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Help with Taylor, ln(1-X), |x|<1

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  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    ln(1-X), |x|<1

    2. Relevant equations
    Could someone verify if it was developed correctly?

    3. The attempt at a solution
    [tex]
    ln(1-x) = \sum_{n=0}^\infty \left (a_nx^n\right )
    [/tex]
    [tex]
    1+a+a^2+a^3+a^4+a^5+a^6... = 1/(1-a)
    [/tex]
    [tex]
    a=x
    [/tex]
    [tex]
    1+x+x^2+x^3+x^4+x^5+x^6... = 1/(1-x)
    [/tex]
    [tex]
    ∫1+x+x^2+x^3+x^4+x^5+x^6...dx = ∫1/(1-x) dx
    [/tex]
    [tex]
    x+(x^2)/2+(x^3)/3+(x^4)/4+(x^5)/5+(x^6)/6... = - ln(x-1)
    [/tex]
    [tex]
    ln(1-x) = - \sum_{n=1}^\infty \left (x^n / n \right )
    [/tex]

    Thanks for your time
     
    Last edited: Mar 21, 2017
  2. jcsd
  3. Mar 21, 2017 #2
    Hi Keven:

    Two issues. One is about your notation of the integrals. There should be a dx.
    More important, you have a problem at the quoted step. Think again about
    ∫1/(1-x)dx = ln(x-1).​
    What is the value of ln(x-1) when |x|<1?

    Hope this helps.

    Regards,
    Buzz
     
  4. Mar 21, 2017 #3
    @Buzz Bloom I made the corrections, thank you
     
  5. Mar 22, 2017 #4
    Hi kevin:

    Thank you for your thanks.

    You still have a problem with -ln(x-1). Think about the value for |x|<1.

    Regards,
    Buzz
     
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