Well I solved it fairly easily, but I had to use my ‘mathematical intuition’ - it seemed ‘obvious’ that the answer was some quadratic in x, then I just crunched through the algebra.
My question is really, is there a more general, systematic way of solving problems of this type? One that doesn’t involve having a lucky guess as to what form the answer might take?
While I think of it, is there a special name for these types of equations (ones where functions are plugged back into themselves)?
Oh that’s three questions. Well my maths ability varies greatly .
I think some of you folks are a tad confused. He’s not looking for a solution to (f(x))^2 = g(x), but of f(f(x))=g(x). Tht is, if you substitute the function f(x) in place of x in the function f(x), then set tha equa to g(x).
As far as I know, there isn’t a general way to solve this but this sort of math isn’t my forte. This kind of thinf became popular and important when computers and especially hand calculaors became available. I note that it wasn’t until that happened that people really started fooling around with such recursive and self-referential formulas. A lot of the work was done by calculators and computers using the brute force method of crunching it through – people wrote programs t draw Mandelbro sets. They didn’t derive formulas for it. So I suspect that there isn’t a general way to do this. (There are probably a few specialized cases that are soluble analytically. I’ll bet you con solve the case for one iteration using the Method of robenius. But it’s probabl more trouble than it’s wortyh.)
There probably is a general solution to f[sup]2[/sup] = g when g is an even-degree polynomial.
Consider the case where g(x) = x. Certainly f = g is a solution. But so is any invertible function on the set where its inverse exists. That should give you some idea of the complexity of the problem.
As for the problem, I don’t know of a general solution for this sort of problem, but I think it should be pretty straightforward when you’re dealing with polynomials. Simply figure out what the degree of f should be, then figure out what the coefficients are.
Yeah, it won’t be so straightforward for polynomials as I first thought. Like ultrafilter mentioned, even degree polynomials might be easier, but even then you may have non-polynomial solutions.
ultrafilter, now that I think about it a bit more, I think you might have been thinking of polynomials with perfect square degree, rather than even degree. Then, it will be easy to find polynomial solutions only if they exist.
For example, f(f(x))=x^9 has the obvious solution f(x)=x^3 (there may be other solutions).
What about f(f(x))=x^6 + 1? I don’t know a solution offhand, but I think it’s clear no solution can be a polynomial in x.
Not necessarily. First of all, if f(x) is a degree-n polynomial, then f(f(x)) is a degree-n[sup]2[/sup] polynomial. Maybe you meant “if g is a polynomial with square degree?”
But even that doesn’t work. If f(x) is of the form ax[sup]2[/sup] + bx + c, then
f(f(x)) = a[sup]2[/sup] x[sup]4[/sup] + 2 a[sup]2[/sup] b x[sup]3[/sup] + a (2ac + b[sup]2[/sup] + b) x[sup]2[/sup] + b (2ac + b) x + c (b + ac + 1)
There aren’t any values of a, b, and c that would correspond to certain values of the coefficients. For example, if the coefficient of the x[sup]4[/sup] is non-zero and the coefficient of the x[sup]3[/sup] is zero, then that means that b must be zero; but if the coefficient of the x[sup]1[/sup] is non-zero, then we’re screwed.
For general n, this won’t work because we have n[sup]2[/sup] + 1 equations that must be satisfied by n + 1 variables (i.e., the coefficients of the polynomial f.)
On preview, I see I’ve conveniently ignored the possibility on non-polynomial solutions. I’ll have to think about this a little more.
When I wrote that, I wasn’t thinking specifically of polynomial solutions. f(x) = x[sup]sqrt(2)[/sup] is fine by me, cause I’m interested in finding all the solutions we can.
According to this page, there is a solution to f[sup]2[/sup] = exp, but it’s not single-valued and is not known to be unique.
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