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?

  • $\begingroup$ A much (much) less elegant method is to use the definition of the Taylor series at a: \begin{align} \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \end{align} i.e. calculate the derivatives. $\endgroup$ – Achal Jun 7 '14 at 1:03

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}


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.

  • $\begingroup$ I think you're missing an x in the second last night in the sum. $\endgroup$ – Achal Jun 7 '14 at 0:40
  • $\begingroup$ Yes thank you!! $\endgroup$ – user63181 Jun 7 '14 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.