1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taylor series question
  1. Taylor series tips (Replies: 4)

  2. Taylor series (Replies: 4)

Loading...