# Why is the Absolute value / modulus function used?

Why is the absolute value function or modulus function $$|x|$$ used ? What are its uses?

For example the square of a modulus number will always be positive, but why is it used when for example the square of any number whether positive or negative is always positive ? For example, $$X^2$$, will give a positive number whether negative or positive where $$X$$ is any number positive or negative.

• $x^2\ne |x|$ so if I want the positive value of $x$ , how would I "just do" $x^2$? Dec 22 '18 at 19:26
• I’m just giving an example. What is the use of the modulus function ?
– Dan
Dec 22 '18 at 19:27
• Obviously: Getting the absolute value of a number. Dec 22 '18 at 19:33
• It has many uses. Have you had Calculus? It is used in definitions where we only care about the distance between two points regardless of which one is greater. e.g. $|x-c|\lt\delta\implies |f(x)-f(c)|\lt\epsilon$. Dec 22 '18 at 19:33
• The absolute value has use in absolutely every application of mathematics, pun intended. Dec 22 '18 at 20:47

One use of it is to define the distance between numbers. For example, in Calculus, you may want to say "the distance between $$x$$ and $$y$$ is less than $$1$$". The way to write that mathematically is $$|x-y|<1$$. And you want to write it mathematically so you can work with it mathematically.

The notation $$\vert x\vert$$ for absolute value of $$x$$ was introduced by Weierstrass in 1841:

K. Weierstrass, Mathematische Werke, Vol. I (Berlin, 1894), p. 67.

Quoted from 

...There has been a real need in analysis for a convenient symbolism for "absolute value" of a given number, or "absolute number," and the two vertical bars introduced in 1841 by Weierstrass, as in $$\vert z\vert$$, have met with wide adoption;...

Extra information: Absolute is from the Latin absoluere, "to free from"; hence suggesting, to free from its sign.

 Florian Cajori, A History of Mathematical Notations (Two volumes bound as one), Dover Publications, 1993.

My take on a usage example of absolute value: $$\min(x,y)=\frac{1}{2}(|x+y|-|x-y|)$$ $$\max(x,y)=\frac{1}{2}(|x+y|+|x-y|)$$

• @HansLundmark me too, that was the reason why I have another reference, let me write it down. Done. Dec 22 '18 at 19:50

In the context of real numbers the absolute value of a number is used in many ways but perhaps very elementarily it is used to write numbers in a canonical form. Every real number $$a\ne 0$$ is uniquely equal to $$\pm \left |a\right|$$. So if we define the sign function $$s\colon \mathbb R\setminus\{0\}\to \{+,-\}$$ given by $$s(a)=+$$ if $$a>0$$ and $$s(a)=-$$ if $$a<0$$, then: for all $$a\ne 0$$ in $$\mathbb R$$ we have $$a=sign(a)\cdot \left | a \right |$$. In a sense this is a way to build all the reals from the positive ones. This is all just a special case of the polar representation of complex numbers, a representation of utmost importance.

## Because both of them are useful.

You explicitly mentioned the square function. Therefore, I want to give some examples. The main idea is that the non-differentiability of $$|\cdot|$$ is useful in minimization problem.

### Estimators

We know that the arithmetic mean $$\hat{\mu}=\sum_{i=1}^n x_i$$ gives

$$\min_{\mu} \,(x_i-\mu)^2$$

but it is less well-known that the median gives

$$\min_{\mu} \, |x_i-\mu|.$$

### Signal Processing

Let's use image processing as an example. Suppose $$g$$ is a given, noisy image. We want to find some smoother image $$f$$ which looks like $$g$$.

The Harmonic L$$^2$$ minimization model solves

$$-\bigtriangleup f + f = g$$

and it turns out to be equivalent to solving a minimization problem:

$$\min_{f} \,(\int_{\Omega} (f(x,y)-g(x,y))^2 dxdy + \int_{\Omega} |\nabla{f(x,y)}|^2 dxdy).$$

An enhanced version is the ROF model. It solves

$$\min_{f} \,(\frac{1}{2} \int_{\Omega} (f(x,y)-g(x,y))^2 dxdy + \lambda \int_{\Omega} |\nabla{f(x,y)}| dxdy).$$

Notice that for appropriate $$\lambda$$, these two models only differ by a square. Another remark is that $$|\cdot|$$ gives the Euclidean norm when the argument is a vector. However, the idea still applies since the norm is non-zero

### Model Selection

In classical model selection problem, we are given a set of predictors and a response (in vector form). We want to decide which predictors are useful. One way is to choose a "good" subset of predictors. Another way is to shrink the regression coefficients.

The classical regression model solves the following minimization problem:

$$\min_{\beta_0,...,\beta_p} \sum_{i=1}^n (y_i-\beta_0-\sum_{j=1}^p \beta_j x_{ij})^2$$

The Ridge Regression solves the following:

$$\min_{\beta_0,...,\beta_p} \sum_{i=1}^n (y_i-\beta_0-\sum_{j=1}^p \beta_j x_{ij})^2+\lambda \sum_{j=1}^p {\beta_j}^2$$

, so that larger $$\beta_j$$ gives penalty.

Another version is Lasso, which solves

$$\min_{\beta_0,...,\beta_p} \sum_{i=1}^n (y_i-\beta_0-\sum_{j=1}^p \beta_j x_{ij})^2+\lambda \sum_{j=1}^p |\beta_j|.$$

In computer science, it can be useful for computing vectors.

Let x be a projection of a vector into 1 dimension such that it has a direction (plus or minus), and a magnitude.

If I want to square the vector but preserve its direction (say, to find air resistance from velocity), then I can take: $$a = x^2 \frac{|x|}{x}$$ or simplified: $$x|x|$$ This gives me the same direction, but the magnitude is squared. Therefore, the vector I am calculating acts in the correct direction.