Find the Maclaurin polynomial of degree $5$ for $f(x) = x\cos(x^2)$.

Question

I'm pretty lost on this one. Can someone help me out?

Find the Maclaurin polynomial of degree $5$ for f(x) = $x\cos(x^2)$.

Answer 1

If $f(x)$ is a power series,the sum of its terms up to and including the term of degree $n$ is the Maclaurin polynomial of $f(x)$ of degree $n.$ $$x\cos(x^2)=x(1-\frac{1}{2!}(x^2)^2+...))$$ $$=x-\frac{1}{2}x^5+...$$. The Maclaurin polynomial of degree 5 is $x-\frac{1}{2}x^5.$

Answer 2

Use $$\cos(t)=\sum_{n=0}^\infty (-1)^n \frac{t^{2n}}{(2n)!}$$ $$\cos(x^2)=\sum_{n=0}^\infty (-1)^n \frac{x^{4n}}{(2n)!}$$ $$x\cos(x^2)=\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{(2n)!}$$ Truncate wherever you need.