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How to solve convergences when dealing with series

  1. Jan 18, 2008 #1
    1. The problem statement, all variables and given/known data

    shows that the sequence [tex]U(n+1) = \sqrt{2U(n) + 5}[/tex] where U(1) = 3 converges to a limit, u, and find the value of u correct to 2 decimal places.

    2. Relevant equations



    3. The attempt at a solution

    I only know how to solve convergences when dealing with series lol.

    Any help please :)

    thanks
     
  2. jcsd
  3. Jan 18, 2008 #2

    Dick

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    If the sequence does have a limit L, then the limit must satisfy L=sqrt(2L+5). Do you see why? As for the proof part can you show the sequence is increasing and bounded?
     
  4. Jan 18, 2008 #3
    ooooo

    yeah I see why L=sqrt(2L+5).

    I still can't go anywhere from here though :S
     
  5. Jan 18, 2008 #4

    Dick

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    Ok. Take the function f(x)=sqrt(2x+5). You get the elements in the sequence by finding f(3), f(f(3)), f(f(f(3))) etc. To show that f(x)>x (that the sequence is increasing), you want to show f(x)-x>0. For what values of x is this true?
     
    Last edited: Jan 18, 2008
  6. Jan 18, 2008 #5

    HallsofIvy

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    You can't solve that for L? L2= 2L+ 5 so L2- 2L= 5. You can solve that by completing the square: L2- 2L+ 1= 5+ 1= 6. (L- 1)2= 6 so [itex]L- 1= \pm\sqrt{6}[/itex] and [itex]L=1\pm\sqrt{6}[/itex]. Here, since the sequence starts at 3 and increases, [itex]L= 1+ \sqrt{6}[/itex]. Of course, we need to prove that the sequnce is increasing and bounded above so it will converge.

    Prove U(n)< U(n+1) for all n, by induction. If n= 1, then U(1)= 3 while U(2)= [itex]2\sqrt{2U(1)+5}= \sqrt{6+ 5}=\sqrt{11}> 3[/itex] (because 11> 9).
    Now, suppose for some k, U(k+1)> U(k). Then U((k+1)+ 1)= [itex]\sqrt{2U(k+1)+ 5}> \sqrt{2U(k)+ 5}= U(k+1)[/itex].

    You will also need to prove that the sequence is bounded above. Since (if the sequence has a limit) the limit is [itex]1+ \sqrt{6}[/itex] which is approximately 3.5so you might try proving, again by induction, that U(n)< 4 for all n.
     
  7. Jan 19, 2008 #6
    o sick

    thanks for help guys :D

    much appreciated

    I see it now :)
     
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