A new definition of $e^x$ for $x\geq 1$ via a limit?

Question

Some month ago and after several attempt to find it I have :

Let $y\geq 1$

$$e^{y}\overset{?}{=}\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}$$

Where we have the composition of Gamma function ($x!!=\Gamma(\Gamma(x+1)+1)$)

As Robjohn answered a previous question Conjecture: $\lim\limits_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots )^{-1/x}\stackrel?=e$ .I have tried the same path without any success .

If true might be this equality have some nice application .

Also I wondering myself if it could be true if $y$ is a complex number .

Some matter :

It seems there is again the binomial coefficient which appears .

Let :

$$f_n(x)=\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!...x!!!...!}}\right|\right)^{\frac{1}{x}}$$

Where we have composing $n$ times the Gamma function with itself .

Then it seems we have :

$$\lim_{x\to 0^+}f_n(x)=e^{y(1-(1-\gamma)^n)}$$

As attempt we can try to use log as the limit is also :

$$\lim_{x\to 0^+}\frac{\ln\left(1-\frac{\left(y-1+x!\right)}{1-\frac{1}{1-x!x!!x!!!\cdot\cdot\cdot}}\right)}{x}=y$$

And then use the linked question with Robjohn's answer.The fraction form can be simplified .

Edit 24/01/2023 :

There is another way if :

$$\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}=\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}$$

Then we can write :

$$\lim_{x \to 0^+}\left|g(x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot)\right|^{\frac{1}{x}}\overset{?}{=} e^y$$

Where :

$$g(x)=1-\frac{y-1+x}{1-\frac{1}{1-x}}=\frac{y-1}{x}-x-y+3$$

How to (dis)prove it ?

If this equation does hold , it barely is useful for anyhing having to do with $e^x$ , in particular not for the calculation of $e^x$. It is too complicated to have any real merit besides of having a nice representation. — Peter, Jan 22 '23 at 16:07
@Peter This is my last question here on MSE but not the least.It's a strange conjecture and perhaps there is some connection with something deep.Have you tried to show it ?I cannot post on overflow so if somebody is agree feel free to ask there.Anyway thanks for your comment Peter. — Miss and Mister cassoulet char, Jan 22 '23 at 16:56
If you don't have a proof of this equation, can you add why you think it's true or how you got it? Did you check it numerically? — Karl, Jan 22 '23 at 17:20
@Karl It's a pure empirical finding and verifying limit for $y\leq 10$ .I have make hundred test to find it perhaps hard to believe but it's a pure derivation starting from Desmos and my imagination . — Miss and Mister cassoulet char, Jan 22 '23 at 17:43
What are the reasons for such a cumbersome equation to be useful and worth proving? — Yalikesifulei, Jan 23 '23 at 07:54

score 1 · Accepted Answer · answered Jan 24 '23 at 10:19

Using the answer @Robjohn and the edit 24/01/2023 we have :

$$\lim_{x \to 0^+}\left|g(x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot)\right|^{\frac{1}{x}}=\lim_{x \to 0^+}\left(g\left(1-x\right)\right)^{\frac{1}{x}}=e^y$$

Where :

$$g(x)=1-\frac{y-1+x}{1-\frac{1}{1-x}}=\frac{y-1}{x}-x-y+3$$

A new definition of $e^x$ for $x\geq 1$ via a limit?

1 Answers1

Linked