3
$\begingroup$

Assume $X = (0,\infty)$, $Y = \mathbb{R}$ and that $f : X \to Y$ is a function with some rule, for example $$ f(x) = ax; \qquad a\in \mathbb{R} := A. $$ These symbols provide the following inference:

($i$) the point $x$ in the domain of $f$ is associated to the point $ax \in Y$ by the relation $f$.

($ii$) the point $x$ in the domain of $f$ is associated to the point $$ \lim_{t \to 0} \frac{1}{t} \left[ a(x + t) - a(x) \right] \in Y $$ by the relation $\partial f/\partial x$.

Two concerns:

  1. I have seen authors switch between the notation $f(x)$ and $f(b)$ carelessly. That is no problem for inference provided above, because in the first instance I know that I should associate $x \mapsto ax$ and in the second $b \mapsto ab$. However, some authors seem to switch between «$f(x)$» and «$f(a)$» as to claim that «the symbol $x$ appears in the rule $f$» and «the symbol $a$ appears in the rule $f$» respectively. Every time this happens, I wonder if the picture in ($i$) and ($ii$) is indeed correct and/or if it is reconcilable with that meaning.

However, this dilemma becomes vivid if you consider the derivative:

  1. Suppose (to my horror) that an author wrote $\partial f/ \partial b = 0$ and I wanted to give meaning to this expression. It seems to me that the formula for the derivative no longer applies because we were never told how to define the limit (the derivative) for any other coordinate than $x$ (the coordinate on the domain of $f$). At best, I can give meaning to $\partial f/ \partial a$ if I redefine the function $f : A \to Y$ with a different domain by $f(a) = ax$ because the symbol $a$ appears in the rule $f$ above. However, I would not know how to define the limit with respect to the symbol $b$, because it does not appear in the product $ax$.

If the view in ($i$) and ($ii$) is indeed correct, can you explain what happens when authors write $f(a)$ «seemingly to say that $f$ contains the symbol $a$» and $\partial f/\partial b = 0$ «seemingly to say that $f$ does not contain the symbol $b$» such that these notions can be reconciled? If ($i$) or ($ii$) is false, please let me know what is correct.

$\endgroup$
-1
$\begingroup$

Some authors highlight the function and pay less attention to using the same designation for the independent variable. In essence, they are implicitly using a function notation similar to the following behind-the-scenes:

$$f(~~) = a(~~)$$

I have even seen some that intentionally do this to ensure that readers are breaking away from the comfort zone of always having $x$ as the independent variable or even $f$ as the function.

For instance, they might define a particular function with $f(x)$ and then in some later text all of a sudden switch to $g(x)$ or $h(k)$.

In the end, it comes down to the trade-off between crystal clear consistency (where $f$ always means the same $f$ and $x$ always meaning the same $x$) and the attempt to write mathematics texts that challenge the reader and not always spelling things out in great detail (thereby opening up for the "personal discovery" aspect of math learning).

Of course, it is always possible that it is due to sheer laziness or that some error has crept in that has not been fixed during editing. After all, many mathematics textbooks have errata lists seemingly no matter how many editions of it that have been published.

Also, not all math texts are equally rigorous. Just look at Thomas versus Spivak for calculus, for instance.

My personal preference is for consistency.

$\endgroup$
6
  • 1
    $\begingroup$ Using the expression for the rule of $f$ as it was defined, is it possible to provide an expression for the limit $\partial f/ \partial b$ provided that $b \in \mathbb{R}$? To me, that would seem like a guessing game. $\endgroup$ – Mikkel Rev Jan 18 '20 at 21:47
  • $\begingroup$ Some authors are struck with what is called "curse of knowledge". So many things are obvious to them that they believe it must be to others as well. Perhaps they are merely assuming that readers understand what they mean, even though it is not spelled out explicitly. $\endgroup$ – MathInferno Jan 18 '20 at 22:02
  • 1
    $\begingroup$ @MikkelRev: Strictly speaking, you should not write $\partial f /\partial b$, but rather $\partial f(b)/\partial b$. Then, the issue vanishes. $\endgroup$ – PhoemueX Jan 19 '20 at 6:40
  • 1
    $\begingroup$ @PhoemueX is $\partial f(b)/\partial b$ to be understood as the derivative of $a b$ with respect to b? $\endgroup$ – Mikkel Rev Jan 19 '20 at 15:48
  • $\begingroup$ Why "challenge the reader"? Is mathematics always a test? If it's a good thing, why can't its virtues be explained without combativeness with the audience? I do understand that there's a great tradition... but I think it is misguided in several ways. $\endgroup$ – paul garrett Jan 20 '20 at 0:02

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.