For two functions described by finite equations, such that $f(x)=g(x)$, must their equations be somehow linked?

Question

A Rephrasing:

Is it possible to have two equations $f$ & $g$, such that $f(x)$ is functionally the same as $g(x)$ for all values of x, but the definition of f does not lead to the definition of g (aside from the fact that they result in the same output)?

The original, more specific question where this stems from:

Given the functional description of a logic circuit, for any possible boolean alegbra/logic diagram representation, is it possible to transform it into all other solutions/representations (i.e by rearranging, using De Morgan's Laws, the relevant Monotone Laws, etc.)?

Given two functions, $f(x)$ & $g(x)$, must those two equations be linked or able to be demonstrated equivalent? Or are there any examples contrary to this?

Examples that follow this rule:

1. • Consider
     $f(x) = g(x)$
     $f(x) = \frac{1}{\sqrt{2}}x$
     $g(x) = \frac{\sqrt{2}}{2}x$
   • Since $\frac{1}{\sqrt{2}}$ can be rearranged as $\frac{\sqrt{2}}{2}$ they can be linked.
2. • Consider
     $f(x) = g(x)$
     $f(x) = \sin(\frac{\pi}{2} - x)$
     $g(x) = \cos(x)$
   • Since $x$ and $\frac{\pi}{2} - x$ are complementary angles they can be linked.
3. • Consider
     $f(x) = g(x)$
     $f(x) = \cos(x)$ $g(x) = \sum_{n=1}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$
   • Since $\sum_{n=1}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$ is the taylor series representation of $\cos(x)$ they can be linked.

Answer 1

Proceeding until I get complete nonsense.

$\begin{array}\\ x &=\ln(e^x)\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{(e^x)^n}{n}\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{e^{nx}}{n}\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}\sum_{k=0}^{\infty}\dfrac{(nx)^k}{k!}\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}(1+\sum_{k=1}^{\infty}\dfrac{(nx)^k}{k!})\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}+\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}\sum_{k=1}^{\infty}\dfrac{(nx)^k}{k!}\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}+\sum_{k=1}^{\infty}\dfrac{x^k}{k!}\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}n^k\\ &=\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}+\sum_{k=1}^{\infty}\dfrac{x^k}{k!}\sum_{n=1}^{\infty} (-1)^{n-1}n^{k-1}\\ \end{array} $

Equating coefficients:

$0 = \sum_{n=1}^{\infty} (-1)^{n-1}\dfrac1{n}$.

$1 = \sum_{n=1}^{\infty} (-1)^{n-1}$.

$0 = \sum_{n=1}^{\infty} (-1)^{n-1}n^{k-1}$ for $k \ge 2$.