How do you obtain these results from $(2\cos(x))^{m}$ In the book "The rise and development of the theory of series up to the early 1820s", the author mentions a paradox that is sometimes known as Poisson Paradox.
$$(2\cos(x))^{m}=\sum_{n=1}^{\infty} \binom{m}{k}\cos(m-2k)x $$
The author wrote that: "In his [1811], Poisson showed that one contradiction  derived from the substitution $m=\frac{1}{3}$ and $x=\pi$, Indeed, from $(2\cos(x))^{m})$ it was possible to obtain the three complex values:
$2^{1/3}\frac{1+\sqrt{-3}}{2}$, $-2^{1/3}$, $2^{1/3}\frac{1-\sqrt{-3}}{2}$
whereas the series gave the value $-2^{1/3}$. According to Poisson, the problem arose from the fact that, when $m$ is a rational number, the series
$$\sum_{n=1}^{\infty} \binom{m}{k}\cos(m-2k)x $$
only represented the real part of the expansion $(2\cos(x))^{m}$
I have consulted the work of Poisson and here is how he expands it.
He lets $u=\cos(x)+i\sin(x)$, $v=\cos(x)-i\sin(x)$
$u+v=2\cos(x)$
$(u+v)^m=(2\cos(x))^{m}$
He then expands the left hand side according to the binomial formula. The expansion is blurred out in the original text so if someone can help me to reconstruct the series, I will appreciate it.
My question is:
1) How does one obtain the binomial expansion of $(2\cos(x))^{m}$
2) How does one obtain those three complex values from the series?
3) How does this represent a paradox in the theory of infinite series?
Poisson's original article is here page 212-217:
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
Thank you!
 A: We define 
$$R_m(x)=\sum_{n\geq0}{m\choose n}\cos\left[(m-2n)x\right]$$
And $$S_m(x)=\sum_{n\geq0}{m\choose n}e^{i(m-2n)x}$$
So we see that $$R_m(x)=\text{Re}\,S_m(x)$$
Then we see that 
$$S_m(x)=e^{imx}\sum_{n\geq0}{m\choose n}e^{-2inx}$$
Then we recall that 
$$(x+1)^a=\sum_{n\geq0}{a\choose n}x^n$$
So we see that 
$$S_m(x)=e^{imx}\left[\left(e^{-2ix}+1\right)^m\right]$$
$$S_m(x)=\left(e^{ix}+e^{-ix}\right)^m$$
$$S_m(x)=2^{m}\cos(x)^m$$
So we have that 
$$R_m(x)=(2\cos x)^m$$
If we plug in $m=1/3$ and $x=\pi$, we get 
$$R_{1/3}(\pi)=\sqrt[3]{2}$$
And I assume that the three values are the three complex roots of $\sqrt[3]{2}$. To find these we recall the formula 
$$z^{1/n}=\sqrt[n]{|z|}\exp\left[\frac{i}{n}(2\pi k+\arg z)\right],\qquad k=0,1,...,n-1$$
Setting $z=2$ and $n=3$, we have that $|z|=2$ and $\arg z=0$ which qives
$$2^{1/3}=\sqrt[3]{2}\exp\frac{2i\pi k}3$$
for every $k=0,1,2$. So, using $e^{ix}=\cos x+i\sin x$,
we have that the three complex roots of $2^{1/3}=\sqrt[3]{2}$ are given by 
$$2^{1/3}\frac{1+i\sqrt3}2,2^{1/3}, -2^{1/3}\frac{1-i\sqrt3}2$$
as desired.
A: Let $z^m$ be defined as $z^m = e^{m \ln z}$, where $\ln$ is the principal value of the logarithm. Then, assuming $x$ and $m$ are real,
$$(e^{i x})^m = e^{i m \arg e^{i x}},$$
which is not necessarily the same as $e^{i m x}$.
For the same reason, assuming that the sums converge,
$$\sum_{k \geq 0} \binom m k a^k b^{m - k} =
b^m \sum_{k \geq 0} \binom m k \left( \frac a b \right)^k =
b^m \left( 1 + \frac a b \right)^m$$
is not necessarily the same as $(a + b)^m$.
We can write
$$(2 \cos x)^m =
(e^{i x} + e^{-i x})^m =
\sum_{k \geq 0} \binom m k e^{i (m - k) x} e^{-i k x} = \\
\sum_{k \geq 0} \binom m k \cos \,(m - 2 k) x +
 i \sum_{k \geq 0} \binom m k \sin \,(m - 2 k) x$$
or
$$(2 \cos x)^m =
(e^{-i x} + e^{i x})^m =
\sum_{k \geq 0} \binom m k e^{-i (m - k) x} e^{i k x} = \\
\sum_{k \geq 0} \binom m k \cos \,(m - 2 k) x -
 i \sum_{k \geq 0} \binom m k \sin \,(m - 2 k) x$$
only if we pick $x$ and $m$ for which the intermediate steps are valid. If $m \not \in \mathbb Z$, then $(e^{i x})^m = e^{i m x}$ can be used if $x \in (-\pi, \pi]$ and $(a b)^m = a^m b^m$ can be used if $\arg a + \arg b \in (-\pi, \pi]$.
