Simplify $(i \theta)^n+(- i \theta)^n$

Question

I need help simplifying this:

$(i \theta)^n+(- i \theta)^n$ with the sum going from $n=0$ to $n=\infty$ This is part of this, sum from n=0 to infinity of (1/4) 2^n((I theta)^n + (-I theta$^n))/n! +(1/2) and I want to simplify this to show the maclaurin series for cos^2(z)

Answer 1

Hint:

• What happens when $n$ is odd? What about when it is even? Simplify the term.
• Geometric series.

Alternatively work on two sums separately and then sum them up later.

Answer 2

Note that

$n=1\implies (i \theta)+(- i \theta)=0$

$n=2\implies (- \theta^2)+(- \theta^2)=-2\theta^2$

$n=3\implies (-i \theta^3)+(i \theta^3)=0$

$n=4\implies (\theta^4)+(\theta^4)=2\theta^4$

...

thus, we can prove by induction, that

• for $n$ odd $(i \theta)^n+(- i \theta)^n =0$

• for $n$ even $(i \theta)^n+(- i \theta)^n =2(-\theta^2)^{n/2}$

thus set n=2k and calculate the

$$\sum_{k=1}^{\infty} 2(-\theta^2)^{k}$$

by geometric series.