Question

The expression $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+}}}}.$$ where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of $$a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}}$$ where \(a\) and \(b\) are positive real numbers.

Short answer

In this problem, we analyzed and manipulated a continued fraction to find its limit, assuming it converges. We first defined the given sequence using the provided recurrence relation. Then, we evaluated the first five terms of the sequence and estimated the limit using computation and graphing. The estimated limit was approximately 1.618. We then determined the exact limit using the quadratic formula, obtaining the golden mean (1 + √5) / 2. Finally, we generalized the expression using two variables (a and b) and found the value to be (a ± √(a^2 + 4ab)) / 2, assuming it converges. Hence, the sequence converges to the golden mean, and its generalized expression converges to (a ± √(a^2 + 4ab)) / 2 when assuming it converges.

Step-by-step solution

Define the sequence using the given recurrence relation

To write the sequence, we simply apply the recurrence relation provided in the problem. Specifically, our sequence is defined as: $$ a_0 = 1 \\ a_{n+1} = 1 + \frac{1}{a_n}, \text{where } n = 0,1,2,3,\ldots $$

Evaluate the first five terms of the sequence

Using the recurrence relation, we can find the first five terms of the sequence by evaluating \(a_{n+1}=1+ \frac{1}{a_{n}}\) iteratively: $$ a_0=1\\ a_1 = 1+\frac{1}{a_0} = 1+\frac{1}{1} = 2 \\ a_2 = 1+\frac{1}{a_1} = 1+\frac{1}{2} = \frac{3}{2} \\ a_3 = 1+\frac{1}{a_2} = 1+\frac{1}{3/2} = \frac{5}{3} \\ a_4 = 1+\frac{1}{a_3} = 1+\frac{1}{5/3} = \frac{8}{5} $$

Estimate the limit of the sequence using computation and/or graphing

By plotting the first several terms of the sequence or using computational tools, we can estimate the limit of the sequence. The sequence appears to be converging towards the value of approximately \(1.618\).

Determine the exact limit using the method of Example 5

Assuming the limit exists, let the limit of the sequence as \(n \to \infty\) be \(L\). This means that as \(n \to \infty\), \(a_n \to L\). By substitution, we have: $$ L = 1 + \frac{1}{L} $$ Solving this equation for \(L\): $$ L^2 - L - 1 = 0 $$ Using the quadratic formula, we find: $$ L = \frac{1 \pm \sqrt{1^2 - 4(-1)}}{2} \\ L = \frac{1 \pm \sqrt{5}}{2} $$ Since the terms of the sequence are positive, we choose the positive root: $$ L = \frac{1 + \sqrt{5}}{2} $$ This is the exact value called the golden mean, confirming our estimation.

Determine the value of the generalized expression with \(a\) and \(b\)

Let's generalize our sequence using the variables \(a\) and \(b\), such that the limit converges towards: $$ S = a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}} $$ Following a method similar to Step 4, we have: $$ S = a + \frac{1}{\frac{a}{b} + \frac{1}{S}} $$ Solve this equation for \(S\): $$ S^2 - aS - ab = 0 $$ By using the quadratic formula, we find: $$ S = \frac{a \pm \sqrt{a^2 + 4ab}}{2} $$ This is the value of the generalized continued fraction when assuming it converges.

Key concepts

Recurrence Relations

A recurrence relation allows us to define a sequence using a formula that relates each term to previous ones. In the given problem, the recurrence relation is crucial for building the sequence of numbers produced by the continued fraction. Here, we started with the initial term:

• \(a_0 = 1\)

Then, we use the recurrence relation for all subsequent terms:

• \(a_{n+1} = 1 + \frac{1}{a_n}\)

This formula highlights the pattern of the continued fraction where each step depends on the last. Such relations are foundational in understanding sequences and can describe natural phenomena, algorithms, and mathematical structures. Recurrence relations are powerful tools that also often appear in computational problems and various algorithm designs.
By applying this method to the initial problem, students can clearly see how each new term is generated from the previous, leading into the general nature of continued fractions.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence "approach" as the term index goes to infinity. In this continued fraction example, we investigate if the sequence \(\{a_n\}\) has a limit.
The process involves observing the sequence:

• \(a_0, a_1, a_2, \ldots\)
• \(1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \ldots\)

Analyzing the terms, it's clear they seem to be settling around some value as more terms are added. This limit, denoted as \(\lim_{n \to \infty} a_n\), captures the essence of where the sequence is heading. If it exists, we're essentially figuring out the "eventual behavior" of our sequence. For many mathematical and real-world applications, understanding this behavior is key. When the sequence \(\{a_n\}\) in the given problem seems to stabilize around \(1.618\), it hints that it's converging towards a limit, and this leads us into defining it precisely.

Golden Mean

The Golden Mean, or Golden Ratio, is a special number approximately equal to 1.6180339887... and represented by the Greek letter \(\phi\). It frequently appears in geometry, art, architecture, and nature, making it a fascinating mathematical constant.
In the context of the continued fraction from the exercise, when we explore the limit of the sequence defined by:

• \(L = 1 + \frac{1}{L}\)

We arrive at the quadratic equation:

• \(L^2 - L - 1 = 0\)

The positive solution of this equation, \(L = \frac{1 + \sqrt{5}}{2}\), is the Golden Mean. What makes it captivating is its aesthetically pleasing properties and appearance in various forms of design and nature. Moreover, mathematically, it showcases a unique property where \(\frac{1}{\phi} = \phi - 1\), contributing to its allure in endless patterns, such as those found in Fibonacci sequences.

Quadratic Formula

Solving quadratic equations is a fundamental skill in mathematics, and the quadratic formula provides a straightforward way to find the roots of any quadratic equation. Quadratic equations generally take the form:

• \(ax^2 + bx + c = 0\)

The solution for \(x\) is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our continued fraction problem, solving for the limit \(L\) by setting up the relation \(L = 1 + \frac{1}{L}\), leads us to solve the quadratic:

• \(L^2 - L - 1 = 0\)

Applying the quadratic formula, we get two potential solutions, but only the positive root makes sense in the sequence context. This positive root \(\frac{1 + \sqrt{5}}{2}\), is not just a root, but a significant number recognized as the Golden Mean. Understanding how to use the quadratic formula not only resolves this mathematical problem but equips students to tackle numerous similar scenarios in algebra and beyond.