Mathematics - 2024-08-03

Jakobian

12:22 AM

@Obliv looks like the definition of an inner product

psie

6:49 AM

I'm studying order statistics. Let there be a sample of just $2$. Then apparently if $X_1$ and $X_2$ are $\text{Ge}(p)$, then $X_{(1)}$ and $X_{(2)}-X_{(1)}$ are independent. How can this be? Seems like $X_{(2)}-X_{(1)}$ depends on $X_{(1)}$...

Here, as always, $X_{(2)}=\max\{X_1,X_2\}$ and $X_{(1)}=\min\{X_{1},X_{2}\}$.

I should add too that $X_1,X_2$ are independent.

Thomas Finley

7:29 AM

0

Q: How to find the second canonical form of a given hyperbolic equation?

Consider the equation $$ y^2 u_{xx} - x^2 u_{yy} = 0. $$ The method used to solve this problem is given below(, and for the real solution given in the book please scroll down): The solution given in the book is as follows: Here $$ A = y^2, \quad B = 0, \quad C = -x^2. $$ Thus, $$ B^2 - 4AC = ...

I need some help with this.

It's very frustrating when authors state something in such a vague way and then does not bother to come back at it.

user 20458579510081670432

8:07 AM

Author and book please.

psie

I've also got a new question on the main site:

This one.

user 20458579510081670432

Add your attempt at a solution please.

Binky McSquigglebottom

9:21 AM

@SineoftheTime Hi !

Sine of the Time

@BinkyMcSquigglebottom hi

Binky McSquigglebottom

Are you free ?

Sine of the Time

yes

you can ask and then ping me, when I'm free I see if I'm able to help

Binky McSquigglebottom

Okay :⁠-⁠)

What does it mean for a function to be equilimited?

Sine of the Time

I'm not familiar with this definition

context?

Binky McSquigglebottom

9:30 AM

I edited the question

Sine of the Time

you mean a family of functions?

like a sequence?

Binky McSquigglebottom

I'll go into more detail

@SineoftheTime yes

Like f_n(x)= 1/nx

in I=(0,1) Why doesn't it converge uniformly?

And it says because it is an equilimited function

Sine of the Time

$(f_n)_n$ is equilimited if $|f_n(x)|\le M$ for all $n\in \mathbb N$ and $x\in [a,b]$

$f_n:[a,b] \to \mathbb R$

@BinkyMcSquigglebottom $f_n=\frac 1{nx}$ or $f_n=\frac 1n x$

Binky McSquigglebottom

The First

Sine of the Time

9:47 AM

$(f_n)$ is not equilimited (?)

Binky McSquigglebottom

....

I read it wrong, it said that they are not equilimited

@SineoftheTime yes

Sine of the Time

are you having trouble for the uniform convergence?

Binky McSquigglebottom

Yes

Sine of the Time

what does $(f_n)$ converge to (pointwise) ?

Binky McSquigglebottom

at f(x)=0

Sine of the Time

10:02 AM

correct

what is $\sup_{x\in ]0,1[} | \frac 1{nx}|$ ?

Binky McSquigglebottom

Do you find that this point is made by trivially making substitutions?

@SineoftheTime 1/n

Sine of the Time

what substitutions?

@BinkyMcSquigglebottom are you sure?

Binky McSquigglebottom

No

In theory I should put 0 but then it comes out 1/0

So I think it doesn't exist

Sine of the Time

it's $+\infty$

Binky McSquigglebottom

How

Sine of the Time

10:06 AM

because the functions are not bounded on $]0,1[$

for $n=1$, you have $f_1=\frac 1x$

Peter

@BinkyMcSquigglebottom This is true , since $\infty$ is not a number. Nevertheless , usually , it is formulated as "sup=$\infty$" meaning that there is no largest value , so no (finite) supremum.

Binky McSquigglebottom

but in this case doesn't the sup occur for x close to 0?

Peter

@BinkyMcSquigglebottom No, because we can beat every finite value by choosing a suitable $x$ close enough to $0$.

Sine of the Time

since $f_n$ are decreasing, the sup is near $x=0$ but the functions blow up

Binky McSquigglebottom

This would be like making a limit lim x->0+ 1/n•0+

And so sup=+∞

Sine of the Time

10:17 AM

so $(f_n)$ can't converge uniformly

Binky McSquigglebottom

Okay

Can I ask one last question?

@SineoftheTime No, because the sup should tend towards 0

Sine of the Time

@BinkyMcSquigglebottom yes

Binky McSquigglebottom

But when I go to see the pointwise/uniform convergence, do I also have to consider the case in which x depends on n?

Sine of the Time

what do you mean?

do you mean when $\sup_x$ is a function of $n$ ?

Binky McSquigglebottom

No

Mmm like I have f_n(x)=x/n, if I consider x in real numbers, it always converges to 0, but if I consider like x=n^2 I find that it doesn't converge, should I do this last thing?

Sine of the Time

10:25 AM

when you consider the pointwise convergence, you have to fix $x_0 \in \Bbb R$ and then take the limit. If you choose $x_0=n^2$, this is not fixed

Binky McSquigglebottom

Right

But what about that uniform?

I don't think there either because if I'm not mistaken it's fixed there too

Sine of the Time

when you study the uniform convergence you first find the sup and then take the limit

Binky McSquigglebottom

Can you go more in detail pls

Sine of the Time

what's not clear?

Binky McSquigglebottom

I didn't understand how to find the sup

Maybe I'm using the wrong method

Sine of the Time

10:35 AM

there's isn't a general rule

you have to use different methods

if the sup is the max, you can use derivatives for example

Binky McSquigglebottom

No, because I have seen that sometimes the derivative is used to find the maximum even when the maximum is not guaranteed

Like in I=(0,+∞)

And from this I wonder why we didn't do it in the exercise above

Sine of the Time

because you don't need it and it does no work

since $f_n$ are decreasing, the sup is as $x\to 0^+$

Binky McSquigglebottom

Correct

So I never have to consider the case where x depends on n?

Sine of the Time

10:55 AM

no

Binky McSquigglebottom

:⁠-⁠)

Thank you very much for helping

🎉🎉🎉

Sine of the Time

np

Soumik Mukherjee

Sine, Binky see this 🤣

Sine of the Time

@SoumikMukherjee that's what happens when opponent tries to humiliate you

there's the summer marathon today btw

Soumik Mukherjee

@SineoftheTime Yeah

@SineoftheTime are you participating?

Sine of the Time

11:05 AM

I played a couple of games

last year I was in top 500

Soumik Mukherjee

Nice

Sine of the Time

but I played like 200-300 games

Soumik Mukherjee

:|

That's a lot

Sine of the Time

yes it's competitive top 500

because then you have a trophy in your profile

Soumik Mukherjee

Ooh

I will play then

Sine of the Time

11:07 AM

but you have to be top 500 :)

last year I scored 545 points

Binky McSquigglebottom

11:45 AM

@SoumikMukherjee ahaha

Soumik Mukherjee

12:19 PM

@SineoftheTime I started, now I am in top 1800

I am just flagging:)

Sine of the Time

@SoumikMukherjee that's the spirit

Soumik Mukherjee

Nf3 Ng5 Nxf7

Jakobian

3:32 PM

glurbingston

Obliv

4:03 PM

by fixed, does that mean $f$ doesn't change at all for any $\phi$?

this theorem implies the existence of this $f$ for any dual space and its vector space?

Xander Henderson

@Obliv Yes?

But I think that you are kind of thinking about it the wrong way. $F$ lives in the dual space. What the theorem says is that there is a unique element of the original space, $f$, such that the action of $F$ on any vector $\varphi$ is the same as the inner product pairing of $f$ with that $\varphi$.

Obliv

the inner product pairing is the isomorphism right?

i.e., $F(\phi+\psi) = (f,\phi +\psi) = F(\phi)+F(\psi)=(f,\phi)+(f,\psi)$ or something?

Xander Henderson

No...? $\langle f, \varphi\rangle$ is always a real (or complex) number, n'est-ce pas?

It does not send bounded linear functionals to vectors (or vice versa).

Jakobian

The theorem says that any bounded linear functional is determined by an element of $V$

Sine of the Time

linear functionals are inner products

Jakobian

4:12 PM

Obviously, $y\mapsto \langle x, y\rangle$ is always a bounded linear functional for $x\in V$, the Riesz theorem gives the converse

Xander Henderson

@SineoftheTime Not by definition. This requires a proof.

Jakobian

@SineoftheTime not how I'd put it. That can cause misunderstandings and is somewhat confusing

Obliv

@Jakobian that every $f \in F$ can be written in the form $\langle x,y\rangle$ for some $x \in V$?

Jakobian

what is $F$

Xander Henderson

As Jakobian says (minus the "obviously"), the map $f \mapsto \langle f, \varphi\rangle$ is a bounded linear functional. But the theorem says that these are the only such bounded linear functionals.

Obliv

4:14 PM

sry i wrote that wrong. should be $F \in V'$

confusing because $F$ can mean functional but also field

Xander Henderson

@Obliv If your book is using $F$ for both a linear functional and a field, find a better book.

Jakobian

@Obliv if anything, I'd write it as, for every $F\in V'$ there is $x\in V$ such that $F(y) =\langle x, y\rangle$ for all $y\in V$

Obliv

@XanderHenderson it was my mistake because I'm looking at different books at the same time. this one says $V'$ is the dual space and $F$ are functionals. Axler says $\textbf{F}$ is the field

Xander Henderson

@Obliv The standard notation would be something like $\langle x, \cdot\rangle$. The point is that feeding some arbitrary vector to $F$ is the same as feeding that arbitrary vector to $\langle x, \cdot \rangle$ (i.e. taking the inner product pairing of that arbitrary vector and $x$).

Obliv

@Jakobian yeah that's best.

Jakobian

4:16 PM

or another, for every $F\in V'$ there is $x\in V$ such that the functions $F$ and $y\mapsto \langle x, y\rangle$ coincide

Xander Henderson

@Obliv Yeah, I've previously advised folk not to try to learn a new topic out of multiple books at the same time. It causes confusion, not least of which is caused by differences in notation.

Jakobian

I think Obliv might be already dealing with distinct notations if they're trying to apply it to something like physics :P

Xander Henderson

@Jakobian Sure, but why make life harder than you have to?

Also, bra-kets can burn in hell.

;)

Jakobian

I actually agree with that

Sine of the Time

I meant that you can express a linear functional on a Hilbert space as an inner product

Xander Henderson

4:24 PM

Weirdly enough, I don't actually believe that bra-kets should burn in hell---they are clearly useful for people doing physics. And I kind of grok how physicists like to view all linear functionals as "nice" linear functionals which come from "nice" inner product pairings, but it always feels so very, very hand-wavy to me. I am not big brained enough to really get bra-kets.

@SineoftheTime Sure, but, again, that is a theorem.

Obliv

I was under the assumption most of quantum theory was inspired by the understanding of linear algebra at the time of the development of the schrodinger equation

since it was a linear eq.

so besides being handwavy with proofs, most of the theory was built from math and deductive reasoning

Jakobian

@SineoftheTime that's too imprecise to convey important information

and I think it's not useful for me, neither for someone who learns the subject

that's what I object to

a linear functional is not an inner product, its an inner product with some fixed vector

I am proposing there is two kinds of hand-waving in physics. The undergraduate hand-waving, and the graduate hand-waving. The latter, still too imprecise, probably makes sense and is educated. The former, not so much

Obliv

is this double arrow the same thing as iff?

or is there some other meaning

Jakobian

it denotes a bijection I'm pretty sure

Obliv

also this notation wasn't confusing when they were separating vectors from linear functionals but now for some reason they're writing linear functionals in both places.

Jakobian

4:35 PM

they mean the map that sends $x\in V$ to a linear functional $y\mapsto \langle x, y\rangle$

Obliv

I thought they'd at least distinguish them by upper and lower case characters. :\

or maybe a super/sub-script character

Jakobian

they're writing $\langle x|$ to mean the functional $y\mapsto \langle x, y\rangle$

and $|x\rangle$ simply to mean the vector $x$

in similar way, if $F$ is a linear functional, then $\langle F|$ will just be $F$, and $|F\rangle$ will be the element of $V$ as per Riesz theorem

Obliv

@Jakobian Got it

@Jakobian I think this is basically what u just said

Jakobian

I don't see how this is the same

they're just saying that you can treat $\langle F|\phi\rangle $ as $\langle F, \phi\rangle$ but there are situations where it matters to not do that

Obliv

oh true I guess it fails to explicitly remind the reader that it's an inner product on a fixed vector on the space of ket vectors.

Jakobian

4:45 PM

I wouldn't say it fails since even though it didn't mention details of Riesz theorem, it still says what it wants to say

psie

> Exercise Suppose $n$ points are chosen uniformly and independently of each other on the unit disc. Compute the expected value of the area of the annulus obtained by drawing circles through the extremes.

I just want to check if I understood the question correctly. We generate $n$ points in the unit circle, and then we look at $X_{(1)}=\min\{|X_1|,\ldots,|X_n|\}$ and $X_{(n)}=\max\{|X_1|,\ldots,|X_n|\}$. Then we draw circles, centered at the origin, and going through $X_{(1)}$ and $X_{(n)}$ respectively. Is that right?

Wait a minute...

Jakobian

@psie the question does seem to talk about $X_{(1)}$ and $X_{(n)}$ you just defined, but the drawing part, of course this is just about the area

psie

"...going through $X_{(1)}$ and $X_{(n)}$..." doesn't make sense.

Jakobian

the circles with radius $X_{(1)}$ and $X_{(n)}$ respectively will go through the points of least and most absolute value

psie

ah ok, thanks! Now this is clearer

Jakobian

4:49 PM

maybe better notation would be to call it $r$ and $R$ because those are not order statistics of $X_1, ..., X_n$

psie

yeah

Obliv

is a linear operator and a vector space automorphism the same thing?

(or module automorphism if vector space isn't an abstract algebra term)

psie

@Jakobian maybe I shouldn't go there (for this problem), but for $X_1,\ldots,X_n$ being two dimensional random vectors, is there such a thing as order statistics? As there is no commonly agreed order in $\mathbb R^2$ I guess talking about $\min$ and $\max$ wouldn't make sense. Feel free to ignore this :)

Obliv

also why are they called operators and not functions? Does 'operator' suggest we're talking about linear vector spaces only?

Jakobian

@Obliv yes

a linear operator is sometimes referred to as a linear function $T:V\to V$ where the domain and codomain are the same

other times its just any linear function $T:W\to V$ between two vector spaces

a linear automorphism is a bijection $T:V\to V$ where domain and codomain is the same

this forces $T^{-1}$ to also be a linear automorphism

the difference between the first definition of linear operator and linear automorphism is that the latter is bijective

@Jakobian my "yes" is supposed to be a "no"

@Obliv operator sometimes means we mean maps $V\to V$

they don't have to be linear in general, although the word is mostly for linear stuff

Hardy–Littlewood maximal function

In mathematics, the Hardy–Littlewood maximal operator M is a significant non-linear operator used in real analysis and harmonic analysis. == Definition == The operator takes a locally integrable function f : Rd → C and returns another function Mf. For any point x ∈ Rd, the function Mf returns the maximum of a set of reals, namely the set of average values of f for all the balls B(x, r) of any radius r at x. Formally, M f ( x ) = sup r > 0...

for example the Hardy-Littlewood maximal operator isn't linear

@psie given that the definition of order statistics depends on the order, I would say no

I'm not an expert on different generalizations of what order statistics means, but I doubt they use them for anything more general than the case of $\mathbb{R}$-valued random variables

(possible, that they generalize them in different ways than just depending on the order)

psie

5:09 PM

Ok. If $r_1=\min\{|X_1|,\ldots,|X_n|\}$ and $r_n=\max\{|X_1|,\ldots,|X_n|\}$, I should compute $E (\pi(r_n^2-r_1^2))$, right?

still having some doubts on what I actually need to do

Jakobian

@psie yep

psie

ok, great

and I think, if one generates a point $X$ uniformly in the unit disc, $|X|$ will not be uniformly distributed in $[0,1]$. Intuitively, this is clear, but it is tempting to think that way

Jakobian

Yes, and you can see this pretty clearly, for $a\leq |X|\leq b$ happens with probability equal to the area of the annulus $a\leq x\leq b$, which varies in a non-uniform way as $a, b$ vary

In fact its clear that $P(|X|\leq t) = t^2$ for $t\in [0, 1]$, and with obvious values otherwise

psie

Indeed

Obliv

5:33 PM

how does operating with $\langle u_i|$ transform the left side into the right

$M$ is a square matrix here? In between a basis vector and the linear functional associated with that basis vector?

by the way as a mathematical aside, do all linear vector spaces have a set of orthonormal basis vectors? Or is there just the guarantee of a set of basis vectors to span the space

I'd imagine only for discrete vector spaces, maybe only by relaxing the definition to 'discrete' modules

Soumik Mukherjee

I started reading The infinite dimensional topology of function spaces by J. van Mill.

Obliv

6:05 PM

@Obliv nvm, it's just an application of riesz theorem

Jakobian

@SoumikMukherjee cool!

Obliv

well I understand how $M|u_j\rangle a_j \vdash |u_k\rangle b_k$ but not how we get the form of a matrix equation in 1.11

how do we represent the inner product $\langle u_i|M|u_k\rangle$ as a matrix?

Soumik Mukherjee

6:45 PM

@Jakobian You suggested this book long ago:), the exposition is very clear, thanks

Obliv

7:06 PM

what is the backward shift operator on $F^{\infty}$?

Is that like mapping each index of the tuple to the one before it?

Or I guess it would be the other way around since idk how i'd map the first index. i.e., $(x_1,x_2,...) \mapsto (x_2,x_3,...)$

leslie townes

obliv: yes, something like that is often called the backward shift

the 'forward shift' or just the 'shift' goes the other way and you add a 0 in the first entry (the only fixed scalar that you could add, to make a linear map)

if S is the forward shift and T is the backward shift defined via your formula then TS is the identity but ST is not, odd little difference from finite dimensional linear algebra

Obliv

7:22 PM

oh okay, thank you. I'm guessing $0$ is just the standard, but you could theoretically put any element of the field in there

@leslietownes because $0$ replaces the first element after ST

?

kind of just deletes $x_1$

leslie townes

well, assuming you wanted a linear map, you couldn't put any other fixed element of F there. the zero vector has to go to the zero vector

yes

Obliv

Oh true

Obliv

8:17 PM

@leslietownes are matrix representations of a linear map unique?

leslie townes

if you specify enough ambient context (e.g. fix bases of the space that the map acts on), which people often do, and which some people do so frequently they don't realize they are doing it, they certainly can be

so in the context of the image like you pasted above, the answer would be, that M is uniquely represented with respect to that chosen orthonormal basis by that matrix

but choose different orthonormal bases and you will generally get different matrix entries M_ij

Obliv

what if the two spaces have differing size of bases, can that happen?

Claudio

@leslietownes but if you prove that your matrix commutes with every invertible matrix then it's also unique

leslie townes

sure, there are interesting special cases like scalar multiples of the identity where the matrix does not change

obliv: can what happen?

Claudio

@leslietownes yep like $a \mathbb{I}, a = const$

@Obliv if you have a map from $\mathbb{R}^n \to \mathbb{R}^n$, and you determined a basis of the space how can that happen?

If I understand what you're trying to say that would mean that the dimension of the space has changed

anyways, I came here to ask a question about a sentence in this paper I'm reading. Please do not make fun of me :P I'm already ashamed to even ask this in the first place

arxiv.org/pdf/2310.19665 It is said that: The prescribed recipe works in all cases except when the Z-axis is in the direction of the south pole, so we exclude this case for the time being

Obliv

8:29 PM

Claudio

I don't see why this could be true. It's clear to me that, at the south pole, I can still make sure that the x-axis is tangent to the 0° meridian, unless the poles are somehow 'ill-defined points' on the sphere, thus creating problems

@Obliv I had this anwer saved btw math.stackexchange.com/a/4774862/1096913

Obliv

Nvm, I figured it out. I was thinking M(T) had to represent T inverse as well

so I was confused if M(T) wasnt a square matrix, then it'd be unable to account for different size bases

@Claudio thanks

Claudio

you're welcome

monoidaltransform

does every compact metric space embeds isometrically into hilbert cube?

Jakobian

8:47 PM

@monoidaltransform what do you mean by isometrically

there is no "the Hilbert cube" as a metric space

Hilbert cube is simply $[0, 1]^{\aleph_0}$ and as a metric space it can be given a lot of different metrics

it can also be treated as a certain subset of $\ell^2$

monoidaltransform

I meant, does there exists a metric on the hilbert cube

compatible with its topology

so that the emebedding is an isometric one?

Jakobian

yes, there exist a lot of metrics on the Hilbert cube

so you mean to extend that metric

monoidaltransform

So that the embedding is isometric..?

yup

Jakobian

I need to think

I believe the answer is yes

monoidaltransform

Could you elaborate on why you think that is?

Jakobian

8:54 PM

because you can extend continuous functions from a compact subspace, and there should be a theorem like that for when you can extend continuous functions, you can also extend pseudometrics

and by obvious adjustment this becomes a metric on the Hilbert cube

eudml.org/doc/264771

according to this article, if we got a metric subspace $(X, d)\subseteq [0, 1]^{\aleph_0}$ of the Hilbert cube, then one can extend $d$ to a continuous pseudometric $\tilde{d}$ on the Hilbert cube

okay my reasoning with the extension can be made a metric was somewhat flawed

actually that there exists such pseudometric is something I even know how to prove

actually no, I don't think I can prove something like that

monoidaltransform

9:13 PM

ofcourse, there exists an infinite dimensional metric space $Z$ such that $X$ embeds isometrically into. But im interested in HC

Jakobian

@monoidaltransform apparently, Hausdorff proved that you can extend a continuous metric from a closed subspace, to a continuous metric on the whole space. This doesn't answer your question however, since the metrics have to be admissible as well

monoidaltransform

Admissible metric? Are these natural @Jakobian

Jakobian

oh no it does, since for compact spaces a continuous metric is admissible

@monoidaltransform it means that the topology it generates coincides with the topology of the topological space

i.e. a metric $d:X\times X\to \mathbb{R}$ on a topological space $X$ is admissible if the open balls with respect to $d$, $B_d(x, r)$, form a basis of $X$

saying that such metric is continuous only gives you that $B_d(x, r)$ must be open

but since we are dealing with a compact space, thankfully we can stop at this and say it must be admissible

monoidaltransform

So since $(X,d)$ embeds into $H$ topologically, as a closed subset, there exists a metric d' on $H$ such that $(X,d)$ isometrically embeds into $(H,d')$ because $d$ is admissible on $X$, since $X$ is compact

?

Jakobian

that $d$ is admissible on $X$ is without a question, its that $d'$ is admissible on $H$ that is important

Hausdorff's theorem above only gives you continuity of $d'$, while compactness gives that its admissible

monoidaltransform

9:19 PM

So there exists an admissible metric $d'$ on $H$ such that $(X,d)$ embeds iso into $(H,d')$ and $d'$ is a metric because $H$ is compact

right?

Jakobian

yes

but not $d'$ is a metric because $H$ is compact

$d'$ is admissible because $H$ is compact. The fact that $d'$ is a metric will always hold, even if $H$ isn't compact

monoidaltransform

I meant $d'$ is a metric compatible with the topology of $H$

Jakobian

then yes

monoidaltransform

okay perfect thank you, if you want to write this as an answer in main, ill accept it without question

(Ive asked it there)

psie

I am working the following problem;

> Exercise: Independent repetitions of an experiment are performed. $A$ is an event that occurs with probability $p$, $0<p<1$. Let $T_k$ be the number of the performance at which $A$ occurs the $k$th time, $k=1,2,\ldots$. Compute (a) $E(T_3\mid T_1=5)$ (b) $E(T_1\mid T_3=5)$.

It is clear $T_1\in \text{Ge}(p)$, but how do I determine the distributions of $T_2$ and $T_3$?

monoidaltransform

9:27 PM

@Jakobian also, do you know if the hilbert cube an ANR?

Yes it is turns out it is true, but what about C(X,\mathbb{R})

Jakobian

@monoidaltransform I think a product of ANR's should be an ANR

monoidaltransform

yup

Jakobian

this holds at least partially

but countable products of AR's are AR's which $[0, 1]$ is

monoidaltransform

Ah I see

Jakobian

which is a stronger property, as AR's are contractible ANR's

monoidaltransform

9:37 PM

what about C(X,\mathbb{R})? continuous functions X->R. Atleast when X is compact

@Jakobian Also, could I ask what reference is this from?

Jakobian

I think its an AR because its a Banach space

@monoidaltransform Infinite-dimensional topology by van Mill

monoidaltransform

Thank you

Jakobian

@monoidaltransform every convex subspace of a locally convex topological vector space is an absolute retract

amazon.com/dp/0444815058

monoidaltransform

Ah interesting

class of ANR's is huge then

Jakobian

here the subspace is $C(X, \mathbb{R})$ itself

@monoidaltransform van Mill does everything for separable metrizable spaces by the way. So it might not exactly hold in this form in general

monoidaltransform

9:46 PM

Two close enough maps into an ANR are homotopic, I was wondering if one could somehow improve this to homotopy equivalence, but it was a random thought maybe its highly false

Jakobian

homotopy equivalence of what?

monoidaltransform

of the domain space of the maps

and the ANR

Sine of the Time

INFINITY GUN

►

zeta space

$I(s)=\zeta(s)+sf(s)$ If I want to analytical continue a function defined by an integral $I(s)$ how would I do so? Assume $\zeta(s)$ is the Riemann zetas function convergent for real $s>1$ and $f(s)$ is some real analytic function convergent for real $s>1$

$I(s)$ is clearly real analytic (as the sum of real analytic functions is real analytic). I thought I might analytically continue $\zeta(s)$ and then $f(s)$ seperately but I am not certain that would work

zeta space

10:28 PM

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A: Is there a closed form for this limit? Is is exactly $1/2$?

By Tonelli's theorem we can move the summation outside of the integral $$ L=\lim_{k\to \infty} \frac{\sum_{n=1}^k \int_0^{e^{-\sqrt{\log n}}}e^{\frac{\log n}{\log r}}~dr}{\sum_{n=1}^k\int_0^1 e^{\frac{\log n}{\log r}}~dr} $$ The numerator can be realized as $$ \sum_{n=1}^k \bigg(\frac{1}{2}e^{-...

a crisp clean proof

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