How to solve a quartic equation

Begin with a general quartic equation

x^4+bx^3+cx^2+dx+e=0,\quad\quad\quad(*)

depress it using the substitution

x = y - \frac{b}{4},

creates a new quartic equation (without the y^3 term) in y:

y^4+qy^2+ry+s=0\quad\quad\quad(**)

where

\begin{cases}q=c-\frac{3b^2}{8}\\r=d-\frac{bc}{2}+\frac{b^3}{8}\\s=e-\frac{bd}{4}+\frac{b^2c}{16}-\frac{3b^4}{256}\end{cases}

This image has an empty alt attribute; its file name is screen-shot-2022-01-07-at-5.40.02-pm.png

When r=0, i.e., d-\frac{bc}{2}+\frac{b^3}{8}=0, the quartic equation in y becomes a quadratic equation in y^2:

(y^2)^2+qy^2+s=0.

It follows that

y^2 = \frac{-q\pm \sqrt{q^2-4s}}{2} \implies y=\pm\sqrt{\frac{-q\pm\sqrt{q^2-4s}}{2}}\implies x=y-\frac{b}{2}.

For example, to solve quartic equation x^4+2x^3-3x^2-4x+4=0, depress it using the substitution x = y-\frac{1}{2}, we obtain

y^4-\frac{9}{2}y^2+\frac{81}{16}=0.

This is a quadratic equation in y^2:

(y^2)^2-\frac{9}{2}y^2+\frac{81}{16}=0.

Therefore,

y^2 = \frac{\frac{9}{2} \pm \sqrt{(\frac{9}{2})^2-4\cdot 1\cdot \frac{81}{16}}}{2}=\frac{9}{4}\implies y = \pm\sqrt{\frac{9}{4}} = \pm\frac{3}{2}.

Consequently,

x_1 = \frac{3}{2}-\frac{1}{2} = 1, x_2=-\frac{3}{2}-\frac{1}{2} = -2.

In general, we solve the depressed quartic equation as follows:

Substituting u+v+w for y in the left side of (**) yields

y^4+qy^2+ry+s

By \begin {cases} y^2 = (u+v+w)^2=u^2+v^2+w^2+2(uv+uw+vw),\\ y^4 = (y^2)^2 = \left((u^2+v^2+w^2)+2(uv+uw+vw)\right)^2 \\= (u^2+v^2+w^2)^2+4(u^2+v^2+w^2)(uv+uw+vw) + 4(uv+uw+vw)^2 \\= (u^2+v^2+w^2)^2+2(u^2+v^2+w^2)\cdot 2(uv+uw+vw) + 4(uv+uw+vw)^2\end {cases}

= \underbrace{(u^2+v^2+w^2)^2 + 4(u^2+v^2+w^2)(uv+uw+vw)+4(uv+uw+vw)^2}_{y^4}+ q\underbrace{(u^2+v^2+w^2+2(uv+uw+vw))}_{y^2}+ r\underbrace{(u+v+w)}_{y}+s

= (u^2+v^2+w^2)^2+4(u^2+v^2+w^2)(uv+uw+vw)+4(uv+uw+vw)^2 + +q(u^2+v^2+w^2)+2q(uv+vw+uw) + r(u+v+w)+s

= (u^2+v^2+w^2)  +q(u^2+v^2+w^2) + 4(u^2+v^2+w^2)(uv+uw+vw) + 2q(uv+uw+vw) +4(uv+uw+vw)^2 + r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2 + 2u^2vw + 2v^2uw +2w^2uv) + q(u^2+v^2+w^2)+r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8u^2vw + 8v^2uw +8w^2uv) + q(u^2+v^2+w^2)+r(u+v+w)+s

= u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8u^2vw + 8v^2uw +8w^2uv + r(u+v+w)+s

=(u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8uvw(u + v + w)  + r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(\underline{2(u^2+v^2+w^2) + q}) +4(u^2v^2+u^2w^2+v^2w^2) + (\underline{8uvw+r})(u + v + w) +s

If \begin{cases} 2(u^2+v^2+w^2) + q=0 \\ 8uvw + r=0\end{cases}\quad\quad\quad(1)

=(u^2+v^2+w^2)^2+q(u^2+v^2+w^2) + 4(u^2v^2+u^2w^2+v^2w^2)+s

Since (1) \implies \begin{cases} u^2+v^2+w^2 = -\frac{q}{2}\\ uvw = -\frac{r}{8}\end{cases}\quad\quad\quad(2)

= (-\frac{q}{2})^2 + q \cdot (-\frac{q}{2}) + 4(u^2v^2+u^2w^2+v^2w^2)+s

= \frac{q^2}{4} - \frac{q^2}{2} + 4(u^2v^2+u^2w^2+v^2w^2)+s

= -\frac{q^2}{4} + s + 4(u^2v^2+u^2w^2+v^2w^2)

= -\frac{q^2-4s}{4} + 4(\underline{u^2v^2+u^2w^2+v^2w^2})

If u^2v^2 + u^2w^2+v^2w^2 = \frac{q^2-4s}{16}\quad\quad\quad(3)

= 0.

It means that u+v+w is a solution of (**) if

\begin{cases} u^2+v^2+w^2=-\frac{q}{2} \\ u^2v^2+u^2v^2+v^2w^2=\frac{q^2-4s}{16}\\ uvw=-\frac{r}{8}\end{cases}\quad\quad\quad(4-1, 4-2, 4-3)

Moreover, squaring (4-3) gives

\begin{cases} u^2+v^2+w^2=-\frac{q}{2} \\ u^2v^2+u^2v^2+v^2w^2=\frac{q^2-4s}{16}\\ u^2v^2w^2=\frac{r^2}{64}\end{cases}\quad\quad\quad(5-1, 5-2, 5-3)

By Vieta’s theorem, u^2, v^2, w^2 satisfying (5-1, 5-2, 5-3) are the three solutions of cubic equation

z^3+\frac{q}{2}z^2+\frac{q^2-4s}{16}z-\frac{r^2}{64}= 0.

(See “How to solve a cubic equation“)

Suppose the three solutions are z_1, z_2, z_3, we have

u^2=z_1 \implies u = \pm\sqrt{z_1}, v^2 = z_2 \implies v = \pm\sqrt{z_2}, w^2=z_3 \implies w=\pm \sqrt{z_3}.

Clearly, there are eight combinations of u,v,w:

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases}

\begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases}

Among them, only four are valid due to constraint (4-3) placed on the product u\cdot v\cdot w.

From (5-3), we see uvw = \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}= \pm\frac{r}{8}.

If \sqrt{z1}\sqrt{z_2}\sqrt{z_3} = +\frac{r}{8}, the valid ones are:

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} since uvw=\sqrt{z_1}\cdot \sqrt{z_2}\cdot -\sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} since uvw=\sqrt{z_1}\cdot -\sqrt{z_2}\cdot \sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} since uvw=-\sqrt{z_1}\cdot \sqrt{z_2}\cdot \sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=-\sqrt{z_1} \\  v=-\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} since uvw=-\sqrt{z_1}\cdot -\sqrt{z_2}\cdot -\sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

Otherwise \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}=-\frac{r}{8}, we have

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases}

Consequently, a solution to the general quartic equation (*) is

x = u+v+w - \frac{b}{4}.

There are four such solutions.


Exercise-1 Show that (5-3) \implies uvw = \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}= \pm\frac{r}{8}.

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Through the Mind’s Eye

Let’s derive Cardano’s formula for the depressed cubic equation x^3+px=q.

Imaging a large cube with length u. It is comprised of six blocks. Each block has its label on the top.

The large cube’s volume, u^3 is the sum of six smaller pieces:

v^3+v^2(u-v) + v(u-v)u + v(u-v)u+(u-v)^2(u-v)+(u-v)^2v= u^3

That is,

(u-v)^3+\underline{2uv(u-v)}+\underline{v^2(u-v)}+\underline{(u-v)^2v}=u^3-v^3.

Factoring u-v from the underlined terms above gives

(u-v)^3+(2uv + v^2 + (u-v)v)(u-v)=u^3-v^3

or simply,

(u-v)^3+\underbrace{3uv}_{p}(u-v)=\underbrace{u^3-v^3}_{q}.\quad\quad\quad(1)

From the cube and (1), we see that for u>v>0, if 3uv = p and u^3-v^3=q, then u-v, a positive quantity, is a solution of x^3 + px =q.

However, (1) is an algebraic identity regardless of the cube that led us to it. i.e.,

(1) is true for all u, v \in R.

Hence,

\forall u,v \in R, (3uv = q, u^3-v^3=p) \implies u-v, not necessary a positive quantity, is a solution of depressed cubic x^3+px=q.

It follows that to find a solution of x^3+px=q, we solve

\begin{cases} 3uv=p \\ u^3-v^3=q \end{cases}

for u, v, after which the Cardano formula emerges:

x=u-v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.\quad\quad\quad(\star)


Exercise-1 Show that

u-v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}

after solving

\begin{cases} 3uv=p \\ u^3-v^3=q\end{cases}

for u, v.

The Lizard and the Fly

An open cylinder garbage can has a height of 4 ft and a circumference of 6 ft. On the inside of the can, 1 ft from the top, is a fly. On the opposite side of the can, 1 ft from the bottom and on the outside, is a lizard. What is the shortest distance that the lizard must walk to reach the fly?


To reach the fly, the lizard must first walk a path on the exterior surface of the garbage can to the opening. From there, it continues walking along a path on the interior surface to catch the prey.

The most salient piece of knowledge we have about minimum distance is that the shortest distance between two points comes from the straight line connecting them. However, it is not immediately apparent how to apply this knowledge in the present problem, since we are constrained to the bent lines on a curved wall. Nonetheless, the curved wall can be cut and flattened, result in a completely flat two dimensional surface. We can then connect the points by a straight line.

Fig. 1

Having flattened out the wall in the manner shown in Fig. 1, we see the lizard’s walking path consists two segments of straight line. The length of the path is

\sqrt{x^2+3^2} + \sqrt{(3-x)^2+1^2}.\quad\quad\quad(1)

Fig. 2

Three centuries before the birth of Christ, Euclid discovered the reflection law of light: if a ray of light is reflected by a mirror from point A to point B, the point of reflection P is such that the rays AP and PB make equal angles with the mirror (\theta_1 = \theta_2 in Fig. 2). However, it was Heron who first observed that if the point P on the mirror were such that \theta_2 \ne \theta_2, then the resulting total path length AP+PB would be longer; i.e.,

AP + PB is minimized when \theta_1 = \theta_2.

For the lizard, when \alpha = \beta, it walks the least (Fig. 1).

From \alpha=\beta, we have \cot(\alpha) = \cot(\beta); i.e.,

\frac{x}{3} = \frac{3-x}{1}\quad\quad\quad(2)

Solving (2) gives

x = \frac{9}{4}

so that (1) yields

\sqrt{(\frac{9}{4})^2 + 3^2} + \sqrt{(3-\frac{9}{4})^2+1^2} = 5.

Therefore, the shortest distance the lizard must walk is 5 ft.


Exercise-1 Find x that minimizes \sqrt{x^2+25} + \sqrt{(x-4)^2+4}.

Exercise-2 If x and N are real numbers such that N=\sqrt{5x+6}+\sqrt{7x+8}, then what is the smallest possible value of N?

Exercise-3 An open cylinder garbage can has a height of 4 ft and a circumference of 6 ft. On the inside of the can, 1 ft from the top, is a fly. On the opposite side of the can, 1 ft from the bottom inside the can, is a lizard. What is the shortest distance that the lizard must walk to reach the fly?

Wallis’ Pi

There is a remarkable expression for the number \pi as an infinite product. Starting with definite integral \int\limits_{0}^{\frac{\pi}{2}} \sin^{m}(x)\;dx, m=0,1,2,3,4..., we derive it as follows:

\int\limits_{0}^{\frac{\pi}{2}} \sin^m(x)\;dx

=\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-1}(x)\cdot\sin(x)\;dx

= \int\limits_{0}^{\frac{\pi}{2}} \sin^{m-1}(x)\cdot(-\cos(x))'\;dx

By \int\limits_{a}^{b} u\cdot v'\;dx = u\cdot  v \bigg|_{b}^{a}- \int\limits_{b}^{a} u' \cdot v\;dx,

= \sin^{m-1}(x)\cdot(-\cos(x))\bigg|_{0}^{\frac{\pi}{2}} - \int\limits_{0}^{\frac{\pi}{2}}(\sin^{m-1}(x))'\cdot(-\cos(x))\;dx

= 0 - \int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cos(x)\cdot(-\cos(x))\;dx

= +\int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cdot\cos^2(x)\;dx

\overset{\cos^2(x) = 1-\sin^2(x)}{=}\int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cdot (1-\sin^{2}(x))\;dx

= \int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x) - (m-1)\sin^{m}(x)\;dx

= (1-m)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x)\;dx + (m-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-2}(x)\;dx.

That is,

m\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x)\;dx = (m-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-2}(x)\;dx.

And so,

\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x) \; dx= \frac{m-1}{m}\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-2}(x)\;dx.\quad\quad\quad(1)

By repeated application of (1) we have the following values for I_{m} = \int\limits_{0}^{\frac{\pi}{2}} \sin^{m}(x)\;dx:

For even m,

I_{2n} =  \left(\prod\limits_{k=0}^{n-1}\frac{2n-2k-1}{2n-2k}\right)I_0\overset{I_0 =  \int\limits_{0}^{\frac{\pi}{2}}\;dx = \frac{\pi}{2}}{\implies}I_{2n}= \left(\prod\limits_{k=0}^{n-1}\frac{2n-2k-1}{2n-2k}\right)\cdot \frac{\pi}{2}.

Similarly, for odd m,

I_{2n+1} = \left(\prod\limits_{k=0}^{n-1}\frac{2n+1-2k-1}{2n+1-2k}\right)\cdot I_{1}\overset{I_1 = \int\limits_{0}^{\frac{\pi}{2}}\sin^{1}(x)\;dx = 1}{\implies} I_{2n+1}= \prod\limits_{k=0}^{n-1}\frac{2n-2k}{2n-2k+1}.

i.e.,

I_{2n} = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdot\frac{2n-7}{2n-6}\cdot...\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}\quad\quad\quad(2)

I_{2n+1} =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdot\frac{2n-4}{2n-3}\cdot\frac{2n-6}{2n-5}\cdot ...\cdot\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}\quad\quad\quad\quad\quad(3)

Since for 0<x<\frac{\pi}{2}, 0< \sin(x) < 1, we have \sin^{2n-1}(x) > \sin^{2n}(x) > \sin^{2n+1}(x). It means

\int\limits_{0}^{\frac{\pi}{2}}\sin^{2n-1}(x) > \int\limits_{0}^{\frac{\pi}{2}}\sin^{2n}(x)\;dx > \int\limits_{0}^{\frac{\pi}{2}}\sin^{2n+1}(x)\;dx>0

or,

I_{2n-1} > I_{2n} > I_{2n+1}>0.

Hence,

\frac{I_{2n-1}}{I_{2n+1}}>\frac{I_{2n}}{I_{2n+1}} > 1.\quad\quad\quad(4)

Moreover, we have

I_{2n+1} \overset{(1)}{=} \frac{(2n+1)-1}{2n+1}I_{(2n+1)-2} = \frac{2n}{2n+1} I_{2n-1},

so that

\frac{I_{2n-1}}{I_{2n+1}} = \frac{2n+1}{2n}.\quad\quad\quad(5)

And,

\frac{I_{2n}}{I_{2n+1}} \overset{(2), (3)}{=} \frac{(2n+1)\cdot(2n-1)^2\cdot (2n-3)^2...7^2\cdot 5^2\cdot 3^2}{(2n)^2\cdot (2n-2)^2\cdot (2n-4)^2\cdot ...\cdot 6^2\cdot 4^2\cdot 2^2}\cdot \frac{\pi}{2}

= \frac{(2n)^2\cdot (2n-1)^2\cdot (2n-2)^2\cdot (2n-3)^2\cdot (2n-4)^2\cdot...\cdot 7^2\cdot 6^2\cdot 5^2\cdot 4^2\cdot 3^2\cdot 2^2}{(2n)^4(2n-2)^4\cdot (2n-4)^4...6^4\cdot 4^4\cdot 2^4}\cdot (2n+1)\cdot\frac{\pi}{2}

= \frac{\left((2n)\cdot (2n-1)\cdot (2n-2)\cdot (2n-3)\cdot (2n-4)\cdot...\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\right)^2}{(2^4\cdot n^4)\cdot (2^4\cdot(n-1)^4) \cdot (2^4\cdot (n-2)^4)\cdot...\cdot (2^4\cdot 3^4)\cdot (2^4\cdot 2^4)\cdot 2^4}\cdot (2n+1)\cdot\frac{\pi}{2}

= \frac{((2n)!)^2}{\underbrace{2^4\cdot 2^4\cdot ...\cdot 2^4}_{n 2^4s}\cdot (n\cdot(n-1)\cdot(n-2)\cdot ...\cdot 3\cdot 2\cdot 1)^4}(2n+1)\cdot\frac{\pi}{2}

= \frac{((2n)!)^2\cdot (2n+1)}{2^{4n}\cdot(n!)^4}\cdot \frac{\pi}{2}

gives

\frac{I_{2n}}{I_{2n+1}} = \frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}\cdot\frac{\pi}{2}.\quad\quad\quad(6)

Substituting (5) and (6) into (4) yields

\frac{2n+1}{2n} >\frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}\cdot\frac{\pi}{2}>1.

Since \lim\limits_{n \rightarrow \infty} \frac{2n+1}{2n} = 1, \lim\limits_{n \rightarrow \infty}1 = 1, by Squeeze Theorem for Sequences,

\lim\limits_{n \rightarrow \infty}\frac{((2n)!)^4(2n+1)}{2^{4n}(n!)^4} \cdot \frac{\pi}{2}= 1\implies \lim\limits_{n \rightarrow \infty}\frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}=\frac{2}{\pi}.

Consequently,

\lim\limits_{n \rightarrow \infty} \frac{2^{4n}(n!)^4}{((2n)!)^2(2n+1)}= \frac{\pi}{2},

i.e.,

\pi = 2\cdot\lim\limits_{n \rightarrow \infty}\frac{2^{4n}(n!)^4}{((2n)!)^2(2n+1)}.

This is Wallis’ product representation for \pi, named after John Wallis who discovered it in 1665.


Maxima knows Wallis’ Pi:

Fig. 1

So does Mathematica:

Fig. 2

Its convergence to \pi is illustrated in Fig. 3:

Fig. 3

A Cautionary Tale of Compute Inverse Trigonometric Functions

From \arcsin(x)‘s definition:

\{(x, y) | \sin(y) = x, -\frac{\pi}{2} \le y \le \frac{\pi}{2}\},

we have

\arcsin(0) = 0, \arcsin(\frac{1}{2})= \frac{\pi}{6},\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}, \arcsin(1) = \frac{\pi}{2}

and,

\arcsin(-\frac{1}{2})= -\frac{\pi}{6},\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}, \arcsin(-1) = -\frac{\pi}{2}.

To obtain other values of \arcsin(x), we may simply solve

\sin(y) - x = 0

for y where -1 \le x \le 1.

For example (see Fig. 1), solving \sin(y) = \frac{1}{\sqrt{2}} for y gives y \approx \frac{\pi}{4}. It is in agreement with the fact that \arcsin(\frac{1}{\sqrt{2}})= \frac{\pi}{4}.

Fig. 1

In Fig. 2, we compute \arcsin(x) from repeatedly solving \sin(y) = x for y where

x=-1+i \cdot\frac{1-(-1)}{n}, i=1,2,...n.

Fig. 2

Since \sin(y) is a periodic function, \sin(y) = x has infinitely many solutions. It is possible that the solution obtained by Newton’s method lies outside of [-\frac{\pi}{2}, \frac{\pi}{2}], the range of \arcsin(x) by its definition. Such solution cannot be considered the value of \arcsin(x).

Fig. 3


Exercise-1 Compute \arccos(x) by solving \cos(y) = x for y.

Exercise-2 Explain Fig. 3.

A Mathematical Allegory

We have defined function \arcsin as a set:

\{(x, y) | \sin(y) =x, \frac{-\pi}{2} \le y \le \frac{\pi}{2}\}.

By definition,

\arcsin(-1) = \frac{-\pi}{2}, \arcsin(0)=0, \arcsin(1) = \frac{\pi}{2}

and

\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.

It means that \arcsin(x) is the unique solution of

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\quad\quad\quad(\star)

where y(-1)=-\frac{\pi}{2}, y(0)=0 and y(1)=\frac{\pi}{2}.

To compute \arcsin(x), we solve (\star) for y(x) as follows:

Integrate \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} from -1 to x gives

\displaystyle\int\limits_{-1}^{x}\frac{dy}{dx} \,dx=\displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\; d\xi\overset{\textbf{FTC}}{\implies}y(x) - y(-1) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\, d\xi.

Therefore,

y(x) = \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt(1-\xi^2}\,d\xi + y(-1) \overset{y(-1)=\frac{-\pi}{2}}{=} \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi - \frac{\pi}{2}.

That is,

\arcsin(x) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\;d\xi - \frac{\pi}{2}.

To obtain \arcsin(x), -1 < x < 1, we numerically evaluate \int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi, using function ‘quad_qags’.

Fig. 1

The result is visually validated in Fig. 2.

Fig. 2

Note: ‘romberg’, another function that computes the numerical integration by Romberg’s method will not work since it evaluates \frac{1}{\sqrt{1-x^2}} at x=-1.

Fig. 3

An alternate approach is to solve \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(0)=0 as an initial-value problem of ODE using ‘rk’ , the function that implements the classic Runge-Kutta algorithm.

Fig. 4 for 0 < x < 1

Fig. 5 -1<x<0

Putting the results together, we have

Fig. 6 -1<x<1

However, we cannot solve \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(-1)=\frac{-\pi}{2} using ‘rk’:

Fig. 7


Exercise-1 Compute \arccos(x) for x \in (-1, 1).

Exercise-2 Explain why ‘rk’ cannot solve \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(-1)=\frac{-\pi}{2}.

Finding Derivative the “Hard” Way

In “Instrumental Flying“, we defined \cosh^{-1}(x), \sinh^{-1}(x) as the inverse of \cosh(x) and \sinh(x) repectively.

To find the derivative of \cosh^{-1}(x), let

y = \cosh^{-1}(x).

We have

x = \cosh(y).

Differentiate it,

\frac{d}{dx} x = \frac{d}{dx} \cosh(y) \implies 1=\frac{d}{dy} \cosh(y)\cdot \frac{dy}{dx},

i.e.,

1 = \sinh(y) \frac{dy}{dx}\implies \frac{dy}{dx} = \frac{1}{\sinh(y)}.

By \cosh(y)^2-\sinh(y)^2=1 (see Exercise-1) and \cosh(y) \ge 1 (see Exrecise-2),

\sinh(y)^2 = \cosh(y)^2-1 \implies |\sinh(y)| = \sqrt{x^2-1} \overset{ (\star) }{\implies} \sinh(y) = \sqrt{x^2-1}.

And so,

\frac{dy}{dx} = \frac{1}{\sqrt{x^2-1}} \implies \boxed{\frac{d}{dx}\cosh^{-1}(x) = \frac{1}{\sqrt{x^2-1}}}.


Similarly, to find \frac{d}{dx}\sinh^{-1}(x), let

y = \sinh^{-1}(x)\implies x=\sinh(y).

Differentiate it,

\frac{d}{dx} x = \frac{d}{dx}\sinh(y) = \frac{d}{dy}\sinh(y)\frac{dy}{dx}\implies 1 = \cosh(y)\cdot\frac{dy}{dx}.

By \cosh(x)^2-\sinh(x)^2=1, \cosh(y) \ge 1,

\cosh(y) = \textbf{+}\sqrt{\sinh(y)^2+1} = \sqrt{x^2+1}.

Therefore,

1 = \sqrt{x^2+1}\frac{dy}{dx} \implies \boxed{\frac{d}{dx}\sinh^{-1}(x) = \frac{1}{\sqrt{x^2+1}}}.


Prove:

\forall x \ge 0, \sinh(x) \ge 0.\quad\quad\quad(\star)

By definition,

\sinh(x) = \frac{e^x-e^{-x}}{2} = \frac{e^{2x}-1}{2 e^{x}} \ge 0, since \forall x>0, e^{x},e^{2x} \ge 1 (see Exercise-3).


Exercise-1 Show that \forall x \in R, \cosh(x)^2 - \sinh(x)^2 =1.

Exercise-2 Show that \forall x \in R, \cosh(x) \ge 1.

Exercise-3 Show that \forall x \ge 0, e^{x} \ge 1.

Exercise-4 Differentiate \cosh^{-1}(x), \sinh^{-1}(x) directly (hint: see”Deriving Two Inverse Functions“).