Date: 06/15/2003 at 14:05:42 From: Khoa Phung Subject: How can I visualize the distance between two functions Dear Dr. Math, What does the distance between two functions mean? Let f(x)=x and g(x)=x^2 on the interval from 0 to 1. Then d(f,g) = 1/sqrt(30). Sqrt( Integral((x-x^2)^2)dx). How can I visually understand the inner product of the distance of those two functions?
Date: 06/20/2003 at 15:40:02 From: Doctor Douglas Subject: Re: How can I visualize the distance between two functions Hi Khoa, thanks for writing to the Math Forum. The quantity that you have computed above is the "L-2 distance" or the "root-mean-square distance" between the two functions f and g on the interval (0,1). More generally, you can define an L-p distance as b (1/p) ||f-g||_p = [ INT |f(x)-g(x)|^p dx ] a . The most common value of p is 2. You may also wish to divide by the length |b-a| of the interval before taking the root. In your example, the distance is the "root-mean-square" of the distance between the function values f(x) and g(x): we take the difference [f(x)-g(x)], square it, take the mean value of the squares by dividing by the length of the interval, and then finally take the square root of that mean to get the distance back in units of the function values (i.e. in units of f or g). Thus, the distance 1/sqrt(30) is the average distance between the two functions f and g in this root-mean-square sense. Now, if we had wanted the average distance between f and g, we could have computed [1/(b-a)] INT f(x)-g(x) dx. For the example above, this would have worked and given us a number that we could have called the average distance. Although it doesn't happen in your example, note that in general the integrand of this last integral can take both positive and negative values, so that the positive difference over one part of the interval can be cancelled by the negative difference (g > f) over another part of the interval. Thus the average distance can be zero even though f and g are not identical to one another. The average distance can also be negative. That doesn't seem like a typical use of the word "distance." This explains why the root-mean-square is useful - all of the discrepancies between f and g contribute to the total distance with the same sign, because the differences are squared. Loosely speaking, a root-mean-square distance of zero implies that f=g over all of the interval, except possibly at a few isolated points. And this is a useful property to have in a definition of distance - if the distance is zero, then the two things are at the same place. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum