Mathematics - 2023-03-23

one potato two potato

12:33 AM

@TedShifrin It's just a meromorphic differential.

$M$ is a compact Riemann surface btw

Xander Henderson

12:45 AM

@mick I have edited the title of your question to remove the very large, vertical-space-stretching expression.

PLease feel free to further edit the title, but note that tall titles are very much frowned upon.

Ted Shifrin

@onepotatotwopotato Yes, so if $\omega$ is a nontrivial meromorphic differential p, then $[\omega]$ is the canonical class and $[\omega^{\otimes k}]$ is the $k$-canonicsl class.

one potato two potato

1:08 AM

@TedShifrin If I write $\omega = f(z)dz$ in local form for meromorphic $f$, then $\omega^{\otimes k} = f^k(z)dz^k$?

Ted Shifrin

Yup.

one potato two potato

Thanks!

Ted Shifrin

Sure thing.

D.C. the III

2:43 AM

Actually being able to apply Cayley Hamilton to questions makes things nice....

one potato two potato

3:20 AM

0

Q: Riemann-Roch theorem for general divisor

$\newcommand{\u}{\mathfrak{U}}$I have a question about the proof of Riemann-Roch theorem in Farkas&Kra Riemann surfaces. Theorem. The Riemann-Roch theorem holds for every divisor on a compact surface $M$. Proof. We write the divisor $\u$ as $$\u = \u_1/\u_2,$$ with $\u_j(j=1,2)$ integral and th...

A minor question about the proof of Riemann-Roch theorem...

Prateek Mourya

3:33 AM

what does this means can anyone explain me with an example

D.C. the III

@PrateekMourya try writing out what the matrix $e_{i,j}$ would look like for a small $m$ and $n$. for example, write out what the matrices look like for $m = 3$ and $n = 2$

Prateek Mourya

3:50 AM

matrix eij?

how i have absolutely no idea

@D.C.theIII

D.C. the III

Did you read the statement? What does the statement say?

Prateek Mourya

if eij belongs to Mmn

but wat does belonging actually mean

D.C. the III

what would the set of matrices $\{ e_{ij}: 1 \leq i \leq 3, 1 \leq j \leq 2 \}$ look like?

belonging means that it is an object in the set

Prateek Mourya

what would be the elemnts of eij

ah yes

1 for ij and 0 for ohter

D.C. the III

in this case the matrices of $e_{ij}$ are objects of the set of matrices $M_{m \times n}$

Poline Sandra

3:55 AM

Just say that you don't have the answer thank you

Prateek Mourya

then to construct any matrix of Mmn we only need eij

D.C. the III

@PolineSandra No...that's not how things work on here. Nobody is entitled to answer your question because you demand it

Prateek Mourya

is this correct?

1 min ago, by Prateek Mourya

then to construct any matrix of Mmn we only need eij

D.C. the III

@PrateekMourya Don't worry about the general $e_{ij}$ just answer my question with $m = 3$ and $n = 2$ tell me what some of these matrices look like. Write out a couple

Poline Sandra

I ask about a composition it means that I don't know from where to start and I ask here to a discussion not to take the answer and go

D.C. the III

3:58 AM

you did not say anything that you tried and you did not write any ideas that you may think are related to the question

Prateek Mourya

a matrix 3 rows and 2 columns and whose element is 1 in ith row and jth column all others are 0

well i got it

how can we construct any matrix of Mmn

D.C. the III

@PrateekMourya good. Now do you see where to go from there?

Prateek Mourya

yess

thanks

D.C. the III

for the general case it does get messy. So don't get too frustrated if you can't present it perfectly. But write out the idea of what is happening and why

Poline Sandra

@D.C.theIII I ask a theoretical question to start a conversation if u is in E and f in E' does the composition can be correct

D.C. the III

4:03 AM

@PolineSandra I was talking about your question posted on MSE not in the chat

but the same idea applies... what have you tried?

Poline Sandra

And I think that we can do this composition...

Ted Shifrin

@D.C.theIII Obliged, not entitled.

D.C. the III

NOw I can't help you specifically with the question because I haven't taken measure theory, but you have to show what you have thought of if you want help @PolineSandra

@TedShifrin 8 hrs of math will have you forgetting proper English

Ted Shifrin

uh huh

Poline Sandra

I know that you don't have idea

leslie townes

4:13 AM

ted: munchkin met a younger child at the pool today. the kid tried to talk to her and she wouldn't respond. after the kid left i asked "why didn't you talk to that kid?" "he was talking wrong." "he was two and a half." "he was saying, bah bah bah mah mah." "he said his name and the name of the toy he had." "he couldn't talk right."

Ted Shifrin

Vindictive as well as judgmental. Where does she get these traits?

leslie townes

her mother is A Lot

Prateek Mourya

5:10 AM

can anyone explain this by giving an exaple

@TedShifrin @D.C.theIII

D.C. the III

Your textbook does not give you examples?

leslie townes

that's not the easiest thing to explain by example. if A' is made by applying row operations to A, then the rows of A' are linear combinations of the rows of A. i would understand that by case analysis, one case for each of the various types of row operation (some books categorize these differently, but there are usually only two or three or four of them).

it also looks kinda like you've even maybe cropped out an attempt at an explanation of the thing you haven't cropped out, but whatever.

Koro

I was able to sleep just fine yesterday :-).

D.C. the III

BAd time to get sick Koro....end of the semester?

Koro

I think my loss of sleep was due to cough.

Poline Sandra

5:24 AM

Hello , if f is in the dual space of a Banach space how we write it's primitive is it $F(t)=\int_0^t f(s) ds$ ?

D.C. the III

Lol...

Koro

@D.C.theIII very bad time, some class tests are on the way.

I feel much better now.

D.C. the III

Whirlwind of stressing about covering all the material before tests come. I'm guessing your professors are not sympathetic to you being sick?

Koro

Ah no, it has nothing to do with the professors really. It's manageable :-).

D.C. the III

Cool

one potato two potato

7:11 AM

If $f:[0,1]\to\Bbb R$ is a smooth function then $\sum_{k=1}^n|f(x_k)-f(x_{k-1})|\leq\int_0^1|f(x)|dx$ where RHS is a Riemann integral? $0<x_0<\cdots<x_n =1$ is a partition of $[0,1]$.

leslie townes

7:28 AM

do you mean 0 = x_0 at the beginning of that? anyway, no. try f(x) = 4x(1-x) and the partition 0 < 1/2 < 1

Vrouvrou

7:40 AM

@leslietownes hello

leslie townes

hi

Vrouvrou

You have some notions on measure right ?

I remember you help me before, please what do you think about this question math.stackexchange.com/questions/4664643/…

Is F exists and satisfies the inequality?

leslie townes

yeah, i dunno. this looks like some differential equations thing. i never really studied that. i don't recognize the names of these function spaces, let alone understand the problem.

someone was asking about something similar earlier, it might be in scrollback. or there was question on MSE about it.

Vrouvrou

Ok thank you

The space are Sobolev spaces

one potato two potato

8:47 AM

@leslietownes It's from math.stackexchange.com/q/2082709/668308 If $f$ is absolutely continuous on $[c,d]$ then $T_f = \int_{c}^d|f'(x)|dx$. $T_f$ is a total variation of $f$.

Oh I should've write $\int_0^1|f'(x)|dx$ Sorry.

Franklin

12:58 PM

Can anyone please help me with this: math.stackexchange.com/questions/4664815/… ?

robjohn

@Franklin can you ever pair an odd with an odd and not pair an even with an even?

Franklin

@robjohn I have edited my solution! Mind taking a look at it ?

Semiclassical

1:16 PM

@TedShifrin got a question for your linear algebra brain: can you think of a good way visualize the action of this matrix? $M=\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{pmatrix}$

It’s easy enough to do this algebraically but I’m trying to find something more geometrically intuitive

Astyx

It permutes two points

But I'm guessing that's not what you want

Is it not a reflection along the hyperplane $x_3-x_4 =0$?

Sine of the Time

it's orthogonal and the determinant is -1

User1974

1:58 PM

I have an extremely dumb question. If there is one boy and three girls, is the ratio of boys to girls 1:3 and the percentage of boys 25%? (1/(1+3))

Franklin

2:16 PM

Can anyone please help me prove : The product of two dedekind cut is a cut as well.

shintuku

@Franklin whats your definition of the product of a dedekind cut

Franklin

@shintuku Well, if $A$ and $B$ are two dedekind cuts then their product $A.B=\{pq| p\in A\space \text{and }\space q\in B, p,q\geq 0\} \cup \Bbb{Q}^-$. This is the definition I use...

This is the definition of cut I use (I am posting this to avoid any confusion):

A set of $\alpha$ of rational numbers is said to be a cut if : (i) $\alpha\neq \emptyset, \alpha\neq \Bbb Q$ (ii) if $p\in\alpha$ and $q\in \Bbb Q$ with $q<p$ then $q\in\alpha$

(iii) if $p\in \alpha$ then $\exists q\in\Bbb Q$ with $q>p$ such that $q\in\alpha.$ The condition (ii) shows that if $\alpha$ contains a rational number then it contains all rational number preceeding it. The third condition shows that the set has no largest rational. If $\alpha=\{x\in \Bbb Q: x<r \space \text{and}\space r\in\Bbb Q\}$ then $\alpha$ is a cut and we denote $\alpha=r^*.$

Now, I want to prove, The product of two cuts is a cut as well, but I am unable to do it

shintuku

why does the product definition need a union with the negative rationals

nevermind

Franklin

@shintuku Actually, that's how we were taught in our college :)

shintuku

2:31 PM

can you show the product set is nonempty

Franklin

Yeah, but surely one might define that in an alternative way, to make it less convoluted...

shintuku

can you show the product set is nonempty and is not equal to the rationals

Franklin

Yes

@shintuku I could do that....but for the rest...I dont know how to proceed...

shintuku

let $p \in A\cdot B$ and $q \in \Bbb Q$ with $ q< p$. Can you show $q \in A\cdot B$?

Suppose $q$ is negative

Franklin

@shintuku No, that's the problem. If q is negative then it is in A.B

But if q>=0

That's the part I am confused about...

shintuku

2:35 PM

suppose $q = 0$

Franklin

If q is negative then it is in A.B as $q\in\Bbb Q^-$

shintuku

yes, now what about $q = 0$

Franklin

@shintuku If q=0, then we know that, p>q, so, p=ab where a\in A and b\in B such that a,b>0. Thus, 0<a and 0<b due to which, 0\in A and 0\in B. So, 0\in A.B

I could only think of this argument for q=0....

shintuku

your definition of product defined $a,b \geq 0$, not $a,b >0$

can you show $0 \in 2 \cdot 5$?

Franklin

@shintuku can't think of any other way 😕?

@shintuku yeah, I missed it...

shintuku

2:50 PM

can you give some elements of $2 \cdot 5$?

Franklin

@shintuku \Bbb {Q}^- is in 2\cdot 5. Some positive elements in it: 1,2.25, ...

0 is in 2.5

That's from the definition of a cut

shintuku

is $0 \in 2$? and is $0 \in 5$?

Franklin

@shintuku yeah surely!

shintuku

so is $0 \times 0$ in $2 \cdot 5$?

Franklin

@shintuku yup

shintuku

2:55 PM

so $0 \in 2\cdot 5$

Franklin

@shintuku all good...

shintuku

so prove for general $A \cdot B$ that $0 \in A \cdot B$

Franklin

@shintuku so the obvious assumtion to be taken is A and B are essentially postive cuts, i.e A,B>0

,right?

shintuku

you've proven in step 1 that $A \cdot B$ is nonempty, this is only possible if $A,B$ are positive

Franklin

@shintuku one thing, that bothers me is that, A and B are non-empty, then following our definition of a product, Q^- is always in A\cdot B, irrespective whether A , B are positive or negative

shintuku

3:00 PM

because you first define positive multiplication before negative multiplication

your definition of multiplication doesn't allow negative multiplication

Franklin

@shintuku I dont get it. Why? What I tried to mean is: since, Q^- is always in A\cdot B, irrespective whether A , B are positive or negative so, A\cdot B is always non-empty.

Koro

0

Q: For a subspace $A$ of $X$, defining relative homology group $H_n(X,A)$.

So we define $H_n(X,A):=C_n(X)/C_n(A)$,the quotient space, where $C_n(X):=$ free Abelian group generated by all singular $n-$maps $\Delta^n\to X$. I don't understand the following: Elements of $H_n(X,A)$ are represented by relative cycles: $n-$ chains $\alpha \in C_n(X)$ such that $∂α \in C_{n−1}...

algebraic-topology

Franklin

Is something wrong with this argument?

shintuku

@Franklin oh my bad, that is right

Franklin

@shintuku no problem. You are helping me, I should be the one grateful to you....

shintuku

3:05 PM

so instead, we know it is a definition for positive multiplication because $p \in A$, $q \in B$, and $p,q \geq 0$

Franklin

@shintuku you are following up from here ?

shintuku

yes, prove $0 \in A\cdot B$ knowing $A,B > 0$

Franklin

@shintuku Yes, so, 0\in A and 0\in B as A and B , both are cuts and they both are positive. So, $0\cdot 0=0\inA.B$

Thus, $0\in A\cdot B$

shintuku

right

now do the general second condition for cuts

Franklin

@shintuku but what if $q>0$ ? That's the major hurdle in my opinion!

I tried to use someratio, but no favorable results was derived

*some ratio

shintuku

3:22 PM

$1 \in 2\cdot 5$?

Franklin

@shintuku yes

shintuku

why

parz

I’d like to say hi but am scared of $dollar signs$ stabbing me.

Franklin

@shintuku as 1\in 2 and 1\in 5

shintuku

$3 \in 2 \cdot 5$?

and $6 \in 2\cdot 5$?

Franklin

3:27 PM

@shintuku yes

shintuku

why

Franklin

@shintuku 1\in 2 and 3\in 5...

shintuku

ok what about $6 \in 2 \cdot 5$

Franklin

@shintuku yes

shintuku

why

Franklin

3:29 PM

As 1.5 \in 2 and 4\in 5

shintuku

ok well now do the general case

Xander Henderson

@parz tinyurl.com/cfqcvpc

Ted Shifrin

3:43 PM

@Semiclassical Astyx gave you the answer I would.

Franklin

@shintuku If this lemma: "if p>q and p=ab such that p,q,ab\in \Bbb Q then there exists c,d \in \Bbb Q such that c<a and d<b and cd=q" holds good, then it's trivial.

I should rephrase that as a conjecture though.

But the question is, does this statement really hold ?

shintuku

edit: evaluating

still evaluating

looks true

Franklin

4:03 PM

@shintuku then it's proven! I mean the multiplication is a cut trivially follows from this , right ?

shintuku

yeah but you have to prove that statement

Franklin

@shintuku now, how to do it ? Any ideas ?

shintuku

maybe get what proportion of $q$ is $p$, and require at least that proportion from combined $cd$

Franklin

@shintuku Hmm...I might right this as a standard result... but is there any way without using this ?

What would your approach be ?

shintuku

that one i just said, but i don't know if it works

Franklin

4:09 PM

@shintuku I see.

I googled this statement for a proof : "multiplication of 2 cuts is a cut"

@shintuku But stupid results!

That's frustrating, you know!

shintuku

if $q$ is $50\%$ of $p$, then try getting $c$ is $50\%$ of $a$, and $d$ is $50\%$ of $b$

Franklin

@shintuku well, that's a nice way to start, but even if we do it, I admit, we do get some sort of an idea of what's going on, but then again we have to prove it for a more general proportion.

I might have a fun experiment with Chat GPT. Let's see...

Whether he knows something 😂😂😂

@shintuku It generates irrelevant results!!!

Koro

4:37 PM

it answered something that has no answer.

Franklin

ChatGPT has become disgusting ! It always was though...

Koro

imagine IVR calls every now and then to annoy us

user4539917

6:09 PM

@User1974 Yes.

Semiclassical

6:28 PM

@Astyx it is. What I have in mind is that, under an appropriate 4d rotation, it should instead be reflection along the $x_1’-x_2’=0$ hyperplane. But said that way it feels pretty weak sauce

Ted Shifrin

@Semiclassic That's way less informative. Any hyperplane would do.

Personally, I say reflection across a hyperplane, rather than along.

Astyx

I look myself along the mirror

Semiclassical

Fair enough

I do have a specific rotation in mind, but my brain is a bit dead at the moment. Here’s an attempt tho

First, I think of the four basis vectors as being the tensor products of the two basis vectors in RR^2

Astyx

Any rotation that sends the normal of the first hyperplane to the normal of the other one will do

Semiclassical

The claim is that, if I apply a 45 degree rotation to these two basis vectors, then I can tensor these to get another orthonormal basis for RR^4

And that M still just permutes two of these new basis vectors

Ted Shifrin

6:35 PM

$(e_1+e_2)\otimes (e_1-e_2) = e_1\otimes e_1 - e_1\otimes e_2 + e_2\otimes e_1 - e_2\otimes e_2$. Say what?

Semiclassical

That corresponds to the vector (1,-1,1,-1) in RR^4, so the M that I cited originally maps it to (1,-1,-1,1). But that’s $e_1\otimes e_1-e_1\otimes e_2-e_2\otimes e_1+ e_2\otimes e_2=(e_1-e_2)\otimes (e_1-e_2).$

By contrast, $M$ instead maps $e_2\otimes e_1$ to $e_2\times e_2$. So in one basis the action is seemingly on the second factor whereas in another it’s seemingly on the first.

Ted Shifrin

6:50 PM

Is there a point to all this?

Semiclassical

Given that this is the CNOT gate in quantum computing, yes, but evidently I can’t find a way to make it interesting

leslie townes

ha ha, snot. haha.

Ted Shifrin

Is there a reason that $CNOT$ should be idempotent?

leslie townes

cnot that i can c, ted.

Semiclassical

Yeah. Controller-not gate: if you run it twice, the target bit gets flipped twice or not at all. But the point is that which bit is the target vs the control bit swaps under the rotation

Ted Shifrin

7:01 PM

Where did rotation come from? We're talking about reflection?

geocalc33

7:17 PM

you know how the surface area and volume of a sphere are related?

Does the relationship hold if you deform the sphere?

that is, the derivative of the volume expression is the surface area expression

for example take the notorious example

S=4pir^2

V=4/3(pir^3)

geocalc33

7:36 PM

I love when an amateur asks a pretty good question and an expert gives an answer saying "it's not rocket science" lol

well it actually is rocket science for the amateur..it's not rocket science for you

PM 2Ring

@geocalc33 Nope. In 2D, the area of an ellipse is trivial, but the perimeter is an elliptic integral.

And of course, ellipse calculus is part of orbital mechanics, hence rocket science. :)

Ted Shifrin

Derivative with respect to what?

PM 2Ring

Elliptic integrals don't have closed-form solutions in terms of the standard elementary functions. However, Gauss & others found expressions for them in terms of the AGM (Arithmetic-Geometric Mean), so they're easy to evaluate to high precision, assuming you can do sqrt. But they were a bit painful to evaluate back in the days of log tables.

With respect to the radius, I assume.

We had a question along those lines a week or so ago:

54

Q: Why isn't the derivative of the volume of the cone its surface area?

The derivative of the volume of a sphere, with respect to its radius, gives its surface area. I understand that this is because given an infinitesimal increase in radius, the change in volume will only occur at the surface. Similarly, the derivative of a circle's area with respect to its radius g...

calculus geometry area volume

Ted Shifrin

7:54 PM

And we all know that the area if a square is $s^2$ and the perimeter is $2s$. I’ve been telling calculus students this for decades!

@PM2Ring what’s radius?

PM 2Ring

@TedShifrin Oh, ok. Not the radius, the semi-major axis.

Or maybe some function involving the semi-major axis and the eccentricity.

That covers ellipses and simple ellipsoids of revolution. Fortunately, in astronomy we rarely need to deal with triaxial ellipsoids. But there are a couple of bodies in the solar system with sufficiently high rotation rates that they are triaxial.

And yes, I realise that there are various ways to define the mean radius of an elliptical orbit. astronomy.stackexchange.com/a/47107/16685

geocalc33

8:18 PM

okay

that's very interesting

PM 2Ring

Gauss & Lagrange did some lovely stuff with ellipsoids. They were trying to figure out how to do geodesy on an ellipsoidal model of the Earth, rather than on the traditional sphere. That paved the way to more accurate surveying & map-making, and we still use an ellipsoid, WGS84, in GPS. en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid

geocalc33

I'm looking to define a new function that is based on the perimeter not of an ellipse, but on the perimeter of $u^2+y^2=R$ where $u=\log x$ and $v=\log y.$ Mainly I just want to explore this via numerical techniques/programming.

with the help of Thorgott I calculated the volume of this

but the surface area is probably radically different

PM 2Ring

Did you mean $u^2+v^2=R$ ?

geocalc33

I was naively thinking that the volume was related to the surface area

@PM2Ring yes

PM 2Ring

8:35 PM

You're going to need differential geometry for that. en.wikipedia.org/wiki/First_fundamental_form Fortunately, Ted knows a thing or two about that. ;)

geocalc33

yes indeed

but say I wanted to plot it

I think I could do it in sage cell

I mean plot the perimeter function

PM 2Ring

Yes, it should be easy to do in Sage. And Sage has lots of nice stuff for dealing with parametric surfaces.

You can give it a surface and it can find geodesics for you.

Give me a few minutes & I'll see if I can do a simple parametric plot...

geocalc33

I'm trying to do it now but this project will take me all day probably

PM 2Ring

Hold on. This is a 2D system, isn't it? If it's 3D, what's z?

geocalc33

yeah 2d is fine

PM 2Ring

8:49 PM

Is R a constant? Can I set it to 1?

geocalc33

yes

PM 2Ring

Cool

In that case, we just have a circle, which is pretty boring.

I guess we can make it 3D by setting z=R

geocalc33

wolframalpha.com/…

trying to calculate the arclength of $\log^2(x)+\log^2(y)=1$

sagecell.sagemath.org

ah it didn't share the link

PM 2Ring

sagecell.sagemath.org/…

robjohn

9:08 PM

@geocalc33 I get $4.4818565$ between $\frac1e$ and $e$.

geocalc33

@PM2Ring having trouble understanding what you plotted there

PM 2Ring

It's the surface
$x = e^u, y = e^v, z^2 = u^2 + v^2$, with u,v running from -1 to 1

robjohn

actually, I only did one side, the whole arclength is $7.5101815753948970173$

geocalc33

$\frac{\pi^2 I_1(\sqrt{2})}{\sqrt{2}}$ is the exact area

robjohn

geocalc33

9:17 PM

so if I integrate from t=1/e to t=x

that's what I'm interested in

it will be sinusoidal

PM 2Ring

How's this?

sagecell.sagemath.org/…

robjohn

@geocalc33 there is a top and a bottom curve as a function of $x$.

geocalc33

@robjohn yeah I'm looking to do something similar to the elliptic integral...but with this curve

not sinusoidal for the arc length up to x

interesting that the complete elliptic integral of the second kind can be expressed in terms of the gauss hypergeometric function

because the enclosed area can be expressed as a regularized hypergeometric function

or equivalently a modified bessel function of the first kind.

robjohn

Area is $3.9952370677480303173$
Length is $7.5101815753948970173$

geocalc33

really?

I'm getting something different for area..

@robjohn you are not doing the total enclosed area

robjohn

9:31 PM

I think so

let me check

geocalc33

I'm getting 6.2757

I think you found area between the rectangle inscribed

the rectangle inscribed in the curve given by $\left[\frac{1}{e},e\right] \times \left[\frac{1}{e},e\right]$

is about 3.995

robjohn

The area and length posted above, should be good for the curve I posted above that

If the length is as I said, then the maximum area (circular area) is $4.4883943842107717230$

geocalc33

9:48 PM

$\int_{B_n}\exp\sum x_i\,d\mu = \int_{B_n}\exp(\sqrt{n} x_1)\,d\mu=\int_{-1}^{1}\exp(\sqrt{n}x)\frac{\pi^{n/2}}{\Gamma(1+n/2)}(1-x^2)^{(n-1)/2}dx$

for $B_n$ denoted by $\{x_1^2+\ldots+x_n^2\leq 1\}$

let $n=2$

This is where I'm getting 6.2757

robjohn

$y_\text{top}=e^{\sqrt{1-\log(x)^2}}$ and $y_\text{bot}=e^{-\sqrt{1-\log(x)^2}}$

So $\int_{1/e}^e\left(e^{\sqrt{1-\log(x)^2}}-e^{-\sqrt{1-\log(x)^2}}\right)\mathrm{d}x$

is the area I get

$\int_{-1}^1\left(e^{\sqrt{1-x^2}}-e^{-\sqrt{1-x^2}}\right)e^x\,\mathrm{d}x$

is the same area

geocalc33

10:09 PM

hmm

you're probably right

robjohn

I just evaluated the second integral and got the same value. If my perimeter is correct, your area is too big.

geocalc33

2

A: Volumes enclosed by $ \ln^2(x)+\ln^2(y)+\ln^2(z)+\cdots=1$

Letting $B_n$ denote $\{x_1^2+\ldots+x_n^2\leq 1\}$ our integral equals $$\int_{B_n}\exp\sum x_i\,d\mu = \int_{B_n}\exp(\sqrt{n} x_1)\,d\mu=\int_{-1}^{1}\exp(\sqrt{n}x)\frac{\pi^{n/2}}{\Gamma(1+n/2)}(1-x^2)^{(n-1)/2}\,dx $$ which has a reasonably simple closed form for any odd $n$. Its maximum is...

@robjohn I guess this answer is incorrect then?

robjohn

10:24 PM

The value is incorrect for $n=2$, I will write an answer and see what I get.

geocalc33

Okay

AMDG

10:42 PM

Can anyone tell me what might be a canonical intuition for the number $\ln(x)$ for integers $x$?

Specifically regarding its numerical representation. I suppose I could ask also for the intuition behind irrational representations and what the pattern is with their numerical representations in general.

AMDG

11:02 PM

I reason that there must be a way to recover the other coefficient from $\ln(x)$ exclusively from its non-integer part since however many factors of the other number are present, this part remains unchanged and is therefore in some way identical with that factor in $x$.

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