# Show that $\Gamma(\Omega)\leq \Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$ on $(0,1]$

Working with Geogebra I have found :

Let $$0 : $$\Gamma(\Omega)\leq f(x)=\Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$$ With the Gamma function and the Lambert's function .

For the RHS the minimum of the difference is $$2-\Gamma\Big(\operatorname{W}\Big(e^{-e^{-1}}\Big)\Big)\simeq0.005167\cdots$$

Using the derivative of $$f(x)$$ it's not hard to show that $$e^{-1}$$ is a maximum on $$(0,1]$$ .

The problem is :

$$f(e^{-1})<2\quad (1)$$

My question :

How to show $$(1)$$ ?

• A very modest gymnopédie for you ! Nov 5, 2020 at 11:11
• Apologies, but what is $\Omega$? Nov 5, 2020 at 11:37
• @K.defaoite. $\Omega=W(1)$ Nov 6, 2020 at 5:04
• Have a look at my new edit for $p=9$ and the "proof" that this is not an overfit. Nov 6, 2020 at 5:05
• @ClaudeLeibovici Digression : If you like Satie's music see the beginning of the 4 gnossienne youtube.com/watch?v=JE8ReAAcZjU . And compare it to the beginning of the Mozart's music kv 397 youtube.com/watch?v=1BkZ8ci8_k4 . Good day cheers :-) Nov 8, 2020 at 10:48

For the time being, consider the function $$f(y)=y\,\Gamma(y)\qquad \text{for} \qquad 0 \leq y \leq 1$$ and, as I did in this question of mine, let me approximate $$f(y)$$ by $$g(y)=1+y(1-y) \sum_{k=0}^p d_k\, y^k$$

Using $$\color{red}{p=3}$$, matching the function and first derivative values at $$x=0$$, $$x=\frac 12$$ and $$x=1$$ , we have $$d_0= -\gamma$$ $$d_1=-17+6 \gamma +4 \sqrt{\pi } (\gamma +2\log (2))$$ $$d_2=4 \left(5-3 \gamma +4 \sqrt{\pi }-3 \sqrt{\pi } (\gamma +2\log (2))\right)$$ $$d_3=-4+8 \gamma -16 \sqrt{\pi }+8 \sqrt{\pi } (\gamma +2\log (2))$$

This gives $$\int \big[f(y)-g(y)\big]^2\,dy =1.28\times 10^{-8}$$ $$\int \Big[\frac{f(y)-g(y)}y\Big]^2\,dy =3.24\times 10^{-7}$$

Now, assuming that we know the exact value of $$W\left(e^{-1/e}\right)$$, we have $$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994799$$ which seems to be quite good for a first approximation when compared to the exact $$1.994832$$.

If we do not know the exact value of $$W\left(e^{-1/e}\right)$$, we can have a reasonable approximation of it using the $$[5,4]$$ Padé approximant of Lambert function $$W(x)$$ built around $$x=0$$. This is $$W(x) \sim x \frac {1+\frac{7430297 }{1597966}x+\frac{1018440443 }{156600668}x^2+\frac{1260595681 }{469802004}x^3+\frac{974868241 }{9396040080}x^4 } {1+\frac{9028263 }{1597966}x+\frac{1668309215 }{156600668}x^2+\frac{3536864687 }{469802004}x^3+\frac{14189787721 }{9396040080}x^4 }$$ which gives

$$W\left(e^{-1/e}\right)\sim 0.444019 \qquad \text{while} \qquad \text{exact}=0.444016$$

Using this approximation leads to $$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994784$$

Using $$\color{red}{p=6}$$, matching the function, first and second derivative values at $$x=0$$, $$x=\frac 12$$ and $$x=1$$, the above norms become $$2.72\times 10^{-12}$$ and $$5.88\times 10^{-11}$$ which is a very significant improvement.

Using the exact value of $$W\left(e^{-1/e}\right)$$, this gives $$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483224 \qquad \text{while} \qquad \text{exact}=1.99483209$$

Using $$\color{red}{p=9}$$, matching the function, first, second and third derivative values at $$x=0$$, $$x=\frac 12$$ and $$x=1$$, the above norms become $$6.19\times 10^{-16}$$ and $$1.23\times 10^{-14}$$ which is another significant improvement.

Using the exact value of $$W\left(e^{-1/e}\right)$$, this gives $$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483209100 \qquad \text{while} \qquad \text{exact}=1.99483209170$$

Remark

As in the linked question, using $$p=9$$ does not correspond to any overfit of function $$f(y)$$. A simple linear regression gives

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} \\ d_0 & -0.577215 & 1.1 \times 10^{-8} \\ d_1 & +0.411823 & 3.7 \times 10^{-7}\\ d_2 & -0.495316 & 5.0 \times 10^{-6} \\ d_3 & +0.482941 & 3.5 \times 10^{-5} \\ d_4 & -0.477680 & 1.4 \times 10^{-4} \\ d_5 & +0.429199 & 3.6 \times 10^{-4}\\ d_6 & -0.325588 & 5.6 \times 10^{-4}\\ d_7 & +0.184385 & 5.3 \times 10^{-4} \\ d_8 & -0.066496 & 2.8 \times 10^{-4} \\ d_9 & +0.011163 & 6.1 \times 10^{-5} \\ \end{array}$$ which show the high statistical significance of the parameters.

• Very nice! a question : it was difficult for you ? Cheers :-). Nov 5, 2020 at 13:38
• @ErikSatie. Not so much since I had in mind my previous problem and I suppose that you noticed that it is excatly the same approach. To me, the most difficult has been the starting point that is to say to consider $f(y)=y\,\Gamma(y)$ and not $\Gamma(y)$ itself. It was amazing. Thanks for the problem. By the way, if I may ask, where are you located ? Cheers -) Nov 5, 2020 at 14:59