Show a conjecture about $\arcsin\left(\frac{9}{10}\right)\cdot\left(\frac{6}{10}\right)!>1$

Question

Recently I found the inequality :

$$\arcsin\left(\frac{9}{10}\right)\cdot\left(\frac{6}{10}\right)!>1$$

I don't ask for a proof .

I ask if someone can give some details on the link with the Gamma function and arcsin .

I ask also because we have a formula on wiki page for the Gamma's function wich gives an equality with the function $\sin$ .

But now why arcsin ?

Can we find a infinite series ?

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I can gives some other terms of the "series" we want :

$$\left(\frac{9}{10}\right)\cdot\frac{\left(\frac{6}{10}\right)!\arcsin\left(\left(\frac{99}{100}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{999}{1000}\right)^{\frac{1}{80}}\right)}\frac{\arcsin\left(\left(\frac{9999}{10000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{99999}{100000}\right)^{\frac{1}{70}}\right)}>1$$

Edit :

I gives more terms :

$$\arcsin\left(\frac{9}{10}\right)\cdot\frac{\left(\frac{6}{10}\right)!\arcsin\left(\left(\frac{99}{100}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{999}{1000}\right)^{\frac{1}{80}}\right)}\frac{\arcsin\left(\left(\frac{9999}{10000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{99999}{100000}\right)^{\frac{1}{70}}\right)}\frac{\arcsin\left(\left(\frac{999999}{1000000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{9999999}{10000000}\right)^{\frac{1}{62}}\right)}\frac{\arcsin\left(\left(\frac{99999999}{100000000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{999999999}{1000000000}\right)^{\frac{1}{85}}\right)}\frac{\arcsin\left(\left(\frac{9999999999}{10000000000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{99999999999}{100000000000}\right)^{\frac{1}{79}}\right)}>1$$

And made a conjecture :

Let $a_k>0$ such that :

$$\arcsin\left(\frac{9}{10}\right)\cdot\frac{\left(\frac{6}{10}\right)!\arcsin\left(\left(\frac{99}{100}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{999}{1000}\right)^{\frac{1}{a_{k}}}\right)}\frac{\arcsin\left(\left(\frac{9999}{10000}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{99999}{100000}\right)^{\frac{1}{a_{k+1}}}\right)}\cdot\cdot\cdot\frac{\arcsin\left(\left(\frac{9999\cdot\cdot\cdot9}{1000\cdot\cdot\cdot00}\right)^{\frac{1}{600}}\right)}{\arcsin\left(\left(\frac{99999\cdot\cdot\cdot9}{10000\cdot\cdot\cdot00}\right)^{\frac{1}{a_{m}}}\right)}\cdot\cdot\cdot=1$$

Then I conjecture that for $a_1=80,a_2=70,a_3=62,a_4=85,a_5=79,a_6=61.75$:

$$a_k<100$$

How to (dis)prove it ?

Answer 1

The solution to

$$\sin^{-1}\left(\frac32x\right)x!=1\implies x=0.59983…$$

is approximated by the series expansion of $x!$ about $x=\frac12$ and of $\sin^{-1}(x)$ about $x=\frac1{\sqrt3}$:

$$\frac{\sqrt\pi}2\left(\frac\pi3+3\left(x-\frac1{\sqrt3}\right)\right)=1\implies x=\frac1{\sqrt3}+\frac2{3\sqrt\pi}-\frac\pi9=0.60441…\approx\frac23$$

More terms give better approximations.

Answer 2

Define $$f(\epsilon)=\sin ^{-1}(1-\epsilon^2 )\,\, \Gamma \left(\frac{3}{2}+\epsilon^2 \right)=\sqrt{\frac{\pi }{2}}\sum_{n=0}^\infty a_n\,\epsilon^n $$ where the first coefficients are

$$\left( \begin{array}{cc} n & a_n \\ 0 & \frac{\pi }{2 \sqrt{2}} \\ 1 & -1 \\ 2 & \frac{\pi \psi ^{(0)}\left(\frac{3}{2}\right)}{2 \sqrt{2}} \\ 3 & -\frac{25}{12}+\gamma +\log (4) \\ 4 & \frac{\pi \left(-8+\pi ^2+2 \psi ^{(0)}\left(\frac{3}{2}\right)^2\right)}{8 \sqrt{2}} \\ 5 & \frac{1}{480} \left(951-120 \pi ^2-40 \psi ^{(0)}\left(\frac{3}{2}\right) \left(1+6 \psi ^{(0)}\left(\frac{3}{2}\right)\right)\right) \\ 6 & \frac{\pi \psi ^{(0)}\left(\frac{3}{2}\right) \left(-24+3 \pi ^2+2 \psi ^{(0)}\left(\frac{3}{2}\right)^2\right)+2 \pi \psi ^{(2)}\left(\frac{3}{2}\right)}{24 \sqrt{2}} \end{array} \right)$$

Now compute the partial sums $$S_p=\sqrt{\frac{\pi }{2}}\sum_{n=0}^p a_n\,10^{-n/2} $$

$$\left( \begin{array}{cc} p & S_p \\ 0 & 1.392081999 \\ 1 & 0.995749269 \\ 2 & 1.000828973 \\ 3 & 0.996079983 \\ 4 & 1.002595858 \\ 5 & 1.000654391 \\ 6 & 1.000485853 \\ 7 & 1.000515895 \\ 8 & 1.000538616 \\ 9 & 1.000532116 \\ 10 & 1.000530898 \\ 11 & 1.000531186 \\ 12 & 1.000531278 \\ 13 & 1.000531254 \\ 14 & 1.000531248 \\ 15 & 1.000531249 \end{array} \right)$$

Otherwise, as @Tyma Gaidash did, starting with $\epsilon_0=0$, the first iterate of Newton method for $f(\epsilon)=1$ is $$\epsilon_{(2)}=\frac{\pi ^{3/2}-4}{2 \sqrt{2 \pi }}$$ and, by Darboux theorem, we know that it is an underestimate of the solution since $$\big[f(0)-1\big]f''(0)=\frac{1}{8} \pi ^{3/2} \left(\pi ^{3/2}-4\right) \psi ^{(0)}\left(\frac{3}{2}\right)~~>~0$$

Similarly, the first iterate of Halley method $$\epsilon_{(3)}=\Bigg[\frac{2 \sqrt{2 \pi }}{\pi ^{3/2}-4}-\frac{\pi \psi ^{(0)}\left(\frac{3}{2}\right)}{2 \sqrt{2}}\Bigg]^{-1}$$ is an overestimate of the solution.

We can expand $$f(\epsilon)=\sum_{n=0}^\infty b_n\,(\epsilon-\epsilon_{(2)})^n $$

Computing again the partial sums

$$\left( \begin{array}{cc} p & S_p \\ 0 & 1.004658787 \\ 1 & 1.000530217 \\ 2 & 1.000531249 \end{array} \right)$$

$$f(\epsilon)=\sum_{n=0}^\infty c_n\,(\epsilon-\epsilon_{(3)})^n $$ using only the first term gives $1.000531214$.

Edit

Pushing more and more $$f\left(\frac{1159+686 \pi -116 \pi ^2}{324+227 \pi +589 \pi ^2}\right)=1.0000000000000000000250068\cdots$$

Update

It is interesting to look at the zero level plot of $$f(x,y)=\sin ^{-1}(x)\,\, \Gamma (1+y)-1$$ (have a look here).

Using the same approach as above, you could prove that $$f\left(\frac{900}{1000},\frac{332}{1000}\right)>0 \qquad \text{and} \qquad f\left(\frac{900}{1000},\frac{596}{1000}\right)>0$$ $$f\left(\frac{901}{1000},\frac{349}{1000}\right)>0 \qquad \text{and} \qquad f\left(\frac{901}{1000},\frac{578}{1000}\right)>0$$ $$f\left(\frac{902}{1000},\frac{369}{1000}\right)>0 \qquad \text{and} \qquad f\left(\frac{902}{1000},\frac{557}{1000}\right)>0$$ $$f\left(\frac{903}{1000},\frac{395}{1000}\right)>0 \qquad \text{and} \qquad f\left(\frac{903}{1000},\frac{530}{1000}\right)>0$$ $$f\left(\frac{904}{1000},\frac{445}{1000}\right)>0 \qquad \text{and} \qquad f\left(\frac{904}{1000},\frac{478}{1000}\right)>0$$