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Binomial Expansion For Rational Index . Please help

  1. Apr 18, 2006 #1

    I wanted to know what is the expansion of (1+x)^n when n is a rational number and |x|<1 ...
    Please let me know as soon as possible..

    Thanks for your help
  2. jcsd
  3. Apr 18, 2006 #2


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    [tex](1+x)^{n} = 1 + nx + \frac{n(n-1)}{1\cdot 2}x^{2} + ... + \frac{n(n-1)...(n-r+1)}{1\cdot 2 ... r}x^{r}[/tex]

    Where [itex]|x|<1[/itex] and n is any real number. This can be derived from the general binomial expansion of [itex](a+b)^n[/itex].

  4. Apr 18, 2006 #3


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    I assume you know that, for n a positive integer
    [tex](1+ x)^n= 1+ nx+ ... + _nC_i x^i+ ...[/tex]
    [tex]_nC_i= \frac{n!}{i!(n-i)!}= \frac{n(n-1)...(n-i+1)}{i!}[/tex]
    For n a rational number, basically the same formula is true. Only now [itex]_nC_i[/itex] is never 0 so we get an infinite sum.

    For example, if n= 1/2 then [itex]_{\frac{1}{2}}C_1= \frac{1}{2}[/itex], [itex]_{\frac{1}{2}}C_2= \frac{\frac{1}{2}(\frac{1}{2}-1)}{2}= -\frac{1}{8}[/itex], [itex]_{\frac{1}{2}}C_3= \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{6}= \frac{1}{16}[/itex], etc. so that
    [tex](1+ x)^{\frac{1}{2}}= 1+ \frac{1}{2}x-\frac{1}{8}x^2+ \frac{1}{16}x^4-...[/tex]
    exactly as you would get from the Taylor's series.
  5. Apr 19, 2006 #4
    Thanks to both .
    The post by HallsofIvy was particularly useful .
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