# $\varphi$, $\pi$, $e$ & $i$ ### Buy Now:

###### David Perkins
Publisher:
MAA Press
Publication Date:
2017
Number of Pages:
176
Format:
Paperback
Price:
50.00
ISBN:
9780883855898
Category:
General
BLL Rating:

The Basic Library List Committee suggests that undergraduate mathematics libraries consider this book for acquisition.

[Reviewed by
Richard Wilders
, on
08/9/2017
]

David Perkins’ $\varphi, \pi, e$ and $i$ is a worthy addition to the MAA Spectrum Series. In keeping with the stated mission of that series, the book should be of interest to a broad range of readers including bright pre-college students. It is aimed at college students and would certainly make a fine text or supplementary text for a senior seminar. Each chapter ends with a section entitled Further Content which could provide nice prompts for student projects.

Each of these celebrated constants has its own chapter, but they also make guest appearances in other chapters. In Chapter 4, they all appear in a remarkable equation: $\varphi = e^{\pi i/5} + e^{-\pi i/5}.$ There are lots of fine books available about these 4 famous numbers. Perkins’ is distinguished by the clarity of writing and his mention of non-western mathematicians who worked with them.

First on stage is $\varphi=\dfrac{1+\sqrt5}{2}$, introduced appropriately as the length of the side of a cube inscribed in a dodecahedron with edges of length $1$. In Plato’s Timaeus, the dodecahedron represents the cosmos and the cube the “element” earth. Perkins then shows that this same number is the ratio of the length to the width of the “golden rectangle” defined by the solution to $\frac{x}{1}=\frac{1}{x-1}.$ $\varphi$ is thus the positive solution to $x^2-x=1$ and so $\varphi^2-\varphi=1$ or $\varphi^2=\varphi+1$. Now the fun starts! We multiply both sides by $\varphi$ and simplify: $\varphi^3=\varphi^2+\varphi=\varphi+1+\varphi=2\varphi+1$. Continuing this process, we obtain the remarkable sequence of equations $\varphi=1\varphi,$$\varphi^2=1\varphi+1,$$\varphi^3=2\varphi+1,$$\varphi^4=3\varphi+2.$

Recalling the Fibonacci Sequence: $0, 1, 1, 2, 3, 5, 8\dots$, we see that the powers of $\varphi$ can be represented as $\varphi^n=F_n\varphi+F_{n-1}$. Perkins mentions that this sequence was discussed by the Indian mathematician Acarya Hemachandra in 1088 in the context of Hemachandra’s discussion of Sanskrit poetry! This is yet another example of a “theorem” I propose in my history of mathematics class: credit for a discovery is most often given to the first white European male who claims credit for it.

Perkins doesn’t mention the fact that the ratio of consecutive terms of the Fibonacci (or Hemachandra) Sequence approach $\varphi$ as a limit but does derive the continued fraction expansion for $\varphi$ as well as providing a nice treatment of the arithmetic-geometric mean inequality. He also provides a really nice derivation of Binet’s formula $F_n=\frac{\varphi^n-\tau^n}{\sqrt{5}},$ where $\tau=\frac{1-\sqrt{5}}{2}$ is the other solution to the quadratic equation that defines $\varphi$ This chapter should be accessible to bright high school students as well as college math majors and should be of interest to both groups.

Next on stage is our old friend $\pi$. The material here is no less interesting than in Chapter 1, but the level is higher. Most of Chapter 2 requires knowledge of calculus through integration techniques. The discussion begins with Liu Hui’s approximation to $\pi$ using inscribed and circumscribed polygons. Perhaps you remember that Archimedes used this same technique?

Next we meet Kilakantha Somayaji (Indian born 1444) who constructed what is now the standard power series for $arctan(x)$ along with power series for other trigonometric functions well before the techniques of calculus were available: $\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5} - \frac{x^7}{7} + \dots$ Evaluating this at $x=1$ yields $\frac{\pi}{4}= 1-\frac{1}{3}+\frac{1}{5} - \frac{1}{7} + \dots,$ a result which was discovered independently by James Gregory and Gottfired Liebnitz over one hundred years later. Perkins then introduces John Machin’s remarkably efficient trick: use $\frac{\pi}{4}=4\arctan\left(\frac{1}{5}\right)-\arctan\left(\frac{1}{239}\right)$ and Somayaji’s formula to approximate $\pi$. Using just three terms of Somayaji’s series to estimate the two arctangent terms yields an estimate of 3.141621.

We then are treated to one of Euler’s solutions to a problem first posed by Pietro Mengoli in 1625: finding the sum of the series of reciprocals of the squares. Finally, Perkins provides a nice proof using calculus that $\pi$ is irrational. Altogether a marvelous tour of (some of) the wonders of $\pi$!

Perkins introduces us to $e$ as the solution to a problem posed to him as a boy by his dad. His dad offered Perkins \$10 or the chance to split the ten and multiply the parts. For example he could obtain $5\times 5=25$. Perkins recounts a wonderful tale of son versus dad culminating in his proposal of using four equal parts to obtain $2.5^4=39.0625$ — at which point his dad, fearing bankruptcy, called a halt to the proceedings. A nice further analysis yields the result that the value using $x$ for each equal part is $x^{10/x}=\left(x^{1/x}\right)^{10}$, so the problem amounts to maximizing $x^{1/x}$. Given the star of the chapter, you shouldn’t be surprised to find that $x=e$ does the job! The resulting payoff from dad would be $\left(e^{1/e}\right)^{10} \approx 39.6$.

Perkins then derives both standard formulas for $e$: $e =\sum_{n=0}^{\infty}\frac{1}{n!} = \lim_{n\to\infty}\left(1+\frac1n\right)^n.$ After giving a proof that $e$ is irrational, Perkins provides a very nice explication of James Stirling’s (Scotland, born 1692) remarkable approximation for $n!$ $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$

Perkins begins his discussion of $i$ with a brief history of the (slow) acceptance of negative numbers due to prior insistence that numbers should represent some geometric property (length, area, volume). The eventual acceptance of these “false positions” led to the possibility of attaching meaning to a number whose square is $-1$.

As Perkins points out, much of the impetus for treating $i$ as a reputable number came from the efforts of multiple mathematicians in multiple cultures to solve the general cubic equation. Through some simple algebra, we can reduce the general cubic to the form $x^3=3px+2q$, after which a clever argument yields the famous cubic formula: $x=\sqrt{q+\sqrt{q^2-p^3}}+\sqrt{q=\sqrt{q^2-p^3}}.$

This is a wonderful result, but the formula does funny things. The equation $x^3=6x+4$, for example, has three real roots $x=-2,1+\sqrt3,1-\sqrt3$. But using the formula yields the apparently nonsensical answer $x=\sqrt{2+\sqrt{-4}}+\sqrt{2-\sqrt{-4}}.$ I asked Maple to evaluate this expression and it returned $2\sqrt{2}\cos(\pi/12)$, otherwise known as $1+\sqrt3$. Perkins does a nice job of explaining how the creation of the number $i$ and the assumptions that the rules of arithmetic carry over into the newly minted set of complex numbers allows one to simplify these expressions. It’s no accident, by the way, that our cubic formula produced an expression involving $i$: as Perkins shows, the equation $x^3=3px+1q$ has three real roots precisely when $q^2-p^3<0$. In other words, complex numbers cannot be avoided.

Perkins then develops the famous formula $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, from which it is only a brief stroll to the equation which opened this review. I thoroughly enjoyed this book and feel certain others will as well. Perkins mixes rigor and intuition in pedagogically sound ways, resulting in a book which enchanting and informative.

Richard Wilders is Professor of Mathematics and Marie and Benice Gantzert Professor of the Liberal Arts and Sciences at North Central College. In addition to the standard undergraduate mathematics curriculum, he teaches courses in the history and philosophy of science. In his spare time, he enjoys watching college football, dance, and musical theater.

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