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 $r=k$,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Finite Difference Approximations of Derivatives

We will derive the finite difference approximations for $y'(x)$ and $y''(x)$.

Let

$y'_i, y''_i$ denotes $y'(x_i)$ and $y''(x_i)$ respectively

and

$h = x_{i+1}-x_i = x_i -x_{i-1} > 0$.

We prove that

[1] $y_i' \approx \frac{y_{i+1} - y_i}{h}$

Let $x=x_{i+1}, a=x_i$, Taylor series

$y(x) = y(a) + y'(a) (x-a) + O((x-a)^2)$

is

$y(x_{i+1}) = y(x_i) + y'(x_i)(x_{i+1}-x_i)+O((x_{i+1} - x_i)^2)$

i.e.,

$y_{i+1} = y_i + y'_i h + O(h^2)$

Hence,

$\frac{y_{i+1} -y_i}{h} = y'_i + O(h)$

[2] $y_i' \approx \frac{y_{i+1}-y_{i-1}}{2h}$

Let $x = x_{i+1}, a = x_i$, Taylor series

$y(x) = y(a) + y'(a) (x-a) + \frac{y''(a)}{2!} (x-a)^2 + O((x-a)^3)$

becomes

$y(x_{i+1}) = y(x_i) + y'(x_i)(x_{i+1}-x_i) + \frac{y''_i}{2!} (x_{i+1}-x_i)^2 + O((x_{i+1}-x_i)^3)$

It follows that

$y_{i+1}= y_i + y'_i h + \frac{y''_i}{2!} h^2 + O(h^3)\quad\quad\quad(1)$

Similarly, let $x=x_{i-1}, a = x_i$,

$y_{i-1}= y_i + y'_i (x_{i-1}-x_1) + \frac{y''_i}{2!} (x_{i-1}-x_i)^2 + O((x_{x-1}-x_i)^3$

Since $x_{i-1}-x_i = -(x_i - x_{i-1}) = -h$, we have

$y_{i-1} = y_i - y'_i h + \frac{y''_i}{2!} + O(h^3)\quad\quad\quad(2)$

(1)-(2) $\implies$

$y_{i+1}-y_{i-1} = 2y'_i h + O(h^3)$

Therefore,

$\frac{y_{i+1}-y_{i-1}}{2h} = y'_i + O(h^2)$

[3] $y_i'' \approx \frac{y_{i+1}-2y_i+y_{i-1}}{h^2}$

Let $x = x_{i+1}, a=x_i$, Taylor series

$y(x) = y(a) + y'(a) (x-a) + \frac{y''(a)}{2!} (x-a)^2 + \frac{y'''(a)}{3!} (x-a)^3 + O((x-a)^4)$

becomes

$y(x_{i+1}) = y(x_i) + y'(x_i) (x_{i+1}-x_i) + \frac{y''(x_i)}{2!} (x_{i+1} - x_i)^2$

$+ \frac{y'''(x_i)}{3!}(x_{i+1} - x_i)^3 + O((x_{i+1}-x_{i})^4)$.

That is,

$y_{i+1} = y_i + y_i' h + \frac{y_i''}{2!} h^2 + \frac{y_i'''}{3!} h^3 + O(h^4)\quad\quad\quad(3)$

Similarly, let $x = x_{i-1}, a=x_i$, we have

$y(x_{i-1}) = y(x_i) + y'(x_i) (x_{i-1}-x_i) + \frac{y''(x_i)}{2!} (x_{i-1} - x_i)^2$

$+ \frac{y'''(x_i)}{3!}(x_{i-1} - x_i)^3 + O((x_{i-1}-x_{i})^4)$.

i.e.,

$y_{i-1} = y_i - y_i' h + \frac{y_i''}{2!} h^2 - \frac{y_i'''}{3!} h^3 + O(h^4)\quad\quad\quad(4)$

(3) + (4) $\implies$

$y_{i+1}+y_{i-1} = 2y_i +y_i'' h^2 + O(h^4)$.

Therefore,

$\frac{y_{i+1}-2y_i+y_{i-1}}{h^2} = y_i'' + O(h^2)$

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 again

We will construct the tangent line of circle differently than what is shown in a previous post.

Start from the equation of circle with radius $r$, centered at the origin of the rectangular coordinate system:

$x^2+y^2=r^2\quad\quad\quad(1)$

Differentiate (1), we have

$2x + 2y { {dy} \over {dx} } = 0$.

At $(x_0, y_0)$,

${x_0 + y_0 {{dy} \over {dx}}} = 0$

i.e.,

$y_0 {{dy} \over {dx}} = -x_0$

If $y_0 \neq 0$,

${{dy} \over {dx} }={ {-x_0} \over {y_0}}$

$y = {{-x_0} \over {y_0}} x + m$

That is

$y_0 y = -x_0 x +my_0$

or

$x_0 x + y_0 y= my_0\quad\quad\quad(2)$

At $(x_0, y_0)$,

$my_0 = x_0^2+y_0^2=r^2$

It follows that by (2),

$x_0 x + y_0 y= r^2$

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