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Homework Help: Adding series sums to one sum

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I have these three series here

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

    what I would like to do is to add the together as one sum representation


    3. The attempt at a solution

    From what I can see the commen sum of three is

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

    What I don't remember do I add the index's together or subtract them?

    If not please direct to a webpage with rules on howto do the above right :) Cause I have a whole in my knowledge then regarding to above :(

    Susanne
     
    Last edited: May 3, 2010
  2. jcsd
  3. May 3, 2010 #2

    Gib Z

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    In Mathematics theres so much stuff to learn, you don't want to fill your head with rules for every thing.

    Why don't you try writing each sum out in full and collect like powers of z, and see what kind of new sum that gives you.
     
  4. May 3, 2010 #3

    HallsofIvy

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    In the second sum, let k= n+1 so that n= k-1. When n= 1, k= 0 so you will have a sum in which k runs from 0 to infinity:
    [tex]\sum_{n=1}^\infty a_n a^{n+ 1}= \sum_{k=0}^\infty a_{k-1}z^k[/tex]

    And since the "k" or "n" in each sum is a dummy index, you can just replace "k" with "n":
    [tex]\sum_{n=0}^\infty a_{n-1}z^n[/tex]


    Now you can add "like powers" in the two sums- although, since the first sum starts at n= 2, you have to treat n=0 and n= 1 separately.

    what I would like to do is to add the together as one sum representation


    3. The attempt at a solution

    From what I can see the commen sum of three is

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

    What I don't remember do I add the index's together or subtract them?

    If not please direct to a webpage with rules on howto do the above right :) Cause I have a whole in my knowledge then regarding to above :(

    Susanne[/QUOTE]
     
  5. May 3, 2010 #4
    [/QUOTE]

    I tried I again and I always end up with [tex]\sum_{n=0}^{\infty} (a_{n+2} - a_{n+1} +a_{n})z^{n+1}[/tex] ??
     
  6. May 3, 2010 #5
    HallsofIvy made a rare mistake above. Let me explain it without intorducing another summation variable. Let's focus on the first summation from n = 2 to infinity. You'll get the same terms in that summation if you let n run from zero to infinity and at the same time you replace n by n + 2. What you are then doing is saying that:


    f(2) + f(3) + f(4) + .....=


    g(0) + g(1) + g(2)+...


    if we define g(n) as f(n+2), so that g(0) = f(2), g(1) = f(3) etc.
     
  7. May 3, 2010 #6
    As I understand you [tex]a_{n+2}z^{n+2}[/tex]
     
  8. May 3, 2010 #7
    Yes, that's correct. You can write out the first few terms and convince yourself that you get the same terms.
     
  9. May 3, 2010 #8
    Any chance you would write the other two sums like the above. Then I could use that to check the rest of my result :)

    Susanne
     
  10. May 3, 2010 #9
    Well, the second summation is of the form:

    f(1) + f(2) + f(3) + ...

    where f(n) = a_n z^(n+1)

    If you want to write

    f(1) + f(2) + f(3) + ..... as

    g(0) + g(1) + g(2)+...

    how do yo have to define g(n) in terms of f(n)?
     
  11. May 3, 2010 #10

    D H

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    You should have offered a correction then -- using summation symbols.

    What you need to do in general, Susanne217, is to first modify the series so that the power terms are of the same form for each of the series. Do not worry about the limits of the series at this stage. So, let's do that, once again picking on the second series.

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

    The series on the left is simply the second series from the original post. From above, the goal here is to express the power term of each of the series in the same form. The canonical form is [itex]z^n[/itex], but here we have terms involving [itex]z^{n+1}[/itex]. So, substitute [itex]k=n+1[/itex], or [itex]n=k-1[/itex]. This substitution may affect other factors in a summand will affect and the sum limits as well. In this case, the factor of [itex]a_n[/itex] becomes [itex]a_{k-1}[/itex] since [itex]n=k-1[/itex] and the lower sum limit changes from [itex]n=1[/itex] to [itex]k=2[/tex] since [itex]k=n+1[/itex]. Thus

    [tex]\sum_{n=1}^\infty a_n z^{n+ 1}= \sum_{k=2}^\infty a_{k-1}z^k[/tex]

    As Halls mentioned, the index is just a dummy variable. There is nothing wrong with changing the k in the series on the right hand side with an n, yielding

    [tex]\sum_{n=1}^\infty a_n z^{n+ 1}= \sum_{n=2}^\infty a_{n-1}z^n[/tex]



    The next stage is to form a single series. Now you do have to worry about the summation limits. Different example from above, suppose you are asked to compute

    [tex]S = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=0}^{\infty} c_n z^{n+1}[/tex]

    After the first step this becomes

    [tex]S = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=1}^{\infty} c_{n-1} z^n[/tex]

    Now we have a problem: The two series don't have the same limits. What you need to do is expand some of the series so that do have series with the same limits. Thus

    [tex]S = b_0 + \sum_{n=1}^{\infty} b_n z^n + \sum_{n=1}^{\infty} c_{n-1} z^n[/tex]

    Finally, the series can be combined. Note that this might well mean a few stray terms that are not a part of the series.

    [tex]S = b_0 + \sum_{n=1}^{\infty} (b_n +c_{n-1})z^n[/tex]
     
  12. May 3, 2010 #11
    See next post...
     
    Last edited: May 4, 2010
  13. May 4, 2010 #12
    Thats so funny cause I get the following result myself

    Let [tex]\frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z_n[/tex]

    where [tex]a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0[/tex] and [tex]a_2=0[/tex] and [tex]a_{n+3} + a_n = 0[/tex]

    By the theorem of Identity

    [tex]1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    If I write out the first sum I get [tex]\sum_{n=3}^{\infty} a_n \cdot z^{n} = a_3 z^3 + a_{4}z^4 + \cdots [/tex]

    I write out a couple terms of the second sum [tex]- \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} = - a_1 \cdot z^{2} - a_2 \cdot z^{3} - \cdots [/tex]

    If I make the commen index n = 0

    it looks like the index increase at n+3 and at a power of n+2 such that

    [tex]a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 +\sum_{n=3}^{\infty} + a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2} = \sum_{n=0}^{\infty} (a_{n+3} +a_n)z^{n+2}[/tex]

    How is that? Better I hope :)

    Sincerely
    Susanne
     
    Last edited: May 5, 2010
  14. May 5, 2010 #13
    In the equation after the sentence "By the theorem of Identity"

    you take out some terms of the summations with seems to be motivated in order to let the summations start with a z^2 term, but then you also take out a2 z^2 from the first summation on theright hand side of that equation. That defeats the purpose of your effort.

    Instead, make sure that the three summations all start with z^2. So, don't take out the a2 z^2 term and instead let the first sumatrion of the r.h.s. be from n = 2 to infinity. Then you can shift the n for all the summations so that they all run from, say, zero to infinity (or from n = 2 to infinity, it doesn't matter what you choose). Then all the summations limits are the same and thus the power of z as a function of n will now be the same, so you can combine the three summations into one and take out the z^(n + something) in the summand.
     
  15. May 5, 2010 #14
    Just understand you correctly Count Iblis,

    I am suppose to end up with a sum that looks someward like this?

    [tex]\sum( a_{n+3} + a_n ) z^{n+2}[/tex]

    because I need it to satisfy the condition [tex]a_{n+3} + a_n = 0[/tex] because [tex]a_{n+3} = -a_{n}[/tex]
     
  16. May 5, 2010 #15
    Yes, that's right. But note that if you rewrite the series in a more standard way, along the lines D.H and I have been suggesting, you find a recurrence relation the expresses a_n in terms of a(n-1} and a_{n-2} and from that you can then derive that
    a_{n+3} = -a_{n}.
     
  17. May 5, 2010 #16

    Count,

    If I subtract the second sum from the first I get (Which is hopefully correct)

    [tex]\sum_{n=1}^{\infty} a_{n-1}z^{n+1}[/tex] if this is correct my next logical step is to add this together with the third sum which produces

    [tex]\sum_{n=0}^{\infty} (??) [/tex]

    But there is a problem with my calculations count :(

    If I add the third sum to the difference between the first two I get

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

    and it sums into on only one term which looks nothing like the intended target.... What am I doing wrong? Why isn't it obvious :(
     
    Last edited: May 5, 2010
  18. May 5, 2010 #17
    You are doing something wrong. Let me start from this point in your posting#12:

    [tex]1 = \left(1 - z + z^2\right) \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + \sum_{n=2}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    Where I put the a_2 z^2 term back inside the summation relative to what you did. What you now have are summations that all start with z^2. If you now simply replace n by n-1 in the second summation, then you need to compensate for that by increasing the lower limit from n= 1 to n = 2. So, the summand in the second summation then contains z^n as in the first summation and it starts at the same lower limit for n, but, of course, you now have a_{n-1} in there.

    You can then rewrite the last summation in the same way, i.e. change n to n-2 and simultaneously increase the lower limit from n=0 to n = 2.
     
    Last edited: May 5, 2010
  19. May 5, 2010 #18
    Okay then by this I simply transformer this into

    [tex] 1 = \left(1 - z + z^2\right) \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z +\sum_{n=2}^{\infty} a_{n}z^{n} - \sum_{n=2}^{\infty} a_{n-1} z^{n+1} + \sum_{n=2}^{\infty} a_{n-2} z^{n+2}[/tex]

    But I can follow this being a recurcive relation half way and its suppose to be n+3 because it has three terms?
     
    Last edited: May 5, 2010
  20. May 5, 2010 #19

    D H

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    You didn't do that right.

    Don't worry about that yet. You will get a different recursive relationship here, and the you will derive the desired one.
     
  21. May 5, 2010 #20
    Good then I'm only semi stupid...

    Anyway please my first step is to squize the two sums 1) and 2) together :)

    If I change the index to n = 0 in the third sum I get

    [tex]\sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    But the difference sum if I do something simular to series 1) and series 2)

    their difference sum is still:

    [tex]\sum_{n=0}^{\infty} a_{n+1} z^{n+2}[/tex] and thats a long way off from

    a_n+3 + a_n ??
     
    Last edited: May 5, 2010
  22. May 5, 2010 #21

    D H

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    Let's change the index in the third sum,

    [tex]\sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    The goal here is to get a sum that involves [itex]z^n[/itex] rather than [itex]z^{n+2}[/itex]. One way to avoid confusion is to invent a new dummy index variable. So, define m=n+2. With this the sum becomes

    [tex]\sum_{m=2}^{\infty} a_{m-2} z^m[/tex]

    Now rename that index variable back to n:

    [tex]\sum_{n=2}^{\infty} a_{n-2} z^n[/tex]
     
  23. May 5, 2010 #22

    D H

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    Now do the same thing to the second sum. When you are done you should have something like

    [tex]1 = \left(1 - z + z^2\right) \sum a_n \,z^n = a_0 + (a_{1}-a_{0})z + \sum_{n=2}^{\infty} a_n \,z^n - \sum_{n=2}^{\infty} b_{n} z^n + \sum_{n=2}^{\infty} c_n z^n[/tex]

    where the [itex]b_n[/itex] and [itex]c_n[/itex] are expressed in terms of some appropriate element from the set [itex]\{a_n\}[/itex].
     
  24. May 5, 2010 #23
    I'm one to something now

    [tex]a_0 + (a_{1}-a_{0})z + \sum_{n=2}^{\infty} (a_n - b_{n} )z^n = - \sum_{n=2}^{\infty} c_n z^n[/tex]

    and for the second sum

    Let's change the index in the third sum,

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

    The goal here is to get a sum that involves [itex]z^n[/itex] rather than [itex]z^{n+1}[/itex]. One way to avoid confusion is to invent a new dummy index variable. So, define m=n+1. With this the sum becomes

    [tex]\sum_{m=2}^{\infty} a_{m-1} z^m[/tex]

    Now rename that index variable back to n:

    [tex]\sum_{n=2}^{\infty} a_{n-1} z^n[/tex]

    How is that ?? and I'm supose to combine sum 1) and 2) into a_n+3 now?
     
    Last edited: May 5, 2010
  25. May 5, 2010 #24

    D H

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    Don't worry about that yet! First get the sum right. Then worry about that. You are putting the cart before the horse.
     
  26. May 5, 2010 #25
    Sorry, but whats my next step then? Cause the whole thing doesn't look obvious to me..
     
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