compute taylor series about $x=0$ of $\arctan(e^x -1 )$

Question

hello I am having some issue and need a little guidance with this taylor expansion

$$f(x)=arctan(e^x -1)$$

the terms i should get are $x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{11 x^4}{24}-\frac{5 x^5}{24}$ but I am having some trouble with the expansion

should I use the definition of $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ and truncate the first two terms $ 1 + x$ or should do three terms instead $1 + x +\frac{x^2}{2}$ after which taking the derivative of $$arctan(x)= \frac{1}{1 + x^2}*\frac{dy}{dx} $$ and then applying the the integration of the geometric series where by $$\int\frac{1}{1-x}= \int\sum_{k=0}^\infty {x^k}$$

or should i approach the question differently all together?

Answer 1

You can definitely do it the way you propose:

\begin{align}\tan^{-1}(x)&=\int\frac1{1+x^2}\text dx=\int\sum_{i=0}^\infty (-x^2)^i\text dx=\sum_{i=0}^\infty \frac{(-1)^ix^{2i+1}}{2i+1}\\ e^x&=\sum_{i=0}^\infty \frac{x^i}{i!}\end{align}

Therefore \begin{align}\tan^{-1}(e^x-1)&\approx (e^x-1)-\frac13(e^x-1)^3+\frac15(e^x-1)^5\\ &\approx (x+\frac12x^2+\frac16x^3+\frac1{24}x^4+\frac1{120}x^5)-\frac13(x+\frac12x^2+\frac16x^3)^3+\frac15x^5\\ &\approx x+\frac12x^2-\frac16x^3+\frac{11}{24}x^4-\frac5{24}x^5\end{align}

Answer 2

All roads lead to rome! Let's proceed by your first idea. We have $$\frac1{1+u^2}=1-u^2+u^4+O(u^6)$$ hence using that $\arctan(0)=0$ we get $$\arctan u=u-\frac{u^3}{3}+\frac{u^5}{5}+O(u^6)$$ moreover we have

$$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O(x^6)$$ now we replace $u$ by $x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}$ and we only retain the terms with degree less or equal $5$ we get the desired result.