# 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?

Thank you!

• You have a typo in the equation. n is summed but does not appear in the summation. – Thomas Feb 7 at 7:26

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.

• Seems OP had a typo and the sum should be $\sum_{\color{red}{n\ge 0}}\dots$ – Mike Earnest Feb 6 at 20:59
• @MikeEarnest Sweet! that actually helps a lot – clathratus Feb 6 at 21:00
• Note that the issue is precisely that $a^m b^m \neq (a b)^m$. Because of this, $S_m(x) \neq 2^m \cos^m x$ (if you're working with single-valued functions). – Maxim Feb 7 at 1:42
• $2^{m}cos^{m}x$ is how it is written in the original document. – James Warthington Feb 7 at 2:34
• @JamesWarthington They're the same: $$(2\cos x)^m=2^m\cos^mx=2^m\cos(x)^m$$ it's just different totation – clathratus Feb 7 at 2:49

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 $$\arg a + \arg b \in (-\pi, \pi]$$.

• I am lost in these comments. I haven't learnt complex exponential so my knowledge is so limited. Thank you though. – James Warthington Feb 7 at 19:11