Cubic equation for closest point to a function

Question

We have $y = x^2 + 1$ and are trying to find the point closest to $(4, 0)$. For context, this is for a high school calculus class.

Clearly, $l = \sqrt{(x - 4)^2 + ((x^2 + 1) - 0)^2}$, from which we get $l^2 = (x - 4)^2 + ((x^2 + 1) - 0)^2$ and so $2 l \frac{\text{d}l}{\text{d}x} = 4x^3 + 6x - 8$. Setting this equal to zero to find maxima and minima, we get $4x^3 + 6x - 8 = 0$. Dividing everything by $2$, we finally have $2x^3 + 3x - 4 = 0$. How do I solve this cubic? It seems that the root is irrational, so the rational root theorem is failing me, and I don't know what else to try. Am I missing something obvious? Did my teacher make a typo?

Answer 1

I don't think there's a high school way to solve this. I am using the standard hyperbolic substitution for depressed cubic with one real root.

$$2x^3 + 3x - 4 = 0 \tag{1}$$

Since one root is real and coefficient of $x>0$, we use: $$\sinh(3\theta) = 4\sinh^3 \theta + 3\sinh \theta$$

Multiplying $(1)$ by $2$ gives $4x^3 + 6x = 8$. Let $x = \sqrt{2}\sinh \theta$. $$ 4(\sqrt{2}\sinh \theta)^3 + 6(\sqrt{2}\sinh \theta) = 8 $$ $$ 8\sqrt{2}\sinh^3 \theta + 6\sqrt{2}\sinh \theta = 8 $$ Dividing by $2\sqrt{2}$: $$ 4\sinh^3 \theta + 3\sinh \theta = \frac{4}{\sqrt{2}} = 2\sqrt{2} $$ $$ \sinh(3\theta) = 2\sqrt{2} \implies \theta = \frac{1}{3}\text{arcsinh}(2\sqrt{2}) $$

Using $\text{arcsinh}(z) = \ln(z+\sqrt{z^2+1})$: $$ \theta = \frac{1}{3}\ln(2\sqrt{2}+3) = \frac{1}{3}\ln((\sqrt{2}+1)^2) = \frac{2}{3}\ln(\sqrt{2}+1) $$ Substituting into $$x = \sqrt{2}\sinh \theta = \frac{\sqrt{2}}{2}(e^\theta - e^{-\theta})$$ $$ x = \frac{1}{\sqrt{2}}\left( (\sqrt{2}+1)^{2/3} - (\sqrt{2}-1)^{2/3} \right) $$

Answer 2

You have been given the exact solution by @匚ㄖㄥᗪ乇ᗪ.

You can have a good approximation of the real root of $$2x^3 + 3x - 4 = 0 $$ knowing (by inspection or plotting) that the solution is "close" to $1$.

Let $x=1-\epsilon$ and expand. Consider now that you look for the zero of $$g(\epsilon)=1-9 \epsilon +6 \epsilon ^2-2 \epsilon^3$$ Neglecting the highest power, solving the quadratic gives $$\epsilon \sim \frac{9-\sqrt{57}}{12}=0.120847 \quad \implies \quad x \sim 0.879153$$ while the exact solution is $0.879615$.

If you want to improve, knowing this result, repeat the same process with $x=1-\frac {12}{100}-\epsilon=\frac {22}{25}-\epsilon$

$$g(\epsilon)=\frac{46}{15625}-\frac{4779 \epsilon}{625}+\frac{132 \epsilon ^2}{25}-2\epsilon ^3$$ and neglect the second and third power. This gives $$\epsilon=\frac{46}{119475}\quad \implies \quad x \sim \frac{105092}{119475}=0.879615$$ which correspond to an absolute error of $10^{-7}$.

Sooner or later, you will also learn that we can approximate functions by rational functions. Around $x=1$ $$2x^3 + 3x - 4 \sim \frac{2 x^2+71 x-64}{2 x^2-10 x+17}$$ To see how decent it is, plot the two functions between $x=0$ and $x=2$.

Solving for zero the numerator gives $$x \sim \frac{3\sqrt{617}-71}{4}=0.879614$$

Edit (for the fun)

Amazing could be to continue with rational approximations, writing $$2x^3 + 3x - 4 \sim \frac {a_nx^2+b_nx+c_n}{1+\sum_{k=1}^n \alpha_k\,x^k }$$ As a function of $n$,below are the quadratic equations to solve and, converted to decimals, the approximate root

$$\left( \begin{array}{ccc} n & \text{quadratic equation} & \text{solution} \\ 0 & 6 x^2-3 x-2 & 0.879152869605896 \\ 1 & 9 x^2+8 x-14 & 0.879597254201444 \\ 2 & 2 x^2+71 x-64 & 0.879613522561331 \\ 3 & 69 x^2-138 x+68 & 0.879614146914231 \\ 4 & 414 x^2-205 x-140 & 0.879614885414259 \\ 5 & 619 x^2+556 x-968 & 0.879614880026944 \\ 6 & 126 x^2+4905 x-4412 & 0.879614879828974 \\ 7 & 4779 x^2-9506 x+4664 & 0.879614879821710 \\ 8 & 28570 x^2-14003 x-9788 & 0.879614879812342 \\ 9 & 42573 x^2+38640 x-66928 & 0.879614879812398 \\ 10 & 7866 x^2+338855 x-304148 & 0.879614879812399 \\ \end{array} \right)$$

How to build these approximations

To approximate a function $f(x)$ around a given point $x_0$, the most typical is a Taylor series expansion. More efficient are the $[m,n]$ Padé approximants which write $$f(x) =\frac {\sum_{k=0}^m \alpha_k\,(x-x_0)^m}{1+\sum_{k=1}^n \beta_k\,(x-x_0)^k }$$ For brevity, I shall use $t=x-x_0$.

In the present case, with $x_0=1$ $$f(x)=2x^3+3x-4 \quad \implies \quad f(t)= 1+9 t+6 t^2+2 t^3$$ and the $[2,3]$ Padé approximants write $$\frac {\sum_{k=0}^2 \alpha_k\,t^m}{1+\sum_{k=1}^3 \beta_k\,t^k }$$ Now, cross multiplying, consider $$\Big(1+\beta_1\,t+\beta_2\,t^2+\beta_3\,t^3\Big)\Big( 1+9 t+6 t^2+2 t^3\Big)=\alpha_0+\alpha_1\,t+\alpha_2\,t^2$$ Put every on the same side, expand and group by powers; this will give a fifth degree polynomial $$\sum_{n=0}^5 \gamma_n\,t^n=0$$ where $$ \gamma_0=1-\alpha_0 \qquad \gamma_1=9-\alpha_1+\beta_1 \qquad \gamma_2= 6-\alpha_2+9\beta_1+\beta_2$$ $$ \gamma_3=2+6\beta_1+9\beta_2+\beta_3\qquad \gamma_4=2\beta_1+6\beta_2+9\beta_3\qquad \gamma_5=2\beta_2+6\beta_3$$

Cancelling all of them is simple (six linear equations for six variables); it gives $$\alpha_0=1 \quad \alpha_1=0 \quad \alpha_2=-69\quad \beta_1=-9 \quad \beta_2=6 \quad \beta_3=-2$$

So, the numerator we need to cancel is $$1-69t^2=1-69(x-1)^2 \qquad \implies \qquad x=1-\frac{1}{\sqrt{69}}=0.879614$$

We could have been more lazy, considering instead the $[1,n]$ Padé approximant; it would generate the sequence of rational numbers $$\left\{\frac{8}{9},\frac{22}{25},\frac{548}{623},\frac{4552}{5175},\frac{548}{623},\frac{314092}{357079}, \frac{2609060}{2966139},\frac{2408068}{2737639},\frac{180027332}{204666083},\cdots\right\}$$

Answer 3

Here is the middle school way for the cubic equation : $$2x^3+3x-4=0 \iff 8x^3-16 + 12x =0\tag{1}$$

We intend to use the known formula $$a^3 +b^3 +c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \tag{2}$$ with $c = 2x$.

We will find $(a,b)$ such that $a^3+b^3 =-16$ and $ab =-2$. It suffices to solve the quadratic equation $$X^2+16X-8=0 \iff(a,b)=\left(\sqrt[3]{-8-6\sqrt{2}}, \sqrt[3]{-8+6\sqrt{2}},\right)$$

As the equation $(2)$ has 1 unique solution $a+b+c=0\iff x=-\frac{a+b}{2}$, therefore, the equation $(1)$ has 1 unique solution $$\color{red}{x = -\frac{a+b}{2}= \frac{1}{2} \left(\sqrt[3]{8+6\sqrt{2}}+ \sqrt[3]{8-6\sqrt{2}} \right)} $$