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Math Help - Epsilon-delta problem

  1. #1
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    Epsilon-delta problem

    Information:
    Given f: [-\sqrt\pi, \sqrt\pi] \rightarrow [-1,1] defined by f(x)= sin(x^2)


    The problem:

    Find a \delta so that |x - y| \leq \delta implicates that |f(x) - f(y)| \leq 0.1 for all x and y in  [-\sqrt\pi, \sqrt\pi]


    I was told that the mean value theorem can be used to find our \delta. Don't know how though. Any ideas on how to proceed?
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  2. #2
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    You are correct about using the mean value theorem.
    From that we know that \left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5 for  - \sqrt \pi   \leqslant a < c < b \leqslant \sqrt \pi  .

    So what is \delta?
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  3. #3
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    Quote Originally Posted by Plato View Post
    From that we know that \left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5
    Where do you get the "5" from? |{f(b) - f(a)}| = |sin(b^2) - sin(a^2)| which is less or equal to 2 (?)
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  4. #4
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    Sorry it should have been \le 5|b-a|.
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  5. #5
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    So that:

    \left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5 |b-a|?


    That would have to mean that: |2c\cos(c^2)| \leqslant 5

    cos(c^2) is always \leqslant 1. And c is a value between -\sqrt\pi and \sqrt\pi. But how is all of this less than, or equal to 5?
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  6. #6
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    Graph \left| {2x\cos (x^2 )} \right| on \left[ { - \sqrt \pi  ,\sqrt \pi  } \right].

    You will see that 5 is good bound.
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  7. #7
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    This is what I did. Please correct any mistakes:


    Given f: [-\sqrt\pi, \sqrt\pi] \rightarrow [-1,1] defined by f(x)= sin(x^2).

    We want to find a \delta so that |x-y| \leq \delta implicates that |f(x) - f(y)| \leq 0.1 for all x and y in [-\sqrt\pi, \sqrt\pi]




    The mean value theorem says that since f is countinous for [-\sqrt\pi, \sqrt\pi], we have:

    \left| {f(x) - f(y)} \right| = \left| {x - y} \right|\left| f'(c) \right|
    where |f'(c)| = |2c cos(c^2)|


    By graphing |2c cos(c^2)| we see that this (absolute) value is never greater than 4 (or 5, or 6...). We now have that:

    | {f(x) - f(y)} | < 4| {x - y} |


    Let's say that \delta = \epsilon / 4, where we have been told that \epsilon = 0.1. Since |x-y| \leq \delta, then |x-y| \leq \epsilon / 4 = 0.1/4.


    So finally:
    | {f(x) - f(y)} | < 4| {x - y} | \leq 4 * (\0.1/4) = 0.1 = \epsilon



    Therefore \delta = 0.1/4 = 1/40 is a value for \delta that makes |x-y| \leq \delta
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  8. #8
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    Quote Originally Posted by jenkki View Post
    So that:

    \left| {f(b) - f(a)} \right| = \left| {b - a} \right|\left| {2c\cos (c^2 )} \right| \leqslant 5 |b-a|?


    That would have to mean that: |2c\cos(c^2)| \leqslant 5

    cos(c^2) is always \leqslant 1. And c is a value between -\sqrt\pi and \sqrt\pi. But how is all of this less than, or equal to 5?
    Yes, -1\le cos(c^2)\le 1 so -2c\le 2c cos(c^2)\le 2c. Since c\le \sqrt{\pi}, 2c\le 2\sqrt{\pi} which is about 2\sqrt{3.14} or 3.5. 4 could have been used but any number that is less than 3.5... is certainly less than 5. I'm not sure why Plato used "5" instead of "4" but it is valid.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    I'm not sure why Plato used "5" instead of "4" but it is valid.
    Because in the original post it required |f(x)-f(y)|<0.1.
    The 5 just makes for nice numbers.
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  10. #10
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    Quote Originally Posted by jenkki View Post
    This is what I did. Please correct any mistakes:
    No one? I'd really like to know if I've got the right answer
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