this, assume that is true, and let . Then, by noting that , for all , we have that as , by the central limit theorem.
Denote by , the Fenchel-Legendre transform [30] of :
Il(z)=supλ∈Rλz−Λl(λ),z∈R. (12)
It can be shown [30] that is nonnegative, strictly convex, , for , and [30]. We now state the result on the centralized detector’s asymptotic performance.
###### Lemma 2
Let Assumption 1 hold, and consider the family of centralized detectors (3) with constant threshold Then, the best (maximal) error exponent:
limk→∞−1klogPe(k,γ)
is achieved for the zero threshold and equals where
The quantity is referred to as the Chernoff information of a single sensor observation Lemma 2 says that the centralized detector’ error exponent is times larger than an individual sensor’s error exponent. We remark that, even if we allow for time-varying thresholds , the error exponent cannot be improved, i.e., the centralized detector with zero threshold is asymptotically optimal over all detectors. We will see that, when a certain condition on the network connectivity holds, the distributed detector is asymptotically optimal, i.e., achieves the best error exponent , and the zero threshold is again optimal. However, when the network connectivity condition is not met, the distributed detector is no longer asymptotically optimal, and the optimal threshold may be non zero.
###### Proof of Lemma 2.
Denote by the LMGF for the log-likelihood ratio for the observations of all sensors at time . Then, , by the i.i.d. in space assumption on the sensors’ observations. The Lemma now follows by the Chernoff lemma (Corollary 3.4.6, [30]):
limk→∞−1klogPe(k,0)=maxλ∈[0,1]{−Λ0,N(λ)}=Nmaxλ∈[0,1]{−Λ0(λ)}=NI0(0).
### Ii-C Distributed detection algorithm
We now consider distributed detection when the sensors cooperate through a randomly varying network. Specifically, we consider the running consensus distributed detector proposed in [20]. Each node maintains its local decision variable , which is a local estimate of the global optimal decision variable in (3). Note that is not locally available. At each time , each sensor updates in two ways: 1) by incorporating its new observation to make an intermediate decision variable ; and 2) by exchanging the intermediate decision variable locally with its neighbors and computing the weighted average of its own and the neighbors’ intermediate variables.
More precisely, the update of is as follows:
xi(k)=∑j∈Oi(k)Wij(k)(k−1kxj(k−1)+1kLj(k)),k=1,2,...xi(0)=0. (13)
Here is the (random) neighborhood of sensor at time (including ), and are the (random) averaging weights. The sensor ’s local decision test at time is:
xi(k)[H0]H1≷γ, (14)
i.e., (respectively, ) is decided when (respectively, ).
Write the consensus+innovations algorithm (13) in vector form. Let and . Also, collect the averaging weights in the matrix , where, clearly, if the sensors and do not communicate at time step . The algorithm (13) becomes:
x(k)=W(k)(k−1kx(k−1)+1kL(k)),k=1,2,...xi(0)=0. (15)
Network model. We state the assumption on the random averaging matrices .
###### Assumptions 3
The averaging matrices satisfy the following:
1. The sequence is i.i.d.
2. is symmetric and stochastic (row-sums equal 1 and ) with probability one, .
3. There exists , such that, for any realization , , , and, whenever , .
4. and are mutually independent over all and .
Condition (c) is mild and says that: 1) sensor assigns a non-negligible weight to itself; and 2) when sensor receives a message from sensor , sensor assigns a non-negligible weight to sensor .
Define the matrices by:
Φ(k,t):=W(k)W(k−1)...W(t),k≥t≥1. (16)
It is easy to verify from (15) that equals:
x(k)=1kk∑t=1Φ(k,t)L(t),k=1,2,... (17)
Choice of threshold . We restrict the choice of threshold to , , , where we recall , Namely, is a stochastic matrix, hence , for all , and thus . Also, , for all , . Now, by iterating expectation:
E[x(k)|Hl]=E[E[x(k)|Hl,W(1),...,W(k)]]=E[1kk∑t=1Φ(k,t)E[L(t)|Hl]]=γl1,l=0,1,
and , for all . Moreover, it can be shown (proof is omitted due to lack of space) that converges in probability to under . Now, a similar argument as with the centralized detector in II-B shows that for , the error probability does not converge to zero. We will show that, for any , the error probability converges to 0 exponentially fast, and we find the optimal that maximizes a certain lower bound on the exponent of the error probability.
Network connectivity. From (17), we can see that the matrices should be as close to as possible for enhanced detection performance. Namely, the ideal (unrealistic) case when for all , corresponds to the scenario where each sensor is equivalent to the optimal centralized detector. It is well known that, under certain conditions, the matrices converge in probability to :
P(∥Φ(k,t)−J∥>ϵ)→0as(k−t)→∞,ϵ>0,
such that vanishes exponentially fast in , i.e., , . The quantity determines the speed of convergence of the matrices . The closer to zero is, the faster consensus is. We refer to as the network connectivity. We will see that the distributed detection performance significantly depends on . Formally, is given by:222It can be shown that the limit in (18) exists and that it does not depend on .
|logr|:=lim(k−t)→∞−1k−tlogP(∥Φ(k,t)−J∥>ϵ). (18)
For the exact calculation of , we refer to [31]. Reference [31] shows that, for the commonly used models of , gossip and link failure (links in the underlying network fail independently, with possibly mutually different probabilities), is easily computable, by solving a certain min-cut problem. In general, is not easily computable, but all our results (Theorem 5, Corollary 6, Corollary 11) hold when is replaced by an upper bound. An upper bound on is given by [31].
The following Lemma easily follows from (18).
###### Lemma 4
Let Assumption 3 hold. Then, for any , there exists a constant (independent of ) such that:
P(∥Φ(k,t)−J∥>ϵ)≤C(δ)e−(k−t)(|logr|−δ),forallk≥t.
## Iii Main results: Asymptotic analysis and error exponents for distributed detection
Subsection III-A states our main results on the asymptotic performance of consensus+innovations distributed detection; subsection III-B proves these results.
### Iii-a Statement of main results
In this section, we analyze the performance of distributed detection in terms of the detection error exponent, when the number of observations (per sensor), or the size of the observation interval tends to . As we will see next, we show that there exists a threshold on the network connectivity such that if is above this threshold, each node in the network achieves asymptotic optimality (i.e., the error exponent at each node is the total Chernoff information equal to ). When is below the threshold, we give a lower bound for the error exponent. Both the threshold and the lower bound are given solely in terms of the log-moment generating function and the number of sensors . These findings are summarized in Theorem 5 and Corollary 6 below.
Let , , and denote the probability of false alarm, the probability of miss, and the error probability, respectively, of sensor for the detector (13) and (14), for the threshold equal to :
αi(k,γ)=P(xi(k)≥γ|H0),βi(k,γ)=P(xi(k)<γ|H1),Pe,i(k,γ)=π0αi(k;γ)+π1βi(k;γ), (19)
where, we recall, and are the prior probabilities.
###### Theorem 5
Let Assumptions 1-3 hold and consider the family of distributed detectors in (13) and (14) with . Let be the zero of the function:
Δl(λ):=Λl(Nλ)−|logr|−NΛl(λ),l=0,1, (20)
and define , by
γ−0 =Λ′0(λs0),γ+0=Λ′0(Nλs0)≥γ−0 (21) γ−1 =Λ′1(Nλs1),γ+1=Λ′1(λs1)≥γ−1. (22)
Then, for every , at each sensor , , we have:
liminfk→∞−1klogαi(k,γ)≥B0(γ),liminfk→∞−1klogβi(k,γ)≥B1(γ), (23)
where
B0(γ) =maxλ∈[0,1]Nγλ−max{NΛ0(λ),Λ0(Nλ)−|logr|}=⎧⎪ ⎪⎨⎪ ⎪⎩NI0(γ),γ∈(γ0,γ−0]NI0(γ−0)+Nλs0(γ−γ−0),γ∈(γ−0,γ+0)I0(γ)+|logr|,γ∈[γ+0,γ1) B1(γ) =maxλ∈[−1,0]Nγλ−max{NΛ1(λ),Λ1(Nλ)−|logr|}=⎧⎪ ⎪⎨⎪ ⎪⎩I1(γ)+|logr|,γ∈(γ0,γ−1]NI1(γ+1)+Nλs1(γ−γ+1),γ∈(γ−1,γ+1)NI1(γ),γ∈[γ+1,γ1).
###### Corollary 6
Let Assumptions 1-3 hold and consider the family of distributed detectors in (13) and (14) parameterized by detector thresholds . Then:
1. liminfk→∞−1klogPe,i(k,γ)≥min{B0(γ),B1(γ)}>0, (24)
and the lower bound in (24) is maximized for the point 333As we show in the proof, such a point exists and is unique. at which
2. Consider , and let:
thr(Λ0,N)=max{Λ0(Nλ∙)−NΛ0(λ∙),Λ0(1−N(1−λ∙))−NΛ0(λ∙)}, (25)
Then, when , each sensor with the detector threshold set to , is asymptotically optimal:
limk→∞−1klogPe,i(k,0)=NCind.
3. When , for , irrespective of the value of (even when .)
Figure 1 (left) illustrates the error exponent lower bounds and in Theorem 5, while Figure 1 (right) illustrates the quantities in (21). ( See the definition of the function in (36) in the proof of Theorem 5.) We consider sensors and a discrete distribution of over a 5-point alphabet, with the distribution under , and under . We set here
Corollary 6 states that, when the network connectivity is above a threshold, the distributed detector in (13) and (14) is asymptotically equivalent to the optimal centralized detector. The corresponding optimal detector threshold is . When is below the threshold, Corollary 6 determines what value of the error exponent the distributed detector can achieve, for any given . Moreover, Corollary 6 finds the optimal detector threshold for a given ; can be found as the unique zero of the strictly decreasing function on , see the proof of Corollary 6, e.g., by bisection on .
Remark. When , for , it can be shown that , and , for all . This implies that the point at which and are equal is necessarily zero, and hence the optimal detector threshold , irrespective of the network connectivity (even when .) This symmetry holds, e.g., for the Gaussian and Laplace distribution considered in Section IV.
Corollary 6 establishes that there exists a “sufficient” connectivity, say , so that further improvement on the connectivity (and further spending of resources, e.g., transmission power) does not lead to a pay off in terms of detection performance. Hence, Corollary 6 is valuable in the practical design of a sensor network, as it says how much connectivity (resources) is sufficient to achieve asymptotically optimal detection.
Equation (24) says that the distribution of the sensor observations (through LMGF) plays a role in determining the performance of distributed detection. We illustrate and explain by examples this effect in Section IV.
### Iii-B Proofs of the main results
We first prove Theorem (5).
###### Proof of Theorem 5.
Consider the probability of false alarm in (19). We upper bound using the exponential Markov inequality [32] parameterized by :
(26)
Next, by setting , with , we obtain:
αi(k,γ) ≤ E[eNkλxi(k)|H0]e−Nkλγ (27) = E[eNλ∑kt=1∑Nj=1Φi,j(k,t)Lj(t)|H0]e−Nkλγ. (28)
The terms in the sum in the exponent in (28) are conditionally independent, given the realizations of the averaging matrices , , Thus, by iterating the expectations, and using the definition of in (4), we compute the expectation in (28) by conditioning first on , :
E[eNλ∑kt=1∑Nj=1Φi,j(k,t)Lj(t)|H0] = E[E[eNλ∑kt=1∑Nj=1Φi,j(k,t)Lj(t)|H0,W(1),…,W(k)]] (29) = E[e∑kt=1∑Nj=1Λ0(NλΦi,j(k,t))].
Partition of the sample space. We handle the random matrix realizations , , through a suitable partition of the underlying probability space. Adapting an argument from [1], partition the probability space based on the time of the last successful averaging. In more detail, for a fixed , introduce the partition of the sample space that consists of the disjoint events , , given by:
As,k={∥Φ(k,s)−J∥≤ϵand∥Φ(k,s+1)−J∥>ϵ},
for , , and . For simplicity of notation, we drop the index in the sequel and denote event by , . for . Intuitively, the smaller is, the closer the product to is; if the event occurred, then the largest for which the product is still -close to equals . We now show that is indeed a partition. We need the following simple Lemma. The Lemma shows that convergence of is monotonic, for any realization of the matrices
###### Lemma 7
Let Assumption 3 hold. Then, for any realization of the matrices :
∥Φ(k,s)−J∥≤∥Φ(k,t)−J∥,for1≤s≤t≤k.
###### Proof.
Since every realization of is stochastic and symmetric for every , we have that and , and, so: . Now, using the sub-multiplicative property of the spectral norm, we get
∥Φ(k,s)−J∥ =∥(W(k)−J)⋯(W(t)−J)(W(t−1)−J)⋯(W(s)−J)∥ ≤∥(W(k)−J)⋯(W(t)−J)∥∥(W(t−1)−J)∥⋯∥(W(s)−J)∥.
To prove Lemma 7, it remains to show that , for any realization of . To this end, fix a realization of . Consider the eigenvalue decomposition , where is the matrix of eigenvalues of , and the columns of are the orthonormal eigenvectors. As is the eigenvector associated with eigenvalue , we have that where . Because is stochastic, we know that , and so
To show that is a partition, note first that (at least) one of the events necessarily occurs. It remains to show that the events are disjoint. We carry out this by fixing arbitrary , and showing that, if the event occurs, then , , does not occur. Suppose that occurs, i.e., the realizations are such that and . Fix any Then, event does not occur, because, by Lemma 7, Now, fix any Then, event does not occur, because, by Lemma 7, Thus, for any , if the event occurs, then , for , does not occur, and hence the events are disjoint.
Using the total probability law over , the expectation (29) is computed by:
E[e∑kt=1∑Nj=1Λ0(NλΦi,j(k,t))]=k∑s=0E[e∑kt=1∑Nj=1Λ0(NλΦi,j(k,t))IAs], (30)
where, we recall, is the indicator function of the event . The following lemma explains how to use the partition to upper bound the expectation in (30).
###### Lemma 8
Let Assumptions 1-3 hold. Then:
1. For any realization of the random matrices , :
N∑j=1Λ0(NλΦi,j(k,t))≤Λ0(Nλ),∀t=1,…,k.
2. Further, consider a fixed in . If the event occurred, then, for : | 2019-12-13 16:03:01 | {"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": 0.8845254182815552, "perplexity": 844.0785177859286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00273.warc.gz"} |
https://www.neetprep.com/question/69635-m-mn-andmp-masses-ofXAZ-nucleus-neutron-proton-respectivelyamAZmnZmpb-mAZmnZmpc-mAZmpZmnd-mAZmnZmp/126-Physics--Nuclei/703-Nuclei | # NEET Physics Nuclei Questions Solved
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If m, ${\mathrm{m}}_{\mathrm{n}}$ and ${\mathrm{m}}_{\mathrm{p}}$ are the masses of ${}_{\mathrm{Z}}\mathrm{X}^{\mathrm{A}}$ nucleus, neutron and proton respectively
(a) $\mathrm{m}<\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{m}}_{\mathrm{n}}+{\mathrm{Zm}}_{\mathrm{p}}$ (b) $\mathrm{m}=\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{m}}_{\mathrm{n}}+{\mathrm{Zm}}_{\mathrm{p}}$
(c) $\mathrm{m}=\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{m}}_{p}+{\mathrm{Zm}}_{\mathrm{n}}$ (d) $\mathrm{m}>\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{m}}_{\mathrm{n}}+{\mathrm{Zm}}_{\mathrm{p}}$
(a) The mass of nucleus formed is always less than the sum of the masses of the constituent protons and neutrons i.e. $\mathrm{m}<\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{m}}_{\mathrm{n}}+{\mathrm{Zm}}_{\mathrm{p}}$.
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Crack NEET with Online Course - Free Trial (Offer Valid Till September 22, 2019) | 2019-09-20 05:31:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.8802827000617981, "perplexity": 6511.068382619743}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00203.warc.gz"} |
https://www.physicsforums.com/threads/how-to-construct-a-table-of-all-the-real-valued-dirichlet-characters.1049776/#post-6852498 | # How to construct a table of all the real-valued Dirichlet characters?
Math100
Homework Statement:
Let ## G ## be the group of reduced residue classes modulo ## 21 ##. Construct a table showing the values of all the real-valued, and one of the complex-valued, Dirichlet characters modulo ## 21 ##. (In your table, use ## \omega ## for ## e^{\pi i/3} ##.)
Relevant Equations:
Definition: Dirichlet characters. Let ## G ## be the group of reduced residue classes modulo ## k ##. Corresponding to each character ## f ## of ## G ##, we define an arithmetical function ## \chi=\chi_{f} ## as follows:
## \chi(n)=f(\hat{n}) ## if ## (n, k)=1 ##,
## \chi(n)=0 ## if ## (n, k)>1 ##.
The function ## \chi ## is called a Dirichlet character modulo ## k ##. The principal character ## \chi_{1} ## has the properties ## \chi_{1}(n)=1 ## if ## (n, k)=1 ##, and ## \chi_{1}(n)=0 ## if ## (n, k)>1 ##.
Corresponding to each character ## f_{i} ## of ## G ##, we define an arithmetical function ## \chi_{i} ## as follows:
## \chi_{i}(n)=f_{i}(\hat{n}) ## if ## (n, k)=1 ##,
## \chi_{i}(n)=0 ## if ## (n, k)>1 ##.
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 2 & 4 & 5 & 8 & 10 & 11 & 13 & 16 & 17 & 19 & 20 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\
\hline \chi_{3}(n) & 1 & 1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & -1 & -1 \\
\hline \chi_{4}(n) & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 \\
\hline \chi_{5}(n) & 1 & \omega & \omega^{2} & -\omega & -1 & -\omega^{2} & -\omega^{2} & -1 & -\omega & \omega^{2} & \omega & 1 \\
\hline \chi_{6}(n) & 1 & \omega^{2} & -\omega & \omega^{2} & 1 & -\omega & -\omega & 1 & \omega^{2} & -\omega & \omega^{2} & 1 \\
\hline \chi_{7}(n) & 1 & -\omega & \omega^{2} & -\omega & 1 & \omega^{2} & \omega^{2} & 1 & -\omega & \omega^{2} & -\omega & 1 \\
\hline \chi_{8}(n) & 1 & -\omega^{2} & -\omega & \omega^{2} & -1 & \omega & \omega & -1 & \omega^{2} & -\omega & -\omega^{2} & 1 \\
\hline \chi_{9}(n) & 1 & \omega & \omega^{2} & \omega & -1 & \omega^{2} & -\omega^{2} & 1 & -\omega & -\omega^{2} & -\omega & -1 \\
\hline \chi_{10}(n) & 1 & \omega^{2} & -\omega & -\omega^{2} & 1 & \omega & -\omega & -1 & \omega^{2} & \omega & -\omega^{2} & -1 \\
\hline \chi_{11}(n) & 1 & -\omega & \omega^{2} & \omega & 1 & -\omega^{2} & \omega^{2} & -1 & -\omega & -\omega^{2} & \omega & -1 \\
\hline \chi_{12}(n) & 1 & -\omega^{2} & -\omega & -\omega^{2} & -1 & -\omega & \omega & 1 & \omega^{2} & \omega & \omega^{2} & -1 \\
\hline
\end{array}
The table above is the answer/solution for this problem.
I know that ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, which means there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. But I do not know how to get ## \chi_{2}(n)... ## except ## \chi_{1}(n) ##. May anyone please tell me how to get these values?
Homework Helper
Gold Member
Your notation has some problems that prevent us making sense of your question:
Corresponding to each character ## f_{i} ## of ## G ##, ....
What do you mean by this? The elements of ##G## are not characters, they are effectively modulus classes, ie elements of ##\mathbb Z_k##. It is the ##\chi## that are characters.
## \chi_{i}(n)=f_{i}(\hat{n}) ## if ## (n, k)=1 ##,
What do you mean by this? You have not defined ##\hat n##, and ##f## is not a function, as you have presented it, but a modulus number.
It depends on what you already know about characters/representations of finite groups (in your case abelian). One thing you can do is find normal subgroups (in the abelian case that is all subgroups). Say ##H## is a normal subgroup of ##G##, then consider the group ##G/H## it is smaller and it is easier to find characters of it. If ##f## is a character of ##G/H##, when you compose it with the projection ##G\rightarrow G/H## you get a character of ##G##, which is trivial on ##H##.
For example look at the group generated by ##5##, then ##H=\langle 5\rangle = \{1,4,5,16,17,20\}##. The the factor ##G/H## has two elements. It has two characters. One is the trivial (all values are 1). This will pull back to ##\chi_1##. The other is the character that gives value -1 to the non-identity element of the group. When you pull back that one you get ##\chi_2##.
Playing around like that you can get many characters, may be all. There are also the orthogonality relations the characters satisfy. So, if you already know some you can use easily these relations to find the others.
Math100
Your notation has some problems that prevent us making sense of your question:
What do you mean by this? The elements of ##G## are not characters, they are effectively modulus classes, ie elements of ##\mathbb Z_k##. It is the ##\chi## that are characters.
What do you mean by this? You have not defined ##\hat n##, and ##f## is not a function, as you have presented it, but a modulus number.
This is my question, too. I do not know what they mean, I just posted them under the relevant equation(s) just because my book has these definitions. Since these definitions are preventing people to make sense of my question, then please ignore them. How should I find those values then, starting from ## \chi_{2}(n)... ##?
Your notation has some problems that prevent us making sense of your question:
What do you mean by this? The elements of ##G## are not characters, they are effectively modulus classes, ie elements of ##\mathbb Z_k##. It is the ##\chi## that are characters.
What do you mean by this? You have not defined ##\hat n##, and ##f## is not a function, as you have presented it, but a modulus number.
The notations are fine. It says that ##f_i## is a character of ##G##. For some reason you think it is a group element, but it doesnt say that.
How should I find those values then, starting from ## \chi_{2}(n)... ##?
I told you! Did you not read my post?
Math100
I told you! Did you not read my post?
I already did. But I still don't understand. How should I find normal subgroups?
My be it would help if you told us what you already know. For example do you know what the structure of finite abelian groups is? Can you figure out how the group of your example decomposes as a product of cyclic groups? Do you know what the characters of cyclic groups are? For instance your group ##G## is isomorphic to ##C_6\times C_2##, two cyclic groups of order 6 and 2. Does this help?
Math100
My be it would help if you told us what you already know. For example do you know what the structure of finite abelian groups is? Can you figure out how the group of your example decomposes as a product of cyclic groups? Do you know what the characters of cyclic groups are? For instance your group ##G## is isomorphic to ##C_6\times C_2##, two cyclic groups of order 6 and 2. Does this help?
No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of ## G ## is isomorphic to ## C_{6}\times C_{2} ##, which are the two cyclic groups of order ## 6 ## and ## 2 ##. But how did you get these?
No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of ## G ## is isomorphic to ## C_{6}\times C_{2} ##, which are the two cyclic groups of order ## 6 ## and ## 2 ##. But how did you get these?
It would be easier if you told us what you do know. Why are looking at this question?
Math100
It would be easier if you told us what you do know. Why are looking at this question?
Since ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. So ## G ## can be generated by order ## 2 ## or ## 6 ##. And we have ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ## and ## \chi(k)^{6}=\chi(k^{6})=\chi(1)=1 ##. But how should I find out which elements are of order ## 2 ## and which ones are of order ## 6 ## from group ## G ##?
Homework Helper
Gold Member
Since ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. So ## G ## can be generated by order ## 2 ## or ## 6 ##. And we have ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ## and ## \chi(k)^{6}=\chi(k^{6})=\chi(1)=1 ##. But how should I find out which elements are of order ## 2 ## and which ones are of order ## 6 ## from group ## G ##?
Any group has only one element of order one, which is its identity. It's easy to see which element of G that is.
Only one element of ##C_6\times C_2## can have order two: ##(1_6,(-1)_2)## where ##1_6## is the identity of ##C_6## and ##(-1)_2## is the sole non-identity element of ##C_2##. So you only need to find one element out of the eleven non-identity elements of G, that has order two. The other ten elements all have order six.
Which of the elements of G gives 1 modulo 21 when you square it?
Math100
Any group has only one element of order one, which is its identity. It's easy to see which element of G that is.
Only one element of ##C_6\times C_2## can have order two: ##(1_6,(-1)_2)## where ##1_6## is the identity of ##C_6## and ##(-1)_2## is the sole non-identity element of ##C_2##. So you only need to find one element out of the eleven non-identity elements of G, that has order two. The other ten elements all have order six.
Which of the elements of G gives 1 modulo 21 when you square it?
## 1, 8, 13, 20 ##
## 1^{2}\equiv 1\pmod {21}, 8^{2}\equiv 1\pmod {21}, 13^{2}\equiv 1\pmod {21}, 20^{2}\equiv 1\pmod {21} ##
Homework Helper
Gold Member
## 1, 8, 13, 20 ##
## 1^{2}\equiv 1\pmod {21}, 8^{2}\equiv 1\pmod {21}, 13^{2}\equiv 1\pmod {21}, 20^{2}\equiv 1\pmod {21} ##
Correct.
BTW, that identifies an error in my post. I should have said that three elements in ##C_6\times C_2## have order two:$$(1_6,g_2),\ \ (b, 1_2),\ \ (b, g_2)$$ where ##g_2## is the sole non-identity element of ##C_2##. and ##b## is the sole non-identity element of ##C_6## whose square is ##1_6##. You can map those three order-two elements of ##C_6\times C_2## to 8, 13 and 20 in G in any way you like.
My apologies.
Math100
Math100
Correct.
BTW, that identifies an error in my post. I should have said that three elements in ##C_6\times C_2## have order two:$$(1_6,g_2),\ \ (b, 1_2),\ \ (b, g_2)$$ where ##g_2## is the sole non-identity element of ##C_2##. and ##b## is the sole non-identity element of ##C_6## whose square is ##1_6##. You can map those three order-two elements of ##C_6\times C_2## to 8, 13 and 20 in G in any way you like.
My apologies.
So now we have that ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ## for all ## k\{8, 13, 20\} ## and ## \chi(k)^{6}=\chi(k^{6})=\chi(1)=1 ## for all ## k\{2, 4, 5, 10, 11, 16, 17, 19\} ##. And this implies that ## \chi(n)=8, 13, 20 ## can either be ## -1, 1 ##. But how should I find out the order of ## -1, 1 ## for each ## \chi_{1}(n),...\chi_{12}(n) ## in ## n=8, 13, 20 ##? For example, why does ## \chi_{1}(8)=1, \chi_{2}(8)=-1, \chi_{3}(8)=1, \chi_{4}(8)=-1, \chi_{5}(8)=-1, \chi_{6}(8)=1, \chi_{7}(8)=1, \chi_{8}(8)=-1, \chi_{9}(8)=-1, \chi_{10}(8)=1, \chi_{11}(8)=1, \chi_{12}(8)=-1 ##? And how should I find other values in ## \chi(2), \chi(4), \chi(5), \chi(10), \chi(11), \chi(16), \chi(17), \chi(19) ##? The question asks to use ## \omega ## for ## e^{\pi i/3} ## but how should I do this? | 2023-03-26 12:00:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6658971309661865, "perplexity": 1181.5519448405546}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00479.warc.gz"} |
https://gamedev.stackexchange.com/questions/184849/image-of-the-room-appearing-incorrectly-for-no-apparent-reason/187833 | # Image of the room appearing incorrectly for no apparent reason
In the game, when I call the room (room_Corrida) from the room of the selection menu, it was for that to appear:
It turns out this is how the room appears:
The effect is as if it had a white rectangle the same size as the room, with less than 100% opacity.
When I start the room immediately (I put it first), it appears normally. The room also appears normally when I have the room called directly after the start menu. It only appears wrongly when I call it after the game selection menu.
Code that makes the room appear wrong (from the selection menu):
if(mouse_x>=1601 && mouse_x<=1917 && mouse_y>=762 && mouse_y<=1077){
room_goto(room_Corrida);
}
Code that makes the room appear correctly (from the home menu):
if(mouse_x>=room_width/7*1 && mouse_x<=room_width/7*2-1 && mouse_y>=room_height/10*1 && mouse_y<=room_height/10*2-1){
room_goto(room_Corrida);
}
Are you using draw? if yes, try using draw_set_colour(c_white) before drawing this whole text. I think the problem could be that you are changing the draw colour somewhere in the code but not changing back to white before drawing the whole ranking on the room. | 2021-04-11 18:59:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32069769501686096, "perplexity": 2000.537843867057}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038064898.14/warc/CC-MAIN-20210411174053-20210411204053-00105.warc.gz"} |
https://math.stackexchange.com/questions/940478/how-can-i-prove-that-sum-k-1-infty-frac2-sqrtkk-1kk4-is-conver | How can I prove that $\sum_{k=1}^{\infty}\frac{(2\sqrt[k]k-1)^k}{k^4}$ is convergent?
I'm trying to prove that the sum $\sum_{k=1}^{\infty}\frac{(2\sqrt[k]k-1)^k}{k^4}$ is convergent.
I've tried Cauchy's root test - but I get the limit to be 1, so the test is inconclusive. I also tried using d'Alemberts ratio test, but that didn't provide any information either. I can't use the integral test, because I can't find the antiderivative. The idea I had was to compare it to $\sum_{k=1}^{\infty} \frac{1}{k^2}$. I know this series converges and I'm pretty confident that $\frac{(2\sqrt[k]k-1)^k}{k^4}\le \frac{1}{k^2}$ for all $k \ge 1$ but I don't know how to prove that.
I would be really grateful for any ideas!
Notice $$\sqrt[k]{k^2} - (2\sqrt[k]{k} - 1) = (\sqrt[k]{k} - 1)^2 \ge 0 \implies \sqrt[k]{k^2} \ge 2\sqrt[k]{k} - 1\tag{*1}$$ This leads to $$\sum_{k=1}^N\frac{\left(2\sqrt[k]{k}-1\right)^k}{k^4} \le \sum_{k=1}^N\frac{\left(\sqrt[k]{k^2}\right)^k}{k^4} = \sum_{k=1}^N\frac{1}{k^2} \le \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$ This means the partial sums are bounded from above. Since the terms are non-negative, the series converges to some number no more than $\displaystyle\;\frac{\pi^2}{6}$.
We have: $$k^{1/k} = e^{\frac{\log k}{k}} \leq \frac{1}{1-\frac{\log k}{k}}$$ hence: $$2k^{1/k}-1 \leq \frac{1+\frac{\log k}{k}}{1-\frac{\log k}{k}}$$ and for any $k$ big enough: $$2k^{1/k}-1 \leq \exp\left(\frac{5}{2}\frac{\log k}{k}\right)$$ so: $$(2k^{1/k}-1)^k \leq k^{5/2}$$ ensuring convergence. | 2019-05-19 20:56:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9694414138793945, "perplexity": 51.51163958861129}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00486.warc.gz"} |
https://math.stackexchange.com/questions/2577643/find-area-of-surface-enclosed-with-given-curve/2577670 | # Find area of surface enclosed with given curve
Find the area of surface with given equation $$\big(x^2 + y^2\big)^2 = a(x^3+y^3)$$
I tried to use polar coordinates $x=r\cos\alpha$ and $y=r\sin\alpha$ and so $$r = a(\cos\alpha + \sin\alpha)\bigg(1-{\sin2\alpha\over{2}}\bigg)$$ Maybe anyone can suggest better way?
• do you mean the area of the region enclosed by the given curve? – daulomb Dec 23 '17 at 10:32
• oh yes, you are right – Spike Bughdaryan Dec 23 '17 at 10:38
• usage of polar coordinates seems okey you need to determine the upper and lover limits of $\theta$. – daulomb Dec 23 '17 at 10:45
• I think it's a best way here. – Michael Rozenberg Dec 23 '17 at 10:47
• @MichaelRozenberg Where "here"? – DonAntonio Dec 23 '17 at 11:07
Okay i solved, with $\alpha \in [-\pi/4, 3\pi/4]$
and $${a^2\over2}\int_{-\pi\over4}^{3\pi\over4}(\sin\alpha + \cos\alpha)^2 (1-{\sin2\alpha \over 2})^2 d\alpha = \bigg(\dfrac{\cos\left(6x\right)+9\sin\left(4x\right)-9\cos\left(2x\right)-48\sin^4\left(x\right)+48\cos^4\left(x\right)-96\cos^2\left(x\right)+60x}{96}\bigg)_{-\pi\over4}^{3\pi\over4} = {5\pi a^2\over 16}$$
Why not directly the double integral (in polar coordinates)?
$$\int_0^{2\pi}\int_0^{a(\cos t + \sin t)\bigg(1-{\sin2t\over{2}}\bigg)}r\,dr\,dt=\frac{a^2}2\int_0^{2\pi}(\cos t+\sin t)^2\left(1-\frac{\sin2t}2\right)^2dt=$$
$$=\frac{a^2}2\int_0^{2\pi}\left(1+\sin2t\right)\left(1-\sin2t+\frac{\sin^22t}4\right)dt=\frac{a^2}2\int_0^{2\pi}\left[1-\frac34\sin^22t+\frac{\sin^32t}4\right]dt=$$
$$=\frac{a^2}4\int_0^{4\pi}\left[1-\frac34\sin^2u+\frac{\sin^3u}4\right]du=$$
$$=\frac{a^2}4\left(4\pi-\frac384\pi+\frac14\overbrace{\int_0^{4\pi}\left(\sin u-\sin u\cos^2u\right)du}^{=0}\right)=\frac{5\pi a^2}8$$
Seeing your solution, at least one of us two is wrong. I wouldn't be surprised at all if it is me...yet in your expression you didn't divide by two $\;\sin2\alpha\;$ ... Check this.
Added. Since it must be $\;t\in\left[-\frac\pi4,\,\frac{3\pi}4\right]\;$ for $\;r\;$ to be positive, the above integral must be changed accordingly to these limits. The outcome indeed is $\;\cfrac{5\pi a^2}{16}\;$
• oh yes, forgot about it, i'm not sure if my solution is all the way right, but the answer is $5\pi a^2 \over 16$ and also we need to chose the $\alpha$ where $sin\alpha + cos\alpha > 0$ and so $[−π/4,3π/4]$ – Spike Bughdaryan Dec 23 '17 at 11:14
• Didn't mean the restriction, but from $r=a(cosα+sinα)(1−{\sin2α\over2})$ we need to chose the angle $\alpha$ , where $r>0$ – Spike Bughdaryan Dec 23 '17 at 11:40
• @SpikeBughdaryan You are completely right, of course... – DonAntonio Dec 23 '17 at 11:42 | 2019-07-19 02:10:32 | {"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": 0.8638865947723389, "perplexity": 483.7088443152698}, "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-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00201.warc.gz"} |
https://titanquestfans.net/index.php?PHPSESSID=ako1eac8m9hfo67ef5b9522qu3&action=profile;area=showposts;u=256 | ### Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
### Messages - koderkrazy
Pages: [1] 2 3 ... 5
1
##### General Discussion / Re: Photo with my favorite monster ^-^
« on: 21 June 2018, 17:41:55 »
img
why do you feel uncomfortable around monsters?
2
##### General Discussion / Photo with my favorite monster ^-^
« on: 21 June 2018, 11:49:20 »
This is my desktop wallpaper
3
##### Other Modifications / Re: [MOD] Battle Nymphs
« on: 20 June 2018, 16:21:48 »
I couldn't see the difference for a minute. Left one kinda fumbles.
BTW how did you make gif? recorded video then ran through some software?
4
##### Anniversary Edition - General discussion / Re: Filter all types of items and more...
« on: 20 June 2018, 06:59:28 »
Quote
2. Invisible as ghost - you can walk pass enemies without getting detected.
How did you achieve this? Is it possible to be tuned to an actual skill in a mod?
I think this is one of the development/debug features(like god mode, invincibility). There are properties in dbr like hiddenFromCombat, invincible but they don't affect pc or monsters.
In the mod made provision to turned it on.
5
##### General Discussion / Re: [AE 1.56 theory] Scatter Shot Alternative Stonespeaker Build
« on: 19 June 2018, 23:14:07 »
edit: i just looked at Hati drop chance and it says "no bosses seem to drop this item".. that explains it.. maybe there's a chance it's given as a quest reward.. but the chance of getting it must be slim
or tq-db isn't updated and Hati now drops in the latest versions
Actually Hati doesn't drop at all!!
It's not there in any loot tables. There are 15 loot tables for ranged weapons in Ragnarok, and Hati is in non of them.
To be sure, I searched entire 1.54 and 1.56 database.arz(57,151 files)
Here is the only way you could get it
Spoiler for Hiden:
@Hector one more use of Blacksmith tool
6
##### Anniversary Edition - General discussion / Re: Filter all types of items and more...
« on: 19 June 2018, 21:04:51 »
Thanks for the support bro! Really helps me keep going
Mods in this collection are simply trying to overcome basic hurdles which players have struggled over many years.
The ideas is to make game more fun to play.
Note: Invisibility doesn't give you much advantage. You can't damage enemies by any means. It's just to have nice quite stroll around the areas.
I would like to add to what @Hector said about Legendary Craftsmanship. There is more unfairness to it...
1) Your weapon could actually get screwed up. It doesn't actually upgrade your weapon! It gives you new weapon of same type with prefix, suffix, and new seed. New seed means your weapon stats change. e.g. when I upgraded my precious Apollo's Will(with 117% pierce!), my upgraded ring got only 96% pierce. I cursed myself... and I never found ring with such good stats again.
2) Also you get to upgrade when the game is almost finished. I mean, you may not find upgraded weapons useful in your next build.
7
##### Anniversary Edition - General discussion / Re: Filter all types of items and more...
« on: 19 June 2018, 15:04:14 »
is this like the cheat TQdefiler used to offer? where whites and yellows don't drop making greens and above quality items drop in abundance? if so, i don't like it
party pooper me
As name suggests it is a filter for dropped items.
Items do drop. It's not TQDefiler type.
Just try it out
8
##### Anniversary Edition - General discussion / Re: Filter all types of items
« on: 19 June 2018, 12:26:30 »
Looks good.
To be honest, it is completely beyond me why green item filtering at the least, was not added to the AE, like in Grim Dawn.
Also the implementation is not hard at all. One line of comparison for each item was enough.
May be policy makers decided to keep farming uneasy.
9
##### Other Modifications / Re: [MOD] TQ Fun - Collection of mods
« on: 19 June 2018, 12:17:26 »
i only want the Typhon Zeus shut up mod.. nothing more
edit: or only talks after last Typhon is dead
Done!
Haven't implemented last typhon dead part though.
Also the second trainer for the pre-AE version ( https://www.cheathappens.com/12762-PC-Titan_Quest_Immortal_Throne_cheats ) have invisibility option.
Using this was much fun, so consider this too.
Done!
10
##### Other Modifications / Re: [Tool] TQ Blacksmith
« on: 18 June 2018, 08:52:36 »
So let me get this straight.. just putting these lines in the text is all that needed? for instance, if I want another prefix/suffix on any part of equipment, I just have to find the related lines for that and put them in the text as well, right?..
Yes you are right. But adding item lines to those data txt files will only work for prefixes and suffixes. For other items you can add only rare, eipc and legendary items' lines, common/broken items will be ignored.
Quote
if that's the case, maybe you can direct me to a place where I can find all the values so that I can put the ones that I need into the text and be done with it.. I don't want you to bear extra workload just for one person asked it, that's why I'm making this post..
There is only one place all the values could be found, <installDir>\Database\database.arz. But compiling data from it not that easy.
tq-db.net is not complete, it doesn't have common and broken items as well.
I've extracted such data by writing specific code.
Here are all the available suffixes and prefixes which are in use(referred by at least one loot table):
https://mega.nz/#F!eOh3mawA!NJBoyOrEvCOvztxijQH3aw
AllPrefixes 928.txt
AllSuffixes 774.txt
You could pick your lines from here to put in TQ_Blacksmith data files.
If you don't find your affix in above files:
Spoiler for Hiden:
Get it from TQValut. It doesn't let you copy text. But just select the path and drag it to notepad or any editor.
Quote
For instance, if I want "impetus" instead , that's tire 3 quality of slow resistance so I have to change 02.dbr to 03.dbr and keep the remaining parts untouched, right?
I would suggest just add a new line for Impetus.
Quote
https://www.tq-db.net/affix/prefixes
There are some prefixes in the list with a note attached as "multiple possibilities". I've never seen a "hardened" prefix offer bleeding resistance instead of pierce resistance for example.. Do you know what the deal with that is? Is it something manually editable like the completion bonuses on charms and relics?
On tq-db.net there is only one page for prefix. And listed prefix doesn't say it belongs to what item type e.g sword, ring, helm, or shield.
So if there are two prefixes:
1) prefix for helm - offers 8% bleeding resist - has name 'Hardened'
2) prefix for shield - offers 8% pierce resist - also has name 'Hardened'
So if these have same name. Then tq-db says "multiple possibilities" and list them in one entry.
These are three prefixes which have same name 'Hardened'
\records\item\lootmagicalaffixes\prefix\default\defensive_resistbleed_01.dbr
\records\item\lootmagicalaffixes\prefix\default\defensive_resistpierce_01.dbr
\records\xpack2\item\lootmagicalaffixes\prefix\default\defensive_resistbleed_01.dbr
I don't know to which item type they get applied, but definitely they are different.
11
##### Other Modifications / Re: [Tool] TQ Blacksmith
« on: 17 June 2018, 22:42:49 »
I enjoyed creating the utility more than using it
If its possible, could you make a little update so that we can choose not to put any prefix or suffix to epic and legendary items? or choose to put only a prefix or a suffix and create the item that way?
Ahhh, good catch. It should allow to create plain version of items as well. I'll change it.
And another thing.. I know that you mentioned common affixes are to many to integrate into the program but some of the suffixes such as "momentum" are very crucial IMO and I would have preferred them over any other godly and rare affixes if I could.. so I'd like to know if you can accept a request about a specific prefix/suffix list to add into your program..
If you are concerned about only common(yellow) prefix and suffixes then I'll add yellow affixes to the data files.
Also there is a quick fix:
Spoiler for Hiden:
These affixes are only for armor.
Code: [Select]
ARM=Rare=15=RECORDS\XPACK\ITEM\LOOTMAGICALAFFIXES\SUFFIX\DEFAULT\DEFENSIVE_RESISTSLOW_02.DBR=xtagSuffix030=of MomentumLEG=Rare=15=RECORDS\XPACK\ITEM\LOOTMAGICALAFFIXES\SUFFIX\DEFAULT\DEFENSIVE_RESISTSLOW_02.DBR=xtagSuffix030=of MomentumHEAD=Rare=15=RECORDS\XPACK\ITEM\LOOTMAGICALAFFIXES\SUFFIX\DEFAULT\DEFENSIVE_RESISTSLOW_02.DBR=xtagSuffix030=of MomentumTORSO=Rare=15=RECORDS\XPACK\ITEM\LOOTMAGICALAFFIXES\SUFFIX\DEFAULT\DEFENSIVE_RESISTSLOW_02.DBR=xtagSuffix030=of Momentum
If you are concerned about all other item types as well then, I'll see if I can add extensive items mode to the utility. But it'll take some time. It's not the implementation, it's compiling data of 8-10K items that takes time.
12
##### Other Modifications / Re: [MOD] TQ Fun - Collection of mods
« on: 17 June 2018, 20:54:19 »
I'll try to add mods in Defiler. Also, let me know if you have any ideas.
How about you bring back all the mods which were in Defiler?
If it's possible of course.
Yeah, actually I meant the same... Cheat Engine is way too powerful than modifying .dlls.
And it's possible and I am on it
13
##### Anniversary Edition - General discussion / Filter all types of items and more...
« on: 17 June 2018, 17:35:53 »
1. Filter dropped items - Following items could be filtered now
Broken
White
Yellow
Green and MI
Epic
Legendary
Potion
Relic
Quest
Artifact
Formula
2. Invisible as ghost - you can walk pass enemies without getting detected.
3. Set/freeze time - you can set game to any time. Also, you can keep time running or freeze it.
Some of the mods which were there in Defiler:
4. Show Damage - shows all damages instead of just critical hits. Only damage over time, and pet damage is not shown.
Enable 'Display critical damage' setting in game.
5. Show Pet Damage - shows damage done by pets.
6. Mute Zeus - no speech or text dialog after you kill Typhon.
7. No gold drops - gold will not drop from chests or monsters.
8. Set potion stack limit to 200 - health and energy potions can stack up to 200.
For more details see https://titanquestfans.net/index.php?topic=504.0
14
##### Other Modifications / [MOD] TQ Fun - Collection of mods
« on: 17 June 2018, 17:29:43 »
All, I've created cheat table(.CT) for following modifications. It is way better than modifying game files
1. Filter dropped items - Following items could be filtered now
Broken
White
Yellow
Green and MI
Epic
Legendary
Potion
Relic
Quest
Artifact
Formula
2. Invisible as ghost - you can walk pass enemies without getting detected.
3. Set/freeze time - you can set game to any time. Also, you can keep time running or freeze it.
Some of the mods which were there in Defiler:
4. Show Damage - shows all damages instead of just critical hits. Only damage over time, and pet damage is not shown.
Enable 'Display critical damage' setting in game.
5. Show Pet Damage - shows damage done by pets.
6. Mute Zeus - no speech or text dialog after you kill Typhon.
7. No gold drops - gold will not drop from chests or monsters.
8. Set potion stack limit to 200 - health and energy potions can stack up to 200.
Spoiler for Hiden:
============================================
http://bit.ly/2I3VavB
File: TQFunV2.ct
Tested Versions 1.54 -1.56
Cheat Engine 6.8 [13.6mb]
Mega Nz: https://mega.nz/#!3fBgmYiC!oxae0l9QULJAGxfF8O9oz4HCIweLYwBkGP1mfb7JcP8
Note: Some antivirus like MalwareBytes block this as PUP.(potentially unwanted program) or RiskWare. You need to add installer as well as installed exe to Exclusions.
[USAGE]
============================================
- Run the GAME.
- Run CE 6.7 or greater.
- Load the game process via CE.
- Load this Table - TQFunV1 .ct .
- Use compact UI mode for convenience. see above image
- Activate scripts by clicking on its box [X].
- You can set time by double clicking on time. To freeze it check the box[X] in front of the cheat.
- For filter dropped items, you can choose any number of item types. Also in game use 'x' key to filter them.
- Go back to the game and have fun.
[Note]
========================================
I am not doing below mods because, either they've already been done here or not feasible.
- infinite player stats e.g. exp, health, energy, gold, skill points etc..
- super flying bodies
- increase chest items
- undo mastery
- set current fountain
- update item stats e.g. bonus on charms
- increase drop chances
Also, let me know if you have any ideas.
15
##### Other Modifications / Re: [MOD] Glorious Darkness
« on: 17 June 2018, 16:43:22 »
BTW I am working on non intrusive solution for such mods, where game files won't be modified.
Waiting for this and of course many thanks for mods.
It's done see TQFun mods.
Pages: [1] 2 3 ... 5
Forum Live Projects TitanQuest Information Live Chat Legion of Champions Challenges Registration Agreement Forum Rules Deities Mod Privacy Policy Discord Hamunaptra Mod Contact Time Zone on Profile | 2018-06-21 22:12:38 | {"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": 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": 0.20425879955291748, "perplexity": 4987.860345380089}, "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-2018-26/segments/1529267864300.98/warc/CC-MAIN-20180621211603-20180621231603-00058.warc.gz"} |
https://math.stackexchange.com/questions/2068291/why-do-we-call-sqrt-1-imaginary-and-1-real?noredirect=1 | # Why do we call $\sqrt{-1}$ imaginary and $-1$ real?
Looking at the numbers I can see in nature only positive real numbers. Because many problems couldn't be solved using only positive real numbers a new number set, called "negative numbers" was introduced. Later on, for similar reasons, complex numbers were introduced and the term of "imaginary".
Can anyone please explain to me why $-1$ is "more real" than $i$? For me both seem to be a convention, a name given to the result of a function or a set of functions to the positive real numbers.
Disclaimer: I'm not a mathematician, I'm just trying to better understand some basic concepts and conventions in math
• This is a philosophical question. See also Does a negative number really exist? and Do complex numbers really exist? – Alex Silva Dec 22 '16 at 9:59
• "For me both seem to be a convention": exactly that. You are adding your own interpretation "more real", which is not intended. Had the qualifier "dense" be used f.i., you would be asking "is $-1$ more dense than $i$ ?" – Yves Daoust Dec 22 '16 at 10:10
• Note that negative integers are defined before reals are defined. – Yves Daoust Dec 22 '16 at 10:12
• It's just an unfortunate naming. And it's extremely narrow minded to say only natural numbers are found in nature. Most natural concepts reflect rationals & reals, a lot of oscillatory phenomena are naturally described with complex numbers, and quantum mechanics is complex on the fundamental physical level. I extremely recommend this entire series of videos: youtube.com/watch?v=T647CGsuOVU – orion Dec 22 '16 at 10:52
• This SMBC comes to mind. – Asaf Karagila Dec 22 '16 at 16:04
In mathematics, every modifier, such as "continuous, real, rational, ...", is reassigned meaning, lest there be unnecessary confusion such as "your 'continuous" is different than mine" that wastes time. So, if you come across a terminology in some math book, find its definition within the book; do not take a dictionary and check the meaning of the term...
I am not sure if you are asking about the origin of the reason why the word "imaginary" was chosen. If not, then leave the "secular" meaning of the word alone. If you do this, then it is crystal clear that "imaginary number" is no more than a word used to name the relatives of $\sqrt{-1}$. Note also that "positive" and "negative" in math have nothing to do with any sentiments.
Logically, you can call the real numbers "imaginary" and the imaginary numbers "real". Then in your book, the theorem "the square of every real number is $\geq 0$" becomes "the square of every imaginary number is $\geq 0$". Noticed? The name thing is rather superficial :), which in math serves as a mnemonic device and a shortcut of reference.
Natural numbers is a very intuitive mathematical construction, corresponding to counting, and this is why they are termed natural. Together with natural numbers, addition appears.
Then the need appears to solve problems like $3+x=7$, how much is $x$ ? This is solved by means of subtraction. But then you face frustrating problems such as $7+x=3$, how much is $x$ ? To cope with these, negative numbers are introduced, forming the integers. (A close friend of entire.)
The next step is to look at multiplication, i.e. repeated addition. All is fine until you want to solve problems like $4\times x=24$, how much is $x$ ?, then $7\times x=17$, how much is $x$ ? This is how division and the rational numbers are introduced. (From ratio.)
The ancient Greeks once discovered that rational numbers are not all, much to their resentment, when they asked the question $x\times x=2$, how much is $x$ ? Then came the real numbers. (Possibly evoking the continuous characteristic of our real world.)
Another step was reached in the Middle-Age when mathematicians started to deal with the equation $x\times x=-1$, also annoyingly unsolvable, and the imaginary and complex numbers were introduced.
You may attach a "romantic" meaning to these terms, but this is not the intent. On the opposite, they are conventional and universally adopted words with a well-defined understanding, and probably a more mnemonic intent.
Even though some of these number categories may look somewhat artificial, there are cases where they come handy just for intermediate results in solving a real-world problem. A famous example is the need for complex numbers to find the three real roots of some cubic equations.
We call them imaginary because people made fun of the person who first used them. They mocked his methods, saying he used "imaginary" numbers. Seems like it stuck.
Ultimately, they're no more fictitious than anything else in mathematics.
• Nor less fictious :) – Yves Daoust Dec 22 '16 at 10:15
For humans, certain abstract objects are more natural than others. For example, natural numbers like $1$, $2$, and $3$ are second nature to us. Rational numbers are useful in the context of "relative" size. Three nickels are $\frac{15}{100}$ as valuable as a dollar. Real numbers are intuitive as well; they can be used to model "continuous" quantities like length and area. Negative numbers can be used to model things like debt and (cold) temperature.
Conversely, complex numbers are not so intuitive. We can't actually "see" $\sqrt{-1}$. Of course, complex numbers are present in the real world; they are used, for example, to mathematically model quantum theory. However, for non-quantum physicists, $\sqrt{-1}$ doesn't mean anything concretely. It is detached from reality, in a sense.
When complex numbers were first introduced, the prevailing ethos in mathematics was that a number is something which lies on the number line. Complex numbers were used, but, for the mathematicians of this time, because these "numbers" did not correspond to anything we could actually imagine, they were regarded as imaginary. For these mathematicians, the complex numbers were only something to be used in formal calculations; they weren't actually "real". Evidently, the name "imaginary" stuck.
Nowadays, whether or not something is "real" in the sense that it corresponds to the real world is largely irrelevant to mathematicians. Mathematics is purely a study of the abstract nowadays. The commutative field $\mathbb{C}$ of complex numbers is the two-dimensional vector space $\mathbb{R}^2$ of real numbers with multiplication defined by $(a,b)(c,d) = (ac - bd, ad + bc)$. "$\sqrt{-1}$" may be identified as the pair $(0, 1)$. This is called the imaginary unit. The name is a relic.
-1is real no because it can represented on real number line but$\sqrt{-1}$ can't be represented on this line
• This meaning has no sentence. – Yves Daoust Dec 22 '16 at 10:17
• @YvesDaoust: In fact, it does: it says that if you define $\sqrt {-1}$ purely algebraically, as one of the roots of $x^2 = -1$, then when you try to picture it geometrically you'll discover that it cannot sit among the real numbers, therefore it is in a way of a different "nature". Notice that in the above answer "no" is a shorthand for "number". While terse and with poor punctuation, this answer is essentially correct. – Alex M. Dec 22 '16 at 10:41
Whenever you have got a question about why we call something so.... refer to the definition.
According to Wikipedia:
A real number is a value that represents a quantity along a line.
An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit $i$, which is defined by its property $i^2 = −1$.
There is no problem in understanding that why $-1$ is a real number as you can easily construct a line and represent $-1$ on that line. The problem lies in the definition of imaginary numbers. It is a real number times $i$. Why $i$ ?? Because imaginary numbers are defined that way and changing the definition is not so good I think. :) :) :)
• "why i?" maybe because imaginary? – Ring Ø Dec 22 '16 at 10:15 | 2019-08-17 11:10:20 | {"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": 0.7368531823158264, "perplexity": 525.4933914349853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00050.warc.gz"} |
http://mathhelpforum.com/pre-calculus/46716-exponential-functions.html | 1. ## Exponential Functions...
Hey guys,
My reading materials didn't really explain how to do these. I have provided an example of one, and would liek ot know hwo to do them. I have a test on them tomorrow, but do not understand them.
Thanks!
2. Originally Posted by mcdanielnc89
Hey guys,
My reading materials didn't really explain how to do these. I have provided an example of one, and would liek ot know hwo to do them. I have a test on them tomorrow, but do not understand them.
Thanks!
this is simply asking for N(10) and N(60), since that gives the number of bacteria after 10 and 60 minutes respectively, as that is how N(t) was defined
can you continue?
3. Originally Posted by Jhevon
this is simply asking for N(10) and N(60), since that gives the number of bacteria after 10 and 60 minutes respectively, as that is how N(t) was defined
can you continue?
So I'd do something like this?
$N(10) = 4000(2)^10/20$
$N(60) = 4000(2)^60/20$
4. Originally Posted by mcdanielnc89
So I'd do something like this?
$N(10) = 4000(2)^{10/20}$
$N(60) = 4000(2)^{60/20}$
5. for pitty sakes. I've been trying to do that the last half hour in the calculator, lol...+
Thank you,. i see what i dind't do, haha
6. Originally Posted by mcdanielnc89
for pitty sakes. I've been trying to do that the last half hour in the calculator, lol...
to be safe, type 2^(10/20) and evaluate first, then multiply the answer by 4000. a similar thing for the 2nd one
half an hour? wow
7. lol, I am not good with calculators, LOOOOOL....
TYSVM
8. Originally Posted by mcdanielnc89
lol, I am not good with calculators, LOOOOOL....
TYSVM
neither am i, but you're worse than i am
not to make you feel bad or anything just two friends messing with each other
9. haha... no worries! | 2017-12-12 01:21:58 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7014830708503723, "perplexity": 1554.6967058060577}, "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-2017-51/segments/1512948514238.17/warc/CC-MAIN-20171212002021-20171212022021-00287.warc.gz"} |
http://dallinger.readthedocs.io/en/v2.7.2/dallinger_with_anaconda.html | # Installing Dallinger with Anaconda¶
If you are interested in Dallinger and use Anaconda, you’ll need to adapt the standard instructions slightly.
## Install psycopg2¶
In order to get the correct bindings, you need to install psycopg2 before you use requirements.txt; otherwise, everything will fail and you will be endlessly frustrated.
conda install psycopg2
## Install Dallinger¶
You’ll follow all of the Dallinger development installation instructions, with the exception of the virtual environment step. Then return here.
## Confirm Dallinger works¶
Now, we need to make sure that Dallinger and Anaconda play nice with one another. At this point, we’d check to make sure that Dallinger is properly installed by typing
dallinger --version
into the command line. For those of us with Anaconda, we’ll get a long error message. Don’t panic! Add the following to your .bash_profile:
export DYLD_FALLBACK_LIBRARY_PATH=$HOME/anaconda/lib/:$DYLD_FALLBACK_LIBRARY_PATH
If you installed anaconda using Python 3, you will need to change anaconda in that path to anaconda3.
After you source your .bash_profile, you can check your Dallinger version (using the same command that we used earlier), which should return the Dallinger version that you’ve installed. | 2017-10-19 21:40:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3639815151691437, "perplexity": 4275.690739769546}, "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/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00554.warc.gz"} |
http://cms.math.ca/cjm/msc/20C30?fromjnl=cjm&jnl=CJM | Canadian Mathematical Society www.cms.math.ca
location: Publications → journals
Search results
Search: MSC category 20C30 ( Representations of finite symmetric groups )
Expand all Collapse all Results 1 - 3 of 3
1. CJM Online first
Ashraf, Samia; Azam, Haniya; Berceanu, Barbu
Representation stability of power sets and square free polynomials The symmetric group $\mathcal{S}_n$ acts on the power set $\mathcal{P}(n)$ and also on the set of square free polynomials in $n$ variables. These two related representations are analyzed from the stability point of view. An application is given for the action of the symmetric group on the cohomology of the pure braid group. Keywords:symmetric group modules, square free polynomials, representation stability, Arnold algebraCategories:20C30, 13A50, 20F36, 55R80
2. CJM 2004 (vol 56 pp. 871)
Schocker, Manfred
Lie Elements and Knuth Relations A coplactic class in the symmetric group $\Sym_n$ consists of all permutations in $\Sym_n$ with a given Schensted $Q$-symbol, and may be described in terms of local relations introduced by Knuth. Any Lie element in the group algebra of $\Sym_n$ which is constant on coplactic classes is already constant on descent classes. As a consequence, the intersection of the Lie convolution algebra introduced by Patras and Reutenauer and the coplactic algebra introduced by Poirier and Reutenauer is the direct sum of all Solomon descent algebras. Keywords:symmetric group, descent set, coplactic relation, Hopf algebra,, convolution productCategories:17B01, 05E10, 20C30, 16W30
3. CJM 1997 (vol 49 pp. 133)
Reeder, Mark
Exterior powers of the adjoint representation Exterior powers of the adjoint representation of a complex semisimple Lie algebra are decomposed into irreducible representations, to varying degrees of satisfaction. Keywords:Lie algebras, adjoint representation, exterior algebraCategories:20G05, 20C30, 22E10, 22E60
© Canadian Mathematical Society, 2014 : https://cms.math.ca/ | 2014-11-27 17:00:31 | {"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": 0.9464932680130005, "perplexity": 1794.2359210860736}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931008919.73/warc/CC-MAIN-20141125155648-00073-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-is-the-slope-of-the-line-perpendicular-to-y-7-4x-3 | # What is the slope of the line perpendicular to y=7/4x -3 ?
Jan 7, 2016
$s l o p e = - \frac{4}{7}$
#### Explanation:
One form of the equation of a straight line is y = mx + c
where m is the slope and c the y-intercept.
the equation here fits this form and by comparison $m = \frac{7}{4}$
Consider 2 lines with slopes ${m}_{1} \mathmr{and} {m}_{2}$
when they are perpendicular to each other then ${m}_{1} \times {m}_{2} = - 1$
$\Rightarrow \frac{7}{4} \times {m}_{2} = - 1$
$\Rightarrow {m}_{2} = - \frac{1}{\frac{7}{4}} = - \frac{4}{7}$ | 2019-11-20 13:00:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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": 0.7233685255050659, "perplexity": 419.0576401625357}, "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/1573496670558.91/warc/CC-MAIN-20191120111249-20191120135249-00113.warc.gz"} |
http://mathhelpforum.com/trigonometry/200739-express-length-x-terms-trigonometric-ratios.html | # Thread: Express the length x in terms of the trigonometric ratios of θ.
1. ## Express the length x in terms of the trigonometric ratios of θ.
Don't know where to start.
2. ## Re: Express the length x in terms of the trigonometric ratios of θ.
Originally Posted by Remriel
Don't know where to start.
To start with, the hypotenuse for the triangle with \displaystyle \displaystyle \begin{align*} x \end{align*} in it is the same as the side opposite angle \displaystyle \displaystyle \begin{align*} \theta \end{align*} in the upper triangle. Call this side \displaystyle \displaystyle \begin{align*} y \end{align*} and we have
\displaystyle \displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}
As for the third side of the triangle with \displaystyle \displaystyle \begin{align*} x \end{align*} in it, call this side \displaystyle \displaystyle \begin{align*} z \end{align*}. If you complete the rectangle, you'll have another right angle triangle with that side opposite \displaystyle \displaystyle \begin{align*} \theta \end{align*} and with a hypotenuse of \displaystyle \displaystyle \begin{align*} 10 \end{align*}. That means
\displaystyle \displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}
So finally, in the triangle with \displaystyle \displaystyle \begin{align*} x \end{align*} in it, we have by Pythagoras
\displaystyle \displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}
I think that's the simplest you can get it
3. ## Re: Express the length x in terms of the trigonometric ratios of θ.
Originally Posted by Remriel
Don't know where to start.
1. I've completed the quadrilateral to a rectangle (see attachment).
2. The side of the rectangle in orange is:
$\displaystyle \displaystyle{s = 10 \cdot \cos(90^\circ - \theta) = 10 \cdot \sin(\theta)}$
3. Then
$\displaystyle \displaystyle{x = s \cdot \tan(\theta) = 10 \cdot \sin(\theta) \cdot \tan(\theta)}$
which is the solution Prove It has posted.
4. ## Re: Express the length x in terms of the trigonometric ratios of θ.
Another person wanted to be able to see the solution, so I am replacing the tex tags with dollar signs:
Originally Posted by Prove It
To start with, the hypotenuse for the triangle with \displaystyle \begin{align*} x \end{align*} in it is the same as the side opposite angle \displaystyle \begin{align*} \theta \end{align*} in the upper triangle. Call this side \displaystyle \begin{align*} y \end{align*} and we have
\displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}
As for the third side of the triangle with \displaystyle \begin{align*} x \end{align*} in it, call this side \displaystyle \begin{align*} z \end{align*}. If you complete the rectangle, you'll have another right angle triangle with that side opposite \displaystyle \begin{align*} \theta \end{align*} and with a hypotenuse of \displaystyle \begin{align*} 10 \end{align*}. That means
\displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}
So finally, in the triangle with \displaystyle \begin{align*} x \end{align*} in it, we have by Pythagoras
\displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}
I think that's the simplest you can get it
5. ## Re: Express the length x in terms of the trigonometric ratios of θ.
Originally Posted by Remriel
Don't know where to start.
Just as a suggestion- it never looks good to say "Don't know where to start"! You can always start by labeling things, drawing new lines that might help, and writing out trig formulas using those thing. For example, here, I see a right triangle with one angle $\theta$ with one leg of length 10. The hypotenuse of that triangle, the line on the left, has length $\frac{10}{cos(\theta)}$. Now I would draw a line from the upper right corner perpendicular to the left side. That creates a right triangle with hypotenuse of length 10 and angle $\theta$ so that the piece cut off by that perpendicular, which I will call "y", has length $y= 10cos(\theta)$. What remains of that left side has length $\frac{10}{cos(\theta)}- 10 cos(\theta)= \frac{10- 10cos^2(\theta)}{cos(\theta)}= \frac{10 sin^2(\theta)}{cos(\theta)}$. That is x.
6. ## Re: Express the length x in terms of the trigonometric ratios of θ.
Originally Posted by HallsofIvy
Just as a suggestion- it never looks good to say "Don't know where to start"! You can always start by labeling things, drawing new lines that might help, and writing out trig formulas using those thing. For example, here, I see a right triangle with one angle $\theta$ with one leg of length 10. The hypotenuse of that triangle, the line on the left, has length $\frac{10}{cos(\theta)}$. Now I would draw a line from the upper right corner perpendicular to the left side. That creates a right triangle with hypotenuse of length 10 and angle $\theta$ so that the piece cut off by that perpendicular, which I will call "y", has length $y= 10cos(\theta)$. What remains of that left side has length $\frac{10}{cos(\theta)}- 10 cos(\theta)= \frac{10- 10cos^2(\theta)}{cos(\theta)}= \frac{10 sin^2(\theta)}{cos(\theta)}$. That is x.
This was a 5 year old post that someone requested the solution be visible for. I resurrected it just to display the LaTex correctly. It is not an active question unless the new user has additional questions.
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# how can i express in term of trigonometric ratio
Click on a term to search for related topics. | 2018-09-19 23:09:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 1.0000100135803223, "perplexity": 1359.4004577982892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156311.20/warc/CC-MAIN-20180919220117-20180920000117-00282.warc.gz"} |
https://ch.mathworks.com/help/econ/moving-average-model.html | ## Moving Average Model
### MA(q) Model
The moving average (MA) model captures serial autocorrelation in a time series yt by expressing the conditional mean of yt as a function of past innovations, ${\epsilon }_{t-1},{\epsilon }_{t-2},\dots ,{\epsilon }_{t-q}$. An MA model that depends on q past innovations is called an MA model of degree q, denoted by MA(q).
The form of the MA(q) model in Econometrics Toolbox™ is
${y}_{t}=c+{\epsilon }_{t}+{\theta }_{1}{\epsilon }_{t-1}+\dots +{\theta }_{q}{\epsilon }_{t-q},$ (1)
where ${\epsilon }_{t}$ is an uncorrelated innovation process with mean zero. For an MA process, the unconditional mean of yt is μ = c.
In lag operator polynomial notation, ${L}^{i}{y}_{t}={y}_{t-i}$. Define the degree q MA lag operator polynomial $\theta \left(L\right)=\left(1+{\theta }_{1}L+\dots +{\theta }_{q}{L}^{q}\right).$ You can write the MA(q) model as
`${y}_{t}=\mu +\theta \left(L\right){\epsilon }_{t}.$`
### Invertibility of the MA Model
By Wold’s decomposition [2], an MA(q) process is always stationary because $\theta \left(L\right)$ is a finite-degree polynomial.
For a given process, however, there is no unique MA polynomial—there is always a noninvertible and invertible solution [1]. For uniqueness, it is conventional to impose invertibility constraints on the MA polynomial. Practically speaking, choosing the invertible solution implies the process is causal. An invertible MA process can be expressed as an infinite-degree AR process, meaning only past events (not future events) predict current events. The MA operator polynomial $\theta \left(L\right)$ is invertible if all its roots lie outside the unit circle.
Econometrics Toolbox enforces invertibility of the MA polynomial. When you specify an MA model using `arima`, you get an error if you enter coefficients that do not correspond to an invertible polynomial. Similarly, `estimate` imposes invertibility constraints during estimation.
## References
[1] Hamilton, J. D. Time Series Analysis. Princeton, NJ: Princeton University Press, 1994.
[2] Wold, H. A Study in the Analysis of Stationary Time Series. Uppsala, Sweden: Almqvist & Wiksell, 1938. | 2020-12-04 16:10:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.920896589756012, "perplexity": 1912.1687052011573}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00480.warc.gz"} |
https://msp.org/agt/2009/9-4/p01.xhtml | #### Volume 9, issue 4 (2009)
Download this article For screen For printing
Recent Issues
The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
The equivariant $J$–homomorphism for finite groups at certain primes
### Christopher P French
Algebraic & Geometric Topology 9 (2009) 1885–1949
##### Abstract
Suppose $G$ is a finite group and $p$ a prime, such that none of the prime divisors of $G$ are congruent to $1$ modulo $p$. We prove an equivariant analogue of Adams’ result that ${J}^{\prime }={J}^{\prime \prime }$. We use this to show that the $G$–connected cover of ${Q}_{G}{S}^{0}$, when completed at $p$, splits up to homotopy as a product, where one of the factors of the splitting contains the image of the classical equivariant $J$–homomorphism on equivariant homotopy groups.
##### Keywords
$J$–homomorphism, Adams operations, equivariant $K$–theory, equivariant fiber spaces and bundles
##### Mathematical Subject Classification 2000
Primary: 19L20, 19L47, 55R91
##### Publication
Received: 7 May 2007
Revised: 17 July 2009
Accepted: 3 August 2009
Published: 3 October 2009
##### Authors
Christopher P French Department of Mathematics and Statistics Grinnell College Grinnell, IA 50112 United States http://www.math.grin.edu/~frenchc/ | 2018-07-23 09:30:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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": 0.3212406039237976, "perplexity": 2257.6091787699975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00335.warc.gz"} |
http://ianfinlayson.net/class/cpsc220/labs/03-grades | ### Objective
To get more practice with using if statements, conditions, and blocks.
For this lab, you will write a program to calculate the letter grade for this class given the numeric grade. The ranges are listed on the syllabus as:
• [92, ∞): A
• [89, 92): A-
• [87, 89): B+
• [82, 87): B
• [79, 82): B-
• [77, 79): C+
• [72, 77): C
• [69, 72): C-
• [67, 69): D+
• [60, 67): D
• [0, 60): F
Input will be the number and output will be the letter grade that would earn.
### Details
• Prompt the user to enter their numeric grade.
• Read in the value using a Scanner.
• Use if and else if statements to check each condition for the user's grade.
• For each one, output the letter grade corresponding to the numeric grade.
• If the grade entered is less than 0, print an error message.
### Sample Run
Below is a sample run of the program. Values entered by the user are shown in bold face. | 2019-03-21 10:07:55 | {"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": 0.45095258951187134, "perplexity": 7931.752421220111}, "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/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00108.warc.gz"} |
https://docs.mantidproject.org/nightly/interfaces/direct/Crystal%20Field%20Python%20Interface.html | $$\renewcommand\AA{\unicode{x212B}}$$
Crystal Field Python Interface¶
The python facilities for Crystal Field calculations are available in Mantid from module CrystalField. The module provides two main classes: CrystalField defines various properties of a crystal field and CrystalFieldFit manages the fitting process.
Theoretical background can be found in the concept page and worked examples in the examples page.
Setting up crystal field parameters¶
A crystal field computation starts with creating an instance of the CrystalField class. The constructor has two mandatory arguments: Ion - the symbolic name of the ion, and Symmetry - the name of the point symmetry group of the field. The rest of the parameters are optional.
Possible values for the Ion argument are:
Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb
These are the trivalent rare-earth ions. For other rare earth ions, use the equivalent trivalent ion based on the number of f-electrons in the outer shell (e.g. for Pr4+ (4f1) use Ce). The rare earth ions sets the correct value of the Lande g-factor. In addition, a pure spin ion with arbitrary (but half-integral) S (or J) values are also supported using the syntax: Ion=S<n> where <n> is an integer or half-integer value, e.g. Ion=S2 or Ion=S1.5. In these cases, the g-factor is set to gJ = 2. The prefix letter can also be J instead of S, and lower case letters are also supported. (e.g. Ion=j1, Ion=s2.5 and Ion=J0.5 are all valid).
Allowed values for Symmetry are:
C1, Ci, C2, Cs, C2h, C2v, D2, D2h, C4, S4, C4h, D4, C4v, D2d, D4h, C3,
S6, D3, C3v, D3d, C6, C3h, C6h, D6, C6v, D3h, D6h, T, Td, Th, O, Oh
The minimum code to create a crystal field object is:
from CrystalField import CrystalField
cf = CrystalField('Ce', 'C2v')
Names of the crystal field parameters have the form Bnn and IBnn where nn are two digits between 0 and 6. Bnn is the real and IBnn is the imaginary part of a complex parameter. If a parameter isn’t set explicitly its default value is 0. To set a parameter pass it to the CrystalField constructor as a keyword argument, e.g.:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770)
An alternative way to set a parameter is to use the square brackets with a CrystalField object:
cf['B40'] = -0.031
Which can also be used to query the value of a parameter:
b = cf['B40']
Calculating the Eigensystem¶
The CrystalField class has methods to calculate the Hamiltonian and its eigensystem:
# Calculate and return the Hamiltonian matrix as a 2D numpy array.
h = cf.getHamiltonian()
# Calculate and return the eigenvalues of the Hamiltonian as a 1D numpy array.
e = cf.getEigenvalues()
# Calculate and return the eigenvectors of the Hamiltonian as a 2D numpy array.
w = cf.getEigenvectors()
It is efficient to call the above methods multiple times as all the outputs are cached and the calculations are repeated only after a parameter changes.
Calculating a Spectrum¶
To calculate a spectrum CrystalField needs to know the sample temperature and the shape of the peaks.
The temperature can be set either via a keyword argument Temperature of the constructor or using the Temperature property:
# Using the keyword argument
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, Temperature=44)
# Using the property
cf.Temperature = 44
Knowing the temperature allows us to calculate a peak list: a list of transition energies and intensities.:
print cf.getPeakList()
Which produces the output:
[[ 0.00000000e+00 2.44006198e+01 4.24977124e+01 1.80970926e+01 -2.44006198e+01]
[ 2.16711565e+02 8.83098530e+01 5.04430056e+00 1.71153708e-01 1.41609425e-01]]
The first row are the energies (in meV) and the second row are the integrated intensities (in milibarn per steradian).
The number of peaks that the function returns is controlled by two tolerance parameters: ToleranceEnergy and ToleranceIntensity. If a peak has an intensity below the value of ToleranceIntensity the peak is ignored. It two peaks have a difference in the energies smaller than ToleranceEnergy they are combined into a single peak.
If we set ToleranceIntensity of the above crystal field object to 1 mb/sr we’ll have only three peaks in the list:
cf.ToleranceIntensity = 1
print cf.getPeakList()
The new output:
[[ 0. 24.40061976 42.49771237]
[ 216.71156467 88.30985303 5.04430056]]
To calculate a spectrum we need to define the shape of each peak (peak profile function) and its default width (FWHM). The width can be set either via a keyword argument or a property with name FWHM. If the peak shape isn’t set the default of Lorentzian is assumed. To set a different shape use the PeakShape property:
cf.PeakShape = 'Gaussian'
cf.FWHM = 0.9
The values of PeakShape are expected to be names of Mantid peak fit functions. At the moment only Lorentzian and Gaussian can be used.
After the peak shape is defined a spectrum can be calculated:
sp = cf.getSpectrum()
The output is a tuple of two 1d numpy arrays (x, y) that can be used with matplotlib to plot:
import matplotlib.pyplot as plt
plt.plot(*sp)
plt.show()
It is possible to change parameters of individual peaks separately. Note though that only the shape parameters can be changed, the peak centre and the integrated intensity are defined by the crystal field parameters. To change the width of a peak use the following syntax:
# If the peak shape is Gaussian
cf.peaks.param[1]['Sigma'] = 2.0
cf.peaks.param[2]['Sigma'] = 0.01
# If the peak shape is Lorentzian
cf.peaks.param[1]['FWHM'] = 2.0
cf.peaks.param[2]['FWHM'] = 0.01
The three peaks now have all different widths. The first peak (index 0) keeps the default value.
If called without arguments getSpectrum() determines automatically the range and number of the x-points. To have more control of how the spectrum is calculated a list (or numpy array) of x-values can be provided as a first argument to getSpectrum. Alternatively, the x-values can be taken from a workspace:
# Use a list for x-values
x = [0, 1, 2, 3, ...]
sp = cf.getSpectrum(x)
# Use the first spectrum of a workspace
sp = cf.getSpectrum(ws)
# Use the i-th spectrum of a workspace
sp = cf.getSpectrum(ws, i)
Plotting¶
To plot a spectrum using the graphing facilities of Mantid CrystalField has method plot. It has the same arguments as getSpectrum and opens a window with a plot, e.g.:
cf.plot()
In addition to plotting, the plot method creates a workspace named CrystalField_<Ion> with the plot data. Subsequent calls to plot for the same CrystalField object will use the same plot window as created by the first call unless this window has been closed in the mean time.
A background has two components: a peak and a general background function. Set a background using the background property:
from CrystalField import CrystalField, CrystalFieldFit, Background, Function
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
Temperature=44.0, FWHM=1.1)
cf.background = Background(peak=Function('Gaussian', Height=10, Sigma=1),
background=Function('LinearBackground', A0=1.0, A1=0.01))
Here is an example of how to access the parameters of the background:
h = cf.background.peak.param['Height']
a1 = cf.background.background.param['A1']
Setting Ties and Constraints¶
Setting ties and constraints are done by calling the ties and constraints methods of the CrystalField class or its components. The Bnn parameters are tied by the CrystalField class directly specifying the tied parameter as a keyword argument:
cf.ties(B20=1.0, B40='B20/2')
The constraints are passed as strings containing expressions:
cf.constraints('1 < B22 <= 2', 'B22 < 4')
For the parameters of the background the syntax is the same but the methods are called on the background property:
cf.background.peak.ties(Height=10.1)
cf.background.peak.constraints('Sigma > 0')
cf.background.background.ties(A0=0.1)
cf.background.background.constraints('A1 > 0')
The names of the peak parameters both in ties and constraints must include the index of the peak to which they belong. Here we follow the naming convention of the CompositeFunction: f<n>.<name>, where <n> stands for an integer index staring at 0 and <name> is the name of the parameter. For example, f1.Sigma, f3.FWHM. Because names now contain the period symbol ‘.’ keyword arguments cannot be used. Instead we must pass a dictionary containing ties. The keys are parameter names and the values are the ties:
cf.peaks.ties({'f2.FWHM': '2*f1.FWHM', 'f3.FWHM': '2*f2.FWHM'})
Constraints are a list of strings:
cf.peaks.constraints('f0.FWHM < 2.2', 'f1.FWHM >= 0.1')
If a parameter of all peaks needs to be tied/constrained with the same expression then the following shortcut methods can be used:
cf.peaks.tieAll('Sigma=0.1', 3)
cf.peaks.constrainAll('0 < Sigma < 0.1', 4)
where the first argument is the general formula of the tie/constraint and the second is the number of peaks to apply to. There is also a version for a range of peak indices:
cf.peaks.tieAll('Sigma=f0.Sigma', 1, 3)
which is equivalent to:
cf.peaks.ties({'f1.Sigma': 'f0.Sigma', 'f2.Sigma': 'f0.Sigma', 'f3.Sigma': 'f0.Sigma'})
Setting Resolution Model¶
A resolution model is a way to constrain the widths of the peaks to realistic numbers which agree with a measured or calculated instrument resolution function. A model is a function that returns a FWHM for a peak centre. The Crystal Field python interface defines the helper class ResolutionModel to help define and set resolution models.
To construct an instance of ResolutionModel one needs to provide up to four input parameters. The first parameter, model, is mandatory and can be either of:
1. A tuple containing two arrays (lists) of real numbers which will be interpreted as tabulated values of the model function. The first element of the tuple is a list of increasing values for peak centres, and the second element is a list of corresponding widths. Values between the tabulated peak positions will be linearly interpolated.
2. A python function that takes a numpy.ndarray of peak positions and returns a numpy array of widths.
If the model is a tuple of two arrays then no additional parameters are required. If it’s a function then the rest of the parameters define how to tabulate this function. xstart and xend define the interval of interpolation which must include all fitted peaks. The last argument is accuracy that defaults to $$10^{-4}$$ and defines an approximate desired accuracy of the approximation. The interval will be split until the largest error of the interpolation is smaller than accuracy. Note that subdivision cannot go on to infinity as the number of points is limited by the class member ResolutionModel.max_model_size.
Example of setting a resolution model using a tuple of two arrays:
from CrystalField import CrystalField, ResolutionModel
rm = ResolutionModel(([1, 2, 3, ...., 100], [0.1, 0.3, 0.35, ..., 2.1]))
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, ..., Temperature=44.0, ResolutionModel=rm)
Or using an arbitrary function my_func:
def my_func(en):
return (25-en)**(1.5) / 200 + 0.1
rm = ResolutionModel(my_func, xstart=0.0, xend=24.0, accuracy=0.01)
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, ..., Temperature=44.0, ResolutionModel=rm)
Finally, the PyChop interface may be used to generate the resolution function for a particular spectrometer:
from pychop.Instruments import Instrument
marires = Instrument('MARI')
marires.setChopper('S')
marires.setFrequency(250)
marires.setEi(30)
rm = ResolutionModel(marires.getResolution, xstart=0.0, xend=29.0, accuracy=0.01)
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, ..., Temperature=44.0, ResolutionModel=rm)
When a resolution model is set, the peak width will be constrained to have a value close to the model. The degree of deviation is controlled by the FWHMVariation parameter. It has the default of 0.1 and is the maximum difference from the value given by the resolution model a width can have. If set to 0 the widths will be fixed to their calculated values (depending on the instant values of their peak centres). For example:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, ..., Temperature=44.0, ResolutionModel=rm, FWHMVariation=0.1)
will allow the peak widths to vary between $$\Delta(E)-0.1$$ and $$\Delta(E)+0.1$$ where $$\Delta(E)$$ is the value of the resolution model at the peak position $$E$$.
Defining Multiple Spectra¶
A CrystalField object can be configured to work with multiple spectra. In this case many of the object’s properties become lists. Here is an example of defining a CrystalField object with two spectra:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
Temperature=[44.0, 50], FWHM=[1.1, 0.9])
cf.PeakShape = 'Lorentzian'
cf.peaks[0].param[0]['FWHM'] = 1.11
cf.peaks[1].param[1]['FWHM'] = 1.12
cf.background = Background(peak=Function('Gaussian', Height=10, Sigma=0.3),
background=Function('FlatBackground', A0=1.0))
cf.background[1].peak.param['Sigma'] = 0.8
cf.background[1].background.param['A0'] = 1.1
Note how Temperature, FWHM, peaks and background become lists. They must have the same size. Ties and constraints similarly change:
# The B parameters are common for all spectra - syntax doesn't change
cf.ties(B20=1.0, B40='B20/2')
cf.constraints('1 < B22 <= 2', 'B22 < 4')
# Backgrounds and peaks are different for different spectra - must be indexed
cf.background[0].peak.ties(Height=10.1)
cf.background[0].peak.constraints('Sigma > 0.1')
cf.background[1].peak.ties(Height=20.2)
cf.background[1].peak.constraints('Sigma > 0.2')
cf.peaks[1].tieAll('FWHM=2*f1.FWHM', 2, 5)
cf.peaks[0].constrainAll('FWHM < 2.2', 1, 4)
The resolution model also needs to be initialised from a list:
x0, y0, x1, y1 = [ ... ], [ ... ], [ ... ], [ ... ]
rm = ResolutionModel([(x0, y0), (x1, y1)])
# or
rm = ResolutionModel([func0, func1], 0, 100, accuracy = 0.01)
cf.ResolutionModel = rm
To calculate a spectrum call the same method getSpectrum but pass the spectrum index as its first parameter:
# Calculate second spectrum, use the generated x-values
sp = cf.getSpectrum(1)
# Calculate third spectrum, use a list for x-values
x = [0, 1, 2, 3, ...]
sp = cf.getSpectrum(2, x)
# Calculate second spectrum, use the first spectrum of a workspace
sp = cf.getSpectrum(1, ws)
# Calculate first spectrum, use the i-th spectrum of a workspace
sp = cf.getSpectrum(0, ws, i)
Note that the attributes Temperature, FWHM, peaks and background may be set separately from the constructor, e.g.:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544)
cf.Temperature = [5, 50]
However, each time that Temperature is set, if it defines a different number of spectra from the previous value (e.g. if Temperature was initially empty or None and is then defined as in the example above, or if Temperature was initially a scalar value but is then redefined to be a list or vice versa), then all Ties, Constraints, FWHM and peaks parameters are cleared. Any crystal field parameters previously defined will be retained, however.
Fitting¶
To fit the crystal field and peak parameters first create a CrystalField object as described above. Then create an instance (object) of the CrystalFieldFit class:
from CrystalField import CrystalFieldFit
# In case of a single spectrum (ws is a workspace)
fit = CrystalFieldFit(Model=cf, InputWorkspace=ws)
# Or for multiple spectra
fit = CrystalFieldFit(Model=cf, InputWorkspace=[ws1, ws2])
Then call fit() method:
fit.fit()
After fitting finishes the CrystalField object updates automatically and contains new fitted parameter values.
The crystal field fit function is derived from the standard Mantid fit function and can be used with all properties described in Fit.
Two step fitting¶
Alternatively, a two step fitting process can be used. Please note that this fitting process is much slower than the standard fitting described above. In this two step process only crystal field parameters are fitted in the first step and only peak parameters in the second step.
Two step fitting is only available for single ions at the moment. It can be used both for a single spectrum or multiple spectra.
There are two versions of two step fitting. One version is entirely based on the standard Mantid fit function and attempts to fit all free field parameters at the same time in the first step. It is used by calling the two_step_fit() method for an instance of the CrystalFieldFit class:
fit.two_step_fit()
The other version, two_step_fit_sc(), applies scipy.optimize.minimize to fit each of the free field parameters sequentially in the first step but uses Mantid fitting for the peak parameters:
fit.two_step_fit_sc()
Both methods allow overwriting the maximal number of iterations both per step and overall as well as the minimizer used for fitting per step. For example:
fit.two_step_fit(OverwriteMaxIterations=[2,10], OverwriteMinimizers=['BFGS', 'Levenberg-Marquardt'], Iterations=30)
runs the first step for up to 2 iterations with the ‘BFGS’ minimizer and then the second step for up to 10 iterations with the ‘Levenberg-Marquardt’ minimizer. The whole fitting process is limited to 30 iterations.
A complete list of minimizers available for scipy.optimize.minimize can be found at: https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.minimize.html
If the minimizer is not overwritten, ‘L-BFGS-B’ is set as a default for scipy.optimize.minimize and ‘Levenberg-Marquardt’ for Mantid fitting.
GOFit fitting¶
The algorithms contained within the GOFit package can also be used from the Crystal Field API. This package is designed for the global optimization of parameters using a non-linear least squares cost function. For more information about the algorithms used in this implementation, please see the related RAL Technical Report.
The GOFit package contains three optimization algorithms called regularisation, multistart and alternating. Please note that the fitting process for multistart and alternating can be slow due to the residuals being evaluated in python.
Before you can use the GOFit package in Mantid, you will need to pip install gofit into your environment because it is an external dependency. Alternatively, run the following code in the Mantid workbench script window:
import subprocess, sys
rv = subprocess.run([sys.executable, '-m', 'pip', 'install', '--user', 'gofit'], capture_output=True)
print(rv.stdout.decode())
print(rv.stderr.decode())
Mantid has to be restarted for the changes to take effect. Please note that the script above requires Python 3.7 or higher.
Once installed, it should be possible to import the package and perform a fit using the regularisation algorithm by passing a GOFit callable into the Crystal Field API:
import gofit
fit.gofit(algorithm_callable=gofit.regularisation, jacobian=True, maxit=500)
The multistart algorithm requires you to pass in parameter_bounds and the number of samples:
parameter_bounds = {'B20': (-0.3013,0.3013), 'B22': (-0.5219,0.5219), 'B40': (-0.004624,0.004624), 'B42': (-0.02068,0.02068), 'B44': (-0.02736,0.02736),
'B60': (-0.0001604,0.0001604), 'B62': (-0.001162,0.001162), 'B64': (-0.001273,0.001273), 'B66': (-0.001724,0.001724),
'IntensityScaling': (0.,10.), 'f0.FWHM': (0.1,5.0), 'f1.FWHM': (0.1,5.0), 'f2.FWHM': (0.1,5.0), 'f3.FWHM': (0.1,5.0), 'f4.FWHM': (0.1,7.0)}
fit.gofit(algorithm_callable=gofit.multistart, parameter_bounds=parameter_bounds, samples=100, jacobian=True, maxit=500, scaling=True)
The alternating algorithm also requires you to pass in parameter_bounds and the number of samples:
fit.gofit(algorithm_callable=gofit.alternating, parameter_bounds=parameter_bounds, samples=100, maxit=500)
A full list of possible arguments for these algorithm can be found here. The output from these fits should be a matrix workspace containing the fitted data, and a table workspace containing the fitted parameters.
Multiple Ions¶
If there are multiple ions you can define CrystalField objects for each ion separately then add them together to create a CrystalFieldMultiSite object:
params = {'B20': 0.377, 'B22': 3.9, 'B40': -0.03, 'B42': -0.116, 'B44': -0.125,
'Temperature': [44.0, 50], 'FWHM': [1.1, 0.9]}
cf1 = CrystalField('Ce', 'C2v', **params)
cf2 = CrystalField('Pr', 'C2v', **params)
cfms = cf1 + cf2
The expression that combines the CrystalField objects also defines the contributions of each site into the overall intensity. The higher the coefficient of the object in the expression the higher its relative contribution. For example:
cf = 2*cf1 + cf2
means that the intensity of cf1 should be twice that of cf2.
Alternatively, you can create a CrystalFieldMultiSite object directly. This takes Ions, Symmetries, Temperatures and peak widths as lists:
from CrystalField import CrystalFieldMultiSite
cfms = CrystalFieldMultiSite(Ions=['Ce', 'Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44.0], FWHMs=[1.1])
Note that Temperature and FWHM (without plural) can also be used in place of the equivalent plural parameters. To access parameters of a CrystalFieldMultiSite object, prefix them with the ion index:
cfms['ion0.B40'] = -0.031
cfms['ion1.B20'] = 0.37737
b = cfms['ion0.B22']
Parameters can be set when creating the object by passing in a dictionary using the parameters keyword:
cfms = CrystalFieldMultiSite(Ions=['Ce', 'Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44.0], FWHMs=[1.1],
parameters={'ion0.B20': 0.37737, 'ion0.B22': 3.9770, 'ion1.B40':-0.031787,
'ion1.B42':-0.11611, 'ion1.B44':-0.12544})
A background can also be set this way, or using cfms.background. It can be passed as a string, a Function object(s), or a CompositeFunction object:
cfms = CrystalFieldMultiSite(Ions='Ce', Symmetries='C2v', Temperatures=[20], FWHMs=[1.0],
Background='name=Gaussian,Height=0,PeakCentre=1,Sigma=0;name=LinearBackground,A0=0,A1=0')
cfms = CrystalFieldMultiSite(Ions=['Ce'], Symmetries=['C2v'], Temperatures=[50], FWHMs=[0.9],
Background=LinearBackground(A0=1.0), BackgroundPeak=Gaussian(Height=10, Sigma=0.3))
cfms = CrystalFieldMultiSite(Ions='Ce', Symmetries='C2v', Temperatures=[20], FWHMs=[1.0],
Background= Gaussian(PeakCentre=1) + LinearBackground())
Ties and constraints are set similarly to CrystalField objects. f prefixes have been changed to be more descriptive:
cfms = CrystalFieldMultiSite(Ions=['Ce','Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44, 50], FWHMs=[1.1, 0.9],
Background=FlatBackground(), BackgroundPeak=Gaussian(Height=10, Sigma=0.3),
parameters={'ion0.B20': 0.37737, 'ion0.B22': 3.9770, 'ion1.B40':-0.031787,
'ion1.B42':-0.11611, 'ion1.B44':-0.12544})
cfms.ties({'sp0.bg.f0.Height': 10.1})
cfms.constraints('sp0.bg.f0.Sigma > 0.1')
cfms.constraints('ion0.sp0.pk1.FWHM < 2.2')
cfms.ties({'ion0.sp1.pk2.FWHM': '2*ion0.sp1.pk1.FWHM', 'ion1.sp1.pk3.FWHM': '2*ion1.sp1.pk2.FWHM'})
Parameters which are not allowed by the specified symmetry will be fixed to be zero, but unlike for the single-site case, all other parameters are assumed to be free (in the single-site case, parameters which are unset are assumed to be fixed to be zero). For the multi-site case, parameters must be fixed explicitly. For example:
params = {'ion0.B20': 0.37737, 'ion0.B22': 3.9770, 'ion1.B40':-0.031787, 'ion1.B42':-0.11611, 'ion1.B44':-0.12544}
cf = CrystalFieldMultiSite(Ions=['Ce', 'Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44.0, 50.0],
FWHMs=[1.0, 1.0], ToleranceIntensity=6.0, ToleranceEnergy=1.0, FixAllPeaks=True,
parameters=params)
cf.fix('ion0.BmolX', 'ion0.BmolY', 'ion0.BmolZ', 'ion0.BextX', 'ion0.BextY', 'ion0.BextZ', 'ion0.B40',
'ion0.B42', 'ion0.B44', 'ion0.B60', 'ion0.B62', 'ion0.B64', 'ion0.B66', 'ion0.IntensityScaling',
'ion1.BmolX', 'ion1.BmolY', 'ion1.BmolZ', 'ion1.BextX', 'ion1.BextY', 'ion1.BextZ', 'ion1.B40',
'ion1.B42', 'ion1.B44', 'ion1.B60', 'ion1.B62', 'ion1.B64', 'ion1.B66', 'ion1.IntensityScaling',
'sp0.IntensityScaling', 'sp1.IntensityScaling')
chi2 = CalculateChiSquared(str(cf.function), InputWorkspace=ws1, InputWorkspace_1=ws2)[1]
fit = CrystalFieldFit(Model=cf, InputWorkspace=[ws1, ws2], MaxIterations=10)
fit.fit()
Calculating a spectrum can be done with CrystalFieldMultiSite in the same way as a CrystalField object.
CrystalFieldMultiSite can also be used in the single-site case to use the CrystalFieldFunction fitting function. It can be used like a CrystalField object in this way, although Temperatures and FWHMs must still be passed as lists:
cfms = CrystalFieldMultiSite(Ions='Ce', Symmetries='C2', Temperatures=[25], FWHMs=[1.0], PeakShape='Gaussian',
BmolX=1.0, B40=-0.02)
Intensity Scaling¶
The IntensityScaling parameter allows applying of a scale factor to the data. This scale factor is not applied immediately but only when a fit is run. As a consequence, a spectrum plotted after setting an IntensityScaling parameter but before running the fit will remain unchanged. Unlike other parameters the default value for IntensityScaling is 1 instead of 0.
There are differences on how to use the IntensityScaling parameter(s) depending on the use-case:
Single ions fitted to a single spectrum¶
When fitting a single ion symmetry to a single spectrum IntensityScaling is set to a single value:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544, Temperature=44.0,
FWHM=1.1, IntensityScaling=3.0)
Instead of initializing IntensityScaling in the constructor this parameter can also be set later:
cf.IntensityScaling = 3.0
In both cases IntensityScaling is initially set to a value of 3.0 but can be varied during the fitting process. Only with a tie on IntensityScaling the value remains fixed:
cf.ties(IntensityScaling = 3.0)
Single ions fitted to multiple spectra¶
In case of multiple spectra the IntensityScaling parameter is replaced by a list of values. The size of this list must match the size of the Temperature list:
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
Temperature=[44.0, 50.0], FWHM=[1.1, 0.9], IntensityScaling=[3.0, 0.05])
The parameters for IntensityScaling can also be initialized later:
cf.IntensityScaling=[3.0, 0.05]
When setting a tie the value for an IntensityScaling parameter can be set directly using the index of the parameter:
cf.ties(IntensityScaling1 = 5.0)
In the example above the initial value of the second IntensityScaling parameter would be ignored and permanently set to 5.
Intensity scaling for multiple ions¶
For multiple ions there are two options for creating a CrystalFieldMultiSite object. Either two CrystalField objects are combined or a CrystalFieldMultiSite object is created directly. The following example:
cf = cf1 + 0.5*cf2
creates a CrystalFieldMultiSite object with cf1 as ion0 and cf2 as ion1. It is also possible to have scaling factors for both CrystalField objects:
cf = 2*cf1 + 3*cf2
The scaling factors are also used as the IntensityScaling setting for the respective ion.
After combining CrystalField objects to a CrystalFieldMultiSite object further changes to the original CrystalField objects are not reflected in the CrystalFieldMultiSite object. Furthermore, the CrystalFieldMultiSite object does not have a set function for IntensityScaling parameters. As a consequence, it is not possible to set these parameters later as for the CrystalField object.
Multiple ions fitted to a single spectrum¶
When combining two CrystalField objects for a single spectrum to a CrystalFieldMultiSite object the original values for IntensityScaling are ignored:
params = {'B20': 0.377, 'B22': 3.9, 'B40': -0.03, 'B42': -0.116, 'B44': -0.125,
'Temperature': [44.0], 'FWHM': [1.1], 'IntensityScaling': [0.2]}
cf1 = CrystalField('Ce', 'C2v', **params)
cf2 = CrystalField('Pr', 'C2v', **params)
cfms = 2*cf1 + cf2
results in the following CrystalFieldMultiSite object and tie:
from CrystalField import CrystalFieldMultiSite
cfms = CrystalFieldMultiSite(Ions=['Ce', 'Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44.0], FWHMs=[1.1], abundances=[2.0, 1.0]
parameters={'ion0.B20':0.377,'ion0.B22':3.9,'ion0.B40':-0.03,'ion0.B42':-0.116,'ion0.B44':-0.125,
'ion1.B20':0.377,'ion1.B22':3.9,'ion1.B40':-0.03,'ion1.B42':-0.116,'ion1.B44':-0.125,
ion0.IntensityScaling:2.0,ion1.IntensityScaling:1.0})
cfms.ties({'ion1.IntensityScaling' : '0.5*ion0.IntensityScaling'})
In addition to creating the equivalent CrystalFieldMultiSite object the coefficient is used to set a tie for the IntensityScaling parameter of ion1 relative to the IntensityScaling parameter of ion0. For the tie the coefficient of the respective ion is divided by the coefficient of the ion with the greatest coefficient. The coefficients from the combining expression are stored as abundances.
Creating the CrystalFieldMultiSite object directly allows for more flexibility. First of all, instead of setting values for abundances it is possible to set the tie directly. Furthermore, if no ties for IntensityScaling are required this can be achieved by not defining any abundances in the constructor. Without a tie the IntensityScaling parameters for each ion can be set individually to an initial value and might vary during the fitting process.
Multiple ions fitted to multiple spectra¶
When fitting multiple ions to multiple spectra the IntensityScaling factor is the product of the IntensityScaling factor of the respective ion and the respective spectrum.
The IntensityScaling factors for the spectra are preserved from the CrystalField objects in the combination. If only one of the CrystalField objects has IntensityScaling values set these are used for the CrystalFieldMultiSite object. In case of different settings for the original CrystalField objects the values for the object defining ion0 are used and a warning about this mismatch displayed.
Creating the CrystalFieldMultiSite object directly allows to set each of the IntensityScaling values individually:
from CrystalField import CrystalFieldMultiSite
cfms = CrystalFieldMultiSite(Ions=['Ce', 'Pr'], Symmetries=['C2v', 'C2v'], Temperatures=[44.0, 50.0], FWHMs=[1.1, 0.9],
parameters={'ion0.B20':0.377,'ion0.B22':3.9,'ion0.B40':-0.03,'ion0.B42':-0.116,'ion0.B44':-0.125,
'ion1.B20':0.377,'ion1.B22':3.9,'ion1.B40':-0.03,'ion1.B42':-0.116,'ion1.B44':-0.125,
'ion0.IntensityScaling':3.0, 'ion1.IntensityScaling':2.0,
'sp0.IntensityScaling':1.5, 'sp1.IntensityScaling':0.007})
As in the single spectra case IntensityScaling values are initial values and default to 1 if there is no setting. The ties can either be added directly or by adding the corresponding abundances with a value per ion in the constructor.
Finding Initial Parameters¶
Using a Monte Carlo estimation method¶
If the initial values of the fitting parameters are not known they can be estimated using estimate_parameters() method. It randomly searches the parameter space in a given region such that the calculated spectra are as close to the fit data as possible. The method uses EstimateFitParameters internally. See algorithm’s description for the available properties. Here is an example of a fit with initial estimation:
from CrystalField.fitting import makeWorkspace
from CrystalField import CrystalField, CrystalFieldFit, Background, Function
# Create some crystal field data
origin = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
Temperature=44.0, FWHM=1.1)
x, y = origin.getSpectrum()
ws = makeWorkspace(x, y)
# Define a CrystalField object with parameters slightly shifted.
cf = CrystalField('Ce', 'C2v', B20=0, B22=0, B40=0, B42=0, B44=0,
Temperature=44.0, FWHM=1.0, ResolutionModel=([0, 100], [1, 1]), FWHMVariation=0)
# Set any ties on the field parameters.
cf.ties(B20=0.37737)
# Create a fit object
fit = CrystalFieldFit(cf, InputWorkspace=ws)
# Find initial values for the field parameters.
# You need to define the energy splitting and names of parameters to estimate.
# Optionally additional constraints can be set on tied parameters (eg, peak centres).
fit.estimate_parameters(EnergySplitting=50,
Parameters=['B22', 'B40', 'B42', 'B44'],
Constraints='20<f1.PeakCentre<45,20<f2.PeakCentre<45',
NSamples=1000)
print 'Returned', fit.get_number_estimates(), 'sets of parameters.'
# The first set (the smallest chi squared) is selected by default.
# Select a different parameter set if required
fit.select_estimated_parameters(3)
print cf['B22'], cf['B40'], cf['B42'], cf['B44']
# Run fit
fit.fit()
Using the point charge model¶
Alternatively, the Point Charge Model may be used to calculate the crystal field parameters. In this case, the crystal field interaction is assumed to be purely electrostatic. At an infinite distance away from an ion, or analogously, at a nonzero distance from an ion of infinitesimal extend (a point charge), the charge in free space is zero, so Gauss’s law becomes $$\nabla^2 V = 0$$ which is Laplace’s equation. The solution of this is a multipole expansion, a sum of spherical harmonic functions: $$V(r,\theta\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^l R_l(r) Y_l^m(\theta,\phi)$$. In the limit of infinite $$r$$, $$R_l(r) = B / r^{l+1}$$. The radial term is the crystal field parameters, and the angular term (spherical harmonics in this case) are the crystal field operators.
One should now note that the quantities noted above are generally complex. In order to have real valued parameters, Stevens chose to use the tesseral harmonics $$Z_l^m(\theta,\phi)$$ instead of the spherical harmonics for the angular part. These functions are simply the hermitian combinations of spherical harmonics of the same rank $$l$$ and opposite signed order $$m$$. (An alternative formulation by Wybourne uses the original spherical harmonics)
In Mantid we use the Stevens convention, as common in the neutron scattering literature. The user should note that the convention amongst optical spectroscopists is that of Wybourne.
A derivation of the point charge energy can be found in many text books (e.g. Morrison), but will not be detailed here, where only the final result is given:
$B_l^m = \frac{4\pi}{2l+1} \frac{| e|^2}{4\pi\epsilon_0} \sum_i \frac{q_i}{r_i^{l+1}} a_0^l \langle r^l \rangle Z_l^m(\theta_i,\phi_i)$
where $$q_i$$, $$r_i$$, $$\theta_i$$ and $$\phi_i$$ are the charge (in units of the elemental charge $$|e|$$) and relative polar coordinates of the $$i^{\mathrm{th}}$$ point charge from the magnetic ion; $$a_0$$ is the Bohr radius, $$\langle r^l \rangle$$ is the $$l^{\mathrm{th}}$$ order expectation value of the radial wavefunction of the magnetic ion and $$\epsilon_0$$ is the permitivity of free space (note this equation is in SI units; many older texts use cgs units, but this does not matter because the value is eventually converted to energy units of meV, rather than Joules or ergs).
In order to calculate the point charge model crystal field parameters a set of charged ligands around the magnetic ion has to be given. This may be done either directly, as a list of 4-element lists [charge, pos_x, pos_y, pos_z]:
from CrystalField import PointCharge
axial_pc_model = PointCharge([[-2, 0, 0, -4], [-2, 0, 0, 4]], 'Nd')
axial_blm = axial_pc_model.calculate()
print(axial_blm)
which represents a simple axial crystal field with charges at $$\pm 4\mathrm{\AA}$$ away from a Nd ion in the z-direction.
Alternatively, the set of ligands may be calculated from a crystal structure and a maximum distance. For example, for a cubic crystal field in the perovskite structure:
from CrystalField import PointCharge
from mantid.geometry import CrystalStructure
perovskite_structure = CrystalStructure('4 4 4 90 90 90', 'P m -3 m', 'Ce 0 0 0 1 0; Al 0.5 0.5 0.5 1 0; O 0.5 0.5 0 1 0')
cubic_pc_model = PointCharge(perovskite_structure, 'Ce', Charges={'Ce':3, 'Al':3, 'O':-2}, MaxDistance=7.5)
The syntax for the CrystalStructure object is given in the Crystal Structure concept page. Instead of the maximum distance, MaxDistance, in Angstrom, the maximum nth neighbour can be specified with:
cubic_pc_model = PointCharge(perovskite_structure, 'Ce', Charges={'Ce':3, 'Al':3, 'O':-2}, Neighbour=2)
note that this might result in a slightly slower calculation, because internally, a maximum distance much greater the nth neighbour is set and then all neighbours up to n are found within this distance.
If a workspace with a defined crystal structure exists, it can be used instead of the CrystalStructure object. Other inputs remain the same. Finally, a CIF file can be given directly:
cif_pc_model = PointCharge('somecompound.cif')
This uses LoadCIF to parse the input CIF file. Note that LoadCIF changes the atom labels, so you should use the getIons() method to get the actual atom labels which PointCharge uses. E.g. using this cif file:
cif_pc_model = PointCharge('AMS_DATA.cif')
print(cif_pc_model.getIons())
gives:
{'O1': [0.125, 0.125, 0.375],
'O2': [0.125, 0.375, 0.375],
'Sm1': [0.25, 0.25, 0.25],
'Sm2': [0.021, 0.0, 0.25],
'Sm3': [0.542, 0.0, 0.25]}
You can then define the charges for each site, the magnetic ion and the maximum distance, and calculate:
cif_pc_model.Charges = {'O1':-2, 'O2':-2, 'Sm1':3, 'Sm2':3, 'Sm3':3}
cif_pc_model.IonLabel = 'Sm2'
cif_pc_model.Neighbour = 1
cif_blm = cif_pc_model.calculate()
print(cif_blm)
Note that only the magnetic structure (as a CrystalStructure object, CIF file name or workspace) is needed to construct a PointCharge object. However, the calculations will return an error unless both IonLabel and Charges are defined. By default a value of 5 $$\mathrm{\AA}$$ for MaxDistance is used if neither MaxDistance nor Neighbour is defined. Whichever of MaxDistance or Neighbour is defined last takes precedent, and if both are defined in the constructor, e.g.:
bad_pc_model = PointCharge('AMS_DATA.cif', MaxDistance=7.5, Neighbour=2)
then the value for MaxDistance will be used regardless of where it appears in the keyword list.
For Charges, instead of listing the charges of each site, you can just give the charge for each element, e.g.:
cif_pc_model.Charges = {'O':-2, 'Sm':3}
cif_blm = cif_pc_model.calculate()
The result of the calculate() method can be put directly into a CrystalField object and used either to calculate a spectrum or as the starting parameters in a fit:
cf = CrystalField('Sm', 'C2', Temperature=5, FWHM=10, **cif_pc_model.calculate())
plot(*cf.getSpectrum())
fit = CrystalFieldFit(cf, InputWorkspace=ws)
fit.fit()
Finally, note that the calculated crystal field parameters are defined with the quantisation axis along the z direction in the Busing-Levy convention (that is, it is perpendicular to the a-b plane). This means that if the particular magnetic ion lies on a higher symmetry site but the highest symmetry rotation axis is not along z (for example, the A or B site in the Pyrochlore lattice, which has a 3-fold axis along [111], whilst z is parallel to c), then the parameters may appear to have a low symmetry (e.g. more m terms are nonzero). You then need to rotate the parameters if you want it quantised along the high symmetry direction.
Calculating Physical Properties¶
In addition to the inelastic neutron spectrum, various physical properties arising from the crystal field interaction can be calculated. These include (but are not necessarily limited to):
• the crystal field contribution to the magnetic heat capacity;
• magnetic susceptibility;
• magnetic moment (and subsequently magnetisation)
• the dipole transition matrix (and individual components).
The calculated values can be invoked using the respective functions:
• getHeatCapacity();
• getSusceptibility();
• getMagneticMoment();
• getDipoleMatrix() (+ getDipoleMatrixComponent(<’X’, ‘Y’ or ‘Z’>)).
To calculate the heat capacity use:
import matplotlib.pyplot as plt
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, Temperature=44.0)
Cv = cf.getHeatCapacity() # Calculates Cv(T) for 1<T<300K in 1K steps (default)
plt.plot(*Cv) # Returns a tuple of (x, y) values
T = np.arange(1,900,5)
Cv = cf.getHeatCapacity(T) # Calculates Cv(T) for specified values of T (1 to 900K in 5K steps here)
plt.plot(T, Cv[1])
# Temperatures from a single spectrum workspace
ws = CreateWorkspace(T, T, T)
Cv = cf.getHeatCapacity(ws) # Use the x-values of a workspace as the temperatures
ws_calc = CreateWorkspace(*Cv)
plot(ws_calc, 0) # Creates workspace from data and plots it (plots the first spectrum, index 0)
# Temperatures from a multi-spectrum workspace
ws = CreateWorkspace(T, T, T, NSpec=2)
Cv = cf.getHeatCapacity(ws, 1) # Uses the second spectrum's x-values for T (e.g. 450<T<900)
plot(*Cv)
All the physical properties methods (excluding dipole matrix functions) returns a tuple of (x, y) values. The heat capacity is calculated in Jmol-1K-1. The theory is described in CrystalFieldHeatCapacity.
The molar susceptibility is calculated using Van Vleck’s formula, and requires in addition knowledge of the applied field direction (default is [0, 0, 1] where the field is along the crystal field quantisation direction):
chi_v = cf.getSusceptibility(T, Hdir=[1, 1, 1])
The field direction is a Cartesian vector with coordinates defined with the z-axis parallel to the quantisation direction of the crystal field parameters (usually taken to be the highest symmetry rotation axis). To calculate for a powder averaged field direction use:
chi_v_powder = cf.getSusceptibility(T, Hdir='powder')
The powder averaging is done by taking the mean of the susceptibility (or magnetisation) along the $$x$$, $$y$$ and $$z$$ directions (e.g. $$\chi^{\mathrm{pow}} = (\chi^x + \chi^y + \chi^z)/3$$).
Note that the function calculates the molar magnetic susceptibility, and by default outputs it in cgs units (cm3/mol or emu/mol). To obtain the result in SI units (m3/mol) use:
chi_v_cgs = cf.getSusceptibility(T, Hdir=[1, 1, 0], Unit='SI')
In addition, “atomic” units ($$\mu_B/\mathrm{T}/\mathrm{ion}$$) can also be obtained using:
chi_v_bohr = cf.getSusceptibility(T, Unit='bohr')
The theory is described in the CrystalFieldSusceptibility function page.
The magnetic moment is calculated by adding a Zeeman interaction to the crystal field Hamiltonian and diagonalising the combined matrix, from which the expectation of the magnetic moment operator is calculated. The moment can be calculated as a function of temperature or applied field magnitude:
moment_t = cf.getMagneticMoment(Temperature=T, Hdir=[1, 1, 1], Hmag=0.1) # Calcs M(T) with at 0.1T field||[111]
H = np.linspace(0, 30, 121)
moment_h = cf.getMagneticMoment(Hmag=H, Hdir='powder', Temperature=10) # Calcs M(H) at 10K for powder sample
By default, the magnetisation is calculated in atomic units of bohr magnetons per magnetic ion. Alternatively, the SI or cgs molar magnetic moments can be calculated:
moment_SI = cf.getMagneticMoment(H, [1, 1, 1], Unit='SI') # M(H) in Am^2/mol at 1K for H||[111]
moment_cgs = cf.getMagneticMoment(100, Temperature=T, Unit='cgs') # M(T) in emu/mol in a field of 100G || [001]
Please note that if cgs units are used, then the magnetic field must be specified in Gauss rather than Tesla (1T == 10000G). Note also that the cgs unit “emu/mol” in this case is “erg/Gauss/mol” quantifying a molar magnetic moment.
Please note that the calculation result is the molar magnetic moment. Thus to get the magnetisation, you should divide this by the molar volume of the material. By default, the calculation temperature is 1K, and the applied magnetic field is 1T along [001]. For further details and a description of the theory, see the CrystalFieldMagnetisation and CrystalFieldMoment pages.
To calculate the dipole transition matrix (and components):
import matplotlib.pyplot as plt
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, Temperature=44.0)
A = cf.getDipoleMatrix() # Calculates the dipole transition matrix, which is equal to the sum of its components::
Ax = cf.getDipoleMatrixComponent('X') # Calculates the component of the dipole transition matrix in the x direction
Ay = cf.getDipoleMatrixComponent('Y') # Calculates the component of the dipole transition matrix in the Y direction
Az = cf.getDipoleMatrixComponent('Z') # Calculates the component of the dipole transition matrix in the Z direction
Fitting Physical Properties¶
Instead of fitting the inelastic neutron spectrum, the physical properties can be fitted using a similar interface to that described above. The main difference is that some experimental setup information has to be given - especially for the susceptibility and magnetisation. This is done by specifying an instance of the PhysicalProperties helper class as the PhysicalProperty attribute of CrystalField, either as a keyword argument in the constructor:
from CrystalField import CrystalField, CrystalFieldFit, PhysicalProperties
# Fits a heat capacity dataset - you must have subtracted the phonon contribution by some method already
# and the data must be in J/mol/K.
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
PhysicalProperty=PhysicalProperties('Cv'))
fitcv = CrystalFieldFit(Model=cf, InputWorkspace=ws)
fitcv.fit()
or separately after construction:
params = {'B20':0.37737, 'B22':3.9770, 'B40':-0.031787, 'B42':-0.11611, 'B44':-0.12544}
cf = CrystalField('Ce', 'C2v', **params)
cf.PhysicalProperty = PhysicalProperties('Cv')
fitcv = CrystalFieldFit(Model=cf, InputWorkspace=ws)
fitcv.fit()
# Fits a susceptibility dataset. Data is the volume susceptibility in SI units
cf = CrystalField('Ce', 'C2v', **params)
cf.PhysicalProperty = PhysicalProperties('susc', Hdir='powder', Unit='SI')
fit_chi = CrystalFieldFit(Model=cf, InputWorkspace=ws)
fit_chi.fit()
# Fits a magnetisation dataset. Data is in emu/mol, and was measured at 5K with the field || [111].
cf = CrystalField('Ce', 'C2v', **params)
cf.PhysicalProperty = PhysicalProperties('M(H)', Temperature=5, Hdir=[1, 1, 1], Unit='cgs')
fit_mag = CrystalFieldFit(Model=cf, InputWorkspace=ws)
fit_mag.fit()
# Fits a magnetisation vs temperature dataset. Data is in Am^2/mol, measured with a 0.1T field || [110]
cf = CrystalField('Ce', 'C2v', **params)
cf.PhysicalProperty = PhysicalProperties('M(T)', Hmag=0.1, Hdir=[1, 1, 0], Unit='SI')
fit_moment = CrystalFieldFit(Model=cf, InputWorkspace=ws)
fit_moment.fit()
Unfortunately only 1D datasets can be fitted (e.g. M(H, T) cannot be fitted as a simultaneous function of field and temperature). Also, note that setting the PhysicalProperty attribute after constructing the CrystalField object (e.g. running cf.PhysicalProperty = PhysicalProperties(‘Cv’)) causes the number of datasets to change and will clear all Ties and Constraints previously set, and also reset all FWHM and peaks to the default values (zero for FWHM and Lorentzian for peaks).
Simultaneous Fitting of Physical Properties and Inelastic Neutron Spectra¶
Finally, physical properties data and neutron spectra may be fitted simultaneously. In this case, all the inelastic neutron spectra must be specified first in the list of input workspaces, with the physical properties dataset(s) following in the same order as specified in the PhysicalProperty attribute, which for multiple physical properties should be a list. E.g.:
# Fits an INS spectrum (at 10K) and the heat capacity simultaneously
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544)
cf.Temperature = 10
cf.FWHM = 1.5
cf.PhysicalProperty = PhysicalProperties('Cv')
fit = CrystalFieldFit(Model=cf, InputWorkspace=[ws_ins_10K, ws_cp])
fit.fit()
# Fits two INS spectra (at 44K and 50K) and the heat capacity, susceptibility and magnetisation simultaneously.
PPCv = PhysicalProperties('Cv')
PPchi = PhysicalProperties('susc', 'powder', Unit='cgs')
PPMag = PhysicalProperties('M(H)', [1, 1, 1], 5, 'bohr')
cf = CrystalField('Ce', 'C2v', B20=0.37737, B22=3.9770, B40=-0.031787, B42=-0.11611, B44=-0.12544,
Temperature=[44.0, 50], FWHM=[1.1, 0.9], PhysicalProperty=[PPCv, PPchi, PPMag] )
fit = CrystalFieldFit(Model=cf, InputWorkspace=[ws_ins_44K, ws_ins_50K, ws_cp, ws_chi, ws_mag])
fit.fit()
Note that PhysicalProperty requires the type of physical property (either ‘Cv’ or ‘Cp’ or ‘heatcap’ for heat capacity; ‘susc’ or ‘chi’ for susceptibility; ‘mag’ or ‘M(H)’ for magnetic moment vs applied field; or ‘mom’ or ‘M(T)’ for moment vs temperature) as the first argument. Subsequent arguments are optional, and are in the following order:
PhysicalProperties('Cp') # No further parameters required for heat capacity
PhysicalProperties('chi', hdir, inverse, unit)
PhysicalProperties('chi', unit)
PhysicalProperties('mag', hdir, temp, unit)
PhysicalProperties('mag', unit)
PhysicalProperties('M(T)', hmag, hdir, inverse, unit)
PhysicalProperties('M(T)', unit)
Or these parameters may be specified using keyword arguments, with the keywords: ‘Hdir’, ‘Hmag’, ‘Inverse’, ‘Unit’, and ‘Temperature’ (note these are case sensitive, and not all parameters apply to all types of physical properties). The default values (Hdir=[0,0,1], Hmag=1, Inverse=False, Unit=’cgs’ and Temperature=1 are used if nothing is specified for a particular attribute.
Categories: Interfaces | Direct | 2022-12-02 03:55:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.5694572329521179, "perplexity": 4386.488961084758}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00638.warc.gz"} |
https://proxies-free.com/convergence-and-divergence-of-series-when-changing-finite-number-of-summands/ | # Convergence and divergence of series when changing finite number of summands
I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:
Let $$sum_{k=n_1}^{infty} a_k$$ and $$sum_{k=n_2}^{infty}b_k$$ be two series with $$(s_n)_{n geq n_1}$$ and $$(t_n)_{t geq n_2}$$ their partial sums. Let’s suppose there exists an $$N$$ so that $$a_k = b_k$$ for alle $$kgeq N$$, than we have
begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and
begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \ &= s_n – left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}
for all $$n geq N$$. Hence, both $$(s_n)_{n geq n_1}$$ and $$(t_n)_ {tgeq n_2}$$ are either convergent or divergent.
Unfortunately I do not see why $$(s_n)_{n geq n_1}$$ and $$(t_n)_{t geq n_2}$$ are either convergent or divergent following this calculation. Moreover I also don’t get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series. | 2021-06-12 23:58:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.991055428981781, "perplexity": 283.935624252465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487586465.3/warc/CC-MAIN-20210612222407-20210613012407-00406.warc.gz"} |
https://www.scienceforums.net/topic/3884-which-form-of-standard-deviation/ | # which form of standard deviation
## Recommended Posts
can someone give me a explernation of when to use the form
$\sigma^2 = \frac{\sum(x-\mu)^2}{n}$
and when to use
$s^2 = \frac{\sum(x-\bar{x})^2}{n-1}$
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Well, it really doesn't make that much of a difference to which one you use. But anyways, here is the explanation:
The first form that you have mentioned needs to be used when you are going to take the standard deviation OF THE WHOLE population.
The second one is when you take a sample of a population ie. only some of the population
{edit}
Got them the wrong way round
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No you didn't.
The n form is the population standard deviation, the n - 1 form is the sample standard deviation. That's why the top one uses Mu, denoting the population mean, and the bottom one uses XBar, denoting the sample mean.
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You didn't understand, the edit WAS the correction that I had made! I put it the other way round, after the edit, it was correct.
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the n-1 is for sample of a population, this is because, since you are estimating a statistic, u want it be unbiased,
which is E(s^2)=sigma^2
s^2 is called the unbiased esimator of the population variance.
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intersting. I never took that formula to that much detail.
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nicked from my stats lectures
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oh wow! It is too complex for me to comprehend ))) !!!!
I'm only at GCSE level for crying out loud! I kind of get the idea of what it is trying to say, but, can't really understand how they get to that answer.
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oh sorry, if u do maths a levels especially the stats modules then u will understand it. you don need to know this now
I had the same problem when doing GCSE cos the course isnt really exact on which formula ur meant to use. Its better to ask you teacher.
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Well, i did, and that's why i know which formula to use. I needed it for the stats project for my coursework. I am planning to do maths A level since I want to go into chemical engineering
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why dont u go to the exam boards website,
if its AQA http://www.aqa.org.uk/qual/gcse.html and select which maths ur doing, and then view the specifications.
and for edexcel
http://www.edexcel.org.uk/qualifications/QualificationSubject.aspx?id=50009
the specs should have everything u need in there. there are should also be some past papers with answers, so if u look at them u should get an idea of which form to use
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Well, thanx for the link, but all the coursework has been finished and all. I have actually sat the non-calculator paper before yesterday!
{Edit}
Just noticed I have 1111 posts!
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You didn't understand, the edit WAS the correction that I had made! I put it the other way round, after the edit, it was correct.
Sorry. Reading your post as it stands, there is no way of knowing that.
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Its ok! I thought that it musht stir up some confusion!
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• Create New... | 2020-08-07 01:19:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41640105843544006, "perplexity": 1485.6184524167804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00342.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/117884/the-specific-rate-constant-for-the-first-order-decomposition-of-n2o5-g-to-no2-g--1 | # Problem: The specific rate constant for the first-order decomposition of N2O5 (g) to NO2(g) and O2(g) is 7.48 x 10-3 s-1 at a given temperature. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise to 0.220 atm .
⚠️Our tutors found the solution shown to be helpful for the problem you're searching for. We don't have the exact solution yet.
###### Problem Details
The specific rate constant for the first-order decomposition of N2O5 (g) to NO2(g) and O2(g) is 7.48 x 10-3 s-1 at a given temperature. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise to 0.220 atm . | 2020-07-15 03:04:25 | {"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": 0.8491494655609131, "perplexity": 1132.4871409212274}, "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/1593657154789.95/warc/CC-MAIN-20200715003838-20200715033838-00477.warc.gz"} |
https://www.mersenneforum.org/showthread.php?p=580993 | mersenneforum.org Getting others to do the work on exponents I like (was: Trial Factoring Progress)
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2021-06-14, 23:27 #133
tuckerkao
"Tucker Kao"
Jan 2020
Head Base M168202123
3·11·13 Posts
Quote:
Originally Posted by LaurV The best case scenario which would really crack me down, would be the following: Some white tuxedo guy comes here and make some crazy prediction about Mxx in 100M-digits being prime. He makes big fuss and convince some other guy to do the work for him. The other guy tests the Mxx. Mxx turns out to be prime. Would love to see the prophet's reaction (especially when the EFF money are awarded)
I understand the rule, if Ben Delo runs the PRPs, the prize will be his if any of them turn out to be the Mersenne Prime.
Quote:
Originally Posted by VBCurtis Mr Delo, I appreciate your sense of humor! I'm sad this kid actually gets his wish instead of doing it himself, but it's really quite something for you to help him out. :)
I've only asked to borrow 3 of Ben Delo's hundreds of fast computational engines. I will finish the P-1 and trial factoring tasks first.
I'll also run some of the PRP tests myself, but it's really not my fault that the new video card models are way above the MSP $. Last fiddled with by tuckerkao on 2021-06-14 at 23:31 2021-06-14, 23:39 #134 chalsall If I May "Chris Halsall" Sep 2002 Barbados 3·72·67 Posts Quote: Originally Posted by tuckerkao Which really makes me thinking there's indeed an unknown Mersenne Prime hidden in 1 of around 1,200 exponents of the 168,***,*23 collection. Billionaires smell the unrevealed money quicker than most other people. This is a bit like watching a James Veitch video backward. Funny in its own quirky way... 2021-06-15, 00:33 #135 moebius Jul 2009 Germany 607 Posts Quote: Originally Posted by tuckerkao I understand the rule, if Ben Delo runs the PRPs, the prize will be his if any of them turn out to be the Mersenne Prime. Is the$ 3000 award for an undefined Mersenne prime still actually? The costs for the Amazon instances for > = 2 years will probably be higher. In the area in which he currently works, he will hardly be able to bag the EFF money award.
2021-06-15, 01:29 #136
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2·32·19·29 Posts
Quote:
Originally Posted by tuckerkao Which really makes me thinking there's indeed an unknown Mersenne Prime hidden in 1 of around 1,200 exponents of the 168,***,*23 collection. Billionaires smell the unrevealed money quicker than most other people.
As has been noted by another poster, he has likely already paid more in fees for compute than the prize money. Also, I would not doubt that he is also responsible for most of the difference between the totals in this post and this one.. Likely not every cent of it, but much of it. Most likely his motivation in throwing you a bone lies somewhere other than money.
2021-06-15, 04:16 #137
slandrum
Jan 2021
California
2·89 Posts
Quote:
Originally Posted by tuckerkao Like LaurV said before, nobody can really see what exactly have been packaged inside each exponent until the box is fully opened, so it'll be a marathon contest between UncWilly and me because he believes that every exponent that I'm going to test will be composite, thus if a Mersenne Prime ever lands under my name, then he will have learn the math in the different ways.
IF this occurs, it will demonstrate exactly NOTHING mathematically. All it will show is that you had a lucky random guess. Just like picking a winning Lotto ticket - there's no skill in that (OK, there's some actual skill in maximizing expectation for your lotto numbers, picking numbers that are less likely to be picked by others will decrease the chance you will have to split the prize if your ticket is a winner - free hint, don't pick 1,2,3,4,5,6 - you'll be sorely disappointed with the payout if it hits).
Until and unless you can show a mathematical basis for your guesses (and not mystic mumbo-jumbo, base 12 is magic, or "this extremely small sample size suggests that..") your guesses actually mean nothing and might as well be random. Now there are a lot of possible candidates for Mersenne Primes, so picking some silly basis for your guesses is probably no worse than just taking the next one in line (except that it takes longer to test than the smaller untested numbers).
Last fiddled with by slandrum on 2021-06-15 at 04:18
2021-06-15, 05:26 #138 axn Jun 2003 140016 Posts Exactly this. Frankly, I expect with the amount of cranks that pass thru this forum, eventually someone will get a lucky hit.
2021-06-15, 05:35 #139 Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 26BE16 Posts
2021-06-15, 17:18 #140
tuckerkao
"Tucker Kao"
Jan 2020
Head Base M168202123
1101011012 Posts
Quote:
Originally Posted by slandrum "this extremely small sample size suggests that..") your guesses actually mean nothing and might as well be random.
UncWilly already said that the sample size of the trial factoring and P-1 have to be at least 300 to advance to the statistic arena, so I won't have any conclusion until I have at least 288 unfactored cases turned in because I already found 12 factors within the M168M range.
https://www.mersenne.ca/status/tf/0/20210201/4/16800
I'm still waiting for the moment when I will have the 1st factor found between 2^75 to 2^78 from a M168M exponent.
There are around 1,200 total available M168,***,*23 exponents, so the sample size should eventually match up the standard, slightly less than 200 of them have been touched as of today.
Last fiddled with by tuckerkao on 2021-06-15 at 17:24
2021-06-15, 18:18 #141
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×32×19×29 Posts
Quote:
Originally Posted by tuckerkao UncWilly already said that the sample size of the trial factoring and P-1 have to be at least 300 to advance to the statistic arena,
That is watermelons vs. bowling balls. Figuring out if your factoring work is tracking what it should be is 1 thing. Finding a new prime is something totally different. Your monomania for baseless base claims has blinded you to reality.
2021-06-17, 18:23 #142
tuckerkao
"Tucker Kao"
Jan 2020
Head Base M168202123
3×11×13 Posts
Quote:
Originally Posted by Viliam Furik If he doesn't want to be advised, let him be and don't waste your time. We've got a saying here, in Slovakia: "Kto chce kam, pomôžme mu tam." Roughly translated as "Who wants (to go) where, let's help them there." If he wants a liquid cooler so much, you're better off recommending him one, than convincing him to choose an air cooler.
Those air cooler units come to the standard size to fit the regular CPU dimensions, but AMD Threadripper is wider, the PC seller doesn't have the dual air cooler units to fit from the top of MSI motherboard: TRX40 PRO WIFI ATX w/ WiFi 6, RGB, Dual LAN, 4 PCIe x16, 1 PCIe x1, 8 SATA3, 2 M.2 SATA/PCIe. Most Threadripper owners install the liquid cooling systems because of the modular dimensions.
I'll patiently wait for the new PC model, I can barely prepare up to 2 to 3 M168M exponents for Ben Delo to run the PRPs per week with my current PC.
M168174323, M168314323 and M168704323 have reached the PRP ready status (TF to 2^78 and P-1 finished), I don't know when Ben Delo will be online again, he can take those when he checks back here the next time.
Last fiddled with by tuckerkao on 2021-06-17 at 18:49
2021-06-17, 19:22 #143
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×32×19×29 Posts
Quote:
Originally Posted by tuckerkao M168174323, M168314323 and M168704323 have reached the PRP ready status (TF to 2^78 and P-1 finished), I don't know when
You don't think that someone as smart as he is could automate the process if he wanted to?
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A copy of the license is included in the FAQ. | 2021-09-19 17:06:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3564724922180176, "perplexity": 4664.456037468429}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00697.warc.gz"} |
https://www.techwhiff.com/issue/find-the-slope-and-y-intercept-of-the-line-3x-4y-7--517263 | # Find the slope and y-intercept of the line. 3x+4y=-7 write your answers in simplest form. slope : y - intercept : (spammers reported) will mark the brainliest.
###### Question:
Find the slope and y-intercept of the line.
3x+4y=-7
slope :
y - intercept :
(spammers reported)
will mark the brainliest.
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Question 4 of 10 2 Points Directions to the Presidio Golf Course & Clubhouse FROM DOWNTOWN Take Geary Blvd. to Arguello Blvd. Turn right on Arguello. Go one-half mile up Arguello through the historic Arguello Gates. Golf course entrance is to the right, 50 yards from the gates. This is an exampl...
### CHEETAHT PEP ABILITY PRACTICE QUESTIONS QUESTION 22 There is an equal number of red, blue and yellow towels in your Aunt's suitcase. How many towels could be in the suitcase? A. 16 B. 29 C. 37 D. 42
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### A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120 v outlet and draws 0.050 a, what are the voltage and current of the secondary coil?1200v,0.0050 a1200v, 0.50a12v,00050a12v,0.50a
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### Riverbed Company sells goods that cost $320,000 to Ricard Company for$407,000 on January 2, 2020. The sales price includes an installation fee, which has a standalone selling price of $38,500. The standalone selling price of the goods is$368,500. The installation is considered a separate performance obligation and is expected to take 6 months to complete. a. Prepare the journal entries (if any) to record the sale on January 2, 2020. (Credit account titles are automatically indented when amount
Riverbed Company sells goods that cost $320,000 to Ricard Company for$407,000 on January 2, 2020. The sales price includes an installation fee, which has a standalone selling price of $38,500. The standalone selling price of the goods is$368,500. The installation is considered a separate performan... | 2022-11-28 09:29:27 | {"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": 0.22238248586654663, "perplexity": 11610.793814301696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00212.warc.gz"} |
https://math.stackexchange.com/questions/3955177/vol-1-shafarevich-two-questions-about-finite-maps | (Vol. 1 Shafarevich) Two questions about finite maps
I am currently reading Basic Algebraic Geometry by Shafarevich. I have two questions about finite maps on affine and quasi-projective varieites.
For a finite map $$f:X\rightarrow Y$$ between affine varieties $$X$$ and $$Y$$, we say that $$f$$ is finite if $$f(X)$$ is dense in $$Y$$ and $$k[X]$$ is integral over $$k[Y]$$ (where $$k[Y]$$ has been identified with a subring of $$k[X]$$).
Question 1
The first question is about the proof of Theorem 1.13. The statement of the theorem is as follows.
If $$f:X\rightarrow Y$$ is a regular map of affine varieties, and every $$x\in Y$$ has an affine neighbourhood $$U\ni x$$ such that $$V = f^{-1}(U)$$ is affine and $$f:V\rightarrow U$$ is finite, then $$f$$ itself is finite.
The proof itself is not so hard, but I find myself unable to understand the first step, which is to assume that each neighbourhood $$U$$ in the statement of the theorem above is a principal open set.
As far as I understand it, the neighbourhood $$U$$ of $$x$$ may be decomposed into a finite union of principal open sets $$U_1,\ldots, U_r$$ and for each $$U_i$$, the set $$V_i := f^{-1}(U_i)$$ is also a principal open set in $$X$$. However, I fail to understand why $$f$$ must be finite when restricted to any of these sets $$V_i$$.
Question 2
The definition of finite maps on quasi-projective varieties follows the "local definition" above. More precisely,
A map $$f: X \rightarrow Y$$ of quasi-projective varieties is finite if any point $$y\in Y$$ has an affine neighbourhood $$U$$ such that $$V = f^{-1}(U)$$ is affine and $$f:V \rightarrow U$$ is a finite map between affine varieties.
This definition makes sense given the theorem above, but it got me wondering regarding the following fairly easy fact about finite maps between affine varieties.
Fact: If $$X,Y$$ are affine varieties and $$f:X\rightarrow Y$$ and $$g:Y\rightarrow Z$$ are finite maps, then so is their composition $$g\circ f$$.
It's not clear to me if the above holds for finite maps between quasiprojective varieties as defined above. The "obvious" proof again runs into a similar obstacle where one needs to show that a restriction of a finite map to some open affine subset is still finite.
Possibly I am missing some simple clean fact about finite maps. If so, I would like to know it.
For question (1), what you are trying to show is that if you have a finite ring map $$A \to B$$ and some element $$f \in A$$ then $$A_f \to B_f$$ is finite (meaning invert the multiplicative subset of powers of $$f$$ in both rings). That is, we have some $$A$$-module surjection $$A^n \twoheadrightarrow B$$. Since localization is exact, and it commutes with direct sums, we get a surjection $$A_f^n \twoheadrightarrow B_f$$, as desired. | 2021-04-20 20:58:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 41, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9553783535957336, "perplexity": 83.3552225369406}, "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-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00470.warc.gz"} |
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# Absolute & Relative References (Excel 97 Macro help, please)
## Absolute & Relative References (Excel 97 Macro help, please)
(OP)
In programming a macro, I would like to have an ABSOLUTE column reference, with a REALITIVE row.
Example,
I have 4 column locations I want to go (ie. C,Z,BB,CD). I want to make 4 seperate macros to go to the 4 seperate locations, with the following
condiditions:
a) I want to mantain my row position (ie. if the ActiveCell is "C11" and I run the "Z" macro, I want my final resting point to be "Z11"
b) I want to be able to run the 4 macros in ANY order (This is the part I am having trouble with)
One possible solution I had was if there were some command (within VB) that would sense where the ActiveCell was and be able to return a
row/column value. I could then use these values in a formula that would end up using the Scroll to move the screen accordingly command.
I am really in a jam on this, so I would appreciate any help or advice that you could provide.
### RE: Absolute & Relative References (Excel 97 Macro help, please)
I think it's not so big a problem. I only wonder if you are programming in VB or you are programming in Excel using VB for Applications. Because if you're programming in VBA then there is an easy solution:
To get the Row or the Column simply change Address in to Row or Column.
The return types of the Adress function is Range. This is just a string. For the Row and Coumn function the return type is a LongInt.
For example the highlighted cell is C11. ActiveCell.Address would return "$C$11", while ActiveCell.Row would give the number 11.
Now I think it's not so hard to select the right column on the right row.
Good luck!
Jonathan
### RE: Absolute & Relative References (Excel 97 Macro help, please)
Cactus13 has explained it very well. I'll just add something on how the macros should be organised:
Say we have the following Macros: MacroC, MacroZ, MacroBB, MacroCD - the names can be anything - I've just named them so their target column would be obvious. You'd write the code something like this:
In the Declaration Section of the module, put the statement:
Dim TargetCell as Range
Then write the macros:
Sub MacroC
Set TargetCell=Activesheet.Cells(Activecell.row,Columns("C:C").Column)
Do what u want with TargetCell in the statements that follow...
End Sub
Sub MacroZ
Set TargetCell=ActiveSheet.Cells(ActiveCell.row,Columns("Z:Z").Column)
Do what u want with TargetCell in the statements that follow...
End Sub
and so on...
Note that there's really no need to refer to the columns as "C:C" or "Z:Z" - simply using "C" or "Z" will also do.
All the best...
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Register now while it's still free! | 2020-04-07 01:25:24 | {"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": 0.24386657774448395, "perplexity": 4393.685952395877}, "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/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00008.warc.gz"} |
http://apmonitor.com/wiki/index.php/Main/Control?action=diff&source=n&minor=y | Main
## Main.Control History
September 19, 2018, at 03:27 PM by 174.148.126.64 -
September 19, 2018, at 03:26 PM by 174.148.126.64 -
Changed lines 58-112 from:
The objective is formulated from Controlled Variables (CVs). The CVs may be controlled to a range, a trajectory, maximized, or minimized. The CVs are an expression of the desired outcome for the controller action.
to:
The objective is formulated from Controlled Variables (CVs). The CVs may be controlled to a range, a trajectory, maximized, or minimized. The CVs are an expression of the desired outcome for the controller action.
#### MPC Example with Python GEKKO
(:source lang=python:) from gekko import GEKKO import numpy as np import matplotlib.pyplot as plt
m = GEKKO() m.time = np.linspace(0,20,41)
1. Parameters
mass = 500 b = m.Param(value=50) K = m.Param(value=0.8)
1. Manipulated variable
p = m.MV(value=0, lb=0, ub=100) p.STATUS = 1 # allow optimizer to change p.DCOST = 0.1 # smooth out gas pedal movement p.DMAX = 20 # slow down change of gas pedal
1. Controlled Variable
v = m.CV(value=0) v.STATUS = 1 # add the SP to the objective m.options.CV_TYPE = 2 # squared error v.SP = 40 # set point v.TR_INIT = 1 # set point trajectory v.TAU = 5 # time constant of trajectory
1. Process model
m.Equation(mass*v.dt() == -v*b + K*b*p)
m.options.IMODE = 6 # control m.solve(disp=False,GUI=True)
import json with open(m.path+'//results.json') as f:
results = json.load(f)
plt.figure() plt.subplot(2,1,1) plt.plot(m.time,p.value,'b-',label='MV Optimized') plt.legend() plt.ylabel('Input') plt.subplot(2,1,2) plt.plot(m.time,results['v1.tr'],'k-',label='Reference Trajectory') plt.plot(m.time,v.value,'r--',label='CV Response') plt.ylabel('Output') plt.xlabel('Time') plt.legend(loc='best') plt.show() (:sourceend:)
June 09, 2017, at 01:01 AM by 10.5.113.159 -
Changed lines 45-49 from:
APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications. The DBS file parameter imode is used to control the simulation mode. This option is set to 6 or 9 for nonlinear control.
nlc.imode = 6 (simultaneous dynamic control)
nlc.imode = 9 (sequential dynamic control)
to:
APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications. The DBS file parameter IMODE is used to control the simulation mode. This option is set to 6 or 9 for nonlinear control.
apm.imode = 6 (simultaneous dynamic control)
apm.imode = 9 (sequential dynamic control)
Changed lines 51-52 from:
apm_option(server,app,'nlc.imode',6);
to:
apm_option(server,app,'apm.imode',6);
Changed line 54 from:
apm_option(server,app,'nlc.imode',9)
to:
apm_option(server,app,'apm.imode',9)
The model1.apm contains a linear first-order differential equation. Other versions are model2.apm (continuous state space) and model3.apm (discrete state space). Each is applied in a model predictive controller to follow a reference trajectory and reach a target value of 7.0.
### Python Model Predictive Control
Continuous and discrete state space models are used in a Python script for Model Predictive Control.
June 16, 2015, at 07:03 PM by 45.56.3.184 -
Changed line 42 from:
apm_option(server,app,'nlc.imode',9);
to:
apm_option(server,app,'nlc.imode',9)
June 16, 2015, at 07:02 PM by 45.56.3.184 -
Changed lines 35-37 from:
• NLC.imode = 6 (simultaneous approach)
• NLC.imode = 9 (sequential approach)
to:
nlc.imode = 6 (simultaneous dynamic control)
nlc.imode = 9 (sequential dynamic control)
% MATLAB example
apm_option(server,app,'nlc.imode',6);
# Python example
apm_option(server,app,'nlc.imode',9);
Deleted lines 46-47:
June 08, 2015, at 10:43 PM by 10.5.113.170 -
Deleted lines 4-7:
## Modeling for Dynamic Optimization
One of the benefits derived from a modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include excessive dependence on empirical formulations.
Deleted lines 7-8:
June 08, 2015, at 03:21 PM by 45.56.3.184 -
Changed lines 41-42 from:
NLC.imode = 6 (simultaneous approach) NLC.imode = 9 (sequential approach)
to:
• NLC.imode = 6 (simultaneous approach)
• NLC.imode = 9 (sequential approach)
June 08, 2015, at 03:20 PM by 45.56.3.184 -
Changed lines 39-41 from:
APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications. The DBS file parameter imode is used to control the simulation mode. This option is set to 6 for nonlinear control.
NLC.imode = 6
to:
APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications. The DBS file parameter imode is used to control the simulation mode. This option is set to 6 or 9 for nonlinear control.
NLC.imode = 6 (simultaneous approach) NLC.imode = 9 (sequential approach)
November 18, 2013, at 11:36 PM by 69.169.137.17 -
Changed line 39 from:
The DBS file parameter imode is used to control the simulation mode. This option is set to 6 for nonlinear control.
to:
APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications. The DBS file parameter imode is used to control the simulation mode. This option is set to 6 for nonlinear control.
November 18, 2013, at 11:36 PM by 69.169.137.17 -
Changed lines 5-7 from:
## Nonlinear Control
One of the benefits derived from a modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical formulations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
to:
## Modeling for Dynamic Optimization
One of the benefits derived from a modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include excessive dependence on empirical formulations.
November 18, 2013, at 11:35 PM by 69.169.137.17 -
(:html:) <iframe width="560" height="315" src="//www.youtube.com/embed/YHAA-uXhI0E?list=PLLBUgWXdTBDgOlvGgbBDvHXdleSgEi9aj" frameborder="0" allowfullscreen></iframe> (:htmlend:)
November 18, 2013, at 06:10 PM by 69.169.137.17 -
November 18, 2013, at 06:09 PM by 69.169.137.17 -
### MATLAB Toolbox for Model Predictive Control
Model Predictive Control (MPC) predicts and optimizes time-varying processes over a future time horizon. This control package accepts linear or nonlinear models. Using large-scale nonlinear programming solvers such as APOPT and IPOPT, it solves data reconciliation, moving horizon estimation, real-time optimization, dynamic simulation, and nonlinear MPC problems.
Three example files are contained in this directory that implement a controller for Linear Time Invariant (LTI) systems:
1. apm1_lti - translate any LTI model into APM format
2. apm2_step - perform step tests to ensure model accuracy
3. apm3_control - MPC setpoint change to new target values
November 18, 2013, at 05:33 PM by 69.169.137.17 -
(:title Model Predictive Control:) (:keywords nonlinear, model, predictive control, moving horizon, differential, algebraic, modeling language:) (:description Tutorial in Excel / Simulink / MATLAB for implementing Model Predictive Control for linear or nonlinear systems.:)
Changed lines 7-8 from:
One of the largest benefits derived from the modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical formulations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
to:
One of the benefits derived from a modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical formulations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
Changed line 17 from:
to:
November 18, 2013, at 05:29 PM by 69.169.137.17 -
Changed lines 3-4 from:
One of the largest benefits derived from the modeling effort is the direct application of the model for optimization. Application of the full nonlinear model directly for control is an elegant but immature field. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical fomulations. These empirical formulations are generally unreliable during abnormal situations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
to:
One of the largest benefits derived from the modeling effort is the direct application of the model for optimization. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical formulations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
### Model Predictive Control Tutorial
A basic Model Predictive Control (MPC) tutorial demonstrates the capability of a solver to determine a dynamic move plan. In this example, a linear dynamic model is used with the Excel solver to determine a sequence of manipulated variable (MV) adjustments that drive the controlled variable (CV) along a desired reference trajectory.
September 12, 2010, at 03:28 AM by 206.180.155.75 -
Changed lines 3-5 from:
Real-time Optimization and Control
The largest benefit derived from the modeling effort is the direct application of the model for optimization. Application of the full nonlinear model directly for control is an elegant but immature field. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical fomulations. These empirical formulations are generally unreliable during abnormal situations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
to:
One of the largest benefits derived from the modeling effort is the direct application of the model for optimization. Application of the full nonlinear model directly for control is an elegant but immature field. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical fomulations. These empirical formulations are generally unreliable during abnormal situations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
September 12, 2010, at 03:27 AM by 206.180.155.75 -
Real-time Optimization and Control
The largest benefit derived from the modeling effort is the direct application of the model for optimization. Application of the full nonlinear model directly for control is an elegant but immature field. When solution robustness or speed are concerns, linearizations of the model may be practical. Other compromises of the model include eccessive dependence on empirical fomulations. These empirical formulations are generally unreliable during abnormal situations. APMonitor enables the use of empirical, hybrid, and fundamental models directly in control applications.
October 02, 2008, at 09:02 PM by 158.35.225.228 -
Nonlinear control adjusts variables that are declared as Manipulated Variables (MVs) to meet an objective. The MVs are the handles that the optimizer uses to minimize an objective function.
The objective is formulated from Controlled Variables (CVs). The CVs may be controlled to a range, a trajectory, maximized, or minimized. The CVs are an expression of the desired outcome for the controller action.
September 30, 2008, at 03:22 PM by 158.35.225.227 - | 2018-09-22 10:39:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35056549310684204, "perplexity": 4393.002905544177}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158320.19/warc/CC-MAIN-20180922103644-20180922124044-00169.warc.gz"} |
https://brilliant.org/problems/easy-but-complicated-2-p/ | # Not too Complicated
Algebra Level 4
Let $$f(x) = \frac{x-1}{x+1}$$ and $$f^{n}(x)$$ denote the $$n-\text{fold}$$ composition of $$f$$ with itself. That is, $$f^1(x) =f(x)$$ and $$f^n(x) = f(f^{n-1}(x))$$. Find $$f^{2007}(2)$$.
× | 2017-05-30 13:22:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9750317335128784, "perplexity": 391.199244789767}, "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-2017-22/segments/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00492.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-p-section-p-1-algebraic-expressions-mathematical-models-and-real-numbers-exercise-set-page-18/126 | ## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter P - Section P.1 - Algebraic Expressions, Mathematical Models, and Real Numbers - Exercise Set: 126
#### Answer
$6x-(-2x)$ $8x$
#### Work Step by Step
The product of 6 and x is simply them multiplied together: $6x$. Two times $x$ is $2x$. Negative two times $x$ is $-2x$. The difference between $6x$ and $-2x$ is the second subtracted from the first: $6x-(-2x)$, which equals $8x$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-07-17 06:20:11 | {"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": 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": 0.6650407910346985, "perplexity": 1457.214070799173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00639.warc.gz"} |
https://nrich.maths.org/1888/solution | ### Tri.'s
How many triangles can you make on the 3 by 3 pegboard?
### Cutting Corners
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Bracelets
Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?
# Triangle Island
##### Age 7 to 11 Challenge Level:
Hugo sent us pictures of his solution. Well done, Hugo!
I know that the area of a triangle is given by the forumula $\frac{1}{2} \times base \times height.$ The original triangle has base $1$ and height $1$, so as long as each triangle on the way also has base $1$ and height $1$ the area will stay the same". | 2019-11-15 02:11:20 | {"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": 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": 0.36171090602874756, "perplexity": 794.4645326936175}, "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/1573496668561.61/warc/CC-MAIN-20191115015509-20191115043509-00015.warc.gz"} |
https://www.physicsforums.com/threads/question-about-an-equation.935909/ | # I Question about an equation
1. Jan 2, 2018
### Arman777
Let's suppose we have an equation,
$$Ω_m+Ω_r+Ω_k=1$$
In this equation whats the values that $Ω_k$ can take ?
Only 1,0 or -1 ?
Also this equation is true for only "now" or any time in the universe ?
2. Jan 2, 2018
### PeroK
Assuming the omegas are positive, then $0 \le \Omega_k \le 1$
If you allow negative values, then $\Omega_k$ can be any real number.
3. Jan 2, 2018
### Arman777
For flat universe we can take $Ω_k=0$. In example for positive curvature universe then what could be the value of $Ω_k$ ?
A negative value ?
Cause $Ω_k=-\frac {κ} {a^2H^2}$
Something seems wrong to me. Either this equation only hold for $Ω_k=0$.
Oh okay I understand
4. Jan 2, 2018
### Arman777
For a negative value of $Ω_k$ corresponds to $Ω>1$ which its positively curved universe
For $Ω_k=0$ corresponds to $Ω=1$ which flat universe
For a postive value of $Ω_k$ corresponds to $Ω<1$ which its negatively curved universe
where $Ω=Ω_m+Ω_r$
5. Jan 2, 2018
### kimbyd
$\Omega_k$ can be any real number that makes the equation above true. Because $\Omega_m$ and $\Omega_r$ must be positive, $\Omega_k$ cannot be greater than 1. But it can be as negative as you like.
This is accurate.
6. Jan 3, 2018
### Arman777
Thanks, its more clear now | 2018-01-21 11:05:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5442320704460144, "perplexity": 1112.9127178242768}, "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-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00719.warc.gz"} |
https://nbviewer.org/github/nell-paquit/learning_predictive_customer_analytics/blob/master/1_customer_propensity_score/LearningPropensity.ipynb | # Predicting Customer Propensity using Classification Algorithms¶
In this notebook, we will show you how to predict customer propensity to buy the product based on his/her interactions on the website. Using that propensity, if a certain threshold are reached we will then decide whether to assign an agent to offer a chat to the customer. We will also be using different classification algorithms to show how each algorithms are coded.
Because we will only be using sample data for exercise purposes, the data set that we'll be using here are very small (500 rows of data). Thus, we might not get a real accurate predictions out of it.
## Call to Action¶
• Analyse real-time customer's actions on website
• Predict the propensity score
• Offer chat once propensity score exceeds threshold
We will load the data file then checkout the summary statistics and columns for that file.
In [1]:
import pandas as pd
import numpy as np
import os
from sklearn.model_selection import train_test_split, cross_val_score
from sklearn.naive_bayes import GaussianNB
from sklearn.linear_model import LogisticRegression, SGDClassifier
from sklearn.neighbors import KNeighborsClassifier
from sklearn.tree import DecisionTreeClassifier
from sklearn.ensemble import RandomForestClassifier
from sklearn.svm import SVC
import sklearn.metrics
# After loading the data, we look at the data types to make sure that the data has been loaded correctly.
customer_data.dtypes
Out[1]:
SESSION_ID int64
IMAGES int64
REVIEWS int64
FAQ int64
SPECS int64
SHIPPING int64
BOUGHT_TOGETHER int64
COMPARE_SIMILAR int64
VIEW_SIMILAR int64
WARRANTY int64
dtype: object
The data contains information about the various links on the website that are clicked by the user during his browsing. This is past data that will be used to build the model.
• Session ID : A unique identifier for the web browsing session
• Other columns : a boolean indicator to show whether the prospect visited that particular page or did the activity mentioned.
In [2]:
# View the top records to understand how the data looks like.
Out[2]:
0 1001 0 0 1 0 1 0 0 0 1 0 0
1 1002 0 1 1 0 0 0 0 0 0 1 0
2 1003 1 0 1 1 1 0 0 0 1 0 0
3 1004 1 0 0 0 1 1 1 0 0 0 0
4 1005 1 1 1 0 1 0 1 0 0 0 0
In [3]:
#Do summary statistics analysis of the data to make sure the data is not skewed in any way.
customer_data.describe()
Out[3]:
count 500.000000 500.000000 500.0000 500.000000 500.0000 500.000000 500.000000 500.000000 500.000000 500.000000 500.000000 500.000000
mean 1250.500000 0.510000 0.5200 0.440000 0.4800 0.528000 0.500000 0.580000 0.468000 0.532000 0.550000 0.370000
std 144.481833 0.500401 0.5001 0.496884 0.5001 0.499715 0.500501 0.494053 0.499475 0.499475 0.497992 0.483288
min 1001.000000 0.000000 0.0000 0.000000 0.0000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
25% 1125.750000 0.000000 0.0000 0.000000 0.0000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
50% 1250.500000 1.000000 1.0000 0.000000 0.0000 1.000000 0.500000 1.000000 0.000000 1.000000 1.000000 0.000000
75% 1375.250000 1.000000 1.0000 1.000000 1.0000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000
max 1500.000000 1.000000 1.0000 1.000000 1.0000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000
### Perform Correlation Analysis¶
In [4]:
customer_data.corr()['BUY']
Out[4]:
SESSION_ID 0.026677
IMAGES 0.046819
REVIEWS 0.404628
FAQ -0.095136
SPECS 0.009950
SHIPPING -0.022239
BOUGHT_TOGETHER -0.103562
COMPARE_SIMILAR 0.190522
VIEW_SIMILAR -0.096137
WARRANTY 0.179156
Name: BUY, dtype: float64
Looking at the correlations above we can see that features like REVIEWS, BOUGHT_TOGETHER, COMPARE_SIMILAR, WARRANTY and SPONSORED_LINKS have medium correlation to the target variable. We will reduce our feature set to only those features that have some good correlation.
In [5]:
#Drop columns with low correlation
## Training and Testing Split¶
We now split the model into training and testing data in the ratio of 70:30
In [6]:
pred_train, pred_test, tar_train, tar_test = train_test_split(predictors, targets, test_size=.3)
print( "Predictor_Training :", pred_train.shape," | ", "Predictor_Testing :", pred_test.shape )
Predictor_Training : (350, 5) | Predictor_Testing : (150, 5)
## Build Model and Check Accuracy¶
Using Naïve Bayes Classifier
Definition: Naïve Bayes algorithm based on Bayes’ theorem with the assumption of independence between every pair of features. Naive Bayes classifiers work well in many real-world situations such as document classification and spam filtering.
Advantages: This algorithm requires a small amount of training data to estimate the necessary parameters. Naive Bayes classifiers are extremely fast compared to more sophisticated methods.
In [7]:
nb = GaussianNB()
nb.fit(pred_train, tar_train)
nb_predictions = nb.predict(pred_test)
#Analyze accuracy of nb predictions
sklearn.metrics.confusion_matrix(tar_test, nb_predictions)
Out[7]:
array([[76, 20],
[24, 30]], dtype=int64)
In [8]:
sklearn.metrics.accuracy_score(tar_test, nb_predictions)
Out[8]:
0.7066666666666667
In [9]:
print(sklearn.metrics.classification_report(tar_test, nb_predictions))
precision recall f1-score support
0 0.76 0.79 0.78 96
1 0.60 0.56 0.58 54
micro avg 0.71 0.71 0.71 150
macro avg 0.68 0.67 0.68 150
weighted avg 0.70 0.71 0.70 150
Using Logistic Regression Classifier
Definition: Logistic regression is a machine learning algorithm for classification. In this algorithm, the probabilities describing the possible outcomes of a single trial are modelled using a logistic function.
Advantages: Logistic regression is designed for this purpose (classification), and is most useful for understanding the influence of several independent variables on a single outcome variable.
Disadvantages: Works only when the predicted variable is binary, assumes all predictors are independent of each other, and assumes data is free of missing values.
In [10]:
lr = LogisticRegression(solver='liblinear')
lr.fit(pred_train, tar_train)
lr_predictions = lr.predict(pred_test)
#Analyze accuracy of predictions
sklearn.metrics.confusion_matrix(tar_test, lr_predictions)
Out[10]:
array([[77, 19],
[28, 26]], dtype=int64)
In [11]:
sklearn.metrics.accuracy_score(tar_test, lr_predictions)
Out[11]:
0.6866666666666666
In [12]:
print(sklearn.metrics.classification_report(tar_test, lr_predictions))
precision recall f1-score support
0 0.73 0.80 0.77 96
1 0.58 0.48 0.53 54
micro avg 0.69 0.69 0.69 150
macro avg 0.66 0.64 0.65 150
weighted avg 0.68 0.69 0.68 150
Definition: Stochastic gradient descent is a simple and very efficient approach to fit linear models. It is particularly useful when the number of samples is very large. It supports different loss functions and penalties for classification.
Advantages: Efficiency and ease of implementation.
Disadvantages: Requires a number of hyper-parameters and it is sensitive to feature scaling.
In [13]:
sgd = SGDClassifier(max_iter=1000, shuffle=True, tol=1e-3, random_state=101)
sgd.fit(pred_train, tar_train)
sgd_predictions = sgd.predict(pred_test)
#Analyze accuracy of sgd predictions
sklearn.metrics.confusion_matrix(tar_test, sgd_predictions)
Out[13]:
array([[84, 12],
[35, 19]], dtype=int64)
In [14]:
sklearn.metrics.accuracy_score(tar_test, sgd_predictions)
Out[14]:
0.6866666666666666
In [15]:
print(sklearn.metrics.classification_report(tar_test, sgd_predictions))
precision recall f1-score support
0 0.71 0.88 0.78 96
1 0.61 0.35 0.45 54
micro avg 0.69 0.69 0.69 150
macro avg 0.66 0.61 0.61 150
weighted avg 0.67 0.69 0.66 150
Using K-Nearest Neighbors Classifier
Definition: Neighbours based classification is a type of lazy learning as it does not attempt to construct a general internal model, but simply stores instances of the training data. Classification is computed from a simple majority vote of the k nearest neighbours of each point.
Advantages: This algorithm is simple to implement, robust to noisy training data, and effective if training data is large.
Disadvantages: Need to determine the value of K and the computation cost is high as it needs to computer the distance of each instance to all the training samples.
In [16]:
knn = KNeighborsClassifier(n_neighbors=5)
knn.fit(pred_train, tar_train)
knn_predictions = knn.predict(pred_test)
#Analyze accuracy of knn predictions
sklearn.metrics.confusion_matrix(tar_test, knn_predictions)
Out[16]:
array([[75, 21],
[20, 34]], dtype=int64)
In [17]:
sklearn.metrics.accuracy_score(tar_test, knn_predictions)
Out[17]:
0.7266666666666667
In [18]:
print(sklearn.metrics.classification_report(tar_test, knn_predictions))
precision recall f1-score support
0 0.79 0.78 0.79 96
1 0.62 0.63 0.62 54
micro avg 0.73 0.73 0.73 150
macro avg 0.70 0.71 0.70 150
weighted avg 0.73 0.73 0.73 150
Using Decision Tree Classifier
Definition: Given a data of attributes together with its classes, a decision tree produces a sequence of rules that can be used to classify the data.
Advantages: Decision Tree is simple to understand and visualise, requires little data preparation, and can handle both numerical and categorical data.
Disadvantages: Decision tree can create complex trees that do not generalise well, and decision trees can be unstable because small variations in the data might result in a completely different tree being generated.
In [19]:
dtree = DecisionTreeClassifier(max_depth=10, min_samples_leaf=15, random_state=101)
dtree.fit(pred_train, tar_train)
dtree_predictions = dtree.predict(pred_test)
#Analyze accuracy of dtree predictions
sklearn.metrics.confusion_matrix(tar_test,dtree_predictions)
Out[19]:
array([[83, 13],
[29, 25]], dtype=int64)
In [20]:
sklearn.metrics.accuracy_score(tar_test, dtree_predictions)
Out[20]:
0.72
In [21]:
print(sklearn.metrics.classification_report(tar_test, dtree_predictions))
precision recall f1-score support
0 0.74 0.86 0.80 96
1 0.66 0.46 0.54 54
micro avg 0.72 0.72 0.72 150
macro avg 0.70 0.66 0.67 150
weighted avg 0.71 0.72 0.71 150
Using Random Forest Classifier
Definition: Random forest classifier is a meta-estimator that fits a number of decision trees on various sub-samples of datasets and uses average to improve the predictive accuracy of the model and controls over-fitting. The sub-sample size is always the same as the original input sample size but the samples are drawn with replacement.
Advantages: Reduction in over-fitting and random forest classifier is more accurate than decision trees in most cases.
Disadvantages: Slow real time prediction, difficult to implement, and complex algorithm.
In [22]:
rf = RandomForestClassifier(n_estimators=100, oob_score=True, n_jobs=-1,
random_state=101, max_features=None, min_samples_leaf=30)
rf.fit(pred_train, tar_train)
rf_predictions = rf.predict(pred_test)
#Analyze accuracy of rf predictions
sklearn.metrics.confusion_matrix(tar_test, rf_predictions)
Out[22]:
array([[74, 22],
[23, 31]], dtype=int64)
In [23]:
sklearn.metrics.accuracy_score(tar_test, rf_predictions)
Out[23]:
0.7
In [24]:
print(sklearn.metrics.classification_report(tar_test, rf_predictions))
precision recall f1-score support
0 0.76 0.77 0.77 96
1 0.58 0.57 0.58 54
micro avg 0.70 0.70 0.70 150
macro avg 0.67 0.67 0.67 150
weighted avg 0.70 0.70 0.70 150
Using Support Vector Machine Classifier
Definition: Support vector machine is a representation of the training data as points in space separated into categories by a clear gap that is as wide as possible. New examples are then mapped into that same space and predicted to belong to a category based on which side of the gap they fall.
Advantages: Effective in high dimensional spaces and uses a subset of training points in the decision function so it is also memory efficient.
Disadvantages: The algorithm does not directly provide probability estimates, these are calculated using an expensive five-fold cross-validation.
In [25]:
svm = SVC(C=1, kernel='linear', gamma='auto')
svm.fit(pred_train, tar_train)
svm_predictions = svm.predict(pred_test)
#Analyze accuracy of svm predictions
sklearn.metrics.confusion_matrix(tar_test, svm_predictions)
Out[25]:
array([[83, 13],
[30, 24]], dtype=int64)
In [26]:
#sklearn.metrics.accuracy_score(tar_test, svm_predictions)
scores = cross_val_score(svm, predictors, targets, cv=5)
scores.mean()
Out[26]:
0.722
In [27]:
print(sklearn.metrics.classification_report(tar_test, svm_predictions))
precision recall f1-score support
0 0.73 0.86 0.79 96
1 0.65 0.44 0.53 54
micro avg 0.71 0.71 0.71 150
macro avg 0.69 0.65 0.66 150
weighted avg 0.70 0.71 0.70 150
## Using probability as propensity score¶
Instead of doing a Yes/No prediction, we want to predict the probability of somebody who wants to buy, and we can do that by using a method called predict_proba.
Let's say we use NB model:
In [28]:
pred_prob=nb.predict_proba(pred_test)
pred_prob[0,1]
Out[28]:
0.10621579226568478
The probability above can be read as 11% chance that the customer will buy the product.
## Real time predictions¶
From the models we examine, let's say the best model to use for our real time prediction is the Naive Bayes.
So when the customer starts visiting the pages one by one, we collect that list and then use it to compute the probability. We do that for every new click that comes in.
So let us start. The prospect just came to your website. There are no significant clicks. Let us compute the probability. The array of values passed has the values for REVIEWS, BOUGHT_TOGETHER, COMPARE_SIMILAR, WARRANTY and SPONSORED_LINKS. So the array is all zeros to begin with
In [29]:
browsing_data = np.array([0,0,0,0,0]).reshape(1, -1)
print("New visitor: propensity :", nb.predict_proba(browsing_data)[:,1] )
New visitor: propensity : [0.04544815]
So the initial probability is 5%. Now, suppose the customer does a comparison of similar products. The array changes to include a 1 for that function. The new probability will be
In [30]:
browsing_data = np.array([0,0,1,0,0]).reshape(1, -1)
print("After checking similar products: propensity :", nb.predict_proba(browsing_data)[:,1] )
After checking similar products: propensity : [0.1082966]
It goes up. Next, he checks out reviews.
In [31]:
browsing_data = np.array([1,0,1,0,0]).reshape(1, -1)
print("After checking reviews: propensity :", nb.predict_proba(browsing_data)[:,1] )
After checking reviews: propensity : [0.5612052]
It shoots up to 50+%. You can have a threshold for when you want to offer chat. You can keep checking this probability against that threshold to see if you want to popup a chat window.
This example shows you how you can use predictive analytics in real time to decide whether a customer has high propensity to convert and offer him a chat with a sales rep/agent.
## References:¶
Next Topic: How to recommend products to customer based on items affinity? | 2021-12-07 06:47:01 | {"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": 0.29533275961875916, "perplexity": 2227.8730794128064}, "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/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00067.warc.gz"} |
https://crypto.stackexchange.com/questions/37259/who-invented-the-1n-notation?noredirect=1 | # Who invented the $1^n$ notation? [duplicate]
I am always wondering about the seemingly odd syntax $1^n$ which designates the security strength of a cryptographic algorithm.
Who invented this (odd) syntax and what was the reason for inventing it?
• It definitely comes from theory, and specifically formal language theory, which means it really comes from mathematical logic. If I had to guess, I'd say S.C. Kleene or Church are likely culprits. – pg1989 Jun 23 '16 at 17:29
• It's not odd at all, it's the string '1' concatenated with itself n times. – fkraiem Jun 23 '16 at 18:02
• @fkraiem That's odd by definition, isn't it? – Maarten Bodewes Jun 23 '16 at 19:51
• @MaartenBodewes Well that depends on your encoding. Usually it's a unary representation of the security parameter, so it's odd if and only if $n$ is odd. – fkraiem Jun 23 '16 at 19:58 | 2020-07-10 15:56:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2522324323654175, "perplexity": 1051.843648217321}, "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/1593655911092.63/warc/CC-MAIN-20200710144305-20200710174305-00034.warc.gz"} |
https://juliapolyhedra.github.io/Polyhedra.jl/stable/generated/Extended%20Formulation/ | # Extended formulation
In this notebook, we show how to work with the extended formulation of a polyhedron. The convex hull of the union of polyhedra that are H-represented can be obtained as the projection of a H-representation [Theorem 3.3, B85]. In order to use the resulting polyhedron as a constraint set in an optimization problem, there is no need to compute the resulting H-representation of this projection. Moreover, other operations such as intersection are also implemented between extended H-representations. We illustrate this with a simple example. We start by defining the H-representation of a square with JuMP.
[B85] Balas, E., 1985. Disjunctive programming and a hierarchy of relaxations for discrete optimization problems. SIAM Journal on Algebraic Discrete Methods, 6(3), pp.466-486.
using JuMP
model = Model()
@variable(model, -1 <= x <= 1)
@variable(model, -1 <= y <= 1)
using Polyhedra
square = hrep(model)
H-representation LPHRep{Float64}:
4-element iterator of HalfSpace{Float64, SparseArrays.SparseVector{Float64, Int64}}:
HalfSpace( [1] = -1.0, 1.0)
HalfSpace( [2] = -1.0, 1.0)
HalfSpace( [1] = 1.0, 1.0)
HalfSpace( [2] = 1.0, 1.0)
Note that the names of the JuMP variables are used as the names of the corresponding dimensions for the polyhedron square.
dimension_names(square)
2-element Vector{String}:
"x"
"y"
In the following, diagonal and antidiag are projections of extended H-representations of dimension 7 hence diamond is a projection of an extended H-representation of dimensions 12.
diagonal = convexhull(translate(square, [-2, -2]), translate(square, [2, 2]))
antidiag = convexhull(translate(square, [-2, 2]), translate(square, [2, -2]))
diamond = diagonal ∩ antidiag
Polyhedra.Projection{LPHRep{Float64}, Base.OneTo{Int64}}(HyperPlane( [1 ] = 1.0
[3 ] = -1.0
[5 ] = -1.0, 0.0) ∩ HyperPlane( [2 ] = 1.0
[4 ] = -1.0
[6 ] = -1.0, 0.0) ∩ HyperPlane( [1 ] = 1.0
[8 ] = -1.0
[10] = -1.0, 0.0) ∩ HyperPlane( [2 ] = 1.0
[9 ] = -1.0
[11] = -1.0, 0.0) ∩ HalfSpace( [3 ] = -1.0
[7 ] = -3.0, 0.0) ∩ HalfSpace( [4 ] = -1.0
[7 ] = -3.0, 0.0) ∩ HalfSpace( [3 ] = 1.0
[7 ] = 1.0, 0.0) ∩ HalfSpace( [4 ] = 1.0
[7 ] = 1.0, 0.0) ∩ HalfSpace( [5 ] = -1.0
[7 ] = -1.0, -1.0) ∩ HalfSpace( [6 ] = -1.0
[7 ] = -1.0, -1.0) ∩ HalfSpace( [5 ] = 1.0
[7 ] = 3.0, 3.0) ∩ HalfSpace( [6 ] = 1.0
[7 ] = 3.0, 3.0) ∩ HalfSpace( [7 ] = -1.0, 0.0) ∩ HalfSpace( [7 ] = 1.0, 1.0) ∩ HalfSpace( [8 ] = -1.0
[12] = -3.0, 0.0) ∩ HalfSpace( [9 ] = -1.0
[12] = 1.0, 0.0) ∩ HalfSpace( [8 ] = 1.0
[12] = 1.0, 0.0) ∩ HalfSpace( [9 ] = 1.0
[12] = -3.0, 0.0) ∩ HalfSpace( [10] = -1.0
[12] = -1.0, -1.0) ∩ HalfSpace( [11] = -1.0
[12] = 3.0, 3.0) ∩ HalfSpace( [10] = 1.0
[12] = 3.0, 3.0) ∩ HalfSpace( [11] = 1.0
[12] = -1.0, -1.0) ∩ HalfSpace( [12] = -1.0, 0.0) ∩ HalfSpace( [12] = 1.0, 1.0), Base.OneTo(2))
Note that the names the first two dimensions are still identical to the names of the JuMP variables and the auxiliary variables have no name.
dimension_names(diamond.set)
12-element Vector{String}:
"x"
"y"
""
""
""
""
""
""
""
""
""
""
We don't need to compute the result of the projection to solve an optimization problem over diamond. For instance, to compute the maximal value that y can take over this polytope with the GLPK solver, we can do as follows. Note that if we use anonymous JuMP variables, the name of the JuMP variables will be the names of the corresponding dimensions of the polyhedron. Therefore, we can retrieve the JuMP variable according to the corresponding dimension name with variable_by_name.
import GLPK
model = Model(GLPK.Optimizer)
@variable(model, [1:2] in diamond)
x = variable_by_name(model, "x")
y = variable_by_name(model, "y")
@objective(model, Max, y)
optimize!(model)
value(x), value(y)
(0.0, 2.0)
In the optimization problem, above, the auxiliary variables of the extended formulation are transparently added inside a bridge. To manipulate the auxiliary variables, one can use the extended H-representation directly instead of its projection. Note that as the auxiliary dimensions have no name, we cannot use variable_by_name to retrieve the corresponding JuMP variables. We can instead catch the returned value of @variable in some variable v in order to use anonymous JuMP variables while still assigning the created JuMP variables to v.
import GLPK
model = Model(GLPK.Optimizer)
v = @variable(model, [1:12] in diamond.set)
y = variable_by_name(model, "y")
@objective(model, Max, y)
optimize!(model)
value.(v)
12-element Vector{Float64}:
0.0
2.0
-0.75
-0.25
0.75
2.25
0.25
-0.75
2.25
0.75
-0.25
0.75 | 2023-03-22 02:26:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6903838515281677, "perplexity": 4047.539084371876}, "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/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00361.warc.gz"} |
https://scikit-survival.readthedocs.io/en/latest/generated/sksurv.metrics.concordance_index_censored.html | # sksurv.metrics.concordance_index_censored¶
sksurv.metrics.concordance_index_censored(event_indicator, event_time, estimate, tied_tol=1e-08)[source]
Concordance index for right-censored data
The concordance index is defined as the proportion of all comparable pairs in which the predictions and outcomes are concordant.
Two samples are comparable if (i) both of them experienced an event (at different times), or (ii) the one with a shorter observed survival time experienced an event, in which case the event-free subject “outlived” the other. A pair is not comparable if they experienced events at the same time.
Concordance intuitively means that two samples were ordered correctly by the model. More specifically, two samples are concordant, if the one with a higher estimated risk score has a shorter actual survival time. When predicted risks are identical for a pair, 0.5 rather than 1 is added to the count of concordant pairs.
See [1] for further description.
Parameters: event_indicator (array-like, shape = (n_samples,)) – Boolean array denotes whether an event occurred event_time (array-like, shape = (n_samples,)) – Array containing the time of an event or time of censoring estimate (array-like, shape = (n_samples,)) – Estimated risk of experiencing an event tied_tol (float, optional, default: 1e-8) – The tolerance value for considering ties. If the absolute difference between risk scores is smaller or equal than tied_tol, risk scores are considered tied. cindex (float) – Concordance index concordant (int) – Number of concordant pairs discordant (int) – Number of discordant pairs tied_risk (int) – Number of pairs having tied estimated risks tied_time (int) – Number of comparable pairs sharing the same time
concordance_index_ipcw() | 2020-07-12 08:26:41 | {"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": 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": 0.32475918531417847, "perplexity": 3243.321977603748}, "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/1593657134758.80/warc/CC-MAIN-20200712082512-20200712112512-00403.warc.gz"} |
https://physics.stackexchange.com/questions/53147/why-is-von-neumann-entropy-maximized-for-an-ensemble-in-thermal-equilibrium | # Why is (von Neumann) entropy maximized for an ensemble in thermal equilibrium?
Consider a quantum system in thermal equilibrium with a heat bath. In determining the density operator of the system, the usual procedure is to maximize the von Neumann entropy subject to the constraint that the ensemble average of the Hamiltonian has a fixed value.
What justifies this assumption?
Sakurai, in his QM text, writes
To justify this assumption would involve us in a delicate discussion of how equilibrium is established as a result of interactions with the environment.
I'd appreciate if someone could shed some light on this. References are welcome as well.
I've heard the suggestion that a thermal equilibrium ensemble is simply defined by that density operator which solves the constrained optimization problem above. If this is the case, then why are real physical systems that are in weak contact with heat baths for long periods well described by such mathematical ensembles, and how would one identify the Lagrange multiplier $\beta$ that arises with inverse temperature of the heat bath?
You need to read this paper by Jaynes. I can't explain it as well as him, but I will try to summarise the main points below.
The first thing is to realise that the entropy is observer-dependent: it depends on what information you have access to about the system. A finite temperature means that you don't have access to all the information about the state of the system; in particular, you cannot keep track of the (infinite) degrees of freedom of the bath. However, suppose that some demon could keep track of all the degrees of freedom of the system and bath: he/she sees zero entropy. For the demon, it looks a bit like the total system is at zero temperature (although really it is better to say that temperature is ill-defined for the demon).
Given that you are ignorant (sorry, but at least I'm not calling you a demon), you need to find a consistent prescription for assigning probabilities to the different microstates. The prescription must be 'honest' about what you do or don't know. The entropy is in some sense a unique measure of ignorance, as proved by Shannon. Therefore you should 'maximise your ignorance', subject to the constraint that you do know certain macroscopic observables, e.g. average energy or average particle number if the system is open, etc.
Maximising the entropy of the system is the most logical way to assign probabilities to the microstates of the system, given access only to a limited subset of observables. The same 'MaxEnt' principle is quite general and applies to all statistical analysis, not only physics. The Lagrange multiplier $\beta$ is identified with inverse temperature by comparing the outcome of this abstract procedure to the experimental facts of phenomenological thermodynamics.
If you are interested in the actual dynamics of equilibration, there has been a lot of literature on this recently, especially in mesoscopic systems. Particular focus is laid on the integrability of the system: non-integrable (chaotic) systems do thermalise, whereas there is a fair bit of evidence that integrable systems do not thermalise properly. Intuitively, this is because integrable systems have a maximal set of locally conserved quantities so that, even when in contact with a heat bath, the memory of the initial conditions is never quite lost.
See, for example: Dynamics of thermalisation in small Hubbard-model systems and Thermalization and ergodicity in many-body open quantum systems, if you search 'thermalization' (sic) on arxiv then you will find many more.
• Thanks Mark. Yeah I actually forgot about the information-theoretic viewpoint for some reason when I was asking this question. I guess I was hoping that the MaxEnt principle could somehow be justified not only from the perspective of statistical inference, but that it would emerge in equilibrium ensembles as a consequence of the process of equilibration (or something to that effect). Do you think such an expectation is unreasonable? I'm not really sure what to think on this point. Also, thanks very much for the references at the end. Feb 6, 2013 at 19:22
• I think this is a very good question, and it's the subject of a lot of research at the moment, although it comes with a caveat. In the context of classical physics, it's important to remember that the system is always really in a pure state. So the emergence of the Gibbs ensemble cannot really be separated from the observer's subjective knowledge about the system. I think of it as ignorance/information/entropy being created dynamically each time the system interacts with degrees of freedom that the observer cannot track. Feb 6, 2013 at 19:30
• (contd.) When the energy/particle etc. distribution is uniform over scales longer than the minimal achievable experimental resolution then further interactions do not increase the observer's entropy, i.e. the probability distribution hits the steady state. Feb 6, 2013 at 19:34
• (contd.) However, in quantum mechanics you can't say the system is always really in a pure state, and the uncertainty of the observer is not subjective, it's required by the laws of physics. You can study equilibration of small quantum systems in a heat bath and you find that the system evolves towards a thermal state, basically due to entanglement with the bath. Likewise, it has been shown that almost all bipartite states for systems with a natural system-environment division are locally thermal, see here. Feb 6, 2013 at 19:39
• Wow thank you so much Mark. I couldn't have asked for better responses; I'll definitely look at all of the references you suggested. Feb 6, 2013 at 19:43
Here is an alternate approach to answering this question (which ignores temperature and lagrange multipliers) given to us by the 1st law of thermodynamics and quantum information.
In short, a weakly interacting ensemble at thermal equilibrium maximises the von Neumann entropy because in doing so it minimises the free energy of the system.
Why can we say this? A weakly interacting ensemble at thermal equilibrium corresponds to a Gibbs state, which is nothing more than a quantum version of a canonical ensemble. We can write this out as $$\hat{\rho}_\beta = \frac{e^{-\beta \hat{H}}}{\mathcal{Z}}$$ where $$\mathcal{Z}$$ is the canonical partition function. Canonical ensembles also have a related quantity known as the Gibbs Free energy given by $$F = U - tS$$ where $$U$$ is the internal energy and $$t$$ is the temperature. We can write out a quantum version of this as $$F = \langle \hat{H} \rangle - tS.$$ For a Gibbs state, this will coincide with the Gibbs Free energy but what if we define this quantity for an arbitrary quantum state modeling an open quantum system taking $$S$$ to be the von Neumann entropy $$S = -\text{tr}\{\hat{\rho}\ln\hat{\rho}\}.$$ In general we have $$F(\hat{\rho}) = \text{tr}\{\hat{\rho}\hat{H}\} + \beta^{-1}\text{tr}\{\hat{\rho}\ln\hat{\rho}\} = \beta^{-1}\text{tr}\{\hat{\rho}(\ln\hat{\rho}+\beta\hat{H})\}$$ by linearity of the trace. The Free energy of a Gibbs state is the Gibbs Free energy $$F(\hat{\rho_\beta}) = -\beta \ln \mathcal{Z} = -\beta^{-1}\ln\left(\text{tr}\{e^{-\beta \hat{H}}\}\right).$$ Considering the relative entropy of this arbitrary state $$\hat{\rho}$$ and the Gibbs state $$\hat{\rho}_\beta$$ we have $$D(\hat{\rho}||\hat{\rho}_\beta) = \text{tr}\{\hat{\rho}\ln\hat{\rho}\} - \text{tr}\{\hat{\rho}\ln\hat{\rho}_\beta\}$$ which can be expressed in terms of the Free Energy as $$D(\hat{\rho}||\hat{\rho}_\beta) = \beta \left(F(\hat{\rho}) - F(\hat{\rho}_\beta)\right).$$ But, the relative entropy is known to be nonnegative $$D(\cdot||\cdot)\geq 0.$$ As a result the Gibbs Free energy is the lowest possible free energy or the free energy of the Gibbs state is minimal.
The first law of thermodynamics can be stated in this context as $$dE = tdS +dF$$ so if Gibbs states minimise the free energy, they maximise the von Neumann entropy, which happens to be the Gibbs Entropy.
Beyond this one can also justify the approach to thermal equilibrium on the grounds of quantum information expanding on this argument. Any quantum channel that preserves the Gibbs state cannot increase the free energy. Rather, the free energy of out-of-equilibrium states is monotonically decreasing to the Gibbs Free Energy giving equilibrium, maximimising von Neumann entropy on long times.
This argument is given by Preskill in his notes on Quantum Shannon Theory. | 2022-05-19 00:07:41 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8252028226852417, "perplexity": 232.9596509236243}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00285.warc.gz"} |
https://www.physicsforums.com/threads/prove-any-cyclic-group-with-more-than-one-two-elements-has-at-least-two-different-gen.287341/ | # Prove any cyclic group with more than one two elements has at least two different gen
1. Jan 25, 2009
### Daveyboy
1. The problem statement, all variables and given/known data
Prove any cyclic group with more than two elements has at least two different generators.
2. Relevant equations
A group G is cyclic if there exists a g in G s.t. <g> = G. i.e all elements of G can be written in the form g^n for some n in Z.
3. The attempt at a solution
Z has 1 and -1.
<i> where i = (-1)^1/2 so i and -i work
now I consider G={e, a, a^2}
All I can think of is a^3 could generate this aside from a, but that is pretty lame. Am I missing somethig?
2. Jan 25, 2009
### Dick
Re: Prove any cyclic group with more than one two elements has at least two different
In G={e,a,a^2}, a^3 doesn't generate. a^(3)=e. But a^2=a^(-1) does. If a is a generator, isn't it's inverse also a generator?
3. Jan 25, 2009
### Symmetryholic
Re: Prove any cyclic group with more than one two elements has at least two different
An infinite cyclic group is isomorphic to Z, and you showed the two generators for it.
A finite cyclic group is isomorphic to $$Z_{n}$$, so an element "a" in $$Z_{n}$$ with gcd(a, n) = 1 generates the whole group.
Now for a finite cyclic group G with more than two elements, we can count how many elements in G can generate the whole group.
Hint) Use the Euler phi function.
Last edited: Jan 25, 2009
4. Nov 6, 2010
### kathrynag
Re: Prove any cyclic group with more than one two elements has at least two different
Hi, I just found this post and noticed this was a homework problem for me.
I'm not really sure I follow this:
In G={e,a,a^2}, a^3 doesn't generate. a^(3)=e. But a^2=a^(-1) does. If a is a generator, isn't it's inverse also a generator?
I follow that a^3 doesn't generate and a^3=e, but afterwards I'm lost
5. Nov 6, 2010
### Dick
Re: Prove any cyclic group with more than one two elements has at least two different
Well, what's a^(-1) in this group? Does it generate the group? You can't be that lost in such a simple group.
6. Nov 7, 2010
### kathrynag
Re: Prove any cyclic group with more than one two elements has at least two different
We want a * a^-1=a^-1*a=e
Can't be e. That'll give a.
Can't be a. That'll give a^2.
Then it must be a^2.
7. Nov 7, 2010
### Dick
Re: Prove any cyclic group with more than one two elements has at least two different
A method like that could take a lot of time if you are dealing with Z100. But, yes, a^(-1)=a^2. Does a^2 generate the group?
8. Nov 7, 2010
### JonF
Re: Prove any cyclic group with more than one two elements has at least two different
If you know x is a generator for G all you need to show is for some n (x-1)^n = x to get x-1 is also a generator. Note this isn't sufficient for your claim, you need to show something else. To see what consider 2 in Z4, 2 = 2-1. So you need to prove why your your generator can't be idempotent.
Also you need to consider infinite groups, so ask your self can an infinite group be generated by a single element?
Last edited: Nov 7, 2010
9. Nov 7, 2010
### kathrynag
Re: Prove any cyclic group with more than one two elements has at least two different
Yes, a^2 generates the group.
Sorry, I've never heard the word idempotent yet.
10. Nov 7, 2010
### Dick
Re: Prove any cyclic group with more than one two elements has at least two different
Ok then can you figure out how this also works for a general cyclic group of order n, {e,a,a^2,...,a^(n-1)} generated by a. Does a^(n-1) (which is a^(-1)) also generate? What's (a^(n-1))^2?
11. Nov 7, 2010
### kathrynag
Re: Prove any cyclic group with more than one two elements has at least two different
Yes a^(n-1) generates. (a^(n-1))^2=a^(2(n-1))=a^(2n)a^-1
12. Nov 7, 2010
### Dick
Re: Prove any cyclic group with more than one two elements has at least two different
Yes, a^(n-1) generates. But your calculation above is quite wrong.
13. Nov 7, 2010
### JonF
Re: Prove any cyclic group with more than one two elements has at least two different
x is called idempotent if x*x=e
14. May 13, 2012
### vohrahul
Re: Prove any cyclic group with more than one two elements has at least two different
Consider a group G = {1,3,5,7,11,13,17} under the multiplication modulo 18 .
Now this group is CYCLIC and have two generators : 5 and 11..
5^1 = 5
5^2 = 7
5^3 = 17
5^4 = 13
5^5 = 11
5^6 = 1
thus giving it order of 6 which is a divisor of order of G. 6(1) = 6 {hence proving lagrange's theorem also }
Similarly,
11^1 = 11
11^2 = 13
11^3 = 17
11^4 = 7
11^5 = 5
11^6 = 1
Again has order 6, a divisor of order of G.
Check and verify!
I know the post is Q-U-I-T-E old.. But I couldn't find any simple answers here.. So posted one! ;) | 2016-10-22 04:07:03 | {"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": 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": 0.6027382016181946, "perplexity": 1096.670435425513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00277-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://clay6.com/qa/32527/ch-equiv-ch-x-2ch-3cooh-y-rightarrow-a-quad-underrightarrow-quad-b-c-b-can- | # $CH\equiv CH(X)+2CH_3COOH(Y)\rightarrow A\quad \underrightarrow{\Delta}\quad B+C$.B can be obtained from X by hydration and c can be obtained from Y by heating with $P_2O_5$.Hence A is
Hence (b) is the correct answer. | 2018-04-24 21:50:57 | {"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": 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": 0.5572612285614014, "perplexity": 589.6054657137838}, "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-2018-17/segments/1524125947328.78/warc/CC-MAIN-20180424202213-20180424222213-00425.warc.gz"} |
https://scicomp.stackexchange.com/questions/23725/the-second-variation-of-displacement-interpolation-function-in-finite-element-me | # The second variation of displacement interpolation function in Finite Element Method
I need to calculate the second variation of displacement interpolation function $u = \sum N_a u_a$ in Finite Element Analysis, where $N_a$ are the shape functions and $u_a$ are the nodal values. Someone told me the second variation is 0. The reason is as shown below. But I don't understand why $\delta u_a$ is a constant.
Could you guys please tell me if the second variation of displacement interpolation function really equals to zero? If it is, could you please explain to me why $\delta u_a$ equals constant? Actually, I think $\delta u$ is arbitrary, so I don't think the second variation of displacement is equal zero.
$$u = \sum_{a=1}^{n} N_a u_a$$
Thus,
$$\delta u = \delta\sum_{a=1}^{n}N_a u_a = \sum_{a=1}^{n}N_a \delta u_a$$
As $\delta u$ is constant (This is what I don't understand). Thus
$$\delta^2 u_a = 0 \enspace .$$
And
$$\delta^2 u = \delta^2\sum_{a=1}^{n}N_a u_a = \sum_{a=1}^{n}N_a \delta^2 u_a = 0 \enspace .$$
• Can you define what you mean by the "variation" of $u$? Do you mean the derivative? Apr 21, 2016 at 16:21
The variation operator applies to the functions, not the constants.
Variations work similarly to derivatives. If your interpolation functions are piecewise linear, the first variation becomes piecewise constant and the second variation becomes zero.
• Hi Paul, Thank you so much for your response. In this variation problem, the variation variable is the nodal value rather than the variable in the shape function. So, we don't see derivatives of the shape function. so I don't think it matters with the order of shape functions or the interpolation functions.
– Joe
Apr 19, 2016 at 17:12
• @Joe: I think you have it backwards. $\delta N_a$ should be constant, not $\delta u_a$.
– Paul
Apr 19, 2016 at 18:11
I am not sure I understand your question. This is how I understand second variation (skipping much of the details). Imagine you have a simple linear spring whose stored energy is $W = {1\over2} k \thinspace u^2$ for spring stiffness $k$ and displacement $u$. Then the first variation gives $\delta W = \delta u \thinspace k \thinspace u$, and the second variation gives $\text{d}\delta W = \delta u \thinspace k \thinspace \text{d}u.$
In this case, the second variation can be expressed as $\text{d}u = N_i \text{d}u_i$ depending on the chosen discretization and shape functions. Thus the quadratic nature of energy expression lends itself to computing second variation. | 2022-08-09 00:53:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.8820458650588989, "perplexity": 218.86256781393843}, "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/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00243.warc.gz"} |
https://math.stackexchange.com/questions/1641044/how-to-configure-simplex-method-to-start-from-a-specific-point | How to configure simplex method to start from a specific point
If I have a linear programming problem e.g. $$\max 2x_1 + x_2$$ with these constraints
$$x_1-2x_2 \leq 14$$ $$2x_1-x_2\leq 10$$ $$x_1-x_2 \leq 3$$
And I want to solve the problem starting from a specific point e.g. $A=(x1=5, x2=0)$ and $B=(x1=0, x2=5)$ .
How should I configure the initial tableau to allow to iterate simplex method starting from these points?
(Note that in standard form when I add the slack variables, the points must be calculated in function of the extra variables.)
• Are these two separate starting points to set up? That is, for the first one you wish to start with $x_1=5,x_2=0$? – coffeemath Feb 4 '16 at 23:35
• A=5,0 is a single starting point – AndreaF Feb 4 '16 at 23:35
• That's what I meant, and also $B=(0,5)$ would be another starting point. What version of tableaus are you to use? Just thought, maybe you mean 5,0 as a decimal like usual 5.0 (some British use comma for decimal) – coffeemath Feb 4 '16 at 23:37
• I usually convert the problem in the standard form and use the 2-phases method. No comma is a separator x1=5 x2=0 – AndreaF Feb 4 '16 at 23:41
One way would be, for a start at (5,0), to set up a new variable $x_1'$ and put the original $x_1$ equal to $5+x_1'$ in the objective and constraints. Then the usual simplex, starting at $(0,0)$ in the $x_1',x_2$ variables, would in effect be starting at $(5,0)$ in the $x_1,x_2$ variables.
• you mean a new constraint $$x6=5+x1$$ that become $$-x1+x6=5$$ to start from $$x1=5$$ and $$x2=0$$ ??? Note: I have used $$x6$$ because $$x3$$, $$x4$$ and $$x5$$ are needed for slack Have I well understand? – AndreaF Feb 4 '16 at 23:51
• Actually I just meant to substitute say $x_1=5+x_6$ into everything in the original objective and constraints, and then work from there, i.e. (since you want other lower index vars for slacks) the new system would be all in terms of $x_6,x_2$ rather than in $x_1,x_2,$ and would after the substitutions have different looking objective and constraints. The way I'm thinking of, there would be no constraint involving $x_1$ and $x_6,$ since $x_1$ would no longer appear in the "new" problem. – coffeemath Feb 5 '16 at 0:08
• So my objective function will be $$2(5+x6)+x2$$, in the same way constraints eg. the first $$x1−2x2≤14$$ becomes $$5+x6−2x2≤14$$. Now is right? And when I have to put objective function in cost row how to deal with the constant introduced? I'm a bit confused about this approach, In addition I have just noted that $$A=(5,0)$$ in the standard form after that I put the slack becomes $$A=(5,0,9,0−2)$$ while the point $$B=(0,5)$$ becomes $$B=(0,5,4,10,8)$$ if you write the first tableau would be helpful. – AndreaF Feb 5 '16 at 2:20 | 2019-08-20 18:16:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7379950881004333, "perplexity": 334.7480349466103}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00202.warc.gz"} |