Monthly Archives: October 2017

Chaos Esthétique

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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}

by Omega CAS Explorer.

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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.

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

 

 

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Knowing Lady L

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.

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Fig. 1

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

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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<x<0,

|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<x<1 \implies 0 < 1+x < 2.

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)

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Fig. 3