Given two real numbers $0 and $0<\delta<1$, I want to find a positive integer $i$ (it is better to a smaller $i$) such that $\frac{a^i}{i!} \le \delta.$
How to solve this inequation
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calculus
sequences-and-series
inequality
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0@draks, how to use Lambert-W-function? I am now obtaining the following inequality: {(a\cdot e)}^i<\sqrt{2\pi}\delta i^{i+1/2} – 2012-05-14
2 Answers
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Here is what I put together from my math toy box:
- Use Stirling's approximation $i!\approx(i/e)^i$ to get $\left( \frac{ae}{i}\right)^i \le \delta$.
- Call $ae=1/b$ and invert to get $(ib)^i\ge \delta^{-1}$.
- Continue with $(ib)^{ib}\ge \delta^{-b}$, define $x:=bi$ to get $x^x\ge\delta^{-b}$
- and then use $ x\ge\frac{\ln(\delta^{-b})}{W(\ln \delta^{-b})}=\frac{\ln(\delta^{-1/ae})}{W(\ln \delta^{-1/ae})}. $
- Resubstitute $x=\frac{i}{ae}$ for the result $i\ge\frac{ae\ln(\delta^{-1/ae})}{W(\ln \delta^{-1/ae})}$.
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1Ask your favorite [math knowlegde base](http://math.stackexchange.com/) and get [this](http://math.stackexchange.com/a/27372/19341): $\displaystyle \log x - \log \log x + \frac{1}{2}\frac{{\log \log x}}{{\log x}} \le W(x) \le \log x - \log \log x + \frac{e}{{e - 1}}\frac{{\log \log x}}{{\log x}}$. Does that help? – 2012-05-16
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Here is a not-very-good answer. Let $i$ be the result of rounding $\log\delta/\log a$ up to the nearest integer. Then $i\ge\log\delta/\log a$, so $i\log a\le\log\delta$ (remember, $\log a\lt0$), so $a^i\le\delta$, so $a^i/i!\le\delta$.
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0@JohnSmith: Stirling doesn't buy you much here: you will get$i$both in the base and in the exponent. – 2012-05-14