# Chaos Esthétique

Drawing fractal by hand is tedious at best and close to impossible at worst. With the help of a CAS program, all you need is a few lines of code with a for-loop operation.

The fractal structure in Fig. 1 is generated from Gumnowski-Mira chaos model

$\begin{cases} F(x)=a x+{{2(1-a)x^2} \over { 1+x^2}} \\ x_{n+1}=y_{n}+F(x_{n}) \\ y_{n+1}=-x_{n} +F(x_{n+1}) \end{cases}$

Fig.1 $a=0.31, (x_0, y_0) = (12, 0)$

Replace the term $x_n$ by $-bx_n$ with b = 0.9998, the result is shown in Fig. 2.

Fig.2 $a=0.7, (x_0, y_0) = (15, 0)$

Let us turn our attention to the numerical calculation of logarithm, introduced in my previous post “Introducing Lady L“.

An example of naively compute $log(x)$, based solely on its definition is shown in Fig. 1.

Fig. 1

However, a more explicit expression is better suited for this purpose.

Fig. 2

From Fig.2, geometrical Interpretation of $\log(1+x)$ as the shaded area reveals that

$\log(1+x) = \int\limits_{1}^{x+1}{1 \over s}\;ds$

$=\int\limits_{1}^{s^*}{1 \over s}\;ds=\int\limits_{0}^{t^*}{1 \over {1+t}}\;dt$

$=\int\limits_{0}^{s^*-1}{1 \over {1+t}}\;dt = \int\limits_{0}^{(x+1)-1}{1 \over {1+t}}\;dt =\int\limits_{0}^{x}{1 \over {1+t}}\;dt$,

i.e.,

$\log(1+x) = \int\limits_{0}^{x}{1 \over {1+t}}\;dt\quad\quad\quad(1)$

Inserting into (1) the well known result

${1 \over {1+x}} =\sum\limits_{i=1}^{n}{(-1)^{i-1}x^{i-1} + {{(-1)^n x^n} \over {1+x}}}$,

we obtain

$\log(1+x)=\int\limits_{0}^{x} \sum\limits_{i=1}^{n}{(-1)^{i-1}t^{i-1} + {{(-1)^n t^n} \over {1+t}}}\; dt$

$= \int\limits_{0}^{x}\sum\limits_{i=1}^{n}{(-1)^{i-1}t^{i-1}\;dt +\int\limits_{0}^{x}{{(-1)^n t^n} \over {1+t}}}\;dt$

$=\sum\limits_{i=1}^{n}{(-1)^{i-1}\int\limits_{0}^{x}{t^{i-1}}\;dt}+\int\limits_{0}^{x}{{(-1)^n t^n} \over {1+t}}\;dt$

$= \sum\limits_{i=1}^{n}{{(-1)^{i-1}x^i} \over {i}} + \int\limits_{0}^{x}{{(-1)^n t^n} \over {1+t}}\;dt$.

Let

$r_n= \int\limits_{0}^{x}{{(-1)^n t^n}\over{1+t}}\;dt$,

we have

$\log(1+x)-r_n=\sum\limits_{i=1}^{n}{{(-1)^{i-1} x^i}\over{i}}$.

If $-1,

$|r_n| = |-\int\limits_{x}^{0}{{(-1)^n t^n}\over{1+t}}\;dt|\leq \int\limits_{x}^{0}|{{(-1)^{n}{t^n}} \over {1+t}}|\;dt=\int\limits_{x}^{0}{{|t|^n} \over {1+t}}\;dt\leq |x|^n\int\limits_{x}^{0}{1 \over {1+t}}\;dt$

otherwise ($0 \leq x < 1)$

$|r_n|\leq \int\limits_{0}^{x}|{{(-1)^{n}{t^n}} \over {1+t}}|\;dt=\int\limits_{0}^{x}{{|t^n|} \over {1+t}}\;dt\leq |x|^n\int\limits_{0}^{x}{1 \over {1+t}}\;dt$.

Therefore, either

$-|x|^n\int\limits_{x}^{0}{1 \over {1+t}}\;dt \leq r_n \leq |x|^n\int\limits_{x}^{0}{1 \over {1+t}}\;dt$

or

$-|x|^n\int\limits_{0}^{x}{1 \over {1+t}}\;dt \leq r_n \leq |x|^n\int\limits_{0}^{x}{1 \over {1+t}}\;dt$.

Since $|x| < 1$,

$\lim\limits_{n \to \infty}-|x|^n\int\limits_{0}^{x}{1 \over {1+t}}\;dt=0$

and

$\lim\limits_{n \to \infty}|x|^n\int\limits_{0}^{x}{1 \over {1+t}}\;dt=0$

We conclude that

$\lim\limits_{n\to\infty}r_n = 0$.

As a consequence,

$\log(1+x) = \log(1+x)-0$

$= \lim\limits_{n\to\infty}\log(1+x)-\lim\limits_{n \to \infty}r_n$

$=\lim\limits_{n\to\infty}(\log(1+x)-r_n)$

$=\lim\limits_{n\to\infty}( \sum\limits_{i=1}^{n}{{(-1)^{i-1}x^i} \over {i}})$,

i.e.,

$\log(1+x) = \sum\limits_{i=1}^{\infty}{{(-1)^{i-1}x^i} \over {i}}\quad\quad(2)$

(2) offers a means for finding the numerical values of logarithm. However, its range is limited to the value of $1+x$ between 0 and 2, since $-1.

To overcome this limitation, we proceed as follows:

$\forall x \in (-1, 1), -x \in (-1, 1)$. By (2),

$\log(1-x)=\log(1+(-x)) = \sum\limits_{i=1}^{\infty}{{(-1)^{i-1}(-x)^i} \over i }= \sum\limits_{i=1}^{\infty}{{-x^i} \over {i}}$

i.e.,

$\log(1-x) = \sum\limits_{i=1}^{\infty}{{-x^i} \over {i}}\quad\quad\quad\quad(3)$

Subtracting (3) from (2) and using the fact that $\log{a}-\log{b}=\log{a \over b}$, we have

$\log{{1+x} \over {1-x}} = \sum\limits_{i=1}^{\infty}{{(-1)^{i-1}x^i} \over {i}} - \sum\limits_{i=1}^{\infty}{{-x^i} \over {i}}$

$= \sum\limits_{i=1}^{\infty}{{(-1)^{i-1}{x^i} \over {i}} + {{x^i} \over {i}}}$

$=\sum\limits_{odd \;i=1}^{\infty}{2{x^i} \over {i}}$.

i.e.,

$\log{{1+x} \over {1-x}} = 2\sum\limits_{i=1}^{\infty}{{x^{2i-1}} \over {2i-1}}\quad\quad\quad(4)$

Solving equation

${{1+x} \over {1-x}} = v$ where $v > 0$,

we find

$x = {{v-1} \over {v+1}}$.

Since this solution can be expressed as

$x = {{-1-v+2v} \over {v+1}} = -1 + {2v \over {v+1}}$

or

$x = {{v+1-2} \over {v+1}} = 1-{{2v} \over {v+1}}$.

It shows that for any $v > 0$, $x \in (-1, 1)$. Therefore,  (4) can be used to obtain the logarithm of any positive number. For example, to obtain $\log(3)$, we solve ${{1+x} \over {1-x}} = 3$ first and then compute a partial sum of (4) with sufficient large number of terms (see Fig. 3)

Fig. 3