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Homework Help: Series involving trig functions

  1. Jun 8, 2007 #1
    1. The problem statement, all variables and given/known data
    http://img87.imageshack.us/img87/6784/dsc07215xt5.jpg [Broken]
    http://img72.imageshack.us/img72/9435/dsc07216aa0.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    Answers for part I are circled, here's parts II-IV.
    http://img485.imageshack.us/img485/8642/dsc07217uq2.jpg [Broken]
    http://img485.imageshack.us/img485/6999/dsc07218qf2.jpg [Broken]

    Any help would be appreciated. This is a very important homework, as you can tell by the title.

    If the pics are too small, download them and zoom in, or use the FF extension ImageZoom.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 8, 2007 #2

    Gib Z

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    2) Derive a taylor series centered at x= pi/3 and just keep summing terms. You will know you have 4 places accuracy when after summing a certain number of terms for 4 places are still the same, or also when the terms become smaller than 0.0001.

    3) You have the right working and series, but it seems you subbed in x=2 instead of x= 0.2.
  4. Jun 9, 2007 #3
    2. I did. I made an Excel thing to find the sums, and it converged to -0.617209760262333. The actual value of cos(63º) is 0.4539905. I can't figure out what's wrong...
    Edit: I was doing it in degrees. I got it now.

    3. Ah, thanks.

    Can anyone help me with Part II B and Part IV?
    Last edited: Jun 9, 2007
  5. Jun 9, 2007 #4

    Gib Z

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    IV) Lets do a and b first.

    A) Just use the definition of a Maclaurin series, not hard to sub in.

    B) Using that series for f(x), change the variable to t, integrate term by term.
  6. Jun 10, 2007 #5
    I did that (see above pics), and it looks right. But part C doesn't work. If I set -2ak=1, and 2a²k/3=1, a comes out to be -1/3 and k=3/2. That doesnt work for the third term...
  7. Jun 10, 2007 #6

    Gib Z

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    Try this approach:

    [itex]f(x)=6e^{-x/3}, f'(x)=-2e^{-x/3}[/itex]

    [tex]kf'(ax) = -2k \cdot e^{-ax/3}[/tex]

    Now for the LHS, you should recognise thats the taylor series for e^x.

    Now you just have to solve [itex]e^x = -2k \cdot e^{-ax/3}[/tex]
  8. Jun 10, 2007 #7
    Ok, with that, I got k=-1/2, and a=-3. Right...?
  9. Jun 10, 2007 #8

    Gib Z

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    Aww come on mate you don't need to be spoon fed like this :P Substituting those into the RHS of what I left you to solve, you can see that it is correct = ] but next time i want something more like:

    " I got k=-1/2 and a=-3 which works out, yay!!"
  10. Jun 10, 2007 #9
    Ok, thanks for the help everyone. Well, actually, just Gib Z, but whatever :).
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