Mathematics - 2016-09-13 (page 1 of 3)

Balarka Sen

00:00

what're you working on?

arctic tern

combinatorial species, root systems

and cooking chicken

Balarka Sen

nice

chicken, i mean :)

It's already the break of a dawn here but I am planning on watching a movie since I'm off tomorrow

arctic tern

00:24

cool

MATH ASKER

00:53

hey

I have a question

In the game of roulette as played in Las Vegas, the wheel has 38 slots. Two slots are numbered 0 and 00, and the rest are numbered 1 to 36. A $1 bet on any number other than 0 or 00 wins $36 ($35 plus the $1 bet). Find the expected value of this game

For this I said 35(36/38)-1(2/38)

but I think thats wrong

can anyone help me with this

arctic tern

what is the expected value of a game?

MATH ASKER

I don't know

It says find the expected value of this game

robjohn

There is 1 way to win and 37 ways to lose. $\frac1{38}35+\frac{37}{38}(-1)=-\frac1{19}$

MATH ASKER

How is it one way to win

Any number beside 0 or 00

so that is 36 ways right, 36 ways a number can be picked

robjohn

@MATHASKER once you choose a number, there is one way to win and 37 ways to lose

MATH ASKER

01:02

How

I'm kind of confused??

robjohn

@MATHASKER once you choose a number if that number comes up you win. if any other number or 0 or 00 comes up, you lose

MATH ASKER

But it says any number other than 0 or 00 wins?

arctic tern

no, it says how much you win on a particular number if you bet on that number and the ball rolled on it

MATH ASKER

Oh

But

The question kind of says any other number, it doesn't say that it can only be one number

Or am I misintrepating the question

Jessy Cat

01:32

@robjohn is in the Halloween spirit.

arctic tern

01:43

@MATHASKER yes, if you bet on 1 and win (i.e. the ball lands on 1) then you get $36. same for betting on 2 and winning (i.e. the ball lands on 2), and any other number.

Semiclassical

02:05

@arctictern Is there a shared context for you being interested in combinatorial species and root systems?I know a little about each, but wouldn't have expected them together

arctic tern

I'm doing root systems to informally lecture to others, I have some combinatorial species stuff I'm thinking about on my own independently

the impetus was: if Lin(X) denotes linear orderings on X, and Perm(X) denotes permutations of X, then |Lin(X)|=|Perm(X)| for every X but Lin and Perm are not naturally isomorphic (if they were, they would be the same as S_X-sets, but Lin(X) is a single orbit while the orbits in Perm(X) are conjugacy classes).

Semiclassical

Ah. I know of root systems from their connection to Lie algebras.

arctic tern

sometimes combinatorial bijections make use of certain choices, which we can view as a "source" of bijections. for instance, if we choose a particular ordering on X, then that induces a bijection Lin(X)->Perm(X). indeed there is a natural iso LinxPerm->LinxLin. so I said: call C a "catalyst" for F and G if there is a natural transformation CxF->CxG that commutes with the projections ->C.

basically I've been thinking about these catalysts

Semiclassical

Mmkay

I'll confess that my knowledge of species is thin. My knowledge of generating functions is far more nitty gritty than absract

arctic tern

turns out a combinatorial species is really just an S_n-set for each n.

0celo7

02:13

@Semiclassical Every time I try to learn about Lie algebras, I get so bored. I'd hoped Helgason would make it tolerable by relating it to the best math, but nah.

Still boring :/

arctic tern

I find the results interesting, but much of the work tedious in lie algebras

Semiclassical

Yeah

arctic tern

that encompasses my impression of much of higher math actually...

Semiclassical

I suspect one thing I'd find interesting, if I could get into it, would be Lie theory as applied to differential equations

But I also can't be arsed to do so :/

0celo7

@Semiclassical I got a book on that, actually.

Completely unreadable.

Semiclassical

02:18

Lol

0celo7

I'll take a crack after a few courses on algebraic geometry lol

Page 5 is Frobenius distributions

Differential ideals and jet spaces

Semiclassical

The one bit of Lie algebra stuff I have used with some frequency is the Baker-Campbell-Hausdorf formula

0celo7

I've used lots of stuff about Lie groups

and some of the algebra spills over

but nothing like root systems or weights

But I hear those are useful in classifying symmetric spaces.

Semiclassical

Sounds right

My knowing about root systems is decidedly a one-off for me

0celo7

I got a book on Lie algebras actually

Was reading a bit during the UTK/VT game last Saturday

Semiclassical

02:23

You see su(2) a lot in condensed matter, but not su(3) so much

0celo7

I got bored quickly

The book defined everything weirdly and the first 60 pages are just showing everything is equivalent to the usual definitions.

Balarka Sen

what's new

0celo7

Lab report approaching 4 pages; no end in sight.

Balarka Sen

excellent

0celo7

Time to start on the procedure ;_;

Intro, equations and data are 4 pages

@Semiclassical Why do they make us do lab reports

Semiclassical

02:35

hell if i know

my version of that is "why do they make us assign lab reports"

0celo7

@Semiclassical wth

who makes you assign them?

I'm blaming my TA

Semiclassical

The department. I'm just a TA in a physics grad program.

The main reason they're assigned, as far as I can tell, is so that the course counts as "writing intensive."

and therefore people are able to take it in order to fulfill a university writing requirement

as compared with taking chemistry, i guess?

0celo7

@Semiclassical I didn't take college chemistry, AP.

Balarka Sen

"Former athlete and sportscaster David Icke has proposed that a race of reptilian, shape-shifting extraterrestrials is bent on world domination and subjugating the human race to slavery. Icke attributes a number of world catastrophes to the aliens, including 9/11."

...

0celo7

It explains a lot of things.

usukidoll

02:50

is $Z \backslash nZ$ the same as $Z_{n}$?

0celo7

depends

Balarka Sen

$\Bbb Z/n\Bbb Z$, you mean? They are isomorphic.

So same in a sense.

usukidoll

crap no chatjax T_T

Semiclassical

should be a bit careful to check that they're using Z_n for Z/nZ

Balarka Sen

Z/nZ generally means the group of integers modulo n. Z_n means the cyclic group of order n, namely, generated by a single element a of order n.

0celo7

02:52

@BalarkaSen $\Bbb Z/n\Bbb Z$ and the $n$-adic integers are isomorphic?

Oh.

Balarka Sen

So in principle not same, but they are isomorphic.

@0celo7 Dude. That's the cyclic group.

Semiclassical

i think you might also sometimes see Z_n as the multiplicative group of integers mod n, but that's not great notation.

Balarka Sen

integers mod n under multiplication is not actually a group. you want to look at the units, the notation for that is Z_n^* or U(Z_n).

Semiclassical

i'll confess, i was lazy and grabbed the title from wikipedia assuming it'd be valid :P

i.e. en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n

Balarka Sen

probably not a good move unless you want to confuse the person asking the question

Semiclassical

02:56

should've said multiplicative group of units mod n

agreed.

not sure why Wikipedia uses that title, though.

Balarka Sen

Integers mod n is a ring. Multiplicative group of that is the multiplicative counterpart of the ring, the group of units.

Semiclassical

hmm, fair enough.

i knew this stuff at one point, but i'm sloppy on the terminology now

0celo7

Very nice result. A Riemannian geodesic ball is isometric to a Euclidean ball.

Balarka Sen

Terminology is confuzzling, yeah.

0celo7

Now, this poses a question

@BalarkaSen Are exotic $\Bbb R^4$s locally diffeomorphic?

Does that even make sense?

Semiclassical

02:59

$(\Bbb Z/n\Bbb Z)^\times$, on the other hand, is perfectly clear

0celo7

What is it that makes one $\Bbb R^4$ different than the rest

Semiclassical

that sounds like something Mike would know

0celo7

So, take an exotic $\Bbb R^4$, give it a metric, then any small ball is diffeomorphic to the standard $\Bbb R^4$.

Balarka Sen

I am not sure what you mean by "locally diffeomorphic". That there is a local diffeomorphism from one exotic R^4 to the other? Related: there are small exotic R^4's, which embed as open sets in R^4 but are not diffeom to R^4.

@0celo7 Don't see why the metric is important.

That seems irrelevant.

0celo7

@BalarkaSen Correct.

Balarka Sen

03:03

Well, an exotic R^4 is after all a smooth 4-manifold.

so just take a smooth chart.

0celo7

Exactly.

I don't know why I said the metric part.

Balarka Sen

so your question is trivial

0celo7

Yeah, I know.

But I am still excited about my result

Balarka Sen

the correct thing to ask or look for along these lines is what I referred to: small exotic R^4's

reference

usukidoll

I'm used to $ Z_{n}$

Not $Z/nZ$

Balarka Sen

03:06

they are more or less the same, so whatever

usukidoll

$Z/5Z$ is the same as mod 5 which is 0,1,2,3,4

Semiclassical

i tend to use Z_n myself, simply out of habit and laziness

Balarka Sen

I do a compromise and use Z/n

Semiclassical

if it's clear in context, it really doesn't matter.

in any case, terminology is boring and so are debates about notation.

usukidoll

What about on this though? http://prntscr.com/chd6iu
I got mod 5 = [0,1,2,3,4]
and then we have 0 or 1,2,4,8
when I take 1,2,4,8 in mod 5 it's 1,2,4,3
so I have 0 or 1,2,4,3
is it because I have retrieved every element in mod 5 the statement is true?

Balarka Sen

03:11

alright, I gotta head to bed

0celo7

Balarka over here spitting bars

Balarka Sen

huh

usukidoll

T___T

ugh I hate the notation

0celo7

@BalarkaSen I will call you Trap Lord

Balarka Sen

no idea what you're talking about. what does "spitting bar" even mean?

0celo7

03:14

rhyming like you Slim Shady

Balarka Sen

oh, that. blah.

I'll quote Stalin while referencing rap: "dog's bark, essentially"

0celo7

I googled that alongside Stalin and didn't find anything

Balarka Sen

Although that was referred to American rock music.

@0celo7 Probably not that much of a famous quote.

0celo7

@BalarkaSen you don't enjoy any modern music

Balarka Sen

maybe

usukidoll

03:22

:O

more like >:O

user116211

03:57

Ah! Thanks @balarka.

user116211

But the price is ridiculously high.

user116211

Nevertheless, when I googled, apart from the amazon page, I came across this:

user116211

Ted Shifrin

39k 1 32 70

user116211

So, the author is in MSE!

arctic tern

and chat regular here...

user116211

03:59

@arctictern ooh!

user116211

Anyways, my search for a good text continues; I found some Berkeley old texts on Calculus III; the exercise is good, I see.

user116211

@arctictern We have authors of books in PSE also.

user116211

04:49

Okay, I did find some good books:

user116211

Classic by Apostol, Calculus III: classic Berkeley text.

user116211

And some advanced books:

user116211

Munkres, Spivak, Baxandall and Lieback, C.H. Edwards and Callahan!!

user116211

05:05

@MAFIA36790 I mean Calculus by Apostol (why in the world, I wrote classic T__T).

0celo7

@Semiclassical 7 pages

user116211

Hey @0celo7, which book did you follow for multivariable calculus?

0celo7

@MAFIA36790 MIT OCW.

user116211

@0celo7 ah! Thanks!!

caub

08:15

not a pure Math question, but I'm coding a visualization, over a circle [0,360[ , but I need to 'emphasize' more the surroundings of the points 60, 180, 300. I'd need to deform the linear function to sketchtoy.com/67453176 roughly, would it a polynomial and how to find its equation?

well, I have 12 equations, 6 with value and 6 with derivatives, that should do

or just x + 0.2*sin(x *PI / 60)

deostroll

If I wanted to understand why the medians of a triangle all pass through one point...where should I start looking?

Secret

08:34

Hey guys I am trying to derive a lebniz integral rule like result for a function f that is discontinous at finite number of points and riemann integrable. However I seemed to get a lot of diverging terms when I tried to partition my integration domain into subintervals in order to make use of mean value theorem. What is the correct way to partition the integration domain?

Consider, where x' is just a dummy variable (where the integral is a function of x, and the integrand is a function of both x and the dummy variable x'), and $f : \mathbb{R}\rightarrow \mathbb{R}$ is any Riemannian integrable function, not necessary continuous at $n$ points, while $u$ is assumed differentiable always. Consider
$$\frac{d}{dx}\int_a^{x} f(x',u(x))dx'$$

Now wrote the differentiation in first principles
$$\lim_{h\rightarrow 0}\frac{\int_a^{x+h} f(x',u(x+h))dx'-\int_a^x f(x',u(x))dx'}{h}$$

Secret

sorry typo

Secret

Consider, where $x'$ is just a dummy variable (where the integral is a function of $x$, and the integrand is a function of both $x$ and the dummy variable $x'$), and $f :\mathbb{R}\rightarrow \mathbb{R}$ is any Riemannian integrable function, not necessary continuous at $n$ points, while $u$ is assumed differentiable always. Consider $$\frac{d}{dx}\int_a^{x} f(x',u(x))dx'$$
Now wrote the differentiation in first principles $$\lim_{h\rightarrow 0}\frac{\int_a^{x+h} f(x',u(x+h))dx'-\int_a^x f(x',u(x))dx'}{h}$$ Use the integral property $\int_a^{b+c}=\int_a^b+\int_b^{b+c}$ $$\lim_{h\rightarrow…

Secret

The issue are the three terms $\frac{(k_1-x) f(c_0,u(x+h))}{h}$, $\frac{(k_{i+1}-k_i) f(c_i,u(x+h))}{h}$, $\frac{(x+h-k_n) f(c_n,u(x+h))}{h}$ in the limit
$$\lim_{h\rightarrow 0}\frac{\int_x^{x+h} f(x',u(x+h))}{h}dx'=\lim_{h\rightarrow 0}\frac{(k_1-x) f(c_0,u(x+h))+\sum_{i=1}^n(k_{i+1}-k_i) f(c_i,u(x+h))+(x+h-k_n) f(c_{n+1},u(x+h))}{h}$$ they seemed to diverge in general when the limit is evaluated, but $f$ restricted to those intervals are given to be continuous, thus the derivatives should be well defined there

Secret

08:56

In order to ensure $f$ is riemann integrable, $f$ is bounded at all points in its domain. There are only at most n discontinuities in the form of jumps in $f$

Therefore the sequence of points $\{k_i\}$ should have nothing to do with the paramter h

Also because of mean value theorem, the terms $f(c_i,u(x))$ should correspond to values of $f$ that are far away from the jumps, thus should be well defined. Therefore why is the limit which is basically computing the derivative, diverges there?

Mathieu K.

09:27

Hi everyone, do you know if a question already exist on mathematical symbols, its representations and definitions? To be more precise, I search an encyclopedia or dictionary of "mathematical language" and a way of simply define the meaning of symbols.

Tobias Kildetoft

@MathieuK. That would make for a poor question, but a fine wikipedia page (and as far as I recall, there is already such a wikipedia page)

though note that a lot of notation is very context dependent

user116211

@MathieuK. You want to study logic?

user116211

Anyways, in this post

user116211

8

A: Strong Newton's third law of action and reaction: Mathematical Interpretation

The full mathematical statement is as follows: Theorem If two particles exert a mutual conservative force $\mathbf{F}_{12}$ and $\mathbf{F}_{21}$ which is independent of any other degree of freedom of any bigger system they're part of, and obeys Newton's third law as $\mathbf{F}_{12}+\ma...

user116211

how did he conclude $$\begin{array}{}
\nabla_{\mathbf{r}_1}=\phantom- \nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}},\\
\nabla_{\mathbf{r}_2}=-\nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}},
\end{array}$$

user116211

09:39

where $\mathbf{R}=\tfrac12\mathbf{r}_1+\tfrac12\mathbf{r}_2$ and $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2\;?$

user116211

In fact, how should I write $\nabla_j\;?$ Basically it means gradient in the direction of $\mathbf r_j$ but how do I write that?

user116211

$$\nabla:= \frac{\partial }{\partial x}~\mathbf{\hat i} + \frac{\partial}{\partial y}~\mathbf{\hat j} + \frac{\partial }{\partial z}~\mathbf{\hat k}\;;$$ how do I make it $\nabla_j\;?$

Lovsovs

09:54

Hey guys, I'm thoroughly confused by something trivial (sorry!) regarding probability: "We have a 47 card deck with two aces and three kings in it. What is the probability that, when drawing two cards, we draw a king and an ace?" If we label all cards as being different, there are 47*46 ways to draw two cards. There are 6 ways to draw an ace and then a king (2 choices of ace, 3 choices of kings), and 6 ways to draw a king and then an ace = 12 ways. So the probability is 12/(47*46).

An recent answer would have me believe that the answer is actually half that - which is correct? Thanks...

Mathieu K.

@TobiasKildetoft list of mathematical symbols still exist on Wikipedia (en.wikipedia.org/wiki/List_of_mathematical_symbols) but, as you said, this depend of context. So, I've also read lot of topic on Wolfram MathWorld but some notation is sometime quite strange. My question would be "is a free/open dictionary or encyclopedia of mathematical exist on internet".

@TobiasKildetoft When I said "free or open" I don't want something like CRC Concise Encyclopedia of Mathematics (around 4 volumes and more than 500pages). I search a more lite and synthetic solution. ;)

Secret

nvm, let me check...

@Lovsovs There are two cases which give your result. Either you drew a king then an ace, or you drew an ace then a king. Since the probability is asking about you drew one of each king and ace without any concern of order, there are $\frac{2}{47}\frac{3}{46}+\frac{3}{47}\frac{2}{46}=\frac{12}{47*46}$ ways to do that

Secret

10:20

You are correct that you start with this expression of gradient. To transform the gradient, leave the unit vectors alone and apply chain rule to the derivatives. You should get e.g. $\hat{e_{1i}}\frac{\partial }{\partial x_{1i}}=\hat{e_{1i}}\left(\frac{\partial X_i}{\partial x_{1i}}\frac{\partial }{\partial X_i}+\frac{\partial x_i}{\partial x_{1i}}\frac{\partial }{\partial x_i}\right)$ which then evaluates to
$$\hat{e_{1i}}\frac{\partial }{\partial x_{1i}}=\hat{e_{1i}}\left(\frac{1}{2}\frac{\partial }{\partial X_i}+\frac{\partial }{\partial x_i}\right)$$. Now multiply the $\hat{e_{1i}}$ in …

Rrjrjtlokrthjji

12:09

I've been wondering why in Stokes, Green and Gauss theorems the integral over the region always involves some kind of derivatives and the integral over the boundary doesn't (in Gauss it's div, in Stokes it's rot etc)

s.harp

@Rrjrjtlokrthjji it is not morally wrong to say that these theorems generalise the fundamental theorem of calculus. They are essentially the following statement:
$$\int_M d\omega = \int_{\partial M}\omega$$
where $d$ is a suitable derivative, $M$ some volume, line or area or whatever and $\partial M$ its boundary.

note that in the case $M$ is an interval $[a,b]$ that $\partial M = \{a,b\}$. In this case the correct result of the formula is
$$\int_a^b \partial_x f(x) = \int_{a,b} f(x) = f(b)-f(a)$$

Rrjrjtlokrthjji

Sweet @s.harp

Is that referred in a book or something

s.harp

yes, you will find it in most books on differential topology or differential geometry

the presentation i like the most is in bott and tu

but I don't know what kind of things you have heard in this direction so such a recommendation may not be the best

Rrjrjtlokrthjji

ok thanks, my calculus book just mentions ffc won't work on multiple integrals and just throws the theorems

on diff geometry I've been studying from kreyszig, you refer to this book? springer.com/us/book/9780387906133

s.harp

yes, thats the on I meant

Rrjrjtlokrthjji

12:18

okay will download the e-book from springer

s.harp

if you know about differential forms and tangent spaces then I think the treatment here for Stokes theorem does not need further knowledge

Rrjrjtlokrthjji

@s.harp just found out that statement $\int_{S} d \omega = \int_{\theta S} \omega $ is on the cover on Zorich's book, I got a copy in my library

s.harp

@Rrjrjtlokrthjji hehe, it is a very nice result, everybody loves it :) thats why its in almost every book and on lots of covers

Rrjrjtlokrthjji

ok, I think I get your point, $\omega$ is a differential form on $S$

Secret

12:36

@s.harp hi can you help me on formulating a problem?

Balarka Sen

@Rrjrjtlokrthjji It's worth exploring the 2 dimensional version (Green's theorem) to understand Stokes'.

Secret

@robjohn Hi can you help me on a integral theorem problem?

s.harp

@Secret what problem? I can look but don't know if I can help

Secret

(Getting details...)

Consider, where $x'$ is just a dummy variable (where the integral is a function of $x$, and the integrand is a function of both $x$ and the dummy variable $x'$), and $f : \mathbb{R}\rightarrow \mathbb{R}$ is any Riemannian integrable function, not necessary continuous at $n$ points, while $u$ is assumed differentiable always. Consider
$$\frac{d}{dx}\int_a^{x} f(x',u(x))dx'$$

Now wrote the differentiation in first principles
$$\lim_{h\rightarrow 0}\frac{\int_a^{x+h} f(x',u(x+h))dx'-\int_a^x f(x',u(x))dx'}{h}$$

I am trying to formulate a lebniz integral rule from first principles, for a function with a discintoniuous number of points, but I don't know why the final limti is not converging

The function f in question can be a piecewise function that has n number of jumps but always bounded

The following are the problematic term: $\frac{(k_1-x) f(c_0,u(x+h))}{h}$, $\frac{(k_{i+1}-k_i) f(c_i,u(x+h))}{h}$, $\frac{(x+h-k_n) f(c_n,u(x+h))}{h}$

Balarka Sen

@Secret Leibniz integral rule does not have a variable in the limit of the integrals.

Indeed, $F(x) = \int_a^x f(t) dt$ is not always differentiable, given $f$ is just Riemann integrable.

Secret

12:48

(Sorry I mean the General version of differentiation under the integral sign)

Balarka Sen

E.g. consider $f(x)$ to be $0$ if $x < 0$ and $1$ if $x \geq 0$. That is Riemann integrable, discontinuous only at $0$.

But the integral is clearly not differentiable (it's not even continuous)

Secret

But isn't according to the mean value theorem applied to the subintervals where the function is continuous, the terms $f(c_i,u(x))$ should correspond to values of $f$ that are far away from the jumps, thus should be well defined. Therefore why is the limit which is basically computing the derivative, diverges there?

Balarka Sen

@Secret You need more hypothesis there than that.

I just gave you an example where the limit would diverge (the function not being differentiable).

s.harp

@Balarka the integral is continuous, and differentiable except at isolated points

Secret

yeah something like this...

Balarka Sen

12:51

@s.harp OK, sorry about that, it's continuous. But not differentiable at $0$.

Secret

so the $k_i$ are where the discontinuities were, thus the integrals of the form $\int_{k_i}^{k_(i+1)}f dx'$ should be over portions of f that are continous, thus I should nto be expecting divergences there

s.harp

I think using the mean value theorem is not the right direction

Secret

But isn't the only requirement for mean value theorem is that the function at some interval [a,b] is continous and bounded then it is vallid?

thus if I dice up my integration domain into n pieces then each piece should be continous and bounded?

s.harp

actually remember that your discontinuities are at isolated points

so you can assume that, upon $h$ becoming small enough eniuogh, that there are either
a) no discontinuties in $[x,x+h]$
b) the only discontinuity in $[x,x+h]$ is at $x$

Balarka Sen

@Secret You wouldn't need to slice up into pieces to use mean value theorem.

s.harp

12:57

at any rate $\lim_{h\to0}\frac{\int_x^{x+h} f(x',u(x+h))dx'}{h}$ should be $\lim_{x'\to x^+} f(x',u(x'))$

ie it should converge to the "continuous from the right" version of $f(x,u(x))$

Secret

@BalarkaSen The section [Mean value theorems for definite integrals] (en.wikipedia.org/wiki/Mean_value_theorem) state the f has to be continuous over the whole interval, thus using mean value theorem on $\int_x^{x+h}f(x',u(x))dx'$ won't there be issues?

@s.harp How do you get rid of that h in the denominator when you convert it into the other limit?

Balarka Sen

If $f$ is continuous at $x$, $f$ is continuous on $[x, x + h]$ for sufficiently small $h$. This is a consequence of $f$ having isolated discontinuities.

Also mentioned by s.harp above.

Secret

Ah I see...

s.harp

@Secret if you replace in $\int_x^{x+h} f(x',u(x+h))$ the $f$ with a function $\tilde f$ that has $\tilde f(x,u(x+h))=\lim_{x'\to x^+}f(x',u(x+h))$ and else $\tilde f = f$ the result will be an integral over a continuous function, since the only discontinuity can be at $x$ as mentioned above, and you are here replacing $f$ essentially with the contiuous extension from $(x,x+h]$ to $[x,x+h]$

for the now continuous function you can use the mean value theorem

Secret

Is the reason this work is because $\int_a^a$ at the discontinuities have measure zero, thus $\tilde{f}$ and $f$ will integrate to the same answer?

Balarka Sen

13:10

Yes, $\tilde{f} - f$ is 0 everywhere except perhaps a point (aka $x$) so integrates to $0$.

s.harp

the Riemann integral does come from a measure, but you are right that it works because if you change a riemann integrable function at a countable amount of isolated points the integral will not change

Secret

Sounds like a new Do Nothing Technique I need to wrap my head around...

(NB Do Nothing Technique are thigns like adding a suitable expression that is zero, 0=x-x and 1=x/x)

s.harp

You could view it as: on $[x,x+h]$ $f$ agrees everywhere except on isolated points with a continuous function. So the integral does not see them as different functions, its "eyes" are not sharp enough to distinguish such features.

since you have nice theorems for continuous functions, use these on the continuous version of $f$

Secret

make sense

Now for the $\int_x^{x+h}$ term, note because $h$ is sufficiently small enough and $f$ has isolated discontinuities, therefore $f$ on any $[x,x+h]$ is continuous except at some isolated discontinuity x. Therefore applying mean value theorem
$$\lim_{h\rightarrow 0}\frac{\int_x^{x+h} f(x',u(x+h))}{h}dx'=\lim_{h\rightarrow 0}\frac{hf(c,u(x+h))}{h}=\lim_{h\rightarrow 0}f(c,u(x+h))=\lim_{c\rightarrow x^+}f(c,u(x))$$

Now if at least [dominant convergence theorem][2] holds then one can conclude the [General Lebniz integral rule][3], as required

ok looks good now, thanks

1

Q: How to integrate $\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx$ and why integration by parts fail?

Th following integral (A,B are constants) $$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx$$ pops up as a curve fitting routine Attempt to integrate it by parts (setting $u=\sqrt{A-bx},v'=e^{x^2}$) gives EDIT: In detail $$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx=\left[\sqrt{A-bx}\left(\int_{...

integration integration-by-parts

Another question:

user116211

13:33

@Secret what's $\hat{e}_{1i}\;?$

Secret

What substitution do I need to do in order to show that this integral is related to Bessel K functions, since none of the forms of BEssel K functions in wikiepedia look remotely like the integral? (and Jack D'Aurizio have stopped replying after some comments)

$\int_{0}^{+\infty}\exp\left(-(c x^{2/3}+d)^2\right)\,dx$

@MAFIA36790 cartesian unit vectors of the $r_1$ gradient

user116211

@Rrjrjtlokrthjji Can be found in any advanced calculus text like Munkres.

Secret

14:02

@Semiclassical Could you help me on this derivation, because there's this extra $\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})$ that I am not sure how to make it disappear?

2

Q: Converse of Euler Homogeneous Thm. How to show that $\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})=\mathbf{0}$?

So basically I read someone else's answer to a question regarding Euler Homogeneous function theorem source: http://quant.stackexchange.com/questions/8911/what-is-exactly-eulers-decomposition Also here http://planetmath.org/sites/default/files/texpdf/40683.pdf I tried to do the proof by hand ...

multivariable-calculus functions tensors

Semiclassical

@Secret absent some context, that integral seems suspicious to me. it becomes complex-valued, after all, when $x>b/A$ (if b>0). so i'd check the derivation of it to see if it's actually the integral you want.

@Secret I'm sort've doubtful I can help with that one.

Probably the place to start is to check that said equation actually makes sense, though, for some choice of $f$

After all, if it doesn't work for a simple $f$ then you're probably asking the wrong question.

Secret

@Semiclassical The main question has already solved, I am just trying to understand how to use a non Mathematica way (perhaps some cleaver substitution and pattern matching) to evaluate the integral $\int_{0}^{+\infty}\exp\left(-(c x^{2/3}+d)^2\right)\,dx$ because Jack D'Aurizo have not explained how it is a bessel K function

Semiclassical

i see. that's too tedious to be interesting to me, if i'm honest.

plus, if I plug it into Mathematica, I don't find that it's a single Bessel K function. I get it in the form of $Ce^{-d^2/2}[K_{3/4}(d^2/2)-K_{1/4}(d^2/2)]$ where $C$ is an overall constant

which falls further into my "too tedious to be interesting" category.

Secret

Let me lookup what $K_{3/4}$ look like...

Semiclassical

note that there's a few different integral representation of Bessel functions, so you may have to look at a few to find one that's relevant.

Secret

14:10

The bessel k function have many names, I am not sure which one to check

Semiclassical

Yeah.

Probably look for an integral representation with the same limits of integration and roughly the same structure as the integral you're considering

Secret

mathworld.wolfram.com/…

This one has some fractions andother scary looking terms besides a exp

Semiclassical

yeah.

the integral representations listed in the DLMF might also be useful: dlmf.nist.gov/10.32

but as you can probably tell I'm happy enough with a "Mathematica tells me so" answer, since they're a bit of a pain to work with.

Secret

The list all contains exp paired with some scary rationals trig and powers, there isnt one where there is a lone exp. Umm if you highlight and lookup the $K_{3/4}(d^2/2)$ term in your mathematica output what function does mathematica said?

Semiclassical

It's definitely the modified Bessel function, of order 3/4, of the second kind.

btw, with regard to the Euler homogeneous Thm question, that statement is definitely false when $f(\mathbf{x})=\mathbf{x}^2.$

Secret

14:20

Ok I will have a closer look, the 3/4 must somehow knock out all the coefficients. I think I might expand the exp(-cx+d)^2 into a series later and doing some pattern matching with the wolfram output and see if there is anything

Semiclassical

possibly. I think there might also be a connection with confluent hypergeometric functions

which is another bad place, usually, but may provide a more direct solution.

Secret

@Semiclassical The f in the Euler homogeneous Thm question is a homogenous function of degree k, as said near the begining. Someone has done a proof there but when I tried to reproduce their proof, the product rule of $\frac{d}{d\lambda}$ somehow gave me an extra term

Semiclassical

and? $f(\mathbf{x})=\mathbf{x}^2$ is homogenous of degree 2, and it doesn't satisfy that equation.

so regardless of what context you're coming from, what you have in the question doesn't work.

Secret

Hmm in that case, let me check that equation with wikipedia...

Given
$$\mathbf{x}\cdot\nabla f(\mathbf{x})=kf(\mathbf{x})$$
$$f(\lambda\mathbf{x})=\lambda^kf(\mathbf{x})$$
Let $f(x)=x^2$ Then for first equation
$$f(\lambda\mathbf{x})=\lambda^kf(\mathbf{x})$$
$$\lambda^2 x^2=\lambda^kx^2$$
Thus
$$k=2$$
Now
$$\mathbf{x}\cdot\nabla f(\mathbf{x})=kf(\mathbf{x})$$
$$x\frac{d}{dx}x^2=2x^2$$
$$2x^2=2x^2$$
Thus $f(x)=x^2$ satisfied Euler Homogenous function thm

en.wikipedia.org/wiki/Homogeneous_function The wikipedia version agrees with the theorem in the question, thus there shoild be no typos

user116211

wait, @secret, I've got one post of Kyle:

user116211

14:33

6

A: A confusion about notation in Goldstein

What Goldstein means by $\nabla_iV_i$ is $$ \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$ which is indeed a vector. Here, $\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$ is th...

Semiclassical

@MAFIA36790 $i$ as particle number index is pretty annouying notation, though there might be no easy alternative in that context.

Secret

yeah, although the V should be able to left indexless

user116211

@Semiclassical Many were confused with the notation, as I've more or less same post many times today at PSE. Thanks to Goldstein ;)

Semiclassical

heh.

user116211

@Secret Yes.

Balarka Sen

14:35

@Secret You seem to have derived that $x \cdot \nabla f(\lambda x) + \lambda x \cdot \frac{d}{d\lambda} \nabla f(\lambda x) = k \phi'(\lambda)$. Why should that mean $\lambda x \frac{d}{d\lambda} \nabla f(\lambda x) = 0$?

user116211

@Secret Goldstein itself retained the index in $V\,,$ however.

user116211

Now, I would be working on your advice, @secret, to deduce $$\left\{\begin{array}{}
\nabla_{\mathbf{r}_1}=\phantom- \nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}},\\
\nabla_{\mathbf{r}_2}=-\nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}}
\end{array}\right..$$

Semiclassical

retaining the index in $V_i$ does make sense, even if it's annouying, because Goldstein wants to specify the external potential experienced by the $i$th particle.

Secret

@BalarkaSen quant.stackexchange.com/questions/8911/…. That extra term is precisely what prevent me from recovering $f(\lambda x)=\lambda^k x$ in the derivation shown in this finance SE link

Balarka Sen

No, your calculations are wrong.

Check them.

user116211

14:39

@MAFIA36790 Goldstein itself .... runs away.

Semiclassical

Amusingly, I've got a copy of Goldstein in the shelf just above my head.

along with a copy of Jackson's Classical Electrodynamics. (now there's a book that's frustrating)

Balarka Sen

Specifically: $$\mathbf{x}\cdot \nabla{f(\lambda \mathbf{x})}+\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})\equiv k\frac{d}{d\lambda}(f(\lambda \mathbf{x}))$$

Why does that mean $\lambda x \cdot \frac{d}{d\lambda} \nabla f(\lambda x) = 0$? It doesn't, actually.

That conclusion is false.

Semiclassical

And, as the counterxample I gave in the question shows, it's moreover a false claim in general. Hence: check thy work.

user116211

Anyways, back to the derivation; $$\nabla_\mathbf{r_1} = \frac{\partial}{\partial x_1}\mathbf{\hat{x}}_1 + \frac{\partial}{\partial y_1}\mathbf{\hat{y}}_1 + \frac{\partial}{\partial z_1}\mathbf{\hat{z}}_1 \;.$$

Semiclassical

dropping $i$ while considering a single particle, though, definitely makes things easier to read.

user116211

14:47

Now, to get $\nabla_\mathbf r,$ where $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$ I have to apply chain rule, is it so?

Semiclassical

yeah, no way around that.

user116211

But what to write....

user116211

In the first case, the differentiation was done with respect to $x_1$ viz. $\frac{\partial}{\partial x_1};$ so would it be now $x_1- x_2\;?$

Semiclassical

sure. you probably want to write $\frac{\partial}{\partial x}$ in terms of $\frac{\partial}{\partial x_1}$ and $\frac{\partial}{\partial x_2}$

and then do the same for $\frac{\partial}{\partial X}$. alternatively, write $r_1$ and $r_2$ in terms of $r,R$ and do it the other way around.

user116211

It's given that $\mathbf{R}=\tfrac12\mathbf{r}_1+\tfrac12\mathbf{r}_2\;.$

Semiclassical

14:58

right. so you can express the COM coordinates in terms of the two-particle coordinates pretty easily.

user116211

@Semiclassical How? Using chain rule?

Secret

@BalarkaSen @semiclassical Actually, on second look of this, in quant.stackexchange.com/questions/8911/…, why when that user compute $\phi'(\lambda)$, that is differentiate wrt $\lambda$, there is no product rule phappening when there is a $\lambda$ dependence in both $\lambda\mathbf{x}$ and $f(\lambda \mathbf{x})$?

Semiclassical

uh, no. just by linear combinations.

user116211

@Semiclassical okay.

Semiclassical

e.g. $\mathbf{r}_1 = \mathbf{R}+\frac12 \mathbf{r}$.

s.harp

15:00

@Secret by the way, there is a mistake in the chain of thought from above. We were always thinking $h>0$, but obviously the limit differential quotient is not restricted to that case. If $h<0$ you get the left sided limit of $f(x,v(x))$, no the right sided limit. So if $f$ is not continuous at $x$ then your function is not differentiable.

Balarka Sen

@Secret There is product rule happening... that's just $\phi'(\lambda) = x \cdot \nabla f(\lambda x)$.

Secret

ok I thnk I see it now, let me qudruple check my calculations again...

0celo7

what's going on here

s.harp

a fistfight

Balarka Sen

nothing particularly interesting

0celo7

15:08

On page 7.

It's not due till 4:30, but I have a damn meeting at 2

I should stop procrastinating

user116211

@Semiclassical okay; but how could I linearly combine those operators?

user116211

Use the given relations?

caub

@deostroll gravity point

*centroid

Secret

15:30

@BalarkaSen I finally located my mistake: I have done all the differentiation correctly and all steps are ok, except at step 1, I differentiate the WRONG equation that should be the starting point of the converse theorem proof

so the weird term is actually there and nonzero, because that is from doing something completely unrelated to the proof

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