User:Ybab321/Sandbox

Feel free to help or even complete whatever proofs appear here.

<!-- = Polynomial Expansion =

Theorem
Let
 * $f(n,k) = \begin{cases}

\{(\{\}, \bigcup^n_{i=1} \{i\})\} & : k = 0\\ \displaystyle \bigcup_{(a,b) \in f(n,k-1)} \bigcup^n_{i=\sup(a)+1} \{(a \cup \{i\},\ b \setminus \{i\})\} & : k \ne 0 \end{cases}$

Then:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) =

\sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i $

Proof by Mathematical Induction
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) = \sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Basis for the Induction
$P \left({1}\right)$ is true, as this just says:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) = \sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Then we need to show:
 * $\displaystyle \prod^{n+1}_{k=1} (a_k x - b_k) = \sum^{n+1}_{k=0} (-1)^{n+1-k} x^k \sum_{(p,q) \in f(n+1,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Induction Step
This is our induction step:

Note that

So $P \left({n}\right) \implies P \left({n+1}\right)$ and the result follows by the Principle of Mathematical Induction. -->

From :


 * $\displaystyle \csc z = \frac 1 z + \sum^\infty_{k=1} \dfrac {(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k} z^{2k-1} } {(2k)!}$


 * $\displaystyle \csc z = \sum^\infty_{k=0} \dfrac {(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k} z^{2k-1} } {(2k)!}$


 * $\displaystyle z \csc z = \sum^\infty_{k=0} \dfrac {(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k} z^{2k} } {(2k)!}$


 * $\displaystyle \int z \csc z\, \mathrm d z = \sum^\infty_{k=0} \dfrac {(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k} z^{2k+1} } {(2k+1)!} + C$

= Cosecant Laurent series =
 * $\displaystyle \csc x = \frac 1 x + \sum^\infty_{n=1} \dfrac {(-1)^{n-1} 2 (2^{2n-1} - 1) B_{2n} x^{2n-1} } {(2n)!}$


 * $\displaystyle \csc x = \frac 1 x + \sum^\infty_{n=2} \dfrac {-i^n 2 (2^{n-1} - 1) B_{n} x^{n-1} } {n!}$


 * $\displaystyle \csc x = \sum^\infty_{n=0} \dfrac {-i^n 2 (2^{n-1} - 1) B_{n} x^{n-1} } {n!}$


 * $\displaystyle \csc x = \dfrac 1 x \sum^\infty_{n=0} \left({\dfrac {-i^n 2^n B_{n} x^n } {n!} - \dfrac {-i^n 2 B_{n} x^n} {n!} }\right)$


 * $\displaystyle \csc x = \dfrac 1 x \sum^\infty_{n=0} \left({-\dfrac {(2ix)^n B_{n} } {n!} + 2 \dfrac {(ix)^n B_{n} } {n!} }\right)$


 * $\displaystyle \csc x = \dfrac 1 x \left({-\dfrac {2ix} {e^{2ix}-1} + 2 \dfrac {ix} {e^{ix}-1} }\right)$


 * $\displaystyle \csc x = \dfrac {-2i} {e^{2ix}-1} + \dfrac {2i} {e^{ix}-1}$


 * $\displaystyle \csc x = -e^{-ix} \dfrac {2i} {e^{ix}-e^{-ix} } + e^{-i \frac x 2} \dfrac {2i} {e^{i \frac x 2}-e^{-i \frac x 2}}$


 * $\displaystyle \csc x = -e^{-ix} \csc x + e^{-i \frac x 2} \csc \dfrac x 2$


 * $\displaystyle \csc x \left({1 + e^{-ix} }\right) = e^{-i \frac x 2} \csc \dfrac x 2$


 * $\displaystyle \csc x = \dfrac {e^{-i \frac x 2} } {1 + e^{-ix} } \csc \dfrac x 2$


 * $\displaystyle \csc x = \dfrac 1 2 \dfrac 2 {e^{i \frac x 2} + e^{-i \frac x 2} } \csc \dfrac x 2$


 * $\displaystyle 2 \sin \dfrac x 2 \cos \dfrac x 2 = \sin x$


 * $\sin x = \sin x$