The identity ∑k=0∞akcos⁡(kx)=1−acos⁡x1−2acos⁡x+a2,|a|<1 can be derived by using the fact that ∑k=0∞akcos⁡(kx)=Re∑k=0∞(aeix)k But can it...

kiki195ms

kiki195ms

Answered

2021-11-19

The identity
k=0akcos(kx)=1acosx12acosx+a2,|a|<1
can be derived by using the fact that k=0akcos(kx)=Rek=0(aeix)k
But can it be derived without using complex variables?

Answer & Explanation

Linda Tincher

Linda Tincher

Expert

2021-11-20Added 14 answers

Here is a very inelegant proof:
(12acosx+a2)×k=0akcos(kx)
=k=0akcos(kx)2k=1akcos((k1)x)cosx+k=2akcos((k2)x)
=1acosx+k=2ak[cos(kx)2cos((k1)x)cosx+cos(k2)x]
=1acosx
Fesion

Fesion

Expert

2021-11-21Added 24 answers

Using the identity,
k=0akcos(kx)=1acosx12acosx+a2,|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:
n=0ancos(nx)=n=0ank=0[n2](1)k(n2k)sin2k(x)cosn2k(x)
=n=0k=0[n2](1)k(n2k)ansin2k(x)cosn2k(x)
=n=0k=0(1)k(2k+n2k)a2k+nsin2k(x)cosn(x)
=k=0(1)ka2ksin2k(x)n=0(2k+n2k)[acos(x)]n
=k=0(1)ka2ksin2k(x)1(1acos(x))2k+1
=11acos(x)k=0(1)k[asin(x)1acos(x)]2k
=11acos(x)11+[asin(x)1acos(x)]2
=11acos(x)(1acos(x))2(1acos(x))2+a2sin2(x)

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