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Math Help - How to prove that the natural logarithm function is real analytic on (0, infinity)?

  1. #1
    CSM
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    Lightbulb [SOLVED]Proving that ln(x) is real analytic on (0, infinity)

    I already know that it is real analytic in 1, with radius of convergence 1 and power series expansion:


    ln(x)=\Sigma_{n=1}^\infty \frac{(-1)^{n+1}}{n} (x-1)^n

    for x \in (0,2)

    Now what?

    I think that I can show that ln is real analytic for all points of (0,2). But I'd like to prove it for (0, + \infty).
    Some thoughts?
    Last edited by CSM; November 15th 2010 at 03:04 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting z=x + i\ y is...

    \displaystyle f(z)= \ln z = \ln \sqrt{x^{2} + y^{2}} + i\ \tan^{-1} \frac{y}{x} = u(x,y) + i\ v(x,y) (1)

    Now necessary condition for f(*) analytic is done by the Cauchy-Riemann relations...

    \displaystyle \frac{\partial{u}}{\partial{x}}=  \frac{\partial{v}}{\partial{y}}<br />

    \displaystyle \frac{\partial{u}}{\partial{y}}= - \frac{\partial{v}}{\partial{x}} (2)

    What are the values of z [if any...] that satisfy (2)?...

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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CSM View Post
    I already know that it is real analytic in 1, with radius of convergence 1 and power series expansion:


    ln(x)=\Sigma_{n=1}^\infty \frac{(-1)^{n+1}}{n} (x-1)^n

    for x \in (0,2)

    Now what?

    I think that I can show that ln is real analytic for all points of (0,2). But I'd like to prove it for (0, + \infty).
    Some thoughts?
    In fact, it's real analytic on (0,2]. Which criteria do you know? The clearest possible one is that \ln can be extended holomorphically to an open set in \mathbb{C} containing \mathbb{R}
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  4. #4
    CSM
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    Thanks for your replies. But I cannot use any knowledge of complex numbers. It's only real analysis, no complex numbers.
    I use the definition from Tao's Analysis II:
    Let E be a subset of R and let f:E--> R be a function. If a is an interior point of E, we say that f is real analytic at a if there exists an open interval (a-r,a+r) in E for some r>0 such that there exists a power series \Sigma_{n=o}^\infty c_n(x-a)^n centered at a which has a radius of convergence greater than or equal to r, and which converges to f on (a-r,a+r). If E is an open set, and f is real analytic at every point a of E, we say that f is real analytic on E.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    I think you should try to prove the following:

    The function f(x)=ln(x) is an infinitely differentiable function such that the Taylor series at any point x0=1 in its domain.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I think you should try to prove the following:

    The function f(x)=ln(x) is an infinitely differentiable function such that the Taylor series at any point x0=1 in its domain.
    Infinite differentiability does not imply real analyticity. Take f(x)=\begin{cases}e^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne 0\\ 0 & \mbox{if}\quad x=0\end{cases}
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  7. #7
    CSM
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    Any ideas?
    How do you in general proof a function to be analytic?
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  8. #8
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    f\in C^{\infty}(U,\mathbb{R}) (where U\subset \mathbb{R} open) is analytic in U if and only if for all x\in U there exist x\in J\subset U an interval, and constants R,C>0 with the property that |f^{(k)}(y)|\leq C \frac{k!}{R^k} for all y\in J. The proof is not difficult, the key is using an appropiate form of the residue in Taylor's formula.

    From this, and noting that (\ln(x))^{(k)}=\frac{(-1)^{k+1}(k-1)!}{x^k} the result follows.
    Last edited by Jose27; November 14th 2010 at 06:16 PM. Reason: wrong letter
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    In fact, it's real analytic on (0,2]. Which criteria do you know? The clearest possible one is that \ln can be extended holomorphically to an open set in \mathbb{C} containing \mathbb{R}
    That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that! The largest possible domain on which a fixed branch of a logarithm can live comfortably is the plane cut along some curve joining 0 and \infty. But you're right in the sense that the existence of a holomorphic function on \mathbb{C}-\mathbb{R}_{\leq 0} which coincides with the usual (real) logarithm on the real line immediately implies that the usual logarithm is analytic there (and everywhere else on \mathbb{C}-\mathbb{R}_{\leq 0}).
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that! The largest possible domain on which a fixed branch of a logarithm can live comfortably is the plane cut along some curve joining 0 and \infty. But you're right in the sense that the existence of a holomorphic function on \mathbb{C}-\mathbb{R}_{\leq 0} which coincides with the usual (real) logarithm on the real line immediately implies that the usual logarithm is analytic there (and everywhere else on \mathbb{C}-\mathbb{R}_{\leq 0}).
    Ah, of course. It must have been late.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Bruno J. View Post
    That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that!...
    That is not fully exact!... the function \ln z is a multivalued function that in z=0 has a singularity classified as branch point. Effectively such type of singularity is, in a certain sense, 'the worst possible' because for an f(z) with an essential singularity in z=0 [like for example f(z)= e^{\frac{1}{z}}...] it is possible to obtain the Laurent expansion around z=0, but for f(z)= \ln z that is impossible...

    Kind regards

    \chi \sigma
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  12. #12
    MHF Contributor chisigma's Avatar
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    For more information about the singularity of \ln z in z=0 see...

    Logarithmic Singularity -- from Wolfram MathWorld

    Kind regards

    \chi \sigma
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