Es Ramanujan la aproximación para el factorial óptimo, o puede ser ajustado? (respuesta a continuación)

Question

Ramanujan la famosa factorial aproximación, $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{30}}$$ es mucho más precisa que la aproximación de Stirling cuando se aplica a los números menores que 1, pero no es muy allá. ¿Qué se podría hacer para ajustar sin necesidad de añadir a su complejidad?

Actualización: Después de experimentar con Ramanujan la aproximación, he descubierto este hecho inquietante: al parecer, el error disminuye más lento que el factorial aumenta, masivos entero errores de nivel superior el factorial se presenta. Me permite demostrar que: $$\begin{array}{r|rr} n&\text{actual }n!&\text{Ramanujan}\\\hline 0&1&1.005513858315898906334685\\ 1&1&1.000283346113497298280502\\ 2&2&2.000066137639113675155990\\ 3&6&6.000048293969899370824935\\ 4&24&24.000067662060676644042510\\ 5&120&120.000147065856635128019467\\ 6&720&720.000442402580258517644894\\ 7&5040&5040.001717876125295382828871\\ 8&40320&40320.008220460028928349902077\\ 9&362880&362880.046912269278701001557543\\ 10&3628800&3628800.311612606631103904381528\\ 11&39916800&39916802.364768173672637433190699\end{array}$$ At first the error reduction is apparent, but then the zeroes start to disappear, until at $11!$ the error overtakes the decimal point. Now I'm not saying the error is increasing—only that it it not decreasing fast enough. What can be done to correct this problem without radically altering the formula? I know that Ramanujan himself anticipated this problem when he wrote this inequality: $$\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{100}}<n!<\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{30}}$$ Podría tener la clave? (Véase la Fórmula 2.1 aquí)

Answer 1

Curiosamente, si hacemos una expansión de la serie en Mathematica

Series[(Gamma[n + 1]/Sqrt[Pi] (n/Exp[1])^-n)^6, {n, Infinity, 6}]

Llegamos $$8 n^3+4 n^2+n+\frac{1}{30}-\frac{11}{240 n}+\frac{79}{3360n^2}+\frac{3539}{201600 n^3}-\frac{9511}{403200 n^4}-\frac{10051}{716800n^5}+\frac{233934691}{6386688000n^6}+O\left(\frac{1}{n}\right)^{13/2}.$$ Esto sugiere que su ad hoc de ajuste de la constante plazo probablemente no es estrictamente justificados.

Answer 2

Hace mucho tiempo, la necesidad de una fórmula explícita para la estadística termodinámica, me estaba enfrentando el mismo problema del cálculo de $n!$$0<n<1$. A partir de Ramanujan la aproximación, lo que hice fue escribir $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+x(n)}$$ What can be easily established is that $$x(0)=\frac{1}{\pi ^3}$$ just as proposed by Brian J. Fink. We can also show that $$x(1/2)=\frac{1}{8} \left(e^3-20\right)$$ and $$x(1)=13-\frac{e^6}{\pi ^3}$$ and I used (this is totally empirical and cannot be justified) a quadratic expansion of $x(n)=a+b n+c n^2$ which leads to $$a=\frac{1}{\pi ^3}$$ $$b=\frac{1}{2} \left(6+e^3-\frac{2 \left(3+e^6\right)}{\pi ^3}\right)$$ $$c=\frac{1}{2} \left(-32-e^3+\frac{4 \left(1+e^6\right)}{\pi ^3}\right)$$ Again, this has never been supposed to be used outside the range $[0,1]$

Answer 3

La fracción $\tfrac{1}{30}$ no puede ser, prácticamente hablando, la mejor cota superior para el término final de la radicando. En efecto, de acuerdo a Ramanujan propias notas, el número de rangos de$\tfrac{1}{30}$$\tfrac{1}{100}$. Observar que cuando se evalúa a $0$, $$\begin{align}0!&\approx\sqrt{\pi}\left(\frac{0}{e}\right)^0\root{\LARGE{6}}\of{8\cdot0^3+4\cdot0^2+0+\frac{1}{30}}\\ &=\sqrt{\pi}\cdot1\cdot\root{\LARGE{6}}\of{\frac{1}{30}}\\ &=\frac{\sqrt{\pi}}{\root{\LARGE{6}}\of{30}}\\ &=\root{\LARGE{6}}\of{\frac{\pi^3}{30}}\\ &\approx1.00551\end{align}$$ Notice the error in the result. However, if we replace the number $30$ in the above formula with $\pi^3$, the result is dead accurate: $$\begin{align}0!&\approx\sqrt{\pi}\left(\frac{0}{e}\right)^0\root{\LARGE{6}}\of{8\cdot0^3+4\cdot0^2+0+\frac{1}{\pi^3}}\\ &=\sqrt{\pi}\cdot1\cdot\root{\LARGE{6}}\of{\frac{1}{\pi^3}}\\ &=\frac{\sqrt{\pi}}{\root{\LARGE{6}}\of{\pi^3}}\\ &=\root{\LARGE{6}}\of{\frac{\pi^3}{\pi^3}}\\ &=1\end{align}$$ Now let's try it on $n=1$: $$\begin{align}1!&\approx\sqrt{\pi}\left(\frac{1}{e}\right)^1\root{\LARGE{6}}\of{8\cdot1^3+4\cdot1^2+1+\frac{1}{\pi^3}}\\ &=\frac{\sqrt{\pi}}{e}\root{\LARGE{6}}\of{13+\frac{1}{\pi^3}}\\ &\approx1.0002695\end{align}$$ For this reason, I recommend this modification to Ramanujan's approximation: $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{\pi^3}}$$ Alternatively, since $\pi^3\approx31.00627668$, a slightly less robust adjustment to the formula can be made by substituting $31$ for $30$, instead of $\pi^3$.

Déjame saber lo que piensa acerca de esto!

Pensamientos adicionales: Basado en el material que he añadido a mi pregunta anterior, quizás $\tfrac{1}{\pi^3}\approx31$ debe reducirse gradualmente hacia la $\tfrac{1}{\pi^4}\approx97$ $n$ aumenta. Un proyecto de la fórmula, a la espera de una mayor investigación: $${\large n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{\pi^{3+c}}}}$$ where $$\left(\frac{n}{n+1}\right)^3>c>\left(\frac{n}{n+1}\right)^2$$ The formula above is constantly evolving as I test and make adjustments. This formula is tricky to adjust. I have to express it in terms of an unknown value $c$ until I can dial it in, so to speak. My current estimate is $$c\approx\left(\frac{n}{n+1}\right)^{{\pi^3}\lower{2pt}/\lower{4pt}{12}}$$ Conclusion: The answer is no and no. According to my own analysis, Ramanujan's formula is only useful for estimating what the value of $n!$ might be for some $n$; and the error reduction rate appears painfully slow, slow enough to begin to accumulate error in the integer range as early as $11!$; as for tweaking it, the need appears to be for an adjustment of a major sort, which perhaps I will be unable to accomplish at this time. Indeed, any adjustment that optimizes the formula over one range appears to diminish its accuracy over another, as with all asymptotic approximations of Factorial and Gamma functions in general. I'm not saying a formula for the factorial function cannot be found that is optimal for all $x\geqslant0$; yo, simplemente, no tienen ni el tiempo ni la experiencia matemática para hacerlo yo mismo.