Binomial expansion of $(2\cos(x))^m$

Question

This is a follow-up from the question:How do you obtain these results from $(2\cos(x))^{m}$

Let $u=\cos(x)+i\sin(x)$

$v=\cos(x)-i\sin(x)$

Then $u+v$ = $2\cos(x)$

$2^{m}\cos^{m}=(u+v)^{m}$

Using the binomial formula, one obtains

$2^{m}\cos^{m}=u^{m}+mu^{m-2}.uv+\frac{m(m-1)}{2}u^{m-4}u^2v^2+...$ (1)

But then we have $uv=[\cos(x)+i\sin(x)][\cos(x)-i\sin(x)]=\cos^2(x)+\sin^2(x)=1$

After the de Moivre formula, we have $u^m=[\cos(x)+i\sin(x)]^m=\cos(mx)+i\sin(mx)$

Substituting all these values to $2^m.cos^m(x)$, we have

$2^m\cos^m(x)=\cos(mx)+m\cos(m-2)x+\frac{m(m-1)}{2}\cos(m-4)_...i(\sin(mx)+m]sin(m-2)x+\frac{m(m-1)}{2}sin(m-4)x+...$

I just copy verbatim from Poisson's article "Note sur le developpement des puissances des sinus et des cosinus, en series de sinus ou de cosinus d'arcs multiples" which can be found here: https://books.google.com/books?id=IZytoPqRRTMC&pg=PA495&lpg=PA495&dq=Correspondance+sur+l%27%C3%89cole+polytechnique+janvier+1811&source=bl&ots=ivbxKUVHqO&sig=ACfU3U2_3gQyUxL9M_SRJCRG_MXdnPKlwA&hl=en&sa=X&ved=2ahUKEwiex8mHpqXgAhXKl-AKHerWDg8Q6AEwDnoECAEQAQ#v=onepage&q&f=false

Page 212-217.

My first question is how do you obtain the binomial series (1) in terms of $u$ and $v$. Isn't the binomial formula for $(u+v)^m$ is: $u^m+mu^{m-1}v+\frac{m(m-1)}{2!}u^{m-2}v^{2}+...muv^{m-1}+v^{m}$.

Poisson expands the series in the form that I cannot recognize instantly.

Answer 1

Yes, your expression is correct. It's just that they've rearranged each term. For example, the $2$nd term is $mu^{m-1}v = mu^{m-2}\left(uv\right)$. They've done this to get powers of $uv$ so that they can simplify it using the $uv = 1$ expression.

Answer 2

Yes, the binomial formula is $(u+v)^m=u^m+mu^{m-1}v+{\frac{m(m-1)}{2!}}u^{m-2}v^2+...$, but this can be arranged to $u^m+mu^{m-2}uv+{\frac{m(m-1)}{2!}}u^{m-4}u^2v^2+...$.