I worked out what the maclaurin of cos x by doing the derivations, plug in 0 and divide by n!. I get thisWrite the summation notation of the maclaurin polynomial of order n for cos x

$\displaystyle 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} ...$

then for the sumation i got this

$\displaystyle \sum_{i = 0}^{??} \frac {(-1)^i x^{2i}} {(2i)!}$

My question is what goes on top of the sigma? The top part is the stopping point when computing the sum, correct? I need to find the Nth so the top is not infinity.

Let's say I want the thrid maclaurin, n =3. would that be

$\displaystyle 1 - \frac {x^2} {2!} + \frac {x^4} {4!}$

or

$\displaystyle 1 + 0 * \frac {x^1} {1!} - \frac {x^2} {2!} + 0 * x^3$

counting on zero but the 0 coeffients remove those odd powered x's so the above is just the same as the previous.

With x^2i that will take into acount the 0 coeffients, for when sine at 0. It out puts 1, x^2, x^4 ..., so that forms the first example of the thrid maclaurin. Does that mean the top of sigma is just n?

The thing is my teacher wrote something like int[n/2 + 1.5] as the top of sigma but said he wasnt sure of that. and after runing different forms n/2 +.5, n/2 +1 it doesnt make sense to me why you would need something like that. the top part is just the stoping point so why the heck do you even need a little equation on the top ?!.

grrr