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Lower Bound of Sequence

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    08112012_54235_0.png
    2. Relevant equations
    3. The attempt at a solution
    This is what I have so far:
    [tex]x_{n+1}=\frac{x^5_n + 1}{5x_n}=1[/tex]

    [tex]x_{n+2}=\frac{x^5_{n+1} + 1}{5x_{n+1}}=\frac{1^5+1}{5}[/tex]

    I think you have to do some sort of repeated substitution but I don't quite see it. Any help? Thanks.
     
  2. jcsd
  3. Nov 14, 2012 #2

    HallsofIvy

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    Why would this be equal to 1? As far as you know to start with, only x1 is equal to 1 and then
    [itex]x_2= 2/5[/itex], [itex]x_3= (32/3125+ 1)/2= 1580/3125= 316/625'[/itex] which is about 0.5056. x1 is the only term that is equal to 1.

    Use induction. You already know that [itex]x_1= 1> 3/11[/itex]. Show that if [itex]x_n> 3/11[/itex] then [itex]x_{n+1}= (x_n^5+ 1)/5x_n> 3/11[/itex].
     
  4. Nov 14, 2012 #3
    I meant to say [tex]x_{1+1}=\frac{x^5_1 + 1}{5x_1}=1[/tex]
    Since the first term of the sequence on the left is [tex]x_{((1)+1)}[/tex]

    won't it be [itex]x_2=1[/itex] and not [itex]x_1=1[/itex]?

    Thanks.
     
  5. Nov 14, 2012 #4

    haruspex

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    No, the sequence is defined from n=1, so the first term is x1. You are confusing that with the first opportunity to use the recurrence relation.
     
  6. Nov 14, 2012 #5
    The notation is confusing me. The problem is equating two different sequences, right? And the sequence on the left is defined as xn+1 from n=1.

    So for n=1: xn+1 = x(1)+1 = x2

    Isn't this the first term?
     
    Last edited: Nov 14, 2012
  7. Nov 14, 2012 #6
    Never mind; silly me. It is the definition of a recursive sequence.

    I don't understand how exactly to prove by induction. Can I do the following?

    [tex]x_{n+1}> ((3/11)^5+ 1)/5(3/11)= 0.74> 3/11[/tex]
     
  8. Nov 14, 2012 #7

    haruspex

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    No, that's not valid. What would happen to the bound on xn+1 if you made xn a bit larger? The numerator is ok, that would just boost xn+1 a bit more, but the denominator goes up too, and it's not obvious what the net effect is.
    You need to show that x > 3/11 implies (x5+ 1)/5x > ((3/11)5+ 1)/5(3/11)
     
  9. Nov 14, 2012 #8
    Unfortunately, I don't understand why that is implied.

    Is this a valid solution?

    [tex]x_n = \frac{x_{n-1}^5 +1}{5x_{n-1}}\\
    f(x) = \frac{x^5+1}{5x}
    [/tex]

    Find the minimum using the first derivative.

    [tex]f '(x)=\frac{4x^5-1}{5x^2}\\
    f '(x)=0\\
    x=(\frac{1}{4})^\frac{1}{5}\\

    f(x) = \frac{((\frac{1}{4})^\frac{1}{5})^5+1}{5((\frac{1}{4})^\frac{1}{5})} \approx 0.33 \\

    \frac{3}{11} \approx 0.273
    [/tex]

    The lowest possible value is greater than 3/11.
     
  10. Nov 14, 2012 #9

    haruspex

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    That proof is valid, but I suspect it is not in the spirit of the question. I'll give it some thought.
     
  11. Nov 14, 2012 #10
    Thank you. That's what I thought too but at least now I've solved it. :) I'd love to find another way to solve it but I don't seem to understand your solution. :( I will look at it again.
     
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