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Powerseries problem issue

  1. May 2, 2006 #1
    HELP: Powerseries problem issue

    Hello People,

    I have this here power series problem which is frustrating me:

    Given the power series

    [tex] (*) f(z) = \sum_{n=0} ^{\infty} (2n+1)z^n[/tex]

    for |z| < 1 and [tex]z \in \mathbb{C}[/tex]

    has the Radius of Convergens R = 1.

    Show that the power series (*) can be written as another power series

    g(z) = f(z) - z f(z)

    which gives the result [tex]g(z) = \frac{1+z}{1-z}[/tex]

    Proof:

    The power series (*) can be written as

    [tex]f(z) = \sum_{n=0} ^{\infty} 2z(n \cdot z^{n-1}) + z^n[/tex]

    Since R = 1

    then

    [tex](1-z) \sum_{n=0} ^{\infty} (2n+1)z^n - z \sum_{n=0} ^{\infty} (2n+1)z^n = 1[/tex]

    But how I proceed from here?

    Sincerely Yours
    Hummingbird25
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 2, 2006 #2

    Physics Monkey

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    Hi Hummingbird25,

    Your first step should be to simply calculate the terms in the series for g(z) defined as f(z) - z f(z), [tex] g(z) = \sum_{n=0} (2n+1) z^n - \sum_{n=0} (2n+1)z^{n+1} [/tex]. If you can't find the pattern directly, just write out a few terms in the series and things will become quite clear. You will find that the result can be easily summed into a function you know.
     
  4. May 2, 2006 #3
    Hello Physics Monkey and thank You for You kind and Quick answer,

    If I write out the sum I get:

    [tex]g(z) = (1-z) + (3z - 3z^2) + (5z^2-5z^3) + \ldots +[/tex]

    But what does that show ?

    Sincerely Hummingbird25
     
    Last edited: May 2, 2006
  5. May 2, 2006 #4

    Physics Monkey

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    Happy to help. Try combining like powers.

    Edit: Also, you made a mistake in writing out the series. There should be plus signs between each of your parenthesis.
     
  6. May 2, 2006 #5

    Curious3141

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    Try rearranging the parantheses so that like powers are bracketed together (z terms together, z^2 together, etc.) :wink:
     
  7. May 2, 2006 #6
    Hello Curious3141, and thank You very much for Your answer,

    If I arrange them together that gives

    [tex]g(z) = (1-z-3z) - (3z - 5z)z + (-5z^3 + 7z^3) + (-7z^4 + 9z^4) + \ldots +[/tex]

    Sincerely Yours Hummingbird25

     
    Last edited: May 2, 2006
  8. May 2, 2006 #7
    But written it out as sums you mean ?

    [tex]g(z) = f(z) - z f(z) = \sum_{n=0} ^{\infty} -(2n+1) \cdot z + 2n +1) z^n[/tex]
     
    Last edited: May 2, 2006
  9. May 2, 2006 #8

    Physics Monkey

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    I would recommend you stick with the approach of writing out terms. Having grouped liked powers together all you need to do is subtract. Do you notice anything? You may need to do a few more terms in the series to see the pattern.
     
  10. May 2, 2006 #9

    Hi

    Then the summation must be

    [tex]g(z) = (1-z+3z) - (3z - 5z)z + (-5z^3 + 7z^3) + (-7z^4 + 9z^4) + \ldots +[/tex]

    Is that what You mean?

    Sincerley Hummingbird25
     
    Last edited: May 2, 2006
  11. May 2, 2006 #10

    Curious3141

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    No, let me illustrate (then you should do the rest yourself).

    [tex]g(z) = (1) + (-z + 3z) + (-3z^2 + 5z^2) + (-5z^3 + 7z^3) + \ldots +[/tex]

    Do you see anything happening within each paranthesis ?
     
  12. May 2, 2006 #11

    Physics Monkey

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    Yes, that is what I mean. Now carry out the subtraction for each term in brackets. What do you notice?
     
  13. May 2, 2006 #12
    Yes the series act [tex]1+ z^n[/tex] ??

    Sincerley
    Hummingbird25
     
  14. May 2, 2006 #13

    Curious3141

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    Ah, no. There's a summation sign and a factor of 2 missing there. Try again ?
     
  15. May 2, 2006 #14
    Okay

    I will try again,

    my surgestion is


    [tex]\sum_{n=0} ^ {z} 1^n = 1+z[/tex]

    That factor of two must I add that within the sum or outside the sum?

    Sincerley Hummingbird

     
    Last edited: May 2, 2006
  16. May 2, 2006 #15

    Curious3141

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    No, that's not right.

    You've gotten to this point right ?

    [tex]g(z) = 1 + 2z + 2z^2 + 2z^3 + ... = 1 + 2(z + z^2 + z^3 + ...)[/tex]

    Put that into summation notation. Also see if you can spot what sort of basic series is represented there.
     
    Last edited: May 2, 2006
  17. May 2, 2006 #16
    That looks like the sum [tex]\sum_{n=0} ^{?} (1+2z^n)[/tex]
     
  18. May 2, 2006 #17

    Curious3141

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    The question mark should be replaced by infinity. Also, when you put the '1' within the summation it means it's going to be added in each term, meaning you'll be doing it an infinite number of times. So that isn't right.

    Also are you sure the lower bound for the summation is zero ?
     
  19. May 2, 2006 #18
    Okay then the sum must be?

    [tex]2 \sum_{n=?} ^{\infty} z^{n+1} = \cdots = \frac{1+z}{1-z}[/tex]

    Sincerely Hummingbird25
     
    Last edited: May 2, 2006
  20. May 2, 2006 #19

    Curious3141

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    You're getting there, it's still not written out quite correctly. Let me show you, I hope you can understand it (it's important to understand more than just remember).

    BTW, I made a mistake in writing f(z) before, I meant g(z).

    [tex]g(z) = 1 + 2(z + z^2 + z^3 + ...) = 1 + 2\sum_{n=1}^{\infty}z^n[/tex]

    Clear so far ? You can start the sum from n = 0 but then the expression in the summation has to become z^(n+1) - you understand ? And don't forget to add the one outside the summation.

    Now, the summation is actually a geometric series with first term z and a common ratio of z. It will converge for |z| < 1. The sum of a geometric series with first term a and common ratio r is :

    [tex]S = \frac{a}{1-r}[/tex]

    So, using that here,

    [tex]g(z) = 1 + 2\sum_{n=1}^{\infty}z^n = 1 + 2(\frac{z}{1-z}) = \frac{1-z}{1-z} + \frac{2z}{1-z} = \frac{1+z}{1-z}[/tex]

    And now, since [tex]g(z) = f(z) - zf(z)[/tex], you can solve for [tex]f(z)[/tex]. Presumably, that's what the question wants you to do at the end even if it's not stated as such.
     
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