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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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not linking to the Wikipedia discussion, which is somewhat brusque, as we have to be with people who want to name their unpublished elementary mathematical discoveries after themselves.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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

$\frac {f(x+1)}{f(x)} = (x+1)\left(\frac {x+1}x \right)^x$. 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 $p(x) = (x+1)e - e/2 - xe + e/2 + O(x^{-1}) = e + O(x^{-1})$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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