9
$\begingroup$

The question is: Let $f:\mathbb R\rightarrow \mathbb R$ and $g:\mathbb R\rightarrow \mathbb R$ be two functions such that $(f\circ g)(x)=4x^2+4x+1$ and $(g\circ f)(X)=x^2+2x+2$. Find $f(x)$ and $g(x)$.

I solved the problem like this.

Because both the compositions are polynomials is $x$, it is clear that $f(x)$ and $g(x)$ also both will be polynomials in $x$. Let $\deg (f)=m, \deg (g)=n$. Then it follows that $m, n\leq 2$ cause otherwise the compositions will be polynomials of degree more than 2.

Now if both $m, n=2$ then $f\circ g, g\circ f$ will be of degree more than 2, contradiction again. Hence one of them will be 2 and the other will be 1. WOLOG let us assume that $m=2, n=1$.

Then we assume $f(x)=ax^2+bx+c, g(x)=dx+e$ we have $f\circ g=a(dx+e)^2+b(dx+e)+c= ad^2x^2+(2ade+bd)x+ae^2+be+c$ and $g\circ f=d(ax^2+bx+c)+e=dax^2+dbx+dc+e$. Equating the coeffcicients of similar terms, we then get the values of $a, b, c, d, e$ and hence the problem is solved.

Please inform me whether my approach to solve this problem was correct or not.

$\endgroup$
13
  • 13
    $\begingroup$ It is not clear that $f,g$ will be polynomials! If $f\circ g$ and $g\circ f$ are the identity (polynomial), then $f$ can be any bijection and $g$ its inverse. $\endgroup$ Jul 29 '14 at 16:31
  • $\begingroup$ but here compositions are not identity functions ! $\endgroup$
    – KON3
    Jul 29 '14 at 16:32
  • 11
    $\begingroup$ You wrote "Since both the compositions are polynomials, it is clear that $f(x)$ and $g(x)$ also will both be polynomials in $x$." I gave an example of nonpolynomial $f,g$ with polynomial compositions. Even though they differ from the given polynomial, this counterexamlpe shows that it is certainly not "clear". $\endgroup$ Jul 29 '14 at 16:35
  • 5
    $\begingroup$ Even assuming that both $f$ and $g$ are polynomials (which I think is probably what is expected in this problem), I don't think you can blithely assume that $f$ is quadratic and $g$ linear, rather than the other way around. It's just as safe to assume both are quadratic, and find out which one is forced to be linear (by having a quadratic coefficient of 0). $\endgroup$
    – Paul Z
    Jul 29 '14 at 16:38
  • 2
    $\begingroup$ @pre-kidney can you please tell us what is the way you meant? $\endgroup$
    – KON3
    Aug 10 '14 at 16:16
4
$\begingroup$

If you disregard the requirement that $f(x)$ be real, there is a non-polynomial solution

$$ f(x) = \dfrac{\sqrt{x-1}}{4} -\dfrac{1}{2},\ g(x) = 1 + 4 (8 x^2 + 8 x + 3)^2$$

$\endgroup$
4
+50
$\begingroup$

If it is given that $f$ and $g$ are polynomials then the product of their degrees must be $2$ (the degree of the composition of two polynomials is the product of their individual degrees), therefore one of the two must be quadratic and the other linear. I do not think that you can assume $\deg f=2$ without loss of generality.

If $\deg f=2$ and $\deg g=1$ then $g$ must be increasing, i.e., its leading coefficient is positive. The graph of $f,$ like the graph of $f\circ g,$ is tangent to the $X$ axis (discriminant $0$) and $f\circ g$ reaches its minimum $0$ when $x=-\frac12.$ The minimum of $g\circ f$ is $g(f(-1))=1$ and, because $g$ is strictly increasing, is reached when $f(x)$ reaches its minimum $0.$ Therefore the $Y$-intercept of $g$ is 1. This also tells us that $f$ reaches its minimum at $x=-1$ and we already know that $g(-\frac12)=-1$. Therefore

$$g(x)=4x+1;\ f(x)=a(x+1)^2$$

Plugging this into $g(f(x))=x^2+2x+2$ gives $a=\frac14,$ and then into $f(g(x))=4x^2+4x+1$ verifies this.

Now suppose $\deg g=2$ and $\deg f=1.$ Then $f$ must increasing (positive leading coefficient). The graph of $g$, like the graph of $g\circ f,$ has minimum $1.$ The function $g\circ f$ reaches this minimum when $x=-1.$ The minimum of $f\circ g$ is $f(g(-\frac12))=0$ and, because $f$ is strictly increasing, is reached when $g(x)$ reaches its minimum $1.$ Therefore $f(1)=0.$ This also tells us that $g$ reaches its minimum at $x=-\frac12$ and we already know that $f(-1)=-\frac12.$ Therefore

$$f(x)=\frac14x-\frac14;\ g(x)=a(2x+1)^2+1$$

Plugging this into $f(g(x))=4x^2+4x+1$ gives $a=4,$ and then into $g(f(x))=x^2+2x+2$ verifies the result.

Without any restriction on $f$ and $g$ the solutions can be very wild. If $(f_1,g_1)$ and $(f_2,g_2)$ are distinct solutions where all 4 functions leave a proper subset $A\subset\mathbb R$ and its complement intact, then two new solutions can be obtained by applying one solution to the numbers in $A$ and the other to its complement. As an example, the two polynomial solutions above map algebraic numbers to algebraic numbers and transcendental numbers to transcendental numbers.

$\endgroup$
2
  • $\begingroup$ Good answer, but I wanted a general solution for all $f,g$. $\endgroup$
    – wythagoras
    Jan 8 '16 at 21:18
  • 2
    $\begingroup$ Working on that right now; no guarantees though :-) Since the original question was, literally, whether the approach was correct, I answered by demonstrating that the OP's approach would necessarily miss at least one good solution. $\endgroup$ Jan 8 '16 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.