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Binomial series vs Binomial theorem, scratching my head for three days on this

  1. Aug 15, 2010 #1
    In my book, it says that the Binomial Series is

    [tex]\sum_{n=0}^{\infty }\binom{n}{r} x^n[/tex]

    Where [tex]\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{n!}[/tex] for [tex]r\geq1[/tex] and [tex]\binom{n}{0} = 1[/tex]

    Now here is where it got to be, I know that the [tex]\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{n!}[/tex] were derived through the power series, but how does it explained the original statement that [tex]\sum_{n=0}^{\infty }\binom{n}{r} x^n[/tex]?

    Since [tex]\binom{n}{r}[/tex] does not actually change. I mean it is still [tex]\frac{n!}{(n-r)!r!}[/tex] and so how can you put negative n into n!?
  2. jcsd
  3. Aug 15, 2010 #2


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    One way to think of it is to rewrite the binomial coefficient in terms of Gamma functions:

    [tex]\left(\begin{array}{c}n \\ r \end{array}\right) = \frac{\Gamma(n+1)}{\Gamma(n-r+1)\Gamma(r+1)}[/tex]

    The Gamma function [itex]\Gamma(z)[/itex] has poles at [itex]z = 0, -1, -2, \dots[/itex], or equivalently, [itex]1/\Gamma(z)[/itex] has zeroes at these points, so when [itex]n-r+1 \leq 0[/itex] the denominator of the binomial coefficient becomes infinite and so the coefficient vanishes. Hence the series terminates and you get a finite expansion for integer powers.

    It is only at the negative integers that the gamma function has poles, so when n - r + 1 is not an integer, the Gamma function doesn't vanish, and you get an infinite series.

    (Note: I don't think this is a rigorous argument, but I think it can be made rigorous by defining the properties of [itex]1/\Gamma(z)[/itex] precisely such that it has zeros where [itex]\Gamma(z)[/itex] has poles. I'll leave it to any more mathematician-minded posters to fill in the gaps in rigor).
  4. Aug 15, 2010 #3

    Gib Z

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    I think the main thing the OP is simply missing is that for Binomial series, n is not necessarily an integer, which is why they write it out in the longer form, as the one with factorials only works with integers.
  5. Aug 15, 2010 #4
    Did they flip the formula of combinatorics?
  6. Aug 15, 2010 #5
    This is incorrect. It should be

    [tex]\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r!}[/tex]
  7. Aug 15, 2010 #6
    Oh right sorry lol, my tex was wrong.
  8. Aug 17, 2010 #7
    I think your original binomial series is wrong too.

    The correct one should be summing about r, not n. i.e.

    [tex] (1+x)^n = \sum_{r=0}^{\infty} \binom{n}{r} x^r [/tex]

    This formula is valid, even for complex n. You are right that this formula can be derived using Taylor series.
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