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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.

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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.

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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, $N_{80}=\exp(81.0000\ldots)$.

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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

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