# Defining the rule of an arbitray function

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1. Jan 6, 2015

### Mr Davis 97

For any $f:\Re \rightarrow \Re$, is the only reason that we typically define the "rule" of the function with the the free variable, $x$, as the argument of the function, e.g., $f(x) = x^2 + 1$, because it's simply easier than having to do something like $f(~) = (~)^2 + 1$? That is, does the use of $x$ not really serve a purpose when defining the function besides that of convenience and readability?

Also, as an extension to this question, if it is the case that the free variable $x$ is just there for convenience, then how would we right the differentiation operator $\frac{\mathrm{d} }{\mathrm{d} x}$? Would you write it as $\frac{\mathrm{d} }{\mathrm{d} (~)}$ or something? Also, would one even be able to define the derivative $\displaystyle\lim_{\Delta x\rightarrow 0}\frac{f(x + \Delta x ) - f(x)}{\Delta x}$ without the use of the free variable $x$?

2. Jan 6, 2015

### jbunniii

Pretty much, yes. The convenience and readability become even more important if you consider functions of multiple variables. E.g. $f(x,y,z) = x^2 y z^3$ vs. something like $f((~), [~], \{~\}) = (~)^2[~]\{~\}^3$.
You could use the notation $D$, e..g $Df$ is the derivative of $f$. If $f$ is a function of $n$ variables, then $D_1, D_2, \ldots D_n$ are standard symbols for the partial derivative operators. See http://en.wikipedia.org/wiki/Differential_operator
$$f'((~)) = \lim_{[~]\rightarrow 0}\frac{f((~) + [~] ) - f((~))}{[~]}$$

3. Jan 8, 2015

### gufiguer

It is not what you have asked, but a symbol for a variable in the "operator" $\frac{d}{dx}$ is just useful for doing long calculations in a draft at most. It is not rigorous, it is not used in the definition and in fact is long time dead.

As a paradox, the same type of variable symbol is largely used in the definition of the limit of a function and is an ancient mistake. A much better notation would be $\displaystyle \lim_{a} f$ for the limit of a function $f$ at an accumulation point $a$ of the domain of $f$. Again, the old notation is great for doing long calculations in a draft.

Now, the derivative of a function $f : A \to \mathbb{R}$ at a point $a \in A \subset \mathbb{R}$ is defined like this: consider the Newton's quocient function of $f$ at the point $a$ given by $\displaystyle Qf_a (t) = \frac{f(a+t) - f(a)}{t}$, with an appropriate domain. Then the derivative of $f$ at the point $a$ is just $\displaystyle \lim_{0} Qf_a$, if it exists. :-)