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Extending Taylor expansions

  1. Oct 29, 2012 #1
    It is well-known that the Taylor series of [itex]\sin x[/itex] about [itex]x=0[/itex] is [tex]\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
    By extension, one might presume that the Taylor series of [itex]\sin x^{3/2}[/itex] about [itex]x=0[/itex] is

    [tex]\sin x^{3/2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{\frac{3}{2}(2n+1)}[/tex]

    This is indeed the case, but why? Finding the Taylor series of [itex]\sin x^{3/2}[/itex] from the definition does not give a convergent series. In general, my question is, if

    [tex] f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n [/tex]
    why is
    [tex] f(g(x)) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}g(x)^n [/tex]
    It seems like this last expression would involve chain differentiation, but does not. Thanks all - I hope I have explained my question properly.
  2. jcsd
  3. Oct 29, 2012 #2


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    Why do you think your last expression is not in conformity with the chain rule??
  4. Oct 30, 2012 #3
    Your series is a valid absolutely convergent series representation of sin(x^(3/2)). However, it is not a power series because it has terms like x^(3/2). Therefore, it cannot be a Taylor series.

    The fact is that sin(x^3/2) is not twice differentiable at x=0, so it does not have a Taylor series representation at 0.

    In general, you can substitute g(x) into the power series representation of f as long as g(x) remains inside the interval of convergence of the power series. You will then have a series representation for f(g(x)). However, the resulting formula won't necessarily be a power series.
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