Mathematics - 2022-11-26

leslie townes

00:53

emphasis and time spent is definitely up to the instructor. i'd imagine div, grad, curl and 'integrating things over other things' including some form of greens and stokes theorem would be covered.

but sometimes the coverage can be very spotty.

monoidaltransform

01:10

math.stackexchange.com/questions/4585132/… @TedShifrin any ideas?

Ted Shifrin

@schn Of course I know it can be proved. You asked, I thought, if it is the definition. Hence my answer.

@SillyGoose There are lots of crummy calc 3 classes that don’t cover what they should. There often isn’t enough time. Some schools have a calc 4 for that reason.

@monoidaltransform First, PRINCIPAL!! Projecting the smooth map onto the second factor is smooth.

I’m confused. The local section maps to $M$?

monoidaltransform

@TedShifrin Okay, so if $\pi_{G}$ is projection onto second factor, then why is $\pi_{G}\circ \psi^{-1}$ smooth? That would be the case if I know $\psi^{-1}$ to be smooth

Yes.

Ted Shifrin

You wrote $M/G$.

That isn’t a section, then.

So show $\psi$ is a local diffeo.

monoidaltransform

To $M$. Sorry. Loud noise came from hall. Got distracted for a seccond

Yeah, that's the idea I had as well. But not sure how

Ted Shifrin

You have to use the free action somewhere.

monoidaltransform

01:23

The free action was used to show $\pi(pg)=\pi(p)$. I don't see how to use it here.

Ted Shifrin

I remember doing this proof in grad geometry and it was a bit technical. How is it used there?

Goku

You know its funny how I feel like the dominant math student in college classes, yet here I feel like an absolute small fry. Weirdly humbling

monoidaltransform

@TedShifrin

This is what Tu has written

I'm just stuck on smoothness of $\phi$

Koro

04:21

I have a small conceptual doubt:

I know that given a commutative ring R with 1 and a left R module M, we have $R\otimes_R M\simeq M$.

Using this, how to get $R/I\otimes R^n\simeq R/I\otimes R^m$ from $ R^n=R^m$?

leslie townes

the longer you do this, the more helpful it will be to replace "\simeq" and its equivalents with "the map ___ from __ to __ is an isomorphism."

focus on maps, not the mere fact of isomorphism.

9 hours ago, by Lukas Heger

if you tensor an isomorphism with the identity on another module, you get another isomorphism

Koro

Leslie, I didn’t understand the “identity on other module part”. We are tensoring a module with another module, not with an identity map.

So we have: there exists an isomorphism i from R^n to R^m. Now I consider the map $f: (r+I)\otimes x\mapsto (r+I)\otimes i(x)$, where x is in R^n.

Looks like this f is an isomorphism.

But this doesn’t help. I should map (r+I) tensor x to ….. for showing isomorphic to (R/I)^n.

Not sure how to do this one. :(

I want to now show: $ R/I \otimes R^n$ is isomorphic to $(R/I)^n$

The map f given earlier works for $R/I\otimes R^m$ isomorphic to $R/I\otimes R^n$.

This is so because one of the answers on showing n=m from R^m isomorphic to R^n did claim that $R/I\otimes R^n$ is isomorphic to (R/I)^n.

Ajay

05:10

@leslietownes Do you know any good references for metric spaces

?

one potato two potato

05:41

0

Q: Compute the integrals $\int_0^\infty \sin(x^2)\ dx$ and $\int_0^{\infty}\cos(x^2)\ dx$

Compute $$\int_0^\infty\sin(x^2)\ dx,\ \int_0^\infty\cos(x^2)\ dx$$ using contour integral. I chose the contour integral over a sector cornered at $0,R$ and $Re^{i\pi/4}$. Let $f(z) = e^{iz^2}$ which is entire. Then, $$\int_0^R e^{ix^2}\ dx+\int_0^{\pi/4}e^{iR^2e^{2i\theta}}\ d\theta -\int_0^R ...

something is wrong...

copper.hat

06:02

there is a lot wrong there

one potato two potato

Lots of missing factors. I know

Deleted the stupid post

Steve

06:29

suppose $X~f$ and $Y~g$ where $f,g$ are two arbitrary distributions and $X|Y ~ N(0,1)$. Can we deduce anything about the distribution $X-Y$? Intuitively I think $X-Y ~ N(0,1)$ but I don't know how to prove it.

copper.hat

What does $X|Y N(0,1)$ mean?

user4539917

06:47

I found this oldie but goldie: Stephen Hawking reveals formula for World Cup success RIP.

one potato two potato

07:48

13

Q: Entire "periodic" function

I am studing for exams and am stuck on this problem. Suppose $f$ is an entire function s.t. $f(z) =f(z+1)$ and $|f(z)| < e^{|z|}$. Show $f$ is constant. I've deduced so far that: a) $f$ is bounded on every horizontal strip b) for every bounded horizontal strip of length greater than 1 a ...

complex-analysis

If I prove that $|f(z)|\leq e^{\epsilon|z|}$ for every $0<\epsilon<1$ then $f$ is constant, then does that imply the question statement?

one potato two potato

08:21

0

A: Entire "periodic" function

Let me try: Let $0<\epsilon<1$. Then Claim: If $f$ is entire with $f(z) =f(z+1)$ such that $$|f(z)|\leq e^{\epsilon|z|},$$ then $f$ is constant. $(\because)$ Since $f$ is $1$-periodc, $g(z) = f\left({\log z\over 2\pi i}\right)$ is a well-defined holomorphic function except for the origin regardle...

ILikeMathematics

09:37

If you want to find out the equation of a line tangent to a curve passing through a point, do you need the point slope form or the formula for a tangent line or can you use $f(x) = mx + k$ in some way?

>$f(x) = 4 - 0.25x^2$, find the equation of the tangent passing through $P(0|5)$ for example

Using $t(x) = mx + k$, you can get $t(x) = -0.5ax + 5$, but after that, can you do anything?

ILikeMathematics

10:29

(I posted it on the forum now, math.stackexchange.com/questions/4585420/…, if you want more details or something)

Silly Goose

10:56

is burgher a common german last name

Koro

12:10

Any idea on how to show this without induction?: $(a_1 + · · · + a_n)^5 \leq n^4(a_1^5 + · · · + a_n^5).$

$a_i$'s are all non negative real numbers.

Thorgott

expanding out the terms and using AM-GM

Koro

not that as well.

It's an exercise problem in a chapter on Lebesgue spaces.

Based on that, I think the exercise wants to use either of the following-Young's, Holder's, Minkowsky inequality.

Thorgott

AM-GM should be an easy corollary of those

Wojciech Kulma

Hi folks, let X be a normally distributed random variable, X ~ N(0, s). I know that the powers of X: X^2, x^3, x^4 and so on aren't normally distributed, but can't derive a generic proof for that. How to do that?

one potato two potato

$$\eta(s) = \sum_{n = 1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \bigl( 1 - 2^{1-s}\bigr)\zeta(s)$$ By Leibniz, $\eta(\sigma) > 0$ for $0 < \sigma$, and since $1 - 2^{1-\sigma} < 0$ for $0 < \sigma < 1$, the result follows. — Daniel Fischer Jan 16, 2017 at 16:54

Why does Leibniz tell $\eta(\sigma)>0$ for $0<\sigma$?

Yai0Phah

12:27

@Koro This is precisely Hölder's inequality for one of the function being 1.

Koro

12:37

@Yai0Phah: Thanks. I was thinking about the same.

But I'm still far from the solution.

I take $f= \sum_{i=1}^n a_i \chi_{\{a_i\}}$, where $\chi$ is a characteristic function.

and the other function to be $1$. And I take counting measure.

Re-writing gives: $\frac{\sum a_i}{n}=(\frac{\sum a_i^5}{n})^\frac 15$

Taking $b_i= a_i/n$ instead of a_i in f, we get $\int f dc=\sum a_i/n=$ LHS

But taking $1$ as the other function makes RHS $\infty$, i.e., I get $\int 1 dc=\infty$. @Yai0Phah.

Koro

13:15

Ha! I managed to solve this-

Set $f=\sum_{i=1}^n a_i \chi_{a_i}$ and to get rid of $\int 1 dc=\infty$, define $g= \chi_{A_n}$, where $A_n= \{a_1,...,a_n\}$. And Holder's inequality on $\int fg$ gives the result :-).

Koro

14:00

any ideas on this one?

9 hours ago, by Koro

I want to now show: $ R/I \otimes R^n$ is isomorphic to $(R/I)^n$

one potato two potato

@Koro This and tensor product commutes with direct sum proves the statement.

$(R/I)\otimes_RM\simeq M/IM$

Koro

Oh, you mean $R^n\simeq R+R+...+R$ and that's why?

@onepotatotwopotato But I am not sure how you got quotient by IM on RHS.

one potato two potato

@Koro Yes. $(R/I)\otimes_R R\simeq R/I$ so $\bigoplus R\otimes (R/I)\simeq bigoplus(R\otimes (R/I))\simeq\bigoplus (R/I)$.

Not Tfue

14:15

They defined span as this $$
\left\langle\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\right\rangle:=\left\{\sum_{i=1}^k c_i \mathbf{v}_i \mid c_1, c_2, \ldots, c_k \in \mathbb{R}\right\}
$$

one potato two potato

@Koro Try to prove the statement via map: $(R/I)\times M\to M/IM$ by $([r],m)\mapsto[rm]$.

Not Tfue

I can only interpret it as $c_1v_1+c_2v_2+...+c_kv_k$.

Koro

@onepotatotwopotato Thanks a lot :-). I've been stuck at it since yesterday.

@onepotatotwopotato I'll try that. Thanks. :-)

This is not as trivial as the post using it makes it out to be.

Not Tfue

Nevermind my brain was having foggy thoughts.

Ignore my message.

Thorgott

14:42

it's trivial insofar as it is immediate from two standard facts

Koro

15:40

0

Q: $\|f_n-f\|_\infty\to 0\implies \|f_n-f\|_p\to 0$ for any $p\in [1,\infty)$.

Suppose that $(X, S, \mu)$ is a measure space and that $f_n:X\to \mathbb R$ is a sequence of functions converging pointwise to $f$ in $L^\infty(\mu)$. It is given that $\mu(X)<\infty$. Then, I want to prove that $f_n\to f$ in $L^p(\mu)$ for any $p\in [1,\infty)$. Remark: This problem has an addi...

real-analysis measure-theory solution-verification

Wojciech Kulma

16:03

X_i are iid random variables with expected value m, |m| < inf. g(x) = x^5. By the weak law of large numbers we know that mean(x) - m converges in probability 0

by applying continuous mapping theorem we can state that g(mean(x) - m) = (mean(x) - m)^5 converges in probability to 0

now I got sqrt(n) * g(mean(x) - m), where does that converge?

the first term goes to inf while the second to 0 in probability
Do I need to consider which one goes faster?

schn

17:01

I have a question about a proof in Spivak's Calculus, see a quote of the proof here. My question is about the sentence:

> Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $\delta > 0$ such that all points $x$ satisfying $b-\delta < x \le b$ are upper bounds for $A$; this also follows from Problem 6-16, since $f(b) > 0$.

I am confused about the interval $b-\delta < x \le b$. When I look at problem 6-16 (6-15 in the third edition I think), it states that if $f(x)$ is right-hand continuous at $a$ and $f(a) > 0$, then there is a number $\delta>0$ such that $f(x)>0$ for all $x$ satisfying $0\leq x-a < \delta$. Shouldn't then the interval be $0\leq x-b < \delta$ in the above quote?

Koro

17:28

@schn

Use continuity of f at b.

@schn f is not defined outside [a,b].

You could try this exercise: If f is a real valued continuous function on [a,b], a<b, and f(b) is non zero, then there exists an r>0 such that (b-r, b] is a subset of [a,b] such that sign of f on (b-r,b] is same as that of f(b).

schn

@Koro thanks for the reply. If f is continuous at b then given $\epsilon > 0$ there exists a $\delta > 0$ such that if $|b - x| < \delta$ then $|f (b) - f (x)| < \epsilon$. How does $b-\delta < x \le b$ follow from that definition?

schn

17:58

@Koro I guess it follows from left-hand continuity. Then $(b-x)$ is negative

Koro

|b-x|<d gives -d<b-x<d so b-d<x<b+d.

schn

@Koro I guess it follows from left-hand continuity. Then $(b-x)$ is positive, and so $-\delta<x-b\leq 0$.

Koro

yes, you could argue like that also.

Continuity at end points is defined such that x does not go outside the interval [a,b]. So continuity at b means 'left continuity' and that at a means 'right continuity'.

schn

Yes

Raffallo

18:27

hello guys :)
maybe someone could me help in this question?
https://math.stackexchange.com/questions/4584985/estimate-variable-using-curve-formula

I've tried so many formulas and still coulnd't get it

Koro

18:49

Are imbedding and embedding the same?

Thorgott

19:05

yes

different spellings of the same word

Koro

Thanks.

copper.hat

19:55

like flammable and inflammable

geocalc33

21:22

Would this be a good result? $\lim_{x\to \infty}\big|\tau(x)-\text{Li}(x)\big|=0$ where tau is some weighted function that measures primes?

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