kiki195ms

Answered

2021-11-19

The identity

$\sum _{k=0}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left(kx\right)=\frac{1-a\mathrm{cos}x}{1-2a\mathrm{cos}x+{a}^{2}},\left|a\right|<1$

can be derived by using the fact that$\sum _{k=0}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left(kx\right)=Re\sum _{k=0}^{\mathrm{\infty}}{\left(a{e}^{ix}\right)}^{k}$

But can it be derived without using complex variables?

can be derived by using the fact that

But can it be derived without using complex variables?

Answer & Explanation

Linda Tincher

Expert

2021-11-20Added 14 answers

Here is a very inelegant proof:

$(1-2a\mathrm{cos}x+{a}^{2})\times \sum _{k=0}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left(kx\right)$

$=\sum _{k=0}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left(kx\right)-2\sum _{k=1}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left((k-1)x\right)\mathrm{cos}x+\sum _{k=2}^{\mathrm{\infty}}{a}^{k}\mathrm{cos}\left((k-2)x\right)$

$=1-a\mathrm{cos}x+\sum _{k=2}^{\mathrm{\infty}}{a}^{k}[\mathrm{cos}\left(kx\right)-2\mathrm{cos}\left((k-1)x\right)\mathrm{cos}x+\mathrm{cos}(k-2)x]$

$=1-a\mathrm{cos}x$

Fesion

Expert

2021-11-21Added 24 answers

Using the identity,

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:

Most Popular Questions