Mathematics - 2022-02-18 (page 1 of 2)

Ted Shifrin

00:02

Matching chains, copper?

robjohn

@copper.hat can you get $-\frac1{12}$ as a product of the positive integers?

copper.hat

The jumpsuit was for the sanatorium not prison hopefully.

Ted Shifrin

@copper.hat surprising!

copper.hat

@robjohn The first time I saw the $-{ 1 \over 12}$ I thought it was a ruse along the lines of an April Fool's.

robjohn

It sort of is, but the fact that $\zeta(-1)=-\frac1{12}$ has given it wings.

copper.hat

00:20

:-) Red Bull $\zeta$.

robjohn

@copper.hat: Note that $$\sum_{k=1}^\infty(-1)^{k-1}kx^k=\frac1{(1+x)^2}$$ Take the limit as $x\to1^-$. We get $$\sum_{k=1}^\infty(-1)^{k-1}k=\frac14$$

Since $$\begin{align}\sum_{k=1}^\infty(-1)^{k-1}k&=\sum_{k=1}^\infty((2k-1)-2k)\\&=\sum_{k=1}^\infty((2k-1)+2k)-\sum_{k=1}^\infty4k\\&=\sum_{k=1}^\infty k-4\sum_{k=1}^\infty k\\&=-3\sum_{k=1}^\infty k\end{align}$$ we get $$\sum_{k=1}^\infty k=-\frac1{12}$$

Ted Shifrin

@robjohn Like Icarus?

robjohn

@TedShifrin indeed

Ted Shifrin

Rearranging lots of infinities is magical.

robjohn

It is, but it gets the same result without using $\zeta$

Ted Shifrin

00:28

Taking non-existent limits.

Still.

But clever enough to fool plenty on MSE.

robjohn

@TedShifrin Of course it is. There is no legal way to get $$\sum_{k=1}^\infty k=-\frac1{12}$$

Ted Shifrin

Aww …

Akiva Weinberger

Hey can someone help me with this 'cause I am very confused

0

Q: A reference to a theorem about L1-bounded Martingales

I'm auditing a course called "Topics in Analysis", and in class we mentioned a theorem that didn't make sense to me. I don't fully remember how it goes, but it was something like: Theorem (half-remembered, probably not the correct formulation): The limit of an $L^1$-bounded Martingale can be uni...

Ted Shifrin

Get notes from classmates?

Nicolás Vilches

00:50

Maybe is relevant the Lebesgue decomposition theorem? A (Radon?) measure on $\mathbb{R}$ can uniquely be decomposed as a sum of Dirac measures, an absolutely continuous part, and a singular continuous part.

Akiva Weinberger

What does "singular continuous" mean?

@NicolásVilches

Nicolás Vilches

Hmmmm... if I recall correctly, a measure is continuous if the cummulative distribution $F(t)=\mu(]-\inf, t])$ is continuous. A nice example is the Cantor staircase: it is continuous, but the associated measure is supported on the Cantor set, hence singular

(As usual, given a continuous non-decreasing $F\colon\mathbb{R} \to [0, 1]$, we can define a measure on $\mathbb{R}$ by putting $\mu([a, b])=F(b)-F(a)$. If $F$ is not continuous this is still possible, but requires extra work!)

Akiva Weinberger

@NicolásVilches I think I get "continuous", but "singular"?

Does it just mean "measure 0 support"?

Nicolás Vilches

In this case, yes. Two measures $\mu_1, \mu_2$ are (mutually) singular if there supported in $A_1, A_2$, such that $\mu_1|_{A_2}=0$ and viceversa

PenAndPaperMathematics

01:16

@geocalc33 hey :)

copper.hat

01:53

@robjohn :-)

geocalc33

@PenAndPaperMathematics hey

robjohn

02:18

@copper.hat I thought I'd add a non-$\zeta$ view.

geocalc33

what's so great about real analytic planar foliations?

leslie townes

nothing. next question

geocalc33

what's so great about hypertopological spaces?

Xander Henderson

@geocalc33 Same answer. Next?

geocalc33

what's so great about isogeny classes of elliptic curves :(

so math is not great?

Xander Henderson

02:29

I'm just going to say that the answer to all of your questions is "nothing", because you are interested in uninteresting things. :P

Anyway, I need to get back to class.

geocalc33

I gotta get interested in interesting math...

Semiclassical

so, here's a silly little thing. consider the following system of difference equations: $$v_{n+1}=v_{n}-r i_n,\quad i_{n+1}=i_n - g v_{n+1}$$

mohan10216

Anyone know good books which study the gamma function in a lot of depth?

leslie townes

mohan10216 you might try this survey paper arxiv.org/abs/1703.05349 and its references.

mohan10216

Thanks Leslie will take a look

I'm not looking for a complex analysis book which studies other topics but one that mainly focuses on the gamma function

leslie townes

02:39

any complex analysis book would likely introduce at least one or two formulas for it and establish its basic properties. articles in the monthly tend to be slightly more expository than textbooks.

and more focused.

mohan10216

I'm trying to study the gamma function to write about it for my EE

Semiclassical

if you take an ansatz solution of the form $(v_n,i_n)=(v_0 \lambda^n,i_0 \lambda^n)$, you end up with the characteristic polynomial $\lambda^2-(2+r_1 g_2)\lambda+1=0$

then the product of the roots is 1 and their sum is larger than 2, so one of the roots will have magnitude larger than 1 and the other will be smaller than 1 (both are positive)

however, on physical grounds I'm only interested in a solution which goes to 0 as $n\to\infty$. so that means i want the root $\lambda\in (0,1)$

and if you work it out, that requires $v_0=R i_0$ where $R=r_1/2+\sqrt{r_1(r_1+4/g_2)}$

so, that's a simple enough exact solution

however, another approach is as follows. the $v_n$'s here denote the potential at nodes in a resistor ladder, each separated by resistors of length $\ell$

so that suggests converting this difference equation to a differential equation: $(v_{n+1}-v_n)/\ell = - (r_1/\ell) i_n\to dv/dx = -(r_1/\ell)i(x)$, $(i_{n+1}-i_n)/\ell = - (g_2/\ell) i_n\to di/dx = -(g_2/\ell) v(x)$

then $v''(x) = (r_1 g_2/\ell^2 v(x)$ has the decaying solution $v(x) = v(0) \exp(-\sqrt{r_1 g_2}x/\ell)$

and therefore $i(x) = -(\ell/r_1)v'(x) = \sqrt{g_2/r_1} v(0)\exp(-\sqrt{r_1 g_2 x/\ell}$

which means $v(0)/i(0) = \sqrt{r_1/g_2}$

now, i had a typo above: the earlier $R$ should've been $R=r_1/2+\sqrt{(r_1/g_2)(1+g_2 r_1/4)}$

so those two answers for $v_0/i_0$ disagree...except when $g_2 r_1\ll 1$, in which case they do agree

so...that's sorta weird

mohan10216

03:12

Anyone know of a math phrase that rhymes? Preferably not a quote.

Koro

03:56

How can I get the relation between: $x_n=\sum_{n=1}^\infty (-1)^{n+1}\frac 1n$ and its rearrangements of the type when one + is followed by k minuses?

For example: 1-1/2-1/4-1/6+1/3-1/8-1/10-1/12+1/5-...

Rajorshi Koyal

04:11

"Pati,Patni and Woh(the three persons) were playing a game. At the beginning of the game Pati and Patni together had 100 % more money than Woh. Patni and Woh together had 300 % more than Pati. By the end of the game Pati and Patni together had 100 % more money than Woh had and Pati had 12.5 % less money than Patni and Woh together had. Finally Pati gained Rs.800 by the end of the game".

How to solve this problem?

I am getting to many variables for solving this problem.

leslie townes

it is cruel and unusual to give two people in a three-person word problem the same first initial

Koro

\begin{align*}
s_{4k}&=1-1/2-1/4-1/6+1/3-...+1/(2k+1)\\
&=1+1/3+...+1/(2k+1)-1/2(1+1/2+...+1/(3k))\\&
=\log (2k+1)+g+o(1)-1/2(\log k+g+o(1))-1/2(\log 3k +g+o(1))
\end{align*}
,where g is Euler-Mascheroni constant.

and this should give me: $\log \frac{2k+1}{\sqrt 3 k}+o(1)\to \log \frac 2{\sqrt 3}$

The only thing left however is to show that this rearrangement $(s_n)$ actually converges. If it does, then the sum is as given above.

Some generalization also seems plausible in terms of number of -ve signs following a + sign.

Oh, the convergence follows from the observation that: $s_{4k+i}-s_{4k}\to 0$ for all $i\in \{0,1,2,3\}$.

Akiva Weinberger

04:52

@leslietownes quietly rewrites problem involving people named Schiff, Schön, and Shah

@mohan10216 "Minimal criminal"

A proof technique equivalent to (strong) induction

To show that a certain property is true of all numbers, consider the smallest number that does not satisfy the property (the minimal criminal), and show that there exists an even smaller criminal

(note also that strong induction and weak induction are equivalent to each other)

leslie townes

Schulz, Scholz, Shultz, and Schuetz are playing a game

Schiff and Shah have a lifelong grudges against Shultz, but only if they're both together, and Schoen and Scheutz cannot be on the same team

Koro

05:08

Schwarz, Sylvestor and Sylow also join the game.

leslie townes

the multiple choice exam used by american law schools has (or had) a section on logic puzzles with people in them. they always named them things like Ann, Bill, Carl, and Diane. they could have ruined the average by naming them Schultz, Schulz, Sholtz, etc

copper.hat

05:30

Saoirse, Tighearnach, Ultan

Random Variable

@robjohn I noticed that the Wikipedia article on the Fourier transform says that Plancherel's theorem holds if $f$ and $g$ are both integrable and square-integrable functions on $\mathbb{R}$.

Isn't the square integrability of both functions and the convergence of $\int_{\mathbb{R}}f(x) \overline{g(x)} \, \mathrm dx $ and $\int_{\mathbb{R}}\hat{f(\xi)} \overline{\hat{g(\xi)}} \, \mathrm d \xi $ enough?

copper.hat

Having $f,g \in L^2$ is sufficient.

Random Variable

@copper.hat That's what my old PDE textbook says, too. But every source seems to say something a little bit different.

copper.hat

The proofs I am aware of show it holds in $L^1 \cap L^2$ and extend to $L^2$ using uniform continuity.

Random Variable

05:51

I think I was misinterpreting what the article meant by "integrable." I now assume they meant in $L^{1}$, which a lot of sources state if they don't want to get into why it can be extend to functions that are only in $L^{2}$.

Ted Shifrin

@leslietownes You left me out!

Random Variable