Question

What is the taylor series in sigma notation of: `sin^2(x)`?

Answer 1

You can start by using the trig identity of `sin^2x = (1 - cos2x)/2`

we know the Maclurin series of `cosx` is `sum_(n=0)^oo (-1)^n (x^(2n))/((2n)!)`

Keep in mind here that 0!=1, so the case of n=0 is still valid.

So we can then sub in `2x` in place of `x` to solve for `cos2x`

`cos2x = sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)`

thus we get:

`sin^2x = (1-cos2x)/2 =1/2- 1/2sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)`