Get to the root of the problem

Relevant wiki: Taylor Series - Problem Solving

Consider the following (J. R. Fielding, pers. comm., May 15, 2002) :

$\begin{aligned} e & = 1 + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \frac 1{4!} + \frac 1{5!} + \cdots \\ & = 1+1+\frac 12 \left(1 +\frac 13 \left(1 + \frac 14 \left(1 + \frac 15 \bigg(1 + \cdots \bigg) \right) \right) \right) \\ x^{e-2} & = \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{\cdots}}}} \\ \implies g(x) & = x^{e-2} \end{aligned}$

Similarly, for $f(x)$ :

$\begin{aligned} \frac 1e & = 1 - \frac 1{1!} + \frac 1{2!} - \frac 1{3!} + \frac 1{4!} - \frac 1{5!} + \cdots \\ & = \frac 12 \left(1 -\frac 13 \left(1 - \frac 14 \left(1 - \frac 15 \bigg(1 - \cdots \bigg) \right) \right) \right) \\ x^\frac 1e & = \sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt[4]{\frac{x}{\ddots}}}}}} \\ \implies f(x) & = x^\frac 1e \end{aligned}$

Therefore, we have:

$\begin{aligned} f(x)g(x) & = 2016 \\ x^{\frac 1e + e-2} & = 2016 \\ \left( \frac 1e + e-2\right) \ln x & = \ln 2016 \\ \ln x & = \frac {\ln 2016}{\frac 1e + e-2} \\ \implies x & \approx 1102.446 \\ \lfloor x \rfloor & = \boxed{1102} \end{aligned}$