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Homework Help: Centre and Radius of Convergence

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the centre and radius of convergence:
    [tex]\stackrel{\infty}{n=1}\sum n.(z+i\sqrt{2})^{n}[/tex]

    2. Relevant equations

    1) Ratio test [tex]\left|\frac{a_{n+1}}{a_{n}}\right|<1[/tex]

    2) Textbook uses [tex]\stackrel{lim}{n-> \infty}\left|\frac{a_{n}}{a_{n+1}}\right|[/tex]

    3. The attempt at a solution

    using 1) and taking the limit as n -> infinity i end up with
    ...but then dont know where to go

    using 2) i end up with

    Last edited: Oct 19, 2009
  2. jcsd
  3. Oct 19, 2009 #2
    Will you please double check the problem statement again and make sure that you typed the problem correctly above? It doesn't look right, because everything but the n is a constant. So your series is just a constant times the summation of n, as n goes from 1 to infinity. A power series is a series of the form
    The an are the coefficients depending on n, and z=c is the center of the power series. You can click on the LaTeX output to see how I entered it.
  4. Oct 19, 2009 #3
    sorry, was meant to be to the power 'n' and i had '2'
  5. Oct 19, 2009 #4
    Okay, then your power series is
    [tex]\sum_{n=1}^\infty n (z+i\sqrt{2})^n.[/tex]
    Comparing this with the power series form I wrote previously, this means that [itex]a_n=n[/itex] are the coefficients and [itex]z=-i\sqrt{2}[/itex] is the center. Now to find the radius of convergence for a complex power series, there is a theorem that the radius of convergence R is
    [tex]R=\lim_{n\to\infty} \left| \frac{a_n}{a_{n+1}} \right|,[/tex]
    if this limit exists. This is what your textbook is using. The first ratio test you listed is something a little different.

    Now, can you find the radius of convergence R from here?
  6. Oct 19, 2009 #5
    I'm not really sure, the textbook seems to ignore the part containing the 'z' but this confuses me because it is to the power 'n' so i would have thought you would need to include it when finding the limit as 'n' goes to infinity. To make things more confusing our lecturer uses the ratio test formula in all the examples he has given us (which only happens to be 2)... and we don't have any tutorials which i think is pathetic.

    If I ignore the brackets to the power 'n' i just get a radius of 1... but this seemed too simple.
    If this is the correct answer, how would i go about solving, do i just ignore the z^2n+1

    [tex]\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1}[/tex] ?
  7. Oct 19, 2009 #6
    Well the point of the theorem I gave, and of what the text no doubts gives, is that it gives a formula for finding the radius of convergence.

    The limit in your first problem is 1, so the radius of convergence is 1. That is all there is to it. This means that the power series converges when [itex]|z+i\sqrt{2}|<1[/itex] and diverges when [itex]|z+i\sqrt{2}|>1[/itex].

    For your other problem, no you don't ignore that part. Look at the theorem I wrote above and the theorem in your book. Note that it is based upon the power series being given in a certain form. Rewrite the series as
    [tex]\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1} = z\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n}[/tex]
    by just factoring out a z.

    Now let [itex]w=z^2[/itex]. Then the series (excluding the lone factored z) becomes
    [tex]\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n} = \sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!} w^n[/tex]
    Find the radius of convergence of this series, say R. Then this series converges for |w|<R, but since w=z2, the original series converges for [itex]|z|<\sqrt{R}[/itex]. The factored out z term doesn't really matter, because if the series directly above converges, then z times that series (the original series) also converges.
  8. Oct 19, 2009 #7


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    Just different ways of looking at the same thing. The ratio test you mention says that a numeric sequence, [itex]\sum a_n[/itex] converges absolutely as long as [itex]|a_{n+1}/a_n|\to a< 1[/itex].

    Now apply that to the power sequence \sum a_n (z-z_0)^n: the ratio is [itex]|a_{n+1}(z-z_0)^{n+1}|/|a_n(z-z_0)^n|= |a_{n+a}/a_n||z- z_0|< 1[/itex].

    Finally solve for [itex]|z- z_0|[/itex] by dividing both sides by [itex]|a_{n+1}/a_n|[/itex]: [itex]|z- z_0|< |a_{n+1}/a_n|[/itex]

    Because you divided both sides by the fraction, in has been inverted.

    No, you don't. In the second form, you are looking only at the limit of the coefficients, without the "[itex](z+ i\sqrt{2})[/itex]". Since the limit of the coefficients is 1, in either case, you get that the radius of convergence is 1.

    And since you want [itex]|z- z_0|= |z+ i\sqrt{2}|= |z- (-i\sqrt{2})|[/itex], the center of convergence is [itex]i\sqrt{2}[/itex].

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