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I have the following: $$e \le \frac{x^n}{n!}.$$

I am trying to get this into format: $$n = \text{something}.$$

Does anyone have any idea how?

Jack M
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XeroZ
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    Is something like my question at http://math.stackexchange.com/questions/1333449/could-this-approximation-be-made-simpler-solve-n-an-10k of any interest for you ? – Claude Leibovici Jan 08 '16 at 11:49
  • @ Claude Leibovici: your question seems similar. I am new to mathematics, is my question impossible to solve? – XeroZ Jan 08 '16 at 11:51
  • Fixing $x$, this can obviously only be true for finitely many $n$. If it's true for $k$ distinct values of $n$, then we have $e^x=\sum x^n/n!\geq ke$, so $k$ is bounded above by $e^{x-1}$. That's all I've got off the top of my head. You might also study the monotonicity of the sequence $x^n/n!$ to get a better picture of the situation. – Jack M Jan 08 '16 at 11:55
  • To solve explicitly, no way. Approximations, yes. Sorry but I must go now. If you are still interested, let me know. I shall be back within 15 hours from now. Is $e$ $e$ or just an arbitrary constant ? – Claude Leibovici Jan 08 '16 at 11:58
  • @Jack M: basically this type calculates the error of taylor. I want, given the order of the taylor to get the error. – XeroZ Jan 08 '16 at 11:59

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(More a comment than an answer):
Perhaps Stirling's approximation to the factorial gives a clue: $$\begin{array} {} e &\le {x^n \over n! } \\ e & \le {x^n \over \sqrt {2 \pi n } (n/e)^n} & \small \text{ ... roughly ...}\\ e \sqrt {2 \pi n } (n/e)^n &\le x^n \\ e \sqrt {2 \pi } \sqrt n n^n &\le (xe)^n \\ \left( \sqrt {2 \pi e^2 }\right)^{1/n} \sqrt {n^{1/n}} \cdot n &\le xe \\ \left( 46.43 n\right) ^{ \frac 1{2n}} \cdot n &\le xe & \small \text{ ... roughly ...}\\ \end{array}$$ After that, $n$ must be smaller than $xe$ but can approach it for large $x,n$

Gottfried Helms
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