# Thread: Taylor Series

1. ## Taylor Series

Use a known Taylor Series to find the Taylor series about c=0 for the given function and find its radius of convergence.

$f(x)= cos x^3$

So, the existing Taylor series for cos(x) is
$\sum \frac{-1^k}{2k!}(x)^{2k}$

So the new Taylor Series for cos x^3 would be
$\sum \frac{-1^k}{2k!}(x)^{3k}$

Am I on the right track? And how do I find its radius of convergence?

2. Originally Posted by kl.twilleger
Use a known Taylor Series to find the Taylor series about c=0 for the given function and find its radius of convergence.

$f(x)= cos x^3$

So, the existing Taylor series for cos(x) is
$\sum \frac{-1^k}{2k!}(x)^{2k}$

So the new Taylor Series for cos x^3 would be
$\sum \frac{-1^k}{2k!}(x)^{3k}$

Am I on the right track? And how do I find its radius of convergence?
\begin{aligned}\cos\left(x\right)&=\sum_{n=0}^{\in fty}\frac{(-1)^nx^{2n}}{(2n)!}\\
&\implies\cos\left(x^3\right)=\sum_{n=0}^{\infty}\ frac{(-1)^n\left(x^3\right)^{2n}}{(2n)!}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n}}{(2n)!}\end{aligned}

And $x\in\mathbb{R}\implies{x^3\in\mathbb{R}}$ because the reals are a field. Therefore this converges $\forall{x}\in\mathbb{R}$