I did it (more on the details later)

[supple wasp]

I have, quite literally, obliterated the even natural number part of the riemann zeta function using a recursive relation and a very funny sum. Basically, off the top of my head, it’s $\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the coefficients of $S_k(n)$ where $S_k(n)=\sum_{k=1}^{n-1} \cot^{2k}(\frac{k\pi}{2n})$. The $S_k$ have an extremely nice recurrence relation in which, where $a_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{a_{n-1}S_{n-1}-a_{n-2}S_{n-2}...}{a_n}$. I am so happy that I was able to do this. Thank you zfn (in server name is I love Linear) for reviewing this with me, as well as the entire helper mod team for introducing this idea to me.

[marsh river]

flying green people eater

[supple wasp]

I have, quite literally, obliterated the even natural number part of the riemann zeta function using a recursive relation and a very funny sum. Basically, off the top of my head, it’s $\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the coefficients of $S_k(n)$ where $S_p(n)=\sum_{k=1}^{n-1} \cot^{2p}(\frac{k\pi}{2n})$. The $S_k$ have an extremely nice recurrence relation in which, where $a_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{a_{n-1}S_{m-1}-a_{n-2}S_{m-2}...}{a_n}$. I am so happy that I was able to do this. Thank you zfn (in server name is I love Linear) for reviewing this with me, as well as the entire helper mod team for introducing this idea to me.

[marsh river]

flying green people eater

[supple wasp]

also this is equal to $\sum_{k=1}^{\infty} \frac{1}{k^{2m}}$

[marsh river]

flying green people eater

[white adder]

dont understand this but cool

i like your magic words magic man

[waxen shadow]

Intresting i may take a look at this in a couple of hours

[glass steppe]

Nice

Reminds me when I found the general formula for zeta(2m) using complex analysis, which is heavily related to the series expansion of tan (with Bernoulli numbers)

[waxen shadow]

Care to share?

[glass steppe]

Sure, but I don’t want to pollute another person’s discussion channel

I can share it in another discussion channel if you want

[waxen shadow]

Respect

[glass steppe]

I’ll type it out later

[waxen shadow]

ok

[supple wasp]

$\sum_{k=1}^{\infty} \frac{1}{k^{2m}}=\pi^{2m}\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the leading coefficients of $S_k(n)$ where $S_p(n)=\sum_{k=1}^{n-1} \cot^{2p}(\frac{k\pi}{2n})$. The $S_k$ have an extremely nice recurrence relation in which, where $a_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{a_{n-1}S_{m-1}-a_{n-2}S_{m-2}...}{a_n}$

[marsh river]

flying green people eater

[supple wasp]

This is the edited and final version.

[peak bloom]

again, you should say "the leading coefficient of Sk(n), which is a polynomial in n defined by .."

[supple wasp]

oop

$\sum_{k=1}^{\infty} \frac{1}{k^{2m}}=\pi^{2m}\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the leading coefficient of $S_k(n)$ where $S_p(n)=\sum_{k=1}^{n-1} \cot^{2p}(\frac{k\pi}{2n})$. The $S_k$ have an extremely nice recurrence relation in which, where $b_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{b_{n-2}S_{m-1}-b_{n-3}S_{m-2}...}{b_{n-1}}$

[peak bloom]

there's only one leading coefficient

that does not seem like an extremely nice recurrence relation, also

[supple wasp]

well it’s a family of functions

[peak bloom]

c(S_k) is one number

[supple wasp]

ok

[peak bloom]

is that an alternating sum

[supple wasp]

it’s technically both

[peak bloom]

i mean an-1Sm-1 - an-2Sm-2 ...

is the next thing +

[supple wasp]

yes

was too lazy to give the general term

[peak bloom]

it is not clear that this sum of cot^2p is polynomial in n, is all

[supple wasp]

?

hiw

[peak bloom]

it's just not

it's also not clear that as polynomials, an divides an-1Sm-1 - an-2Sm-2 + ...

[supple wasp]

i mean i could prove that $\sum_{k=1}^{n-1} \cot^{2}\left(\frac{k\pi}{2n}\right)=\frac{(2n-1)(n-1)}{3}$

[marsh river]

flying green people eater

[supple wasp]

it does becuase the a_n have factorials in them.

$a_k$ is defined as $(-1)^k\binom{2n}{2k+1}$

[peak bloom]

ak has degree 2k+1
so an-1/an has degree -2
this seems to imply that degree of Sm is less than the degree of S(m-1)

[supple wasp]

huh?

[peak bloom]

i think the letters are causing more issues

ak = (-1)^k (2n choose 2k+1)
then what's an

a is a function of k and n here

[supple wasp]

deg(a_n)=k+1

wait

[peak bloom]

an is really (-1)^n * (2n choose 2n + 1)

but that's 0

[marsh river]

flying green people eater

[supple wasp]

uh ok lemme change the letters

gimme a few mins

[marsh river]

flying green people eater

[supple wasp]

i think that’s fine.

[peak bloom]

i would be hesitant to believe this it seems like you're just moving letters around until it works

[supple wasp]

no i realized it’s the wrong relation

[peak bloom]

ok so b(n-1) = (-1)^(n-1) (2n) choose (2n - 1)

[supple wasp]

yes

[peak bloom]

which is (-1)^(n-1) * (2n)

[supple wasp]

in the polynomial i used to get this the a_n were misaligned

[peak bloom]

b_(n-2) = (-1)^(n-2) (2n) choose (2n - 3)

[supple wasp]

yes

is that a little clearer?

[peak bloom]

desmos suggests Sm is degree 2m

[supple wasp]

?

[peak bloom]

k <=> n-1-k
so bk has degree
2q - 2|x-q|, where q = (n-1)/2

[supple wasp]

bruh

[peak bloom]

n - 1 - |n - 1 - 2k|

[supple wasp]

ok

so that’s what you’re trying to prove

[peak bloom]

im just trying to find which is the term of your alternating sum that has the leading coefficient in it

[supple wasp]

oh ok

the whole point of the “leading coefficient” thing is because i needed it for the squeeze theorem limit approximation

[supple wasp]

$\sum_{k=1}^{\infty} \frac{1}{k^{2m}}=\frac{\pi^{2m}}{2^{2m}}\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the leading coefficient of $S_k(n)$ where $S_p(n)=\sum_{k=1}^{n-1} \cot^{2p}(\frac{k\pi}{2n})$. The $S_k$ have an extremely nice recurrence relation in which, where $b_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{b_{n-2}S_{m-1}-b_{n-3}S_{m-2}...}{b_{n-1}}$

[marsh river]

flying green people eater

[supple wasp]

$\sum_{k=1}^{\infty} \frac{1}{k^{2m}}=\frac{\pi^{2m}}{2^{2m}}\sum_{k=1}^{m} (-1)^{k+1}\frac{4^k c(S_{m-k})}{(2k+1)!}$, where $c(S_k)$ represents the leading coefficient of $S_k(n)$ where $S_p(n)=\sum_{k=1}^{n-1} \cot^{2p}(\frac{k\pi}{2n})$ and $c(S_0)=m$. The $S_k$ have an extremely nice recurrence relation in which, where $b_k=(-1)^k\binom{2n}{2k+1}$, and $S_m=\frac{b_{n-2}S_{m-1}-b_{n-3}S_{m-2}...}{b_{n-1}}$

[marsh river]

dark matter

[normal wren]

Is that a polynomial in n??

[supple wasp]

yeah

[normal wren]

I need a proof

How

In

Fucking tarnation

Is that a polynomial in n

[supple wasp]

S_k is a polynomial in n

[supple wasp]

you can ask coffey

dead serious

i can prove for p=1 that it is a polynomial

[normal wren]

Wait what

[supple wasp]

and then p=k

[normal wren]

Try inducting on p??

[supple wasp]

becuase it’s literally just a choose function

[normal wren]

What

[supple wasp]

it’s the sum of the roots of $Q(x^2)=\sum_{k=1}^{n-1} \binom{2n}{2k+1}(-1)^k x^{2(n-1-k)}$

wtf

purple role

it fits

[normal wren]

jumps off the Empire State Building

[marsh river]

dark matter

[supple wasp]

like i said, ask coffey.

i didn’t make the original problem.

i just made a generalization using newton’s sums

[normal wren]

Hm

Try proving that cot(whatever) actually makes it 0??

[supple wasp]

wdym

“makes it zero”

[waxen shadow]

zeta(s) = 0

Find

[supple wasp]

@waxen shadow just like, ignore the first section

and i want you to analyze the last part

of the paper

[supple wasp]

i will update the paper once again.