kiki195ms

2021-11-19

The identity
$\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)=\frac{1-a\mathrm{cos}x}{1-2a\mathrm{cos}x+{a}^{2}},|a|<1$
can be derived by using the fact that $\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)=Re\sum _{k=0}^{\mathrm{\infty }}{\left(a{e}^{ix}\right)}^{k}$
But can it be derived without using complex variables?

Linda Tincher

Here is a very inelegant proof:
$\left(1-2a\mathrm{cos}x+{a}^{2}\right)×\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)$
$=\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)-2\sum _{k=1}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(\left(k-1\right)x\right)\mathrm{cos}x+\sum _{k=2}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(\left(k-2\right)x\right)$
$=1-a\mathrm{cos}x+\sum _{k=2}^{\mathrm{\infty }}{a}^{k}\left[\mathrm{cos}\left(kx\right)-2\mathrm{cos}\left(\left(k-1\right)x\right)\mathrm{cos}x+\mathrm{cos}\left(k-2\right)x\right]$
$=1-a\mathrm{cos}x$

Fesion

Using the identity,
$\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)=\frac{1-a\mathrm{cos}x}{1-2a\mathrm{cos}x+{a}^{2}},|a|<1$
the infinite series in question may be rewritten as a double infinite series over a triangle. Changing the order of summation (if you're like me and the transformation gymnastics with multiple indices makes you dizzy, here's a very handy cheat-sheet), we're left with fairly elementary summations:
$\sum _{n=0}^{\mathrm{\infty }}{a}^{n}\mathrm{cos}\left(nx\right)=\sum _{n=0}^{\mathrm{\infty }}{a}^{n}\sum _{k=0}^{\left[\frac{n}{2}\right]}\left(-1{\right)}^{k}\left(\begin{array}{c}n\\ 2k\end{array}\right){\mathrm{sin}}^{2k}\left(x\right){\mathrm{cos}}^{n-2k}\left(x\right)$
$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\left[\frac{n}{2}\right]}\left(-1{\right)}^{k}\left(\begin{array}{c}n\\ 2k\end{array}\right){a}^{n}{\mathrm{sin}}^{2k}\left(x\right){\mathrm{cos}}^{n-2k}\left(x\right)$
$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\left(\begin{array}{c}2k+n\\ 2k\end{array}\right){a}^{2k+n}{\mathrm{sin}}^{2k}\left(x\right){\mathrm{cos}}^{n}\left(x\right)$
$=\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}{a}^{2k}{\mathrm{sin}}^{2k}\left(x\right)\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}2k+n\\ 2k\end{array}\right)\left[a\mathrm{cos}\left(x\right){\right]}^{n}$
$=\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}{a}^{2k}{\mathrm{sin}}^{2k}\left(x\right)\frac{1}{{\left(1-a\mathrm{cos}\left(x\right)\right)}^{2k+1}}$
$=\frac{1}{1-a\mathrm{cos}\left(x\right)}\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}{\left[\frac{a\mathrm{sin}\left(x\right)}{1-a\mathrm{cos}\left(x\right)}\right]}^{2k}$
$=\frac{1}{1-a\mathrm{cos}\left(x\right)}\cdot \frac{1}{1+{\left[\frac{a\mathrm{sin}\left(x\right)}{1-a\mathrm{cos}\left(x\right)}\right]}^{2}}$
$=\frac{1}{1-a\mathrm{cos}\left(x\right)}\cdot \frac{{\left(1-a\mathrm{cos}\left(x\right)\right)}^{2}}{{\left(1-a\mathrm{cos}\left(x\right)\right)}^{2}+{a}^{2}{\mathrm{sin}}^{2}\left(x\right)}$

Do you have a similar question?