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Math Help - Riemann Sums

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Riemann Sums

    Although we implicitly use the definition of the Riemann Sums every time we cacluate a definite integral is it actually possible to calculate the majority of integrals via Riemann Sums?

    Example:

    \int_0^1 x~dx this is easily able to be calculated using the rieman formula \lim_{n\to\infty}\sum_{i=1}^{n}f\left(M_i\right)\d  elta x_i (I omit the details)

    But what about another fairly elementary integral? How would we go about using the above defintion to find say

    \int_1^e \ln(x)~dx?

    Is it actually feasible using just summing techniques (besides obviously integrating) to calculate \lim_{n\to\infty}\sum_{i=1}^{n}\ln\left(1+\frac{e-1}{n}i\right)\cdot\frac{e-1}{n}?

    Even if someone could do the above sum (which Im sure someone can) what about a harder one like \int_0^1 e^x\sin(x)~dx. And what about all of this using the formal Riemann-Stieltjes definitions?


    A similar argument brings us to ask if we can practically find say

    \lim_{x\to 0}\frac{\tan(x)-\sin(x)}{x^3} using only \delta-\varepsilon defintions?


    Note: I am not asking whether or not we should use these defintions to find integrals and limits. What I am asking is if someone said compute \int_0^1 e^x\sin(x)~dx using only Riemann sums would it be feasible?


    Any input/discussion would be appreciated
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  2. #2
    Eater of Worlds
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    I think it is indeed possible, but probably so difficult in cases like these it is

    impractical to do. That is why easy ones are always given as examples when illustrating the technique.


    Using the right endpoint method:

    \frac{(e-1)\left(\displaystyle\sum_{k=1}^{n}ln\left(\frac{k  \cdot (e-1)}{n}\right)\right)}{n}. Which is what you have as well.

    Just for fun, I ran this through Maple to see what it would give.

    This sum equals....check this out:

    \frac{(e-1)\left((n+1)ln(\frac{e-1}{n})+ln({\Gamma}(n+1+\frac{n}{e-1}))\right)}{n}-\frac{(e-1)\left(ln(\frac{e-1}{n})+ln({\Gamma}(1+\frac{n}{e-1}))\right)}{n}

    Now, taking the limit, it is 1...as hoped/expected.

    I wasn't about to attempt this by hand.

    Now, for the even more difficult \int_{0}^{1}e^{x}sin(x)dx

    Whose sum is \frac{\displaystyle\sum_{k=1}^{\infty}e^{\frac{k}{  n}}sin(\frac{k}{n})}{n}

    I am not about to type out what Maple gave back for this. But taking the limit gives:

    \frac{1}{2}(1-ecos(1)+esin(1))\approx .90933....

    My point being, yes it is possible, but it is so laborious to be impractical.

    Give it a go if you have that much free time.
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  3. #3
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    You do not need to use equally subdivided partitions.
    Let P_n = \{ e^{k/n} : k=0,1,2,...,n\} where n\geq 1.
    Then, \Delta_k = x_k - x_{k-1} = e^{k/n} - e^{(k-1)/n} = e^{1/n} (e^k - e^{k-1})
    And, M_k = \sup \{ \log x : x \in [x_{k-1},x_k] \} = \log  \left( e^{k/n} \right) = \tfrac{k}{n}
    With, m_k = \inf \{ \log x : x\in [x_{k-1},x_k ] \} = \log \left( e^{(k-1)/n} \right) = \tfrac{k-1}{n}

    The upper sum is, U_n = \sum_{k=1}^n M_k \Delta_k = e^{1/n} \sum_{k=1}^n \tfrac{k}{n} \left( e^k - e^{k-1} \right)
    The lower sum is, L_n = \sum_{k=1}^n m_k \Delta_k = e^{1/n} \sum_{k=1}^n \tfrac{k-1}{n} \left( e^k - e^{k-1} \right)

    Perhaps these are more convient for you to deal with.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    You do not need to use equally subdivided partitions.
    The classic example here is to find

    \int_0^1 \sqrt{x}\,dx

    Instead of choosing the right endpoint, i.e, x_i = \frac{i}{n} for the Riemann sum giving

    \lim_{n \rightarrow \infty}\sum_{i=1}^n \sqrt{\frac{i}{n}}\, \frac{i}{n}

    choose x_i = \left( \frac{i}{n} \right)^2
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