Riemann sum with circles!

Question

My friend and I were talking about Riemann sum and friend suggested that all rectangles would be replaced by circles this is the formula I came up with: $$ \lim _{n\ \to\ \infty}\sum_{i\ =\ 1}^{n}\pi\,{\rm f}^{2}\left(\, i\,\right) $$ Is this formula right and if so does this formula work for all Riemann integrable functions ?. I can visualize it failing for a straight line. What is the error bound for this formula in terms of the number of circles ?. I am very confident that it is related to the maximum curvature somehow. I should note here that circles may not stack on each other. This would mean that for a straight line it would probably fail.

Answer 1

I've just written a paper on this subject for class. My answer is a bit different from yours, because I am allowing circles to stack on each other.

Rectangular Riemann sums take the form $$\sum_{i=0}^n f(a+i\Delta x)\Delta x $$ when given a function $f(x)$, divided into $n$ partitions, bounded between $ a \text{ and } b. \Delta x = \frac{b-a}{n} $. Multiplying the function value and the width of the partition gives us the area of the rectangle that is being created.

For circles, we can set the diameter of the circle equal to the partition width, and then stack the circles up to the height of the function value. We can then get the area of the stack by finding the number of circles which will fit under the curve and then multiplying by the area of a partition-diameter circle.

$$ A_{stack} = \frac{f(a+n_i\Delta x)}{\Delta x} \times \frac{\pi \Delta x^2}{4} $$

Substituting $ \Delta x = \frac{b-a}{n} $ back in, and summing up to the number of partitions, we eventually come to this simplification:

$$ A_{total} = \sum_{i=0}^n \frac{f(a+i(\frac{b-a}{n}))(b-a)\pi}{4n} $$

Any area generated by this sum is, interestingly, 78% of the value given by a rectangular sum. This is because a circle has 78% of the area of a square with the same radius (a line from the middle of the rectangle to a side, which is perpendicular to that side). This calculation is with givens $ f(x)=x^2+1, [a,b] = [0,1], n = 10 $.

$$ A_{circle} = \displaystyle \sum_{i=0}^{10} \frac{f(\frac{i}{10})\pi}{40} = \frac{\pi}{40} + \frac{1.01\pi}{40} + \dots + \frac{2\pi}{40} = \frac{297}{800}\pi \approx 1.166 $$

$$ A_{rectangle} = \displaystyle \sum_{i=0}^{10} \frac{f(\frac{i}{10})}{10} = \frac{1}{10} + \frac{1.01}{10} + \dots + \frac{2}{10} = 1.485 $$

$$ \frac{1.166}{1.485} \approx 0.7851 $$

When you take the sum to infinity, it converges at a value that is 78% of the real value.

$$ A_{total} = \displaystyle \sum_{i=0}^n \frac{f\left(\frac{i}{n}\right)\pi}{4n} = \frac{(\left(\frac{0}{n}\right)^2+1)\pi}{4n} + \dots + \frac{(\left(\frac{n}{n}\right)^2+1)\pi}{4n} $$ $$ = \frac{\pi}{4n}\left[\left(\left(\frac{0}{n}\right)^2 + 1 \right) + \left(\left(\frac{1}{n}\right)^2 + 1 \right) + \dots + \left(\left(\frac{n}{n}\right)^2 + 1 \right)\right] $$ $$ = \frac{\pi}{4n}\left[\left(\left(\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)^2 + \dots + \left(\frac{n}{n}\right)^2\right) + n \right] $$ $$ = \frac{\pi}{4n}\left[\left(\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)^2 + \dots + \left(\frac{n}{n}\right)^2 \right] + \frac{\pi}{4} $$ $$ = \frac{\pi}{4n^3}\left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{\pi}{4} $$ $$ = \frac{8\pi n^2 + 3\pi n + \pi}{24n^2} = \frac{8\pi}{24} + \frac{3\pi}{24n} + \frac{\pi}{24n^2} $$ $$ = \displaystyle\lim_{n \to \infty}\left(\frac{8\pi}{24} + \frac{3\pi}{24n} + \frac{\pi}{24n^2}\right) = \frac{1}{3}\pi \approx 1.0472 $$