quality:corruption
Versions suitable for calculators

The approximation Γ (z) ≈ 2 π z (z e z sinh 1 z + 1 810 z 6) z {\displaystyle \Gamma (z)\approx {\sqrt {\frac {2\pi }{z}}}\left({\frac {z}{e}}{\sqrt {z\sinh {\frac {1}{z}}+{\frac {1}{810z^{6}}}}}\right)^{z}} and its equivalent form 2 ln Γ (z) ≈ ln (2 π) − ln z + z (2 ln z + ln (z sinh 1 z + 1 810 z 6) − 2) {\displaystyle 2\ln \Gamma (z)\approx \ln(2\pi)-\ln z+z\left(2\ln z+\ln \left(z\sinh {\frac {1}{z}}+{\frac {1}{810z^{6}}}\right)-2\right)} can be obtained by rearranging Stirling's extended formula and observing a coincidence between the resultant power series and the Taylor series expansion of the hyperbolic sine function. This approximation is good to more than 8 decimal digits for z with a real part greater than 8. Robert H. Windschitl suggested it in 2002 for computing the gamma function with fair accuracy on calculators with limited program or register memory.

Gergő Nemes proposed in 2007 an approximation which gives the same number of exact digits as the Windschitl approximation but is much simpler: Γ (z) ≈ 2 π z (1 e (z + 1 12 z − 1 10 z)) z, {\displaystyle \Gamma (z)\approx {\sqrt {\frac {2\pi }{z}}}\left({\frac {1}{e}}\left(z+{\frac {1}{12z-{\frac {1}{10z}}}}\right)\right)^{z},} or equivalently, ln Γ (z) ≈ 1 2 (ln (2 π) − ln z) + z (ln (z + 1 12 z − 1 10 z) − 1). {\displaystyle \ln \Gamma (z)\approx {\tfrac {1}{2}}\left(\ln(2\pi)-\ln z\right)+z\left(\ln \left(z+{\frac {1}{12z-{\frac {1}{10z}}}}\right)-1\right).}

An alternative approximation for the gamma function stated by Srinivasa Ramanujan in Ramanujan's lost notebook is Γ (1 + x) ≈ π (x e) x (8 x 3 + 4 x 2 + x + 1 30) 1 6 {\displaystyle \Gamma (1+x)\approx {\sqrt {\pi }}\left({\frac {x}{e}}\right)^{x}\left(8x^{3}+4x^{2}+x+{\frac {1}{30}}\right)^{\frac {1}{6}}} for x ≥ 0. The equivalent approximation for ln n ! has an asymptotic error of 1 / 1400 n and is given by ln n ! ≈ n ln n − n + 1 6 ln (8 n 3 + 4 n 2 + n + 1 30) + 1 2 ln π. {\displaystyle \ln n!\approx n\ln n-n+{\tfrac {1}{6}}\ln(8n^{3}+4n^{2}+n+{\tfrac {1}{30}})+{\tfrac {1}{2}}\ln \pi.}

The approximation may be made precise by giving paired upper and lower bounds; one such inequality is π (x e) x (8 x 3 + 4 x 2 + x + 1 100) 1 / 6 < Γ (1 + x) < π (x e) x (8 x 3 + 4 x 2 + x + 1 30) 1 / 6. {\displaystyle {\sqrt {\pi }}\left({\frac {x}{e}}\right)^{x}\left(8x^{3}+4x^{2}+x+{\frac {1}{100}}\right)^{1/6}<\Gamma (1+x)<{\sqrt {\pi }}\left({\frac {x}{e}}\right)^{x}\left(8x^{3}+4x^{2}+x+{\frac {1}{30}}\right)^{1/6}.}