## Tuesday, June 21, 2016

### Illustration of an Integral

The 'slices' of the x-axis are non-zero:

How can we calculate this area? There are many ways, varying in sophistication, efficiency, and accuracy. The easiest way is to use a Newton-Cotes method of order zero.
That's a fancy way to say, break up the interval into small pieces calculate the value of the function at the location of each piece multiply the width of the piece by its height to find its area add up all the areas...
How small should the intervals be? The smaller you make the pieces, the better the approximation. How small should they be? It depends on your purpose.

Bob said...

Why would you want to calculate the area?
This is a signal of some sort, not a raster. In analog to digital conversion, the quality would depend on the sampling rate and bit size. For example, an 8 bit integer would chop the y-axis into 256 parts.

Brian Romanchuk said...

That's what we teach to (non-math major) undergraduates about integrals, and doesn't work.

Try integrating: f(x) on the interval [0,1], where:

f(x) = { 0, if x is rational (can be written as p/q, where p,q are integers),
{ 1, if x is irrational.

Note: an any non-zero interval, there is always a rational and irrational number in it. So if we divide f(x) into slices, what is the height of the slice - 0 or 1?

Whee, isn't math fun!

(The answer is 1; we need to use the notion of Lebesgue intervals to get there.)

Matt Franko said...

Well Brian couldnt we just shift the period from 2 to 3; or 4 to 5 or something... iow dont start at 0 ?

Brian Romanchuk said...

Shifting the interval does not help, since rationals/irrationals are mixed up everywhere on the real line. (Needless to say, the derivative exists nowhere.)

This is a classic example from real analysis. There's some good arguments that functions like that are meaningless, but it's hard to exclude them from consideration. You need to do something like get rid of the real line; the problem is - what do you replace it with?

In the real world, the described method is pretty much how we calculate integrals numerically. However, we know that these numerical calculations are approximations, and they do not converge to the right answer if we run into degenerate functions like the one I gave above.