Is simplifying a rational function considered as a continuous extension?

Question

Given the rational function $f(x)=\frac{x^2-1}{x+1}$. The expression can be simplified to $g(x)=x-1$ and thus the singularity at $x=-1$ is removed.

I would personally claim that $f$ and $g$ are the same function and that $f$ is defined on all $\mathbb R$ with no discontinuities.

However, in some text book I read, the author stated that by doing so, one obtained the continuous extension of $f$ and would remove the discontinuity.

I am aware that this question is somehow opinion based, but: Where does the creation of a new (extended) function start and where are we just refactoring terms while maintaining a given function?

Answer 1

I would say they are definitely not the same function. Let's take an even simpler example:

$$h(x)=\frac{x}{x}$$

This function is defined for all $x \ne 0$, and at such points it is equal to $1$. But it is not the same function as the constant function $$k(x)=1$$ because the two functions have different domains. See also my answer at https://math.stackexchange.com/questions/1525054/why-are-removable-discontinuities-even-discontinuities-at-all/1525156#1525156.

Answer 2

The expression $f(x) = \frac{x^2 - 1}{x + 1}$ is a formula (with two free variables, $f$ and $x$), not a function. It is not the same formula as $f(x) = x - 1$, but neither is $1 + 1 = 2$ the same formula as $2 = 1 + 1$.

By convention, such formulas are sometimes taken to represent functions in contexts where the domain and codomain can be inferred from context. In this case, one could say more precisely that $f: \mathbb{R} \setminus \{-1\} \to \mathbb{R}$ is the function defined for each $x \in \mathbb{R} \setminus \{-1\}$ by $f(x) = \frac{x^2 - 1}{x + 1}$. This is a continuous function. (The "missing point" in the domain, $-1$, is a (removable) singularity, not a discontinuity; it makes no sense to talk about continuity of a function outside its domain.)

If we instead tried to say that $f: \mathbb{R} \to \mathbb{R}$ is the function defined for each $x \in \mathbb{R}$ by $f(x) = \frac{x^2 - 1}{x + 1}$, we have spoken nonsense, because the expression $\frac{x^2 - 1}{x + 1}$ has no meaning if $x = -1$. However, we could instead take the domain to be literally any subset of $\mathbb{R} \setminus \{-1\}$, or even a subset excluding $-1$ of some other field (who said we're working in the real numbers?), provided we change the codomain to match.

Similarly, given any $D \subseteq \mathbb{R}$, one can say $g: D \to \mathbb{R}$ is the function defined for each $x \in D$ by $g(x) = x - 1$. Of course, this is a different function for each choice of $D$; from a structural standpoint, it doesn't even make sense to compare functions with different domains or codomains (and if one works in a framework where doing so does make sense, they're never equal anyway).

So, yes, if we choose $D = \mathbb{R}$, then $f$ and $g$ are not the same function. But they are related, in that $f$ is the restriction of $g$ to the domain of $f$; in other words, $g$ is an extension of $f$ (and, as it happens, a continuous extension).

On the other hand, if we choose $D = \mathbb{R} \setminus \{-1\}$, then $f$ and $g$ are literally the same function, because they have the same domain and codomain, and $x - 1$ and $\frac{x^2 - 1}{x + 1}$ are the same number for every $x$ in that domain.

Also, if we interpret $f$ and $g$ not as functions, but as "rational functions in one indeterminate $x$", i.e., elements of the field $\mathbb{R}(x)$, then $f = g$. Note, however, that "rational functions" in this sense are not functions, but equivalence classes of formal fractions of polynomials.

Answer 3

Given the rational function $f(x)=\frac{x^2-1}{x+1}$. The expression can be simplified to $g(x)=x-1$ and thus the singularity at $x=-1$ is removed.

I would personally claim that $f$ and $g$ are the same function and that $f$ is defined on all $\mathbb R$ with no discontinuities.

This is false (and not a matter of opinion). When $x = -1$, then $f(x)$ is undefined (not any number), but $g(x) = -2$. They are clearly not the same function.

One can only make the simplification $\frac{x^2-1}{x+1} = x-1$ so long as $x$ is not $-1$. You sort of admit as much when you use the language "the singularity at $x=-1$ is removed", which recognizes that some change has been made.

This may the most common error at the level of precalculus?

Whenever one learns a new mathematical operation, it is imperative also to learn the limitations under which the operation may be performed. Lack of this additional knowledge can lead to the employment of the new operation in a blindly formal manner in situations where the operation is not properly applicable, perhaps resulting in absurd and paradoxical conclusions. Instructors of mathematics see mistakes of this sort made by their students almost every day... (Howard Eves, Great Moments in Mathematics, Lecture 32)

Answer 4

This is really a matter of convention. But a usual convention is that for polynomials $g,h$ one considers the rational mapping
$$\frac{g(x)}{h(x)}$$ to have domain $\mathbb{R}\setminus \{x \colon h(x)\neq 0\}$ and likewise in the case one is working over the complex numbers.

One could also set things up so that one reduces $$\frac{g(x)}{h(x)}= \frac{\overline{g}(x)}{\overline{h}(x)}$$ where $\overline{g}, \overline{h}$ are co-prime and then have domain $\mathbb{R}\setminus \{x \colon \overline{h}(x)\neq 0\}$ but this is just not what is common.

Not only but also as generalizing from polynomials $g,h$ to more general functions there is no clear-cut analog for the co-prime condition. Say already with $$\frac{x}{\sin(x)}$$ it feels more natural to consider this as defined on $\mathbb{R} \setminus \pi \mathbb{Z}$ rather than to special case $0$.

It is however true that as elements of the algebraic structure field of rational functions $\mathbb{R}(X)$ the two expressions $X-1$ and and $\frac{X^2-1}{X+1}$ denote the very same element.

To sum it up: the convention is to read such expressions as quotient of the "visible" (polynomial) maps and not as representing an elements of the field of rational functions to which then a map is assigned to.