According to this survey paper on Bloom Filters, we can do the following approximation

$$(1 - \frac{1}{m})^{kn} \approx e^{\frac{-kn}{m}}$$

which allegedly is a $O(\frac 1m)$ approximation. I am curious as to why this is so. I know that:

  • $(1 - \frac{1}{x})^x$ eventually converges to $e^{-1}$ because of L'Hospital's rule.

  • $(1 - \frac{1}{x})^x$ can be expressed as $e^{-1} \prod_{i=1}^{\infty} e^{\frac{-1}{(i+1)x^i}} $ using a Taylor expansion.

However, I don't see a clean way of showing that it is a $O(\frac 1m)$ approximation.


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We have: $$ \left(1 - \frac{1}{x}\right)^x = e^{x\log\left(1-\frac{1}{x}\right)} = e^{x\left(-\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right)} = e^{\left(-{1}-\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)}$$

by Taylor expansion of $\log(1-t)$ at $0$. Then by expanding the exponential : $$ e^{\left(-\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)} = 1-\frac{1}{2x}+o\left(\frac{1}{x}\right) + o\left(\frac{1}{x}\right) = 1+O(1/x) $$

So that plugging everything back: $$\left(1 - \frac{1}{x}\right)^x = e^{-1}(1+O(1/x)) = e^{-1}+O(1/x)$$

  • $\begingroup$ Ahh I should have discarded the higher order terms for an asymptotic. Thank you! $\endgroup$ – iart Oct 17 '17 at 22:59
  • $\begingroup$ @iart I am not sure what you did, but the asymptotic should be exactly the same even if you keep higher-order terms. That is always the case with O-notation arithmetic. $\endgroup$ – Federico Poloni Oct 18 '17 at 6:45
  • $\begingroup$ If you have more terms in the bigO you can collapse them, for instance, $O(1/x^2)+O(1/x^3)$ is a $O(1/x^2)$. Doing the computations with an arbitrary long Taylor expansion should yield the same result in the end. If you're happy with this answer, please mark it as Solved! $\endgroup$ – user70925 Oct 18 '17 at 8:51
  • $\begingroup$ Oh, wow, changing sign inside the brackets changes sign in power of $e$. $\endgroup$ – rus9384 Oct 18 '17 at 9:01
  • $\begingroup$ Yup marked it as solved. $\endgroup$ – iart Oct 19 '17 at 23:10

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