From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: A question about polynomials Date: 22 Oct 1997 18:07:24 GMT In article <62falj$1cm@vidar.diku.dk>, Asger Grunnet wrote: >Consider two polynomials f and g, both of degree n, with f(0)=g(0)=0. >Futhermore assume that f and g are increasing on the interval [0,infinity). > >These will of course be bijective functions on [0,infinity), and so it is >possible to define a function h(x) = g^{-1}(f(x)). >With this definition we have : f(x) = g(h(x)). > >The derived function h'(x) can be found to be f'(x)/g'(h(x)), >and the twice derived h''(x) can be calculated likewise. > >My question now is : Is h'(x) monotonous on some interval [a,infinity) ? >By monotonous I mean either increasing, decreasing or constant. > >Equivalently one can ask : Is it true that either h'' = 0 or h'' >has only finitely many zeros on [0,infinity) ? The answer to your question is 'yes', and you don't need all the assumptions; moreover, you can even compute your interval explicitly. The graph of h consists of the points (x,y) with h(x) = y, i.e., it's the zero locus of the function H(x,y) = f(x)-g(y). The implicit function theorem allows us to compute the derivative of the implied functional relationship at points on curves: dy/dx = - (dH/dx) / (dH/dy) = f'(x)/g'(y). Differentiating also gives d^y/dx^2 as a rational function of x and y: it's d/dx ( f'(x)/g'(y) ) = f''(x)/g'(y) - f'(x)g''(y)(dy/dx)/(g'(y))^2 = f''(x)/g'(y) - (f'(x))^2 g''(y)/(g'(y))^3 Thus the points where h''(x)=0 are the points on the graph where some other function of x and y vanishes, i.e., they are the intersection points of two algebraic curves in the x-y plane. Now it's simple algebraic geometry to conclude that the intersection is either itself a curve (i.e., h''(x) is identically zero along the graph of h) or a zero-dimensional variety, i.e., a finite set of points. For example, if f(x) = 2x^3+x+1 and g(x)=x^3-x^2-4, then we have h''(x) = 12x/(3y^2-2y) - (6x^2+1)^2(6y-2)/(3y^2-2y)^3, which vanishes where 12x(3y^2-2y)^2=(6x^2+1)^2(6y-2), i.e. on the zero locus of 4 3 2 4 4 2 2 108 x y - 144 x y + 48 x y - 216 x y + 72 x - 72 x y + 24 x - 6 y + 2 The graph of h itself is the zero locus of 3 3 2 2 x + x + 5 - y + y We can determine the intersection of these curves by eliminating y: the only x which can lead to a point of intersection are the roots of 9 8 7 6 5 4 - 12528 x - 532656 x - 8108208 x - 41251248 x - 4052016 x - 41044104 x 3 2 - 103207788 x + 3417876 x - 37509 x + 137 which are all complex or negative except .0105... So for these two curves, h' is indeed increasing past .011. Can we generalize? If f and g are any two functions, we can look at the zero locus of H(x,y)=f(x)-g(y). At almost all points of this curve, the curve is the graph of a locally-defined function, on which we can ask about h', h'' etc. As noted above, the points where h''=0, for example, are the points in another function's zero locus. We can draw the appropriate finiteness conclusion as long as the curves are algebraic. It is sufficient that f and g be rational functions. dave