Question

If $f\left(x\right)=(\cos x+i\sin x)(cos3x+i\sin 3x)\dots$$\left(\cos \right(2n-1)x+i\sin (2n-1\left)x\right)$, then $f^{\prime \prime }\left(x\right)$ =

• A. $n^{2}f\left(x\right)$
• B. $-n^{4}f\left(x\right)$
• C. $-n^{2}f\left(x\right)$
• D. $n^{4}f\left(x\right)$

Answer 1

Answer: (B)

$-n^{4}f\left(x\right)$

$f\left(x\right)=\left(e^{ix}\right)\left(e^{i3x}\right)\left(e^{i5x}\right)\cdots e^{i(2n-1)}x$

$f\left(x\right)=e^{\sum i(2k-1)x}$

$f\left(x\right)=e^{i(\frac{2n(n+1)}{2}-n)x}$

$f\left(x\right)=e^{i{n}^{2}x}$

$f^{\prime }\left(x\right)=i{n}^2{e}^{i{n}^{2}x}$

$f^{\prime \prime }\left(x\right)=-n^4{e}^{i{n}^{2}x}$

$f^{\prime \prime }\left(x\right)=-n^{4}f\left(x\right)$