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Math Help - convergence radius question..

  1. #1
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    convergence radius question..

    \sum \frac{n^2}{3^nz^n}\\

    \frac{1}{R}=\lim_{n->\infty}\sqrt[n]{|\sum \frac{n^2}{3^nz^n}|}\\

    what to do now?
    Last edited by transgalactic; October 13th 2009 at 10:43 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    \sum \frac{n^2}{3^nz^n}\\

    \frac{1}{R}=\lim_{n->\infty}\sqrt[n]{|\sum \frac{n^2}{3^nz^n}|}\\

    what to do now?
    I don't know why you have the summation inside the square root. Correct application of the nth root test means that the given series will converge for values of z that satisfy \lim_{n \rightarrow +\infty} \left| \frac{n^{2/n}}{3z}\right| < 1 \Rightarrow \frac{1}{3|z|} \lim_{n \rightarrow +\infty} n^{2/n} < 1.

    Then you have to test each endpoint.
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  3. #3
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    i cant understand the numenator
    we have n^{\frac{n}{2}}
    so we get "infinty by power 0"

    what is the result of this expression

    i think its undefined form

    ?
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    Quote Originally Posted by transgalactic View Post
    i cant understand the numenator
    we have n^{\frac{n}{2}}
    so we get "infinty by power 0"

    what is the result of this expression

    i think its undefined form

    ?

    No, you don't have n^(n/2) but n^(2/n) and the limit of this thing is 1, so the lmit you're looking for is (1/3)(1/|z|) = 1/R and etc.

    Tonio
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  5. #5
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    sorry i ment that
    n^(2/n)

    i cant understand why it equals 1
    we have (infinity)^0
    why its 1?
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    Quote Originally Posted by transgalactic View Post
    sorry i ment that
    n^(2/n)

    i cant understand why it equals 1
    we have (infinity)^0
    why its 1?

    Since power series and stuff belongs at least to calculus II , I assume you've already had calculus I: limits, continuity and stuff...right?
    Then you must know that n^(1/n) --> 1 when n --> oo, but then
    n^(2/n) = [n^(1/n)]^2 --> 1*1 = 1 by arithmetic of limits.

    Tonio
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  7. #7
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    ok now i have
    \frac{1}{3|z|}


    i dont know how to deside if it converge or diverge

    so what if its less then 1

    by what law we decided that it converged?
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    Quote Originally Posted by transgalactic View Post
    ok now i have
    \frac{1}{3|z|}


    i dont know how to deside if it converge or diverge

    so what if its less then 1

    by what law we decided that it converged?

    Hmmm...rechecking your question I realize that what you posted is NOT a power series in z since z appears in the denominator, so this is at most a Laurent series, a very different stuff belonging to complex analysis.
    What did you really mean to ask here? Is z a variable or just a parameter?

    Tonio
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  9. #9
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    z is a complex number
    here it acts as a parameter

    what is the convergence radius
    ?
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    Quote Originally Posted by transgalactic View Post
    z is a complex number
    here it acts as a parameter

    what is the convergence radius
    ?

    Oh, I see: so z is a constant. Well, since it MUST be (1/3)(1/|z|) < 1, from here we see that the series converges whenever |z| > 1/3.
    Now just check the end points 1/3 and -1/3 (hint: divergence at both)

    Tonio
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  11. #11
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    why it converges
    whenever |z| > 1/3

    ??
    by what law?

    i know that a series converges if the limit is a constant
    it diverges if its infinity
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    Quote Originally Posted by transgalactic View Post
    why it converges
    whenever |z| > 1/3

    ??
    by what law?
    Inequalities...??! If 1/(3|z|) < 1 then multiplying by |z| you get |z| > 1/3...

    Tonio
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  13. #13
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    ok i understand
    but the formula is
    1/R=1/(3|z|)
    R=3|z|
    so if going stricktly by the formula i get undefined expression 3|z|

    i understand your logic
    that it uniformly coverges from -1/3<z<1/3

    but i get a different answer by the formula
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  14. #14
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    Quote Originally Posted by transgalactic View Post
    ok i understand
    but the formula is
    1/R=1/(3|z|)
    R=3|z|
    so if going stricktly by the formula i get undefined expression 3|z|

    i understand your logic
    that it uniformly coverges from -1/3<z<1/3

    but i get a different answer by the formula

    True, because as I pointed out before what you ngave is not a power series! What you gave is a palin series with a parameter (constant) z, as you said. There's no convergence radius here, then.
    For something to be a proper power series the powers of the VARIABLE have to be positive integers.

    Tonio
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  15. #15
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    Quote Originally Posted by transgalactic View Post
    ok i understand
    but the formula is
    1/R=1/(3|z|)
    R=3|z|
    so if going stricktly by the formula i get undefined expression 3|z|

    i understand your logic
    that it uniformly coverges from -1/3<z<1/3

    but i get a different answer by the formula
    Instead of blindly applying formulas you would be well advised to understand the principles of where those formulas come from and then applying those same principles to the questions you get. The question you have posted and the subsequent exchanges of replies is a classic example of why I give that advice.
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