Benedict Irwin edited untitled.tex  over 9 years ago

Commit id: 3939b9f1ba5c8d9c9239301fa367fa03c0e1f97b

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\end{array}  \end{equation}  Above we have used (or discovered depending on your point of view!) that $(z-1)\Gamma(z)-\Gamma(z+1)=-\Gamma(z)$. I.e $(n-1)(n-1)!-n!=-(n-1)!$...  Then reused this again an used/discovered that $(1-2z)\Gamma(z+1)+\Gamma(z+2)+(z-1)^2\Gamma(z)=\Gamma(z)$...  Altohugh if one knows the inner workings of the gamma function this is perhaps obvious, we could potentially using this method, discover identities for horrendously complicated functions without knowing what they were... I will demonstrate below later on.