Mathematics - 2022-06-20 (page 1 of 3)

geocalc33

00:30

what are the benefits of a pdf on compact support vs. a pdf on semi compact support or on the entire real line

geocalc33

00:44

there aren't many well known pdfs whose inverse is itself. why is this?

Koro

01:08

Is there a real valued function f from R to R such that 1) f is differentiable but has no power series, 2) f’=f, 3) f(0)=1

?

If f has power series then, f(x)= e^x.

leslie townes

you don't need to assume that f is real analytic to deduce that f(x) = e^x. you might need to assume that f is real analytic to use a "series method" to solve the equation.

for example, let g(x) = f(x)/e^x and use the chain rule to see that g'(x) = 0 identically. and g(0) = 1

Akiva Weinberger

Was just gonna say that

If $f$ is differentiable and $f'=f$, then you can prove $fe^{-x}$ is constant

(by showing it, too, is differentiable and has zero derivative)

There are also more general existence-and-uniqueness theorems for differential equations

leslie townes

more generally, often, solutions to [nice enough] differential equations are going to be real analytic, so assuming that for purposes of guessing at a potential solution is not a a loss of generality, just a convenient assumption that helps you find solutions

e.g. you could do it here, recognize e^x from its series, and then use the argument above (which does draw upon having the idea of e^x as something to compare f to) to finish the job

sometimes you can also learn stuff if e.g. you assume there's a series and then solve a recurrence or get partial info about it, and it turns out its coefficients grow too fast to give a series that converges everywhere. might be a sign that something interesting is going on about solutions to the equation

Calvin Khor

01:37

@Koro +1, but I think you can directly show u(x) > u(0) without splitting into cases. If at any point u(x) ≤ u(0), by continuity of u, there is y in [0,x] such that 0 < u(y) ≤ u(0), hence u(y)^2 ≤ u(0)^2, which is a contradiction

leslie townes

slick

Koro

Hi @CalvinKhor et @Leslie.

@CalvinKhor Thanks for the review :). How do you know though that u(y) is not negative?

Calvin Khor

@Koro f'=f implies that f is C^\infty and has a taylor series which coincides with the exponential. You can show its real analytic (def: equal to the taylor series, hence equal to exp) by checking the gevrey type estimates but that's a lot more work than what they said

@Koro u(0) > 0 and continuity

that was given right? let me check...

Koro

u(0)>0 + continuity is given, yes.

Calvin Khor

right, so either u(x)>0, then take y=x, or by ivt, u takes all values between u(0) and u(x)≤0

@leslietownes please consider pegging lesliecoin to rep

Koro

01:47

@CalvinKhor yes, while writing the answer I thought of eliminating the need of mentioning that u(y) could be negative too (after assuming suppose on the contrary) by introducing max().

Akiva Weinberger

@leslietownes @Koro A similar thing: suppose we want to show that the only solutions to $f''=-f$ are $A\cos x+B\sin x$

You can show directly that $f\cos x-f'\sin x$ has zero derivative and that $f\sin x+f'\cos x$ has zero derivative

Calvin Khor

@Koro yeah, but i think ivt is cleaner :) anyway +1

Akiva Weinberger

so we can call those $A$ and $B$

and then we directly see that $A\cos x+B\sin x=f(\cos^2x+sin^2x)=f$

leslie townes

akiva: that's too cute by half but i love it

these are the kinds of fun things you always seem to have in your bag of puzzles and games

Akiva Weinberger

Similarly: just from knowing $\sin'=\cos$ and $\cos'=-\sin$, you can see $\cos^2+\sin^2$ is constant

and in general if $f''=-f$ then $f^2+f'^2$ is constant

Calvin Khor

01:54

this reminds me that i haven't seen a full derivation of all of sin and cos' properties (...that's an apostrophe) from the ode/taylor series but i know it exists somewhere

Akiva Weinberger

@CalvinKhor Well what I just said tells you that $\cos^2+\sin^2=1$ from Taylor

because the derivative properties are clear from the Taylor series

Calvin Khor

yup

hm how do you get periodic

Koro

@AkivaWeinberger :-) I studied an another way to show that.

for any linear homogeneous ode with constant coefficients.

Akiva Weinberger

@CalvinKhor Show $\cos1>0$ and $\cos2<0$ because they're alternating series so you can do that directly with enough (not many) terms

IVT says there's a zero

Call the first zero $\pi/4$

Calvin Khor

oh nice

Akiva Weinberger

01:59

then from angle addition properties (also doable) you can work your way up to $\sin(x+2\pi)=\sin x$, $\cos(x+2\pi)=\cos x$

Calvin Khor

that sounds like a fun exercise

might work out the details when im done staring at the abyss for today

the first zero should be pi/2 = 1.57... but wtv

Akiva Weinberger

@CalvinKhor That's to see if you're paying attention

(joking)

(that was an actual mistake, sorry)

Calvin Khor

:')

Akiva Weinberger

@CalvinKhor For the addition properties, it's probably easier to deal with the properties of exponentials: if $f'=cf$ and $f(0)=1$, then $f(x+y)/f(x)$ is constant in $x$

Plug it $x=0$, conclude $f(x+y)=f(x)f(y)$

Finally, note that if $f=\cos x+i\sin x$, then $f'=if$

All the addition formulas fall out from that

Calvin Khor

right ok. translation of f also solves the ode

Akiva Weinberger

02:06

Or find $\frac d{dx}\frac{f(x+y)}{f(x)}$ directly if you want

Calvin Khor

ah right

Akiva Weinberger

Actually, even better, if you want to do this from the Taylor series:

Calvin Khor

and i do

Akiva Weinberger

if $f=\sum(cx)^n/n!$, then $f(x+y)=f(x)f(y)$ actually follows from the binomial theorem directly

(This also counts as a proof of the binomial theorem)

Calvin Khor

sicc

Akiva Weinberger

02:09

and then find zero via IVT and proceed as before, and bam, periodicity

Really, all the pure trig properties are just sparkling exponential properties

Calvin Khor

p sweet. feels like i should have known all this a while ago

Akiva Weinberger

@CalvinKhor Here's a neat thing

Suppose we have the indefinite integral $\int f(x)\sin xdx$

Wait I'm wrong

Calvin Khor

still on f''=-f? or exp?

Akiva Weinberger

No just any ol' (infinitely differentiable) function

I claim that this equals $-f(x)\cos x+f'(x)\sin x+f''(x)\cos x-f^{(3)}(x)\sin x-\cdots+C$

Can you see why? (and/or can you check that I haven't messed up)

Calvin Khor

02:26

integration by parts

well in spirit, havent checked

ok it seems right

Akiva Weinberger

Arright

Now let $f(x)=x^n(\pi-x)^n$

Calvin Khor

niven?

Akiva Weinberger

I want to think about $\int_0^\pi f(x)\sin xdx$

Oh damn you know where I'm going with this!

Yeah I was gonna show $\pi$ is irrational

Calvin Khor

:D sorry hahaha its the only time i've seen pi in a poly

but i didnt recall the earlier bit

Akiva Weinberger

OK so $\int_0^\pi x^n(\pi-x)^n\sin xdx>0$ because the integrand is positive

Calvin Khor

02:29

yup

Akiva Weinberger

On the other hand, plugging in the thing from earlier, it's $f(0)-f''(0)+\cdots+(-1)^nf^{(2n}(0)+f(\pi)-f''(\pi)+\cdots+(-1)^nf^{(2n)}(\pi)$

Oh also $\int_0^\pi f(x)\sin x<\pi(\pi/2)^{2n}$

I'm trying to remember the next step

I think if $\pi=a/b$ then that's all a multiple of $n!/b^n$

Right, yeah. @CalvinKhor If $f(x)=x^n(\pi-x)^n$, then $f^{(k)}(0)$ is either $0$ (if $k<n$) or equal to $k!\binom n{k-n}\pi^{k-n}$ (if $k\ge n$), and these are both integer multiples of $n!/b^n$ if $\pi=a/b$

Calvin Khor

ok, bear with me, i dont work with integers often...

Akiva Weinberger

and since $f(x)=f(\pi-x)$, then $f^{(k)}(\pi)=(-1)^kf^{(k)}(0)$

Calvin Khor

yuuuuup

Akiva Weinberger

@CalvinKhor Let's write it out explicitly. $(\pi-x)^n=\pi^n-\binom n1\pi^{n-1}x+\binom n2\pi^{n-2}x^2-\dotsb+x^n$, so $x^n(\pi-x)^n=\pi^nx^n-\binom n1\pi^{n-1}x^{n+1}+\binom n2\pi^{n-2}x^{n+2}-\dotsb+x^{2n}$

Right yeah so anyway $\int_0^\pi x^n(\pi-x)^n\sin xdx$ is positive and an integer multiple of $n!/b^n$, so it must be greater than or equal to $n!/b^n$

but on the other hand $x^n(\pi-x)^n\sin x$ peaks at $x=\pi/2$, where it equals $(\pi/2)^{2n}$, so $\int_0^\pi x^n(\pi-x)^n\sin xdx$ must be less than or equal to $\pi(\pi/2)^{2n}$

Factorials grow larger than exponentials, so eventually $n!/b^n>\pi(\pi/2)^{2n}$

Contradiction, QED

Calvin Khor

02:45

right yeah

p nice

where does this x^n(pi - x)^n come from tho

apparently niven is slightly different

Akiva Weinberger

Oh yeah he did $x^n(a-bx)^n/n!$

That's mine times $b^n/n!$

Calvin Khor

yup

Akiva Weinberger

He wanted the ending to be an integer

I just thought $x^n(\pi-x)^n$ looked nicer

Calvin Khor

yeah

Akiva Weinberger

Wait so I got an explicit thing

$\int_0^\pi x^n(\pi-x)^n\sin xdx=2\sum_{k=0}^n(n+k)!\binom nk\pi^k$

Wait, no

Only every other term

and some minus signs

Akiva Weinberger

03:04

Maybe it's $2\sum_{k=0}^{\lfloor n/2\rfloor}(n+2k)!\binom n{2k}(-\pi^k)$

Doesn't work numerically

Maybe $2\sum_{k=0}^{\lfloor n/2\rfloor}(2n-2k)!\binom n{2k}(-\pi^2)^k$

This seems correct

Calvin Khor

....im gonna trust you :D

Akiva Weinberger

Yeah I dunno lol

Thought it was gonna be nicer than it ended up being

Calvin Khor

theres a correct version, just leave it as an exercise

Akiva Weinberger

It also shows $\pi^2$ is irrational, as well, since it works out to only have even exponents of $\pi$

(basically because we're only doing even derivatives)

Calvin Khor

ooh ok sure

Akiva Weinberger

03:19

Is this the simplest polynomial $f(x)$ for which $\int_0^\pi f(x)\sin xdx$ is nice?

Well, "nice"

copper.hat

0

Calvin Khor

lol

Akiva Weinberger

Ah

lol

Calvin Khor

just some of the serious work that goes on in mathoverflow

one potato two potato

03:35

That seal drawing is essential in the proof

copper.hat

wow, things have changed

Akiva Weinberger

Somewhat confused why write it in terms of a weight $d\omega(y)=dy/y^2$ rather than directly write $dy/y^2$ in the integral

Why write $\int_1^{\infty}\frac{d\omega(y)}{1+(y-1)^{\alpha}}$ rather than more simply $\int_1^{\infty}\frac{dy/y^2}{1+(y-1)^{\alpha}}$? — Akiva Weinberger 16 secs ago

Koro

05:08

@JoeShmo .

leslie townes

akiva: all i can think is it's a special case of a more general/complicated result about more general weights

self-answered questions that seem to come outta nowhere often have the form "special case of something i noticed while doing something else that's a secret"

Gwyn

Assume $(a_k)_k \to 0$ as $k \to \infty$ and that $a_k>0$ for any $k \in \mathbb{N}$. Can I say that $\lim_{n \to \infty} \sum_{k=0}^n a_k>0$ because: I know that the limit of positive numbers is nonnegative, hence assuming by contradiction that it is $\lim_{n \to \infty} \sum_{k=0}^n a_k=0$, from $a_k>0$ it is $\sum_{k=0}^n a_k=a_0+\dots +a_n>a_0$ and hence limit $0=\lim_{n\to\infty} \sum_{k=0}^n a_k \ge a_0>0$ implies that $0 \ge a_0>0$, and this is a contradiction because it implies $0>0$.

Calvin Khor

if $\lim_{n\to\infty} \sum_{k=0}^n a_k$ exists, which is not implied by your assumptions, then yes @Gwyn

leslie townes

one side note, there's no reason under the given hypotheses that the limit exists. but that's the right idea. you might be able to simplify the proof.

it's not just a limit of positive numbers, but a limit of positive numbers $\geq a_0$, for example.

so it's just "if c_n > k for all n, then lim c_n, if it exists, is >= k". no reason to give special attention to the result with k = 0. unless for some reason you don't have it.

one generalization with the same proof is that under those hypotheses, liminf [that sequence of sums], which does exist, is positive (in fact \geq a_0)

Gwyn

05:43

@CalvinKhor thank you for your help, the limit should be implied from the hypothesis $a_k>0$ for any $k\in\mathbb{N}$ and so $\sum_{k=0}^{n+1} a_k -\sum_{k=0}^n a_k =a_{n+1}>0$ implies that $\sum_{k=0}^n a_k$ is increasing, or am I wrong?

leslie townes

the sequence is strictly increasing, but usually people avoid writing the limit when it might not be finite. the usual definition of the limit assumes finiteness. if you're working in the 'extended reals' and the reader understands that the limit might be infinity, this is not an issue.

although it still might help to put somewhere, under these hypotheses there is no reason for the sequence to tend to a finite limit, but in those cases it will tend to +infty.

if you are in the middle of a book of some other form of exposition there might even be a canned theorem somewhere, saying that a monotone sequence of real numbers has a limit (in the extended real number system) and maybe even includes an inequality of the form "if a_n >= k for all n then lim a_n >= k". which still makes sense there.

what wouldn't make sense is appealing to "if a_n >= k for all n then lim a_n >= k" for a more general sequence might not be guaranteed to converge even in the extended real number system. (e.g. a_n = (-1)^n and k = -1, which does not arise in this problem)

Calvin Khor

@Gwyn leslie backed by lesliecoin speaks the truth

leslie townes

khor: SE should consider pegging rep to lesliecoin.

monoidaltransform

06:00

Do we have

$(F\times F')_{}(T_pM\times T_p'M')= F_{}T_pM \times F_{*}T_pM'$?

Koro

$f(x)=\sum_{n=1}^\infty \frac 1{1+n^2x}$.
1) On what intervals does the above converge absolutely?
On R+ and on R-. I think that this is so because if $x>0$ then the series can be bounded by $\frac a{n^2}$, where a is a constant.

Calvin Khor

@Koro on (-\infty , 0) there are points where some summands = \infty

Koro

Here is how a comes: if 0<x<1, then multiplying it by $10^k$ where k is large enough makes $10^k x>1$. Setting a:=$10^kx, \frac 1{1+n^2x}=\frac a{a+n^2 (ax)}\le \frac a{a+n^2}$

@CalvinKhor Oh, the points of the form 1/k^2.

hmm, and at those points the series won't even be series anymore.

Calvin Khor

yes and please consult a physicist for what i mean by = \infty

Koro

Thanks. For now, I'll interpret \infty as not defined.

So the series should converge absolutely on R+ and on $(-\infty, 0)\setminus \{-1/k^2:k\in \mathbb N\}$

i.e., on $R-(\{0\}\cup \{-1/k^2: k\in \mathbb N\})$

Calvin Khor

06:12

brackets

sounds right

Koro

2) on what intervals does the series converge uniformly?
On any interval properly contained in R+, and on (-\infty, -1).

Calvin Khor

what about the other intervals that don't touch 0 or -1/k^2?

like (-1+0.001,-1/2-0.001)

Koro

06:27

yeah, I think that will work too. :)

intervals of the form $(-\frac 1{k^2}, -\frac 1{(k+1)^2})$

Calvin Khor

:) probably easier to just say intervals properly contained in the set you identified earlier (R minus 0 and the -1/k^2s)

Koro

haha, yes.

Silent