# How do I see that $1-(1-\frac 1 M)^Q \approx \frac Q M$ (provided that $Q$ is small compared to $M$), where both $Q$ and $M$ are integers?

How do I see that $$1-(1-\frac 1 M)^Q \approx \frac Q M$$ (provided that $Q$ is small compared to $M$), where both $Q$ and $M$ are integers ?

The approximation is stated in a book without proof.

I've tried looking into some analysis and using the binomial expansion theorem, but haven't come to a conclusion.

Also, how "small" should $Q$ be compared to $M$ before the approximation is good ?

• Using the Taylor expansions of $-\ln(1-x)$ and $e^x$ around $0$, you get \begin{align} 1-\left(1-\frac 1 M\right)^Q &= 1-e^{Q\ln (1-\frac 1 M)} = 1-e^{-\frac{Q}{M} + o\!\left(\frac{Q}{M}\right)} = 1-\left(1-\frac{Q}{M} + o\!\left(\frac{Q}{M}\right)\right) \\&= \frac{Q}{M} + o\!\left(\frac{Q}{M}\right) \end{align} – Clement C. Dec 31 '14 at 10:59
• You buy an item and you get 1% discount. And because you are the first customer of 2015, another 1% of the remaining amount. Then, because 1% is really tiny anyway, this is almost the same thing as getting 1% of the total amount instead. So $(1-\tfrac1{100})^2 \simeq 1 - \tfrac{2}{100}$. – Myself Dec 31 '14 at 12:13

Another way to look at it compared with clement nice comment/answer is $$(x+y)^n = \sum_{k=0}^{n}\left(\matrix{n\\k}\right)x^{k}y^{n-k}$$ Putting in $x= -1/M$ and $y=1$ for convience. We find $$(-\frac{1}{M} +1)^Q= \sum_{k=0}^{Q}\left(\matrix{Q\\k}\right)\left(-\frac{1}{M}\right)^{k}$$ The first few terms are $$\left(\matrix{Q\\0}\right)\left(-\frac{1}{M}\right)^{0} + \left(\matrix{Q\\1}\right)\left(-\frac{1}{M}\right)^{1} + ...$$ Which yields your result. The size of $x$ in this case determines you can truncate the binomal expansion to first order terms.
Consider the geometric progression: $$\overbrace{\left(1{}+{}\left(1-\frac{1}{M}\right){}+{}\ldots{}+{}\left(1{}-{}\frac{1}{M}\right)^{Q-1}\right)}^{Q\mbox{ terms}}{}={}\frac{\left(1{}-{}\left(1{}-{}\frac{1}{M}\right)^Q\right)}{\left(1{}-{}\left(1{}-{}\frac{1}{M}\right)\right)}{}={}M\left(1{}-{}\left(1{}-{}\frac{1}{M}\right)^Q\right)$$
For $M$ sufficiently large, each of the $Q$ terms on the l.h.s is approximately $1$. Therefore, their sum in this case is approximately $Q$, giving $$\frac{Q}{M}\approx\left(1{}-{}\left(1{}-{}\frac{1}{M}\right)^Q\right)\,.$$ For $Q$ and $M$ not integers, the argument required is more sophisticated.