1. Not finding help here? Sign up for a free 30min 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!

Boas: Mathematical Methods for Phys Sci Pr.1.13.25

  1. Feb 22, 2014 #1
    Mclaurin Series with Division by Zero?

    Boas: Mathematical Methods for Phys Sci Pr.1.13.25

    1. The problem statement, all variables and given/known data

    Using the methods of this section:
    (a) Find the first few terms of the Maclaurin series for each of the following functions.
    (b) Find the general term and write the series in summation form.
    (c) Check your results in (a) by computer.
    (d) Use a computer to plot the function and several approximating partial sums of the
    series.

    [tex]f(x)=\frac { 2x }{ { e }^{ 2x }-1 } [/tex]


    2. Relevant equations

    [tex]f(x)=\sum_{n=0}^{\infty}f^{n}(0)x^{n}/n![/tex]


    3. The attempt at a solution

    Since f(0) is division by zero, how do you find a Maclaurin series for it?

    Thanks,
    Chris Maness
     
    Last edited: Feb 22, 2014
  2. jcsd
  3. Feb 22, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the function, as written, is not defined. But it is a "removable" discontinuity. I suspect they intend you to use the limit there. You can simplify by letting u= 2x. Then the limit is [itex]\lim_{u\to 0}\frac{u}{e^u- 1}[/itex]. And now you can take that limit in a number of ways:
    1) L'Hopitals rule: differentiating both numerator and denominator gives [itex]\frac{1}{e^u}[/itex] which has limit 1 as u goes to 0.

    2) Since you are dealing with MacLaurin series, the MacLaurin series of [itex]e^u[/itex] is [itex]1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot[/itex]. So [itex]e^u- 1= u+ u^2/2+u^3/6+ \cdot\cdot\cdot[/itex] and [itex]\frac{u}{e^u- 1}= \frac{u}{u+ u^2/2+ u^3/6+ \cdot\cdot\cdot}[/itex][itex]= \frac{1}{1+ u/2+ u^3/6+ \cdot\cdot\cdot}[/itex] which again has limit 1 as u goes to 0.

    (It would have been better if the problem had said "[itex]f(x)= \frac{2x}{e^{2x}- 1}[/itex] if [itex]x \ne 0[/itex], [itex]f(0)= 1[/itex]".)
     
    Last edited: Feb 22, 2014
  4. Feb 22, 2014 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Find the limit, if it exist, as x goes to 0.
     
  5. Feb 22, 2014 #4
    Yes, they don't mention using the limit if f(0)=D.N.E, and I should have use L'Hosp.

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Boas: Mathematical Methods for Phys Sci Pr.1.13.25
Loading...