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Thread: Epsilon/Delta.

  1. #1
    Junior Member pearlyc's Avatar
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    Epsilon/Delta.

    Just needed a double check on my working to see if I am on the right track!

    Prove that $\displaystyle f(x) = x^4$ is continuous at $\displaystyle x = 0$.

    $\displaystyle | f(x) - f(0) | = | x^4 |$

    Given $\displaystyle \epsilon > 0$, we have $\displaystyle | f(x) - f(0)| < \epsilon$, provided that $\displaystyle | x^4 | < \epsilon$

    Therefore, $\displaystyle \delta = \epsilon$.

    Is this correct?

    And I have another question about continuity!

    Let $\displaystyle {Q}$ denotes the set of rational numbers, consider the function

    $\displaystyle g(x)=\left\{\begin{array}{cc}sin x, &\mbox{ } x \in {Q}\\x, &\mbox{ } x \notin {Q}\end{array}\right.$

    (i) Prove that g(x) is continuous at x=0.
    (ii) State the least upper bound for g on the interval [0,1] and state whether g attains a maximum on [0,1]. Give clear explanations to justify your statements.
    Last edited by pearlyc; Aug 12th 2008 at 02:30 AM.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    Just needed a double check on my working to see if I am on the right track!

    Prove that $\displaystyle f(x) = x^4$ is continuous at $\displaystyle x = 0$.

    $\displaystyle | f(x) - f(0) | = | x^4 |$

    Given [math\epsilon > 0[/tex], we have $\displaystyle | f(x) - f(0)| < \epsilon$, provided that $\displaystyle | x^4 | < \epsilon$

    Therefore, $\displaystyle \delta = \epsilon$.

    Is this correct?
    For a given $\displaystyle \epsilon > 0$, you have to obtain a $\displaystyle \delta$ such that whenever $\displaystyle |x - 0| < \delta, |f(x) - f(0)| < \epsilon$

    And you have claimed that $\displaystyle \delta = \epsilon$ will work for all epsilon...

    So check if your choice works for $\displaystyle \epsilon = 3$.

    Hints:

    Hint 1: So does $\displaystyle |x| < 3 \implies |x^4| < 3$?

    Hint 2: What about x=2?

    Think about it again. You have to break your argument into two parts. For one part choosing $\displaystyle \delta = \epsilon$ will work. Figure out the other part and we can continue.
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  3. #3
    Junior Member pearlyc's Avatar
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    I guess it does not satisfy all epsilons :\ But I don't know what went wrong. I am really confused with this whole epsilon/delta business but I really want to know how to nail it as I know it's a very fundamental concept to understand if I wanna further my degree in Math.

    Is there like a guide or something :\?
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  4. #4
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    Because $\displaystyle \left| x \right| < 1\quad \Rightarrow \quad x^4 < 1$
    so you want to make sure that $\displaystyle \delta = \min \left\{ {1,\varepsilon } \right\}$.
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  5. #5
    Lord of certain Rings
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    To elaborate on:
    Quote Originally Posted by Isomorphism View Post
    For a given $\displaystyle \epsilon > 0$, you have to obtain a $\displaystyle \delta$ such that whenever $\displaystyle |x - 0| < \delta, |f(x) - f(0)| < \epsilon$

    And you have claimed that $\displaystyle \delta = \epsilon$ will work for all epsilon...

    So check if your choice works for $\displaystyle \epsilon = 3$.

    Hints:

    Hint 1: So does $\displaystyle |x| < 3 \implies |x^4| < 3$?

    Hint 2: What about x=2?

    Think about it again. You have to break your argument into two parts. For one part choosing $\displaystyle \delta = \epsilon$ will work. Figure out the other part and we can continue.
    Observe that for $\displaystyle \epsilon < 1, \epsilon ^4 < 1$ and thus choosing $\displaystyle \delta = \epsilon$ will work.

    However when $\displaystyle \epsilon \ge 1$, the above argument fails.

    To understand what we need to do with this specific case, lets work with a concrete example. How will you choose $\displaystyle \delta$ if i give $\displaystyle \epsilon = 3$?

    I want a $\displaystyle \delta$ such that $\displaystyle |x| < \delta \implies |x^4| < 3 $.

    Note that it suffices if I find x's such that $\displaystyle |x^4| < 1$, since 1 < 3.

    But for all $\displaystyle |x| < 1, |x^4| < 1$. Thus if

    a) $\displaystyle \epsilon < 1$ then choose $\displaystyle \delta = \epsilon$

    b) $\displaystyle \epsilon \ge 1$ then choose $\displaystyle \delta = 1$

    You are actually done with the problem. But generally there is a tricky way of combining both parts and writing $\displaystyle \delta = \text{min}\{1,\epsilon\}$.

    NOTE:If fourth root is defined to you, you dont need all this circus. The problem is straightforward then.
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  6. #6
    Junior Member pearlyc's Avatar
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    therefore i just have to change my final part of the working to $\displaystyle \delta = \text{min}\{1,\epsilon\}$?

    how about the 2nd question? i don't know where to start. i reckon the approach is almost similar. but what is L in this case?
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  7. #7
    Junior Member pearlyc's Avatar
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    wait, should i let $\displaystyle \delta = \epsilon^{\frac{1}{4}}$ instead?
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by pearlyc View Post
    wait, should i let $\displaystyle \delta = \epsilon^{\frac{1}{4}}$ instead?
    yes, that is better..
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  9. #9
    Junior Member pearlyc's Avatar
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    would the answer still be the same ?

    or would it be $\displaystyle \delta = min \{1,\epsilon^{\frac{1}{4}}\}$?
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  10. #10
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by pearlyc View Post
    would the answer still be the same ?

    or would it be $\displaystyle \delta = min \{1,\epsilon^{\frac{1}{4}}\}$?
    no, it is just the same, it can be $\displaystyle \delta = \epsilon^{\frac{1}{4}}$ since you can take any positive epsilon.

    for example, if you take $\displaystyle \epsilon = 16$, taking $\displaystyle \delta = \epsilon^{\frac{1}{4}} = 2$ such that if $\displaystyle |x| < \delta=2$, then $\displaystyle |x^4| < \delta^4 =16=\epsilon$

    the more important thing is, you should be able to come up with $\displaystyle |f(x) - f(a)| < \epsilon$

    unlike if you only take $\displaystyle \delta = \epsilon = 16$, if $\displaystyle |x| < \delta=16$, then $\displaystyle |x^4| < \delta^4 =16^4$ and after that, you are not sure whether it will still be less that $\displaystyle 16$ which is your epsilon.
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  11. #11
    Junior Member pearlyc's Avatar
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    okay. so my working above is sufficient? all i have to do is just change the delta value right? how about the 2nd question? how do i start with it?
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