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Radius of convergence

  1. Jul 21, 2014 #1
    Hello.
    I need explanation on why the answer for this problem is R = ∞.

    Here's the question and the solution.

    Expand the function into maclaurin series and find the radius of convergence.
    $zsin(z^2)$

    Solution:
    $$zsin(z^2)=z\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

    Divide both sides by z,

    $$sin(z^2)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

    So here's the calculation but I don't know how to get the radius of convergence. Answer is ∞.
     
  2. jcsd
  3. Jul 21, 2014 #2
  4. Jul 21, 2014 #3

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    Last edited: Jul 21, 2014
  5. Jul 21, 2014 #4
    I think the ratio test should start off like this

    [itex]lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} | [/itex]

    You should find that the limit is 0, so R = ∞.
     
  6. Jul 21, 2014 #5
    That's what I did, only that I use the formula to find R directly, instead of L because R is just 1/L.
    Could you show me the steps until you get 0, please? Because as you see mine, I got 4 :/
     
  7. Jul 21, 2014 #6
    Oh, I get it now! I just googled about limits and learned it again. I think I understand now why the lmit is infinity. Thanks!
     
  8. Jul 21, 2014 #7

    HallsofIvy

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    For others who might be wondering here is your mistake: near the end you have
    [tex]\lim_{n\to\infty}\left|-4n^2+ 10n+ 6\right|[/tex]
    and then immediately have
    [tex]\lim_{n\to\infty}\left|-(4- \frac{10}{n}- \frac{6}{n^2})\right|[/tex]
    where you have divided by n^2. Obviously you cannot do that! You have essentially factored out "[itex]n^2[/itex]" and should have
    [tex]\left(\lim_{n\to\infty}n^2\right)\left(\lim_{n\to\infty}\left|4- \frac{10}{n}- \frac{6}{n^2}\right|\right)[/tex]

    The limit on the right is 4 but the limit on the left is [itex]\infty[/itex]. Their product is "[itex]\infty[/itex]".
     
  9. Jul 21, 2014 #8
    I think there's some confusion about the ratio test. For a series [itex] \sum a_n [/itex], the ratio test is [itex] lim_{n \to \infty} \frac {a_{n+1}}{a_n} [/itex].

    This is exactly what I started off with when I saw that you had simplified the series. I start off with

    [itex]lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} | [/itex]

    Since this is an absolute value, the [itex] -1 [/itex] terms can be ignored, and simplifying the exponents of the first fraction, you should see

    [itex] lim_{n \to \infty} | \frac {z^{4n+6}}{(2n+3)!} \cdot \frac {(2n+1)!}{z^{4n+2}} | [/itex]

    To simplify, observe that [itex] (2n+3)! = (2n+3) \cdot (2n+2) \cdot (2n+1)! [/itex], [itex] z^{4n+6} = z^{4n} \cdot z^6 [/itex] and that [itex] z^{4n+2} = z^{4n} \cdot z^2 [/itex].

    So we have

    [itex] lim_{n \to \infty} | \frac {z^{4n} \cdot z^6}{(2n+3) \cdot (2n+2) \cdot (2n+1)!} \cdot \frac {(2n+1)!}{z^{4n} \cdot z^2} | [/itex]

    The [itex] z^{4n} [/itex]'s and [itex] (2n+1)! [/itex]'s cancel out. You can also factor the remaining z terms. This will leave you with

    [itex] \frac {z^6}{z^2} lim_{n \to \infty} \frac 1{(2n+3)(2n+2)} [/itex].

    The limit is 0. Interpret this to mean, it does not matter what values you choose for the variable z. Any real number will work. So we have [itex] -\infty < z < \infty [/itex], so the radius of convergence is ∞.

    If you like my explanation, be sure to "thank" me (hit the Thanks button). I have 0 "thank"s thus far, and I would like some.
     
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