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Power Series Equality Proof

  1. Jun 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Given [itex]\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}[/itex] and [itex]\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n} [/itex] that are in R. Then, [itex]\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}=\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n} [/itex] if and only if [itex]a_{n}=b_{n} [/itex] for every [itex] n=0,1,2,... [/itex]

    The attempt at a solution
    (<<) Assume an=bn for every n=0,1,2,...
    Then [itex]\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}=a_{0}+a_{1}(x-a)+...=b_{0}+b_{1}(x-a)+...=\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n} [/itex].
    (>>) Now with this direction I am having some issues as the series may not necessarily converge. My attempts have been feeble at best. The problem I have had is that the sums are infinite so I don't think I can use that polynomials are equal if and only if their coefficients are equal. At this point I am thinking about looking at the nth partial sums but I am not sure and just would like any advice.
     
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  3. Jun 21, 2012 #2

    SammyS

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    How can it be said that [itex]\sum_{n=0}^\infty a_{n}(x-a)^{n}=\sum_{n=0}^\infty b_{n}(x-a)^{n}\ , [/itex] unless both sums converge?
     
  4. Jun 21, 2012 #3
    That is the thing. This was something I was going to ask my professor that I am going to be doing research with in p-adic analysis. I am starting to think this may just be a definition since if I recall my professor he was saying how a p-adic integer which is a power series may not be a convergent power series in the ring of p-adic integers. My professor is busy right now. This might not make sense.
     
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