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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!

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

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    $\begingroup$ Seems OP had a typo and the sum should be $\sum_{\color{red}{n\ge 0}}\dots$ $\endgroup$ – Mike Earnest Feb 6 at 20:59
  • $\begingroup$ @MikeEarnest Sweet! that actually helps a lot $\endgroup$ – clathratus Feb 6 at 21:00
  • $\begingroup$ 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). $\endgroup$ – Maxim Feb 7 at 1:42
  • $\begingroup$ $2^{m}cos^{m}x$ is how it is written in the original document. $\endgroup$ – James Warthington Feb 7 at 2:34
  • $\begingroup$ @JamesWarthington They're the same: $$(2\cos x)^m=2^m\cos^mx=2^m\cos(x)^m$$ it's just different totation $\endgroup$ – clathratus Feb 7 at 2:49
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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 $\arg a + \arg b \in (-\pi, \pi]$.

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

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