2
$\begingroup$

A common Calculus exercise I keep running upon is a variation of the following:

When $f(x) = \frac{p(x)}{q(x)}$, define $f(x_0)$ such as that $f$ is continuous at $x_0$.

Usually, $x_0$ is a common zero of both $p$ and $q$, so the question is asking the reader to factor out that common root and proceed by taking the limit of the now defined, continuous quotient. Then, for $f$ to be continuous at $x_0$, it's value must agree with the limit. My question is, before factoring out the common zero, was $f$ really undefined? If a function is a relation between sets, and we exchanged it's formula by an equivalent one, how comes this relation is changed? What exactly means to give a function by a formula?


I try to make sense of it this way: let $K$ be a field, and $K[X]$ it's polynomial ring. Then, a polinomial function $p(x)$ can be identified with an element in $K[X]$ that is simply evaluated at the morphism $X \mapsto x$. But rational functions, functions given by a formula which is a quotient of polynomials, must be identified with an element in the field of fractions of $K[X]$. But this field is a quotient set, and any proper morphism from it needs to be properly defined. One way to do so is to take the irreducible representation of each fraction. Then, we get rid of the common zero problem, and our function $f$ was never undefined at $x_0$ in the first place, as it's value is given by the irreducible version of the formula. Given that Calculus textbooks neglect this, how is this usually handled in the mainstream?


EDIT: While I understand and appreciate the comments so far on the topic, I think I haven't made the point of the question clear enough. I comprehend that a function is not only compromised by the relation itself, but also by the domain. Thus, if I take a real function given by a formula such as $f(x) = \frac{1}{x}$, then this function is characterized by the subset $f \subset \mathbb{R}\times\mathbb{R}$ defined by $f = \{(x, \frac{1}{x}) \mid x \neq 0\}$. Somewhat formally, a function defined by $g(x) = \frac{x+1}{x(x+1)}$ will be then different from $f$ because it's domain would explicitly exclude $-1$, but it's unique continuous extension will coincide with $f$.

This definition seems ok intuitively, but when I try to make it more rigorous is where I ran into the "problems" I described. First, it's necessary to define properly what a "formula" is, as a mathematical object. It's redundant to say that a formula is an expression, because "expression" would then need to be properly defined as a mathematical object and constructed. The very fact that the difference between $\frac{1}{x}$ and $\frac{x+1}{x(x+1)}$ is subtle shows that the concept of a "formula" deserves a proper, rigorous definition. Then, once defined what a formula $f^*(x)$ is (denoted with a $*$ to separate it from the function itself), we need to define what is it's "domain" $D(f^*)$, the set of points where it applies. Then, in the reals for example, the function will be the subset of $D(f^*)\times \mathbb{R}$ given by $f = \{(x, f^*(x))\}$.

In the paragraph before this edit, I argued in favor of a definition of "formula", at least for rational functions, in which $f^*$ and $g^*$ given as above wouldn't be different formulas, and would define the same function over the base field. Points in which I think it deserves credit are that it works over any base field, not only $\mathbb{R}$, and for transcendental formulas, it can be easily generalized to the algebraic set of power series over $K$, given that $K$ is a topological field and convergence questions are properly addressed. I understand that such approach must not be canon, given the usual terminology in textbooks and the comments on this topic so far - what I ask is, if this is not the usual definition of "formula", what is it then?

I'm inclined to think that, in the mainstream, a formula is simply a fraction in the congruence class of the fraction field of $K[X]$. Then, even equivalent fractions in the usual sense could wield different functions, so $f$ and $g$ above would be different. But this goes against the usual modus operandi in mathematics, where we normally take proper care in checking that a function from a quotient set is properly defined and doesn't depend on representatives, when this can be done, as it usually simplifies a problem. Why would we go against that current?

$\endgroup$
  • $\begingroup$ Note that the domain of a rational function is the set of all points for which the denominator does not vanish. Now if $x_0$ is a point for which $q(x_0)=0$ then the function $f(x)=\frac{p(x)}{q(x)}$ is not defined at $x_0$. Only if $\lim_{x\to x_0} f(x)$ exists can we define the function $f$ at $x=x_0$ so as to make it continuous at that point. In that case we define $f(x_0)$ to be the value of the limit. $\endgroup$ – Naive Jul 18 '17 at 11:42
  • $\begingroup$ @Naive then, if we define two rational functions, each one with a different, but equivalent, fraction of polynomials, they are different functions? Take for instance any rational function $f$ with denominator vanishing at $x_0$, and multiply both denominator and numerator by $x - x_0$. It's domain has not changed, because $x_0$ was already excluded, but by such definition it must be a different function. Both still agree at every point on the domain, though. $\endgroup$ – Henrique Augusto Souza Jul 18 '17 at 11:49
  • $\begingroup$ @Naive and if we set that both functions are the same, then we are saying that the function is invariant to changing a fraction with an equivalent one. If we say it's domain excludes the points where the denominator vanishes, which equivalent fraction should we choose for defining the domain? By choosing an appropriate equivalent fraction, we could have any set of finite points excluded. $\endgroup$ – Henrique Augusto Souza Jul 18 '17 at 11:55
  • $\begingroup$ In your scenario (with the original rational function having $x_0$ as a zero of the denominator and the new function obtained by multiplying the numerator and denominator by $x-x_0$ ) both the functions are one and the same. $\endgroup$ – Naive Jul 18 '17 at 11:56
  • $\begingroup$ @Naive my argument is that the finite limit exists if and only if the zero on the denominator is also a zero on the numerator with at least the same multiplicity. Thus, a function "apparently" undefined at $x_0$ is no different than the function obtained by multiplying both terms of the fraction by $x -x_0$. If on one case we say that the functions are the same and nothing has changed, why doesn't it applies to the other? $\endgroup$ – Henrique Augusto Souza Jul 18 '17 at 11:59
2
$\begingroup$

Let us examine the - interesting - questions you pose as follows:

My question is, before factoring out the common zero, was f really undefined? If a function is a relation between sets, and we exchanged it's formula by an equivalent one, how comes this relation is changed? What exactly means to give a function by a formula?

Let us now, at first, rememeber the definition of a function between two sets $A$ and $B$:

Let $A,B$ be two sets and let $f\subseteq A\times B$ be a relation from $A$ to $B$. Then, $f$ is a function from $A$ to $B$ if and only if for every $a\in A$ exists exactly one $b\in B$ such that $(a,b)\in f$.

So, let us now see two simple functions $f\subseteq\mathbb{R}\setminus\{1\}\times\mathbb{R}$ and $g\subseteq\mathbb{R}\times\mathbb{R}$, as follows: $$f=\{(x,x+1):x\in\mathbb{R}\setminus\{1\}\}$$ $$g=\{(x,x+1):x\in\mathbb{R}\}$$ So, $f$ consists of all the arranged pairs of the form $(x,x+1)$ except for the pair $(1,2)$ and $g$ consists of all the arranged pairs of the form $(x,x+1)$ - with no exceptions. Then, it is clear that, by definition, $f\neq g$ and, more precisely, $f\underset{\neq}{\subset} g$.

So, we could easily say that $$g(x)=x+1$$ But, what about $f$?

Well, we can write, "inspired" by its set definition: $$f(x)=x+1,\ x\neq1$$ But, since $x\neq1$ we directly see that $x-1\neq0$ so we can divide by $x-1$, with no fear. Actually, we can divide by $(x-1)^n$ for every $n=1,2,\dots$, and we can get, for $x\neq1$: $$f(x)=x+1=\frac{(x-1)^n(x+1)}{(x-1)^n}$$ for every $n=1,2,\dots$. So, if we set: $$f_n(x)=\frac{(x-1)^n(x+1)}{(x-1)^n}$$ we can easily see that $f_n\equiv f$, for every $n=1,2,\dots$

By the above example, it is made clear that $\displaystyle f(x)=\frac{x^2-1}{x-1}$ and $g(x)=x+1$ are not only different with respect to their formulas, but also with respect to their domain. It has to be made clear that a function is defined by both its formula and its domain. So, $f$ and $g$ are two different relations between two different sets! (actually, $f$ is a restriction of $g$ over $\mathbb{R}\setminus\{1\}$).

With respect to your last question, defining a function only by its formula means that we, of course give the general formula of $f(x)$, as well as the restrictions over the domain implied by the formula, in order to be well defined.

For instance, $f(x)=2\ln(x)$ and $g(x)=\ln(x^2)$ are two entirely different functions, since $f$ is defined over $(0,+\infty)$ but $g$ is defined over $(-\infty,0)\cup(0,+\infty)$. However, we can see that $$f(x)=2\ln x=\ln(x^2)=g(x)\mbox{, for }x>0$$ In terms of our set definition: $$f=\{(x,\ln(x^2)):x>0\}$$ and $$g=\{(x,\ln(x^2)):x\neq0\}$$ So, it is again clear that $f\neq g$.

Lastly, it is clear that we cannot identify the functions $\displaystyle f_n=\frac{(x-1)^n(x+1)}{(x-1)^n}$ and $g(x)=x+1$ for any $n=1,2,\dots$, since they have different domains. We can, however, identify $f_n$ with $f_m$ for every $n,m=1,2,\dots$.

Note: By all this discussion, I feel that the most important thing to remain, is that the domain of a function is part of its definition (it is exactly the set $A$, of the definition, as it is given in the beginning of this answer).

Hope this helped! :)

$\endgroup$
  • $\begingroup$ Thanks for the careful and detailed answer! But I would like to cite one particular passage: With respect to your last question, defining a function only by its formula means that we, of course give the general formula of $f(x)$, as well as the restrictions over the domain implied by the formula, in order to be well defined. - my point is precisely that "formula" is undefined yet in our context. I've never seen a formal definition of a "formula" in a textbook, though the concept is widely used, and I see two different, clashing candidates for a definition that could work. See my edit. $\endgroup$ – Henrique Augusto Souza Jul 21 '17 at 19:37
  • 1
    $\begingroup$ As I see it, there might be no need for a strict mathematic definition of formula, other than the "linguistic" one, since formula is not a mathematical concept, but an intuitively commonly accepted one. It is just a means of representation of - a part of - functions. Moreover, there do exist functions that may not be represented by formulas - think of the Implicit function Theorem. In the same "mood" there does not a mathematically strict definition of the table of the values of a function. $\endgroup$ – Βασίλης Μάρκος Jul 22 '17 at 8:38
  • $\begingroup$ I see. But don't you find not having a solid and rigorous definition of formula, or in that same mood, of a table of values of a function, or of anything else as widely used as those, a bit unsettling? My feeling is that basically everything in mathematics gets formalized in some point, if used enough - even if it's formalism is one of those that you verify once in your life, in an introductory course, check that it mimics the intuitive, "linguistic" reality and then forgets entirely about it! For example, the formal definition of an ordered pair. $\endgroup$ – Henrique Augusto Souza Jul 22 '17 at 16:31
  • $\begingroup$ I see your point. Shall we refer to someone with expertise on logic and recursive languages, then? $\endgroup$ – Βασίλης Μάρκος Jul 22 '17 at 18:30
1
$\begingroup$

The definition of expression of a rational function can be carried out in a way that is similar to what one usually does for propositions in propositional logic.

Let us fix a variable $\mathsf x$ and an alphabet $C$ for real constants (or coefficients) together with a function $V \colon C \to \mathbb R$ that assigns to each constant its value.

We define inductively a set $E$ of expressions for rational functions of one variable by the following rules:

  • (variable) $\mathsf x \in E$;
  • (constant) if $c \in C$, then $c \in E$;
  • (sum) if $e, e' \in E$, then $(e + e') \in E$;
  • (product) if $e, e' \in E$, then $(e \cdot e') \in E$;
  • (quotient) if $e, e' \in E$, then $(e / {e'}) \in E$.

Each expression is simply a string in the alphabet $\{ \mathsf x, {(}, {)}, {+}, {\cdot}, {/} \} \cup C$.

Then, we simultaneously define two functions $D \colon E \to \mathcal P (\mathbb R)$ and $I \colon E \to \bigcup_{A \subseteq R} \mathbb R^A$ by structural induction on $E$ that associate to each expression $e$ a subset $D(e)$ of $\mathbb R$ called its natural domain and a function $I(e) \colon D(e) \to \mathbb R$ called its interpretation by the following rules:

  • (variable) $D(\mathsf x) = \mathbb R$ and $I(\mathsf x)(x) = x$ for any $x \in D(\mathsf x)$;
  • (constant) if $c \in C$, then $D(c) = \mathbb R$ and $I(c)(x) = V(c)$ for any $x \in D(c)$;
  • (sum) if $e, e' \in E$, then $D((e + e')) = D(e) \cap D(e')$ and $I((e + e'))(x) = I(e)(x) + I(e')(x)$ for any $x \in D((e + e'))$;
  • (product) if $e, e' \in E$, then $D((e \cdot e')) = D(e) \cap D(e')$ and $I((e \cdot e'))(x) = I(e)(x) I(e')(x)$ for any $x \in D((e \cdot e'))$;
  • (quotient) if $e, e' \in E$, then $D((e / e')) = D(e) \cap \{x \in D(e') \mid I(e')(x) \neq 0 \}$ and $I((e / e'))(x) = \dfrac {I(e)(x)} {I(e')(x)}$ for any $x \in D((e / e'))$.

One can easily check that $D$ and $I$ are well defined.

Then of course one might want to make expressions more "natural" by introducing some kind of syntactic equality: a binary relation $\equiv$ on $E$ such that, for example, $$((e + e') + e'') \equiv (e + (e' + e'')) \qquad \text{for any } e, e', e'' \in E$$ and then prove that both $D$ and $I$ are well defined on the quotient $E/{\equiv}$.

Also, it is desirable to have rules for syntactic sugar: for example, we might get rid of the outermost parentheses; write $\dfrac e {e'}$ for $(e / e')$; write $\mathsf x^3$ for $((\mathsf x \cdot \mathsf x) \cdot \mathsf x)$; and so on.

It is a bit harder to define equivalence of fractions and irreducible fractions in this context, but you should see now that all of the usual constructions can be properly defined with strings and symbols.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.