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So, I was just exploring the limit definition of e and seeing what I could create from it and after some time I landed at these 2 approximations for the power tower function - "$x^x$".

$ e(x^{x-1}+(x-1)^x) \approx x^x$

$ \frac{e(x-1)^{x+1}}{x-e} \approx x^x$

(I know they don't make much sense, but you can't separate out the x^x. Instead, they could used as large number approximations, or limits for e). I will explain how I got the first one - (as the second is just a re-arrangement fo the first using $x^{x-1}$ as the subject).

Also, for this I will be using the limit for the reciprocal of e --> $\lim_{x\to\infty} (1-\frac1x)^x = 1/e$

Proof for this is here


$\lim_{x\to\infty} (1-\frac1x)^x = 1/e$

As 1/x approaches 0 as x goes to infinity, it can just be added on with changing the limit.

$\lim_{x\to\infty} (1-\frac1x)^x +\frac1x = 1/e$

$\lim_{x\to\infty} (1-\frac1x)^x +\frac{x^{x-1}}{x^x} = 1/e$

Then rewrite the inside of the bracket

$\lim_{x\to\infty} (\frac{x-1}{x})^x +\frac{x^{x-1}}{x^x} = 1/e$

$\lim_{x\to\infty} \frac{(x-1)^x}{x^x} +\frac{x^{x-1}}{x^x} = 1/e$

$\lim_{x\to\infty} \frac{{(x-1)^x} + {x^{x-1}}}{x^x} = 1/e$

Then just do some simple re-arrangement to get and discard the limit to get:

$ e(x^{x-1}+(x-1)^x) \approx x^x$


I haven't seen this anywhere before, so I assume it's new.

My question is whether this could somehow be used to solve the integral of $x^x$ or define the function of $x^x$ in terms of an integral, just like the gamma function does for the factorial (after all, the factorial and power tower are similar functions)

I will also try using this to make an infinite continued fraction for $x^x$.

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  • $\begingroup$ Going to the last line in your proof, you have separated the limit of a quotient to quotient of limits. However, both of the individual limits go to infinity and thus you aren’t guaranteed to be able to separate the limit. And in any case, $\lim_{x\to \infty} x^x=\infty$. $\endgroup$ – imas145 Dec 27 '19 at 11:38
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    $\begingroup$ A limit such as $\lim_{x\to a} f(x)$ cannot be an expression with $x$. After all, (very loosely speaking) the limit "replaces" $x$ with $a$. ... In particular, it makes no sense to write $\lim_{x\to a}\left(\frac{f(x)}{g(x)}\right) = \lim_{x\to a}f(x) = g(x)$. $\endgroup$ – Blue Dec 27 '19 at 11:39
  • $\begingroup$ Ok, I see what you mean. However , it could still be used as an approximation for for larger numbers $\endgroup$ – Jakub Skop Dec 27 '19 at 11:41
  • $\begingroup$ Also, you could move some terms over and use them as limits for e $\endgroup$ – Jakub Skop Dec 27 '19 at 11:41
  • $\begingroup$ You started the “derivation” using one definition for $e$, so any rearrangements should just result in where you started. And as others have said, your steps are not even valid. $\endgroup$ – imas145 Dec 27 '19 at 11:43
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The fact of the matter that, just because two functions seem to have the same limit as $x\rightarrow\infty$ does not mean that the functions are necessarily close to each other. Because the function $x^x$ grows really quickly, and there is a lot of deviation you can have, while retaining the same limit.

For an example, the function ${\pi}(x)$ gives the number of primes less than or equal to $x$. One long-known fact is that the limit as $x\rightarrow\infty$ of $\pi(x)$ is the same as that of ${\rm li}(x):=x/\ln (x)$, in the sense that $\displaystyle\lim_{x\rightarrow\infty}\frac{\pi(x)}{{\rm li}(x)}=1$.

However, if you introduce a special metric to measure how good a certain approximation is, then you can see that ${\rm li}(x)-\frac 12{\rm li}(\sqrt x)$ is in fact a better approximation to $\pi(x)$. If you allow discontinuous functions, then in fact there is a whole range of other better estimates.

In other words, what I am saying is that, just because both functions seem to go to infinity in the same way does not mean that there is a special relationship between the two functions.

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  • $\begingroup$ Are you referring to the factorial and power tower functions? If you are, then they are similar in more than what you said. For example, they both include repeated multiplication x times. The only way they differ is that x! is off by an x number of terms from x^x. $\endgroup$ – Jakub Skop Dec 27 '19 at 11:54
  • $\begingroup$ No, I have references in my proof as to what I'm referring to. Just click those links, and they should take you to read about what I'm referring to. $\endgroup$ – Alex Dec 27 '19 at 11:57
  • $\begingroup$ No, I understand what you mean, but what I'm asking is whether your initial point that "two functions with the same limit are not neccesarily close to each other" was meant to be attributed to my question on the factorial and tetration functions $\endgroup$ – Jakub Skop Dec 27 '19 at 12:00
  • $\begingroup$ I see. So I guess I meant that just because what you've shown in your conclusion is true (that the two functions at the beginning of your post have the same limit) does not necessarily mean that there is a further relationship between these two functions, which can be applied to solve new problems. $\endgroup$ – Alex Dec 27 '19 at 12:04
  • $\begingroup$ I'm not asking whether the relationships of the functions can be used to solve problems, but whether they can independently be used to solve tetration problems, such as the ones I discussed in my question $\endgroup$ – Jakub Skop Dec 27 '19 at 12:06
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None of these identities make sense. The right-hand sides $x^x$ are meaningless as $x$ is not defined (or at best could be interpreted as $\infty^\infty$).


Note that

$$\lim_{x\to\infty}\frac{e(x^{x-1}+(x-1)^x)}{x^x}=e\lim_{x\to\infty}\frac1x+e\lim_{x\to\infty}\left(\frac{x-1}x\right)^x=0+e\cdot e^{-1}$$ so that $$e(x^{x-1}+(x-1)^x)\sim x^x$$

but one may wonder if the first term has any relevance.

One can also wonder what use is to replace an expression by an approximation that is more complicated.

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  • $\begingroup$ Yeah, someone already pointed that. I would just use them as approximations. $\endgroup$ – Jakub Skop Dec 27 '19 at 11:44
  • $\begingroup$ @JakubSkop: fix your question then. $\endgroup$ – Yves Daoust Dec 27 '19 at 11:44
  • $\begingroup$ What do you mean - my question is whether there is any application $\endgroup$ – Jakub Skop Dec 27 '19 at 11:46
  • $\begingroup$ @JakubSkop: as I said, the "identities" are meaningless. Change the notation. $\endgroup$ – Yves Daoust Dec 27 '19 at 11:47
  • $\begingroup$ How can I change whether it's meaningless or not - That doesn't make any sense $\endgroup$ – Jakub Skop Dec 27 '19 at 11:49

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