Final Project 1: Fabulous Fibonacci Numbers

by Elizabeth Gieseking

Leonardo of Pisa (c. 1170-1250), commonly known as Fibonacci, posed and solved the following problem involving an idealized rabbit population in his book Liber Abaci.

Suppose a newly-born pair of rabbits, one male and one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits. Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month from the second month on. How many pairs will there be in one year? 

The table below illustrates what is happening.  When the rabbits are first placed in the field, their age is 0 months.  At this point they are immature and cannot mate, so at the end of month 1, there is still one pair of rabbits.  At that point they are mature and mate, so at the end of the second month there are 2 pairs of rabbits, one mature and one immature.  In each month after the first one, the number of immature rabbit pairs is equal to the number of mature rabbit pairs in the previous month.  Likewise the number of mature rabbit pairs is equal to the total number of rabbit pairs in the previous month.

 

Month	Mature Rabbit Pairs	Immature Rabbit Pairs	Total Rabbit Pairs
0	0	1	1
1	1	0	1
2	1	1	2
3	2	1	3
4	3	2	5
5	5	3	8
6	8	5	13
7	13	8	21
8	21	13	34
9	34	21	55
10	55	34	89
11	89	55	144
12	144	89	233
13	233	144	377
14	377	233	610
15	610	377	987
16	987	610	1597
17	1597	987	2584
18	2584	1597	4181
19	4181	2584	6765
20	6765	4181	10946
21	10946	6765	17711
22	17711	10946	28657
23	28657	17711	46368
24	46368	28657	75025
25	75025	46368	121393

The resulting sequence of number of rabbit pairs is the famous Fibonacci sequence.  The Fibonacci sequence is defined recursively as: .  By varying the starting values, we can create many similar sequences.  The most famous of these is the Lucas sequence in which .  Another variation is to the change the number of values being summed at each step.  For example, in a tribonacci sequence the three previous terms are summed at each step: .

A spreadsheet can be a useful tool for studying the Fibonacci sequence and other related sequences.  We simply enter the starting values and our formula and let the computer do the computation.  This is especially useful if we wish to examine ratios of Fibonacci numbers.  In the following table, we are examining the ratios of a Fibonacci number to a previous Fibonacci number.  In the first ratio, we are calculating , the ratio of the Fibonacci number to the previous term.  In the second ratio, we are calculating , the ratio of the Fibonacci number to the number two spots earlier in the sequence.  Likewise, in the third ratio, we are using terms three apart, and so on.

 

n
0	1
1	1	1
2	2	2	2
3	3	1.5	3	3
4	5	1.6666667	2.5	5	5
5	8	1.6000000	2.6666667	4	8	8
6	13	1.6250000	2.6000000	4.3333333	6.5	13
7	21	1.6153846	2.6250000	4.2000000	7	10.5
8	34	1.6190476	2.6153846	4.2500000	6.8	11.3333333
9	55	1.6176471	2.6190476	4.2307692	6.875	11
10	89	1.6181818	2.6176471	4.2380952	6.8461538	11.125
11	144	1.6179775	2.6181818	4.2352941	6.8571429	11.076923
12	233	1.6180556	2.6179775	4.2363636	6.8529412	11.095238
13	377	1.6180258	2.6180556	4.2359551	6.8545455	11.088235
14	610	1.6180371	2.6180258	4.2361111	6.8539326	11.090909
15	987	1.6180328	2.6180371	4.2360515	6.8541667	11.089887
16	1597	1.6180344	2.6180328	4.2360743	6.8540773	11.090277
17	2584	1.6180338	2.6180344	4.2360656	6.8541114	11.090128
18	4181	1.6180341	2.6180338	4.2360689	6.8540984	11.090185
19	6765	1.6180340	2.6180341	4.2360676	6.8541033	11.090163
20	10946	1.6180340	2.6180340	4.2360681	6.8541014	11.090172
21	17711	1.6180340	2.6180340	4.2360679	6.8541022	11.090169
22	28657	1.6180340	2.6180340	4.2360680	6.8541019	11.090170
23	46368	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
24	75025	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
25	121393	1.6180340	2.6180340	4.2360680	6.8541020	11.090170

We see that these values go up and down but converge fairly quickly.  The number for the first ratio looks familiar – it is  or the golden ratio.  The golden ratio is defined as: .  This can be rewritten as:   We can solve this using the quadratic formula to get:

We also quickly see that the second ratio  converges to .  Although the other ratios converge, it is not yet obvious how these values relate to   If we again examine the equations, we see that   Thus our second ratio is .  We can then verify that our third ratio is , our fourth ratio is , and our fifth ratio is

Now we know that the n^th ratio converges to , but we do not have a reason yet.  Let’s go back to our equation for  and use it to generate other powers of .

We see two relationships between the powers of  and the Fibonacci series.

1.   The sequence of powers of  has the same structure as the Fibonacci sequence. 
Just as , we see that

2.   When we express  as a function of , the coefficients are the Fibonacci numbers.

Interestingly, no matter what values we choose for  and  , the nth ratio will still converge to   The table below shows the Lucas numbers in which .

n
0	1
1	3	3
2	4	1.3333333	4
3	7	1.7500000	2.3333333	7
4	11	1.5714286	2.7500000	3.6666667	11
5	18	1.6363636	2.5714286	4.5000000	6	18
6	29	1.6111111	2.6363636	4.1428571	7.25	9.666667
7	47	1.6206897	2.6111111	4.2727273	6.7142857	11.750000
8	76	1.6170213	2.6206897	4.2222222	6.9090909	10.857143
9	123	1.6184211	2.6170213	4.2413793	6.8333333	11.181818
10	199	1.6178862	2.6184211	4.2340426	6.8620690	11.055556
11	322	1.6180905	2.6178862	4.2368421	6.8510638	11.103448
12	521	1.6180124	2.6180905	4.2357724	6.8552632	11.085106
13	843	1.6180422	2.6180124	4.2361809	6.8536585	11.092105
14	1364	1.6180308	2.6180422	4.2360248	6.8542714	11.089431
15	2207	1.6180352	2.6180308	4.2360845	6.8540373	11.090452
16	3571	1.6180335	2.6180352	4.2360617	6.8541267	11.090062
17	5778	1.6180342	2.6180335	4.2360704	6.8540925	11.090211
18	9349	1.6180339	2.6180342	4.2360671	6.8541056	11.090154
19	15127	1.6180340	2.6180339	4.2360683	6.8541006	11.090176
20	24476	1.6180340	2.6180340	4.2360678	6.8541025	11.090168
21	39603	1.6180340	2.6180340	4.2360680	6.8541018	11.090171
22	64079	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
23	103682	1.6180340	2.6180340	4.2360680	6.8541019	11.090170
24	167761	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
25	271443	1.6180340	2.6180340	4.2360680	6.8541020	11.090170

We can even choose non-integer values for  and  and still get the same ratios.

n
0	0.7
1	9.2	13.1428571
2	9.9	1.0760870	14.1428571
3	19.1	1.9292929	2.0760870	27.2857143
4	29	1.5183246	2.9292929	3.1521739	41.4285714
5	48.1	1.6586207	2.5183246	4.8585859	5.2282609	68.714286
6	77.1	1.6029106	2.6586207	4.0366492	7.7878788	8.380435
7	125.2	1.6238651	2.6029106	4.3172414	6.5549738	12.646465
8	202.3	1.6158147	2.6238651	4.2058212	6.9758621	10.591623
9	327.5	1.6188828	2.6158147	4.2477302	6.8087318	11.293103
10	529.8	1.6177099	2.6188828	4.2316294	6.8715953	11.014553
11	857.3	1.6181578	2.6177099	4.2377657	6.8474441	11.119326
12	1387.1	1.6179867	2.6181578	4.2354198	6.8566485	11.079073
13	2244.4	1.6180521	2.6179867	4.2363156	6.8531298	11.094414
14	3631.5	1.6180271	2.6180521	4.2359734	6.8544734	11.088550
15	5875.9	1.6180366	2.6180271	4.2361041	6.8539601	11.090789
16	9507.4	1.6180330	2.6180366	4.2360542	6.8541562	11.089934
17	15383.3	1.6180344	2.6180330	4.2360732	6.8540813	11.090260
18	24890.7	1.6180338	2.6180344	4.2360660	6.8541099	11.090135
19	40274	1.6180340	2.6180338	4.2360687	6.8540989	11.090183
20	65164.7	1.6180340	2.6180340	4.2360677	6.8541031	11.090165
21	105438.7	1.6180340	2.6180340	4.2360681	6.8541015	11.090172
22	170603.4	1.6180340	2.6180340	4.2360679	6.8541021	11.090169
23	276042.1	1.6180340	2.6180340	4.2360680	6.8541019	11.090170
24	446645.5	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
25	722687.6	1.6180340	2.6180340	4.2360680	6.8541020	11.090170

Even negative values give ratios that converge to the powers of

n
0	-3
1	-2.815	0.9383333
2	-5.815	2.0657194	1.9383333
3	-8.63	1.4840929	3.0657194	2.8766667
4	-14.445	1.6738123	2.4840929	5.1314387	4.8150000
5	-23.075	1.5974386	2.6738123	3.9681857	8.1971581	7.691667
6	-37.52	1.6260022	2.5974386	4.3476246	6.4522786	13.328597
7	-60.595	1.6150053	2.6260022	4.1948771	7.0214368	10.420464
8	-98.115	1.6191930	2.6150053	4.2520043	6.7923157	11.369061
9	-158.71	1.6175916	2.6191930	4.2300107	6.8780065	10.987193
10	-256.825	1.6182030	2.6175916	4.2383860	6.8450160	11.130011
11	-415.535	1.6179694	2.6182030	4.2351832	6.8575790	11.075027
12	-672.36	1.6180586	2.6179694	4.2364060	6.8527748	11.095965
13	-1087.895	1.6180246	2.6180586	4.2359389	6.8546090	11.087958
14	-1760.255	1.6180376	2.6180246	4.2361173	6.8539083	11.091015
15	-2848.15	1.6180326	2.6180376	4.2360491	6.8541759	11.089847
16	-4608.405	1.6180345	2.6180326	4.2360752	6.8540737	11.090293
17	-7456.555	1.6180338	2.6180345	4.2360652	6.8541128	11.090123
18	-12064.96	1.6180341	2.6180338	4.2360690	6.8540978	11.090188
19	-19521.515	1.6180340	2.6180341	4.2360676	6.8541035	11.090163
20	-31586.475	1.6180340	2.6180340	4.2360681	6.8541014	11.090173
21	-51107.99	1.6180340	2.6180340	4.2360679	6.8541022	11.090169
22	-82694.465	1.6180340	2.6180340	4.2360680	6.8541019	11.090170
23	-133802.45	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
24	-216496.92	1.6180340	2.6180340	4.2360680	6.8541020	11.090170
25	-350299.37	1.6180340	2.6180340	4.2360680	6.8541020	11.090170

Yet another connection exists between  and the Fibonacci numbers.  We can use  to write a closed form expression for   Remember  was one of the solutions to the quadratic equation:   There is also a negative solution to this equation.  .

Fibonacci Numbers and Pythagorean Triples

We will conclude this exploration with one more interesting observation about Fibonacci numbers.  We can take any four consecutive Fibonacci numbers and use them to generate Pythagorean triples – solutions to the Pythagorean Theorem: . 

Let .

First we will test this with the first two sets of four consecutive Fibonacci numbers, and then we will determine why this is always true.

                                               

This is the well-known Pythagorean triple (3, 4, 5).

                                    13

This yields the Pythagorean triple (5, 12, 13).

Because A, B, C, and D are four consecutive Fibonacci numbers,  and    We can substitute these values into the equation.

Thus, no matter what values are chosen for A and B, we will get a Pythagorean triple.  This means a spreadsheet is a great means of generating Pythagorean triples.  The following values were obtained from the standard Fibonacci sequence.  The highlighted values are the numbers in the Pythagorean triples.

A	B	C	D	AD	2BC
1	1	2	3	3	4	5		25	25
1	2	3	5	5	12	13		169	169
2	3	5	8	16	30	34		1156	1156
3	5	8	13	39	80	89		7921	7921
5	8	13	21	105	208	233		54289	54289
8	13	21	34	272	546	610		372100	372100
13	21	34	55	715	1428	1597		2550409	2550409
21	34	55	89	1869	3740	4181		17480761	17480761
34	55	89	144	4896	9790	10946		119814916	119814916
55	89	144	233	12815	25632	28657		821223649	821223649
89	144	233	377	33553	67104	75025		5628750625	5628750625

 

When we vary the starting values of A and B, we can generate other Pythagorean triples.

A	B	C	D	AD	2BC
1	3	4	7	7	24	25		625	625
3	4	7	11	33	56	65		4225	4225
4	7	11	18	72	154	170		28900	28900
7	11	18	29	203	396	445		198025	198025
11	18	29	47	517	1044	1165		1357225	1357225
1	4	5	9	9	40	41		1681	1681
4	5	9	14	56	90	106		11236	11236
5	9	14	23	115	252	277		76729	76729
9	14	23	37	333	644	725		525625	525625
14	23	37	60	840	1702	1898		3602404	3602404
1	5	6	11	11	60	61		3721	3721
5	6	11	17	85	132	157		24649	24649
6	11	17	28	168	374	410		168100	168100
11	17	28	45	495	952	1073		1151329	1151329
17	28	45	73	1241	2520	2809		7890481	7890481
1	6	7	13	13	84	85		7225	7225
6	7	13	20	120	182	218		47524	47524
7	13	20	33	231	520	569		323761	323761
13	20	33	53	689	1320	1489		2217121	2217121
20	33	53	86	1720	3498	3898		15194404	15194404

 

This exploration highlighted two useful ways that spreadsheets can be used to explore properties of the Fibonacci numbers.  The ability to quickly generate data based on a simple formula allows us to test conjectures and more easily make mathematical connections.

         

 

 

 

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