TITLE: Convert $f(x)=(\cos(x))^3$ to powers of x and find if converges.
QUESTION [1 upvotes]: I started out by writing the Taylor series for $x_0=0$ (Maclaurin series) of $f(x)=(\cos(x))^3$.
If my calculations are correct $$f(x)=1-\frac{3x^2}{2!}+\frac{21x^4}{4!}-\frac{183x^2}{6!}+...$$
and after some simplifications $$f(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}a_n$$ where $a_n$ is a sequence of $a_0=1, a_1=3, a_2=21,a_3=183, ..$ 
Is there any way to find the $a_n$ sequence or a better way to solve the whole problem?

REPLY [0 votes]: Note that $\cos^3x=\frac{1}{4}(3\cos(x)+\cos(3x))$. But
$\cos(x)=\sum_{k=0}^\infty\dfrac{(-1)^kx^{2k}}{(2k)!}$
$\cos(3x)=\sum_{k=0}^\infty\dfrac{(-9)^kx^{2k}}{(2k)!}$
Then 
$\begin{eqnarray}
4\cos^3(x)&=&3\cos(x)+\cos(3x)\\
&=&3\sum_{k=0}^\infty\dfrac{(-1)^kx^{2k}}{(2k)!}+\sum_{k=0}^\infty\dfrac{(-9)^kx^{2k}}{(2k)!}\\
&=&\sum_{k=0}^\infty\dfrac{[3(-1)^k+(-9)^k]x^{2k}}{(2k)!}\\
&=&\sum_{k=0}^\infty\dfrac{a_kx^{2k}}{(2k)!}
\end{eqnarray}$
where $a_0=4$ and $a_{n+1}=-10a_n-9a_{n-1}$.
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