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Math Help - [SOLVED] Logarithm real analytic

  1. #1
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    [SOLVED] Logarithm real analytic

    I want to proof that the natural logarithm is real analytic.
    For  x \in (0,2) I have ln(x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n

    Is it satisfied when I say when x \in (2,\infty),
    I have ln(a)=-ln(\frac{1}{a})
    Because \frac{1}{a} \in (0,2) it is also real analytic when x \in (2,\infty) ?
    Thanks
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  2. #2
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    Definition Real Analytic;

    F is real analytic at a if there excist an open interval (a-r, a+r) in E for some r>0 such that there exist a power series \sum_{n=0}^{\infty} c_n (x-a)^n centered at a which has radius of convergence greater than or equal to r, and which converges to f on (a-r, a+r).
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  3. #3
    MHF Contributor chisigma's Avatar
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    The 'natural logarithm' function is analytic in x=1 so that the Taylor expansion around this point exists and it is...

    \ln x = -\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n} , |1-x|<1 (1)

    If You want to 'expand' the range of definition of the function, there are two possibilities...

    a) because is \ln \frac{1}{x}= - \ln x, writing \frac{1}{x} instead to x in the second term of (1) and changing sign You obtain...

    \ln x = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n} , |1-\frac{1}{x}|<1 (2)

    b) because is \ln \frac{x}{a} = \ln x - \ln a , writing \frac{x}{a} instead of x in the second term of (1) You obtain…

    \ln x = \ln a - \sum_{n=1}^{\infty} \frac{(1-\frac{x}{a})^{n}}{n} , |1-\frac{x}{a}|<1 (3)

    Kind regards

    \chi \sigma


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  4. #4
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    Quote Originally Posted by chisigma View Post
    The 'natural logarithm' function is analytic in x=1 so that the Taylor expansion around this point exists and it is...

    \ln x = -\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n} , |1-x|<1 (1)

    If You want to 'expand' the range of definition of the function, there are two possibilities...

    a) because is \ln \frac{1}{x}= - \ln x, writing \frac{1}{x} instead to x in the second term of (1) and changing sign You obtain...

    \ln x = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n}, |1-\frac{1}{x}|<1 (2)

    b) because is \ln \frac{x}{a} = \ln x - \ln a , writing \frac{x}{a} instead of x in the second term of (1) You obtain…

    \ln x = \ln a - \sum_{n=1}^{\infty} \frac{(1-\frac{x}{a})^{n}}{n} , |1-\frac{x}{a}|<1 (3)

    Kind regards

    \chi \sigma

    Thanks a lot Chisigma.
    The Taylor expansion is \ln x = -\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n
    so if i'm right the answer should be something like this;
    \ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(\frac{1}{x}-1)^n
    Or am I missing something?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by bram kierkels View Post
    ... the Taylor expansion is \ln x = -\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n
    so if i'm right the answer should be something like this;
    \ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(\frac{1}{x}-1)^n
    All right!... because is...

    (x-1)^{n}= (-1)^{n} \cdot (1-x)^{n} (1)

    ... we have...

    \ln x = - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\cdot (x-1)^{n} = - \sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n} = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n} (2)

    Kind regards

    \chi \sigma
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