User:Ybab321/Sandbox

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

= Cosine Multiple Angle Formula =

Theorem

 * $\displaystyle \cos n \theta = n \sum^{\left\lfloor{\frac n 2}\right\rfloor}_{k \mathop = 0} \frac{\left({-1}\right)^k \left({n-k-1}\right)! 2^{n-2k-1} \cos^{n-2k} \theta}{k!\left({n-2k}\right)!}$

Proof 2
= Partial Fraction Decomposition =


 * $\dfrac {ax - b} {(c_1 x - d_1)(c_2 x - d_2)} = \left(\dfrac {a d_1 - b c_1} {c_2 d_1 - c_1 d_2}\right) \left(\dfrac 1 {c_1 x - d_1}\right) - \left(\dfrac {a d_2 - b c_2} {c_2 d_1 - c_1 d_2}\right) \left(\dfrac 1 {c_2 x - d_2}\right)$


 * $\dfrac {(a_1 x - b_1) (a_2 x - b_2)} {(c_1 x - d_1) (c_2 x - d_2) (c_3 x - d_3)} =

\left(\dfrac {(a_1 d_1 - b_1 c_1) (a_2 d_1 - b_2 c_1)} {(c_1 d_2 - d_1 c_2) (c_1 d_3 - d_1 c_3)}\right) \left(\dfrac 1 {c_1 x - d_1}\right) + \left(\dfrac {(a_1 d_2 - b_1 c_2) (a_2 d_2 - b_2 c_2)} {(c_2 d_3 - d_2 c_3) (c_1 d_2 - d_1 c_2)}\right) \left(\dfrac 1 {c_2 x - d_2}\right) + \left(\dfrac {(a_1 d_3 - b_1 c_3) (a_2 d_3 - b_2 c_3)} {(c_3 d_1 - d_3 c_1) (c_3 d_2 - d_3 c_2)}\right) \left(\dfrac 1 {c_3 x - d_3}\right)$

Algebra is too unmanageable to expand now

= Polynomial Expansion =

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

\{(\{\}, \bigcup^n_{i=1} \{i\})\} & : k = 0\\ \displaystyle \bigcup_{(a,b) \in f(k-1,n)} \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(k,n)} \prod_{i \in p} a_i \prod_{i \in q} b_i $

Proof by Mathematical Induction
Note that
 * $\displaystyle \bigcup_{(a,b) \in f(n-1,k)} \{(a, b \cup \{n\})\} \cup \bigcup_{(a,b) \in f(n-1,k-1)} \{(a \cup \{n\}, b)\}$

Proof by induction:

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

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

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

Induction Step
This is our induction step:

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