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Homework Help: Function approximation near a given point

  1. Sep 5, 2007 #1
    I've came up to a problem, where I would like to prove that a differentiable function f(x) can be approximated by

    [tex]f(x) = f(x_0) \left(\frac{x}{x_0}\right)^{\alpha}[/tex]


    [tex]\alpha = \frac{d \ln f(x)}{d \ln x} \Big |_{x=x_0}[/tex]

    But I'm not sure this is true. The problem and solution can be found at:
    http://crydee.sai.msu.su/~konon/Book/Book.html [Broken]
    see Problem 8.2, the book is in Russian. If you can't find it, the problem and solution are respectively at pages
    http://crydee.sai.msu.su/~konon/Book/ch3L/node12.html [Broken]
    http://crydee.sai.msu.su/~konon/Book/ch4L/node9.html [Broken]

    The idea depicted in that book is to use a Taylor expansion for a function [tex]\ln f(x)[/tex], but use a variable [tex]\ln x[/tex] instead of [tex]x[/tex], as if the function realy was [tex]\ln [f(\ln x)][/tex], but that's not shown. Anyway, say I Taylor expand this function:

    [tex]\ln [f(\ln x)] = \ln [f(\ln x_0)] + \frac{d \ln [f(\ln x)]}{d \ln x} \Big |_{x=x_0} (\ln x-\ln x_0) + \cdots = \ln[f(\ln x_0)] + \ln \left(\frac{x}{x_0}\right)^{\alpha} + \cdots[/tex]

    OK, now I put those two terms together and obtain

    [tex]f(\ln x) = f(\ln x_0) \left(\frac{x}{x_0}\right)^{\alpha}[/tex]

    Looks like it, but if we let [tex]\ln x = z[/tex], then

    [tex]f(z) = f(z_0) e^{\alpha(z-z_0)} \approx f(z_0) [1 + \alpha(z-z_0)][/tex]

    after approximating exponent for small [tex]\Delta z[/tex] and this is just Taylor series, i.e. it gives nothing new and no expected new awesome cool approximation. Am I missing something, or the textbook is rubbish?

    Ironically, the author complains that no textbook gives this approximation, but it is used everywhere in astrophysics.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 5, 2007 #2


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    There's not really anything to prove. The real question is, what is a good approximation to f(x) near x0? This depends on f. It might be that a power-law approximation, f = f0(x/x0)a, is better than a linear approximation, f = f0 + a(x-x0), or it might not. Or there might be some other 1-paramater approximation that's better, like f = f0 exp(a(x-x0)) or f = f0 + a ln(x/x0).

    In astrophysics, it often happens that you get power-law relations between two quantities over some range, so power-law functions are often good approximations. But there's nothing mathematically fundamental about this.
    Last edited: Sep 5, 2007
  4. Sep 7, 2007 #3
    Your advice was of help. What I learned is that you may choose any functional form of the approximation you like, and then choose the parameters to best fit the real thing. That way I was able to solve the problem. And it's 8.3, I made a mistake :))
  5. Sep 7, 2007 #4


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    Yes, exactly. Of course, some may be better than others (over some range), and you could do something like see which form (of those you tried) minimizes the average squared variation to pick the best one.
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