# A functional approach for solving equations

## Contents

Solving or interpreting equations is a fundamental topic, usually taught in young ages, when students are not very eager to capture the essence of the equations or do not know some more advanced concepts that can reinforce their understanding of them.

When students are presented with an equation $x+4=13$, usually they are asked about the value of $x$. As a student I was shown a number of manipulations that I had to learn in isolation, making the task cumbersome initially but then mechanic. Equations though have an underlying symmetry that allows you to infer most properties by intuition only.

In this post we explore equations and we will develop a different but useful perspective on the subject.

# An initial clarification about “two types of equation”

There are two cases which its critical to distinguish before we proceed

$$ \exists x \in \mathcal{D}: x+4=13$$ $$ \forall x \in \mathcal{D}: x+4=13 \Leftrightarrow x+4 \equiv 13$$

The first one corresponds to problems of the form “find $x$”. The question is: what is the value that satisfies the equation, that makes the two parts indeed equal.

The second, usually is a statement which helps you to reason about the set $\mathcal{D}$ or other parameters that accompany the equivalence.

In this post we will only discuss the first case.

# Describing the conventional solution of an example

When you are given an equation, $9x+10=91$, you are told that the left side has the same value as the right side.

Since those two parts have the same value, modifying them in the same way will maintain the equality. The fact that you don’t know the particular value of $x$ yet, is irrelevant.

Now, let’s try to describe the steps involved in solving the equation.

First we will subtract $10$ from each side. This is allowed since subtracting a number from two values that are equal, creates two new values that are also equal.

That step was useful since the two new values still contain $x$, but they are also simpler from the initial values.

Then, we have $9x=81$. Since we need to isolate x, by dividing the two equal values ($9x$ and $81$) by the same number $9$, we end up with two new values, for which we have the information of being equal.

Thus $ x=9 $.

# A function-oriented approach

Essentially what we have done before is applying transformations. In mathematics, we have a name for deterministic (same input same output) transformations. We call them functions. The need for a same-input, same-output mechanism asserts that since the left and the right part of the equation were initially the same (same input), after applying the function we will end up with equal values (same output), resulting recursively in a new equation.

Suppose we have a function $f(x)=x-10$ at our disposal. This function performs a transformation. It takes a number and it reduces it by 10. Also consider another function $g(x)=x/9$. These two functions are one-to-one, they have the property that $f(x)=f(y) \Rightarrow x=y$.

Returning to our initial problem, we apply the two functions one by one:

$9x+10=91 \Rightarrow f(9x+10) = f(91) \Rightarrow 9x+10-10=91-10 \Rightarrow 9x = 81$.

Repeating the same method:

$9x=81 \Rightarrow g(9x)=g(81) \Rightarrow \frac {9x} {9} = \frac {81} {9} \Rightarrow x = \frac {81}{9}$

## Inverse of a function

In mathematics the inverse of a function $f_1$ is a function $f_2 \equiv f_1^{-1}$ so that $f_1(f_2(x)) = x$. For instance the inverse of $f_1(x) = x/2$ is $f_2(x)=2x$ since $f_1(f_2(x)) = f_1(2x) = \frac {2x}{2} = x$.

Inverse functions “cancel” the effect of the original function.

## Equation in terms of Inverse functions.

In the equation $9x+10=91$, if we define $f(x)=9x+10$, then we essentially have to solve the equation $f(x)=91$, or to simply find the inverse of $f,\ f^{-1}$ and apply it to both sides:

$$f(x)=91 \Rightarrow f^{-1}(f(x))=f^{-1}(91) \Rightarrow x = f^{-1}(91)$$

## Composability

The fact that we can solve the equation the way we do (through a number of transformations) means that the function is decomposable into a few non-trivial functions.

Indeed, in the case of $f(x)=9x+10$, we can decompose the function in two functions: $f_1(x)=9x$, $f_2(x)=x+10$: $f(x)=f_2(f_1(x))$

Thus we can simply find the inverse function of $f_2$ and then the inverse of $f_1$.

This process is like unwrapping a gift. First your friend wraps it into multiple layers, and then you will have to unwrap it, one layer at a time, in reverse order to recover it.

## Not every function has an inverse

$f(x)=x^2$ doesn’t have a regular inverse. But we can partition the set of $x$ in which we are interested so that the individual sets give us invertible functions. For example if we are interested for solutions in $\mathbb R$, partition that space to $(-\infty,0], (0,\infty)$. In these intervals the function $x^2$ has inverses.

## Not every function takes every value

if you have the equation $f(x)=35$, $f$ doesn’t always take the value $35$. That means that the equation doesn’t have a solution; there is no $x$ that leads to that value.

## Not every function can be decomposed nicely

If our function is $f(x)=e^x + lnx$, and we need to solve the equation $f(x)=e$, Can you find an obvious decomposition? There is no one. That means that we have to perform the task of inversion in one step instead of decomposing. Turns out the inverse of this function is not even expressed in terms of common functions.

## Not every decomposition is obvious

If the equation in question is: $x^2+2x+3=0$

$f(x)=x^2+2x+3 = x^2 + 2x +1 + 2 = (x+1)^2+2 =f_3(f_2(f_1(x)))$

Where

$f_1(x)=x+1$,$f_2(x)=x^2$, $f_3(x)=x+2$.

# Extension to Inequalities

This process extends naturally to inequalities, and we might explore this issue in the future.