Find T(10) (x): the Taylor polynomial of degree 10 of the function f(x)=arctan(x^3) at a=0.
Can someone show me the steps to finding this, please?
note that if
$\displaystyle f(x)=\tan^{-1}(x)$ then
$\displaystyle f'(x)=\frac{1}{1+x^2}$
by the def of a geometric series we get...
$\displaystyle f'(x)=\sum_{n=0}^{\infty}(-1)^nx^{2n}$ so if integrate both sides we get
$\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$
so then $\displaystyle f(x^3)=\tan^{-1}(x^3)=\sum_{n=0}^{\infty}\frac{(-1)^n(x^3)^{2n+1}}{2n+1}$
so we get...
$\displaystyle tan^{-1}(x^3)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+3}}{2n+1}$
Just write out as many terms as you need.