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Math Help - The limit of R_n as n-->inf is...

  1. #1
    Member dokrbb's Avatar
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    The limit of R_n as n-->inf is...

    It seems that I have some issues with the sums...

    I have the function  f(x) = \sqrt{4 - x^2}

    I need to evaluate it \displaystyle \int_0^2 (\sqrt{4 - x^2})dx by using the definition of Riemann sum for R_n - right end-points,

     \sum_{n=1}^{n} (\sqrt{4 - x^2})

    \delta x = \frac{2-0}{n} = \frac{2}{n}

    then,

     a) R_n = \sum_{n=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =
     b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =
     c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =
     d) R_n = (\frac{2}{n})(2n) \sum_{n=1}^{n} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} = ...

    I'm not sure I did right so far, and how should I proceed further...

    thanks for hints,
    dokrbb
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  2. #2
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by dokrbb View Post
    It seems that I have some issues with the sums...

    I have the function  f(x) = \sqrt{4 - x^2}

    I need to evaluate it \displaystyle \int_0^2 (\sqrt{4 - x^2})dx by using the definition of Riemann sum for R_n - right end-points,

     \sum_{n=1}^{n} (\sqrt{4 - x^2})

    \delta x = \frac{2-0}{n} = \frac{2}{n}

    then,

     a) R_n = \sum_{n=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =
    The sum is over i not n.

     b) R_n = (\frac{2}{n}) \sum_{i=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =
     c) R_n = (\frac{2}{n}) \sum_{i=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =
     d) R_n = (\frac{2}{n})(2n) \sum_{n=1}^{n} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} = ...
    At this point you have already done the sum. You should not have summation sign.

    I'm not sure I did right so far, and how should I proceed further...

    thanks for hints,
    dokrbb
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  3. #3
    Member dokrbb's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by HallsofIvy View Post
    The sum is over i not n.


    At this point you have already done the sum. You should not have summation sign.
    ok,

     a) R_n = \sum_{i=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =
     b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =
     c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =

     (\frac{2}{n})(2n) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =

    is that right...  d) (4) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} = what can I do with all this under the radical? I have no idea
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  4. #4
    Member dokrbb's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by dokrbb View Post
    ok,

     a) R_n = \sum_{i=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =
     b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =
     c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =

     (\frac{2}{n})(2n) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =

    is that right...  d) (4) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} = what can I do with all this under the radical? I have no idea
    someone???
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  5. #5
    Junior Member FelixFelicis28's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by dokrbb View Post
    someone???
    Could you post what the exact question is, as you found it? To compute \int_0^2 \sqrt{4 - x^2} \ dx from first principles would involve deriving a closed form for this partial sum:

    \sum_{i=1}^n \sqrt{n^2 - i^2}

    which seems to be very difficult, I've got nothing...
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  6. #6
    Member dokrbb's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by FelixFelicis28 View Post
    Could you post what the exact question is, as you found it? To compute \int_0^2 \sqrt{4 - x^2} \ dx from first principles would involve deriving a closed form for this partial sum:

    \sum_{i=1}^n \sqrt{n^2 - i^2}

    which seems to be very difficult, I've got nothing...
    I'll try, the question is like this:

    The following sum  \sqrt{4 - (\frac{2}{n})^2}*(\frac{2}{n}) + \sqrt{4 - (\frac{4}{n})^2}*(\frac{2}{n}) + ...+ \sqrt{4 - (\frac{2n}{n})^2}*(\frac{2}{n})

    is a Right Riemann sum for the definite integral \displaystyle \int_0^b (f(x))dx

    where b = ...

    and f(x) = ...

    I checked my answer for b=2 and f(x) = (\sqrt{4 - x^2}) and they are correct; The next question is: the limit of these Riemann sums as n-> infinity is....

    does that change my approach?
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  7. #7
    Junior Member FelixFelicis28's Avatar
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    Re: The limit of R_n as n-->inf is...

    does that change my approach?
    Yes, this changes everything.

    Quote Originally Posted by dokrbb View Post
    I'll try, the question is like this:

    The following sum  \sqrt{4 - (\frac{2}{n})^2}*(\frac{2}{n}) + \sqrt{4 - (\frac{4}{n})^2}*(\frac{2}{n}) + ...+ \sqrt{4 - (\frac{2n}{n})^2}*(\frac{2}{n})

    is a Right Riemann sum for the definite integral \displaystyle \int_0^b (f(x))dx

    where b = ...

    and f(x) = ...
    We use the definition of the Riemann sum to be \sum_{i=1}^n f(x_i) \Delta x

    Clearly:

    \Delta x = \frac{b-a}{n} = \frac{b}{n} = \frac{2}{n} \implies b = 2

    as you've done.

    And x_i = 0 + i\Delta x = \frac{2i}{n} \implies f(x_i) = \sqrt{4 - \left(\frac{2i}{n}\right)^2}, thus f(x) = \sqrt{4 - x^2} which you've also done correctly.

    The next question is: the limit of these Riemann sums as n-> infinity is....
    When they ask for this, they're not asking you to compute this integral \int_0^2 \sqrt{4 - x^2} \ dx by evaluating this limit:

    \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}

    But rather, what they're asking you to do is evaluate this limit:

    \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}

    by using this relation:

    \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x = \int_a^b f(x) \ dx

    where \Delta x = \frac{b-a}{n}, \ x_i = a + i\Delta x.

    In other words, they're not asking you to compute the integral by using the Riemann sum, they're asking you to compute the Riemann sum by using the integral.
    Last edited by FelixFelicis28; September 22nd 2013 at 06:23 PM.
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  8. #8
    Member dokrbb's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by FelixFelicis28 View Post
    Yes, this changes everything.



    We use the definition of the Riemann sum to be \sum_{i=1}^n f(x_i) \Delta x

    Clearly:

    \Delta x = \frac{b-a}{n} = \frac{b}{n} = \frac{2}{n} \implies b = 2

    as you've done.

    And x_i = 0 + i\Delta x = \frac{2i}{n} \implies f(x_i) = \sqrt{4 - \left(\frac{2i}{n}\right)^2}, thus f(x) = \sqrt{4 - x^2} which you've also done correctly.



    When they ask for this, they're not asking you to compute this integral \int_0^2 \sqrt{4 - x^2} \ dx by evaluating this limit:

    \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}

    But rather, what they're asking you to do is evaluate this limit:

    \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}

    by using this relation:

    \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x = \int_a^b f(x) \ dx

    where \Delta x = \frac{b-a}{n}, \ x_i = a + i\Delta x.

    In other words, they're not asking you to compute the integral by using the Riemann sum, they're asking you to compute the Riemann sum by using the integral.
    I'm not sure I understand really how it should be done, do I really need in this case to evaluate it as

    \lim_{n\to\infty} \sum_{i=1}^n \left[ \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =
     \left[ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] = and so on, without using the actual values of b and a ?
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  9. #9
    Junior Member FelixFelicis28's Avatar
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    Re: The limit of R_n as n-->inf is...

    Quote Originally Posted by dokrbb View Post
    I'm not sure I understand really how it should be done, do I really need in this case to evaluate it as

    \lim_{n\to\infty} \sum_{i=1}^n \left[ \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =
     \left[ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] = and so on, without using the actual values of b and a ?
    Hmmm? I'm not quite sure what you're doing...

    Your nth Riemann sum is:

    R_n = \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}.
    And they're asking you to evaluate \lim_{n \to \infty} R_n.

    We proceed with the definition that:

    \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = \int_a^b f(x) \ dx.
    Thus:

    \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n} = \int_0^2 \sqrt{4 - x^2} \ dx.

    All that's left to do is compute the integral on the RHS and you're done.
    Last edited by FelixFelicis28; September 23rd 2013 at 04:30 PM.
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