APPENDIX II (To Chapters IX, X)

A Note on Double Limit Problems

In the course of Chapters IX and X we came on several occasions into contact with problems of a kind which invariably puzzle beginners and are indeed, when treated in their most general forms, problems of great difficulty and of the utmost interest and importance in higher mathematics.

Let us consider some special instances. In § 213 we proved that $$\log (1+x)=x–\frac{1}{2}{x}^2+\frac{1}{3}{x}^3–\dots ,$$ where $-1<x\le 1$, by integrating the equation $$1/(1+t)=1–t+t^2–\dots$$ between the limits $0$ and $x$. What we proved amounted to this, that $${\int }_0^x\frac{dt}{1+t}={\int }_0^{x}dt–{\int }_0^{x}tdt+{\int }_0^x{t}^{2}dt–\dots ;$$ or in other words that the integral of the sum of the infinite series $1–t+t^2–\dots$, taken between the limits $0$ and $x$, is equal to the sum of the integrals of its terms taken between the same limits. Another way of expressing this fact is to say that the operations of summation from $0$ to $\mathrm{\infty }$, and of integration from $0$ to $x$, are commutative when applied to the function $(-1{)}^n{t}^n$, that it does not matter in what order they are performed on the function.

Again, in § 216, we proved that the differential coefficient of the exponential function $$\exp x=1+x+\frac{x^2}{2!}+\dots$$ is itself equal to $\exp x$, or that $$D_x(1+x+\frac{x^2}{2!}+\dots )=D_{x}1+D_{x}x+D_x\frac{x^2}{2!}+\dots ;$$ that is to say that the differential coefficient of the sum of the series is equal to the sum of the differential coefficients of its terms, or that the operations of summation from $0$ to $\mathrm{\infty }$ and of differentiation with respect to $x$ are commutative when applied to $x^n/n!$.

Finally we proved incidentally in the same section that the function $\exp x$ is a continuous function of $x$, or in other words that $$\underset{x\to \xi }{lim}(1+x+\frac{x^2}{2!}+\dots )=1+\xi +\frac{{\xi }^2}{2!}+\cdots =\underset{x\to \xi }{lim}1+\underset{x\to \xi }{lim}x+\underset{x\to \xi }{lim}\frac{x^2}{2!}+\dots ;$$ that the limit of the sum of the series is equal to the sum of the limits of the terms, or that the sum of the series is continuous for $x=\xi$, or that the operations of summation from $0$ to $\mathrm{\infty }$ and of making $x$ tend to $\xi$ are commutative when applied to $x^n/n!$.

In each of these cases we gave a special proof of the correctness of the result. We have not proved, and in this volume shall not prove, any general theorem from which the truth of any one of them could be inferred immediately. In Ex. XXXVII. 1 we saw that the sum of a finite number of continuous terms is itself continuous, and in § 113 that the differential coefficient of the sum of a finite number of terms is equal to the sum of their differential coefficients; and in § 160 we stated the corresponding theorem for integrals. Thus we have proved that in certain circumstances the operations symbolised by $$\underset{x\to \xi }{lim}\dots ,D_x\dots ,\int \dots dx$$ are commutative with respect to the operation of summation of a finite number of terms. And it is natural to suppose that, in certain circumstances which it should be possible to define precisely, they should be commutative also with respect to the operation of summation of an infinite number. It is natural to suppose so: but that is all that we have a right to say at present.

A few further instances of commutative and non-commutative operations may help to elucidate these points.

(1) Multiplication by $2$ and multiplication by $3$ are always commutative, for $$2\times 3\times x=3\times 2\times x$$ for all values of $x$.

(2) The operation of taking the real part of $z$ is never commutative with that of multiplication by $i$, except when $z=0$; for $$i\times \mathrm{\Re }(x+iy)=ix,\mathrm{\Re }\{i\times (x+iy)\}=-y.$$

(3) The operations of proceeding to the limit zero with each of two variables $x$ and $y$ may or may not be commutative when applied to a function $f(x,y)$. Thus $$\underset{x\to 0}{lim}\{\underset{y\to 0}{lim}(x+y)\}=\underset{x\to 0}{lim}x=0,\underset{y\to 0}{lim}\{\underset{x\to 0}{lim}(x+y)\}=\underset{y\to 0}{lim}y=0;$$ but on the other hand $$\begin{array}{rlrl}2\underset{x\to 0}{lim}\left(\underset{y\to 0}{lim}\frac{x–y}{x+y}\right)& =\underset{x\to 0}{lim}\frac{x}{x}& & =\underset{x\to 0}{lim}1=1,\\ \underset{y\to 0}{lim}\left(\underset{x\to 0}{lim}\frac{x–y}{x+y}\right)& =\underset{y\to 0}{lim}\frac{-y}{y}& & =\underset{y\to 0}{lim}(-1)=-1.\end{array}$$

(4) The operations $\sum _1^{\mathrm{\infty }}\dots$, $\underset{x\to 1}{lim}\dots$ may or may not be commutative. Thus if $x\to 1$ through values less than $1$ then $$\begin{array}{rlrl}2\underset{x\to 1}{lim}\left\{\sum _1^{\mathrm{\infty }}\frac{(-1{)}^n}{n}{x}^n\right\}& =\underset{x\to 1}{lim}\log (1+x)& & =\log 2,\\ \sum _1^{\mathrm{\infty }}\left\{\underset{x\to 1}{lim}\frac{(-1{)}^n}{n}{x}^n\right\}& =\sum _1^{\mathrm{\infty }}\frac{(-1{)}^n}{n}& & =\log 2;\end{array}$$ but on the other hand $$\begin{array}{rl}\underset{x\to 1}{lim}\{\sum _1^{\mathrm{\infty }}(x^n–x^{n+1})\}& =\underset{x\to 1}{lim}\{(1–x)+(x–x^2)+\dots \}=\underset{x\to 1}{lim}1=1,\\ \sum _1^{\mathrm{\infty }}\{\underset{x\to 1}{lim}(x^n–x^{n+1})\}& =\sum _1^{\mathrm{\infty }}(1–1)=0+0+0+\cdots =0.\end{array}$$

The preceding examples suggest that there are three possibilities with respect to the commutation of two given operations, viz.: (1) the operations may always be commutative; (2) they may never be commutative, except in very special circumstances; (3) they may be commutative in most of the ordinary cases which occur practically.

The really important case (as is suggested by the instances which we gave from Ch. IX) is that in which each operation is one which involves a passage to the limit, such as a differentiation or the summation of an infinite series: such operations are called limit operations. The general question as to the circumstances in which two given limit operations are commutative is one of the most important in all mathematics. But to attempt to deal with questions of this character by means of general theorems would carry us far beyond the scope of this volume.

We may however remark that the answer to the general question is on the lines suggested by the examples above. If $L$ and $L^{\prime }$ are two limit operations then the numbers $L{L}^{\prime }z$ and $L^{\prime }Lz$ are not generally equal, in the strict theoretical sense of the word ‘general’. We can always, by the exercise of a little ingenuity, find $z$ so that $L{L}^{\prime }z$ and $L^{\prime }Lz$ shall differ from one another. But they are equal generally, if we use the word in a more practical sense, viz. as meaning ‘in a great majority of such cases as are likely to occur naturally’ or in ordinary cases.

Of course, in an exact science like pure mathematics, we cannot be satisfied with an answer of this kind; and in the higher branches of mathematics the detailed investigation of these questions is an absolute necessity. But for the present the reader may be content if he realises the point of the remarks which we have just made. In practice, a result obtained by assuming that two limit-operations are commutative is probably true: it at any rate affords a valuable suggestion as to the answer to the problem under consideration. But an answer thus obtained must, in default of a further study of the general question or a special investigation of the particular problem, such as we gave in the instances which occurred in Ch. IX, be regarded as suggested only and not proved.

Detailed investigations of a large number of important double limit problems will be found in Bromwich’s Infinite Series.

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