7

I came across the following fact in a paper and am having trouble understanding why it is true:

Consider the error made when truncating the expansion for $e^a$ at the $K$th term. By choosing $K = O(\frac{\log N}{\log \log N})$, we can upper bound the error by $1/N$, in other words $$\sum_{j={K+1}}^\infty \frac{a^j}{j!} \leq \frac{1}{N}.$$

Here, the $O$ is "big-O" notation. I'm thinking a Stirling approximation probably has to be used in the denominator but still can't reproduce this result. Maybe this is some well-known result that I'm not aware of?

J. Doe
  • 71

1 Answers1

0

What we are asked to prove is this: there is a constant $C$ such that if $$ K \ge C \frac{\log N}{\log \log N} $$ then $$ \sum_{j=K+1}^\infty \frac{a^j}{j!} \le \frac1N. \tag1$$ Now clearly the left hand side of (1) grows to infinity as $a \to \infty$, so it must be that $C$ depends upon $a$. Suppose $K \ge 2a$ (which follows if $C \ge 2a$). Then the left hand side of (1) can be bounded above by a geometric series $$ \sum_{j=K+1}^\infty \frac{a^{K}}{K!} 2^{j-K} = \frac{a^K}{K!} .$$ Now use Stirling's formula, noting that if $$ K \ge C \frac{\log N}{\log \log N} $$ then $$ \log(K^K) = K \log K \ge C \frac{\log N}{\log\log N} (\log\log N - \log\log\log N) $$ and that if $N$ is sufficiently large then $$ \log\log N - \log\log\log N \ge \tfrac12 \log\log N .$$

By the way, if you want a much sharper result, use the approximation of the Poisson distribution by the normal distribution, and Proof of upper-tail inequality for standard normal distribution, to get $$ K \ge C \sqrt{\log N} .$$

Stephen Montgomery-Smith
  • 26,965