User:X7q/MathSandbox

Please don't erase
$$ \begin{array}{l} S_1 = \sum_{i=1}^n c[i] s1[i], \\ S_2 = \sum_{i=1}^n c[i] s2[i], \\ S_3 = \sum_{i=1}^n c[i] s3[i], \\ S_{21} = \sum_{i=1}^n c[i] (s2[i] - s1[i]), \\ S_{31} = \sum_{i=1}^n c[i] (s3[i] - s1[i]). \end{array} $$

$$S_{21} = \sum_{i=1}^n c[i] (s2[i] - s1[i]) = \sum_{i=1}^n (c[i] s2[i] - c[i] s1[i]) = \sum_{i=1}^n c[i] s2[i] - \sum_{i=1}^n c[i] s1[i] = S_2 - S_1.$$

$$\Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^{\mathrm z}}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^{\mathrm z}}{1+\frac{\mathrm z}{n}}$$

$$\Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^{z}}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^{z}}{1+\frac{z}{n}}$$