Can the Binomial Series Exist When α < n?

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    Alpha Binomial Series
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Discussion Overview

The discussion centers around the existence of the binomial series when the parameter α is less than the index n. Participants explore the implications of this condition on the binomial coefficient and the validity of the series.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions how the binomial series can exist when α < n, suggesting that the binomial coefficient should only be valid for α > n.
  • Another participant states that the binomial coefficients are identically zero for n > α, assuming α is an integer.
  • A participant expresses confusion about the factorial definition of the binomial coefficient when α < n, particularly regarding the factorial of a negative number.
  • Another participant clarifies that the binomial coefficient can be defined in a way that is valid for non-integer α, which allows for the series to be defined even when α < n.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the binomial coefficient when α < n, with some asserting it is zero while others question the implications of this definition.

Contextual Notes

There is a lack of consensus on the definitions and implications of the binomial coefficient for non-integer values of α, particularly in cases where α < n. The discussion highlights the need for clarity on these definitions.

steven187
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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
 
Last edited:
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they are identically zero for n>alpha (assuming alpha is an integer)
 
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
 
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.
 
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
 
Last edited:

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