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Find the sum of following series:

$$1 + \cos \theta + \frac{1}{2!}\cos 2\theta + \cdots$$

where $\theta \in \mathbb R$.

My attempt: I need hint to start.

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    $\begingroup$ Hint: $\cos(n\theta) = \Re( e^{i n\theta})$ for $\theta \in \mathbb{R}$. $\endgroup$ Aug 14, 2015 at 15:00

3 Answers 3

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Hint: $$ 1+\cos x + \frac{1}{2!}\cos 2x + \ldots = \Re(e^{0ix} + e^{1ix} + \frac{1}{2!}e^{2ix} + \ldots)=\Re e^{e^{ix}} $$

$$ e^{ix}=\cos x+i\sin x\\\Longrightarrow e^{e^{ix}} = e^{\cos x}e^{i\sin x}=e^{\cos x}(\cos(\sin x)+i\sin(\sin x)) $$ Your sum is $$ e^{\cos x}\cos(\sin x) $$

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  • $\begingroup$ Spoilers with \$\$ doesn't work properly $\endgroup$ Aug 14, 2015 at 15:19
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HINTS: You have $$\sum_{k=0}^\infty\frac{\cos(k\theta)}{k!}$$

Remember that: $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$ And: $$\cos(\theta)=\Re{e^{i\theta}}$$

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  • $\begingroup$ Is $\cos\theta$ really equal to $e^{i\theta}$? $\endgroup$ Aug 14, 2015 at 15:04
  • $\begingroup$ Then what is $\cos\theta+i\sin\theta$? $\endgroup$ Aug 14, 2015 at 15:05
  • $\begingroup$ @AdityaAgarwal typo fixed $\endgroup$
    – entrelac
    Aug 14, 2015 at 15:05
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Hint to start $$ 1+\cos \theta + \frac{1}{2!}\cos(2\theta)+\cdots = \mathcal{Re}\sum_{k=0}^n\frac{\mathrm{e}^{ik\theta}}{k!} = \mathcal{Re}\sum_{k=0}^n\frac{\left(\mathrm{e}^{i\theta}\right)^k}{k!} $$ Also remember that in the $\lim_{n\to\infty}$ we have $$ \lim_{n\to\infty}\sum_{k=0}^n\frac{x^k}{k!}=\mathrm{e}^x $$

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