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Maclaurin series homework help

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data

    the maclaurin series for f(x) is given by 1/2! - x2/4! + x4/6! - x6/8! + ... + (-1)nx2n/(2n+2)! + ...

    a) Let g'(x) = 1-x2 * f(x)
    Write the Maclaurin series for g'(x), showing the first three nonzero terms and the general term.

    b) write g'(x) in terms of a familiar function without using series. then write f(c) in terms of the same familiar function.

    c) given that g(0)=3 write g(x) in terms of a familiar function without series.


    3. The attempt at a solution
    the solution is:

    a) 1 - x2/2! - x4/4! -...+ (-1)nx2n/(2n)!

    b) g'(x) = cosx, f(x) = { (1-cosx)/x2 if x≠0
    { 1/2 if x=0

    c) g(x) = sinx+3


    i tried doing part A but could not figure out how to find the maclaurin series for g'(x)=1-x2 * f(x)
    first you have to find the g'(0), g''(0), g'''(0), etc., then continue to find the maclaurin series by multiplying the terms with xn and dividing by n!
    but i couldnt figure out how to do it. any help will be appreciated, thanks!
     
  2. jcsd
  3. Apr 16, 2014 #2

    micromass

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    Let me do a different example:

    The Taylor series of ##e^x## is ##1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...##.

    If we substitute in ##-x## instead of ##x##, then we get the Taylor series of ##e^{-x}##:

    [tex]1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ...[/tex]

    Multiply by ##x## to find the Taylor series of ##x*(e^{-x})##:

    [tex]xe^{-x} = x(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ... ) = x - x^2 + \frac{x^3}{2}- \frac{x^4}{6} + ...[/tex]

    We can also find ##(1+x)e^{-x}##:

    [tex](1+x)e^{-x} = e^{-x} + xe^{-x} = (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ...) + (x - x^2 + \frac{x^3}{2}- \frac{x^4}{6} + ...) = (1 - \frac{x^2}{2} + \frac{x^3}{3} + ...)[/tex]

    Does that help you a bit?
     
  4. Apr 16, 2014 #3

    BiGyElLoWhAt

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    just multiply g'(x) through, and you'll see you're mistake, and hopefully you'll see the general solution as well =]
    [itex]g'(x)=(1-x^2)f(x)=\frac{1}{2!}(1-x^2) - \frac{x^2}{4!}(1-x^2)... = \frac{1-x^2}{2!} +\frac{-x^2+x^4}{4!}...[/itex]
     
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