# Introducing Operator Delta

The $r^{th}$ order finite difference of function$f(x)$ is defined by

$\Delta^r f(x) = \begin{cases} f(x+1)-f(x), r=1\\ \Delta(\Delta^{r-1}f(x)), r > 1\end{cases}$

From this definition, we have

$\Delta f(x) = \Delta^1 f(x) = f(x+1)-f(x)$

and,

$\Delta^2 f(x) = \Delta (\Delta^{2-1} f(x))$

$= \Delta (\Delta f(x))$

$= \Delta( f(x+1)-f(x))$

$= (f(x+2)-f(x+1)) - (f(x+1)-f(x))$

$= f(x+2)-2f(x)+f(x+1)$

as well as

$\Delta^3 f(x) = \Delta (\Delta^2 f(x))$

$= \Delta (f(x+2)-2f(x)+f(x+1))$

$= (f(x+3)-2f(x+1)+f(x+2)) - (f(x+2)-2f(x)+f(x+1))$

$= f(x+3)-3f(x+2)+3f(x+1)-f(x)$

The function shown below generates $\Delta^r f(x), r:1\rightarrow 5$ (see Fig. 1).

delta_(g, n) := block(
local(f),

define(f[1](x),
g(x+1)-g(x)),

for i : 2 thru n do (
define(f[i](x),
f[i-1](x+1)-f[i-1](x))
),

return(f[n])
);


Fig. 1

Compare to the result of expanding $(f(x)-1)^r=\sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x)^{r-i}, r:1\rightarrow 5$ (see Fig. 2)

Fig. 2

It seems that

$\Delta^r f(x) = \sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x+r-i)\quad\quad\quad(1)$

Lets prove it!

We have already shown that (1) is true for $r= 1, 2, 3$.

Assuming (1) is true when $r=k-1 \ge 4$:

$\Delta^{k-1} f(x) = \sum\limits_{i=0}^{k-1}(-1)^i \binom{r}{i} f(x+k-1-i)\quad\quad\quad(2)$

When $r=k$,

$\Delta^k f(x) = \Delta(\Delta^{k-1} f(x))$

$\overset{(2)}{=}\Delta (\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i))$

$=\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+1+k-1-i)-\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$=(-1)^0 \binom{k-1}{0}f(x+k-0)$

$+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$-(-1)^{k-1}\binom{k-1}{k-1}f(x+k-1-(k-1))$

$\overset{\binom{k-1}{0} = \binom{k-1}{k-1}=1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$=f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)+\sum\limits_{i=0}^{k-2}(-1)^{i+1}\binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$\overset{j=i+1, i:0 \rightarrow k-2\implies j:1 \rightarrow k-1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{j=1}^{k-1}(-1)^j \binom{k-1}{j-1}f(x+k-j)-(-1)^{k-1}f(x)$

$= f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i-1}f(x+k-i)+(-1)^k f(x)$

$= f(x+k) + \sum\limits_{i=1}^{k-1}(-1)^i f(x+k-i) (\binom{k-1}{i} + \binom{k-1}{i-1})+(-1)^k f(x)$

$\overset{\binom{k-1}{i} + \binom{k-1}{i-1}=\binom{k}{i}}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k f(x)$

$= (-1)^0 \binom{k}{0}f(x+k-0)+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k \binom{k}{k} f(x+k-k)$

$= \sum\limits_{i=0}^{k}(-1)^i \binom{k}{i} f(x+k-i)$

# A pair of non-identical twins

A complex number $x + i y$ can be plotted in a complex plain where the $x$ coordinate is the real axis and the $y$ coordinate the imaginary.

Let’s consider the following iteration:

$z_{n+1} = z_{n}^2 + c\quad\quad\quad(1)$

where $z, c$ are complex numbers.

If (1) are started at $z_0 = 0$ for various values of $c$ and plotted in c-space, we have the Mandelbrot set:

When $c$ is held fixed and points generated by (1) are plotted in z-space, the result is the Julia set:

# Constructing the tangent line of circle without calculus

The tangent line of a circle can be defined as a line that intersects the circle at one point only.

Put a circle in the rectangular coordinate system.

Let $(x_0, y_0)$ be a point on a circle. The tangent line at $(x_0, y_0)$ is a line intersects the circle at $(x_0, y_0)$ only.

Let’s first find a function $y=kx+m$ that represents the line.

From circle’s equation $x^2+y^2=r^2$, we have

$y^2=r^2-x^2$

Since the line intersects the circle at $(x_0, y_0)$ only,

$r^2-x^2=(kx+m)^2$

has only one solution.

That means

$k^2x^2+x^2+2kmx+m^2-r^2 =0$

has only one solution. i.e., its discriminant

$(2km)^2-4(k^2+1)(m^2-r^2)=0\quad\quad\quad(1)$

By the definition of slope,

$kx+m-y_0 = k(x-x_0)$.

It follows that

$m =y_0-kx_0\quad\quad\quad(2)$

Substitute (2) into (1) and solve for $k$ gives

$k = \frac{-x_0}{y_0}\quad\quad\quad(3)$

The slope of line connecting $(0, 0)$ and $(x_0, y_0)$ where $x_0 \neq 0$ is $\frac{y_0}{x_0}$.

Since $\frac{-x_0}{y_0}\cdot \frac{y_0}{x_0} = -1$, the tangent line is perpendicular to the line connecting $(0, 0)$ and $(x_0, y_0)$.

Substitute (3) into $y = k x +m$, we have

$y=-\frac{x_0}{y_0} x + m\quad\quad\quad(4)$.

The fact that the line intersects the circle at $(x_0, y_0)$ means

$y_0 = -\frac{x_0^2}{y_0} + m$

or

$y_0^2=-x_0^2+ my_0$.

Hence,

$m =\frac{x_0^2+y_0^2}{y_0} = \frac{r^2}{y_0}$.

It follows that by (4),

$x_0 x +y_0 y = r^2\quad\quad\quad(5)$

(5) is derived under the assumption that $y_0 \neq 0$. However, by letting $y_0 =0$ in (5), we obtain two tangent lines that can not be expressed in the form of $y=kx+m$:

$x=-r, x=r$

# Constructing the tangent line of quadratic without calculus

The tangent line of a quadratic function at $(x_0, y_0)$is a line $y=kx+m$ that intersects $y=ax^2+bx+c$ at $(x_0, y_0=ax_0^2+bx_0+c)$ only.

The presence of function $y=kx+m$ immediately excludes the vertical line $x=x_0$ which also intersects $y=ax^2+bx+c$ at $(x_0, ax_0^2+bx_0+c)$ only (see Fig. 1).

Fig. 1

Let’s find $k$.

Line $y = kx+m$ intersects $y=ax^2+bx+c$ at $(x_0, ax_0^2+bx_0+c)$ only means quadratic equation

$ax^2+bx +c =kx +m$

has only one solution. That is, the discriminant of $ax^2+bx+c-kx-m =0$ is zero:

$(b-k)^2-4a(c-m) = 0\quad\quad\quad(1)$

Fig. 2

And, by the definition of slope (see Fig. 2),

$(x-x_0)k = (kx+m)-(ax_0^2+bx_0+c)$.

It follows that

$m = (ax_0^2+b_0+c)-x_0 k\quad\quad\quad(2)$

Substituting (2) into (1), we have

$(b-k)^2-4a(c-((a_0 x^2+b x_0 + c)-x_0 k)=0$.

Solve it for $k$ gives

$k = 2 a x_0 +b$.

Fig. 3

# A Proof without Calculus

Now, solve $\sin(x)=x$.

By inspection, $x=0$.

Is this the only solution?

Visually, it is difficult to tell (see Fig. 1 and Fig. 2)

Fig. 1

Fig. 2

However, we can prove that $0$ is the only solution:

Fig. 3

If $0 < x \leq 1 < \frac{\pi}{2}$ then from Fig. 3, we have

Area of triangle OAB < Area of circular sector OAB.

That is,

$\frac{1}{2}\cdot 1 \cdot \sin(x)< \frac{1}{2}\cdot 1 \cdot x$.

Hence,

$\forall 0< x \leq 1, \sin(x) < x\quad\quad\quad(1)$

If $x>1$ then

$\sin(x) \leq 1 \overset{x>1}{\implies} \sin(x) \leq 1< x \implies \forall x>1, \sin(x)

Put (1) and (2) together, we have

$\forall x > 0, \sin(x) < x\quad\quad\quad(3)$

If $x < 0$ then

$-x > 0 \overset{(3)}\implies \sin(-x) < -x \overset{\sin(-x)=-\sin(x)}\implies -\sin(x) < -x \implies \sin(x) > x$

i.e.,

$\forall x < 0, \sin(x) > x\quad\quad\quad(4)$

Therefore by (3) and (4), we conclude that

only when $x = 0, \sin(x)=x$.

# From proof to simpler proof

Solve $\sin(x)=2x$ for $x$.

By mere inspection, we have $x=0$.

Visually, it appears that $0$ is the only solution (see Fig.1 or Fig. 2)

Fig. 1

Fig. 2

To show that $0$ is the only solution of $\sin(x)=2x$ analytically, let

$f(x) = \sin(x)-2x$

$0$ is a solution of $\sin(x)=2x$ means

$f(0) = 0\quad\quad\quad(1)$

Suppose there is a solution

$x^* \neq 0\quad\quad\quad(2)$

then,

$f(x^*)=\sin(x^*)-2x^*=0\quad\quad\quad(3)$

Since $f(x)$ is an function continuous and differentiable on $(-\infty, +\infty)$,

by Lagrange’s Mean-Value Theorem (see “A Sprint to FTC“), there $\exists c \in (0, x^*)$ such that

$f(x^*)-f(0)=f'(c)(x^*-0)$

$\overset{(1), (3)}{\implies} 0 = f'(c) x^*\quad\quad\quad(4)$

$\overset{(2)}{\implies} f'(c)=0\quad\quad\quad(5)$.

We know

$f'(x) = \cos(x)-2$.

From (5), we have

$\cos(c)-2=0$.

i.e.,

$\cos(c)=2$.

This is not possible since $\forall x \in R, -1 \le \cos(x) \le 1$.

A simpler alternative without direct applying Lagrange’s Mean-Value Theorem is:

$f(x) = \sin(x)-2x \implies f'(x) = \cos(x) - 2 \overset{-1 \le \cos(x) \le 1}{\implies} f'(x) < 0\implies$

$f(x) =\sin(x)-2x$ is a strictly decreasing function.

Since $f(0) = 0, \forall x<0, f(x) > 0$ and $\forall x>0, f(x) < 0$.

Therefore, $0$ is the only solution of $f(x)=0$. i.e.,

$0$ is the only solution of $\sin(x) = 2x$.