The logarithmic scale. Let us consider once more the series of functions $x,\quad \sqrt{x},\quad \sqrt{x},\ \dots,\quad \sqrt[n]{x},\ \dots,$ which possesses the property that, if $$f(x)$$ and $$\phi(x)$$ are any two of the functions contained in it, then $$f(x)$$ and $$\phi(x)$$ both tend to $$\infty$$ as $$x \to \infty$$, while $$f(x)/\phi(x)$$ tends to $$0$$ or to $$\infty$$ according as $$f(x)$$ occurs to the right or the left of $$\phi(x)$$ in the series. We can now continue this series by the insertion of new terms to the right of all those already written down. We can begin with $$\log x$$, which tends to infinity more slowly than any of the old terms. Then $$\sqrt{\log x}$$ tends to $$\infty$$ more slowly than $$\log x$$, $$\sqrt{\log x}$$ than $$\sqrt{\log x}$$, and so on. Thus we obtain a series $x,\quad \sqrt{x},\quad \sqrt{x},\ \dots,\quad \sqrt[n]{x},\ \dots\quad \log x,\quad \sqrt{\log x},\quad \sqrt{\log x},\ \dots\quad \sqrt[n]{\log x},\ \dots$ formed of two simply infinite series arranged one after the other. But this is not all. Consider the function $$\log\log x$$, the logarithm of $$\log x$$. Since $$(\log x)/x^{\alpha} \to 0$$, for all positive values of $$\alpha$$, it follows on putting $$x = \log y$$ that $(\log\log y)/(\log y)^{\alpha} = (\log x)/x^{\alpha} \to 0.$ Thus $$\log\log y$$ tends to $$\infty$$ with $$y$$, but more slowly than any power of $$\log y$$. Hence we may continue our series in the form $\begin{gathered} x,\quad \sqrt{x},\quad \sqrt{x},\ \dots\qquad \log x,\quad \sqrt{\log x},\quad \sqrt{\log x},\ \dots\\ \log\log x,\quad \sqrt{\log\log x},\ \dots\quad \sqrt[n]{\log\log x},\ \dots;\end{gathered}$ and it will by now be obvious that by introducing the functions $$\log\log\log x$$, $$\log\log\log\log x$$, … we can prolong the series to any extent we like. By putting $$x = 1/y$$ we obtain a similar scale of infinity for functions of $$y$$ which tend to $$\infty$$ as $$y$$ tends to $$0$$ by positive values.1

Example LXXXIV
1. Between any two terms $$f(x)$$$$F(x)$$ of the series we can insert a new term $$\phi(x)$$ such that $$\phi(x)$$ tends to $$\infty$$ more slowly than $$f(x)$$ and more rapidly than $$F(x)$$. [Thus between $$\sqrt{x}$$ and $$\sqrt{x}$$ we could insert $$x^{5/12}$$: between $$\sqrt{\log x}$$ and $$\sqrt{\log x}$$ we could insert $$(\log x)^{5/12}$$. And, generally, $$\phi(x) = \sqrt{f(x) F(x)}$$ satisfies the conditions stated.]

2. Find a function which tends to $$\infty$$ more slowly than $$\sqrt{x}$$, but more rapidly than $$x^{\alpha}$$, where $$\alpha$$ is any rational number less than $$1/2$$. [$$\sqrt{x}/(\log x)$$ is such a function; or $$\sqrt{x}/(\log x)^{\beta}$$, where $$\beta$$ is any positive rational number.]

3. Find a function which tends to $$\infty$$ more slowly than $$\sqrt{x}$$, but more rapidly than $$\sqrt{x}/(\log x)^{\alpha}$$, where $$\alpha$$ is any rational number. [The function $$\sqrt{x}/(\log\log x)$$ is such a function. It will be gathered from these examples that incompleteness is an inherent characteristic of the logarithmic scale of infinity.]

4. How does the function $f(x) = \{x^{\alpha} (\log x)^{\alpha’} (\log\log x)^{\alpha”}\}/ \{x^{\beta} (\log x)^{\beta’} (\log\log x)^{\beta”}\}$ behave as $$x$$ tends to $$\infty$$? [If $$\alpha \neq \beta$$ then the behaviour of $f(x) = x^{\alpha-\beta} (\log x)^{\alpha’-\beta’} (\log\log x)^{\alpha”-\beta”}$ is dominated by that of $$x^{\alpha-\beta}$$. If $$\alpha = \beta$$ then the power of $$x$$ disappears and the behaviour of $$f(x)$$ is dominated by that of $$(\log x)^{\alpha’-\beta’}$$, unless $$\alpha’ = \beta’$$, when it is dominated by that of $$(\log\log x)^{\alpha”-\beta”}$$. Thus $$f(x) \to \infty$$ if $$\alpha > \beta$$, or $$\alpha = \beta$$, $$\alpha’ > \beta’$$, or $$\alpha = \beta$$, $$\alpha’ = \beta’$$, $$\alpha” > \beta”$$, and $$f(x) \to 0$$ if $$\alpha < \beta$$, or $$\alpha = \beta$$, $$\alpha’ < \beta’$$, or $$\alpha = \beta$$, $$\alpha’ = \beta’$$, $$\alpha” < \beta”$$.]

5. Arrange the functions $$x/\sqrt{\log x}$$, $$x\sqrt{\log x}/\log\log x$$, $$x\log\log x/\sqrt{\log x}$$, $$(x\log\log\log x)/\sqrt{\log\log x}$$ according to the rapidity with which they tend to infinity as $$x \to \infty$$.

6. Arrange $\log\log x/(x\log x),\quad (\log x)/x,\quad x\log\log x/\sqrt{x^{2} + 1},\quad \{\sqrt{x + 1}\}/x(\log x)^{2}$ according to the rapidity with which they tend to zero as $$x \to \infty$$.

7. Arrange $x\log\log(1/x),\quad \sqrt{x}/\{\log(1/x)\},\quad \sqrt{x\sin x\log(1/x)},\quad (1 – \cos x)\log(1/x)$ according to the rapidity with which they tend to zero as $$x \to +0$$.

8. Show that $D_{x}\log\log x = 1/(x\log x),\quad D_{x}\log\log\log x = 1/(x\log x\log\log x),$ and so on.

9. Show that $D_{x}(\log x)^{\alpha} = \alpha/\{x(\log x)^{1-\alpha}\},\quad D_{x}(\log\log x)^{\alpha} = \alpha/\{x\log x(\log\log x)^{1-\alpha}\},$ and so on.

1. For fuller information as to ‘scales of infinity’ see the author’s tract ‘Orders of Infinity’, Camb. Math. Tracts, No. 12.↩︎