Ordinary Newton binomial theorem in high school before, is the time when the power of positive integers, but sometimes need to use cases (such as generating function) is not necessarily a positive integer, and so will need scores or negative, so Newton generalized binomial theorem came out.
First, we introduce Newton binomial coefficients $ {r \ choose n} $.
Newton binomial coefficient is defined:
Let r is a real number, is an integer of n-introduced in the form of symbols
$${r \choose n}=
\begin{cases}
0， & n<0\\
1， & n=0\\
\frac{r(r-1)\cdots (r-n+1)}{n!}， & n>0
\end{cases}$$
Newton generalized binomial theorem:
$(x+y)^{\alpha}=\sum_{n=0}^\infty{\alpha \choose n}x^{n}y^{\alpha-n}$
Wherein x, y, α is a real number, and $ \ mid \ frac {x} {y} \ mid <1 $
Proof:
Demonstrate the need to use knowledge of mathematical analysis, never learned, you should read 2333.
Order $ z = \ frac {x} {y} $ there is:
$(x+y)^{\alpha}=y^{\alpha}(1+z)^{\alpha}，\mid z\mid <1$
Provided $ f (z) = (1 + z) ^ {\ alpha} $, so there are:
$f^{(n)}(z)=\alpha (\alpha -1)\cdots (\alpha -n+1)(1+z)^{\alpha -n}$
Thus, when Z 0 = 0, the Taylor formula function (in this case should be referred to McLaughlin Expandable) has the following form:
$(1+z)^{\alpha} =1+\frac{\alpha}{1!}z+\frac{\alpha (\alpha -1)}{2!}z^{2}+\cdots +\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}z^{n}+r_n(0;z)$
That is:
$(1+z)^{\alpha}=1+{\alpha \choose 1}z+{\alpha \choose 2}z^{2}+\cdots +{\alpha \choose n}z^{n}+r_n(0;z)$
The remainder will have to use the Cauchy formula:
$r_n(0;z)=\frac{\alpha (\alpha -1)\cdots (\alpha -n)}{n!}(1+ξ)^{\alpha -n}(z-ξ)^{n}z$
Wherein ξ is between 0 to z.
The remainder deformation can be obtained at:
$r_n(0;z)=\alpha (1-\frac{\alpha}{1})\cdots (1-\frac{\alpha}{n})(1+ξ)^{\alpha}(\frac{z-ξ}{1+ξ})^{n}z$
Because when $ \ mid z \ mid <1 $ time, from between 0 and z ξ This condition can be introduced:
$\mid \frac{z-ξ}{1+ξ}\mid =\frac{\mid z\mid -\mid ξ\mid}{\mid 1+ξ\mid}\leq \frac{\mid z\mid -\mid ξ\mid}{1-\mid ξ\mid}=1-\frac{1-\mid z\mid}{1-\mid ξ\mid}\leq 1-(1-\mid z\mid)=\mid z\mid$
于是$\mid r_n(0;z)\mid \leq\mid \alpha (1-\frac{\alpha}{1})\cdots (1-\frac{\alpha}{n})\mid (1+ξ)^{\alpha}\mid z\mid ^{n+1}$
Because $ \ mid r_ {n + 1} (0; z) \ mid = \ mid r_n (0; z) \ mid \ times \ mid (1- \ frac {\ alpha} {n + 1}) z \ mid and because $ $ \ mid z \ mid <1 $, so if $ \ mid z \ mid <q <1 $, regardless of $ \ alpha $ how the value for n will be sufficiently large $ \ mid (1- \ frac {\ alpha} {n + 1}) z \ mid <q <1 $, which means that when $ n \ rightarrow \ infty $, there $ r_n (0; z) \ rightarrow 0 $, whereby described when $ \ mid z \ mid <1 $ when infinite series $ 1 + {\ alpha \ choose 1} z + {\ alpha \ choose 2} z ^ {2} + \ cdots + {\ alpha \ choose n} z ^ {n} + \ cdots (*) $ converges to $ (1 + z) ^ {\ alpha} $.
For this case equation $ y ^ {\ alpha} (1 + z) ^ {\ alpha}, \ mid z \ mid <1 $ to the left and then expanded to the infinite series $ y ^ \ alpha $ get multiplied our $ (x + y) ^ {\ alpha} = \ sum_ {n = 0} ^ \ infty {\ alpha \ choose n} x ^ {n} y ^ {\ alpha-n} $
When $ \ mid z \ mid> $ 1, by the d'Alembert ratio test method can be obtained, as long as $ \ alpha \ notin \ mathbb { N} $, stages (*) always diverge.
Derivative particular, when $ \ alpha \ in \ mathbb {N} $ when a function $ f (z) $, the above arbitrary order n are 0, 0 remainder is directly launched on the bin, expand the binomial theorem is obtained through high school.
"Mathematical Analysis" References Zhuoli Qi and Qu Wanling "Discrete Mathematics"