# Math Help - Taylor Series

1. ## Taylor Series

I have come across the following in regards to a taylor series. I know what the solution is but not how it was arrived at.

1-summation n=0->inf ((-1)^n*4^n*x^2n)/2n!

It's the 1-summation part that has me stumped.

2. Originally Posted by p75213
I have come across the following in regards to a taylor series. I know what the solution is but not how it was arrived at.

1-summation n=0->inf ((-1)^n*4^n*x^2n)/2n!

It's the 1-summation part that has me stumped.
I have no idea what "1- summation" means! Is it possible that the "1" is just the problem number? Also "x^2n" could be (x^2)n or x^(2n)- I suspect the latter- and 2n! correctly is 2(n!) but I suspect you mean (2n)!

If I am correct about what you mean then this is $\sum_{n=0}^\infty \frac{(-1)^n 4^n x^{2n}}{(2n)!}= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(4x^2)^n= sin(4x^2)$

3. The equation you have shown is correct. It is the Taylor series for sine^2 x. By 1-summation I meant 1 minus summation.