Sir hutch, sorry to post an image in your server, I know that spends a lot of bandwidth, but without this image this will be hard to explain. Well, I suppose can be usefull to your eletrical deals.

I am tired, after a long journey of study I see only disappointments. I will reveal the secret behind this topic. The best I can do is keep weeding and growing food.

Public domain.

I could feel arithmetic compression but couldn't explain it. I tried mental regressions but in vain. Take your beer, your cigarette, sit down comfortably.

PART 1: PHYSICS, ELECTRICITY

I find it strange that the electron charge is 1.6. Being Catholic and believing in the golden or divine number, I wondered if there was any relationship between the electron charge and Fibonacci. And here came the crack.

In electrical we study circuits with resistors in series, parallel and mixed. I think it is Thevenin who said that every resistive circuit can be shortened with only one voltage source (voltage) and only one resistor called the equivalent resistance.

No matter how many resistors I have in a circuit, no matter if they are in series and parallel, it just matters that we can represent the entire circuit with only 1 resistor.

The formula used is: Voltage = Resistance * Amps

If the resistance is equal to 1ohm, then the voltage is equal to the intensity of the electric current (amps).

PART 2: FIBONACCI

The fibonacci sequence is known to everyone:

1,1,2,3,5,8,13,21,34,55,89, ....

If we take two sequential numbers and divide them we tend to find the golden number.

55/34 = 1.6176470588235294117647058823529

89/55 = 1.618181818181818181818181818181818

But something curious happens when we reverse the division, see:

34/55 = 0.61818181818181818181818181818182

Did you notice that the ratio between 89/55 and 34/55 is the same? Same floating points except for one detail: One starts with 0 and the other with 1 for the entire part.

Joining mixed electrical circuits with fibonacci. If all resistors are 1 ohm, strangely I can fit fibonacci into the electrical circuit.

PART 3: ARITHMETIC DATA COMPRESSION - fibonacci compression or sunflower compression.

Joining the useful with the pleasant. If we think of equivalent resistance as a form of data compression and the electrical circuit (resistors) as an abstract representation of the fibonacci sequence then it can be done. But in digital circuits we deal with the binary base. Then I should combine the 3 themes into one. Mathematicians do not compress data, mathematicians model data.

Imagine the following electrical circuit of figure 3:

we have arranged the resistors: 3,4,2,3,1

Series = numerator, parallel = denominator

Series = 1, parallel = 0

34231

Converting the circuit to binary:

1110000110001

Starting from right to left we do:

1/3

3 * 2 + 1 = 7/3

7 * 4 + 3 = 7/31

31 * 3 + 7 = 100/31

The equivalent pseudo resistance or the compressed circuit or the compressed binary digits can be represented by the division 100/31 = 3,2258064516129032258064516129032

Note that the nonzero integer (the number 3) corresponds to the circuit starting with ones instead of zeros (starting in series), and changing to 0 (parallel), then to 1, to 0, to 1, .. .

As seen in fibonacci, the data are sequences of parallels and series, or series and parallels, or if viewed in binary, a sequence of 01010101 ... or 10101010 ...

How will we have the digits again only with 3,2258064516129032258064516129032? Through inversion and subtraction of whole parts:

We already know that the circuit has 3 equal elements repeated and that starts with one's, let's subtract the whole part 3 and continue:

3.2258064516129032258064516129032 - 3 = 0.22580645161290322580645161290323

Now we invert the result:

1 / 0.22580645161290322580645161290323 = 4.4285714285714285714285714285714

subtract the whole part 4 and invert

1 / 0.42857142857142857142857142857143 = 2.3333333333333333333333333333333

subtract the whole part 2 and invert

1 / 0.33333333333333333333333333333333 = 3

and here we reach the end. If you wonder why we didn't get to resistor 1 or the first resistor was because I started the sequence from 1 and not 2 which would be the proper fibonacci number to be the initial one. I mean, I didn't do the math operation for the first resistor, just from the second resistor.

Reducing equivalent resistance at its lowest lossless representation? This was the question I raised in this topic.

3,2258064516129032258064516129032

trying by brute force the operations have to 3.2258 can represent the lossless circuit (i'm ignoring the 1st ohm resistor).

3.2258 in binary equals 11.001110011100111000000 ... in single words 11.00111 (times 3)

Well, if I enter as information that the circuit starts in series or parallel, just one bit, then I can reverse the number 3,2258 and get:

1 / 3.2258 = 0.31000062000124000248000496000992

Now the number is more palatable and compact: 0.31 is the number representing that electrical circuit, or the sequence 34231

Continue inverting and subtracting the entire part and you will see:

1 / 0.31 = 3.2258064516129032258064516129032

1 / 0.22580645161290322580645161290323 = 4.428571428

...