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Math Help - Epsilon - Delta Proof with Squares

  1. #1
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    Talking Epsilon - Delta Proof with Squares

    Hey i've been working on this problem for a few days now, just seeing if anyone else has some insight.

    ∀x ∈ R, ∀ε ∈ R+, ∃δ ∈ R+, ∀y ∈ R, |x – y| < δ ==> |x^2 – y^2| < ε

    prove using epsilon-delta structured proof

    ive got to the point where

    |x^2 – y^2| < δ|x + y|

    but am not completely stuck, I think I have to replace y with a constant somehow? any help would be great! thanks in advance!
    Last edited by kale3193; February 22nd 2010 at 10:42 PM. Reason: latex error
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  2. #2
    DBA
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    I cannot read that very well. It also would help if you would add the entire question.

    You mentioned that
    |x-y| < \delta and |x^{2}-y^{2}| < \epsilon

    From there I would rewrite
    |x^{2}-y^{2}| < \epsilon
    |(x-y) (x+y)| < \epsilon
    |(x-y) * C| < \epsilon

    Then
    |(x-y)| < \frac{\epsilon}{C} = \delta

    To find C, I would look at:
    if  0< |x-a| < \delta then  |f(x) - L| < \epsilon

     a- \delta < x< a+ \delta
    In this case y would be a

    But to be sure I would need the entire question. If you still need this problem, maybe you can submit the question asked.
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  3. #3
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    Quote Originally Posted by kale3193 View Post
    Hey i've been working on this problem for a few days now, just seeing if anyone else has some insight.

    ∀x ∈ R, ∀ε ∈ R+, ∃δ ∈ R+, ∀y ∈ R, |x – y| < δ ==> |x^2 – y^2| < ε

    prove using epsilon-delta structured proof

    ive got to the point where

    |x^2 – y^2| < δ|x + y|

    but am not completely stuck, I think I have to replace y with a constant somehow? any help would be great! thanks in advance!
    if i am not wrong this is just the prove that f(x) = x^2 is continuous for all x on R.
    if we take delta = min{ |x|/2, 2epsilon/5|x| } for all x from R \ {0} the formula will work

    and for x = 0 we can take delta = sqrt( epsilon )

    you choose x and epsilon first and than you choose delta, so delta must depend just on x and epsilon, not on y
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  4. #4
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    Okay but

    How can you write < epsilon first?
    I thought we had to assume |x-y| < delta first and show that it leads to |x^2 - y^2 | < Epsilon
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  5. #5
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    Quote Originally Posted by kale3193 View Post
    How can you write < epsilon first?
    I thought we had to assume |x-y| < delta first and show that it leads to |x^2 - y^2 | < Epsilon
    i didnt wrote < epsilon first...
    you need to find delta ( based on epsilon and x ) such that the statment in first post is true! Dont know if you learnd about continuosity of functions ( maybe you did, but didnt realise that very good, because this is just an example of that )
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