October 2017

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.

Screen Shot 2017-10-07 at 6.12.01 PM.png

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

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

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

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

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

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

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

Month: October 2017

Chaos Esthétique

Knowing Lady L