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Suppose the function $f:\Bbb R\to\Bbb R$ satisfies the following conditions:

$$\begin{align} f(4xy)&=2y[f(x+y)+f(x-y)] \\[4pt] f(5)&=3 \end{align}$$

Find the value of $f(2015)$.

I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.

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Firstly, it’s trivial that $$f(0) = f(4\cdot 0\cdot 0) = 2\cdot 0 \cdot [f(0) + f(0)] = 0$$

Next, we see that $$0 = f(0) = f(4\cdot 0 \cdot y) = 2y\cdot [f(y) + f(-y)]\text{,}$$ which implies that $f(-y) = -f(y)$ for all $y\neq 0$.

Afterwards, one notices that $$f(4xy) = f(4yx)$$ Hence, $$2y\cdot [f(x+y) + f(x-y)] = 2x\cdot [f(x+y) + f(y-x)]$$ $$\Leftrightarrow (x-y)\cdot f(x+y) = (x+y)\cdot f(x-y)$$ $$\Leftrightarrow f(x+y) = \frac{x+y}{x-y}\cdot f(x-y)$$

If we substitute now $x=1010$ and $y=1005$, we get that $$f(2015) = \frac{2015}{5}\cdot 3 = 1209$$

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  • $\begingroup$ Very nice! It's a more direct way to get the function value than what I did in my answer. $\endgroup$ – John Omielan Jan 1 at 1:22
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With the provided function,

$$f\left(4xy\right) = 2y\left[f\left(x + y\right) + f\left(x - y\right)\right] \tag{1}\label{eq1}$$

First, substitute $y = 0$ to get

$$f\left(0\right) = 0 \tag{2}\label{eq2}$$

Next, substitute $x = 0$ and use \eqref{eq2} to get

$$0 = 2y\left[f\left(y\right) + f\left(-y\right)\right] \tag{3}\label{eq3}$$

Thus, for all values of $y$ other than $0$, dividing both sides by $2y$ gives

$$f\left(y\right) = -f\left(-y\right) \tag{4}\label{eq4}$$

In other words, $f$ is an odd function. Next, substituting $y = 1$ into \eqref{eq1} gives

$$f\left(4x\right) = 2\left[f\left(x + 1\right) + f\left(x - 1\right)\right] \tag{5}\label{eq5}$$

Now, using $x = 1$ in \eqref{eq5} gives

$$f\left(4\right) = 2\left[f\left(2\right) + f\left(0\right)\right] \tag{6}\label{eq6}$$

Thus, using \eqref{eq2} gives

$$f\left(4\right) = 2f\left(2\right) \tag{7}\label{eq7}$$

Similarly, using $x = 3$ in \eqref{eq5} gives

$$f\left(12\right) = 2\left[f\left(4\right) + f\left(2\right)\right] \tag{8}\label{eq8}$$

Thus, using \eqref{eq7} gives

$$f\left(12\right) = 6f\left(2\right) \tag{9}\label{eq9}$$

Note that \eqref{eq2}, \eqref{eq4}, \eqref{eq7} and \eqref{eq9} all satisfy

$$f\left(nx\right) = nf\left(x\right) \forall \; n \in N \tag{10}\label{eq10}$$

In particular, for \eqref{eq2}, $n = 0 \; \forall \; x \in R$; for \eqref{eq4}, $n = -1 \; \forall \; x \in R$; and for \eqref{eq7} & \eqref{eq9}, $x = 2$ only, with $n = 2$ & $n = 6$, respectively. This doesn't prove \eqref{eq10} always works, but it suggests that $f$ is a linear multiple of $x$, i.e, $f\left(x\right) = kx \text{ for a non-zero constant } k \in R$, which from the given condition of

$$f\left(5\right) = 3 \tag{11}\label{eq11}$$

gives a value of $k = \frac{3}{5}$ so the function would be

$$f\left(x\right) = \cfrac{3x}{5} \tag{12}\label{eq12}$$

To confirm this, substitute \eqref{eq12} into \eqref{eq5} to give on the left side

$$\cfrac{12x}{5} \tag{13}\label{eq13}$$

with the right side becoming

$$2\left[\cfrac{3\left(x + 1\right)}{5} + \cfrac{3\left(x - 1\right)}{5} \right] = 2\left[\cfrac{\left(3x + 3 + 3x - 3\right)}{5} \right] = \cfrac{12x}{5} \tag{14}\label{eq14}$$

Similarly, using \eqref{eq12} in \eqref{eq1} gives a left & right hand side of $\frac{12xy}{5}$, confirming it's also a solution of the original equation. I'm not quite sure offhand how to prove it's unique, but I assume it is. We thus get a final answer of

$$f\left(2015\right) = \cfrac{3 \times 2015}{5} = 1209 \tag{15}\label{eq15}$$

Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did. In particular, I tried using small constant values for $x$ and $y$, checking what information this offered, and then seeing if I could leverage this when using other values. For example, I started with $x = 0$ and $y = 0$, then used $y = 1$, $x = 1$ and $x = 3$. Each time, I used previous details to help simplify and learn more about the new values, until I saw the pattern which I then confirmed worked.

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  • $\begingroup$ I am trying to give a reason for the 10th step. $\endgroup$ – yuanming luo Dec 27 '18 at 2:08
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    $\begingroup$ @yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago. $\endgroup$ – John Omielan Dec 27 '18 at 2:17
  • $\begingroup$ Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best. $\endgroup$ – yuanming luo Dec 27 '18 at 3:14
  • $\begingroup$ I don't see how (10) is implied by the equations you claim imply it. $\endgroup$ – Henning Makholm Dec 31 '18 at 22:38
  • $\begingroup$ @HenningMakholm Thanks for your question. ($10$) works with all the equations I list. With ($2$), $f\left(0x\right) = 0x \forall \; x \in R$; with ($4$), $f\left(-x\right) = -x \forall \; x \in R$; with ($7$), $f\left(2 \times 2\right) = 2\left(2\right)$; & with ($9$), $f\left(6 \times 2\right) = 6f\left(2\right)$. I have updated my solution so this is more clear. This obviously doesn't prove that ($10$) works for all values of $x$ & $n$, but it would have to be a quite unusual function for it not to. I just used the apparent pattern to determine a proposed solution that I confirmed works. $\endgroup$ – John Omielan Dec 31 '18 at 22:55

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