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Math Help - Proving the definition of the integral

  1. #1
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    Proving the definition of the integral

    Suppose f \in R[0, \pi]. Prove that the sequence

    a_{n}=\frac{\pi}{n}\sum f(\frac{k\pi}{n})

    converges to \int f.

    Note that the sum is from k=1 to n, and the integral is from 0 to pi.

    Can anyone at LEAST give me a starting point for this? I have no idea where to even begin this...
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  2. #2
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    Note that the sum you have, is a Riemann sum of the form:

    \dfrac{b-a}{n}\displaystyle\sum_{i=0}^nf\left(a+i\dfrac{b-a}{n}\right)

    Where \dfrac{b-a}{n} is a partition of the interval, a+i\dfrac{b-a}{n} is a selection of points of each interval defined by the partition. The norm of the partition is regular, that means that every interval defined by the partition has the same lenght. This Riemann sum is defined by a aritmetic progression. In more abstract sence, a Riemann sum just have to look like this:

    \sum_{i=0}^nf(a_i)\Delta x_i

    The Riemann integration tell us that if a Riemann sum converges when the norm of the partition tends to 0, the function is Riemann-integrable. So, can you tell me which is the partition in this particular case? can you express your sum like a Riemann sum?
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  3. #3
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    I have no idea what you mean by prove.
    I assume the by f\in R[0,\pi] you mean the function is Riemann integrable of the interval.
    Then observe that each a_n is a right-hand approximating sum over a regular partition of the interval.
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  4. #4
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    Quote Originally Posted by felper View Post
    Note that the sum you have, is a Riemann sum of the form:

    \dfrac{b-a}{n}\displaystyle\sum_{i=0}^nf\left(a+i\dfrac{b-a}{n}\right)

    Where \dfrac{b-a}{n} is a partition of the interval, a+i\dfrac{b-a}{n} is a selection of points of each interval defined by the partition. The norm of the partition is regular, that means that every interval defined by the partition has the same lenght. This Riemann sum is defined by a aritmetic progression. In more abstract sence, a Riemann sum just have to look like this:

    \sum_{i=0}^nf(a_i)\Delta x_i

    The Riemann integration tell us that if a Riemann sum converges when the norm of the partition tends to 0, the function is Riemann-integrable. So, can you tell me which is the partition in this particular case? can you express your sum like a Riemann sum?
    So the partition is \frac{\pi}{n}
    So \sum_{i=0}^nf(a_i)\Delta x_i=\frac{\pi}{n}\sum_{i=0}^nf(a_i)? But this is where I started...
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