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I want to prove the following;

$$\sum_{k=0}^{\infty}a^k\cos{kx}=\frac{1-a\cos{x}}{1-2a\cos{x}+a^2}, \; |a|<1 \tag{1}$$

I know that;

$$\text{S}(r)=\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}, \; |r|<1$$

We can use this expression by rewriting (1);

$$\sum_{k=0}^{\infty}a^k\cos{kx}=\text{Re}\left[\sum_{k=0}^{\infty}(a\text{e}^{\pm ix})^k\right]=\text{Re}\left[\text{S}(a\text{e}^{\pm ix})\right]$$

Evaluating this sum we get;

$$\text{Re}\left[\text{S}(a\text{e}^{\pm ix})\right]=\text{Re}\left[\frac{1}{1-[a \text{e}^{\pm ix}]}\right]=\text{Re}\left[\frac{1-a\text{e}^{\pm ix}}{\left(1-a\text{e}^{\pm ix}\right)^2}\right]=\text{Re}\left[\frac{1-a\text{e}^{\pm ix}}{1+(a\text{e}^{\pm ix})^2-2a\text{e}^{\pm ix}}\right] \tag{3}$$

Now this would, in fact, be eq. (1) if; $1+(a\text{e}^{\pm ix})^2-2a\text{e}^{\pm ix}=1-2a\cos{x}+a^2$, however, I cannot see how we get rid of the two terms containing $\text{e}^{\pm ix}$, after all one term depends on $a$ while the other depends on $a^2$. Do anyone have an idea/a trick to use to derive eq. (1) from my current work?

Rasmus Andersen
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    To find the rael part of $\frac 1 z$ you don't multiply numerator and denominator by $z$. You multiply numerator and denominator by $\overline z$. – geetha290krm Jul 24 '23 at 09:44
  • Helpful info here: https://math.stackexchange.com/questions/685278/ – David H Jul 24 '23 at 09:46
  • Thank you. I successfully derived eq. (1) by multiplying the numerator and denominator of $\frac{1}{z}$ by $\bar{z}$. – Rasmus Andersen Jul 24 '23 at 10:02
  • $$\sum_{k=1}^\infty\left(ae^{ix}\right)^k=\frac{1}{1-ae^{ix}}\frac{1-ae^{-ix}}{1-ae^{-ix}}=\frac{1-ae^{-ix}}{1-a(e^{-ix}+e^{ix})+a^2}=\frac{1-ae^{-ix}}{1-2a\cos(x)+a^2}.$$ – pshmath0 Jul 25 '23 at 17:45

4 Answers4

3

The usual trick of introducing $\cos(x) =\Re(e^{ix}) $.

$\begin{array}\\ s(x, a) &=\sum_{k=0}^{\infty}a^k\cos{kx}\\ &=\sum_{k=0}^{\infty}a^k\Re(e^{ikx})\\ &=\Re(\sum_{k=0}^{\infty}a^ke^{ikx})\\ &=\Re(\sum_{k=0}^{\infty}(ae^{ix})^k)\\ &=\Re(\dfrac{1}{1-ae^{ix}})\\ &=\Re(\dfrac{1}{1-a\cos(x)-ai\sin(x)})\\ &=\Re(\dfrac{1}{1-a\cos(x)-ai\sin(x)})\dfrac{1-a\cos(x)+ai\sin(x)}{1-a\cos(x)+ai\sin(x)})\\ &=\Re(\dfrac{1-a\cos(x)+ai\sin(x)}{(1-a\cos(x))^2+a^2\sin^2(x)})\\ &=\Re(\dfrac{1-a\cos(x)+ai\sin(x)}{1+2a\cos(x)+a^2\cos^2(x)+a^2\sin^2(x)})\\ &=\Re(\dfrac{1-a\cos(x)+ai\sin(x)}{1+2a\cos(x)+a^2})\\ &=\dfrac{1-a\cos(x)}{1+2a\cos(x)+a^2})\\ \\ \text{also}\\ s(x, a) &=\sum_{k=0}^{\infty}a^k\sin{kx}\\ &=\sum_{k=0}^{\infty}a^k\Im(e^{ikx})\\ &=\Im(\dfrac{1-a\cos(x)+ai\sin(x)}{1+2a\cos(x)+a^2})\\ &=\dfrac{a\sin(x)}{1+2a\cos(x)+a^2})\\ \end{array} $

marty cohen
  • 107,799
1

An elementary proof not using complex numbers.

$\begin{array}\\ d(x, a) &=(1-2a\cos{x}+a^2)\sum_{k=0}^{\infty}a^k\cos{kx}\\ &=\sum_{k=0}^{\infty}a^k\cos{kx}-2a\cos{x}\sum_{k=0}^{\infty}a^k\cos{kx}+a^2\sum_{k=0}^{\infty}a^k\cos{kx}\\ &=\sum_{k=0}^{\infty}a^k\cos{kx}-2\cos{x}\sum_{k=0}^{\infty}a^{k+1}\cos{kx}+\sum_{k=0}^{\infty}a^{k+2}\cos{kx}\\ &=1+a\cos(x)+\sum_{k=2}^{\infty}a^k\cos{kx}-2\cos{x}\sum_{k=1}^{\infty}a^{k}\cos{(k-1)x}+\sum_{k=2}^{\infty}a^{k}\cos{(k-2)x}\\ &=1+a\cos(x)+\sum_{k=2}^{\infty}a^k\cos{kx}-2\cos{x}(a+\sum_{k=2}^{\infty}a^{k}\cos{(k-1)x})+\sum_{k=2}^{\infty}a^{k}\cos{(k-2)x}\\ &=1+a\cos(x)-2a\cos(x)+\sum_{k=2}^{\infty}a^k\cos{kx}-2\cos{x}(\sum_{k=2}^{\infty}a^{k}\cos{(k-1)x})+\sum_{k=2}^{\infty}a^{k}\cos{(k-2)x}\\ &=1-a\cos(x)+\sum_{k=2}^{\infty}a^k(cos(kx)-2\cos(x)\cos((k-1)x)+\cos((k-2)x)\\ &=1-a\cos(x) \qquad\text{(since }\cos(kx)=2\cos(x)\cos((k-1)x)-\cos((k-2)x))\\ \end{array} $

marty cohen
  • 107,799
0

In eq. (3), instead of multiplying $\frac{1}{z}$ by $z$ in the numerator and denominator, we multiply by $\bar{z}$ in the numerator and denominator. This yields;

$$\text{Re}\left[\frac{1}{1-a\cos{x}-ai\sin{x}}\right]=\text{Re}\left[\frac{1-a\cos{x}+ai\sin{x}}{(1-a\cos{x})^2+(a\sin{x})^2}\right]=\text{Re}\left[\frac{1-a\cos{x}+ai\sin{x}}{1+a^2-2a\cos{x}}\right]=\frac{1-a\cos{x}}{1+a^2-2a\cos{x}}$$

$\textbf{Note}$; For a complex number, $z=a+ib$, we have $|z|^2=\left[\text{Re}(z)\right]^2+\left[\text{Im}(z)\right]^2$

Which proves that; $\sum_{k=0}^{\infty}a^k\cos{kx}=\frac{1-a\cos{x}}{1-2a\cos{x}+a^2}$

Rasmus Andersen
  • 279
0

A more general case, with an obvious relation to the power series for a linear recurrence relation.

If $\sum_{j=0}^m g_j(x)f_{k-j}(x) =0$ then $\sum_{k=0}^{\infty}a^kf_k(x) =\dfrac{\sum_{k=0}^{m-1}a^k(\sum_{j=0}^k g_j(x)f_{k-j}(x))}{\sum_{j=0}^m a^jg_j(x)} $.

Standard examples with $f_k(x)=f(kx)$:

$(f(x), m, recurrence, g_j(x), result)\\ (\cos(x). 2, 0=\cos(kx)-2\cos(x)\cos((k-1)x)+\cos((k-2)x), (1, -2\cos(x), 1),\dfrac{1-a\cos(x)}{1-2a\cos(x)+a^2})\\ (\sin(x). 2, 0=\sin(kx)-2\cos(x)\sin((k-1)x)+\sin((k-2)x), (1, -2\cos(x), 1), \dfrac{a\sin(x)}{1-2a\cos(x)+a^2}) $

The derivation.

$\begin{array}\\ \sum_{j=0}^m a^jg_j(x)\sum_{k=0}^{\infty}a^kf_k(x) &=\sum_{j=0}^m g_j(x)\sum_{k=0}^{\infty}a^{k+j}f_k(x)\\ &=\sum_{j=0}^m g_j(x)\sum_{k=j}^{\infty}a^{k}f_{k-j}(x)\\ &=\sum_{j=0}^m g_j(x)(\sum_{k=j}^{m}a^{k}f_{k-j}(x)+\sum_{k=m+1}^{\infty}a^{k}f_{k-j}(x))\\ &=\sum_{j=0}^m g_j(x)\sum_{k=j}^{m}a^{k}f_{k-j}(x)+\sum_{j=0}^m g_j(x)\sum_{k=m+1}^{\infty}a^{k}f_{k-j}(x)\\ &=\sum_{k=0}^{m}\sum_{j=0}^k g_j(x)a^{k}f_{k-j}(x)+\sum_{k=m+1}^{\infty}a^{k}\sum_{j=0}^m g_j(x)f_{k-j}(x)\\ &=\sum_{k=0}^{m}a^{k}\sum_{j=0}^k g_j(x)f_{k-j}(x)\\ \end{array} $

so $\sum_{k=0}^{\infty}a^kf_k(x) =\dfrac{\sum_{k=0}^{m}a^{k}\sum_{j=0}^k g_j(x)f_{k-j}(x)}{\sum_{j=0}^m a^jg_j(x)} $.

Coefficients of $a^k$:

$\begin{array}\\ k &\sum_{j=0}^k g_j(x)f_{k-j}(x)\\ 0 &g_0(x)f_0(x)\\ 1 &g_0(x)f_1(x)+g_1(x)f_0(x)\\ 2 &g_0(x)f_2(x)+g_1(x)f_1(x)+g_2(x)f_0(x)\\ \end{array} $

marty cohen
  • 107,799