To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:
$e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$
$\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$
$\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$
$\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$
In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$
However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$.
What did I do wrong? I can't seem to find what or where.
$\endgroup$ 12 Answers
$\begingroup$$$=e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{(x-x^3/6)^3}{6}+\dfrac{(x-x^3/6)^4}{24}+O(x^5)$$
and this term $$\dfrac{(x-x^3/6)^4}{24}=\dfrac{x^4}{24}+o(x^5)$$ so $$e^{\sin{x}}=1+x+\dfrac{x^2}{2}+\left(-\dfrac{x^4}{6}+\dfrac{x^4}{24}\right)+o(x^5)$$
$\endgroup$ $\begingroup$HINT: You have to consider all summands up to $O(x^5)$.
$\endgroup$
