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Help with power series representation

  1. Dec 4, 2006 #1
    1. The problem statement, all variables and given/known data

    find a power series representation for the function and determine the radius of convergence.

    f(t)= ln(2-t)


    2. Relevant equations



    3. The attempt at a solution

    i first took the derivative of ln(2-t) which is 1/(t-2)

    then i tried to write the integral 1/(t-2) in the form of a power series i got the sum for n=0 to infinity of (t/2)^n i don't know if this is write if someone can please help me
     
  2. jcsd
  3. Dec 4, 2006 #2
    If [itex] f(t) = \ln(2-t) [/tex] then write this as follows:


    [itex] -\ln(2-t) = \int \frac{1}{2-t} \; dt [/tex]

    How would you transform [itex] \frac{1}{2-t} [/tex] into a more recognizable form: i.e. [itex] \frac{1}{1-t} [/itex]?
     
  4. Dec 4, 2006 #3
    i would factor out a 1/2 so i would have (1/2)*1/(1-(1/2t))
     
  5. Dec 4, 2006 #4
    ok then what would you do?
     
  6. Dec 4, 2006 #5
    i would write that as a power series:
    (1/2)*the sum for n=0 to infinity of (1/2t)^n
     
  7. Dec 4, 2006 #6
    correct you have:

    [tex] \frac{1}{2}\sum_{n=0}^{\infty} (\frac{t}{2})^{n} [/tex]
     
  8. Dec 4, 2006 #7
    so is that the answer??
     
  9. Dec 4, 2006 #8
    yes that is the answer.

    or write it like this:

    [tex] \sum_{n=0}^{\infty} \left(\frac{t^{n}}{2^{n+1}}\right) [/tex]
     
  10. Dec 4, 2006 #9
    ok thank you very much
     
  11. Mar 31, 2009 #10

    I'm confused... What happens to the integral?
     
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