ASSIGNMENT PROBLEM 4 Generate a Fibonnaci sequence in the first column using f 0 =, f 0 =, = f n f n a. Construct the ratio of each pair of adjacent terms in the Fibonnaci sequence. What happens as n increases? What about the ratio of every second term? etc. b. Explore sequences where f 0 and f are some arbitrary integers other than. If f 0 = and f =3, then your sequence is a Lucas Sequence. All such sequences, however, have the same limit of the ratio of successive terms. PART ONE Here is the first 0 terms of the Fibonacci Sequence in the first column of a spreadhseet: f(n) 3 5 8 3 34 55 89 44 33 377 60 987 597 584 48 6765
Question: Construct the ratio of each pair of adjacent terms in the Fibonnaci sequence. f(n) f(n)/f(n-).0000000000.0000000000 3.5000000000 5.6666666667 8.6000000000 3.650000000.65384654 34.69047690 55.676470588 89.68888 44.6797758 33.680555556 377.680575 60.68037353 987.68037869 597.680344478 584.68033834 48.680340557 6765.68033963 Question: What happens as n increases? Answer: As n increases, the ratio f n converges to the irrational number 5 =.68033988749894848045868343656 Proof: We are given that = f n f n. Now dividing each side of the equation by f n, we get: f n = f n f n f n f n. Simplifying we get: f n = f n f n.
Here is the trick. Let us define the ratio f n =x as n. f n Therefore, we can say that the ratio = x as n. Moreover, the ratio f n f n = as n as well. x Therefore our original equation becomes: x= x which can be written as x =x Rewriting this equation in the a x b x c=0 form, we get x x =0. This is a quadratic equation. In Assignment, we learned that the roots of a quadratic equation by are given by using the quadratic formula by: x= b± b 4 ac a In our case, a=,b=, c=. Plugging these in the equation above, we get: x= ± 4 Simplification yields: x= ± 5 The negative root is not acceptable because all the terms of the fibonacci sequence are positive. Therefore the solution is x= 5 =.68033988749894848045868343656 This number is called golden ratio. Now let us make sure that we understand what this means: This means that the ratio of the consecutive terms converges to x= 5 as n. In other words, the f n ratio converges to the irrational number f n.68033988749894848045868343656 as n.
Question: What about the ratio of every second term? f(n) f(n)/f(n-) f(n)/f(n-).0000000000.0000000000.0000000000 3.5000000000 3.0000000000 5.6666666667.5000000000 8.6000000000.6666666667 3.650000000.6000000000.65384654.650000000 34.69047690.65384654 55.676470588.69047690 89.68888.676470588 44.6797758.68888 33.680555556.6797758 377.680575.680555556 60.68037353.680575 987.68037869.68037353 597.680344478.68037869 584.68033834.680344478 48.680340557.68033834 6765.68033963.680340557 f n converges to the square of the golden ratio:.68033988749894848045868343656 =.68033988749894848045868343656..68033988749894848045868343656 = x. x = x Namely as n goes to infinity, the ratio f n = f n
Question: What about the ratio of every third term? f(n) f(n)/f(n-) f(n)/f(n-) f(n)/f(n-3).0000000000.0000000000.0000000000 3.5000000000 3.0000000000 3.000000000 5.6666666667.5000000000 5.000000000 8.6000000000.6666666667 4.000000000 3.650000000.6000000000 4.333333333.65384654.650000000 4.00000000 34.69047690.65384654 4.50000000 55.676470588.69047690 4.307693 89.68888.676470588 4.3809538 44.6797758.68888 4.35948 33.680555556.6797758 4.36363636 377.680575.680555556 4.35955056 60.68037353.680575 4.36 987.68037869.68037353 4.360550 597.680344478.68037869 4.360747 584.68033834.680344478 4.36065574 48.680340557.68033834 4.36068896 6765.68033963.680340557 4.3606767 f n 3 converges to the cube of the golden ratio, namely: f n 3 =x3 as n. In fact, this ratio is seen to be: x 3 =x x x 3 =x x x 3 =x x x 3 = x x x 3 = x =4.3606797749978969640973668733. Question: What about the ratio of every forth term? f n 4 converges to the forth power of the golden ratio, namely: f n 4 =x4 as n. In fact, this ratio is seen to be:
x 4 =x x 3 x 4 =x x x 4 = x x x 4 = x x x 4 =3 x Question: What about the ratio of every fifth term? f n 5 converges to the fifth power of the golden ratio, namely: f n 5 =x5 as n. In fact, this ratio is seen to be: x 5 =x x 4 x 5 =x 3 x x 5 =3 x x x 5 =3 x x x 5 =5 x 3 Question: What about the ratio of every sixth term? f n 6 converges to the fifth power of the golden ratio, namely: f n 6 =x6 as n. In fact, this ratio is seen to be: x 6 =x x 5 x 6 =x 5 x 3 x 6 =5 x 3 x x 6 =5 x 3 x x 6 =8 x 5
Question: Can you generalize this? What about the ratio of every nth term? Answer: It converges to the nth power of the golden ratio. x n =x x n x n =x [ f n x f n ] x n = f n x f n x x n = f n x f n x x n =[ f n f n ] x f n x n = x f n PART TWO Explore sequences where f 0 and f are some arbitrary integers other than. If f 0 = and f =3, then your sequence is a Lucas Sequence. All such sequences, however, have the same limit of the ratio of successive terms. Here is the first 0 terms of the Lucas Sequence generated in spreadsheet: f(n) 3 4 7 8 9 47 76 3 99 3 5 843 364 07 357 5778 9349 57
Question: Construct the ratio of each pair of adjacent terms in the Lucas sequence. f(n) f(n)/f(n-) 3 3.000000000 4.333333333 7.750000000.574857 8.636363636 9.6 47.60689655 76.67077 3.684053 99.6788679 3.6809045 5.6804 843.68046 364.6803084 07.680359 357.68033530 5778.6803464 9349.680339 57.6803404 Therefore, we see that as n increases, the ratio f n converges to golden ratio again. This is expected because the recursive definition of the sequence is still valid and therefore, my proof is still valid.
Question: What about the ratio of every second term? f(n) f(n)/f(n-) f(n)/f(n-) 3 3.000000000 4.333333333 4.000000000 7.750000000.333333333.574857.750000000 8.636363636.574857 9.6.636363636 47.60689655.6 76.67077.60689655 3.684053.67077 99.6788679.684053 3.6809045.6788679 5.6804.6809045 843.68046.6804 364.6803084.68046 07.680359.6803084 357.68033530.680359 5778.6803464.68033530 9349.680339.6803464 57.6803404.680339 The ratio f n converges to the square of the golden ratio. My proof is valid for any recursively defined sequence of the form: = f n f n with arbitray first two nonzero and nonnegative terms f 0 = f 0, f = f The ratio of consecutive terms of all such sequences converges to golden ratio irrespective of how we define the first two terms f 0 = f 0 0, f = f 0. Therefore, Fibonnacci Sequence is a special case with f 0 =, f = as well as the Lucas Sequence, which is a special case with f 0 =, f =3 (See proof on pages -3)