ALRIGHT!
I FINALLY FUCKING HAVE IT!
THIS IS THE MOST BAFFLING FUCKING FUNCTION MAN!
I AM HERE TO REDEEM MYSELF!
I AM GOING TO GO VERY SLOW!
Let \(f(x)\) be known as Caleb's function; it is written as:
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
The theorem we are going to present; is that for \( k \ge 1\) the value:
\[
F_k = \frac{1}{2\pi i} \int_{|z| = 2} f(z)\frac{dz}{z^{k+1}}\,dz = 0\\
\]
And that \(F_0 = 3/2\).
I have confirmed this with pari-gp; and am finally confident enough to answer this stupid annoying ass question, lmao.
We additionally, get a more complex formula for \(k \le -1\).
We start by showing that:
\[
a_{kn} = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{1+z^n}\frac{dz}{z^{k+1}} = 0\\
\]
For \(k \ge 1\).
This value just equals:
\[
b_{kn} + \frac{1}{n}\sum_{q^n = -1} q^{-k}\\
\]
If \(n \not | k\) then this value is just zero. This is because \(b_{kn}\) is zero. And because the sum on the right is zero.
If \(n | k\) then something unbelievable happens; then \(k = nm\) and:
\[
(-1)^{m+1} + (-1)^m = 0\\
\]
Like holy fuck, how did I miss this.
The only time this doesn't happen is when \(k=0\), upon which the sum on the right is \(1\), and \(b_{kn} = 0\). So we end up with:
\[
F_0 = f(0) + \sum_{n=1}^\infty \frac{1}{2^n} = 3/2\\
\]
And for \(F_k\) we just get:
\[
F_k = \sum_{n=0}^\infty \frac{a_{kn}}{2^n} = 0\\
\]
This means that:
\[
B(x) = \frac{1}{2\pi i} \int_{|z| =2} f(z) \frac{dz}{z-x} = \frac{3}{2}\\
\]
Which is exactly as you saw Caleb.
When \(k \le -1\) we basically get the exact same thing. Except the term \(b_{kn} = 0\). So there's no cancellation and we get:
\[
F_k = -\frac{f^{(-k)}(0)}{(-k)!}\\
\]
Okay for fucks SAKES MAN!!!
I'm taking a break now. Been yelling at pari for the past 2 hours, lmao!!!!!
The cool formula I actually wanted is that:
\[
2-f(x) = \sum_{k\in \mathbb{Z}} F_k x^{-k}\\
\]
We should also be able to generalize this. Assume that \(\sum_{n=0}^\infty |a_n| < \infty\); then if:
\[
G(x) = \sum_{n=0}^\infty a_n \frac{x^n}{1+x^n}\\
\]
Then:
\[
F^G_k = \frac{1}{2\pi i} \int_{|z| = 2} G(z) \frac{dz}{z^{k+1}}\\
\]
We have
\[
F^G_0 = \frac{1}{2}a_0+ \sum_{n=1}^\infty a_n\\
\]
And
\[
F^G_k = 0\,\,\text{for}\,\,k \ge 1\\
\]
And for \(k \le -1\); we have:
\[
F^G_k = - \frac{G^{(-k)}(0)}{(-k)!}\\
\]
So that the general formula is for \(|x| > 1\):
\[
C-G(1/x) = \sum_{k\in \mathbb{Z}} F^G_k x^k\\
\]
where \(C\) is an easy constant.
I FINALLY FUCKING HAVE IT!
THIS IS THE MOST BAFFLING FUCKING FUNCTION MAN!
I AM HERE TO REDEEM MYSELF!
I AM GOING TO GO VERY SLOW!
Let \(f(x)\) be known as Caleb's function; it is written as:
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
The theorem we are going to present; is that for \( k \ge 1\) the value:
\[
F_k = \frac{1}{2\pi i} \int_{|z| = 2} f(z)\frac{dz}{z^{k+1}}\,dz = 0\\
\]
And that \(F_0 = 3/2\).
I have confirmed this with pari-gp; and am finally confident enough to answer this stupid annoying ass question, lmao.
We additionally, get a more complex formula for \(k \le -1\).
We start by showing that:
\[
a_{kn} = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{1+z^n}\frac{dz}{z^{k+1}} = 0\\
\]
For \(k \ge 1\).
This value just equals:
\[
b_{kn} + \frac{1}{n}\sum_{q^n = -1} q^{-k}\\
\]
If \(n \not | k\) then this value is just zero. This is because \(b_{kn}\) is zero. And because the sum on the right is zero.
If \(n | k\) then something unbelievable happens; then \(k = nm\) and:
\[
(-1)^{m+1} + (-1)^m = 0\\
\]
Like holy fuck, how did I miss this.
The only time this doesn't happen is when \(k=0\), upon which the sum on the right is \(1\), and \(b_{kn} = 0\). So we end up with:
\[
F_0 = f(0) + \sum_{n=1}^\infty \frac{1}{2^n} = 3/2\\
\]
And for \(F_k\) we just get:
\[
F_k = \sum_{n=0}^\infty \frac{a_{kn}}{2^n} = 0\\
\]
This means that:
\[
B(x) = \frac{1}{2\pi i} \int_{|z| =2} f(z) \frac{dz}{z-x} = \frac{3}{2}\\
\]
Which is exactly as you saw Caleb.
When \(k \le -1\) we basically get the exact same thing. Except the term \(b_{kn} = 0\). So there's no cancellation and we get:
\[
F_k = -\frac{f^{(-k)}(0)}{(-k)!}\\
\]
Okay for fucks SAKES MAN!!!
I'm taking a break now. Been yelling at pari for the past 2 hours, lmao!!!!!
The cool formula I actually wanted is that:
\[
2-f(x) = \sum_{k\in \mathbb{Z}} F_k x^{-k}\\
\]
We should also be able to generalize this. Assume that \(\sum_{n=0}^\infty |a_n| < \infty\); then if:
\[
G(x) = \sum_{n=0}^\infty a_n \frac{x^n}{1+x^n}\\
\]
Then:
\[
F^G_k = \frac{1}{2\pi i} \int_{|z| = 2} G(z) \frac{dz}{z^{k+1}}\\
\]
We have
\[
F^G_0 = \frac{1}{2}a_0+ \sum_{n=1}^\infty a_n\\
\]
And
\[
F^G_k = 0\,\,\text{for}\,\,k \ge 1\\
\]
And for \(k \le -1\); we have:
\[
F^G_k = - \frac{G^{(-k)}(0)}{(-k)!}\\
\]
So that the general formula is for \(|x| > 1\):
\[
C-G(1/x) = \sum_{k\in \mathbb{Z}} F^G_k x^k\\
\]
where \(C\) is an easy constant.
