## [solved]Commuting fucntions problem

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MrMonkey
Posts: 60
Joined: Sun Jan 31, 2010 6:55 am UTC

### [solved]Commuting fucntions problem

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

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

ThomasS
Posts: 585
Joined: Wed Dec 12, 2007 7:46 pm UTC

### 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].

Token
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### 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: $(ax^3 + bx^2 + cx + d)^2 - 2 = a(x^2-2)^3 + b(x^2-2)^2 + c(x^2-2) + d$ Try expanding this out and seeing what you can say about the possible values of a, b, c and d.
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DavCrav
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Location: Oxford, UK

### 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.

tendays
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### 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:
Spoiler:
Since p and q commute, [imath]2[q[p(x)]]^4 + 2 = 2[p[q(x)]]^4 + 2[/imath]

Spoiler:
Assume X is a (necessarily real) q-fixed point. So: [imath]2[p[q(X)]]^4 + 2 = 2[p[X]]^4 + 2 = [p(X)]^4 + [p(X)]^3[/imath]

Spoiler:
Set [imath]y=p(X)[/imath]: [imath]2y^4 + 2 - y^4 - y^3 = 0 \Rightarrow y^4 - y^3 + 2 = 0[/imath]

Spoiler:
If that [imath]f(y) = y^4 - y^3 + 2[/imath] polynomial has a real zero then there must be a value Y such that [imath]f(Y)<0[/imath] and [imath]f'(Y)=0[/imath]

Spoiler:
[imath]f'(Y)=4y^3 - 3y^2=0 \Rightarrow y \in \{\frac 3 4 , 0\}[/imath]

Spoiler:
[imath]f(0)=2>0[/imath] so [imath]y\neq 0[/imath]. [imath]\left(\frac 3 4\right)^3 - \left(\frac 3 4\right)^4<2[/imath] so [imath]y \neq \frac 4 3[/imath] either. Contradiction, so [imath]q[/imath] has no fixed point.
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