Mathematics Underfoot: The Formulas That Came to Würzburg from New Haven

Appendices

Appendix A: The First Formula

It is convenient to rewrite the formulas defining the inner and outer patterns as

$$(A)\quad F(y)+H(x)-b=0 \;\text{and} \;(B)\quad G(y)=a H(x),$$

(A1)

where

$$\begin{aligned} &F(y)=\sin (y)\;\sin (y/3),\;G(y)=\sin (y)\;\sin (y/2), \\ &\qquad \qquad \quad H(x) =\sin (x)\;\sin (x/3).\end{aligned}$$

(A2)

The graphs of the periodic functions F, G, and H are shown in Figure 7.

Figure 7

On the left, graph of H(x) and F(y) over a period from $-3\pi /2$ to $3\pi /2$. On the right, graph of G(y) over a period from $-2\pi$ to $2\pi$

Newton and Phillips used $b=-0.44$ and $a=1.37$ to construct their Figures 94 and 145, respectively, whereas Strouhal put $b=-0.4375$ and $a=1.3685$ in his formulas (1). Here and in Appendix B we consider slightly more general values.

If we consider a chain of figures in the $XY-$plane determined by formula (A) in (A1), with b values decreasing from $b=-0.40$ to $b=-0.48$ for example, as shown from L to R in Figure 8, we see in the middle figure that the curves close to form an especially attractive pattern when $b=b_0\approx -0.4375\approx -0.44$, as chosen in (1) and [8]. To determine this critical value exactly, we focus attention on $b>b_0$ and in particular on the two points where $dy/dx=0$, at $x=x_0\approx 2$ and $y\approx 5$.

Figure 8

Patterns determined by Formula (A) with, from L to R, $b=-0.40$, $b=-0.4375$, and $b=-0.48$

First, we find $x_0$ from

$$F^{\prime}(y)\;\frac{dy}{dx}=-\,H^{\prime}(x)=\cos (x)\;\sin (x/3)+\frac{1}{3}\sin (x)\;\cos (x/3)=0,$$

(A3)

giving

$$3\tan (x/3)=-\tan (x)$$

(A4)

and hence, with $t=\tan (x/3)$,

$$3t=-\,(3t-t^3)/(1-3t^2),\;(\Rightarrow t = 0, \sqrt{3/5} \; \text{or} \; - \sqrt{3/5})$$

(A5)

The root of interest is $t=\sqrt{3/5}$, giving

$$x_0=3\;\arctan \left( \sqrt{3/5}\right) \approx 1.9772.$$

(A6)

Now we find

$$H(x_0)=\sin (x_0/3)\;\sin (x_0)=s_0(3s_0-4s_0^3),\; s_0=\sin (x_0/3),$$

(A7)

and use $t_0=s_0/\sqrt{1-s_0^2}$ to get

$$s_0^2=t_0^2/(1+t_0^2)=3/8,$$

(A8)

and hence

$$H(x_0)=9/16.$$

(A9)

From formula (A) we then have at $x=x_0$

$$\sin (y)\;\sin (y/3)=-\,H(x_0)+b=\gamma,\;\text{say},$$

(A10)

or

$$S\;(3S-4S^3)=\gamma, \quad S=\sin (y/3),$$

(A11)

implying

$$S^2=(3\pm \sqrt{9-16\gamma })/8.$$

(A12)

With $b<0$, we have $\gamma <0$ from (A9) and (A10), and it follows that only the upper sign is relevant in (A12). Because $S^2\le 1$, it then follows using (A9) that

$$3+\sqrt{9-16\gamma }\le 8\quad \left(\Rightarrow -16\gamma \le 16\;\Rightarrow b\ge - 7/16\right).$$

(A13)

For each $b>-7/16$, it now follows that $S^2=\sin (y/3)^2<1$, leading to two possible values of y with $dy/dx=0$ and with $0<y<3\pi$. But when $b=-7/16$, it follows that $y=3\pi /2$ is the only possible value in this interval. Thus the critical value $b_0$, when the two y values with $dy/dx=0$ collapse to one, is the value given by Strouhal,

$$b_0=-7/16 =- 0.4375.$$

(A14)

Appendix B: The Second Formula

Now we consider a chain of figures determined by formula (B) in (A1), with a values increasing from 1.33 to 1.41 as shown in Figure 9, and this time concentrate on the points near $x=2$, $y=2$ where $dy/dx=0$.

Figure 9

Patterns determined by Formula (B) with, from L to R, $a=1.30$, $a=1.36853$, and $a=1.44$

We see again a coincidence of these two points, this time when $a = a_0\approx 1.3685\approx 1.37$, which are the values chosen in (1) and [8]. To determine the exact value of $a_0$, we first note that for general a, the x-coordinate of these points is again $x=x_0$ as in (A6), so the y-coordinates of these points are determined by

$$G(y)=a\;H(x_0)=9a/16=\tau,\;\text{say}.$$

(B1)

If we consider the graph of G(y) as in Figure 7, we see that there is a single maximum in the range of y-values of interest (say $0<y<\pi$), occurring at a value $y=y_0$ to be determined. Denoting $G(y_0)$ by $G_{\rm max.}$, we see that there are two roots of (B1) in the designated range provided $\tau < G_{\rm max.}$, and just one root if $\tau = G_{\rm max.}$.

To calculate $G_{\rm max.}$, we use

$$G^{\prime}(y)=\cos (y)\;\sin (y/2)+\frac{1}{2}\sin (y)\;\cos (y/2)=0 \; \text{at}\;y=y_0,$$

(B2)

giving

$$(1-2\sigma ^2)\sigma +\sigma (1-\sigma ^2) =0,\; \sigma =\sin (y_0/2),$$

(B3)

and hence (noting the allowed range of y-values)

$$\sigma =\sqrt{2/3}.$$

(B4)

Now $\cos (y_0/2)=\sqrt{1-\sigma ^2}=1/\sqrt{3}$, and hence

$$G_{\rm max.}=G(y_0)=2\sin^2(y_0/2)\;\cos (y_0/2)=4/3\sqrt{3}\approx 0.770.$$

(B5)

Because the critical value of $\tau$ is given by $G_{\rm max.}$, it now follows from (B1) that the critical value of a is given by

$$a_0= 64\sqrt{3}/81=1.36853\ldots$$

(B6)