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Revision difference : binomial formula
Version 7 Version 6
The binomial formula gives the power series expansion of the The binomial formula gives the power series expansion of the
$p\supth$ power function. The power $p$ can be an integer, $p\supth$ power function. The power $p$ can be an integer,
rational, real, or even a complex number. The formula is rational, real, or even a complex number. The formula is
\begin{align*} \begin{align*}
(1+x)^p &= \sum_{n=0}^\infty \frac{p^{\underline{n}}}{n!} \, x^n\\ (1+x)^p &= \sum_{n=0}^\infty \frac{p^{\underline{n}}}{n!} \, x^n\\
&= \sum_{n=0}^\infty \binom{p}{n} x^n &= \sum_{n=0}^\infty \binom{p}{n} x^n
\end{align*} \end{align*}
where $p^{\underline{n}}= p(p-1)\ldots (p-n+1)$ denotes the falling where $p^{\underline{n}}= p(p-1)\ldots (p-n+1)$ denotes the falling
factorial, and where $\binom{p}{n}$ denotes the generalized binomial factorial, and where $\binom{p}{n}$ denotes the generalized binomial
coefficient. coefficient.
For $p=0,1,2,\ldots$ the power series reduces to a polynomial, and we For $p=0,1,2,\ldots$ the power series reduces to a polynomial, and we
obtain the usual binomial theorem. For other values of $p$, the obtain the usual binomial theorem. For other values of $p$, the
radius of convergence of the series is $1$; the right-hand series radius of convergence of the series is $1$; the right-hand series
converges pointwise for all complex $|x|<1$ to the value on the left converges pointwise for all complex $|x|<1$ to the value on the left
side. Also note that the binomial formula is valid at $x=\pm 1$, but side. Also note that the binomial formula is valid at $x=\pm 1$, but
for certain values of $p$ only. Of course, we have convergence if $p$ for certain values of $p$ only. Of course, we have convergence if $p$
is a natural number. Furthermore, for $x=1$ and real $p$, we have is a natural number. Furthermore, for $x=1$ and real $p$, we have
absolute convergence if $p>0$, and conditional convergence if absolute convergence if $p>0$, and conditional convergence if
$-1<p<0$. For $x=-1$ we have absolute convergence for $p>0$. $-1<p<0$. For $x=-1$ we have absolute convergence for $p>0$.