Mysterious Function

$f^{ 1 }\left( x \right) +f^{ 2 }\left( x \right) +...=f\left( x \right)$ --- (1)

Differentiating w.r.t. $x$ , $f^{ 2 }\left( x \right) +f^{ 3 }\left( x \right) +...=f^{ 1 }\left( x \right)$ --- (2)

Substituting (2) into (1), $2f^{ 1 }\left( x \right) =f\left( x \right)$

$\frac { 2f^{ 1 }\left( x \right) }{ f\left( x \right) } =1$

Integrating w.r.t. $x$ , $2\ln { f\left( x \right) } =x+C$

$f(0)=1\Rightarrow C=0$

$f\left( x \right) ={ e }^{ \frac { x }{ 2 } }$

$\therefore \ln { f\left( 2 \right) =1 }$

Yo Shao Hong, long time no see huh! This is how I did it. I just randomly guessed $f(x) = e^{kx}$ because $ln f(2)$ gives you an integer so $f(x)$ is probably an exponential function. We have:

$f'(x) + f''(x) + f'''(x).... = f(x)$

$ke^{kx} + k^2 e^{kx} +k^3 e^{kx}+ ... = e^{kx}$

Dividing by $e^{kx}$ throughout, we have:

$k+k^2 +k^3 ... = 1$

$k(1+k+k^2.... )= 1$

$k(\frac{1}{1-k}) =1$ by sum to infinity

$k = 1-k$

$2k = 1$

$k = \frac{1}{2}$ so $f(x) = e^{\frac{x}{2}}$ .

So this means you will get an integer for $ln (2)$ , $ln f(4)$ , $ln f(6)$ and so on and so forth...