Chapter 4
Infinite Purely Periodic Continued fraction expansions
4.1 Decomposition of Rectangle never ends
So far the rectangles were decomposed completely into an exact number of squares. Are there rectangles that cannot be decomposed into a finite number of squares? What does that mean in terms of the ratio of the sides of the rectangles? What is the implication to the continued fraction expansion?
Leftover space
If the decomposition of a rectangle into squares never ends, it will imply that the continued fraction expansion is infinite, since the number of squares represents the quotients in the continued fraction expansion.
Are the squares in the rectangle arranged in any specific way? Do the numbers in the list of the continued fraction expansion or quotients show any patterns or periodicity?
The squares inside of the rectangle could be arranged in one of the following forms as shown in the chart.
In the following sections, we will consider the different possibilities of arrangements of squares inside the rectangle and discover what kind of numbers the ratios of the sides represent.
4.2 Decomposition of a rectangle with the Golden Proportions
The simplest way to imagine that the decomposition never stops is if we have a rectangle and are able to fit only one square. In the left over space, we are able to fit only another smaller square. We continue this process fitting only one square in the leftover space and are always left with a smaller rectangle which again allows to fit only one square and so on.
In the situation where a remaining rectangle allows fitting one square and so on, what is the relationship of the smaller rectangle to the original one? If both were similar, then the same pattern of decomposition we see in the original rectangle will be seen in the leftover rectangle. Therefore, similarity of the original rectangle with the leftover rectangle will result in periodicity.
The simplest case would be:
Similar rectangles
The decomposition, after a few steps, would look like this:
The continued fraction expansion would be [1; 1, 1, 1….]
Since the first leftover rectangle is similar to the original rectangle, then using algebra we can find the ratio of the sides of the rectangles.
If the make the length of the shorter side of the original rectangle 1, then the longer side would be 1 + x. We can make the shorter side 1 since we are only interested in the simplified ratio of the sides. The leftover rectangle will have a long side of length 1 and a short side of length x.
Since the rectangles are similar, the proportion of longer sides to shorter sides is:
Cross-multiplying, we get a quadratic equation:
Using the quadratic formula, and taking into account only the positive answer (since length cannot be negative)
Then the longer length of the original rectangle is
This number is the golden ratio. The golden ratio is an algebraic irrational number of degree 2. (Refer to the section titled “definitions” in the first part for the definition of algebraic irrational numbers).
We would want to verify this discovery using DGS. Since the golden ratio is an irrational number, the construction of a segment with that length requires some steps as follows:
To construct the Golden rectangle using DGS:
- Construct segment AB
- Construct a perpendicular line to segment AB through endpoint A.
- Using the circle tool, divide the perpendicular line in two sections equal to the length of the segment AB. Name the upper point of intersection of the circle and the perpendicular line C. If we think of a right triangle ABC with legs ratio of 1:2, then the hypotenuse BC will have length.
- Draw ray AB.
- Using the circle tool draw a circle centered at B with radius BC.
- Label the point of intersection of the last circle with the ray AB point D. The distance from point A to D is
- Find the midpoint between A and D. Label the point E. The distance between points A and E will be or the Golden ratio.
- Construct a perpendicular line to segment AE through point E and another perpendicular line to AF through point F. The resulting rectangle will have sides in the golden ratio.
Decomposing the golden rectangle into squares, we can see that we are able to fit only one square in each rectangle.
The decomposition of the golden rectangle suggests that the short notation of the continued fraction expansion is [1; 1, 1, 1…].
4.3Continued fraction expansion and Algorithmic Similarity for the Golden Ratio
In Chapter 2 we saw that the Golden ratio is the positive solution of the equation:
Rearranging this equation, we get:
Which means that x has a value that is 1 plus a fractional part.
If we use the equationand substitute the right part of the equation in the x of the denominator, we have:
The result is:. We can keep doing this substitution:
.
The process of substituting the right side of the equation into the x at the bottom of the continued fraction expansion can be said to have “algorithmic similarity”. We could keep doing it forever.
The short notation for the continued fraction expansion is indeed [1; 1, 1, 1, …] as we saw after the decomposition of the golden rectangle.
The decimal expansion of the golden ratio is 1.618033988749895…. Irrational numbers, like the golden ratio, are such that their decimal expansion is not periodic or eventually periodic.
In this example we can see the beauty of continued fraction expansions. We are able to see periodicity in the continued fraction expansion and none could be seen in the decimal notation.
Since the periodicity in the continued fraction expansion can be seen from the beginning, this type of continued fraction expansion is called periodic and the short notation is . In some literatures this type of continued fraction expansion is also called “purely periodic”.
4.4 Periodicity and similar rectangles
In the previous example we saw that in the decomposition of the golden rectangle, we had similar rectangles and the continued fraction expansion showed periodicity. At this point, we can make the following proposition:
The decomposition of a rectangle into squares will show similar rectangles if and only if the continued fraction expansion is periodic or eventually periodic.
- If the decomposition of a rectangle shows similar rectangles, then the continued fraction expansion has periodicity. For this we can have two situations:
- If the continued fraction expansion has periodicity, then the decomposition of the rectangle will show similar rectangles.
If the continued fraction expansion is periodic or eventually periodic it means that it has a sequence of quotients that will repeat over and over again
Periodic:.
Eventually periodic:
Since each quotient represent squares in the decomposition, showing this fact geometrically means that the decomposition of the rectangle will also have a repetitive pattern. For this pattern to repeat it means that after doing a series of decomposition steps, the space left must be similar to the original rectangle or the rectangle where the pattern started.
4.5 Continued fraction expansions, the Golden ratio, and the Fibonacci Numbers
The first numbers in the Fibonacci sequence are: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
If we assume that each Fibonacci number represents the length of the side of a square, and we place them next to each other, we can construct rectangles like this:
The construction of the rectangles using the Fibonacci numbers has a close resemblance to the decomposition of the golden rectangle.
Golden RectangleFibonacci rectangles
The Golden rectangle is made of single squares that decrease in size and has a leftover rectangle similar to the original rectangle. The Fibonacci rectangles are a series of rectangles made of squares that increase in size and these rectangles do not have any leftover rectangles.
Since the Golden Rectangle has a leftover space, which is similar to the original rectangle, it means that the continued fraction expansion is infinite.
The Fibonacci rectangles do not have any empty spaces, which means that the continued fraction is finite. In the case of the picture above, the continued fraction expansion is. It only has 7 terms since only 7 different sizes squares that make the rectangle.
Furthermore, the difference between these two rectangles is that the ratio of the sides of the Golden rectangle represents an irrational number and the ratio of the sides of the Fibonacci rectangle represents a rational number.
Johannes Kepler proved that the ratio of consecutive Fibonacci numbers converges to the golden ratio [10].
Proof: assume that the ratio of successive terms of the Fibonacci numbers converges to a limit x.
In this proof we assume that the sequence converges. We know that this assumption is true because are the convergents of x. (look at the next page)
The last equation of the proof () was the one used to develop the continued fraction expansion of the golden ratio. The equation also yields the quadratic equation for which is the root. (Refer to page 56)
The continued fraction expansion of the golden ratio is an infinite continued fraction expansion. Truncating the continued fraction expansion will give rational approximations to the golden ratio. These rational approximations are called the convergents of the continued fraction expansion.
Let’s consider the first few rational approximations or convergents of the golden ratio:
The rational approximations of the golden ratio are:
These approximations are ratios of consecutive Fibonacci numbers. We can conclude that the convergents of the golden ration are . Therefore the limit of is which we know is .
Using EXCEL, we can see how the ratio of the consecutive Fibonacci numbers converges to the Golden Ratio.
How fast are these convergents approaching the golden ratio? In theorem 2, conclusion 4, we saw that the difference between the number x and the convergent is:
The nth convergent is
In the case of the Golden ratio, some of the convergents using this recursive relation are (reestablishing what we know already):
Knowing that are the ratios of the Fibonacci numbers, we can use Excel to figure out and in that way we can see how close the convergents get to the golden ratio confirming that
The red line represents. The blue line represents . We can see clearly that
By the 10th convergent, the difference is less than 1/100, it is about 0.00006.
Another interesting fact we can see from Theorem 2:
Conclusion 3: convergents do not simply increase or decrease. We have;
Some convergents are greater than the previous convergent, some are smaller. The even ones increase and the odd ones decrease.
In the diagram, the red segment has slope .If we let the convergents be slope of lines, we can see that the odd convergents (C1 and C3) are above the red segments and the even convergent (C0 and C2) are under the red segment.
It is interesting to see with only these 4 convergents how fast they get closer to the red line that represent the value of the golden ratio.
4.6 More examples of periodic continued fraction expansions
The Golden ratio is an example of a periodic continued fraction expansion of the form [n; n, n, …] or. The definition of periodic says that the general form is:. The quotients do not need to be always the same, but after a few quotients, they start repeating.
For example, the decomposition of the rectangle could be done as follows:
The decomposition of this rectangle can be described as one pink square, two smaller blue squares, one smaller tan square, and two even smaller green squares.
The decomposition suggests a continued fraction expansion of the form: [1; 2, 1, 2, 1, 2, 1…].
Since the quotient that represents the integral part of this continued fraction expansion is 1, we can think about the periodicity in two ways
The pattern of getting one square and then two smaller ones starts from the first quotient, suggesting the periodicity of the continued fraction expansion is. This notation is not the notation used throughout this paper, but it can found in some literatures.
In this section we will do the work using the idea that the periodicity starts from the first quotient and that the continued fraction expansion is .
What is the ratio of the length and width of this rectangle?
We saw before that after going through one cycle of the decomposition, the leftover rectangle is similar to the first rectangle if the decomposition is periodic. Imagine the width of the original rectangle is x and the length is x+y, after going through one cycle of the decomposition, the rectangle looks like the rectangle below and the colored rectangle is similar to the original rectangle:
Since the colored rectangle is similar to the original one, we can write the following equations:
Using cross-multiplication:
We are only interested in the ratio, so we can let x be equal to 1. The equation then becomes:
The solution to the equation is:
And the ratio of the sides of the rectangle becomes:
This number is an algebraic irrational number. From the set up for the decomposition problem we conjecture that
Let’s see how we can get the continued fraction expansion from the quadratic equation.
In order to obtain the continued fraction expansion, we manipulate the quadratic equation:
Substituting the y on the denominator by the expression of y we get:
We can use the idea of algorithmic similarity and substituting the whole expression into the variable in the denominator.
In this case we can see that the quotients alternate between the numbers 1 and 2. The notation for the continued fraction expansion of y will be:
y=[0; 2, 1, 2, 1, 2, …]
The ratio of the sides of the rectangle is:
(x + y):x = (1 + y):1 =1 + y =
[1, 2, 1, 2, 1, 2, … ]
The decimal expansion of the ratio is:
The continued fraction expansion for this irrational number is:
We can see periodicity in the continued fraction expansion but none in the decimal expansion of the irrational number.
Second example:
If the colored rectangle is similar to the original rectangle, what continued fraction expansion does the following decomposition represent and what would be the equation used to find the continued fraction expansion?
Since the colored rectangle is similar to the original rectangle, the following proportion can be written:
Using cross-multiplication:
We can make x = 1 (we are only interested in the simplified ratio), the equation becomes:
The positive solution is:
This decomposition represents the irrational number:
We conjecture the continued fraction expansion is
To find the continued fraction expansion we need to factor the variable y as follow:
We will use the last equation to find the continued fraction expansion by replacing the variable y in the denominator by its own expression.
Using the idea of algebraic similarity:
The quotients 3 and 1 alternate in the continued fraction expansion
The short notation for the continued fraction expansion of y is:
y = [0; 3, 1, 3, 1, 3, 1, …]
The ratio of the sides of the rectangle is 1+ y : 1
1+y = [1; 3, 1, 3, 1, 3, 1, …]
The periodicity of this continued fraction expansion starts from the first quotient of the continued fraction expansion. The continued fraction expansion is periodic with period of length 2 .
The ratio of the sides of the rectangle is the irrational number with decimal expansion
Summarizing the findings from the examples on purely periodic continued fraction expansions:
Quadratic Equation / Irrational Number / Equation for CF / CF / Decomposition/ / / [] /
To verify the last generalization, we need to check three situations:
- Continued fraction expansion is known, how to find the equation and its solution.
- Equation is known, how to find its solution and the continued fraction expansion.
- Irrational number is known, how to find the continued fraction expansion and the equation.
a. Continued fraction expansion is known, how to find the equation and its solution.
If the short notation of the continued fraction expansion is, the expanded form of the continued fraction expansion, without the integral part, is:
Simplifying this fraction from the bottom up:
Solving the equation for y, we get the positive root of the equation:
Adding 1 to this last number gives us the ratio of the length and width of the rectangle. The irrational number y +1 is:
b. Equation is known, how to find its solution and the continued fraction expansion.
Knowing the equation, we can find the solution using the quadratic formula as shown above.
Rewriting the formula:
Substituting the expression for y by itself:
At this point, we can see algorithmic similarity:
The periodicity can be seen and the continued fraction expansion becomes: [0; n, 1, n, 1, ….]
Adding the integral part 1 back:
c. Irrational number is known, how to find the continued fraction expansion and the equation.
The value or irrational number since:
Therefore:
The number can then be written as 1 plus a fractional part 1/x:
Doing some algebra to solve for x:
If we divide the last expression by n, we get again the irrational number
Since
and