Exercise 1.3.14

Proof.

(a)

Let $f(x)=x^3+2x^2+10x$.

For all $x\in ℝ$, $f^{\prime }(x)=3x^2+4x+10=2x^2+{(x+2)}^2+6>0$, thus $f$ is strictly increasing, $f$ is continuous, and $\underset{y\to +\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(y)=+\infty$, $\underset{y\to -\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(y)=-\infty$. By the intermediate values theorem, $f:ℝ\to ℝ$ is a bijection. Therefore there exists $x\in ℝ$ such that $f(x)=20$, so there exists a unique real root of

$$g(x)=x^3+2x^2+10x-20.$$

As $g(0)=-20$ and $\underset{x\to +\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}g(x)=+\infty$, this root is positive.

(b)

To remove the coefficient of $x^2$, let $h(y)=g\left(y-\frac{2}{3}\right)$. $$\begin{array}{llll}\hfill h(y)& =g\left(y-\frac{2}{3}\right)& \hfill & \\ \hfill & ={\left(y-\frac{2}{3}\right)}^3+2{\left(y-\frac{2}{3}\right)}^2+10\left(y-\frac{2}{3}\right)-20& \hfill & \\ \hfill & =y^3-2y^2+\frac{4}{3}y-\frac{8}{27}+2y^2-\frac{8}{3}y+\frac{8}{9}+10y-\frac{20}{3}-20& \hfill & \\ \hfill & =y^3+\frac{26}{3}y-\frac{704}{27}.& \hfill & \\ \hfill & & \hfill \end{array}$$

Define $p,q$ by

$$\frac{p}{3}=\frac{26}{9},\frac{q}{2}=-\frac{352}{27}$$

$$\begin{array}{llll}\hfill {\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{3}\right)}^3& =\frac{123904}{3^6}+\frac{17576}{3^6}& \hfill & \\ \hfill & =\frac{141480}{3^6}& \hfill & \\ \hfill & =\frac{5240}{27}& \hfill & \end{array}$$

Consequently, the real solution of the equation is given by

$$\begin{array}{llll}\hfill x_1& =-\frac{2}{3}+\sqrt[3]{\frac{352}{27}+\sqrt{\frac{5240}{27}}}+\sqrt[3]{\frac{352}{27}-\sqrt{\frac{5240}{27}}}& \hfill & \\ \hfill & =\frac{1}{3}\left(-2+\sqrt[3]{352+6\sqrt{3930}}+\sqrt[3]{352-6\sqrt{3930}}\right)& \hfill & \\ \hfill & =\frac{1}{3}\left(-2+\sqrt[3]{6\sqrt{3930}+352}-\sqrt[3]{6\sqrt{3930}-352}\right).& \hfill & \\ \hfill & & \hfill \end{array}$$

With a calculator, an approximation at $10^{-10}$ of the real root $x_1$ is $x_1\simeq 1.3688081078$.

The two other roots are given by

To verify the Fibonacci’s result in 1225, we compute a decomposition of $x_1$ with basis 60 with the following Maple program (or Sage program):

 x1:=1/3*(-2+(6*sqrt(3930)+352)^(1/3)-(6*sqrt(3930)-352)^(1/3)):
 l:=[1]:Digits:=20:
 u:=evalf(x1):
 u:=u-1:
 for i to 6 do a:=floor(60*u):u:=60*(u-a/60.0):l:=l,a: od:
 [l];

           [[1], 22, 7, 42, 33, 4, 38]

which gives

Only the last fraction is slightly different in Fibonacci’s result $\frac{40}{60^6}$, the difference being $2∕60^6\simeq 4.10^{-11}$.

Idem with Sage :

x1 = 1/3*((6*sqrt(3930) + 352)^(1/3) - (6*sqrt(3930) - 352)^(1/3) - 2)
R = RealField(200)
u = R(x1)
n = floor(u)
l = [[n]]; u = u-n
for i in range(6):
    a = floor(60*u)
    u = 60*(u-a/60)
    l.append(a)
print l
[[1], 22, 7, 42, 33, 4, 38]

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