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$$f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$$

Given the functional equation above, I am trying to find the value of $f(3)$.

I do not remember the exact statement of the problem precisely, so I am not sure whether an initial value of the form $f(a) = p$ was provided.

I started off by finding the degree of $f(x)$ on both sides, which must be equal. If the degree of $f(x)$ is $n$, then the right side clearly has degree $2n$, while the left side seems to have a degree of the form $3(n-1)$.

Equating this, I got that $f(x)$ is a polynomial of degree $3$.

Now I am stuck — I thought of setting $f(x) = a x^3 + b x^2 + cx + d$ but I don't believe that will help much.

EDIT: Ok I did that, and substituted back into the functional equation to get:

$$a x^9-a \left(x^3-2\right)^3+b x^6-b \left(x^3-2\right)^2+c x^3-c \left(x^3-2\right)=\left(a x^3+b x^2+c x+d\right)^2+12$$

Ew. I guess I could expand and equate coefficients on both sides (and indeed, churching through that with Mathematica gives $6x^3 - 6$), but seeing as this is an AIME-esque problem, there should probably be an easier way. Is there one?

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    $\begingroup$ WA seems to say the answer is $f(x) \rightarrow 6 (x^3-1)$ $\endgroup$ – Brevan Ellefsen Nov 12 '15 at 0:18
  • $\begingroup$ @BrevanEllefsen Yea I followed through with expanding coefficients and solving the resulting system in Mathematca and got the same, though it is a very inelegant approach. I am hoping there is a clever way (or at least something that can be done by hand). $\endgroup$ – 1110101001 Nov 12 '15 at 0:19
  • $\begingroup$ Ya, I was just sharing the answer in case someone wants to work backwards or check their proof. $\endgroup$ – Brevan Ellefsen Nov 12 '15 at 0:21
  • $\begingroup$ Is [x] the floor function? $\endgroup$ – fleablood Nov 12 '15 at 0:24
  • $\begingroup$ Traditionally one starting point would be to try and conflate coefficients, but here that isn't as useful as it might be - you can write $f(1)$ in terms of $f(-1)$ and $f(0)$ in terms of $f(-2)$ and $f(-1)$ in terms of $f(-3)$, but $x=x^3-2$ (which is really the equation you'd like to be able to conflate) doesn't have a rational solution and I don't see any clean way of finding values to bootstrap with. $\endgroup$ – Steven Stadnicki Nov 12 '15 at 0:24
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Note that the left side depends only on $x^3$, so the right side should also. Thus we look for a function of the form $f(x) = g(x^3)$. As you say, $f$ should have degree $3$, so $g$ should have degree $1$. If $f(x) = a x^3 + b$, then taking $x = 0$ we get $b - (-8a + b) = b^2 + 12$, i.e. $8 a = b^2 + 12$. Substitute $a= (b^2+12)/8$ in to the equation, expand, and subtract right from left: I get $$ \left(-\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4}\right) x^6 + \left(-\frac{b^3}{4} - \frac{3}{2} b^2 - 3 b - 18 \right) x^3 = 0 $$ so both the coefficients of $x^6$ and $x^3$ must be $0$. Factoring the first, which is a quadratic in $b^2$: $$ -\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4} = - \frac{1}{64} (b^2 - 36)(b^2 + 12) = 0 $$ so if we're looking for real solutions, $b = \pm 6$. Of those only $b = -6$ makes the other coefficient $0$. Thus the only polynomial solution with real coefficients is $b = -6$, $a = 6$: $f(x) = 6 x^3 - 6$.

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