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Math Help - rigor versus "formally"

  1. #1
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    rigor versus "formally"

    My text makes the statement:

    "Note that expression (2) can be obtained formally by differentiating the integral in equation (1) with respect to z".

    Equation (1) is:
    f(z) = \frac 1{2\pi i} \int _C \frac{f(s)ds}{s-z}

    Equation (2) is:
    f'(z) = \frac 1{2\pi i} \int _C \frac{f(s)ds}{(s-z)^2}

    the text then follows with a longer proof (in terms of limits etc) that the derivative of (1) exists and that it is (2).

    My question is just what does "formally" mean in this context? Why can't I just differentiate (1) and say that (2) is proven?
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  2. #2
    Super Member PaulRS's Avatar
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    In that case "formally" means that you treat it as an algebraic object, ignoring convergence issues.

    See here and here

    Say, pick the generating function f(z)=\sum_{k=1}^{\infty}{k!\cdot z^k} ; as a series it converges for z=0 only and thus it'd make no sense to talk about f in general, however one can treat it as an algebraic object. It'd be more like a symbolic manipulation, you define certain operations over these ojects and manipulate them, but you can't actually evaluate them at a given number.

    Of course, having convergence adds some nice things.

    The reason you can't just differentiate is that exchanging limit and integral doesn't always yield the same result, but under suitable conditions this happens.
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