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Use geometric series to write power series representation

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Complete the proof that ln (1+x) equals its Maclaurin series for -1< x ≤ 1 in the following steps.

    Use the geometric series to write down the powe series representation for 1/ (1+x) , |x| < 1




    This is the part (b) of the question where in part (a)I proved that ln (1+x) equals its Maclaurin series for 0< x ≤ 1by showing the limit of the errom is zero (hense converges).
    2. Relevant equations


    3. The attempt at a solution
    The solution is actually given, I just couldn't understand it. Shown below

    1. 1/(1-x) =∑ x^k for |x|<1
    2. 1/(1+x) =∑ (-x)^k
    3. = ∑(-1)^k x^k
    4. = 1 - x + x^2 - x^3 ........for |-x| < 1 which is for |x|<1

    To be clear, I only dont understand step one, the rest is just rearranging. Specifically I dont see how the expression on the left equals the summation on the right.

    Note - all the summations are to infinity starting with k= 0
     
  2. jcsd
  3. Mar 29, 2016 #2

    blue_leaf77

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    Do you know the formula for a convergent, infinite geometric series starting from unity as the first term and with the ratio ##x##?
     
    Last edited: Mar 29, 2016
  4. Mar 29, 2016 #3

    Ray Vickson

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    By DEFINITION,
    [tex] \sum_{k=0}^{\infty} x^k = \lim_{n \to \infty} \sum_{k=0}^n x^k [/tex]
    if that limit exists.

    You can apply high-school formulas for the finite sum, and so examine in detail what happens when ##n \to \infty##. You will see immediately why having ##|x| < 1## is important. Of course, the summation ##\sum (-1)^k x^k## is obtained from that of ##\sum y^k## just by putting ##y = -x## (in case that was what was bothering you).
     
  5. Mar 29, 2016 #4
    Ahh now that you mention it...... totally get it now. Thank you
     
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