# 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

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?

• What do you want to show exactly the sum identity or the integral? Commented Feb 21, 2014 at 21:28
• This can also be seen as a mild restatement of the generating function for Chebyshev polynomials of the first kind. Commented Feb 21, 2014 at 21:47
• @Mhenni Benghorbal The sum identity. That's just a link to the other thread. Commented Feb 21, 2014 at 21:54
• You can use this technique. Commented Feb 22, 2014 at 2:08
• @MhenniBenghorbal: doesn't that method use complex variables?
– robjohn
Commented Feb 25, 2014 at 23:44

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

• There may be some... I agree with you that it would be much better! Commented Feb 21, 2014 at 22:07

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!

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.

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)