# How do you expand this function around infinity?

How do you expand this function at n= $\infty$? \begin{align*} f(x) = \exp \left( - \left( \frac{x}{n + \frac{1}{3} + \frac{0.1}{n+1}} \right)^2 \left( n+ \frac{1}{6} \right) \right) \end{align*} From Wolfram alpha I got \begin{align*} f(x)= 1 - \frac{x^2}{n} + \frac{x^4+x^2}{2n^2} + \mathcal{O} (n^{-3}) \end{align*} $\\$ When I expand the exponent using Wolfram alpha I got:

\begin{align*} - \frac{x^2}{n} + \frac{x^2}{2n^2} + \mathcal{O} (n^{-3}) \end{align*}

Then by taking the exponential on this term and using Taylor expansion $\left(\exp(x) = 1+ x + \frac{x^2}{2} +\mathcal{O}(n^{-3}) \right)$ I get : \begin{align*} f(x) &= 1- \frac{x^2}{n} + \frac{x^2}{2n^2} + \frac{1}{2} \left( \frac{x^4}{n^2} + \frac{x^4}{4n^4} - \frac{x^4}{n^3} \right) + \mathcal{O} (n^{-3}) \\ &= 1- \frac{x^2}{n} + \frac{x^4+x^2}{2n^2} + \mathcal{O} (n^{-3}) \end{align*} Where I in the second equality have collected the same order terms under the big-O.

• Can you write what did you try? Dec 12, 2017 at 0:49
• I tried using $\exp(x) = 1 + x + \frac{x^2}{2} + ...$ Dec 12, 2017 at 0:51
• When expanding the exponent using Wolfram alpha you get: $- \frac{x^2}{n} + \frac{x^2}{2n^2} + \mathcal{O} (n^{-3})$. Dec 12, 2017 at 1:05
• Then by taking the exponential on this term and using Taylor expansion I get: \begin{align*} f(x) &= 1- \frac{x^2}{n} + \frac{x^2}{2n^2} + \frac{1}{2} \left( \frac{x^4}{n^2} + \frac{x^4}{4n^4} - \frac{x^4}{n^3} \right) + \mathcal{O} (n^{-3}) \\ &= 1- \frac{x^2}{n} + \frac{x^4+x^2}{2n^2} + \mathcal{O} (n^{-3}) \end{align*} Dec 12, 2017 at 1:13
• you can edit your question, it would be simpler to understand there Dec 12, 2017 at 1:14

$$f = \exp \left( - \left( \frac{x}{n + \frac{1}{3} + \frac{1}{10(n+1)}} \right)^2 \left( n+ \frac{1}{6} \right) \right)$$ $$\log(f)=- \left( \frac{x}{n + \frac{1}{3} + \frac{1}{10(n+1)}} \right)^2 \left( n+ \frac{1}{6} \right)=-x^2\frac {n+\frac{1}{6}}{\left(n + \frac{1}{3} + \frac{1}{10(n+1)} \right)^2}$$ Now, using long division or Taylor series $$\frac {n+\frac{1}{6}}{\left(n + \frac{1}{3} + \frac{1}{10(n+1)} \right)^2}=\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{45 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(f)=-\frac{x^2}{n}+\frac{x^2}{2 n^2}-\frac{x^2}{45 n^3}+O\left(\frac{1}{n^4}\right)$$ Now, using $f=e^{\log(f)}$ and Taylor again $$f=1-\frac{x^2}{n}+\frac{x^2+x^4}{2 n^2}-\frac{2 x^2+45 x^4+15 x^6}{90 n^3}+O\left(\frac{1}{n^4}\right)$$