Assignment # 12
Fibonnaci and Phi
Let's look at our sequence for the first 30 terms. As we can
see, the ratio of subsequent terms approaches The Golden Ratio
closer and closer.
| 1 |
|
| 1 |
1 |
| 2 |
2 |
| 3 |
1.5 |
| 5 |
1.66666666666667 |
| 8 |
1.6 |
| 13 |
1.625 |
| 21 |
1.61538461538462 |
| 34 |
1.61904761904762 |
| 55 |
1.61764705882353 |
| 89 |
1.61818181818182 |
| 144 |
1.61797752808989 |
| 233 |
1.61805555555556 |
| 377 |
1.61802575107296 |
| 610 |
1.61803713527851 |
| 987 |
1.61803278688525 |
| 1597 |
1.61803444782168 |
| 2584 |
1.61803381340013 |
| 4181 |
1.61803405572755 |
| 6765 |
1.61803396316671 |
| 10946 |
1.6180339985218 |
| 17711 |
1.61803398501736 |
| 28657 |
1.6180339901756 |
| 46368 |
1.61803398820532 |
| 75025 |
1.6180339889579 |
| 121393 |
1.61803398867044 |
| 196418 |
1.61803398878024 |
| 317811 |
1.6180339887383 |
| 514229 |
1.61803398875432 |
What is the Golden Ratio?
Phi, or the Golden Ratio, is equal to the following
constant:
|
From this, we can see that the successive ratios are getting
closer and closer to the true value. Let's take a look at the
variance from the actual value as well as the ratio of our values
to the true value.
| Fibonnaci |
f(n-1)/f(n) |
variance |
ratio |
| 1 |
|
|
|
| 1 |
1 |
-0.618033988749895 |
0.618033988749895 |
| 2 |
2 |
0.381966011250105 |
1.23606797749979 |
| 3 |
1.5 |
-0.118033988749895 |
0.927050983124842 |
| 5 |
1.66666666666667 |
0.0486326779167718 |
1.03005664791649 |
| 8 |
1.6 |
-0.0180339887498948 |
0.988854381999832 |
| 13 |
1.625 |
0.0069660112501051 |
1.00430523171858 |
| 21 |
1.61538461538462 |
-0.00264937336527948 |
0.998362597211369 |
| 34 |
1.61904761904762 |
0.00101363029772417 |
1.00062645797602 |
| 55 |
1.61764705882353 |
-0.000386929926365465 |
0.999760864154242 |
| 89 |
1.61818181818182 |
0.000147829431923263 |
1.00009136361347 |
| 144 |
1.61797752808989 |
-0.000056460660007307 |
0.999965105393088 |
| 233 |
1.61805555555556 |
0.0000215668056606777 |
1.00001332901893 |
| 377 |
1.61802575107296 |
-8.23767693347577E-06 |
0.999994908835667 |
| 610 |
1.61803713527851 |
3.14652861965747E-06 |
1.00000194466163 |
| 987 |
1.61803278688525 |
-1.20186464891425E-06 |
0.999999257206797 |
| 1597 |
1.61803444782168 |
4.59071787028975E-07 |
1.00000028372197 |
| 2584 |
1.61803381340013 |
-1.75349769593325E-07 |
0.999999891627882 |
| 4181 |
1.61803405572755 |
6.69776591966098E-08 |
1.00000004139447 |
| 6765 |
1.61803396316671 |
-2.55831884565794E-08 |
0.99999998418872 |
| 10946 |
1.6180339985218 |
9.77190839357434E-09 |
1.00000000603937 |
| 17711 |
1.61803398501736 |
-3.73253694618825E-09 |
0.999999997693165 |
| 28657 |
1.6180339901756 |
1.4257022229458E-09 |
1.00000000088113 |
| 46368 |
1.61803398820532 |
-5.44569944693762E-10 |
0.999999999663437 |
| 75025 |
1.6180339889579 |
2.08007167046276E-10 |
1.00000000012856 |
| 121393 |
1.61803398867044 |
-7.94517784896698E-11 |
0.999999999950896 |
| 196418 |
1.61803398878024 |
3.03477243335237E-11 |
1.00000000001876 |
| 317811 |
1.6180339887383 |
-1.15918386001113E-11 |
0.999999999992836 |
| 514229 |
1.61803398875432 |
4.42756942220512E-12 |
1.00000000000274 |
We can also see that our sequence is oscillating around the
correct value. Let's examine how this functions behaves:
 |
|
By the 8th term, we are pretty much constant
... |
Looking at terms other than f(n+1)/f(n)
...
| Fibonnaci |
f(n+1)/f(n) |
f(n+2)/f(n) |
f(n+3)/f(n) |
f(n+4)/f(n) |
| 1 |
|
|
|
|
| 1 |
1 |
2 |
3 |
5 |
| 2 |
2 |
3 |
5 |
8 |
| 3 |
1.5 |
2.5 |
4 |
6.5 |
| 5 |
1.66666666666667 |
2.66666666666667 |
4.33333333333333 |
7 |
| 8 |
1.6 |
2.6 |
4.2 |
6.8 |
| 13 |
1.625 |
2.625 |
4.25 |
6.875 |
| 21 |
1.61538461538462 |
2.61538461538462 |
4.23076923076923 |
6.84615384615385 |
| 34 |
1.61904761904762 |
2.61904761904762 |
4.23809523809524 |
6.85714285714286 |
| 55 |
1.61764705882353 |
2.61764705882353 |
4.23529411764706 |
6.85294117647059 |
| 89 |
1.61818181818182 |
2.61818181818182 |
4.23636363636364 |
6.85454545454545 |
| 144 |
1.61797752808989 |
2.61797752808989 |
4.23595505617978 |
6.85393258426966 |
| 233 |
1.61805555555556 |
2.61805555555556 |
4.23611111111111 |
6.85416666666667 |
| 377 |
1.61802575107296 |
2.61802575107296 |
4.23605150214592 |
6.85407725321888 |
| 610 |
1.61803713527851 |
2.61803713527851 |
4.23607427055703 |
6.85411140583554 |
| 987 |
1.61803278688525 |
2.61803278688525 |
4.23606557377049 |
6.85409836065574 |
| 1597 |
1.61803444782168 |
2.61803444782168 |
4.23606889564336 |
6.85410334346505 |
| 2584 |
1.61803381340013 |
2.61803381340013 |
4.23606762680025 |
6.85410144020038 |
| 4181 |
1.61803405572755 |
2.61803405572755 |
4.23606811145511 |
6.85410216718266 |
| 6765 |
1.61803396316671 |
2.61803396316671 |
4.23606792633341 |
6.85410188950012 |
| 10946 |
1.6180339985218 |
2.6180339985218 |
4.23606799704361 |
6.85410199556541 |
| 17711 |
1.61803398501736 |
2.61803398501736 |
4.23606797003472 |
6.85410195505207 |
| 28657 |
1.6180339901756 |
2.6180339901756 |
4.23606798035119 |
6.85410197052679 |
| 46368 |
1.61803398820532 |
2.61803398820533 |
4.23606797641065 |
6.85410196461598 |
| 75025 |
1.6180339889579 |
2.6180339889579 |
4.2360679779158 |
6.85410196687371 |
| 121393 |
1.61803398867044 |
2.61803398867044 |
4.23606797734089 |
6.85410196601133 |
| 196418 |
1.61803398878024 |
2.61803398878024 |
4.23606797756049 |
|
| 317811 |
1.6180339887383 |
2.6180339887383 |
|
|
| 514229 |
1.61803398875432 |
|
|
|
We can see that taking the f(n+2)/f(n) sequence yields us ...
phi + 1. The other terms give us successive powers of phi, i.e.:
This pattern is bound to continue ...
Looking starting points other than f(1)
= 1 and f(2) = 1 ...
| Fibonnaci |
f(n+1)/f(n) |
| 1 |
|
| 3 |
3 |
| 4 |
1.33333333333333 |
| 7 |
1.75 |
| 11 |
1.57142857142857 |
| 18 |
1.63636363636364 |
| 29 |
1.61111111111111 |
| 47 |
1.62068965517241 |
| 76 |
1.61702127659574 |
| 123 |
1.61842105263158 |
| 199 |
1.61788617886179 |
| 322 |
1.61809045226131 |
| 521 |
1.61801242236025 |
| 843 |
1.61804222648752 |
| 1364 |
1.61803084223013 |
| 2207 |
1.61803519061584 |
| 3571 |
1.6180335296783 |
| 5778 |
1.61803416409969 |
| 9349 |
1.61803392177224 |
| 15127 |
1.61803401433308 |
| 24476 |
1.61803397897799 |
| 39603 |
1.61803399248243 |
| 64079 |
1.61803398732419 |
| 103682 |
1.61803398929446 |
| 167761 |
1.61803398854189 |
| 271443 |
1.61803398882935 |
| 439204 |
1.61803398871955 |
| 710647 |
1.61803398876149 |
| 1149851 |
1.61803398874547 |
|
| Fibonnaci |
f(n+1)/f(n) |
| 2 |
|
| 5 |
2.5 |
| 7 |
1.4 |
| 12 |
1.71428571428571 |
| 19 |
1.58333333333333 |
| 31 |
1.63157894736842 |
| 50 |
1.61290322580645 |
| 81 |
1.62 |
| 131 |
1.61728395061728 |
| 212 |
1.61832061068702 |
| 343 |
1.61792452830189 |
| 555 |
1.61807580174927 |
| 898 |
1.61801801801802 |
| 1453 |
1.61804008908686 |
| 2351 |
1.6180316586373 |
| 3804 |
1.61803487877499 |
| 6155 |
1.61803364879075 |
| 9959 |
1.61803411860276 |
| 16114 |
1.61803393915052 |
| 26073 |
1.61803400769517 |
| 42187 |
1.61803398151344 |
| 68260 |
1.61803399151397 |
| 110447 |
1.61803398769411 |
| 178707 |
1.61803398915317 |
| 289154 |
1.61803398859586 |
| 467861 |
1.61803398880873 |
| 757015 |
1.61803398872742 |
| 1224876 |
1.61803398875848 |
| 1981891 |
1.61803398874662 |
|
| Fibonnaci |
f(n+1)/f(n) |
| 7 |
|
| 59 |
8.42857142857143 |
| 66 |
1.11864406779661 |
| 125 |
1.89393939393939 |
| 191 |
1.528 |
| 316 |
1.6544502617801 |
| 507 |
1.60443037974684 |
| 823 |
1.6232741617357 |
| 1330 |
1.61603888213852 |
| 2153 |
1.6187969924812 |
| 3483 |
1.6177426846261 |
| 5636 |
1.61814527706001 |
| 9119 |
1.6179914833215 |
| 14755 |
1.61805022480535 |
| 23874 |
1.61802778719078 |
| 38629 |
1.61803635754377 |
| 62503 |
1.61803308395247 |
| 101132 |
1.61803433435195 |
| 163635 |
1.61803385674168 |
| 264767 |
1.61803403917255 |
| 428402 |
1.61803396949016 |
| 693169 |
1.61803399610646 |
| 1121571 |
1.61803398593994 |
| 1814740 |
1.6180339898232 |
| 2936311 |
1.61803398833993 |
| 4751051 |
1.61803398890649 |
| 7687362 |
1.61803398869008 |
| 12438413 |
1.61803398877274 |
| 20125775 |
1.61803398874117 |
|
It looks like it doesn't matter which values we choose as our
starting point - the sequence is bound to converge to phi sooner
or later. Let's look at why that's the case ...
| a |
| b |
| a+b |
| a + 2b |
| 2a + 3b |
| 3a + 5b |
| 5a + 8b |
| 8a + 13b |
| 13a + 21b |
| 21a + 34b |
| 34a + 55b |
| etc É |
|
The coefficients of both the 'a' and 'b' terms are consecutive
numbers from the Fibonnaci sequence. Dividing f(n+1)/f(n) will
start approaching phi. Why is that the case?
Here goes our proof ...
The positive value will give us Phi.
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