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

The identity
$$\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({k}{x}\right)}}}={\frac{{{1}-{a}{\cos{{x}}}}}{{{1}-{2}{a}{\cos{{x}}}+{a}^{{2}}}}},{\left|{a}\right|}{<}{1}$$</span>
can be derived by using the fact that $$\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({k}{x}\right)}}}={R}{e}{\sum_{{{k}={0}}}^{{\infty}}}{\left({a}{e}^{{{i}{x}}}\right)}^{{k}}$$
But can it be derived without using complex variables?

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Linda Tincher
Here is a very inelegant proof:
$$\displaystyle{\left({1}-{2}{a}{\cos{{x}}}+{a}^{{2}}\right)}\times{\sum_{{{k}={0}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({k}{x}\right)}}}$$
$$\displaystyle={\sum_{{{k}={0}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({k}{x}\right)}}}-{2}{\sum_{{{k}={1}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({\left({k}-{1}\right)}{x}\right)}}}{\cos{{x}}}+{\sum_{{{k}={2}}}^{{\infty}}}{a}^{{k}}{\cos{{\left({\left({k}-{2}\right)}{x}\right)}}}$$
$$\displaystyle={1}-{a}{\cos{{x}}}+{\sum_{{{k}={2}}}^{{\infty}}}{a}^{{k}}{\left[{\cos{{\left({k}{x}\right)}}}-{2}{\cos{{\left({\left({k}-{1}\right)}{x}\right)}}}{\cos{{x}}}+{\cos{{\left({k}-{2}\right)}}}{x}\right]}$$
$$\displaystyle={1}-{a}{\cos{{x}}}$$
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Fesion

Using the identity,
$$\sum_{k=0}^{\infty} a^k \cos(kx)=\frac{1-a \cos x}{1-2a \cos x+a^2}, |a|<1$$
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:
$\sum_{n=0}^{\infty} a^n \cos(nx)=\sum_{n=0}^{\infty}a^n \sum_{k=0}^{[\frac{n}{2}]}(-1)^k \begin{pmatrix}n \\2k \end{pmatrix} \sin^{2k}(x)\cos^{n-2k}(x)$
$=\sum_{n=0}^{\infty}\sum_{k=0}^{[\frac{n}{2}]}(-1)^k \begin{pmatrix}n \\2k \end{pmatrix}a^n \sin^{2k} (x)\cos^{n-2k}(x)$
$=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^k \begin{pmatrix}2k+n \\2k \end{pmatrix} a^{2k+n}\sin^{2k}(x)\cos^n(x)$
$=\sum_{k=0}^{\infty}(-1)^k a^{2k} \sin^{2k}(x) \sum_{n=0}^{\infty} \begin{pmatrix}2k+n \\2k \end{pmatrix}[a \cos(x)]^n$
$$\displaystyle={\sum_{{{k}={0}}}^{{\infty}}}{\left(-{1}\right)}^{{k}}{a}^{{{2}{k}}}{{\sin}^{{{2}{k}}}{\left({x}\right)}}{\frac{{{1}}}{{{\left({1}-{a}{\cos{{\left({x}\right)}}}\right)}^{{{2}{k}+{1}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{1}-{a}{\cos{{\left({x}\right)}}}}}}{\sum_{{{k}={0}}}^{{\infty}}}{\left(-{1}\right)}^{{k}}{\left[{\frac{{{a}{\sin{{\left({x}\right)}}}}}{{{1}-{a}{\cos{{\left({x}\right)}}}}}}\right]}^{{{2}{k}}}$$
$$\displaystyle={\frac{{{1}}}{{{1}-{a}{\cos{{\left({x}\right)}}}}}}\cdot{\frac{{{1}}}{{{1}+{\left[{\frac{{{a}{\sin{{\left({x}\right)}}}}}{{{1}-{a}{\cos{{\left({x}\right)}}}}}}\right]}^{{{2}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{1}-{a}{\cos{{\left({x}\right)}}}}}}\cdot{\frac{{{\left({1}-{a}{\cos{{\left({x}\right)}}}\right)}^{{2}}}}{{{\left({1}-{a}{\cos{{\left({x}\right)}}}\right)}^{{2}}+{a}^{{2}}{{\sin}^{{2}}{\left({x}\right)}}}}}$$
$$\displaystyle={\frac{{{1}-{a}{\cos{{\left({x}\right)}}}}}{{{1}-{2}{a}{\cos{{\left({x}\right)}}}+{a}^{{2}}{{\cos}^{{2}}{\left({x}\right)}}+{a}^{{2}}{{\sin}^{{2}}{\left({x}\right)}}}}}$$
$$=\frac{1-a \cos (x)}{1-2a \cos (x)+a^2}$$