Mathematics - 2022-11-29

geocalc33

12:00 AM

$(3,2+1/3)$ is an example of a point

$(5,2+1/3+5)$ another example

just kidding

reverse those coordinates and then it's accurate

desmos.com/calculator/ypvwlxigiw

one potato two potato

2:34 AM

I already proved (a) and (b) but can't conclude (c). All I can conclude is that $\zeta$ has infinitely many zeros in $\Bbb C$ not in the critical strip.

Lukas Heger

@onepotatotwopotato I'm not very good with analytical stuff, but the zeroes of $\xi$ should be exactly the non-trivial zeroes of $\zeta$, because the Gamma function has poles that cancel with the trivial zeroes (i.e. the ones outside the critical strip)

(this is assuming that you know about the poles of the Gamma function and the location of the zeroes of $\zeta$ outside the critical strip)

one potato two potato

2:49 AM

@LukasHeger Oh so $G$ has infinitely many zeros implies $F$ has infinitely many zeros implies $\xi$ has infinitely zeros so the statement follows.

Lukas Heger

yes

one potato two potato

Thank you!

Lukas Heger

np

Jack Ohara

3:33 AM

@LukasHeger If I want to show that subsets of Q that has more than 1 point are not connected , can I argue like this? we can find x<z<y and take two nbhd that of right radius that do not intersect , hence our subset can be written as union of two seperated subsets that are nonempty

I notice the question has been posted in the forum but they argue that z is irrational

Lukas Heger

you need to take z irratoinal

Jack Ohara

i dont see the need for this , a rational z would also work?

Lukas Heger

otherwise the argument doesn't work

how do you get two neighborhoods that don't intersect if z is rational?

in R you can also find x<y<z, but R is connected

Jack Ohara

well it works like this take radius to be less than half the distance between (z,x) and ( z,y)

this can always be done

Ted Shifrin

Focus on what exactly you are doing.

Lukas Heger

3:36 AM

but what if z is contained in the subset? and I agree with Ted

you could use more clarity

Jack Ohara

the goal to have that the subset has to be disconnected

Ted Shifrin

What is your subset of $\Bbb Q$?

Jack Ohara

I see that if z is irrational it does work but why does it fail if z is rational ( this i need to understand )

Ted , so far i am taking two element set

Lukas Heger

you can't just take a two element set

Jack Ohara

but once i do that it will work for any subsets that has more than two elements

Lukas Heger

3:38 AM

you need to take any subset that contains at least two element

@JackOhara no that's just not true

any two element subset of R is disconnected

but R has plenty of subsets with more than one point that are connected

Jack Ohara

Right, I see the flaws in my argument now

it should be any subset that contains at least two elements

Lukas Heger

right

Jack Ohara

so we make a sorta CUT at irrational point

to make sure the seperation is clear

Lukas Heger

yes, that's the point

Ted Shifrin

Valid, not clear

Jack Ohara

3:40 AM

Oh yes valid is the correct word!

thanks Ted and Lukas!

Ted Shifrin

You’re most welcome.

Lukas Heger

no problem

one potato two potato

0

Q: $\hat{f}(\xi) = \Gamma(a+i\xi)$ for $f(z) =e^{az}e^{-e^z}$

The following is Exercise 6.11 in Stein Complex analysis. Let $f(z) = e^{az}e^{-e^z}$ where $a>0$. Observe that in the strip $\{x+iy:|y|<\pi\}$ the function $f(x+iy)$ is exponentially decreasing as $|x|$ tends to infinity. Prove that $$\hat{f}(\xi) = \Gamma(a+i\xi),\quad\text{for all}\ \xi\in\Bb...

The problem has several typos I think.

Lukas Heger

have you googled for errata of the book? Plenty of popular books have errata lists (official or inofficial) online

one potato two potato

assets.press.princeton.edu/releases/m7563c.pdf

The problem is Exercise 11 in P.177

Ted Shifrin

3:48 AM

I didn’t cover that part of the book, but I found errors — some substantive — when I taught out of the book years ago.

Errata from 2006 are not up-to-date.

But Stein is dead now, I believe.

leslie townes

yes

Lukas Heger

yeah, these errata don't say much, unfortunately

some books have more detailed errata

Ted Shifrin

Mine are constantly updated, but when I’m dead the webpages will probably go poof.

Lukas Heger

@Ted you could make sure that they are archived on the wayback machine: web.archive.org

Ted Shifrin

I have a feeling it’ll be too late to worry about that when I head out the door :)

Lukas Heger

3:54 AM

yeah true

but I mean you could make sure they are archived regularly

Ted Shifrin

I could put you in charge :)

Lukas Heger

I think I can handle that :)

okay I'm archiving your three errata pages now

done

leslie townes

munchkin and i just went through my fourth grade yearbook. she didn't recognize me but had very funny things to say about how people looked in photos.

Ted Shifrin

Thanks, Lukas!

Cotton Headed Ninnymuggins

5:35 AM

@Peter Best doesn't imply good

one potato two potato

5:47 AM

1

A: Elementary bound on the Riemann zeta function

Let $s = \sigma+ it$. For $\sigma > 1$ and by analytic continuation for $Re(s) > 0$ : $$\zeta(s) - \sum_{n < N} n^{-s} = s \int_N^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1} N^{1-s} +s \underbrace{\int_N^\infty (\lfloor x \rfloor-x) x^{-s-1}dx}_{\textstyle< \int_N^\infty x^{-1-\sigma}dx <...

Need an explanation for this answer...

Koro

7:40 AM

Suppose that f is complex valued and is integrable. Then how to prove that $|\int f|\le \int |f|$?

I have a proof but it is weird.

Naturally, one would write $f=Re(f)+i Im (f)\implies \int f= \int Re f+i\int Im f\implies |\int f|^2=(\int Re (f))^2+(\int Im (f))^2$

$\int |f|=\int \sqrt{(Re f)^2+(Im (f))^2}$

I'm not sure how to go from here.

I don't see how, unfortunately.

Lukas Heger

nevermind

I mean, there's a really pedestrian approach using Riemann sums and the triangle inequality

but there's probably also something more elegant

Koro

suppose that f is defined on some measure space that doesn't have to be the familiar R with Lebesgue measure.

I got it math.stackexchange.com/a/2152367/266435

It's such an amazing solution.

Lukas Heger

certainly very concise

I mean you can replace Riemann sums by approximation with simple functions for the measure-theoretic integral

then it's still just triangle inequality

that's not as elegant as the proof you have linked, but it's something one might come up with more easily

Koro

7:56 AM

The weird one that I had was: let $c=\int f$. Define $x=\frac {\bar c}{|c|}$ so that $xc=|c|\implies |c|= x\int f= \int (xf)=\int Re (xf)\le \int |xf|\le \int |f|$

the linked one is intuitive because one knows that $z= |z| e^{i\theta}$

one potato two potato

8:37 AM

The above answer I shared here, I think it's wrong.

Raffallo

@PM2Ring oh, thats why i couldn't calculate it using Newton's method :) I will try it now and will give info after that if it works

copper.hat

8:59 AM

@Koro that solution is in Rudin afaik

Koro

Ok. thanks.

@copper.hat

Given a Cauchy sequence $(x_n)$, can we find a subsequence $(x_{n_k})$ such that
$\sum_k |x_{n_{k+1}}- x_{n_k}|<\infty$?

PM 2Ring

9:16 AM

@Raffallo $\cosh(x) = \cosh(-x)$, so there will generally be two solutions for $p$. Using your numbers & my equation, I get $p\approx$ 225.2939 and -71.0939

Raffallo

@PM2Ring the function is symmetric so I know there will be 2 answers :) But I'm not sure I can figure out the equation for p

one potato two potato

9:37 AM

0

Q: $\varphi(x)\sim x$ then $\psi_1(x)\sim x^2/2$

The problem is from Exercise 7.11 in Stein complex analysis Let $$\varphi(x) =\sum_{p\leq x}\log p$$ where the sum is taken over all primes $\leq x$. Prove the following are equivalent as $x\to\infty$: (i) $\varphi(x)\sim x,$ (ii) $\pi(x)\sim x/\log x,$ (iii) $\psi(x)\sim x,$ (iv) $\psi_1(x)\sim...

complex-analysis analytic-number-theory

tbh, I don't like this problem.

Raffallo

@PM2Ring also I'm not sure i I know x value. I know destination delta y and delta x, and value of a. Now I need to calculate parameter p to know how much my curve needs to be moved in the X axis direction to meet my delta x and delta y requirements

PM 2Ring

10:03 AM

@Raffallo In that case, I don't understand your question. I thought you knew x, and p is basically the delta x. If you just have delta y, then x could be anywhere, and there are two delta x values that will correspond to the given delta y. Maybe you need to draw a diagram.

Raffallo

I have two points, where one is always on Y axis and I calculated delta x and delta y between them

also I have my curve without p parameter, which will always have her minimum value at Y axis and also on point 1. Now I need to add p parameter to move my curve along X axis to fit it to my two points. @PM2Ring

https://www.desmos.com/calculator/3xsgjkzpy7
There is and diagram in the online viewer. When you will turn on drawing second curve you will see, that he is almost fitting.
The shift on the Y axis is negligible, because I can easily calculate it.

Not Tfue

10:50 AM

the relationship between x and w should be -1/tan(w)=x. I don't think I am wrong.

PM 2Ring

@Raffallo Oh, ok. Then you need x+49.506713 or x-203.706713

Not Tfue

I don't see the x=tan(w) here lol.

And my professor said book is always right.

PM 2Ring

11:06 AM

@Raffallo So your point 1 is $x=0$, $y=a$. And so you want to solve for $\Delta x$ in $a+\Delta y = a \cosh\left(\frac{x+\Delta x}{a}\right)$. That gives us $\Delta x = -x \pm a \cosh^{-1}\left(1+\frac{\Delta y}{a}\right)$

BTW, $\cosh^{-1}(x)\equiv \ln(x+\sqrt{x^2-1})$

@NotTfue It looks like there's a $\pi/2$ error. If two lines are perpendicular, the product of their slopes is $-1$.

Raffallo

I've tried to use x+49.506713 and x-203.706713 in the viewer and its not looking good.
When you check curve intersection point with x=0 and x=77.10, the delta y is not equal to 7.05.
https://www.desmos.com/calculator/3xsgjkzpy7

Not Tfue

@PM2Ring which 2 lines?

PM 2Ring

11:26 AM

@NotTfue Any two perpendicular lines.

@Raffallo Huh? I checked my values on your Desmos graph before I posted them. Both the resulting curves go through your point 2.

Raffallo

PM 2Ring

@Raffallo sagecell.sagemath.org/…

Raffallo

@PM2Ring I've used a*cosh((x+49.506713)/a) and here is the result. Curve passes through the point 2, but not through the point 1, so the delta y is wrong

@PM2Ring yes, you are right, that Y value is correct for point 2, but in the point 1 curve is no longer have correct y

PM 2Ring

@Raffallo I thought you agreed that y is 1138 & delta y is 7.05. And you want the curve to go through point 2.

Raffallo

11:42 AM

@PM2Ring i want the curve to go through both of the points

but maybe not exactly points, i want that |f(77.10) - f(0.0)| will be equal to 7.05

PM 2Ring

12:00 PM

@Raffallo I think that means that you want to solve for $p$ such that $a\cdot\cosh((77.1+p)/a) - a\cdot\cosh(p/a) = 7.05$. Does that sound correct?

Raffallo

12:11 PM

@PM2Ring hmmm, yea, sounds correct. It can be even the absolute value there, since both results for p should be the same number but one negative and one possitive

I've used wolframalpha.com to calculate p and I've got 65.3441, which looks correct in my example data :)
How would look like my final equation to solve? wolfram is giving me very complex formula :/

one potato two potato

Let $\mathcal{F}$ be a family of functions of the form $\sum_{n=0}^\infty a_nz^n$ where $|a_n|\leq n$. Then $\mathcal{F}$ is a normal family such that every uniform limit of a given arbitrary sequence $\{f_n\}$ in $\mathcal{F}$ is contained in $\mathcal{F}$.

PM 2Ring

@Raffallo I get ~65.3440832006 using Newton's method.

I'm pretty sure there's only 1 solution. The graph of $a\cdot\cosh((77.1+p)/a) - a\cdot\cosh(p/a) - 7.05$ looks like this (temporary image file):

sagecell.sagemath.org/kernel/…

Raffallo

12:34 PM

so, I had right that I have to use Newton's method:)
where you are calculating it?

PM 2Ring

@Raffallo Yes. Newton's method works ok. Sage can't find a symbolic solution.

one potato two potato

12:50 PM

I wrote something wrong. If $\{f_n(z) = \sum_{k=1}^\infty a_{n,k}z^k\}$ is a sequence of functions contained in $\mathcal{F}$ and $f_n(z)\to f(z) = \sum_{k=1}^\infty a_kz^k$ uniformly, then $\lim_{n\to\infty}a_{n,k} = a_k$ and $|a_{n,k}|\leq k$ for all $n$ implies $|a_k|\leq k$. The family being normal is easy.

PM 2Ring

@Raffallo Here's my Newton's method code, in Sage / Python. Sage can also do arbitrary precision arithmetic, if you need it.

sagecell.sagemath.org/…

one potato two potato

If I suddenly change the family to be $|a_n|<n$ or $n+a_n$ then suddenly becomes hard.

Yai0Phah

@LukasHeger Maybe it is community related. I don't think that Bourbaki is hated among French mathematicians.

PM 2Ring

@Raffallo Here's a few more digits: 65.3440832006162642220406756199669063780

Raffallo

1:06 PM

@PM2Ring I've used it in my c++ code and it works! Thank you very much man!

Jakobian

don't you love it when the code you've downloaded off the internet just doesn't work

PM 2Ring

@Raffallo No worries.

@Jakobian It can be very annoying & frustrating. Especially if you're tempted to debug it.

geocalc33

1:44 PM

$\sum_{p* \le x} p^*$ where $p^*$ denotes alternating terms are primes and reciprocals of primes

any idea how to code this in sage cell or write it in better notation?

Jakobian

@PM2Ring this is actually what I had but it wasn't that annoying, I just didn't have any program to compile LaTeX and it didn't use TinyTeX by default

geocalc33

I can code it for $\sum_{p \le x} 1/p$

Jakobian

which I did have

PM 2Ring

1:59 PM

@geocalc33 How big is your $x$? If it's not that large it probably won't matter if it's not that efficient. ;)

geocalc33

@PM2Ring Want to plot it for say x<1000

and also I wanted to ask if this sort of counts as a kind of prime counting function

PM 2Ring

@geocalc33 Ok. That's easy. Even up to a million (or ten million) a naive method would be fine.

Summing primes is kinda related to counting primes. But (obviously) the sum of primes is significantly larger than the number of primes.

There's actually a fairly nice sieve-like algorithm for prime sums, so you can get the sum of primes < x without actually having to generate all the primes < x. I've often used it as a way of testing my prime generating code.

But we don't need that here.

geocalc33

@PM2Ring the sum of those primes and reciprocals (alternating) looks like it's distributed like the logarithmic integral at least for some data

PM 2Ring

Ok. Give me a minute & I'll write you some code.

geocalc33

Okay

I was trying for a while to get 2 functions on the same plot in sage cell

PM 2Ring

2:20 PM

We can create the sequence by zipping the primes with an endless cycle of (1, -1). And to make a plot, we don't just want the final sum of the sequence, we want all the cumulative sums of the series, so we use accumulate. Like this:

sagecell.sagemath.org/…

However, it's a bit more efficient to do the summing with floats, rather than rationals. I'll make a plot.

Here you go. It's pretty fast for m<100000. It takes about 15 secs for m=1000000

sagecell.sagemath.org/…

@geocalc33 It's easy to plot multiple functions. But we aren't plotting continuous functions here, we're just plotting points.

geocalc33

2:40 PM

can i switch the ordered pairs easily in that code?

I was gonna compare to Li(x), forgot that part

think i got it

PM 2Ring

You can combine plots by simply adding plot objects together. But here's a simple example of plotting multiple functions in a single plot object.

sagecell.sagemath.org/…

Here's the same thing, but using the "add plot objects together" technique. We could also do that using the sum function & a generator expression.

sagecell.sagemath.org/…

Koro

4:26 PM

2

Q: Suppose $f \in L^{p_0} \cap L^{\infty}$ for some $p_0 < \infty$. Prove that: $\lim_{p \rightarrow \infty} \|f\|_p=\|f\|_{\infty}$

Suppose $f \in L^{p_0} \cap L^{\infty}$ for some $p_0 < \infty$. Prove that: $\lim_{p \rightarrow \infty} \|f\|_p=\|f\|_{\infty}$ My attempt: So the first step seems to be showing that the limit exists at all. I'm having trouble showing this, I feel like I'm getting a massive case of tunnel ...

real-analysis measure-theory

Can anyone please tell how to show existence of this limit?

Thanks.

Ted Shifrin

My recollection from grad school is that it all falls out of Hölder. Finite measure space, right?

Thorgott

I believe this shouldn't need finiteness

Semiclassical

o/

Ted Shifrin

I don't berember.

Koro

@TedShifrin not necessarily finite.

Given p>q. I want to prove that: $\|f\|_p\le \|f\|_q^{\frac qp} \|f\|_\infty^{1-\frac qp}$. Is this result even true?

I think the correct should be (by Ho:lder's: inequality) $\|f\|_p\le \|f\|_q^{\frac qp} \||f|^{p-q}\|_\infty^\frac 1p$$

Ted Shifrin

4:39 PM

You have an unnecessary absolute value in there.

Thorgott

this looks way too complicated

Koro

$\|f\|_\infty$ denotes essential sup.

Ted: the absolute sign can be removed, yes.

hmm, I don't think the absolute value sign can be removed. f could be negative and q-p could be say 1/2.

Ted Shifrin

Isn't $\|f^r\|_\infty = \|f\|_\infty^r$?

But the $L^p$ norms all work with $|f|$ to start with.

Koro

$\int |f|^p =\int (|f|^q |f|^{p-q})\le \||f|^q\|_1 \||f|^{q-p}\|_\infty$ (Because Ho:lder's inequality is true here). But not sure how to get what I want to prove.

@TedShifrin Let me check that.

We can for sure say that $\||f|^r\|_\infty \le \|f\|_\infty^r$.

Ted Shifrin

4:56 PM

Remember that we’re often talking about complex-valued functions, so $|f|$ is needed everywhere.

Koro

5:12 PM

yeah

geocalc33

@PM2Ring thanks for helping, i was able code this cool plot

Semiclassical

what'd you plot

geocalc33

I plotted prime counting function in green, log integral in red

Semiclassical

neat

geocalc33

blue things and then the average of the blues are that yellow one

Semiclassical

5:25 PM

what's the blue things

@PM2Ring does your knowledge of Sagemath include any expertise in GAP

geocalc33

@Semiclassical ill share my code one second. It times out for higher values unfortunately

Semiclassical

not surprising

geocalc33

It's cool how the yellow is nestled between the li(x) and pi(x)

Semiclassical

yeah, tho i expect that wouldn't last forever given that pi and li eventually do cross

(and then cross again, etc)

geocalc33

yeah they probably all cross eventually

Semiclassical

5:35 PM

the word "eventually" is the annoying part

geocalc33

5:49 PM

yes

it definitely looks like the yellow and green are avoiding eachother

Oliver Díaz

6:13 PM

@Koro This question has been asked several times before and has many answers. Here is one by yours truly. I think robjohn has another one somewhere

Ted Shifrin

6:27 PM

@OliverDíaz I assume you meant $|f|/\|f\|_\infty \le 1$?

Oliver Díaz

@TedShifrin: sure, I will edit shortly (if there is a typo)

Ted Shifrin

Not a huge deal, but it might confuse some folks.

copper.hat

7:09 PM

@Koro You can always choose the next point so that any subsequent points are less than ${1 \over 2^{n+1}}$ away, if yoyu see what I mean, so the sum is bounded.

Koro

7:20 PM

@OliverDíaz thanks. I managed to get the proof after this-

3 hours ago, by Koro

We can for sure say that $\||f|^r\|_\infty \le \|f\|_\infty^r$.

@copper.hat yes, I'd understood that as well. Thanks.

We can take any convergent series of positive numbers $c_n$ and can get a subsequence $(x_{n_j})$ of Cauchy $(x_n)$ such that $|x_{n_{j+1}}-x_{n_j}|\lt c_j$

for every $j\in \mathbb N$.

anak

9:16 PM

@TedShifrin I had a question about some notation. In a paper I was reading there was a situation where they wrote $\mathcal{L}_u(f\,dx) = (\nabla f\cdot u + f\nabla \cdot u)dx = (\nabla\cdot fu)dx$, where $u$ is a vector field, $f$ a non-vanishing function, and $dx$ the volume form on $\mathbb R^n$. I am unsure how they did the first equality. The second one is a product rule, but I don't quite understand their use of $\nabla$ alone. I thought the cdot was contraction with u, but I am unsure.

I tried going about it with Cartan's magic formula a few ways, but that gets me $\mathcal L_u(f\,dx) = \iota_u d(f\,dx) + d(\iota_u f\,dx) = \iota_u(\nabla f \wedge dx) + d(f\,\iota_udx)$.

The first term looks like something I'd call $(\nabla f\cdot u)\,dx$, but the rest seems harder to reconcile.

Koro

9:31 PM

Suppose that F:[a,b]-->R is monotonic. Then F is absolutely continuous iff $\lambda_F<<\lambda$ How do I prove this?

Jakobian

@Koro one direction is obvious, no?

the other one follows from some considerations using regularity

Koro

I am not comfortable with using regularity :(.

Ted Shifrin

@anak it is product rule. First term gives directional derivative of $f$. Second term is Lie derivative of the volume form, which is div $u$ times volume form. This is often called the divergence theorem. You can get it out of Cartan. $d(\iota_u dx)$ is divergence.

Koro

The other direction from right to left seems doable but I have no idea.

Jakobian

Let $A$ be a Borel set such that $\lambda(A) = 0$, $\delta > 0$ as in definition of absolute continuity and fix $\varepsilon > 0$. There is a decreasing sequence of open sets $U_n$ with $\lambda(U_n) < \delta$ such that $\lambda_F(U_n)\to \lambda_F(A)$

Ted Shifrin

9:40 PM

I have no idea what your $\nabla f\wedge dx$ means.

Jakobian

Now use that $U_n$ are a countable union of open intervals and estimate $\lambda_F(A)$.

you might wonder why I took the limit, and that's because I was thinking about complex $F$ of bounded variation

but yeah, that's how the proof goes, and I didn't come up with it on the spot, I just remember it

anak

@TedShifrin df\wedge dx, but I this notation was towards trying to reconcile it with their notation

Jakobian

do I need to learn differential topology

say, if I'm going into analysis direction, probably

Ted Shifrin

@anak That’s 0. $dx$ is a top form.

anak

Ah! You are right.

Let me try re-writing it neatly now and see if I understand the div thing.

Ted Shifrin

9:51 PM

Everyone should learn differential forms and Stokes’s Theorem. That’s undergraduate analysis. Manifolds are ubiquitous. Differential topology with bundles, intersection theory, degree …. Depends on what analysis.

Koro

@Jakobian :-)

I checked the proof given in a book.

It's similar to what you suggested.

Goku

We seem to have a bit of a rivalry with humanities and med school/departments right across our department

Brownian_Motion

What is a good multiple shooting solver for BVP?

anak

@TedShifrin Okay so I seem to be getting not what I want still. With the div suggestion, clearly I get the $f(\nabla\cdot u)dx$ part. However, the other part was $df\wedge(\iota_udx)$, which seems different from $(\nabla f\cdot u)dx$ in general?

Or does it not matter where you contract a vector field with the interior product?

(I feel like it definitely does)

Ted Shifrin

Huh?

Are you doing $L_u(f\,dx) = (L_u f)\,dx + f(L_u\,dx)$ as we started out?

anak

10:03 PM

I use Cartan to get: $L_u(f\,dx) = \iota_ud(f\,dx) + d(\iota_u f\,dx) = 0 + d(f\,\iota_udx) = df\wedge(\iota_udx) + f(\nabla\cdot u)dx.$

Unless I am doing something stupid?

Ted Shifrin

So you're granting that $d(\iota_u dx) = \text{div}\, u\,dx$.

anak

Yeah, after doing cartan again and using the def of divergence

Ted Shifrin

It's easier to see that $L_uf = \nabla f\cdot u$ without Cartan.

But you can work out the first term. Extend $u$ to a basis $u=v_1,\dots,v_n$ and let $\theta^1,\dots,\theta^n$ be the dual basis. Then write $dx = \phi \,\theta^1\wedge\dots\wedge\theta^n$.

$\iota_u dx = \phi \,\theta^2\wedge\dots\wedge \theta^n$.

Write $df = \sum f_i\theta^i$. Etc.

Note that $f_1 = df(u) = \nabla f\cdot u$.

Ted Shifrin

10:20 PM

Alternatively, you don't have to change basis. Just write it out in the standard coordinates and you'll see $\nabla f\cdot u$ appearing.

Sonozaki

10:51 PM

Consider the relation on $\mathbb{R}$ given by $x \sim y$ if $xy \ge 0$. It is correct to say that this is reflexive because $x \sim x$ if $x^2 \ge 0$ (which is true for any $x\in\mathbb{R}$), it is symmetric because $x \sim y$ implies $xy \ge 0$, hence $yx \ge 0$ (because commutativity of product on $\mathbb{R}$) but it is not transitive because for $x=1, y=0$ and $z=-1$ we have $xy=0 \ge 0$, $yz=0 \ge 0$ but $xz=-1<0$?

PM 2Ring

@Semiclassical Sorry, I don't know GAP. I have run GAP and Pari/GP in SageMathCell, but I can't remember what it was.

Ted Shifrin

@Sonozaki Looks fine to me.

KZ-Spectra

11:37 PM

Hello, when I preview the code mathb.in/73434 in the question I am writing, it is not properly formatted. Why? It works in the mathbin link

Transcript for

$Mathematics$ Mathematics