Recent content by jonas.hall

I can try... but I'm not sure I'm up to showing by induction something I can't show i the special cases. Though maybe you're saying: I have started by assuming the highest order term = 1, instead I should start by the lowest term? I'll think about it anyway.

Thanks for both answers - they gave me clues I can work on but unfortunately MrJB has done some mistakes. The equation x^2 + 1 = x has no real solutions so a large part of the reasoning fails. If one evaluate the derivatives at x=0 however some progress can be made but every even derivative is =...

I think Martin Gardner has written in one of his many books that the problem is unsolved. That was a while ago though, back in the 60ies/70ies and a lot has happened since then. Typically though, most problems of this sort are difficult to solve.

Are there any ploynomials p(x) such that p(x)^2 -1 = p(x^2+1) for all x? To cut it short: With CAS software I have verified that there are no solutions except p(x) = 1.618... and p(x) = -0.618... (constants) up to order 53 or so, but I have to prove this (or find the other solutions)...

Here are two very similar questions about polynomials that I feel may have deeper roots (excuse pun). a) Does anyone know of any interesting theory related to them that I could read up upon? b) How would one start solving them? Here are the problems: 1) Show that there are...

I tried that and the fencing needed is then exactly 0.5 of the total perimeter. Using a circle to cut of the corners lowers this to approximately 0.475. I have som argument that points towards thie circle being best, but no proofs or counterexamples. Your example is actually the special case...

I do admit magic is wonderful, but I don't shirk hard work - If I know what to do. My problem here is I hardly know where to begin. What, for instance do you mean by picking some different y's and seeing what happens? I guess you mean doing the multiplication but for what X? Numeric example or...

Condition 1 Condition 1 specifically states that for all x,y in S and a,b in R ax+by should be in S. As far as I understand this is a condition that states that the subsets S have to be linear subspaces like straight lines, flat planes etc. My solutions so far seem to be just that but there...

...? I like to visualise. Do offer an alternative before discarding other peoples mode of thinking. By thinking like that I can visualize what x1 = -x4 means. Admittedly rotations are difficult :-) but I find I can understand problems better if I can visualise them as well as numbercrunch...

Some findings... Well to be perfectly true I did find something when I did the multiplication. If You let z = ax + by, I the tried to multiply z with x to see if that would yield anything and I found that z*x = by * x and z*y = ax*y. It is as if the product takes just the part...

Missing something...? OK, so I've tried your advice. But setting x and y to x1... and y1... and doing the multiplication didn't really give anything more than the original definition of the product. What am I missing here? Please remember that I'm liable to miss simple observations simply...

I take it an ideal is a concept in linear algebra...?

I just realized a typo in Condition 1. it shoud read ...and all a,b in R (not R4).

So I have an equilateral triangle an I want to divide it in 4 parts, all having the same area. This can be done in a multitude of ways of course. But assuming it's a garden and the division is about putting up a fence, which division uses the least fencing? Now I have two alternatives so...

Here's a difficult problem in an area I'm not at all familiar with. A product * is defined on R^{4} in the following way: (a, b, c, d)*(a', b', c', d') = (cd' - c'd, ac' - a'c + cb' - c'b, a'd - ad' + bd' - b'd, c'd - cd') Find all subsets S of R^{4} that satisfies the following two...