Question

Question #b9e60

Answer 1

`f''(x)=-n^4 e^(i n^2x)`

Explanation:

By Euler's Formula: `cos nx+i sin nx=e^(i nx)`,

`f(x)=e^(ix)cdot e^(i 3x)cdot e^(i 5x)cdots e^(i (2n-1)x)=e^(i[1+3+5+cdots (2n-1)]x)`

By the sum formula: `1+3+5+cdots+(2n-1)=n^2`,

`f(x)=e^(i n^2x)`

By differentiating w.r.t. `x`,

`f'(x)=i n^2 e^(i n^2x)`

By differentiating w.r.t.`x` one more time,

`f''(x)=i^2 n^4 e^(i n^2x)=-n^4e^(i n^2x)`

I hope that this was clear.