For a Calculus course i am taking, we are going through successive differentiation($n^{th}$-derivatives if you will). I came across this technique, given below, for successive derivatives of functions of the form $cos^px,sin^px ,sin^pxcos^qx$:
- First,| we let z=cosx+isinx $\implies z^{-1}$=cosx-isinx | so we have that $$z+z^{-1}=2cosx,and z-z^{-1}=2i*sinx....(1) $$ Further, we have: $ z^r+z^{-r}=2cos(rx),z^r-z^{-r}=2i*sin(rx)$......(2)
this first step was okay to understand. But then they did something i didnt quite understand:
- let $y=cos^px \implies 2^p*y=(2cosx)^p$ $$=(z+z^{-1})^p=(z^p+z^{-p})+C^p_1(z^{p-2}+z^{-(p-2)})+C^p_2(z^{p-4}+z^{-(p-4)})+\cdots$$ then by using (1)&(2) we will go on to evaluate the successive derivatives.
My question is, how did they expand out the $y=cos^px \implies 2^p*y=(2cosx)^p$ term into a seemingly binomial theorem-esque expansion?. What theorem(s) did they employ to write it out like that? I haven't taken a complex analysis course yet and what i do know about complex numbers is from Intro. ODE courses for my major requirement. so any further resources from which i could learn more about this would be appreciated.
I think it's just the binomial theorem. $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$. When $b = a^{-1}$ then this is $\sum_{k=0}^n \binom{n}{k} a^{n-k} a^{-k} = \sum_{k=0}^n \binom{n}{k} a^{n-2k}$. Because $\binom{x}{y} = \binom{x}{x-y}$, when $n$ is odd, you can perfectly pair each positive power of $a$ (e.g. $a^3$) in this sum with the corresponding same negative power of $a$ (e.g. $a^{-3}$) having the same coefficient. When $n$ is even you can also do this but there is also a constant term.