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I Heavyside step function chain rule

  1. Mar 21, 2017 #1
    Hi,

    I have a probably very stupid question:

    Suppose that there is an expression of the form $$\frac{d}{da}ln(f(ax))$$ with domain in the positive reals and real parameter a. Now subtract a fraction ##\alpha## of f(ax) in an interval within the interval ##[ x_1, x_2 ]##, i.e.

    $$f(ax) \rightarrow f(ax) \left[\theta(a(x-x_1)) \theta(a(x_2-x))+ (1-\alpha) \theta(a(x_1-x)) \theta(a(x-x_2)) \right]$$

    Now in the case that ##a>0## one can use ##\theta(ax) = \theta(a)\theta(x)## and after doing a case analysis for ##x \in [x_1,x_2]## and ##x\not\in [x_1,x_2]## above expression becomes

    $$\frac{d}{da} ln(f(ax)) \rightarrow \frac{d}{da}ln(f(ax)) + \frac{d}{da} ln(\theta(a)) + \frac{d}{da} ln(\theta(a)) \theta(x-x_1) \theta(x_2-x) $$

    I have two questions:
    i) Are the steps above legitimate, as one is dealing with functionals?
    It does not seem right to me that the result does not depend on ##\alpha##...
    ii) How can one evaluate ##\frac{d}{da} ln(\theta(a))## ? Is the typical chain rule applicable there (which would yield 0 for a > 0)?

    Thank you and best greetings :)
     
    Last edited: Mar 21, 2017
  2. jcsd
  3. Mar 21, 2017 #2

    andrewkirk

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    The derivative does not depend on ##\alpha## because the expression does not depend on it.

    The expression written above gives ##\log f(ax)## for ##x\in[x_1,x_2]## and 0 elsewhere.

    That's because the product of Heaviside functions in the second term is always zero, since it is the indicator function of the set
    $$\{x\ :\ x<x_1\wedge x>x_2\}$$
    which is the empty set, since ##x_1\leq x_2##.

    Have another go at writing the coefficient with Heaviside functions so that it gives ##(1-\alpha)## for ##x\in[x_1,x_2]## and 1 elsewhere.
     
  4. Mar 22, 2017 #3
    Thanks for your reply!

    Yep, I made an editing error...Of course it should read:

    $$f(ax) \rightarrow f(ax) \left{(1-\alpha) [\theta(a(x-x_1)) \theta(a(x_2-x))]+ \theta(a(x_1-x)) + \theta(a(x-x_2)) \right}$$

    Then the second equation should also make sense.

    But I am still not sure if one could proceed in such a fashion. Shouldn't it be equivalent to just do a bunch of case analyses (dropping all the Heaviside functions) or to actually take deriviatives of the Heaviside function applying the usual rules?
     
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