Linearization of arbitrary equation If we define linearization of equation as here:
http://academic.macewan.ca/physlabs/Linearization.pdf,
is it possible to linearize any arbitrary equation?
If so, is there an algorithm, I could code let's say in Mathematica, that could always take some equation and output, equation for $x$ axis, and equation for $y$ axis?
 A: It is always possible to linearize equations (given some technical requirements), but not always useful to do so. To understand the how and why, note that any equation in $x$ can be put into the form $F(x)=0$. If the equation has more than one variable in it, then $F$ would depend on these other variables as well. The simplest way to see that all equations can be put into this form is to write the equation as $(something)=(something else)$ and move the right hand side to the left: $(something)-(something else)=0$ so $F=(something)-(something else)$.
Given this, and assuming $F$ is an analytic function (meaning its Taylor series exists), then we may Taylor expand about any point $x_0$:
$$
0=F(x)=F(x_0)+F^\prime(x_0)(x-x_0)+\frac{1}{2}F^{\prime\prime}(x_0)(x-x_0)^2+\cdots.
$$
Truncating this expansion at first order (it is interesting to think about when this is reasonable to do) we would find
$$
F(x_0)+F^\prime(x_0)(x-x_0)=0
$$
but since $x_0$ is some fixed number, $F(x_0)$ and $F^\prime(x_0)$ are constants with respect to $x$, and so this is a linear equation. We would say that this is the linearization of the original equation.
I will note that what I have described here shows linearization to, in general, be an approximation to the true equation, and hence any solutions thereof will only be approximate.
The link in the question only considers examples where $F(x)$ is an invertible function. In this special case, there exists a function $F^{-1}(x)$ such that $F^{-1}(F(x))=x$. For example if $F(x)=\ln x$, then $F^{-1}(x)=e^x$. With this we are able to apply $F^{-1}$ to both sides of $F(x)=0$ and obtain
$$
F^{-1}(0)=F^{-1}(F(x))=x,
$$
which is not only linear, but is in fact the solution.
