Assignment 12
For this assignment, I chose to write up problem number nine:
4. Generate a Fibonnaci sequence in the first column using f(0) = 1, f(1) = 1, f(n) = f(n1) + f(n2).
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(1) are some arbitrary integers other than 1. If f(0)=1 and f(1) = 3, then your sequence is a Lucas Sequence. All such sequences, however, have the same limit of the ratio of successive terms.
I constructed the table to the left using Microsoft Excel. Column A shows the first 22 fibonacci numbers, while in column B, row n shows the ratio (n+1)/n. These ratios seem to be converging towards a number near 1.61903399. You might recall that the golden ratio Phi = = also happens to be very near 1.61903399. In fact, the sequence displayed in column B does converge to Phi. A geometric interpretation of this convergence follows. Note that .
, Since Phi is a solution to this equation, we get that , which means that if we have a rectangle with , then from our equation above, we find that . Therefore, if we have a golden rectangle (a rectangle in which the ratio of the length to the width is the golden ratio), .
In the picture above all angles are right angles, all red segments have the same length, and all blue segments have the same length. Rectangle ABCD is a golden rectangle, and by the equation above, so is rectangle ABEF! Since ABEF is a golden rectangle we can continue the process indefinitely:: ... This GSP file contains a script tool that constructs this spiral of golden rectangles. To increase or decrease the number of rectangles, press '+' or '' immediately after using the tool. If we repeat this construction starting with a square of length 1, rather than a golden rectangle, we get a spiral of squares with lengths equal to the fibonacci numbers:

What happens if we look at the ratio of every second term, every third, every fourth, and so on? The following table displays these numbers and shows that all of these ratios seem to converge. The entries in table C, the ratios of every second term, seem to converge to Phisquared. The entries in table D, the ratios of every third term, seem to converge to Phicubed. In general, it appears that the ratios of every nth term converge to Phi to the (n1)th power.
If we make similar tables with f(0) and f(1) not both equal to one, we see that the sequence of successive ratios still seems to converge to Phi. Amazing, isn't it?