Here's my first problem for you all... Recursively define the class of "ASU functions" of x: 1. f(x)=x is ASU. 2. If f(x) is an ASU function and c is a (complex) constant, then c+f(x), c-f(x), c*f(x), c/f(x) all are ASU functions. 3. If m is a positive integer constant and f(x) is an ASU function then f(x)^m is an ASU function. Prove or disprove: The "Weierstrass approximation theorem" for ASU functions. That is, is it true that an arbitrary continuous function of x on [0,1] can be approximated to within epsilon by an ASU function (for any epsilon>0)? The usual Weierstrassian proof techniques cannot handle this problem; they depend on the function class being closed under linear combinations. ASU class is not. I proved the following partial result. Given any finite subset of [0,1] and any set of real data on that set.... that is, given any finite set of 2-tuples (x1,y1), (x2,y2), ... (xN,yN) with the xj distinct... there exists an ASU function which fits the data with |additive error|<epsilon. However, that does not settle the problem for infinite datasets. Possible way to disprove: just find a function which cannot be approximated... e.g. the fact the real ASU functions are monotone except for rule 3, seems rather restrictive... Possible way to prove: I suspect that there is some sense in which you can keep on using the recursive definition randomly each time (for some very carefully chosen notion of "random"!) for N iterations. Then you can argue that in the limit N-->infinity, the resulting function will be "locally universal." That is, with probability-->1, if you zoom in with your microscope to look at its graph-picture, and set the (N-dependent) horizontal and vertical magnifications right (and high enough), then you will see any desired curve, accurate to epsilon, on your screen if the place you zoom in on was chosen right and if N(epsilon) was large enough. I think it would be quite interesting to prove or disprove this.