Showing that $\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}}$ without using complex variables

Question

The identity $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} \ , \ |a| <1$$

can be derived by using the fact that $ \displaystyle \sum_{k=0}^{\infty} a^{k} \cos(kx) = \text{Re} \sum_{k=0}^{\infty} (ae^{ix})^{k}$.

But can it be derived without using complex variables?

Answer 1

Here is a very inelegant proof: \begin{eqnarray} (1-2a\cos x+a^2)&\times&\sum_{k=0}^\infty a^k\cos(kx) \\&=&\sum_{k=0}^\infty a^k\cos(kx)-2\sum_{k=1}^\infty a^k\cos((k-1)x)\cos x+\sum_{k=2}^\infty a^k\cos((k-2)x)\\ &=&1-a\cos x+\sum_{k=2}^\infty a^k\left[\cos(kx)-2\cos((k-1)x)\cos x+\cos(k-2)x) \right]\\ &=&1-a\cos x\, . \end{eqnarray}

Edit. I just realize that this is LutzL's answer

Answer 2

Using the identity,

$$\cos{\left(nx\right)}=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)},$$

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:

$$\begin{align} \sum_{n=0}^{\infty}a^{n}\cos{\left(nx\right)} &=\sum_{n=0}^{\infty}a^{n}\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}a^{n}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^k\binom{2k+n}{2k}a^{2k+n}\sin^{2k}{\left(x\right)}\cos^{n}{\left(x\right)}\\ &=\sum_{k=0}^{\infty}(-1)^ka^{2k}\sin^{2k}{\left(x\right)}\sum_{n=0}^{\infty}\binom{2k+n}{2k}\left[a\cos{\left(x\right)}\right]^n\\ &=\sum_{k=0}^{\infty}(-1)^ka^{2k}\sin^{2k}{\left(x\right)}\frac{1}{\left(1-a\cos{\left(x\right)}\right)^{2k+1}}\\ &=\frac{1}{1-a\cos{\left(x\right)}}\sum_{k=0}^{\infty}(-1)^k\left[\frac{a\sin{\left(x\right)}}{1-a\cos{\left(x\right)}}\right]^{2k}\\ &=\frac{1}{1-a\cos{\left(x\right)}}\cdot\frac{1}{1+\left[\frac{a\sin{\left(x\right)}}{1-a\cos{\left(x\right)}}\right]^{2}}\\ &=\frac{1}{1-a\cos{\left(x\right)}}\cdot\frac{\left(1-a\cos{\left(x\right)}\right)^2}{\left(1-a\cos{\left(x\right)}\right)^2+a^2\sin^2{\left(x\right)}}\\ &=\frac{1-a\cos{\left(x\right)}}{1-2a\cos{\left(x\right)}+a^2\cos^2{\left(x\right)}+a^2\sin^2{\left(x\right)}}\\ &=\frac{1-a\cos{\left(x\right)}}{1-2a\cos{\left(x\right)}+a^2}.~~\blacksquare\\ \end{align}$$

Happy holidays!

Answer 3

This has a standard approach: You multiply with the denominator of the right fraction, sort the product with the series on the left by powers of $a$ and apply trigonometric identities to the resulting coefficients of $a^n$. Everything should cancel except the lowest order terms that constitute the numerator of the right side.

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One should also note that this is the Poisson kernel of the Abel summation method for Fourier series, see for instance Carl Offner: "A little harmonic analysis" (Online PDF article)