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Taylor series question

  1. Jan 13, 2013 #1
    The Taylor Series of f(x) = exp(-x^2) at x = 0 is 1-x^2...

    Why is this?

    The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ...

    and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0???

    Can someone please explain why it is 1-x^2???
     
  2. jcsd
  3. Jan 13, 2013 #2
    Although f'(0) = 0, yet f''(0) will NOT be zero.
     
  4. Jan 13, 2013 #3
    The formula is (x^n)/n! (with the sigma crap in front)
    0! is 1 and x^0 is 1

    Just substitute x=-t^2 and you get the series for e^-t^2
     
  5. Jan 13, 2013 #4
    Then why doesn't the second derivative term have an x^4?
     
  6. Jan 17, 2013 #5
    Here's what HomogenousCow is trying to say. Consider the expansion of ##\exp(t)## where ##t=-x^2##. The Taylor expansion of ##\exp(t)## is $$\sum_{n=0}^\infty \frac{t^n}{n!},$$ so substituting gives you $$e^{-x^2}=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}.$$

    The second derivative term does not have an ##x^4## by its definition: Each term is given by ##\frac{f^{(n)}(0)x^n}{n!}## (for MacLaurin; for Taylor about ##c##, it's ##\frac{f^{(n)}(c)(x-c)^n}{n!}##). When you plug in 2, you get $$\frac{f''(0)x^2}{2!}.$$ In this case, ##f''(0)=-2##. This cancels with the ##2!## in the denominator, so you end up with ##1-x^2## as the first two nontrivial terms of the expansion.
     
    Last edited: Jan 17, 2013
  7. Jan 18, 2013 #6

    HallsofIvy

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    It looks like you are trying to combine two different ways of finding the Taylor series of [itex]e^{-x^2}[/itex]

    1) Using the basic definition: the "[itex]x^2[/itex]" term has coefficient f''(0)/2 so that the second derivative term, by definition, involves [itex]x^2[/itex], not [itex]x^4[/itex].

    2) Replacing x in the Taylor's series for [itex]e^x[/itex] with [itex]-x^2[/itex]. In that case, the term that you get from the second derivative of [itex]e^x[/itex] has [itex]x^4[/itex] but that has nothing to do with the second derivative of [itex]e^{-x^2}[/itex]. That term comes from replacing x with [itex]-x^2[/itex] in the [itex]f'(0) x[/itex] term of the Taylor series for [itex]e^x[/itex].
     
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