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Math Help - Proving this using Maclaurin's.

  1. #1
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    Proving this using Maclaurin's.

    Can someone show me how you'd prove:



    using Maclaurin's theorem please?
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  2. #2
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    First, can you tell me what is the MacLaurin series for ln(x+1)?
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  3. #3
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    For any 0\not = x>-1 we can find y between 0 and x so that:
    \log (1+x) - x = \frac{f''(y)}{2!}x^2 < 0
    Since the second derivative is negative.

    Therefore, \log (1+x) < x for -1<x\not = 0.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    For any 0\not = x>-1 we can find y between 0 and x so that:
    \log (1+x) - x = \frac{f''(y)}{2!}x^2 < 0
    Since the second derivative is negative.

    Therefore, \log (1+x) < x for -1<x\not = 0.
    I think it's important to say that

    f(x) = \ln (x+1)

    Obvious to most but not all
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  5. #5
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    Another proof:

    For each x>-1 it's \ln (x+1)=\int_{1}^{x+1}{\frac{dy}{y}}. Hence \frac{1}{x+1}\le \frac{1}{y}\le 1\implies \frac{x}{x+1}\le \ln (x+1)\le x.\quad\blacksquare
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