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Thread: Formal Power Series

  1. #1 Formal Power Series 
    Forum Junior Vroomfondel's Avatar
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    Hello. I was working on my complex analysis work, and I came across a problem that I am having significant difficulties with. The problem is basically to show that one formal power series is the inverse of another, given both of the power series. The ones I am working with now are actually exp(T) and ln(T). I know that these two series extend from the reals to the complex numbers uniquely, but I am pretty sure that the author wants it purely in the formal power series methods. Thanks!

    Edit: I would much prefer hints, or at least a commentary on what I am doing than a correct answer. My method so far is:

    I need to show that exp(log(1+T)) = 1+T. I am truncating log and exp at the N term, calling these exp<sub>N</sub>(T) and ln<sub>N</sub>(T). All that I need to show then is that exp<sub>N</sub>(ln<sub>N</sub>(1+T)) = 1+T+o(T<sup>N</sup>). Now I am going to try to expand (ln<sub>N</sub>(1+T))<sup>n</sup> and see if things cancel out when put into exp<sub>N</sub>(T).

    Any pointers? Am I going down a dead end? Thanks.


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  3. #2  
    Forum Ph.D. Hanuka's Avatar
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    eek! math!!


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  4. #3  
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    This is my method using calculus.

    Note that you can differentiate both power series to show that d[exp(x)]/dx = exp(x) and d[log(1+x)]/dx = 1*⁄*(1+x).

    Let y = exp(log(1+x)). Note that y = 1 when x = 0.

    Differentiating using the chain rule, dy/dx = exp(log(1+x))*⁄*(1+x) = y*⁄*(1+x).

    Now solve this variables-separable differential equation. The general solution is

    y = C(1+x)

    As y = 1 when x = 0, C = 1. Hence

    y = exp(log(1+x)) = 1+x

    I think this method is okay. If it isn’t, hopefully Serpicojr will say so.
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  5. #4  
    Forum Junior Vroomfondel's Avatar
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    Well, let's see. That does the trick for the reals, and let's us expand log(x) around any point along the positive x axis. Using analytic continuation, let us expand log(1+z) around any z on the positive real axis. Say we expand it around z<sub>0</sub>. Then convergence is assured for |z - z<sub>0</sub>| < |z<sub>0</sub>|, the open disk of radius |z<sub>0</sub>| centered at z<sub>0</sub>. But this disk has complex numbers with only positive real parts, and does not assure that exp(log(1+z)) = 1+z for numbers with negative real parts. Not quite enough, I need the punctured plane (that is, C - {0})!

    Now, I had an idea. What if we start covering the entire plane with overlapping open disks. All that we require of these disks is that they do not touch z = -1, so that absolute convergence of log(1+z) is assured. we can cover the punctured plane with these, and by analytic continuation we have that exp(log(1+z)) = 1+z everywhere but z = -1. equivalence of these two series (exp(log(1+z)) and 1+z) implies equivalence of their formal series, as was to be shown. What do you think?

    I still think the author wanted an argument based purely on formal series techniques though (ignoring convergence and just looking at the series themselves... any ideas?)
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  6. #5  
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    Oops. I thought the functions were for real numbers only. I missed the part where you mentioned complex numbers. Sorry.
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  7. #6  
    Forum Junior Vroomfondel's Avatar
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    No problem at all, your derivation helped me see the whole problem from another point of view. Thanks!
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  8. #7  
    Forum Professor serpicojr's Avatar
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    I think Jane's proof is fine, even formally and without calculus, because you can formally define differentiation of power series in the obvious way. I imagine the chain rule isn't too hard to show, but you'd need to show it for this proof, as we're not using calculus. And then the differential equation can be interpreted as a condition on the coefficients of the power series, from which you can show, indeed, all solutions are C(1+x).

    Just plugging and chugging--i.e., literally computing the coefficients of exp(log(1+x))--is probably doable, although you'll likely have to manipulate some weird looking sums.
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  9. #8 Re: Formal Power Series 
    Forum Professor serpicojr's Avatar
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    Quote Originally Posted by Vroomfondel
    All that I need to show then is that exp<sub>N</sub>(ln<sub>N</sub>(1+T)) = 1+T+o(T<sup>N</sup>). Now I am going to try to expand (ln<sub>N</sub>(1+T))<sup>n</sup> and see if things cancel out when put into exp<sub>N</sub>(T).
    This will work if the calculations don't get too ugly!
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  10. #9  
    Forum Junior Vroomfondel's Avatar
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    Quote Originally Posted by serpicojr
    I think Jane's proof is fine, even formally and without calculus, because you can formally define differentiation of power series in the obvious way. I imagine the chain rule isn't too hard to show, but you'd need to show it for this proof, as we're not using calculus. And then the differential equation can be interpreted as a condition on the coefficients of the power series, from which you can show, indeed, all solutions are C(1+x).
    This seems like a good idea, but I can't help but think that it would take more work than expected to prove the chain rule, working in a purely formal environment. I'll give it a shot though, thanks for the idea. Oh, and yes, my original method did get quite messy.
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