ok great so how do we now show the next bit converges to zero

the next thing is to show that $x^{p-1}\equiv1\bmod p$, which is equivalent to saying that the $p$-adic expansion of $x^{p-1}$ begins as $\color{Green}{1}+\square p+\square p^2+\cdots$.

after that, we will show that if $z\equiv1\bmod p$ (where $z$ is any $p$-adic integer), then $z^{p^n}-1\to0$.

|-O

the reason these two things gets us what we want is because $x^{p^n(p-1)}-1=(x^{p-1})^{p^n}-1=(1+\square p+\cdots)^{p^n}-1\to0$ will follow

i get you

that's very interesting

Now, $x\bmod p$ is just some integer residue mod $p$, so we can apply Fermat's little theorem and obtain $x^{p-1}\equiv1\bmod p$.

@anon well isn't $p^n(p-1)=\varphi(p^{n+1})$?

One question, if I have a measurable function f with bounded support, exists M such that the lebesgue measure of the set K = {x such that f(x) > M} is less than epsilon?

@robjohn yes, that is even quicker, let's go that route

so then, by Fermat's little theorem, we have $x^{p^n(p-1)}=x^{\varphi(p^n)}\equiv1\bmod p^n$ (when $x\not\equiv0$ mod $p$, which we handle as a special case), which is to say that $|x^{p^n(p-1)}-1|_p\le\frac{1}{p^n}$

@robjohn I should edit that into my answer, which used the binomial theorem :(

@anon Speaking of the binomial theorem, I just came up with a simple way to show that $\frac1{2^n}\sum\limits_{k=0}^{\alpha n}\binom{n}{k}\to0$ exponentially for $\alpha\lt\frac12$ today.

It might not be new, but I had not seen it before

that isn't too surprising, but haven't seen it before and I don't have the tools for that sort of thing

@robjohn Share! I command you!

@PeterTamaroff Can you help me now please?

how did you get that inequality?

@MathproofP. do you know how to respond to chat messages? when you hover over the message, there is a small gray arrow in the right-hand corner of it that you can click.

Hey guys, I got some brain lag - how do you call it when the function is the inverse of its inverse? :(

@MathproofP. I got it from $a\equiv b\bmod p^n\iff p^n\mid(a-b)\iff |a-b|_p\le\frac{1}{p^n}$, where with $a=x^{(p-1)p^n}=x^{\varphi(p^n)}$ and $b=1$.

@WojciechMorawiec I would call it a compositional involution

Involution! Right, thank you so much XD

@WojciechMorawiec injective?

@anon ok i got it

@rob Pardon? Whether my function is injective?

@WojciechMorawiec every injective function is the inverse of its inverse (defined as a function on the original function's image). an involution is something that is its own inverse (this is how I read your question)

@robjohn

@WojciechMorawiec I was asking if injective was the term you were looking for.

Nope, my wording was kinda wrong - I meant its own inverse ;)

Words are bad - f(f(x)) = x is what I was looking for and that should be an involution

@WojciechMorawiec ah, then what anon said :-)

@MathproofP. so, we have now that $x^{p^n}$ converges (when $x\not\equiv0$) $p$-adically since the difference between consecutive terms tends to zero. If $x\equiv0\bmod p$, then $|x|\le\frac{1}{p}$, hence $|x^{p^n}|\le\frac{1}{p^{p^n}}\to0$, so $x^{p^n}\to0$ in such a case. Follow?

@anon yes excellent the x=pk case is a lot easier.

@Carpediem With what?

so we've done part (a). it's actually easier at this point, in my opinion to do part (c) before we do part (b). @MathproofP. ready?

@PeterTamaroff With this: math.stackexchange.com/questions/360793/…

@Carpediem I have no idea what you're asking. Could you define what you're talking about?

@anon ok great yes lets go

that part a proof was great thank you

@MathproofP. In ${\bf Z}_p$, addition and multiplication are continuous, so more generally every polynomial is continuous. In particular, $x\mapsto x^p$ is continuous. As a consequence, it preserves limits. So, defining $\omega(x):=\lim_{n\to\infty}x^{p^n}$, we must have $$\omega(x)^p=(\lim_{n\to\infty}x^{p^n})^p=\lim_{n\to\infty}x^{p^{n+1}}=\omega(x‌​)$$

@robjohn This question of mine is dead

@Carpediem OK, good.

@anon ok

$But I need to find p(D)=D^2+aD+b$

Why not write $$\frac{100}{(s-3)^2+1}$$ though?

@PeterTamaroff sos440 killed it?

Makes things more orderedly.

@PeterTamaroff Because p(D) is in a developped form

@Carpediem What is p(D)?

When I have W(s)=\frac{1}{s^2+as+b}

Then p(D) is of the above form

@anon yes offcourse

@Carpediem is $p(D)$ a constant coefficient operator?

@PeterTamaroff p(D) is the differential operator

@Carpediem yes, but are $a$ and $b$ constants?

@robjohn Oh, right. It is an old question of mine, so I didn't really read the comments.

a and b are constants

@PeterTamaroff why do you say that it is dead?

@Carpediem Oh, OK.

@robjohn I meant it has no answer and is really old.

@robjohn I'll add one.

@PeterTamaroff It seems hard to read through... quite long.

@robjohn What do you suggest?

I was merely describing the context.

@PeterTamaroff I don't know... I haven't read through it yet :-)

@robjohn LOL ok

I'm a man without conviction.

@MathproofP. now, $x\equiv x^p\bmod p$, so $x\equiv x^p\equiv x^{p^2}\equiv x^{p^3}\equiv\cdots\bmod p$. Hence $x\equiv\omega(x)\bmod p$. (This is hard to explain without saying "the reduction-mod-$p$ ring homomorphism ${\bf Z}_p\to{\bf F}_p$ is continuous, where ${\bf F}_p$ has the discrete topology.")

@JonasTeuwen three convictions and you have mandatory jail time here.

That's not so wicked if you have two of those already!

@JonasTeuwen You were flagged?

@robjohn @PeterTamaroff What do you suggest ?

@Carpediem I still don't know what you're asking =)

I'm a man who doesn't know.

Well what will be the form of p(D)

@PeterTamaroff

@anon $x\equiv x^p\pmod{p}$ uses fermat thm right? then you go inductively

@MathproofP. the parenthetical I just put forth is a fancy way of saying that reducing-mod-$p$ and taking limits in the $p$-adics can be done in either order

@MathproofP. correct

@Carpediem Isn't $p(D)$ fixed? I still don't understand what you're asking.

@robjohn I need to find p(D) basically

Here $D=\frac d{dx}$ yes?

Yes and we have p(D) of the form D^2+aD+b

And I know that when W(s)=\frac{1}{s^2+as+b}

Then p(D) is of the above form

but in this case we have: \frac{100}{s^2+as+b}

so what is p(D) ?

@anon ok

the reason $x\equiv\omega(x)\bmod p$ is useful is because then we know that $\omega(x)\equiv0\bmod p\implies x\equiv0\bmod p\implies\omega(x)=0$, so either $\omega(x)$ is zero or it is a unit in ${\bf Z}_p$. If it's not zero, then $\omega(x)^p=\omega(x)\implies\omega(x)^{p-1}=1$, i.e. $\omega(x)$ is a $(p-1)$st root of unity.

@Carpediem $p(D)$ would be the operator that works on a one the yield your function, yes?

For example

If $y=\sin z$, then $y''+y=0$ is $(D^2+1)y=0$

And $\mathcal L(\sin z)(s)=\frac 1{1+s^2}$

Is that what you're asking?

Yes

In said case, constants do no harm since $Dcy=cDy$ for any function $y$.

So your differential equation will remain unchanged.

....

@anon ow that's excellent, that proves it then.

I have to leave now.

So it's: p(D)=100(D^2-6D+10) ?

Be back in 10'. Munch time.

@MathproofP. And since $1,2,\cdots,p-1$ are all distinct mod $p$, $\omega(1),\omega(2),\cdots,\omega(p-1)$ are all distinct mod $p$, hence they are distinct as $p$-adic integers, hence we have at least $p-1$ different $(p-1)$st roots of unity. But a field cannot contain more than $n$ different $n$th roots of unity, so then we know that the number of $(p-1)$s roots of unity in ${\bf Z}_p$ is $(p-1)$. And ${\bf Z}_p\subseteq{\bf Q}_p$, so this proves the claim.

@PeterTamaroff

@Carpediem No, just the same.

The constant $100$ plays to role in the ODE.

Oh so it's just D^2-6D+10

@PeterTamaroff

@Carpediem I thinks so. But I'm not sure about what your definitions are.

See you later.

see you

thanks

@MathproofP. Note that the problem said "the" roots of unity, which meant we had to show it had a full set (i.e. precisely p-1 in count) roots of unity.

@anon yes exactly, that's great is there any way for me to save this chat? I think I need to go over it in more rigour before we attempt the last bit and it's awfully late and I need to get up early to study tomorrow.

@MathproofP. the last bit follows pretty easily from (c) and the stuff we've shown so far. you can Ctrl+PrntScrn to take a screenshot (it will be in your "clipboard," meaning you can then Paste it into mspaint say). I use a google chrome extension to save long webpages as images.

actually, I could probably screenshot our discussion and post it for you, hang on

thank you ever so much for the help it made things much clearer for me

p

ok

well, my extension is buggy, you'll have to figure out out yourself

yeah sure I'm print screening it. goodnight. I will also try do c when I wake up tomnorrow and get back to you.