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Limit of series.

  1. Dec 14, 2015 #1
    1. The problem statement, all variables and given/known data
    If we have a number sequence such that: a0, a1 are given, and every other element is given as ##a_n=\frac{(a_{n-1} + a_{n-2})}{2} then express an in terms of a0, a1 and n , and fin the limit of an

    2. Relevant equations


    3. The attempt at a solution
    If i try to express a3 in terms of a1 and a0 and then a4 in terms of a3 and a2 (using the a2 i previously expressed in terms of a1 and a0 ) and going so on with next elements up to an element, that isn't actually a good idea, since i am supposed to express it only using a1 and a0, no an-1 and stuff. So if i try to find a2 - a1, then a3 - a2 all the way to the an - an-1 i would have something like this:

    ## a_2 - a_1= - \frac{a_1 - a_0}{2} \\ a_3 - a_2= \frac{a_1 - a_0}{2^2} \\ \vdots \\ a_n - a_{n-1} = (-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}} ##

    If i sum up all of these, i actually have:
    ## a_n - a_1 = -\frac{a_1 - a_0}{2} + \frac{a_1 - a_0}{2^2} - \frac{a_1 - a_0}{2^3} + \cdots +(-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}} ##

    Which is what i need to have, i should just "move" a_1 to the RHS but i don't know how to simplify that series so that i could find a limit of it.
     
  2. jcsd
  3. Dec 14, 2015 #2

    HallsofIvy

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    The "[itex]a_1[/itex]" on the left is not really important. What is important is that the right side is a geometric series, [itex]\sum_{i=0}^n ar^i[/itex] with [itex]a= a_1- a_0[/itex] and common factor [itex]-\frac{1}{2}[/itex]. After you have taken that sum, then add [itex]a_1[/itex] to both sides.
     
  4. Dec 14, 2015 #3

    BvU

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    But you want it in terms of a1 and a0 , not a2 etc.
    So:
    $$a_0, \ a_1, \ {a_0 + a_1 \over 2}, \ { a_1 + {a_0 + a_1 \over 2} \over 2} , \ {{a_0 + a_1 \over 2} + \ { a_1 + {a_0 + a_1 \over 2} \over 2} \over 2 }. \ ... $$and that you can easily generalize, which is exactly what is asked from you.

    :smile: I see Ivy has already spoiled the thing for you. Pity. :frown:
     
  5. Dec 14, 2015 #4
    Well, i think i understand the idea, but the fact that sings are switching is confusing me, i mean, when i have something like this: ## \sum\limits_{k=1}^n \frac{1}{2k}## (which is almost exactly the same as in this case, except for the (-1)k part) then it's simple:
    ##
    \sum\limits_{k=1}^{n-1} \frac{1}{2k} = \frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}## So, what can i do about that ##(-1)^k## ?
     
  6. Dec 14, 2015 #5

    micromass

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    Do you know matrix diagonalization?
     
  7. Dec 14, 2015 #6
    Matrix diagonalization? I know that if i have upper or lower triangular matrix that determinant of such matrix is product of elements on the diagonal, if that's what you meant.
     
  8. Dec 14, 2015 #7

    Ray Vickson

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    The solution of the recurrence can be obtained using a standard method for such a "constant coefficient" problem. If we assume a solution of the form ##a_n = c r^n## (with constant ##c##), then the recursion ##a_{n+2} = (a_{n+1} + a_n)/2## yields ##r^n \,r^2 = r^n \,(1+r)/2 ##, hence ##r^2 = (1+r)/2## and so ##r = 1,-1/2##. That implies ##a_n = A 1^n + B(-1/2)^n = A + B\,(-1/2)^n##. Find ##A## and ##B## by using the given conditions at ##n = 0## and ##n = 1##.

    Note: you can easily check that for any constants ##A,B## the form ##a_n = A + B \,(-1/2)^n## satisfies the recursion.
     
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