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

$$(1+i)^{n+k}\approx(1-k)(1+i)^n+k(1+i)^{n+1}=(1+i)^n(1+ki)$$

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 $(1+i)^{n}+ik(1+i)^{n}$.

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    @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
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    @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 $$ f(k)\approx (1-k)f(0) + k f(1). $$

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