Question #b9e60

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Answer 1 · 2017-04-15T09:06:35

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

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.