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Math Help - Help needed. Writing ln(1 + x) as a convergent series

  1. #1
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    Help needed. Writing ln(1 + x) as a convergent series

    I have a problem in expressing ln (1 + x) as a convergent series. Can people show me the steps of doing this? thanks in advance!!!
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  2. #2
    Member HappyJoe's Avatar
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    You could expand ln(1+x) in its Taylor series around x=0.

    For this, you need all the derivative of ln(1+x) evaluated at x=0.

    Let f(x)=ln(1+x). (note first that f(0) = 0).

    Then
    f'(x) = 1/(1+x).
    f''(x) = -1/(1+x)^2.
    f'''(x) = 2/(1+x)^3.
    f''''(x) = -6/(1+x)^4
    and so on.

    In general, we have that the n'th derivative of f is (-1)^{n+1}\frac{(n-1)!}{(1+x)^n} (which you may prove by induction). Then

    f^{(n)}(0) = (-1)^{n+1}(n-1)!,

    and so the Taylor series is:

    \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n,

    which converges for |x|\leq 1 (except for x=-1).
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  3. #3
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    Thanks! Your response is extremely lucid and helpful! It does not only answer the question perfectly, but it also gives me some very good intuition to understand Taylor series expansion! I have become more confident in real analysis now!

    Thank you very much
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  4. #4
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    \displaystyle \frac{1}{1+x} = \frac{1}{1-(-x)} = 1 - x + x^2 - x^3 + ...<br />

    integrate ...

    \displaystyle \ln(1+x) = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...

    x = 0 \implies C = 0

    \displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}
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