# Finding a Function Given a Condition

1. Feb 22, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
I am interested in working on this problem: https://www.physicsforums.com/threads/unknown-composite-function.902469/

From I gather, it seems this problem has been solved already. I was considering joining the thread, but I didn't want to flood songoku's inbox; so I started a new one. I hope it is okay of me to start this thread. The problem reads:

Find $f(x)$ given that $f(f(x) - x^2) = x^2 - 5x + 3$

2. Relevant equations

3. The attempt at a solution

Following some of the suggestions there, I assumed that $f(x) - x^2 = ax + b$ or that $f(x) = x^2 + ax + b$. Hence, the condition becomes $f(ax+b) = x^2 - 5x + 3$. Then, taking $f(x) = x^2 + ax + b$ and substituting in $ax+b$, it seems we need to require that

$f(ax+b) = x^2 - 5x + 3$

or

$(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3$

to hold for all $x$.

However, this doesn't have the appearance of an elegant solution...

My next question is, are polynomials the only functions that satisfy the condition? How do I know whether there exists some intricate combination of trig functions, exponentials, etc that satisfy the condition?

2. Feb 22, 2017

### SammyS

Staff Emeritus
Why the need for an elegant solution?​
alternatively:

3. Feb 22, 2017

### haruspex

Following my suggestion, you only need to assume f is a polynomial.
You could try relaxing that to any power series convergent for |x|<1, say.

4. Feb 23, 2017

### Bashyboy

Here is one way of solving it; although I didn't fully carry out the other method, this seems easier. Let $f(x) = x^2 + ax + b$. Then the condition $f(f(x)-x^2) = x^2 - 5x + 3$ becomes $f(ax+b) = x^2 -5x + 3$. Taking the derivative of each, we get

$f'(x) = 2x + a$

$af'(ax+b) = 2x-5$

Thus, multiplying the first by $a$ and substituting in $ax+b$, we require that $af'(ax+b) = 2a(ax+b)+a = 2a^2x + 2ab + a^2$ must equal $2x-5$ for every $x$; i.e.,

$2a^2 x + 2ab + a^2 = 2x-5$ for every $x$.

Taking $x=0$ and $x=1$ gives us a system which, when solved, yields the solutions $(a,b) = (1,-3)$ and $(a,b) = (-1,-3)$. However, $(a,b) = (1,-3)$ is the only solution that insures $f(f(x)-x^2) = x^2 - 5x + 3$ is satisfied.

Are there any methods that does not appeal to calculus, besides the one in my first post? This problem is no fun without Calculus---too many nasty multivariable equations and ways of messing up a calculation. Anyone who knows me knows that I am terrible at calculating. I had to do the above solution twice, because I didn't properly distribute a $2$---a stinkin' $2$!!

I wonder, are these the only $a$ and $b$ that work? Obviously by taking $f(x)$ to have the form $f(x) = x^2 + ax + b$, we are considerably narrowing the search space; but it also seems that also requiring the derivatives to be equal further narrows the search space since we are loosing quadratic terms (not sure about this last point, though).

Last edited: Feb 23, 2017
5. Feb 23, 2017

### haruspex

I do not see this as any advance on your original method. And as I mentioned, you do not need to assume f is a quadratic. If you just assume it is a polynomial the proof is not hard. Relaxing the assumption to allow any power series would be the challenge.

6. Feb 23, 2017

### Bashyboy

Really? I find that rather surprising. The constraint $(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3$ becomes

$a^2(x^2+x) +ab(2x+1)+b^2+b = x^2 - 5x + 3$ for every $x$. The easiest values to work with would be $x=0$ and $x = - \frac{1}{2}$, yielding the equations

$3a^2 + 4b^2 + 4b = 23$

$ab + b^2 + b = 3$

I tried to solve for $a$ and $b$ by multiplying the second equation by $-4$ and adding the result to the first equation, but had no luck. How should I proceed from here?

7. Feb 23, 2017

### haruspex

It is not necessary to consider specific values. Just equate terms with the same power of x. That is how you can solve it on the assumption that it is a polynomial, and maybe extend to power series.

8. Feb 25, 2017

### SammyS

Staff Emeritus
Expand the left side and collect like terms.
$a^2x^2+(2ab+a^2)x+ab+b^2+b$
Equating coefficients leads to finding $\ a\ \text{and}\ b\$ without difficulty.
$a^2=1\; \rightarrow \; \left | a \right |=1$

Plugging that into $\ (2ab+a^2)=-5\$ gives a result for $\ ab\,.$

From this you get two possible combinations for $\ (a,\,b)\,.$

Equating constant terms eliminates one of those two.