Prove $e^{x \cos(x)}=1+x+\frac{x^2}{2} - \frac{x^3}{3}-\frac{11x^4}{24}- \frac{x^5}{5} + \cdots$

Question

How do we do this question using Maclaurin's Series. I tried expanding it by putting $x \cos x$ in place of $x$ in Maclaurin's expansion of $e^x$, and then using multinomial theorem to open the squares, cubes etc of $\cos x$. Is my approach correct?

Answer 1

Since $x\cos{x} \to 0$ as $x\to 0$, you could do that yeah. You can also use this formula : $$f(x)\underset{\mathrm{0}}{=}\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k$$ where $$f^{(k)}=\frac{d^kf(x)}{dx^k}$$ Generally, if you want to find the expansion of some composite function $f(g(x))$ at some point $a$, you do the expansion of $f$ at $g(a)$ and that of $g$ at $a$ (let's call it $h$) and then you replace $x$ with $h$ in $f$'s expansion.

Answer 2

There's a sign error in your $x^3$ coefficient. With $\equiv$ denoting equality up to $x^5$ terms, $$\exp(x\cos x)\equiv\exp\left(x-\frac{x^3}{2}+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1-\frac{x^3}{2}\right)\left(1+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{29x^5}{120}\right)\left(1+\frac{x^5}{24}\right)\\\equiv 1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{x^5}{5}.$$