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

### 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:

• 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.
• 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.
• 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.
• In all of your examples, the natural domains of the functions on both sides are the same and they take the same values, so they are not only equivalent, but just the same functions, only expressed in different ways. – user539887 May 4 '18 at 18:48
• I think you'll need a more rigorous definition of "linked". For instance, does it disregard substitution? Cause if we know $f=g$, then we can directly substitute $f$ for $g$ and it's always true. – Dando18 May 4 '18 at 18:50

## 1 Answer

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 $k \ge 2$.