Results 1 to 13 of 13

Math Help - Real Analysis (Sequences)

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    65

    Real Analysis (Sequences)

    Here is the problem. How would I prove this problem?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Is it clear that \sqrt{x_n}-\sqrt{a}=\dfrac{x_n-a}{\sqrt{x_n}+\sqrt{a}}?

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    Do I go with a direct proof?

    like:

    Given \epsilon > 0

    Then |\sqrt(x_n)-\sqrt(a)| < \epsilon

    Thus |{\frac{x_n-a}{\sqrt(x_n)+\sqrt(a)}| < \epsilon

    but what do I do next? Do I split it up into two?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Quote Originally Posted by Phyxius117 View Post
    Given \epsilon > 0
    Then |\sqrt(x_n)-\sqrt(a)| < \epsilon
    Thus |{\frac{x_n-a}{\sqrt(x_n)+\sqrt(a)}| < \epsilon
    That is pure wishful thinking.
    First show that \left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow x_n  > \frac{a}{2}} \right]

    From that you show that
    \frac{1}{{\sqrt {x_n }  + \sqrt a }} < \frac{{\sqrt 2 }}<br />
{{\sqrt a  + \sqrt {2a} }}

    You have a lot of work to do.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    I don't quite understand. how did you get x_n > a/2 ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Quote Originally Posted by Phyxius117 View Post
    I don't quite understand. how did you get x_n > a/2 ?
    I did not prove that!
    I said that you must prove that as part of this problem.

    Look we are not going to do this for you.
    It is your problem. You do it.
    I told you how it is done.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    Thank you for all the help.

    after looking at your hints on how to approach this problem, i'm trying hard to figure out how to solve it but I just don't understand what I am suppose to do.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    I happened on this problem and I was wondering if the following reasoning is enough to prove that x_n>\frac{a}{2}.

    If x_n is monotonically increasing or non-decreasing then it must eventually pass \frac{a}{2} in order to get to a. If it is monotonically decreasing or non-incresing then all values of x_n must be greater then a. If it is none of those we can still choose a monotonic subsequence because all sequences have a monotonic subsequence and apply the logic previously stated.

    Is this enough just for x_n>\frac{a}{2}?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Quote Originally Posted by Plato View Post
    That is pure wishful thinking.
    First show that \left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow x_n  > \frac{a}{2}} \right]

    From that you show that
    \frac{1}{{\sqrt {x_n }  + \sqrt a }} < \frac{{\sqrt 2 }}<br />
{{\sqrt a  + \sqrt {2a} }}

    You have a lot of work to do.
    Hi. I like these assignments, they can be challenging. I am curios Plato if your first hint can be proved like this:

    \lim x_n =a \Rightarrow \forall \epsilon >0, \exists n_\epsilon such that  \forall n>n_\epsilon is |x_n-a|<\epsilon. Setting \epsilon =\frac{a}{2} then set N=n_\epsilon=n_{\frac{a}{2}} and for all n>N we'd have |x_n-a|<\epsilon=\frac{a}{2}.

    Now we get
    |x_n-a|<\frac{a}{2},
    -\frac{a}{2}<x_n-a<\frac{a}{2}, (add a to all)
    \frac{a}{2}<x_n<\frac{3a}{2}, and the left part of this extended inequality gives us x_n>\frac{a}{2}.

    I guess this is ok, Plato can you respond. And am trying to get the second part. Still, I don't see were this leads. Thx in advance.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Continuing on my previous post. If my getting to x_n>\frac{a}{2} is ok then:

    \sqrt{x_n}>\sqrt{\frac{a}{2}}, (now add \sqrt{a})
    \sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a}, (compare the reciprocals)
    \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt  {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\  sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr  t{2a}}

    So we get the \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}}{\sqr  t{a}+\sqrt{2a}}. Now what?

    EDIT: see my next post, I almost got it
    Last edited by MathoMan; September 20th 2010 at 03:18 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    I thought I should put it all in one place.

    \lim x_n =a \Rightarrow \forall \epsilon >0, \exists n_\epsilon such that  \forall n>n_\epsilon is |x_n-a|<\epsilon. Setting \epsilon =\frac{a}{2} then set N=n_\epsilon=n_{\frac{a}{2}} and for all n>N we'd have |x_n-a|<\epsilon=\frac{a}{2}.

    Now we get
    |x_n-a|<\frac{a}{2},
    -\frac{a}{2}<x_n-a<\frac{a}{2}, (add a to all)
    \frac{a}{2}<x_n<\frac{3a}{2}, and the left part of this extended inequality gives us x_n>\frac{a}{2}.
    Then,
    \sqrt{x_n}>\sqrt{\frac{a}{2}}, (now add \sqrt{a})
    \sqrt{x_n}+\sqrt{a}>\sqrt{\frac{a}{2}}+\sqrt{a}, (compare the reciprocals)
    \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{1}{\frac{\sqrt  {a}}{\sqrt{2}}+\sqrt{a}}=\frac{1}{\frac{\sqrt{a}+\  sqrt{2a}}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{a}+\sqr  t{2a}}

    So we get the \frac{1}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}}{\sqr  t{a}+\sqrt{2a}}.

    If we were to multiply that with x_n-a
    finally I see why \sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}} comes in handy.

    \sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}

    Forcing limit on the above should lead towards solution. Well almost. Strict inequality is bothering me here. Anyway Physix117 that started the thread should work out the fine details. I gave my shot at the rough stuff and I think I did just fine. Waiting for feedback.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,977
    Thanks
    1643
    Physix117, one problem with people not showing any work in their first post is that there are often many different ways to solve a problem, of differing difficulty and mathematical sophistication- and we have no way of knowing which is appropriate for you.

    Plato chose to show you the most basic- and therefore, most difficult, method, assuming the least he could about your knowledge- that if you are asked to find limits, you at least know the definition of "limit". If I were doing a problem like this, I think I would just observe that \sqrt{x}, for x> 0, is a continuous function and so \limit_{n\to\infty} \sqrt{x_n}= \sqrt{\lim_{n\to\infty} x_n}= \sqrt{a}.

    Mathoman, "strict inequality" should not bother you. If a_n< B for all n, it is NOT true that \lim_{n\to\infty} a_n< B. What is true is that \lim_{n\to\infty}\le B.

    Also, x_n> \frac{a}{2} is not always true nor is it claimed to be. For "sufficiently large n", x_n> \frac{a}{2}. That is, "there exist a positive integer N such that if n> N, then x_n> \frac{a}{2}". That follows directly from the definition of "limit", taking \epsilon= \frac{a}{2}> 0. Since \lim_{n\to\infty} x_n= a, given any \epsilon> 0 there exist N such that if n> N, then |x_n- a|< \epsilon. Taking \epsilon= \frac{a}{2} that becomes |x_n- a|< \frac{a}{2} which is the same as -\frac{a}{2}< x_n- a< \frac{a}{2}. Adding a to each part, a- \frac{a}{2}= \frac{a}{2}< x_n< a+ \frac{a}{2}= \frac{3a}{2}. Of course, it is only the left part you need here.
    Last edited by HallsofIvy; September 20th 2010 at 05:06 AM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Halls of Ivy and Mathoman. I can see now where plato was leading but there's something that bothers me about the rhs. It involves x_n and I can't get it in a form leading to |\sqrt{x_n}-\sqrt{a}| so I was wondering if something different would work.

    Starting from
    \sqrt{x_n}-\sqrt{a}=\frac{x_n-a}{\sqrt{x_n}+\sqrt{a}}<\frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}

    lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq lim\limits_{n\to\infty} \frac{\sqrt{2}(x_n-a)}{\sqrt{a}+\sqrt{2a}}

    lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq  \frac{\sqrt{2}lim\limits_{n\to\infty}(x_n-a)}{\sqrt{a}+\sqrt{2a}}

    lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq  \frac{0}{\sqrt{a}+\sqrt{2a}}
    lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq  0

    Is this valid and does it help any? I figure that from the original frac{a}{2}<x_n<\frac{3a}{2} I could find a lower bound with a (x_n-a) term in the numerator that would make the limit of it equal 0 and then since 0 \leq lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} \leq  0

    We'd know that  lim\limits_{n\to\infty}\sqrt{x_n}-\sqrt{a} =0

    Is this at all valid?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Real Analysis - Sequences 3
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 16th 2010, 02:18 PM
  2. Real Analysis - Sequences 2
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 15th 2010, 06:43 PM
  3. Real Analysis - Sequences
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 15th 2010, 06:36 PM
  4. Real Analysis - Sequences #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 20th 2009, 02:57 PM
  5. Real Analysis - Sequences #3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 20th 2009, 01:35 PM

Search Tags


/mathhelpforum @mathhelpforum