The identity ∑k=0∞akcos⁡(kx)=1−acos⁡x1−2acos⁡x+a2,|a|&lt;1 can be derived by using the fact that ∑k=0∞akcos⁡(kx)=Re∑k=0∞(aeix)k But can it be derived without using complex variables?

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The identity   ∑k=0∞akcos⁡(kx)=1−acos⁡x1−2acos⁡x+a2,|a|&amp;lt;1   can be derived by using the fact that  ∑k=0∞akcos⁡(kx)=Re∑k=0∞(aeix)k    But can it be derived without using complex variables?

Linda Tincher · Accepted Answer

Here is a very inelegant proof:  (1−2acos⁡x+a2)×∑k=0∞akcos⁡(kx)  =∑k=0∞akcos⁡(kx)−2∑k=1∞akcos⁡((k−1)x)cos⁡x+∑k=2∞akcos⁡((k−2)x)  =1−acos⁡x+∑k=2∞ak[cos⁡(kx)−2cos⁡((k−1)x)cos⁡x+cos⁡(k−2)x]  =1−acos⁡x

Fesion · Accepted Answer

Using the identity, ∑k=0∞akcos⁡(kx)=1−acos⁡x1−2acos⁡x+a2,|a|&amp;lt;1 the infinite series in question may be rewritten as a double infinite series over a triangle. Changing the order of summation (if you&#039;re like me and the transformation gymnastics with multiple indices makes you dizzy, here&#039;s a very handy cheat-sheet), we&#039;re left with fairly elementary summations: ∑n=0∞ancos⁡(nx)=∑n=0∞an∑k=0[n2](−1)k(n2k)sin2k⁡(x)cosn−2k⁡(x) =∑n=0∞∑k=0[n2](−1)k(n2k)ansin2k⁡(x)cosn−2k⁡(x) =∑n=0∞∑k=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(1−acos⁡(x))2k+1 =11−acos⁡(x)∑k=0∞(−1)k[asin⁡(x)1−acos⁡(x)]2k =11−acos⁡(x)⋅11+[asin⁡(x)1−acos⁡(x)]2 =11−acos⁡(x)⋅(1−acos⁡(x))2(1−acos⁡(x))2+a2sin2(x)

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

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