Riemann Summation Notation

sponsoredwalk

Okay I've seen how crazy Riemann sums can get in real analysis and I've noticed a heirarchy of notation.

The Stewart/Thomas etc... kinds of books use;

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x[/latex]

Where;

[latex]\Delta x = \frac{b - a}{n} \ and \ x_i = a + i\Delta x[/latex]

Then the books like Spivak and Bartle's real analysis use;

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(x_i) (x_i - x_i_-_1)[/latex]


and what I'd like to know is how to calculate the [latex](x_i - x_i_-_1)[/latex] for some equation like [latex]\int_{2}^{8} x^2\,dx [/latex]

I'll stick to using a Right endpoint sum, not an upper or lower or a left or midpoint etc...


If I were to use the normal easy notation at the very top of this thread I'd do it like so;


[latex]\int_{2}^{8} x^2\,dx [/latex]

[latex]\Delta x = \frac{8 - 2 }{n} = \frac{6}{n} [/latex]

By a Right Riemann sum it would be;

[latex]\lim_{x \to \infty} \sum_{k=1}^{n} (2 + \frac{6k}{n} )^2 \cdot ( \frac{6}{n} ) [/latex]

By a Left Riemann sum;

[latex]\lim_{x \to \infty} \sum_{k=1}^{n} (2 + \frac{6(k - 1)}{n} )^2 \cdot ( \frac{6}{n} ) [/latex]

By a Midpoint sum;

[latex]\lim_{x \to \infty} \sum_{k=1}^{n} (2 + \frac{6(k - \frac{1}{2})}{n} )^2 \cdot ( \frac{6}{n} ) [/latex]


That's all fine. But, I would assume that using the more advanced notation I would set up the sum like so (but I'm probably wrong!);


[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(x_i) (x_i - x_i_-_1)[/latex]

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(2 + i \Delta x) (x_i - x_i_-_1)[/latex]

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(2 + \frac{6i}{n}) (\frac{6i}{n} - \frac{6(i - 1)}{n})[/latex]

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} f(2 + \frac{6i}{n}) (\frac{6i}{n} - \frac{6i - 6}{n})[/latex]


[latex]\lim_{x \to \infty} \sum_{i=1}^{n} (2 + \frac{6i}{n})^2 (\frac{6i}{n} - ( \frac{6i}{n} - \frac{6}{n}))[/latex]

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} (2 + \frac{6i}{n})^2 (\frac{6i}{n} - \frac{6i}{n} + \frac{6}{n})[/latex]

[latex]\lim_{x \to \infty} \sum_{i=1}^{n} (4 + \frac{24i}{n} + \frac{36i^2}{n^2}) ( \frac{6}{n} )[/latex]

[latex]\lim_{x \to \infty} ( \frac{6}{n} ) [ \sum_{i=1}^{n} 4 + \sum_{i=1}^{n} \frac{24i}{n} + \sum_{i=1}^{n} \frac{36i^2}{n^2} ] [/latex]

[latex] \lim_{x \to \infty} ( \frac{6}{n} ) [ \sum_{i=1}^{n} 4 + \frac{24}{n} \sum_{i=1}^{n} i + \frac{36}{n^2} \sum_{i=1}^{n} i^2 ] [/latex]

[latex] \lim_{x \to \infty} ( \frac{6}{n} ) [ 4n + \frac{24}{n} \frac{n(n + 1)}{2}+ \frac{36}{n^2} \frac{n(n + 1)(2n + 1)}{6} ] [/latex]

[latex] \lim_{x \to \infty} [24 + \frac{144[n^2 + n]}{2n^2} + \frac{216[2n^3 + 3n^2 + n]}{6n^3}] [/latex]

[latex] \lim_{x \to \infty} [24 + 72 + \frac{1}{2n} + 72 + \frac{108}{n} + \frac{36}{n^2} [/latex]

[latex] 168 [/latex]

Okay, I was already told that this method was wrong but I'm getting the right answer. I've spent four days working on getting this method right so please let me know if I'm wrong in any particular, especially the way I set up [latex](x_i - x_i_-_1)[/latex] . Is the method I've used the general method to set this up for a right sum?

seandoiler

I'm not sure which type of riemann sum you are trying to calculate, from what you have typed in you are calculating the RIGHT riemann sum, this is because you choose [latex]x_{i}[/latex] to be the right end point for each interval. Is this what you were trying to do? The [latex](x_{i}-x_{i-1})[/latex] is just the interval width regardless.

If you want to be more general then I would say [latex]\sum_{i=1}^{n} f(y_{i})(x_{i}-x_{i-1})[/latex]where now [latex]x_{i-1},x_{i}[/latex] are the end-points of the interval and [latex]y_{i}[/latex] is the point in the interavl that you select to use for the Riemann sum, if [latex]y_{i}=x_{i-1}[/latex] this is the LEFT sum, if [latex]y_{i}=x_{i}[/latex] this is the RIGHT sum, if [latex]y_{i}=\frac{x_{i-1}+x_{i}}{2}[/latex] this is the MID-POINT sum....or if we choose
[latex]y_{i}[/latex] to be the supremum of the function f on the interval then this is the UPPER sum and similarly if we choose the infinum this is then the LOWER sum

hope this helps...if not you can pm me

sponsoredwalk

Yeah I was actually doing a right Riemann sum, my main concern was actually using the notation for [latex](x_i - x_i_-_1)[/latex] in practice.

Just looking at your notation you've clarified everything for me.

The interval stays the same regardless, [latex](x_i - x_i_-_1)[/latex] but to change from a right sun to a left sum to a midpoint sum all you do is insert the appropriate position into the function. I think I have it down now, thank you.