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Taylor expansion - imaginary coefficients?

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the first two non-zero terms in the Taylor expansion of [itex]\frac{x}{\sqrt{x^2-a^2}}[/itex] where a is a real constant


    2. Relevant equations
    [tex]
    f(x)=f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{f^{\prime\prime}(x_0)}{2!}(x-x_0)^2+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n
    [/tex]


    3. The attempt at a solution
    If a=0 then f(x)=1 (in the case a=0,x=0, use l'Hopital's rule to find the limit of the ratio as [itex]x[/itex] approaches 0 - is this correct?)

    then [tex]
    f(0)=0

    f^{\prime}(0)=\frac{-ai}{a}

    f^{\prime\prime}(0)=\frac{-i}{2a^{13}}
    [/tex]

    Is it OK that there are imaginary terms here? I guess [itex]f(x)[/itex] is imaginary if [itex]a^2>x^2[/itex] anyway?
     
  2. jcsd
  3. Apr 5, 2012 #2
    If you really need to expand it at x=0, then your function is imaginary, so of course you expect to get imaginary coefficients. Usually though in expressions like this, x is limited to values larger than a.

    I'm not sure if your coefficients are correct though. Maybe it's easier to expand [itex] g(x)=(x^2-a^2)^{-1/2} [/itex]
     
  4. Apr 7, 2012 #3
    Thanks! I hadn't thought of expanding [itex]\frac{1}{\sqrt{x^2-a^2}}[/itex] and then multiplying my [itex]x[/itex] - that's really cool - thanks!
     
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