Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume $\lambda =80$, $\frac{e ^{\lambda x}}{x!}>0.9$ find $x$. I know how to use normal approximations but i want other method other than normal approximation.

share|improve this question

2 Answers 2

You can use Stirling's approximation where $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+\frac{1}{12n}+\frac{1}{288n^2}+o\left(\frac{1}{n^3}\right)\right)$$

First apply $\ln(\cdot)$ to both side of your inequality, then solve $x$ numerically.

share|improve this answer

Let $u_n=\mathrm e^{\lambda n}/n!$, then $u_{n+1}/u_n=\mathrm e^{\lambda}/(n+1)$ hence the sequence $(u_n)_{n\geqslant0}$ is increasing on $n\leqslant\mathrm e^\lambda$ and decreasing on $n\geqslant\mathrm e^\lambda$. Since $u_0=1\gt0.9$, this indicates that $u_n\gt0.9$ if and only if $n\leqslant N_\lambda$, for some $N_\lambda\geqslant\mathrm e^\lambda$. Furthermore, one can show that, when $\lambda\to\infty$, $N_\lambda\sim\mathrm e^{\lambda+1}$.

Edit: Refined forms of Stirling's approximation yield the following approximation. Call $x_\lambda(a)$ the solution of $(\lambda+1)x-(x+\tfrac12)\log(x)=\log(0.9a)$. Then $x_\lambda(\mathrm e)-1\lt N_\lambda\lt x_\lambda(\sqrt{2\pi})+1$. Finally, first-order estimates of the derivative show that, when $\lambda$ is large, the order of the width $x_\lambda(\sqrt{2\pi})-x_\lambda(\mathrm e)$ is $\log(\mathrm e/\sqrt{2\pi})=0.08$ hence a good heuristics is to take for $N_\lambda$ the closest integer to $x_\lambda(\sqrt{2\pi})$ and/or $x_\lambda(\mathrm e)$.

These considerations yield $N_{80}=1.5061\ldots\cdot10^{35}$, that is, $N_{80}=\exp(81.0000\ldots)$.

share|improve this answer
    
i still don't quite get how to solve the inequality. $\lambda =80$ and i am not sure how to proceed to the step $\lambda \rightarrow \infty$ –  Mathematics Oct 28 '12 at 11:30
    
What do you call solve the inequality? We proved that $\mathrm e^{80n}/n!\gt0.9$ if and only if $n\leqslant N_{80}$ for some $N_{80}\geqslant\mathrm e^{80}$ and the equivalent at the end of my answer suggests that $\log N_{80}$ might be close to $81$, or, loosely speaking, $N_{80}$ close to $1.5\cdot10^{35}$. Exact bounds might be deduced from refined forms of Stirling's approximation, if this is your point. See Edit. –  Did Oct 28 '12 at 12:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.