Mathematics - 2018-09-23 (page 1 of 2)

$\sum_{n=1}^{\infty} \dfrac{\sin nz}{s^n} = \sum_{n=1}^{\infty} \dfrac{e^{inz}-e^{-inz}}{i \cdot 2^{n+1}} = \sum_{n=1}^{\infty} \dfrac{1}{i \cdot 2^{n+1}}[e^{inz}-e^{-inz}]$

Semiclassical

12:28 AM

yep. At that point it's useful to split the series in two

\cdot, maybe?

For instance, can you compute $\sum_{n=1}^\infty e^{i n z}/2^n$ ?

That's a good deal less painful to deal with

Joe Shmo

and and that point ^ im stuck

Semiclassical

How about if you write that as $\sum_{n=1}^\infty (e^{iz}/2)^n$?

See if you can remember anything about series of that form.

(back in a bit)

Joe Shmo

geometric series

Joe Shmo

12:44 AM

hm, so perhaps the region is $-\ln2 < |iz| < \ln2$

and the sum evaluates to $\dfrac{1}{2i} \Bigg[\dfrac{1}{1-\dfrac{e^{iz}}{2}}-\dfrac{1}{1-\dfrac{e^{-iz}}{2}}\Bigg]$

Semiclassical

Second looks right. But for the first, you’d have z=pi lying outside that region

Joe Shmo

hrmp

i get $|e^{iz}| < 2$

Semiclassical

more generally, the series should vanish term-by-term for any multiple of pi

Joe Shmo

true. can i get more specific than $|e^{iz}| < 2$? take $\ln$ of both sides?

Semiclassical

Sure. But that equals |e^x e^(iy)|

Joe Shmo

12:50 AM

yeah, more specific than that?

Semiclassical

yes.

Joe Shmo

how?

taking $\ln$ of both sides doesn't work (?)

Semiclassical

Well, x and y are real. So what can you say about e^(i y)

Joe Shmo

complex

Semiclassical

What about it’s modulus?

Joe Shmo

12:52 AM

drawing blanks

Semiclassical

So you don’t know the absolute value of e^(i pi/3)? (Picking an arbitrary example)

Joe Shmo

oh

its $1$

by Euler, and then Pythagoras

Semiclassical

Right. That applies to any $e^{iy}$ with $y$ real.

Alternatively, you've got $|e^{iz}|^2=e^{iz}e^{-i \overline{z}}=e^{-2y}$

Joe Shmo

right

Semiclassical

Oh. I wrote |e^x e^(i y)|=|e^z| earlier when it should have been |e^(ix)e^(-y)|=|e^(iz)|

same idea, but slightly different algebra

So what does that inequality become?

Joe Shmo

12:59 AM

yes yes

Jalapeno Nachos

is it unwise to concentrate in abstract algebra for a math PhD?

Joe Shmo

ok good. that bothered me acutally

Jalapeno Nachos

seems so many people pick applied topics for their PhD these days

Joe Shmo

drumroll

The region is $-2 < \ln\Im{z} < 2$

Semiclassical

still not right :(

that'd be e^(-2) < Im(z)<e^2

but both of those are positive, i.e. you'd be talking about a region of positive imaginary part

which again would exclude real z

Joe Shmo

1:04 AM

$|e^{-y}| < 2$ for ${y\in \mathbb{R}}$

specifies the region right?

Semiclassical

Right. But $e^{-y}$ is positive for all real y

so you can drop the absolute value sign

Joe Shmo

yes

and we get $y > -\ln 2$

Semiclassical

Yep. So the imaginary part must be larger than -ln(2)

That definitely includes all real z, so it passes that check

Joe Shmo

:)

thanks

Semiclassical

Hmm, though

The region we get this way isn't symmetric across the real line

Joe Shmo

1:07 AM

what do you mean

Semiclassical

Suppose $\sum_{n=1}^\infty \frac{\sin n z}{2^n}$ converges. What can we say about $\sum_{n=1}^\infty \frac{\sin n\overline{z}}{2^n}$?

Joe Shmo

ought to converge as well

Semiclassical

Right. So the region in which the series converges out to be symmetric across the real line

since otherwise you'd have z for which the series would converge but not z-bar

Joe Shmo

im not following what you mean by symmetric across the real line?

for real z, z = z-bar

Semiclassical

Draw the region in the complex plane. Is the real line a line of symmetry?

Joe Shmo

1:11 AM

no

ohh

Semiclassical

Yeah

I think the resolution is simple, though. the requirement that |e^(iz)/2|<1 is the condition for $\sum_{n=1}^\infty (e^{iz}/2)^n$ to converge

Joe Shmo

right

Semiclassical

But you actually had two series. The second was $\sum_{n=1}^\infty (e^{-i z}/2)^n$. What's the condition for that to converge?

(It's not quite the same!)

Joe Shmo

$y < \ln 2$

Semiclassical

Yep. So between the two you get $-\ln 2 <y <\ln 2$

Joe Shmo

1:15 AM

yes

which is why i had $-2 < \ln\Im{z} < 2$ except my $\ln$'s were misplaced

Semiclassical

Ah.

In any case, I think that's the final correct domain of convergence

Joe Shmo

looks good. thanks again!

Semiclassical

np

Strikers

Hi

For solving proofs, in general, (started proofs now), is getting to a solution more on based on doing many different proofs and luck, or...?

Jalapeno Nachos

1:51 AM

@Strikers both luck and practice -- maybe practice on how to use the given assumptions

Ultradark

2:03 AM

there are two solutions to $e^{1/lnx}=x$ the solutions are: $x=e$ and $x=1/e$

Semiclassical

those are the two positive real solutions, at any rate

Ultradark

what is pythagoras's constant?

ohh

it's $\sqrt{2}$

Ultradark

2:26 AM

what kind of number is $e^{-\sqrt{2}}$

Lucas Henrique

4:05 AM

One of those that are less than < 1/2

Secret

5:11 AM

4

Q: Uncountable orderings

Let $P$ be an uncountable linear ordering. Is it true that either $P$ contains an order-copy of $\omega_1$ or there is $x_0\in P$ such that there exist uncountably many distinct $y\in P$ with $y< x_0$? If so, where can I find a reference for this? Thank you.

This link provide an example of constructing a singular $\omega_1$ in ZF

Secret

6:05 AM

[Random]

a is 0, a is positive, a is negative, a is fractional, a is decimal, a is continuous, a is undefined, a is unknown, a is ill defined

user400188

6:31 AM

@Semiclassical thank you. I mistyped and wrote y(b) instead of y(a).

Piyush Divyanakar

7:03 AM

Is anyone familiar with abstract algebra / groups here

If I take product of two normal groups is the result group abelian?

Fargle

7:14 AM

@PiyushDivyanakar Well, a group can't be normal by itself, it's only normal as a subgroup of another group.

But if you take the product of two abelian groups, the result will indeed be abelian. You should check this yourself; i.e. compute $(g_1, h_1) \cdot (g_2, h_2)$ and compare it with $(g_2, h_2) \cdot (g_1,h_1)$, keeping in mind that each group is abelian by itself.

Leaky Nun

hi @Gargle

Fargle

ouch

Leaky Nun

digaonal is normal iff group is abelian right

Fargle

7:31 AM

Yeah. In particular, for any $g,h \in G$, $(g,1)(h,h)(g^{-1}, 1) = (k,k)$ for some $k \in G$, so that $ghg^{-1} = k = h$.

Piyush Divyanakar

7:42 AM

@Fargle let me rephrase that. If we have H,K as normal subgroups of G then is the product set HK which is pairwise product of all elements in H and K. Then is HK abelian.

Fargle

Oh, that's my bad, I misread you.

I think that's false: if you let $H$ be the subgroup of the quaternion group generated by i, and $K$ the subgroup generated by j, then $HK = G$, but the group is nonabelian.

Leaky Nun

@Fargle no that's not your bad

he clearly said "normal group" instead of "normal subgroup"

Secret

1

Q: Why is the group G a normal subgroup of itself?

Apparently $g^{-1}Gg=G$ for all $g$ in $G$. I understand that by closure if you multiply on the right by $g$ and the left by $g^{-1}$ you will get an element of the group $G$. However, how do i know that some of the elements of $g^{-1}Gg$ will not be duplicate and hence I will not be able to gene...

abstract-algebra

"self normal subgroup"

Piyush Divyanakar

8:12 AM

@Fargle thanks I am getting it now.

Secret

9:20 AM

[Random]

How big is the leap between 0 and 1

Given any definable function $f$, $f(\varnothing)=\varnothing$ always

Jasper Loy

@Secret 1-0=1

Secret

so that means 1 is effectively inaccessible from 0 (because to even get to 1 we need 1 itself)

yet $\mathcal{P}(\varnothing) = \{\varnothing\} = 1$

so somehow powersets (in the absence of inaccessibles) can reach a higher hierarchy of numbers from below

Is it even possible to have $1$ if we don't have:

Axiom of infinity
Axiom of pairing
Axiom of power set
?

In fact, one can argue the gap between $0$ and $1$ is much much bigger than the gap between any two infinite cardinals (however big they are). This is because for any cardinal $\kappa$, $0\kappa = 0$ and since cardinals are initial ordinals, it follows this is actually a well ordered sum of zeros which gets nowhere other than zero itself.

Secret

11:12 AM

Therefore, one important observation that the zero ordinal is the lowest level of the hierarchy of "unreachable" quantities is the following:

> Any sequence formed by the arithmetic combination of zeros will always be the zero sequence, hence converging to zero for any topology $\tau$

So the zero ordinal is special. It is not an unreachable thing like anything else higher up, but it is "very hard to escape from"

The zero ordinal, in fact, is the opposite of any limit ordinal: It is something that everything will eventually converge to, give enough time for things to wander around in the class of ordinals

Math geek

12:00 PM

0

Q: Doubt on Theorem 4.4.4 (Tutte [181]; Nash-Williams [145])

I understood that $T$ is one of the spanning trees which has disjoint edges compared to other $k-1$ edges. then, $T$ has $|V(G)|-1$ edges. What does the author say in the underlined statement? Isn't it a trivial thing?we know that $|\mathscr P-1|\leq|V(G)|-1$. Is it follow from this inequality...

graph-theory

please help me to find the converse

user193319

1:33 PM

Problem: If $b \neq 0$ and $b_n \to b$, then $b_n \neq 0$ except for possibly a finite number of integers $n$. I discovered a simpler solution but I just want to verify that my original one is correct. If am not mistaken, the contrapositive states "If for every finite set $F$ in $\Bbb{N}$ there is some $n \notin F$ s.t. $b_n = 0$, then $b=0$."

With that in mind, assuming the hypothesis, there exists some $n_1 > 1$ such that $b_{n_1} = 0$. And given the finite $\{1,...,n_1\}$, there is some $n_2 > n_1$ such that $b_{n_2}=0$. From this we can inductively construct a subsequence $\{b_{n_k}\}$ such that $b_{n_k}=0$ for all $k \in \Bbb{N}$. Clearly $b_{n_k} \to 0$, so $b =0$, since the limit of the parent sequence must agree with the limit of any one of its subsequence.

Does this sound right? Did I get the contrapositive of the original statement right?

Akiva Weinberger

"The 3 groups of ppl most in conflict in israel are the Jews, the Arabs, and the Bus Drivers" - my sister

2

@user193319 Seems good

Yeah, you can't have infinitely many 'cause then you'd get an infinite subsequence of zeroes

Unrelatedly:

(I wonder if there's a book or article on the Riemann Hypothesis with the title "From Zeroes to Heroes")

user193319

@AkivaWeinberger Thanks!

Leaky Nun

1:50 PM

@AkivaWeinberger because people don't thank the bus drivers enough?

Soham Chowdhury

because the bus drivers have to thank each other

Leaky Nun

did you type that in 10 seconds

Soham Chowdhury

i'm ... pretty fast at typing?

Akiva Weinberger

No because they're just horrible

I JUST missed a bus three times in a row today

three separate busses

Leaky Nun

lol

Akiva Weinberger

1:52 PM

Oh also they changed the law on how you have to pay them

You can't use cash, you have to have a card, and there are very few places where you can actually refill the cards

It's annoying

Not like a credit card, a special card specifically for public transport

Leaky Nun

can you pay them by thanks?

Soham Chowdhury

cash, card, or fortnite

Akiva Weinberger

I do yell "Todah!" before stepping off but I think I'm the only one in the country who does this

Leaky Nun

why is cos(nx) n>=0 and sin(nx) for n>=1 a countable basis for continuous functions on [-1,1]?

Akiva Weinberger

I'm too ~~Canadian~~ American

Soham Chowdhury

1:56 PM

IME spending over a decade in a Catholic school also does this to you

Akiva Weinberger

'Cause Fourier? @LeakyNun

Leaky Nun

why does Fourier work?

Akiva Weinberger

Should really be an interval of length $2\pi$ or something

Leaky Nun

[-pi,pi] then

Akiva Weinberger

The proofs work a lot easier if you use $\{e^{inx}\}$ instead

and it's equivalent

Leaky Nun

1:59 PM

ok why does {e^inx} work?

Akiva Weinberger

But uh you find a formula that says, if it is a sum of the thingies, this formula gives you the coefficients. And then you can go the other way and prove the formula always gives you things that add up to the function

Leaky Nun

well I understand the first part because they're all orthogonal to each other

Akiva Weinberger

And if you use sin/cos then you need the formula for $\sum^N\sin nx$ or something

Leaky Nun

and int fg dx is just the inner product <f,g> in this Hilbert space

Akiva Weinberger

which makes me think if you use $e^{inx}$ you just need the geometric series which is easier

Soham Chowdhury

2:00 PM

i mean, $\sum \sin nx$ is just two copies of that, just a bit messier

Leaky Nun

I think you're just transforming my question into equivalent statements

Akiva Weinberger

I don't remember the details but work it out, if $f=\sum a_ne^{inx}$ for $n$ from $-\infty$ to $\infty$ then what's the formula for $a_n$?

Leaky Nun

multiply it by e^{-inx} and integrate

Akiva Weinberger

Ye

Leaky Nun

because they're orthogonal

it's like extracting a coefficient using the dot product

Akiva Weinberger

2:02 PM

So then define define $\bar f$ to be $\sum a_ne^{inx}$ where $a_n$ is defined by that formula, and then show $\bar f=f$

I honestly don't remember any more details on this

Leaky Nun

and that's my question...

Akiva Weinberger

Well, a good start would be to write out what $\bar f$ is

That's probably not the official notation or whatever

Leaky Nun

$\displaystyle \overline f = \frac1{2\pi} \sum_{n=-\infty}^\infty e^{inx} \int_{-\pi}^{\pi} f(z) e^{-inz} \ \mathrm dz$

Akiva Weinberger

Any way you can switch the integral and summation signs?

Leaky Nun

M?

Akiva Weinberger

2:05 PM

?

Leaky Nun

LDC?

Weierstrass M-test or Lebesgue Dominated Convergence

Akiva Weinberger

Uh probably

I don't remember

Leaky Nun

maybe @MikeMiller will know

Akiva Weinberger

$f$ is bounded anyway

'cause it's a continuous function on a compact interval

Leaky Nun

sure

Akiva Weinberger

2:06 PM

so that probably helps

Leaky Nun

ok...

Akiva Weinberger

I don't remember the criteria

Mike Miller

@LeakyNun Do you understand why they are L^2 orthogonal?

Leaky Nun

yes

Mike Miller

You need to show that the L^2 norm of \sum a_n sin(nx) decays as you start the sum from larger n

Akiva Weinberger

2:08 PM

Oh, am I over the Green Line right now?

I think I might be

Mike Miller

You know those are orthogonal. So the L^2 norm is just the sum of a_n^2

Akiva Weinberger

I'll cross it again before I get off the bus

but right now I think I'm in the West Bank

Mike Miller

The assumption that sum a_n^2 converges implies that the sum from N to infinity goes to 0 as N goes to infinity

(basically just definition of convergence of sums)

Akiva Weinberger

Yep, I am

Leaky Nun

@MikeMiller would it be easier to do with e^inx?

Mike Miller

2:12 PM

The argument would be the same - the essential point is orthogonality

Akiva Weinberger

It would actually be incredibly easy for me to go into Ramallah at any point, just take the 218 bus

Obviously I'm not gonna do that, but I could

Mike Miller

I would probably do that using the Fourier transform as it takes multiplication to convolution and vice versa

Akiva Weinberger

(Would probably not be a fun bus ride)

Leaky Nun

@MikeMiller and then?

Akiva Weinberger

Oh, a soldier with a huge gun just went inside briefly, I assume that means we crossed the border again

Yup

Also I missed my stop 'cause I was distracted by checking the map for that

Mike Miller

2:19 PM

@LeakyNun I'm confused - the argument is finished. What do you mean?

Leaky Nun

you showed that the sum converges

I don't see why it would converge to the original function

Mike Miller

Aha I now see what you mean

Akiva Weinberger

Like, if we removed an element from our list, it would no longer be a basis

but they would still all be orthogonal

Mike Miller

Yes I got it :)

Gimme a minute to fix my thoughts

Akiva Weinberger

17 mins ago, by Leaky Nun

$\displaystyle \overline f = \frac1{2\pi} \sum_{n=-\infty}^\infty e^{inx} \int_{-\pi}^{\pi} f(z) e^{-inz} \ \mathrm dz$

Wait, so start here, swap the thingies

Mike Miller

2:22 PM

The problem is that this is a nontrivial theorem and I was trying to prove it with trivial facts...

Akiva Weinberger

The $e$s combine to $e^{in(x-z)}$ and you're summing over $n$

Mike Miller

The nontrivial input is Parseval's theorem, which says that Fourier transform preserves $L^2$ norm

Leaky Nun

that's not very non-trivial right, that's just swapping integrals and sums

let's assume we can swap them

Mike Miller

I haven't thought about the proof of that theorem in a long time

So I will have to take your word of it

Leaky Nun

it's still only saying that the things are orthogonal

maybe orthonormal

Akiva Weinberger

2:25 PM

Wait, hold on, I'm confused again

because summing a geometric series from $-\infty$ to $\infty$ never makes sense

I guess you interpret it as the limit of the sum from $-N$ to $N$ as $N\to\infty$?

Mike Miller

the coefficients are not geometric?

Akiva Weinberger

$e^{in(x-z)}$?

Mike Miller

there is an integral term...

Akiva Weinberger

Yeah, we're swapping them

You know what

Let's think of the finite sum for now

Define $\bar f_N$ to be $\sum_{-N}^Na_ne^{inx}$ 'cause I forget the actual notation

Mike Miller

OK I'm too far behind so I'm going to go do something else, sorry I wasn't much help. Deserves more thought than I remembered.

Akiva Weinberger

2:28 PM

And we just need $\bar f_N\to f$, and then we don't need to worry about weirdness involving swapping sums and integrals

Leaky Nun

ok

Akiva Weinberger

$\displaystyle \overline f_N = \frac1{2\pi} \sum_{n=-N}^N e^{inx} \int_{-\pi}^{\pi} f(z) e^{-inz} \ \mathrm dz$

$\displaystyle {}= \frac1{2\pi} \int_{-\pi}^\pi f(z) \sum_{n=-N}^N e^{in(x-z)} \ \mathrm dz$

Mike Miller

this really shouldn't be that hard but i am utterly blanking lol

i've seen this proof so many times

Akiva Weinberger

I mean clearly the next step is to simplify that sum, no?

Mike Miller

i know i said i was doing something else but then i got embarassed that i couldn't figure out why the standard basis of L^2 is a basis

Akiva Weinberger

2:36 PM

Get rid of the $\sum$

Leaky Nun

@AkivaWeinberger what are you going to do, turn it into dirac delta?

Akiva Weinberger

$\dfrac{e^{i(N+1)(x-z)}-e^{-iN(x-z)}}{e^{i(x-z)}-1}$, right?

Ugh, maybe doing the real version would have been simpler

What are the real and imaginary parts of that?

Or

Leaky Nun

I don't really think doing the real case would be simpler

anyway, the common ratio of the geometric series has norm 1 so it doesn't converge

Akiva Weinberger

What's $\left(e^{iN(x-z)}-1\right)\left(e^{-iN(x-z)}-1\right)$?

Mike Miller

terrible

Leaky Nun

2:44 PM

I really feel some dirac delta business going on there

Akiva Weinberger

Never mind, doesn't equal what I thought it would

What's $\left(e^{i(N+1)(x-z)}-1\right)\left(e^{-iN(x-z)}-1\right)$?

Mike Miller

@LeakyNun That's definitely what goes on in Parseval

Akiva Weinberger

Hm, the signs aren't right

Leaky Nun

@MikeMiller what's parseval's theorem?

Mike Miller

The Fourier transform is an isometry $L^2 \to L^2$. One should be able to use this to then show that the Fourier series is an isometry $L^2 \to \ell^2$. Unfortunately I am a doofus and do not see how

Akiva Weinberger

2:46 PM

$\left(1+e^{i(N+1)(x-z)}\right)\left(1-e^{-iN(x-z)}\right)$?

Mike Miller

Alternatively I am scrunched on a plane with 4h sleep so that's my excuse

Akiva Weinberger

Yeah that's what I want

Leaky Nun

maybe my notes stated a bogus version of Parseval

no that's the same as what wiki says

Parseval's theorem

In mathematics, Parseval's theorem usually refers to the result that the Fourier transform is unitary; loosely, that the sum (or integral) of the square of a function is equal to the sum (or integral) of the square of its transform. It originates from a 1799 theorem about series by Marc-Antoine Parseval, which was later applied to the Fourier series. It is also known as Rayleigh's energy theorem, or Rayleigh's Identity, after John William Strutt, Lord Rayleigh.Although the term "Parseval's theorem" is often used to describe the unitarity of any Fourier transform, especially in physics, the most...

so is wiki's version also bogus?

Akiva Weinberger

$\displaystyle \frac{e^{i(N+1)(x-z)}-e^{-iN(x-z)}}{e^{i(x-z)}-1}=\\ 1+\frac{\left(1+e^{i(N+ 1)(x-z)}\right)\left(1-e^{-iN(x-z)}\right)}{e^{i(x-z)}-1}$

I think I made it worse

Mike Miller

definitely not, I'm probably just misremembering names

Oh I see, your point is that it assumes the basis already

I am about 80% sure that the theorem is not supposed to assume that

wwwf.imperial.ac.uk/~jdg/eeft3.pdf

No series in this just transforms

So still need to figure out why the map to $\ell^2$ is also an isometry

Leaky Nun

2:51 PM

@MikeMiller hey that's not discrete

Mike Miller

If I parse correctly that's what I meant by my last two lines

thanks for your patience with me this morning lol

Leaky Nun

@MikeMiller indeed

so I now have one more equivalent statement

$\displaystyle \overline f = \frac1{2\pi} \sum_{n=-\infty}^\infty e^{inx} \int_{-\pi}^{\pi} f(z) e^{-inz} \ \mathrm dz$

so I guess the idea is that after you switch it, the terms become zero except around z=0 and when n=0

GENESECT

If a continious function is increasing from [a, b], does it mean it will also increase from (a, b)

Mike Miller

3:08 PM

@GENESECT What is the definition of increasing function on a subset of R? Let's call the subset I for convenience (I stands for interval, but we don't need to restrict to intervals).

GENESECT

Okay.. I would say that if a, b €I and a<=b then f(a) <=f(b)

Mike Miller

Let's use different letters than you used above, so maybe x, y instead of a,b

GENESECT

Okay. If x, y €I and x<= y then f(x) <=f(y)

Mike Miller

Okay. So you know that's true for x, y in [a,b] by assumption.

Is it true for (a,b)?

GENESECT

Going by the graph... Yes

But at the end points the function is not differentiable

Mike Miller

3:14 PM

I don't know anything about the graph, though. We never said anything about that.

I just know that when x <= y are both elements of [a,b], then f(x) <= f(y).

I am asking if when x and y are both elements of (a,b), then f(x) <= f(y).

GENESECT

Yes

Mike Miller

Why is that true?

GENESECT

Because it falls under the closed interval and we know the closed interval is increasing.......?

Mike Miller

yeah!

So that's the answer to your question

If one interval sits inside another, and your function is increasing on the bigger interval, it's definitely increasing on the smaller interval too

GENESECT

Okay got it. Thanks for your help sir

Mike Miller

3:18 PM

Sure thing. There's no need to call me sir. I am no knight.

Leaky Nun

@MikeMiller that's answering a question by citing a more general fact...

Mike Miller

@LeakyNun We answered the question first and the same logic implied that statement

I certainly did not cite that, but rather said the statement in an appropriate level of generality

Leaky Nun

alright

Leaky Nun

3:41 PM

Riemann's lemma: if $g$ is integrable then $\lim_{\lambda \to \infty} \int_a^b g(x) \sin(\lambda x) \ \mathrm dx = 0$

@MikeMiller maybe this is the key

also do you know the complex version of this lemma?

Leaky Nun

4:28 PM

$\begin{array}{cl} & \displaystyle \frac1{2\pi} \sum_{n=-N}^N e^{inx} \int_{-\pi}^{\pi} f(z) e^{-inz} \ \mathrm dz \\=& \displaystyle \frac1{2\pi} \int_{-\pi-x}^{\pi-x} f(z+x) \sum_{n=-N}^N e^{-inz} \ \mathrm dz \\=& \displaystyle \frac1{2\pi} \int_{-\pi-x}^{\pi-x} \frac {f(z+x) [\exp(i(N+1)z) - \exp(-iNz)]} {\exp(iz)-1} \ \mathrm dz
\\\approx& \displaystyle \frac1{2\pi} \int_{-\delta}^{\delta} \frac {f(z+x) [\exp(i(N+1)z) - \exp(-iNz)]} {\exp(iz)-1} \ \mathrm dz \\\approx& \displaystyle \frac1{2\pi} \int_{-\delta}^{\delta} \frac {f(x) [i(2N+1)z]} {iz} \ \mathrm dz \\=& \dfrac {2\delta (2n+…

parts of this is clearly bogus

Luigi M

Hi all, can you please help me in proving this statement? let $L: H\to H'$ be a continuous linear Fredholm map between hilbert spaces, and let $c : H\to H'$ be a compact map. prove that L+c is proper when restricted to any closed bounded subset $A \subset H$.

The idea of proof given was to consider the following factorisation of the map. First we have $A \to H' \times \overline{c(A)}\times \overline{\rho(A)}$ $a\mapsto (L(a),c(a),\rho(a)$ where $\rho(A)$ is the orthogonal projection $H\to ker(L)$. The claim is that this map is injective and CLOSED. Injectiviness is clear, I've problems proving that this map is closed, in particular that the map $a \mapsto c(a)$ is closed

Leaky Nun

@LuigiM I may be spewing complete nonsense, but does it help that compact subset of hausdorff space is closed?

Luigi M

@LeakyNun I thought about it, but A is not necessarily compact if the hilbert spaces are infinite dimensional :(

Leaky Nun

I just saw "c, compact, c, closed"

Luigi M

nono, c is just a compact map, i.e. images of bounded are precompact

Leaky Nun

4:43 PM

maybe if you care to define the terms for me then I can help you think :P

Luigi M

mmh I think I'll write a question then, so everything will be written in a more clear way

Mark S.

5:20 PM

Does anyone have an example (or just experience) of how to refer to a result in a talk that is not online yet. Like if Jane Smith is giving the talk but hasn't posted things on the arXiv yet, should it just say "(Smith, unpublished)" near the statement? Is "(Smith)" presumptuous/implying too much?

user193319

5:47 PM

Problem: Let $T_X$ and $T_Y$ be topologies on $X$ and $Y$, respectively. If $T = \{U \times V \mid U \in T_X \mbox{ and } V \in T_Y\}$ is a topology on $X \times Y$, then either $T_X$ or $T_Y$ is trivial.

I could use a hint on this. I tried a proof-by-contradiction, trying to show that the union axiom fails, but I had no luck...

Mark S.

5:58 PM

@user193319 Draw a picture of simple examples (e.g. what if X and Y have only two points?) to get an idea of how your proof must go. Then carefully work out the logical consequence of "neither T_X nor T_Y is trivial" and write the proof based on the intuition your picture(s) gave you.

user193319

Okay. I can try that again.

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$Mathematics$ Mathematics