User:Turtlemanman

Physics Regents Grade


 * $$\frac{2}$$

The gamma function's connection to the factorial function.


 * $$\Gamma(z) = \int_{0}^{\infty} t^{z-1}e^{-t}\,dt$$     definition


 * $$\Gamma(z) = [-t^{z-1}e^{-t}]_{0}^{\infty} - \int_{0}^{\infty} -e^{-t}(z-1)t^{z-2}\,dt$$     integration by parts


 * $$\Gamma(z) = 0 + (z-1)\int_{0}^{\infty} e^{-t}t^{(z-1)-1}\,dt $$

Hence


 * $$\Gamma(z+1) = z\int_{0}^{\infty} e^{-t}t^{z-1}\,dt = z\Gamma(z) $$

It follows that, since $$\textstyle \Gamma(1) = 1$$,


 * $$\textstyle \Gamma(z+1) = z\Gamma(z) = z(z-1)\Gamma(z-1) = z(z-1)(z-2)\Gamma(z-2) = ... = z!\Gamma(1) = z!$$

Thus, in summary,


 * $$\textstyle \Gamma(z+1) = z!$$

Which shows the gamma function, $$\textstyle \Gamma(z)$$, to be an extension of the factorial function, $$\textstyle z!$$ (for all natural numbers z).