## cos(i) using taylor series

Is there a way to find the cosine of i, the imaginary unit, by computing the following infinite sum?

$$cos(i)=\sum_{n=0}^\infty \frac{(-1)^ni^{2n}}{(2n)!}$$

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

$$\sum_{n=0}^\infty \frac{(-1)^n(-1)^n}{(2n)!}$$

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

$$cos(i)=\sum_{n=0}^\infty \frac{1}{(2n)!}$$

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

Edit: got the inline equations working

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 By the way, can someone remind me how to do inline equations? single dollar signs are not working for some reason. Also, why won't my last equation come out?

Recognitions:
 Quote by piercebeatz By the way, can someone remind me how to do inline equations? single dollar signs are not working for some reason. Also, why won't my last equation come out?
The forum can use "tags" inclosed in square brackets (with no spaces between the tag and the surrounding brackets). The tag for beginning an expression in-line with text is itex. The tag for ending it is /itex. For writing latex on a separate line, the tags are tex and /tex.

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Mentor

## cos(i) using taylor series

Looks correct so far. cos(i)=cosh(1) which can be expressed with exponential functions. If you write them as series and simplify, you should get the same result.

You can use ## for inline TeX.

 Quote by mfb Looks correct so far. cos(i)=cosh(1) which can be expressed with exponential functions. If you write them as series and simplify, you should get the same result. You can use ## for inline TeX.
I am aware that it is possible to extend trigonometric functions to imaginary values using hyperbolic trig functions. However, I am wondering if this sum can be evaluated without doing so

 Quote by piercebeatz I am aware that it is possible to extend trigonometric functions to imaginary values using hyperbolic trig functions. However, I am wondering if this sum can be evaluated without doing so
$$\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac12 \sum_{m=0}^{\infty} \left(\frac{1}{m!} + \frac{(-1)^m}{m!}\right) = \frac12(e + e^{-1}).$$

 Quote by pasmith $$\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac12 \sum_{m=0}^{\infty} \left(\frac{1}{m!} + \frac{(-1)^m}{m!}\right) = \frac12(e + e^{-1}).$$
How did you change the first sum to the second?

 Mentor Clever guessing (based on the known result). Write down the first terms of the second sum and you'll see how it works. For odd m, both summands cancel each other, for even m, they stay (and get a factor of 2 to cancel the 1/2).

 Tags taylor series