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| Vroomfondel |
 Posted: Sat Apr 19, 2008 12:51 pm Post subject: Formal Power Series |
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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 expN(T) and lnN(T). All that I need to show then is that expN(lnN(1+T)) = 1+T+o(TN). Now I am going to try to expand (lnN(1+T))n and see if things cancel out when put into expN(T).
Any pointers? Am I going down a dead end? Thanks.
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| Hanuka |
| Posted: Sat Apr 19, 2008 4:23 pm Post subject: |
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eek! math!!   |
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| JaneBennet |
| Posted: Sat Apr 19, 2008 5:01 pm Post subject: |
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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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| Vroomfondel |
| Posted: Sat Apr 19, 2008 5:42 pm Post subject: |
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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 z0. Then convergence is assured for |z - z0| < |z0|, the open disk of radius |z0| centered at z0. 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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| JaneBennet |
| Posted: Sat Apr 19, 2008 5:53 pm Post subject: |
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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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| Vroomfondel |
| Posted: Sat Apr 19, 2008 6:07 pm Post subject: |
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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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| serpicojr |
| Posted: Sat Apr 19, 2008 7:48 pm Post subject: |
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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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| serpicojr |
| Posted: Sat Apr 19, 2008 7:50 pm Post subject: Re: Formal Power Series |
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| Vroomfondel wrote: |
| All that I need to show then is that expN(lnN(1+T)) = 1+T+o(TN). Now I am going to try to expand (lnN(1+T))n and see if things cancel out when put into expN(T). |
This will work if the calculations don't get too ugly! |
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| Vroomfondel |
| Posted: Sat Apr 19, 2008 10:18 pm Post subject: |
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| serpicojr wrote: |
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).
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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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