And how do you show that it really is a isometry ?

Anyway I guess I'll give it some sleep before buggind you

Thanks for the help @Akiva

"Find 5 different positive integers such that the sum of any three terms divides the sum of the other two." Any ideas?

@Astyx Maybe use the dot-product definition of isometry in vector spaces

What do you mean by that ?

Actually, how to prove that

To give you some broader context, I'm trying to find a norm on $\Bbb R^2$ such that the only isometries are the identity and its opposite. To do this I showed that given any convex compact containing a neighbourhood of 0 and that has a central symmetry through the origin is the unit ball of a norm

And I want to argue that therefore if we find such a compact that has no other symmetry than the one stated above, then the norm induced by it satisfies the property

So I just have to draw such a compact to find the norm

But as I said, I need sleep :)

@FrankScience I am sorry I got repelled from the library because it is closing. Just arrived home. Thanks for the advice.

@ZachHauk it says that there exist a diagonalizing orthonormal basis of eigenvectors to be precise

If |r_n | is summable, how is it that \sum^{inf}_{n=1} |r_n|= \sum |r_n| = \sum r+_n -\ sumr-_n ?

If you pick random linearly independent eigenvectors they won't necessarily be an orthonormal basis

@JingWeng You're omitting the $ symbols so the Latex doesn't show

@FrankScience You can always study what you deem useful by yourself.

And I believe that this is by deifnition @JingWeng

@FrankScience My experience so far has been that courses serve to bait me into something interesting, and then it is up to me to really learn to content. With very few exceptions, I never learned anything substantial from courses.

hi chat

how to find the number of zeros of complex valued function

if you say "infinitely many" you'll be right most of the time

Hello there !

hi

What is the pattern in the infinite sum of $\pi$ ?

Algebraic Geometry theorem generator

hi guys

How to determnine the number of zeros and poles of a function ?

@AkivaWeinberger I haven't understood any of the questions that algorithm randomized, the more random it actually is, the more I'll feel stupid ..

There is so many formulas for $\pi$, is this because it's important, or just a consequence of the hype around it, in the mathematical culture ?

@Mahmoud It's meant to be gibberish. The person who wrote that page is convinced that algebraic geometry is gibberish.

Hi, @Mahmoud and DogAteMy.

Hello Sir @TedShifrin

hi chat

@TedShifrin no hello for poor old kasmir Ted ?

Hi semi

Hello @Semiclassical

@Zach: Good teaching takes skill, caring, and experience.

Oh, and hi @Kasmir.

:)

@Mahmoud I'd wager it primarily has to do with the connection to sine and its good analytic properties

So, one advantage to being a University of Minnesota student: One of the things that the University has is the IMA (Institute for Mathematics and its Applications)

which among other things hosts quite a few workshops/conferences during the year.

I'll go now, G'Night (sorry if it's morning for you, I can't be in multiple places at once), Bye all.

Night, @Mahmoud.

Yes, @Semiclassic. It's well known.

$:)$

How to determnine the number of zeros and poles of a function eg f (z) = z^5 sin(Pi *z ) / (( 1-z^2) ( 2-z^2) (3-z^4))

In particular, they host the annual Abel Conference in celebration of the year's Abel Prize winner.

And it was just announced that Andrew Wiles won said prize.

Ah cool.

I think I'm going to try to attend that :)

Find where the numerator is $0$ and where the denominator is $0$, @Kasmir. Not rocket science.

should I care about multiplicities?

the question is not so clear for me

You can say what the multiplicities of each are, yes.

The main thing to watch out for is whether some zeros of the numerator are also zeros of the denominator. I can see two instances of that.

okay :D

hmm

Why is that a problem ?

@Semiclassic makes an excellent point. You do have to decide if certain singularities are removable.

oh okay =p and it is only for abs(z) < 5.5

this is from an old exam am trying to do

Start by finding the zeros of the numerator and denominator respectively, and see if any roots are in both lists.

well the top is zero at z= 0

multiplicity is 6

Right. But there are other zeros in the numerator.

oh yeah

0, 2,4

makes sin pi z = 0

More than that. Ahem.

You're missing some. You want sin(pi*z)=0.

yes but I added the condition later

we looking for abs (z) < 5.5

Doesn't matter. There are more roots than that in that interval.

><

yes yes

0,1,2,3,4,5

Okay, that's (a bit more than) half of them :)

and on the bottom they are 1,-1 , -sqrt2 , +sqrt 2 and 3^(1/4) , - 3^(1/4)

Still missing some from the top.

I should mention that this is my first question of this kind so be easy on me :D

oh

yes yes

negative those works too

Right.

-1 , -2, -3 ,-4 - 5 :D

So -5 through 5.

yes :)

zero has multiplicty 7 now

if i got this correct

Actually, no.

-0 = 0, so it only shows up once from sin(pi*z)=0.

z^5 has 6 right?

or wait am mixing stuff up

Imagine you change it from z^5=0 to z^5 = 0.0001. How many roots does the latter equation have?

is must have 5 zeros

fundamental theorem of algebra

Right. And as you change the right-hand side back to zero, those five zeros all move to the origin.

So z^5=0 has multiplicity 5.

oh thats a good argument :D

Okay. So, back to what I was saying. Are any of the 6 roots of the denominator also a root of the numerator?

only +1 and -1

Prove $a^{p^2 - p} \equiv 1 \pmod{p^2}$, where $p$ is a prime number. Has anyone seen a problem like this before/know how to get it started?

Yeah. Those are the two you have to think about.

if they are removable they cancel out right ?

Right.

they are both removable

An easy check is whether they have the same multiplicity in both lists.

And they do, so they're removable.

oh nice

is that allways the case?

By contrast, sin(z)/z^2 would have a pole at zero (z=0 is a single zero on top and a double zero on bottom)

and vice versa for z^2/sin(z)

I got it but sinz / z is removable

Right.

:)

So let me count the zeros

@MathematicsStudent1122 "Hey there, it seems we have a lot in common. What say we have a one-to-one correspondence, to see if the isomorphism is natural?"

...there's a reason I don't go on many dates.

To see why it works, remember what it means to have a root of multiplicity n: $f(z)$ has a root of multiplicity $n$ at $z=z_0$ if $f(z)=(z-z_0)^n \tilde{f}(z)$ where $\tilde{f}(z)$ is analytic and nonzero at $z=z_0$.

Got it :)

So if $f(z)$ and $g(z)$ both have a zero of multiplicity $n$ at $z=z_0$, then you can write $f(z)/g(z)=\tilde{f}(z)/\tilde{g}(z)$ which is analytic and finite at $z=z_0$.

Anyways.

We did not go that deep yet but it is always good to know more thanks! :D

@Semiclassical I may be misremembering something I've read, but I thought a root of multiplicity $n$ at $z_0$ meant that the function and its first $n-1$ derivatives are zero at $z_0$, but the $n$th derivative is non-zero. I know these are equivalent for polynomials, but are they equivalent otherwise? They don't seem like they would be, and my characterization might not be true for non-polynomials, hence "I may be misremembering".

Ehm what answer did you get for the number of zero and poles?

i mean how do we proceed

I got 11 zeros on top we deleted 2 of them

so 9

and z=0 with multiplicity of 5

14 right ?

Hmm. I think the above works. If $\tilde{f}(z)$ is analytic and finite at $z=z_0$, then $f^{(k)}(z_0)=0$ for any $k<n$ but $f^{(n)}(z_0)\neq 0.$

Okay, yeah. Product rule. You're right.

5+11-2=14 on top, yeah.

For a second, I thought that trig functions might be a problem, but they aren't--all of $\sin z$'s roots are multiplicity 1, as $\cos z \neq 0$ whenever $\sin z = 0$. It doesn't skip a derivative. Oh well.

How many poles, just to check?

hmm I dont know how to count them

Well, how many zeros in the denominator?

if I understood right they should be 4

Yep. 6-2=4.

but I really have many gaps

I should read the chapter again

So you have 14 zeros and 4 poles, counting multiplicity, and 2 removable singularities.

its alot of stuff to take in

Yes it is easy when you help me

but I should learn this in better ways :D

Okay thank you Semi !: D

Well, the basic tack is simple: Write down the zeros of the numerator, and write down the zeros of the denominator to get the poles.

multiplicities and removable singularities complicate that, but not too much.

Oh yeah =p but the computation is alot in some of these questins

we have to do them by hand on exam

need to practice alot =p

@Ted oh, I'm surprised you saw that message :P

im back everyone

I find residue computations to be more tedious, tbh.

lol ><

I was about to ask you about that just now

hah.

The easiest scenario is when you can organize the function as $f(z)=g(z)/(z-z_0)$ where $g(z)$ is analytic and nonzero at $z=z_0$. In that case the residue is just $g(z_0)$.

@Semiclassical doesn't that formula use the $n$-th derivative of some rational function?

I think what we are doing right now should be easy for you

(for pole of order $n$)

its undergrad complex analysis

For finding residues, yeah.

Zack focus on that math camp and let semi help me :D

But it can still be a right pain to compute by hand.

We do all by papper and pen on exams

nothing is allowed

Yeah.

Some of online lectures the teacher let students have formulae sheet

that could be handy in many exams