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Math Help - Proving continuity

  1. #1
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    Proving continuity

    Use the given definition of continuity to prove f is continuos at 0.

    Definition: For all >0, there exists a k such that abs(f(x)-f(a))< for all x such at
    abs(x-a)<k

    F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0

    I have trouble understanding the implications of the definition of continuity.

    I have that abs(f(x)-f(0))=abs(f(x)-1) but don't what to do next.
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  2. #2
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    Quote Originally Posted by poirot View Post
    Use the given definition of continuity to prove f is continuos at 0.
    Definition: For all >0, there exists a k such that abs(f(x)-f(a))< for all x such at
    abs(x-a)<k
    F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0
    I have trouble understanding the implications of the definition of continuity.
    I have that abs(f(x)-f(0))=abs(f(x)-1) but don't what to do next.
    You have to consider two cases:
    \left| {F(x) - F(0)} \right| = \left\{ {\begin{array}{rl}   {x^2 ,} & {x > 0}  \\   {-x^3 ,} & {x < 0}  \\ \end{array} } \right..
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  3. #3
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    Can you show me how to do it?
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  4. #4
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    Quote Originally Posted by poirot View Post
    Can you show me how to do it?
    Of course I could. But it is your problem.
    You have shown no effort on your part.
    Now what do you know about the \epsilon/\delta method of proof?
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  5. #5
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    Nothing, I haven't encountered it before.
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  6. #6
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    Quote Originally Posted by poirot View Post
    Nothing, I haven't encountered it before.
    How can you say with any sincerity?
    Quote Originally Posted by poirot View Post
    Use the given definition of continuity to prove f is continuos at 0. Definition: For all >0, there exists a k such that abs(f(x)-f(a))< for all x such at
    abs(x-a)<k
    F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0
    That is what this entire question is about. You posted it.
    Did you post a question that you know absolutely nothing about?
    If so, that is exceedingly odd.
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  7. #7
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    I know the definition but I don't really get it and don't know how to apply it to questions. I meant I have not done a question like this from first principles. So if you could post half of the method I will probably get it. I would prefer if you posted it all but you seem averse to that.
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  8. #8
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    Quote Originally Posted by poirot View Post
    I know the definition but I don't really get it and don't know how to apply it to questions. I meant I have not done a question like this from first principles. So if you could post half of the method I will probably get it. I would prefer if you posted it all but you seem averse to that.
    If that is the truth, the show us.
    Suppose \varepsilon  > 0 then let \delta  = \min \left\{ {1,\varepsilon } \right\}.
    Now YOU show that \left| {x - 0} \right| < \delta \, \Rightarrow \,\left| {F(x) - 1} \right| < \varepsilon .
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  9. #9
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    I don't really understand what the definition intuitively means. An explanation would be welcomed.

    In post 2,how have got that equation?

    Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)

    abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.

    Probably total rubbish.
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  10. #10
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    Quote Originally Posted by poirot View Post
    I don't really understand what the definition intuitively means. An explanation would be welcomed.
    In post 2,how have got that equation?
    Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)
    abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.
    Probably total rubbish.
    Yes, it is rubbish.
    Over and out
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  11. #11
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    Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)

    abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.
    I think this is pretty correct. I would expand it a little as follows. Let δ (it is called delta) be min(1,ε) and suppose that 0 < -x < δ. Then |f(x) - 1| = |x^3| = (-x)^3 < δ^3 <= δ (since δ <= 1) <= ε (since δ = min(1,ε)).

    The case x >= 0 is very similar but simpler since |x| = x.

    I don't really understand what the definition intuitively means.
    Looks like you need a tutorial on epsilon-delta. I don't have good links, but this one I found just now seems all right. Look especially at this intuitive explanation using Flash.
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  12. #12
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    eminently more helpful. How did you know to set delta to be min(1,epsilon)?
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  13. #13
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    In this example, this was Plato's suggestion. The intuition is that we must make x^2 to be < ε when x < δ. Since δ^2 < δ when 0 < δ < 1, it is sufficient to choose such δ that both δ < 1 and δ < ε hold. It is also possible to choose \delta = \min(\sqrt{\varepsilon},\sqrt[3]{\varepsilon}). And of course, any smaller value of δ also works, such as min(1,ε) / 2.

    I general, there is no algorithm for choosing δ given ε; it completely depends on the function. The definition of the continuity just requires that for every ε some δ exists. When proving continuity from the definition, one can use any means to find a δ. One becomes better at this with practice. It is even possible to prove the existence of δ by contradiction instead of coming up with a specific example, i.e., to show that the assumption that no suitable δ exists is absurd.
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