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Shortcut to taylor series of f, given taylor series of g

  1. May 3, 2012 #1
    So, I have the series of [itex]g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!} [/itex]

    and I am asked to find the series of [itex]f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}}[/itex] for x [itex]\neq[/itex] 1 and f(1) = 1. The Taylor series is centered about x = 1

    I am told that there is an easy way to do this, but I don't see it.

    any help would be appreciated, thanks
  2. jcsd
  3. May 3, 2012 #2


    Staff: Mentor

    1. What do you get for e(x - 1)2 - 1?
    2. What do you get if you divide the result from #1 by (x - 1)2?

    Is this a homework problem?
  4. May 3, 2012 #3
    I see it now, thanks

    It's an old AP problem that I am doing to prepare for the test
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