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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?

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    $\begingroup$ These issues involve conventions used when functions are defined by formulas. It is standard practice today (and for at least the past 100 years) to regard $f$ as undefined at $x=-1.$ Also, it is typically the case that one would not call $x=-1$ a point of discontinuity of $f,$ since this point is not in the domain of $f$ (any more than one would say that $f$ is discontinuous at some quaternion value, or at some matrix value, or at some geometrical triangle value, or at some well-formed formula (in logic) value, etc.). $\endgroup$ – Dave L Renfro Dec 29 '15 at 14:50
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    $\begingroup$ You are right. I should have written singularity instead of discontinuity (as I did in the introduction). However, I somehow remember being taught that instead of having $f$ undefined at $x=-1$, one should rather simplify the fraction and thus avoid that singularity. $\endgroup$ – Philipp Imhof Dec 29 '15 at 14:54
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    $\begingroup$ I think it depends a lot on the level of rigor desired and what one wants to do with the concepts. In an elementary algebra course the emphasis is (mostly) on algebraic manipulation, and not on set-theoretic/logical analysis of functions, but in an undergraduate real analysis course the emphasis gets reversed. $\endgroup$ – Dave L Renfro Dec 29 '15 at 14:58
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    $\begingroup$ Your question explains very well (one reason) why I dislike the secondary education view of defining functions by a formula, rather than by the specification of domain, range, and of the image of each element of the domain. $\endgroup$ – Benoît Kloeckner Dec 29 '15 at 16:46
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    $\begingroup$ @BenoîtKloeckner Doesn't the formula for $f$ above (and formulae used to define functions in general) neatly specify exactly the domain, range, and image of each element? That's how we know $g$ is not a valid alternate formula for $f$. Furthermore, I'm hard pressed to come up with a reasonable and rigorous way to define any function with an infinite domain without using a formula. $\endgroup$ – Todd Wilcox Dec 29 '15 at 18:50
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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.

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    $\begingroup$ That's a very good point indeed. $\endgroup$ – Philipp Imhof Dec 29 '15 at 16:39
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    $\begingroup$ $h(x) = \frac{x}{x}$ is a formula, not a function; you haven't specified a function because you haven't given the domain and codomain of $h$. If you do specify the domain as all real $x \neq 0$, you could just as well write $h(x) = 1$ and it'd say the same thing. $\endgroup$ – Daniel Hast Dec 29 '15 at 17:00
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    $\begingroup$ @DanielHast In secondary education the standard convention is that unless otherwise specified any formula implicitly defines a real-valued function whose domain is the maximal subset of $\mathbb{R}$ for which the formula is well-defined. $\endgroup$ – mweiss Dec 29 '15 at 18:01
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    $\begingroup$ That convention is probably a large part of why many students come out of high school with so many misconceptions and so much confusion about what functions are, which becomes readily apparent as soon as they take calculus or any higher mathematics in college. $\endgroup$ – Daniel Hast Dec 29 '15 at 23:42
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    $\begingroup$ Agreeing with @mweiss, this is also standard in college calculus texts, e.g.: Stein and Barvellos, Calculus and Analytic Geometry, 5E (1992) Sec 2.1: "The domain consists of all $x$ such that $f(x)$ is defined. So exclude those $x$ for which the formula makes no sense (requiring, for instance, division by 0 or the square root of a negative number)." It's pretty reasonable that if say, a course concerns only the context of real numbers, that be taken as the domain for the whole course and not have to be repeated for every individual example and exercise. $\endgroup$ – Daniel R. Collins Dec 30 '15 at 3:28
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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.

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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)

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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.

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  • $\begingroup$ The rationals in $\mathbb{R}(X)$ are most definitely not a "set of functions", they are (or can be represented as) a set of quotients of formal polynomials. The equality between such quotients is defined by formal manipulations, not by looking what happens when the indeterminate $X$ takes a value. $\endgroup$ – vonbrand Dec 30 '15 at 16:10
  • $\begingroup$ I did not say and did not intend to imply it is a set of functions, I merely used "field of rational functions" as name for the algebraic structure (also note that I purposefully used a captial $X$ and at the start of the post said mapping not function), which as you said is the quotient field of the ring of polynomials (in one variable) or equivalently the purely transcendental field extension of transcedence degree $1$. I believe to call it field of rational functions or rational function field (in one variable) is not uncommon. $\endgroup$ – quid Dec 30 '15 at 20:48
  • $\begingroup$ Tangentially as the reals are infinite it'd be isomorphic in a natural way to a set of functions/maps, namely the rational ones without removable singularity. And in some sense this is the main point of my answer: it is a convention to interpret the expression by first considering the two polynomials as definining maps and taking quotients then (where possible), yet one could also take the quotient in a formal way first and assign a function/map to the formal quotient. $\endgroup$ – quid Dec 30 '15 at 20:51

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