1. Intro
  2. A simple approximation
  3. Further remarks
  4. Epilogue

Intro

One of the previous posts touched upon the importance of rescaling in simple perturbation problems. As the example discussed in that post involved a straightforward polynomial equation that contained explicitly a small parameter, a question remains: is rescaling relevant only in situations that involve such small parameters? The answer might come as a surprise to the “uninitiated”, but this is definitely not true. Employing rescaling strategies proves advantageous across a range of different scenarios that initially appear unrelated to perturbation theory. One such situation is considered next.

A simple approximation

We are interested in finding (without a calculator or Cardano’s formulae) reasonable approximations for the roots of the cubic

\[x^3- 21x^2 + 35x-7 = 0. \tag{1}\]

The crucial step is to conveniently re-scale this equation in order to get a tight bracketing interval for one of its roots. By making the substitution $y=10x$, we are led to the new equation

\[P(y)\equiv y^3-210y^2+3500y-7000 = 0. \tag{2}\]

It might seem that not much progress was made in passing from (1) to (2). However, this is not entirely true. Note that $P(2)<0$ and $P(3)>0$, so by the intermediate value theorem we know that there is a root $y_0$ of (2) such that $y_0\in(2,3)$. In terms of the original equation, there is a root $x_0=y_0/10\in (0.2,0.3)$.

Set $x = 0.2 + \varepsilon$, with $\varepsilon$ a small number (we already know that $0 < \varepsilon < 0.1$). Substituting this into (1) we find another cubic in $\varepsilon$:

\[(\varepsilon+0.2)^3 -21(\varepsilon+0.2)^2+35(\varepsilon+0.2) - 7 = 0, \tag{3}\]

and since $\varepsilon^3\simeq 10^{-3}=0.001$ we can safely ignore the cubic term when expanding (3). This leaves us with the quadratic

\[-20.4{\varepsilon^2} + 26.72{\varepsilon}- 0.832 = 0.\]

The root of interest is

\[\varepsilon_1\simeq \dfrac{13.36-12.71}{20.4}\simeq 0.032\quad\Longrightarrow\quad x_1\simeq 0.2 + \varepsilon_1 = 0.232.\]

To find approximations for the remaining roots, we use Viete’s relations for the original cubic

\[x_1+x_2+x_3 = 21\quad\mbox{and}\quad x_1x_2x_3 = 0.\]

We already have an idea about $x_1$ (see above), so

\[\begin{aligned}[t] &a\equiv x_2+x_3 = 21-x_1\simeq 20.768\\ &{}\\ &b\equiv x_2x_3 = 7/x_1\simeq 30.17. \end{aligned}\]

The quadratic that has roots $x_2$ and $x_3$ is simply $x^2-ax+b=0$ or $x^2-20.768x+30.17=0$, whence $x_2\simeq 1.57$ and $x_3 \simeq 19.20$.

To summarise, we have found that the roots of the original cubic are

\[x_1\simeq 0.232,\quad x_2\simeq 1.57,\quad x_3\simeq 19.20.\]

Using one of the many online calculators we discover the following values (rounded to 4 d.p. accuracy)

\[x_1\simeq 0.2319,\quad x_2\simeq 1.5724,\quad x_3\simeq 19.1957.\]

Clearly, our approximations are pretty decent.

Further remarks

Equation (1) is related to trigonometry, in particular the value of $\tan(2\pi/7)$. To see the link with the above developments, we need some basic trigonometric identities. Let’s start with the well-known result

\[\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}\,\]

which is valid for any $\alpha$, $\beta\in\mathbb{R}$ such that all the expressions in this formula exist.

Setting $\alpha\to{\theta}$ and $\beta\to{\theta}$ gives

\[\tan(2\theta) = \dfrac{2\tan \theta}{1-\tan^2\theta}, \tag{4}\]

while for $\alpha\to{2\theta}$ and $\beta\to{\theta}$ we find

\[\tan(3\theta) = \dfrac{\tan(2\theta)+\tan \theta}{1-\tan(2\theta)\tan \theta}=\dots =\dfrac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}. \tag{5}\]

Furthermore, with $\theta\to 2\theta$ in (4),

\[\tan(4\theta) = \dfrac{2\tan(2\theta)}{1-\tan^2(2\theta)} = \dots =\dfrac{4\tan\theta(1-\tan^2\theta)}{1-6\tan^2\theta + \tan^4\theta}. \tag{6}\]

We are now set to go. If $\theta=2\pi/7$ then $7\theta=2\pi$; taking the tangent of both sides,

\[\begin{aligned}[1] &0 = \tan(2\pi) = \tan(7\theta)=\tan(4\theta + 3\theta)\\ &{}\\ &{}\qquad\Longrightarrow\quad 0 = \dfrac{\tan(4\theta)+\tan(3\theta)}{1-\tan(4\theta)\tan(3\theta)}\\ &{}\\ &{}\qquad\Longrightarrow\quad \tan(4\theta)+\tan(3\theta)= 0. \end{aligned}\]

The last equation above can be transformed with the help of (5) and (6), so that everything is expressed in terms of $\tan\theta$. Carrying out the relevant algebraic manipulations, it transpires that

\[\tan^6\theta-21\tan^4\theta+35\tan^2\theta - 7 = 0. \tag{7}\]

Finally, by making the substitution $x=\tan^2\theta$ in (7) results in the original cubic (1). We remark in passing that our calculations in the previous section have shown that

\[\tan\dfrac{2\pi}{7}\simeq\sqrt{1.57}\simeq 1.25.\]

The value $\pi/7$ is the hero of many other trigonometric stories; for example, the following interesting identity

\[\cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7}= -\dfrac{1}{2}. \tag{8}\]

This is easily proved with the help of the standard formula

\[\sin\alpha\cos\beta =\dfrac{1}{2}\left[ \sin(\alpha+\beta)-\sin(\beta-\alpha) \right]. \tag{9}\]

Let $S$ be the left-hand side of (8), and then multiply $S$ by $1\equiv \sin(\pi/7)/\sin(\pi/7)$. The result can be re-arranged in the more convenient form

\[S\times \dfrac{\sin(\pi/7)}{\sin(\pi/7)} = \dfrac{1}{\sin(\pi/7)} \left[ \sin\dfrac{\pi}{7}\cos\dfrac{2\pi}{7} + \sin\dfrac{\pi}{7}\cos\dfrac{4\pi}{7}+\sin\dfrac{\pi}{7}\cos\dfrac{6\pi}{7} \right].\]

To evaluate the terms in the square bracket in this last equation, we need to apply (9) for suitably chosen values for $\alpha$ and $\beta$:

\[\begin{aligned}[t] &\sin\dfrac{\pi}{7}\cos\dfrac{2\pi}{7} = \dfrac{1}{2}\left[ \sin\dfrac{3\pi}{7}-\sin\dfrac{\pi}{7} \right],\\ &{}\\ &\sin\dfrac{\pi}{7}\cos\dfrac{4\pi}{7} = \dfrac{1}{2}\left[ \sin\dfrac{5\pi}{7}-\sin\dfrac{3\pi}{7} \right],\\ &{}\\ & \sin\dfrac{\pi}{7}\cos\dfrac{6\pi}{7} = \dfrac{1}{2}\left[ \sin\dfrac{7\pi}{7}-\sin\dfrac{5\pi}{7} \right]. \end{aligned}\]

Adding together the last three results and taking into account that $\sin(7\pi/7)=\sin(\pi)=0$, we eventually recover the identity (8).

Epilogue

Here are a few other interesting polynomial equations related to trigonometry:

  1. The cubic $x^3-21x^2+35x-7=0$ has the roots: $\tan^2\dfrac{\pi}{7}$, $\tan^2\dfrac{2\pi}{7}$, $\tan^2\dfrac{3\pi}{7}$.
  2. $x=\tan\dfrac{\pi}{20}$ is one of the roots of the quartic $x^4-4x^3-14x^2-4x+1=0$.
  3. The quartic $x^4-6x^2+1=0$ has the roots: $\tan\dfrac{\pi}{8}$, $\tan\dfrac{3\pi}{8}$, $\tan\dfrac{5\pi}{8}$, $\tan\dfrac{7\pi}{8}$.
  4. The cubic $x^3-5x^2+6x-1=0$ has the roots: $4\cos^2\dfrac{\pi}{7}$, $4\cos^2\dfrac{2\pi}{7}$, $4\cos^2\dfrac{3\pi}{7}$.
  5. A regular heptagon $A_1A_2A_3A_4A_5A_6A_7$ is inscribed in a circle of radius $R=1$. Then, the lengths of the chords $A_1A_2$, $A_1A_3$, $A_1A_4$ are the roots of the cubic $x^3-7x^2+14x-7=0$.
  6. $x=2\sin\dfrac{\pi}{18}$ is one of the roots of the cubic $x^3-3x-1=0$.