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Binomial expansion with n=1/2

  1. Dec 16, 2013 #1
    Is it possible to do a binomial expansion of [itex](x+y)^{1/2}[/itex]? I tried to compute it with the factorial expression for the binomial coefficients, but the second term already has n=1/2 and k=1, which makes the calculation for the binomial coefficient (n 1) weird, I think.

    Any advice?
  2. jcsd
  3. Dec 16, 2013 #2


    Staff: Mentor

  4. Dec 16, 2013 #3
  5. Dec 16, 2013 #4


    Staff: Mentor

  6. Dec 16, 2013 #5
    You may want to look at something like

    Assuming neither x or y are zero (and both are positive), I would recommend factoring out the larger of x or y and let your task reduce to that of finding [itex](1+z)^{1/2}[/itex] with [itex]z<1[/itex].

    For example, assume [itex]y < x[/itex], then your expression would be

    [tex] f = \sqrt{x}\,(1+z)^{1/2}[/tex]

    Expand [itex](1+z)^{1/2}[/itex] using the binomial series. The expansion will be an infinite series due to the non-integer exponent.
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