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In this biology textbook I found the following approximation:

$$\frac{1}{2N}\left( 1-\frac{1}{2N} \right)^t ≈ \frac{1}{2N}e^{\frac{-t}{2N}}$$

Can you help me to understand this approximation and help me to understand what assumption are needed for this approximation to be useful.

Antonio Vargas
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Remi.b
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  • Have you heard of the Taylor/McLaurin series of logarithm in 0 ? – T_O Apr 15 '14 at 10:29
  • @T_O I know what is a Taylor series expansion. I don't know about McLauring series but from your comment I'd guess they are two names for the same series. I understand what is a taylor series of log(x) but I don't understand what we mean by "…of logarithm in o". Thanks! – Remi.b Apr 15 '14 at 10:32
  • Sorry I worded that poorly I will try to give a detailed answer – T_O Apr 15 '14 at 10:33
  • Ok others have given good enough answers. The key point is that you can replace a function by its first k terms in the Taylor/McLaurin (I don't know how it is called in English) as they are equivalent : their ratio will be 1, they behave the same way. Edit : for more information, see http://en.wikipedia.org/wiki/Asymptotic_analysis – T_O Apr 15 '14 at 10:41

2 Answers2

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One way to define $e^x$ is via $$ e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n $$

By re-arranging your forumula a bit, you get $$ \frac{1}{2N}\left(1 - \frac{1}{2N}\right)^t = \frac{1}{2N}\underbrace{\left(1 + \frac{\frac{-t}{2N}}{t}\right)^t}_{\text{Compare to $e^x$ for $x=\frac{-t}{2N}$}} \approx \frac{1}{2N}e^{-\frac{-t}{2N}} \text{,} $$ at least for large enough $t$.

fgp
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  • Thanks a lot, you answer made things clear up to the understanding that $e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$. So here is my following question ;) – Remi.b Apr 15 '14 at 11:10
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Let us compare Taylor expansions of $\left( 1-\frac{1}{x} \right)^t$ and $e^{\frac{-t}{x}}$ around $t=0$. For the first one, we have $$\left( 1-\frac{1}{x} \right)^t \simeq 1+t \log \left(1-\frac{1}{x}\right)+\frac{1}{2} t^2 \log ^2\left(1-\frac{1}{x}\right)+\frac{1}{6} t^3 \log ^3\left(1-\frac{1}{x}\right)+O\left(t^4\right)$$ while for the second one, we have $$e^{\frac{-t}{x}}\simeq 1-\frac{t}{x}+\frac{t^2}{2 x^2}-\frac{t^3}{6 x^3}+O\left(t^4\right)$$ Now, in the first one, assume that $x$ is large compared to $1$ and use the Taylor expansion of $\log(1+y)$ for small values of $y$. Replace $y$ by $1/x$.

I am sure that you can take from here and conclude about the validity of the approximation.

Claude Leibovici
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