Talk:Binomial Theorem/Hurwitz's Generalisation


 * The source (Knuth) is incorrect.
 * Based on Binomial Theorem/Abel's Generalisation, (and Hurwitz himself), that lone y does not belong there.


 * Abel's Generalisation


 * $\displaystyle \left({x + y}\right)^n = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}$


 * Hurwitz's Generalisation (incorrect - including the lone y)
 * $\displaystyle \left({x + y}\right)^n = \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} y \left({y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}$

Source 1 - Exercise 30: * : $\S 1.2.6$: Binomial Coefficients: Exercise $51$


 * Hurwitz's Generalisation (corrected - lone y removed)
 * $\displaystyle \left({x + y}\right)^n = \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} \left({y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}$


 * Letting Hurwitz speak for himself (page 202)


 * "Diese vereinfacht sich noch, wenn man $\nu $ durch $\nu - \paren {x_1 + x_2 + \cdots + x_n } $ ersetzt. Dadurch erhalt man namlich"


 * $\displaystyle \sum_{k } \paren {\mu + x_{\alpha 1 } + x_{\alpha 2 } + \cdots + x_{\alpha \lambda }}^{\lambda - 1 } \paren {\nu - x_{\alpha 1 } - x_{\alpha 2 } - \cdots - x_{\alpha \lambda }}^{n - \lambda } = \frac 1 {\mu } \paren {\mu + \nu }^n$


 * I saw your humorous comment on Binomial Theorem/Abel's Generalisation/Proof 3 and that's what generated this.


 * --Robkahn131 (talk) 13:58, 21 June 2020 (EDT)