Functional means are a way of inducing a mean Mf from a differentiable function f.
If f(x) = x2, Mf(x, y) = (x + y)/2 is the arithmetic mean.
If f(x) = 1/x, Mf(x, y) = √(xy) is the geometric mean.
Seems like this should be useful for something.
Related post: Means and inequalities
I don’t see an obvious or intuitive relationship between the two functions and the arithmetic and geometric means. Given M and the goal of finding the two means, is it straightforward to find the functions? Or is it trial and error? This seems a bit magical.
I think you need more than differentiability, since the first derivative has to be invertible.
Differentiable and strictly convex or concave should do it.
In fact, now that I think about it, continuously differentiable and strictly convex gives you a Bregman divergence, which is a “functional distance” in the same manner. latex$$f(x) = x^2$$ in a Bregman divergence gives you squared Euclidean distance.
Do you know of any other instance, besides the two you listed here, in which the function is something manageable and the “mean” is one that anyone cares about?
If we take f(x) = x^3 we get sqrt((xx+xy+yy)/3), and likewise for other powers x^k; the cases k=-1 and k=2 seem like the only really interesting ones. (Maybe one can make something of k=1/2, which produces AM(AM,GM), not a million miles from the famous AGM iteration.)
I basically had the same reaction as ‘g’.. I’m not sure it results in anything useful in other cases.
I think that the meaning is less mysterious if you think about it geometrically…plot f(x) and construct the secant through (x, f(x)) and (y,f(y)). The functional mean is the point on the curve where the tangent is parallel to this secant.
What flashed through my head was “Jensen’s inequality”, but I think the foregoing comments may already be deeper than that.
I agree with GlennF about the geometric view. Or look at it this way: The difference quotient provides the average rate of change. If the function is continuous on [x,y] and differentiable within, then mean value theorem guarantees existence of c such that f'(c) gives this average rate of change. If in f’ is invertible then this c is unique and functional mean just “extracts” the value c.
I agree with the above comments. The mean value theorem from Calculus I tells us that for smooth f there is an M in the interval (x,y) such that f'(M)=(f(y)-f(x))/(y-x). If M is to be a well-defined function of x,y and f, you’d next some other condition on f, like convexity or concavity. If f(x)=exp(x), you get M(x,y)=log((exp(y)-exp(x))/(y-x)), and M(x,x)=x.
Also, for f(x)=x^p, this is known as the Stolarsky mean, apparently:
http://en.wikipedia.org/wiki/Stolarsky_mean
An additional observation about this: The function leading to a particular mean is not unique. In fact any scale or translation has no effect on this mean. As an example f(x)=k(x-a)^2 leads to the same mean as f(x)=x^2.
I think the “magic” Lawrence noted is in the connection between distance and mean, Euclidean squared distance to standard mean being the most well known. If there are other interesting examples, they probably stem from other distance functions.
Dan: Just to expand on/clarify your comment: The mean is invariant under non-zero scaling of the function f and invariant under adding any linear function to f. It is *not* invariant under a translation of the argument of f, i.e. f(z) -> f(z-a), except in the case where f is a quadratic function.