Improving Formulae - Writing Well

Mathematical Writing - Vivaldi Franco 2014

Improving Formulae
Writing Well

When the physicist Stephen Hawking was writing his book A Brief History of Time, an editor warned him that every new equation in the book would halve the readership. So he included a single equation. When writing for the general public, one should expect diffidence—even hostility—towards formulae. Even though mathematicians are trained to deal with formulae, it is still safe to assume that no one likes to struggle with too many symbols. In this section we explore techniques to improve the clarity of formulae, through presentation, notation, and layout.

Formulae may be embedded in the text or displayed, and in a document one usually finds both arrangements. In either case a formula must obey standard punctuation rules. The following passage features embedded formulae with appropriate punctuation.

For each $$x\in X$$ we have the decomposition $$x=\xi +\lambda $$, with $$\xi \in \Xi $$ and $$\lambda \in \varLambda $$; accordingly, we define the function $$P:X\rightarrow \varXi ,\,\,x\mapsto \xi $$, which extracts the first component of $$x$$.

The punctuation generates rests as it would in an ordinary English sentence. The two items comprising the function definition are separated by a comma, which would not be necessary in a displayed formula, see (2.16). The formula defining $$P$$ is echoed by a short sentence, for added clarity.

This is a fully punctuated displayed formula:

$$ a_1=1;\quad a_{k+1}={\left\{ \begin{array}{ll}a_k^2-1,&{} 1\leqslant k< 10;\\ a_k^2, &{} k\geqslant 10.\\ \end{array}\right. } $$

If full punctuation seems heavy (it may do so here), then we may replace some punctuation by increased spacing or by words:

$$ a_1=1;\qquad a_{k+1}={\left\{ \begin{array}{ll}a_k^2-1&{}\,\,\mathrm{{if}}\, 1\leqslant k< 10;\\ a_k^2 &{}\,\,\mathrm{{if}}\, k\geqslant 10.\\ \end{array}\right. } $$

A displayed formula is normally embedded within a sentence, and the punctuation at the end of a formula must be appropriate to the structure of the sentence. In particular, if a sentence terminates at a formula—as in the example above—then the full stop at the end of the formula must always be present.

EXAMPLE. The following untidy formula

$$ f(x)=\frac{14x-2x^3-2x^2+14}{-2x-4} $$

could represent a typical unprocessed output of a computer algebra system. It contains redundant information (a common factor between numerator and denominator), the monomials at numerator are not ordered, and there are too many negative signs. The properties of the rational function $$f(x)$$ are not evident from it. There are several ways to improve the formula’s layout:

$$\begin{aligned} f(x)&= \frac{x^3+x^2-7x-7}{x+2}\\ f(x)&= \frac{(x+1)(x^2-7)}{x+2}\\ f(x)&= x^2-x-5+\frac{3}{x+2}. \end{aligned}$$

If the degree or the coefficients of $$f$$ are important, then the first version is appropriate; the second version makes it easy to solve the equation $$f(x)=0$$; the third version is a preparation for integrating $$f$$.

EXAMPLE. The following subset of the rationals

$$ A=\biggl \{y\in \mathbb {Q}\,:\,y=\frac{x}{x^2+1},\,\,x\in \mathbb {Z},\,\,x<0 \biggr \} $$

is defined with a disproportionate volume of notation. To save symbols we switch from the Zermelo to the standard set definition, and consider only elements of the required form. We also remove the inequality by introducing a negative sign, so we can replace the integers by the natural numbers:

$$ A=\biggl \{\frac{-n}{n^2+1}\,:\,n\in \mathbb {N}\biggr \}. $$

Now the formula is transparent.

EXAMPLE. To simplify a complex formula, we apply the ’divide and conquer’ method. In the cluttered expression

$$ R=x(ad-bc)-y(ad-cb)^2+z(ad-cb)^3 $$

the sub-expression $$ad-bc$$ appears as a unit. We exploit this fact to improve the layout:

$$\begin{aligned} R=x\varDelta -y\varDelta ^2+z\varDelta ^3\qquad \quad \varDelta =ad-bc. \end{aligned}$$

The formula is tidier, and the structure of $$R$$ is clearer.

EXAMPLE. Our next challenge is to improve an intricate double sum:

$$\begin{aligned} z(y_1,y_2,\ldots )=\sum _{i=1}^\infty \sum _{y=0}^{y_i-1}(y+1)x^{i-1}. \end{aligned}$$

(6.5)

The meaning of $$z$$ is not at all evident. This quantity depends on $$x$$, but its dependence does not appear explicitly; a poor choice of symbols obscures matters further. We note that the parameters $$y_i$$ are integers, since they are the upper limit of the inner summation; accordingly, we replace the symbol $$y$$ with $$n$$—see (6.1). Then we adopt the ’divide and conquer’ method, splitting up the sum as follows:

$$ z(x,n_1,n_2,\ldots )=\sum _{i=1}^\infty d_i x^{i-1} \qquad \quad d_i=\sum _{k=0}^{n_i-1}(k+1). $$

We now see that $$z$$ is a power series in $$x$$; its coefficients are finite sums, determined by the elements of an integer sequence. These are sums of arithmetic progressions, which can be evaluated explicitly:

$$ \sum _{0\leqslant k\leqslant n-1}(k+1) = \sum _{1\leqslant k\leqslant n}k=\frac{n(n+1)}{2}. $$

(In this passage we have dropped the subscript $$i$$, since the association $$n\leftrightarrow n_i$$ is clear, and we have switched to the standard sigma-notation to change variable—cf. Eq. (3.7)). Our original double sum (6.5) can now be written as

$$ z(x,\mathbf{n})\,=\,\frac{1}{2}\sum _{i=1}^\infty n_i(n_i+1) x^{i-1} \qquad \mathbf{n}=(n_1,n_2,\ldots ) $$

where we have used again the ’divide and conquer’ method. The dependence of $$z$$ on the variable $$x$$ and the sequence $$\mathbf{n}$$ is now clear.