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12:53 AM
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.
 
@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 $M$?
 
@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.
 
You wrote $M/G$.
That isn’t a section, then.
So show $\psi$ is a local diffeo.
 
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
 
You have to use the free action somewhere.
 
1:23 AM
The free action was used to show $\pi(pg)=\pi(p)$. I don't see how to use it here.
 
I remember doing this proof in grad geometry and it was a bit technical. How is it used there?
 
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
 
@TedShifrin
This is what Tu has written
I'm just stuck on smoothness of $\phi$
 
 
3 hours later…
4:21 AM
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$?
 
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
 
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.
 
5:10 AM
@leslietownes Do you know any good references for metric spaces
?
 
5:41 AM
0
Q: Compute the integrals $\int_0^\infty \sin(x^2)\ dx$ and $\int_0^{\infty}\cos(x^2)\ dx$

one potato two potato 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...
 
6:02 AM
there is a lot wrong there
 
Lots of missing factors. I know
Deleted the stupid post
 
6:29 AM
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.
 
What does $X|Y N(0,1)$ mean?
 
 
1 hour later…
7:48 AM
13
Q: Entire "periodic" function

MykieI 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 ...

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?
 
8:21 AM
0
A: Entire "periodic" function

one potato two potatoLet 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...

 
 
1 hour later…
9:37 AM
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?
 
10:29 AM
(I posted it on the forum now, math.stackexchange.com/questions/4585420/…, if you want more details or something)
 
10:56 AM
is burgher a common german last name
 
 
1 hour later…
12:10 PM
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.
 
expanding out the terms and using AM-GM
 
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.
 
AM-GM should be an easy corollary of those
 
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?
 
$$\eta(s) = \sum_{n = 1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \bigl( 1 - 2^{1-s}\bigr)\zeta(s)$$ 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 $0<\sigma$?
 
12:27 PM
@Koro This is precisely Hölder's inequality for one of the function being 1.
 
12:37 PM
@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.
 
1:15 PM
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 :-).
 
2:00 PM
any ideas on this one?
9 hours ago, by Koro
I want to now show: $ R/I \otimes R^n$ is isomorphic to $(R/I)^n$
 
@Koro This and tensor product commutes with direct sum proves the statement.
$(R/I)\otimes_RM\simeq M/IM$
 
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.
 
@Koro Yes. $(R/I)\otimes_R R\simeq R/I$ so $\bigoplus R\otimes (R/I)\simeq bigoplus(R\otimes (R/I))\simeq\bigoplus (R/I)$.
 
2:15 PM
They defined span as this $$
\left\langle\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\right\rangle:=\left\{\sum_{i=1}^k c_i \mathbf{v}_i \mid c_1, c_2, \ldots, c_k \in \mathbb{R}\right\}
$$
 
@Koro Try to prove the statement via map: $(R/I)\times M\to M/IM$ by $([r],m)\mapsto[rm]$.
 
I can only interpret it as $c_1v_1+c_2v_2+...+c_kv_k$.
 
@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.
 
Nevermind my brain was having foggy thoughts.
Ignore my message.
 
2:42 PM
it's trivial insofar as it is immediate from two standard facts
 
3:40 PM
0
Q: $\|f_n-f\|_\infty\to 0\implies \|f_n-f\|_p\to 0$ for any $p\in [1,\infty)$.

KoroSuppose 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...

 
4:03 PM
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?
 
5:01 PM
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?
 
5:28 PM
@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).
 
@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?
 
5:58 PM
@Koro I guess it follows from left-hand continuity. Then $(b-x)$ is negative
 
|b-x|<d gives -d<b-x<d so b-d<x<b+d.
 
@Koro I guess it follows from left-hand continuity. Then $(b-x)$ is positive, and so $-\delta<x-b\leq 0$.
 
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'.
 
Yes
 
6:27 PM
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
 
6:49 PM
Are imbedding and embedding the same?
 
7:05 PM
yes
different spellings of the same word
 
Thanks.
 
7:55 PM
like flammable and inflammable
 
 
1 hour later…
9:22 PM
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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