text:
# BAT.jl Documentation

BAT.jl is a Bayesian Analysis Toolkit in Julia. It is a high-performance tool box for Bayesian inference with statistical models expressed in a general-purpose programming language instead of a domain-specific language.

Typical applications for this package are parameter inference given a model (in the form of a likelihood function and prior), the comparison of different models in the light of a given data set, and the test of the validity of a model to represent the data set at hand. BAT.jl provides access to the full Bayesian posterior distribution to enable parameter estimation, limit setting and uncertainty propagation. BAT.jl also provides supporting functionality like plotting recipes and reporting functions.

BAT.jl is implemented in pure Julia and allows for a flexible definition of mathematical models and applications while enabling the user to code for the performance required for computationally expensive numerical operations. BAT.jl provides implementations (internally and via other Julia packages) of algorithms for sampling, optimization and integration. BAT's main focus is on the analysis of complex custom models. It is designed to enable parallel code execution at various levels (running multiple MCMC chains in parallel is provided out-of-the-box).

It's possible to use BAT.jl with likelihood functions implemented in languages other than Julia: Julia allows for calling code in C and Fortran, C++, Python and several other languages directly. In addition, BAT.jl provides (as an experimental feature) a very lightweight binary RPC protocol that is easy to implement, to call non-Julia likelihood functions written in another language and running in separate processes.

BAT.jl originated as a rewrite/redesign of BAT, the Bayesian Analysis Toolkit in C++. BAT.jl now offer a different set of functionality and a wider variety of algorithms than it's C++ predecessor.

Note

BAT.jl requires Julia >= v1.3, we recommend to use Julia >= v1.5.

## Citing BAT.jl

When using BAT.jl for research, teaching or similar, please cite

Schulz et al. "BAT.jl - A Bayesian Analysis Toolkit in Julia" (2020) arXiv:2008.03132

@article{Schulz:2020ebm,
author = "Schulz, Oliver and Beaujean, Frederik and Caldwell, Allen and Grunwald, Cornelius and Hafych, Vasyl and Kröninger, Kevin and La Cagnina, Salvatore and Röhrig, Lars and Shtembari, Lolian",
title = "{BAT.jl -- A Julia-based tool for Bayesian inference}",
eprint = "2008.03132",
archivePrefix = "arXiv",
primaryClass = "stat.CO",
month = "8",
year = "2020"
}

If you use BAT.AHMIntegration as an important part of your work, please also cite

Caldwell et al. "Integration with an Adaptive Harmonic Mean Algorithm" (2020) arXiv:1808.08051

@article{caldwell2020integration,
author = "Caldwell, Allen and Eller, Philipp and Hafych, Vasyl and Schick, Rafael and Schulz, Oliver and Szalay, Marco",
title = "{Integration with an adaptive harmonic mean algorithm}",
journal = "International Journal of Modern Physics A",
volume = "35",
number = "24",
pages = "2050142",
year = "2020",
publisher = "World Scientific",
doi = "10.1142/S0217751X20501420"
}

If you use BAT.PartitionedSampling (experimental feature) as an important part of your work, please also cite

Hafych et al. "Parallelizing MCMC Sampling via Space Partitioning" (2020) arXiv:2008.03098

@article{Hafych:2008.03098,
author = "Hafych, Vasyl and Eller, Philipp and Caldwell, Allen and Schulz, Oliver",
title = "{Parallelizing MCMC Sampling via Space Partitioning}",
eprint = "2008.03098",
archivePrefix = "arXiv",
primaryClass = "stat.CO",
month = "8",
year = "2018"
}


text:
mersenneforum.org Simple Diophantine equation
 Register FAQ Search Today's Posts Mark Forums Read

 2009-04-10, 01:20 #1 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×3×1,571 Posts Simple Diophantine equation $a^3 = b^5 + 100$ Is there more than one solution? ________ P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35. If so, then of course I am convinced, too. Sorry, then it was ...too simple. Last fiddled with by Batalov on 2009-04-10 at 02:12
 2009-04-10, 01:56 #2 CRGreathouse     Aug 2006 32×5×7×19 Posts Doubtful. No small solutions (|a^3| < 10^35) other than 7^3 = 3^5 + 100, and powers are rarely close together. S. S. Pillai conjectures that a positive integer can be expressed as the difference of powers only finitely many ways, which suggests that finite checking is meaningful. Last fiddled with by CRGreathouse on 2009-04-10 at 02:36
2009-04-10, 02:38   #3
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by Batalov P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35. If so, then of course I am convinced, too. Sorry, then it was ...too simple.
I just checked -10 million to 10 million, raised to the fifth power and added 100, and checked if the numbers were cubes (using modular restrictions to avoid taking too many cube roots).

2009-04-10, 11:32   #4
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Batalov $a^3 = b^5 + 100$ Is there more than one solution? ________ P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35. If so, then of course I am convinced, too. Sorry, then it was ...too simple.

Note that Faltings proof of the Mordell Conjecture shows that there
are only finitely many solutions. Actually, this is like hitting a thumbtack
with a sledgehammer. Siegel's Theorem suffices to show the same thing.

Unfortunately, neither is effective. Nor would an application of the
ABC conjecture be effective.

 Similar Threads Thread Thread Starter Forum Replies Last Post Nick Number Theory Discussion Group 2 2016-12-18 14:49 jasong jasong 4 2012-02-20 03:33 flouran Math 7 2009-12-12 18:48 grandpascorpion Math 11 2009-09-23 03:30 philmoore Puzzles 8 2008-05-19 14:17

All times are UTC. The time now is 06:18.

Wed May 12 06:18:12 UTC 2021 up 34 days, 59 mins, 0 users, load averages: 1.60, 1.66, 1.70

format: plain


text:
How to call a java method that takes a boolean (not Boolean)?

I'm trying to call a method on a java object that takes a single boolean argument - note boolean (native or primative) not Boolean (Object or Reference type).

object@method[True]

Goes bang with...

Java::argx1: Method named method defined in class blah.blah.blah.Object was called with an incorrect number or type of arguments. The argument was True.

Various experiments with MakeJavaObject fail for the obvious reason they build True on the Mathematica side of JLink.

• You may try tocall Boolean.TRUE – s.s.o Sep 16 '14 at 18:36
• @Ymareth Can you look at the latest edit of my answer and specify which operating system you are working on? – halirutan Sep 1 '16 at 10:56

Let's make a test by ourselves. The class java.lang.Boolean has a static method

public static String toString(boolean b)

that takes a primitive boolean and returns a string. Here we go:

<< JLink

and now

BooleantoString[True]
(* "true" *)

So my educated guess is that your method does not take a single argument as the warning already tells you.

Aftermath

I considered this a simple mistake of the OP, but @Picket gave a minimal example in a comment, that fails on his machine

I took the time to download the JSoup library and tried it myself. Mysteriously, it works on my machine

One explanation is the difference in operating systems. While Picket is using OS X, I'm working on Linux here. I tried the above example in Mathematica version 11.

• I have the same problem as the OP: screenshot. I have no problems with any of the other ~40 methods that I've used from jsoup, and this is the only one with a boolean. Is your minimal example really conclusive? Could there not be problems introduced with overloaded methods for example, which I suppose are more difficult for JLink to handle? – C. E. Sep 1 '16 at 0:18
• @C.E. Hmm, strange. I just tried a non-trivial example with Matcher but it worked. Do we find an example that works in a standard java (without Jsoup)? Can you, just for for fun create a string object for your first argument? Like s = JavaNew["java.lang.String", "this is a string"]. – halirutan Sep 1 '16 at 0:42
• That didn't work. I will try to reproduce the problem without jsoup tomorrow. – C. E. Sep 1 '16 at 0:50


text:
# How does fed cutting federal fund rate affect student loan

If fed cuts rate, it means it's cutting federal fund rate right? then why do loans like credit card and stuend loan? those aren't like Mortgage rate which the fed cuts its rate by buying MBS

• The Fed doesn't cut mortgage rates by buying MBS. MBS didn't even exist until maybe a decade ago. The fed funds rates affects the rate at which banks can borrow from the fed, and hence the rate banks are willing to lend to people. Only in the case of – Mox Nov 4 '20 at 0:15

This affects among others also student loans (when they are provided by commercial banks - some places have government provided and subsidized $$0\%$$ interest student loans). However, of course if you already had a loan in a past it will only be affected if you took it at variable interest rate or if you are taking out a new loan, the fixed interest rate loans won't be changed retroactively.


text:
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text:
• ms@ms.lt
• +370 607 27 665
• My work is in the Public Domain for all to share freely.

Introduction E9F5FC

Understandable FFFFFF

Questions FFFFC0

Notes EEEEEE

Software

• Compare: projective - real numbers, conformal - complex numbers, symplectic - quaternions.
• Relate the existence of a quaternionic derivative df/dq with a symplectic map preserving oriented area.

Reals, Complexes, Quaternions, Octonions, Cayley-Dickson construction.

The Cayley-Dickson construction

The construction makes use of complex conjugation, which is an important duality: (ab)* = b*a* and also a** = a.

The construction is based on working with pairs of numbers, where multiplication is defined (a,b)(c,d) = (ac-d*b, da+bc*)

The construction can also be thought in terms of multiplicative conjugation: aba^-1

Analysis is based on the "looseness" by which a local property (the slope locally) may not maintain globally. And this looseness is of different kinds:

• nonlooseness of path - discrete (integer, rational) affine
• looseness of line - reals - projective
• looseness of angle - complexes - conformal
• looseness of orientation (cross product) - quaternions - symplectic

So looseness is the flip-side of invariance. We see the role of equivalence as based on limits. And also we see the qualitative distinction based on the nature of the limiting process - so taking the limit in all directions for the complexes relates to preserving angles.

What would be the notion of differentiation for a function on the quaternions?

Study Dyson's analysis of the random walk matrix - the generic matrix:

• the most general matrix ensemble, defined with a symmetry group which may be completely arbitrary, reduces to a direct product of independent irreducible ensembles each of which belongs to one of the three known types: complex Hermitian, real symmetric, and quaternion self-dual.

Symp(M,ω) vs. Vol(M,ω) . In dimension 2, the concepts of area preserving and symplectic maps coincide. In higher dimensions, it turns out that volume preserving maps may or may not be symplectic. A distinction is given by Gromov's famous Non-squeezing Theorem: in R2n,n>1 one cannot embed a unit ball inside an appropriately constructed, narrow enough cylinder, whereas this is always possible with a volume preserving diffeomorphism. The proof of this theorem gave rise to the notion of "symplectic width" and "capacities". Scholarpedia: Symplectic maps

Given: T T* = T* T = 1 for n x n real matrices T where * is the relevant conjugate transpose and consider the relevant (real, complex, quaternion) linear transformations of the relevant (real, complex, quaternion) n-dimensional Hilbert space that preserve the inner product. Then:

• The orthogonal matrices O(n) are real.
• The unitary matrices U(n) are complex.
• The symplectic matrices Sp(n) are quaternion.

Baez (see his proof): Given an irreducible unitary representation H of some group, and suppose H is isomorphic to its dual. Then there is a conjugate-linear isomorphism j:H->H and either:

• j2=1, so that j is a real structure and H is its complexification, H is equipped with an orthogonal structure where there is a nondegenerate symmetric bilinear pairing, intertwining operator f: H tensor H -> C.
• j2=-1, so we have the usual complex i, and j, and ij=k, yielding the quaternions. H is equipped with an orthogonal structure where there is a nondegenerate antisymmetric bilinear pairing, intertwining operator f: H tensor H -> C.

Complex derivative: In regions where the first derivative is not zero, holomorphic functions are conformal in the sense that they preserve angles and the shape (but not size) of small figures.

the only left and right holomorphic quaternion functions (with domain all of H) are the affine functions qa+b and aq+b respectively Stack Exchange q2 has no derivative because the value of hqh−1 is a rotation by 2 theta and thus depends on the direction in which h goes to zero.

References

Symplectic geometry

#### Numbers

Naujausi pakeitimai

 Puslapis paskutinį kartą pakeistas 2018 sausio 21 d., 14:37


text:
herlp voc :)... - QuestionCove

herlp voc :)

3 days ago

@Vocaloid

3 days ago

Sal’s Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and The profit on every wrap is$3. Sal made a profit of $1,470 from lunch specials last month. The equation 2x + 3y = 1,470 represents Sal’s profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold. Change the equation into slope-intercept form. Identify the slope and y-intercept of the equation. Be sure to show all of your work below: 3 days ago Vocaloid: 2x + 3y = 1,470 solve this for y 3 days ago cookiemonster: 2x + 3y = 1,470 -2x -2x 3y/3= (-2x+1470)/3 y=(-⅔)x + 490 3 days ago cookiemonster: what do you mean copy and pasting... why did you just first assume that. im working on this assignment in a doc.. i copy and pasted it here to make sure i was correct. 3 days ago Vocaloid: looks fine then. 3 days ago Vocaloid: now find the slope and y-intercept. 3 days ago cookiemonster: is this it? Slope-intercept form: y=mx+b, where “m” is the slope of the line and “b” is the y-value at which the line intercepts the y-axis of the graph, the y-value when x=0. m=(-⅔) b= 490 3 days ago Vocaloid: good, that's it 3 days ago cookiemonster: awesome :) thx 3 days ago cookiemonster: @Vocaloid ok now following the same like problem about sarah its just a different question but i answered it and i wanted you to check if its correct 3 days ago cookiemonster: Describe how you would graph this line using the slope-intercept method. Be sure to write in complete sentences. I would begin to graph this line by first putting appropriate numerical values on the y-axis and x-axis of my graph. For example, I know that when x is zero the y-value is huge. The y-intercept is close to 500. So, I would label my y-axis in 10 increments of 100 up to 1000. Next, I appropriately label the x-axis from 0-1000 in 10 increments 100. I would not include either the negative x or y values because it is not possible to sell negative sandwiches or wraps.The line on the graph represents the points at which the profit is$1, 470 for x amount of sandwiches sold and the corresponding y number of wraps that would have to be sold in order to reach that profit point exactly. The first point I would groph would be the y-intercept because I know, from the equation, (0,490) is the point at which the line intercepts the y-axis. The negative slope tells me that the y-values will be decreasing from the y-intercept to y being equal to zero. The -⅔ slope tells me that the y-values will decrease at a rate of 2 while the x-values increase at a rate of 3. I know this because I know slope is equal to the change in rise or fall in the y-direction over run in the x-direction.

3 days ago
Vocaloid:

Looks fine to me Very detailed but good

3 days ago

ok great now i do need your help with the rest of the questions

3 days ago

i answered this one as well please check Write the equation in function notation. Explain what the graph of the function represents. Be sure to use complete sentences. Let f(x) be equal to the number of wraps sold to reach a total profit of $1,470 for which the number of sandwiches is equal to x sandwiches. Therefore, the function notation of this equation would be f(x)= (-⅔)x+490. The function f(x) represents the points along the y-axis at which the profit is$1, 470 for x amount of sandwiches sold. The y-axis values are number of wraps that would have to be sold to reach that profit.

3 days ago

@Vocaloid

3 days ago
Vocaloid:

Good

3 days ago

ok now i need help with this: Graph the function using one of the following two options below. On the graph, make sure to label the intercepts. ~You may graph your equation by hand on a piece of paper and scan your work. ~You may graph your equation using graphic technology. Paste your graph below:

3 days ago

@Vocaloid

3 days ago
Vocaloid:

I mean, you have pretty detailed instructions in this post here

3 days ago
Vocaloid:

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @cookiemonster Describe how you would graph this line using the slope-intercept method. Be sure to write in complete sentences. I would begin to graph this line by first putting appropriate numerical values on the y-axis and x-axis of my graph. For example, I know that when x is zero the y-value is huge. The y-intercept is close to 500. So, I would label my y-axis in 10 increments of 100 up to 1000. Next, I appropriately label the x-axis from 0-1000 in 10 increments 100. I would not include either the negative x or y values because it is not possible to sell negative sandwiches or wraps.The line on the graph represents the points at which the profit is $1, 470 for x amount of sandwiches sold and the corresponding y number of wraps that would have to be sold in order to reach that profit point exactly. The first point I would groph would be the y-intercept because I know, from the equation, (0,490) is the point at which the line intercepts the y-axis. The negative slope tells me that the y-values will be decreasing from the y-intercept to y being equal to zero. The -⅔ slope tells me that the y-values will decrease at a rate of 2 while the x-values increase at a rate of 3. I know this because I know slope is equal to the change in rise or fall in the y-direction over run in the x-direction. $$\color{#0cbb34}{\text{End of Quote}}$$ 3 days ago Vocaloid: Just follow the directions you wrote here and you should be good to go 3 days ago Vocaloid: Try sketching this on paper and take a pic with your phone or something 3 days ago cookiemonster: but how do i like make the graph... i have to use a graphing website? 3 days ago Vocaloid: It says you can draw it by hand and then scan it 3 days ago cookiemonster: ok.... so from what number to what number? 3 days ago Vocaloid: Again, read your directions 3 days ago cookiemonster: i have to do it from 0-1000? 3 days ago Vocaloid: That's what it says 3 days ago Vocaloid: You don't need to calculate every single value, once you have two points you can just connect them in a straight line 3 days ago Vocaloid: I would recommend graphing your x and y intercepts first 3 days ago Vocaloid: (Once you have set up your axes that is) 3 days ago Vocaloid: Need to take a break After you finish graphing, you can double check by plugging your function into wolframalpha.com 3 days ago cookiemonster: @Vocaloid ok can you atleast tell me what the function is so i can graph it 3 days ago Vocaloid: ??? It's the same function you've already been working on 3 days ago Vocaloid: The one about the sandwiches You know, the one that you wrote that entire huge paragraph explaining how to graph? 3 days ago Vocaloid: Your answers so far have been great, I don't know why you're suddenly struggling with this 3 days ago Vocaloid: Just follow the instructions (you know, the ones that you wrote) and you should be fine 3 days ago Bearclaws72: Good demonstration jesus 3 days ago Vocaloid: Cookie what do you have graphed so far? 3 days ago cookiemonster: nothing im confused... what would be my function? 3 days ago cookiemonster: im using this website http://www.wolframalpha.com/input/?i=graphing 3 days ago Vocaloid: The exact same function we have been working with for this ENTIRE problem 3 days ago Vocaloid: The one about the sandwiches 3 days ago cookiemonster: yes... is this it? y=(-⅔)x + 490 3 days ago Vocaloid: Yes good 3 days ago cookiemonster: so let me see hold on let me insert that into where it says function 3 days ago cookiemonster: http://www.wolframalpha.com/input/?i=graphing&rawformassumption=%22FSelect%22+-%3E+%7B%7B%22Plot%22%7D%7D&rawformassumption=%7B%22F%22,+%22Plot%22,+%22plotfunction%22%7D+-%3E%22y%3D(-%E2%85%94)x+%2B+490%22&rawformassumption=%7B%22FVarOpt%22%7D+-%3E+%7B%7B%22Plot%22,+%22plotvariable%22%7D,+%7B%22Plot%22,+%22plotlowerrange%22%7D,+%7B%22Plot%22,+%22plotupperrange%22%7D%7D&rawformassumption=%7B%22C%22,+%22graphing%22%7D+-%3E+%7B%22Calculator%22%7D this is what it shows for the graph^^ 3 days ago Vocaloid: Take a screen grab and save it 3 days ago cookiemonster: is that what the graph should look like? 3 days ago Vocaloid: Then paste it into your problem 3 days ago cookiemonster: is the graph correct? 3 days ago Vocaloid: Take a screenshot of the graph not the URL 3 days ago cookiemonster: i knoww 3 days ago Vocaloid: Does the slope and y-intercept match the equation? 3 days ago cookiemonster: well i think so the graph goes through the number 40 but then the x-axis ends at 10? 3 days ago Vocaloid: If the slope and y-intercept match the equation, then you should get good to go 3 days ago Vocaloid: *be good to go 3 days ago cookiemonster: right but how do i make sure that the match/ 3 days ago Vocaloid: Try adding "from x = 0 to x = 1000" to the end of the function line 3 days ago cookiemonster: im so confused i literally just want to know what the graph will look like 3 days ago cookiemonster: i dont know what to input in the blank parts .... 3 days ago Vocaloid: Type your function into the input section Then write "from x = 0 to x = 1000" Then press enter 3 days ago cookiemonster: it doesnt let me put that 3 days ago cookiemonster: ughhhh 3 days ago Vocaloid: Wolframalpha.com 3 days ago cookiemonster: ill just draw it then how would i draw it 3 days ago Bearclaws72: Why did I click on this 3 days ago Bearclaws72: Im forever notified 3 days ago Vocaloid: Again... if you want to do this by hand you have that entire paragraph of instructions 3 days ago cookiemonster: |dw:1507940495057:dw| 3 days ago cookiemonster: like please just help me create it... 3 days ago Vocaloid: $$\color{#0cbb34}{\text{Originally Posted by}}$$ @Vocaloid $$\color{#0cbb34}{\text{Originally Posted by}}$$ @cookiemonster Describe how you would graph this line using the slope-intercept method. Be sure to write in complete sentences. I would begin to graph this line by first putting appropriate numerical values on the y-axis and x-axis of my graph. For example, I know that when x is zero the y-value is huge. The y-intercept is close to 500. So, I would label my y-axis in 10 increments of 100 up to 1000. Next, I appropriately label the x-axis from 0-1000 in 10 increments 100. I would not include either the negative x or y values because it is not possible to sell negative sandwiches or wraps.The line on the graph represents the points at which the profit is$1, 470 for x amount of sandwiches sold and the corresponding y number of wraps that would have to be sold in order to reach that profit point exactly. The first point I would groph would be the y-intercept because I know, from the equation, (0,490) is the point at which the line intercepts the y-axis. The negative slope tells me that the y-values will be decreasing from the y-intercept to y being equal to zero. The -⅔ slope tells me that the y-values will decrease at a rate of 2 while the x-values increase at a rate of 3. I know this because I know slope is equal to the change in rise or fall in the y-direction over run in the x-direction. $$\color{#0cbb34}{\text{End of Quote}}$$ $$\color{#0cbb34}{\text{End of Quote}}$$

3 days ago

ok voc, i get that i have that paragraph doesnt mean i know how to draw it

3 days ago


text:
# Tikz - How not style nodes

New to using Tikz, i created tikz pictures I'd like to include in my latex document but latex keeps giving me a ! Package pgfkeys Error: I do not know the key '/tikz/none' and I am going to ignore it. error because I don't want to color my nodes (no fill or draw color). I just wanted to make a line in my image but to make a line you need to connect two nodes so I don't really need the nodes and would prefer they stay invisible.

\begin{tikzpicture}
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (1) at (7.5, 0) {};
\node [style=none] (2) at (-1.5, 1.75) {};
\node [style=none] (6) at (0, 3) {};
\node [style=none] (7) at (-0.5, 0) {};
\node [style=none] (8) at (0, 0) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw (0.center) to (1.center);
\draw (6.center) to (8.center);
\end{pgfonlayer}
\end{tikzpicture}

How do I get around this? I made a seperate style file as well but still didn't help.

(Note I'm using a mac app tikzit to make these things)

• if you not like some style for nodes, just dropout this option: node (0) at (-1.5, 0) {}; Mar 12, 2020 at 22:36
• seems better but getting now: ! Package pgf Error: Sorry, the requested layer 'nodelayer' is not part of the even though i have this: \pgfdeclarelayer{edgelayer} \pgfdeclarelayer{nodelayer} \pgfsetlayers{edgelayer, nodelayer}
– guy
Mar 12, 2020 at 22:47
• This works: \begin{tikzpicture} \pgfdeclarelayer{nodelayer} \pgfdeclarelayer{edgelayer} \pgfsetlayers{main,nodelayer,edgelayer} \begin{pgfonlayer}{nodelayer} \node (0) at (-1.5, 0) {}; \node (1) at (7.5, 0) {}; \node (2) at (-1.5, 1.75) {}; \node (6) at (0, 3) {}; \node (7) at (-0.5, 0) {}; \node (8) at (0, 0) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (0.center) to (1.center); \draw (6.center) to (8.center); \end{pgfonlayer} \end{tikzpicture}
– user194703
Mar 12, 2020 at 22:58
• ... if you wrap it into \documentclass[tikz,border=3mm]{standalone} \begin{document} and \end{document}, say.
– user194703
Mar 12, 2020 at 22:59
• @Schrödinger'scat yep that worked, thanks!
– guy
Mar 12, 2020 at 23:09

To provide an official answer to the problem. As Zarko is saying, if you do not want to add a style, do not add a style, but use none only if you have defined a style of that name. Also, if you use layers, you need to define them.

\documentclass[tikz,border=3mm]{standalone}
\begin{document}
\begin{tikzpicture}
\pgfdeclarelayer{nodelayer}
\pgfdeclarelayer{edgelayer}
\pgfsetlayers{main,nodelayer,edgelayer}
\begin{pgfonlayer}{nodelayer}
\node  (0) at (-1.5, 0) {};
\node  (1) at (7.5, 0) {};
\node  (2) at (-1.5, 1.75) {};
\node  (6) at (0, 3) {};
\node  (7) at (-0.5, 0) {};
\node  (8) at (0, 0) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw (0.center) to (1.center);
\draw (6.center) to (8.center);
\end{pgfonlayer}
\end{tikzpicture}
\end{document}

The ordering of the layers has been guessed here, a common problem when the code in the question is incomplete.


text:
# Classical theory fails to explain quantization of motions?

Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT \sim \hbar \omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.

The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $\propto \text{exp}(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.

For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.

Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.

By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).


text:
# Homework Help: (DC motor derive ω expression with respect time

1. May 10, 2014

### tcuay

1. The problem statement, all variables and given/known data
I try to derive the expression of ω with respect to t,
but the problem is the armature current Ia or the flux is unknown. I dont know how to solve it.
Anyone could give me ideas? Thanks a lot

2. Relevant equations

τ=J(ω/t)
Vs=IaRa+Ea
T=KEøIa
Ea=KEøω

#### Attached Files:

• ###### DC.png
File size:
22.6 KB
Views:
116
2. May 10, 2014

### Simon Bridge

Either (a) you are mistaken, and there is some way to find Ia, or (b) you can make a substitution for Ia so the term no longer appears in the equation - like you do with simultaneous equations.

3. May 12, 2014

### tcuay

I did in the following steps:
1.J(ω/t)=KøIa;
2. and then substitute Ia from Vs=IaRa+Ea into 1;
finally, i got:
ω=Vs/(Kø+JRa/Køt)

But it turns out the term ø cannot be eliminated and it is unknown.
Where am i wrong?

4. May 12, 2014

### Simon Bridge

Really?
What physical property does "ø" represent?

... isn't that supposed to be: $$\tau=J\frac{d\omega}{dt}$$...??

how did you get this from what went before?

The fact they give you a bunch of initial values suggests to me you should look for a DE.

Last edited: May 12, 2014
5. May 13, 2014

### tcuay

ø is the magnetic flux.
i have double checked for it, the deriving process should be right.
and it seems reasonable if t approaches infinity
Vs=Køω quite similar to Ea=Køω

6. May 13, 2014

### Simon Bridge

Derivation:
... So long as you are sure.

You don't appear to have used the fact that the armature resistance is "negligible".

What does that do to the derivation?


text:
# Matrix difference equation explained

A matrix difference equation is a difference equation in which the value of a vector (or sometimes, a matrix) of variables at one point in time is related to its own value at one or more previous points in time, using matrices.[1] [2] The order of the equation is the maximum time gap between any two indicated values of the variable vector. For example,

xt=Axt-1+Bxt-2

is an example of a second-order matrix difference equation, in which is an vector of variables and and are matrices. This equation is homogeneous because there is no vector constant term added to the end of the equation. The same equation might also be written as

xt+2=Axt+1+Bxt

or as

xn=Axn-1+Bxn-2

The most commonly encountered matrix difference equations are first-order.

## Nonhomogeneous first-order case and the steady state

An example of a nonhomogeneous first-order matrix difference equation is

xt=Axt-1+b

with additive constant vector . The steady state of this system is a value of the vector which, if reached, would not be deviated from subsequently. is found by setting in the difference equation and solving for to obtain

x*=[I-A]-1b

where is the n×n identity matrix, and where it is assumed that is invertible. Then the nonhomogeneous equation can be rewritten in homogeneous form in terms of deviations from the steady state:

\left[xt-x*\right]=A\left[xt-1-x*\right]

## Stability of the first-order case

The first-order matrix difference equation is stable - that is, converges asymptotically to the steady state - if and only if all eigenvalues of the transition matrix (whether real or complex) have an absolute value which is less than 1.

## Solution of the first-order case

Assume that the equation has been put in the homogeneous form . Then we can iterate and substitute repeatedly from the initial condition, which is the initial value of the vector and which must be known in order to find the solution:

\begin{align} y1&=Ay0\\ y2&=Ay

 2y 0

\\ y3&=Ay

 3y 0 \end{align}

and so forth, so that by mathematical induction the solution in terms of is

 t y t=A

y0

Further, if is diagonalizable, we can rewrite in terms of its eigenvalues and eigenvectors, giving the solution as

yt=PDtP-1y0,

where is an matrix whose columns are the eigenvectors of (assuming the eigenvalues are all distinct) and is an diagonal matrix whose diagonal elements are the eigenvalues of . This solution motivates the above stability result: shrinks to the zero matrix over time if and only if the eigenvalues of A are all less than unity in absolute value.

## Extracting the dynamics of a single scalar variable from a first-order matrix system

Starting from the -dimensional system, we can extract the dynamics of one of the state variables, say . The above solution equation for shows that the solution for is in terms of the eigenvalues of . Therefore the equation describing the evolution of by itself must have a solution involving those same eigenvalues. This description intuitively motivates the equation of evolution of, which is

y1,t=a1y1,t-1+a2y1,t-2+...+any1,t-n

where the parameters are from the characteristic equation of the matrix :

λn-a1λn-1-a2λn-2-...-anλ0=0.

Thus each individual scalar variable of an -dimensional first-order linear system evolves according to a univariate th-degree difference equation, which has the same stability property (stable or unstable) as does the matrix difference equation.

## Solution and stability of higher-order cases

Matrix difference equations of higher order - that is, with a time lag longer than one period - can be solved, and their stability analyzed, by converting them into first-order form using a block matrix (matrix of matrices). For example, suppose we have the second-order equation

xt=Axt-1+Bxt-2

with the variable vector being and and being . This can be stacked in the form

\begin{bmatrix}xt\xt-1\\end{bmatrix}=\begin{bmatrix}A&B\I&0\\end{bmatrix}\begin{bmatrix}xt-1\xt-2\end{bmatrix}

where is the identity matrix and is the zero matrix. Then denoting the stacked vector of current and once-lagged variables as and the block matrix as, we have as before the solution

zt=Ltz0

Also as before, this stacked equation, and thus the original second-order equation, are stable if and only if all eigenvalues of the matrix are smaller than unity in absolute value.

## Nonlinear matrix difference equations: Riccati equations

In linear-quadratic-Gaussian control, there arises a nonlinear matrix equation for the reverse evolution of a current-and-future-cost matrix, denoted below as . This equation is called a discrete dynamic Riccati equation, and it arises when a variable vector evolving according to a linear matrix difference equation is controlled by manipulating an exogenous vector in order to optimize a quadratic cost function. This Riccati equation assumes the following, or a similar, form:

Ht-1=K+A'HtA-A'HtC\left[C'H

 -1 tC+R\right]

C'HtA

where,, and are, is, is, is the number of elements in the vector to be controlled, and is the number of elements in the control vector. The parameter matrices and are from the linear equation, and the parameter matrices and are from the quadratic cost function. See here for details.

In general this equation cannot be solved analytically for in terms of ; rather, the sequence of values for is found by iterating the Riccati equation. However, it has been shown[3] that this Riccati equation can be solved analytically if and, by reducing it to a scalar rational difference equation; moreover, for any and if the transition matrix is nonsingular then the Riccati equation can be solved analytically in terms of the eigenvalues of a matrix, although these may need to be found numerically.[4]

In most contexts the evolution of backwards through time is stable, meaning that converges to a particular fixed matrix which may be irrational even if all the other matrices are rational. See also .

A related Riccati equation[5] is

Xt+1=-\left[E+BXt\right]\left[C+AX

 -1 t\right]

in which the matrices,,,, and are all . This equation can be solved explicitly. Suppose, which certainly holds for with and with . Then using this in the difference equation yields

\begin{align} Xt+1&=-\left[E+BNtD

 -1 t

\right]DtD

 -1 t

\left[C+ANtD

 -1 t

\right]-1\\ &=-\left[EDt+BNt\right]\left[\left[C+ANt

 -1 D t
 -1 \right]D t\right]

\\ &=-\left[EDt+BNt\right]\left[CDt+AN

 -1 t\right]

\\ &=Nt+1

 -1 D t+1

\end{align}

so by induction the form holds for all . Then the evolution of and can be written as

\begin{bmatrix}Nt+1\Dt+1\end{bmatrix}=\begin{bmatrix}-B&-E\A&C\end{bmatrix}\begin{bmatrix}Nt\Dt\end{bmatrix}\equivJ\begin{bmatrix}Nt\Dt\end{bmatrix}

Thus by induction

\begin{bmatrix}Nt\Dt\end{bmatrix}=Jt\begin{bmatrix}N0\D0\end{bmatrix}


text:
## Seminars and Colloquia Schedule

Monday, October 1, 2018 - 13:55 , Location: Skiles 005 , Dr. Andre Wibisono , Georgia Tech CS , Organizer: Molei Tao
Accelerated gradient methods play a central role in optimization, achieving the optimal convergence rates in many settings. While many extensions of Nesterov's original acceleration method have been proposed, it is not yet clear what is the natural scope of the acceleration concept. In this work, we study accelerated methods from a continuous-time perspective. We show there is a Bregman Lagrangian functional that generates a large class of accelerated methods in continuous time, including (but not limited to) accelerated gradient descent, its non-Euclidean extension, and accelerated higher-order gradient methods. We show that in continuous time, these accelerated methods correspond to traveling the same curve in spacetime at different speeds. This is in contrast to the family of rescaled gradient flows, which correspond to changing the distance in space. We show how to implement both the rescaled and accelerated gradient methods as algorithms in discrete time with matching convergence rates. These algorithms achieve faster convergence rates for convex optimization under higher-order smoothness assumptions. We will also discuss lower bounds and some open questions. Joint work with Ashia Wilson and Michael Jordan.
Monday, October 1, 2018 - 14:00 , Location: Skiles 006 , Lenny Ng , Duke University , Organizer: Caitlin Leverson
I'll describe a way to construct an A-infinity category associated to a contact manifold, analogous to a Fukaya category for a symplectic manifold. The objects of this category are Legendrian submanifolds equipped with augmentations. Currently we're focusing on standard contact R^3 but we're hopeful that we can extend this to other contact manifolds. I'll discuss some properties of this contact Fukaya category, including generation by unknots and a potential application to proving that augmentations = sheaves''. This is joint work in progress with Tobias Ekholm and Vivek Shende.
Monday, October 1, 2018 - 15:30 , Location: Skiles 006 , Jason Joseph , UGA , Organizer: Caitlin Leverson
The knot group has played a central role in classical knot theory and has many nice properties, some of which fail in interesting ways for knotted surfaces. In this talk we'll introduce an invariant of knotted surfaces called ribbon genus, which measures the failure of a knot group to 'look like' a classical knot group. We will show that ribbon genus is equivalent to a property of the group called Wirtinger deficiency. Then we will investigate some examples and conclude by proving a connection with the second homology of the knot group.
Wednesday, October 3, 2018 - 12:20 , Location: Skiles 005 , Justin Lanier , Georgia Tech , Organizer: Trevor Gunn
After briefly describing my research interests, I’ll speak on two results that involve points moving around on surfaces. The first result shows how to “hear the shape of a billiard table.” A point bouncing around a polygon encodes a sequence of edges. We show how to recover geometric information about the table from the collection of all such bounce sequences. This is joint work with Calderon, Coles, Davis, and Oliveira. The second result answers the question, “Given n distinct points in a closed ball, when can a new point be added in a continuous fashion?” We answer this question for all values of n and for all dimensions. Our results generalize the Brouwer fixed point theorem, which gives a negative answer when n=1. This is joint work with Chen and Gadish.
Wednesday, October 3, 2018 - 12:55 , Location: Skiles 006 , Xingyu Zhu , Georgia Institute of Technology , , Organizer: Galyna Livshyts
 The n-dimensional L^p Brunn-Minkowski inequality for p<1 , in particular the log-Brunn-Minkowski inequality, is proposed by Boroczky-Lutwak-Yang-Zhang in 2013, based on previous work of Firey and Lutwak . When it came out, it promptly became the major problem in convex geometry. Although some partial results on some specific convex sets are shown to be true, the general case stays wide open.  In this talk I will present a breakthrough on this conjecture due to E. Milman and A Kolesnikov, where we can obeserve a beautiful interaction of PDE, operator theory, Riemannian geometry and all sorts of best constant estimates. They showed the validity of the local version of this inequality for orgin-symmtric convex sets with a C^{2} smooth boundary and strictly postive mean curvature, and for p between 1-c/(n^{3/2}) and 1. Their infinitesimal formulation of this inequality reveals the deep connection with the poincare-type inequalities. It turns out, after a sophisticated transformation, the desired inequality follows from an estimate of the universal constant in Poincare inequality on convex sets.
Wednesday, October 3, 2018 - 13:55 , Location: Skiles 005 , Allysa Genschaw , University of Missouri , , Organizer: Michael Lacey
We prove a criterion for nondoubling parabolic measure to satisfy a weak reverse H¨older inequality on a domain with time-backwards ADR boundary, following a result of Bennewitz-Lewis for nondoubling harmonic measure.
Wednesday, October 3, 2018 - 14:00 , Location: Skiles 006 , Stephen Mckean , GaTech , Organizer: Anubhav Mukherjee
Many problems in algebraic geometry involve counting solutions to geometric problems. The number of intersection points of two projective planar curves and the number of lines on a cubic surface are two classical problems in this enumerative geometry. Using A1-homotopy theory, one can gain new insights to old enumerative problems. We will outline some results in A1-enumerative geometry, including the speaker’s current work on Bézout’s Theorem.
Wednesday, October 3, 2018 - 16:00 , Location: Skiles 005 , Rupert Frank , LMU Munich/Caltech , , Organizer: Michael Loss
We generalize the Lp spectral cluster bounds of Sogge for the Laplace-Beltrami operator on compact Riemannian manifolds to systems of orthonormal functions. We show that these bounds are optimal on any manifold in a very strong sense. These spectral cluster bounds follow from Schatten-type bounds on oscillatory integral operators and their optimality follows by semi-classical analysis.
Wednesday, October 3, 2018 - 16:30 , Location: Skiles 006 , James Anderson , Georgia Tech , Organizer: Xingxing Yu
Erdős and Nešetřil conjectured in 1985 that every graph with maximum degree Δ can be strong edge colored using at most (5/4)Δ^2 colors. In this talk we discuss recent progress made in the case of Δ=4, and go through the method used to improve the upper bound to 21 colors, one away from the conjectured 20.
Thursday, October 4, 2018 - 13:30 , Location: Skiles 006 , Daniel Minahan , Georgia Tech , Organizer: Trevor Gunn
We discuss the construction of spectral sequences and some of their applications to algebraic geometry, including the classic Leray spectral sequence.  We will draw a lot of diagrams and try to avoid doing anything involving lots of indices for a portion of the talk.
Friday, October 5, 2018 - 13:05 , Location: Skiles 005 , , ISyE, Georgia Tech , , Organizer: He Guo
Abstract: Queueing systems are studied in various asymptotic regimes because they are hard to study in general. One popular regime of study is the heavy-traffic regime, when the system is loaded very close to its capacity. Heavy-traffic behavior of queueing systems is traditionally studied using fluid and diffusion limits. In this talk, I will present a recently developed method called the 'Drift Method', which is much simpler, and is based on studying the drift of certain test functions. In addition to exactly characterizing the heavy-traffic behavior, the drift method can be used to obtain lower and upper bounds for all loads. In this talk, I will present the drift method, and its successful application in the context of data center networks to resolve a decade-old conjecture. I will also talk about ongoing work and some open problems.   Bio: Siva Theja Maguluri is an Assistant Professor in the School of Industrial and Systems Engineering at Georgia Tech. Before that, he was a Research Staff Member in the Mathematical Sciences Department at IBM T. J. Watson Research Center. He obtained his Ph.D. from the University of Illinois at Urbana-Champaign in Electrical and Computer Engineering where he worked on resource allocation algorithms for cloud computing and wireless networks. Earlier, he received an MS in ECE and an MS in Applied Math from UIUC and a B.Tech in Electrical Engineering from IIT Madras. His research interests include Stochastic Processes, Optimization, Cloud Computing, Data Centers, Resource Allocation and Scheduling Algorithms, Networks, and Game Theory. The current talk is based on a paper that received the best publication in applied probability award, presented by INFORMS Applied probability society.
Friday, October 5, 2018 - 14:00 , Location: Skiles 005 , , University of Iowa , Organizer: Philipp Jell
In this talk, we introduce rather exotic algebraic structures called hyperrings and hyperfields. We first review the basic definitions and examples of hyperrings, and illustrate how hyperfields can be employed in algebraic geometry to show that certain topological spaces (underlying topological spaces of schemes, Berkovich analytification of schemes, and real schemes) are homeomorphic to sets of rational points of schemes over hyperfields.
Friday, October 5, 2018 - 15:05 , Location: Skiles 156 , Adrian P. Bustamante , Georgia Tech , Organizer: Adrian Perez Bustamante
In this talk I will present a proof of a generalization of a theorem by Siegel, about the existence of an analytic conjugation between an analytic map, $f(z)=\Lambda z +\hat{f}(z)$, and a linear map, $\Lambda z$, in $\mathbb{C}^n$. This proof illustrates a standar technique used to deal with small divisors problems. I will be following the work of E. Zehnder. This is a continuation of last week talk.


text:
## WeBWorK Main Forum

### Can "Email instructor" be directed to different emails depending on the student's section?

by Debbie Yuster -
Number of replies: 10
Hello,

I am managing several sections of a course in one WeBWorK course. Each section has its own instructor. Rather than having all instructors receive all student emails, is there a way to direct emails from students in Section 1 to Instructor 1, emails from students in Section 2 to Instructor 2, etc?

It doesn't look like this can be done in WeBWorK...might have to send all the emails to a certain account and somehow filter and redirect them to the appropriate instructor.

Any better ideas?

Thanks,
Debbie

### Re: Can "Email instructor" be directed to different emails depending on the student's section?

This is a good question. To my knowledge the answer is no.

Our initial solution was to have a separate course for each
discussion section.

That was cumbersome since we have 4 discussion sections per
discussion leader. So now we have a separate course for

This is still awkward because homework is assigned for the class
as a whole, but it seems like the right level of compromise for
our setup.

-Hal

### Re: Can "Email instructor" be directed to different emails depending on the student's section?

by Gavin LaRose -
Hi Debbie & Hal,

I think this is explicitly possible by setting the $feedback_by_section variable in global.conf (or, for an individual course, in the course.conf file). Setting that to a true value should result in only the instructors with the same section number as the student submitting the feedback request receiving the feedback e-mail. Gavin In reply to Gavin LaRose ### Re: Can "Email instructor" be directed to different emails depending on the student's section? by Hal Sadofsky - Thanks Gavin - I didn't realize that. We'll try it. In reply to Hal Sadofsky ### Re: Can "Email instructor" be directed to different emails depending on the student's section? by Michael Schroeder - Debbie, We have one course with 24 sections, and we found a way to distribute the emails. We put the following flag in the course.conf file: $feedback_by_section = 1;

Then if a student submits an email, it will go to all those TA or higher with the student's section in their section name. For example, if I have my section as "001,002,003" and a student from section "002" sends an email, I will receive it, but I will not from a student in section "004".

If you have someone wanting to receive all emails, setting their section to ".*" did the trick for us.

Hope this helps!

Mike Schroeder

### Re: Can "Email instructor" be directed to different emails depending on the student's section?

by Debbie Yuster -
Thanks everyone! This is an easy way to do exactly what I wanted.

Best,
Debbie

### Re: Can "Email instructor" be directed to different emails depending on the student's section?

by Zach Teitler -

Does this still work in 2021-2022?

by Glenn Rice -


text:
The Book Stacking Problem - Maple Programming Help

Home : Support : Online Help : Math Apps : Discrete Mathematics : MathApps/TheBookStackingProblem

The Book Stacking Problem

Main Concept

If you were to stack n identical books on the edge of a table, what is the maximum amount of overhang that can be achieved without the stack falling over?

 Solution Simulated below is the single-wide case, where there is only one block at every given level (no counter-balancing). The maximum amount of overhang, in terms of book lengths, is given by half of the ${n}^{\mathrm{th}}$ partial sum of the Harmonic series.   So, for a stack of n books, the largest possible distance from the edge of the table to the leading edge of the ${n}^{\mathrm{th}}$ book is:   Why? Let ${d}_{n}$ be the maximum overhang for a stack of n identical books. The key to obtaining the largest overhang is to have the center of gravity of n books lie directly above the table's edge, while the center of gravity of the top  books lies directly over the leading edge of the bottom book. By assigning M to be the mass of a single book, we can compute the total moment of n books with respect to the leading edge (the right edge in this simulation): Solving for ${d}_{n}$ , we obtain the following recurrence relation: It is obvious that . Since the center of gravity of a single book lies at the center of its length, the optimal way to balance it would be to have one half rest on the table, while the other half hangs over the edge. Now, we can see that: and so on... The general formula for this relation can be written as: Recognizing the sum $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}$ as the Harmonic series, we can simplify this formula:  , where ${H}_{n}$ is the ${n}^{\mathrm{th}}$ Harmonic number   Since the Harmonic series diverges as , the maximum amount of overhang will become arbitrarily large as the number of books grows.

Use the slider below to choose the number of books you would like to stack. Move the books around by clicking and dragging on the plot. Then, click "Release" to check if your stack is still balanced. Try to have your stack reach the bright green line, which represents the maximum overhang possible. Click "Reset" to restart the activity.

Check the "Show Current Centers of Gravity" box to see the centers of gravity of various sub-stacks. Each dashed line shows the center of gravity of the substack of books from the top book to the book directly above that line. The complete stack is balanced when all of these dashed lines occur over a book rather than over empty space.

Click "Show the Solution!" to see the solution of the Book Stacking Problem.

 Number of Books, n =     Current total overhang = book lengths Maximum total overhang = book lengths Current overhang for this book = book lengths Maximum overhang for this book = book lengths

 More MathApps


text:
## GIF Animations / GIFアニメーション

1. Euler's formula / オイラーの公式

LEFT: The complex exponential function $e^z=\exp z$ is defined by the series $\displaystyle e^z = 1 + \frac{z}{1!}+\frac{z^2}{2!}+\cdots$ for each $z \in \mathbb{C}$. When we let $z = it$ with $t \in \mathbb{R}$, the celebrated Euler's formula $e^{it}=\cos t + i \sin t$ follows. Indeed, the sequence $\{z_n\}$ defined by the recursive formula $\displaystyle z_0=0; z_{n+1}=z_{n} + \frac{(it)^n}{n!}$ converges to $\cos t + i \sin t$, a point on the unit circle. This GIF animation depicts the convergence of the sequence $\{z_n\}$ (blue heavy dots joined by blue segments) with angular parameter $t$ increasing from 0 to $4 \pi$.
RIGHT: The same idea applied for $z=\log 2 + i t~(0 \le t \le 4 \pi)$.
: 複素数 $z \in \mathbb{C}$ に対し， 複素指数関数 $e^z=\exp z$ は 級数 $\displaystyle e^z := 1 + \frac{z}{1!}+\frac{z^2}{2!}+\cdots$ によって定義される．とくに，$z = it$（$t \in \mathbb{R}$）を代入したとき， 有名な オイラーの公式 $e^{it}=\cos t + i \sin t$ が得られる. 実際，数列 $\{z_n\}$ を漸化式 $\displaystyle z_0=0; z_{n+1}=z_{n} + \frac{(it)^n}{n!}$ によって定義するとき，これは単位円周上の点 $\cos t + i \sin t$ に収束する． この GIF アニメーションは， 角度のパラメーター $t$ を 0 から $4 \pi$ まで動かしたとき， 数列 $\{z_n\}$ （青い線で結ばれる青い点たち） が収束する様子を表現したものである．
:同様に $z=\log 2 + i t~(0 \le t \le 4 \pi)$ としたもの.

2. Differentiation and linear approximation / 微分と1次近似

LEFT: We say a function $y=f(x)$ is differentiable at $x=a$ if there exists a constant $A \in \mathbb{R}$ such that $\displaystyle f(x)= f(a)+A (x-a) + o(x-a)~~(x \to a).$ This means that $f(x)$ is approximated by the tangent line $y = f(a) + A (x-a)$ and its relative precision gets better as $x \to a$. In other words, if we magnify the graph of $y=f(x)$ at $(a, f(a))$, it becomes harder to distinguish it from the tangent line. In this GIF animation $f(x)=\sin 3x + \sin^2 5x$, $a=0$, and $A=3$.
RIGHT: We say a function $z=f(x, y)$ in two variable is differentiable at $(x,y)=(a,b)$ if there exists a vector $(A, B) \in \mathbb{R}^2$ such that $\displaystyle ~~~~~~~~f(x,y)= f(a,b)+A (x-a) + B(y-b)+o(\{(x-a)^2+(y-b)^2\}^{1/2})$
as $(x,y) \to (a,b).$ Instead of considering the tangent plane, we use contour graphs to understand what it means. The contours (level curves) of $z=f(a,b)+A (x-a) + B(y-b)$ are straight lines perpendicular to the gradient vector $(A,B)$. Hence the differentiability means the contours of $z=f(x,y)$ are approximated by those of $z=f(a,b)+A (x-a) + B(y-b)$. In this GIF animation $f(x,y)=\sin(x + y)\cos( y - x^2/3 + 2)$, $(a,b)=(0,0)$, and $(A,B)= (\cos 2,\cos 2)$.
: 関数 $y=f(x)$ が $x=a$ で 微分可能 であるとは， ある定数$A \in \mathbb{R}$ が存在し， $\displaystyle f(x)= f(a)+A (x-a) + o(x-a)~~(x \to a).$ が成り立つことをいう．すなわち， $f(x)$ は接線 $y = f(a) + A (x-a)$ によって近似され，その相対的な精度は $x \to a$ のときより高くなる． 言い換えると，$y=f(x)$ のグラフを点 $(a, f(a))$ を中心に 拡大すればするほど，接線との区別が難しくなるということである． この GIF アニメーションでは， $f(x)=\sin 3x + \sin^2 5x$, $a=0$, $A=3$ としている．
: 2変数関数 $z=f(x, y)$ が $(x,y)=(a,b)$ において 微分可能 であるとは，あるベクトル $(A, B) \in \mathbb{R}^2$ が存在し，$(x,y) \to (a,b)$ のとき $\displaystyle ~~~~~~~~f(x,y)= f(a,b)+A (x-a) + B(y-b)+o(\{(x-a)^2+(y-b)^2\}^{1/2})$
となることをいう． その意味するところを理解するには，接平面を考えるよりも， 等高線グラフを考えるほうがよい． 関数 $z=f(a,b)+A (x-a) + B(y-b)$ の等高線はその勾配ベクトル $(A,B)$ と垂直な直線たちになる． よって，$z=f(x,y)$ の微分可能性とは， その等高線が $z=f(a,b)+A (x-a) + B(y-b)$ の等高線によって近似される， ということである． この GIF アニメーションでは， $f(x,y)=\sin(x + y)\cos( y - x^2/3 + 2)$, $(a,b)=(0,0)$, and $(A,B)= (\cos 2,\cos 2)$ としている.

3. Integrations / 積分

LEFT: In the 19th century Riemann introduced a classical definition of the integration which is still fundamental in college mathematics and numerical analysis. The idea is to approximate the region between the graph of a function and the axis by thin rectangles. In this animation the approximation starts with 5 rectangles (with randomly chosen thickness) and it gets finer as the number of rectangles increases up to 100.
MIDDLE and RIGHT: The same idea is applied for multivariable functions. For functions with two variables we use thin cuboids (like square columns) for approximation.
: 19世紀にリーマンが与えた積分の古典的な定義は， 現在も大学教養課程における数学や，数値解析学における基礎となっている． そのアイディアは，関数のグラフと軸で囲まれる部分を細長い長方形によって近似する， というものである．このアニメーションでは，5つの長方形（太さはランダムに選んでいる） による近似からスタートし，長方形の数が100まで増加する過程で近似が改善される様子を表している．

3'. Surface area / 曲面積

Defining the area" of a given surface is quite non-trivial. If the surface is smooth, then we have a natural definition by integration. For example, let $S$ be a surface parametrized by a vector-valued smooth (or $C^1$) function $\vec{p}=\vec{p}(s,t) \in \mathbb{R}^n$ where $(s,t)$ ranges over a domain $D$ of $\mathbb{R}^2$. Then its area is defined by $\displaystyle \iint_D \left ||\vec{p}_s\times\vec{p}_t \right || \, ds \, dt$. The integration is a limit of the total areas of tiny tangent planes (panels).

4. Fourier Series / フーリエ級数

Let $f:\mathbb{R} \to \mathbb{R}$ be a square wave function, that is, a periodic function of period $2 \pi$ such that
$f(x)=-1~(-\pi < x \le 0)$ and $f(x)=1~(0< x \le \pi)$. By Dirichlet's theorem, a classic result on Fourier series, it follows that that the equation $f(x)=\frac{4}{\pi}\Big( \sin x + \frac{\sin 3x}{3}+ \frac{\sin 5x}{5}+\cdots\Big) =\lim_{N \to \infty} \sum_{n=-N}^N \frac{-2i}{(2n-1)\pi}\big\{e^{(2n-1) xi} - e^{-(2n-1) xi}\big\}$ holds unless $f$ is discontinuous at $x$. The middle one is the Fourier expansion and the right-most one is its complex form. This animation shows the motion of the finite sum $S_N(x):=\frac{-2i}{\pi} \Big\{e^{xi} -e^{-xi} + \frac{e^{3xi}}{3} - \frac{e^{-3xi}}{3} +\cdots +\frac{e^{(2N-1)xi}}{2N-1} - \frac{e^{-(2N-1)xi}}{2N-1} \Big\}$ for sufficiently large $N$. Since $e^{\pm (2N-1)xi}$ rotates along the unit circle $\pm(2N-1)$ times for each period $2 \pi$, $S_N(x)$ is a composition of rotations with distinct weights. The animations below shows the case for $N=1,3$ and $5$.

5. $L^p$ convergence / $L^p$ 収束

It is hard to imagine the topology of the $L^p$ space $L^p([0,1])$ for $p \ge 1$, that is, the vector space of functions $f(x)$ on $[0,1]$ whose Lebesgue norm $\displaystyle ||f(x)||=\Big (\int_0^1 |f(x)|^p \, dx \Big)^{1/p}$ is finite. The animation depicts a sequence of functions $\{f_n\}$ such that it converges to $f \equiv 0$ (the zero vector) in $L^p$ but the sequence of numbers $\{f_n(x)\}$ diverges for every $x \in [0,1]$.
$L^p$ 空間 $L^p([0,1])$（$p \ge 1$）とは， $[0,1]$区間上の関数 $f(x)$ で，ルベーグのノルム $\displaystyle ||f(x)||=\Big (\int_0^1 |f(x)|^p \, dx \Big)^{1/p}$ が有限となるもの全体からなるベクトル空間だが， その位相を思い描くのはなかなか難しい． このアニメーションでは，関数列 $\{f_n\}$ で，$L^p$ において $f \equiv 0$ （0ベクトル）に収束するが， 任意の点 $x \in [0,1]$ で数列 $\{f_n(x)\}$ が発散するようなものを表現している．

Copyright (C) 2001-2020 by Tomoki Kawahira


text:
# How do you graph y=sqrt(x+4)?

Jun 20, 2015

You get hal parabola but along the x axis...have a look:

#### Explanation:

First of all you need to discover the allowed $x$ values for your function, the domain .
The square root can accept all $x$ as long as its argument doesn't get negative (you cannot find a real number as solution of a negative square root). You need all the $x$ such that:
$x + 4 \ge 0$ or:
$x \ge - 4$
Your domain is then all the x bigger or equal to $- 4$:

Next we choose "good" values for $x$ to be used into our function (are good because they allow to evaluate the square root easily):

$x = - 4$ then $y = \sqrt{0} = 0$
$x = - 3$ then $y = \sqrt{1} = 1$
$x = 0$ then $y = \sqrt{4} = 2$
$x = 5$ then $y = \sqrt{9} = 3$
$x = 12$ then #y=sqrt(16)=4

Plotting these points you get:


text:
## Developed counties not using SI units

So I understand changing up your system of measurement could be a bit of a chore. And I completely understand clinging on to a temperature scale based on fish brine smoothies for nostalgia. I get it, its quirky. Like totally makes sense. But even the kings of holding onto archaic traditions, the UK, have manage to wriggle free of its irrational grip and have put in a decent effort. Yes, we’re all looking at you USA…


text:
• BERT uses many layers of bidirectional transformers. Source: Adapted from Devlin et al. 2019, fig. 3.
• Encoder self-attention distribution for the word 'it' in different contexts. Source: Uszkoreit 2017.
• Asynchronous memory copy overlaps with computation. Source: Mukherjee et al. 2019, fig. 3.
• An example showing how BERT improves query understanding. Source: Nayak 2019.
• Fine-tuning BERT for various NLP tasks. Source: Devlin et al. 2019, fig. 4.
• BERT does basic token masking whereas ERNIE uses three types of masking. Source: Liu 2019, fig. 2.
• For sentiment analysis, Target-Dependent BERT uses tokens at target positions rather than [CLS]. Source: Gao et al. 2019, fig. 2.
• Input embeddings in BERT. Source: Devlin et al. 2019, fig. 2.
• BERT has spawned many variants. Source: thunlp 2019.

# BERT (Language Model)

arvindpdmn
1237 DevCoins
Last updated by arvindpdmn
on 2019-12-03 14:01:26
Created by arvindpdmn
on 2019-11-30 09:41:03

## Summary

NLP involves a number of distinct tasks each of which typically needs its own set of training data. Often each task has only a few thousand samples of labelled data, which is not adequate to train a good model. However, there's plenty of unlabelled data readily available online. This data can be used to train a baseline model that can be reused across NLP tasks. Bidirectional Encoder Representations from Transformers (BERT) is one such model.

BERT is pre-trained using unlabelled data on language modelling tasks. For specific NLP tasks, the pretrained model can be fine-tuned for that task. Pre-trained BERT models, and their variants, have been open sourced. This makes it easier for NLP researchers to fine-tune BERT and quickly advance the state of the art for their tasks.

## Milestones

Jun
2017

Vaswani et al. propose the transformer model in which they use a seq2seq model without RNN. The transformer model relies only on self-attention, although they're not the first to use self-attention. Self-attention is about attending to different tokens of the sequence. This would later prove to be the building block on which BERT is created.

Feb
2018

Peters et al. use many layers of bidirectional LSTM trained on a language model objective. The final embeddings are based on all the hidden layers. Thus, their embeddings are deeply contextual. They call it Embeddings from Language Models (ELMo). They show that higher-level LSTM states capture semantics while lower-level states capture syntax.

Oct
2018

Devlin et al. from Google publish on arXiv a paper titled BERT: Pre-training of Deep Bidirectional Transformers forLanguage Understanding.

Nov
2018

Google open sources pre-trained BERT models, along with TensorFlow code that does this pre-training. These models are for English. Later in the month, Google releases multilingual BERT that supports about 100 different languages. The multilingual model preserves case. The model for Chinese is separate. It uses character-level tokenization.

Apr
2019

NAACL announces the best long paper award to the BERT paper by Devlin et al. The annual NAACL conference itself is held in June.

May
2019

Whole Word Masking is introduced in BERT. This can be enabled with the option --do_whole_word_mask=True during data generation. For example, when the word 'philammon' is split into sub-tokens 'phil', '##am' and '##mon', then either all three are masked or none at all. Overall masking rate is not affected. As before, each sub-token is predicted independent of the others.

Aug
2019

Since a BERT model has 12 or 24 layers with multi-head attentions, using it in a real-time application is often a challenge. To make this practical for applications such conversational AI, NVIDIA releases TensorRT optimizations for BERT. In particular, the transformer layer has been optimized. Q, K and V are fused into a single tensor, thus locating them together in memory and improving model throughput. Latency is 2.2ms on T4 GPUs, well below the 10ms acceptable latency budget.

Oct
2019

Google Search starts using BERT for 10% of English queries. Since BERT looks at the bidirectional context of words, it helps in understanding the intent behind search queries. Particularly for conversational queries, prepositions such as "for" and "to" matter. BERT's bidirectional self-attention mechanism takes these into account. Because of the model's complexity, for the first time, Search uses Cloud TPUs.

## Discussion

• What's the typical process for using BERT?

BERT is an evolution of self-attention and transformer architecture that's becoming popular for neural network models. BERT is an encoder-only transformer. It's deeply bidirectional, meaning that it uses both left and right contexts in all layers.

BERT involves two stages: unsupervised pre-training followed by supervised task-specific fine-tuning. Once a BERT model is pre-trained, it can be shared. This enables downstream tasks to do further training on a much smaller dataset. Different NLP tasks can thus benefit from a single shared baseline model. In some sense, this is similar to transfer learning that's been common in computer vision.

While pre-training takes a few days on many Cloud TPUs, fine-tuning takes only 30 minutes on a single Cloud TPU.

For fine-tuning, one or more output layers are typically added to BERT. Likewise, input embeddings reflect the task. For question answering, an input sequence will contain the question and the answer while the model is trained to learn the start and end of answers. For classification, the [CLS] token at the output is fed into a classification layer.

• Could you describe the tasks on which BERT is pre-trained?

BERT is pre-trained on two tasks:

• Masked Language Model (MLM): Given a sequence of tokens, some of them are masked. The objective is then to predict the masked tokens. Masking allows the model to be trained using both left and right contexts. Specifically, 15% of tokens are randomly chosen for masking. Of these, 80% are masked, 10% are replaced with a random word, 10% are retained.
• Next Sentence Prediction (NSP): Given two sentences, the model predicts if the second one logically follows the first one. This task is used for capturing relationship between sentences since language modelling doesn't do this.

Unlike word embeddings such as word2vec or GloVe, BERT produces contextualized embeddings. This means that BERT produces multiple embeddings of a word, each representing the context around the word. For example, word2vec embedding for the word 'bank' would not differentiate between the phrases "bank account" and "bank of the river" but BERT can tell the difference.

• Which are some possible applications of BERT?

In October 2019, Google Search started using BERT to better understand the intent behind search queries. Another application of BERT is to recommend products based on a descriptive user request. Use of BERT for question answering on SQuAD and NQ datasets is well known. BERT has also been used for document retrieval.

BERT has been used for aspect-based sentiment analysis. Xu et al. use BERT for both sentiment analysis and comprehending product reviews so that questions on those products can be answered automatically.

Among classification tasks, BERT has been used for fake news classification and sentence pair classification.

To aid teachers, BERT has been used to generate questions on grammar or vocabulary based on a news article. The model frames a question and presents some choices, only one of which is correct.

BERT is still new and many novel applications might happen in future. It's possible to use BERT for quantitative trading. BERT can be applied to specific domains but we would need domain-specific pre-trained models. SciBERT and BioBERT are two examples.

• Which are the essential parameters or technical details of BERT model?

BERT pre-trained models are available in two sizes:

• Base: 12 layers, 768 hidden size, 12 self-attention heads, 110M parameters.
• Large: 24 layers, 1024 hidden size, 16 self-attention heads, 340M parameters.

Each of the above took 4 days to train on 4 Cloud TPUs (Base) or 16 Cloud TPUs (Large).

For pre-training, a batch size of 256 sequences was used. Each sequence contained 512 tokens, implying 128K tokens per batch. The corpus for pre-training BERT had 3.3 billion words: 800M from BooksCorpus and 2500M from Wikipedia. This resulted in 40 epochs for 1M training steps. Dropout of 0.1 was used on all layers. GELU activation was used.

For fine-tuning, batch sizes of 16 or 32 are recommended. Only 2-4 epochs are needed for fine-tuning. Learning rate is also different from what's used for pre-training. Learning rate is also task specific. Dropout used was same as in pre-training.

• How do I represent the input to BERT?

BERT input embeddings is a sum of three parts:

• Token: Tokens are basically words. BERT uses a fixed vocabulary of about 30K tokens. To handle rare words or those not in token vocabulary, they're broken into sub-words and then mapped to tokens. The first token of a sequence is [CLS] that's useful for classification tasks. During MLM pre-training, some tokens are masked.
• Segment/Sentence: An input sequence of tokens can be a single segment or two segments. A segment is a contiguous span of text, not an actual linguistic sentence. Since two segments are packed into the same sequence, each segment has its own embedding. Each segment is terminated by [SEP] token. For example in question answering, question is the first segment and answer is the second.
• Position: This represents the token's position within the sequence.

In practice, input embeddings can also contain an input mask. Since sequence length is fixed, the final sequence may involve padding. Input mask is used to differentiate between actual inputs and padding.

Beginners may wish to look at a visual explanation of BERT input embeddings.

• What are some variants of BERT?

BERT has inspired many variants: RoBERTa, XLNet, MT-DNN, SpanBERT, VisualBERT, K-BERT, HUBERT, and more. Some variants attempt to compress the model: TinyBERT, ALERT, DistilBERT, and more. We describe a few of the variants that outperform BERT in many tasks:

• RoBERTa: Showed that the original BERT was undertrained. RoBERTa is trained longer, on more data; with bigger batches and longer sequences; without NSP; and dynamically changes the masking pattern.
• ALBERT: Uses parameter reduction techniques to yield a smaller model. To utilize inter-sentence coherence, ALBERT uses Sentence-Order Prediction (SOP) instead of NSP.
• XLNet: Doesn't do masking but uses permutation to capture bidirectional context. It combines the best of denoising autoencoding of BERT and autoregressive language modelling of Transformer-XL.
• Could you share some resources for developers to learn BERT?

Developers can study the TensorFlow code for BERT. This follows the main paper by Devlin et al. (2019). This is also the source for downloading BERT pre-trained models.

Google has shared TensorFlow code that fine-tunes BERT for Natural Questions.

McCormick and Ryan show how to fine-tune BERT in PyTorch. HuggingFace provides transformers Python package with implementations of BERT (and alternative models) in both PyTorch and TensorFlow. They also provide a script to convert a TensorFlow checkpoint to PyTorch.

IBM has shared a deployable BERT model for question answering. An online demo of BERT is available from Pragnakalp Techlabs.

## Milestones

Jun
2017

Vaswani et al. propose the transformer model in which they use a seq2seq model without RNN. The transformer model relies only on self-attention, although they're not the first to use self-attention. Self-attention is about attending to different tokens of the sequence. This would later prove to be the building block on which BERT is created.

Feb
2018

Peters et al. use many layers of bidirectional LSTM trained on a language model objective. The final embeddings are based on all the hidden layers. Thus, their embeddings are deeply contextual. They call it Embeddings from Language Models (ELMo). They show that higher-level LSTM states capture semantics while lower-level states capture syntax.

Oct
2018

Devlin et al. from Google publish on arXiv a paper titled BERT: Pre-training of Deep Bidirectional Transformers forLanguage Understanding.

Nov
2018

Google open sources pre-trained BERT models, along with TensorFlow code that does this pre-training. These models are for English. Later in the month, Google releases multilingual BERT that supports about 100 different languages. The multilingual model preserves case. The model for Chinese is separate. It uses character-level tokenization.

Apr
2019

NAACL announces the best long paper award to the BERT paper by Devlin et al. The annual NAACL conference itself is held in June.

May
2019

Whole Word Masking is introduced in BERT. This can be enabled with the option --do_whole_word_mask=True during data generation. For example, when the word 'philammon' is split into sub-tokens 'phil', '##am' and '##mon', then either all three are masked or none at all. Overall masking rate is not affected. As before, each sub-token is predicted independent of the others.

Aug
2019

Since a BERT model has 12 or 24 layers with multi-head attentions, using it in a real-time application is often a challenge. To make this practical for applications such conversational AI, NVIDIA releases TensorRT optimizations for BERT. In particular, the transformer layer has been optimized. Q, K and V are fused into a single tensor, thus locating them together in memory and improving model throughput. Latency is 2.2ms on T4 GPUs, well below the 10ms acceptable latency budget.

Oct
2019

Google Search starts using BERT for 10% of English queries. Since BERT looks at the bidirectional context of words, it helps in understanding the intent behind search queries. Particularly for conversational queries, prepositions such as "for" and "to" matter. BERT's bidirectional self-attention mechanism takes these into account. Because of the model's complexity, for the first time, Search uses Cloud TPUs.

Author
No. of Edits
No. of Chats
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0
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2
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## Cite As

Devopedia. 2019. "BERT (Language Model)." Version 4, December 3. Accessed 2020-01-28. https://devopedia.org/bert-language-model
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text:
## Journal of Mathematics Research

Journal of Mathematics Research (ISSN: 1916-9795;  E-ISSN 1916-9809) is an open-access, international, double-blind peer-reviewed journal published by the Canadian Center of Science and Education. This journal, published quarterly (March, July, September, and December) in both print and online versions, keeps readers up-to-date with the latest developments in all aspects of mathematics.

The scopes of the journal include, but are not limited to, the following topics: statistics, approximation theory, numerical analysis, operations research, dynamical systems, mathematical physics, theory of computation, information theory, cryptography, graph theory, algebra, analysis, probability theory, geometry and topology, number theory, logic and foundations of mathematics.

This journal accepts article submissions online or by e-mail (jmr@ccsenet.org).

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b) Peer review. We use double-blind system for peer-review; both reviewers and authors’ identities remain anonymous. The paper will be peer-reviewed by three experts; two reviewers from outside and one editor from the journal typically involve in reviewing a submission. The review process may take 2-3 weeks.

c) Notification of the result of review by E-mail.

d) The authors revise paper and pay publication fee (300USD).

e) After publication, the corresponding author will receive two copies of printed journals, free of charge.

f) E-journal in PDF is available on the journal’s webpage, free of charge for download.

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The publisher and journal have a policy of “Zero Tolerance on the Plagiarism”. We check the plagiarism issue through two methods: reviewer check and plagiarism prevention tool (ithenticate.com).

All submissions will be checked by iThenticate before being sent to reviewers.

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We are calling for submission of papers. Please find the journal’s profile at http://www.ccsenet.org/jmr and submit your manuscripts online.

Authors are strongly encouraged to use the LaTeX style files to prepare their manuscript. You may submit your manuscript in the format of PDF or MS-Word to be reviewed, but if the paper is accepted to be published, you are required to provide the LaTeX source files or MS-Word.

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### Vol 6, No 4 (2014)

Journal of Mathematics Research, Vol. 6, No. 4, 2014, e-Version FirstTM

#### Articles

 Remarks on the Continuity of the Local Minimizer of Scalar Integral Functionals With Nonstandard General Growth Conditions PDF Tiziano Granucci p1
 Cordiality of a Star of the Complete Graph and a Cycle Graph $C(N\cdot K_{N})$ PDF V. J. Kaneria, H. M. Makadia, Meera Meghpara p18
 Fit States on Girard Algebras PDF Remigijus Petras Gylys p29
 Backward Ornstein-Uhlenbeck Transition Operators and Mild Solutions of Non-Autonomous Hamilton-Jacobi Equations in Banach Spaces PDF Rafael Serrano p55
 One-Parameter Equations of Spherical Conics and Its Applications PDF Bülent Altunkaya, Yusuf Yayli, H. Hilmi Hacisalihoglu, Fahrettin Arslan p77
 Diversity soliton solutions for the (2+1)-Dimensional Boiti-Leon-Manna-Pempinelli Equation PDF Xichao Deng, Hanlin Chen, Zhenhui Xu p85
 Effect of Some Geometric Transfers on Homology Groups PDF Samy M. Mostafa, Abdelaziz E. Radwan, Fayza Abelhalim Ibrahem, Fatema F.Kareem p90


text:
# FAST-SCAN, PULSED-JET SUBMILLIMETER WAVE ABSORPTION SPECTROSCOPY

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/19367

Files Size Format View
1999-RG-05.jpg 121.3Kb JPEG image

 Title: FAST-SCAN, PULSED-JET SUBMILLIMETER WAVE ABSORPTION SPECTROSCOPY Creators: Melnik, Dmitry G.; Gopalakrishnan, Sandhya; Miller, Terry A.; De Lucia, Frank C. Issue Date: 1999 Publisher: Ohio State University Abstract: A newly developed pulsed jet spectrometer utilizing an extension of the previously reported FASSST $technique^{a}$ is discussed. The output beam from a backward wave oscillator (BWO) is focused on a pulsed supersonic jet and rapidly scanned in frequency (up to $10^{5}$ Doppler linewidths per second), covering the entire accessible spectrum range. The absolute frequency calibration and system synchronization is performed in real time by linking the output signal to a computer running appropriate software. The sensitivity of the apparatus is such that $^{13}C^{18}O$ can be seen in natural abundance with modest integration time. The apparatus is optimized for searching for weak molecular lines. The method and apparatus has been developed for use in the spectroscopy of low lying vibrational modes corresponding to conformational, torsional, and psecudorotational motion of non-rigid molecules, including Van der Waals clusters. Initial experiments have involved complexes of the CO molecule. These and related results will be discussed. Description: $^{a}$ Petkie D.T.; Goyette T.M.; Bettens R.P.A; Belov S.P.; Albert S.; Helminger P; and Dc Lucia F.C., Rev. Sci. Instrum., 1997, 68(4), 1675 Author Institution: The Ohio State University, Dept. of Chemistry, Laser Spectroscopy Facility; The Ohio State University, Department of Physics URI: http://hdl.handle.net/1811/19367 Other Identifiers: 1999-RG-05