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Here are a few fun mathematical puzzles having to do with unknown functions.

If $f(x)=xf(x^2-7)+x$, then what is $f(2)$?

If $g(x)=g(2x-3)+g(3x-2)$, then what is $g(-1)+g(-5)+g(7)$?

If $h(x)=\frac{h(x^2-4)}{h(x-4)}$, then what is $h(\frac{-7+\sqrt{17}}{2})$?

If $j(x)=j(3x)+\frac{2}{3}x$ and j(36)=1, then what is $j^{-1}(j(54)-18)$?

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  • $\begingroup$ +1 Great puzzle. I saw there was a vote to close as off-topic, but this is more puzzle than mathematics despite how it may look, IMHO $\endgroup$ – Dr Xorile May 20 '17 at 22:35
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A solution to the third question under an assumption:

The functional equation reads \begin{align*}h(x) = \frac{h(x^2 - 4)}{h(x-4)},\end{align*} which can be rewritten to \begin{align*}h(x)h(x-4) = h(x^2-4).\end{align*} If we solve the equation $x^2-4=x$, we get a solution $x_0$ with the property that \begin{align*}x_0 = \frac{1+\sqrt{17}}{2}\quad\text{and}\quad x_0-4 = \frac{-7+\sqrt{17}}{2}.\end{align*} So if we fill in $x_0$ to the functional equation we get \begin{align*}h(x_0)\cdot h\left(\frac{-7+\sqrt{17}}{2}\right) = h(x_0).\end{align*} If we assume that $h(x_0)$ is nonzero, we can divide this equation by $h(x_0)$ to obtain \begin{align*}\boxed{h\left(\frac{-7+\sqrt{17}}{2}\right) = 1.}\end{align*} I think it might implicitly be given in the original functional equation that $h(x-4)$ is never zero, otherwise $\frac{h(x^2-4)}{h(x-4)}$ is technically not defined, although a limit might still exist ofcourse. Now $h(x-4)$ never being zero is equivalent to saying that $h(x)$ is never zero. This means that our assumption that $h(x_0) \neq 0$ is not an unreasonable one.

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If $f(x)=xf(x^2-7)+x$, then what is $f(2)$?

$f(2) = 2f(2^2-7)+2 = 2(-3f(2)-3)+2$ If $f(2) = x$ $x=2(-3x-3)+2=-6x-6+2=-6x-4 \implies x=-6x-4 \implies 7x = -4 \implies x = -\frac{4}{7}$.

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  • $\begingroup$ Nice! Please hide your answer unless it is moused over, though, so others who might want to try the problem don't see it right away. $\endgroup$ – Franklin Pezzuti Dyer May 20 '17 at 20:55
  • $\begingroup$ How do you do that? (I'm kind of new) $\endgroup$ – Eric Lee May 20 '17 at 20:57
  • $\begingroup$ I'll edit it into your answer for you. $\endgroup$ – Franklin Pezzuti Dyer May 20 '17 at 21:00
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Second question (g):

First of all, set x=3; we get g(3)=g(3)+g(7). So g(7)=0. Next, set x=1; we get g(1)=g(-1)+g(1). So g(-1)=0. Finally, set x=-1; we get g(-1)=g(-5)+g(-5). Since we already know g(-1)=0, this tells us that also g(-5)=0. Hence the required quantity is 0.

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Possibly-incorrect answer to fourth question (j):

We have j(54)=j(18)-12, and j(18)=j(6)-4, and j(6)=j(2)-4/3, etc. Adding up any number of these and telescoping, we have $j(54)=j(54\cdot3^{-k})-(18-6\cdot3^{1-k})$. So if j is continuous at 0 then it follows that $j(54)=j(0)-18$ which almost tells us that the answer we're looking for is 0 -- but the sign is wrong. So this would yield a solution if we were told that j is continuous and if the recurrence for j had a $+$ instead of a $-$.

However,

with the question as it stands I claim the answer is not uniquely determined by the available information. Say that $x$ and $y$ are "in the same class" if their ratio is an integer power of 3. Notice that our recurrence only ever relates numbers in the same class. So we may choose independently what happens in each: e.g., define j(x) arbitrarily for $1\leq x<3$, and then extend via the recurrence. By doing this we may pick any value at all for $j(54)$ (note in particular that 54 and 36 are in different classes), and then we may arrange for pretty much anything to map to $j(54)-18$.

I suppose

one might argue that the use of the notation $j^{-1}$ implies that $j$ has a well-defined inverse everywhere, or at least that exactly one thing maps to $j(54)-18$; that might be nontrivial to arrange with a construction of the sort above (though I bet it's possible). But it does look to me as if the question may perhaps be in error.

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  • $\begingroup$ Continuity is enough of an assumption (you don't need to swap signs), because then j(x) = j(0) - x/3, and from j(36) you can solve for the whole function. $\endgroup$ – ffao May 20 '17 at 22:05
  • $\begingroup$ Oh, good point. You might want to make an answer saying that. $\endgroup$ – Gareth McCaughan May 20 '17 at 23:09

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