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Let $f(x) = x^x$. Then, let us define a function $p(x)$ such that: $$p(x) = \frac {f(x+1)}{f(x)} - \frac {f(x)}{f(x-1)}$$

As the value of $x$ increases, $p(x)$ approaches $e$, or (I think), $$\lim_{x\rightarrow \infty} p(x) = e$$

I have no idea why this occurs as I'm no advanced math student, but could someone explain the reason to me? I just found this out playing with a calculator.

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  • $\begingroup$ It's a good observation. may come handy in some calculations. Though may be not new definition of $e$. $\endgroup$
    – user45099
    Mar 4 '13 at 10:26
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    $\begingroup$ It's not new; a guy named Felix Keller tried repeatedly to insert it into the Wikipedia article about $e$ under the name "Keller's expression". At the time I said that if it had not been published before, that was probably because it was not hard to find. I don't want to discourage you; It is a good observation and you should be pleased to have discovered it; this is why I am not linking to the Wikipedia discussion, which is somewhat brusque, as we have to be with people who want to name their unpublished elementary mathematical discoveries after themselves. $\endgroup$
    – MJD
    Mar 4 '13 at 14:24
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    $\begingroup$ It also appears in H. J. Brothers and J. A. Knox, "New closed-form approximations to the Logarithmic Constant $e$The Mathematical Intelligencer, Vol. 20, No. 4, 1998; pp 25–29; this formula is at the bottom right of page 26. $\endgroup$
    – MJD
    Mar 4 '13 at 14:32
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    $\begingroup$ Is there any motivation for calling it $p$? If It's random then you should have choosen something else, e.g. $\bar f$, $F$, $E$, $fe$ or $\eta$. $\endgroup$
    – Nikolaj-K
    May 15 '13 at 11:39
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$\frac {f(x+1)}{f(x)} = (x+1)\left(\frac {x+1}x \right)^x$. Define $g(x) = (1+1/x)^x$.

Then $p(x) = (x+1)g(x+1) - xg(x)$.
It is well known that $\lim_{x \to \infty} g(x) = e$, but we need a more precise development of $g(x)$ as $x \to \infty$ :

$g(x) = \exp(\log(g(x))) = \exp(x \log(1+x^{-1})) \\ = \exp(x (x^{-1} - x^{-2}/2 + O(x^{-3})) \\ = \exp(1 - x^{-1}/2 + O(x^{-2})) = e - (e/2)x^{-1} + O(x^{-2})$.

Now we can get $xg(x) = xe - e/2 + O(x^{-1})$
And finally $p(x) = (x+1)e - e/2 - xe + e/2 + O(x^{-1}) = e + O(x^{-1})$

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You are right. Use the fact $$\lim_{x\rightarrow\infty}\bigg(1+\frac{1}{x}\bigg)^x = e$$

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