Can the Binomial Series Exist When α < n?

[steven187]

hello all

I thought this might be an interesting question to ask, consider the following series
\sum_{n=0}^{\infty}\left(\begin{array}{cc}\alpha\\n \end{array}\right)x^{n}=(1+x)^{\alpha}
this is known as the binomial series, what's confusing me is that how could this series exist when \alpha&lt; n especially when its a series that adds infinitely
number of terms, from my understanding this \left(\begin{array}{cc}\alpha\\n \end{array}\right) can only be evaluated when \alpha&gt;n please help

thanxs

 

[matt grime]

they are identically zero for n>alpha (assuming alpha is an integer)

 

[steven187]

hello Matt
well I am still confused, what I am not understanding is that if

\left(\begin{array}{cc}\alpha\\n \end{array}\right)=\frac{\alpha !}{n!(\alpha-n)!} and \alpha&lt; n then \alpha-n is negetive how could you find the factorial of a negetive

number, and if they were identically equal to zero i couldn't see how this
\frac{\alpha !}{n!(\alpha-n)!}=0

 

[matt grime]

They are *DEFINED* to be zero. Who said that that formula holds for the cases you have a problem with?

there is another way to define the binomial coefficient for non-integer alpha too, it's


a(a-1)(a-2)..(a-n+1)/n!

that is defined for all a and all integer n.

 

[steven187]

yeah I get what you mean, its just that I couldn't find any site that tells what happens in that sanario so i got really confused anyway

Thanxs Matt