I found this passage in my book which is not very clear to me as to how it was obtained. Given $n\in\mathbb{N},\,k\in(0,1)$, then


up vote 3 down vote accepted

Is $i$ supposed to be small? If so, the derivative of $(1+i)^{n}$ with respect to $n$ is $(1+i)^n\log(1+i)$. Thus a first order approximation of $(1+i)^{n+k}$ is $$ (1+i)^{n}+k(1+i)^{n}\ln(1+i). $$ If $0<i<1$ then $i\approx\ln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^{n}+ik(1+i)^{n}$.

  • 1
    @user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$) – TomGrubb Sep 19 at 4:57
  • but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$? – user372003 Sep 19 at 4:59
  • 1
    @user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^{n+k}$, and taking $n$ to be a constant. You'll arrive at the same destination either way. – TomGrubb Sep 19 at 5:04

Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)\cdot 1 + x\cdot a.$

As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.

Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^{n+k}$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently, $$ \frac{f(k)-f(0)}{k-0}\approx\frac{f(1)-f(0)}{1-0} $$ Rearranging this gives $$ f(k)\approx (1-k)f(0) + k f(1). $$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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