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Macluarin series and radius of convergence

  1. Jul 21, 2014 #1
    Hello.
    I am stuck on this question. I'd appreciate if anyone could help me on how to do this.

    The question:
    Expand the following into maclaurin series and find its radius of convergence.

    $$\frac{2-z}{(1-z)^2}$$

    I know that we can use geometric series as geometric series is generally
    $$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$

    So,
    \begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}

    $$\frac{d}{dz}\frac{1}{(1-z)}=\frac{1}{(1-z)^2}$$

    Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
    $$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$

    How do I finish this? The answer I was given is :

    $$\sum_{n=0}^{\infty}(n+2)z^n$$
     
  2. jcsd
  3. Jul 21, 2014 #2

    HallsofIvy

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    You have started off very well. Yes, [itex]\frac{2- z}{(1- z)^2}= \frac{1+ 1- z}{(1- z)^2}= \frac{1}{(1- z)^2}+ \frac{1}{1- z}[/itex]. And [itex]\frac{1}{1- z}= \sum z^n[/itex], the "geometric series".

    Now, note that [itex]d(1/u)/dx= -1/u^2[/itex]. So [itex]1/(1- z)^2= d/dx(1/(1- z))[/itex]. Differentiating the geometric sequence "term by term" gives a power series for [itex]1/(1- z)^2[/itex]
     
  4. Jul 21, 2014 #3
    I already did that in my solution. Look,

    Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
    $$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$

    How do I finish this? The answer I was given is :

    $$\sum_{n=0}^{\infty}(n+2)z^n$$
     
  5. Jul 21, 2014 #4

    HallsofIvy

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    You have your first sum starting with [itex]z^{0- 1}= z^{-1}[/itex]. Because it is multiplied by n= 0, it doesn't change the sum but I think it is causing confusion. Instead write it as [itex]\sum_{n=1}^\infty nz^{n-1}[/itex] which is, as I said, exactly the same thing.

    If you meant $$\sum_{n=1}^\infty nz^{n-1}+ \sum_{n=0}^\infty z^n$$
    then you need to change the indexing on the first sum. Let j= n- 1 so that n= j+ 1. When n= 1 j= 0. The formula becomes [tex]\sum_{j= 0}^\infty (j+ 1)z^j+ \sum_{n=0}^\infty z^n[/tex].
    Now use the fact that "j" is just a "dummy index"- replace j with n in the first sum and combine.
     
  6. Jul 21, 2014 #5
    I don't get it, how can I just replace j with n, when j=n-1? But if I do that, then I'll go back to square one.
     
  7. Jul 21, 2014 #6

    Ray Vickson

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    You have ##S_1 + S_2##, where
    [tex] S_1 = \sum_{n=1}^{\infty} n z^{n-1} \; \text{ and } \; S_2 = \sum_{n=0}^{\infty} z^n [/tex]
    In ##S_1## the power of ##z## ranges from ##n-1 = 0## to ##\infty##, so if we call that power ##j## (instead of ##n-1##) the general term is ##(j+1) z^j##, for ##j = 0,1,2, \ldots##. Now you can re-name ##j## freely, because you really have something of the form
    [tex] (\text{index-name}+1)\, z^{\text{index-name}}[/tex]
    and we are currently using ##\text{index-name} = j## because that is a lot easier to write. However, "index-name" can be anything we want, so we can easily call it ##n## if we want.

    If you remain unconvinced, just use the alternative of re-naming the summation index from ##n## to ##j## in the second sum ##S_2## to get
    [tex] \sum_{j=0}^{\infty} (j+1) z^j + \sum_{j=0}^{\infty} z^j . [/tex]
    Do you see it now?
     
  8. Jul 21, 2014 #7
    Yes, thank you!
     
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