This has been bothering me for a while (it was on a math competition):

Suppose f and g are functions.

We say the real number c is a real fixed point of f if [imath]f(c) = c[/imath]

We say that f and g commute if [imath]f(g(x)) = g(f(x))[/imath] for all real numbers x

a) if [imath]f(x) = x^2-2[/imath] determine all real fixed points of f

(this ones easy, just solve [imath]x^2-2 = x[/imath])

b)if [imath]f(x) = x^2-2[/imath] find all cubic polynomials g that commute with f

I know that an inverse would commute, but its not cubic . Cant do this one.

c)Suppose that p and q are real-valued functions that commute.

if [imath]2[q[p(x)]]^4 + 2 = [p(x)]^4 + [p(x)]^3[/imath] for all real numbers, prove q has no fixed points

No idea about this one.

## [solved]Commuting fucntions problem

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### [solved]Commuting fucntions problem

Last edited by MrMonkey on Sun Jan 31, 2010 9:00 pm UTC, edited 1 time in total.

### Re: Commuting fucntions problem

Here is an idea which might be relevant.... First of all, if [imath]x[/imath] is a fixed point of [imath]f[/imath], and [imath]g[/imath] commutes with [imath]f[/imath], can we say anything about [imath]g(x)[/imath]? Let's see [imath]f(g(x)) = g(f(x)) = g(x)[/imath]. But by definition, this means that [imath]g(x)[/imath] is a fixed point of [imath]f[/imath].

### Re: Commuting fucntions problem

Part (b): Suppose [imath]g(x) = ax^3 + bx^2 + cx + d[/imath]. If [imath]f(g(x)) = g(f(x))[/imath], then we have that: [math](ax^3 + bx^2 + cx + d)^2 - 2 = a(x^2-2)^3 + b(x^2-2)^2 + c(x^2-2) + d[/math] Try expanding this out and seeing what you can say about the possible values of a, b, c and d.

All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

### Re: Commuting fucntions problem

For part (c), first apply the fact that p and q commute, and then specialize x to be a fixed point under q. Do a bit of cancelling and see what happens.

### Re: Commuting fucntions problem

MrMonkey wrote:Suppose that p and q are real-valued functions that commute.

if [imath]2[q[p(x)]]^4 + 2 = [p(x)]^4 + [p(x)]^3[/imath] for all real numbers, prove q has no fixed points

Ninja'd, but I typed all that so I'll submit.

Spoilering each step of my solution separately:

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