How to find the cosine of i using Taylor series?

[pierce15]

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

 

[pierce15]

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?

 

[Stephen Tashi]

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.

(Begin a reply-with-quote to someone's post that has the type of expression you want and you'll see the details of the LaTex.)

If you right click on a an expression you can get some menus and one of them gives options to pick the "math renderer". Picking a different renderer can fix some problems with how formulas are displayed.

 

[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.

 

[pierce15]

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

 

[pasmith]

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

<br /> \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}).<br />

 

[pierce15]

<br /> \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}).<br />

How did you change the first sum to the second?

 

[mfb]

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).