Functional equations game

\(\begin{array} {} P(x,y): & f(x+y) + f(x-y) = 2f(x) g(y) & ...(0) \\ P(0,x): & f(x) + f(-x) = 2f(0) g(x) & ...(1) \\ P(0,-x): & f(-x) + f(x) = 2f(0) g(-x) & ...(2) \end{array} \)

From $(1)$ and $(2)$ , we note that $g(x) = g(-x)$ . This means that $g(x)$ is even.

\(\begin{array} {} P(x,x): & f(2x) + f(0) = 2f(x) g(x) & ...(3) \\ P(-x,-x): & f(-2x) + f(0) = 2f(-x) \color{blue} g(x) & ...(4) & \small \color{blue} \text{Since }g(x) = g(-x) \end{array} \)

Then $(3)+(4):$

$\begin{aligned} {\color{#3D99F6}f(2x) + f(-2x)} + 2f(0) & = 2g(x) ({\color{#3D99F6} f(x) + f(-x)}) & \small \color{#3D99F6} (1): \ f(x) + f(-x) = 2f(0) g(x) \\ {\color{#3D99F6}2f(0)g(2x)} + 2f(0) & = 2g(x) \cdot \color{#3D99F6}2f(0)g(x) \\ g(2x) + 1 & = 2(g(x))^2 \\ \implies g(2x) & = 2(g(x))^2 - 1 \quad ...(5) \end{aligned}$

It would appear that $g(x) = \cos x$ , so that $\cos (2x) = 2\cos^2 x - 1$ . But $g''(x) = - \cos x$ and $g''(0) = -1 \ne 1$ . Therefore $g(x) \ne \cos x$ .

Instead, it is $g(x) = \cosh x = \dfrac {e^x + e^{-x}}2$ . Then $2(g(x))^2 - 1 = 2\left(\dfrac {e^x+e^{-x}}2\right)^2 - 1 = \dfrac {e^{2x}+e^{-2x}}2 = g(2x)$ ; and $g''(x) = \cosh x$ , $\implies g''(0) = 1$ . And $f(x) = e^x$ . Then $P(x,y) = f(x+y) + f(x-y) = e^{x+y} + e^{x-y} = 2e^x \left(\dfrac {e^y+e^{-y}}2 \right) = 2f(x) g(y)$ , proving $f(x) = e^x$ and $g(x) = \cosh x$ satisfy the functional equation.

Therefore, $\lfloor \ln (g(2019)) \rfloor = \lfloor \ln (\cosh 2019) \rfloor = \boxed {2018}$ .

First, let's differentiate twice with respect to $y$ . We obtain: $f''(x+y) + f''(x-y) =2 f(x)g''(y)$

By making $y = 0$ and $g''(0)=1$ , we obtain the differential equation $f''(x)+f''(x)=2f(x)$ or $f''(x)-f(x) = 0$ , with solutions of $f(x) = C_1e^x + C_2e^{-x}$ (1).

Substituting (1) into the original equation gives us: $C_1e^{x+y} + C_2e^{-(x+y)} + C_1e^{x-y} + C_2e^{-(x-y)} = 2(C_1e^x + c_2e^{-x})g(y)$

Using direct calculations, we obtain $\frac{1}{2}(e^y + e^{-y} ) = g(y)$ or $g(y) = cosh(y)$ .

Finally, we evaluate the function at $2019$ and find $\lfloor\ln(g(2019))\rfloor$ .

We obtain that $\lfloor\ln(g(2019))\rfloor = 2018$ .

Let us first differentiate the original functional equation twice with respect to $y$ :

$f''(x+y) + f''(x-y) = 2f(x)g''(y) \Rightarrow g''(y) = \frac{f''(x+y)+f''(x-y)}{2f(x)}$ (i)

and if $g''(0) = 1$ , then we have:

$1 = \frac{f''(x) + f''(x)}{2f(x)} \Rightarrow f''(x) - f(x) = 0 \Rightarrow f(x) = Ae^{x} + Be^{-x}$ (ii)

Substituting (ii) back into the original functional equation now gives:

$Ae^{x+y} + Be^{-(x+y)} + Ae^{x-y} +Be^{-(x-y)} = 2(Ae^{x} + Be^{-x}) \cdot g(y)$ ;

or $Ae^{x}(e^{y} + e^{-y}) + Be^{-x}(e^{y} +e^{-y}) = (Ae^{x} + Be^{-x}) \cdot 2g(y);$

or $(Ae^{x} + B^{-x})(e^{y} + e^{-y}) = (Ae^{x} + Be^{-x}) \cdot 2g(y)$ ;

or $e^{y} + e^{-y} = 2g(y)$ ;

or $g(y) = \frac{e^{y} + e^{-y}}{2} = cosh(y).$

Finally, $\lfloor ln(g(2019)) \rfloor = \lfloor ln(\frac{e^{2019} + e^{-2019}}{2}) \rfloor;$ ;

or $\lfloor ln(\frac{e^{2019}(1 + e^{-4038})}{2}) \rfloor$ ;

or $\lfloor ln(e^{2019}) + ln(1 + e^{-4038}) - ln(2) \rfloor \approx \lfloor 2019 - ln(2) \rfloor = \boxed{2018}.$