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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cos(i) using taylor series

**Physics Forums | Science Articles, Homework Help, Discussion**