Is there a way to find the cosine of i, the imaginary unit, by computing the following infinite sum?(adsbygoogle = window.adsbygoogle || []).push({});

[tex]cos(i)=\sum_{n=0}^\infty \frac{(-1)^ni^{2n}}{(2n)!}[/tex]

Since the value of ##i^{2n}## alternates between -1 and 1 for every ##n\in\mathbb{N}##, it can be rewritten as ##(-1)^n##.

[tex]\sum_{n=0}^\infty \frac{(-1)^n(-1)^n}{(2n)!}[/tex]

##(-1)^n(-1)^n=(-1)^{2n}##, which is equal to one for all ##n\in\mathbb{N}##. Thus,

[tex]cos(i)=\sum_{n=0}^\infty \frac{1}{(2n)!}[/tex]

Is all of this correct? If so, how would the final sum be calculated?

Edit: got the inline equations working

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# Cos(i) using taylor series

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