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Math Help - square root function and convergence

  1. #1
    Member pberardi's Avatar
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    square root function and convergence

    This one is giving me such a hard time. I guess this is important to know because it is saying that the square root of the function translates to the square root of the limit. Similar to mutliplication and addition. The hint they give is to use the fact that if all convergent subsequences have the same limit, then the sequence is convergent. I need help on this one. Thanks.

    Suppose <x_n> --->x for each natural number n, where x_n >= 0. Prove that <sqrt(x_n)> ---> to sqrt(x)
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  2. #2
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    Perhaps this is not what you're looking for, but sqrt is continuous, so the limit of f(x_n)=f(lim x_n). Or are you confused about epsilon delta?
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  3. #3
    Member pberardi's Avatar
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    Quote Originally Posted by nqramjets View Post
    Perhaps this is not what you're looking for, but sqrt is continuous, so the limit of f(x_n)=f(lim x_n). Or are you confused about epsilon delta?
    I have to prove this for all x_n>= 0. I probably have to use a convergence proof or use the squeeze theorem. Any suggestions?
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  4. #4
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    I think I'm confused by what you mean... Do you mean that you have a sequence (x_n)\rightarrow x and you're trying to show that (\sqrt{x_n})\rightarrow \sqrt{x} or something else?
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  5. #5
    Member pberardi's Avatar
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    Quote Originally Posted by nqramjets View Post
    I think I'm confused by what you mean... Do you mean that you have a sequence (x_n)\rightarrow x and you're trying to show that (\sqrt{x_n})\rightarrow \sqrt{x} or something else?
    Exactly that. Nothing else.
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  6. #6
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    This isn't so bad:
    First show that the square root is continuous:
    Let \varepsilon >0, then given some x\in(0,\infty) choose \delta < \varepsilon. Now, consider y\in(0,\infty) such that |x-y|<\delta, then
    |\sqrt{x}-\sqrt{y}|\leq |\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|=|x-y|<\delta<\varepsilon, So square root is continuous.
    Now, by continuity, if x_n\rightarrow x for x_n\in(0,\infty) with x\in(0,\infty) you have
    \lim_{n\rightarrow \infty}\sqrt{x_n}=\sqrt{\lim_{n\rightarrow \infty}x_n}=\sqrt{x}
    as desired. Hope this helps!
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  7. #7
    Member pberardi's Avatar
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    Helps alot. Thanks bud.
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  8. #8
    Super Member Random Variable's Avatar
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    The following proof is valid if x>0. You can prove the case when x=0 separately.


    Assume ( x_{n}) converges to x and that  x_{n} \ge 0

     | \sqrt{x_n} - \sqrt{x} | = \frac { |\sqrt{x_n} - \sqrt{x}||\sqrt{x_n} + \sqrt{x}|}{\sqrt{x_n} + \sqrt{x}}

    = \frac {|x_{n}-x|}{ \sqrt{x_{n}} + \sqrt{x}}

    \le \frac {1}{\sqrt{x}}|x_{n}-x|

    And since lim (x_{n}-x) = 0 , and since  \frac {1}{\sqrt{x}} > 0 , you can conclude that lim  \sqrt{x_{n}} = \sqrt{x}
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  9. #9
    Member pberardi's Avatar
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    This is awesome RandomVariable. Pardon me but how would one begin the case for zero? And also, what does this proof have anything to do with the fact that x_n converges to x?
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  10. #10
    Super Member Random Variable's Avatar
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    And also, what does this proof have anything to do with the fact that x_n converges to x?
    It's used at the end.

    I need to show that  | \sqrt{x_n} - \sqrt{x} | \le k|a_{n}| for some k>0 and for some series a_{n} that converges to zero.
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  11. #11
    Super Member Random Variable's Avatar
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    I think you should be able to prove the the case for when x=0 straight from the definition. And I emphasize "think" because I can't seem to prove it.
    Last edited by Random Variable; June 1st 2009 at 10:41 PM.
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  12. #12
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    I'm pretty sure the proof I posted above actually works just fine for x=0. That affects nothing in the proof. I mean, sqrt is certainly continuous at x=0, so you are good.
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  13. #13
    Super Member Random Variable's Avatar
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    I think I came up with a proof for the case when x=0.

    Assume  x_{n} \ge 0

     (x_{n}) \to 0 \Longrightarrow \forall \epsilon >0, \ \exists N : \forall n>N, \ |x_{n}-0| = |x_{n}| = x_{n} < \epsilon^{2}

    then  |\sqrt{x_{n}} -0| = |\sqrt{x_{n}}| = \sqrt{x_{n}} < \sqrt{ \epsilon^{2}} = \epsilon
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  14. #14
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    You can use the continuous inverse theorem on f(x)=x^2 on [0,\infty) to get continuity for the whole thing. This result follows from the preservation of interval theorem.

    By the way, I think you need to follow the logic on the second proof provided here, because mine actually has a fault in it. I didn't even think about the fact that \sqrt{x}+\sqrt{y} < 1 in which case that inequality isn't true... So yea, go with the mult/div by that expression to fix it. Sorry!
    Last edited by nqramjets; June 2nd 2009 at 02:31 PM. Reason: WRONG!
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  15. #15
    Super Member Random Variable's Avatar
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    So my proof for the case when x=0 is valid?
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