The sum of the series C = 1 + frac{cos x}{1!} | vedclass

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Question

(C) Consider the complex series $C + iS$:$C + iS = 1 + \frac{(\cos x + i\sin x)}{1!} + \frac{(\cos 2x + i\sin 2x)}{2!} + \frac{(\cos 3x + i\sin 3x)}{3

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$\exp[\exp(ix)]$

Vedclass · Answer

(C) Consider the complex series $C + iS$:$C + iS = 1 + \frac{(\cos x + i\sin x)}{1!} + \frac{(\cos 2x + i\sin 2x)}{2!} + \frac{(\cos 3x + i\sin 3x)}{3!} + \dots$Using Euler's formula $e^{i	heta} = \cos 	heta + i\sin 	heta$,we get:$C + iS = 1 + \frac{e^{ix}}{1!} + \frac{e^{i2x}}{2!} + \frac{e^{i3x}}{3!} + \dots$$C + iS = 1 + \frac{(e^{ix})}{1!} + \frac{(e^{ix})^2}{2!} + \frac{(e^{ix})^3}{3!} + \dots$This is the expansion of the exponential series $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ where $z = e^{ix}$.Therefore,$C + iS = e^{(e^{ix})} = \exp[\exp(ix)]$.

The sum of the series $C = 1 + \frac{\cos x}{1!} + \frac{\cos 2x}{2!} + \frac{\cos 3x}{3!} + \dots$ and $S = \frac{\sin x}{1!} + \frac{\sin 2x}{2!} + \frac{\sin 3x}{3!} + \dots$ is equal to

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