A Mathematical Allegory

October 20, 2021October 20, 2021 / Michael Xue / Leave a comment

We have defined function $y = \arcsin(x)$ 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

October 8, 2021October 14, 2021 / Michael Xue / Leave a comment

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

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

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 $\cos^{-1}(x), \sinh^{-1}(x)$ directly (hint: see”Deriving Two Inverse Functions“).

Oasis

October 2, 2021October 11, 2021 / Michael Xue / Leave a comment

An oasis awaits

The above image is created by Omega CAS Explorer:

The governing equations are:

$x_{n+1} = \rho + c_2(x_n\cos(t_n)-y_n\sin(t_n),$

$y_{n+1} = c_2(x_n\sin(t_n) + y_n\cos(t_n))$

where $t_n = c_1-\frac{c_3}{1+x_n^2 + y_n^2}.$

Polar plot

September 28, 2021September 27, 2021 / Michael Xue / Leave a comment

The polar coordinates $r$ and $\theta$ can be converted to the Cartesian coordinates $x$ and $y$ using the trigonometry functions:

$\begin{cases} x=r\cdot\cos(\theta) \\ y=r\cdot\sin(\theta)\end{cases}$

It follows that a figure specified in $(r, \theta)$ can be plotted by ‘plot2d’ as a parametric curve:

Fig. 1 $r = \cos(5\theta)$

It is possible to plot two or more parametric curves together:

Fig. 2 $r=\theta$ and $r=\cos(5\theta)$

An alternate is the ‘draw2d’ function, it draws graphic objects created by the ‘polar’ function:

Fig. 3 $r=\theta$ and $r=\cos(5\theta)$

Fig. 4 shows a graceful geometric curve that resembles a butterfly. Its equation is expressed in polar coordinates by

$r = e^{\sin(\theta)} - 2\cos(4\theta)+\sin(\frac{\theta}{12})^5$

Fig. 4

Let’s combine them!

September 25, 2021September 24, 2021 / Michael Xue / Leave a comment

It is possible to combine two or more plots into one picture.

For example, we solve the following initial-value problem

$\begin{cases} \frac{dy}{dx} = (x-y)/2 \\ y(0)=1 \end{cases}\quad\quad\quad(\star)$

and plot the analytic solution in Fig. 1.

Fig. 1

We can also solve $(\star)$ numerically and plot the discrete data points:

Fig. 2

Fig. 3 is the result of combining Fig.1 and Fig. 2.

Fig.3

It validates the numerical solution obtained by ‘rk’: the two figures clearly overlapped.

An Alternate Solver of ODEs

September 22, 2021September 22, 2021 / Michael Xue / Leave a comment

Besides ‘ode2’, ‘contrib_ode’ also solves differential equations.

For example,

$\frac{d^2y}{dx^2} - \frac{1+x}{x} \cdot \frac{dy}{dx}+ \frac{1}{x}\cdot y=0.$

While ‘ode2’ fails:

‘contrib_ode’ succeeds:

This is an example taking from page 4 of Bender and Orszag’s “Advanced Mathematical Methods for Scientists and Engineers“. On the same page, there is another good example:

$\frac{dy}{dx} = \frac{A^2}{x^4}-y^2, \quad\quad A$ is a constant.

Using ‘contrib_ode’, we have

It seems that ‘contrib_ode’ is a better differential equation solver than ‘ode2’:

Even though it is not perfect:

From the examples, we see the usage of ‘contrib_ode’ is the same as ‘ode2’. However, unlike ‘ode2’, ‘contrib_ode’ always return a list of solution(s). It means to solve either initial-value or boundary-value problem, the solution of the differential equation is often lifted out of this list first:

Exercise Solve the following differential equations without using a CAS:

$\frac{dy}{dx}= \frac{A^2}{x^4} - y^2$ (hint: Riccati Equation)
$\frac{d^2 y}{dx^2} = \frac{y \frac{dy}{dx}} {x}$

DIY

September 17, 2021September 17, 2021 / Michael Xue / Leave a comment

As far as I know, ‘bc2’, Maxima’s built-in function for solving two-point boundary value problem only handles the type:

$\begin{cases} y'' = f(x, y, y')\\ y(a)=\alpha, y'(b) = \beta \end{cases}\quad\quad\quad(*)$

For example, solving $\begin{cases} y'' + y (y')^3=0 \\y(0) = 1, y(1)=3 \end{cases}:$

But ‘bc2’ can not be applied to

$\begin{cases} -y'' + 9 y =12 \sin(t)\\ y(0)=y(2\pi), y'(0) = y'(2\pi) \end{cases}$

since it is not the type of (*). However, you can roll your own:

An error occurs on the line where the second boundary condition is specified. It attempts to differentiate the solution with respect to $t$ under the context that $t=0$ . i.e.,

diff(rhs(sol), 0);

which is absurd.

The correct way is to express the boundary conditions using ‘at’ instead of ‘ev’:

The following works too as the derivative is obtained before using ‘ev’:

Nonetheless, I still think using ’at’ is a better way:

Solving x = a x + b

August 27, 2021August 28, 2021 / Michael Xue / Leave a comment

Problem: Solving $x = ax +b$ for $x$ .

Solution:

Choose $x_0$ arbitrarily, we generate the sequence $\{x_i\}$ recursively from $x_i = a x_{i-1} + b, i=1, 2, ...$ :

$x_1= a x_0 + b,$

$x_2 = a x_1 +b =a(a x_0+b)+b =a^2 x_0 +ab + b,$

$x_3 = a x_2 + b = a (a^2 x_0 + ab +b) +b = a^3 x_0 +a^2b +ab +b,$

$\ddots$

$x_n = a x_{n-1} +b = a^n x_0 + b \sum\limits_{i=0}^{n-1} a^i=a^n x_0 + b\cdot\frac{1-a^{n}}{1-a}.$

It follows that for $|a| <1$ ,

$x = \lim\limits_{n \rightarrow \infty} x_n = \lim\limits_{n\rightarrow \infty}a^n x_0+\frac{b(1-a^n)}{1-a}= \lim\limits_{n\rightarrow \infty}a^n x_0+ \lim\limits_{n \rightarrow \infty}\frac{b(1-a^n)}{1-a}\overset{(\star)}{=}0\cdot x_0 + \frac{b(1-0)}{1-a}.$

i.e.,

$x = \frac{b}{1-a}.$

Fig. 1 shows a CAS -aided solution using Omega CAS Explorer:

Fig. 1

For $|a|> 1$ , we rewrite $x=ax +b$ as

$x = \frac{1}{a}x - \frac{b}{a} \overset{A=\frac{1}{a}, B=-\frac{b}{a}}{=} Ax +B.$

Since $|a| >1 \implies \frac{1}{|a|} = |A| <1\implies \lim\limits_{n\rightarrow \infty}A^n = 0$ ,

$x =\lim\limits_{n \rightarrow \infty}A^n x_0 + B\cdot \frac{1-A^n}{1-A}\frac{}{} = \frac{B}{1-A} = \frac{-\frac{b}{a}}{1-\frac{1}{a}} = \frac{-b}{a-1} = \frac{b}{1-a}.$

We have used fact that

$|a| < 1 \implies \lim\limits_{n \rightarrow \infty} a^n = 0.\quad\quad\quad(\star)$

A proof is as follows:

Since

$|a^n| = |\underbrace{a\cdot a\cdot a ... a}_{n\;a's}| =\overbrace{|a||a|...|a|}^{n\;|a|'s} = |a|^n,$

if $a=0$ then $a^n = 0\cdot 0 ... 0 = 0.$ It means

$(\forall \epsilon >0, \forall n \in N^+, |a^n-0|<\epsilon)\implies \lim\limits_{n \rightarrow \infty} a^n = 0.$

Otherwise ( $a \ne 0$ ),

$\forall \epsilon >0, |a^n-0| =|a|^n < \epsilon \implies n \log(|a|)<\log(\epsilon).$

For $|a|<1,$

$n\log(|a|)<\log(\epsilon) \overset{\log(|a|) < 0}{\implies} n > \frac{\log(\epsilon)}{\log(|a|)}.$

And so,

$(\forall \epsilon > 0, \exists n^* = \lceil\frac{\log(\epsilon)}{\log(|a|)}\rceil \ni n > n^*, |a^n-0| < \epsilon) \implies \lim\limits_{n \rightarrow \infty}a^n = 0.$

Exercise-1 Prove by mathematical induction:

$(x_{k} = a x_{k-1} + b, k=1,2,...,n ) \implies x_n = a^n x_0 + b\sum\limits_{i=0}^{n-1}a^i.$

“Instrumental Flying”

August 14, 2021August 14, 2021 / Michael Xue / Leave a comment

Consider the following set

$S_1 = \{(x,y) | x=\frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1-1)$

For all $(x, y_1), (x, y_2) \in S_1$ , we have

$\frac{e^{y_1}-e^{-y_1}}{2}=x, \frac{e^{y_2}-e^{-y_2}}{2}=x.$

That is,

$(x, y_1), (x, y_2) \in S_1 \implies \frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0.\quad\quad\quad(1-2)$

Since $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$ (Exercise-1),

$\frac{e^{y_1}-e^{-y_1}}{2}-\frac{e^{y_2}-e^{-y_2}}{2}=0 \implies y_1 = y_2.\quad\quad\quad(1-3)$

From (1-2) and (1-3), we conclude:

$\forall (x, y_1), (x, y_2) \in S_1 \implies y_1 = y_2.$

i.e., $S_1$ defines a function.

Alternatively, (1-1) can be expressed as

$S_1 = \{(x, y) | x = \sinh(y)\}.$

It shows that $S_1$ is the inverse of $\sinh(x)$ . Therefore, we name the function defined by (1-1) $\sinh^{-1}$ and write it as

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

Let’s look at another set:

$S_2 = \{(x,y)|x=\frac{e^y+e^{-y}}{2}\}.\quad\quad\quad(2-1)$

For a pair $(x, y_1>0) \in S_2$ , we have

$x=\frac{e^{y_1}+e^{-y_1}}{2}.\quad\quad\quad(2-2)$

For another pair $(x, y_2=-y_1)$ ,

$\frac{e^{y_2} + e^{-y_2}}{2} = \frac{e^{-y_1}+e^{-(-y_1)}}{2} = \frac{e^{-y_1} + e^{y_1}}{2} \overset{(2-2)}{=} x\implies (x, y_2=-y_1) \in S_2.$

Since $y_1>0, y_2 = -y_1 \neq y_1, (x, y_1), (x, y_2) \in S_1$ does not implie $y_1 = y_2$ . It means $S_2$ does not define a function.

However, modification of $S_2$ gives

$S_3 = \{(x, y) | x=\frac{e^y+e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2-3)$

It defines a function.

Rewrite (2-3) as

$S_3 = \{(x, y) | x = \cosh(y), y \ge 0\}.\quad\quad\quad(2-4)$

Then $\forall (x, y_1), (x, y_2) \in S_3$ , we have

$x=\cosh(y_1), x=\cosh(y_2)\implies \cosh(y_1) = \cosh(y_2).\quad\quad\quad(2-5)$

Notice that by (2-3), $y_1, y_2 \ge 0.$

Cleary, (2-5) is true if and only if $y_1 =y_2$ . For if $y_1 \ne y_2$ , by $(\star)$ , $\cosh(y_1) \ne \cosh(y_2).$

Therefore,

$(x, y_1), (x, y_2) \in S_3 \implies y_1 = y_2.$

We name the function defined by (2-3) $\cosh^{-1}$ as (2-4) shows that it is the inverse of $\cosh(x).$

See “Deriving Two Inverse Functions” for the explicit expressions of $\sinh^{-1}$ and $\cosh^{-1}$ .

Prove

$t_1 \ge 0, t_2 \ge 0, t_1 \ne t_2 \implies \cosh(t_1) \ne \cosh(t_2).\quad\quad\quad(\star)$

Without loss of generality, we assume that $t_2 > t_1.$ By Lagrange’s mean-value theorem (see “A Sprint to FTC“),

$\cosh(t_2) -\cos(t_1) =\cosh'(\xi) (t_2-t_1)$

where $\xi \in (t_1, t_2).$

We have

$t_2 > t_1 \implies t_2 - t_1 >0$

and

$\forall t > 0, (\cosh(t))' = (\frac{e^t+e^{-t}}{2})' = \frac{e^t-e^{-t}}{2} \overset{t>0\implies t >-t \implies e^t >e^{-t}}{>} 0.$

Consequently,

$\cosh(t_2) - \cosh(t_1) = \cosh'(\xi) (t_2-t_1) > 0.$

i.e.,

$\cosh(t_2) \ne \cosh(t_1).$

Exercise-1 Show that $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$ .

Deriving Two Inverse Functions

July 22, 2021September 9, 2021 / Michael Xue / Leave a comment

In “Instrumental Flying“, we defined function $y=\sinh^{-1}(x)$ as

$\{(x, y) | x = \frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1)$

From $x = \frac{e^y-e^{-y}}{2}$ , we obtain

$(e^y)^2-2x\cdot e^y-1=0.$

It means either $e^y = x-\sqrt{x^2+1}$ or $e^y = x+\sqrt{x^2+1}.$

But $e^y = x-\sqrt{x^2+1}$ suggests $e^y < 0$ (see Exercise-1), contradicts the fact that $\forall t \in R, e^t > 0$ (see “Two Peas in a Pod, Part 2“).