Binomial Theorem/Abel's Generalisation

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Theorem

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

for $n \in \Z_{\ge 0}$ and $x \in \R_{\ne 0}$.


Special Case: $x + y = 0$

Consider Abel's Generalisation of Binomial Theorem:

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


This holds in the special case where $x + y = 0$.


Negative $n$

Abel's Generalisation of Binomial Theorem:

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

does not hold for $n \in \Z_{< 0}$.


Proof 1

By admitting $y = \left({x + y}\right) - x$, we have that:

$\left({x + y}\right)^n = \left({x + \left({x + y}\right) - x}\right)^n$


Expanding the right hand side in powers of $\left({x + y}\right)$:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({x + \left({x + y}\right) + k z}\right)^{n - k}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \sum_j \left({x + y}\right)^j \left({-x + k z}\right)^{n - k - j} \binom {n - k} j\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_j \binom n j \left({x + y}\right)^j \sum_k \binom {n - j} {n - j - k} x \left({x - k z}\right)^{k - 1} \left({-x + k z}\right)^{n - k - j}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop \le n} \binom n j \left({x + y}\right)^j 0^{n - j}\) $\quad$ Abel's Generalisation of Binomial Theorem: Special Case $x + y = 0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x + y}\right)^n\) $\quad$ Binomial Theorem $\quad$

$\blacksquare$


Proof 2

From this formula:

$(1): \quad \displaystyle \sum_{k \mathop \in \Z} \binom n k x \left({x + k z}\right)^{k - 1} y \left({y + \left({n - k}\right) z}\right)^{n - k - 1} = \left({x + y}\right) \left({x + y + n z}\right)^{n - 1}$

The given formula:

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

can then be transformed into $(1)$ by


Proof 3

From Hurwitz's Generalisation of Binomial Theorem:

$(1): \quad \left({x + y}\right)^n = \displaystyle \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} y \left({y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}$

Setting $z = z_1 = z_2 = \cdots z_n$ we have:

\(\displaystyle \) \(\) \(\displaystyle \sum x \left({x + \epsilon_1 z + \cdots + \epsilon_n z}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} y \left({y - \epsilon_1 z - \cdots - \epsilon_n z}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum \binom n k x \left({x + k z}\right)^{k - 1} y \left({y - k z}\right)^{n - k}\) $\quad$ where $\epsilon_1 + \cdots + \epsilon_n = k$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum \binom n {n - k} x \left({x + k z}\right)^{k - 1} y \left({y - k z}\right)^{n - k}\) $\quad$ Symmetry Rule for Binomial Coefficients $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum \binom n k x \left({x - k z}\right)^{k - 1} y \left({y + k z}\right)^{n - k}\) $\quad$ $\quad$



Hence the result.


Source of Name

This entry was named for Niels Henrik Abel.


Historical Note

Niels Henrik Abel presented this generalisation of the Binomial Theorem in $1826$ in the first volume of August Leopold Crelle's Journal für die reine und angewandte Mathematik.


Sources