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Radius of convergence problem

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    how to prove that radius of convergence of a sum of two series is greater or equal to the minimum of their individual radii

    i don't know how to begin, can someone give me some ideas?
     
  2. jcsd
  3. Feb 25, 2012 #2

    kai_sikorski

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    Hi Jay,

    Welcome to the Forums!

    I wouldn't try to do this by working with the coefficients of the series directly. Instead fix some x within the radius of convergence of both sequences. Define fn(x) to be the quantity that x gets mapped to by the first n terms of the first power series, and gn(x) the same thing except for the second series. What do you know about the sequences

    f1(x), f2(x), f3(x), ... and
    g1(x), g2(x), g3(x), .... ?

    Using that what can you say about the sequence (g + f)n(x)?
     
  4. Feb 25, 2012 #3
    now , i got some ideas. can i say that cause fn(X) gn(x) are both converges to somthing ,assume T ,S .then by definition of convergent we have |fn(x)-T|<esillope/2 for |x|<R1,and |gn(x)-S|<esillope/2 for |x|<R2 (R1 R2 are the radius of f(X) and g(X)) then choose the min{R1,R2} then we can get |fn(X)+gn(x)-(T+S)|<esillope which is convergent for |x|<min{R1,R2},then the new radius K=min{R1,R2} but how to conclude that it is greater than min{R1,R2}??and is the proof above correct?
     
  5. Feb 25, 2012 #4

    kai_sikorski

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    Above proof is correct.

    I don't see how the statement that the radius of convergence of the sum is strictly greater than the min(R1,R2) is even true. I mean take the taylor series for g(x)=0 as one of the series. Clearly the radius of convergence of the sum must be exactly equal to the radius of convergence of the other series. So greater than or equal to is best you can hope for.
     
  6. Feb 25, 2012 #5

    kai_sikorski

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    Oh if you're asking why it might be greater, then it's simple because your proof excluded the possibility that

    R1 + 2 < min(R1, R2)

    So it must be true that

    R1 + 2 ≥ min(R1, R2)

    That means that it might be greater maybe, or it might be equal. And in fact you could find examples for both.
     
  7. Feb 25, 2012 #6
    sorry, i still cannot understand. from my proof i concluded that |x|<min{R1,R2}then the new radius =min{R1,R2}=K. then the sum of series convergent.but if K can be greater than min{R1,R2}then |x|<min{R1,R2}=<K ,but how can i get this from the proof?
     
  8. Feb 25, 2012 #7
    sorry, i think i get it . because from my proof that |x|<min{R1,R2} then it means that the new radius K cannot smaller than {R1,R2}so K>=min{R1,R2} is this the idea?
     
  9. Feb 25, 2012 #8

    kai_sikorski

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    Yes exactly.

    :smile:
     
  10. Feb 25, 2012 #9

    kai_sikorski

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    I guess here technically you should say that there exists an N, such that this is true for all fn(x) with n>N.
     
  11. Feb 25, 2012 #10

    ok thank you so much
     
  12. Feb 26, 2012 #11

    sorry, can use give me one example that the T>min{R1,R2}(inequality is strict) cause i cannot find one case that satisfies.
     
  13. Feb 26, 2012 #12

    kai_sikorski

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    just take a taylor series for some function f(x) that has finite radius of convergence. make g(x) the taylor series for -f(x).
     
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