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Taylor Series Rewriting

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data
    I understand the whole concept of Taylor Series and Maclaurin series but I don't know how to rewrite them in sigma notation.

    I'll use this generic example. Find the Maclaurin series of the function [tex]\ f(x)=e^{x}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    [tex]\ e^{x}+\frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3}\cdot\cdot\cdot[/tex]

    I know the answer is [tex]\sum^{\infty}_{n=0} \frac{x^{n}}{n!}[/tex] but I don't know how to get it from the expanded form.
     
  2. jcsd
  3. Feb 15, 2010 #2

    Mark44

    Staff: Mentor

    You're missing the first term. What you have above should be
    [tex]\ e^{x} = \frac{1}{0!}(x - 0)^0 + \frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3} + \cdot\cdot\cdot[/tex]

    When simplified, the expression on the right is
    [tex]\ e^{x} = \frac{1}{0!}x^0 + \frac{1}{1!}x+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+ \cdot\cdot\cdot[/tex]

    What will be the general term in the sum above? Do you notice anything that's common to each of the terms above?

     
  4. Feb 15, 2010 #3
    I noticed that there should be an [tex]x^{n}[/tex] in there and there should be an n!. I guess I need to use a slightly more complicated example. How about sin(x)
    The expanded form is (after the 0s are dropped out) [tex]sin(x)+\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+\cdot\cdot\cdot[/tex].

    I'll try and get an answer for this one now since I don't know the answer.

    [tex]sin(x)+\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex] I know that that can't possibly be correct but that's my attempt.
     
    Last edited: Feb 15, 2010
  5. Feb 15, 2010 #4

    Mark44

    Staff: Mentor

    You're making the same mistake you made earlier. The Maclaurin series for sin(x) should not start off with sin(x) + ... What you should be saying is that sin(x) = <the series>.

    Other than that, your summation is exactly right. Here's what you have with the correction.
    [tex]sin(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex]
     
  6. Feb 15, 2010 #5
     
  7. Feb 15, 2010 #6

    Mark44

    Staff: Mentor

    What you're doing with Taylor's series (and in this case Maclaurin series) is writing a function f(x) as a power series. For a Maclaurin series,

    [tex]f(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + ... + \frac{f^{(n)}(0)x^n}{n!} + ...[/tex]
     
  8. Feb 16, 2010 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As Mark44 told you, this should be
    [tex]sin(x)= \frac{1}{1!}x^2- \frac{1}{3!}x^3+ \frac{1}{5!}x^5- \frac{1}{7!}x^7+ \cdot\cdot\cdot[/tex]

    The first thing I would not is that the powers of x are all odd. You can write any even number as "2n" and any odd number as "2n+1", so I would try "2n+1" as as the powers and not that if n=0, 2n+1= 1, if n= 1, 2n+1= 2+ 1= 3, etc.

    I would then notice that the signs alternate and think of (-1) to some power. That will alternate + and - for even and odd powers. In some cases, you might need to experiment with [itex](-1)^n[/itex] or [itex](-1)^{n+1}[/itex] to get the right parity but here I see that the first term corresponds to n= 0 and (-1)0= 1 is the right sign: I want (-1)n.

    Finally, I see that the denominator is the factorial of the power- since I am writing the powers as "2n+1" I want to do the same in the denominators- each term is of the form
    [tex]\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
    and the sum is
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]

    I'll try and get an answer for this one now since I don't know the answer.

    [tex]sin(x)+\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex] I know that that can't possibly be correct but that's my attempt.[/QUOTE]
     
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