8 - Florida Atlantic University - Charles E. Schmidt ...

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Chapter 6

Continued Fraction of Transcendental Numbers

6.1 Decomposition of rectangles does not show any periodicity or patterns

In this chapter I will show how the decomposition of a rectangle looks like when the length of one side is a non-quadratic irrational number like[pic].

The cube root of two is the subject of a Greek legend.

On the island of Delos, the birthplace of Apollo and Artemis was an oracle. At one time, when there was a plague in Athens, an emissary went to the oracle to ask what to do. The oracle replied that the plague would cease if the altar to Apollo were doubled in size. The altar was a perfect cube, one cubit on each side. So the people of Athens built a new altar that was a cube two cubits long on each side. But the plague continued, and when asked, the oracle explained that the new altar was eight times as large as the old one. So they built an altar that was two cubits long in one direction, but one cubit in the other two directions, but the plague continued, the altar was no longer a cube. Apollo was only to be pleased by an exact cube that was twice the volume of the original —[pic]cubits on each side.(From Conway and Guy, The Book of Numbers page 190, paraphrased.)

A real number has one real cube root and two other cube roots, which form a complex conjugate pair. This implies that [pic]has one real solution.

How could we construct a segment of length [pic] using dynamic geometry software like CABRI?

The construction is not a Euclidean construction, meaning it cannot be done using a compass and a straightedge only.

1. Star with a segment AO = 1

2. Construct a perpendicular line through point O

3. Construct point B anywhere on the perpendicular line

4. Rotate line OB 30( clockwise (using a negative angle in CABRI)

5. Using the compass tool, construct a circle in point B with radius AO.

6. Mark the point of intersection of the circle and the rotated line C

7. Construct line BC

8. Adjust point B so that line BC passes through point A.

9. Segment AB is of the desired length.

How can we prove that AB is[pic]?

Construct ray AO and construct a perpendicular line through point C. Segment BC has length 1 because it is the radius of the circle constructed at B with radius AO.

Using similar triangles:

[pic]

Also:

[pic]

Since xy =1, then y = 1/x

[pic]

Using the Pythagorean Theorem:

[pic]

Multiplying by the denominator:

[pic]

Now that we can construct a segment with length[pic], we can construct a rectangle with sides 1 and[pic], and find the continued fraction expansion of[pic]?

Decomposition a rectangle with sides’ ratio of [pic]to 1, we can see a big square, followed by 3 squares of smaller size, followed by one even smaller and 5 very small squares. The decomposition does not end here.

The continued fraction will have, so far, the short notation of:

[1; 3, 1, 5…]

The method of decomposing a square to find the continued fraction is not very effective in this case. It might be better to use another tool like Excel to find the quotients:[pic]

The decimal expansion of [pic] can be expressed as 1 + 0.259921049895…. The decimal part is the same as [pic] which equals[pic]. We can star the same process with the new decimal: remove the integer part and find the reciprocal of the fractional part. Write this answer as the denominator of a fraction with numerator 1.

This algorithm can be done very easily using software like Excel.

Cell A2: the formula to get the value of [pic] is “=2^(1/3)”

Cell B2: The integer part of the number in cell A2 is “=int(A2)”.

Cell C2: Subtracted the integer part from the number in cell A2 “=A2 - B2”.

Cell A3: found the reciprocal of the number in cell C2 “=1/C2”.

After copying down the formulas. This is what we get:

The short notation of the irrational number[pic]is:

[pic]= [1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14, 3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1, 1, 1, 3, 7, 2, 2, 9, 4, 1, 3, 7, 6, 1, 1, 2, 2, 9, 3, 1, 1, 69, 4, 4, 5, 12, 1, 1, 5, 15, 1, 4, 1, 1, 1, 1, 1, 89, 1, 22, 186, 5, 2, 4, 3, 3, 1….]

Since this method of finding the continued fraction depends on the decimal expansion of [pic], it will be as accurate as the original number of decimal we start with. However, we can see that this expansion does not show any periodicity or pattern.

Lagrange proved that the simple continued fraction of an irrational number is periodic if and only if the number is a quadratic irrational. If the number is not a quadratic irrational, but is a real algebraic number of higher degree, like[pic], then its continued fraction expansion is not periodic.

Construction with the ruler and the compass of the quadratic irrationals on the real line is possible because the Euclidean Algorithm gives periodicity, and because of its periodicity there is an algorithmic approximation for every quadratic irrational.

No such algorithmic approximation exists for higher degree irrationals.

6.2 Decomposition shows patterns but not periodicity

What kind of numbers will show patterns in their continued fraction expansion?

We have seen that rational numbers always give finite continued fractions. Irrational numbers give us infinite continued fractions.

Irrational numbers are made of algebraic and transcendental numbers. We saw that algebraic numbers that were quadratic produced continued fractions with periodicity; non-quadratic algebraic did not show any periodicity. We have only left to consider transcendental numbers.

[pic]

6.3 The number e and its continued fraction

One number that is not an algebraic number and has a nice pattern in its continued fraction is the number e. The number e is the base of the natural logarithms and it is defined as

[pic]

e is a transcendental number, proved by Hermite in 1873. Its value with 15 decimal places is:

2.718281828459045…

To find the continued fraction of it, we can express it as 2 + 0.718281828459045. The decimal part is the same as [pic] which equals[pic]. We can star the same process with the new decimal: remove the integer part and find the reciprocal of the fractional part. Write this answer as the denominator of a fraction with numerator 1.

This algorithm can be done very easily using software like Excel.

Cell A2: the formula to get the value of e is “=exp(1)”

Cell B2: I found the integer part of the number in cell A2.

Cell C2: I subtracted the integer part from the number in cell A2.

Cell A3: found the reciprocal of the number in cell C2.

After copying down the formulas. This is what we get:

In the middle column we can see the notation of the simple continued fraction that displays a nice pattern:

e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12,…..1, 2n, 1…] [11]

Leonhard Euler (1707-1783) found another continued fraction expressions for e and found a pattern in the expansion

[pic]

In this beautiful continued fraction each quotient increases by adding 4 each time. Euler did not give a proof that the patterns he saw continue but he knew that if such a proof were given it would prove that e is irrational. If the continued fraction for (e - 1)/2 were to follow the pattern shown in the first few terms, 1, 6, 10, 14, 18, 22, 26, ... then it will never terminate so (e - 1)/2 (and so e) cannot be rational. Some people see this as the first attempt to prove that e is not rational.

This pattern can be reproduced in Excel up to a certain value. After a few steps of the iteration, the integral part does not follow the expected pattern.

There are other continued fractions for e that are not simple; they do not always have 1 as the numerator. [11]

Other expression using e

The number e is involved in many expressions that have interesting patterns in their continued fractions.

One interesting pattern is[pic]. The continued fraction pattern can be seen using Excel. The algorithm is the same as before, the only difference is that the beginning number is the square root of the number e.

[pic]

The continued fraction is [pic]= [1; 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, …, 4n+1, 1, 1, …]. It follows a pattern of 4n+1, 1, 1 for[pic]. [11]

Two other expressions that have interesting patterns in their continued fractions are

[pic]

Which is a special case of the more general form

[pic]

This last expression is known as the hyperbolic function [pic]

If in the expression of the hyperbolic function, we let k=1, then we have:

[pic]

While working these last two functions in Excel, the pattern stopped after a few steps. Only the numbers inside the red rectangle follow the pattern. I believe the difference is because Excel only uses 15 decimal places for the value of e.

6.4 The number [pic]and its continued fraction

Pi can also be expressed as a simple continued fraction. The algorithm for changing its decimal expansion is the same as for changing the number e.

Using Excel:

[pic]

The simple continued fraction for ( is given by:

[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]

Excel does not show this sequence of integers because of the limitations in the amount of decimals used. We cannot see any pattern in the integers representing the quotients in the simple continued fraction.

The first few convergent, or approximations of ( using the quotients of the continued fraction are

3 = 3 (with accuracy of 0 decimal digits)

[pic] (with accuracy of 2 decimal digits)

[pic] (with accuracy of 4 decimal digits)

[pic]

(with accuracy of 6 decimal digits)

[pic]

(with accuracy of 9 decimal digits)

The very large term 292 means that the convergent 355/113 is an extremely good approximation good to six decimal places that was first discovered by astronomer Tsu Ch'ung-Chih in the fifth century A.D. (Gardner 1966, pp. 91-102). []

We can see why the convergent [pic] is a good approximation by looking at the excel graph. The fractional part at the fourth step of the iteration is almost 0. At this point ( wants to become a rational number, with a very small fractional part. When we take the reciprocal of this small decimal, we get a large number, and that explains why we see a large number like 292. This happens every time a large number appears in the continued fraction, the convergent before this large quotient is a good approximation of (. If the number gets larger then the convergent is more accurate.

[pic]

One of the interesting things about the continued-fraction expansion of irrational numbers is that they are base-independent. Instead of repeating digits of the base in which we are representing the number (digits 0 - 9 in base ten), we get whole numbers as large as you wish. A relatively large 292 is found fairly early in the simple continued fraction form of pi. What other large numbers are there?

In 2003, E.W. Weisstein calculated [pic]terms of the continued fraction using Mathematica. The smallest integer that does not occur in the first 20 million terms is 2297 and some numbers are as large as 878,783,625. [}

There are other continued fraction expressions for ( that are not simple continued fractions (numerator can be any number). William Brouncker found more general forms of continued fractions in 1655 [11].

Since[pic], pi can be expressed as a complex continued fraction (one where the numerators are not 1) as follows:

6.5 Liouville Numbers and continued fractions

In 1844 Joseph Liouville proved that if a number x is algebraic of degree n, there exists a constant M such that for all rational numbers p/q, with[pic], the distance from x to p/q is greater than M/qn, where n is the degree of the minimal polynomial satisfied by x. He also defined a set of numbers that failed to have that property.

A Liouville number is a real number x with the property that for any positive integer n, exists integers p and q (q>1) such that:

[pic]

Joseph Liouville showed that numbers with this property are not just irrational, but are always transcendental. [12]

A Liouville number can be approximated by sequences of rational numbers.

Examples of Liouville numbers are:

[pic]

The last example is known as the Liouville’s constant. Its value is the sum of the sequence of rational numbers:

[pic]

This value can also be calculated using Excel, but the denominators get too large fast, and it is difficult to display the value.

Excel can also be used to figure out the continued fraction expansion of this Liouville number. The steps are very similar to the ones used to figure out the numbers e and (

[pic]

The numbers in the middle column represent the quotients of the continued fraction. The numbers inside the red rectangle are accurate, the rest of the numbers might have some errors due to the limitations of excel.

Another example of a Liouville number figured out in Excel is[pic].

The decimal expansion would be:

The quotients of the continued fraction are in the middle column.

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