03/01/2023, 03:26 AM
(02/28/2023, 11:20 PM)Caleb Wrote: Hmm, I did a little bit of numerical testing and I've found that
\[\frac{1}{2\pi i} \int_C \frac{f(z)}{z-x} dz = -\frac{3}{2}\]
if the contour \(C\) encloses the whole natural boundary. In particular, I think we actually have that
\[\frac{1}{2\pi i} \int_C \frac{x^n}{x^n+1} \frac{1}{z-x} = 1 \]
As long as \(C\) encloses all the poles of \(x^n+1\).
Let me know if any of these computations are wrong, I did these in a rush and I'll check them later
Oh My GOD!
The plot thickens!
No, you are right!
I wrote:
\[
\frac{1}{n}\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{2\pi i \frac{jkm}{n}} = 0\\
\]
But the actual formula is:
\[
\frac{1}{n}\sum_{j=0}^{n-1} \sum_{m=1}^\infty \sqrt[n]{-1}^{-m} e^{2\pi i \frac{jkm}{n}} = 0\\
\]
When \(m=0\), the formula produces a \(1\)!
\[
\frac{1}{n}\sum_{j=0}^{n-1} 1 = 1\\
\]
....Okay so for \(|x| > 1\) the Cauchy integral formula doesn't hold.
But it does hold for \(|x| < 1\)...
God this shit makes my head spin, lmao!
I apologize, Caleb
