f(x) = (cos x + i sin x) cdot (cos 3x + i sin | vedclass

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(B) Using De Moivre's Theorem,we know that $(\cos 	heta + i \sin 	heta) = e^{i	heta}$.Thus,$f(x) = e^{ix} \cdot e^{i3x} \cdot e^{i5x} \cdots e^{i(2

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$-n^4 f(x)$

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(B) Using De Moivre's Theorem,we know that $(\cos 	heta + i \sin 	heta) = e^{i	heta}$.Thus,$f(x) = e^{ix} \cdot e^{i3x} \cdot e^{i5x} \cdots e^{i(2n-1)x}$.$f(x) = e^{i(1 + 3 + 5 + \cdots + (2n-1))x}$.The sum of the first $n$ odd numbers is $n^2$,so $f(x) = e^{i(n^2)x}$.Now,find the first derivative: $f'(x) = i n^2 e^{i(n^2)x} = i n^2 f(x)$.Find the second derivative: $f''(x) = i n^2 f'(x) = i n^2 (i n^2 f(x)) = i^2 n^4 f(x)$.Since $i^2 = -1$,we get $f''(x) = -n^4 f(x)$.

$f(x) = (\cos x + i \sin x) \cdot (\cos 3x + i \sin 3x) \cdots [\cos(2n-1)x + i \sin(2n-1)x]$,$n \in N$. Then $f''(x) = ?$ (Where $i = \sqrt{-1}$)

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