Sequences and Their Properties: Apply It

Giselle Miller

Sequences and Their Properties: Apply It

Find the pattern in a sequence and write a formula for its terms
Determine what value a sequence approaches (if it approaches any value at all)
Figure out if a sequence converges or diverges

The Fibonacci Sequence: From Recursion to the Golden Ratio

The Fibonacci sequence is one of mathematics’ most fascinating patterns, appearing everywhere from flower petals to spiral galaxies. Named after Leonardo Fibonacci, who introduced it to Western mathematics in 1202, this sequence demonstrates how simple recursive rules can produce profound mathematical relationships.

The Fibonacci numbers are defined recursively by the sequence [latex]\left\{{F}_{n}\right\}[/latex] where [latex]{F}_{0}=0[/latex], [latex]{F}_{1}=1[/latex] and for [latex]n\ge 2[/latex], [latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[/latex].

In this exploration, you’ll discover how to move from the recursive definition of Fibonacci numbers to a closed formula, and ultimately connect this ancient sequence to the golden ratio—a number that has captivated mathematicians, artists, and architects for centuries.

Write out the first twenty Fibonacci numbers.

Find a closed formula for the Fibonacci sequence by using the following steps.

Consider the recursively defined sequence [latex]\left\{{x}_{n}\right\}[/latex] where [latex]{x}_{o}=c[/latex] and [latex]{x}_{n+1}=a{x}_{n}[/latex]. Show that this sequence can be described by the closed formula [latex]{x}_{n}=c{a}^{n}[/latex] for all [latex]n\ge 0[/latex].
Using the result from part a. as motivation, look for a solution of the equation

[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[/latex]

of the form [latex]{F}_{n}=c{\lambda }^{n}[/latex]. Determine what two values for [latex]\lambda[/latex] will allow [latex]{F}_{n}[/latex] to satisfy this equation.
Consider the two solutions from part b.: [latex]{\lambda }_{1}[/latex] and [latex]{\lambda }_{2}[/latex]. Let [latex]{F}_{n}={c}_{1}{\lambda }_{1}{}^{n}+{c}_{2}{\lambda }_{2}{}^{n}[/latex]. Use the initial conditions [latex]{F}_{0}[/latex] and [latex]{F}_{1}[/latex] to determine the values for the constants [latex]{c}_{1}[/latex] and [latex]{c}_{2}[/latex] and write the closed formula [latex]{F}_{n}[/latex].

For [latex]n = 0[/latex]: [latex]x_0 = c[/latex] (given) [latex]ca^0 = c \cdot 1 = c[/latex] So the formula holds for [latex]n = 0[/latex].

Assume [latex]x_k = ca^k[/latex] for some [latex]k \geq 0[/latex]. We need to show [latex]x_{k+1} = ca^{k+1}[/latex].

From the recurrence relation: [latex]x_{k+1} = ax_k[/latex] Using: [latex]x_{k+1} = a(ca^k) = ca^{k+1}[/latex]

Therefore, by mathematical induction, [latex]x_n = ca^n[/latex] for all [latex]n \geq 0[/latex].
If [latex]F_n = c\lambda^n[/latex], then:
- [latex]F_{n-1} = c\lambda^{n-1}[/latex]
- [latex]F_{n-2} = c\lambda^{n-2}[/latex]
Substituting into the Fibonacci recurrence relation: [latex]c\lambda^n = c\lambda^{n-1} + c\lambda^{n-2}[/latex]

Dividing by [latex]c\lambda^{n-2}[/latex] (assuming [latex]c \neq 0[/latex] and [latex]\lambda \neq 0[/latex]): [latex]\lambda^2 = \lambda + 1[/latex]

Rearranging: [latex]\lambda^2 - \lambda - 1 = 0[/latex]

Using the quadratic formula: [latex]\lambda = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}[/latex]

Therefore, the two values are: [latex]\lambda_1 = \frac{1 + \sqrt{5}}{2}[/latex] and [latex]\lambda_2 = \frac{1 - \sqrt{5}}{2}[/latex]
We have [latex]F_n = c_1\lambda_1^n + c_2\lambda_2^n[/latex] where:
- [latex]\lambda_1 = \frac{1 + \sqrt{5}}{2}[/latex]
- [latex]\lambda_2 = \frac{1 - \sqrt{5}}{2}[/latex]
Using the initial conditions:

For [latex]n = 0[/latex]: [latex]F_0 = 0[/latex] [latex]0 = c_1\lambda_1^0 + c_2\lambda_2^0 = c_1 + c_2[/latex] So [latex]c_1 + c_2 = 0[/latex], which means [latex]c_2 = -c_1[/latex]

For [latex]n = 1[/latex]: [latex]F_1 = 1[/latex] [latex]1 = c_1\lambda_1 + c_2\lambda_2[/latex]

Substituting [latex]c_2 = -c_1[/latex]: [latex]1 = c_1\lambda_1 - c_1\lambda_2 = c_1(\lambda_1 - \lambda_2)[/latex]

Now, [latex]\lambda_1 - \lambda_2 = \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}[/latex]

Therefore: [latex]1 = c_1\sqrt{5}[/latex], so [latex]c_1 = \frac{1}{\sqrt{5}}[/latex] and [latex]c_2 = -\frac{1}{\sqrt{5}}[/latex]

The closed formula for the Fibonacci sequence is: [latex]F_n = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n[/latex]

This can also be written as: [latex]F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right][/latex]

Use the answer in the third part of the previous question to show that:

[latex]\underset{n\to \infty }{\text{lim}}\frac{{F}_{n+1}}{{F}_{n}}=\frac{1+\sqrt{5}}{2}[/latex].

From part 2(c), we have: [latex]F_n = \frac{1}{\sqrt{5}}\left[\lambda_1^n - \lambda_2^n\right][/latex]

where [latex]\lambda_1 = \frac{1 + \sqrt{5}}{2}[/latex] and [latex]\lambda_2 = \frac{1 - \sqrt{5}}{2}[/latex]

Therefore: [latex]F_{n+1} = \frac{1}{\sqrt{5}}\left[\lambda_1^{n+1} - \lambda_2^{n+1}\right][/latex]

Computing the ratio: [latex]\frac{F_{n+1}}{F_n} = \frac{\frac{1}{\sqrt{5}}[\lambda_1^{n+1} - \lambda_2^{n+1}]}{\frac{1}{\sqrt{5}}[\lambda_1^n - \lambda_2^n]} = \frac{\lambda_1^{n+1} - \lambda_2^{n+1}}{\lambda_1^n - \lambda_2^n}[/latex]

Factoring out [latex]\lambda_1^n[/latex] from both numerator and denominator: [latex]\frac{F_{n+1}}{F_n} = \frac{\lambda_1^n(\lambda_1 - \lambda_2(\frac{\lambda_2}{\lambda_1})^n)}{\lambda_1^n(1 - (\frac{\lambda_2}{\lambda_1})^n)} = \frac{\lambda_1 - \lambda_2(\frac{\lambda_2}{\lambda_1})^n}{1 - (\frac{\lambda_2}{\lambda_1})^n}[/latex]

Now we need to examine [latex]\frac{\lambda_2}{\lambda_1}[/latex]: [latex]\frac{\lambda_2}{\lambda_1} = \frac{\frac{1 - \sqrt{5}}{2}}{\frac{1 + \sqrt{5}}{2}} = \frac{1 - \sqrt{5}}{1 + \sqrt{5}}[/latex]

Rationalizing the denominator: [latex]\frac{\lambda_2}{\lambda_1} = \frac{1 - \sqrt{5}}{1 + \sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} = \frac{(1 - \sqrt{5})^2}{1 - 5} = \frac{1 - 2\sqrt{5} + 5}{-4} = \frac{6 - 2\sqrt{5}}{-4} = \frac{\sqrt{5} - 3}{2}[/latex]

Since [latex]\sqrt{5} \approx 2.236[/latex], we have [latex]\frac{\lambda_2}{\lambda_1} = \frac{\sqrt{5} - 3}{2} \approx \frac{-0.764}{2} \approx -0.382[/latex]

Therefore [latex]\left|\frac{\lambda_2}{\lambda_1}\right| < 1[/latex], which means [latex]\left(\frac{\lambda_2}{\lambda_1}\right)^n \to 0[/latex] as [latex]n \to \infty[/latex].

Taking the limit: [latex]\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \lim_{n \to \infty} \frac{\lambda_1 - \lambda_2(\frac{\lambda_2}{\lambda_1})^n}{1 - (\frac{\lambda_2}{\lambda_1})^n} = \frac{\lambda_1 - \lambda_2 \cdot 0}{1 - 0} = \lambda_1 = \frac{1 + \sqrt{5}}{2}[/latex]

Therefore, [latex]\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \frac{1 + \sqrt{5}}{2}[/latex], which is indeed the golden ratio [latex]\varphi[/latex].

The number [latex]\varphi =\frac{\left(1+\sqrt{5}\right)}{2}[/latex] is known as the golden ratio (Figures 7 and 8).

This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number. — Figure 7. The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)

This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple’s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof. — Figure 8. The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)