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http://mathhelpforum.com/calculus/21262-hard-mathematics-c-questions.html
|
1. Hard Mathematics C Questions
Hey,
This is an advanced mathematics C problem in Australia. Anyone who can solve the attached question is a GOD!!!!!!! Please gimme a hand, I cant quite get it worked out.
Cheers,
Ned
2. Originally Posted by Ned Hunter
Hey,
This is an advanced mathematics C problem in Australia. Anyone who can solve the attached question is a GOD!!!!!!! Please gimme a hand, I cant quite get it worked out.
Cheers,
Ned
You have the following initial value problem to solve:
$
\frac{dN}{dt}=0.0004~N~(1000-N),\ \ N(0)=1
$
This is of variables seperable type, so:
$
\int \frac{1}{N~(1000-N)}~dN = \int 0.0004~dt = 0.0004t+C
$
The integral on the left can be done by using partial fractions.
RonL
3. Hey,
do you think you can show me how to do that? i cannot remember for the life of me and my textbook is very sketchy on it. that would be a great help.
Cheers
4. Originally Posted by Ned Hunter
Hey,
do you think you can show me how to do that? i cannot remember for the life of me and my textbook is very sketchy on it. that would be a great help.
Cheers
$
\frac{1}{N(1000-N)}=\frac{A}{N}+ \frac{B}{1000-N}=\frac{A(1000-N)+BN}{N(1000-N)}
$
So $A=B$ and $A=0.001$
Hence:
$
\int \frac{1}{N(1000-N)}~dN=\int \frac{0.001}{N}+ \frac{0.001}{1000-N}~dN=0.001 \left[ \ln(N) - \ln(1000-N) \right]$
$=0.001 \ln \left(\frac{N}{1000-N}\right)
$
So:
$
0.001 \ln \left(\frac{N}{1000-N}\right)=0.0004 t +C
$
Now rearrange:
$
\left(\frac{N}{1000-N}\right)=A \exp(0.4 t)
$
Now change the subject to get this in the form: $N=f(t)$ and
choose the constant so that the initial condition is satisfied.
RonL
|
2016-09-30 00:50:16
|
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|
https://savosh.com/medical-instrumentation-webster-solution-manual-zip-verified/
|
# Medical Instrumentation Webster Solution Manual.zip VERIFIED
Medical Instrumentation Webster Solution Manual.zip
Medical Instrumentation / Medical Instruments, · ۱۱۲۰۵۰۳۹۲۶ – Maths and Science – Assessment, Key Stage 2 – Teach your children about maths and science – Find maths and science games and activities for children – Play, explore, and learn with a range of games, activities and puzzles – Get flexible and extend the learning effect across the curriculum. · ۱۰۶۶۰۹۷۸۲۲ – Maths and Science – Assessment, Key Stage 1 – Maths and science – Find maths and science games and activities for children – Play, explore, and learn with a range of games, activities and puzzles – Get flexible and extend the learning effect across the curriculum. · ۱۳۷۸۴۵۶۱۵۶ – Maths and Science – Assessment, Key Stage 1 Maths and Science – Activities, Activities, Games, Puzzles and Games, Maths and science – Play, explore, and learn with a range of games, activities and puzzles – Get flexible and extend the learning effect across the curriculum. Math and Science: Key Stage 1. Key Stage 1 – Maths and Science – Activities, Activities, Games, Puzzles and Games .
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The method of using ether-alcohol solution in the determination of total cholesterol in serum by a kinetic reaction with cholesterol esters catalyzed by lecithin : Part I.Esterification products of cholesterol in the serum and the influence of sterols upon the formation of esters: Part II.Biosynthetic origin of cholesterol esters and sterols: Part III.An improved method for the determination of cholesterol esters and cholesterol in serum and other biological fluids by the divalent alkali. The simple technique of using ether-alcohol solution in the determination of total cholesterol in serum by a kinetic reaction with cholesterol esters catalyzed by lecithin: Part I.the method for the determination of the sterol composition of plasma lipoproteins : Part II.Alissa 2
Alissa 2 is the second studio album by British progressive metal band Alissa. It was released on 15 November 2014 through Candlelight Records.
Alissa 2 is notable for the inclusion of an alternate version of the song “Smile”, which was released in July 2014 as a free download from the band’s official website.
Track listing
Personnel
Alissa – vocals, guitars
Andy St. John – bass guitar
Paul Crivelli – guitars
Tom Mason – drums
Simon Collins – piano on “I Won’t Let You Go”
John McKean – violin on “Smile”
Production
Pavel Civelli – producer, engineer, mixing, mastering
Nicola Renga – executive producer
Rick Watson – co-mixing and mastering engineer
References
Alissa, Alissa 2 at Encyclopaedia Metallum
Category:2014 albums
Category:Alissa (band) albumsQ:
AngularJS & AJAX
Im creating a dynamic website that consists of some apps.
These apps are basically divs in html,with their own Controller.
The idea is to give the user the ability to change some style parameters from the main page without reloading the entire page
I have no experience in JavaScript.
I took a look on W3schools but I couldn’t get to understand to structure the HTML and Javascript.
This is my idea of how to do it:
JS / Jquery / Bootstrap:
function MainController($scope,$http) {
function changeStyle() {
0cc13bf012
Attention professionals and researchers in the scientific, medical, and health care industries.
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– The Book of Mormon is a religious and historical book, believed to contain.
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The Book of Mormon is a religious and historical book, believed to contain an account of the experiences of the. It is the original text in several languages of the “American. Book of Mormon The Book of Mormon, written by Joseph Smith, Jr. is a religious and historical book, believed to contain an account of the experiences of the ancient “American. The Book of Mormon originally.
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By this device, the instruments with a very similar identification code to that used in inventory and distribution management or financial management can be easily. an early fully automatic instrument, the Micro-Matic was a small, very low-cost product. In, 80 more free medical instruments available, micro-patent.
MR.. S. Webster, F.J. Anderson, M.A. Webster, and E.M. Kochkin. British Micro-Electronic Laboratories Co. The automatic volume dispensing unit, part, medical instrumentation webster solution manual.zip was one of the first refillable instruments which made it suitable for a wide range of laboratory applications and could be designed for use with a large variety of products ( Hibbs, 1999, p.
MANUAL OF PHOTOGRAPHY—۲. PROCESS OF THERMAL STRETCHING A. Imaging and Image Analysis, Handbook of Biomedical Engineering,.
, industry is the separation of the organic substances from the ( 4) To control the dose, we used a delivery rate of $\unit[63.5]{{\mu s}}$ and a ratio of $\unit[15.5]{{\mu J/pulse}}$. The incident laser pulses were $\unit[4.6]{{\mu J}}$. We delivered a laser pulse every $\unit[160]{ms}$ and scanned over the plane of the liquid. The focus was $\unit[50]{\upmu m}$ and the scan was $\unit[1.3]{mm}$ per point.. The reconstruction algorithm. To obtain the images, the light of the fluorescence (of the specimens in suspension) was collected with a 20/20 BSG20/BSG20-A CCD camera (made by Hamamatsu, Japan) and imaged with an iXon EM+-CCD camera (resolution of $\unit[1320\times 1024]{pixels}$). To avoid blurring due to slow camera response, it was required to measure the specimen fluorescence with a dwell time of $\unit[24]{\upmu s}$.. The maximum diameter of the object illuminated by the X-ray, is given by the Rayleigh criterion.. The results obtained with these two different devices are illustrated in Fig.. These results are obtained for a specimen having an absorption coefficient of $\unit[1]{cm^{ -1}}$ and a refractive index of
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2022-09-24 22:14:02
|
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https://physics.stackexchange.com/questions/617059/missing-factor-1-2-when-using-generalized-stokes-theorem
|
# Missing factor 1/2 when using generalized Stokes theorem
I'm doing the following homework question:
By invoking Stokes' theorem, according to which the integral of a vector field (which equals the field strength) over any two-dimension surface S that is bounded by that closed path (see 1.57), $$\Phi(x_1;x_1)=\exp\left[\oint-ieA_\mu(x)\mathrm d x^\mu\right]=\exp\left[\int_S-\tfrac{1}{2}ie F_{\mu\nu}\mathrm{ d} S^{\mu\nu}\right]$$ where $$\int_S F_{\mu\nu}\mathrm d S^{\mu\nu}$$ denotes twice the total electromagnetic flux through the surface $$S$$. The quantity $$\Phi(x_1;x_2)$$ is known as a non-integrable phase factor [305].
Equation 1.57 refers to the vector version of Stokes' theorem i.e. $$\int_S \mathbf B\cdot \mathrm d \mathbf S=\oint_C\mathbf A\cdot \mathrm d \mathbf l$$. Using this formula naively I'm able to find the answer but I'm not sure how to handle the fact that the dot product is using the Minkowski metric. I want to solve this using the theory of one-forms because it's also a nice practice. To do this I'm using the following definitions (which might be wrong)
\begin{align} A&=A_\mu\mathrm d x^\mu\\ F&=\mathrm d A\\ (\mathrm d A)_{\mu\nu}&=\partial_\mu A_\nu-\partial_\nu A_\mu \\ \mathrm d A&=(\mathrm d A)_{\mu\nu}\,\mathrm d x^\mu\wedge\mathrm d x^\nu \end{align}
Then I get the following \begin{align} \oint A_\mu\mathrm d x^\mu&=\int_{\partial S}A\\ &=\int_S\mathrm d A&\text{(Generalized Stokes)}\\ &=\int_S F\\ &=\int_S F_{\mu\nu}\,\mathrm d x^\mu\wedge\mathrm d x^\nu\\ &=\int_S F_{\mu\nu}\,\mathrm d S^{\mu\nu} \end{align} So I get an overall factor 1/2 wrong. Where does this factor come from?
If $$A=A_\nu\, dx^{\nu}$$ then $$dA= \partial_{\mu} A_{\nu} \,dx^{\mu} \wedge dx^{\nu}\\ = \frac 12 (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})dx^{\mu} \wedge dx^{\nu}\\ = \frac 12 F_{\mu\nu} dx^{\mu} \wedge dx^{\nu}$$
• Thanks that explains the discrepancy. On wikipedia's article for the exterior derivative en.wikipedia.org/wiki/… I found the definition $(\mathrm d\omega)_{\mu\nu}=\partial_\mu\omega_\nu-\partial_\nu\omega_\mu$. Does that definition agree with the one you're using? Or are those two different conventions? Feb 25 at 12:51
• It's not a notation that I would use precisely because of the factor of two. I Always write $\omega= \frac 1{p!} \omega_{i_1\ldots i_p} dx^{{i_1}}\cdots dx^{i_p}$, where the $p!$ eliminates the overcounting. Feb 25 at 13:14
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2021-12-05 04:56:36
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http://mathoverflow.net/revisions/79342/list
|
2 added 135 characters in body
The answer of following classical problem is surely known, but I can't find a reference
For which positive integer $n$ is the set $S_n$ of primes of the form $x^2+n y^2$ ($x$, $y$ integers) determined by congruences?
A set of prime $S$ is said determined by congruences if there is a positive integer $m$ and a set $A \subset (\mathbb{Z}/m\mathbb{Z})^\ast$ such that a prime $p$ not dividing $m$ is in $S$ if and only if $p$ modulo $m$ is in $A$. There is a natural place to look for this question: the book by Cox "primes number of the form $x^2+ny^2"$". Unfortunately I don't have it, my library doesn't have it, and I can't find it on the internet, except for some preview at Amazon and Google. From the table of content and the preview it seems that the book does not contain the answer to my question (otherwise I wouldn't ask) but it is still possible that the answer be hidden precisely in one of the sporadic pages that amazon doesn't want me to see.
From that book one knows that a prime $p$ is in $S_n$ if and only if it splits in the ring class field $L_n$ of the order $\mathbb{Z}[\sqrt{-n}]$ in the quadratic imaginary field $K_n:=\mathbb{Z}[\sqrt{-n}]$. Therefore the question becomes: is $L_n$ abelian over $\mathbb{Q}$? Now $H_n:=Gal(L_n/K_n)$ is the ring class group of $\mathbb{Z}[\sqrt{-n}]$, hence abelian, and $Gal(L_n/\mathbb{Q})$ is a semi-direct extension of $\mathbb{Z}/2\mathbb{Z}$ by $H_n$, the action of the non-trivial element of $\mathbb{Z}/2\mathbb{Z}$ on $H_n$ being $x \mapsto x^{-1}$. Hence, if I am not mistaken (am I?), the question is equivalent to
For which $n$ is the ring class group $H_n$ killed by $2$?
Thanks for any clue or reference. I am especially interested in the case $n=32$.
1
# Primes of the form $x^2+ny^2$ and congruences.
The answer of following classical problem is surely known, but I can't find a reference
For which positive integer $n$ is the set $S_n$ of primes of the form $x^2+n y^2$ ($x$, $y$ integers) determined by congruences?
A set of prime $S$ is said determined by congruences if there is a positive integer $m$ and a set $A \subset (\mathbb{Z}/m\mathbb{Z})^\ast$ such that a prime $p$ not dividing $m$ is in $S$ if and only if $p$ modulo $m$ is in $A$. There is a natural place to look for this question: the book by Cox "primes number of the form $x^2+ny^2"$". Unfortunately I don't have it, my library doesn't have it, and I can't find it on the internet, except for some preview at Amazon and Google. From the table of content and the preview it seems that the book does not contain the answer to my question (otherwise I wouldn't ask) but it is still possible that the answer be hidden precisely in one of the sporadic pages that amazon doesn't want me to see.
From that book one knows that a prime $p$ is in $S_n$ if and only if it splits in the ring class field $L_n$ of the order $\mathbb{Z}[\sqrt{-n}]$ in the quadratic imaginary field $K_n:=\mathbb{Z}[\sqrt{-n}]$. Therefore the question becomes: is $L_n$ abelian over $\mathbb{Q}$? Now $H_n:=Gal(L_n/K_n)$ is the ring class group of $\mathbb{Z}[\sqrt{-n}]$, hence abelian, and $Gal(L_n/\mathbb{Q})$ is a semi-direct extension of $\mathbb{Z}/2\mathbb{Z}$ by $H_n$, the action of the non-trivial element of $\mathbb{Z}/2\mathbb{Z}$ on $H_n$ being $x \mapsto x^{-1}$. Hence, if I am not mistaken (am I?), the question is equivalent to
For which $n$ is the ring class group $H_n$ killed by $2$?
Thanks for any clue or reference. I am especially interested in the case $n=32$.
|
2013-05-21 07:08:47
|
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https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-approach-4th-edition/chapter-3-exercises-page-133/118
|
## Chemistry: Molecular Approach (4th Edition)
a) ethane b) $CH_3CH_2CH_2CH_2CH_3$ c) hexane d) $CH_3CH_2CH_2CH_2CH_2CH_2CH_3$
a) 2 carbon atoms in the alkane. So, its name is ethane b) pentane has 5 carbon atoms and 12 hydrogen atoms. Its formula is $CH_3CH_2CH_2CH_2CH_3$ c) 6 carbon atoms in the alkane. So, its name is hexane d) heptane has 7 carbon atoms and 16 hydrogen atoms. Its formula is $CH_3CH_2CH_2CH_2CH_2CH_2CH_3$
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2018-10-21 11:14:50
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http://gmatclub.com/forum/timeline-of-year-2-recruiting-61042.html
|
Find all School-related info fast with the new School-Specific MBA Forum
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# Timeline of Year 2 recruiting
Author Message
Senior Manager
Joined: 23 Jan 2008
Posts: 498
Followers: 1
Kudos [?]: 52 [0], given: 0
Timeline of Year 2 recruiting [#permalink]
### Show Tags
09 Mar 2008, 13:40
So, I have been unable to find a straightforward answer to this question. Hope my fellow GMAT-clubbers can help.
When do I need to be in the country for 2nd year (full time) recruiting for the various positions?
I ask because I had wanted to do a semester abroad (fall of 2nd year). This means I would be out of the country through december of my 2nd year. What sorts of job/industry/company (regional, local, national, etc). Would be possible/impossible to get if I do this? what sorts of things would I miss out on, etc? Just trying to see how this would alter my 2nd year job search! (I am a 2010 graduate, so haven't started yet. This means I can't give info about my internship)
VP
Joined: 11 Dec 2006
Posts: 1430
Location: New York, NY
Schools: NYU Stern 2009
Followers: 42
Kudos [?]: 220 [0], given: 6
Re: Timeline of Year 2 recruiting [#permalink]
### Show Tags
09 Mar 2008, 15:18
I would think that you are better set to look to be out for the final semester if you are recruiting banking or consulting, maybe earlier if you are doing more boutique stuff.
It would appear to me that the bigger the firm, the earlier in the fall/winter period they want their recruiting tidied up. Most the MBA2's I have spoken to felt that having (or not having) a job sorted by the holidays was a key point for the banks and consulting.
This is just a feel for things though. You can study abroad, but will have to be prepared to travel for interviews to meet people etc. I can't see why it should be a critical point as recruiters understand the MBA lifestyle, options and activity, but it may add to the stress.
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Director
Joined: 18 Dec 2007
Posts: 983
Location: Hong Kong
Concentration: Entrepreneurship, Technology
Schools: Hong Kong University of Science and Technology (HKUST) - Class of 2010
Followers: 13
Kudos [?]: 134 [0], given: 10
Re: Timeline of Year 2 recruiting [#permalink]
### Show Tags
09 Mar 2008, 15:26
westsider wrote:
So, I have been unable to find a straightforward answer to this question. Hope my fellow GMAT-clubbers can help.
When do I need to be in the country for 2nd year (full time) recruiting for the various positions?
I ask because I had wanted to do a semester abroad (fall of 2nd year). This means I would be out of the country through december of my 2nd year. What sorts of job/industry/company (regional, local, national, etc). Would be possible/impossible to get if I do this? what sorts of things would I miss out on, etc? Just trying to see how this would alter my 2nd year job search! (I am a 2010 graduate, so haven't started yet. This means I can't give info about my internship)
Also, if youre looking to work abroad, you can build up potential networks in those countries. Some of the current students Ive spoken to at international universities have said that some of the excahnge students who arrive on campus build up networks and connections whilst they are abroad, and some even get offers at a later date
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09 Mar 2008, 17:07
Good info. Seems to back up my vague notion that ibanking and consulting are started in the fall. what about other areas?
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10 Mar 2008, 11:18
Will I miss the domestic recruiting season for large national companies in positions other than ibanking/consulting if I am out of the country September - December.
Thanks!
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10 Mar 2008, 21:03
westsider wrote:
Will I miss the domestic recruiting season for large national companies in positions other than ibanking/consulting if I am out of the country September - December.
Thanks!
I cant speak for other schools, but at the GSB, if you leave Sept to December, you basically miss 98% of full time recruiting options as I understand it. IB, Consulting, GM, IM, finance, whatever... it all happens in the fall.
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11 Mar 2008, 14:02
i would agree with rhyme, but people DO leave during that period. it can be done.
i would just plan to get and take an offer from your internship....then you can do whatever you want!
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11 Mar 2008, 18:24
Quote:
IB, Consulting, GM, IM, finance
GM= General Management
IM = ?
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11 Mar 2008, 19:08
Investment management
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12 Mar 2008, 13:42
Fair enough. So, for my purposes I would simplify that list to "financial positions (i banks, etc) and General Management".
What about ops/marketing/product management (e.g. consumer products, etc)
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12 Mar 2008, 14:35
I think I know why you have struggled to get a straight forward answer to this. There isn't one.
Big companies = fall. Smaller companies = later. Though that is no hard and fast rule. You will be counting out a lot of the companies that you typically see in the fall (the big list companies), and maybe even all of those that look to recruit from summer interns and top-up their staff.
It depends on how you want to do things. Would you like to be away in the fall knowing you don't have a job and have to return to desperately recruit? Or be sure you have a job before you go away (I am pretty certain you could back out if needs be on a spring study abroad). Or do you not want to recruit for MBA standard stuff anyway, at which point it is mostly individual work anyway?
What I do know is that people who don't have jobs January of their year of graduation look really scared and worried. I am sure they sort it out, but for the most part that seems to be a signal.
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12 Mar 2008, 15:18
Quote:
What I do know is that people who don't have jobs January of their year of graduation look really scared and worried
Nicely stated.
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31 Mar 2008, 07:49
I have a question related to internship recruiting: what happens in December? is it relatively quiet or are students busy applying for jobs? Is it wise to take a 2-3 week break during this period?
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31 Mar 2008, 10:49
at Duke, it's completely nuts. you're getting your resume and cover letters together, making last-minute contacts, going to final recruitment activities (consulting companies give workshops on case prep, for example) and also busting your butt because it's the end of the semester. i was wiped by the time that was done.
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31 Mar 2008, 12:51
Thanks aau- that was helpful. What about the vacation period? Does it get real busy during the 3 week break as well?
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31 Mar 2008, 14:15
dosa_don wrote:
Thanks aau- that was helpful. What about the vacation period? Does it get real busy during the 3 week break as well?
that's up to you. Some students do approximatley a million cases, and are still dropping resumes/cover letters, and are networking their tails off.
I did minimal job prep and was just really busy travelling a lot. I recommend scheduling in some down time. You'll need it coming off your first semester. And your mom coming to visit does not count for down time.
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Re: Timeline of Year 2 recruiting [#permalink]
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31 Mar 2008, 18:25
I spoke to students and alumni at HKUST... Their timeline is slightly different.
Recruiting for internships starts in Autumn, but not many people secure one at the start. Most of the action for internships happen after Chinese New Year when lots of internship openings appear.
For recruitment, a similar cycle happens (although quite a few people are offered a role after the internship)
Re: Timeline of Year 2 recruiting [#permalink] 31 Mar 2008, 18:25
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Display posts from previous: Sort by
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2016-07-27 02:16:12
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https://me.gateoverflow.in/201/gate-mechanical-2014-set-3-question-5
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The definite integral $\int_{1}^{3}\dfrac{1}{x}$ is evaluated using Trapezoidal rule with a step size of $1$. The correct
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2023-02-07 02:27:07
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https://petsc.org/release/docs/manualpages/Mat/MatSubMatrixVirtualUpdate/
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# MatSubMatrixVirtualUpdate#
Updates a MATSUBMATRIX virtual submatrix
## Synopsis#
#include "petscmat.h"
PetscErrorCode MatSubMatrixVirtualUpdate(Mat N, Mat A, IS isrow, IS iscol)
Collective
## Input Parameters#
• N - submatrix to update
• A - full matrix in the submatrix
• isrow - rows in the update (same as the first time the submatrix was created)
• iscol - columns in the update (same as the first time the submatrix was created)
## Note#
Most will use MatCreateSubMatrix() which provides a more efficient representation if it is available.
MATSUBMATRIX, MatCreateSubMatrixVirtual()`
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2022-12-06 18:16:09
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https://quantumprogress.wordpress.com/2010/12/15/perfectionism-and-sbg/
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The last few days of the term can be hell for most any teacher, but under Standards Based Grading (SBG), where every kid has the “opportunity to earn an A,” and thinks all they need to do is reassess their way to an A, which often translates into hanging out the teacher’s classroom working on problems I give them until they are satisfied, or give up in despair, and along the entire focus on learning we’ve been building all year long can fall away.
Here are a collection of random observations and thoughts that, because I’m in the middle of dealing with all my students’ perfection seeking, I haven’t had time to really hammer out into a deep connected understanding.
## The problem of perfection seeking in general
This past weekend, I got the following email:
I would like to enter the exam having showed understanding on absolutely everything. Is there any way that you could let me know the concepts that I have not yet showed understanding on so that I could look them over, make sure that I understand them, and then re-assess my understanding of them so that I could not only show understanding on everything, but also review any dicey subject matters that I need to go over. If you could please let me know the subjects I don’t have 3’s on, that’d be great.
Five years ago, I would have thought—”wow, this student really cares about physics. Let me look in my trusty gradebook and send them a detailed list of their scores on every concept and questions they can use to reassess them.” (This isn’t completely true. 5 years ago I would have said something, like “your sores on the previous exams were X, Y, and Z; they are fixed and cannot change, they show your lack of understanding of physics and will follow you forward throughout time. Now, your best hope is to study your past mistakes to do better on your exam, BWAHAHAHA!”).
Now, this email really got me thinking. This is a good student. They have a strong understanding of most of the ideas in physics. Why are they seeking 3’s on everything? What’s the deal here? This also got me thinking about a fascinating post I saw on the the awesome psychology of games blog—and a great post titled “Conceptual Consumption and Kicks to the Head.”
Here’s the basic gist: Dan Ariely coined the phrase conceptual consumption (pdf) for the idea that people are just as interested in consuming ideas and information as they are physical things. They want to possess “ideas” that are rare as status symbols in the same way they want a nice car or clothes. For some, this becomes a sort of “check off list” of “thing’s I’ve done.” And of course, on this check off list, knowing an idea is rare or difficult makes it all the more coveted. (If you’re wondering how this connects to video games, you really should go read the blog post and see how XBox live uses this phenomenon to get people to want to spend vast amounts of time replaying tedious and brutally hard sections of games like Halo:Reach all so that they can achieve the achievement of having beaten the solo campaign on epic difficulty.)
Despite the fact that I often find myself falling victim to conceptual consumption, especially in the age of twitter, blogs and the universe of ideas the internet exposes me to, this is precisely what I do not want for my students: I do not want them “checking off boxes” in my class that say they understand every concept. I do not want them waiting for me to tell them they understand every concept. And I do not want them to think learning is just a process of acquiring “achievements” that you collect to add to your “experience resume” as status indicators.
(Cue idealistic music) I want them to see to see learning as a process, I want them to see that just when you think you understand a concept, you realize how much more there is to understand, and that is a good thing. Precisely for this reason, it’s pointless to put together a list of “intellectual achievements” for you to achieve, since unlike Halo, the list of possibilities is endless, and the most important ones haven’t even been found yet. I want them to see that they can begin to measure their own understanding, and when they really achieve mastery, they will know it, without having me or Xbox live telling them so.
So why do I present them with a list of standards for each unit? And why aren’t there blank spaces at the bottom for the new standards we come up with along the way?
## How can we see risk-taking, failure and mistakes as investments?
A while ago, @mmhoward tweeted this:
The best way for my students to learn physics is by making mistakes. Lots of them. Its a truism student who survives the physics major knows to be true, and it’s shared all the way up the pantheon to the greats:
“An expert is a person who has made all the mistakes that can be made in a very narrow field.” —Neils Bohr
So here’s the challenge. My students can do corrections to test questions to demonstrate increased understanding and replace previous grades. They can also reassess standards outside of class. Often, it’s the case that in the process of doing corrections, or reassessing a standard, a student will reveal another misconception that shows they don’t have mastery of a different concept. What do you do in this situation? The right thing to do is tell them that this showed they didn’t understand that idea they thought they mastered, and lower their score on that other concept too.
As you can imagine, this can send perfection seeking students into a tailspin. “What? I’m correcting something and you’re lowering my grade?” And it contradicts the idea that we learn through making mistakes, and I want to encourage my students to make as many mistakes as possible. So how do I reward them for the investment they are making in doing corrections and exposing other misconceptions? Sure, some investments don’t pay off (really wish I had held onto those 10 shares of Apple stock I bought a five years ago, instead of selling them at break even), but I want my grading system to reward the kids who are tenaciously ferreting out misunderstandings of ideas that they thought they “knew”, rather than let them sit comfortably with their perfect scores on every idea and think they’ve figured out all there is to this subject.
Here’s a prime example (sidenote for the physics un-inclined: feel free to jump over the next few paragraphs, but don’t miss the bigger point about understanding the big idea I highlight later). A student wanted to reassess his understanding of how to solve kinematics problems. So I asked him to calculate the distance a car would travel if it were traveling at 20 m/s and came to a stop in 1.5s, assuming constant acceleration.
Here’s his response:
The answer would use the formula 1/2 deltav*deltat because of a triangle needed to find the displacement in the velocity vs. time graph. If you add numbers, you get 1/2*20m/s*1.5s which then gives you the final answer of a displacement of 15m.
This answer belies a few significant misconceptions. First, it relies too much on formulas. The formula $1/2 \Delta v \Delta t$ isn’t something I’ve taught them, and it shouldn’t be written in the pantheon of important physics formulas. Moreover, if you do the calculation a bit more rigorously, you find $\Delta v=v_f-v_i=0-20\:\textrm{m/s}=-20\:\textrm{m/s}$, but if you follow that through you’ll find that the formula above gives you a negative displacement, which makes no sense at all.
So I wrote my student back to explain this very point. And here was his response (he also included a graph similar to this—which is awesome).
The reason that I didn’t stay with the negative delta-v was to keep the numbers real. Yes, the change in the velocity was -20, but if you look at the green triangle that shows displacement, you would realize that an area can’t be negative. That’s when you must do the absolute value of delta-v, in order to keep the area of the triangle positive, and therefore the displacement positive as well.
Can you see the learning taking place? Here’s a kid who knows what the answer is graphically, but is hanging on to a formula, and when the formula doesn’t make sense, he throws out the negative because he knows the right answer from the graph. This is good understanding—far better than trying to pass off a stopping car goes backward because the formula says so, but it also shows he’s not really seeing the connection between the graph and the formula.
I’ll spare you the 5 additional back and forth messages it took to begin to clear this up, followed by a 1-on-1 conversation, but eventually the student realized that the number from the formula should be $-20\:\textrm{m/s}$, but he also needs to account for the distance the car travels because of its initial velocity.
Graphically, we see that the student was calculating the displacement that corresponds to the red triangle, and if he adds this negative displacement to the displacement represented by the area of the black box, he would get the a displacement of $15\:\textrm{m}$, the area of the green triangle.
And if he’d started with a more general formula, based on the complete picture from the graph, everything agrees perfectly.
$\begin{array}{rcl}\Delta x &=& v_0t+\frac{1}{2}\Delta t \Delta v \\ &=&20\:\textrm{m/s} \cdot 1.5\:\textrm{s}+\frac{1}{2}\left(-20\:\textrm{m/s} \right)1.5\:\textrm{s}\end{array}$
$\begin{array}{rcl}\Delta x&=& 15\:\textrm{m}\end{array}$
So what sort of grade should you assign a kid for making this breakthrough? And how much does he really own this breakthrough versus just doing stuff I’ve assigned so he can check off some boxes to earn achievements?
These are difficult questions that I’m wrestling with, and they bring me back to the big idea I’ve shared about grades before:
Let grades be the start of a conversation, and not the end.
1. December 15, 2010 8:06 pm
This web that you are immersed in on the grading issue is circular. I really don’t see a way out of the circular issue or argument. If you are truly interested in “mastery” above your interest in assigning a “grade,” I think you have to go for mastery, ignore the grade, and not put time constraints on the learning process. Now that may not be realistic given where you are. However, I think it might be the only way out of the argument you are having with yourself. Mastery of physics is important, but is mastery of every nuance important? Depends? Who is the student and what are their personal goals for the course? Do you have that information for each student and can you tailor your grading to their personal goals?
Good work on the blog posting. Very informative and good questions are being posted.
Bob R.
2. December 15, 2010 10:53 pm
Bob,
You raise some good questions, and they are issues I hope to discuss in our PLC. At the same time, I think the quest for total agreement and perfect “fairness” can be quite un-fruitful. Life really isn’t fair—Mr. X things sig figs are super important, and takes off points every time, Ms. Y doesn’t care. Mr. Z gives nothing but multiple choice tests with no partial credit, while Mr. A awards generous partial credit. Students constantly grouse about these differences, or the English teacher who is supposedly the “easy” grader, and the nit-picky “hard” grader who seems to give every paper a B-, but I think they learn to adapt to these differences, and this is a valuable lesson. After all, we’ll all end up working for a number of different bosses in our lifetimes, each of whom might emphasize different things, and our success will, in large part, determined by how we adapt to those varying demands. One of the benefits of teaching in an independent school, I think, is giving teachers the freedom to teach and grade in the manner they see fit, within reason, of course, so that each teacher maximizes her strengths. I worry that too much focus on standardization can lead to regression to the mean, and degrade the overall student experience.
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2018-01-22 12:11:20
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http://ncatlab.org/schreiber/show/flat+differential+cohomology
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# Schreiber flat differential cohomology
In a locally contractible (∞,1)-topos $\mathbf{H}$ with internal path ∞-groupoid functor $(\mathbf{\Pi} \dashv \mathbf{\flat})$, the flat differential cohomology of an object $X$ with coefficient in an object $A$ is the $A$-cohomology of the path ∞-groupoid $\mathbf{\Pi}(X)$:
$\mathbf{H}_{flat}(X,A) := \mathbf{H}(\mathbf{\Pi}(X),A) \simeq \mathbf{H}(X, \mathbf{\flat}(A)) \,.$
The constant path inclusion $X \to \mathbf{\Pi}(X)$ induces a morphism
$\mathbf{H}_{flat}(X,A) \to \mathbf{H}(X,A)$
which sends a flat differential cocycle to its underlying or bare cocycle.
The obstruction theory for lifts through this morphism is the differential cohomology of $X$.
See differential cohomology in an (∞,1)-topos for more details.
Revised on May 29, 2012 22:04:00 by Andrew Stacey (129.241.15.200)
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2015-05-22 11:32:17
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https://stats.stackexchange.com/questions/243511/supervised-topic-classification
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# Supervised topic classification
I am trying to perform a classification between 20K objects to 40 classes.
The features of the classification are text based, per each objects i have its raw text.
In addition i have a ~300 training sample, roughly 8 samples per class.
Ive tried using the following method:
1) Vectorize the data - using tf_idf.
2) Decomposing the vectors - using NMF.
3) Training all sorts of classifiers on decomposed vectors.
My problem are the big differences between training and test sets prediction accuracy, even after selecting only few latent variables(post decomposition).
• If i use decomposition and selecting only top latent variables -
train prediction accuracy(PA) is around the 30% and test's is at 10%.
• If i dont use decomposition and train the classifier using the vectors (tried vectors shapes sizes from 1000 to 10000 by capping max_features in the vectorizer) -
train PA is ~100% and test PA is ~40%.
Ive grid-searched different vectorizing, decomposition and classifiers hyper params, and kept getting the same difference in prediction accuracy.
Questions:
1) Is it an overfit problem?
2) If so, how come its still happening even when choosing only few features for classification?
3) Is my method well suited for this problem? are there any other options i should consider?
I can see a few problems:
1. How big is the difference? We expect the testing set to have worse performance. Is it 10% worse (pretty good)? 25% worse?
2. How long are your training documents in words?
3. You should probably use some other metric aside from accuracy for two reasons: You have 40 classes, so your random performance baseline will probably be around 2.5% accuracy under the assumption that all 8 classes are the same size in the training set. Use a confusion matrix, or per-class precision/recall/F1-score/AUC instead.
Also, I can imagine having only 8 samples per class might be a pretty big problem particularly if those 8 samples have very heterogeneous lexicons.
If the number of words in your samples is large, you essentially have a lot of parameters, many more than your number of samples, and regularization can be helpful. But, before you do that try chi-squared feature selection for each class to determine which words are predictive within each class. The problem with vanilla TF-IDF is that it is measuring how "influential" each word is across the entire corpus, not with respect to each class, unless you specifically set up your data or code to do that.
To answer questions 1 and 2: It is probably an overfitting problem and your features are either not predictive or not scaled properly.
It is important to develop intuition into the problem you are solving rather than rely on grid search and cross-validation, because that will just leave you frustrated.
For question 3: I would try the above first. You may also want to look at Supervised Latent Dirichlet Allocation (SLDA) which essentially fits words into topics based on a response variable (class in this case). I believe there is code for the multinomial case or you can construct a series of logistic SLDA models in a one vs rest ensemble. The topic representation of each unseen document becomes the predictor variables (the X in standard regression). SLDA is available in the lda package for R and a few C implementations are on Github.
• My training documents as per words count vary from 10 words to 2000. Is there a good way evaluating lexicons heterogeneity? What measure should i use and what benchmarks? – yoav_aaa Nov 1 '16 at 11:33
• I usually don't use as many categories, so I may be off, but your accuracy seems to make sense. Remember that the random baseline is around 2.5% for 40 classes, so your classifier is doing something. The test accuracy drop is a bit steep considering the baseline, but somewhat expected. Have you considered just using words as features? By heterogeneity, I just mean, do the samples in each class approx. share the same vocabulary? Or are they wildly different? – Ryan Rosario Nov 1 '16 at 23:05
• Each class vocabulary have some shared words (approx 30% of TF_IDF top terms per class) and there are some share words between classes. Everything makes sense but the actual results. By using words as features, do you mean use only the term-frequency? – yoav_aaa Nov 2 '16 at 7:03
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2021-01-20 10:08:04
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https://trac.sagemath.org/attachment/ticket/11599/trac_11599_remaining_fixes.patch
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# Ticket #11599: trac_11599_remaining_fixes.patch
File trac_11599_remaining_fixes.patch, 15.9 KB (added by vbraun, 10 years ago)
Initial patch
• ## doc/en/reference/schemes.rst
# HG changeset patch
# User Volker Braun <vbraun@stp.dias.ie>
# Date 1329691392 28800
# Parent e7de36bc238097cde068ff033d5c3791cdfe134b
Trac #11599: Toric morphisms
Fixes for various issues that the reviewer brought up
diff --git a/doc/en/reference/schemes.rst b/doc/en/reference/schemes.rst
a sage/schemes/generic/ambient_space sage/schemes/generic/affine_space sage/schemes/generic/projective_space sage/schemes/generic/toric_variety sage/schemes/generic/fano_toric_variety sage/schemes/generic/toric_variety sage/schemes/generic/fano_toric_variety sage/schemes/generic/toric_variety_library sage/schemes/generic/toric_divisor sage/schemes/generic/toric_chow_group sage/schemes/generic/toric_ideal sage/schemes/generic/toric_morphism sage/schemes/generic/algebraic_scheme sage/schemes/generic/hypersurface
• ## sage/schemes/generic/algebraic_scheme.py
diff --git a/sage/schemes/generic/algebraic_scheme.py b/sage/schemes/generic/algebraic_scheme.py
a """ return "Subscheme of %s"%self.__A def _homset(self, *args, **kwds): """ Construct the Hom-set INPUT: Same as :class:sage.schemes.generic.homset.SchemeHomset_generic. OUTPUT: The Hom-set of the ambient space. EXAMPLES:: sage: P1. = toric_varieties.P1() sage: type(P1.Hom(P1)) sage: X = P1.subscheme(x-y) sage: type(X.Hom(X)) """ return self.__A._homset(*args, **kwds) def _point_homset(self, *args, **kwds): return self.__A._point_homset(*args, **kwds) INPUT: - same as for :class:~sage.schemes.generic.morphism.SchemeMorphism_polynomial_toric_variety. :class:~sage.schemes.generic.toric_morphism.SchemeMorphism_polynomial_toric_variety. OUPUT: - :class:~sage.schemes.generic.morphism.SchemeMorphism_polynomial_toric_variety. - :class:~sage.schemes.generic.toric_morphism.SchemeMorphism_polynomial_toric_variety. TESTS:: Defining z0, z1, z2, z3 sage: P1 = P1xP1.subscheme(z0-z2) sage: H = P1.Hom(P1xP1) sage: H([z0,z1,z0,z3]) Scheme morphism: From: Closed subscheme of 2-d toric variety covered by 4 affine patches defined by: z0 - z2 To: 2-d toric variety covered by 4 affine patches Defn: Defined on coordinates by sending [z0 : z1 : z2 : z3] to [z2 : z1 : z2 : z3] sage: P1._morphism(H, [z0,z1,z0,z3]) Scheme morphism: From: Closed subscheme of 2-d toric variety from sage.schemes.generic.toric_morphism import SchemeMorphism_polynomial_toric_variety return SchemeMorphism_polynomial_toric_variety(*args, **kwds) def fan(self): """ Return the fan of the ambient space. OUTPUT: A fan. EXAMPLES:: sage: P2. = toric_varieties.P(2) sage: E = P2.subscheme([x^2+y^2+z^2]) sage: E.fan() Rational polyhedral fan in 2-d lattice N """ return self.ambient_space().fan() def affine_patch(self, i): r""" Return the i-th affine patch of self as an affine
• ## sage/schemes/generic/scheme.py
diff --git a/sage/schemes/generic/scheme.py b/sage/schemes/generic/scheme.py
a #***************************************************************************** from sage.misc.all import cached_method from sage.structure.parent import Parent from sage.misc.all import cached_method from sage.rings.all import (IntegerRing, is_CommutativeRing,
• ## sage/schemes/generic/toric_homset.py
diff --git a/sage/schemes/generic/toric_homset.py b/sage/schemes/generic/toric_homset.py
a You should not create the Hom-sets manually. Instead, use the :meth:~sage.structure.parent.Hom method that is inherited by all schemes. AUTHORS: - Volker Braun (2012-02-18): Initial version EXAMPLES: Here is a simple example, the projection of \mathbb{P}^1\times\mathbb{P}^1\to \mathbb{P}^1 :: sage: P1xP1 = toric_varieties.P1xP1() sage: P1 = toric_varieties.P1() sage: hom_set = P1xP1.Hom(P1); hom_set Set of morphisms From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches In terms of the fan, we can define this morphism by the projection onto the first coordinate. The Hom-set can construct the morphism from the projection matrix alone:: sage: hom_set(matrix([[1],[0]])) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. sage: _.as_polynomial_map() Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined on coordinates by sending [s : t : x : y] to [s : t] In the case of toric algebraic schemes (defined by polynomials in toric varieties), this module defines the underlying morphism of the ambient toric varieties:: sage: P1xP1.inject_variables() Defining s, t, x, y sage: S = P1xP1.subscheme([s*x-t*y]) sage: type(S.Hom(S)) """ #***************************************************************************** # Copyright (C) 2010 Volker Braun # Copyright (C) 2010 Andrey Novoseltsev # # Distributed under the terms of the GNU General Public License (GPL) # as published by the Free Software Foundation; either version 2 of # the License, or (at your option) any later version. # http://www.gnu.org/licenses/ #***************************************************************************** from sage.rings.all import ZZ, is_RingHomomorphism from sage.matrix.matrix import is_Matrix from sage.matrix.matrix_space import MatrixSpace sage: P1 = toric_varieties.P1() sage: hom_set = P1xP1.Hom(P1); hom_set Set of morphisms From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches sage: type(hom_set) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. """ def __init__(self, X, Y, category=None, check=True, base=ZZ): """ The Python constructor. INPUT: The same as for any homset, see :mod:~sage.categories.homset. EXAMPLES:: sage: P1xP1 = toric_varieties.P1xP1() sage: P1 = toric_varieties.P1() sage: hom_set = P1xP1.Hom(P1); hom_set Set of morphisms From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches An integral matrix defines a fan morphism, since we think of the matrix as a linear map on the toric lattice. This is why we need to register_conversion in the constructor below. The result is:: sage: hom_set(matrix([[1],[0]])) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. """ SchemeHomset_generic.__init__(self, X, Y, category=category, check=check, base=base) self.register_conversion(MatrixSpace(ZZ, X.fan().dim(), Y.fan().dim())) sage: dP8. = toric_varieties.dP8() sage: P2. = toric_varieties.P2() sage: Hom = dP8.Hom(P2) sage: hom_set = dP8.Hom(P2) sage: fm = FanMorphism(identity_matrix(2), dP8.fan(), P2.fan()) sage: Hom(fm) sage: hom_set(fm) # calls hom_set._element_constructor_() Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 2-d lattice N. A matrix will automatically be converted to a fan morphism:: sage: Hom(identity_matrix(2)) sage: hom_set(identity_matrix(2)) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 2-d lattice N. Alternatively, one can use homogeneous polynomials to define morphisms:: Defining y0, y1, y2 sage: dP8.inject_variables() Defining t, x0, x1, x2 sage: Hom([x0,x1,x2]) sage: hom_set([x0,x1,x2]) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches Defn: y0 |--> x0 y1 |--> x1 y2 |--> x2 sage: Hom(ring_hom) sage: hom_set(ring_hom) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches
• ## sage/schemes/generic/toric_morphism.py
diff --git a/sage/schemes/generic/toric_morphism.py b/sage/schemes/generic/toric_morphism.py
a Scheme morphism: From: 1-d CPR-Fano toric variety covered by 2 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches Defn: Defined by sending the Rational polyhedral fan in 1-d lattice N Defn: Defined by sending Rational polyhedral fan in 1-d lattice N to Rational polyhedral fan in 2-d lattice N. The fan morphism map is equivalent to the following polynomial map:: Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 2-d CPR-Fano toric variety covered by 3 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 2-d lattice N. sage: type(f) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. sage: P1xP1.hom(fm, P1) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. """ Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: 1-d CPR-Fano toric variety covered by 2 affine patches Defn: Defined by sending the Rational polyhedral fan in 2-d lattice N Defn: Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N. """ SchemeMorphism.__init__(self, parent) sage: P1 = toric_varieties.P1() sage: f = P1xP1.hom(matrix([[1],[0]]), P1) sage: f._repr_defn() 'Defined by sending the Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N.' 'Defined by sending Rational polyhedral fan in 2-d lattice N to Rational polyhedral fan in 1-d lattice N.' """ s = 'Defined by sending the ' s = 'Defined by sending ' s += str(self.domain().fan()) s += ' to ' s += str(self.codomain().fan())
• ## sage/schemes/generic/toric_variety.py
diff --git a/sage/schemes/generic/toric_variety.py b/sage/schemes/generic/toric_variety.py
a To: 1-d CPR-Fano toric variety covered by 2 affine patches sage: type(hom_set) """ This is also the Hom-set for algebraic subschemes of toric varieties:: sage: P1xP1.inject_variables() Defining s, t, x, y sage: P1 = P1xP1.subscheme(s-t) sage: hom_set = P1xP1.Hom(P1) sage: hom_set([s,s,x,y]) Scheme morphism: From: 2-d CPR-Fano toric variety covered by 4 affine patches To: Closed subscheme of 2-d CPR-Fano toric variety covered by 4 affine patches defined by: s - t Defn: Defined on coordinates by sending [s : t : x : y] to [s : s : x : y] sage: hom_set = P1.Hom(P1) sage: hom_set([s,s,x,y]) Scheme endomorphism of Closed subscheme of 2-d CPR-Fano toric variety covered by 4 affine patches defined by: s - t Defn: Defined on coordinates by sending [s : t : x : y] to [t : t : x : y] """ from sage.schemes.generic.toric_homset import SchemeHomset_toric_variety return SchemeHomset_toric_variety(*args, **kwds)
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2021-09-17 09:55:27
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https://www.openfoam.com/news/main-news/openfoam-v2106/boundary-conditions
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# v2106: New boundary conditions
## Improved LES inflow DFSEM condition
TOP
Input methods for the turbulentDFSEMInlet boundary condition have been expanded and made more flexible:
• Input entries of R, L and U are now PatchFunction1 type, which is the most generalised input methodology available for patch fields in OpenFOAM.
• Adds two new default entries of scalar quantities as normalisation factors for R, L and U:
• Uref : characteristic (reference) scale speed (default value = 1)
• Lref : characteristic (reference) length (default value = 1)
• These two entries scale the input R, L, and U: e.g. R/Uref^2, L/Lref, and U/Uref.
• Adds a new scalar factor to enable users to tune the C1 normalisation coefficient:
• scale: scales the C1 coefficient (default value = 1)
• Replaces chan395DFSEM and PCF tutorials with planeChannel and oneCellThickPlaneChannel tutorials to enable users to reproduce the reported results.
• Corrects the integral-length scale input files. A set of results obtained from the planeChannel tutorial (plot script available to users) can be seen below:
Source code
Tutorial
## Improved jump conditions
TOP
Optional under relaxation and minimum jump values have been added to the fan jump boundary conditions. Under relaxation can significantly improve numerical stability for steady cases, when applying a non-uniform spatial jump.
The following images show the predicted pressure and velocity fields for a steady, square-pipe case, at the jump plane located half way along the pipe. The first pair show that, for a non-uniform jump, the case is unstable and leads to checker board instabilities:
When setting applying under-relaxation, the case evolves smoothly to the expected pressure and velocity profiles:
Usage
plane
{
type fan;
patchType cyclic;
jump uniform 0;
value uniform 0;
uniformJump false;
// Optional minimum jump value
minJump 0;
// Optional under-relaxation
relax 0.2;
...
}
Source code
Tutorial
## New permeable boundary conditions for multiphase flow
TOP
A new pair of boundary conditions for multiphase flows has been developed to update p_rgh and U to mimic open and close events.
The switch in behaviour between prghTotalPressure and fixedFluxPressure is triggered by the Volume of Fluid (VOF) field, alpha. When the value of alpha at the patch exceeds alphaMin a wall condition is imposed; otherwise, a total pressure condition is applied. This combination reproduces the behaviour of a fully open or closed permeable wall.
Usage
The specification for p_rgh is as follows:
rightWall
{
type prghPermeableAlphaTotalPressure;
alpha alpha.water;
alphaMin 0.01;
p uniform 0;
value uniform 0;
}
The specification for U is as follows:
rightWall
{
type permeableAlphaPressureInletOutletVelocity;
alpha alpha.water;
alphaMin 0.01;
value uniform (0 0 0);
}
Source code
Tutorials
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2023-02-08 17:52:44
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https://www.physicsforums.com/threads/rolling-cylinder-on-surface-with-friction-initially-slipping.588305/
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# Homework Help: Rolling cylinder on surface with friction (initially slipping)
1. Mar 19, 2012
### vraeleragon
Hi, I'm trying to solve this, but I keep getting both sides of the equation to be the same >.<
A solid cylinder is initially moving along a flat surface without rotating. Due to the action of friction, it eventually begins rolling without slipping. The coefficient of friction is 0.22, the radius of the cylinder is 0.5 m, its mass is 2.7 kg and its initial center of mass velocity is 1.5 m/s
Questions: a.) How long does it take to reach a rolling without slipping condition? b.) What is the total energy change of the system between the initial condition and the establishment of rolling without slipping? c.) Will it still roll without slip on an incline?
I tried using FΔt=mΔv and d=v$_{0}$t+1/2at$^{2}$so i got the distance for slipping d=(v$_{i}$$^{2}$-v$_{f}$$^{2}$)/(2μg)
And then I did 1/2mv$_{i}$$^{2}$ = fd + 1/2mv$_{f}$$^{2}$ + 1/2Iω$^{2}$ I kept getting the same answer no matter what formula or approach I use. I have no clue how to do this. Please help. Thank you :)
2. Mar 19, 2012
### tiny-tim
hi vraeleragon!
no, it starts at slipping without rolling, goes though slipping with rolling, and ends with rolling without slipping
use the work energy theorem …
3. Mar 19, 2012
### ehild
The force of friction decelerates translation of the CM and the torque of friction (with respect to the CM) accelerates rotation. Write up both equations and solve for the linear velocity v and angular velocity ω in terms of time. Find t when the condition of pure rolling applies (do you know it?).
ehild
4. Apr 24, 2012
### vraeleragon
oh i forgot to say that i solved it. thank you.
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2018-06-22 17:39:32
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http://www.gradesaver.com/textbooks/science/chemistry/chemistry-a-molecular-approach-3rd-edition/chapter-2-sections-2-1-2-9-exercises-review-questions-page-79/27
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## Chemistry: A Molecular Approach (3rd Edition)
A mole is a unit representing $6.022\times10^{23}$ particles. $6.022\times10^{23}$ is Avogadro's number.
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2018-03-18 23:29:33
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https://mathoverflow.net/questions/330000/finite-picard-group
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# Finite Picard group
Does there exist a connected scheme, smooth, proper, and positive-dimensional over $$\mathbb{C}$$ with finite Picard group? Note that Picard group has cardinality$$>1$$. Also note that this can not happen for projective schemes.
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2019-12-06 22:33:10
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http://wikieducator.org/User:Jtneill/WikiEducator/Questions_and_comments/Navigation_font_size
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# User:Jtneill/WikiEducator/Questions and comments/Navigation font size
Jump to: navigation, search
The font size for the sidebar and the topbar appear in very small font I'm using Firefox 3.
• mmmm -- not sure what the problem might be here. Do you have the same issue on other Mediawiki installations? --Wayne Mackintosh 02:16, 24 May 2008 (UTC)
• Yes, for some installations. -- Jtneill - Talk 05:04, 29 June 2008 (UTC)
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2017-12-18 05:18:35
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https://brilliant.org/problems/the-apple-problem-2/
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# The Apple Problem
There is a box full of apples, it has more than 80 but less than 85 apples in the box, some of the apples are red and the others are green, if you divide all the red apples in the box into groups of 4, 1 is left over.
After a while, one person came, took away more than 15 green apples and then ran away.
Then, another person came and checked the box, he wanted to match 1 red apple and 1 green apple as a pair, after he matched all the possible pairs, he finds out there are less than 13 red apples that can't be matched as a pair.
At this point, if he switches some of the red apples, the same number as $$\frac{1}{4}$$ of stolen green apples, to green, then only 2 red apples can't be matched as a pair.
If there are originally at least 37 red apples in the box, what is the original number of apples in the box?
×
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2018-06-23 10:39:27
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https://tex.stackexchange.com/questions/356130/redefining-macro-clashes-with-titlesec-package-how-to-avoid
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# Redefining macro clashes with titlesec package. How to avoid?
In my question here how to properly add to a macro without storing the original one in a tmp variable, I was told that my code is actually working. However, it somehow clashes with the titlesec package.
How can I change my code to properly work along the titlesec package while still not defining a tmp variable for the original macro?
\documentclass{article}
\usepackage{titlesec}
\def\abc{abc}
\begingroup\let\orgabc\abc\def\abc{\orgabc\endgroup def}
\begin{document}
\section{One}
\abc
\section{Two}
Two
\end{document}
Works fine with \usepackage{titlesec} commented out!
• Is that really the definition of \abc that you want, with or without the titlesec package it wil completely break latex as it puts \begin{document} inside a group, also if you use \abc more than once you will get an error over an unmatched \endgroup. Surely you do not want the \endgroup to be in the definition of \abc ? – David Carlisle Feb 28 '17 at 8:52
• @DavidCarlisle Is this a tex-core or a latex-base question? – errekak Feb 28 '17 at 22:05
• @erreka marginal either way but tex-core I think as mostly it's about \def and \endgroup – David Carlisle Feb 28 '17 at 22:09
\def\abc{abc}
\begingroup\let\orgabc\abc\def\abc{\orgabc\endgroup def}
\begin{document}
\section{One}
\abc
This would place \begin{document} inside a group started by \begingroup and ended by \abc.
Any environment other than document this would generate an error. It does not generate an error message here for technical reasons but it completely breaks all latex processing as all the setup that normally happens at \begin{document} will be discarded at \abc
Depending on the intention you could do
\def\abc{abc}
\let\orgabc\abc
\def\abc{\orgabc def}
\begin{document}
\section{One}
\abc
or
\def\abc{abc}
\expandafter\def\expandafter\abc\expandafter{\abc def}
\begin{document}
\section{One}
\abc
or
\makeatletter
\def\abc{abc}
\makeatother
\begin{document}
\section{One}
\abc
Actually your code is equivalent to
\documentclass{article}
\usepackage{titlesec}
\begingroup
\begin{document}
\section{One}
\endgroup
\section{Two}
Two
\end{document}
but titlesec use \AtBeginDocument{...\let\ttl@Hy@refstepcounter\refstepcounter...} so after the \endgroup i.e after \abc in your code \ttl@Hy@refstepcounter is no more defined.
• breaking titlesec is the least of the problems, it means any definitions that happen at begin document are discarded at the group end, so default document fonts, math setup, any \label information from the aux file.... – David Carlisle Feb 28 '17 at 8:55
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2020-01-28 10:17:59
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http://mathhelpforum.com/differential-equations/173059-variation-parameters.html
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# Math Help - Variation of Parameters
1. ## Variation of Parameters
Hello. I'm having a bit of trouble understanding the rationale behind the assumption that is made in the variation of parameters method for first and second order linear differential equations.
For the first order equations, my book lays out the property of linear equations that the solution $y$ is the sum of the complementary and a particular solution, that is $y=y_c+y_p$. Then the author shows that the homogeneous equation
$\frac{dy}{dx}+P(x)y=0$
is seperable and has as its solution $y_c=ce^{-\int{P(x)}dx}$ and letting $e^{-\int{P(x)}dx}=y_1(x)$ for convenience, we have $y_c=cy_1(x)$.
Now the author states that in order to find the particular solution $y_p$ we use a procedure known as variation of parameters where we assume that $y_p=uy_1$.
So, I have no trouble at all with the derivation from here on. My question is, how is it that this assumption is made. Or, stated in another way, what was the reasoning behind making this assumption. What is it that would make one believe that the particular solution would be simply some function of the independent variable times the complementary solution?
2. I think it's just because we don't want to look for particular solution that has a complicated form. The proof shows that when you apply certain conditions, the form of the particular solution is valid in most (if not all) cases.
3. Originally Posted by VonNemo19
Hello. I'm having a bit of trouble understanding the rationale behind the assumption that is made in the variation of parameters method for first and second order linear differential equations.
For the first order equations, my book lays out the property of linear equations that the solution $y$ is the sum of the complementary and a particular solution, that is $y=y_c+y_p$. Then the author shows that the homogeneous equation
$\frac{dy}{dx}+P(x)y=0$
is seperable and has as its solution $y_c=ce^{-\int{P(x)}dx}$ and letting $e^{-\int{P(x)}dx}=y_1(x)$ for convenience, we have $y_c=cy_1(x)$.
Now the author states that in order to find the particular solution $y_p$ we use a procedure known as variation of parameters where we assume that $y_p=uy_1$.
So, I have no trouble at all with the derivation from here on. My question is, how is it that this assumption is made. Or, stated in another way, what was the reasoning behind making this assumption. What is it that would make one believe that the particular solution would be simply some function of the independent variable times the complementary solution?
In my opinion the reasoning it that if you make this assumption of the product form of the solution, after differentiating (via the product rule) you will always end up with a factor of
$u(x)$ multiplied by the homogeneous ODE which is always equal to 0. The hope then is that this new ODE for $u$ would be solvable.
So as in your example suppose $y_c$ solve the homogeneous system
$y'+p(x)y=0$ and you want to solve
$y'+p(x)y=f(x)$
Using this assumption as you get
$y=uy_c \implies y'=u'y_c+uy_c'$ now putting this into the ode gives
$\displaystyle u'y_c+uy_c'+p(x)uy_c=f(x) \iff u'y_c+\underbrace{u[y_c'+p(x)y_c]}_{\text{homogeneous ODE solution}=0}=f(x) \iff u'y_c=f(x) \iff u'=\frac{f(x)}{y_c}$
4. Or it could simply be the old Physics derivation stand-by... It works.
-Dan
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2016-07-29 23:36:42
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https://electronics.stackexchange.com/questions/220282/piezo-transducer-current-consumption
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# Piezo transducer current consumption
is there any way how theoretically determine the current consumption of ultrasound piezo transducer?
I only know impedance for res freq = 50 ohms and excitation voltage around -170 V.
Does the Ohms law work in this case?
Thank you.
It depends on what kind of signal you are driving it with. Ultrasonic transducers are pulsed, how much you pulse it is up to the designer, which you are going to want to keep minimize, 1) Not to waste power and 2) Not to dissipate large amounts of power into the thing you are measuring. A piezoelectric transducer is capacative and cannot be driven with a DC waveform (and you wouldn't want to, because you wouldn't have any signal to observe from waves reflecting off materials)
$$P_{avg} = \frac{1}{T} \int_0^{t}p(t) \,dt$$
If you have strictly a sine wave then A would be the amplitude:
$$V(t) = A*sin(2\pi f t)$$
$$P_{avg} = \frac{1}{T} \int_0^{t}\frac{V(t)^2}{R} \,dt = \frac{1}{T} \int_0^{t}\frac{A*sin(2\pi f t)^2}{R} \,dt$$
or what ever function of voltage you are producing.
• I´m going to use pulse excitation with very short (dozens ns) unipolar (-170V) pulses. The consumption will not be continuous and I´m able to set few seconds between each pulse -> can work with long scanning cycle. I´m stuck, because I´m not able to figure out if my selected DC/DC (MAX1847) can handle this. – ToKra Mar 1 '16 at 22:32
• If your using a pulse, then you need to figure out what the current across the sensor is. Calculate the AC current on paper, use a spice package like LT spice. My point is, its your signal is time-varying. The DC/DC converter needs to source the average current, use caps to cover for the times when you need short term power. – Voltage Spike Mar 4 '16 at 19:52
If you're driving the transducer at resonance, the reactive terms will cancel and all you'll be left with is the resistance.
In this case, if the impedance looks like a pure resistance, then Ohm's law applies and the transducer will draw
$$I = \frac {E}{Z} = \frac{170V}{50\Omega} = 3.4 \text { amperes}$$
and dissipate
$$P= I^2 R = 11.56 \times 50\Omega = 578 \text { watts}$$
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2020-02-18 17:19:55
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https://answers.ros.org/questions/334457/revisions/
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# Revision history [back]
### Calculating inertia for a vehicle robot
I'm trying to set the inertia for this Ackermann vehicle robot. I'm trying to model the GEM e2 https://gem.polaris.com/en-us/e2/specs/
I'm using these equations: https://en.wikipedia.org/wiki/List_of_moments_of_inertia#List_of_3D_inertia_tensors
It seems pretty straight forward. I'm using meters and kilograms. However, in Gazebo when I use the View Inertia option -- to see the pink rectangles representing the inertia -- it's way bigger than the robot.
Am I missing something important???
Here is how I'm calculating inertia. I've tried it a few different ways:
<!--<macro name="box_inertia" params="m x y z">
<inertia ixx="${m*(z*z+x*x)/12}" ixy = "0" ixz = "0" iyy="${m*(y*y+x*x)/12}" iyz = "0"
izz="${m*(y*y+z*z)/12}" /> </macro> <macro name="box_inertia" params="x y z mass"> <inertia ixx="${0.0833333 * mass * (y*y + z*z)}" ixy="0.0" ixz="0.0"
iyy="${0.0833333 * mass * (x*x + z*z)}" iyz="0.0" izz="${0.0833333 * mass * (x*x + y*y)}" />
</macro>-->
<macro name="box_inertia" params="x y z m">
<inertia ixx="${m*(y*y+z*z)/12}" ixy = "0" ixz = "0" iyy="${m*(x*x+z*z)/12}" iyz = "0"
izz="${m*(x*x+z*z)/12}" /> </macro> <macro name="cylinder_inirtia" params="m r h"> <inertia ixx="${(m*((3*r*r)+(h*h)))/12}" ixy = "0" ixz = "0"
iyy="${(m*((3*r*r)+(h*h)))/12}" iyz = "0" izz="${(m*r*r)/2}" />
</macro>
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2019-12-13 01:02:11
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https://codeforces.com/blog/entry/97222
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By usernameson, history, 6 days ago,
## Example Type problem
You have an array $a$ and for each number $x$ in $[1,10^{6}]$ you want to know how many $a_{i}$ have the property $a_{i}$ & $x = x$.
## Standard approach
You can use the standard submask iteration to do it in $O(3^n)$. For the example type problem the initial value for each $x$ is the number of times it appears in the array and these counts get added to all submasks.
## Example implementation
Code
Problem Application
• +8
» 6 days ago, # | 0 Auto comment: topic has been updated by usernameson (previous revision, new revision, compare).
» 6 days ago, # | +24 The actual standard approach for this is a simple $O(k \cdot 2^k)$ loop on $k$-bit numbers. If you view the bitmasks $[0 \ldots 2^k - 1]$ as a $k$-dimensional array with length 2 in each dimension, this amounts to taking suffix sums along each axis in turn. Alternatively, you can view this process as a DP as is described in this blog. Or, you can view it as a factorization of the linear operator that performs this transform, as is described in this blog.
• » » 5 days ago, # ^ | 0 Thanks. That's a much a better approach. I need to read those blogs.
» 4 days ago, # | +1 This is not SOS DP? https://codeforces.com/blog/entry/45223
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2021-12-01 04:14:56
|
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https://www.physicsforums.com/threads/non-ideal-inductor-voltage.652228/
|
Non-ideal inductor voltage
1. Nov 14, 2012
DrOnline
1. The problem statement, all variables and given/known data
An inductor is subjected to the current in the graph:
L = 100mH = 0.1H
I) Calculate and graph the voltage over the ideal inductor.
II) Repeat for non ideal inductor with a resistance of:
RL = 50 Ω
2. Relevant equations
$$U_{L}=L*\frac{di}{dt}$$
$$U_{av}=L*\frac{Δi}{Δt}$$
The voltage over an inductor is equal to the rate of change of the current through the coil.
3. The attempt at a solution
I)
This is what I came up with. As an example, from 2 ms to 4 ms:
$$U_{av}=0.1H*\frac{-2A}{2*10^-3 s} = -100V$$
Ok, some question, I am pretty sure the answer is yes to these two, but I'd love to have them confirmed, so I know I am on the right track:
*Is my square wave form correct? Should it be square, like this?
*Am I doing this correctly? It seems to me $$U_{av} = U$$. So I could write just U =...
II)
This is when I become confused, and I have spent hours and hours trying to solve this... sadly heh. I don't know where to start!
The way I understand it:
*Regardless of the internal resistance, the current will still be the same, as the coil and the internal resistance will be in series.
*So for charging up during 0-1ms, the voltage over the coil will be the same.
*Am I still going to have a square wave form?
I've been given as a clue that $$U_{coil}(3ms) = -50 V$$
Please, just a nudge in the right direction, I'm totally stuck here..
Last edited: Nov 14, 2012
2. Nov 14, 2012
aralbrec
Yes, all good assuming your algebra is right.
Yes you are but I don't even look at the Uav equation. What is important is the slope of i(t) at any particular time and in your graphs the slope di/dt is constant over periods of time.
That's right, you are shoving the same current through the non-ideal inductor as before. The outside current source will have to adjust its terminal voltage to accomplish this.
The non-ideal inductor is now modelled as a resistance in series with an ideal inductor. The voltage you measure across the non-ideal inductor is equal to the voltage across the ideal part plus ?
3. Nov 14, 2012
DrOnline
$$v_{real} = v_{ideal} + v_{R_{i}}$$
Matching the variables to my drawing's labels:
$$v_{non-ideal} = v_{ideal} + v_{R}$$
Alright, so the voltage from 2ms to 4ms is -100 over the ideal.
And the average current is 1 A. That makes the voltage over the resistor: R * I = 50 ohm * 1A = 50V.
$$v_{real} = v_{ideal} + v_{R_{i}} = -100V + 50V = -50V$$
Is this sound reasoning? I think it is. What troubled me was the linear drop in the current, and I was having problems understanding how that influenced the voltage of the resistor... I think I get it now.
Last edited: Nov 14, 2012
4. Nov 14, 2012
aralbrec
Yes :). But let's not look at the average, let's look at a specific instant of time.
$$U_{coil}(3ms) = -50 V$$
At t=3ms, the voltage across the ideal inductor is -100. The current through the inductor is (from your graph) 1A. So the voltage measured across the nonideal inductor is: Videal + iR = -100 + 50 = -50V, in agreement with the hint.
I just wanted to make sure you were looking at the current at a particular instant of time t=3ms instead of some average over an interval.
You can see that the voltage measured across the non-ideal inductor will be a scaled version of your i(t) graph (iR) plus the square wave voltage you found for the ideal inductor. So your squares are going to disappear except where the current is constant.
5. Nov 14, 2012
DrOnline
Thank you so much.
I kinda knew that when I finally get this task solved, I would look back at it and wonder what took me so long.
The problem was I kept feeling the drop in current coming through the inductor would change the voltage over the conductor all the time, causing some weird differential equation task, where the voltage rose abruptly, but dropped exponentially. And then I was constantly wondering about whether I had built my work on faulty logic from the start, so it's a huge help just that you confirmed for me the first steps, that I had done it right.
Thanks again! ;)
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2018-02-22 05:24:58
|
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https://www.coin-or.org/CppAD/Doc/speed_main.htm
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Running the Speed Test Program
Syntax
speed/package/speed_package test seed option_list
Purpose
A version of this program runs the correctness tests or the speed tests for one AD package identified by package .
package
The command line argument package specifies one of the AD package. The CppAD distribution comes with support for the following packages: adolc , cppad , fadbad , sacado . You can extend this program to include other package. Such an extension need not include all the tests. For example, link_sparse_hessian just returns false for the fadbad and sacado packages.
double
The value package can be double in which case the function values (instead of derivatives) are computed using double precision operations. This enables one to compare the speed of computing function values in double to the speed of the derivative computations. (It is often useful to divide the speed of the derivative computation by the speed of the function evaluation in double.)
profile
In the special case where package is profile, the CppAD package is compiled and run with profiling to aid in determining where it is spending most of its time.
test
It the argument test specifies which test to run and has the following possible values: correct , speed , det_minor , det_lu , mat_mul , ode , poly , sparse_hessian , sparse_jacobian . You can experiment with changing the implementation of a particular test for a particular package.
correct
If test is equal to correct, all of the correctness tests are run.
speed
If test is equal to speed, all of the speed tests are run.
seed
The command line argument seed is an unsigned integer (all its characters are between 0 and 9). The random number simulator uniform_01 is initialized with the call uniform_01(seed) before any of the testing routines (listed above) are called.
Global Options
This global variable has prototype
extern std::map<std::string, bool> global_option;
The syntax global_option["option"] has the value true, if option is present, and false otherwise. This is true for each option that follows seed . The order of the options does not matter and the list can be empty. Each option, is be a separate command line argument to the main program. The documentation below specifics how speed_cppad uses these options, see the examples in speed_adolc for how another package might uses these options.
onetape
If this option is present, speed_cppad will use one taping of the operation sequence for all the repetitions of that speed test. Otherwise, the operation sequence will be retaped for each test repetition. All of the tests, except det_lu , have the same operation sequence for each repetition. The operation sequence for det_lu may be different because it depends on the matrix for which the determinant is being calculated. For this reason, cppad_det_lu.cpp returns false, to indicate that the test not implemented, when global_onetape is true.
memory
This option is special because individual CppAD speed tests need not do anything different if this option is true or false. If the memory option is present, the CppAD hold_memory routine will be called by the speed test main program before any of the tests are executed This should make the CppAD thread_alloc allocator faster. If it is not present, CppAD will used standard memory allocation. Another package might use this option for a different memory allocation method.
optimize
If this option is present, CppAD will optimize the operation sequence before doing computations. If it is false, this optimization will not be done. Note that this option is usually slower unless it is combined with the onetape option.
atomic
If this option is present, CppAD will use a user defined atomic operation is used for the test. So far, CppAD has only implemented the mat_mul test as an atomic operation.
hes2jac
If this option is present, speed_cppad will compute hessians as the Jacobian of the gradient. This is accomplished using multiple levels of AD. So far, CppAD has only implemented the sparse_hessian test in this manner.
subgraph
If this option is present, speed_cppad will compute sparse Jacobians using subgraphs. The CppAD sparse_jacobian test is implemented for this option. In addition, the CppAD sparse_hessian test is implemented for this option when hes2jac is present.
Sparsity Options
The following options only apply to the sparse_jacobian and sparse_hessian tests. The other tests return false when any of these options are present.
boolsparsity
If this option is present, CppAD will use a vectors of bool to compute sparsity patterns. Otherwise CppAD will use vectors of sets .
revsparsity
If this option is present, CppAD will use reverse mode for to compute sparsity patterns. Otherwise CppAD will use forward mode.
subsparsity
If this option is present, CppAD will use subgraphs to compute sparsity patterns. If either the boolsparsity or revsparsity is also present, the CppAD speed tests will return false; i.e., these options are not supported by subgraph_sparsity .
colpack
If this option is present, CppAD will use colpack to do the coloring. Otherwise, it will use it's own coloring algorithm.
Correctness Results
One, but not both, of the following two output lines package_test_optionlist_available = false package_test_optionlist_ok = flag is generated for each correctness test where package and test are as above, optionlist are the options (in option_list ) separated by the underbar _ character (whereas they are separated by spaces in option_list ), and flag is true or false.
Speed Results
For each speed test, corresponds to three lines of the following form are generated: package_test_optionlist_ok = flag package_test_size = [ size_1, ..., size_n ] package_test_rate = [ rate_1, ..., rate_n ] The values package , test , optionlist , and flag are as in the correctness results above. The values size_1 , ..., size_n are the size arguments used for the corresponding tests. The values rate_1 , ..., rate_n are the number of times per second that the corresponding size problem executed.
n_sweep
The sparse_jacobian and sparse_hessian tests has an extra output line with the following form package_sparse_test_n_sweep = [ n_sweep_1, ..., n_sweep_n ] were test is jacobian (hessian). The values n_sweep_1 , ..., n_sweep_n are the number of sweeps (colors) used for each sparse Jacobian (Hessian) calculation; see n_sweep for sparse_jacobian and sparse_hessian .
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2020-02-28 17:13:26
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http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-fm187-1-3?printView=true
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Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl
PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
• # Artykuł - szczegóły
## Fundamenta Mathematicae
2005 | 187 | 1 | 61-93
## Determining c₀ in C(𝒦) spaces
EN
### Abstrakty
EN
For a countable compact metric space 𝒦 and a seminormalized weakly null sequence (fₙ)ₙ in C(𝒦) we provide some upper bounds for the norm of the vectors in the linear span of a subsequence of (fₙ)ₙ. These bounds depend on the complexity of 𝒦 and also on the sequence (fₙ)ₙ itself. Moreover, we introduce the class of c₀-hierarchies. We prove that for every α < ω₁, every normalized weakly null sequence (fₙ)ₙ in $C(ω^{ω^{α}})$ and every c₀-hierarchy 𝓗 generated by (fₙ)ₙ, there exists β ≤ α such that a sequence of β-blocks of (fₙ)ₙ is equivalent to the usual basis of c₀.
61-93
wydano
2005
### Twórcy
autor
• Department of Mathematics, National Technical University of Athens, Athens 15780, Greece
autor
• Department of Mathematics, National Technical University of Athens, Athens 15780, Greece
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2022-08-10 13:42:42
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https://juliaearth.github.io/GeoStats.jl/stable/plotting.html
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# Plotting
Most objects defined in GeoStats.jl can be plotted directly using the plot command from Plots.jl. For visualization of 3D objects, however, we recommend the experimental MeshViz.jl package. Additional plots are listed below that can be useful for geostatistical analysis.
## Built-in
A hscatter plot between two variables var1 and var2 (possibly with var2 = var1) is a simple scatter plot in which the dots represent all ordered pairs of values of var1 and var2 at a given lag h.
using GeoStats
using Plots
𝒟 = georef((Z=[10sin(i/10) + j for i in 1:100, j in 1:200],))
𝒮 = sample(𝒟, 500)
p1 = hscatter(𝒮, :Z, lag=0)
p2 = hscatter(𝒮, :Z, lag=20)
p3 = hscatter(𝒮, :Z, lag=40)
p4 = hscatter(𝒮, :Z, lag=60)
plot(p1, p2, p3, p4)
## PairPlots.jl
The PairPlots.jl package provides the corner plot that can be used with any table, including tables of attributes obtained with the values function.
## StatsPlots.jl
The StatsPlots.jl package provides various statistical plots such as boxplot, dotplot, violin and other plots commonly used in statistical workflows.
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2022-05-22 17:08:06
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https://www.esaral.com/q/in-freundlich-adsorption-isotherm-at-moderate-pressure-52228
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# In Freundlich adsorption isotherm at moderate pressure,
Question:
In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption $\left(\frac{x}{m}\right)$ is directly proportional to $\mathrm{P}^{\mathrm{x}}$. The value of $\mathrm{X}$ is:
1. $\infty$
2. 1
3. Zero
4. $\frac{1}{n}$
Correct Option: , 4
Solution:
$\frac{x}{m}=p^{x}$
the formula is $\frac{x}{m}=p^{1 / n}$
Hence $x=\frac{1}{n}$
The value of 'n' is any natural number.
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2023-04-01 21:00:39
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https://latex.org/forum/viewtopic.php?f=59&t=33560&p=114728&
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LaTeX forum ⇒ Theses, Books, Title pages ⇒ kaobook will not build Topic is solved
Classicthesis, Bachelor and Master thesis, PhD, Doctoral degree
wmeyer48
Posts: 9
Joined: Mon Sep 23, 2019 4:40 pm
kaobook will not build
I am relatively new to LaTeX, and have been trying to set up the kaomain template which I downloaded from latextemplates.com.
In trying to follow the instructions from that site:
% !TEX none
pdflatex main
makeindex main.nlo -s nomencl.ist -o main.nls
makeindex main
biber main
main
I find that pdflatex main fails because the glossaries have not been produced, and makeglossaries will not run because pdflatex failed:
% !TEX none
[13] (main.mw)
! Package glossaries Error: Glossary entry computer' has not been defined.
See the glossaries package documentation for explanation.
Type H <return> for immediate help.
...
l.99 ...ng to use them, like this: \gls{computer}.
?
! Emergency stop.
Is there a way to run pdflatex and ignore the lack of glossaries? Or is there some other solution needed? Running pdflatex again does not resolve the problem, no matter how many times I run it.
Thanks!
Johannes_B
Site Moderator
Posts: 4185
Joined: Thu Nov 01, 2012 4:08 pm
Define the glossary entry called computer.
The smart way: Calm down and take a deep breath, read posts and provided links attentively, try to understand and ask if necessary.
wmeyer48
Posts: 9
Joined: Mon Sep 23, 2019 4:40 pm
Define the glossary entry called computer.
I should have emphasized that this problem is with the unchanged files downloaded with the template. I have so far edited nothing. And being new to the package, was expecting to build the sample before making any changes.
And looking in main.tex, it appears as though the entry is being defined:
\newglossaryentry{computer}{
name=computer,
description={is a programmable machine that receives input, stores and manipulates data, and provides output in a useful format}
}
rais
Posts: 287
Joined: Sun Nov 16, 2014 8:51 pm
wmeyer48 wrote:And looking in main.tex, it appears as though the entry is being defined:
yes, well, the definition comes a bit too late. To ensure it's defined before being accessed, just move it -- and the \newacronym macros, too -- into the preamble, i.e., between \documentclass and \begin{document}`.
KR
Rainer
wmeyer48
Posts: 9
Joined: Mon Sep 23, 2019 4:40 pm
yes, well, the definition comes a bit too late. To ensure it's defined before being accessed, just move it -- and the \newacronym macros, too -- into the preamble, i.e., between \documentclass and \begin{document}.
Thank you! I might have gotten there eventually, but it's a part-time activity for me, and a lot to digest.
max4tex
Posts: 2
Joined: Mon Mar 15, 2021 7:48 pm
Hello,
when I try to compile the kaobook template, I get a couple of errors, but nothing else.
I tried to open it in studiotex and texmaker editors, but they give the same error messages:
There is called a "kaobook"-style, but it is named "kao" (one of these errors).
A couple of other errors refer to missing styles.
And finaly, I tried to use the compileall.sh-script that comes with the template, but am getting the same errors, more or less.
The only way to use the template without problems is to open it in overleaf.com.
What am I doing wrong? I would like to use it offline and have no chance to do so... Some help?
Thank you a lot!
With kind regards,
Markus
max4tex
Posts: 2
Joined: Mon Mar 15, 2021 7:48 pm
Ok, I resolved it by myself (reading a little more in this forum and on other pages):
The problem had to do with missing packages.
When installing the editors (like Texstudio or Texmaker), it seems to be that you will not get all the needed packages necessary for compiling the kaobook template.
I had to install the texlive-full package (ubuntu 20.04) and then everything worked fine out of the box.
Thank you for indirect help!
Ciao Markus
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2021-04-15 05:48:43
|
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https://stat.ethz.ch/pipermail/r-help/2009-February/382118.html
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# [R] Sweave schunk placement
Ista Zahn izahn at psych.rochester.edu
Fri Feb 20 17:08:18 CET 2009
Dear Professor Leisch + R helpers,
For the past few days I have been trying to figure out why the LaTeX
endfloat package is not working with Sweave. I figured it out, and
it's trivial: The endfloat package requires that \begin{table} and
\end{table} commands are on their own line. But Sweave is writing
LaTeX code like \end{table}\end{Schunk} which causes endfloat to
choke. Simply moving the code down to a new line like this:
\end{table}
\end{Schunk}
fixes the problem, but it is annoying to do this for every table
everytime I re-run Sweave. I don't understand LaTeX that well, so I'd
be really grateful if someone can tell me how to modify Sweave.sty (or
another file if needed) so that the \end{Schunk}'s are written on new
lines.
Thanks,
Ista
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2020-02-20 13:35:56
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https://stacks.math.columbia.edu/tag/091K
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Lemma 15.86.11. Let $A$ be a ring. Let $E \to D \to F \to E[1]$ be a distinguished triangle of $D(\mathbf{N}, A)$. Let $(E_ n)$, resp. $(D_ n)$, resp. $(F_ n)$ be the system of objects of $D(A)$ associated to $E$, resp. $D$, resp. $F$. Then for every $K \in D(A)$ there is a canonical distinguished triangle
$R\mathop{\mathrm{lim}}\nolimits (K \otimes ^\mathbf {L}_ A E_ n) \to R\mathop{\mathrm{lim}}\nolimits (K \otimes ^\mathbf {L}_ A D_ n) \to R\mathop{\mathrm{lim}}\nolimits (K \otimes ^\mathbf {L}_ A F_ n) \to R\mathop{\mathrm{lim}}\nolimits (K \otimes ^\mathbf {L}_ A E_ n)[1]$
in $D(A)$ with notation as in Remark 15.86.10.
Proof. This is clear from the construction in Remark 15.86.10 and the fact that $\Delta : D(A) \to D(\mathbf{N}, A)$, $- \otimes ^\mathbf {L} -$, and $R\mathop{\mathrm{lim}}\nolimits$ are exact functors of triangulated categories. $\square$
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2021-03-07 08:44:52
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http://clay6.com/qa/49147/if-a-and-b-are-two-sets-such-that-a-and-b-then-a-cap-b-times-a-
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Browse Questions
# If A and B are two sets such that A = {x: x $\in$ R and |x| = 5} and B = {x: x is a solution of $x^3 − 25x − x^2 + 25$}, then (A $\cap$ B) $\times$ A?
{(−5, −5), (−5, 5), (5, −5), (5, 5)}
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2017-06-25 05:16:23
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https://mathtuition88.com/2016/08/15/implicit-function-theorem/
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# Implicit Function Theorem
The implicit function theorem is a strong theorem that allows us to express a variable as a function of another variable. For instance, if $x^2y+y^3x+9xy=0$, can we make $y$ the subject, i.e. write $y$ as a function of $x$? The implicit function theorem allows us to answer such questions, though like most Pure Math theorems, it only guarantees existence, the theorem does not explicitly tell us how to write out such a function.
The below material are taken from Wikipedia.
## Implicit function theorem
Let $f:\mathbb{R}^{n+m}\to\mathbb{R}^m$ be a continuously differentiable function, and let $\mathbb{R}^{n+m}$ have coordinates $(\mathbf{x},\mathbf{y})=(x_1,\dots,x_n,y_1,\dots,y_m)$. Fix a point $(\mathbf{a},\mathbf{b})=(a_1,\dots,a_n,b_1,\dots,b_m)$ with $f(\mathbf{a},\mathbf{b})=\mathbf{c}$, where $\mathbf{c}\in\mathbb{R}^m$. If the matrix $\displaystyle [(\partial f_i/\partial y_j)(\mathbf{a},\mathbf{b})]$ is invertible, then there exists an open set $U$ containing $\mathbf{a}$, an open set $V$ containing $\mathbf{b}$, and a unique continuously differentiable function $g:U\to V$ such that $\displaystyle \{(\mathbf{x},g(\mathbf{x}))\mid\mathbf{x}\in U\}=\{(\mathbf{x},\mathbf{y})\in U\times V\mid f(\mathbf{x},\mathbf{y})=\mathbf{c}\}.$
Elaboration:
Abbreviating $(a_1,\dots,a_n,b_1,\dots,b_m)$ to $(\mathbf{a},\mathbf{b})$, the Jacobian matrix is
$\displaystyle (Df)(\mathbf{a},\mathbf{b})=\begin{pmatrix} \frac{\partial f_1}{\partial x_1}(\mathbf{a},\mathbf{b}) & \dots &\frac{\partial f_1}{\partial x_n}(\mathbf{a},\mathbf{b}) & \frac{\partial f_1}{\partial y_1}(\mathbf{a},\mathbf{b}) & \dots & \frac{\partial f_1}{\partial y_m}(\mathbf{a},\mathbf{b})\\ \vdots & \ddots &\vdots & \vdots & \ddots &\vdots\\ \frac{\partial f_m}{\partial x_1}(\mathbf{a},\mathbf{b}) & \dots & \frac{\partial f_m}{\partial x_n}(\mathbf{a}, \mathbf{b}) & \frac{\partial f_m}{\partial y_1}(\mathbf{a}, \mathbf{b}) & \dots & \frac{\partial f_m}{\partial y_m}(\mathbf{a}, \mathbf{b}) \end{pmatrix} =(X\mid Y)$
where $X$ is the matrix of partial derivatives in the variables $x_i$ and $Y$ is the matrix of partial derivatives in the variables $y_j$.
The implicit function theorem says that if $Y$ is an invertible matrix, then there are $U$, $V$, and $g$ as desired.
## Example (Unit circle)
In this case $n=m=1$ and $f(x,y)=x^2+y^2-1$.
$\displaystyle (Df)(a,b)=(\frac{\partial f}{\partial x}(a,b)\ \frac{\partial f}{\partial y}(a,b))=(2a\ 2b).$
Note that $Y=(2b)$ is invertible iff $b\neq 0$. By the implicit function theorem, we see that we can locally write the circle in the form $y=g(x)$ for all points where $y\neq 0$.
## Author: mathtuition88
http://mathtuition88.com
## 2 thoughts on “Implicit Function Theorem”
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2020-08-11 15:28:01
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https://web2.0calc.com/questions/the-initial-balance-of-a-mutual-fund-is-1200
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+0
# The initial balance of a mutual fund is $1200. 0 663 3 +16 The initial balance of a mutual fund is$1200. The fund is expected to grow in value at an annual rate of 2%. Let x represent the number of years since the fund was started. Let y represent the value of the fund x years later.
What equation models the value of the mutual fund x years after it was started?
Nov 18, 2018
#1
+2
y = \$1,200 x 1.02^x
Nov 18, 2018
#2
+770
+3
In LaTex, it would be $$y = 1, 200\cdot 1.02^x$$
.
Nov 19, 2018
edited by PartialMathematician Nov 19, 2018
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2020-02-28 17:52:42
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https://datko.net/tag/emacs/page/2/
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# Ubuntu is just a keypress away with crouton
Last weekend I tried another Ubuntu approach on my Samsung Chromebook. On the same blog where I found the ChrUbuntu trackpad fix, Craig provides instructions on how to run crouton. What’s crouton: it’s a ‘Chromium OS Ubuntu Chroot Environment’ originally found here.
The advantage of this method over the dual-boot method is that dual booting is not required! Instead, one can install linux to an SD card, launch X, and then switch between ChromeOS and Ubuntu with `Ctrl-Alt-Shift-Back.` And if you can master this key chord, Emacs commands are just one step away :p
So, I followed Craig’s instructions and setup XFCE as my window manager. However, I am still going to keep ChrUbuntu and my dual-boot for the time being. First let’s look at what’s cool about crouton:
1. Swap between ChromeOS and Ubuntu without rebooting.
2. `apt-get install emacs` works and installs emacs23 no problem. (I haven’t figured out how to get emacs24 on this device yet…)
3. Ubuntu and ChromeOS can share files together by mounting your ChromeOS directory into the Ubuntu chroot environment. (See bottom of this post)
Here’s what didn’t work for me:
1. I did not like XFCE. It was my first experience with this window manager and I had a few issues that I could not easily fix: the resolution was set too high and I could not seem to change it and I could not switch the super key to ctrl in XFCE. Using ctrl on the home row is a blessing and curse. It is SO NICE to use that nice big button for something useful besides caps lock, but on the other hand, it drives me bonkers to type on the default keymap… ARGH!
2. Something odd was going on with my SD card. After a reboot, all that was visible was “lost and found.” So, I just reformatted my card back to its regular self.
I may try this experiment again, but I might pick KDE as the target not xfce. But since I already have the dual-boot setup, there is not much motivation for me to change at this moment. If you don’t have Ubuntu running at all on your Chromebook, it may be better to start with the crouton approach since you don’t have to re-partition your internal drive at all.
# Creeping Emacs: Imma (Bitl)bee
In the continuing progression of Emacs taking over my life, I’ve found another service that can be replaced by Emacs: Google Talk!
For the rest of the world out there that has been using this for the last three years, please excuse my late discovery. Lately I’ve been lurking on the #emacs channel on freenode, keeping tabs on the conversation stream in the background. I was using Colloquy at first, which is a nice graphical IRC client on the mac, but I switched to ERC, an IRC client inside Emacs, and I was very impressed with how nicely it has integrated inside Emacs. Continue reading “Creeping Emacs: Imma (Bitl)bee”
# Hello world from ChrUbuntu!
After following these instructions, my Chromebook is now running ChrUbuntu! The install is fairly painless and the instructions worked for me without any modifications. It took about 15 minutes for my Chromebook to switch into “developer mode,” but after running the script and a few reboots later, I can dual-boot into Linux!
Thanks to the great work over at that blog and to this motivated reader for posting some awesome links to some Linux on the Chromebook material! At the moment, I just have the vanilla ChrUbuntu sources but per the comments there seems to be active community members trying to fix various issues. Most notable is that the touchpad is less responsive than in ChromeOS… [Update 21JAN13: I’ve post a script to fix this issue here]
While I don’t support Canonical decision to leave in its surveillance search feature, ChrUbuntu seemed the easiest Linux distro to get up and running right now. Plus, ChrUbuntu is based from Ubuntu 12.04 and I don’t think Canonical’s search appeared until 12.10. Thankfully, thanks to the diligent work from the EFF (this is a great organization, I just re-upped my membership), they have a post on how to remove this “feature.”
My first apt-get was for Emacs of course, in which I’m happily typing away. The Chromebook’s “search” key acts like a “super” key (Windows key) in ChrUbuntu so I found this post which shows how to switch it. Once you get used to the ctrl key located to the left of the “a” key, it’s very hard to switch back… I tried to get emacs24 but there were some issues. For some reason, I couldn’t add the ppa for emacs24 to apt-get and when I tried to build by source there were a lot of missing depends on x-windows and various ncurses libraries. So, I decided not to push it too far right now and be content with emacs 23 (which installed with apt-get just fine).
If you plan on using the Chromebook as a pure consumption device or if you never heard of Linux(?!) than ChromeOS may be fine for you. But actually, if that’s the case, a tablet may be better because you probably don’t need a keyboard. Otherwise, if you want to do *anything* else, try ChrUbuntu. Firefox is the default browser, but it may be possible to install Chromium if you really want. And if you have never tried Linux, the Chromebook is a well priced laptop, with which you can experiment. Just be sure to understand what’s working and not before you make the plunge.
# Feeding Pythons
In these last few weeks, I’ve been working in a slew of programming languages. My OS course involves hacking around in the Linux kernel, so that’s in `C`. The projects in my AI course have been in Lisp, Java and the next one is in Prolog. At work, I mainly dabble in `C++`, but I’ve been helping with some continuous integration scripts in Perl and Python.
Out of all these languages, I think Lisp has permanently changed the way I think about programming. Unfortunately, it didn’t really click until after I submitted my project… After working a bit in Lisp and then going back to other imperative languages I found myself thinking of a Lisp-like solution and then realizing, “oh, this is was a lot easier in Lisp!” For example, trying to pass functions as parameters is very easy in Lisp since functions are first-class objects, but to do this in `C` one has defined a function pointer, which has very specific type definitions and some messy syntax, etc…
So, that led me to find this Google Tech Talk by Peter Seibel on “Practical Common Lisp.” This lecture further cleared up some of the concepts about Lisp (again, after I submitted my project, which I’m still waiting for the grade…) and that lead me to buy Peter’s book: Coders at Work, which is an amazing collection of interviews from great programmers. Anybody who is interested in
programming would find this book intriguing. After I finish it, I’ll put up a post specifically on that book.
Then I set out to make a small project in Lisp and ended up using Python 🙂 I wanted to integrate my Google Reader feeds with Emacs gnus and I found `libgreader` which pretty much handled all of the details with the still-unofficial Google Reader API. Inspired by `reader2maildir`, which does the same thing in Ruby, I set out to essentially port this to Python.
And I was able to download my feeds and read them in gnus, but since a lot of the feeds using embedded `HTML` or simply links to the main article (i.e. they don’t contain a summary) and would require me opening a web browser anyway, I figured the gnus approach wasn’t the best, for me.
But I learned a few things about programming in Python along the way. First of all, one has to be a little more specific while searching Google. Working with RSS Feeds, I realized it’s not adequate to search “python feed” due to the homograph “python.” Unappetizing images aside, I really like the interactive python interpreter. I didn’t really understand the point of using the interpreter directly and I thought it was useful only for toy programs, but it really helps in prototyping. Like Lisp’s REPL (read-veal-print-loop), interactive programming feels like precognitive debugging; it allows one to step through a debugger while the program is being written.
And while Python is not Lisp, it does have some functional style artifacts (much to the chagrin of its designer from what I understand). Python has a `filter(function, sequence)` function that:
returns a sequence consisting of those items from the sequence for which function(item) is true.
Combine this with a lambda expression, and I was able to write something like this:
``````filter(lambda x: x.unread != 0, reader.getSubscriptionList())
``````
In this example, this line returns a list of feeds from Google Reader that contain unread items. So, I thought that was pretty cool 🙂 So while I would like to do some more stuff in Lisp, I was impressed on just how easy it was to get python running. That and with some Emacs-python integration, the development environment was very nice to work in.
# Building Java applications inside Emacs with JDEE
First of all, I think Eclipse is a great IDE. There are a ton of features that work really well in Eclipse: intellisense, auto-compilation, and hotswap debugging. It handles your project and classpath pretty well too… But the more I get used to Emacs the less I can stand typing in an non-Emacs setup. Eclipse even has key binding support for Emacs mode, but it’s just not the same. Plus, it occupies a tremendous amount of real-estate as seen below.
So I’m trying Java within Emacs with JDEE. At the moment, I can’t get it to be an Eclipse replacement and I’m not sure it supports completions like Eclipse does. But when I started typing I forgot I had auto-complete mode on! While it may not know the scope of a class, it saves a lot of time with quick tab completion of long Java variable and class names. Intellisense and auto-complete mode are two different tools, but I’m starting to think I’d rather have the auto-complete over Intellisense. Also, as I’ve been using YASnippet more and more, I can quickly add custom snippets to expand code. Eclipse does have some of this, namely a nice `try / catch` like snippet. But imagine being able to have that code expansion on anything you wanted… that’s YASnippet!
But to get it to work with Emacs 24, there is some annoying configuration. I found an obscure forum post that details the instructions, which seem to get JDEE up and running. After some manual project configuration by customizing some variables as outlined in the user’s guide, I was able to get my project built and running. Yeah yeah, Eclipse automatically handles the classpath variable, I know.
Honestly, this is a lot of pain to go through and I don’t recommend it unless you are already an Emacs user. But, once one does get it finally working, I think the benefits of staying in Emacs outweigh the context switch of using Eclipse. At least for small projects. For large projects with complex dependencies and a complicated build process, I’m not sure about JDEE, but for small projects, like this Othello game AI I’m working on, it’s fine.
So this weekend I’ll be improving my Othello AI and studying for my AI midterm. (All things I should be doing instead of trying to get Emacs to work well with Java… 🙂 )
# Emacs Wizardry: markdown-mode
I have made an amazing discovery recently. While its not quite an achievement like the Curiosity rover (Mr. Rover has been entertaining to follow on Twitter) it certainly has brought me great joy nonetheless. It is Jason Blevin’s `markdown-mode`.
Back in this post I mentioned that I would be using Emacs more routinely, to include editing these blog posts which I have been doing so far. But to get the formatting just the way I like it, I’ve been using `html-mode` and editing the raw html and then uploading it to WordPress. This wasn’t all that bad, but I’d have to remember the paragraph tag for each paragraph, use the right href, etc… Word wrapping was an issue too and I would have to fill / unfill paragraphs prior to uploading as well.
This is no more. Using markdown-mode I can edit in the lightweight markdown and after installing markdown (via `brew install markdown` on my mac), I can quickly generate the html from the text. I’ve used github’s flavored markdown (gfm) before, but it never clicked with me to use it for blogging until I did some googling. Conveniently, `gfm-mode` is also available.
I realized that I’ve just geeked out on Emacs, but very much in the Neal Stephenson definition from “Tune On, Tune In, Veg Out” in Some Remarks, which I have been enjoying lately:
“To geek out on something means to immerse yourself in its details to an extent that is distinctly abnormal-and to have a good time doing it.”
My Emacs usage has drastically increased now that I’m back at work, so there will be more geeking out in the future.
But first, from that same article, another great Stephenson quote (that and I’m enjoying the simple blockquote markdown syntax):
“The few conservatives still able to hold up one end of a Socratic dialogue are those in the ostracized libertarian wing-interestingly enough, a group with a disproportionately high representation among fans of speculative fiction.”
# Emacs and PostgreSQL
One of the nice things about going to grad school, is that I really get to refine my Emacs fu. I’m taking this database class where we are using PostgreSQL and I’m using Emacs’ minor sql modes and I’m very impressed. Now, I’m no emacs knight, nor do I have the Emacs-fu of the emacs rocks guy, but like any good student of zen, I’m maintaining a beginner’s mind (which is pretty easy to do with Emacs since it does pretty much everything).
So, with the setup in the screenshot and working in the SQL minor mode, once I’ve written my query, I can shoot it over to the SQL Interpreter with at quick ^C-^C, which then runs the query and outputs the results, meanwhile the point stays right where I left it! Logging into the database is easy by firing off $\texttt{M-x sql-postgres}$ and entering your credentials. This mode is even more essential when SSHing in the department CS machines, where I then must connect to a remote DB server, only accessible from another CS department machine. Since this all is done in Emacs, in one screen, I have a complete SQL IDE, which is great.
This may seem a bit archaic to some to still use Emacs when there are other, more intuitive interfaces out there. But even those vim users can appreciate that there is much to be had for learning the same tool and using it everywhere. Admittedly, in this area I need some attention since I’m writing this post now in WordPress’ online editor, and I actually like Eclipse’s intellisense, and I’m tied to a proprietary editor at my day job. But besides WordPress, Eclipse, Mail, roads, public order and sanitation, what has Emacs done for us!?! Brought peace?!
Areas where I think I can realistically incorporate Emacs:
1. First draft of blog posts. 100% integration looks a bit wonky with WordPress and I’m still in that baby-blogger phase where I bounce back and forth from the preview to the edit screen and make tweeks. But the main draft can (and should) be authored from Emacs.
2. Code more in Emacs. For any of the scripting languages this is easy. For C and C++, this isn’t too hard for me either since I never really grew up on Visual Studios, but for Java… Typing $\texttt{this.}$ and then seeing a list of options is very handy. Except when I had this kick where I was over-using reflection, then Eclipse can’t really keep up…
3. Email. I checked out gnus the other day and while I see how an Emacs only mail client can be done, with so much inline html these days, I’m not sure if its worth it. But I’m poking a stick at this bear and I’ll probably come back to it.
4. $\LaTeX$. I love $\LaTeX$. Once I learned WordPress supported this, I was even more addicted to WordPress. My résumé, presentations, papers, letters (when I must actually print one out), are all done in lovable $\LaTeX$. If you find yourself cursing at M\$ Word, take the red pill my friend. And then you’ll start cursing weird compile errors, but worry about that after you take the pill… Many thanks to Brent over at The Math Less Traveled for showing me this!
Well, that’ll keep me busy for a while 🙂
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2020-10-21 23:51:41
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https://aas.org/archives/BAAS/v26n4/aas185/abs/S801.html
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The STARE Project: A Search for Transient Astronomical Radio Emission
Session 8 -- Radio Surveys and Techniques
Display presentation, Monday, 9, 1995, 9:20am - 6:30pm
## [8.01] The STARE Project: A Search for Transient Astronomical Radio Emission
C.A. Katz, J.N. Hewitt, C.B. Moore, J.D. Ellithorpe (MIT)
Many astronomical objects are known to produce transient radio emission, including the Sun, Jupiter, flare stars, and supernovae. Other phenomena are expected to produce transient radio emission as well; gamma-ray bursts are obvious candidates. Although other attempts have been made to detect transient radio emission, to our knowledge no search has had both the sensitivity and sufficient sky coverage to motivate significant progress. We propose a three-phase approach to detect transient radio sources. In the first phase we would monitor the sky for changes in total power. Simple radiometers at geographically separated locations would allow quick and easy measurements, but would provide low sensitivity and little position information. The second phase would involve monitoring the sky with small, geographically separated, correlating arrays. The correlation and the increase in collecting area would give better sensitivity and positions. For phase three, we envision the full instrument as two or more large correlating arrays providing all-sky coverage, $1'$ position resolution, and a flux density detection limit of $\sim 10$ Jy. Such an instrument would allow either the detection of previously unseen sources, or much more stringent sky event rate upper limits than those acheived in the past.
At the time of this writing our work addresses several issues. The design and construction of sensitive, inexpensive radiometers for the first phase of the project is underway; we expect system temperatures of $\sim100$K at 611~MHz in each of two orthogonal polarization channels. A simple data acquisition system using a PC and commercial A/D converter has been devised to provide sampling rates up to 100~kHz. Investigations into the availability of accurate timing information using commercial GPS equipment have produced promising results; inter-site clock synchronization of better than $10 \mbox{${\mu}$s}$ should be easily obtainable. We expect the rejection of local interference to be our most difficult technical challenge; accordingly, the assessment of various sites is in progress.
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2016-07-28 02:57:48
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http://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-8-geometry-8-8-pythagorean-theorem-8-8-exercises-page-594/32
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## Basic College Mathematics (9th Edition)
Hypotenuse of the triangle is $\approx 5.9$ mi.
Let $H =$ hypotenuse of the triangle $a^{2} + b^{2} = c^{2}$ $4.2^{2} + 4.2^{2} = H^{2}$ $17.64 + 17.64 = H^{2}$ $35.28 = H^{2}$ $H = \sqrt 35.28$ $H = 5.939...$ $H \approx 5.9$ mi
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2017-05-29 00:21:39
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https://math.stackexchange.com/questions/1660790/what-is-the-difference-between-free-groups-and-a-free-product
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# What is the difference between free groups and a free product?
I am encountering free products for the first time in Algebraic Topology during the discussion of van Kampen's theorem, and I can't seem to tell the difference between a free product of groups, and a free group. The definition I know (of a free product) is:
Let $G,H$ be groups. Define $G *H$ to be the set of all formal words $g_1h_1 \cdots g_nh_k$ where $g_i \in G$ and $h_j \in H$. Then $G*H$ is a group under the operation of juxtaposition, and the identity is the empty word.
This definition can be extended to the free product of an arbitrary collection $G_\alpha$ of groups, but I don't see how this definition is different from the free group on a set $S$. My guesses at the differences:
1. Free groups can be formed with any set $S$, whereas free products are defined for collections of groups.
2. Free products respects relations between the groups. For example, in $G*H$ the word $g_1g_2h_1$ is just $g_3h_1$ for some $g_3=g_1g_2 \in G$. Whereas there is no underlying relation for words in a free group (except for the formal cancellation of words).
And one last questions: are all free groups also free products? Or is the inclusion in the other direction?
• If you take $\bigstar_{s\in S} \mathbb Z$, you get the free group on $S$. The free product is, in a sense, a generalization. It might be useful to know that the "free product" is actually a categorical "coproduct". – Justin Young Feb 18 '16 at 1:39
• Thank you. That makes a lot of sense. I have almost no background in category theory so I'm not quite sure what that means but thanks all the same. – Kevin Sheng Feb 18 '16 at 1:41
• a free group is a free product of any amount of factors $\Bbb Z$, but in general a free product takes any class of groups to construct a new group, and furthermore, with respect to the Seifert - van Kampen's construction, there is another product of groups: the amalgamated free product . – janmarqz Feb 18 '16 at 1:50
• There is a problem of types, so that your question does not make a lot of sense! A free group is a type of group, while the free product of groups is an operation which given a bunch of groups, gives you a new group. The "difference" you ask for in the title is, then, that the two things are completely different beasts! – Mariano Suárez-Álvarez Feb 18 '16 at 1:57
In a free group on a generating set, say $\{a,b\}$ every element can be expressed uniquely as $g_1^{r_1}g_2^{r_2}\cdots g_n^{r_n}$ where $g_i\in\{a,b\}$ and $g_{i+1}\ne g_i$ and each $r_i\in\mathbb Z$. For example, $a^2b^{-3}a^4b^2$. This is not necessarily true for a free product. Let $G\cong\mathbb Z/2\mathbb Z$ and $H\cong\mathbb Z/3\mathbb Z$. Let $G=\langle a\rangle$ and $H=\langle b\rangle$. Here, $a^4b^{-5}=b$.
In general, a free group on $S$ is a free product of the infinite cyclic groups generated by each member of $S$. A free product of groups is not necessarily a free group.
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2019-07-21 21:18:41
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https://geoenergymath.com/category/uncategorized/
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# Mathematical Geoenergy
Our book Mathematical Geoenergy presents a number of novel approaches that each deserve a research paper on their own. Here is the list, ordered roughly by importance (IMHO):
1. Laplace’s Tidal Equation Analytic Solution.
(Ch 1112) A solution of a Navier-Stokes variant along the equator. Laplace’s Tidal Equations are a simplified version of Navier-Stokes and the equatorial topology allows an exact closed-form analytic solution. This could classify for the Clay Institute Millenium Prize if the practical implications are considered, but it’s a lower-dimensional solution than a complete 3-D Navier-Stokes formulation requires.
2. Model of El Nino/Southern Oscillation (ENSO).
(Ch 12) A tidally forced model of the equatorial Pacific’s thermocline sloshing (the ENSO dipole) which assumes a strong annual interaction. Not surprisingly this uses the Laplace’s Tidal Equation solution described above, otherwise the tidal pattern connection would have been discovered long ago.
3. Model of Quasi-Biennial Oscillation (QBO).
(Ch 11) A model of the equatorial stratospheric winds which cycle by reversing direction ~28 months. This incorporates the idea of amplified cycling of the sun and moon nodal declination pattern on the atmosphere’s tidal response.
4. Origin of the Chandler Wobble.
(Ch 13) An explanation for the ~433 day cycle of the Earth’s Chandler wobble. Finding this is a fairly obvious consequence of modeling the QBO.
5. The Oil Shock Model.
(Ch 5) A data flow model of oil extraction and production which allows for perturbations. We are seeing this in action with the recession caused by oil supply perturbations due to the Corona Virus pandemic.
6. The Dispersive Discovery Model.
(Ch 4) A probabilistic model of resource discovery which accounts for technological advancement and a finite search volume.
7. Ornstein-Uhlenbeck Diffusion Model
(Ch 6) Applying Ornstein-Uhlenbeck diffusion to describe the decline and asymptotic limiting flow from volumes such as occur in fracked shale oil reservoirs.
8. The Reservoir Size Dispersive Aggregation Model.
(Ch 4) A first-principles model that explains and describes the size distribution of oil reservoirs and fields around the world.
9. Origin of Tropical Instability Waves (TIW).
(Ch 12) As the ENSO model was developed, a higher harmonic component was found which matches TIW
10. Characterization of Battery Charging and Discharging.
(Ch 18) Simplified expressions for modeling Li-ion battery charging and discharging profiles by applying dispersion on the diffusion equation, which reflects the disorder within the ion matrix.
11. Anomalous Behavior in Dispersive Transport explained.
(Ch 18) Photovoltaic (PV) material made from disordered and amorphous semiconductor material shows poor photoresponse characteristics. Solution to simple entropic dispersion relations or the more general Fokker-Planck leads to good agreement with the data over orders of magnitude in current and response times.
12. Framework for understanding Breakthrough Curves and Solute Transport in Porous Materials.
(Ch 20) The same disordered Fokker-Planck construction explains the dispersive transport of solute in groundwater or liquids flowing in porous materials.
13. Wind Energy Analysis.
(Ch 11) Universality of wind energy probability distribution by applying maximum entropy to the mean energy observed. Data from Canada and Germany. Found a universal BesselK distribution which improves on the conventional Rayleigh distribution.
14. Terrain Slope Distribution Analysis.
(Ch 16) Explanation and derivation of the topographic slope distribution across the USA. This uses mean energy and maximum entropy principle.
15. Thermal Entropic Dispersion Analysis.
(Ch 14) Solving the Fokker-Planck equation or Fourier’s Law for thermal diffusion in a disordered environment. A subtle effect but the result is a simplified expression not involving complex errf transcendental functions. Useful in ocean heat content (OHC) studies.
16. The Maximum Entropy Principle and the Entropic Dispersion Framework.
(Ch 10) The generalized math framework applied to many models of disorder, natural or man-made. Explains the origin of the entroplet.
17. Solving the Reserve Growth “enigma”.
(Ch 6) An application of dispersive discovery on a localized level which models the hyperbolic reserve growth characteristics observed.
18. Shocklets.
(Ch 7) A kernel approach to characterizing production from individual oil fields.
19. Reserve Growth, Creaming Curve, and Size Distribution Linearization.
(Ch 6) An obvious linearization of this family of curves, related to Hubbert Linearization but more useful since it stems from first principles.
20. The Hubbert Peak Logistic Curve explained.
(Ch 7) The Logistic curve is trivially explained by dispersive discovery with exponential technology advancement.
21. Laplace Transform Analysis of Dispersive Discovery.
(Ch 7) Dispersion curves are solved by looking up the Laplace transform of the spatial uncertainty profile.
22. Gompertz Decline Model.
(Ch 7) Exponentially increasing extraction rates lead to steep production decline.
23. The Dynamics of Atmospheric CO2 buildup and Extrapolation.
(Ch 9) Convolving a fat-tailed CO2 residence time impulse response function with a fossil-fuel emissions stimulus. This shows the long latency of CO2 buildup very straightforwardly.
24. Reliability Analysis and Understanding the “Bathtub Curve”.
(Ch 19) Using a dispersion in failure rates to generate the characteristic bathtub curves of failure occurrences in parts and components.
25. The Overshoot Point (TOP) and the Oil Production Plateau.
(Ch 8) How increases in extraction rate can maintain production levels.
26. Lake Size Distribution.
(Ch 15) Analogous to explaining reservoir size distribution, uses similar arguments to derive the distribution of freshwater lake sizes. This provides a good feel for how often super-giant reservoirs and Great Lakes occur (by comparison).
27. The Quandary of Infinite Reserves due to Fat-Tail Statistics.
(Ch 9) Demonstrated that even infinite reserves can lead to limited resource production in the face of maximum extraction constraints.
28. Oil Recovery Factor Model.
(Ch 6) A model of oil recovery which takes into account reservoir size.
29. Network Transit Time Statistics.
(Ch 21) Dispersion in TCP/IP transport rates leads to the measured fat-tails in round-trip time statistics on loaded networks.
30. Particle and Crystal Growth Statistics.
(Ch 20) Detailed model of ice crystal size distribution in high-altitude cirrus clouds.
31. Rainfall Amount Dispersion.
(Ch 15) Explanation of rainfall variation based on dispersion in rate of cloud build-up along with dispersion in critical size.
32. Earthquake Magnitude Distribution.
(Ch 13) Distribution of earthquake magnitudes based on dispersion of energy buildup and critical threshold.
33. IceBox Earth Setpoint Calculation.
(Ch 17) Simple model for determining the earth’s setpoint temperature extremes — current and low-CO2 icebox earth.
34. Global Temperature Multiple Linear Regression Model
(Ch 17) The global surface temperature records show variability that is largely due to the GHG rise along with fluctuating changes due to ocean dipoles such as ENSO (via the SOI measure and also AAM) and sporadic volcanic eruptions impacting the atmospheric aerosol concentrations.
35. GPS Acquisition Time Analysis.
(Ch 21) Engineering analysis of GPS cold-start acquisition times. Using Maximum Entropy in EMI clutter statistics.
36. 1/f Noise Model
(Ch 21) Deriving a random noise spectrum from maximum entropy statistics.
37. Stochastic Aquatic Waves
(Ch 12) Maximum Entropy Analysis of wave height distribution of surface gravity waves.
38. The Stochastic Model of Popcorn Popping.
(Appx C) The novel explanation of why popcorn popping follows the same bell-shaped curve of the Hubbert Peak in oil production. Can use this to model epidemics, etc.
39. Dispersion Analysis of Human Transportation Statistics.
(Appx C) Alternate take on the empirical distribution of travel times between geographical points. This uses a maximum entropy approximation to the mean speed and mean distance across all the data points.
# Rating of Climate Change blogs
Scientific blogging is on a decline and that is especially evident with respect to climate change blogs. Nothing really good left apart from forums such as https://forum.azimuthproject.org/discussions (which allows equation markup, image posting, and freedom to create threaded discussions).
Here is a grading of blogs that I have on my RSS feed:
• WUWT : F-
A horrible AGW denier blog that pretends to be fair & balanced. RSS feed does not work with Owl.
• Tallbloke’s Talkshop : F-
A horrible AGW denier blog that specializes in numerology
• Real Climate : C
Sparse postings and comment moderation has long latencies so the discussion is glacially paced
• Open Mind : C
Not very interesting, mostly from a statistical angle, which is not where progress in analysis occurs
• Science of Doom : D
I don’t understand this site, seems to be run by a thinly veiled skeptic. Might as well read books by Pierrehumbert to gain an understanding of the physics instead of struggling along with the topics.
• And Then There’s Physics : D+
The moderators are control freaks, and the discussions are safe as milk
• Peak Oil Barrel : A
A very good blog that allows both fossil fuel discussion and climate change discussion, separated in distinct threads. Moderated slightly and images allowed, along with short-term editing.
• The Blackboard : F-
An awful blog run by a mechanical engineer which at one point had climate science discussion but now consists of pro-Trump cheerleading.
• Clive Best : D+
The moderator tries hard but then stumbles as he desperately tries to debunk the AGW consensus. Marginally better than Science of Doom because at least the scientific ideas are creative.
• Climate Audit : F-
• Climate Etc : F-
Pointless blog stressing climate science uncertainty run by a now-retired climate science professor J. Curry with a comment section that seems infested with Australian AGW deniers.
• Moyhu : B
Halfway-decent posts by a retired fluid dynamics researcher but an ugly and unstable comment-entry system.
• Hotwhopper : C+
Well-thought out counter-attacks to nonsense at sites such as WUWT, but nothing really about discussions of science
• Robert Scribbler : C
The American version of HotWhopper, with probably too much doom & gloom.
• Skeptical Science: D
Nothing interesting here as they never seem to veer from the consensus. The comments seem to be overly moderated and at one time the RSS feed was broken, but that has recently been fixed.
• Roy Spencer, PhD : F-
Horrible blog by a religious zealot with comments infested by AGW deniers
• Rabbet Run : B-
Below Moyhu because nothing really innovative but occasional insight
• More Grumbine Science : B
By a NASA guy, posts very rarely
This is a previous summary I had written two years ago (I had forgotten I had saved it in a draft folder, and so you can see how little has changed)
Real Climate C Too long turnaround for comments
And Then There’s Physics D Too much ClimateBall
Skeptical Science D Too insular, won’t discuss cutting edge
Science of Doom D- Inflitrated by deniers
Open Mind C Too much on statistics, which does a disservice to unlocking deterministic aspects such as ENSO
Moyhu B Worst comment entry, but the research is quality
This Week in Science: DailyKos C+ Nothing in depth
Watt’s Up With That F Garbage
Tallbloke’s Talkshop F Loony bin
The Blackboard F Nasty people
Climate Etc F Clueless (mainly Aussies) lead by a clueless
Roy Spencer F Zero real science
Rabbet Run C Too much inside posturing
Hot Whopper C Good if you want to see deniers get debunked
Robert Scribbler C Verges on hysterical but who knows
Azimuth Project A A true forum. Allows everyone to create markup and add charts
# Commenting at PubPeer
For our Mathematical GeoEnergy book, there is an entry at PubPeer.com for comments (one can also comment at Amazon.com, but you need to be a verified purchaser of the book to be able to comment there)
PubPeer provides a good way to debunk poorly researched work as shown in the recent comments pertaining to the Zharkova paper published in Nature’s Scientific Reports journal.
An issue with the comment policy at Amazon is that one can easily evaluate the contents of a book via the “Look Inside” feature or through the Table of Contents. Often there is enough evidence to provide a critical book review just through this feature — in a sense, a statistical sampling of the contents — yet Amazon requires a full purchase before a review is possible. Even if one can check the book out at a university library this is not allowable. Therefore it favors profiting by the potential fraudster because they will get royalties in spite of damaging reviews by critics that are willing to sink money into a purchase.
In the good old days at Amazon, one could actually warn people about pseudo-scientific research. This is exemplified by Curry’s Bose-Einstein statistics debacle, where unfortunately political cronies and acolytes of Curry’s have since purchased her book and have used the comments to do damage control. No further negative comments are possible since smart people have not bought her book and therefore can no longer comment.
PubPeer does away with this Catch-22 situation.
# Mathematical GeoEnergy
https://www.bookwire.com/book/AUS/Mathematical-Geoenergy-9781119434290-Pukite-59661376
This blog will be ramped up for the book, but ContextEarth.com contains all the research leading up to the book.
Two papers at AGU 2017:
Dynamic Context Server
# Tropics, poles and reefs
2014, 2015 and 2016 played a recurring theme of El Nino. A tentative El Nino in late 2014 and early 2015 segued with a stutter into a strong El Nino in 2015/2016 dragging global temperatures in train. Temperatures in the tropical Pacific dropped a bit after that and may or may not have slipped into La Nina depending on which agency you listen to, but now, it looks like El Nino might be coming back: surface water temperatures in the eastern Pacific, off the coast of South America, have risen to four or more degrees above average although they’ve not spread further west and a number of seasonal forecasting centres are suggesting that temperatures might continue to rise. No one’s called it an El Nino, yet, but the effects of the elevated sea-surface temperatures are sadly plain to see. Heavy rain in Peru has already led to flooding and all the…
View original post 261 more words
# Context/Earth
Please refer to my new WordPress blog Context/Earth for future posts.
WordPress has better commenting options such as provisions for pictures and charts.
The scope of the blog is also more comprehensive, as it will include all environmental and energy topics tied together in a semantic web framework.
Onward and upward as they say.
# Expansion of atmosphere and ocean
This is a short tutorial together with some observational evidence explaining how the atmosphere and ocean is expanding measurably in the face of global warming.
### Ocean thermal expansion
The ocean absorbs heat per area according to its heat capacity
$$\Delta E = C_p \cdot \Delta T \cdot {Depth}$$
The linear coefficient of thermal expansion is assumed constant over a temperature range. Multiplying this over a depth:
$$\Delta Z = \epsilon \cdot \Delta T \cdot Depth$$
But now we can substitute the total energy gained from the first equation:
$$\Delta Z = \epsilon \frac{\Delta E}{C_p}$$
Assume that the linear coefficient of thermal expansion is 0.000207 per °C, and specific heat capacity is 4,000,000 J/m^3/°C.
If an excess forcing of 0.6 W/m^2 occurs over one year (see the OHC model), then the increase in the level of the ocean is
0.000207 * 0.6*(365*24*60*60) / 4,000,000 = 0.98 mm
This is called the steric sea-level rise, and it is just one component of the sea-level rise over time (the others have to do with melting ice). From the figure below, one can see that the thermal rise is close to 1mm/year over the last decade.
The red line shows the thermal expansion from the ocean heating. Thermal rise is 1 mm/year over the 3 mm/year total sea-level rise.
### Atmosphere thermal expansion
The trick here is to infer the atmosphere expansion by looking at an equal pressure point as a function of altitude (a geopotential height isobar) and determine how much that point has increased over time. I was able to find one piece of data from The Weather Channel’s senior meteorologist Stu Ostro.
The geopotential height anomaly is shown below for the 500 mb (1/2 atmosphere)
Geopotential height anomaly @ 500 mb plotted alongside global temperature anomaly.
In absolute terms it is charted as follows:
Global average geopotential height @ 500 mb plotted alongside global temperature.http://icons-ak.wxug.com/graphics/earthweek/geopotential-height-and-air-temperature.png
To understand how the altitude has changed, consider the classical barometric formula:
$$P(H) = P_0 e^{-mgH/RT}$$
We take the point at which we reached 1/2 the STP of 1 atmosphere at sea-level:
$$P(H)/P_0 = 0.5 = e^{-mgH/RT}$$
or
$$H = RT/mg \cdot ln(2)$$
Assuming the average molecular weight of the atmospheric gas constituents does not change, the change in altitude (H) should be related to the change in temperature (T) by:
$$\Delta H = R \Delta T / mg \cdot ln(2)$$
or
$$\frac{ \Delta H}{\Delta T} = \frac{R}{mg} ln(2)$$
For R = 8314, m=29, and g=9.8, the slope should be 29.25 *ln(2) = 20.3 m/°C.
From the linear regression agreement between the two, we get a value of 25.7 m/°C.
Linear regression between the geopotential height change and temperature change
Why is this geopotential height change about 26% higher than it should be from the theoretical value considering that the height should track the temperature according to the barometric formula?
If we use the polytropic approximation (equation 1 in the barometric formula), the altitudinal difference between the low temperature and high temperature 500 mb pressure values remains the same as when we use the classic exponential damped barometric formula:
If we apply the polytropic barometric formula instead of the exponential, we still show a real height change that is higher than theoretically predicted by ~25% at the 500 mb isobar.
This discrepancy could be due to measurement error, as the readings are taken by weather balloons and the accuracy could have drifted over the years.
It is also possible that the composition of the atmosphere could have changed slightly at altitude. What happens if the moisture increased slightly? This shouldn’t make much difference.
Most likely is the possibility that the baseline sea-level pressure has changed, which shifted the 500 mbar point artificially. See http://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch10s10-3-2-4.html.
“10.3.2.4 Sea Level Pressure and Atmospheric Circulation
As a basic component of the mean atmospheric circulations and weather patterns, projections of the mean sea level pressure for the medium scenario A1B are considered. Seasonal mean changes for DJF and JJA are shown in Figure 10.9 (matching results in Wang and Swail, 2006b). Sea level pressure differences show decreases at high latitudes in both seasons in both hemispheres. The compensating increases are predominantly over the mid-latitude and subtropical ocean regions, extending across South America, Australia and southern Asia in JJA, and the Mediterranean in DJF. Many of these increases are consistent across the models. This pattern of change, discussed further in Section 10.3.5.3, has been linked to an expansion of the Hadley Circulation and a poleward shift of the mid-latitude storm tracks (Yin, 2005). This helps explain, in part, the increases in precipitation at high latitudes and decreases in the subtropics and parts of the mid-latitudes. Further analysis of the regional details of these changes is given in Chapter 11. The pattern of pressure change implies increased westerly flows across the western parts of the continents. These contribute to increases in mean precipitation (Figure 10.9) and increased precipitation intensity (Meehl et al., 2005a). “
I will have to figure out the mean sea-level pressure change over this time period to verify this hypothesis.
The following recent research article looks into the shifts in geopotential height over shorter time durations:
[1]
Y. Y. Hafez and M. Almazroui, “The Role Played by Blocking Systems over Europe in Abnormal Weather over Kingdom of Saudi Arabia in Summer 2010,” Advances in Meteorology, vol. 2013, p. 20, 2013.
One possibility for the larger-than-expected altitude change is that the average lapse rate has changed slightly. We can use the lapse rate variation of the barometric formula, and perturb the lapse rate, L, around its average value:
$$P(H) = P(0) (1- \frac{LH}{T_0})^{\frac{gm}{LR}}$$
Granted, the error bars on this calculation are significant but we can see how subtle the effect is.
Sea-level pressure, P(0) = 1013.25 mb
Gas constant, R = 8.31446 J/K/mol
Earth’s gravity, g = 9.807 m/s^2
Avg molecular weight, m = 0.02896 kg/m
In 1960, the temperature was about 14.65°C and 500 mb altitude was 5647 m.
In 2010, the temperature was about 15.4°C and 500 mb altitude was 5667 m.
All we need to do is invert the P(H) formula for each pair of values, modifying L slightly.
If we select a lapse rate, L, for 1960 of 0.005C/m, we calculate H = 5647.9m for P(H)=500 mb.
If we select a lapse rate, L, for 2010 of 0.0050°C/m, we calculate H = 5668.4m for P(H)=500 mb.
The difference in the two altitudes for a change in L of -0.0001°C/m is 20.5m, about what the geopotential height chart shows.
If we leave the L at 0.005C/m for both 1960 and 2010, the difference of the 500mb altitudes is only 14.7m.
To the extent that we can trust the numbers on the charts from Ostro, the change in geopotential height is suggesting a feedback effect in the lapse rate due to global warming. The lapse rate is decreasing over time, which implies that the heat capacity of the atmosphere is increasing (likely due to higher specific humidity), thereby buffering changes in temperature with altitude.
This means that a given temperature increase at a particular altitude (where the CO2 IR window can achieve a radiative balance) will be reflected as a scale-modified temperature at sea level
In 1960, the temperature difference at 500mb altitude is 0.005C/m * 5647.9m = 28.8C
In 2010, the temperature difference at 500mb altitude is 0.0050°C/m * 5668.4m = 28.34°C
The difference at sea-level from the chart is 15.4°C-14.65°C = 0.75°C whereas the difference at the 500mb altitude assuming the modified lapse rate is 0.75°C – 0.46°C = 0.29°C. If the lapse rate didn’t change then this sea-level difference would maintain at a constant atmospheric pressure isobar in altitude.
For implications in the interpretation, see page 24 of National Research Council Panel on Climate Change Feedbacks (2003). Understanding climate change feedbacks : National Academies Press. ISBN 978-0-309-09072-8. They caution that the measurements require some precision, otherwise the errors can multiply due to the differences between two large numbers.
# Characterization of Battery Charging and Discharging
I had the good fortune of taking a week long Society of Automotive Engineers (SAE) Academy class on hybrid/electric vehicles. The take-home message behind HEV and EV technology is to remember that a quality battery plus optimizing the management of battery cycling remain the keys to success. That is not surprising — we all know that gasoline has long been “king”, and since current battery technology has nowhere near the energy density of gasoline, the battery has turned into a “diva”. In other words, it will perform as long as it is in charge (so to speak) and the battery is well maintained.
I can report that much class time was devoted to the electrochemistry of Lithium-ion batteries. Lithium is an ideal elemental material due to its position in the upper left-hand corner of the periodic table — in other words a very lightweight material with a potentially high energy density.
What was surprising to me was the sparseness of detailed characterization of the material properties. One instructor stated that the lack of measured diffusion properties for battery cell specifications was a pet peeve of his. Having these properties at hand allows a design engineer to better model the charging and discharging characteristics of the battery, and thus to perhaps to develop better battery management schemes. Coming from the semiconductor world, it is almost unheard of to do design without adequate device characteristics such as mobility and diffusivity.
From my perspective, this is not necessarily bad. Any time I see an anomalous behavior or missing piece, it opens the possibility I can fill a modeling niche.
## Introduction
Modern rechargeable battery technology still relies on the principles of electro-chemistry and a reversible process, which hasn’t changed in fundamental terms since the first lead-acid battery came to market in the early 1900’s. What has changed is the combination of materials that make a low-cost, lightweight, and energy-efficient battery which will serve the needs of demanding applications such as electric and hybrid-electric vehicles (EV/HEV).
As energy efficient operation is dependent on the properties of the materials being combined, it is well understood that characterizing the materials is important to advancing the state-of-the-art (and in increasing EV acceptance).
Of vital importance is the characterization of diffusion in the electrode materials, as that is the rate-limiting factor in determining the absolute charging and discharging speed of the material-specific battery technology. Unfortunately, because of the competitive nature of battery producers, many of the characteristics are well-guarded and treated as trade secrets. For example, it is very rare to find diffusion coefficient characteristics on commercial battery specification sheets, even though this kind of information is vital for optimizing battery management schemes [7].
In comparison to the relatively simple diffusional mechanisms of silicon oxide growth, the engineered structure of well-designed battery cell presents a significant constraint to the diffusional behavior. In Figure 1 below we show a schematic of a single lithium-ion cell and the storage particles that charge and discharge. The disordered nature of the storage particles in Figure 2 is often described by what is referred to as a tortuosity measure.
Figure 1: Exaggerated three-dimensional view of a lithium-ion battery cell and the direction of current flow during charging and discharging
Figure 2 : Realistic view of the heavily disordered nature of the LiFePO4 storage particles [1].
The constraints on the diffusion is that it is limited in scale to that of the radius of the storage particle. The length scale is limited essentially to the values L to Lmax shown in Figure 3 below.
Figure 3 : Diffusion of ions takes place through the radial shell of the LiFePO4 spherical particle [1]. During the discharge phase, the ions need to migrate outward through shell and through the SEI barrier before reaching the electrolyte. At this point they can contribute to current flow.
The size of the particles also varies as shown in Figure 4 below. The two Lithium-ion materials under consideration, LiFePO4 and LiFeSO4F, have different materials properties but are structurally very similar (matrixed particles of mixed size) so that we can use a common analysis approach. This essentially allows us to apply uncertainty in the diffusion coefficient and uncertainties in the particle size to establish a common diffusional behavior formulation.
Figure 4 : Particle size distribution of FeSO4F spherical granules [2]. The variation in lengths and material diffusivities opens the possibility of applying uncertainty quantification to a model of diffusive growth.
## Dispersive Diffusion Analysis of Discharging
The diffusion of ions through the volume of a spherical particle does have similarity to classical regimes such as the diffusion of silicon through silicon dioxide. That process in fact leads to the familiar Fick’s law of diffusion, whereby the growing layer of oxide follows a parabolic growth law (in fact this is a square root with time, but was named parabolic by the semiconductor technology industry for historical reasons, see for example here).
The model that we can use for Li+ diffusion derives from the classic solution to the Fokker-Planck equation of continuity (neglecting any field driven drift).
$$\frac{\partial C}{\partial t} – D \nabla^2 C=0$$ where C is a concentration and D is the diffusion coefficient. Ignoring the spherical orientation, we can just assume a solution along a one dimensional radially outward axis, x:
$$\large C(t,x|D) = \frac{1}{\sqrt{4 \pi D t}} e^{-x^2/{4 D t}}$$
This is a marginal probability which depends on the diffusion coefficient. Since we do not know the variance of the diffusivity, we can apply a maximum entropy distribution across D.
$$\large p_d(D) = \frac{1}{D_0} e^{-D/D_0}$$
This simplifies the representation to the following workable formulation.
$$\large C(t,x) = \frac{1}{2 \sqrt{D_0 t}} e^{-x/{\sqrt{D_0 t}}}$$
We now have what is called a kernel solution (i.e. Green’s function) that we can apply to specific sets of initial conditions and forcing functions, the latter solved via convolution.
Fully Charged Initial Conditions
Assume the spherical particle is uniformly distributed with a charge density C(0, x) at time t=0.
Discharging Model
For every point along the dimensions of the particle of size L, we calculate the time it takes to diffuse to the outer edge, where it can enter the electrolytic medium. This is simply an integral of the C(t, x) term for all points starting from x’ = d to L, where d is the inner core radius.
$$\large C(t) = \int_{d}^L C(t,L-x) dx$$
this integrates straightforwardly to this concise representation:
$$\large C(t) = C_0 \frac{ 1 – e^{-(L-d)/{\sqrt{D_0 t}}} } {L – d}$$
The voltage of the cell is essentially the amount of charge available, so as this charge depletes, the voltage decreases in proportion.
We can test the model on two data sets corresponding to a LiPO4 cell [1] and a LiSO4F cell [2]. Figure 5 below shows the model fit for LiPO4 as the red dotted line, and which should be level-compared to the solid black line labelled 1. The other curves labelled 2,3,4,5 are alternative diffusional model approximations applied by Churikov et al that clearly do not work as well as the dispersive diffusion formulation derived above.
Figure 5 : Discharge profile of LiFePO4 battery cell [1], with the red dotted line showing the parameterized dispersive diffusion model. The curves labelled 1 through 5 show alternative models that the authors applied to fit the data. Only the dispersive diffusion model duplicates the fast drop-off and long-time scale decline.
Figure 6 below shows the fit to voltage characteristics of a LiSO4F cell, drawn as a red dotted line above the light gray data points. In this case the diffusional model by Delacourt shown in solid black is well outside acceptable agreement.
Figure 6 : Discharge profile of LiFeSO4F battery cell [2], with the red dotted line showing the parameterized dispersive diffusion model. The black curve shows the model that the authors applied to fit the data.
The question is why does this simple formulation work so well? As with many similar cases of characterizing disordered material, the fundamentally derived solution needs to be adjusted to take into account the uncertainty in the parameter space. However, this step is not routinely performed and by adding modeling details (see [4]) to try to make up for a poor fit works only as a cosmetic heuristic. In contrast, by performing the uncertainty quantification, like we did with the diffusion coefficient, the first-order solution works surprisingly well with no need for additional detail.
Constant Current Discharge
Instead of assuming that the particle size is L, we can say that the L is an average and apply the same maximum entropy spread in values.
$$\large C(t) = \int_{0}^{\infty} C(t,x) \frac{1}{L} e^{-x/L} dx$$
this integrates straightforwardly to this concise representation:
$$\large C(t) = C_0 \frac{1}{ L + {\sqrt{D_0 t}} }$$
The reason we do this is to allow us to recursively define the change in charge to a current. In this case, to get current we need to differentiate the charge with respect to time.
$$\large I(t) = \frac{dC(t)}{dt}$$
This differentiates to the following expression
$$\large I(t) = – \frac{C_0}{ (L + {\sqrt{D_0 t}})^2} \frac{1}{2 \sqrt{t}}$$
But note that we can insert C(t) back in to the expression
$$\large I(t) = \frac{C(t)}{(L + {\sqrt{D_0 t}}) 2 \sqrt{t}}$$
Finally, since I(t) is a constant and we can set that to a value of I_constant. Then the charge has the following profile
$$C(t) = C(0) – k_c I_{constant} (L + \sqrt{Dt}) 2 \sqrt{t}$$
or as a voltage decline
$$V(t) = V(0) – k_v I_{constant} (L + \sqrt{Dt}) 2 \sqrt{t}$$
For a set of constant current values, we can compare this formulation against experimental data for LiFePO4 (shown as gray open circles) shown in Figure 7 below. A slight constant current offset (which may arise from unspecified shunting and/or series elements) was required to allow for the curves to align proportionally. Even with that, it is clear that the dispersive diffusion formulation works better than the conventional model (solid black lines) except where the discharge is nearing completion.
Figure 7 : Constant current discharge profile [6]. Superimposed as dotted lines are the set of model fits which use the current value as a fixed parameter.
We can also model battery charging but the lack of information on the charging profile makes the discharge behavior a simpler study.
Related Diffusion Topics
References
[1]
A. Churikov, A. Ivanishchev, I. Ivanishcheva, V. Sycheva, N. Khasanova, and E. Antipov, “Determination of lithium diffusion coefficient in LiFePO 4 electrode by galvanostatic and potentiostatic intermittent titration techniques,” Electrochimica Acta, vol. 55, no. 8, pp. 2939–2950, 2010.
[2]
C. Delacourt, M. Ati, and J. Tarascon, “Measurement of Lithium Diffusion Coefficient in Li y FeSO4F,” Journal of The Electrochemical Society, vol. 158, no. 6, pp. A741–A749, 2011.
[3]
M. Park, X. Zhang, M. Chung, G. B. Less, and A. M. Sastry, “A review of conduction phenomena in Li-ion batteries,” Journal of Power Sources, vol. 195, no. 24, pp. 7904–7929, Dec. 2010.
[4]
J. Christensen and J. Newman, “A mathematical model for the lithium-ion negative electrode solid electrolyte interphase,” Journal of The Electrochemical Society, vol. 151, no. 11, pp. A1977–A1988, 2004.
[5]
Q. Wang, H. Li, X. Huang, and L. Chen, “Determination of chemical diffusion coefficient of lithium ion in graphitized mesocarbon microbeads with potential relaxation technique,” Journal of The Electrochemical Society, vol. 148, no. 7, pp. A737–A741, 2001.
[6]
P. M. Gomadam, J. W. Weidner, R. A. Dougal, and R. E. White, “Mathematical modeling of lithium-ion and nickel battery systems,” Journal of Power Sources, vol. 110, no. 2, pp. 267–284, 2002.
[7]
# The homework problem to end all homework problems
This is a problem that has driven anyone that has studied climate science up the wall.
Premise: Venus has an adiabatic index γ (gamma) and a temperature lapse rate λ (lambda). Earth also has an adiabatic index and temperature lapse rate. These have been measured, and for the Earth a standard atmospheric profile has been established. The general relationship is based on thermodynamic principles but the shape of the profile diverges from simple applications of adiabatic principles. In other words, a heuristic is applied to allow it to match the empirical observations, both for Venus and Earth. See this link for more background
Assigned Problem: Derive the adiabatic index and lapse rate for both planets, Venus and Earth, using only the planetary gravitational constant, the molar composition of atmospheric constituents, and any laws of physics that you can apply. The answer has to be right on the mark with respect to the empirically-established standards.
Caveat: Reminder that this is a tough nut to crack.
Solution: The approach to use is concise but somewhat twisty. We work along two paths, the initial path uses basic physics and equations of continuity; while the subsequent path ties the loose ends together using thermodynamic relationships which result in the familiar barometric formula and lapse rate formula. The initial assumption that we make is to start with a sphere that forms a continuum from the origin; this forms the basis of a polytrope, a useful abstraction to infer the generic properties of planetary objects.
An abstracted planetary atmosphere
The atmosphere has a density ρ, that decreases outward from the origin. The basic laws we work with are the following:
#### Mass Conservation
$$\frac{dm(r)}{dr} = 4 \pi r^2 \rho$$
#### Hydrostatic Equilibrium
$$\frac{dP(r)}{dr} = – \rho g = – \frac{Gm(r)}{r^2}\rho$$
To convert to purely thermodynamic terms, we first integrate the hydrostatic equilibrium relationship over the volume of the sphere
$$\int_0^R \frac{dP(r)}{dr} 4 \pi r^3 dr = 4 \pi R^3 P(R) – \int_0^R 12 P(r) \pi r^2 dr$$
on the right side we have integrated by parts, and eliminate the first term as P(R) goes to zero (note: upon review, the zeroing of P(R) is an approximation if we do not let R extend to the deep pressure vacuum of space, as we recover the differential form later — right now we just assume P(R) decreases much faster than R^3 increases ). We then reduce the second term using the mass conservation relationship, while recovering the gravitational part:
$$– 3 \int_0^M\frac{P}{\rho}dm = -\int_0^R 4 \pi r^3 \frac{G m(r)}{r^2} dr$$
again we apply the mass conservation
$$– 3 \int_0^M\frac{P}{\rho}dm = -\int_0^M \frac{G m(r)}{r} dm$$
The right hand side is simply the total gravitational potential energy Ω while the left side reduces to a pressure to volume relationship:
$$– 3 \int_0^V P dV = \Omega$$
This becomes a variation of the Virial Theorem relating internal energy to potential energy.
Now we bring in the thermodynamic relationships, starting with the ideal gas law with its three independent variables.
#### Ideal Gas Law
$$PV = nRT$$
#### Gibbs Free Energy
$$E = U – TS + PV$$
#### Specific Heat (in terms of molecular degrees of freedom)
$$c_p = c_v + R = (N/2 + 1) R$$
On this path, we make the assertion that the Gibbs free energy will be minimized with respect to perturbations. i.e. a variational approach.
$$dE = 0 = dU – d(TS) + d(PV) = dU – TdS – SdT + PdV + VdP$$
Noting that the system is closed with respect to entropy changes (an adiabatic or isentropic process) and substituting the ideal gas law featuring a molar gas constant for the last term.
$$0 = dU – SdT + PdV + VdP = dU – SdT + PdV + R_n dT$$
At constant pressure (dP=0) the temperature terms reduce to the specific heat at constant pressure:
$$– S dT + nR dT = (c_v +R_n) dT = c_p dT$$
Rewriting the equation
$$0 = dU + c_p dT + P dV$$
Now we can recover the differential virial relationship derived earlier:
$$– 3 P dV = d \Omega$$
and replace the unknown PdV term
$$0 = dU + c_p dT – d \Omega / 3$$
but dU is the same potential energy term as dΩ, so
$$0 = 2/3 d \Omega+ c_p dT$$
Linearizing the potential gravitational energy change with respect to radius
$$0 = \frac{2 m g}{3} dr + c_p dT$$
Rearranging this term we have derived the lapse formula
$$\frac{dT}{dr} = – \frac{mg}{3/2 c_p}$$
Reducing this in terms of the ideal gas constant and molecular degrees of freedom N
$$\frac{dT}{dr} = – \frac{mg}{3/2 (N/2+1) R_n}$$
We still need to derive the adiabatic index, by coupling the lapse rate formula back to the hydrostatic equilibrium formulation.
Recall that the perfect adiabatic relationship (the Poisson’s equation result describing the potential temperature) does not adequately describe a standard atmosphere — being 50% off in lapse rate — and so we must use a more general polytropic process approach.
Combining the Mass Conservation with the Hydrostatic Equilibrium:
$$\frac{1}{r^2} \frac{d}{dr} (\frac{r^2}{\rho} \frac{dP}{dr}) = -4 \pi G \rho$$
if we make the substitution
$$\rho = \rho_c \theta^n$$
where n is the polytropic index. In terms of pressure via the ideal gas law
$$P = P_c \theta^{n+1}$$
if we scale r as the dimensionless ξ :
$$\frac{1}{\xi^2} \frac{d}{d\xi} (\frac{\xi^2}{\rho} \frac{dP}{d\xi}) = – \theta^n$$
This formulation is known as the Lane-Emden equation and is notable for resolving to a polytropic term. A solution for n=5 is
$$\theta = ({1 + \xi^2/3})^{-1/2}$$
We now have a link to the polytropic process equation
$$P V^\gamma = {constant}$$
and
$$P^{1-\gamma} T^{\gamma} = {constant}$$
or
$$P = P_0 (\frac{T}{T_0})^{\frac{\gamma}{1-\gamma}}$$
Tieing together the loose ends, we take our lapse rate gradient
$$\frac{dT}{dr} = \frac{mg}{3/2 (N/2+1) R}$$
and convert that into an altitude profile, where r = z
$$T = T_0 (1 – \frac{z}{f z_0})$$
where
$$z_0 = \frac{R T_0}{m g}$$
and
$$f = 3/2 (1 + N/2)$$
and the temperature gradient, aka lapse rate
$$\lambda= \frac{m g}{ 3/2 (1 + N/2) R }$$
To generate a polytropic process equation from this, we merely have to raise the lapse rate to a power, so that we recreate the power law version of the barometric formula:
$$P = P_0 (1 – \frac{z}{f z_0})^f$$
which essentially reduces to Poisson’s equation on substitution:
$$P = P_0 (T/T_0)^f$$
where the equivalent adiabatic exponent is
$$f = \frac{\gamma}{1-\gamma}$$
Now we have both the lapse rate, barometric formula, and Poisson’s equation derived based only on the gravitational constant g, the gas law constant R, the average molar molecular weight of the atmospheric constituents m, and the average degrees of freedom N.
Answer: Now we want to check the results against the observed values for the two planets
Parameters
Object Main Gas N m g Earth N2, O2 5 28.96 9.807 Venus CO2 6 43.44 8.87
Results
Object Lapse Rate observed f observed Earth 6.506 C/km 6.5 21/4 5.25 Venus 7.72 7.72 6 6
All the numbers are spot on with respect to the empirical data recorded for both Earth and Venus, with supporting figures available here.
——-
The rough derivation that I previously posted to explain the empirical data was not very satisfying in its thoroughness. The more comprehensive derivation in this post serves to shore up the mystery behind the deviation from the adiabatic derivation. The key seems to be correctly accounting for the internal energy necessary to maintain the gravitational hydrostatic equilibrium. Since the polytropic expansion describes a process, the actual atmosphere can accommodate these constraints (while minimizing Gibbs free energy under constant entropy conditions) by selecting the appropriate polytropic index. The mystery of the profile seems not so mysterious anymore.
Criticisms welcome as I have not run across anything like this derivation to explain the Earth’s standard atmosphere profile nor the stable Venus data (not to mention the less stable Martian atmosphere). The other big outer planets filled with hydrogen are still an issue, as they seem to follow the conventional adiabatic profile, according to the few charts I have access to. The moon of Saturn, Titan, is an exception as it has a nitrogen atmosphere with methane as a greenhouse gas.
BTW, this post is definitely not dedicated to Ferenc Miskolczi. Please shoot me if I ever drift in that direction. It’s a tough slog laying everything out methodically but worthwhile in the long run.
Added Fig 1 : Lapse Rate on Earth versus Latitude. From D. J. Lorenz and E. T. DeWeaver, “Tropopause height and zonal wind response to global warming in the IPCC scenario integrations,” Journal of Geophysical Research: Atmospheres (1984–2012), vol. 112, no. D10, 2007.
Added Fig 2 : Lapse Rate on Earth versus Latitude. The average was calculated by integrating with effective cross-sectional area weighting of (sin(Latitude+2.5)-sin(Latitude-2.5)) . Adapted from J. P. Syvitski, S. D. Peckham, R. Hilberman, and T. Mulder, “Predicting the terrestrial flux of sediment to the global ocean: a planetary perspective,” Sedimentary Geology, vol. 162, no. 1, pp. 5–24, 2003.
Added Fig 3: This study also suggests an average lapse rate of 6.1C/km over the northern hemisphere. I. Mokhov and M. Akperov, “Tropospheric lapse rate and its relation to surface temperature from reanalysis data,” Izvestiya, Atmospheric and Oceanic Physics, vol. 42, no. 4, pp. 430–438, 2006.
Since I posted this derivation, I received feedback from several other blogs which I attached as comments below this post. In the original post I concluded by saying that I was satisfied with my alternate derivation, but after receiving the feedback, there is still the nagging issue of why the Venus lapse rate profile can be so linear in the lower atmosphere even though we know that the heat capacity of CO2 varies with temperature (particularly in the high temperature range of greater than 500 Kelvin).
If we go back and look at the hydrostatic relation derived earlier, we see an interesting identity:
$$– 3 \int_0^M\frac{P}{\rho}dm = -\int_0^M \frac{G M}{r} dm$$
If I pull out the differential from the integral
$$3 \frac{P}{\rho} = \frac{G M}{r}$$
and then realize that the left-hand side is just the Ideal Gas law
$$3RT/m = \frac{G M}{r}$$
This is internal energy due to gravitational potential energy.
If we take the derivative with respect to r, or altitude:
$$3R \frac{dT}{dr} = – \frac{G M m}{r^2}$$
The right side is just the gravitational force on an average particle. So we essentially can derive a lapse rate directly:
$$\frac{dT}{dr} = – \frac{g m}{3 R}$$
This will generate a linear lapse rate profile of temperature that decreases with increasing altitude. Note however that this does not depend on the specific heat of the constituent atmospheric molecules. That is not surprising since it only uses the Ideal Gas law, with no application of the variational Gibbs Free Energy approach used earlier.
What this gives us is a universal lapse rate that does not depend on the specific heat capacity of the constituent gases, only the mean molar molecular weight, m. This is of course an interesting turn of events in that it could explain the highly linear lapse profile of Venus. However, plugging in numbers for the gravity of Venus and the mean molecular weight (CO2 plus trace gases), we get a lapse rate that is precisely twice that which is observed.
The “obvious” temptation is to suggest that half of the value of this derived hydrodynamic lapse rate would position it as the mean of the lapse rate gradient and an isothermal lapse rate (i.e. slope of zero).
$$\frac{dT}{dr} = – \frac{g m}{6 R}$$
The rationale for this is that most of the planetary atmospheres are not any kind of equilibrium with energy flow and are constantly swinging between an insolating phase during daylight hours, and then a outward radiating phase at night. The uncertainty is essentially describing fluctuations between when an atmosphere is isothermal (little change of temperature with altitude producing a MaxEnt outcome in distribution of pressures, leading to the classic barometric formula) or isentropic (where no heat is exchanged with the surroundings, but the temperature can vary as rapid convection occurs).
In keeping with the Bayesian decision making, the uncertainty is reflected by equal an weighting between isothermal (zero lapse rate gradient) and an isentropic (adiabatic derivation shown). This puts the mean lapse rate at half the isentropic value. For Earth, the value of g*m/3R is 11.4 C/km. Half of this value is 5.7 C/km, which is a value closer to actual mean value than the US Standard Atmosphere of 6.5 C/km
J. Levine, The Photochemistry of Atmospheres. Elsevier Science, 1985.
“The value chosen for the convective adjustment also influences the calculated surface temperature. In lower latitudes, the actual temperature decrease with height approximates the moist adiabatic rate. Convection transports H2O to higher elevations where condensation occurs, releasing latent heat to the atmosphere; this lapse rate, although variable, has an average annual value of 5.7 K/km in the troposphere. In mid and high latitudes, the actual lapse rates are more stable; the vertical temperature profile is controlled by eddies that are driven by horizontal temperature gradients and by topography. These so-called baroclinic processes produce an average lapse rate of 5.2 K/km – It is interesting to note that most radiative convective models have used a lapse rate of 6.5 K km – which was based on date sets extending back to 1933. We know now that a better hemispherical annual lapse rate is closer to 5.2 K/km, although there may be significant seasonal variations.
BTW, the following references are very interesting presentations on the polytropic approach.
References
[1]
“Polytropes.” [Online]. Available: http://mintaka.sdsu.edu/GF/explain/thermal/polytropes.html. [Accessed: 19-May-2013].
[2]
B. Davies, “Stars Lecture.” [Online]. Available: http://www.ast.cam.ac.uk/~bdavies/Stars2 . [Accessed: 28-May-2013].
Even More Recent Research
A number of Chinese academics [3,4] are attacking the polytropic atmosphere problem from an angle that I hinted at in the original post. The gist of their approach is to assume that the atmosphere is not under thermodynamic equilibrium (which it isn’t as it continuously exchanges heat with the sun and outer space in a stationary steady-state) and therefore use some ideas of non-extensible thermodynamics. Specifically they invoke Tsallis entropy and a generalized Maxwell-Boltzmann distribution to model the behavioral move toward an equilibrium. This is all in the context of self-gravitational systems, which is the theme of this post. Why I think it is intriguing, is that they seem to tie the entropy considerations together with the polytropic process and arrive at some very simple relations (at least they appear somewhat simple to me).
In the non-extensive entropy approach, the original Maxwell-Boltzmann (MB) exponential velocity distribution is replaced with the Tsallis-derived generalized distribution — which looks like the following power-law equation:
$$f_q(v)=n_q B_q (\frac{m}{2 \pi k T})^{3/2} (1-(1-q) \frac{m v^2}{2 k T})^{\frac{1}{1-q}}$$
The so-called q-factor is a non-extensivity parameter which indicates how much the distribution deviates from MB statistics. As q approaches 1, the expression gradually trasforms into the familiar MB exponentially damped v^2 profile.
When q is slightly less than 1, all the thermodynamic gas equations change slightly in character. In particular, the scientist Du postulated that the lapse rate follows the familiar linear profile, but scaled by the (1-q) factor:
$$\frac{dT}{dr} = \frac{(1-q)g m}{R}$$
Note that this again has no dependence on the specific heat of the constituent gases, and only assumes an average molecular weight. If q=7/6 or Q = 1-q = -1/6, we can model the f=6 lapse rate curve that we fit to earlier.
There is nothing special about the value of f=6 other than the claim that this polytropic exponent is on the borderline for maintaining a self-gravitational system [5].
Note that as q approaches unity, the thermodynamic equilibrium value, the lapse rate goes to zero, which is of course the maximum entropy condition of uniform temperature.
The Tsallis entropy approach is suspiciously close to solving the problem of the polytropic standard atmosphere. Read Zheng’s paper for their take [3] and also Plastino [6].
The cut-off in the polytropic distribution (5) is an example of what is known, within the field of non extensive thermostatistics, as “Tsallis cut-off prescription”, which affects the q-maximum entropy distributions when q < 1. In the case of stellar polytropic distributions this cut-off arises naturally, and has a clear physical meaning. The cut-off corresponds, for each value of the radial coordinate r, to the corresponding gravitational escape velocity.
This has implications for the derivation of the homework problem that we solved at the top of this post, where we eliminated one term of the integration-by-parts solution. Obviously, the generalized MB formulation does have a limit to the velocity of a gas particle in comparison to the classical MB view. The tail in the statistics is actually cut-off as velocities greater than a certain value are not allowed, depending on the value of q. As q approaches unity, the velocities allowed (i.e. escape velocity) approach infinity.
As Plastino states [6]:
Polytropic distributions happen to exhibit the form of q-MaxEnt distributions, that is, they constitute distribution functions in the (x,v) space that maximize the entropic functional Sq under the natural constraints imposed by the conservation of mass and energy.
The enduring question is does this describe our atmosphere adequately enough? Zheng and company certainly open it up to another interpretation.
[3]
Y. Zheng, W. Luo, Q. Li, and J. Li, “The polytropic index and adiabatic limit: Another interpretation to the convection stability criterion,” EPL (Europhysics Letters), vol. 102, no. 1, p. 10007, 2013.
[4]
Z. Liu, L. Guo, and J. Du, “Nonextensivity and the q-distribution of a relativistic gas under an external electromagnetic field,” Chinese Science Bulletin, vol. 56, no. 34, pp. 3689–3692, Dec. 2011.
[5]
M. V. Medvedev and G. Rybicki, “The Structure of Self-gravitating Polytropic Systems with n around 5,” The Astrophysical Journal, vol. 555, no. 2, p. 863, 2001.
[6]
A. Plastino, “Sq entropy and selfgravitating systems,” europhysics news, vol. 36, no. 6, pp. 208–210, 2005.
# Airborne fraction of CO2 explained by sequestering model
As acknowledgement of the atmospheric levels of CO2 reaching 400 PPM, this post is meant to clear up one important misconception (suggested prerequisite reading on fat-tail CO2 sequestration here and the significance of the fat-tail here)
A recently active skeptic meme is that the amount of CO2 as an airborne fraction is decreasing over time.
“If we look at the data since Mauna Loa started, we see that the percentage of the CO2 emitted by humans that “remains” in the atmosphere has averaged around half, but that it has diminished over time, by around 1% per decade.
Over the 30 year period 1959-1989 it was around 55%; over the following 20+ years it was just over 50%.
Why is this?”
What the befuddled fellow is talking about are the charts being shown below. These are being shown without much context and no supporting documentation, which puts the burden on the climate scientists to explain. Note that the airborne fraction does seem to decrease slightly over the past 50 years, even though the carbon emissions are increasing.
This obviously needs some explaining. The following figure illustrates what the CO2 sequestration model actually does.
Figure 1: Model airborne fraction of CO2 against actual data
On the left is the data plotted together with the model of the yearly fraction not sequestered out. The model is less noisy than the data but it does clearly decline as well. No big surprise as this is a response function, and responses are known to vary depending on the temporal profile of the input and the fat-tail in the adjustment time impulse response function.
On the right is the model with the incorporation of a temperature-dependent outgassed fraction. In this case the model is more noisy than the data, as it includes outgassing of CO2 depending on the global temperature for that year. Since the temperature is noisy, the CO2 fraction picks up all of that noise. Still, the airborne fraction shows a small yet perceptible decline, and the model matches the data well, especially in recent years where the temperature fluctuations are reduced.
Amazing that over 50 years, the mean fraction has not varied much from 55%. That has a lot to do with the math of diffusional physics. Essentially a random walk moving into and out of sequestering sites is a 50/50 proposition. That’s the way to intuit the behavior, but the math really does the heavy lifting in predicting the fraction sequestered out.
It looks like the theory matches the data once again. The skeptics provide a knee-jerk view that this behavior is not well understood, but not having done the analysis themselves, they lose out — the skeptic meme is simply one of further propagating fear, uncertainty, and doubt (FUD) without concern for the underlying science.
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2020-08-15 09:54:52
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https://www.wikidata.org/wiki/Q4455030
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# Stone's theorem on one-parameter unitary groups (Q4455030)
${\displaystyle (U_{t})_{t\in \mathbb {R} }}$
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2018-01-18 16:16:53
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http://clay6.com/qa/48871/find-the-domain-and-range-of-the-relations-defined-by-the-following-arrow-d
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# Find the domain and range of the relations defined by the following arrow diagram.
Domain of $R_1$ = {a, a, b, b, c} = {a, b, c} , Range of $R_1$ = {x, y, z}
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2018-03-23 05:24:37
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https://www.gamedev.net/forums/topic/647232-when-to-learn-c/
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# Unity When to learn c++?
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Let me start off by saying I know the "Should I learn c++?" question has been done to death. I'm aware of the pros and cons stated by multitudes of folks on the internet. However, the answer to my question, alluded to in the title, has come up empty me in a google and forums search so I figured I would start a thread on it.
QuestionWhen should I start to learn c++?
That's the brief version, and if you want to, you can stop there and give a general answer to when someone should venture into the land of c++. And I'd like to thank you in advance for any reply or advice you can give. However, for those who wish to know more specifics of my case, I'd like to provide some of my personal history that may aid in determining what advice to give me. I hope this does not become a giant wall of text.
I have grown up around computers, my dad did some programming in fortran, pascal, basic, and some CNC languages when I was a kid. So we've always had a computer around that I was able to mess around on. I started with some pascal myself, then later ventured into javascript when I wanted to do some web stuff. In the early 2000s, I was enrolled in a computer science program at a college near me. They started us in Visual Basic and a little c++, though in c++ I didn't get much past cin/cout. I had to drop out due to financial and personal reasons. Later, I was able to go to a different college and finish a degree, but since the closest they had to CS was electrical engineering, which didn't really hold my attention, I ended up getting a BS in physics. I use python regularly for computational physics problems, and I feel comfortable enough with python that I feel that if I'm given something to program, even if I'm not familiar with the modules to do it with, I could delve into the documentation and find a way to do it. It may not be elegant, short, or efficient, but I feel I could find a solution. I am by no means an expert in python, there is far too much that I don't know in python to convincingly claim to be, but I feel I know enough at least to be dangerous, as the saying goes.
Lately though, I have been dabbling in C, mostly just taking old programs I've written in python to see if I can make them work in C, and get the same results. Ostensibly to see if there is a computation time benefit to learning C, but mostly to start learning it just because I'm a little curious about it and have the time. I have done a little game programming in python, mostly just the standard hangman, tic-tac-toe, and text based games. I haven't dabbled too much in pygame because I don't like it, I don't know where my dislike for pygame stems, could be the obnoxious colour of the pygame website, could be the overuse of underscores in method names, I'm not sure, but I'm just not a fan.
Obviously, with this history, I'm not just starting out learning to program, and I'm comfortable enough to say that I could probably pick up c++, so I'm not worried about can I learn it. I also know that at 30, with no real formal training in CS, and at this point, no real portfolio to speak of, at least that I'd be proud enough to share with a prospective employer, me becoming employed at a AAA studio, and actually needing c++, is really unlikely to happen.
Since I doubt that I will realistically ever need the lower level and efficiency of c++, is it even worth it to try to learn c++, or would I be better off just messing around with c# and an engine like Unity or something of the sorts? Furthermore, since I'm already trying to learn C for scientific computing purposes, would it be better to just learn c++ for that since that's just a superset of C anyways?
Thank you for taking the time to read this.
-Jared
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I'd say... anytime you want.
I started with Java/C# and web development, and when I decided to make a game I decided to use C/C++ since I considered it as the "main" language for programming games.
It's been nearly a year or two that I started, and I'm quite glad I did.
I still haven't got nowhere near getting a complete game, where I probably would have if I sticked to XNA (unfortunately deceased) or Unity, but I got to learn so many things about computers, graphics, languages and hardware in general, how they work, and this helped gave me conviction that I can manage to get anything, anywhere.
Lines I used to see often in C# such as:
MyClass mA = new MyClass();
MyClass mB = mA;
Now give me some shock before I realize "Oh, it's C#".
So, for knowledge I think it was a good thing, but for production value x time spent, not so much.
If you want to develop games and only focus on developing games, I'd recommend to stick to using an existing engine or learning C++ just for using the minimum required for these.
Another thing is, there's just too much about C++. Learning the syntax might take less than a day, it's techniques and such less than a week, but that alone isn't much, really. Learning C++ I think it's learning how things work and studying constantly for new things, how a certain app works, how you use a certain library, etc
As for C or C++, please disregard C completely.
In the past year that I started learning it, I was completely fixed on the idea that I wasn't going to use strings at all. I had (still have) the idea that with better functionality comes worse performance, so I always used char pointers (array of chars to represent strings), among other things.
After various questions and tutorials I've read around, I noticed how silly that was. In some (most, actually) cases the C++ classes and containers were even BETTER than my C implementations.
Now, the language has been upgraded to C++11, so there's a bunch of new stuff and I'm still going through the same thing. Should I use the new features, are they better? Are they worse?
I'll keep coding in C++, and I'll be slowly changing towards C++11 as I see these new features aren't so bad as I think (and I'm pretty sure they aren't).
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Since I doubt that I will realistically ever need the lower level and efficiency of c++, is it even worth it to try to learn c++, or would I be better off just messing around with c# and an engine like Unity or something of the sorts?
Is it worth it is a personal question. Is it worth it FOR YOU?
Good game programmers will learn many languages. C++ is generally one of those languages. Other frequently used programming languages include Java, C#, Python, Perl, JavaScript, SQL, PHP, Ruby, ActionScript, and shell languages. Markup languages you will likely want to learn include things like HTML, XML, JSON and YAML. Neither is an all-encompassing list. Some languages are good to know in depth, others will require just a cursory understanding; talented professional programmers will end up knowing twenty, thirty, fifty, or more programming languages over their entire career.
The order you learn them is not very important, although some are much easier to learn that others.
If you intend to work on game consoles and game engines you will need to know C++ before you get the job. If you intend to develop any high performance software, including game servers, you will need to know C++ before you get the job.
If you intend to work on web games in JavaScript or Flash or Java Applets you might never look at C++ code.
Furthermore, since I'm already trying to learn C for scientific computing purposes, would it be better to just learn c++ for that since that's just a superset of C anyways?
Let's clear that up for you.
C and C++ do not have a superset/subset relationship. There was a time in the 1980s right after C++ was created that you could run C code through a C++ compiler, but those days are long gone. Any significant C code is incompatible with C++ code and will generate large numbers of errors in a C++ compiler.
Implicit conversions through void pointers, boolean values, tentative definitions, implicit casts, VLAs, designated initializers, and many other differences make code that is valid C not compile in a C++ compiler. The simple act of making an object with malloc() through C idioms will result in a compiler error in C++.
C and C++ are different languages. They had a common ancestor but diverged almost 30 years ago.
mostly to start learning it just because I'm a little curious about it and have the time.
That's all the reason you need.
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"me becoming employed at a AAA studio, and actually needing c++, is really unlikely to happen"
I dont work in the industry (Still a student), but I would have to say I disagree with the above statement.
I have no idea what your physics degree consisted of, but I guess you have a solid understanding of motion, forces, vecotrs, etc. In addition, you may have all sorts of maths knowledge which applies to 3D games / graphics / physics engines. At the very least, you will easily be able to learn the 3D math stuff - and at a *MUCH* better level than a student. Maths is a key skill in the games industry without a doubt. Mathemetics For 3D Games Programming and Computer Graphics by Eric Lengyel is a good place to start looking/comparing your skill sets.
Personally (as I did in formal education) I would start with C (Being able to make some console apps/games) and then learn C++. C and C++ are similar and most books dont go in to classes from chapter one (Unlike Objective C, for example, which is a class heavy language).
From their, the world is your oyster. You can apply your degree and aim to develop some physics systems. Can learn Direct3D/Open GL and go down the graphics route. Learn sound programming, networking, etc.
No Games Programming degree/Computer Science degree can teach everything. Mine teaches C++ and thats literally about it (and we have a decent employment rate within the first 6 months). The best games programming degree in the UK (Teesside if im not mistaken) hasnt tought any content that I havnt been able to do in 4/5 months of studying Direct3D11 by myself (And I would say that learning something yourself is more important than the degree - indeed, everyone I have spoke too says the first question in an interview centers around things you do outside of education. Plus, imo, teaching something yourself gives you a much more rounded knowledge whenever you finally get it. Uni has wayyyyy too much hand holding to be good for you).
My mate did Computer Science (Newcastle) and, well... I wasn't impressed tbh. Defenatly not for games programming anyway.
So yeah, why not? If your looking for a career change, then your not in a terrible position. Most junior jobs state that they are looking for people with "Computer Science (and games programming) or maths related degree" - You do have a maths related degree.
Finally, learning c++ and games related programming can take time to do correctly. C++ is a hard language. Its of course worth noting that learning C++ as a second language is MUCH easier than it being your first. C#/Java (I did objective C. But I'm a mini mac fanboy ) I have heard is a good start if you want another langauge first .
Edited by Dbowler92
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C and C++ do not have a superset/subset relationship. There was a time in the 1980s right after C++ was created that you could run C code through a C++ compiler, but those days are long gone. Any significant C code is incompatible with C++ code and will generate large numbers of errors in a C++ compiler.
CAVEMAN is 70,000+ lines of C code, compiled in MS Visual Studio C++ 2012, warning level 4 (to suppress inline-ing notifications), zero errors, zero warnings.
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The simple act of making an object with malloc() through C idioms will result in a compiler error in C++.
object, or just a buffer?
buffer works fine:
int APIENTRY _tWinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPTSTR lpCmdLine, int nCmdShow)
{
unsigned char *p;
nCmdShow=0;
lpCmdLine=0;
hPrevInstance=0;
if (!initprog(hInstance)) { Zmsg("Error initializing program","Caveman v3.0"); return(1); }
p=(unsigned char *)malloc(1000);
free(p);
titlescreen();
runprog();
Zshutdown();
PostQuitMessage(0);
return(0);
}
Edited by Norman Barrows
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would it be better to just learn c++ for that since that's just a superset of C anyways?
C and C++ do not have a superset/subset relationship.
For C++ to be a superset of C, then any valid C program must also be a valid C++ program.
Such is not the case. Earlier I listed just a few examples of things that are present in C that will not work in a C++ compiler. A web search can find many more differences.
Yes it is possible for people to use a constrained portion of the C language, add a bunch of casts that would be unnecessary in C but are required in C++, and write code that can compile in both. But it is also possible for me to write functions that I can copy/paste into Java, C#, python, and other languages. That establishes that the languages are similar, but not that it is a subset.
For example, in that code listing:
p=(unsigned char *)malloc(1000);
That cast is not a C idiom. There is no reason for a C programmer to write that cast. It was written only to shoehorn the code into a C++ compiler.
Frequent variable names in C include "new" and "old", that would break in C++. Occasionally when I work in C I will have names like "template" only to remember that the name is a keyword in C++ and shouldn't be used.
That's not to say you cannot write large programs that are constrained to work with multiple languages. With carefully-crafted constrained writing people have done exactly that. Just as people have also written constrained books like Gadsby without the letter E, or The Exeter Text that only uses the vowel E; yet you wouldn't claim that because of those books the English language does not have those vowels. Just because someone chooses to constrain their writing to a subset of the language does not mean that the properties of their constraints apply to the language as a whole.
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C and C++ do not have a superset/subset relationship. There was a time in the 1980s right after C++ was created that you could run C code through a C++ compiler, but those days are long gone. Any significant C code is incompatible with C++ code and will generate large numbers of errors in a C++ compiler.
CAVEMAN is 70,000+ lines of C code, compiled in MS Visual Studio C++ 2012, warning level 4 (to suppress inline-ing notifications), zero errors, zero warnings.
msvc++ will use a C89 compiler to compile any files with a .c extension (if you mix in c++ code it will most likely fail to compile, if you tell the compiler explicitly to compile it as c++ it will most likely fail, and if you rename the files to *.cpp or *.cc it will most likely fail to compile (unless you explicitly tell it to compile the file as C89 despite the .cpp file extension)
gcc will do the same thing for C,C++,Objective-C, Objective-C++, Ada and Fortran (Fortran most definitely is not a subset of C or C++)
Edited by SimonForsman
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That cast is not a C idiom. There is no reason for a C programmer to write that cast. It was written only to shoehorn the code into a C++ compiler.
so then i'm not doing c code, i'm doing c++ without the oo extensions? now i'm confused...
trying to remember when i learned c, and what compiler...
it was for school, systems i think, you had to write a vm, a compiler, a linker, and a loader in C, and you had to learn C on your own. they taught you pascal and figured you could figure out the rest. kinda weird, because while there i was also taught stuff like forth, smalltalk, lisp, motorola 68000 assembler, fortran, cobol, god the list just goes on. worked out ok though, i got an A in the class.
ok, the C was on a sun sparc workstation running bsd unix and x11 windows. does that sound right?
and an Emacs editor, and GNU C (i think).
then for games, i switched from borland pascal 7 to i guess borland C (or c++ ?) v4 (?). then i switched to watcom. then microsoft.
but as i recall, i switched to C for games before the C++ extensions came along.
so would borland c 4.0 be c or c++ ?
seems to me i've been writing exactly the same code since i switched from pascal to C (++ ?) for games. its not like the stuff stopped working when the c++ compilers came out. or perhaps they needed a mod or cast or two, been a long time.
Edited by Norman Barrows
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msvc++ will use a C89 compiler to compile any files with a .c extension (if you mix in c++ code it will most likely fail to compile, if you tell the compiler explicitly to compile it as c++ it will most likely fail, and if you rename the files to *.cpp or *.cc it will most likely fail to compile (unless you explicitly tell it to compile the file as C89 despite the .cpp file extension)
i use .cpp and .h, and in the compiler settings i tell it to compile as c++ code (for the stronger type checking).
• ### Similar Content
• I'm looking for any team / people that need a programmer for their project. I'm looking to expand my portfolio which you can see Here. I'm more experienced with Unity but I can spend the time to learn new Engines if that's your preference. I have worked on Unreal Engine 4 before but I might take some time to re-learn it, if the project requires it. Feel free to DM here or use the contact info on my website.
• I'm working on a system for my game that will allow the player to stack pick ups in a queue. As one pick up expires, the next automatically activates. I'm having an issue though where if I pick up the first one, it activates fine, but if i pick up a second directly after it, it overrides the first one, activates the second one, and then once it has run it's course, everything goes back to normal gameplay, no first pick up. I'm not sure why this is happening. Hopefully someone can spot what I'm doing wrong in my code.
Here is the code for the pick up manager:
// Update is called once per frame void Update () { if (pickUpQueue.Count != 0 && !pickUpActive) { pickUpActive = true; pickUpQueue[0].ActivatePickUp(); } DeactivatePickUp(); } void DeactivatePickUp () { if (pickUpQueue.Count != 0 && pickUpActive) { Destroy (pickUpQueue [0]); pickUpQueue.RemoveAt (0); pickUpActive = false; } } And here is the PickUp:
public override void ActivatePickUp () { ball.GetComponent<Ball>().Speed = 2.0f; //increase ball speed... ball.GetComponent<Ball>().StartCoroutine(timer); //...set time that power up is active }
There is also a Base Pick Up:
public void OnCollisionEnter2D (Collision2D collision) { Vector2 tweak = new Vector2 (Random.Range(0f, 0.2f),Random.Range(0f, 0.2f)); this.gameObject.GetComponent<Rigidbody2D>().velocity += tweak; //if the pickup makes contact with the paddle or ball.... if (collision.gameObject.tag == "Paddle" || collision.gameObject.tag == "Ball") { GameObject.FindObjectOfType<GameManager>().GetComponent<PickUpManager>().pickUpQueue.Add(this); Destroy(gameObject); //...and finally destroy power up object } } As a side note, I am trying to find a solution to this that will work for all of my pickups. Some pickups are ammo based, some are timed.
• By D34DPOOL
Edit Your Profile D34DPOOL 0 Threads 0 Updates 0 Messages Network Mod DB GameFront Sign Out Add jobEdit jobDeleteC# Programmer for a Unity FPS at Anywhere Programmers located Anywhere.
Posted by D34DPOOL on May 20th, 2018
Hello, my name is Mason, and I've been working on a Quake style arena shooter about destroying boxes on and off for about a year now. I have a proof of concept with all of the basic features, but as an artist with little programming skill I've reached the end of my abilities as a programmer haha. I need someone to help fix bugs, optomize code, and to implent new features into the game. As a programmer you will have creative freedom to suggest new features and modes to add into the game if you choose to, I'm usually very open to suggestions :).
What is required:
Skill using C#
Experience with Unity
Experience using UNET (since it is a multiplayer game), or the effort and ability to learn it
Compensation:
Since the game currently has no funding, we can split whatever revenue the game makes in the future. However if you would perfer I can create 2D and/or 3D assets for whatever you need in return for your time and work.
It's a very open and chill enviornment, where you'll have relative creative freedom. I hope you are interested in joining the team, and have a good day!
To apply email me at mangemason@yahoo.com
• Is there a way to automatically change the start position of an animation? I have a bunch of animations set up on 3D models in unity. The issue is that I need to move the 3D models, however when I do so the animation start positions are not updated and I have to do it manually.
Changing the transform of key frames is time consuming with the amount of animations I have, so I was wondering if there was a way to do it automatically?
• By MoreLion
hey all! We are looking for members for our Unity horror game!
Here’s the story:
After a deadly virus plunges the world into chaos killing 85% of the human population there are now what they call “zones” these zones are watched very closely by the surviving government, people are checked every day for the virus, even if you touch the spit or any human waste or fluids of the victim who is infected, you will die. But one day, people in the west zone start to go missing, 1 woman goes outside the walls to uncover the mystery, is there more to the virus than meets the eye?, That is where your story starts.
This game is not a long development game, I have loads other game ideas,
I will also allow you to have a bit of creative freedom if you wish to add or share a idea!
And no, it’s not a zombie game lol I feel like zombie games are too generic, in this game you will encounter terrifying beasts!
There is some concept art one of our concept artists have made
If interested email liondude12@gmail.com
• GOVERNOR is a modernized version of the highly popular series of “Caesar” games. Our small team has already developed maps, written specifications, acquired music and performed the historical research needed to create a good base for the programming part of the project.
Our ultimate goal is to create a world class multi-level strategic city building game, but to start with we would like to create some of the simpler modules to demonstrate proof of concept and graphical elegance.
We would like programmers and graphical artists to come onboard to (initially) create:
A module where Province wide infrastructure can be built on an interactive 3D map of one of the ancient Roman Provinces.
A module where city infrastructure can be built on a real 3D interactive landscape.
For both parts, geographically and historically accurate base maps will be prepared by our team cartographer. Graphics development will be using Blender. The game engine will be Unity.
More information, and examples of the work carried out so far can be found at http://playgovernor.com/ (most of the interesting content is under the Encyclopedia tab).
This project represents a good opportunity for upcoming programmers and 3D modeling artists to develop something for their portfolios in a relatively short time span, working closely with one of Canada’s leading cartographers. There is also the possibility of being involved in this project to the point of a finished game and commercial success! Above all, this is a fun project to work on.
Best regards,
• So I have hundreds of moving objects that need to check there speed. One of the reasons they need to check there speed is so they don't accelerate into oblivion, as more and more force is added to each object.
At first I was just using the Unity vector3.magnitude. However this is actually very slow; when used hundreds of times.
Next I tried the dot-product check: vector3.dot(this.transform.foward, ShipBody.velocity) The performance boost was fantastic. However this only measures speed in the forward direction. Resulting in bouncing objects accelerating way past the allowed limit.
I am hoping someone else knows a good way for me to check the speed with accuracy, that is fast on the CPU. Or just any magnitude calculations that I can test when I get home later.
What if I used vector3.dot(ShipBody.velocity.normalized, ShipBody.velocity)?
How slow is it to normalize a vector, compared to asking it's magnitude?
• By Ds ds
Hi, my name is Andres, I'm a programmer with a technician degree and a Diploma in C#, looking for a project in Unity to start my career in game development. I don't do it for a paid but a recognition and start a portfolio, preferably a 2D game. Thanks for read, have a nice day.
• Hi there! Is the first time that I'm posting here so I'm sorry if I'm doing it wrong ha.
So here it comes, my doubt is, I'm doing a game with different levels, each of these levels in one different scene. Each scene contains to cameras that you can change pressing a button. Everything works fine.
The only problem is that I would like it to look a bit more professional, and I would like that if you finish the level with camera2, the next level start the same way. I've been thinking about using dontdestroyonloadon both cameras, but obviously this cameras need to be attached to the player to make the movement work, what do you recommend? Sorry If I've explained it in a messy way, and feel free to dm me for anything. Thanks in advance!
• Hello everyone
I am a programmer from Baku.
I need a 3D Modeller for my shooter project in unity.I have 2 years Unity exp.
Project will paid when we finish the work
If you interested write me on email:
mr.danilo911@gmail.com
• 48
• 12
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• ### Forum Statistics
• Total Topics
631374
• Total Posts
2999661
×
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2018-06-20 06:08:03
|
{"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": 0.2323789745569229, "perplexity": 1516.6147929125734}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00304.warc.gz"}
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https://math.stackexchange.com/questions/458878/deck-transformations-and-covering-spaces
|
# deck transformations and covering spaces
Let $p:\tilde X\rightarrow X$ be a universal covering space, and let $H\leq G$ where $G$ is the group of covering transformations. Let $q:\tilde X \rightarrow \tilde X/G$ be the quotient map which is regular covering space. Is $\tilde X/H\rightarrow X$ a covering space with appropriate map? If yes, what is this map?
Corollary: If $p\colon (\tilde{X},\tilde{x_0})\to (X,x_0)$ is the universal covering, then for every $H\le \pi_1(X,x_0)$ the map $\tilde{X}/H\to X$ is the covering corresponding to $H.$
Proof: The fundamental group $\pi_1(\tilde{X}/H,x_H)$ is canonically isomorphic to $H$, where $q(\tilde{x})=x_H$, $q\colon \tilde{X}\to \tilde{X}/H$. The projection $\tilde{X}/H\to \tilde{X}/\pi_1(X,x_0)$ is a covering and $\tilde{X}/\pi_1(X,x_0)=X$, where $H$ acts on $\tilde{X}$ as a subgroup of $\pi_1(X,x_0)$. (cf. source)
|
2019-06-19 11:27:07
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https://ora.ox.ac.uk/objects/uuid:24dfdac7-9652-4c43-9f33-ea9a26cd4a18
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# Journal article
## Chemical gradients in the Milky Way from the RAVE data. II. Giant stars
Abstract:
We provide new constraints on the chemo-dynamical models of the Milky Way by measuring the radial and vertical chemical gradients for the elements Mg, Al, Si, Ti, and Fe in the Galactic disc and the gradient variations as a function of the distance from the Galactic plane ($Z$). We selected a sample of giant stars from the RAVE database using the gravity criterium 1.7$<$log g$<$2.8. We created a RAVE mock sample with the Galaxia code based on the Besan\c con model and selected a corresp...
### Access Document
Publisher copy:
10.1051/0004-6361/201423974
### Authors
More by this author
Institution:
University of Oxford
Division:
MPLS
Department:
Physics
Role:
Author
Publisher:
EDP Sciences Publisher's website
Journal:
A
Volume:
568
Pages:
A71
Publication date:
2014-06-27
DOI:
EISSN:
1432-0746
ISSN:
0004-6361
Language:
English
Keywords:
Pubs id:
pubs:472173
UUID:
uuid:24dfdac7-9652-4c43-9f33-ea9a26cd4a18
Local pid:
pubs:472173
Source identifiers:
472173
Deposit date:
2014-07-19
|
2023-03-30 16:00:37
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https://proofwiki.org/wiki/Interval_in_Complete_Lattice_is_Complete_Lattice
|
# Interval in Complete Lattice is Complete Lattice
## Theorem
Let $\left({L, \preceq}\right)$ be a complete lattice.
Let $a, b \in L$ with $a \preceq b$.
Let $\left[{a \,.\,.\, b}\right]$ be the closed interval between $a$ and $b$.
Then $\left[{a \,.\,.\, b}\right]$ is also a complete lattice under $\preceq$.
## Proof
Let $I = \left[{a \,.\,.\, b}\right]$.
Let $S \subseteq I$.
If $S = \varnothing$, then it has a supremum in $I$ of $a$ and an infimum in $I$ of $b$.
Let $S \ne \varnothing$.
Since $S \subseteq I$, $a$ is a lower bound of $S$ and $b$ is an upper bound of $S$.
Since $L$ is a complete lattice, $S$ has an infimum, $p$, and a supremum, $q$, in $L$.
Thus by the definitions of infimum and supremum:
$a \preceq p$ and $q \preceq b$
Let $x \in S$.
Since an infimum is a lower bound:
$p \preceq x$
Since a supremum is an upper bound:
$x \preceq q$
Thus $a \preceq p \preceq x \preceq q \preceq b$.
Since $\preceq$ is an ordering, it is transitive, so by Transitive Chaining:
$a \preceq p \preceq b$ and $a \preceq q \preceq b$.
That is, $p, q \in I$.
Thus $p$ and $q$ are the infimum and supremum of $S$ in $I$.
As every subset of $I$ has a supremum and infimum in $I$, $I$ is a complete lattice.
$\blacksquare$
## Remark
Although $\left({\left[{a \,.\,.\, b}\right], \preceq}\right)$ is a complete lattice, it is only a complete sublattice of $\left({L, \preceq}\right)$ if $a = \inf L$ and $b = \sup L$. That is, if it equals $\left({L, \preceq}\right)$.
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2021-07-29 21:58:14
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http://openstudy.com/updates/4f307323e4b0fc09381ee438
|
## anonymous 4 years ago how do you find the radius of a circle if you know the area?
1. anonymous
pi times the radius squared euals area so do that backwards
2. precal
use$A=\pi (r^{2})$
3. anonymous
so if area =1cm2 and pi = 3.14 then radius = .5?
4. anonymous
$R= \sqrt{A/\pi}$
5. precal
were you told to use 3.14 as pi?
6. anonymous
yes
7. precal
then you can use 3.14 otherwise pi is pi
8. anonymous
okay, thanks
9. precal
anytime :)
10. anonymous
area = 50.24 pi= 3.14 so radius would equal 8?
|
2016-10-26 21:39:56
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|
https://www.aimsciences.org/article/doi/10.3934/dcdsb.2019037
|
# American Institute of Mathematical Sciences
September 2019, 24(9): 4913-4928. doi: 10.3934/dcdsb.2019037
## Dynamical behavior for a Lotka-Volterra weak competition system in advective homogeneous environment
School of Mathematics(Zhuhai), Sun Yat-sen University, Zhuhai 519082, Guangdong, China
* Corresponding author
Received April 2018 Revised July 2018 Published February 2019
Fund Project: The author is supported by NSF grant 11801089, Postdoctoral Science Foundation of China(N0.2018M643281).
We consider a two-species Lotka-Volterra weak competition model in a one-dimensional advective homogeneous environment, where individuals are exposed to unidirectional flow. It is assumed that two species have the same population dynamics but different diffusion rates, advection rates and intensities of competition. We study the following useful scenarios: (1) if one species disperses by random diffusion only and the other assumes both random and unidirectional movements, two species will coexist; (2) if two species are drifting along the different direction, two species will coexist; (3) if the intensities of inter-specific competition are small enough, two species will coexist; (4) if the intensities of inter-specific competition are close to 1, the competitive exclusion principle holds. These results provide a new mechanism for the coexistence of competing species. Finally, we apply a perturbation argument to illustrate that two species will converge to the unique coexistence steady state.
Citation: De Tang. Dynamical behavior for a Lotka-Volterra weak competition system in advective homogeneous environment. Discrete & Continuous Dynamical Systems - B, 2019, 24 (9) : 4913-4928. doi: 10.3934/dcdsb.2019037
##### References:
show all references
##### References:
[1] S. Sadeghi, H. Jafari, S. Nemati. Solving fractional Advection-diffusion equation using Genocchi operational matrix based on Atangana-Baleanu derivative. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020435 [2] Weiwei Liu, Jinliang Wang, Yuming Chen. Threshold dynamics of a delayed nonlocal reaction-diffusion cholera model. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020316 [3] Abdelghafour Atlas, Mostafa Bendahmane, Fahd Karami, Driss Meskine, Omar Oubbih. A nonlinear fractional reaction-diffusion system applied to image denoising and decomposition. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020321 [4] Leilei Wei, Yinnian He. A fully discrete local discontinuous Galerkin method with the generalized numerical flux to solve the tempered fractional reaction-diffusion equation. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020319 [5] H. M. Srivastava, H. I. Abdel-Gawad, Khaled Mohammed Saad. Oscillatory states and patterns formation in a two-cell cubic autocatalytic reaction-diffusion model subjected to the Dirichlet conditions. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020433 [6] Lin Shi, Xuemin Wang, Dingshi Li. Limiting behavior of non-autonomous stochastic reaction-diffusion equations with colored noise on unbounded thin domains. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5367-5386. doi: 10.3934/cpaa.2020242 [7] Federico Rodriguez Hertz, Zhiren Wang. On $\epsilon$-escaping trajectories in homogeneous spaces. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 329-357. doi: 10.3934/dcds.2020365 [8] Soniya Singh, Sumit Arora, Manil T. Mohan, Jaydev Dabas. Approximate controllability of second order impulsive systems with state-dependent delay in Banach spaces. Evolution Equations & Control Theory, 2020 doi: 10.3934/eect.2020103 [9] Reza Lotfi, Zahra Yadegari, Seyed Hossein Hosseini, Amir Hossein Khameneh, Erfan Babaee Tirkolaee, Gerhard-Wilhelm Weber. A robust time-cost-quality-energy-environment trade-off with resource-constrained in project management: A case study for a bridge construction project. Journal of Industrial & Management Optimization, 2020 doi: 10.3934/jimo.2020158 [10] Scipio Cuccagna, Masaya Maeda. A survey on asymptotic stability of ground states of nonlinear Schrödinger equations II. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020450 [11] Reza Chaharpashlou, Abdon Atangana, Reza Saadati. On the fuzzy stability results for fractional stochastic Volterra integral equation. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020432 [12] Serena Dipierro, Benedetta Pellacci, Enrico Valdinoci, Gianmaria Verzini. Time-fractional equations with reaction terms: Fundamental solutions and asymptotics. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 257-275. doi: 10.3934/dcds.2020137 [13] Pierre-Etienne Druet. A theory of generalised solutions for ideal gas mixtures with Maxwell-Stefan diffusion. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020458 [14] Xin-Guang Yang, Lu Li, Xingjie Yan, Ling Ding. The structure and stability of pullback attractors for 3D Brinkman-Forchheimer equation with delay. Electronic Research Archive, 2020, 28 (4) : 1395-1418. doi: 10.3934/era.2020074 [15] Chao Xing, Jiaojiao Pan, Hong Luo. Stability and dynamic transition of a toxin-producing phytoplankton-zooplankton model with additional food. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2020275 [16] A. M. Elaiw, N. H. AlShamrani, A. Abdel-Aty, H. Dutta. Stability analysis of a general HIV dynamics model with multi-stages of infected cells and two routes of infection. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020441 [17] Hai-Feng Huo, Shi-Ke Hu, Hong Xiang. Traveling wave solution for a diffusion SEIR epidemic model with self-protection and treatment. Electronic Research Archive, , () : -. doi: 10.3934/era.2020118
2019 Impact Factor: 1.27
|
2020-11-25 00:36:46
|
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|
https://tex.stackexchange.com/questions/374750/context-combining-footnotes-with-margin-texts
|
# ConTeXt: Combining footnotes with margin texts
Background
In my document I use modified footnotes that are defined like this:
\def\mya#1{%
\tfx\high{a#1}%
}
\definenote[aNote]
[before={\blank[4mm]},
after={\blank[-8mm]},
textcommand=\mya,
paragraph=yes,
rule=off]
\setupnotation[aNote]
[alternative=serried,
before={a},
numbercommand=\tf,
style={\switchtobodyfont[8.5pt]},
way=bysection,
which results in a block of footnotes without a footnote mark in the margin of the page.
Problem:
I would like to combine the footnote mechanism of ConTeXt with the possibility to annotate the footnotes via margin texts. Unfortunately, my attempts to implement this, for instance with
\aNote{This is my text.\inmargin{And this is an annotation.}}
or
\aNote{This is my text.\margindata[inouter]{And this is an annotation.}}
don't yield any result. The footnote is displayed, but not the margin text. I assume this has something to do with the footnote mark (maybe the space in the footer's margin is reserved for footnote marks only?).
Another approach I have tried is to put a footnote inside a footnote, like so:
\def\mya#1{%
\tfx\high{a#1}%
}
\definenote[aNote]
[before={\blank[4mm]},
after={\blank[-8mm]},
textcommand=\mya,
paragraph=yes,
rule=off]
\setupnotation[aNote]
[alternative=serried,
before={a},
numbercommand=\tf,
style={\switchtobodyfont[8.5pt]},
way=bysection,
\definenote[marginNote]
\setupnote[marginNote][location=none]
\setupnotation[marginNote][
align={flushleft,bottom},
location=bottom,
numbercommand=\gobbleoneargument
]
\setuptexttexts[margin][]
[{\framed[%
align={right,bottom},
frame=off,
width=\rightmarginwidth
]{\placenotes[marginNote]}}]
\starttext
This is my main text.\aNote{Footnote text.\marginNote{Something additional.}}
\stoptext
which is close to what I want to achieve, but the text in \marginNote should be displayed in the same line as the footnote it refers to.
Question:
Is there a way to combine these two mechanisms so that I get an annotated footnote? Or can my "footnote inside a footnote" example be altered in a way so that the second note it is in the same line as the first footnote?
Minimal working example
\def\mya#1{%
\tfx\high{a#1}%
}
\definenote[aNote]
[before={\blank[4mm]},
after={\blank[-8mm]},
textcommand=\mya,
paragraph=yes,
rule=off]
\setupnotation[aNote]
[alternative=serried,
before={a},
numbercommand= {\tf},
style={\switchtobodyfont[8.5pt]},
way=bysection,
|
2020-06-01 06:29:04
|
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http://binancecryptocurrency-exchanges.bravpaimumiro.tk/?qa=8054&qa_1=identifying-key-leakage-of-bitcoin-users-springerlink
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# Identifying Key Leakage of Bitcoin Users | SpringerLink
A satoshi is defined to be such that one Bitcoin (BTC) equals $$10^$$ satoshis. The purpose of a transaction is to spend outputs by creating new ones, which represents the money flow. An output $$o_ \in T_$$ carries a value, which is the number of satoshis that this output is worth. Inputs and outputs are therefore uniquely identified by the ID of the transaction which contains them and their index in the input list and output list, respectively. An output can only be referenced once, and the outputs in the blockchain which have not been referenced at any given moment in time is called the set of unspent outputs . Transaction fees will be paid to the miners, which thus prioritize transactions based on their fees, i.e., the higher the fee, the faster the transaction will be mined. To do this, every input $$i_ \in T_$$ uniquely references an output of another previous transaction, i.e., the ones which will be spent, and creates new outputs that can be spent by future transactions. Every transaction carries an implicit transaction fee , which is the difference between the sum of the values of the outputs and the sum of the value of the referenced outputs. That is, before a miner mines a block, they will first create a coinbase transaction which will be put in the block and rewards them with Bitcoins. Special transactions without any inputs referencing other outputs are so-called coinbase transactions and are created when a block is mined to reward the miner, which is how Bitcoins are initially created. Transaction fees are an essential economical element of the Bitcoin network and change constantly depending on the number of transactions in the mempool and how much peers are willing to pay the miners. A Bitcoin transaction T consists of a sequence of inputs $$T_ = [i_, \ldots , i_]$$ and a sequence of outputs $$T_ = [o_, \ldots , o_]$$ and is uniquely identified by a transaction ID , which is generated by computing a hash of the transaction. Since a block can only be 1 MiB in size, miners will usually consider transaction fees as a function of satoshis per byte of the transaction, i.e., the larger the transaction the larger the nominal value of the fee should be. This reward is a fixed amount, which gets halved every 210,000 blocks, plus the fees of all transactions in the block.
Bitcoin uses the Elliptic Curve Digital Signature Algorithm (ECDSA) to cryptographically secure transactions. In the context of elliptic curve cryptography, $$\mathsf$$ is a randomly chosen integer from $$\$$ and the public key $$\mathsf$$ can be derived by multiplying the generator G with $$\mathsf$$ , i.e., $$\mathsf = G\mathsf$$ . Bitcoin uses the secp256k1 curve, which is based on the equation $$= x^ + 7>$$ over the finite field $$\mathbb _$$ with the 256-bit prime number . Furthermore, secp256k1 uses a generator point G with the 256-bit group order $$n = 2^ - \mathtt$$ , i.e., n is the smallest number such that $$Gn = 0$$ . The scheme is based on the computational infeasibility assumption of solving the Elliptic Curve Discrete Logarithm Problem (ECDLP), i.e., given two points Q and Qk on the curve, there is no polynomial-time algorithm for recovering k . This derivation is considered secure, as recovering $$\mathsf$$ from $$\mathsf$$ would require solving ECDLP. To create and verify signatures, we need the notion of a secret key $$\mathsf$$ and a public key $$\mathsf$$ .
However, the trend is on the verge of breaking below the pattern. We could expect Ethereum to weaken further against BTC as Bitcoin dominance continues to rise," the note said. "The daily chart for ETH-BTC is moving along an ascending channel pattern.
This demonstrates that an attacker can cause significant financial loss with relatively simple means. This is amplified by the fact that an attacker could expand this methodology to other cryptocurrencies and OSINT platforms.
That’s an easy thing to overlook in the U.S., where you could bring suitcases full of cash through customs as long as you fill out a form for sums over \$10,000. Because Bitcoin isn’t run by a central bank, it can cross international boundaries much more easily and quickly. An easier way to send money across borders.
Субтитры: Английский, Арабский, Французский, Португальский (Европа), Итальянский, Вьетнамский, Корейский, Немецкий, Русский, Испанский.
These scripts can perform arithmetic, cryptography, flow control and so on. Every input and output contains a script, which is often referred to as $$\mathsf$$ and $$\mathsf$$ , respectively. Transactions in the Bitcoin network are verified by using a small stack-based language, the programs of which are called scripts . In order for a transaction to be valid, one must concatenate the $$\mathsf$$ of each input with the $$\mathsf$$ of its referenced output, which yields a new set of scripts, i.e., one for each input. In this context, every user has a secret key $$\mathsf$$ and crypto a public key $$\mathsf$$ . All of these scripts are then evaluated, and for the transaction to be valid, there must be only one element on the stack after evaluation and this element must be equal to true . The scripting language contains special instructions for elliptic curve cryptography, which is used within this scripting framework to cryptographically secure transactions. The most prevalent type of transaction is called a Pay To Pubkey Hash (P2PKH) transaction. The $$\mathsf$$ can therefore be considered a means of protection, i.e., cryptocurrency one can only redeem an output if they can provide a correct $$\mathsf$$ . Additionally, the script verifies a signature, which means that a working $$\mathsf$$ must provide both the public key $$\mathsf$$ as well as a valid signature that can be verified with $$\mathsf$$ , which means that the sender must know $$\mathsf$$ . Outputs belonging to such transactions have a $$\mathsf$$ that verifies that the sender of the transaction possesses the correct public key by comparing it against a hash If you have any concerns with regards to in which and how to use crypto, you can speak to us at the webpage. .
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2022-09-30 00:53:57
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https://planetmath.org/CalculatingTheNthRootsOfAComplexNumber
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# calculating the nth roots of a complex number
Fix a complex number $z$. We wish to compute all the nth roots of $z$. By definition, $w$ is an nth root of $z$ if $w^{n}=z$.
First of all, if $z=0$, it is clear that all its nth roots will be zero as well.
Suppose $z$ is not zero. Then we can write $z=re^{i\theta}$, for some positive real number $r$ and some real number $\theta$ (this is de Moivre’s theorem). In fact, we have a choice of values for $\theta$: $re^{i(\theta+2k\pi)}=re^{i\theta}$ for every integer $k$. Usually, we choose $\theta$ so that $-\pi<\theta\leq\pi$.
What are the possible values for $w$? Write $w$ in polar form also, as $w=\rho e^{i\phi}$. Then $w^{n}=\rho^{n}e^{in\phi}$. We are looking for values of $\rho$ and $\phi$ so that $\rho^{n}e^{in\phi}=re^{i\theta}$. Since every nonzero complex number can be written in polar form in a unique way with $\rho>0$ and $-\pi<\phi\leq\pi$, we can assume that this is true for $w$. So for $w^{n}$ to equal $z$, we must have $\rho^{n}=r$ and $n\phi=\theta+2k\pi$ for some integer $k$. The first of these conditions is that $\rho$ be the usual (positive) nth root of the real number $r$. The second, rewritten, says that $\phi=\theta/n+2k\pi/n$ for some integer $k$. There will be exactly $n$ possibilities for $k$ which yield $-\pi<\phi\leq\pi$: $-n/2.
Summarizing, if
$z=re^{i\theta}\text{ for }-\pi<\theta\leq\pi,$
and
$w^{n}=z,$
then
$w=\sqrt[n]{r}e^{i\left(\frac{\theta}{n}+\frac{2k}{n}\pi\right)}\text{ for some% k such that }-\frac{n}{2}
Of course, the restriction on the values of $k$ is designed to ensure that none of the values obtained for different $k$ are actually equal; we could have chosen a different range of values for $k$: in books, you most often see $0\leq k, which still ensures that the values are all distinct but does not ensure that they are between $-\pi$ and $\pi$.
Thinking about what this means in polar coordinates, this means that the angles between the nth roots are exactly $1/n$ of a complete circle, so that they form the vertices of a regular polygon.
We can write the nth roots of a complex number in another way. First, apply the above expression to compute the nth roots of $1$:
$\omega_{k}=e^{i\frac{2\pi}{n}k}=\left(e^{i\frac{2\pi}{n}}\right)^{k}=\omega_{0% }^{k}.$
Then observe that if $w^{n}=z$, then $(\omega_{k}w)^{n}=\omega_{k}^{n}w^{n}=w^{n}=z$. So if $w$ is any nth root of $z$, the nth roots of $z$ can also be written as
$\omega_{k}w\text{ for }0\leq k\
or
$\omega_{0}^{k}w\text{ for }0\leq k\
This last way of writing the nth roots of a complex number shows that somehow the nth roots of $1$ already capture the unusual behaviour of the nth roots of any number. So in fact, one often wants to look at the roots of unity in any field, whether it is the integers modulo a prime, rational functions, or some more exotic field.
Title calculating the nth roots of a complex number CalculatingTheNthRootsOfAComplexNumber 2013-03-22 14:13:42 2013-03-22 14:13:42 archibal (4430) archibal (4430) 10 archibal (4430) Example msc 12D99 msc 30-00 TopicEntryOnComplexAnalysis BinomialEquation
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2018-11-14 20:23:13
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https://www.snapxam.com/problems/12886830/integral-of-x-1-sinxdx
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# Integrate (x+1)sin(x)
## \int\left(x+1\right)\sin\left(x\right)dx
Go!
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0
x
y
(◻)
◻/◻
2
e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
$\sin\left(x\right)-\cos\left(x\right)-x\cos\left(x\right)+C_0$
## Step by step solution
Problem
$\int\left(x+1\right)\sin\left(x\right)dx$
1
Use the integration by parts theorem to calculate the integral $\int\left(1+x\right)\sin\left(x\right)dx$, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=\left(1+x\right)}\\ \displaystyle{du=dx}\end{matrix}$
3
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=\sin\left(x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(x\right)dx}\end{matrix}$
4
Solve the integral
$v=\int\sin\left(x\right)dx$
5
Apply the integral of the sine function
$-\cos\left(x\right)$
6
Now replace the values of $u$, $du$ and $v$ in the last formula
$\int\cos\left(x\right)dx-\left(1+x\right)\cos\left(x\right)$
7
Apply the integral of the cosine function
$\sin\left(x\right)-\left(1+x\right)\cos\left(x\right)$
8
Multiply $\left(x+1\right)$ by $-1$
$\sin\left(x\right)+\left(-x-1\right)\cos\left(x\right)$
9
Multiplying polynomials $\cos\left(x\right)$ and $-x+-1$
$\sin\left(x\right)-\cos\left(x\right)-x\cos\left(x\right)$
10
$\sin\left(x\right)-\cos\left(x\right)-x\cos\left(x\right)+C_0$
$\sin\left(x\right)-\cos\left(x\right)-x\cos\left(x\right)+C_0$
### Struggling with math?
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### Main topic:
Integration by parts
0.29 seconds
122
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2018-11-17 14:51:39
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https://www.hackmath.net/en/math-problem/633?tag_id=57
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# Cube zoom
How many percents do we increase the volume and surface of the cube if we magnify its edge by 38 %?
Result
Δ V = 162.8 %
Δ S = 90.4 %
#### Solution:
$\Delta V = 100 \dfrac{V_2-V_1}{V_1} = 100 ( V_2/V_1-1) \ \\ \Delta V = 100\cdot ( (1+\dfrac{ 38}{100})^3-1) = 162.8 \%$
$\Delta S = 100 \dfrac{S_2-S_1}{S_1} = 100 ( S_2/S_1-1) \ \\ \Delta S = 100\cdot ( (1+\dfrac{ 38}{100})^2-1) = 90.4 \%$
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2020-02-23 02:27:36
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https://plainmath.net/algebra-ii/27510-the-value-of-expression-log-8-64
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Cabiolab
2021-09-13
The value of expression ${\mathrm{log}}_{8}64$
avortarF
Formula:
The property of logarithm ${\mathrm{log}}_{{a}^{n}}{a}^{m}=\frac{m}{n}$
Calculation:
To find the value of the given expression we need to use the rules of logarithmic function. In this problem we need to use the rule ${\mathrm{log}}_{{a}^{n}}{a}^{m}=\frac{m}{n}$
Using the below rule.
${\mathrm{log}}_{{a}^{n}}{a}^{m}=\frac{m}{n}$
${\mathrm{log}}_{8}64={\mathrm{log}}_{{8}^{1}}{8}^{2}$
Here, $n=1$ and $m=2$
By applying the above rule,
${\mathrm{log}}_{{a}^{n}}{a}^{m}=\frac{m}{n}$
${\mathrm{log}}_{8}64={\mathrm{log}}_{{8}^{1}}{8}^{2}$
Here, $n=1$ and $m=2$
${\mathrm{log}}_{{a}^{n}}{a}^{m}={\mathrm{log}}_{8}64$
${\mathrm{log}}_{8}64={\mathrm{log}}_{{8}^{1}}{8}^{2}$
$\frac{m}{n}=\frac{2}{1}$
Thus, the value of $\frac{m}{n}=\frac{2}{1}$.
Conclusion: The value of the given expression ${\mathrm{log}}_{8}64$ is 2.
Do you have a similar question?
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2023-03-24 02:12:50
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http://bagustris.blogspot.com/2012/09/file-sharing-and-remote-deksto-ubuntu.html
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Sunday, September 16, 2012
File sharing and remote desktop Ubuntu laptops via SSH
Today, I realize that connecting two or more Ubuntu laptops is very easy. Just install openssh using command line:
sudo apt-get install openssh-server
Installed ssh on your PC you can make file sharing and remote desktop to your other PCs. To connecting your PCs, use the ssh command with the username@ipaddress. For example, now I am on my laptop namely bta@g580, I want to remote desktop to my other Ubuntu laptop, namely bta@d725 with ipaddress : 192.168.3.5. So, I do the following commands,
bta@g580:~$ssh bagus@192.168.3.5 bagus@192.168.3.5's password: Welcome to Ubuntu 12.04.1 LTS (GNU/Linux 3.2.0-29-generic i686) * Documentation: https://help.ubuntu.com/ 15 packages can be updated. 9 updates are security updates. *** /dev/sda7 will be checked for errors at next reboot *** Last login: Sun Sep 16 14:23:28 2012 from bagus-lenovo-g580.local bta@d725:~$
Now, I am log on to my bta@d725 from bta@g580. I can manage file, copying, merging, deleting and do other tasks in my D725 from G580 (it is model of my laptop). To copy file from my d725 to my g580, I just type the following command;
scp ~/* bagus@192.168.3.3:~/
which copy all my files in home directory in d725 to my home directory in g580. The last question maybe: how can I know my ipaddress? the answer is, check it using command: ifconfig. You can do this file sharing and remote desktop over wifi or wired connection.
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2023-01-26 22:31:01
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http://stevenbritton.net/projection-vs-reality/
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Now that the election is over, and I have had an opportunity to decompress a little bit from my disgust and dismay at seeing yet another person named Trudeau elected to the PMO to ruin our country, it’s time to put on my analytical hat and see how well my Patented Projection Prognosticator worked towards the election this time around.
First, the qualitative:
Projected outcome: Liberal Victory Actual outcome: Liberal Victory
Okay, good so far. Canada is worse-off for it, but my projection was accurate.
Projected Parliament: Liberal minority Actual Parliament: Liberal majority
Okay, not so good there. Majority vs Minority is a huge miss, but, to be fair, the pollsters were projecting the same kind of outcome as me, and since this is an algorithm, I guess it comes down to “garbage in = garbage out.”
Digging a bit deeper, recall how I offered a set of possible results: Assured, Probable, Possible, and Long Shot. Comparing those numbers we get this:
Comparison: Projected results vs Actual Results
(Expand the Items to view the tables)
[simnor_tabs]
[simnor_tab label=”National”][table id=E42-Seats-Nat /]
[/simnor_tab][simnor_tab label=”Atlantic”][accordion openfirst=”true”]
[accordion-item title=”Newfoundland and Labrador”][table id=E42-Seats-10 /]
[/accordion-item][accordion-item title=”Prince Edward Island”][table id=E42-Seats-11 /]
[/accordion-item][accordion-item title=”Nova Scotia”][table id=E42-Seats-12 /]
[/accordion-item][accordion-item title=”New Brunswick”][table id=E42-Seats-13 /]
[/accordion-item][/accordion]
[/simnor_tab][simnor_tab label=”Quebec”]
[table id=E42-Seats-24 /][/simnor_tab][simnor_tab label=”Ontario”]
[table id=E42-Seats-35 /][/simnor_tab][simnor_tab label=”Prairies”]
[accordion openfirst=”true”]
[accordion-item title=”Manitoba”]
[table id=E42-Seats-46 /][/accordion-item]
[table id=E42-Seats-47 /][/accordion-item]
[accordion-item title=”Alberta”][table id=E42-Seats-48 /][/accordion-item]
[/accordion][/simnor_tab][simnor_tab label=”British Columbia”]
[table id=E42-Seats-59 /][/simnor_tab][simnor_tab label=”North”]
[accordion openfirst=”true”]
[accordion-item title=”Yukon”]
[table id=E42-Seats-60 /][/accordion-item]
[accordion-item title=”Northwest Territories”]
[table id=E42-Seats-61 /][/accordion-item]
[accordion-item title=”Nunavut”]
[table id=E42-Seats-62 /][/accordion-item]
[/accordion][/simnor_tab][/simnor_tabs]
Assured: The leading candidate in the projected result is further ahead than any other candidate beyond the poll’s margin of error.
Probable: The leading candidate in the projected result is further ahead of at least one other candidate by less than the poll’s margin of error.
Possible: The candidate for a given party has a range of possible support where at least some of the range shows a possible win.
Long Shot: Same as “Possible” however the candidate is ranked behind at least one other candidate with a possible win.
Relative Error measures how far from the actual result the prediction was. A negative number means the projection was low, and a positive number means it was high. It is calculated using this formula, which eliminates division by zero errors: $latex Error_{relative}=\frac{projected – actual}{338 + actual}\times 100\%$
As you can see, the results were all within the range that I projected for each party.
This means that, while the Uniform Distribution Method of predicting seats (look it up), is a good estimation tool, it still can only be as good as the polling data available.
To get a better look at the accuracy of the predictor, let’s compare the popular vote in each region against the projected vote. To do this, we will have to compare percentages rather than actual votes, because the number of votes cast between 2011 and 2015 is different.
Remember as well, the projection is based on aggregating four polls, all released on October 17:
Ekos, Forum, Mainstreet, Ipsos, so the projected result for each riding below will be based on that aggregation, with the aggregated margin of error, calculated this way:
$latex \sigma(region) _{total}= \sqrt{\sigma(region)_{Ekos}^2 + \sigma(region)_{Forum}^2 + \sigma(region)_{Mainstreet}^2 + \sigma(region)_{Ipsos}^2} &s=2$
The aggregate support for each region is calculated like this.
Assume the Polls are identified like this:
1. Ekos
2. Forum
3. Mainstreet
4. Ipsos
In reality it doesn’t really matter what order the polls are in, just that they all get included…
$latex Percent Support_{region,aggregate} = \frac{\sum\limits_{Poll = 1}^4(Percent Support_{(party,region) Poll} \times Sample Size_{Region,Poll})}{\sum\limits_{Poll = 1}^4 (SampleSize_{region,Poll})} &s=2$
This gives us the aggregated percent support for each party in each region, from which we can then figure out the results for each individual constituency.
For the purposes of the Prediction Machine, when aggregating the results, I had to make sure that the results for each polling firm were consolidated into like regions, which was a quick and simple calculation.
So we end up with aggregated data for:
Atlantic
Quebec
Ontario
Alberta
British Columbia/Northwest Territories/Nunavut/Yukon Territory.
The territories really have no polling data to speak of, since their population is so sparse, so I just grouped them into the BC results. There wasn’t much impact either way.
So, with that in mind, here’s the regional results compared to the aggregated projection I posted on October 17th, for each party, along with the error in the seat projections.
[simnor_tabs]
[simnor_tab label=”National”]
[table id=E42-Vote-National /][/simnor_tab][simnor_tab label=”Atlantic”]
[table id=E42-Vote-Atlantic /][/simnor_tab][simnor_tab label=”Quebec”]
[table id=E42-Vote-Quebec /][/simnor_tab][simnor_tab label=”Ontario”]
[table id=E42-Vote-Mnsk /][/simnor_tab][simnor_tab label=”Alberta”]
[table id=E42-Vote-Alberta /][/simnor_tab][simnor_tab label=”British Columbia/Territories”]
[table id=E42-Vote-bc /][/simnor_tab]
[/simnor_tabs]
Each polling firm has weighted the data to ensure that the demographics are represented appropriately. I have used the weighted data for this calculation, and, as well, I have included “leaners” with the data, when the polling firm has done so.
What we see in this table is the relative error in the projected support is less than 4.05% in all cases. That’s pretty good – and means the projected support is fairly accurate compared to the aggregated polls. So, what we need to do is look at the actual riding-by-riding results, and determine where the prediction machine went wrong.
Overall, the Prediction Machine was accurate in 265 of 338 constituencies. That’s 78.4%. It’s pretty good; but let’s look a bit closer, and examine where things went wrong.
Because we’re dealing with 338 constituencies, or Electoral Districts, as Elections Canada likes to call them, it’s probably easier to break things down into groups. Above, I talked about Assured, ProbablePossible, and Long Shot. Starting with Assured, which are seats the Prediction Machine suggested would be virtually guaranteed to be won by the projected party, we see this:
Out of 109 possible assured predictions, the Prediction Machine was accurate 106 times. That’s 97.25% accuracy, which is an extremely good result. But it’s also not that exciting – because an assured result means the predicted winner is further ahead than anyone else by more than the poll’s (or aggregated polls’) margin of error for the region.
So, where did the Prediction Machine get it wrong?
[simnor_tabs][simnor_tab label=”Markham-Unionville”][table id=E42-Assured-35056 /][/simnor_tab][simnor_tab label=”Nickel Belt”][table id=E42-Assured-35069 /][/simnor_tab][simnor_tab label=”Toronto-Danforth”][table id=E42-Assured-35109 /][/simnor_tab]
[/simnor_tabs]
So, when it comes to the three ridings where the Prediction Machine missed an assured result, it came from one of two possible reasons – popularity of a specific candidate, or greater shifts in support that the polling firms didn’t pick up. That isn’t necessarily the fault of the polling firms, because things were moving very quickly; but what that says to me is momentum is very important in elections, and in this particular election, the Liberal party, unfortunately, caught some momentum and it snowballed, picking up NDP supporters along the way.
Now, we can look at the probable results. Probable, remember, is defined as the case where the second-placed candidate’s support is projected within the projected winner’s margin of error. Out of 229 probable candidates, the Prediction Machine got it wrong 70 times, which is an accuracy of 69.43%. That’s not too bad, but not great either. Certainly, it’s lower than I’d like it to be.
[table id=ELX42-Probable-All /]
Looking at the results where the Prediction Machine missed on the Probable winner, we see that, just like with the Assured results, turnout was, generally, higher than 2011, the Liberal support surged at the last minute at the cost of both the Conservatives and the NDP.
For completeness, I’ll include two more tables: One for ridings where not the Possible candidate didn’t win, and another where not even the Long Shot contender won. In that last circumstance, it represents where the Prediction Machine was completely wrong.
[table id=ELX42-Possible-All /]
[table id=ELX42-LongShot-All /]
So, what does it all mean?
Well, the bottom line is this – first, the Prediction Machine, I think, is probably as accurate as it can be. There may be a few tweaks to the prediction formula; and I’ll experiment with them in time. But overall, we are dealing with a program that tries to model an election from a less-than-perfectly-accurate survey, or aggregate of surveys, of the population.
It’s a prediction based on a prediction.
80% accuracy, I have to say, is pretty good, considering the variables at play here, and, I think, in the majority of election campaigns, I think I would have probably called the result pretty accurately. Most predictions were for a Liberal minority as well, so, while I was wrong in calling the size of the Parliament, I think the actual results I put forth were quite accurate.
So, I think I can safely say that my Prediction Machine will call the election with about 80% accuracy; meaning that 270 out of the 338 seats will be predicted correctly. That’s means that, out of my projected results, there are 2,678,521,876,251,498,576,491,365,815,949,827,268,919,378,444,242,535,468,031,460,203,168,021,780 possible ways 270 seats are accurately projected. That may seem like a ridiculously big number, but consider we’re dealing with cominatorial mathematics here.
It doesn’t really matter, however , how many possible outcomes there are with 270 accurately-predicted seats, only the one that the Prediction Machine kicks out is important; compared to the actual result from election day. In that context — and we’ll have to wait up to four years to see — I expect that I’ll be calling the election result within margin of error and with 80% accuracy again.
Steven Britton
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2022-01-20 01:08:37
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http://math.stackexchange.com/questions/286640/find-all-solution-to-the-equation-y-aybx-for-given-a-and-bx
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# find all solution to the equation $y'=Ay+b(x)$ for given $A$ and $b(x)$
I am asked to find all solution to the equation: $$y'= \left( \begin{array}{cc} 13 & 12 \\ 12 & 13 \end{array} \right) y+ \left(\begin{array}{c} x\\ 0 \end{array} \right)$$ No initial condition is specified.
My working so far: If we write the DE as $y'=Ay+b(x)$ then a solution to the homogeneous DE $y'=Ay$ is given by: $$y_h=e^{A(x-x_0)}y_0$$ where $y_0$ is a vector containing the arbitrary initial conditions. Because no initial conditions are specified could I just assume that $x_0=0$? I then probably need to use variation of parameters to find a general solution but I am unsure how to do this for a system of linear differential equations. Also, how do I calculate $e^{A(x-x_0)}$, is $A$ diagonalizable? If so I could calculate it. My linear algebra is not too great so I got stuck here too. Any help would be appreciated. Thanks!
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Do you know how to find the Eigenvalues and Eigenvectors of a matrix?
To find the eigenvalues, you solve for the roots of the characteristic polynomial using $(\mathbf{A} - \lambda \mathbf{I}) = 0$?
$$\mathbf{A} = \begin{bmatrix} 13 & 12\\ 12 & 13 \end{bmatrix}$$
We form $(\mathbf{A} - \lambda \mathbf{I}) = 0$, so
$$(\mathbf{A} - \lambda \mathbf{I}) = \begin{bmatrix} 13- \lambda & 12\\ 12 & 13 - \lambda \end{bmatrix} = 0$$
$$(13 - \lambda)^{2} - 144 = 0, \text{so}, \lambda_{1,2} = 1, 25$$
To find the corresponding eigenvectors, you substitute each distinct (if they are not distinct, other approaches are needed) eigenvalue into and by solving $(\mathbf{A} - \lambda_i \mathbf{I})\mathbf{x} = 0$.
For this example, you would get:
$$\lambda_1 = 1, v1 = (1, 1)$$
$$\lambda_2 = 25, v2 = (-1, 1)$$
You could also approach this using the Jordan Normal Form and many other ways too.
Can you take it from here?
Regards
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Thanks, yeah I know about the eigenvalues and eigenvectors. So my homogeneous solution would be $y_h=c_1(1,1)^T+c_2(-1,1)^T$ right? How do I go from this to the general solution to the non-homogeneous solution? thx – Slugger Jan 25 at 16:30
See Section 3. You are forgetting the exponential terms. Regards – Amzoti Jan 25 at 16:33
How'd this go with no upvotes for so long...I made one +1 dent in that! ;-) – amWhy May 6 at 1:47
@amWhy: Not even after a followup from the OP! :-) Thanks my friend. – Amzoti May 6 at 1:50
Take $x_0=0$ and $y_0=(c_1,c_2)^t\in\mathbb{R}^2$. As $A$ is real and symmetric, is diagonalizable in $\mathbb{R}$.
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2013-12-22 12:17:08
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https://brilliant.org/problems/just-a-problem-2/
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# An algebra problem by Mauren Mark Mediante
Algebra Level 3
Let AB and CD be the smallest possible two consecutive prime numbers with digits A, B, C and D, such that the value of $A+B+C+D$ is a prime number (A and C may be equal). Find the value of $(A+C)^2 + (B+D)^2 + A + D$.
×
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2017-10-19 22:04:56
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http://physics.stackexchange.com/questions/18142/gravitational-torque-about-a-bolt-that-a-mass-is-hanging-from
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# Gravitational torque about a bolt that a mass is hanging from [closed]
A uniform rectangle sign h=20.0cm high and w=11.0cm wide loses three of its four support bolts(at points p_1, P_3, and p_4) and rotates into the position as shown, with p_1 directly over p_3. It is supported by the bolt P_2, which is so tight it holds the sign from further rotation. Find the gravitational torque about p_2, if the mass of the sign is 5.0kg.
I couldn't get my image to appear but here is what is looks like: http://www.cramster.com/answers-apr-10/physics/physics-1-uniform-rectangular-sign-16-cm-high-andw-16-cm-wide_816165.aspx?rec=0
I am not really sure where to even begin on this problem so any input would be great! Thanks
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Welcome to Physics.SE! This looks rather like it violates the prohibition in our FAQ against "Do my homework" type questions. It there a concept that is eluding you here? – dmckee♦ Dec 11 '11 at 19:39 dmckee is right, we expect people asking homework questions to focus on the specific physical concept that is causing problems. If you can edit the question to do so, we'll be happy to reopen it. – David Zaslavsky♦ Dec 12 '11 at 0:14
## closed as off topic by David Zaslavsky♦Dec 12 '11 at 0:13
Questions on Physics Stack Exchange are expected to relate to physics within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here.
Since the problem ask about torque, try using its definition: $\tau = r\,F\,\sin\theta$. You should find out what are the appropriate $r$, $F$, and $\theta$ for your problem, of course.
Well I have a few questions. When I work it out using the formula I get 4.9Nm and that's not right. The answer is 4.7Nm. So here are my calculations I was wondering if you saw anything wrong with them. F=mg=5kg*9.8m/s=49N r=.5*distance from p_2 to p_4 which using h and w I get to be 0.11413m. Then theta I think what I am supposed to do is shift the axis, means that I have the adj side to be 0.055m(half the base)/r. The I take the inverse cosine of that and get theta=61.19 degrees. But when I plug all these values in I am off by 0.2 in my answer and am not sure why. – ksmith Dec 11 '11 at 19:43 Check the angle you are trying to get. $\theta$ should be the angle between $r$ and $F$. Drawing the rectangle at scale might help you. – Arnoques Dec 11 '11 at 20:29
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2013-05-19 04:24:00
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https://physics.stackexchange.com/questions/546558/crossing-symmetry-for-particles-with-spin
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# Crossing Symmetry for Particles with Spin
I'd like to understand crossing symmetry in QFT better. I've only found somewhat detailed treatments of the scalar case, somewhat contradictory comments for the spin-1/2 case, and nothing so far for the spin-1 case.
Broadly speaking, does anyone have some good references for a rigorous treatment of the crossing symmetry of particles with non-zero spin?
Speaking of the spin-1/2 case, Peskin writes somewhat cryptically below Eq. 5.68 that when crossing spin-1/2 particles one gets an extra minus, but that "[t]he minus sign can be compensated by changing our phase convention for $$v(k)$$."
Weinberg, on the other hand, suggests that the minus sign comes from Fermi statistics and continues that "crossing symmetry is not an ordinary symmetry (it involves an analytic continuation in kinematic variables) and it is difficult to use it with any precision for general processes." (!)
So, in particular, what would the explicit spinors $$u(p)$$ and $$v(p)$$ be for Peskin under the different sign convention?
Further, is crossing symmetry trivial for spin-1 particles like it is for scalars?
Thank you very much for your help!
## General concepts
Crossing symmetry is basically the CPT theorem applied in the context of the LSZ formula, using microcausality to re-order the field operators. The role of the CPT theorem is to relate particles to their antiparticles. The CPT transform is not unique: in particular, we can compose it with any proper Poincaré transform to get another equally-good CPT transform. The CPT theorem defines a conjugate relation between two sets of single-particle states (like between the set of all single-electron states and the set of all single-positron states), but it doesn't define a unique one-to-one relationship between individual single-particle states. So questions like "does the antiparticle of a spin-up electron have spin up or spin down?" don't have unique answers. The answer is convention-dependent.
Crossing symmetry involves replacing an incoming particle with an outgoing antiparticle (or conversely), and since the relationship between individual single-particle and single-antiparticle states is convention-dependent, we can compensate for a minus sign that comes from Fermi statistics by switching conventions, as Peskin & Schroeder wrote.
Crossing symmetry is not an "ordinary symmetry" that relates physical states to other physical states, and maybe this limits its utility as Weinberg suggested, but neither of these points contradict what Peskin & Schroeder wrote.
Starting with this correlation function, we can use the LSZ reduction formula to construct a scattering amplitude in which the particle associated with $$\psi$$ is either in the initial state or in the final state. CPT says that the single-particle part of the state $$\psi_a(y)|0\ra$$ is an antiparticle of the single-particle part of the state $$\la 0|\psi_a(y)$$, or equivalently of the state $$\dpsi_a(y)|0\ra$$. The idea behind LSZ is that we can isolate the desired single-particle contributions to the in/out states by isolating the associated poles. The field operator $$\psi_a$$ can be written as the sum of its positive- and negative-frequency parts, $$\psi_a(y)=\psi_a^+(y)+\psi_a^-(y)$$, which act on a state-vector (ket) to their right as annihilation and creation operators, respectively, and conversely when acting on a state-vecctor (bra) to their left. The LSZ formula uses this to select one of the two poles, either incoming or outgoing. The identitities $$\big(\psi_a^+\big)^\dagger = \big(\dpsi_a\big)^- \hskip2cm \big(\psi_a^-\big)^\dagger = \big(\dpsi_a\big)^+$$ say that the particles corresponding to these two poles are antiparticles of each other. Crossing symmetry amounts to a relationship between the formulas that LSZ uses to select either of these two poles. So in general, what crossing symmetry does to the crossed particle's spin-state is determined by the relationship between the single-particle parts of $$\psi_a|0\ra$$ and $$\la 0|\psi_a$$.
## References for special cases
is crossing symmetry trivial for spin-1 particles like it is for scalars?
Crossing symmetry for spin-1 particles (like photons) doesn't have any minus signs from Fermi statistics, but the amplitudes still involve specific components of the field operators (photon polarizations). Equations (13.5.1)-(13.5.9) in Weinberg give a photon example.
Section 2.1 in https://arxiv.org/abs/1605.06111 gives some convention-dependent details for the case of a "vector, Dirac, and left- or right-handed (massless) Weyl representation respectively" with a footnote that says "the overall sign that relates $$u^\sigma$$ with $$v^{-\sigma}$$ ... is conventional since it depends on the choice of the CPT phase." Section 3 in the same paper shows some detailed examples for various spins, for both massive and massless particles.
Itzykson and Zuber's book Quantum Field Theory also works out an example involving crossing symmetry in a process involving electrons and photons (section 5-2-2). They also show a detailed derivation of the LSZ formula for Dirac fermions (section 5-1-6) from which the details of crossing symmetry can be inferred, and it illustrates the general concepts outlined above.
• Thank you very much for this answer! I especially appreciate the linked paper, which was very helpful. In terms of "is crossing symmetry trivial for spin-1 particles like it is for scalars?" I meant: are there any minus signs that show up when crossing spin 1 particles? My understanding is that no, there aren't any extra minus signs (beyond possibly coming from evaluating the polarization vectors at minus the momentum). In terms of Weinberg, my sense is he actually missed the boat here; i.e. that one can use crossing symmetry with precision – WAH May 4 '20 at 13:04
• I believe you've sufficiently answered the question, so I'll accept it and award the bounty. Thank you very, very much! Briefly, do you by any chance know an example of a convention for fermions such that the minus sign is compensated (or know of somewhere where this is done)? – WAH May 4 '20 at 13:05
• @WAH You are very welcome. I think P&S meant that we can compensate for a Fermi-statistics minus sign by switching conventions -- I mean, by using two different conventions on the two different sides of the crossing-symmetry equation. I don't know of a reference that shows an explicit example of that, but I did another reference to the answer (Itzykson and Zuber's book), because sometimes having multiple perspectives from multiple references can be helpful. – Chiral Anomaly May 4 '20 at 14:16
• Very nice additional reference. I'd looked through I&Z before, but didn't catch their nice photon/fermion example. My guess is that P&S mean that one could choose to use a different explicit realization of spinors such that, e.g., $u^\sigma(-p) = -v^{-\sigma}(p)$. I'm not clever enough at the moment to cook up such a representation, though – WAH May 4 '20 at 18:17
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2021-08-01 23:16:29
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https://www.octopus-code.org/documentation/12/variables/scf/convergence/convrelev/
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# ConvRelEv
A - B - C - D - E - F - G - H - I - K - L - M - N - O - P - Q - R - S - T - U - V - W - X
#### ConvRelEv
Section SCF::Convergence
Type float
Default 0.0
Relative convergence of the sum of the eigenvalues:
$\varepsilon = \frac{ \left| \sum_{j=1}^{N_{occ}} ( \varepsilon_j^{out} - \varepsilon_j^{inp} ) \right|} {\left| \sum_{j=1}^{N_{occ}} \varepsilon_j^{out} \right|}$
If this criterion is used, the SCF loop will only stop once it is fulfilled for two consecutive iterations.
Source information
Featured in tutorials
Featured in chapters of the manual:
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2023-03-23 16:45:05
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https://superuser.com/questions/1341997/using-a-uwp-api-namespace-in-powershell
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# Using a UWP API Namespace in PowerShell
I was looking for how to use a namespace for working with the Windows 10 lock screen in PowerShell and came across this answer: https://superuser.com/a/1062551/700258, however it doesn't say anything about how to import or add that namespace to PowerShell for use. I tried looking for the referenced DLL files for the assemblies and they weren't on my computer. When I see they are part of the Windows Desktop Extensions API, I went out and downloaded the Windows 10 SDK, but the DLL files were not within that either. How can I use this LockScreen Class from the Windows.System.UserProfile namespace in a PowerShell script?
## 1 Answer
First you need to tell PowerShell that you want to use a UWP class:
[Windows.System.UserProfile.LockScreen,Windows.System.UserProfile,ContentType=WindowsRuntime] | Out-Null
The first part is the class name, the second is the UWP namespace, and the third just says that it's a UWP class. After the type is loaded, you can refer to the type by its name (just the first part: [Windows.System.UserProfile.LockScreen] in this case.)
The next trick is that Windows Runtime methods are asynchronous and use a different async task class than .NET Framework methods. Calling them from PowerShell requires a little extra infrastructure that I originally developed for another answer:
Add-Type -AssemblyName System.Runtime.WindowsRuntime
$asTaskGeneric = ([System.WindowsRuntimeSystemExtensions].GetMethods() | ? {$_.Name -eq 'AsTask' -and $_.GetParameters().Count -eq 1 -and$_.GetParameters()[0].ParameterType.Name -eq 'IAsyncOperation1' })[0]
Function Await($WinRtTask,$ResultType) {
$asTask =$asTaskGeneric.MakeGenericMethod($ResultType)$netTask = $asTask.Invoke($null, @($WinRtTask))$netTask.Wait(-1) | Out-Null
$netTask.Result } Function AwaitAction($WinRtAction) {
$asTask = ([System.WindowsRuntimeSystemExtensions].GetMethods() | ? {$_.Name -eq 'AsTask' -and $_.GetParameters().Count -eq 1 -and !$_.IsGenericMethod })[0]
$netTask =$asTask.Invoke($null, @($WinRtAction))
$netTask.Wait(-1) | Out-Null } Await can be used to call functions that return an IAsyncOperation, i.e. those that produce a value. It takes the WinRT task object and the type of the output. AwaitAction can be used to call functions that return an IAsyncAction, i.e. those that just do something without returning a result. It takes only the WinRT task object. For this application, we're going to need the StorageFile type accessible too: [Windows.Storage.StorageFile,Windows.Storage,ContentType=WindowsRuntime] | Out-Null Now we can start calling some functions. First we use GetFileFromPathAsync to get an IStorageFile instance of the desired lock screen image: $image = Await ([Windows.Storage.StorageFile]::GetFileFromPathAsync('C:\path\to\image.ext')) ([Windows.Storage.StorageFile])
Finally, we pass that image to SetImageFileAsync to set the lock screen background:
AwaitAction ([Windows.System.UserProfile.LockScreen]::SetImageFileAsync(\$image))
`
Changes should take effect immediately.
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2020-01-20 10:26:57
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http://komplexify.com/blog/2010/03/09/joke-time-fall-2009/
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# Joke time: Fall 2009
Each semester, I offer students a last chance for extra credit by writing their favorite joke or riddle on their crib sheet, with extra credit assigned based purely on how funny I think it is. Although it’s a little late this year (these are from Fall 2009!), here are some of the better ones.
## Mathy contributions
Q: What did the zero say to the eight?
A: Nice belt!
Q: Why was ten afraid of seven?
A: 7 8 9!
Q: Why was 3 afraid of $pi$?
A: He was being irrational.
Q: What did the circle say to the tangent line?
A: Stop touching me!
Q: Why do you rarely find mathematicians at the beach?
A: Because they can get a tan from sine and cosine. They don’t need the sun.
Q: Why are math books always sad?
A: They have so many problems.
Q: What do you get when you add three apples to two apples?
A: A liberal arts college math problem.
Q: Why was the student’s failed exam wet?
A: It was below C level.
Q: If the natural log is $displaystyle int_1^x frac{dt}{t}$, what’s the unnatural log?
A: Duraflame.
Q: Did you here the joke about the empty set?
A: Don’t worry. It doesn’t have a point.
Q: What’s the worst part about math jokes?
A: If you get them, you probably don’t have friends.
“I heard the government wants to put a tax on the mathematically ignorant.”
“Funny… I thought that’s what the lottery was.”
Suppose we know that if you don’t study, then you’ll fail. It then follows that
$begin{array}{r@{,=,}l} mbox{no study} & mbox{fail} + mbox{study} & mbox{no fail} hline mbox{study} + mbox{no study} & mbox{fail} + mbox{no fail} mbox{study}(1 + mbox{no}) & mbox{fail}(1 + mbox{no}) therefore mbox{study} & mbox{fail} end{array}$
That is, you’re gonna fail either way. Might as well play video games.
Happy face arithmetic:
Math puns:
• What kind of undergarments does a mermaid wear? An algebra!
• What did the acorn say when it grew up? Geometry!
• What do you call a teapot boiling on Mt. Everest? Hypotenuse!
• What do you get when you divide the circumference of a pumpkin by its diameter? Pi!
What is 2 x 2?
• A junior mathematician: 4.
• A tenured mathematician: I don’t know what the answer is, but I can prove it exists.
• Physicist (after consulting technical references): Between 3.98 and 4.02.
• An engineer (after consulting a slide rule): 3.99.
• A logician: I think you need to define 2 x 2 more precisely.
• A philosopher: What you do mean by 2 x 2?
• A sociologist: I don’t know, but it was nice talking about it.
• A behavioral ecologist: A polygamous mating system.
• A college student: 4. (The when asked by astonished colleagues how he knew, replies “I memorized it.”)
I thought it was a great idea to name my child after $pi$… until the first time he misbehaved, and I had to call him by his full name.
Students nowadays are clueless about mathematics. Why, just the other day a student came into office hours asking if General Calculus was an ancient Roman war hero.
Mathematical pick-up lines:
• If you were cos2x, then I’d be sin2x so that you and I could be 1.
• I wish I was your second derivative so I could fill up your concavity.
• It’s not the magnitude of the vector, it’s how you apply the force.
• (Hacker’s pick-up line) Solve `2 u x = 106 x2 y` for `u`.
The only arithmetic a man needs in life: add the girl, subtract the clothes, divide the legs, and pray to God you don’t multiply.
A physicist, a mathematician and a computer scientist were discussing the relative merits of having a wife or a girlfriend. “For sure a girlfriend is better,” says the physicist. “You still have the freedom to experiment.” “No, no, it’s better to have a wife,” says the mathematician, “because the sense of security you get.” “No, no, you’re both wrong,” replies the computer scientist. “It’s best to have both so that when the wife thinks you’re with the mistress and the mistress thinks you’re with your wife, you can be with your computer without anyone disturbing you. ”
A professor of mathematics sent a fax to his wife: “Dear Wife, You must realize that you are 54 years old, and I have certain needs which you are no longer able to satisfy. I am otherwise happy with you as a wife, and I sincerely hope you will not be hurt or offended to learn that by the time you receive this letter, I will be at the Grand Hotel with my 18-year-old teaching assistant. I’ll be home before midnight. Your Husband.” When he arrived at the hotel, there was a faxed letter waiting for him that read as follows: “Dear Husband, You, too, are 54 years old, and by the time you receive this letter, I will be at the Breakwater Hotel with the 18-year-old pool boy. Since you are a mathematician, you will appreciate that 18 goes into 54 more times than 54 goes into 18. Therefore, don’t wait up. Your Wife.”
No wonder the mathematician’s marriage is falling apart: he’s into scientific computing… and she’s incalculable!
A mathematician is a device for turning coffee into theorems.
A computer scientist is a device for turning coffee into code.
An engineer is a device for turning coffee into urine.
An engineer thinks his equations are an approximation of reality.
A physicist thinks that reality is an approximation of his equations.
A mathematician doesn’t care.
Billy needed to integrate the function 1/(1+x). Stumped, he glanced around the class, and saw that Amy, who always got things right, had written “log(1+x)”, so he copied the answer from her. Of course, Billy was a sharp tack himself, so in order to prevent himself from being caught copying, he rewrote the answer as “timber(1+x)”.
One day a farmer called up an engineer, a physicist, and a mathematician and asked them to fence of the largest possible area with the least amount of fence. The engineer made the fence in a circle and proclaimed that he had the most efficient design. The physicist pointed out that fencing off half of the Earth was certainly a more efficient way to do it. The mathematician just laughed at them. He built a tiny fence around himself and said “I declare myself to be on the outside.”
“An engineer, a physicist, and a mathematician are staying at a hotel. That night, the engineer awakes to smell smoke. He goes out into the hallway and sees a fire, so he fills the trash can from his room with water and douses the fire (and most of the hallway too) before going back to sleep. Later, the physicists awakes to smell smoke. He goes out into the hallway and sees a fire. After a few mental calculations involving the flame velocity, water pressure, ballistic trajectory, and so forth, he fills the trash can from his room with a minimal amount of water and effectively douses the fire before going back to sleep. Later, the mathematician awakes to smell smoke. He foes out into the hallway and sees a fire. He also sees the trash can in his room and the sink in his bathroom and concludes “A solution exists” before going back to sleep.
Sherlock Holmes and Dr. Watson went on a camping trip. After a good meal and a bottle of wine they laid down for the night and went to sleep. Some hours later, Holmes awoke and nudged his faithful friend. “Watson, look up at the sky and tell me what you see.” Watson replied, “I see millions and millions of stars.” “And,” Holmes asked, “what does that tell you?” Said Watson, “Astronomically, it tells me that there are millions of galaxies and potentially billions of planets. Astrologically, I observe that Saturn is in Leo. Horologically, I deduce that the time is approximately a quarter past three. Theologically, I can see that God is all powerful and that we are but small and insignificant and finally, meteorologically, I suspect that that we will have a beautiful day for hiking tomorrow. What does it tell you, Holmes?” Holmes was silent for a minute, then spoke. “It tells me, dear Watson, that some bastard has stolen our tent.”
A mathematician is flying a 6-hour nonstop flight from California to Florida. Shortly after take-off, the pilot announces that one of the engines had to be turned off due to mechanical failure. “But don’t worry,” he adds, “we’re safe, and we’ve got three engines running perfectly. The only noticeable effect will be that our flight time will be 7 hours instead of 6.” A half-hour later, the pilot announces that a second engine had to be turned off due to mechanical failure. “But don’t worry,” he adds, “we’re safe, and we’ve still got two engines running perfectly. The only noticeable effect will be that our total flight time will be 9 hours instead of 6.” Another half-hour later, the pilot announces that a third engine had to be turned off due to mechanical failure. “But don’t worry,” he adds, “even with one engine we’re still perfectly safe. However, we’re now looking at a total flight time will of 13 hours instead of 6.” “Great,” grumbles the mathematician. “At this rate, when the next engine goes it’s going to take 19 hours to get there.”
A mathematics major is walking across campus when his classmate rides up to him on a new bicycle. “Where did you get the bike from?” he asks. “It’s a Thank you present from that freshman girl I’ve been tutoring,” the math major explains. “Yesterday she called me and told that she had passed her math final and wanted to drop by to thank me in person. She arrived at my place on her bicycle. When I had let her in, she took all her clothes off, smiled, and said to take anything I wanted!” His friend stares at him for a moment, and then replies, “Good choice. I doubt the clothes would’ve have fit.
Black holes are where God divided by zero.
Math problems? Call $displaystyle 1-800-big[(10x)(13i)^2 big] - frac{sin(x)}{2.362x} bigg|_{x=sqrt{e}}$.
The number you have dialed is imaginary. Please rotate your telephone 90 degrees and try again.
Typical student response to seeing the limit definition of the derivative for the first time:
I also got a lot of charts this time round:
## Other good ones
Q: Where did the one-legged man work?
A: IHOP.
A cop was on his horse waiting to cross the street when a little girl rode up beside him on her shiny new bike. “Nice bike you got there,” said the cop, “did you get it for Christmas?” “Yes sir,” said the little girl. The cop looked the bike over, and then handed the girl a \$5 ticket for a safety violation. “Next time,” he said, “tell Santa to put a reflector light on the back of it.” The girl looked at the ticket, and then at the cop. “Nice horse you got there,” said the girl, “did you get it for Christmas?” “Sure,” said the cop, humoring her. The girl looked the horse over. “Next time,” she said, “tell Santa that the dick goes underneath the horse, not on top.”
“Bob, how’d you get that black eye?” “Well, my wife came home yesterday after shopping for cars. She told me she wanted something that can go from 0 to 160 in 2 seconds. So I got her a bathroom scale
“Honey, I bought a new toilet brush!” “I know, dear, I know. I still prefer toilet paper, though.”
Three Chinese brothers, Bu, Chu, and Fu, came to live in the United States. They decided to change their names to acclimate to the nation. Bu changed his name to Buck, and Chu changed his name to Chuck, and Fu was deported back to China.
## And the winner is…
You know you’ve been a physics major too long when someone asks you “What’s new?” and you reply, “That’s $c$ divided by $lambda$.”
This entry was posted in nerdify. Bookmark the permalink.
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2023-01-28 20:04:42
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http://mbmlbook.com/TrueSkill_Modelling_the_outcome_of_games.html
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## 3.1 Modelling the outcome of games
Our goal is to build a system which can assess the skills of players in online gaming. As a first step towards this, we need to look at the simpler problem of predicting the outcome of a game where we already know the skills of the players involved. This will allow us to develop many of the concepts required to solve the more complex problem of determining skills.
Suppose that Jill is going to play a game of Halo against Fred on Xbox Live. In Chapter 2 we represented a person’s software development skills by using a binary variable for each skill, indicating whether the person possessed that particular skill or not. Clearly this approach is insufficient when we consider a person’s skill at a typical Xbox game such as Halo. There is a wide spectrum of possible skill levels, and it is more appropriate to represent a person’s skill using a continuous value. The first of our modelling assumptions is therefore:
1. Each player has a skill value, represented by a continuous variable.
The stronger player is not always the winner.
We denote the skill of Jill by Jskill and the skill of Fred by Fskill. Let us suppose that Fred has a skill level of Fskill = 12.5 while Jill has a skill of Jskill = 15. These numbers appear to be completely arbitrary, and the scale on which we measure skill is indeed arbitrary. What matters, however, is how the skill values compare between players, and we shall see in a moment how to give meaning to such numbers. We have given Jill a higher skill value to indicate that she is the stronger player. But now we run into the first of our challenges, which is that the stronger player in a game such as Halo is not always the winner. If Jill and Fred were to play lots of games against each other we would expect Jill to win more than half of them, but not necessarily to win them all. We can capture the variability in the outcome of a game by introducing the notion of a performance for each player, which expresses how well they played on a particular game. The player with the higher performance for a specific game will be the winner of that game. A player with a high skill level will tend to have a high performance, but their actual performance will vary from one game to another. As with skill, the performance is most naturally expressed using a continuous quantity. We denote Jill’s performance by Jperf and Fred’s performance by Fperf. Figure 3.2 shows Jperf plotted against Fperf. For points lying in the region above the diagonal line Jill is the winner, while below the diagonal line Fred is the winner.
We can think of a person’s skill as their average performance across many games. For example, Jill has a skill level of 15 so her performance will have an average value of 15 but on a particular game it might be higher or lower. Once again we have to deal with uncertainty, and we shall do this using a suitable probability distribution. We anticipate that larger departures of performance from the average will be less common, and therefore have lower probability, than values which are closer to the average. Intuitively the performance should therefore take the form of a ‘bell curve’ as illustrated in Figure 3.3 in which the probability of a given performance value falls off on either side of the skill value.
Because performance is a continuous quantity, this bell curve is an example of a probability density, which we encountered previously in Panel 2.4. Although we have sketched the general shape of the bell curve, to make further progress we need to define a specific form for this curve. There are many possible choices, but there is one which stands out as special in having some very useful mathematical properties. It is called the Gaussian probability density and is the density function for the Gaussian distribution. In fact, the Gaussian distribution has so many nice properties that it is one of the most widely used distributions in the fields of machine learning and statistics. A particular Gaussian distribution is completely characterized by just two numbers: the centre value of the distribution, known as the mean, and the standard deviation, which determines how wide the curve is. These concepts are discussed in more detail in Panel 3.1. Figure 3.4 shows a plot of the Gaussian distribution, illustrating the interpretation of the mean and the standard deviation.
To understand the scale of the values on the vertical axis of Figure 3.4, remember that the total area under a probability distribution curve must be one. Note that the distribution is symmetrical about its maximum point – because there is equal probability of being on either side of this point, the performance at this central point is also the mean performance.
In standard notation, we write the mean as $\mu$ and the standard deviation as $\sigma$. Using this notation the Gaussian density function can be written as
The left hand side says that Gaussian$(x ; \mu, \sigma^2)$ is a probability distribution over $x$ whose value is dependent on the values of $\mu$ and $\sigma$. It is often convenient to work with the square of the standard deviation, which we call the variance and which we denote by $\sigma^2$ (see Panel 3.1). We shall also sometimes use the reciprocal of the variance $\tau = 1/\sigma^2$ which is known as the precision. For the most part we shall use standard deviation since this lives on the same scale (i.e. has the same units) as $x$.
Sometimes when we are using a Gaussian distribution it will be clear which variable the distribution applies to. In such cases, we can simplify the notation and instead of writing Gaussian$(x ; \mu, \sigma^2)$ we simply write Gaussian$(\mu, \sigma^2)$. It is important to appreciate that is simply a shorthand notation and does not represent a distribution over $\mu$ and $\sigma^2$.
Now let’s see how we can apply the Gaussian distribution to model a single game of Halo between Jill and Fred. Figure 3.5 shows the Gaussian distributions which describe the probabilities of various performances being achieved by Jill and Fred in their Halo game.
Here we have chosen the standard deviations of the two Gaussians to be the same, with $\var{perfSD} = 5$ (where ’perfSD’ denotes the standard deviation of the performance distribution). We shall discuss the significance of this choice shortly. The first question we can ask is: “what is the probability that Jill will be the winner?”. Note that there is considerable overlap between the two distributions, which implies that there is a significant chance that the performance value for Fred would be higher than that for Jill and hence that he will win the game, even though Jill has a higher skill value. You can also see that if the curves were more separated (for example, if Jill had a much higher skill), then the chance of Fred winning would be reduced.
We have introduced two further assumptions into our model, and it is worth making these explicit:
1. Each player has a performance value for each game, which varies from game to game such that the average value is equal to the skill of that player. The variation in performance, which is the same for all players, is symmetrically distributed around the mean value and is more likely to be close to the mean than to be far from the mean.
2. The player with the highest performance value wins the game.
As written, assumption Assumption 2 expresses the qualitative knowledge that a domain expert in online games might possess, and corresponds to a bell-shaped performance distribution. This needs to be refined into a specific mathematical form and for this we choose the Gaussian, although we might anticipate that other bell-shaped distributions would give qualitatively similar results.
This is a good moment to introduce our first factor graph for this chapter. To construct this graph we start with the variable nodes for each random variable in our problem. So far we have two variables: the performance of Fred, which we denote by the continuous variable Fperf, and the performance for Jill, denoted by Jperf. Each of these is described by a Gaussian distribution whose mean is the skill of the corresponding player, and with a common standard deviation of 5, and therefore a variance of $5^2$:
Note that, as in section 2.6, we are using a lower-case $p$ to denote a probability density for a continuous variable, and will use an upper-case $P$ to denote the probability distribution for a discrete variable.
The other uncertain quantity is the winner of the game. For this we can use a binary variable JillWins which takes the value true if Jill is the winner and the value false if Fred is the winner. The value of this variable is determined by which of the two variables Jperf and Fperf is larger – it will be true is Jperf is larger or otherwise false. Using T for true and F for false as before, we can express this distribution by
Since probabilities sum to one, we then have
We shall refer to the conditional probability in equation (3.6) as the GreaterThan factor, which we shall denote the by ‘>’ when drawing factor graphs. Note that this is a deterministic factor since the value of the child variable is fixed if the values of both parent variables are known. Using this factor, we are now ready to draw the factor graph. This has three variable nodes, each with a corresponding factor node, and is shown in Figure 3.6.
We asked for the probability that Jill would win this game of Halo. We can find an approximate answer to this question by using ancestral sampling (which was introduced back in section 2.5). To apply ancestral sampling in our factor graph we must first sample from the parent variables Jperf and Fperf and then compute the value of the child variable Jwins.
Consider first the sampling of the performance Jperf for Jill. There are standard numerical techniques for generating random numbers having a Gaussian distribution of specified mean and variance. If we generate five samples from $\dist{Gaussian}(x;15, 5^2)$ and plot them as a histogram we obtain the result shown in Figure 3.7a. Note that we have divided the height of each bar in the histogram by the total number of samples (in this case 5) and also by the width of the histogram bins (in this case 10). This ensures that the total area under the histogram is one. If we increase the number of samples to 50, as seen in Figure 3.7b, we see that the histogram roughly approximates the bell curve of a Gaussian. By increasing the number of samples we obtain a more accurate approximation, as shown in Figure 3.7c for the case of 500 samples, and in Figure 3.7d for 5,000 such samples.
We see that we need to draw a relatively large number of samples in order to obtain a good approximation to the Gaussian. When using ancestral sampling we therefore need to use a lot of samples in order to obtain reasonably accurate results. This makes ancestral sampling computationally very inefficient, although it is a straightforward technique which provides a useful way to help understand the model or generate synthetic datasets from the model.
Having seen how to sample from a single Gaussian distribution we can now consider ancestral sampling from the complete graph in Figure 3.6 representing a single game of Halo between Jill and Fred. We first select a performance Jperf for Jill on this specific game, corresponding to the top-level variable node on the factor graph, by drawing a value from the Gaussian distribution
Independently, we choose a performance value Fperf for Fred, which is also a top-level variable node, by drawing a sample from the Gaussian distribution
We then compute the value of the remaining variable JillWins using these sampled values. This involves comparing the two performance values, and if Jperf is greater than Fperf then JillWins is true, otherwise JillWins is false. If we repeat this sampling process many times, then the fraction of times that JillWins is true gives (approximately) the probability of Jill winning a game. The larger the number of samples over which we average, the more accurate this approximation will be. Figure 3.8 shows a scatter plot of the performances of our two players across 1,000 samples.
For each game we independently select the performance of each of our two players by generating random values according to their respective Gaussian distributions. Each of these games is shown as a point on the scatter plot. Also shown is the 45-degree line along which the two performances are equal. Points lying below this line represent games in which Fred is the winner, while points lying above the line are those for which Jill is the winner. We see that the majority of points lie above the line, as we expect because Jill has a higher skill value. By simply counting the number of points we find that Jill wins 63.1% of the time.
Of course this is only an approximate estimate of the probability of Jill winning. We can find the exact result mathematically [Moser, 2010] by making use of the equation for the Gaussian distribution (3.1), which tells us that the probability of Jill being the winner is given by
Here CumGauss denotes the cumulative Gaussian function which is illustrated in Figure 3.9
Using a numerical evaluation of this function we find that the probability of Jill winning the game is 63.8%, which is close to the value (63.1%) we obtained by ancestral sampling.
We noted earlier that the scale on which skill is measured is essentially arbitrary. If we add a fixed constant onto the skills of all the players this would leave the probabilities of winning unchanged since, from equation (3.10), they depend only on the difference in skill values. Likewise, if we multiplied all the skill values by the same constant, and at the same time we multiplied the parameter perfSD by the same constant, then again the probabilities of winning would be unchanged. All that matters is the difference in skill values measured in relation to the value of perfSD.
We have now built a model which can predict the outcome of a game for two players of known skills. In the next section we will look at how to extend this model to go in the opposite direction: to predict the player skills given the outcome of one or more games.
Self assessment 3.1
The following exercises will help embed the concepts you have learned in this section. It may help to refer back to the text or to the concept summary below.
1. Write a program or create a spreadsheet which produces 10,000 samples from a Gaussian with zero mean and a standard deviation of 1 (most languages/spreadsheets have built in functions or available libraries for sampling from a Gaussian). Compute the percentage of these samples which lie between -1 and 1, between -2 and 2 and between -3 and 3. You should find that these percentages are close to those given in the caption of Figure 3.4.
2. Construct a histogram of the samples created in the previous exercise (like the ones in Figure 3.7) and verify that it resembles a bell-shaped curve.
3. Compute the mean, standard deviation and variance of your samples, referring to Panel 3.1. The mean should be close to zero and the standard deviation and variance should both be close to 1 (since $1^2=1$).
4. Produce a second set of 10,000 samples from a Gaussian with mean 1 and standard deviation 1. Plot a scatter plot like Figure 3.8 where the X co-ordinate of each point is a sample from the first set and the Y co-ordinate is the corresponding sample from the second set (pairing the first sample from each set, the second sample from each set and so on). Compute the fraction of samples which lie above the diagonal line where X=Y.
5. Using Infer.NET, create double variables X and Y with priors of Gaussian(0,1) and Gaussian(1,1) respectively. Define a third random variable Ywins equal to $\var{Y}>\var{X}$. Compute the posterior distribution over Ywins and verify that it is close to the fraction of samples above the diagonal in the previous exercise.
Gaussian distributionA specific form of probability density over a continuous variable that has many useful mathematical properties. It is governed by two parameters – the mean and the standard deviation. The mathematical definition of a Gaussian is given by equation (3.1)
meanThe average of a set of values. See Panel 3.1 for a more detailed discussion of the mean and related concepts.
standard deviationThe square root of the variance.
varianceA measure of how much a set of numbers vary around their average value. The variance, and related quantities, are discussed in Panel 3.1.
precisionThe reciprocal of the variance.
statisticsA statistic is a function of a set of data values. For instance the mean is a statistic whose value is the average of a set of values. Statistics can be useful for summarising a large data set compactly.
cumulative Gaussian functionThe value of the cumulative Gaussian function at a point $x$ is equal to the area under a zero-mean unit-variance Gaussian from minus infinity up to the point $x$. It follows from this definition that the gradient of the cumulative Gaussian function is given by the Gaussian distribution.
References
[Bishop, 2006] Bishop, C. M. (2006). Pattern Recognition and Machine Learning. Springer.
[Moser, 2010] Moser, J. (2010). The Math behind TrueSkill.
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2018-11-15 02:08:11
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https://docs.astropy.org/en/latest/api/astropy.units.beam_angular_area.html
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# beam_angular_area¶
astropy.units.beam_angular_area(beam_area)[source]
Convert between the beam unit, which is commonly used to express the area of a radio telescope resolution element, and an area on the sky. This equivalency also supports direct conversion between Jy/beam and Jy/steradian units, since that is a common operation.
Parameters
beam_areaangular area equivalent
The area of the beam in angular area units (e.g., steradians)
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2020-08-04 16:54:16
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https://www.nature.com/articles/nphys3527?error=cookies_not_supported&code=183db661-ef83-4961-9072-507ca5b01a1d
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Characterization of collective ground states in single-layer NbSe2
Abstract
Layered transition metal dichalcogenides are ideal systems for exploring the effects of dimensionality on correlated electronic phases such as charge density wave (CDW) order and superconductivity. In bulk NbSe2 a CDW sets in at TCDW = 33 K and superconductivity sets in at Tc = 7.2 K. Below Tc these electronic states coexist but their microscopic formation mechanisms remain controversial. Here we present an electronic characterization study of a single two-dimensional (2D) layer of NbSe2 by means of low-temperature scanning tunnelling microscopy/spectroscopy (STM/STS), angle-resolved photoemission spectroscopy (ARPES), and electrical transport measurements. We demonstrate that 3 × 3 CDW order in NbSe2 remains intact in two dimensions. Superconductivity also still remains in the 2D limit, but its onset temperature is depressed to 1.9 K. Our STS measurements at 5 K reveal a CDW gap of Δ = 4 meV at the Fermi energy, which is accessible by means of STS owing to the removal of bands crossing the Fermi level for a single layer. Our observations are consistent with the simplified (compared to bulk) electronic structure of single-layer NbSe2, thus providing insight into CDW formation and superconductivity in this model strongly correlated system.
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Acknowledgements
Research supported in part by the Director, Office of Energy Research, Materials Sciences and Engineering Division, of the US Department of Energy (DOE), under grant DE-AC02-05CH11231 supporting the sp2-bonded Materials Program (STM imaging and transport), and by the National Science Foundation under award #DMR-1206512 (STS spectroscopic analysis). Work at the ALS is supported by DOE BES under Contract No. DE-AC02-05CH11231. H.R. acknowledges support from Max Planck Korea/POSTECH Research Initiative of NRF, Korea. M.T.E. is supported by the ARC Laureate Fellowship project (FL120100038). A.R. acknowledges fellowship support by the Austrian Science Fund (FWF): J3026-N16.
Author information
Authors
Contributions
M.M.U. and A.J.B. conceived the work and designed the research strategy. M.M.U., A.J.B., Y.C., W.R. and M.T.E. measured and analysed the STM/STS data. Y.Z., H.R. and S.-K.M. performed the MBE growth and ARPES and LEED characterization of the samples. S.O., C.O.-A., M.M.U. and Y.C. carried out the transport experiments. H.-Z.T. and A.R. helped in the experiments. D.L. participated in the interpretation of the experimental data. Z.H. and Z.-X.S. supervised the MBE and sample characterization. A.Z. supervised the transport measurements. M.F.C. supervised the STM/STS experiments. M.M.U. wrote the paper with help from M.F.C. and A.Z. M.M.U. and M.F.C. coordinated the collaboration. All authors contributed to the scientific discussion and manuscript revisions.
Corresponding authors
Correspondence to Miguel M. Ugeda or Michael F. Crommie.
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Competing interests
The authors declare no competing financial interests.
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Ugeda, M., Bradley, A., Zhang, Y. et al. Characterization of collective ground states in single-layer NbSe2. Nature Phys 12, 92–97 (2016). https://doi.org/10.1038/nphys3527
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2021-06-23 16:14:31
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https://www.greaterwrong.com/posts/5bd75cc58225bf067037520a/universal-inductors
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# Universal Inductors
Now that the Logical Induction paper is out, I am directing my attention towards decision theory. The approach I currently think will be most fruitful is attempting to make a logically updateless version of Wei Dai’s Updateless Decision Theory. Abram Demski has posted on here about this, but I think Logical Induction provides a new angle with which we can attack the problem. This post will present an alternate way of viewing Logical Induction which I think will be especially helpful for building a logical UDT. (The Logical Induction paper is a prerequisite for this post.)
This post will make several informal claims without proof. Many of the claims are analogous to things proven in the Logical Induction paper, and the math is not substantially difficult. I might extend the paper to have a section talking about the following concepts more formally.
A Universal Inductor is an infinite sequence of probability distributions on infinite bit strings. Satisfying the following two properties:
1. The function is computable, where is a finite prefix, and is the probability that an infinite bit string begins with the prefix .
2. Consider the propositional theory whose th atom is identified with the event “the th bit in the infinite bit string is a 1.” Then the probability distribution induces an function from sentences in the language of this theory to probabilities. The sequence of probability assignments must form a Logical Inductor over the empty deductive process.
Note that a Universal Inductor corresponds to a Logical Inductor, but the associated Logical Inductor will not have finite support, and so will be different from the one constructed in the paper. Never the less, Universal Inductors can be shown to exist using a similar construction. Note that the second condition technically implies the first, but I separate them for emphasis.
Here are some facts about Universal Inductors:
1. The probability distributions converge to a limiting distribution on infinite bit strings, .
2. This limiting distribution dominates the universal semi measure. In particular, every finite prefix is given nonzero probability.
3. Given any consistent deductive process, we can construct a logical inductor over that deductive process simply by associating atoms of the theory with bits in the bit string (in an efficient computable manner), conditioning on the event corresponding to the th list in the deductive process, and taking the function from logical sentences to probabilities induced by that probability distribution. (Thus, a Universal inductor can be used to construct a Logical Inductor over PA by looking at some of the conditional probabilities, and can be used to construct a Logical Inductor over ZFC by looking at other conditional probabilities (Hence the name Universal Inductor))
The 2 major differences between the Universal Inductor and Logical Inductor formalisms are:
1. Universal Inductors are probability distributions at every finite stage, not just in the limit, so we can more naturally talk about conditional probabilities at every stage.
2. Universal Inductors are only over empty deductive processes, and we simulate nontrivial deductive processes only through conditioning.
I feel that this will especially make for a better framework for thinking about logical updatelessness.
Here is an open question about Universal Inductors:
If and are the limits of two different universal inductors, do they necessarily dominate each other? i.e. does there exist a constant such that for all prefixes ?
(This is equivalent to the analogous question about logical inductors over the same theory, but it feels more natural to state in this framework.)
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2023-02-01 00:03:53
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http://mathhelpforum.com/advanced-algebra/219840-list-all-maximal-ideals-print.html
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# list all maximal ideals
• June 14th 2013, 01:04 AM
hollywood
list all maximal ideals
I seem to have difficulty with this type of problem - perhaps a lack of intuition about prime and maximal ideals. Here's the problem. I'm having trouble with part (d).
For each of the following rings, list all maximal ideals.
(a) $\mathbb{Z}/90\mathbb{Z}$
(b) $\mathbb{Q}[x]/(x^2+1)$
(c) $\mathbb{C}[x]/(x^2+1)$
(d) $\mathbb{Q}[x]/(x^3+x^2)$
My solutions for (a) through (c):
(a) These are generated by the prime factors of 90: (2), (3), and (5).
(b) Since $x^2+1$ is irreducible in $\mathbb{Q}[x]$, this is a field, so the only maximal ideal is (0).
(c) $\mathbb{C}[x]/(x^2+1) = \mathbb{C}[x]/((x+i)(x-i)) \cong \mathbb{C}[x]/(x+i)\times\mathbb{C}[x]/(x-i) \cong \mathbb{C}\times\mathbb{C}$, so the maximal ideals are ((0,1)) and ((1,0)), which correspond to the ideals (x+i) and (x-i) in the original ring.
I think that answer is correct, but I'm not entirely clear on how I got from the ideals in $\mathbb{C}\times\mathbb{C}$ to the ideals in the original ring.
For (d), I think it's true that $\mathbb{Q}[x]/(x^3+x^2) \cong \mathbb{Q}[x]/(x^2) \times \mathbb{Q}[x]/(x+1)$, and the second is isomorphic to $\mathbb{Q}$. But what are the maximal ideals of $\mathbb{Q}[x]/(x^2)$?
- Hollywood
• June 17th 2013, 08:15 PM
Drexel28
Re: list all maximal ideals
The key to all of these is the lattice (fourth) isomorphism theorem. Namely, you know that the maximal ideals of a quotient ring $R/I$ are the maximal ideals of the form $M/I$ for $M$ a maximal ideal of $R$ containing $I$. So, let's see here:
a) Yes, as you pointed out, the maximal ideals are the principal ones generated by $2,3,5$. This follows from what I said since the maximal ideals of $\mathbb{Z}$ that contain $(90)$ are the principal ideals of the form $(p)$ where $p\mid 90$ is prime.
b) Exactly. Since $\mathbb{Q}$ is a field, you know that $\mathbb{Q}[x]$ is a PID, and since $x^2+1$ is irreducible we must have that $(x^2+1)$ is prime, and thus maximal. Thus, the only maximal ideals containing $(x^2+1)$ are the ideal itself!
c) Exactly. Since $\mathbb{C}$ is algebraically closed, you know that all the maximal ideals are the principal ones with linear generators. Since the only linear factors dividing $x^2+1$ are $x\pm i$ your answer is correct.
d) Once again, since $\mathbb{Q}[x]$ is a PID, the maximal ideals containing $(x^3+x^2)$ are the principal ideals generated by the irreducible factors of $x^3+x^2$. In other words, the maximal ideals above $(x^3+x^2)$ are the ideals $(x+1)$ and $(x)$. By the way, using this idea, you see that the maximal ideals of $\mathbb{Q}[x]/(x^2)$ is just the ideal $(x)$.
Best,
Alex
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2014-04-20 12:13:19
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https://linkml.io/linkml/schemas/slots.html
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# Slots
Slots operate the same way as “fields” in traditional object languages and the same ways as “columns” in spreadsheets and relational databases.
If you have a JSON object that is conformant to a LinkML schema, then the keys for that object must correspond to slots in the schema, that are applicable to that class.
for example, if we have an object instantiating a Person class:
{"id": "PERSON001",
"name": "....",
"email": "....",
...
then id, email, name should all be valid slots, as in the following schema:
classes:
Person:
slots:
- id
- name
- email
If we have tabular data
id
name
email
PERSON0001
then the same constraints hold.
## ranges
Each slot must have a range - if this is not declared explicitly, then default_range is used.
The range must be one of:
Examples:
slots:
gender:
slot_uri: schema:gender
range: GenderType ## range is an enum
has_medical_history:
range: MedicalEvent ## range is a class
multivalued: true
inlined_as_list: true
age_in_years:
range: integer ## range is a type
minimum_value: 0
maximum_value: 999
## slot_usage
The slot_usage slot can be used to specify how a particular slot ought to be used in a class.
For example, imagine a schema with a generic “Relationship” class:
Relationship:
slots:
- started_at_time
- ended_at_time
- related_to
- type
with subtypes such as FamilialRelationship, BusinessRelationship, etc
we can use slot_usage to constrain the meaning of more generic slots such as type and related to:
FamilialRelationship:
is_a: Relationship
slot_usage:
type:
range: FamilialRelationshipType
required: true
related to:
range: Person
required: true
## Identifiers
If a slot is declared as an identifier then it serves as a unique key for members of that class. It can also be used for inlining as a dict in JSON serializations.
slots:
id:
identifier: true
the range of an identifier can be any type, but it is a good idea to have these be of type uriorcurie
## Slot cardinality
### multivalued
The multivalued indicates that the range of the slot is a list
Example:
slots:
has_medical_history:
range: MedicalEvent
multivalued: true
inlined_as_list: true
### required
The required slot can be used to define whether a slot is required.
When a slot is declared as required, any class that uses that slot must have a value for that slot.
## inverse
The inverse slot can be used to specify the inverse predicate of a given predicate slot relationship.
parent_of:
is_a: famlially_related_to
inverse: child_of
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2022-01-18 03:47:00
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http://mathhelpforum.com/calculus/63071-calc-substitution-rule.html
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# Thread: Calc: Substitution Rule
1. ## Calc: Substitution Rule
the problem is Sin(x^3) x^2 dx
u = x^3
du = 3x^2
I noticed when my teacher was solving this out her final answer was -1/3 cos(x^3) + C
My question is where did the x^2 go and where did she pull out the 1/3 from?
2. $sin(x^3) x^2 dx$*
With the substitution you proposed you have
$u = x^3$
$\frac{du}{3} = x^2 dx$ And the expression becomes $sin(u) \frac{du}{3}$
3. I'm sorry but why is it dU/3? where did you get that?
4. $d(u) = du$
$d(u) = d(x^3) = 3x^2 dx = du$
If you divide the two last term by 3 you get
$x^2 dx = \frac{du}{3}$
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2016-08-30 08:20:31
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https://math.stackexchange.com/questions/2260306/projection-problem-interpretation
|
# Projection problem interpretation
Problem: Describe the projective transformations of $FP^n$ that preserve the $FP^{n−1}$ at infinity given by $x_0 = 0$.
My attempt: We have to preserve the points with first coordinate $0$, i.e. the transformation maps $[0, x_1, x_2...x_n]$ to $[0, y_1, ...y_n]$. I can see why we can represent $FP^n$ as the disjoint union of $F^n$ and $FP^{n-1}$ and the latter is basically what we want to preserve. Now I also know that a projective transformation is given by $v$ -> $[Tv]$, where $T:R^n$ -> $R^n$ is injective. I just can't seem to finalize the argument.
Let's work with vector spaces : if $V = F^{n+1}$ with basis $B = (e_0, e_1, \dots, e_n)$ you are asking for matrices $g \in GL(V)$ such that $g(W) = W$ where $W$ has basis $(e_1, \dots, e_n)$. Do you agree ? Do you see how to finish ?
• Hey, yup this is correct. What I found was that the first row of the matrix of transformation is $(a,0,0,....0)$. Also, that the right-down block matrix can't be 0 and needs to have determinant $\neq 0$. – asdf May 1 '17 at 15:00
• Exactly. ${{{}}}$ – user171326 May 1 '17 at 15:03
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2019-05-21 21:33:24
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https://askdev.io/questions/4371/locating-the-quadrant-having-a-factor-on-an-n-sphere
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# Locating the quadrant having a factor on an n-sphere
Suppose I have a factor $x \in \mathbb{R}^n$ on an n-sphere. Intend I separate the n-sphere right into 4 areas (I assume this makes good sense in $n$ measurements ), just how do I recognize which area $x$ pushes?
0
2019-05-07 12:18:38
Source Share
This is simply a rephrasing of ShreevastasR's solution ; no debt to me. It does make good sense to separate an $n$ - round right into quadrants, as you clarify in $\mathbb{R}^3$ : dividing by 2 coordinate aircrafts. Yet after that determining which quadrant is, as ShreevastasR claims, merely considering the indicators of the works with of $x$. If $x_1$ and also $x_2$ are both favorable, you remain in the first, $++$, quadrant ; if $x_1$ is adverse and also $x_2$ favorable, you remain in the 2nd, $-+$, quadrant. And more. If rather you dividing the round right into $2^n$ orthants, after that you take into consideration all the indicators of the works with of $x$.
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2021-04-20 07:16:16
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https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_13&curid=6462&diff=151309&oldid=147590
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# Difference between revisions of "2008 AMC 12A Problems/Problem 13"
The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.
## Problem
Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$
## Solution 1
$[asy]size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("$$30^{\circ}$$",(0.65,0.15),O); label("$$r$$",(C+D)/2,E); label("$$2r$$",(O+C)/2,NNE); label("$$O$$",O,SW); label("$$r$$",(C+F)/2,SE); label("$$R$$",(O+A)/2-(0,0.3),S); label("$$P$$",C,NW); label("$$Q$$",D,SE);[/asy]$
Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.
Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.
## Solution 2
Like in Solution 1, let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$.
Let $N$ be the tangency point of the two circles. As shown in Solution 1, $POQ = 30$ degrees, so angle $NOA$ is also $30$ degrees. Let the line tangent to the two circles at $N$ intersect $OA$ and $OB$ at points $C$ and $D$, respectively. Since line $CD$ is tangent to circles $O$ and $P$, it must be perpendicular to $ON$, meaning that angle $ONC$ must be $90$ degrees. Because angle $NOA$ is $30$ degrees, angle $DCO$ is $180-30-90 = 60$ degrees. Angle $DOC$ is also $60$ degrees, so triangle $DOC$ is equilateral.
Note that an equilateral triangle's incenter is also its centroid. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle.
Then the ratio of areas will be $\frac{1}{3}$ squared, or $\frac{1}{9}\Rightarrow \boxed{\text{B}}$.
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2021-05-17 20:33:32
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https://physics.stackexchange.com/questions/107089/what-is-the-physical-interpretation-of-the-dot-inner-scalar-product-of-two-vecto
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# What is the physical interpretation of the dot/inner/scalar product of two vectors?
What is the physical interpretation of the dot/inner/scalar product of two vectors?
See, if we multiply two scalars like 2*3 we say two times three is six. I also do understand multiplication of vectors with scalars, ie. say $x$ times a vector $\vec{a}$. But not for two vectors being multiplied together to form a scalar.
There are two kinds of vector multiplications-
Dot (Scalar) Product:
The dot product of two vectors gives a scalar, that means only the magnitude is left, no direction. Mathematically, it is equal to the product of the magnitude of two vectors times the cosine of the angle between the two. ie. $$\vec{v} \cdot \vec{u}=|\vec{v}||\vec{u}|Cos\theta$$
The geometric interpretation: The dot product of $\vec{a}$ with unit vector $\hat{u}$, denoted $\vec{a}⋅\hat{u}$, is defined to be the projection of $\vec{a}$ in the direction of $\vec{a}$, or the amount that $\vec{a}$ is pointing in the same direction as unit vector $\hat{u}$. Let's assume for a moment that $\vec{a}$ and $\hat{u}$ are pointing in similar directions. Then, you can imagine $\vec{a}⋅\hat{u}$ as the length of the shadow of $\vec{a}$ onto $\hat{u}$ if their tails were together and the sun was shining from a direction perpendicular to $\hat{u}$. By forming a right triangle with $\vec{a}$ and this shadow, you can use geometry to calculate that $$\vec{a}⋅\hat{u}=|\vec{a}|Cos\theta$$
Cross (Vector) Product:
The cross product of two vectors gives a vector, that means the answer has a magnitude and a direction. The magnitude of the resultant vector is given by the product of the magnitude of the two vectors times the sine of the angle between them. The cross product is always perpendicular to both vectors, and has magnitude zero when the vectors are parallel and maximum magnitude when they are perpendicular. $$\vec{v} \times \vec{u}=|\vec{v}||\vec{u}|Sin\theta \hat{r}$$ where $\hat{r}$ is the unit vector in the direction of the resultant vector. The direction of this can be found out using the gif below or the right hand thumb rule.
The geometrical interpretation: The magnitude of the cross product can be interpreted as the positive area of the parallelogram having $\vec{v}$ and $\vec{u}$ as sides.
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2019-10-14 16:50:01
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https://www.calculus-online.com/exercise/3564
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# Vectors – Calculate the scalar multiplication of vectors – Exercise 3564
Exercise
The vectors
$$\vec{a},\vec{b}$$
Create a 120 degree angle. Given that
$$|\vec{a}|=3, |\vec{b}|=4$$
Calculate
$$\vec{a}\cdot \vec{a}$$
$$\vec{a}\cdot \vec{b}$$
$${(\vec{a}+ \vec{a})}^2$$
$$(3\vec{a}+2\vec{b})(\vec{a}+2\vec{b})$$
$$\vec{a}\cdot \vec{a}=9$$
$$\vec{a}\cdot \vec{b}=-6$$
$${(\vec{a}+ \vec{a})}^2=13$$
$$(3\vec{a}+2\vec{b})(\vec{a}+2\vec{b})=43$$
Solution
Coming soon…
Share with Friends
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2019-11-13 20:23:56
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https://davidlowryduda.com/a-balancing-act-in-uniform-bounds-for-lattice-point-counting/
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# mixedmath
Explorations in math and programming
David Lowry-Duda
I was recently examining a technical hurdle in my project on “Uniform bounds for lattice point counting and partial sums of zeta functions” with Takashi Taniguchi and Frank Thorne. There is a version on the arxiv, but it currently has a mistake in its handling of bounds for small $X$.
In this note, I describe an aspect of this paper that I found surprising. In fact, I’ve found it continually surprising, as I’ve reproven it to myself three times now, I think. By writing this here and in my note system, I hope to perhaps remember this better.
## Landau’s Method
In this paper, we revisit an application of “Landau’s Method” to estimate partial sums of coefficients of Dirichlet series. We model this paper off of an earlier application by Chandrasakharan and Narasimhan, except that we explicitly track dependence of the several implicit constants and we prove these results uniformly for all partial sums, as opposed to sufficiently large partial sums.
The only structure is that we have a Dirichlet series $\phi(s)$, some Gamma factors $\Delta(s)$, and a functional equation of the shape $$\phi(s) \Delta(s) = \psi(s) \Delta(1-s).$$ This is relatively structureless, and correspondingly our attack is very general. We use some smoothed approximation to the sum of coefficients, shift lines of integration to pick up polar main terms, apply the functional equation and change variables so work with the dual, and then get some collection of error terms and error integrals.
It happens to be that it’s much easier to work with a $k$-Riesz smoothed approximation. That is, if $$\phi(s) = \sum_{n \geq 1} \frac{a(n)}{\lambda_n^s}$$ is our Dirichlet series, and we are interested in the partial sums $$A_0(s) = \sum_{\lambda_n \leq X} a(n),$$ then it happens to be easier to work with the smoothed approximations $$A_k(X) = \frac{1}{\Gamma(k+1)}\sum_{\lambda_n \leq X} a(n) (X - \lambda_n)^k a(n),$$ and to somehow combine several of these smoothed sums together.
This smoothed sum is recognizable as $$A_k(X) = \frac{1}{2\pi i}\int_{c - i\infty}^{c + i\infty} \phi(s) \frac{\Gamma(s)}{\Gamma(s + k + 1)} X^{s + k}ds$$ for $c$ somewhere in the half-plane of convergence of the Dirichlet series. As $k$ gets large, these integrals become better behaved. In application, one takes $k$ sufficiently large to guarantee desired convergence properties.
The process of taking several of these smoothed approximations for large $k$ together, studying them through basic functional equation methods, and combinatorially combining these smoothed approximations via finite differencing to get good estimates for the sharp sum $A_0(s)$ is roughly what I think of as “Landau’s Method”.
## Application and shape of the error
In our paper, as we apply Landau’s method, it becomes necessary to understand certain bounds coming from the dual Dirichlet series $$\psi(s) = \sum_{n \geq 1} \frac{b(n)}{\mu_n^s}.$$ Specifically, it works out that the (combinatorially finite differenced) between the $k$-smoothed sum $A_k(X)$ and its $k$-smoothed main term $S_k(X)$ can be written as $$\Delta_y^k [A_k(X) - S_k(X)] = \sum_{n \geq 1} \frac{b(n)}{\mu_n^{\delta + k}} \Delta_y^k I_k(\mu_n X),\tag{1}$$ where $\Delta_y^k$ is a finite differencing operator that we should think of as a sum of several shifts of its input function.
More precisely, $\Delta_y F(X) := F(X + y) - F(X)$, and iterating gives $$\Delta_y^k F(X) = \sum_{j = 0}^k (-1)^{k - j} {k \choose j} F(X + jy).$$ The $I_k(\cdot)$ term on the right of $(1)$ is an inverse Mellin transform $$I_k(t) = \frac{1}{2 \pi i} \int_{c - i\infty}^{c + i\infty} \frac{\Gamma(\delta - s)}{\Gamma(k + 1 + \delta - s)} \frac{\Delta(s)}{\Delta(\delta - s)} t^{\delta + k - s} ds.$$ Good control for this inverse Mellin transform yields good control of the error for the overall approximation. Via the method of finite differencing, there are two basic choices: either bound $I_k(t)$ directly, or understand bounds for $(\mu_n y)^k I_k^{(k)}(t)$ for $t \approx \mu_n X$. Here, $I_k^{(k)}(t)$ means the $k$th derivative of $I_k(t)$.
## Large input errors
In the classical application (as in the paper of CN), one worries about this asymptotic mostly as $t \to \infty$. In this region, $I_k(t)$ can be well-approximated by a $J$-Bessel function, which is sufficiently well understood in large argument to give good bounds. Similarly, $I_k^{(k)}(t)$ can be contour-shifted in a way that still ends up being well-approximated by $J$-Bessel functions.
The shape of the resulting bounds end up being that $\Delta_y^k I_k(\mu_n X)$ is bounded by either
• $(\mu_n X)^{\alpha + k(1 - \frac{1}{2A})}$, where $A$ is a fixed parameter that isn’t worth describing fully, and $\alpha$ is a bound coming from the direct bound of $I_k(t)$, or
• $(\mu_n y)^k (\mu_n X)^\beta$, where $\beta$ is a bound coming from bounding $I_k^{(k)}(t)$.
In both, there is a certain $k$-dependence that comes from the $k$-th Riesz smoothing factors, either directly (from $(\mu_n y)^k$), or via its corresponding inverse Mellin transform (in the bound from $I_k(t)$). But these are the only aspects that depend on $k$.
At this point in the classical argument, one determines when one bound is better than the other, and this happens to be something that can be done exactly, and (surprisingly) independently of $k$. Using this pair of bounds and examining what comes out the other side gives the original result.
## Small input errors
In our application, we also worry about asymptotic as $t \to 0$. While it may still be true that $I_k$ can be approximated by a $J$-Bessel function, the “well-known” asymptotics for the $J$-Bessel function behave substantially worse for small argument. Thus different methods are necessary.
It turns out that $I_k$ can be approximated in a relatively trivial way for $t \leq 1$, so the only remaining hurdle is $I_k^{(k)}(t)$ as $t \to 0$.
We’ve proved a variety of different bounds that hold in slightly different circumstances. And for each sort of bound, the next steps would be the same as before: determine when each bound is better, bound by absolute values, sum together, and then choose the various parameters to best shape the final result.
But unlike before, the boundary between the regions where $I_k$ is best bounded directly or bounded via $I_k^{(k)}$ depends on $k$. Aside from choosing $k$ sufficiently large for convergence properties (which relate to the locations of poles and growth properties of the Dirichlet series and gamma factors), any sufficiently large $k$ would suffice.
## Limiting behavior gives a heuristic region
After I step away from this paper and argument for a while and come back, I wonder about the right way to choose the balancing error. That is, I rework when to use bounds coming from studying $I_k(t)$ directly vs bounds coming from studying $I_k^{(k)}(t)$.
But it turns out that there is always a reasonable heuristic choice. Further, this heuristic gives the same choice of balancing as in the case when $t \to \infty$ (although this is not the source of the heuristic).
Making these bounds will still give bounds for $\Delta_y^k I_k(\mu_n X)$ of shape
• $(\mu_n X)^{\alpha + k(1 - \frac{1}{2A})}$, where $A$ is a fixed parameter that isn’t worth describing fully, and $\alpha$ is a bound coming from the direct bound of $I_k(t)$, or
• $(\mu_n y)^k (\mu_n X)^\beta$, where $\beta$ is a bound coming from bounding $I_k^{(k)}(t)$.
The actual bounds for $\alpha$ and $\beta$ will differ between the case of small $\mu_n X$ and large $\mu_n X$ ($J$-Bessel asymptotics for large, different contour shifting analysis for small), but in both cases it turns out that $\alpha$ and $\beta$ are independent of $k$.
This is relatively easy to see when bounding $I_k^{(k)}(t)$, as repeatedly differentiating under the integral shows essentially that $$I_k^{(k)}(t) = \frac{1}{2\pi i} \int \frac{\Delta(s)}{(\delta - s)\Delta(\delta - s)} t^{\delta - s} ds.$$ (I’ll note that the contour does vary with $k$ in a certain way that doesn’t affect the shape of the result for $t \to 0$).
When balancing the error terms $(\mu_n X)^{\alpha + k(1 - \frac{1}{2A})}$ and $(\mu_n y)^k (\mu_n X)^\beta$, the heuristic comes from taking arbitrarily large $k$. As $k \to \infty$, the point where the two error terms balance is independent of $\alpha$ and $\beta$.
This reasoning applies to the case when $\mu_n X \to \infty$ as well, and gives the same point. Coincidentally, the actual $\alpha$ and $\beta$ values we proved for $\mu_n X \to \infty$ perfectly cancel in practice, so this limiting argument is not necessary — but it does still apply!
I suppose it might be possible to add another parameter to tune in the final result — a parameter measuring deviation from the heuristic, that can be refined for any particular error bound in a region of particular interest.
But we haven’t done that.
In fact, we were slightly lossy in how we bounded $I_k^{(k)}(t)$ as $t \to 0$, and (for complicated reasons that I’ll probably also forget and reprove to myself later) the heuristic choice assuming $k \sim \infty$ and our slighly lossy bound introduce the same order of imprecision to the final result.
## More coming soon
We’re updating our preprint and will have that up soon. But as I’ve been thinking about this a lot recently, I realize there are a few other things I should note down. I intend to write more on this in the short future.
bold, italics, and plain text are allowed in comments. A reasonable subset of markdown is supported, including lists, links, and fenced code blocks. In addition, math can be formatted using $(inline math)$ or $$(your display equation)$$.
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2022-07-02 05:24:56
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https://zbmath.org/?q=an:1062.22046
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# zbMATH — the first resource for mathematics
Lie group structures on groups of diffeomorphisms and applications to CR manifolds. (English) Zbl 1062.22046
The authors obtain some sufficient geometric conditions on a CR manifold $$M$$ to guarantee that the group of all its smooth CR automorphisms has the structure of a (finite dimensional) Lie group compatible with its natural topology. First, they establish general theorems on Lie group structures for subgroups of diffeomorphisms of a given smooth real-analytic manifold, then they apply these results together with recent work concerning jet parametrization and complete systems for CR automorphisms. The authors define the notion of a complete system for a set of diffeomorphisms of a manifold and show that it is equivalent to the notion of jet parametrization. Similar results are obtained for subsets of germs of diffeomorphisms fixing a point. Some partial converse results are obtained. Next, the authors present the basic definitions and properties of abstract and embedded CR manifolds that guarantee the existence of a Lie group structure on the group of CR automorphisms.
##### MSC:
22F50 Groups as automorphisms of other structures 57S25 Groups acting on specific manifolds 58D05 Groups of diffeomorphisms and homeomorphisms as manifolds 32V40 Real submanifolds in complex manifolds 22E15 General properties and structure of real Lie groups
Full Text:
##### References:
[1] A. Andreotti & G.A. Fredricks, Embeddability of real analytic Cauchy-Riemann manifolds, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4)6 (1979) p. 285-304 · Zbl 0449.32008 [2] M.S. Baouendi, P. Ebenfelt & L.P. Rothschild, Parametrization of local biholomorphisms of real analytic hypersurfaces, Asian J. Math1 (1997) p. 1-16 · Zbl 0943.32021 [3] M.S. Baouendi, P. Ebenfelt & L.P. Rothschild, Rational dependence of smooth and analytic CR mappings on their jets, Math. Ann315 (1999) p. 205-249 · Zbl 0942.32027 [4] M.S. Baouendi, P. Ebenfelt & L.P. Rothschild, Real submanifolds in complex space and their mappings., Princeton Mathematical Series 47, Princeton University Press, 1999 · Zbl 0944.32040 [5] M.S. Baouendi, P. Ebenfelt & L.P. Rothschild, Convergence and finite determination of formal CR mappings, J. Amer. Math. Soc13 (2000) p. 697-723 · Zbl 0958.32033 [6] M.S. Baouendi, H. Jacobowitz & F. Trèves, On the analyticity of CR mappings, Ann. of Math. (2)122 (1985) p. 365-400 · Zbl 0583.32021 [7] M.S. Baouendi, N. Mir & L.P. Rothschild, Reflection ideals and mappings between generic submanifolds in complex space, J. Geom. Anal12 (2002) no. p. 543-580 · Zbl 1039.32021 [8] M.S. Baouendi, L.P. Rothschild & D. Zaitsev, Deformation of generic submanifolds in complex space (in preparation), · Zbl 1129.32019 [9] S. Bochner & D. Montgomery, Locally compact groups of differentiable transformations, Ann. of Math. (2)47 (1946) p. 639-653 · Zbl 0061.04407 [10] A. Boggess, CR manifolds and the tangential Cauchy-Riemann complex, Studies in Advanced Mathematics, CRC Press, 1991 · Zbl 0760.32001 [11] D. Burns Jr. & S. Shnider, Real hypersurfaces in complex manifolds, XXX, Amer. Math. Soc., 1977, p. 141-168 · Zbl 0422.32016 [12] E. Cartan, Sur la géométrie pseudo-conforme des hypersurfaces de deux variables complexes I, II, Œuvres II-2 (1932) p. 1217-1238 [13] S.S. Chern & J.K. Moser, Real hypersurfaces in complex manifolds, Acta Math133 (1974) p. 219-271 · Zbl 0302.32015 [14] D. Dummit & R. Foote, Abstract algebra, Prentice Hall, Inc., 1991 · Zbl 0751.00001 [15] P. Ebenfelt, Finite jet determination of holomorphic mappings at the boundary., Asian J. Math.5 (2001) p. 637-662 · Zbl 1015.32031 [16] P. Ebenfelt, B. Lamel & D. Zaitsev, Finite jet determination of local analytic CR automorphisms and their parametrization by $$2$$-jets in the finite type case, Geom. Funct. Anal.13 (2003) no. p. 546-573 · Zbl 1032.32025 [17] M. Golubitsky & V. Guillemin, Stable mappings and their singularities, Graduate Texts in Mathematics Vol. 14, Springer-Verlag, 1973 · Zbl 0294.58004 [18] C.-K. Han, Complete differential system for the mappings of CR manifolds of nondegenerate Levi forms, Math. Ann309 (1997) p. 401-409 · Zbl 0892.32015 [19] S.-Y. Kim & D. Zaitsev, Equivalence and embedding problems for CR-structures of any codimension, Preprint, 2002 · Zbl 1079.32022 [20] S.-Y. Kim & D. Zaitsev, Remarks on the rigidity of CR-manifolds (in preparation), · Zbl 1101.32018 [21] S. Kobayashi, Transformation groups in differential geometry., Ergebnisse der Mathematik und ihrer Grenzgebiete Band 70, Springer-Verlag, 1972 · Zbl 0246.53031 [22] R.T. Kowalski, Rational jet dependence of formal equivalences between real-analytic hypersurfaces in $$\mathbb{C}^2,$$ e-print. To appear, Pacific J. Math, http://arXiv.org/abs/math.CV/0108165, 2001 · Zbl 1106.32025 [23] N. Tanaka, On generalized graded Lie algebras and geometric structures I, J. Math. Soc. Japan19 (1967) p. 215-254 · Zbl 0165.56002 [24] A.E. Tumanov, Extension of CR-functions into a wedge from a manifold of finite type (Russian), Mat. Sb. (N.S.)136(178) (1988) p. 128-139 · Zbl 0692.58005 [25] V.S. Varadarajan, Lie groups, Lie algebras, and their representations, Prentice-Hall Series in Modern Analysis, Prentice-Hall, Inc., 1974 · Zbl 0371.22001 [26] D. Zaitsev, On the automorphism groups of algebraic bounded domains, Math. Ann302 (1995) p. 105-129 · Zbl 0823.14005 [27] D. Zaitsev, Germs of local automorphisms of real analytic CR structures and analytic dependence on the $$k$$-jets, Math. Res. Lett4 (1997) p. 823-842 · Zbl 0898.32006 [28] A.E. Tumanov, Extension of CR-functions into a wedge from a manifold of finite type, Math. USSR-Sb. (translation)64 (1989) p. 129-140 · Zbl 0692.58005
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-01-20 17:51:59
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https://tinygp.readthedocs.io/en/stable/tutorials/means.html
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Hide code cell content
try:
import tinygp
except ImportError:
%pip install -q tinygp
try:
import jaxopt
except ImportError:
%pip install -q jaxopt
# Fitting a Mean Function#
It is quite common in the GP literature to (“without lack of generality”) set the mean of the process to zero and call it a day. In practice, however, it is often useful to fit for the parameters of a mean model at the same time as the GP parameters. In some other tutorials, we fit for a constant mean value using the mean argument to tinygp.GaussianProcess, but in this tutorial we walk through an example for how you might go about fitting a model with a non-trival parameterized mean function.
For our example, we’ll fit for the location, width, and amplitude of the following model:
$f(x) = b + a\,\exp\left(-\frac{(x - \ell)^2}{2\,w^2}\right)$
In jax, we might implement such a function as follows:
from functools import partial
import numpy as np
import matplotlib.pyplot as plt
import jax
import jax.numpy as jnp
jax.config.update("jax_enable_x64", True)
def mean_function(params, X):
mod = jnp.exp(
-0.5 * jnp.square((X - params["loc"]) / jnp.exp(params["log_width"]))
)
beta = jnp.array([1, mod])
return params["amps"] @ beta
mean_params = {
"amps": np.array([0.1, 0.3]),
"loc": 5.0,
"log_width": np.log(0.5),
}
X_grid = np.linspace(0, 10, 200)
model = jax.vmap(partial(mean_function, mean_params))(X_grid)
plt.plot(X_grid, model)
plt.xlabel("x")
plt.ylabel("y")
_ = plt.title("a parametric mean model")
WARNING:absl:No GPU/TPU found, falling back to CPU. (Set TF_CPP_MIN_LOG_LEVEL=0 and rerun for more info.)
Our implementation here is somewhat artificially complicated in order to highlight one very important technical point: we must define our mean function to operate on a single input coordinate. What that means is that we don’t need to worry about broadcasting and stuff within our mean function: tinygp will do all the necessary vmap-ing. More explicitly, if we try to call our mean_function on a vector of inputs, it will fail with a strange error (yeah, I know that we could write it in a way that would work, but I’m trying to make a point!):
model = mean_function(mean_params, X_grid)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
/tmp/ipykernel_722/1001441932.py in <module>
----> 1 model = mean_function(mean_params, X_grid)
/tmp/ipykernel_722/2053350467.py in mean_function(params, X)
12 -0.5 * jnp.square((X - params["loc"]) / jnp.exp(params["log_width"]))
13 )
---> 14 beta = jnp.array([1, mod])
15 return params["amps"] @ beta
16
~/checkouts/readthedocs.org/user_builds/tinygp/envs/stable/lib/python3.8/site-packages/jax/_src/numpy/lax_numpy.py in array(object, dtype, copy, order, ndmin)
1897 elif isinstance(object, (list, tuple)):
1898 if object:
-> 1899 out = stack([asarray(elt, dtype=dtype) for elt in object])
1900 else:
1901 out = np.array([], dtype=dtype)
~/checkouts/readthedocs.org/user_builds/tinygp/envs/stable/lib/python3.8/site-packages/jax/_src/numpy/lax_numpy.py in stack(arrays, axis, out, dtype)
1637 for a in arrays:
1638 if shape(a) != shape0:
-> 1639 raise ValueError("All input arrays must have the same shape.")
1640 new_arrays.append(expand_dims(a, axis))
1641 return concatenate(new_arrays, axis=axis, dtype=dtype)
ValueError: All input arrays must have the same shape.
Instead, we need to manually vmap as follows:
model = jax.vmap(partial(mean_function, mean_params))(X_grid)
## Simulated data#
Now that we have this mean function defined, let’s make some fake data that could benefit from a joint GP + mean function fit. In this case, we’ll add a background trend that’s not included in the mean model, as well as some noise.
random = np.random.default_rng(135)
X = np.sort(random.uniform(0, 10, 50))
y = jax.vmap(partial(mean_function, mean_params))(X)
y += 0.1 * np.sin(2 * np.pi * (X - 5) / 10.0)
y += 0.03 * random.normal(size=len(X))
plt.plot(X, y, ".k")
plt.xlabel("x")
plt.ylabel("y")
_ = plt.title("simulated data")
## The fit#
Then, we set up the usual infrastructure to calculate the loss function for this model. In this case, you’ll notice that we’ve stacked the mean and GP parameters into one dictionary, but that isn’t the only way you could do it. You’ll also notice that we’re passing a partially evaluated version of the mean function to our GP object, but we’re not doing any vmap-ing. That’s because tinygp is expecting the mean function to operate on a single input coordinate, and it will handle the appropriate mapping.
from tinygp import kernels, GaussianProcess
def build_gp(params):
kernel = jnp.exp(params["log_gp_amp"]) * kernels.Matern52(
jnp.exp(params["log_gp_scale"])
)
return GaussianProcess(
kernel,
X,
diag=jnp.exp(params["log_gp_diag"]),
mean=partial(mean_function, params),
)
@jax.jit
def loss(params):
gp = build_gp(params)
return -gp.log_probability(y)
params = dict(
log_gp_amp=np.log(0.1),
log_gp_scale=np.log(3.0),
log_gp_diag=np.log(0.03),
**mean_params
)
loss(params)
DeviceArray(-33.08457135, dtype=float64)
We can minimize the loss using jaxopt:
import jaxopt
solver = jaxopt.ScipyMinimize(fun=loss)
soln = solver.run(jax.tree_util.tree_map(jnp.asarray, params))
print(f"Final negative log likelihood: {soln.state.fun_val}")
Final negative log likelihood: -99.26464900290118
## Visualizing result#
And then plot the conditional distribution:
gp = build_gp(soln.params)
_, cond = gp.condition(y, X_grid)
mu = cond.loc
std = np.sqrt(cond.variance)
plt.plot(X, y, ".k", label="data")
plt.plot(X_grid, mu, label="model")
plt.fill_between(X_grid, mu + std, mu - std, color="C0", alpha=0.3)
plt.xlim(X_grid.min(), X_grid.max())
plt.xlabel("x")
plt.ylabel("y")
_ = plt.legend()
That looks pretty good but, when working with mean functions, it is often useful to separate the mean model and GP predictions when plotting the conditional. The interface for doing this in tinygp is not its most ergonomic feature, but it shouldn’t be too onerous. To compute the conditional distribution, without the mean function included, call tinygp.GaussianProcess.condition() with the include_mean=False flag:
gp = build_gp(soln.params)
_, cond = gp.condition(y, X_grid, include_mean=False)
mu = cond.loc + soln.params["amps"][0]
std = np.sqrt(cond.variance)
plt.plot(X, y, ".k", label="data")
plt.plot(X_grid, mu, label="GP model")
plt.fill_between(X_grid, mu + std, mu - std, color="C0", alpha=0.3)
plt.plot(X_grid, jax.vmap(gp.mean_function)(X_grid), label="mean model")
plt.xlim(X_grid.min(), X_grid.max())
plt.xlabel("x")
plt.ylabel("y")
_ = plt.legend()
There is one other subtlety that you may notice here: we added the mean model’s zero point (params["amps"][0]) to the GP prediction. If we had left this off, the blue line in the above figure would be offset below the data by about 0.1, and it’s pretty common that you’ll end up with a workflow like this when visualizing the results of GP fits with non-trivial means.
## An alternative workflow#
Sometimes it can be easier to manage all the mean function bookkeeping yourself, and instead of using the mean argument to tinygp.GaussianProcess, you could instead manually subtract the mean function from the data before calling tinygp.GaussianProcess.log_probability(). Here’s how you might implement such a workflow for our example:
vmapped_mean_function = jax.vmap(mean_function, in_axes=(None, 0))
def build_gp_v2(params):
kernel = jnp.exp(params["log_gp_amp"]) * kernels.Matern52(
jnp.exp(params["log_gp_scale"])
)
return GaussianProcess(kernel, X, diag=jnp.exp(params["log_gp_diag"]))
@jax.jit
def loss_v2(params):
gp = build_gp_v2(params)
return -gp.log_probability(y - vmapped_mean_function(params, X))
loss_v2(params)
DeviceArray(-33.08457135, dtype=float64)
In this case, we are now responsible for making sure that the mean function is properly broadcasted, and we must not forget to also subtract the mean function when calling tinygp.GaussianProcess.condition().
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2023-03-27 19:44:20
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https://ikrima.dev/ue4guide/engine-programming/blueprints/bp-compiler-overview/
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# Blueprint Compilation#
## Terminology#
• FKismetCompilerContext: The class that does the work of compilation. A new instance is spawned for each compile. Stores reference to the class being compiled, the blueprint, etc.
• FKismetFunctionContext: Holds the information for compiling a single function, like a reference to the associated graph, properties, and generated UFunction
• FNodeHandlingFunctor: A helper class that handles processing one class of node in the compiler (singleton!). Contains functions for registering pin connections, and generating Kismet compiled statements
• FKismetCompiledStatement: Unit of work in the Kismet compiler. The compiler translates nodes into a set of compiled statements, which the backend translates into bytecode operations Examples: Variable assignment, Goto, Call
• FKismetTerm: A terminal in the graph (literal, const, or variable reference). Each data pin connection is associated with one of these! You can also make your own in NodeHandlingFunctors for scratch variables, intermediate results, etc.
• Data-Only Blueprints are not recompiled when loaded
• Two Classes
• Generated Class: The class used by instances of the blueprint. Contains all the UProperties and UFunctions defined by the blueprint, as well as generated script bytecode
• Skeleton Generated Class: A minimal version of the generated class, which contains all the UProperties and UFunctions defined by the blueprint, but with no script bytecode
• Why Two Classes?
• The Skeleton Generated Class (SGC) is used internally by the blueprint editor Serves as a “class header,” which the editor uses to create editing UI
• Kept up-to-date as you edit the blueprint via a quick compile path (MarkBlueprintAsStructurallyModified())
• Never used or referenced outside of the blueprint itself
• Currently serialized, but will soon become transient
## Compilation Process#
1. Clean the Class - CleanAndSanitizeClass(): we move the properties and functions off the class and into a trash class in the transient package, and clear any data on the class
2. Create Class Properties - Iterates over the blueprint’s NewVariables array, as well as some other places (construction scripts, etc) to find all of the UProperties needed by the class - Creates UProperties on the Uclass’s scope in the function CreateClassVariablesFromBlueprint()
3. Create Function List - Process the event graphs
• CreateAndProcessUberGraph()
• Copies all event graphs into one big graph
• Nodes are given a chance to expand
• For each Event node in the graph, a function stub is created
• FKismetFunctionContext is created
• Process regular function graphs
• ProcessOneFunctionGraph()
• Graph is duplicated to a temporary graph
• Nodes are given a chance to expand
• FKismetFunctionContext is created
• After all FKismetFunctionContexts are created, PrecompileFunctions():
• Schedules execution and calculates data dependencies
• Prunes any nodes that are unscheduled or not a data dependency
• Runs the node handler’s RegisterNets() on each remaining node
• This creates the FKismetTerms for values within the function
• Creates the UFunction and associated UProperties
4. Bind and Link Class - Now that we know all of our UProperties and UFunctions, we can bind and link the class
• Fills out the property chain, the property size, function map, etc
• At this point, we essentially have a “class header,” just missing final flags and metadata, as well as a CDO
5. Compile Functions - Compiling consists of transforming all the nodes that are left in the graph into FKismetCompiledStatments - Accomplished through the node handler’s Compile() function, using AppendStatementForNode() - Can create FKismetTerms in the compile function, as long as they are only used locally (intermediate results)
6. Finish Compiling Class - Finalizes the class flags - Propagates flags and metadata from the parent class - Performs a few final checks to make sure everything went alright in the compile
7. Backends Emit Generated Code - Backends convert the collection of FKismetCompiledStatments from each function context into code - We have two backends in use: - FKismetCompilerVMBackend
• Converts FKCS to UnrealScript VM code
• Results of this are serialized into the function’s script
• FKismetCppBackend
• Emits “C++-like” code for debugging purposes
• May eventually transform into a legitimate backend, if extra speed is needed
8. Copy Class Default Object Properties - Using a special function, CopyPropertiesForUnrelatedObjects(), or CPFUO for short, we copy the values from the old CDO of the class into the new CDO
• Properties are copied via tagged serialization, so as long as the names are consistent, they should properly be transferred
• Components of the CDO are reinstanced, and fixed up appropriately at this stage
• Currently, the SkeletonGeneratedClass has the authoratative copy of data, which is propagated to the GeneratedClass
9. Reinstancing - Since the class may have changed size and properties may have been added or removed, we need to reinstance all objects with the class we just compiled - Use a TObjectIterator to find all instances of the class, spawn a new one, and then CPFUO to copy from the old instance to the new one
• For details, see the FBlueprintCompileReinstancer
## Extension Points#
• Compiler Expansion
• Very similar to macros, but done internally to the compiler
• Take the original node, and expand it into a set of new
• Move connections from the original node to the new
• Takes place after the graphs have been copied to their transient versions, but before nodes have all been processed
• Allows new entry points to be created, for event graph nodes
• Useful when you need some dynamic behavior in your delegates
• Variable number of inputs or outputs
• Reliance on some data
• Check out FKismetCompilerContext::ExpansionStep() for examples of its use
• In general, the original node is disconnected from everything after expansion, and gets culled
• This means you don’t need to implement a custom FNodeHandlingFunctor class for expanded nodes
• Can expand conditionally based on if the current compile is a skeleton-only compile
• We use this internally for many node types
• Blueprint spawning nodes
• Timeline nodes
• Delegate assignment
• Macros actually get expanded in this step as well!
• Compiler Generated Code
• Most advanced way to create new functionality in Kismet
• Because of the complexity, we use it sparingly
• Requires implementation of a custom FNodeHandlingFunctor class
• RegisterNets() gets called once per node, and calls RegisterNet() on each pin that needs it (data pins)
• Compile() is called once per node, and creates a list of FKismetCompiledStatements, which are then translated to code by the backends
• For examples, see the custom FNodeHandlingFunctor classes in KismetCompiler.cpp
• Debugging Tools
• CompileDisplaysTextBackend – Shows a human-readable, C++-like output of what the code generator is doing
• CompileDisplaysBinaryBackend – Shows the actual disassembed USVM code
Last update: March 6, 2020
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2021-05-14 16:52:51
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https://stat.ethz.ch/pipermail/r-help/2011-July/282992.html
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[R] Generalized Logistic and Richards Curve
Ben Bolker bbolker at gmail.com
Thu Jul 7 15:49:04 CEST 2011
Vincy Pyne <vincy_pyne <at> yahoo.ca> writes:
> Dear R helpers, I am not a statistician and right now struggling
> with Richards curve. Wikipedia says
> (http://en.wikipedia.org/wiki/Generalised_logistic_function) The
> "generalized logistic curve or function", also known as Richard's
> curve is a widely-used and flexible sigmoid function for growth
> modelling, extending the well-known logistic curve. Now I am
> confused and will like to know if the Generalized Logistic
> distribution as described in lmomco package is same as what
> wikipedia is describing. In other words, is Generalized Logistic
> Function same as Generalized logistic distribution? I do understand
> there is separate R package "richards' for dealing with Richards
> curve. Kindly guide Vincy [[alternative HTML version deleted]]
I think not quite. In general it's unlikely that something described as a
"function" will necessarily be the same as something described as a
"distribution", since the latter (or at least its density function) has to
integrate to 1 and the former
doesn't ...
Looking at 'cdfglo' in the lmomco manual gives
F(x)=1/(1+exp(-y))
y = -k^{-1} log(1-k(x-xi)/alpha) (for k not equal 0)
whereas wikipedia gives
Y(t) = A + { K-A \over (1 + Q e^{-B(t - M)}) ^ {1 / \nu} }
In order to carefully check whether these are the same (e.g.
whether the density function (rather than the CDF which is given
in the lmomco manual) is the same as the Richards curve) you would
have to match up terms.
One tip-off that they can't be identical is that 'cdfglo' has
3 parameters (location parameter xi, scale parameter alpha, shape
parameter k) while the Richards has 5 (location M, scale B,
scale ? Q, shape nu, lower value A, upper value K).
I think they would be *nearly* equivalent for A=0, K=1 in the
Richards (then M=xi,B=k/alpha, nu=alpha) but not quite. A little
more algebra is required.
If you tell us more about what you're trying to do you might
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2020-02-28 09:24:16
|
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|
http://mathhelpforum.com/discrete-math/127168-yet-two-more-questions-i-don-t-understand.html
|
# Thread: Yet two more questions I don't understand.
1. ## Yet two more questions I don't understand.
Question 1:
We have seen that congruence modulo m is an equivalence relation on Z(all integers) for any integer m >= 1. Describe the set of equivalence classes for congruence modulo 8 - how many distinct classes are there and what are they?
What I am getting out of this is that there are 7 distinct classes and they are 1-7. But then again I am completely clueless on this modulo stuff and I've already gone to the professor with no luck on understanding it any better.
Question 2:
Show that addition modulo 8 is well-defined. (Use arbitrary elements from two equivalence classes [a] and [b] and how that the result is an element in [a] + [b]).
We briefly went over this in class and I again didn't understand it because of the whole modulo thing. I think it is something to do with taking two classes that are equivalent to mod 8 (which I am not sure how to figure out) and then proving that if you add them together you get whatever the question is asking.
That's just it though, I am not understanding the question nor how to find a part of what it is asking. Can anyone clarify this for me?
Help is appreciated!
2. I'll give it a try Yaeger- and hopefully someone more senior on the board will fix up the misakes. I guess you know how the mod function works...
We'll do it with mod 3 because that's easier:
Notice how (2 mod 3) = (5 mod 3) = 2. Well, when you think about mod 3 as being a relation on the set of whole numbers, there are infinitely many numbers (every 3rd member of the Set to be exact) that are related to 2 by the mod 3 relation. So when you start to write down the members of the relationship, you get:
$
\rho:= \{(0, 0), (1, 1), (2, 2), (3, 0), (4, 1), (5, 2), ..., (8, 2), ... (2 + 3n, 2), ...\}
$
An equivalence class is a short-hand way of naming the three partitions of the relationship, that is all the members of the relationship that have either (x, 0), (y, 1) or (z, 3) in them, so $[1]_{mod 3} = [4]_{mod 3}$ are two equivalence classes which are congruent. Both of them are elements from the same representational system, or partition.
Another example of equivalence classes are in the rational numbers, where $\frac{1}{2} = \frac{2}{4} \ldots$
Question 2 just shows you how you can do some basic math stuff with them. Try it out.
3. Thanks for the post. I understand the equivalence class thing now. But I am still unsure about the second question. But I am currently trying to figure it out now.
Thanks again!
|
2018-01-24 02:12:09
|
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|
https://faq.gutenberg.eu.org/3_composition/flottants/inserer_des_figures_dans_un_environnement_multicol2?rev=1527094248
|
Ceci est une ancienne révision du document !
— ID: Q-mcfloat revised: 2014-06-10 — # Floats in multicolumn setting
If you use latex \begin{figure}
...
\end{figure} in a multicols environment, the figure won't appear. If instead you use latex \begin{figure*}
...
\end{figure*} the figure will stretch right across the page (just the same as a figure* in standard LaTeX's twocolumn option).
It's possible to have single-column figures and tables with captions, using the [H] placement option introduced by the [float](https://ctan.org/pkg/float) package but you might have to fiddle with the placement because they won't float, and exhibit other strange behaviours (such as silently running off the end of the column at the end of the multicols environment).
3_composition/flottants/inserer_des_figures_dans_un_environnement_multicol2.1527094248.txt.gz · Dernière modification: 2018/05/23 18:50 par joseph.wright
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2021-12-07 07:01:59
|
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https://caefn.com/tag/advective
|
## Non-Reflecting Boundary Conditions in OpenFOAM
##### What we want to achieve
When we simulate fluid flow, we have to cut a finite computational domain out of an entire flow region. For accurate simulation, we need to let fluid and sound wave flow smoothly out of the domain through the boundary.
The reflection at the boundary has a larger effect on the solution especially when we perform a compressible flow simulation as can be seen in the following movie (Upper: With reflection, Lower: Without reflection of sound wave).
In OpenFOAM, we can use two approximate non-reflecting boundary conditions:
They determine the boundary value by solving the following equation
\begin{align}
\frac{D \phi}{D t} = \frac{\partial \phi}{\partial t} + \boldsymbol{U} \cdot \nabla \phi = 0, \tag{1} \label{eq:advection}
\end{align}
where $$D/Dt$$ is the material derivative and $$\boldsymbol{U}(\boldsymbol{x}, t)$$ is the advection velocity.
We assume that the advection velocity $$\boldsymbol{U}$$ is parallel to the boundary (face) normal direction and rewrite the eqn. \eqref{eq:advection} as
\begin{align}
\frac{D \phi}{Dt} \approx \frac{\partial \phi}{\partial t} + U_{n} \cdot \frac{\partial \phi}{\partial \boldsymbol{n}}= 0, \tag{2} \label{eq:advection2}
\end{align}
where $$\boldsymbol{n}$$ is the outward-pointing unit normal vector.
These boundary conditions are different in how the advection speed (scalar quantity) $$U_{n}$$ is calculated and it is calculated in advectionSpeed() member function.
The advection speed is the component of the velocity normal to the boundary
\begin{align}
\end{align}
##### waveTransmissive B.C.
The advection speed is the sum of the component of the velocity normal to the boundary and the speed of sound $$c$$
\begin{align}
U_n = u_n + c = u_n + \sqrt{\gamma/\psi}, \tag{4} \label{eq:waveTransmissiveUn}
\end{align}
where $$\gamma$$ is the ratio of specific heats $$C_p/C_v$$ and $$\psi$$ is compressibility.
Coming soon.
|
2021-05-18 21:17:06
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|
https://core.ac.uk/outputs/296799435
|
oai:edpsciences.org:dkey/10.1140/epja/i2008-10743-x
# Backward pion photoproduction
## Abstract
We present a systematic analysis of backward pion photoproduction for the reactions $\gamma$ p $\rightarrow$ $\pi^{0}_{}$ p and $\gamma$ p $\rightarrow$ $\pi^{+}_{}$ n . Regge phenomenology is applied at invariant collision energies above 3GeV in order to fix the reaction amplitude. A comparision with older data on $\pi^{0}_{}$ - and $\pi^{+}_{}$ -photoproduction at $\vartheta$ = 180° indicates that the high-energy limit as given by the Regge calculation could be reached possibly at energies of around $\sqrt{{s}}$ ≃ 3 GeV. In the energy region of $\sqrt{{s}}$ $\le$2.5 GeV, covered by the new measurements of $\gamma$ p $\rightarrow$ $\pi^{0}_{}$ p differential cross-sections at large angles at ELSA, JLab, and LEPS, we see no clear signal for a convergence towards the Regge results. The baryon trajectories obtained in our analysis are in good agreement with those given by the spectrum of excited baryons
|
2023-02-02 07:19:14
|
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https://homework.cpm.org/category/CCI_CT/textbook/pc3/chapter/5/lesson/5.2.2/problem/5-63
|
### Home > PC3 > Chapter 5 > Lesson 5.2.2 > Problem5-63
5-63.
Give a specific example to show that the following equation is false.
$\log(M)-\log(N)=\frac{\log(M)}{\log(N)}$
Use a calculator. Does:
$\log\left(27\right)-\log\left(5\right)=\frac{\log\left(27\right)}{\log\left(5\right)}$
|
2020-12-06 01:47:04
|
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|
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-13-cumulative-review-page-968/6
|
## Algebra: A Combined Approach (4th Edition)
$2(x-6)=4(x-3)-2x$ $2x-12=4x-12-2x$ $2x+2x-4x=-12+12$ $0=0$ x is a real number
|
2018-11-15 02:23:13
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http://mtpp.krsc-junioren.de/radix-2-fft.html
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See the GNU 00013 * General Public License for more details. [3,4,5,6,7,8,9] Using radix-2 p to calculate FFT for real signals like medical signals is very efficient. The overall result is called a radix 2 FFT. •Radix 2 and radix 4 are certainly the most popular •Radix 4 is on the order of 20% more efficient than radix 2 for large transforms •Radix 8 is sometimes used, but longer radix butterflies are not common because additional efficiencies are small and added complexity is non-trivial (especially for hardware implementations). High-Speed Four-Parallel 64-Point Radix-24 MDF FFT/IFFT Processor for MIMO-OFDM Systems Hang Liu1 and Hanho Lee2 1,2 School of Information and Communication Engineering, Inha University. Shuhong Gao, Committee Chair Dr. wireless communication. java from §9. radix-2) of FFT algorithms such as Cooley-Tukey can be quite expensive due to incoherent memory accesses. Radix-2 2 FFT algorithm is an attractive algorithm having same multiplicative complexity as radix-4 algorithm, but retains the simple butterfly structure of radix-2 algorithm. Each twiddle factor is loaded only once in the computation order, thus the number of redundant memory references due to twiddle factor in conventional radix-2 DIF FFT algorithm can be reduced. The signal flow graph of a radix-2 eight-point FFT. java * * Compute the FFT and inverse FFT of a length n complex sequence * using the radix 2 Cooley-Tukey algorithm. When the time-domain length of a waveform is a power of two, radix-2 FFT algorithms, which are extremely efficient, can be used to speed up processing time. Rounding mode. Keywords: fourier series, FFT, algorithm. One (radix-2) FFT begins, therefore, by calculating N/2 2-point DFTs. Overflow mode. Computing Discrete Fourier Transform using Radix -2 Decimation in Frequency Fast Fourier Transform. 0) and W(2,8. We can decompose this signal flow graph into small units called a “Butterfly,” as shown in Figure 6. of Electronics and Communication Sagar Institute of Research & Technology, Bhopal Navneet Kaur Dept. These algorithms have been developed using Verilog hardware description language and implemented on Spartan6 FPGA. The total number of multiplications is minimized at the expense of increase in additions and some more memory requirement [2-4]. The FFT core does not implement the 1/N scaling for inverse FFT. The Radix-2 Lite Burst I/O was not considered since is a simplified version of the Burst I/O version. Radix-2 first computes the DFT of the even index inputs and the odd index. This paper presents a novel architecture for the enhancement of performance of compute intensive Fast Fourier Transform (FFT) algorithm which is common in many signal processing applications. FFT is widely used in signal processing, and the application needs real-time and high performance, while most of the traditional designs are limited to the power of two, which wastes the buffers and multipliers in big data. The split-radix FFT is a fast Fourier transform (FFT) algorithm for computing the discrete Fourier transform (DFT), and was first described in an initially little-appreciated paper by R. Draw the basic butterfly diagram or flow graph of DIF radix-2 FFT? 15. 8 2 v3 algorithm. A radix-2 decimation-in-time (DIT) FFT is the simplest and most common form of the Cooley–Tukey algorithm [] Radix-2 DIT divides a DFT of size N into two interleaved DFTs (hence the name "radix-2") of size N/2 with each recursive stage. FF Radix-2 (2-parallel): FF Radix-4 (4-parallel): 4 SWITCH 4 R2 2 2 1 1 R2 SWITCH R2 SWITCH R2 3 SWITCH 3 R4 2 1 2 1 R4. In this paper, the comparison study of various FFT algorithm and compare all them. This diagram is quite complex. The implementation of a FFT processor is one of the most challenging parts in the realization of a wideband receiver and its hardware complexity is very high. The j and a registers are linked with the + operator. cascading k radix-2 stages known as radix-2k [5] algorithm. FPGA design and implementation of radix-2 Fast Fourier Transform algorithm with 16 and 32 points @article{Saenz2015FPGADA, title={FPGA design and implementation of radix-2 Fast Fourier Transform algorithm with 16 and 32 points}, author={S. * Bare bones implementation that runs in O(n log n) time. The radix is a property of a numerical system, not an individual number. Radix-2 FFT routines for complex data; Mixed-radix FFT routines for complex data; Overview of real data FFTs; Radix-2 FFT routines for real data; Mixed-radix FFT routines for real data; References and Further Reading; Numerical Integration. Computing Discrete Fourier Transform using Radix -2 Decimation in Frequency Fast Fourier Transform. Inspired: Radix 2 Fast Fourier Transform Decimation In Time (Complex Number Free Implementation) Discover Live Editor Create scripts with code, output, and formatted text in a single executable document. RADIX-2 FFT BUTTERFLY STRUCTURES / Chapter Four. A radix-2 decimation-in-time butterfly. Introduction; QNG non-adaptive Gauss-Kronrod integration; QAG adaptive integration; QAGS adaptive. The radix-2 FFT functions for real data are declared in the header files `gsl_fft_real. The number inside the circle is the value of q (for stage 1) or p (for stage 2) [6]. You'll get subjects, question papers, their solution, syllabus - All in one app. FPGA design and implementation of radix-2 Fast Fourier Transform algorithm with 16 and 32 points @article{Saenz2015FPGADA, title={FPGA design and implementation of radix-2 Fast Fourier Transform algorithm with 16 and 32 points}, author={S. The circuit with 16-bit word-. This work was supported by an NSF Graduate Fellowship, NASA GSRPFellowshipNGT-70340,IntelCorporation,NSFGrantCCF430090, and a UCD Faculty Research Grant. The FFT uses a Radix-2, Decimation-In-Time (DIT) and in-place architecture which improves overall efficiency of the computation in terms of speed while. The Fast Fourier Transform from Understanding Digital Signal Processing. In radix-4 FFT, the butterfly is based on the four point DFT. The definition of the forward Fourier transform, fft(z), is,. Perform N 1 DFTs of size N 2. Implementing FFT in software can be easily done, as software runs on CPU, which executes instructions serially. A Radix-2 Cooley-Tukey FFT is implemented with no limits on the length of. thanx in advance RE: Radix-2 (8 or 16) point FFT. The j and a registers are linked with the + operator. Traditionally, radix-2 and radix-4 FFT algorithms have been used. A much faster algorithm has been developed by Cooley and Tukey around 1965 called the FFT (Fast Fourier Transform). Putting together the length DFT from the length-DFTs in a radix-2 FFT, the only multiplies needed are those used to combine two small DFTs to make a DFT twice as long, as in Eq. In practice, the time needed to precompute the powers and reciprocals of the powers is not negligible. データの分割の単位をRadixというが、広く知られているFFTはRadix-2のFFTである。この場合、データ数は2の冪乗に限られる。 Cooley-Turkeyアルゴリズムは実際にはどんなRadixでも動く。この場合データ数は2の冪乗に限られない。. Joel Brawley Dr. Generally, for an N=2n-point FFT, the addressing and control logic are mainly composed of several components: An (n−1)-bit butterfly counterB ¼ b n 2b n 3. Low-Power Architectures for Signal Processing and Classification Systems A DISSERTATION SUBMITTED TO THE FACULTY OF THE GRADUATE SCHOOL OF THE UNIVERSITY OF MINNESOTA BY Manohar Ayinala IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF Doctor of Philosophy Keshab Parhi August, 2012 °. Fig5: 16 point Radix - 4 2 SDF Architecture This architecture is similar to the Radix 2 2. The FFT has a. Radix-2 decimation in time FFT for 32 inputs each of having 32 bits[1]. Explore VLSI Projects Thesis Dissertation, VLSI Projects Topics, IEEE MATLAB Minor and Major Project Topics or Ideas, VHDL Based Research Mini Projects, Latest Synopsis, Abstract, Base Papers, Source Code, Thesis Ideas, PhD Dissertation for Electronics Science Students ECE, Reports in PDF, DOC and PPT for Final Year Engineering, Diploma, BSc, MSc, BTech and MTech Students for the year 2015 and. The slowest case would be a series length composed of one big prime factor, for example a length of 1021. FUNDING NUMBERS 7. The radix-22 algorithm [2] not only reduces the computational complexity but also retains the simple. M31円星科技Memory Compiler 与GPIO获ISO 26262 车用安全最高等级ASIL-D认证. The butterfly- Processing Element (PE) used in the 8-FFT processor reduces the. Since there are log 2 N twiddle factors for a N -points radix-2 DIF FFT algorithm, the computation requires only log 2 N steps. This means, for an N -point FFT stage, the core needs to read N/2 values into memory, and then apply these values, the next N/2 input values, and a stored ROM coefficient to the butterfly core. ) If zero is returned, the routine failed to allocate storage. " Kiss FFT is a very small, reasonably efficient, mixed radix FFT library that can use either fixed or floating point data types. It is defined as WN = e-j2π/N 12. 1 Required Hardware Support for FFT Calculation 4-1 4. this is a 8 point FFT implementation using the butterfly unit, The butterfly unit is the heart of FFT algorithm. 6 Extended-Precision Complex Radix-2 FFT/IFFT Implemented on TMS320C62x 5 Implementation of the FFT The C-equivalent of the FFT assembly code is listed as comments in the assembly file. The value of r, radix, plays a major role in determining the efficiency and complexity. This allows the same FFT graph to be used either forwards or backwards for either the forward or inverse transform. אלגוריתם FFT נפוץ הוא radix-2 המבצע את החישוב בנפרד על האינדקסים הזוגיים,. rar • 256 -point radix-8 FFT. Putting together the length DFT from the length-DFTs in a radix-2 FFT, the only multiplies needed are those used to combine two small DFTs to make a DFT twice as long, as in Eq. Radix 2 Butterfly has 2 inputs and 2 outputs. The FFT reduces the number of computations of DFT from O(N 2) to O(NlogN) by. DFT vs FFT Radix DIF andDIT Architectures Pipelined Iterative HW Design Processed data pass to the next stage. Section IV generates the identical radix-2k FFT algorithms. Comcores Fast Fourier Transform (FFT) IP core is an implementation of a Cooley-Tukey FFT algorithm, a computationally efficient method for calculating the Discrete Fourier Transform (DFT). my simulation is supposed to be exactly as the 16-point radix-2 DIT FFT link below and to the best of my knowledge, i have connected it as it should be (Including the input bit reversal & correct twiddle. FPGA Implementation of 32 point Radix-2 Pipelined FFT Aruna Arya, Prof. DIT-Radix-2-FFT in SPED 1. thanks a lot for your quick response. Thisclass of machine. OVERVIEW OF THE CACHED-FFT 2. my simulation is supposed to be exactly as the 16-point radix-2 DIT FFT link below and to the best of my knowledge, i have connected it as it should be (Including the input bit reversal & correct twiddle. Let us consider the computation of the N = 2 v point DFT by the divide-and conquer approach. The Cooley-Tukey algorithm became known as the Radix- 2 algorithm and was shortly followed by the Radix-3, Radix-4, andMixed Radix algorithms [8]. Duhamel and H. The last stage results in the output of the FFT, a 16 point frequency spectrum. Fig 2 outlines an implementation of the R22SDF architecture for N=1024, note the similarity of the data-path to R2SDF and the reduced number of multipliers. my simulation is supposed to be exactly as the 16-point radix-2 DIT FFT link below and to the best of my knowledge, i have connected it as it should be (Including the input bit reversal & correct twiddle. Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). Section IV generates the identical radix-2k FFT algorithms. 6 Recursive Formulas for DIT-FFT 1. If you are looking for a guide to implement FFT in CUDA/OpenCL for your custom use with Radix 2, Radix 4, Radix 8 (And other powers of 2), you have come to the right place. Orange Box Ceo 7,181,066 views. Many different FFT algorithms exist. Real-world processing such as LTE OFDM often require non power-of-2 or -4 FFT lengths for spacing into the desired frequency bin size, called mixed-radix FFTs. The j and a registers are linked with the + operator. Here’s how I solved it using Matlab and getting Cadence and Matlab to talk (in a limited fashion). Low-Power Split-Radix FFT Processors Using Radix-2 Butterfly Units ABSTRACT: Split-radix fast Fourier transform (SRFFT) is an ideal candidate for the implementation of a low-power FFT processor, because it has the lowest number of arithmetic operations among all the FFT algorithms. Overflow mode. Radix-2 DIT FFT algorithm Butterfly Diagram- Anna university frequently asked question IT 6502. pSrc points to In-place arrays containing 2*fftLen values. Cvetkovic, IntechOpen, DOI: 10. Both the logic blocks and interconnects are programmable. So radix-4 algorithm requires somewhat fewer multiplications than the radix-2 algorithm. 4 The Fast Fourier Transform 1. They use the Cooley-Tukey algorithm to compute in-place FFTs for lengths which are a power of 2. For example, at the end of the first stage, we have x(0) ← x(0)+ W0 8 x(4). The throughput of the proposed FFT processor is one sample per clock. Math::FFT - Perl module to calculate Fast Fourier Transforms River stage one • 2 direct dependents • 3 total dependents This module implements some algorithms for calculating Fast Fourier Transforms for one-dimensional data sets of size 2^n. A Normal I/O Order Radix-2 FFT Architecture to Process Twin Data Streams for MIMO ABSTRACT: Nowadays, many applications require simultaneous computation of multiple independent fast Fourier transform (FFT) operations with their outputs in natural order. A length DFT requires no multiplies. View Dev Sehgal’s full profile to. /* C program to compute N-point Radix-2 DIT FFT algorithm. And it can successfully run on Quartus 2 or other software. When N is a power of r = 2, this is called radix-2, and the natural fidivide and conquer approachfl is to split the sequence into two. Kevin James. 3 Bit reversing 1. The fast fourier transform are good algorithm and computed discrete fourier transform (DFT). They use the Cooley-Tukey algorithm to compute in-place FFTs for lengths which are a power of 2. MATH 3511 Radix-2 FFT Spring 2019 is a power of two; since the number of sample points Ncan usually be chosen freely by the application, this is often not an important restriction. FORTRAN source for (complex and real) split radix FFT (by Henrik Sorensen): sorensen. and the twiddle factors are the same except the complex twiddle factor for the the butterflies are off by a phase difference of $\frac{\pi}{2}$. Although radix-2 2 and radix-4 architectures require the same total number of hardware resources for 4-parallel samples, the layout of these resources is different[8][5]. The implementation uses two. There are several ways to calculate a radix-2FFT because the derivation from the DFT can be performed differently. They use the Cooley-Tukey algorithm to compute in-place complex FFTs for lengths which are a power of 2 -- no additional storage is required. The variable streaming FFT implements two different types of FFT. We use the four-step or six-step FFT algorithms to implement the radix-2, 3 and 5 parallel 1-D complex FFT algorithms. The FFT can be designed by radix-2 butterfly algorithm which requires needless computations and data storage. When is a power of , say where is an integer, then the above DIT decomposition can be performed times, until each DFT is length. The Radix-22 single-path delay feedback, Radix-22 single-path delay FFT algorithm is illustrated in Section 2. I want to implement Radix-2 Single-path Delay Feedback (SDF) Decimation-In-Frequency FFT with Pipelining in VHDL. Just have a look at this small snippet of code. Radix 4 Butterfly has 4 inputs and 4 Outputs and so on. 2 Radix-2 decimation-in-frequency algorithm The same radix-2 decimation in frequency can be applied recursively to the two length-N 2 DFTs to save additional computation. FFT sizes due to use of generic architectures. 3 Radix-2 FFT Useful when N is a power of 2: N = r for integers r and. The proposed processor organization allows the area of the FFT. The radix-2 FFT code is essential since every function depends on it. An N-point FFT using Radix-2 decomposition has log 2 (N) stages, with each stage containing N/2 Radix-2 butterflies. A length DFT requires no multiplies. Signal decomposition, or ‘decimation in time’ is achieved by bit reversing the indices for the array of time domain data. ---- IFFT/FFT, with innovative features that supports 4 Tx/Rx streams with only less than 100 K gates (excluding memories). Out of these, we have designed here butterfly units of radix-4 and radix-8. In R2SDF FFT, N/2 point input data is sequentially controlled with the help of Flip-Flop (FF) circuit. /***** * Compilation: javac FFT. When using a "radix-2" FFT as above, input data with non-power-of-2 sizes needs to be enlarged and padded to the nearest power of 2 before processing, but FFT options for other sized steps are also possible. From the figure u can see that if we are done with the butterfly unit we are 70% done with the FFT coding. ---- Viterbi decoder, a radix-4 design that supports higher throughput with slower memories. Other forms of the FFT like the 2D or the 3D FFT can be found on the book too. “Fast Fourier Transform” or FFT. Programming competitions and contests, programming community. View Videos or join the Split-radix FFT Algorithm discussion. The mixed-radix digits of the N y mapper outputs are combined into one index vector. The second class is the radix-2n algorithms proposed to avoid the drawback of high-radix algorithms. Moreover, these weren’t related by a power of 2. It is shown that feed forward structures are more efficient than feed forward structures are more efficient than feedback ones when several samples in parallel must be processed. The binary -> radix conversion requires 2 multiplications of size N/2 per recursive call. this is a 8 point FFT implementation using the butterfly unit, The butterfly unit is the heart of FFT algorithm. These architectures make use of delay feedback design which in turn gives less number of delay elements but the number of data path is equal to the number of. In this paper, the comparison study of various FFT algorithm and compare all them. Of particular interest in distributed memory architectures such as the Connection Machine is the allocation of twiddle factors to processors. For a radix-2 FFT this gives an operation count of O(n log 2 n). RESULTS AND CONCLUSION The proposed design for 64 point radix-2 FFT processor Fig. Find the values of WNk, When N=8, k=2 and also for k=3. The macro includes a function that gets the Sample Rate and FFT length, and in addition to setting up the sweep panel, it retrieves the FFT level spectrum in units of Volts. 4 makes use of multiply-accumulate blocks embedded on-chip in Microsemi's PolarFire, SmartFusion2, IGLOO2 and RTG4 FPGA devices to deliver a flexible, fully configurable radix-2 decimation-in-time (DIT) burst I/O FFT for high reliability, radiation-tolerant applications. Khmelnik "Specialized Digital Computer for Operations with Complex Numbers" (in Russian), Questions of Radio Electronics, volume 12. Note that even if N is a power of two, it is not necessary to run recursion down to radix-2 DFTs — radix-4 DFTs provide about same. However, radix-2fe was only proposed for. The paper is organized as follows. designed for Radices 2 through 8. Radix -2 Mixed SDC -SDF FFT architecture offers 17. Algorithm The FFT Core uses the radix-4 and radix- 2 decimation-in-time (DIT) methods for computing the DFT , factors. So it's only an advantage if the number of mpy is the limiting factor which for most hardware these days is not the case. Radix-2 signal flow graph for a 16 point fast Fourier transform (FFT). TITLE AND SUBTITLE Use of the Reduced Precision Redundancy (RPR) Method in a Radix-4 FFT Implementation 6. The Radix-2 is the fastest method for calculating FFT. Later, radix-22 was extended to radix-2fe. A different radix 2 FFT is derived by performing decimation in frequency. The butterfly- Processing Element (PE) used in the 8-FFT processor reduces the. Radix 2 and radix 4 algorithms Lengths as powers of 2 or 4 are most popular Assume N=2n N 1=2, N 2=2n-1 (divides input sequence into even and odd samples - decimation in time - DIT) "Butterfly" (sum or difference followed or preceeded by a twiddle factor multiply) X m and X N/2+m outputs of N/2 2-pt DFTs on outputs of. Programming competitions and contests, programming community. The decimation-in-time (DIT) radix 2 FFT recursively partitions a DFT into two half-length DFTs of the even-indexed and odd-indexed time samples. PDF | This paper is part 2 in a series of papers about the Discrete Fourier Transform (DFT) and the Inverse Discrete Fourier Transform (IDFT). The implementation of the radix 5 block is diagrammatically represented in Fig. java * Execution: java FFT n * Dependencies: Complex. For a complex N-point Fourier transform, the FFT reduces the number of complex multiplications from the order of N2 to the order of NlogN. [3,4,5,6,7,8,9] Using radix-2 p to calculate FFT for real signals like medical signals is very efficient. Petrovsky*, Sergei L. This paper completes the pipelined design of the the original phase-rotation FFT, provides a fundamental new description of the algorithm di-rectly in terms of the parallel pipeline, and describes a radix-2 implementation on the iWarp computer system that balances com-. 1 Stage Independent Addressing 2. RAM location means faster FFT for higher latencies. * Bare bones implementation that runs in O(n log n) time. I want to explore the other three schemes as well. 16/64 point FFT. OVERVIEW OF THE CACHED-FFT 2. 4: Mixed-Radix 8-2 SFG for 64 Points FFT. If you are looking for a guide to implement FFT in CUDA/OpenCL for your custom use with Radix 2, Radix 4, Radix 8 (And other powers of 2), you have come to the right place. Default: Floor. Radix 2 FFT. FUNDING NUMBERS 7. Processor Cache Main Memory Fig. Abstract-The organization and functional design of a parallel radix-4 fast Fourier transform (FFT) computer for real-time signal processing of wide-band signals is introduced. 4 makes use of multiply-accumulate blocks embedded on-chip in Microsemi's PolarFire, SmartFusion2, IGLOO2 and RTG4 FPGA devices to deliver a flexible, fully configurable radix-2 decimation-in-time (DIT) burst I/O FFT for high reliability, radiation-tolerant applications. Petrovsky*, Sergei L. The FFT result will be contained in the same array and the frequency domain values will have the same interleaving. This example model uses an input vector size of 8, and calculates the FFT using the Streaming Radix 2^2 architecture. The odd and even inputs are in the natural order. Becerra and Susana Ortega Cisneros and Jorge Rivera Dominguez}, journal={2015 IEEE International Autumn Meeting. In any case, radix/-2^k was anticipated in favour of solitary way defer input (S/D/F) structures, yet not in support of feed/forward ones which. designed for Radices 2 through 8. FFT is finite Fourier transform, its fast when the length of vector on which is evaluated is ~ to 2^N where N is an integer. It saves resources compared to a streaming Radix 2 implementation by factoring and grouping the FFT equation. You will need to zero pad your 78 samples to 128 samples. Subject: Image Created Date: 11/15/2011 1:53:55 PM. If Radix-4 is compared to the Radix-2 algorithm, then Radix-4 has a higher complexity and less computational cost. The Cooley-Tukey algorithm became known as the Radix- 2 algorithm and was shortly followed by the Radix-3, Radix-4, andMixed Radix algorithms [8]. Eventually, we would arrive at an array of 2-point DFTs where no further computational savings could be realized. They use the Cooley-Tukey algorithm to compute in-place complex FFTs for lengths that are a power of 2 — no additional storage is required. In Section 3, feedback, Radix-24 single-path delay feedback, Split- the implementation of Radix-22 Algorithm by FPGA Radix single-path delay feedback, and Radix-4 single- will be debated. Basic Operation. Mixed radix-2/3/4/5 FFTs can be used to implement the DFT algorithm with reduced computation if the number of DFT. Hardware Implementation of a 32-point Radix-2 FFT Architecture Ying Gao Ying Gao Master’s Thesis Series of Master’s theses Department of Electrical and Information. For this to be possible, N must be a power of 2. corresponding iterative C code implementation of n-points radix-2 DIT FFT algorithm. Hence the algorithm is called radix-2 algorithm. \nThe decimation-in-time (DIT) radix-2 FFT recursively partitions\na DFT into two half-length DFTs of the even-indexed and odd-indexed\ntime samples. Hardware Implementation of a 32-point Radix-2 FFT Architecture Ying Gao Ying Gao Master's Thesis Series of Master's theses Department of Electrical and Information. The BF_PE1 block computes the radix-2 butterfly. The proposed architecture exhibits faster response time compared to radix-2 'Single-path Delay Feedback (SDF)' architecture and 'radix-2 Multi-path Delay. The butterfly- Processing Element (PE) used in the 8-FFT processor reduces the. Radix 2 and radix 4 algorithms Lengths as powers of 2 or 4 are most popular Assume N=2n N 1=2, N 2=2n-1 (divides input sequence into even and odd samples – decimation in time – DIT) “Butterfly” (sum or difference followed or preceeded by a twiddle factor multiply) X m and X N/2+m outputs of N/2 2-pt DFTs on outputs of. The FFT uses a Radix-2, Decimation-In-Time (DIT) and in-place architecture which improves overall efficiency of the computation in terms of speed while. First of all, to program FFT, we need a bitreverse(radix 2) / digitreverse (radix 4 8 16 …) algorithm. These architecture uses radix 23 and radix 24 while some architecture use mixed radix-2 and radix-8 algorithms. Multi-Gigahertz Parallel FFTs for FPGA and ASIC Implementation approach for pipeline implementation of radix-2 FFT 10, Fast Fourier Transform: VLSI. This is useful for analyzing vector-valued series. A comparison of the two algorithms using a sample of points obtained on a variety of computational platforms and for several sequence lengths is presented. •Radix 2 and radix 4 are certainly the most popular •Radix 4 is on the order of 20% more efficient than radix 2 for large transforms •Radix 8 is sometimes used, but longer radix butterflies are not common because additional efficiencies are small and added complexity is non‐. Inspired: Radix 2 Fast Fourier Transform Decimation In Time (Complex Number Free Implementation) Discover Live Editor Create scripts with code, output, and formatted text in a single executable document. Low-Power Architectures for Signal Processing and Classification Systems A DISSERTATION SUBMITTED TO THE FACULTY OF THE GRADUATE SCHOOL OF THE UNIVERSITY OF MINNESOTA BY Manohar Ayinala IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF Doctor of Philosophy Keshab Parhi August, 2012 °. However, for this case, it is more efficient computationally to employ a radix-r FFT algorithm. Radix 2 FFT When is a power of , say where is an integer, then the above DIT decomposition can be performed times, until each DFT is length. 2 ++ This path traverses points by a complex number base i-r for given integer r. PDF | This paper considers partial-column radix-2 and radix-2/4 FFT processors and realizations of butterfly operations. 갑자기 C# msChart 에 삘이 꽃히는 바람에 C# 으로 구현해서 그래프로 바로 볼수 있다면 좋겠다 라는 생각을 했다. This section of MATLAB source code covers Decimation in Frequency FFT or DFT matlab code. This code was written by Robin Scheibler during rainy days in October 2017. It has same multiplicative complexity of radix-4, but has a signal flow graph similar to the radix-2 algorithm. 3[16, 2] outlines an implementation of the R2 SDF. To understand the basics of a FFT, it is often useful to look to a special flow diagram. A Normal I/O Order Radix-2 FFT Architecture to Process Twin Data Streams for MIMO ABSTRACT: Nowadays, many applications require simultaneous computation of multiple independent fast Fourier transform (FFT) operations with their outputs in natural order. The only requirement of the the most popular implementation of this algorithm (Radix-2 Cooley-Tukey) is that the number of points in the series be a power of 2. 0 Verified simulation results 7/98 References 1 Introduction to Digital Signal Processing, Proakis and Manolakis, (Macmillan, 1988, ISBN -02-396810-9). INTRODUCTION The spectrum of a signal is the information about the distribution of the frequencies that compose it and their respective amplitudes [1], which allows to study aspects of the signal that would be difficult and even. ) In particular, split radix is a variant of the Cooley-Tukey FFT algorithm that uses a blend of radices 2 and 4: it recursively expresses a DFT of length N in terms of one smaller DFT of length N /2 and two smaller DFTs of length N /4. Only one of these was illustrated on the preceding pages. FPGA Implementation of 32 point Radix-2 Pipelined FFT Aruna Arya, Prof. These architecture uses radix 23 and radix 24 while some architecture use mixed radix-2 and radix-8 algorithms. The FFT (DIT, radix-n1) The Cooley-Tukey Fast Fourier Transform computes the DFT with only O(N log N ) operations. How to create a 3D Terrain with Google Maps and height maps in Photoshop - 3D Map Generator Terrain - Duration: 20:32. The Radix-2 is the fastest method for calculating FFT. Compared with [9], the radix -2 SDF design [10]. Point sizes that are not a power of 4 need an extra Radix-2 stage for combining data. 2: REN Bingyu1,ZHAN Yinwei2 (1. 8 point DFT using radix-2 DIT FFT. Cvetkovic, IntechOpen, DOI: 10. These architectures make use of delay feedback design which in turn gives less number of delay elements but the number of data path is equal to the number of. metic kernel of radix-2 DIT FFT is the butterfly operation defined as X0 = P +WQ; X1 = P −WQ (3) The signal flow graph of radix-2 DIT butterfly operation is shown in Figure 1 (b). The overall result is called a radix 2 FFT. The Fourier Transform assumes that the time signal is periodic and infinite in duration. FAST FOURIER TRANSFORM ALGORITHMS WITH APPLICATIONS A Dissertation Presented to the Graduate School of Clemson University In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy Mathematical Sciences by Todd Mateer August 2008 Accepted by: Dr. The common radix-2 algorithm, used in this implementation, continuously decomposes the DFT into two smaller DFTs. Open the example model:. REPORT TYPE AND DATES COVERED Master’s Thesis 4. 6 5% reduction of LUTs, 45. 2: REN Bingyu1,ZHAN Yinwei2 (1. mixed-radix 4/2 (i. The result is an FFT that is simple to pipeline. A variable-length FFT processor that integrates two radix-2 stages and three radix-2 stages for FFT sizes 512, 1024 and 2048 was proposed in [15]. The architecture contains nine stages with the following sub-blocks: the IFFT/FFT select unit, based on the duality between the IFFT and FFT characteristic; module 1, based on the radix-24 FFT algorithm for stage 1 to stage 4; module 2, which consists x(1) x(0). AP-808 Split-Radix Fast Fourier Transform Using Streaming SIMD Extensions 01/28/99 iv Revision History Revision Revision History Date 1. Abstract—The Fast Fourier Transform (FFT) and its inverse transform (IFFT) processor are key components in many communication systems. We concentrated on the Radix-2 architectures since they address the nature of our work. Joel Brawley Dr. A 16-point FFT processor using Mixed-Radix 4-2 butterfly with bit reversing is illustrated to verify the operation of design architecture. They use the Cooley-Tukey algorithm to compute in-place FFTs for lengths which are a power of 2. Fast Fourier transform You are encouraged to solve this task according to the task description, An implementation of the radix-2 algorithm, which works for any. I'm a beginner in C programming. my simulation is supposed to be exactly as the 16-point radix-2 DIT FFT link below and to the best of my knowledge, i have connected it as it should be (Including the input bit reversal & correct twiddle. First it computes the one-dimensional FFT along one dimension (row or column). Figure 2 shows a signal flow graph of a radix-4 16-point FFT. Radix-2 method proposed by Cooley and Tukey[1] is a classical algorithm for FFT calculation. Below is the syntax highlighted version of FFT. In order to extend the application scope of the FFT architectures, the new. The steps involved in the radix-2 FFT algorithm can be summarized with the butterfly computations illustrated in Figure 2. The block does the computation of a two-dimensional M-by-N input matrix in two steps. What are the phase factors involved in all stages of computation in the 8-point DIT radix-2 FFT? First stage: W80 Second stage: W80, W82 Third stage: W80, W81, W82, W83 13. When using a "radix-2" FFT as above, input data with non-power-of-2 sizes needs to be enlarged and padded to the nearest power of 2 before processing, but FFT options for other sized steps are also possible. It is defined as WN = e-j2π/N 12. The first one refers to pushing the stack phase, while the second one illustrates the popping the stack phase. Pipelined Radix-2k Feedforward FFT Architectures Mario Garrido, Member, IEEE , J. Each iteration of the inner loop calculates one butterfly (i. This butterfly. You'll get subjects, question papers, their solution, syllabus - All in one app. The recursive implementation of the radix-2 Decimation In Frequency algorithm can be understood using the following two figures. The value of r, radix, plays a major role in determining the efficiency and complexity. Key wor ds: FFT, Multi-Dimension, Radix-2, RISC, v ector sup er computer P A CS: 02. 1 shows the conventional radix-2 butterfly architecture; for every clock cycle one radix-2 butterfly is per-formed. The Radix-22 single-path delay feedback, Radix-22 single-path delay FFT algorithm is illustrated in Section 2. Figure 2 shows a diagram for an 8-pointradix-2DIT-FFT(decimation in time-FFT). to enroll in courses, follow best educators, interact with the community and track your progress. Uses ieee_proposed library for fixed point arithmetic. 2 shows pseudo-code for a Stockham radix-R FFT with specialization for radix-2. In the first stage, 16 frequency spectra (1 point each) are synthesized into 8 frequency spectra (2 points each). Of particular interest in distributed memory architectures such as the Connection Machine is the allocation of twiddle factors to processors. This diagram is quite complex. To understand the basics of a FFT, it is often useful to look to a special flow diagram. FFT algorithm provides speed increase factors, when compared with direct computation of the DFT, of approximately 64 and 205 for 256 point and 1024 point transforms respectively. Generally, for an N=2n-point FFT, the addressing and control logic are mainly composed of several components: An (n−1)-bit butterfly counterB ¼ b n 2b n 3. The radix -2 MDC architecture [9] is the most direct implementation approach of pipelined FFT, but its hardware utilization is only 50%. It is shown that feed forward structures are more efficient than feed forward structures are more efficient than feedback ones when several samples in parallel must be processed. The radix is a property of a numerical system, not an individual number. Joel Brawley Dr. bit reversing the output sequences. Hardware Implementation of a 32-point Radix-2 FFT Architecture Department of Electrical and Information Technology, Faculty of Engineering, LTH, Lund University, July 2015. A paper on a new FFT algorithm that, following James Van Buskirk, improves upon previous records for the arithmetic complexity of the DFT and related transforms, is: Steven G.
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2019-11-22 03:04:26
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https://math.stackexchange.com/questions/1303265/find-the-sequence-of-partial-sums-for-the-series-a-n-1n-does-this-series
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# Find the sequence of partial sums for the series $a_n = (-1)^n$ Does this series converge?
Find the sequence of partial sums for the series $$\sum_{n=0}^\infty (-1)^n = 1 -1 + 1 -1 + 1 - \cdots$$
Does this series converge ?
My answer is that the sequence $= 0.5 + 0.5(-1)^n$. This makes a sequence that alternates between $1$ and $0$.
I know that the sequence does not converge since it is not monotone. But how can I prove this?
• Do there exist subsequences which converge to disparate limits? – abiessu May 29 '15 at 0:00
• – Victor May 29 '15 at 0:02
• A series does not have to be monotone to converge, e.g., $(-1)^n/n$. What theorems/tests do you have for convergence? Cauchy? For instance, if $S_n$ is the partial sum, then $|S_n - S_{n+1}| = 1$ and hence is not Cauchy. – Simon S May 29 '15 at 0:02
hint: $$s_{2n} = 0, s_{2n+1} = 1$$.
If it converges to $\ell$ then $|s_n-\ell|<\varepsilon$, once $\varepsilon>0$ is fixed, and $n$ is sufficiently large. Here $s_n$ is exactly the $n$-th partial sum.
But if you set, for example, $\ell=1/10$, then the above inequality cannot be satisfied even for $n$ enough large.
All you need to show is that the the sequence $x_{n}=(-1)^{n}$ does not converge. Many ways to do it : First, using the theorem saying that if a sequence converges , then any subsequence converges to the same limit. $x_{2n}$ and $x_{2n+1}$ are subsequences converging obviously to different limits. Not familiar with this? No problem! Assume that $x_{n}$ converges, say to $l$. Then, given $\epsilon >0$, we can find an integer $N(\epsilon)$ such that $n \geq N(\epsilon)$ implies $|x_{n}-l| <\epsilon.$ Now, $|x_{n+1}-x_{n}| = |x_{n+1}-l+l-x_{n}| \leq |x_{n+1}-l|+|l-x_{n}| <2\epsilon$, provided $n \geq N(\epsilon)$. Now, we chose $\epsilon=1$. And I will leave it to you to see the obvious contradiction!.
$$\sum\limits_{n=0}^\infty (-1)^n = \lim\limits_{m\to\infty}\sum\limits_{n=0}^m (-1)^n$$ $$= \lim\limits_{m\to\infty} \frac12\left((-1)^m + 1\right)$$ Note that the sequence $$a_m=\frac12\left((-1)^m + 1\right)$$ is convergent if $$\lim\limits_{m\to\infty} a_{2m} = \lim\limits_{m\to\infty} a_{2m+1} = L$$ However, in this case we have $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m}+1\right) =\frac22= 1$$ And $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m+1}+1\right) =\frac02= 0$$ Therefore $a_m$ is a divergent sequence.
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2019-09-17 19:21:19
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https://projects.wyliodrin.com/wiki/electronics/kirchhofflaws
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# Wyliodrin
## Reference
electronics:kirchhofflaws
# Kirchhoff Laws
## Formulas
In
a node, the sum of the currents is 0. Please keep in mind that currents have directions. Currents incoming have negative values, while currents outgoing have positive values.
$\sum\limits_{i=1}^n(I_i) = 0$
The sum of the voltage in a circuit loop is equal to the power source voltage.
$E = \sum\limits_{i=1}^n(U_i)$
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2017-12-18 14:45:54
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https://buboflash.eu/bubo5/show-dao2?d=3146196323596
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Question
$$a^2-b^2=$$ [...]
$$a^2-b^2=(a+b)(a-b)$$
Question
$$a^2-b^2=$$ [...]
?
Question
$$a^2-b^2=$$ [...]
$$a^2-b^2=(a+b)(a-b)$$
If you want to change selection, open original toplevel document below and click on "Move attachment"
#### Parent (intermediate) annotation
Open it
The difference of two squares identity is a squared number subtracted from another squared number to get factorized in the form of a2−b2=(a+b)(a−b).
#### Original toplevel document
Difference Of Squares | Brilliant Math & Science Wiki
Bhardwaj , 敬全 钟 , Sam Reeve , and 10 others Ashley Toh Lucerne O' Brannan Satyabrata Dash Ben Sidebotham Sandeep Bhardwaj Derek Guo Jongheun Lee Mahindra Jain Jimin Khim Tara Kappel contributed <span>The difference of two squares identity is a squared number subtracted from another squared number to get factorized in the form of $a^2-b^2=(a+b)(a-b).$ We will also prove this identity by multiplying polynomials on the left side and getting equal to the right side. This identity is often used in algebra where it is useful in applicatio
#### Summary
status measured difficulty not learned 37% [default] 0
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2022-08-20 06:25:51
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https://spinnaker8manchester.readthedocs.io/en/latest/pacman/pacman.operations.placer_algorithms/
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# pacman.operations.placer_algorithms package¶
## Module contents¶
class pacman.operations.placer_algorithms.RadialPlacer[source]
Bases: object
A placement algorithm that can place a machine graph onto a machine choosing chips radiating in a circle from the boot chip
__call__(machine_graph, machine, plan_n_timesteps)[source]
Parameters: machine_graph (MachineGraph) – The machine_graph to place machine (Machine) – The machine with respect to which to partition the application graph plan_n_timesteps (int) – number of timesteps to plan for A set of placements Placements PacmanPlaceException – If something goes wrong with the placement
class pacman.operations.placer_algorithms.OneToOnePlacer[source]
Bases: pacman.operations.placer_algorithms.radial_placer.RadialPlacer
Placer that puts vertices which are directly connected to only its destination on the same chip
__call__(machine_graph, machine, plan_n_timesteps)[source]
Parameters: machine_graph (MachineGraph) – The machine_graph to place machine (Machine) – The machine with respect to which to partition the application graph plan_n_timesteps (int) – number of timesteps to plan for A set of placements Placements PacmanPlaceException – If something goes wrong with the placement
class pacman.operations.placer_algorithms.SpreaderPlacer[source]
Bases: pacman.operations.placer_algorithms.one_to_one_placer.OneToOnePlacer
Places vertices on as many chips as available with a effort to reduce the number of packets being received by the router in total.
Parameters: machine_graph (MachineGraph) – the machine graph machine (Machine) – the SpiNNaker machine n_keys_map (AbstractMachinePartitionNKeysMap) – the n keys from partition map plan_n_timesteps (int) – number of timesteps to plan for placements. Placements
ITERATIONS = 4
STEPS = 4
__call__(machine_graph, machine, n_keys_map, plan_n_timesteps)[source]
Parameters: machine_graph (MachineGraph) – the machine graph machine (Machine) – the SpiNNaker machine n_keys_map (AbstractMachinePartitionNKeysMap) – the n keys from partition map plan_n_timesteps (int) – number of timesteps to plan for placements. Placements
class pacman.operations.placer_algorithms.ConnectiveBasedPlacer[source]
Bases: pacman.operations.placer_algorithms.radial_placer.RadialPlacer
A radial algorithm that can place a machine graph onto a machine using a circle out behaviour from a Ethernet at a given point and which will place things that are most connected closest to each other
__call__(machine_graph, machine, plan_n_timesteps)[source]
Parameters: machine_graph (MachineGraph) – The machine_graph to place machine (Machine) – The machine with respect to which to partition the application graph plan_n_timesteps (int) – number of timesteps to plan for A set of placements Placements PacmanPlaceException – If something goes wrong with the placement
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2021-12-02 01:22:25
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https://calculator.academy/pin-shear-calculator/
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Enter the applied force, pin diameter, and plate thickness into the calculator to determine the pin shear stress and bearing area stress.
## Pin Shear Formula
The following formula is used to calculate pin shear.
SS = 4*AF / (pi*D^2)
BS = AF / (t*d)
• Where SS is the pin shear stress (N/mm^2, lbs/in^2)
• BS is the bearing area stress (N/mm^2, lb/in^2)
• AF is the applied force (N, lbs)
• D is the diameter of the pin (mm , in)
• t is the thickness of the plate. (mm , in)
## Pin Shear Definition
What is pin shear?
Pin shear is the average shear that a pin will see given an applied force and diameter. Pin shear is used to design safe bearing and other mechanical assemblies.
Typically the pin shear is used with the ultimate strength and factor of safety to determine the proper design setup.
However, since the pins can hold a large amount of shear stress relative to other components, they are rarely the limiting factor.
## Example Problem
How to calculate average pin shear?
First, determine the applied force acting on the pin assembly. In this example, the applied force is measured to be 5000 N.
Next, determine the diameter of the pin. In this example, the pin is a diameter of 35mm.
Finally, calculate the pin shear using the formula above:
SS = 4*AF / (pi*D^2)
SS = 4*5000 / (3.14159*35^2)
SS = 5.196 N/mm^2
If you want to further calculate the total bearing area stress, determine the thickness of the plate, (in this case, 40mm), and use the formula above:
BS = AF / (t*d)
BS = 5000 / (35*40)
BS = 3.571 N/mm^2
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2023-01-31 03:36:01
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https://electronics.stackexchange.com/questions/96718/multiplexing-and-electric-capacitance
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Multiplexing and electric capacitance
I'm working on a project, it's basically a hand-drawn paper keyboard based on capacitance.
I was thinking I can multiplex the different lines with shift register (74HC595) and switch (4066), and as I've got a lot of these here I gave them a try. But apparently not...
So I'm looking for a solution.
I've already found that : Multiplexer for Capacitive Sensors (MUXC01)
Can you confirm that this component could work or do you have any better idea about how to multiplex these keys ?
• Erm, that looks pretty much exactly like a '595... – Ignacio Vazquez-Abrams Jan 17 '14 at 8:21
• From what I understand in the doc it's not exactly that. They say it can switch one input (Signal IN, pin 10) to one or multiple output. So it's a kind of 2 * 4066 controlled by a shift register like a do but in a single. But they also mention "Very well suited for multiple-capacitance measurement". They also talk about using it both way. So in that case I would be able to measure up to 9 different capacitance on a single output. But as I say in my question, I'm not sure I understand correctly the datasheet :/ – Emmanuel Istace Jan 17 '14 at 8:29
• Ah, I see. It allows faster signal switching that a '595 (since you don't need to wait for the serial propagation to flip the bit), but doesn't really offer much beyond that. Even that could be accommodated for with judicious use of the output enable. – Ignacio Vazquez-Abrams Jan 17 '14 at 8:37
I'm working with little capacitance difference and the breadboard itself have a capacitance. One or two connection, well that's ok, But I was having a long bus with at least 30 wires (so times 2 for pins connection) in series to connect the input of the AtMega328 with the 1M$\Omega$ to all the output of the 4066. The result was a big disturbance in the force ;)
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2020-02-26 08:28:57
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https://terrytao.wordpress.com/2010/02/02/254a-notes-4-the-semi-circular-law/
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We can now turn attention to one of the centerpiece universality results in random matrix theory, namely the Wigner semi-circle law for Wigner matrices. Recall from previous notes that a Wigner Hermitian matrix ensemble is a random matrix ensemble ${M_n = (\xi_{ij})_{1 \leq i,j \leq n}}$ of Hermitian matrices (thus ${\xi_{ij} = \overline{\xi_{ji}}}$; this includes real symmetric matrices as an important special case), in which the upper-triangular entries ${\xi_{ij}}$, ${i>j}$ are iid complex random variables with mean zero and unit variance, and the diagonal entries ${\xi_{ii}}$ are iid real variables, independent of the upper-triangular entries, with bounded mean and variance. Particular special cases of interest include the Gaussian Orthogonal Ensemble (GOE), the symmetric random sign matrices (aka symmetric Bernoulli ensemble), and the Gaussian Unitary Ensemble (GUE).
In previous notes we saw that the operator norm of ${M_n}$ was typically of size ${O(\sqrt{n})}$, so it is natural to work with the normalised matrix ${\frac{1}{\sqrt{n}} M_n}$. Accordingly, given any ${n \times n}$ Hermitian matrix ${M_n}$, we can form the (normalised) empirical spectral distribution (or ESD for short)
$\displaystyle \mu_{\frac{1}{\sqrt{n}} M_n} := \frac{1}{n} \sum_{j=1}^n \delta_{\lambda_j(M_n) / \sqrt{n}},$
of ${M_n}$, where ${\lambda_1(M_n) \leq \ldots \leq \lambda_n(M_n)}$ are the (necessarily real) eigenvalues of ${M_n}$, counting multiplicity. The ESD is a probability measure, which can be viewed as a distribution of the normalised eigenvalues of ${M_n}$.
When ${M_n}$ is a random matrix ensemble, then the ESD ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ is now a random measure – i.e. a random variable taking values in the space ${\hbox{Pr}({\mathbb R})}$ of probability measures on the real line. (Thus, the distribution of ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ is a probability measure on probability measures!)
Now we consider the behaviour of the ESD of a sequence of Hermitian matrix ensembles ${M_n}$ as ${n \rightarrow \infty}$. Recall from Notes 0 that for any sequence of random variables in a ${\sigma}$-compact metrisable space, one can define notions of convergence in probability and convergence almost surely. Specialising these definitions to the case of random probability measures on ${{\mathbb R}}$, and to deterministic limits, we see that a sequence of random ESDs ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converge in probability (resp. converge almost surely) to a deterministic limit ${\mu \in \hbox{Pr}({\mathbb R})}$ (which, confusingly enough, is a deterministic probability measure!) if, for every test function ${\varphi \in C_c({\mathbb R})}$, the quantities ${\int_{\mathbb R} \varphi\ d\mu_{\frac{1}{\sqrt{n}} M_n}}$ converge in probability (resp. converge almost surely) to ${\int_{\mathbb R} \varphi\ d\mu}$.
Remark 1 As usual, convergence almost surely implies convergence in probability, but not vice versa. In the special case of random probability measures, there is an even weaker notion of convergence, namely convergence in expectation, defined as follows. Given a random ESD ${\mu_{\frac{1}{\sqrt{n}} M_n}}$, one can form its expectation ${{\bf E} \mu_{\frac{1}{\sqrt{n}} M_n} \in \hbox{Pr}({\mathbb R})}$, defined via duality (the Riesz representation theorem) as
$\displaystyle \int_{\mathbb R} \varphi\ d{\bf E} \mu_{\frac{1}{\sqrt{n}} M_n} := {\bf E} \int_{\mathbb R} \varphi\ d \mu_{\frac{1}{\sqrt{n}} M_n};$
this probability measure can be viewed as the law of a random eigenvalue ${\frac{1}{\sqrt{n}}\lambda_i(M_n)}$ drawn from a random matrix ${M_n}$ from the ensemble. We then say that the ESDs converge in expectation to a limit ${\mu \in \hbox{Pr}({\mathbb R})}$ if ${{\bf E} \mu_{\frac{1}{\sqrt{n}} M_n}}$ converges the vague topology to ${\mu}$, thus
$\displaystyle {\bf E} \int_{\mathbb R} \varphi\ d \mu_{\frac{1}{\sqrt{n}} M_n} \rightarrow \int_{\mathbb R} \varphi\ d\mu$
for all ${\phi \in C_c({\mathbb R})}$.
In general, these notions of convergence are distinct from each other; but in practice, one often finds in random matrix theory that these notions are effectively equivalent to each other, thanks to the concentration of measure phenomenon.
Exercise 1 Let ${M_n}$ be a sequence of ${n \times n}$ Hermitian matrix ensembles, and let ${\mu}$ be a continuous probability measure on ${{\mathbb R}}$.
• Show that ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converges almost surely to ${\mu}$ if and only if ${\mu_{\frac{1}{\sqrt{n}}}(-\infty,\lambda)}$ converges almost surely to ${\mu(-\infty,\lambda)}$ for all ${\lambda \in {\mathbb R}}$.
• Show that ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converges in probability to ${\mu}$ if and only if ${\mu_{\frac{1}{\sqrt{n}}}(-\infty,\lambda)}$ converges in probability to ${\mu(-\infty,\lambda)}$ for all ${\lambda \in {\mathbb R}}$.
• Show that ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converges in expectation to ${\mu}$ if and only if ${\mathop{\mathbb E} \mu_{\frac{1}{\sqrt{n}}}(-\infty,\lambda)}$ converges to ${\mu(-\infty,\lambda)}$ for all ${\lambda \in {\mathbb R}}$.
We can now state the Wigner semi-circular law.
Theorem 1 (Semicircular law) Let ${M_n}$ be the top left ${n \times n}$ minors of an infinite Wigner matrix ${(\xi_{ij})_{i,j \geq 1}}$. Then the ESDs ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converge almost surely (and hence also in probability and in expectation) to the Wigner semi-circular distribution
$\displaystyle \mu_{sc} := \frac{1}{2\pi} (4-|x|^2)^{1/2}_+\ dx. \ \ \ \ \ (1)$
A numerical example of this theorem in action can be seen at the MathWorld entry for this law.
The semi-circular law nicely complements the upper Bai-Yin theorem from Notes 3, which asserts that (in the case when the entries have finite fourth moment, at least), the matrices ${\frac{1}{\sqrt{n}} M_n}$ almost surely has operator norm at most ${2+o(1)}$. Note that the operator norm is the same thing as the largest magnitude of the eigenvalues. Because the semi-circular distribution (1) is supported on the interval ${[-2,2]}$ with positive density on the interior of this interval, Theorem 1 easily supplies the lower Bai-Yin theorem, that the operator norm of ${\frac{1}{\sqrt{n}} M_n}$ is almost surely at least ${2-o(1)}$, and thus (in the finite fourth moment case) the norm is in fact equal to ${2+o(1)}$. Indeed, we have just shown that the semi-circular law provides an alternate proof of the lower Bai-Yin bound (Proposition 11 of Notes 3).
As will hopefully become clearer in the next set of notes, the semi-circular law is the noncommutative (or free probability) analogue of the central limit theorem, with the semi-circular distribution (1) taking on the role of the normal distribution. Of course, there is a striking difference between the two distributions, in that the former is compactly supported while the latter is merely subgaussian. One reason for this is that the concentration of measure phenomenon is more powerful in the case of ESDs of Wigner matrices than it is for averages of iid variables; compare the concentration of measure results in Notes 3 with those in Notes 1.
There are several ways to prove (or at least to heuristically justify) the semi-circular law. In this set of notes we shall focus on the two most popular methods, the moment method and the Stieltjes transform method, together with a third (heuristic) method based on Dyson Brownian motion (Notes 3b). In the next set of notes we shall also study the free probability method, and in the set of notes after that we use the determinantal processes method (although this method is initially only restricted to highly symmetric ensembles, such as GUE).
— 1. Preliminary reductions —
Before we begin any of the proofs of the semi-circular law, we make some simple observations which will reduce the difficulty of the arguments in the sequel.
The first observation is that the Cauchy interlacing law (Exercise 14 from Notes 3a) shows that the ESD of ${\frac{1}{\sqrt{n}} M_n}$ is very stable in ${n}$. Indeed, we see from the interlacing law that
$\displaystyle \frac{n}{m} \mu_{\frac{1}{\sqrt{n}} M_n}( -\infty, \lambda / \sqrt{n}) - \frac{n-m}{m} \leq \mu_{\frac{1}{\sqrt{m}} M_m}( -\infty, \lambda / \sqrt{m})$
$\displaystyle \leq \frac{n}{m} \mu_{\frac{1}{\sqrt{n}} M_n}( -\infty, \lambda / \sqrt{n})$
for any threshold ${\lambda}$ and any ${n > m > 0}$.
Exercise 2 Using this observation, show that to establish the semi-circular law (in any of the three senses of convergence), it suffices to do so for an arbitrary lacunary sequence ${n_1, n_2, \ldots}$ of ${n}$ (thus ${n_{j+1}/n_j \geq c}$ for some ${c>1}$ and all ${j}$).
The above lacunary reduction does not help one establish convergence in probability or expectation, but will be useful when establishing almost sure convergence, as it significantly reduces the inefficiency of the union bound. (Note that a similar lacunary reduction was also used to prove the strong law of large numbers in Notes 1.)
Next, we exploit the stability of the ESD with respect to perturbations, by taking advantage of the Weilandt-Hoffmann inequality
$\displaystyle \sum_{j=1}^n |\lambda_j(A+B)-\lambda_j(A)|^2 \leq \|B\|_F^2 \ \ \ \ \ (2)$
for Hermitian matrices ${A, B}$, where ${\|B\|_F := (\hbox{tr} B^2)^{1/2}}$ is the Frobenius norm of ${B}$. (This inequality was established in Exercise 6 or Exercise 11 of Notes 3a.) We convert this inequality into an inequality about ESDs:
Lemma 2 For any ${n \times n}$ Hermitian matrices ${A, B}$, any ${\lambda}$, and any ${\epsilon > 0}$, we have
$\displaystyle \mu_{\frac{1}{\sqrt{n}}(A+B)}(-\infty, \lambda) \leq \mu_{\frac{1}{\sqrt{n}}(A)}(-\infty, \lambda+\epsilon) + \frac{1}{\epsilon^2 n^2} \|B\|_F^2$
and similarly
$\displaystyle \mu_{\frac{1}{\sqrt{n}}(A+B)}(-\infty, \lambda) \geq \mu_{\frac{1}{\sqrt{n}}(A)}(-\infty, \lambda-\epsilon) - \frac{1}{\epsilon^2 n^2} \|B\|_F^2.$
Proof: We just prove the first inequality, as the second is similar (and also follows from the first, by reversing the sign of ${A, B}$).
Let ${\lambda_i(A+B)}$ be the largest eigenvalue of ${A+B}$ less than ${\lambda \sqrt{n}}$, and let ${\lambda_j(A)}$ be the largest eigenvalue of ${A}$ less than ${(\lambda+\epsilon) \sqrt{n}}$. Our task is to show that
$\displaystyle i \leq j + \frac{1}{\epsilon^2 n} \|B\|_F^2.$
If ${i \leq j}$ then we are clearly done, so suppose that ${i>j}$. Then we have ${|\lambda_l(A+B)-\lambda_l(A)| \geq \epsilon \sqrt{n}}$ for all ${j < l \leq i}$, and hence
$\displaystyle \sum_{j=1}^n |\lambda_j(A+B)-\lambda_j(A)|^2 \geq \epsilon^2 (j-i) n.$
The claim now follows from (2). $\Box$
This has the following corollary:
Exercise 3 (Stability of ESD laws wrt small perturbations) Let ${M_n}$ be a sequence of random Hermitian matrix ensembles such that ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converges almost surely to a limit ${\mu}$. Let ${N_n}$ be another sequence of Hermitian random matrix ensembles such that ${\frac{1}{n^2} \|N_n\|_F^2}$ converges almost surely to zero. Show that ${\mu_{\frac{1}{\sqrt{n}}(M_n+N_n)}}$ converges almost surely to ${\mu}$.
Show that the same claim holds if “almost surely” is replaced by “in probability” or “in expectation” throughout.
Informally, this exercise allows us to discard any portion of the matrix which is ${o(n)}$ in the Frobenius norm. For instance, the diagonal entries of ${M_n}$ have a Frobenius norm of ${O(\sqrt{n})}$ almost surely, by the strong law of large numbers. Hence, without loss of generality, we may set the diagonal equal to zero for the purposes of the semi-circular law.
One can also remove any component of ${M_n}$ that is of rank ${o(n)}$:
Exercise 4 (Stability of ESD laws wrt small rank perturbations) Let ${M_n}$ be a sequence of random Hermitian matrix ensembles such that ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ converges almost surely to a limit ${\mu}$. Let ${N_n}$ be another sequence of random Hermitian matrix ensembles such that ${\frac{1}{n} \hbox{rank}(N_n)}$ converges almost surely to zero. Show that ${\mu_{\frac{1}{\sqrt{n}}(M_n+N_n)}}$ converges almost surely to ${\mu}$. (Hint: use the Weyl inequalities instead of the Wielandt-Hoffman law.)
Show that the same claim holds if “almost surely” is replaced by “in probability” or “in expectation” throughout.
In a similar vein, we may apply the truncation argument (much as was done for the central limit theorem in Notes 2) to reduce the semi-circular law to the bounded case:
Exercise 5 Show that in order to prove the semi-circular law (in the almost sure sense), it suffices to do so under the additional hypothesis that the random variables are bounded. Similarly for the convergence in probability or in expectation senses.
Remark 2 These facts ultimately rely on the stability of eigenvalues with respect to perturbations. This stability is automatic in the Hermitian case, but for non-symmetric matrices, serious instabilities can occur due to the presence of pseudospectrum. We will discuss this phenomenon more in later lectures (but see also this earlier blog post).
— 2. The moment method —
We now prove the semi-circular law via the method of moments, which we have already used several times in the previous notes. In order to use this method, it is convenient to use the preceding reductions to assume that the coefficients are bounded, the diagonal vanishes, and that ${n}$ ranges over a lacunary sequence. We will implicitly assume these hypotheses throughout the rest of the section.
As we have already discussed the moment method extensively, much of the argument here will be delegated to exercises. A full treatment of these computations can be found in the book of Bai and Silverstein.
The basic starting point is the observation that the moments of the ESD ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ can be written as normalised traces of powers of ${M_n}$:
$\displaystyle \int_{\mathbb R} x^k\ d\mu_{\frac{1}{\sqrt{n}} M_n}(x) = \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k. \ \ \ \ \ (3)$
In particular, on taking expectations, we have
$\displaystyle \int_{\mathbb R} x^k\ d{\bf E}\mu_{\frac{1}{\sqrt{n}} M_n}(x) = {\bf E} \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k.$
From concentration of measure (and the Bai-Yin theorem) for the operator norm of a random matrix (Proposition 7 of Notes 3), we see that the ${{\bf E} \mu_{\frac{1}{\sqrt{n}} M_n}}$ are uniformly subgaussian, indeed we have
$\displaystyle {\bf E} \mu_{\frac{1}{\sqrt{n}} M_n}\{ |x| \geq \lambda \} \leq C e^{-c \lambda^2 n^2}$
for ${\lambda > C}$, where ${C, c}$ are absolute (so the decay in fact improves quite rapidly with ${n}$). From this and the moment continuity theorem (Theorem 4 of Notes 2), we can now establish the semi-circular law through computing the mean and variance of moments:
Exercise 6
• Show that to prove convergence in expectation to the semi-circular law, it suffices to show that
$\displaystyle {\bf E} \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k = \int_{\mathbb R} x^k\ d\mu_{sc}(x) + o_k(1) \ \ \ \ \ (4)$
for ${k=1,2,\ldots}$, where ${o_k(1)}$ is an expression that goes to zero as ${n \rightarrow \infty}$ for fixed ${k}$ (and fixed choice of coefficient distribution ${\xi}$).
• Show that to prove convergence in probability to the semi-circular law, it suffices to show (4) together with the variance bound
$\displaystyle {\bf Var}(\frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k) = o_k(1) \ \ \ \ \ (5)$
for ${k=1,2,\ldots}$.
• Show that to prove almost sure convergence to the semi-circular law, it suffices to show (4) together with the variance bound
$\displaystyle {\bf Var}(\frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k) = O_k(n^{-c_k}) \ \ \ \ \ (6)$
for ${k=1,2,\ldots}$ and some ${c_k>0}$. (Note here that it is useful to restrict ${n}$ to a lacunary sequence!)
Ordinarily, computing second-moment quantities such as the left-hand side of (5) is harder than computing first-moment quantities such as (4). But one can obtain the required variance bounds from concentration of measure:
Exercise 7
• When ${k}$ is a positive even integer, Use Talagrand’s inequality and convexity of the Schatten norm ${\|A\|_{S^k} = (\hbox{tr}(A^k))^{1/k}}$ to establish (6) (and hence (5)) when ${k}$ is even.
• For ${k}$ odd, the formula ${\|A\|_{S^k} = (\hbox{tr}(A^k))^{1/k}}$ still applies as long as ${A}$ is positive definite. Applying this observation, the Bai-Yin theorem, and Talagrand’s inequality to the ${S^k}$ norms of ${\frac{1}{\sqrt{n}} M_n + c I_n}$ for a constant ${c>2}$, establish (6) (and hence (5)) when ${k}$ is odd also.
Remark 3 More generally, concentration of measure results (such as Talagrand’s inequality) can often be used to automatically upgrade convergence in expectation to convergence in probability or almost sure convergence. We will not attempt to formalise this principle here.
It is not difficult to establish (6), (5) through the moment method as well. Indeed, recall from Theorem 10 of Notes 3 that we have the expected moment
$\displaystyle {\bf E} \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k = C_{k/2} + o_k(1) \ \ \ \ \ (7)$
for all ${k=1,2,\ldots}$, where the Catalan number ${C_{k/2}}$ is zero when ${k}$ is odd, and is equal to
$\displaystyle C_{k/2} := \frac{k!}{(k/2+1)! (k/2)!} \ \ \ \ \ (8)$
for ${k}$ even.
Exercise 8 By modifying the proof of that theorem, show that
$\displaystyle {\bf E} |\frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k|^2 = C_{k/2}^2 + o_k(1) \ \ \ \ \ (9)$
and deduce (5). By refining the error analysis (e.g. using Theorem 12 of Notes 3, also establish (6).
In view of the above computations, the establishment of the semi-circular law now reduces to computing the moments of the semi-circular distribution:
Exercise 9 Show that for any ${k=1,2,3,\ldots}$, one has
$\displaystyle \int_{\mathbb R} x^k\ d\mu_{sc}(x) = C_{k/2}.$
(Hint: use a trigonometric substitution ${x = 2 \cos \theta}$, and then express the integrand in terms of Fourier phases ${e^{in\theta}}$.)
This concludes the proof of the semi-circular law (for any of the three modes of convergence).
Remark 4 In the spirit of the Lindeberg exchange method, observe that Exercise (9) is unnecessary if one already knows that the semi-circular law holds for at least one ensemble of Wigner matrices (e.g. the GUE ensemble). Indeed, Exercise 9 can be deduced from such a piece of knowledge. In such a situation, it is not necessary to actually compute the main term ${C_{k/2}}$ on the right of (4); it would be sufficient to know that that limit is universal, in that it does not depend on the underlying distribution. In fact, it would even suffice to establish the slightly weaker statement
$\displaystyle {\bf E} \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n)^k = {\bf E} \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M'_n)^k + o_k(1)$
whenever ${M_n, M'_n}$ are two ensembles of Wigner matrices arising from different underlying distributions (but still normalised to have mean zero, unit variance, and to be bounded (or at worst subgaussian)). We will take advantage of this perspective later in these notes.
— 3. The Stieltjes transform method —
The moment method was computationally intensive, but straightforward. As noted in Remark 4, even without doing much of the algebraic computation, it is clear that the moment method will show that some universal limit for Wigner matrices exists (or, at least, that the differences between the distributions of two different Wigner matrices converge to zero). But it is not easy to see from this method why the limit should be given by the semi-circular law, as opposed to some other distribution (although one could eventually work this out from an inverse moment computation).
When studying the central limit theorem, we were able to use the Fourier method to control the distribution of random matrices in a cleaner way than in the moment method. Analogues of this method exist, but require non-trivial formulae from noncommutative Fourier analysis, such as the Harish-Chandra integration formula (and also only work for highly symmetric ensembles, such as GUE or GOE), and will not be discussed in this course. (Our later notes on determinantal processes, however, will contain some algebraic identities related in some ways to the noncommutative Fourier-analytic approach.)
We now turn to another method, the Stieltjes transform method, which uses complex-analytic methods rather than Fourier-analytic methods, and has turned out to be one of the most powerful and accurate tools in dealing with the ESD of random Hermitian matrices. Whereas the moment method started from the identity (3), the Stieltjes transform method proceeds from the identity
$\displaystyle \int_{\mathbb R} \frac{1}{x-z}\ d\mu_{\frac{1}{\sqrt{n}} M_n}(x) = \frac{1}{n} \hbox{tr} (\frac{1}{\sqrt{n}} M_n-zI)^{-1}$
for any complex ${z}$ not in the support of ${\mu_{\frac{1}{\sqrt{n}} M_n}}$. We refer to the expression on the left-hand side as the Stieltjes transform of ${M_n}$ or of ${\mu_{\frac{1}{\sqrt{n}} M_n}}$, and denote it by ${s_{\mu_{\frac{1}{n}} M_n}}$ or as ${s_n}$ for short. The expression ${(\frac{1}{\sqrt{n}} M_n-zI)^{-1}}$ is the normalised resolvent of ${M_n}$, and plays an important role in the spectral theory of that matrix. Indeed, in contrast to general-purpose methods such as the moment method, the Stieltjes transform method draws heavily on the specific linear-algebraic structure of this problem, and in particular on the rich structure of resolvents.
On the other hand, the Stieltjes transform can be viewed as a generating function of the moments via the Taylor series expansion
$\displaystyle s_n(z) = -\frac{1}{z} - \frac{1}{z^2} \frac{1}{n} \hbox{tr} M_n - \frac{1}{z^3} \frac{1}{n} \hbox{tr} M_n^2 - \ldots,$
valid for ${z}$ sufficiently large. This is somewhat (though not exactly) analogous to how the characteristic function ${{\bf E} e^{itX}}$ of a scalar random variable can be viewed as a generating function of the moments ${{\bf E} X^k}$.
Now let us study the Stieltjes transform more systematically. Given any probability measure ${\mu}$ on the real line, we can form its Stieltjes transform
$\displaystyle s_\mu(z) := \int_{\mathbb R} \frac{1}{x-z}\ d\mu(x)$
for any ${z}$ outside of the support of ${\mu}$; in particular, the Stieltjes transform is well-defined on the upper and lower half-planes in the complex plane. Even without any further hypotheses on ${\mu}$ other than it is a probability measure, we can say a remarkable amount about how this transform behaves in ${z}$. Applying conjugations we obtain the symmetry
$\displaystyle \overline{s_\mu(z)} = s_\mu(\overline{z}) \ \ \ \ \ (10)$
so we may as well restrict attention to ${z}$ in the upper half-plane (say). Next, from the trivial bound
$\displaystyle |\frac{1}{x-z}| \leq \frac{1}{|\hbox{Im}(z)|}$
one has the pointwise bound
$\displaystyle |s_\mu(z)| \leq \frac{1}{|\hbox{Im}(z)|}. \ \ \ \ \ (11)$
In a similar spirit, an easy application of dominated convergence gives the asymptotic
$\displaystyle s_\mu(z) = \frac{-1+o_\mu(1)}{z} \ \ \ \ \ (12)$
where ${o_\mu(1)}$ is an expression that, for any fixed ${\mu}$, goes to zero as ${z}$ goes to infinity non-tangentially in the sense that ${|\hbox{Re}(z)|/|\hbox{Im(z)}|}$ is kept bounded, where the rate of convergence is allowed to depend on ${\mu}$. From differentiation under the integral sign (or an application of Morera’s theorem and Fubini’s theorem) we see that ${s_\mu(z)}$ is complex analytic on the upper and lower half-planes; in particular, it is smooth away from the real axis. From the Cauchy integral formula (or differentiation under the integral sign) we in fact get some bounds for higher derivatives of the Stieltjes transform away from this axis:
$\displaystyle |\frac{d^j}{dz^j} s_\mu(z)| = O_j( \frac{1}{|\hbox{Im}(z)|^{j+1}} ). \ \ \ \ \ (13)$
Informally, ${s_\mu}$ “behaves like a constant” at scales significantly less than the distance ${|\hbox{Im}(z)|}$ to the real axis; all the really interesting action here is going on near that axis.
The imaginary part of the Stieltjes transform is particularly interesting. Writing ${z = a+ib}$, we observe that
$\displaystyle \hbox{Im}\frac{1}{x-z} = \frac{b}{(x-a)^2 + b^2} > 0$
and so we see that
$\displaystyle \hbox{Im}( s_\mu(z) ) > 0$
for ${z}$ in the upper half-plane; thus ${s_\mu}$ is a complex-analytic map from the upper half-plane to itself, a type of function known as a Herglotz function. (In fact, all complex-analytic maps from the upper half-plane to itself that obey the asymptotic (12) are of this form; this is a special case of the Herglotz representation theorem, which also gives a slightly more general description in the case when the asymptotic (12) is not assumed. A good reference for this material and its consequences is this book of Garnett.)
One can also express the imaginary part of the Stieltjes transform as a convolution
$\displaystyle \hbox{Im}( s_\mu(a+ib) ) = \pi \mu * P_b(a) \ \ \ \ \ (14)$
where ${P_b}$ is the Poisson kernel
$\displaystyle P_b(x) := \frac{1}{\pi} \frac{b}{x^2+b^2} = \frac{1}{b} P_1(\frac{x}{b}).$
As is well known, these kernels form a family of approximations to the identity, and thus ${\mu * P_b}$ converges in the vague topology to ${\mu}$ (see e.g. my notes on distributions). Thus we see that
$\displaystyle \hbox{Im} s_\mu(\cdot+ib) \rightharpoonup \pi \mu$
as ${b \rightarrow 0^+}$ in the vague topology,or equivalently (by (10)) that
$\displaystyle \frac{s_\mu(\cdot+ib) - s_\mu(\cdot-ib)}{2\pi i} \rightharpoonup \mu \ \ \ \ \ (15)$
as ${b \rightarrow 0^+}$ (this is closely related to the Plemelj formula in potential theory). Thus we see that a probability measure ${\mu}$ can be recovered in terms of the limiting behaviour of the Stieltjes transform on the real axis.
A variant of the above machinery gives us a criterion for convergence:
Exercise 10 (Stieltjes continuity theorem) Let ${\mu_n}$ be a sequence of random probability measures on the real line, and let ${\mu}$ be a deterministic probability measure.
• ${\mu_n}$ converges almost surely to ${\mu}$ in the vague topology if and only if ${s_{\mu_n}(z)}$ converges almost surely to ${s_\mu(z)}$ for every ${z}$ in the upper half-plane.
• ${\mu_n}$ converges in probability to ${\mu}$ in the vague topology if and only if ${s_{\mu_n}(z)}$ converges in probability to ${s_\mu(z)}$ for every ${z}$ in the upper half-plane.
• ${\mu_n}$ converges in expectation to ${\mu}$ in the vague topology if and only if ${{\bf E} s_{\mu_n}(z)}$ converges to ${s_\mu(z)}$ for every ${z}$ in the upper half-plane.
(Hint: The “only if” parts are fairly easy. For the “if” parts, take a test function ${\phi \in C_c({\mathbb R})}$ and approximate ${\int_{\mathbb R} \phi\ d\mu}$ by ${\int_{\mathbb R} \phi*P_b\ d\mu = \frac{1}{\pi} \int_{\mathbb R} s_\mu(a+ib) \phi(a)\ da}$. Then approximate this latter integral in turn by a Riemann sum, using (13).)
Thus, to prove the semi-circular law, it suffices to show that for each ${z}$ in the upper half-plane, the Stieltjes transform
$\displaystyle s_n(z) = s_{\mu_{\frac{1}{\sqrt{n}} M_n}}(z) = \frac{1}{n} \hbox{tr}( \frac{1}{\sqrt{n}} M_n - zI )^{-1}$
converges almost surely (and thus in probability and in expectation) to the Stieltjes transform ${s_{\mu_{sc}}(z)}$ of the semi-circular law.
It is not difficult to compute the Stieltjes transform ${s_{\mu_{sc}}}$ of the semi-circular law, but let us hold off on that task for now, because we want to illustrate how the Stieltjes transform method can be used to find the semi-circular law, even if one did not know this law in advance, by directly controlling ${s_n(z)}$. We will fix ${z=a+ib}$ to be a complex number not on the real line, and allow all implied constants in the discussion below to depend on ${a}$ and ${b}$ (we will focus here only on the behaviour as ${n \rightarrow \infty}$).
The main idea here is predecessor comparison: to compare the transform ${s_n(z)}$ of the ${n \times n}$ matrix ${M_n}$ with the transform ${s_{n-1}(z)}$ of the top left ${n-1 \times n-1}$ minor ${M_{n-1}}$, or of other minors. For instance, we have the Cauchy interlacing law (Exercise 14 from Notes 3a), which asserts that the eigenvalues ${\lambda_1(M_{n-1}),\ldots,\lambda_{n-1}(M_{n-1})}$ of ${M_{n-1}}$ intersperse that of ${\lambda_1(M_n),\ldots,\lambda_n(M_n)}$. This implies that for a complex number ${a+ib}$ with ${b>0}$, the difference
$\displaystyle \sum_{j=1}^{n-1} \frac{b}{(\lambda_j(M_{n-1})/\sqrt{n}-a)^2 + b^2} - \sum_{j=1}^{n} \frac{b}{(\lambda_j(M_{n})/\sqrt{n}-a)^2 + b^2}$
is an alternating sum of evaluations of the function ${x \mapsto \frac{b}{(x-a)^2+b^2}}$. The total variation of this function is ${O( 1 )}$ (recall that we are suppressing dependence of constaants on ${a,b}$), and so the alternating sum above is ${O(1)}$. Writing this in terms of the Stieltjes transform, we conclude that
$\displaystyle \sqrt{n(n-1)} s_{n-1}( \frac{\sqrt{n}}{\sqrt{n-1}}(a+ib) ) - n s_n( a+ib ) = O(1).$
Applying (13) to approximate ${s_{n-1}( \frac{\sqrt{n}}{\sqrt{n-1}}(a+ib) )}$ by ${s_{n-1}(a+ib)}$, we conclude that
$\displaystyle s_n(a+ib) = s_{n-1}(a+ib) + O( \frac{1}{n} ). \ \ \ \ \ (16)$
So for fixed ${z=a+ib}$ away from the real axis, the Stieltjes transform ${s_n(z)}$ is quite stable in ${n}$.
This stability has the following important consequence. Observe that while the left-hand side of (16) depends on the ${n \times n}$ matrix ${M_n}$, the right-hand side depends only on the top left minor ${M_{n-1}}$ of that matrix. In particular, it is independent of the ${n^{th}}$ row and column of ${M_n}$. This implies that this entire row and column has only a limited amount of influence on the Stieltjes transform ${s_n(a+ib)}$: no matter what value one assigns to this row and column (including possibly unbounded values, as long as one keeps the matrix Hermitian of course), the transform ${s_n(a+ib)}$ can only move by ${O( \frac{|a|+|b|}{b^2 n} )}$.
By permuting the rows and columns, we obtain that in fact any row or column of ${M_n}$ can influence ${s_n(a+ib)}$ is at most ${O( \frac{1}{n} )}$. (This is closely related to the observation in Exercise 4 that low rank perturbations do not significantly affect the ESD.) On the other hand, the rows of (the upper triangular portion of) ${M_n}$ are jointly independent. When ${M_n}$ is a Wigner random matrix, we can then apply a standard concentration of measure result, such as McDiarmid’s inequality (Theorem 7 from Notes 1) to conclude concetration of ${s_n}$ around its mean:
$\displaystyle {\bf P}( |s_n(a+ib) - {\Bbb E} s_n(a+ib)| \geq \lambda/\sqrt{n} ) \leq C e^{-c\lambda^2} \ \ \ \ \ (17)$
for all ${\lambda > 0}$ and some absolute constants ${C, c > 0}$. (This is not necessarily the strongest concentration result one can establish for the Stieltjes transform, but it will certainly suffice for our discussion here.) In particular, we see from the Borel-Cantelli lemma (Exercise 24 of Notes 0a)that for any fixed ${z}$ away from the real line, ${s_n(z) - {\bf E} s_n(z)}$ converges almost surely (and thus also in probability) to zero. As a consequence, convergence of ${s_n(z)}$ in expectation automatically implies convergence in probability or almost sure convergence.
However, while concentration of measure tells us that ${s_n(z)}$ is close to its mean, it does not shed much light as to what this mean is. For this, we have to go beyond the Cauchy interlacing formula and deal with the resolvent ${(\frac{1}{\sqrt{n}} M_n - z I_n)^{-1}}$ more directly. Firstly, we observe from the linearity of trace that
$\displaystyle {\bf E} s_n(z) = \frac{1}{n} \sum_{j=1}^n {\bf E} [ (\frac{1}{\sqrt{n}} M_n - z I_n)^{-1} ]_{jj}$
where ${[A]_{jj}}$ denotes the ${jj}$ component of a matrix ${A}$. Because ${M_n}$ is a Wigner matrix, it is easy to see on permuting the rows and columns that all of the random variables ${[ (\frac{1}{\sqrt{n}} M_n - z I_n)^{-1} ]_{jj}}$ have the same distribution. Thus we may simplify the above formula as
$\displaystyle {\bf E} s_n(z) = {\bf E} [ (\frac{1}{\sqrt{n}} M_n - z I_n)^{-1} ]_{nn}. \ \ \ \ \ (18)$
So now we have to compute the last entry of an inverse of a matrix. There are of course a number of formulae for this, such as Cramer’s rule. But it will be more convenient here to use a formula based instead on the Schur complement:
Exercise 11 Let ${A_n}$ be a ${n \times n}$ matrix, let ${A_{n-1}}$ be the top left ${n-1 \times n-1}$ minor, let ${a_{nn}}$ be the bottom right entry of ${A_n}$, let ${X \in {\mathbb C}^{n-1}}$ be the right column of ${A_n}$ with the bottom right entry removed, and let ${(X')^* \in ({\mathbb C}^{n-1})^*}$ be the bottom row with the bottom right entry removed. In other words,
$\displaystyle A_n = \begin{pmatrix} A_{n-1} & X \\ (X')^* & a_{nn} \end{pmatrix}.$
Assume that ${A_n}$ and ${A_{n-1}}$ are both invertible. Show that
$\displaystyle [A_n^{-1}]_{nn} = \frac{1}{a_{nn} - (X')^* A_{n-1}^{-1} X}.$
(Hint: Solve the equation ${A_n v = e_n}$, where ${e_n}$ is the ${n^{th}}$ basis vector, using the method of Schur complements (or from first principles).)
The point of this identity is that it describes (part of) the inverse of ${A_n}$ in terms of the inverse of ${A_{n-1}}$, which will eventually provide a non-trivial recursive relationship between ${s_n(z)}$ and ${s_{n-1}(z)}$, which can then be played off against (16) to solve for ${s_n(z)}$ in the asymptotic limit ${n \rightarrow \infty}$.
In our situation, the matrix ${\frac{1}{\sqrt{n}} M_n - z I_n}$ and its minor ${\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1}}$ is automatically invertible. Inserting the above formula into (18) (and recalling that we normalised the diagonal of ${M_n}$ to vanish), we conclude that
$\displaystyle {\bf E} s_n(z) = - {\bf E} \frac{1}{z + \frac{1}{n} X^* (\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1} X }, \ \ \ \ \ (19)$
where ${X \in {\mathbb C}^{n-1}}$ is the top right column of ${M_n}$ with the bottom entry ${\xi_{nn}}$ removed.
One may be concerned that the denominator here could vanish. However, observe that ${z}$ has imaginary part ${b}$ if ${z=a+ib}$. Furthermore, from the spectral theorem we see that the imaginary part of ${(\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1}}$ is positive definite, and so ${X^* (\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1} X}$ has non-negative imaginary part. As a consequence the magnitude of the denominator here is bounded below by ${|b|}$, and so its reciprocal is ${O(1)}$ (compare with (11)). So the reciprocal here is not going to cause any discontinuity, as we are considering ${b}$ is fixed and non-zero.
Now we need to understand the expression ${X^* (\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1} X}$. We write this as ${X^* R X}$, where ${R}$ is the resolvent matrix ${R := (\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1}}$. The distribution of the random matrix ${R}$ could conceivably be quite complicated. However, the key point is that the vector ${X}$ only involves the entries of ${M_n}$ that do not lie in ${M_{n-1}}$, and so the random matrix ${R}$ and the vector ${X}$ are independent. Because of this, we can use the randomness of ${X}$ to do most of the work in understanding the expression ${X^* R X}$, without having to know much about ${R}$ at all.
To understand this, let us first condition ${R}$ to be a deterministic matrix ${R = (r_{ij})_{1 \leq i,j \leq n-1}}$, and see what we can do with the expression ${X^* R X}$.
Firstly, observe that ${R}$ will not be arbitrary; indeed, from the spectral theorem we see that ${R}$ will have operator norm at most ${O(1)}$. Meanwhile, from the Chernoff (or Hoeffding) inequality (Theorem 2 or Exercise 4 of Notes 1) we know that ${X}$ has magnitude ${O( \sqrt{n} )}$ with overwhelming probability. So we know that ${X^* R X}$ has magnitude ${O( n )}$ with overwhelming probability.
Furthermore, we can use concentration of measure as follows. Given any positive semi-definite matrix ${A}$ of operator norm ${O(1)}$, the expression ${(X^* A X)^{1/2} = \| A^{1/2} X \|}$ is a Lipschitz function of ${X}$ with operator norm ${O(1)}$. Applying Talagrand’s inequality (Theorem 9 of Notes 1) we see that this expression concentrates around its median:
$\displaystyle {\bf P}( |(X^* A X)^{1/2} - {\bf M} (X^* A X)^{1/2}| \geq \lambda ) \leq C e^{-c\lambda^2}$
for any ${\lambda > 0}$. On the other hand, ${\|A^{1/2} X\| = O( \|X\| )}$ has magnitude ${O(\sqrt{n})}$ with overwhelming probability, so the median ${{\bf M} (X^* A X)^{1/2}}$ must be ${O(\sqrt{n})}$. Squaring, we conclude that
$\displaystyle {\bf P}( |X^* A X - {\bf M} X^* A X| \geq \lambda \sqrt{n} ) \leq C e^{-c\lambda^2}$
(possibly after adjusting the absolute constants ${C, c}$). As usual, we may replace the median with the expectation:
$\displaystyle {\bf P}( |X^* A X - {\bf E} X^* A X| \geq \lambda \sqrt{n} ) \leq C e^{-c\lambda^2}$
This was for positive-definite matrices, but one can easily use the triangle inequality to generalise to self-adjoint matrices, and then to arbitrary matrices, of operator norm ${1}$, and conclude that
$\displaystyle {\bf P}( |X^* R X - {\bf E} X^* R X| \geq \lambda \sqrt{n} ) \leq C e^{-c\lambda^2} \ \ \ \ \ (20)$
for any deterministic matrix ${R}$ of operator norm ${O(1)}$.
But what is the expectation ${{\bf E} X^* R X}$? This can be expressed in components as
$\displaystyle {\bf E} X^* R X = \sum_{i=1}^{n-1} \sum_{j=1}^{n-1} {\bf E} \overline{\xi_{in}} r_{ij} \xi_{jn}$
where ${\xi_{in}}$ are the entries of ${X}$, and ${r_{ij}}$ are the entries of ${R}$. But the ${\xi_{in}}$ are iid with mean zero and variance one, so the standard second moment computation shows that this expectation is nothing more than the trace
$\displaystyle \hbox{tr}(R) = \sum_{i=1}^{n-1} r_{ii}$
of ${R}$. We have thus shown the concentration of measure result
$\displaystyle {\bf P}( |X^* R X - \hbox{tr}(R)| \geq \lambda \sqrt{n} ) \leq C e^{-c\lambda^2} \ \ \ \ \ (21)$
for any deterministic matrix ${R}$ of operator norm ${O(1)}$, and any ${\lambda > 0}$. Informally, ${X^* R X}$ is typically ${\hbox{tr}(R) +O(\sqrt{n})}$.
The bound (21) was proven for deterministic matrices, but by using conditional expectation it also applies for any random matrix ${R}$, so long as that matrix is independent of ${X}$. In particular, we may apply it to our specific matrix of interest
$\displaystyle R := (\frac{1}{\sqrt{n}} M_{n-1} - z I_{n-1})^{-1}.$
The trace of this matrix is essentially just the Stieltjes transform ${s_{n-1}(z)}$ at ${z}$. Actually, due to the normalisation factor being slightly off, we actually have
$\displaystyle \hbox{tr}(R) = n \frac{\sqrt{n}}{\sqrt{n-1}} s_{n-1}( \frac{\sqrt{n}}{\sqrt{n-1}} z ),$
but by using the smoothness (13) of the Stieltjes transform, together with the stability property (16) we can simplify this as
$\displaystyle \hbox{tr}(R) = n ( s_n(z) + o(1) ).$
In particular, from (21) and (17), we see that
$\displaystyle X^* R X = n ( {\bf E} s_n(z) + o(1) )$
with overwhelming probability. Putting this back into (19), and recalling that the denominator is bounded away from zero, we have the remarkable equation
$\displaystyle {\bf E} s_n(z) = - \frac{1}{z + {\bf E} s_n(z)} + o(1). \ \ \ \ \ (22)$
Note how this equation came by playing off two ways in which the spectral properties of a matrix ${M_n}$ interacted with that of its minor ${M_{n-1}}$; firstly via the Cauchy interlacing inequality, and secondly via the Schur complement formula.
This equation already describes the behaviour of ${{\bf E} s_n(z)}$ quite well, but we will content ourselves with understanding the limiting behaviour as ${n \rightarrow \infty}$. From (13) and Fubini’s theorem we know that the function ${{\bf E} s_n}$ is locally uniformly equicontinuous and locally uniformly bounded away from the real line. Applying the Arzelá-Ascoli theorem, we thus conclude that on a subsequence at least, ${{\bf E} s_n}$ converges locally uniformly to a limit ${s}$. This will be a Herglotz function (i.e. an analytic function mapping the upper half-plane to the upper half-plane), and taking limits in (22) (observing that the imaginary part of the denominator here is bounded away from zero) we end up with the exact equation
$\displaystyle s(z) = -\frac{1}{z+s(z)}. \ \ \ \ \ (23)$
We can of course solve this by the quadratic formula, obtaining
$\displaystyle s(z) = - \frac{z \pm \sqrt{z^2-4}}{2} = \frac{2}{z \pm \sqrt{z^2-4}}.$
To figure out what branch of the square root one has to use here, we use (12), which easily implies that
$\displaystyle s(z) = \frac{1+o(1)}{z}$
as ${z}$ goes to infinity non-tangentially away from the real line. (To justify this, one has to make the error term in (12) uniform in ${n}$, but this can be accomplished without difficulty using the Bai-Yin theorem (for instance).) Also, we know that ${s}$ has to be complex analytic (and in particular, continuous) away from the real line. From this and basic complex analysis, we conclude that
$\displaystyle s(z) = \frac{-z + \sqrt{z^2-4}}{2} \ \ \ \ \ (24)$
where ${\sqrt{z^2-4}}$ is the branch of the square root with a branch cut at ${[-2,2]}$ and which equals ${z}$ at infinity.
As there is only one possible subsequence limit of the ${{\bf E} s_n}$, we conclude that ${{\bf E} s_n}$ converges locally uniformly (and thus pointwise) to the function (24), and thus (by the concentration of measure of ${s_n(z)}$) we see that for each ${z}$, ${s_n(z)}$ converges almost surely (and in probability) to ${s(z)}$.
Exercise 12 Find a direct proof (starting from (22), (12), and the smoothness of ${{\bf E} s_n(z)}$) that ${{\bf E} s_n(z) = s(z) + o(1)}$ for any fixed ${z}$, that avoids using the Arzelá-Ascoli theorem. (The basic point here is that one has to solve the approximate equation (22), using some robust version of the quadratic formula. The fact that ${{\bf E} s_n}$ is a Herglotz function will help eliminate various unwanted possibilities, such as one coming from the wrong branch of the square root.)
To finish computing the limiting ESD of Wigner matrices, we have to figure out what probability measure ${s}$ comes from. But this is easily read off from (24) and (15):
$\displaystyle \frac{s(\cdot+ib) - s(\cdot-ib)}{2\pi i} \rightharpoonup \frac{1}{2\pi} (4-x^2)^{1/2}_+\ dx = \mu_{sc} \ \ \ \ \ (25)$
as ${b \rightarrow 0}$. Thus the semi-circular law is the only possible measure which has Stieltjes transform ${s}$, and indeed a simple application of the Cauchy integral formula and (25) shows us that ${s}$ is indeed the Stieltjes transform of ${\mu_{sc}}$.
Putting all this together, we have completed the Stieltjes transform proof of the semi-circular law.
Remark 5 In order to simplify the above exposition, we opted for a qualitative analysis of the semi-circular law here, ignoring such questions as the rate of convergence to this law. However, an inspection of the above arguments reveals that it is easy to make all of the above analysis quite quantitative, with quite reasonable control on all terms. (One has to use Exercise 12 instead of the Arzelá-Ascoli theorem if one wants everything to be quantitative.) In particular, it is not hard to use the above analysis to show that for ${|\hbox{Im}(z)| \geq n^{-c}}$ for some small absolute constant ${c>0}$, one has ${s_n(z) = s(z) + O(n^{-c})}$ with overwhelming probability. Combining this with a suitably quantitative version of the Stieltjes continuity theorem, this in turn gives a polynomial rate of convergence of the ESDs ${\mu_{\frac{1}{\sqrt{n}} M_n}}$ to the semi-circular law ${\mu_{sc}}$, in that one has
$\displaystyle \mu_{\frac{1}{\sqrt{n}} M_n}( -\infty, \lambda ) = \mu_{sc}(-\infty,\lambda) + O(n^{-c})$
with overwhelming probability for all ${\lambda \in {\mathbb R}}$.
A variant of this quantitative analysis can in fact get very good control on this ESD down to quite fine scales, namely to scales ${\frac{\log^{O(1)} n}{n}}$, which is only just a little bit larger than the mean spacing ${O(1/n)}$ of the normalised eigenvalues (recall that we have ${n}$ normalised eigenvalues, constrained to lie in the interval ${[-2-o(1), 2+o(1)]}$ by the Bai-Yin theorem). This was accomplished by Erdös, Schlein, and Yau (under some additional regularity hypotheses on the distribution ${\xi}$, but these can be easily removed with the assistance of Talagrand’s inequality) by using an additional observation, namely that the eigenvectors of a random matrix are very likely to be delocalised in the sense that their ${\ell^2}$ energy is dispersed more or less evenly across its coefficients. We will return to this point in later notes.
— 4. Dyson Brownian motion and the Stieltjes transform —
We now explore how the Stieltjes transform interacts with the Dyson Brownian motion introduced in Notes 3b. We let ${n}$ be a large number, and let ${M_{n}(t)}$ be a Wiener process of Hermitian random matrices, with associated eigenvalues ${\lambda_{1}(t),\ldots,\lambda_{n}(t)}$, Stieltjes transforms
$\displaystyle s(t,z) := \frac{1}{n} \sum_{j=1}^n \frac{1}{\lambda_{j}(t)/\sqrt{n} - z} \ \ \ \ \ (26)$
and spectral measures
$\displaystyle \mu(t,z) := \frac{1}{n} \sum_{j=1}^n \delta_{\lambda_j(t)/\sqrt{n}}. \ \ \ \ \ (27)$
We now study how ${s}$, ${\mu}$ evolve in time in the asymptotic limit ${n \rightarrow \infty}$. Our computation will be only heuristic in nature.
Recall from Notes 3b that the eigenvalues ${\lambda_i = \lambda_i(t)}$ undergo Dyson Brownian motion
$\displaystyle d\lambda_i = dB_i + \sum_{j \neq i} \frac{dt}{\lambda_i-\lambda_j}. \ \ \ \ \ (28)$
Applying (26) and Taylor expansion (dropping all terms of higher order than ${dt}$, using the Ito heuristic ${dB_i = O(dt^{1/2})}$), we conclude that
$\displaystyle ds = - \frac{1}{n^{3/2}} \sum_{i=1}^n \frac{dB_i}{(\lambda_i/\sqrt{n}-z)^2} - \frac{1}{n^2} \sum_{i=1}^n \frac{|dB_i|^2}{(\lambda_i/\sqrt{n}-z)^3}$
$\displaystyle - \frac{1}{n^{3/2}} \sum_{1 \leq i,j \leq n: i \neq j}\frac{dt}{(\lambda_i - \lambda_j) (\lambda_j/\sqrt{n}-z)^2}.$
For ${z}$ away from the real line, the term ${\frac{1}{n^2} \sum_{i=1}^n \frac{|dB_i|^2}{(\lambda_i/\sqrt{n}-z)^3}}$ is of size ${O( dt / n )}$ and can heuristically be ignored in the limit ${n \rightarrow \infty}$. Dropping this term, and then taking expectations to remove the Brownian motion term ${dB_i}$, we are led to
$\displaystyle {\bf E} ds = - {\bf E} \frac{1}{n^{3/2}} \sum_{1 \leq i,j \leq n: i \neq j}\frac{dt}{(\lambda_i - \lambda_j) (\lambda_j/\sqrt{n}-z)^2}.$
Performing the ${i}$ summation using (26) we obtain
$\displaystyle {\bf E} ds = - {\bf E} \frac{1}{n} \sum_{1 \leq j \leq n} \frac{s(\lambda_j/\sqrt{n}) dt}{(\lambda_j/\sqrt{n}-z)^2}$
where we adopt the convention that for real ${x}$, ${s(x)}$ is the average of ${s(x+i0)}$ and ${s(x-i0)}$. Using (27), this becomes
$\displaystyle {\bf E} s_t = - {\bf E} \int_{\mathbb R} \frac{s(x)}{(x-z)^2}\ d\mu(x) \ \ \ \ \ (29)$
where the ${t}$ subscript denotes differentiation in ${t}$. From (15) we heuristically have
$\displaystyle s(x \pm i0) = s(x) \pm \pi i \mu(x)$
(heuristically treating ${\mu}$ as a function rather than a measure) and on squaring one obtains
$\displaystyle s(x \pm i0)^2 = (s(x)^2 - \pi^2 \mu^2(x)) \pm 2 \pi i s(x) \mu(x).$
From this the Cauchy integral formula around a slit in real axis (using the bound (11) to ignore the contributions near infinity) we thus have
$\displaystyle s^2(z) = \int_{\mathbb R} \frac{2s(x)}{x-z}\ d\mu(x)$
and thus on differentiation in ${z}$
$\displaystyle 2 s s_z(z) = \int_{\mathbb R} \frac{2s(x)}{(x-z)^2}\ d\mu(x).$
Comparing this with (29), we obtain
$\displaystyle {\bf E} s_t + {\bf E} s s_z = 0.$
From concentration of measure, we expect ${s}$ to concentrate around its mean ${\overline{s} := {\bf E} s}$, and similarly ${s_z}$ should concentrate around ${\overline{s}_z}$. In the limit ${n \rightarrow \infty}$, the expected Stieltjes transform ${\overline{s}}$ should thus obey Burgers’ equation
$\displaystyle s_t + s s_z = 0. \ \ \ \ \ (30)$
To illustrate how this equation works in practice, let us give an informal derivation of the semi-circular law. We consider the case when the Wiener process starts from ${M(0) = 0}$, thus ${M_t \equiv \sqrt{t} G}$ for a GUE matrix ${G}$. As such, we have the scaling symmetry
$\displaystyle s(t,z) = \frac{1}{\sqrt{t}} s_{GUE}(\frac{z}{\sqrt{t}})$
where ${s_{GUE}}$ is the asymptotic Stieltjes transform for GUE (which we secretly know to be given by (24), but let us pretend that we did not yet know this fact). Inserting this self-similar ansatz into (30) and setting ${t=1}$, we conclude that
$\displaystyle -\frac{1}{2} s_{GUE} - \frac{1}{2} z s'_{GUE} + s s'_{GUE} = 0;$
multiplying by two and integrating, we conclude that
$\displaystyle z s_{GUE} + s_{GUE}^2 = C$
for some constant ${C}$. But from the asymptotic (12) we see that ${C}$ must equal ${-1}$. But then the above equation can be rearranged into (23), and so by repeating the arguments at the end of the previous section we can deduce the formula (24), which then gives the semi-circular law by (15).
As is well known in PDE, one can solve Burgers’ equation more generally by the method of characteristics. For reasons that will be come clearer in the next set of notes, I will solve this equation by a slightly different (but ultimately equivalent) method. The idea is that rather than think of ${s=s(t,z)}$ as a function of ${z}$ for fixed ${t}$, we think of ${z=z(t,s)}$ as a function of ${s}$ for fixed ${t}$. (This trick is sometimes known as the hodograph transform, especially if one views ${s}$ as “velocity” and ${z}$ as “position”.) Note from (12) that we expect to be able to invert the relationship between ${s}$ and ${z}$ as long as ${z}$ is large (and ${s}$ is small).
To exploit this change of perspective, we think of ${s, z, t}$ as all varying by infinitesimal amounts ${ds, dz, dt}$ respectively. Using (30) and the total derivative formula ${ds = s_t dt + s_z dz}$, we see that
$\displaystyle ds = - s s_z dt + s_z dz.$
If we hold ${s}$ fixed (i.e. ${ds=0}$), so that ${z}$ is now just a function of ${t}$, and cancel off the ${s_z}$ factor, we conclude that
$\displaystyle \frac{dz}{dt} = s.$
Integrating this, we see that
$\displaystyle z(t,s) = z(0,s) + ts. \ \ \ \ \ (31)$
This, in principle, gives a way to compute ${s(t,z)}$ from ${s(0,z)}$. First, we invert the relationship ${s=s(0,z)}$ to ${z=z(0,s)}$; then we add ${ts}$ to ${z(0,s)}$; then we invert again to recover ${s(t,z)}$.
Since ${M_t \equiv M_0 + \sqrt{t} G}$, where ${G}$ is a GUE matrix independent of ${M_0}$, we have thus given a formula to describe the Stieltjes transform of ${M_0 + \sqrt{t} G}$ in terms of the Stieltjes transform of ${M_0}$. This formula is a special case of a more general formula of Voiculescu for free convolution, with the operation of inverting the Stieltjes transform essentially being the famous ${R}$-transform of Voiculescu; we will discuss this more in the next section.
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2022-01-21 11:18:02
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https://www.r-bloggers.com/non-negative-least-squares/
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# Non-negative least squares
November 27, 2019
By
[This article was first published on R – Statistical Odds & Ends, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
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Imagine that one has a data matrix $X \in \mathbb{R}^{n \times p}$ consisting of $n$ observations, each with $p$ features, as well as a response vector $y \in \mathbb{R}^n$. We want to build a model for $y$ using the feature columns in $X$. In ordinary least squares (OLS), one seeks a vector of coefficients $\hat{\beta} \in \mathbb{R}^p$ such that
\begin{aligned} \hat{\beta} = \underset{\beta \in \mathbb{R}^p}{\text{argmin}} \quad \| y - X\beta \|_2^2. \end{aligned}
In non-negative least squares (NNLS), we seek a vector coefficients $\hat{\beta} \in \mathbb{R}^p$ such that it minimizes $\| y - X \beta\|_2^2$ subject to the additional requirement that each element of $\hat{\beta}$ is non-negative.
There are a number of ways to perform NNLS in R. The first two methods come from Reference 1, while I came up with the third. (I’m not sharing the third way Reference 1 details because it claims that the method is buggy.)
Let’s generate some fake data that we will use for the rest of the post:
set.seed(1)
n <- 100; p <- 10
x <- matrix(rnorm(n * p), nrow = n)
y <- x %*% matrix(rep(c(1, -1), length.out = p), ncol = 1) + rnorm(n)
Method 1: the nnls package
library(nnls)
mod1 <- nnls(x, y)
mod1x # [1] 0.9073423 0.0000000 1.2971069 0.0000000 0.9708051 # [6] 0.0000000 1.2002310 0.0000000 0.3947028 0.0000000 Method 2: the glmnet package The glmnet() function solves the minimization problem \begin{aligned} \hat{\beta} = \underset{\beta \in \mathbb{R}^p}{\text{argmin}} \quad \frac{1}{2n} \| y - X\beta \|_2^2 + \lambda \left[ \frac{1-\alpha}{2}\|\beta\|_2^2 + \alpha \|\beta\|_1 \right], \end{aligned} where $\alpha$ and $\lambda$ are hyperparameters the user chooses. By setting $\alpha = 1$ (the default) and $\lambda = 0$, glmnet() ends up solving the OLS problem. By setting lower.limits = 0, this forces the coefficients to be non-negative. We should also set intercept = FALSE so that we don’t have an extraneous intercept term. library(glmnet) mod2 <- glmnet(x, y, lambda = 0, lower.limits = 0, intercept = FALSE) coef(mod2) # 11 x 1 sparse Matrix of class "dgCMatrix" # s0 # (Intercept) . # V1 0.9073427 # V2 . # V3 1.2971070 # V4 . # V5 0.9708049 # V6 . # V7 1.2002310 # V8 . # V9 0.3947028 # V10 . Method 3: the bvls package NNLS is a special case of bounded-variable least squares (BVLS), where instead of having constraints $\beta_j \geq 0$ for each $j = 1, \dots, p$, one has constraints $a_j \leq \beta_j \leq b_j$ for each $j$. BVLS is implemented in the bvls() function of the bvls package: library(bvls) mod3 <- bvls(x, y, bl = rep(0, p), bu = rep(Inf, p)) mod3x
# [1] 0.9073423 0.0000000 1.2971069 0.0000000 0.9708051
# [6] 0.0000000 1.2002310 0.0000000 0.3947028 0.0000000
In the above, bl contains the lower limits for the coefficients while bu contains the upper limits for the coefficients.
References:
1. Things I Thought At One Point. Three ways to do non-negative least squares in R.
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|
2019-12-06 03:48:42
|
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|
https://stats.stackexchange.com/questions/571157/if-there-is-no-censoring-can-be-the-naive-3rd-quantile-different-from-the-one-c
|
# If there is no censoring, can be the naive 3rd quantile different from the one calculated with from the Kaplan-Meier?
I know, that the median survival time calculated from the Kaplan-Meier estimator is equal to the "naive" descriptive median of the survival time when no censoring in data occurs.
Does it apply also to the other quantiles, like the 3rd quantile? Is this possible that they differ, if no censoring occurs and all subjects experience the event?
> quantile(d$time) 0% 25% 50% 75% 100% 2.5000 3.1375 4.3050 11.3700 71.4200 > km <- survfit( Surv(time, event) ~ 1, data = d, conf.type = "log-log") > > quantile(km)$quantile
25 50 75
3.120 4.305 12.140
No censoring:
> d %>% count(event)
event n
1 1 86
EDIT: OK, got it, thanks to @Frank Harrell
I should use the empirical CDF with averaging at discontinuities:
> quantile(km)$quantile 25 50 75 3.120 4.305 12.140 > quantile(d$time, type=2)
0% 25% 50% 75% 100%
2.500 3.120 4.305 12.140 71.420
|
2022-10-06 19:31:33
|
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|
http://ibpsexamguide.org/quantitative-aptitude/quantitative-aptitude-course/online-average-aptitude-test/exercise-5-2.html
|
# Exercise : 5
1. A family consists of grandparents, parents and three grand children. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
(a) 2847 years
(b) 3157 years
(c) 3217 years
(d) 2712 years
(e) None of these
#### View Ans & Explanation
Ans.b
Required average = $\left ( \frac{67 \times 2 + 35 \times 2 + 6 \times 3}{2 + 2 + 3} \right )$
= $\left ( \frac{134 + 70 + 18}{7} \right ) = \frac{222}{7}$
= 3157 years
2. In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all of them are correct in their estimation, what is the average of different probable weights of Arun?
(a) 67 kg
(b) 68 kg
(c) 69 kg
(d) 66.5 kg
(e) None of the above
#### View Ans & Explanation
Ans.d
Let Arun’s weight be X kg.
According to Arun, 65 < X < 72.
According to Arun’s brother, 60 < X < 70.
According to Arun’s mother, X < 68.
The values satisfying all the above conditions are 66 and 67.
∴ Required average = $\left ( \frac{66 + 67}{2} \right ) = \left ( \frac{133}{2} \right ) = 66.5 \; kg.$
3. The average age of a board of 8 functional directors in a company is the same as it was 3 years ago, a younger man having been substituted for one of the directors. How much younger was the new man than the director whose place he took.
(a) 24 years
(b) 26 years
(c) 28 years
(d) 27 years
(e) None of the above
#### View Ans & Explanation
Ans.a
Let the new man was younger than the director = x years and 3 years ago, the sum of ages of board of directors
= S – 8 × 3 = S – 24
Then, 3 years ago, average age of board of directors
= $\frac{S - 24}{8}$
Now, $\frac{S - 24}{8} = \frac{S - x}{8}$
⇒ x = 24 years
Shortcut Method : If the new young director would have been not substituted, then total age would have increased at present by 8 × 3 = 24 years.
Therefore, the new man is 24 years younger keeping the average at present same as 3 years ago.
4. A batsman makes a scores of 98 runs in his 19th inning and thus increases his average by 4. What is his average after 19th inning ?
(a) 22
(b) 24
(c) 28
(d) 26
(e) None of the above
#### View Ans & Explanation
Ans.d
Let the average score of 19 innings be x.
Then, $\frac{18x + 98}{19} = x + 4$
The average score after 20th innings
= x + 4 = 22 + 4 = 26
5. The average weight of 45 students in a class is 52 kg. 5 of them whose average weight is 48 kg leave the class and other 5 students whose average weight is 54 kg join the class. What is the new average weight (in kg) of the class ?
(a) 5113
(b) 5223
(c) 5213
(d) 43.42
(e) None of these
#### View Ans & Explanation
Ans.b
Total weight of 45 students
= 45 × 52 = 2340 kg
Total weight of 5 students who leave
= 5 × 48 = 240 kg
Total weight of 5 students who join
= 5 × 54 = 270 kg
Therefore, new total weight of 45 students
= 2340 – 240 + 270 = 2370
⇒ New average weight = 237045 = 5223 kg
6. The average of 10 numbers is 40.2. Later it is found that two numbers have been wrongly copied. The first is 18 greater than the actual number and the second number added is 13 instead of 31. Find the correct average.
(a) 40.2
(b) 40.4
(c) 40.6
(d) 40.8
(e) None of the above
#### View Ans & Explanation
Ans.a
Sum of 10 numbers = 402
Corrected sum of 10 numbers
= 402 – 13 + 31 – 18 = 402
Hence, new average = 40210 = 40.2
|
2018-05-25 18:32:31
|
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|
https://web2.0calc.com/questions/need-help_28574
|
+0
# need help
0
76
2
Let x, y and z be positive real numbers such that x + y + z = 1. Find the minimum value of (x + y + z)/(xyz).
Jan 25, 2022
#1
+516
+3
x + y + z = 1. Since xyz is in the denominator, we want xyz to be the largest value possible.
The largest value possible of xyz is only if x = y = z, so x = y = z = 1/3.
$$1\over{({1\over3})^2}$$ = $$(x + y + z)\over{xyz}$$
Thus, the minimum value of $$(x + y + z)\over{xyz}$$ is 27.
Jan 25, 2022
#2
+360
0
I think you made a mistake on format you wrote $$\frac{1}{(\frac{1}{3})^2}$$ instead of $$\frac{1}{(\frac{1}{3})^3}$$
Jan 26, 2022
|
2022-05-25 15:27:44
|
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|
https://tex.stackexchange.com/questions/346773/im-confused-about-latex-3-returning-values-from-functions
|
# I'm confused about LaTeX 3 returning values from functions
The below code works as expected and prints on the terminal "answer is 5". How can I change the code below so I can replace lines B and C with, in pseudocode,
set \l_width_int to \width:N \l_test_seq
I tried many things around lines A, B, C and other places and couldn't get anything to work. I'm confused. Thanks.
\documentclass{article}
\usepackage[check-declarations]{expl3}
\begin{document}
\ExplSyntaxOn
\int_new:N \l_width_int
\seq_new:N \l_test_seq
\seq_set_from_clist:Nn \l_test_seq {alpha,beta,gamma}
% This function sets \l_width_int to the number of characters
% in the longest element of the sequence argument.
\cs_new:Npn \width:N #1
{
\int_new:N \l__width_curwidth_int
\int_new:N \l__width_maxwidth_int
\tl_new:N \l__width_cur_tl
\int_set:Nn \l__width_maxwidth_int {-1}
\seq_map_inline:Nn {#1}
{
\tl_set:Nn \l__width_cur_tl {##1}
\int_set:Nn \l__width_curwidth_int {\tl_count:N \l__width_cur_tl}
\int_compare:nT
{\l__width_curwidth_int > \l__width_maxwidth_int}
{\int_set:Nn \l__width_maxwidth_int \l__width_curwidth_int}
}
\int_set:Nn \l_width_int \l__width_maxwidth_int % line A
}
\width:N \l_test_seq % line B
\ExplSyntaxOff
\end{document}
To be able to set \l_width_int (or any variable) to the result of a function (really a macro) in TeX, we have to arrange for the calculation to be done by expansion. In expl3 terms, that means you can only use functions which are shown with a star in interface3. To do that, we need to rearrange the code. I'd go with something like
\cs_new:Npn \MS_width:N #1
{
\seq_map_function:NN {#1} \__MS_width:n
\__MS_result:n { -1 }
}
\cs_new:Npn \__MS_result:n #1 {#1}
\cs_new:Npn \__MS_width:n #1 #2 \__MS_result:n #3
{
\__MS_width:fnn
{ \int_eval:n { \tl_count:n {#1} } } {#2} {#3}
}
\cs_new:Npn \__MS_width:nnn #1#2#3
{
\int_compare:nNnTF {#1} > {#3}
{ #2 \__MS_result:n {#1} }
{ #2 \__MS_result:n {#3} }
}
\cs_generate_variant:Nn \__MS_width:nnn { f }
The idea is to set up the mapping such that the result is kept in the input stream all of the time: I use \__MS_result:n as a marker to know where it is and to allow the mapping tokens to be 'moved around'. I also arrange for this to be a simple 'do nothing' function to finally dump the result at the end of the mapping. This would then work in
(I've given your public sequence a 'proper' name here.)
Notice that I've 'forced' evaluation of the length of the token list using f-type expansion, so the value is worked out exactly rather than potentially at every pass (if the first item in the list is the longest).
I'm using the fact here that I know that I can 'move stuff around' with the sequence mapping. I don't want to rely on a particular internal form for the sequence so have to stick to \seq_map_function:NN. One might argue that it would be better to work directly from the comma list and to therefore know exactly what is going on in all of the code:
\cs_new:Npn \MS_width_comma:N #1
{
\exp_after:wN \__MS_width:w #1 , \q_nil , \q_stop
{ -1 }
}
\cs_new:Npn \__MS_width:w #1 , #2 \q_stop #3
{
\quark_if_nil:nTF {#1}
{ #3 }
{
\__MS_width:fnn { \int_eval:n { \tl_count:n {#1} } }
{#2} {#3}
}
}
\cs_new:Npn \__MS_width:nnn #1#2#3
{
\int_compare:nNnTF {#1} > {#3}
{ \__MS_width:w #2 \q_stop {#1} }
{ \__MS_width:w #2 \q_stop {#3} }
}
\cs_generate_variant:Nn \__MS_width:nnn { f }
The basic idea is of course the same: depending on the nature of your input, you might go either way. (I'm not clear if you start with a user comma list or you might have a sequence with empty items, etc.)
• I see egreg has mentioned various other things in his answer, so I don't need to ;) – Joseph Wright Jan 2 '17 at 22:59
• I've kept width as it's the term you've used, but normally that is a typesetting concept: you are really talking about length. – Joseph Wright Jan 3 '17 at 7:35
You're making things more complicated than needed. ;-)
\documentclass{article}
\usepackage[check-declarations]{expl3}
\ExplSyntaxOn
\int_new:N \l_senn_width_int
\seq_new:N \l_senn_test_seq
\seq_set_from_clist:Nn \l_senn_test_seq {alpha,beta,gamma}
% This function sets \l_senn_width_int to the number of characters
% in the longest element of the sequence argument.
\cs_new_protected:Npn \senn_width:N #1
{
\int_set:Nn \l_senn_width_int {-1}
\seq_map_inline:Nn #1
{
\int_set:Nn \l_senn_width_int
{
\int_max:nn { \l_senn_width_int } { \tl_count:n { ##1 } }
}
}
}
\senn_width:N \l_senn_test_seq
\ExplSyntaxOff
Some points to note:
1. Use prefixes.
2. Never allocate variables in functions, unless the functions are specifically designed to allocate variables for future usage.
3. You can set a variable using its current value.
4. The function \tl_count:n is expandable and can be used in the context of an integer.
5. The main function \senn_width:N should be protected, as it does assignments to variables.
\int_set:Nn \l_senn_width_int { \senn_width:N \l_senn_test_seq }
requires \senn_width:N to be fully expandable, which I don't see easily possible.
What you can do is defining a two argument macro:
\senn_width_set:NN \l_senn_width_int \l_senn_test_seq
so problems with expandability are overcome:
\cs_new_protected:Npn \senn_width_set:NN #1 #2
{
\int_set:Nn #1 {-1}
\seq_map_inline:Nn #2
{
\int_set:Nn #1
{
\int_max:nn { #1 } { \tl_count:n { ##1 } }
}
}
}
\senn_width_set:NN \l_senn_width_int \l_senn_test_seq
|
2020-04-04 13:06:05
|
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|
https://ericphanson.com/teaching/qit/es4/exercise_9/
|
# Exercise 9
## from Example Sheet 4
### Exercise 9.
Continuity of the von Neumann entropy (Fannes’ inequality): Suppose $$\rho, \sigma \in \mathcal{D}( \mathcal{H})$$ are states such that their trace distance $$D(\rho, \sigma)$$ satisfies the bound $2 D(\rho, \sigma) = \|\rho - \sigma\|_1 \le 1/e.$ Then $|S(\rho) - S(\sigma)| \le \|\rho - \sigma\|_1 \log d + \eta\left( \|\rho - \sigma\|_1\right),$(1) where $$d = \dim \mathcal{H}$$, and $$\eta(x) := - x \log x.$$
Let us prove this theorem in steps:
1. Let $$r_1 \ge r_2 \ge \ldots \ge r_d$$ be the eigenvalues of $$\rho$$ arranged in non-increasing order, and let $$s_1 \ge s_2 \ge \ldots \ge s_d$$ be the eigenvalues of $$\sigma$$ arranged in non-increasing order. Then prove that: $\|\rho - \sigma\|_1 \ge \sum_{j=1}^d |r_j - s_j|$(2)
2. Check (using elementary calculus) that if $$|r-s| \le 1/2$$, then $|\eta(r) - \eta(s)| \le \eta(|r-s|),$ where $$\eta(x) := - x \log x.$$
3. Use Step 2 and the triangle inequality to prove that $|S(\rho) - S(\sigma)|\le \sum_j \eta(|r_j - s_j|).$
4. Let $$\varepsilon_j := |r_j - s_j|$$, $$\forall\, j=1,2,\ldots, d$$, and $$\varepsilon:= \sum_j \varepsilon_j$$. Let $$\lambda_j := \varepsilon_j/\varepsilon$$ and note that $$\{\lambda_j\}$$ forms a probability distribution. Use this fact and Step 3 to prove that $|S(\rho) - S(\sigma)|\le \varepsilon\log d + \eta(\varepsilon).$
5. Note that $$\eta(\varepsilon)$$ is a monotonically increasing function of $$\varepsilon$$ for $$0 \le \varepsilon\le 1/e$$. Use this to finally arrive at the statement (1).
1. I could not find a short proof of this fact (without assuming something that directly implies it). This subquestion is then nonexaminable; see e.g. the excellent book Matrix Analysis by Bhatia1 for several (more involved) proofs.
2. Without loss of generality, take $$r > s$$ and set $$\delta := r-s$$. Then \begin{aligned} \eta(s+\delta) - \eta(s) - \eta(\delta) &= - (s+\delta)\log (s+\delta) + s \log s + \delta\log(\delta) \\ &= - s\log (s+\delta) - \delta \log(s+\delta) + s \log s + \delta\log(\delta)\\ &= s\log\frac{s}{s+\delta} + \delta \log \frac{\delta}{s+\delta} \\ &\leq 0 \end{aligned} since both $$\frac{s}{s+\delta}\leq 1$$ and $$\frac{\delta}{s+\delta} \leq 1$$. Thus, $\eta(r) - \eta(s) \leq \eta(r-s).$ On the other hand, set $f(\delta) := \eta(s) - \eta(s+\delta) - \eta(\delta).$ It remains to show that $$f(\delta)\leq 0$$ for $$\delta \leq \frac{1}{2}$$ (and $$r=s+\delta\leq 1$$), which implies $\eta(s) - \eta(s+\delta) \leq \eta(\delta)$ meaning that we have proven $|\eta(r) - \eta(s)|\leq \eta(|r-s|).$
First, $$f(0) = 0$$, and $f'(\delta) = - \eta'(s+\delta) -\eta'(\delta)$ while $\eta'(x) = \frac{\mathrm{d}}{\mathrm{d}x}(- x \log x) = - \log x - \frac{1}{\ln 2},$ so $f'(\delta) = \log ( s + \delta) + \frac{1}{\ln 2} + \log (\delta) + \frac{1}{\ln 2}$ and $f''(\delta) = \frac{1}{\ln 2} \left[ \frac{1}{s+\delta} + \frac{1}{\delta}\right] > 0$ for $$\delta > 0$$. Thus, $$f$$ is convex. We wish to prove $$f(\delta)\leq 0$$ for $$\delta \in [0, \min(1-s, \frac{1}{2})]$$, and by convexity, we only need to check the boundaries (since $$f$$ will be less than either boundary point between the two). We have immediately $$f(0)=0$$. Next, consider $$s \geq \frac{1}{2}$$. Then let’s check $$g(s) := f(1-s) = \eta(s) - \eta(1-s) \leq 0$$ for $$s\geq \frac{1}{2}$$. We can see at $$s= \frac{1}{2}$$, $$g(s)=0$$, and $$g'(s) = -\log(s)+\log(1-s) \leq 0$$ since $$1-s \leq s$$. Thus, $$g$$ is decreasing for all $$s\in [ \frac{1}{2},1]$$, proving indeed, $$g(s)\leq 0$$ on that interval.
Lastly, we need to check when $$s < \frac{1}{2}$$ that $$f(\frac{1}{2}) \leq 0$$. In this case, $h(s) := f(\frac{1}{2} ) = \eta(s) - \eta(s+\tfrac{1}{2}) - \eta(\tfrac{1}{2}).$ Then $$h(0) = - 2 \eta(\frac{1}{2})\leq 0$$, and $$h'(s) = - \log(s) + \log(s + \frac{1}{2}) > 0$$, so $$h$$ is increasing. Thus, $$h$$ takes a maximum at the end of the interval we wish to test, at $$s= \frac{1}{2}$$. But here, $$h(\tfrac{1}{2}) = \eta(\tfrac{1}{2} ) - \eta(1) - \eta(\tfrac{1}{2}) = 0$$. Thus, $$f(\tfrac{1}{2}) \leq 0$$ for $$s\in [0, \tfrac{1}{2}]$$.
3. We have that $$S(\rho) = \sum_j \eta(r_j)$$ and $$S(\sigma) = \sum_j \eta(s_j)$$. Therefore, \begin{aligned} |S(\rho) - S(\sigma)| &= \left| \sum_j \eta(r_j) - \eta(s_j)\right| \\ &\leq \sum_j |\eta(r_j) - \eta(s_j)| \\ &\leq \sum_j \eta(|r_j-s_j|). \end{aligned}
4. Note that $\eta(xy) = - xy \log (xy) = - xy (\log x + \log y) = - xy \log x - xy \log y = y \eta(x) + x \eta(y)$ Thus, $$\eta(\varepsilon_j) = \eta(\varepsilon\lambda_j) = \varepsilon\eta(\lambda_j) + \lambda_j\eta(\varepsilon)$$, and \begin{aligned} |S(\rho) - S(\sigma)| &\leq \sum_j \eta(\varepsilon_j) \\ &= \varepsilon\sum_j \eta(\lambda_j) + \eta(\varepsilon) \\ &= \eta(\varepsilon) + H(\{ \lambda_j\}_{j})\\ &\leq \eta(\varepsilon)+ \varepsilon\log d \end{aligned} using that $$\{ \lambda_j\}_{j}$$ is a probability distribution, and the entropy of any probability distribution with $$d$$ elements is bounded by $$\log d$$.
5. By step 1, $$\varepsilon\leq \|\rho-\sigma\|_1$$. Since $$\varepsilon\mapsto \eta(\varepsilon)$$ is monotonically increasing on the range $$[0,1/e]$$, if $$\|\rho-\sigma\|_1\leq 1/e$$, then \begin{aligned} |S(\rho) - S(\sigma)| &\leq\eta(\varepsilon) + \varepsilon\log d \\ &\leq \eta(\|\rho-\sigma\|_1) + \|\rho-\sigma\|_1 \log d \end{aligned} as desired.
1. DOI: 10.1007/978-1-4612-0653-8. In particular, equation IV.62 of that book is a restatement of this question.
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https://www.greenwich143.com/2yw9c6j/47aa8c-given-these-points-example
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# given these points example
Step 2: Join Aâ to O. Using y-y1=m (x-x1) to write the equation of a line. Slope intercept vs Point Slope Form. When we substitute -1 for x in the standard form equation, the y value will be 3. Write the equation of the line in slope-intercept form with a slope of - \,5 and a y-intercept of 3. Again, remember that while the derivative doesnât exist at $$w = 3$$ and $$w = - 2$$ neither does the function and so these two points are not critical points for this function. You can use either (4,5) or (8,7), Substitute b, 3, into the equation from step 2, Find the equation of a line through the following the points: (-6,7) and (-9,8), Substitute the slope for 'm' in the slope intercept equation, Substitute either point into the equation. (-6,7) and (-9,8), y − y1 = m(x −x1) Visit the Big Ideas Math Algebra 2: Online Textbook Help page to learn more. We get 3 = a(-1)2 + b(-1) + c. This simplifies to 3 = a - b + c. Do the same with the other points and we'll get a total of three equations: We can take any two of these equations, subtract one from the other, and the c will cancel leaving us with two unknowns. Create an account to start this course today. y − 7 = 2(x− 3), using (5,11): In the next scenario we are given the x-intercepts and one point on the curve. The first half of this page will focus on writing the equation in slope intercept form like example 1 below. Substitute the slope for 'm' in the point slope equation, y − y1 = m(x −x1) You can use either (3,7) or (5,11), Solve for b, which is the y-intercept of the line, Substitute b, -1, into the equation from step 2, You can use the calculator below to find the equation of a line from any two points. There are a few different ways to find the equation of line from 2 points.. You can use either (3,7) or (5,11), using (3,7): So you are going to move from Point 1 to Point 2. Substituting for x and y: 3 = a (-1 - 3) 2 - 1 => 3 = a (-4) 2 - 1. If the sl⦠The reason is the statement given above - any three points in 3-dimensional space determine a plane. Then substitute these into the equation for slope: We get the same answer as before! Find the y intercept given two points. Real World Math Horror Stories from Real encounters. If has fixed point at , then = â ( ) has a zero at . Be careful of the signs! Two cloud shapes down and one to go. The angle of rotation is 62Ë anticlockwise or +62Ë Manual rotation of a polygon about a given point at a given angle Example 1. Points P, Q, X, and W, for example, are coplanar; the plane that contains them is the left side of the box.Each of the six faces of the box contains four coplanar points, but these are not the only groups of coplanar points. We are either given: Three forms, three scenarios, and you the only mathematician! Then you describe what you see. All other trademarks and copyrights are the property of their respective owners. So, together we are going to learn how to: Basic Definitions. We are done! How to find y=mx+b with two points. first two years of college and save thousands off your degree. There are a few different ways to find the equation of line from 2 points. In the given choices, when we plug in (1, 10) we get 10 = 7 + 2, which is false, making this is the desired answer. Just type numbers into the boxes below and the calculator (which has its own page here) will automatically calculate the equation of line in point slope and slope intercept forms, ( Try this 'equation from 2 points' calculator on its own page here . Use the given point (-1, 3), which says y is 3 for x equal to -1. There are three typical scenarios when writing a quadratic equation from points. This will happen on occasion. What Is the Rest Cure in The Yellow Wallpaper? So this is my x-axis. Step 2: Use the slope to find the y-intercept. share. Example 3: Determine the point-slope form of the line passing through the points \left( {2,10} \right) and \left( {5,1} \right). In those cases, you might use a low-order polynomial fit (which tends to be smoother between points) or a different technique, depending on the problem. You can test out of the With three points, we use the standard form for the quadratic equation: Get access risk-free for 30 days, {{courseNav.course.mDynamicIntFields.lessonCount}} lessons If the slope is positive, the line slants upward to the right. That means p is 1 and the q is 5. This is a pretty nice feature of these equations. In this example we are given two points, (15, 8) and (10, 7), on a straight line. y = ½x +b, Substitute either point into the equation. Not sure what college you want to attend yet? In fact, all points along the straight line connecting B and C are on the same âequipotential line.â A more complete discussion of equipotential will be given in Section 3.5. So, or In the example below, youâll see that the line has two points each indicated as an ordered pair. y − y1 = ½(x −x1), using (4,5): We should also not expect this integral to be the same for all paths between these two points. From a - b + c = 3: 1/4 - (-3/2) + c = 3 => c = 5/4. Plus, get practice tests, quizzes, and personalized coaching to help you Quadratic equations appear in all types of science and engineering applications. As a member, you'll also get unlimited access to over 83,000 Those are the x-intercept values. succeed. All rights reserved. Many translated example sentences containing "given these points" â Spanish-English dictionary and search engine for Spanish translations. An error occurred trying to load this video. You need information to write the quadratic equation. credit by exam that is accepted by over 1,500 colleges and universities. Hmmm, three major shapes: related but different. This means h = 3 and k = -1. The point is (6, 5/4) and the x-intercepts are located at (1, 0) and (5, 0). So notice, these are just the negatives of these values from when we swapped them. For example, given numbers from a table of loga-rithms, estimate the logarithm of a number xnot in the table. Slope is sometimes described as rise over run. What Is a Bachelor of Professional Studies Degree? Note that this time, unlike the line integral we worked with in Examples 2, 3, and 4 we got the same value for the integral despite the fact that the path is different. ⢠Given a set of data points {(xi,yi)}, ï¬nd a curve passing thru these points that is âpleasing to the eyeâ. Thus, our quadratic equation in vertex form is y = 0.25(x - 3)2 - 1. So let me draw our two points. However, if you are comfortable using the point slope form of a line, then skip to the second part of this page because writing the equation from 2 points is easier with point slope form . study All of these are true. Substituting the 1 and 5 in the equation gives us y = a(x - 1)(x - 5). Try refreshing the page, or contact customer support. Free Algebra Solver ... type anything in there! Enrolling in a course lets you earn progress by passing quizzes and exams. Already registered? \\ F of two, well, two is less than or equal to two, and it's greater than zero. flashcard set{{course.flashcardSetCoun > 1 ? The first half of this page will focus on writing the equation in slope intercept form like example 1 below.. In the next example that you can see here, we're given three points located at (-1, 3), (4, -3/4) and (6, 5/4). No x-intercepts and no vertex. Example: Step-by-Step Guide to Writing a Great Reading Response Paper, Step-by-Step Guide to Writing Compare and Contrast Essays, Make Your Writing Shine: Tips for Perfect Usage, Learn Writing in the Blogosphere: Top 10 Writing Blogs, Break that Block: Five Fun Writing Prompts, Tips for Writing an Effective Cover Letter, Best Bachelor's in Health Science Degrees, Best Online Bachelor's in Sociology Programs. Let's say we are given a point on the curve at (-1, 3) and the vertex located at (3, -1). Study.com has thousands of articles about every We wish to interpolate Æ(x) = x 2 over the range 1 ⤠x ⤠3, given these three points: x 0 = 1 f ( x 0 ) = 1 x 1 = 2 f ( x 1 ) = 4 x 2 = 3 f ( x 2 ) = 9. Example sentences with the word given. Step 1: Find the slope of the line. Find more ways to say given, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. ... (10, 7) as the point (x2, y2). For this example, NaCl completely dissociates into the two ions, Na + and Cl-. Thus, an equation of this plane is 0(x 1)+0(y 2)+1(z 3) = 0 or z 3 = 0 Example 2. And it's still a nice day! Quiz & Worksheet - Writing Quadratic Equations, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Axis of Symmetry of a Parabola: Equation & Vertex, Using Quadratic Models to Find Minimum & Maximum Values: Definition, Steps & Example, Big Ideas Math Algebra 2: Online Textbook Help, Biological and Biomedical Let's do this substitution. It indicates the direction in which a line slants as well as its steepness. | {{course.flashcardSetCount}} From (1, 0) we want the x value which is 1. 10 Great Study Abroad Locations for Art and Architecture Students, Serving the Community Education Portal Speaks with the University of Colorado Boulder, University of Alabama Students Finally Graduate Amid Tornado Wreckage, Big Ideas Math Algebra 2 - Chapter 1: Linear Functions, Writing Quadratic Equations for Given Points, Big Ideas Math Algebra 2 - Chapter 2: Quadratic Functions, Big Ideas Math Algebra 2 - Chapter 3: Quadratic Equations and Complex Numbers, Big Ideas Math Algebra 2 - Chapter 4: Polynomial Functions, Big Ideas Math Algebra 2 - Chapter 5: Rational Exponents and Radical Functions, Big Ideas Math Algebra 2 - Chapter 6: Exponential and Logarithmic Functions, Big Ideas Math Algebra 2 - Chapter 7: Rational Functions, Big Ideas Math Algebra 2 - Chapter 8: Sequences and Series, Big Ideas Math Algebra 2 - Chapter 9: Trigonometric Ratios and Functions, Big Ideas Math Algebra 2 - Chapter 10: Probability, Big Ideas Math Algebra 2 - Chapter 11: Data Analysis and Statistics, Introduction to Statistics: Tutoring Solution, High School Geometry: Homework Help Resource, The Role of Probability Distributions, Random Numbers & the Computer in Simulations, Static Vs Dynamic Simulation in Quantitative Analysis, Developing Linear Programming Models for Simple Problems, The Importance of Extreme Points in Problem Solving, Quiz & Worksheet - Find a Taylor Polynomial, Quiz & Worksheet - Derivatives of Square Root Functions Practice, Worksheet & Practice - Trig Function Derivatives & the Chain Rule, Worksheet & Practice - Solving Derivatives of Trig Functions, CSET Math: Integrals in Geometry and Trigonometry, CPA Subtest IV - Regulation (REG): Study Guide & Practice, CPA Subtest III - Financial Accounting & Reporting (FAR): Study Guide & Practice, ANCC Family Nurse Practitioner: Study Guide & Practice, Advantages of Self-Paced Distance Learning, Advantages of Distance Learning Compared to Face-to-Face Learning, Top 50 K-12 School Districts for Teachers in Georgia, Finding Good Online Homeschool Programs for the 2020-2021 School Year, Those Winter Sundays: Theme, Tone & Imagery. Those x values are the p and the q. Step 3: Measure the angle AOAâ. CONDITION 4: A point P moves so that it is always equidistant from two intersecting lines AB and CD. Find the equation of a line through the points (3,7) and (5,11), Substitute the slope for 'm' in the slope intercept form of the equation, y = mx +b Substituting for a, b and c in the standard quadratic form gives: That last cloud shape was fun and you see clear skies ahead. How do we connect a set of points to make a smooth curve? The slope of a line refers to the ratio of the vertical change in y over the horizontal change in xbetween any two points on a line. Select a subject to preview related courses: How do we find the a, b and c? Repeat step 2 until there is only one point. It's looking good but we still need the value for a. Therefore, all of the following groups of points are coplanar: A, B, E Example 1: Find the equation of the line passing through the points (â1, â2) and (2, 7). Log in or sign up to add this lesson to a Custom Course. Instead of 5 steps, you can find the line's equation in 3 steps, 2 of which are very easy and require nothing more than substitution! Typically, this information is available in one of three scenarios. Cloud art! y − y1 = 2(x −x1), Substitute either point as x1, y1 in the equation. In order to write the equation of a line in point-slope form, we will need two essential things here which are the slope of the two given points and any point found on the line. lessons in math, English, science, history, and more. courses that prepare you to earn Services. As you can see in this image, we use the vertex form of the quadratic equation: See the vertex at (3, -1)? imaginable degree, area of When two points (x1, x2), (y1, y2) are given and the equation contains these two points, the first step is to find the slope of the line. y − y1 = (x −x1), using (-6,7): y − 5 = ½(x − 4), using (5,11) : We are given a point and the vertex, a point and the x-intercepts, or three points. y = 7x + 2 (2, 16) gives 16 = 7(2) + 2 = 14 + 2 = 16 (â1, â5) gives â5 = 7(â1) + 2 = â7 + 2 = â5 (0, 2) gives 2 = 7(0) + 2 = 0 + 2 = 2 (â2, â12) gives â12 = 7(â2) + 2 = â14 + 2 = â12. Given these two points, the PIs decided to combine results from "1" with "2" and "4" with "5," resulting in a 3 level ordinal response metric, "Unprofessional", "Neutral", ⦠Create your account. Anyone can earn DONE! So once again, this is equal to negative 2. y + 6 = (x − 15 ), If you read this whole page and looked at both methods (slope intercept form and point slope, you can see that it's substantially quicker to find the equation of line through 2 points by means of point slope. y − 6 = (x + 3), using (15, -6): We are given a point in the plane. The main advantage, in this case, is that you do not have to solve for 'b' like you do with slope intercept from. y − 11 = 2(x − 5), y − y1 = m(x −x1) ⢠A number is a fixed point for a given function if = ⢠Root finding =0 is related to fixed-point iteration = . To find the slope of the line passing through these two points we need to use the slope formula: ( ) ( ) So the slope of the slope of the line passing through these two points is 3. Locus formed: Angle bisectors of angles between lines AB and CD. These points make the curve. Solution: Step 1: Join A to O. That is my y-axis. , Find the equation of a line through the following the 2 points: (-3,6) and (15,-6), Substitute either point into the equation. You look at clouds and you see a work of art. And the vertex is given, we still need the value two is less than or equal two...: step 1: find the y-intercept the x-intercepts, or in the example below, see... These equations in a better fit a quick graph here just to show you what a slope... Bisectors of angles given these points example lines AB and CD in fact, this is a point! This lesson you must be a Study.com Member y=m ( x-x1 ) + c 3! ), the only calculation, that you 're going to make a smooth curve Ideas math Algebra:... Fact, this is what is done continually with computer graphics -3/4 with =. Of their respective owners Na + and Cl- get practice tests,,... Engine for Spanish translations top, point slope form is the Rest Cure the! K = -1 a downward slope would look like just to show you what a downward would... C = 5/4 the data points, leading to a poorer fit to the data subject to related... That moves a constant distant of 2 cm from AB statement given above - any three points line has points... Slants upward to the xy-plane ; 3 ) and is parallel to the Community on the! Polyfit does not always result in a better fit wanted to evaluate at... Y is 3 when x is -1 in fact, this is to! Y when x=24 a, b and c, two is less than or equal to two, well two! From 3a + b = -3/4 with a = 1/4: 3/4 + b = -3/2 are the... 'Ll look at clouds and you see a work of art scenarios relate directly to a Custom...., 6 ) as the point ( 4,5 ) and slope of the line slants upward to the xy-plane points! ) ( x - 5 ) we find the equation for slope: we connect them to initially... To evaluate it at two, and you the only calculation, that you going... Parallel lines 2 cm from a straight line AB by passing quizzes and exams which is 1 and vertex... ) or ( 15, -6 ) only calculation, that you 're going make. Find an equation of the plane that passes through the points ( â1, â2 ) and (,! Or three points in 3-dimensional space determine a plane would look like of cm. We are going to learn how to: Another word for given of age or education level leading to particular. Given function if = ⢠Root finding =0 is related to fixed-point iteration.! Slants upward to the data points, but will need to identify two points y-intercept.: three forms, three scenarios for writing quadratic equations when given points on the curve,... Form is the Rest Cure in the equation of a line slants as well as its steepness an. Another word for given or x 2 â x 1 1: find the a run! X values are the property of their respective owners ways to find slope! Are going to make a smooth curve straight line AB substituting the 1 and q! Earning Credit page, youâll see that the line slants upward to the xy-plane: get! Construct the locus of a line from 2 points line in slope-intercept form a. Get the unbiased info you need to find the slope to find the y-intercept example. High-Order polynomials can be oscillatory between the data points, increasing the degree of the line has two from... Of line from 2 points the problem reduces to finding two unknowns with equations! Them to get initially N-1 segments means P is 1 and the q is 5 three typical scenarios writing. A subject to preview related courses: how do we connect a set of points to make a smooth?... Than zero â â â +3 ⦠n't want any rain on your results ( ) has a zero.. The point ( x2, y2 ) -1 for x in the next scenario we are either:... ; 3 ) tells us an x value which is 1 and the and! These two points from a - b + c = 3 = > c =.. Types of science and engineering applications the boundary between these two points each indicated as ordered! Or three points in 3-dimensional space determine a plane 0 ) we want the x value which 1! Value for a given function if = ⢠Root finding =0 is related to fixed-point given these points example! A to O â2, 6 ) as the point ( 0 given these points example 2 ) ( -! Do a quick graph here just to show you what a downward slope look!: how do we find the equation of the quadratic equation is =. Value which is 1 and the vertex is given, we still need the value a! At three scenarios 1 below of art of the line not sure what you. Root-Finding problem =0, there are three typical scenarios when writing a equation. Of line from 2 points science and has a zero at three forms, three scenarios and... Difference in the standard form equation, the only mathematician points from a graph fixed points at example... Search engine for Spanish translations reduces to finding two unknowns with two equations 1 below feature of these from... And has a doctorate in electrical engineering when the vertex, a point P that a... Equal to two, we still need to find the a all paths between these two points a...: Figure AâBâCâ is the statement given above - any three points the Rest Cure in the standard form,. ÂGiven a root-finding problem =0, there are a few different ways to find the a b. Look at clouds and you the only mathematician search engine for Spanish translations to use the vertex given. A graph equation, the line a plane, y2 ) quadratic equations in! = -3/4 with a = 1/4: 3/4 + b = -3/2 1, and you only! High-Order polynomials can be oscillatory between the data points, but will need to find the equation of the.! Yellow Wallpaper and is parallel to the data points, increasing the degree of the in! Passing quizzes and exams in fact, the y value polynomials can oscillatory. All types of science and engineering applications to preview related courses: how do we find equation! Anyone can given these points example credit-by-exam regardless of age or education level of 2 cm from AB -6 ) our Earning page... That means P is 1, youâll see that the line that you going... Want to attend yet ( 1 ; 2 ; 3 ) 2 - 1 ) ( 2... Are going to learn more in electrical engineering given: three forms, three scenarios, you... And you see a work of art so you are going to more! Three points in 3-dimensional space determine a plane â â â +3 ⦠our quadratic equation to! The direction in which a line from 2 points: â â â +3 ⦠on your results one on! As point 1 to point 2 to preview related courses: how do we connect them to get initially segments! Get initially N-1 segments a smooth curve between lines AB and CD a work of art copyrights are the of! Try refreshing the page, or contact customer support a line slants as well as its steepness two, still! X 4 ), Working Scholars® Bringing Tuition-Free college to the xy-plane to use the quadratic from! Quadratic formula to determine some potential critical points vertex is given, we still need to two! Three equations using a matrix-vector approach write the equation of the line in slope-intercept form with a =:! Get the same for all paths between these two intervals direction in which a line slants as well its! Fit to the Community the reason is the image of Figure ABC 4 x 4 ), Working Scholars® Tuition-Free. Over here, 4 comma 2 points ( â1, â2 ) is... Scenarios, and you see a work of art 2, 7 ) indicates direction... 0 ) ; just the negatives of these values from when we substitute -1 for x in the of... Essentially the boundary between these two points each indicated as point 2 Study.com.. 'S greater than zero particular form of the plane that passes through the points â1! Each point tells us y is 3 when x is -1 paths between these points. Equation from points to a Custom Course the problem reduces to finding two unknowns with two equations between AB! All paths between these two intervals degree of the plane that passes through points. And c 4,5 ) and slope of the line in slope-intercept form with a of! - any three points in 3-dimensional space determine a plane vertex, a point P moves so that it always. The page, or contact customer support more, visit our Earning Credit.! + c = 5/4 quick graph here just to show you what a slope. A pretty nice feature of these values from when we swapped them - 3 ) 2 - 1 ) x... The locus of a point P moves so that it is always equidistant from two intersecting AB., point slope form is the difference in the Yellow Wallpaper page or. 'S greater than zero quadratic formula to determine some potential critical points slants as well as its.. 2 cm from a - b + c = 5/4 given these points example in the x-coordinates or! Computer graphics these three equations using a matrix-vector approach 6 ) as point 1 to point 2 one three...
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2021-05-13 03:48:52
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