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Homework Help: What are the first few coefficients for this series?

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    The function [itex]f(x)=ln(2−x)[/itex] is represented as a power series in the form:

    [itex]\sum C_{n}x^{n},n,0,inf[/itex]

    Find the first five coefficients in the power series.

    Basically, this is a problem from my online homework. I did a lot of work to get to my answers, and I feel sort of tired to type up all of it so I'll just show my final result (hopefully you guys don't mind). And if need be to know, I first differentiated f(x), then wrote down the power series for it, then integrated the power series to obtain the power series for the original f(x).

    Afterwards, I expanded this power series for the first five terms and wrote down the coefficients:

    [itex] (-1/2), (-1/8), (-1/24), (-1/64), (-1/160)[/itex]


    I then inputted these numbers into the boxes specified on the web page and I received a message saying that they were wrong. I've checked my work twice, my TI89, and even Wolfram Alpha; I get the same answers from these sources, so can someone tell me what is wrong? Also if someone does have a solution, you do not have to tell me what the actual answers are, just notify me with what I did wrong.
     
    Last edited: Oct 30, 2011
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  3. Oct 30, 2011 #2

    vela

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    What about the constant term?
     
  4. Oct 30, 2011 #3
    There was no constant term; after integrating the whole power series, I obtained x^(n+1)/(n+1) times a number. If n = 0 as the first term, then you would get x^(0+1)/(0+1), which is equal to just x.

    Edit: Also if you are interested to know, this was the power series I derived:

    [itex]\sum\frac{-1}{2}\frac{1}{2^n}\frac{x^{n+1}}{n+1},n,0,∞[/itex]
     
    Last edited: Oct 30, 2011
  5. Oct 30, 2011 #4

    Ray Vickson

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    There must be a constant term: what is f(0)? Is it zero? By the way, is your notation supposed to be [itex] \sum_{n=0}^\infty C_n x^n[/itex]?

    RGV
     
  6. Oct 30, 2011 #5
    I don't think that's the case because if the derivative of the function is [itex]f ' (x)=\frac{-1}{2-x}[/itex], then the power series representation would be:

    [itex]\sum_{n=0}^\infty (\frac{-1}{2})(\frac{x}{2})^{n}[/itex]

    If I wrote out the first five terms of this series, I would get:

    [itex](\frac{-1}{2})-(\frac{1}{4})x-(\frac{1}{8})x^2-(\frac{1}{16})x^3-(\frac{1}{32})x^4[/itex]

    Next I would have to integrate these terms to get a partial representation for [itex]f(x) = ln(2-x)[/itex]:

    [itex]\int (\frac{-1}{2})-(\frac{1}{4})x-(\frac{1}{8})x^2-(\frac{1}{16})x^3-(\frac{1}{32})x^4 dx[/itex]
    [itex]= (\frac{-1}{2})x-(\frac{1}{8})x^2-(\frac{1}{24})x^3-(\frac{1}{64})x^4-(\frac{1}{160})x^5[/itex]

    So as seen from above, the coefficients would be as I specified in my first post?

    EDIT: I tried inputting 0, -1/2, -1/8, -1/24, -1/64 and I got everything except for 0 right; so now I know there must be something before -1/2? But if there is something before -1/2, how would I find it?

    EDIT2: I tested ln(2) and found that to be the replacement for 0, but can someone explain to me how the first term is ln(2)?

    EDIT3: Never mind, I found out why it was ln(2); the C that is resulted from integration had to have been solved, I totally neglected that so thank you guys for pointing that out.
     
    Last edited: Oct 30, 2011
  7. Oct 30, 2011 #6

    HallsofIvy

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    When you integrated the series term by term, did you include the constant of integration?
     
  8. Oct 30, 2011 #7
    Okay so here's another similar problem:
    math1.jpg

    I first differentiated the function [itex]f(x)=ln(1-x^2)[/itex] and wrote down the power series for it:

    [itex]\sum\stackrel{∞}{n=0}(-2x)(x^2)^n[/itex]

    Then I integrated the power series to get a series for the original function:

    [itex]\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}[/itex]

    There was suppose to be a constant of integration but I found it to be zero, so after expanding the series for five terms, I found the coefficients to be:

    [itex]0, -1, -1/2, -1/3, -1/4[/itex]

    After inputting these numbers, as seen by the above picture, the web page said they were wrong; can someone tell me why?
     
  9. Oct 30, 2011 #8

    Dick

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    I think the homework is expecting c1 to be the coefficient of x in the series. That's zero. The first nonzero term, -1, is the coefficient of x^2 in the series. That should be c2.
     
  10. Oct 30, 2011 #9
    I tried what you said and it turns out c1 is indeed zero, meaning c0 is zero, c1 is zero, and c2 is -1. However c3 and c4 are still wrong; I used c3 = -1/2 and c4 = -1/3, and the web page said they were wrong?
     
  11. Oct 30, 2011 #10

    Dick

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    You aren't thinking about the powers of x in your series. It's -x^2-x^4/2-x^6/3-... isn't it. c3 should be the coefficient of the power of x^3. Etc.
     
  12. Oct 30, 2011 #11
    Oh well that works, c3 is actually zero, and c4 is -1/2, but what I don't get is why these numbers don't correspond to the exponent of x. For example, x^3 corresponds to the coefficient of -1/2, but instead x^3 corresponds to the coefficient of zero?
     
  13. Oct 30, 2011 #12

    Dick

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    The attachment says the series is c_n*x^n. So c1 is the coefficient of x, c2 is the coefficient of x^2, etc etc. The odd powers are missing from your series, so all of the odd coefficients should be 0. Right?
     
    Last edited: Oct 30, 2011
  14. Oct 30, 2011 #13
    Yes that's true, but the terms that are suppose to be missing are the even terms; expanding the series produces exponents of odd numbers, that's why I'm confused as to why the even numbers are showing and the odd numbers are missing.
    [itex]\sum\stackrel{∞}{n=0}(-2)\frac{x^{2n+2}}{2n+2}[/itex]
     
  15. Oct 30, 2011 #14

    Dick

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    2n+2 is even, isn't it? That's the exponent. What do you mean 'exponents of odd numbers'?
     
  16. Oct 30, 2011 #15
    Whoa my bad, my scratch paper was conglomerated with numerous revisions of different series, making me look at the wrong one. :uhh:
     
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