Taylor Series- Quick Math Help?

Start with the geometric series for 1/(1-x)

1/(1-x)=sum(x^n) n=0 to oo

If I replace x with "-x^2"

1/(1+x^2)=sum((-1)^n*x^(2n))

After integration with respect to x, we get:

arctan(x)+C=sum([(-1)^n*x^(2n+1)]/(2n+1)) n=0 to oo

Setting x=0, we find that C=0. For the 4th degree Taylor polynomial, we need the first 2 terms.

T4(x)=[(-1)^(0)*x^(2(0)+1)]/(2(0)+1]+[(-1)^1*x^(2(1)+1)]/(2(1)+1)

T4(x)=x+x^3/3

The 4th degree Taylor polynomial is actually 3rd degree. This is due to the fact that arctan(x) is an odd function. It shouldn't be too hard to replace x with "8x" and simplify

i ought to anticipate your hassle arises from calculating the derivatives that produce the words contained in the series. Why not use a finite large difference code to approximate the words? It gained't take a lot attempt because you're already using a computer, and then you mustn't want to remedy them analytically.