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Math Help - 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 f(x) = x^4 is continuous at x = 0.

    | f(x) - f(0) | = | x^4 |

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

    Therefore, \delta = \epsilon.

    Is this correct?

    And I have another question about continuity!

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

    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; August 12th 2008 at 02:30 AM.
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  2. #2
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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 f(x) = x^4 is continuous at x = 0.

    | f(x) - f(0) | = | x^4 |

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

    Therefore, \delta = \epsilon.

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

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

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

    Hints:

    Hint 1: So does |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 \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 \left| x \right| < 1\quad  \Rightarrow \quad x^4  < 1
    so you want to make sure that \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 \epsilon > 0, you have to obtain a \delta such that whenever |x - 0| < \delta, |f(x) - f(0)| < \epsilon

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

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

    Hints:

    Hint 1: So does |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 \delta = \epsilon will work. Figure out the other part and we can continue.
    Observe that for \epsilon < 1, \epsilon ^4 < 1 and thus choosing \delta = \epsilon will work.

    However when \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 \delta if i give \epsilon = 3?

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

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

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

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

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

    You are actually done with the problem. But generally there is a tricky way of combining both parts and writing \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 \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 \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 \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 \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 \delta = min \{1,\epsilon^{\frac{1}{4}}\}?
    no, it is just the same, it can be \delta = \epsilon^{\frac{1}{4}} since you can take any positive epsilon.

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

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

    unlike if you only take \delta = \epsilon = 16, if |x| < \delta=16, then |x^4| < \delta^4 =16^4 and after that, you are not sure whether it will still be less that 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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