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Math Help - mathematical analysis problem. please help.

  1. #1
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    mathematical analysis problem. please help.

    suppose {x(n)} is a sequence such that
    x(1)=1 and
    x(n+1)=1+x(n)/(1+x(n+1)) ; n>=1

    Prove that this sequence converges and find the limit.


    I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
    Last edited by Kat-M; January 1st 2009 at 07:07 PM.
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    suppose {x(n)} is a sequence such that
    x(1)=1 and
    x(n+1)=1+x(n)/(1+x(n+1)) ; n>=1

    Prove that this sequence converges and find the limit.


    I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
    First, multiplying the series by 1 + x_{n+1} gives

    x_{n+1} (1+x_{n+1})=1 + x_{n+1}+x_n or x^2_{n+1} =1 + x_n

    To prove that the sequence converges we will prove two things

    (i) the sequence is increasing and

    (ii) it is bounded above


    First off, I think it's clear that x_n > 0

    (i) writing out the first few terms suggests its increasing. To prove this assume that

     x_{k} < x_{k+1}

    thus,

     x_{k} + 1 < x_{k+1} + 1\;\;\; \Rightarrow\;\;\;x^2_{k+1} < x^2_{k+2}

    so

     x_{k+1} < x_{k+2}

    and by induction,  x_{n} < x_{n+1}\;\; \forall n

    (ii) It is bounded by 2

    The first few terms suggest this. Assume that

     x_{k} < 2

    then

     x_{k} + 1 < 2 + 1 = 3 < 4

    thus,

     x^2_{k+1} < 4\;\;\; \Rightarrow \;\;x_{k+1} < 2

    so again by induction, it is ture for all n. Since the sequence is increasing and bounded above, it must converge.
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  3. #3
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    Here's one very similar that I think can be solved exactly

    2 x^2_{n+1} = x_n + 1,\;\;\;x_0 = 0.

    Any ideas?
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    thanks

    i made a typo in the original question and i fixed it which is posted below.
    Last edited by Kat-M; January 1st 2009 at 07:07 PM.
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    oops i had a typo in the question.

    here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).

    suppose {x(n)} is a sequence such that
    x(1)=1 and
    x(n+1)=1+x(n)/(1+x(n)) ; n>=1

    Prove that this sequence converges and find the limit.


    I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
    Last edited by Kat-M; January 1st 2009 at 07:10 PM.
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  6. #6
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    Quote Originally Posted by Kat-M View Post
    here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).
    I wondered about that! What you had originally seemed a strange way to write the formula.
    suppose {x(n)} is a sequence such that
    x(1)=1 and
    x(n+1)=1+x(n)/(1+x(n)) ; n>=1

    Prove that this sequence converges and find the limit.


    I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
    x(n+1)= 1+ x(n)/(1+ x(n)) which we can rewrite as [tex]\frac{1+ 2x(n)}{1+ x(n)}= 2- 1/(1+ x(n)).

    Assuming that the limit exists and is "x", we must have lim x(n+1)= 2- 1/(1+ lim x(n) or x= 2- 1/(x+1). Multiplying on both sides by x+1, x^2+ x= 2x+ 2- 1 or x^2- x- 1= 0. The quadratic formula gives two possible values for x, \frac{1- \sqrt{5}}{2} and \frac{1+\sqrt{5}}{2}. Since x(n) is positive for all x, the limit must be \frac{1+ \sqrt{5}}{2}. I presume that is what you got.

    That is assuming the limit exists. We still need to prove that. Since x(1)= 1 is less than \frac{1+ \sqrt{5}}{2}, the sequence has to increase to it and it seems reasonable to show that the sequence is increasing and has an upper bound- that is sufficient to show that it converges.

    So, first we want to prove that the sequence is increasing: that x(n+1)> x(n) for all n. x(n+1)= 2- x(n)/(1+ x(n))> x(n), multiplying on both sides by the positive number 1+ x(n), is equivalent to 2+ 2x(n)- x(n)> x(n)^2+ x(n) or x(n)^2- x(n)- 1< 0. We have already seen that x^2- x- 1= 0 only for x= \frac{1- \sqrt{5}}{2} and x= \frac{1+ \sqrt{5}}{2}. If x= 0, which is between those two values, then (0)^2- 0 - 1= -1< 0 so x^2- x- 1< 0 for all x between those values. x(n) is positive for all n so if we can prove that x(n)< \frac{1+ \sqrt{5}}{2} for all n, we will have proved both that the sequence is increasing and that it has an upper bound and so converges.

    Let's try induction. Clearly x(1)= 1< \frac{1+\sqrt{5}}{2}, which is about 1.47. Suppose x(k)< \frac{1+\sqrt{5}}{2}
    Then x(k)+ 1< \frac{3+ \sqrt{5}}{2}, \frac{1}{x(k)+1}> \frac{2}{3+ \sqrt{5}}, and x(k+1)= 2- 1/(x(k)+1)< 2- \frac{2}{3+ \sqrt{5}}= \frac{4+\sqrt{5}}{3+ \sqrt{5}}

    Rationalizing the denominator, we have x(n+1)< \frac{4+ 2\sqrt{5}}{3+ \sqrt{5}}\frac{3-\sqrt{5}}{3-\sqrt{5}}= \frac{2+ 2\sqrt{5}}{4}= \frac{1+ \sqrt{5}}{2}, exactly what was needed.

    That is, since x(n) lies between \frac{1- \sqrt{5}}{2} and \frac{1+\sqrt{5}}{2}, it is, as we saw above, an increasing sequence. Further, since it has \frac{1+ \sqrt{5}}{2} as upper bound, and is increasing, it is a convergent sequence.
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  7. #7
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    Quote Originally Posted by Kat-M View Post
    here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).

    suppose {x(n)} is a sequence such that
    x(1)=1 and
    x(n+1)=1+x(n)/(1+x(n)) ; n>=1

    Prove that this sequence converges and find the limit.


    I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
    Another way is to solve your problem directly. The substitution

    x_n = \frac{a y_n}{y_n + 1}

    for suitable a (guess what number see HallsofIvy reply) the difference equation will become a linear difference equation which can be solved exactly.
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  8. #8
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    Thank you very much

    HollsofIvy and danny arrigo. you two helped me a lot. Thank you so much.
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