Christian Remling has put me to shame.
He's expressed your qualms, Caleb. And done so, to show that my Cauchy formula is not correct!!!
There is a residue factor which gets added in. He uses the \(f(0) = 1/2\) and my formula equals \(3/2\) as an example.
I definitely screwed up some part of my formula. But it's probably just a couple indicator values which get compounded...
In the case of \(f(0) = 1/2\); my formula is \(f(0) + 1\). And for further derivatives \(f^{(k)}(0)\)..
There's some small error I forgot. And it's probably \(n=k\) kind of \(\delta[n=k] = 1\) value.
This'll probably point even more towards your work!
There's some small rational function that:
\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z^{k+1}}\,dz = B_k + d_k\\
\]
Where now:
\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z-x}\,dz = f(x) + \sum_{k=0}^\infty d_k x^k\\
\]
Where \(d_k\) is a small collection of \(1\)'s and \(-1\)'s I missed out on...
Fuckkkkkkkk
He's expressed your qualms, Caleb. And done so, to show that my Cauchy formula is not correct!!!
There is a residue factor which gets added in. He uses the \(f(0) = 1/2\) and my formula equals \(3/2\) as an example.
I definitely screwed up some part of my formula. But it's probably just a couple indicator values which get compounded...
In the case of \(f(0) = 1/2\); my formula is \(f(0) + 1\). And for further derivatives \(f^{(k)}(0)\)..
There's some small error I forgot. And it's probably \(n=k\) kind of \(\delta[n=k] = 1\) value.
This'll probably point even more towards your work!
There's some small rational function that:
\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z^{k+1}}\,dz = B_k + d_k\\
\]
Where now:
\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z-x}\,dz = f(x) + \sum_{k=0}^\infty d_k x^k\\
\]
Where \(d_k\) is a small collection of \(1\)'s and \(-1\)'s I missed out on...
Fuckkkkkkkk
