Mathematics - 2022-02-03

Jakobian

00:06

When is C_b(X) separable if X is separable metrizable?

16

Q: reference for "X compact <=> C_b(X) separable" (X metric space)

I know (and am able to prove via Stone-Čech compactification) that the following is correct: Theorem: A metric space is compact if and only if its space of bounded, continuous, real-valued functions is separable in the uniform topology. I use it in a paper for readers who are presumably not...

Alex

04:39

@TedShifrin Hello professor, I'm sorry for the late response to your question. I've been away for a while. The problem was a calculation everywhere.

Hi everyone

Ted Shifrin

@Alex I had totally forgotten. A yucky mess, but doable.

dtn

10:51

Matrices are known to be positive-definite if all their eigenvalues are positive. Gershgorin circle theorem can be used to localize eigenvalues of a matrix (including symmetric). Let $A$ be a real symmetry $n × n$ matrix, with entries $a_{ij}$. For $i∈{1,…,n}$ let $R_i$ be the sum of the absolute values of the non-diagonal entries in the $i$-th row:

$R_{i}=\sum _{j\neq {i}}\left|a_{ij}\right|$

Let $D(a_{ii},R_{i})\subseteq \mathbb {C}$ be a closed disc centered at $a_{ii}$ with radius $R_{i}$. Such a disc is called a Gershgorin disc.

How, using this theorem, to choose the parameters of a symmetric matrix in such a way that all its eigenvalues are positive? The first idea that came to my mind was to observe the following condition. $R_{i}<\sum _{j\neq {i}}\left|a_{ij}\right|$ If the circles lie in the right half-plane, then the eigenvalues must also have a positive real part? What additional conditions should be used for the imaginary part of the eigenvalues to be strictly zero?

Such situations are also possible

Koro

12:01

3

Q: How to Count Homomorphism from any finite group G to infinite cyclic group?

Question How to Count Homomorphism from any finite group G to infinite cyclic group MY Approach I Know I also know that a homomorphism is completely determined by its action on unit element (like 1)

abstract-algebra group-theory group-isomorphism group-homomorphism

In the above linked question, how is existence of an onto homomorphism possible?

If $A$ and $B$ are two sets such that $|A|<|B|$, where |.| denote cardinality then $f$ can't be onto.

So how can there be an onto homomorphism from identity group {e} to an infinite cyclic group?

Jakobian

maybe it's a "choose one" type of question

anyway, i) is correct the rest is in general wrong

Koro

@Jakobian no, it's not.

Jakobian

it is

the only normal subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ for $n\in\mathbb{N}_0$

Koro

because in my coaching centre's book, it is there.

I would call that a printing mistake.

Jakobian

ah

that's what you meant

Koro

12:08

if what I said above about cardinality is correct. :)

there is an answer also to the post.

I think I should drop a comment there as well.

Jakobian

the answer understood the question as a "choose one" type of question as well it seems

and it'd make sense since only one of those is correct

Koro

Oh, I suppose you're right.

Jakobian

if you have a homomorphism $f:A\to \mathbb{Z}$ where $A$ is a finite group, then $f(A)$ is a subgroup of $\mathbb{Z}$, so since the only finite subgroup of $\mathbb{Z}$ is $\{0\}$ we need to have $f(A) = \{0\}$.

I think this is conceptually easier than the answer

I'm learning about set-valued functions and Michael selection theorem, cool stuff

Koro

@Jakobian makes sense :).

there are many examples. I was wondering why that condition was put.

But it seems it was an mcq.

thanks @Jakobian. :)

Jakobian

12:23

np

AMDG

16:04

Morning!

I'm almost ready to rewrite the draft. Just need to see what more I can squeeze out of division before I'm ready to rewrite.

This one also won't be nearly as long to write since it's simpler... or it will take twice as long because it's simpler... not sure which.

In particular, I'm trying to raise the value in the numerator for computing floored quotients of the form $\lfloor\frac{2^x}{y}\rfloor$. Currently, I can do that up to $2^4$ which is pretty nice, but if I can manage the same cost or better using $2^8$ or $2^{16}$, that would just be ideal for 64-bit floored quotients.

$$\lfloor\frac{2^4}{x}\rfloor \equiv \lfloor 2^{3 - \lfloor\log_2(x)\rfloor} \left(\frac{3}{2} +\sqrt{2} - 2^{-\lfloor\log_2(x)\rfloor} x \right)\rfloor + \lfloor\frac{2^{\lfloor\log_2(x)\rfloor + 1} }{x}\rfloor$$

for natural $0 < x \leq 16$

robjohn

17:01

@TedShifrin: I was able to use that integral I posted in chat a while ago in a comment.

AMDG

Is it valid to say that a function is periodic for a given subset of the codomain?

Eminem

Given that $f(x)$ is differentiable, is it true to say that $|f(x)|^2$ is also differentiable?

robjohn

@AMDG how would that look?

@Eminem is $f$ real or complex?

AMDG

@robjohn Suppose, for instance, I have a piecewise function where $y = 0$ for $x<1$ and $y=1$ for $x\geq 1$. The codomain can be said to be periodic with respect to a particular domain, but not both: for all reals $k$ in $x + k < 1$ and in $x + k \geq1$, but not for the entire domain of $x$ as ${x + k} \in \Bbb{R}$.

Eminem

$f$ is complex and the derivative is also continouse

$f$ is a quantom wave function for that matter

robjohn

17:12

@Eminem Then no. $|x|^2$ is not a differentiable function on $\mathbb{C}$

Differentiability on $\mathbb{C}$ brings a lot of structure with it.

AMDG

Now, consider in my particular case, we have the modular reduction of $2^x$ in $y$ for $\lbrace x,y\rbrace \in \Bbb{N}$ as $2^x \bmod y$.

Modular reductions of this form exhibit a periodic behavior in a subset of the domain of $x$.

Eminem

hmmm.. I will phrase the full question:
Given a complex wave function $f(x)$, is it true that for any operator $y$ we can calculate $<y> = \int_{-\infty}^{\infty}f(x)^*yf(x)dx$

AMDG

Plot of $2^x\bmod y$: desmos.com/calculator/wdzvcammxk

robjohn

@Eminem any operator? That seems a bold claim.

Eminem

Well its a physics question, so a quantom operator

So for example differentiation is also ok

AMDG

17:18

I'm interested in this specifically because I'm interested in finding a function which describes the rate at which we must add one every time we double the fractional part of a quotient, and this is directly related to its modular reduction.

robjohn

@Eminem It looks as if you are applying the operator only to $f$

Eminem

Sorry this is the question
Given a complex wave function $f(x)$, is it true that for any operator $y$ we can calculate $<y> = \int_{-\infty}^{\infty}y|f(x)|^2dx$

I though that if $|f|^2$ is not differentiable than the statement is false

robjohn

@Eminem There are obviously restrictions related to something being a quantum operator. I don't know what those are, so I have no idea.

Eminem

Since y can be any quantom operator - therefor differentiation is ok - than the statement is nonsense

Ted Shifrin

@robjohn You actually found an elementary antiderivative? Or just estimated?

Eminem

17:20

Given that y can be differentiation, is this false?

robjohn

@Eminem It looks as if you are only integrating over $\mathbb{R}$, so you are doing real integration, not complex.

Eminem

Yes but $f$ is a complex wave function

Ted Shifrin

@robjohn Since he's doing wave functions, I suspect he means real differentiable, not holomorphic.

AMDG

Yes, but you are integrating with $|f(x)|$

Which is real

Ted Shifrin

But it seems irrelephant.

robjohn

17:22

yes, but you are not looking at complex differentiation, just real differentiation. If so, then it is possible, but not necessary that the derivative of $|f|^2$ would be integrable.

Eminem

hmm yes that is also true. The derivative can be non integrable so the statement is false

robjohn

@Eminem but $f:\mathbb{R}^n\to\mathbb{C}$?

Eminem

f is a wave function - no other data is provided.

robjohn

If that is true, then $|f|^2$ would be differentiable as a function on $\mathbb{R}^n$

context is important here

Ted Shifrin

Why are we caring about differentiability if the question is convergence of the integral? Are you trying to integrate by parts and use something about behavior at infinity?

robjohn

17:26

The original question was about the differentiability.

the integral came after

Ted Shifrin

The question was to Eminem :)

robjohn

sorry

Eminem

Well my initial idea to contradict the statement is to say that |f|^2 done not have to be differentiability so y can not operate on it

Ted Shifrin

What does "operator $y$" mean?

Eminem

Quantom operator, with regards to quantum mechanics

Ted Shifrin

17:30

What does that have to do with differentiation?

Eminem

Because the operation can be differentiation

So applying y on f will be to differentiate f

Ted Shifrin

But this is often meant in the distributional sense when you're doing this stuff.

The math of quantum mechanics is not beginning calculus.

Eminem

Im not sure what you mean

Ted Shifrin

Probably you don't know the math required to make sense of this rigorously. Quantum mechanics works with $L^2$ (square integrable) functions that are differentiable in a "generalized functions" sense, not according to beginning calculus.

We can't guess how your course is teaching this.

Eminem

You are right, I dont know the math for this

robjohn

17:35

Here is a car. It can take you from place to place. Have fun!

Ted Shifrin

@robjohn But it's a 5-speed. Who here knows how to drive a standard shift car?

robjohn

@TedShifrin who in the physics class can differentiate in the distributional sense?

I can drive a standard transmission

Eminem

The real issue here is that im not a physics student nor a math student, and uni is still making me taking this class :)

Ted Shifrin

@robjohn But they do know how to play with delta functions, Heaviside functions, etc. They aren't doing it the way we would in a graduate math course, no.

@robjohn Me too. My current car is my first with automatic transmission.

robjohn

I know, but it's like giving someone a car without proper training, imo

Ted Shifrin

17:40

Well, we don't expect undergraduate physics (or applied math) students to learn Sobolev spaces.

@Eminem How can a uni make you take a course you do not have prerequisites for?

If you do have the prerequisites, then the course should have taught you more or less what you need to answer the homework questions.

AMDG

I find many things straightforward in theory. It's the practices that make me cringe. That said: stick-shift is straightforward in theory, but a pain in practice. Never have, never will, drive a stick on the road. Although, driving a stick on the road sounds like fun.

Ted Shifrin

"On the road" (i.e., on the highway) is easy. It's mountains and stop-and-go traffic that are a test.

I miss my standard transmission, but it was horrible (especially with a bad back) in Atlanta stop-and-go traffic.

AMDG

Yeah, but what consumer needs that level of control? You really only need that control if you're into racing.

Ted Shifrin

Not true.

AMDG

Alright, I'm listening

(Also, I've never driven anything besides a fast go-kart lol)

Random Variable

17:45

@robjohn It looks like the developers fixed those remaining MathJax issues.

Ted Shifrin

Better gas mileage. Better control in any sort of bad weather conditions (especially snow). Better engine braking and acceleration in hilly conditions ...

AMDG

My argument is that to have to think about gear shifts, especially in a stressful moment, is more to think about and thus more of a chance to lead to an accident. With auto trans, you only have to focus on where you're going, and how fast you're going. Also torque converters controlled by a chip doing calculations has more potential for gas mileage than a human. I don't think anyone can argue with that.

Gears are just an implementation detail.

Ted Shifrin

Well, data disagree with you, assuming a competent driver. And after a year of experience (and some of us have 50) one doesn't even think about gear shifts.

Jakobian

I thought $n^2+n+41$ was remarkable because it give you $40$ primes for $n = 0, 1, ..., 39$, but apparently the primes are consecutive as well?

Ted Shifrin

@Jakobian That's an increasing function for $n>0$, no?

Oh, consecutive.

Is that really true?

Jakobian

17:50

it's true for the first 8 of them

Ted Shifrin

So the consecutive primes are $2n+2$ apart?

AMDG

Then my question is then why, after 50 years, you can do better than a computer in this regard given that a computer, for example, given a few sensors and a bit of math, can determine the optimal amount of gas to accelerate, and the perfect nanosecond to switch gears or change torque to minimize effort.

Ted Shifrin

It's not true for the first 8. We go from 53 to 61 to 71. Aren't we missing a few?

@AMDG You're naive about how automatic transmission cars actually work, and you're forgetting that they too have drivers.

robjohn

@RandomVariable yes. Thanks for pointing them out. Of course, the rest of the layout issues are still there

AMDG

I'm naive concerning existing implementations as I'm not a car mechanic. I make my judgements based on a charitable assumption that certain contingencies have been met which thus give in this case a physical certainty provided that the contingencies are true.

Ted Shifrin

17:54

@robjohn My apologies for interrupting your differentiability discussion.

robjohn

@TedShifrin no problem. The question he wants to talk about was actually about the integral, so the differentiability was of secondary importance.

Ted Shifrin

@Jakobian Did you see my question? I'm no expert on primes, but shouldn't 59 and 67 be on the list?

Jakobian

yeah, in the video I'm watching, they said "none of these is prime" and crossed over 59

https://youtu.be/n4VWgquP8D0
at 7:10

AMDG

So, rephrasing what I said: if there were a few sensors, and if there were a bit of math, and if it can determine the optimal amount of gas to accelerate, etc. according to what I imagine the implementation would be based on what I know about cars in principle, then such and such would happen.

Hope that explains my judgements better.

Ted Shifrin

@Jakobian You need to be more critical when you watch videos. Most of us make mistakes :P

robjohn

17:59

@TedShifrin they are even Gaussian primes.

Ted Shifrin

All the more shameful :P

robjohn

indeed

AMDG

18:11

~~So now if we consider what a car is in principle...~~

hyper-neutrino

18:27

full disclosure this is related to a hw problem so i won't ask the question i'll just give the part i'm stuck on and i'd appreciate any pointers
so we have a combinatorial object where $f_n$ is the number of combinations for each $n$ and i've already calculated $\sum_{n \geq 0} f_n x^n$
the next part is to prove something about $\lim_{n \to \infty} \frac{f_n}{n^3}$ being equal to something (that thing is in the problem statement) and the hint provided was about the form of the partial fraction decomp of the sum from the earlier part

Jakobian

that's very cryptic

all I can tell you it means a limit of something divided by $n^3$

Ted Shifrin

Why shouldn't it be meaningful? Presumably, $f_n$ grows like some constant times $n^3$ plus lower order terms.

Partial fractions on a power series doesn't make sense, however.

hyper-neutrino

well, I mostly don't want to give too many details cuz policies lol

Ted Shifrin

I think the point is that $\sum f_n x^n$ is the power series expansion of something like $P(x)/Q(x)$, and they mean the partial fraction decomposition of that.

hyper-neutrino

^ yes exactly that

Ted Shifrin

18:33

So, do it :)

hyper-neutrino

the hint says that we're supposed to find the limit in terms of the decomp first and then only compute values of the decomp that we need, lol

the rational function has 16 terms in its denominator so i'm not a huge fan of doing that directly

Ted Shifrin

LOL

Well, it seems you have pretty explicit hints and I cannot guess any more than I've already said.

hyper-neutrino

yeah i don't think i can really say more. anyway i'll just have to think this through more but thanks

Jakobian

I think the only way for $f_n/n^3$ to exist and be non-zero is for $Q(x)$ to have a triple root at $x = 1$

Ted Shifrin

Wow. That's some insight. I have no idea.

I'm not quite connecting the dots in my head.

Jakobian

18:37

well, it's going to be sum of terms of the form $an^{k-1} z^n$ where $z$ is the root and $k$ is the number of occurrences of that root

for some constants $a$

hyper-neutrino

18:48

oh, looking more carefully at the hint, it mentions the partial frac decomp will be a double sum (the roots and each of their multiplicities) of $\frac{A_{ij}}{(1 - x / \zeta_i)^j}$ for some constant $A_{ij}$ and we're supposed to compute the limit in terms of $A_{ij}$ first and then only grab the values of $A$ that we need
(where $\zeta$ are the roots of the denominator)

i've... never seen that form for partial fractions before...

Koro

19:05

If $\bar{\mathbb N}=\mathbb N\cup\{\infty\}$ and $d(m,n):=|1/m-1/n|, d(n,\infty):=1/n$, where d is metric space. Suppose $f:\bar{\mathbb N}\to \mathbb R$ is a continuous function, then is $\bar{\mathbb N}$ compact?

I could solve a question on Sylow's theorem which I thought I'd never be able to solve :).

The problem is that f is not given to be onto. Had it been onto, the answer would have been in negative.

I think that the only way in this case is to proceed by definition. So let's say $B_i, i\in I$, where I is an index set (it doesn't have to be countable) form an open cover for N bar. I don't know how to existence of finite cover of this covering.

Alex

19:25

$\mathbf{N}\cup \{\infty\}$ is homeomorphic to $\{0\}\cup \{1/n: n\in \mathbf{N}\}$, then $\tilde{\mathbf{N}}$ is compact set.

AMDG

Ah, the set of Spanish naturals.

Jakobian

@Koro why is there an $f$ there?

Koro

@Alex In my case, $\mathbb N$ doesn't contain $0$.

@Jakobian it's part of an MCQ question.

Jakobian

$x\mapsto \frac{1}{x}$ where we understand $1/\infty = 0$ is a homeomorphism between those two spaces

Koro

I'm afraid, I don't know what homeomorphism is.

Jakobian

19:35

it's a continuous map such that its inverse is continuous

compactness is preserved under homemorphisms

Koro

So far, I know topology only from chapter 2 of Rudin's PMA.

Jakobian

it's really basic topology though

Koro

@Jakobian I see. I didn't know that.

what's this result called?

Jakobian

it's not called anything, most properties in topology are preserved under homeomorphisms

because homeomorphisms are basically maps such that they give a bijection between open sets and the underlying sets simultaneously

since compactness is defined in terms of open covers, this instantly shows they are preserved under homeomorphisms

Koro

just quickly looked at Apostol's and 'topological mapping' aka homeomorphism has been defined there :).

Jakobian

19:41

if you defined it in terms of sequential compactness instead then I guess it takes some more flexing, but it's basically the same process, there's nothing really clever about it, it's really just set theory at this point

homeomorphism is the commonly accepted term

Koro

@Jakobian hence compactness is 'topological property'.

:)

Alex

one-point compactification or Alexandroff compactification

Jakobian

I still don't see how $f$ relates to this all @Koro

Koro

Like I said, it's part of an MCQ:
other options are as follows: 2) f is unbounded
3) $f^{-1}\{x_0\}$ is compact.
and I don't recall the 4th option.

yeah, 4)th option was: $f$ is uniformly continuous.

The question was: which of the following is necessarily false?

Jakobian

2) because image of f is compact, so bounded

Koro

19:47

yeah, exactly.

UnderMathUate

Is it annoying to explain what you think an answer might be on an assignment, with a caveat that you couldn't complete the proof? I know an answer mainly by intuition, but I'm struggling to put the last piece together.

I was just going to provide this explanation on my write-up

By annoying, I mean annoying for the professor.

Koro

That's why I asked about proving compactness of N bar.

Alex

@UnderMathUate Mathematics consists of making any recipient familiar with that area understand your reasoning. Personally, I think that if the idea is clear, it should be well received, even if you have not completed your support.

Koro

@Jakobian about 3) however, we can for sure say that $f^{-1}\{x_0\}$ is closed.

Ah, okay.

Every closed subset of a compact set is compact.

@Alex I'm afraid I haven't studied those yet.

Jakobian

Alexandroff compactification, of Alexandroff extension is a way to make a space compact by adding one point to it

A good visualization of this is how $S^n$ is a one-point compactification of $\mathbb{R}^n$ using the stereographic projection

Koro

19:57

@Jakobian I understood. Homeomorphisms map open sets (resp. closed sets) to open sets (resp. closed sets).

Jakobian

yes, bijectively

Koro

@Jakobian for example: adding 0 to {1/n: n =1,2,3,...} in R.

Jakobian

yep

Koro

@Alex I understand it now. :)

Thanks a lot @Jakobian and @Alex.

Jakobian

well, Alexandroff compactification is a specific construction, and those spaces are homeomorphic to it, because every locally compact Hausdorff space has a unique Hausdorff one-point compactification, but it can have multiple non-Hausdorff compactfications

which extend it by a point

Ted Shifrin

22:28

@hyper-neutrino Sure you have. This is just like writing $\displaystyle{\sum_i\sum_{j=1}^{n_i} \frac{A_{ij}}{(x-a_i)^j}}$ when the multiplicity of root $a_i$ is $n_i$.

They wrote $1-x/a_i$ instead for power series purposes.

hyper-neutrino

yeah, after examining it a bit more thoroughly i noticed it was just the same form as usual but slightly modified

also thanks to jakobian's hint and an approach i found it turns out the only root i need care about is 1 so it... doesn't even make a difference if it's x-a or 1-x/a :P

thanks

Ted Shifrin

I still haven't understood where Jakobian's observation came from. But I haven't worked on it.

robjohn

I should probably keep my nose out. I just got back.

Ted Shifrin

Well, hyper was being very careful about rules for getting help on his homework.

hyper-neutrino

@TedShifrin TBH, I didn't get it either. after using an approach via negative binomial theorem i have found that their conclusion is pretty much right (although it actually needs a quadruple root at x=1, at least if i haven't done something wrong)

Ted Shifrin

22:35

Why does the limit of $f_n/n^k$ tells us something about a root at $1$?

So $\sum f_n x^n$ is the power series for this rational function we've written down.

Oh, do we know all the other roots are larger than $1$?

hyper-neutrino

@TedShifrin not sure lol

all of the roots happen to be roots of unity

since the rational function is a product of terms 1/(1-x^k)

Ted Shifrin

Oh, I didn't know you'd told us that.

This is a very complicated problem.

You just said you had a generating function for a combinatorial problem :)

hyper-neutrino

i might not've, in an effort to not give away too much / get too much help, :P

a classmate gave me the hint to just "use neg binom theorem" and that's gotten me through almost the whole problem i think

Ted Shifrin

Well, I'm now curious enough. If you type up your homework problem, I'd love to see your write-up.

Well, you don't need the negative binomial theorem to do $1/(1-x)^k$, but it doesn't hurt.

You can think of it as $\left(\sum x^n\right)^k$ and pick out the coefficient you need.

hyper-neutrino

oh. well i'm never sure the solutions i do are optimal or necessary anyway :P so long as it's valid

Ted Shifrin

22:42

Anyhow, I'm sure @robjohn's curiosity has intensified by now.

robjohn

@TedShifrin looks hard that way

@TedShifrin I'd just take some derivatives of $\frac1{1-x}$

or use the negative binomial theorem

Ted Shifrin

Yeah, taking derivatives is the way I've usually done it.

robjohn

$\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$

So I need to read back about this partial fraction hyper is talking about

Ted Shifrin

It all came from a generating function $\sum f_n x^n$ and wanting the limit $\lim_{n\to\infty} f_n/n^3$.

robjohn

the third integral of the generating function would be the generating function of $\frac{f_{n-3}}{n(n-1)(n-2)}$

robjohn

23:04

if that limit exists and is non-zero, then I'd say that the function has a degree four pole on the unit circle, but no higher.

and possibly it needs to be at $1$

hyper-neutrino

alright, thanks for the insight :) yep, I came to that result as well so that checks out

robjohn

what is it you need to show? or would I need to kill you if you told me?

Xander Henderson

So... I have this dead body... er... animal that I want to keep in my freezer. Is that gonna be a problem?

hyper-neutrino

Hm, I probably shouldn't say more, just to be cautious, so thanks for the tips you've given and I'll just have to work this through; if I don't fully solve it then so be it

@XanderHenderson o_O

Derivative

23:23

Is there a holomorphic function $f:D\to D$ where $D$ is the open unit disk and $f(0)=f'(0)=\dots =f^{(n-1)}(0)=0$ and $f^{(n)}(0)\neq 0$ which is not of the form $\lambda z^n$?

I've been banging my head on this problem all afternoon

robjohn

$z^n+z^{n+1}$

Derivative

has to be surjective, sorry forgot to mention

robjohn

Oh, injective

and surjective, too?

Derivative

only surjective, not injective

Ted Shifrin

Yours doesn’t map to $D$, robjohn.

robjohn

23:26

$f:D\to D$

Um not injectivity, but it maps into $D$

Ted Shifrin

No injectivity?

Derivative

no, it only has to map to $D$

only surjectivity

Ted Shifrin

This sounds like the Schwarz lemma to me.

It cannot be injective if $n>1$.

robjohn

maximum modulus applied to $\frac{f(z)}{z^n}$ perhaps

Ted Shifrin

Yup.

Derivative

23:29

?

leslie townes

or z^{n-1} if you want the modified function to have a zero at 0

robjohn

@TedShifrin yes, I misspoke. I meant into.

Ted Shifrin

Review the proof of the Schwarz lemma, Derivative.

Derivative

hmmm what about it?

Ted Shifrin

Modify it.

Derivative

23:32

okay, in the previous exercise I managed to prove that if $f$ is a function satisfying those conditions then $|f(z)|\leq z^n$ for all $z\in D$ and $|f^{(n)}(0)|\leq n!$ with equality in either iff $f(z)=\lambda z^n$

but I don't know how to construct a surjective function from that

Ted Shifrin

There isn’t a different one.

Derivative

the book asks to produce an example...but it's an error-prone book

geocalc33

does anyone know about deforming a path of integration making it encircle a line so that you can sum up the residues at the poles of the integrand?

does the integrand need decay conditions in order to be able to deform the path/

robjohn

@geocalc33 can you be more specific?

@Derivative an example of a function other than $\lambda z^n$?

Derivative

yes

robjohn

23:42

what is the next chapter about?

Derivative

"sequences, series and products of holomorphic and meromorphic functions"

robjohn

do they talk about Blaschke products?

geocalc33

@robjohn $\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\zeta(sz)\,dz.$

For $0<s<1$, the integrand tends to $0$ rapidly enough when $z\to\infty$ in the half-plane $\Re z\leqslant c$ and out of a neighborhood of the line $L=\{z : \Im z=0\wedge\Re z\leqslant 1/s\}$. This allows us to deform the path of integration, making it encircle $L$

Derivative

idk I haven't read it

geocalc33

we see that $\Phi(s)$ is equal to the (infinite) sum of residues of the integrand at its poles (which are $z=1/s$ and $z=-n$ for nonnegative integers $n$)

Derivative

23:44

but it's not in the index

geocalc33

$c>1/s$

robjohn

@Derivative Hmm... it could be a cruel exercise so that you see that $\lambda z^n$ is the only such function.

Derivative

the exact wording is:
"10. Let $f:D\to D$ be a holomorphic function so that $f(0)=f'(0)=\dots f^{(n-1)}(0)=0$ and $f^{(n)}(0)\neq 0$.
(...)
(c) Give an example of such a function $f$ so that $f(D)=D$ but $|f(z)|<|z|^n$ for all $z\in D$, $z\neq 0$."

robjohn

@geocalc33 do they cover the size of $\Gamma(z)\zeta(z)$ for $\operatorname{Im}(z)\to\infty$?

@Derivative It seems to be impossible, but let me think more.

geocalc33

23:59

@robjohn I don't think so, what's written is everything I have

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