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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?

Random Variable
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  • What do you want to show exactly the sum identity or the integral? – Mhenni Benghorbal Feb 21 '14 at 21:28
  • This can also be seen as a mild restatement of the generating function for Chebyshev polynomials of the first kind. – Eugene Shvarts Feb 21 '14 at 21:47
  • @Mhenni Benghorbal The sum identity. That's just a link to the other thread. – Random Variable Feb 21 '14 at 21:54
  • You can use this technique. – Mhenni Benghorbal Feb 22 '14 at 02:08
  • @MhenniBenghorbal: doesn't that method use complex variables? – robjohn Feb 25 '14 at 23:44
  • @RandomVariable I was happy to see my response below sufficiently satisfied Pranav for him to award his bounty (though he may have done so somewhat by default since I was the only user to attempt a response after he posted the bounty). But since this was originally you're question, I would also be interested in feedback from you. Is there any way in which you find it unsatisfying or think it might be improved? The reason I take such an interest here is (as you likely noticed) this is a key lemma in solving many integrals by purely real methods, and I anticipate citing it in the future. Thanks! – David H Dec 30 '14 at 06:30
  • @DavidH If I were asked to prove the multiple angle formula for cosine, I would use Euler's formula. But I guess you could prove it by induction without using Euler's formula. – Random Variable Dec 30 '14 at 09:15

3 Answers3

16

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

Etienne
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12

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!

David H
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7

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)

Lutz Lehmann
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