1

I tried to solve this problem; and I looked at the answer. It said that $e^{\frac{-r}{n}} =\sum_{0}^{\infty}{\frac{(\frac{-r}{n})^k}{k!}}\approx 1-\frac{r}{n} $ I can't under stand the summation approximation.

Thank you so much for your reply.

Henry Cai
  • 641
  • Since $r$ is much smaller than $n$, the higher order terms of $\frac{r}{n}$ are ignored, yielding the result – Lukas Rollier Jan 08 '20 at 12:57
  • They just keep the term with $k=0$, which is $1$, and the term with $k=1$, which is $-r/n$. They assume that for higher powers $(r/n)^k <<1$ – Andrei Jan 08 '20 at 12:58
  • Thank you very much for you guys' replies.

    This seems quite difficult for me to understand. Since I thought $\frac{-r}{n} <<1 $ also. Why don't we just abandon all the terms and leave only '1'.

     – Henry Cai Jan 08 '20 at 13:10
  • The approximation is getting worse, obviously. If $r$ is the interest rate, $r$ is near $0$. Then $\frac{r}{n}$ is becomes very fast small if it is raised to power. – callculus42 Jan 08 '20 at 13:21
  • @callculus Thank you. May I ask whether it is a normal practice to abandon the terms with ''power' when doing approximation? Thank you so much, and sorry for any inconvenience causedd. – Henry Cai Jan 08 '20 at 17:05
  • Sure, especially if the term which is raised to power is near $0$. – callculus42 Jan 08 '20 at 17:28
  • Here is an example for such an approximation. Here the interest rate plays a role as well. – callculus42 Jan 08 '20 at 17:31
  • thank you very much – Henry Cai Jan 08 '20 at 19:56

0 Answers0