Editing Binomial theorem (section)

=== Newton's generalized binomial theorem ===
{{Main|Binomial series}}
Around 1665, [[Isaac Newton]] generalized the binomial theorem to allow real exponents other than nonnegative integers. (The same generalization also applies to [[complex number|complex]] exponents.) In this generalization, the finite sum is replaced by an [[infinite series]]. In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials.  However, for an arbitrary number {{mvar|r}}, one can define
<math display="block">{r \choose k}=\frac{r(r-1) \cdots (r-k+1)}{k!} =\frac{(r)_k}{k!},</math><!--This is not the same as \frac{r!}{k!(r−k)!}. Please do not change it.-->
where <math>(\cdot)_k</math> is the [[Pochhammer symbol]], here standing for a [[falling factorial]].  This agrees with the usual definitions when {{mvar|r}} is a nonnegative integer.  Then, if {{mvar|x}} and {{mvar|y}} are real numbers with {{math|{{abs|''x''}} > {{abs|''y''}}}},<ref name=convergence group=Note>This is to guarantee convergence. Depending on {{mvar|r}}, the series may also converge sometimes when {{math|1={{abs|''x''}} = {{abs|''y''}}}}.</ref> and {{mvar|r}} is any complex number, one has
<math display="block">\begin{align}
   (x+y)^r & =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \\
   &= x^r + r x^{r-1} y + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \cdots.
 \end{align}</math>

When {{mvar|r}} is a nonnegative integer, the binomial coefficients for {{math|1=''k'' > ''r''}} are zero, so this equation reduces to the usual binomial theorem, and there are at most {{math|1=''r'' + 1}} nonzero terms. For other values of {{mvar|r}}, the series typically has infinitely many nonzero terms.

For example, {{math|1=''r'' = 1/2}} gives the following series for the square root:
<math display="block">\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots.</math>

Taking {{math|1=''r'' = &minus;1}}, the generalized binomial series gives the [[Geometric series#Sum|geometric series formula]], valid for {{math|{{abs|''x''}} < 1}}:
<math display="block">(1+x)^{-1} = \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots.</math>

More generally, with {{math|1=''r'' = −''s''}}, we have for {{math|{{abs|''x''}} < 1}}:<ref name=wolfram2>{{cite web| url=https://mathworld.wolfram.com/NegativeBinomialSeries.html|title=Negative Binomial Series|website=Wolfram MathWorld|last=Weisstein|first=Eric W.}}</ref>
<math display="block">\frac{1}{(1+x)^s} = \sum_{k=0}^\infty {-s \choose k} x^k = \sum_{k=0}^\infty {s+k-1 \choose k} (-1)^k x^k.</math>

So, for instance, when {{math|1=''s'' = 1/2}},
<math display="block">\frac{1}{\sqrt{1+x}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + \cdots.</math>

Replacing {{mvar|x}} with {{mvar|-x}} yields:
<math display="block">\frac{1}{(1-x)^s} = \sum_{k=0}^\infty {s+k-1 \choose k} (-1)^k (-x)^k = \sum_{k=0}^\infty {s+k-1 \choose k} x^k.</math>

So, for instance, when {{math|1=''s'' = 1/2}}, we have for {{math|{{abs|''x''}} < 1}}:
<math display="block">\frac{1}{\sqrt{1-x}} = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \frac{35}{128}x^4 + \frac{63}{256}x^5 + \cdots.</math>