Magic behind the Fibonacci sequence

Let’s take a look at patterns that can be discovered in Fibonacci numbers and how we can find them around us.

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In mathematics, there are countless sequences such as arithmetic sequences, geometric sequences, and many more. The Fibonacci sequence is one of them, but it is different from other sequences in that it can be easily found in everyday life. Let’s take a look at patterns that can be discovered in Fibonacci numbers and how we can find them around us.

In a Fibonacci sequence, every number after the first two numbers is the sum of the two preceding ones.

0, 1, 1, 2, 3, 5, 8, 13,…

This is a Fibonacci sequence because 2 is found by adding up the two former numbers, 1 and 1. Likewise, 3 is found by adding 1 and 2, which are the preceding numbers. Based on this rule, we can figure out that the next number in this sequence would be 8 + 13 = 21.

Numbers that appear in the Fibonacci sequence such as 1, 2, 3, 5, 8, and so on are referred to as Fibonacci numbers. Here is a list of Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, … (A000045)

We can write this sequence by using symbols instead of numbers:

\begin{align*}
u_n &= \text{term number} (n)\\
u_{n-1} &= \text{former term} (n-1)\\
u_{n-2} &= \text{term before} (n-2)
\end{align*}

$u_n= u_{n-1} + u_{n-2}$

$n$ 0 1 2 3 4 5 6 7 8
$u_n$ 0 1 1 2 3 5 8 13 21

The above table shows Fibonacci numbers for the $n$th term in the Fibonacci sequence. Let’s find out the next term, $u_9$, by using this equation.

\begin{align*}
u_9 &= u_8 + u_7\\
{} &= 21 + 13\\
{} &= 34
\end{align*}

Fibonacci sequence formula

There is another way to figure out Fibonacci numbers, other than adding up two former terms. In fact, there is a formula that easily lets us find Fibonacci numbers, just by substituting numbers. Following Nevil Hopley‘s derivation of the Fibonacci sequence’s general term, we proceed as follows:

Let’s define a certain Fibonacci number as $u_n$. We will assume that $u_n = k x^n$ because Fibonacci numbers grow exponentially, and an exponential function can be described as $k x^n$. Since $u_n$ is a Fibonacci number, it is the sum of two former Fibonacci terms.

$u_{n+1} = u_n + u_{n-1}$

This can also be written as:

$k x^{n+1} = k x^n + k x^{n-1}$

We can eliminate $k$ from the equation above by dividing all terms by $k$.

$x^{n+1} = x^n + x^{n-1}.$

This can be simplified by dividing all the terms by $x^{n-1}$.

$x^2 = x + 1$

The value of $x$ can be measured with the quadratic formula.

$\displaystyle x = \frac{1 \pm \sqrt{5}}{2}$

$\displaystyle x_1 = \frac{1 + \sqrt{5}}{2}, x_2 = \frac{1 – \sqrt{5}}{2}$

Therefore, $\displaystyle u_n = k x^n = k_1  \left(\frac{1 + \sqrt{5}}{2}\right)^n$ or $\displaystyle u_n = k x^n = k_2  \left(\frac{1 – \sqrt{5}}{2}\right)^n$.

The sum of two possible values of $u_n$ would also work.

$\displaystyle u_n = k_1  \left(\frac{1 + \sqrt{5}}{2}\right)^n + k_2  \left(\frac{1 – \sqrt{5}}{2}\right)^n$

Since $u_n$ is a formula for Fibonacci numbers, $u_0$ should be 0 and $u_1$ should be 1.
\begin{align*}
n&=0&&\implies& u_0 = k_1  \left(\frac{1 + \sqrt{5}}{2}\right)^0 + k_2  \left(\frac{1 – \sqrt{5}}{2}\right)^0 &= 0\\
&&&&k_1+k_2 &= 0\\
&&&&k_2 &= -k_1
\end{align*}
\begin{align*}
n &= 1 &&\implies&u_1 = k_1  \left(\frac{1 + \sqrt{5}}{2}\right)^1 + k_2  \left(\frac{1 – \sqrt{5}}{2}\right)^1 &= 1\\
&&&&k_1  \left(\frac{1 + \sqrt{5}}{2}\right) + k_2  \left(\frac{1 – \sqrt{5}}{2}\right) &= 1\\
\end{align*}
Multipling all terms by $2$ in order to eliminate fractions, we get:

$$k_1  \left(1 + \sqrt{5}\right) + k_2  \left(1 – \sqrt{5}\right) = 2$$

From the previous calculation when we substituted $0$ for $n$, we derived the equation $k_2 = -k_1$. Thus we will rewrite $k_2$ as $-k_1$ in the equation above.
\begin{align*}
k_1  \left(1 + \sqrt{5}\right) – k_1  \left(1 – \sqrt{5}\right) &= 2\\
k_1 + \sqrt{5} k_1 – k_1 + \sqrt{5} k_1 &= 2\\
2\sqrt{5} k_1 &= 2
\end{align*}
So we have found that:
\begin{align*}
k_1 &= \frac{1}{\sqrt{5}}\\
k_2 &= -\frac{1}{\sqrt{5}}
\end{align*}

Therefore:
$$_n = k_1  \left(\frac{1 + \sqrt{5}}{2}\right)^n + k_2  \left(\frac{1 – \sqrt{5}}{2}\right)^n = \frac{1}{\sqrt{5}}  \left(\frac{1 + \sqrt{5}}{2}\right)^n + \frac{1}{\sqrt{5}} \left(\frac{1 – \sqrt{5}}{2}\right)^n $$

What happens when the numbers are squared?

In fact, something much more interesting happens when the numbers are squared.

$u_n$ 1 1 2 3 5 8 13 21 34 55
$(u_n )^2$ 1 1 4 9 25 64 169 441 1156 3025

When we look at this new sequence closely, we can find a pattern.

$\displaystyle (u_n)^2 + (u_{n+1})^2 = u_{2n+1}$

This sequence can be derived not only by observation, but also by using algebra. Previously, we derived the $n$th term of the Fibonacci sequence:

$$u_n = \frac{1}{\sqrt{5}}  \left(\frac{1 + \sqrt{5}}{2}\right)^n + \frac{1}{\sqrt{5}} \left(\frac{1 – \sqrt{5}}{2}\right)^n$$

In order to justify $\displaystyle (u_n)^2 + (u_{n+1})^2 = u_{2n+1}$ in a simpler way, I will write $\displaystyle \frac{1 + \sqrt{5}}{2}$ as $A$ and $\displaystyle \frac{1 – \sqrt{5}}{2}$ as $B$.

In other words, I will write $\displaystyle u_n = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^n – \left(\frac{1 – \sqrt{5}}{2}\right)^n}{\sqrt{5}} = \frac{A^n – B^n}{\sqrt{5}}$

$\displaystyle u_n^2$ and $\displaystyle u_{n+1}^2$ can be written by using $A$ and $B$:
\begin{align*}
u_n^2 &= \left( \frac{A^n – B^n}{\sqrt{5}} \right)^2 = \frac{A^{2n} – 2 A^n B^n + B^{2n}}{5}\\
u_{n+1}^2 &= \left( \frac{A^{n+1} – B^{n+1}}{\sqrt{5}} \right)^2 = \frac{A^{2n+2} – 2 A^ {n+1}B^{n+1} + B^{2n+2}}{5} \\&= \frac{A^{2n} \times A^2 – 2 A^ {n}B^{n} \times A \times B + B^{2n} \times B^2}{5}
\end{align*}

Here $A \times B$ equals $-1$, so:
\begin{align*}
u_{n+1}^2 &=  \frac{A^{2n} \times A^2 + 2 A^ {n}B^{n} + B^{2n} \times B^2}{5}\\
u_n^2 + u_{n+1}^2 &=  \frac{A^{2n} + B^{2n} + A^{2n} \times A^2 + 2 A^ {n}B^{n} + B^{2n} \times B^2}{5} \\&= \frac{A^{2n}\left( 1 + A^2\right) + B^{2n}\left( 1 + B^2\right)}{5}
\end{align*}

Here,
$$1 + A^2 = 1 +  \left(\frac{1 + \sqrt{5}}{2}\right)^2 = \frac{5 + \sqrt{5}}{2}$$
and
$$1 + B^2 = 1 +  \left(\frac{1 – \sqrt{5}}{2}\right)^2 = \frac{5 – \sqrt{5}}{2}.$$
This means that:
\begin{align*}
u_n^2 + u_{n+1}^2 &= \frac{A^{2n}\left( \frac{5 + \sqrt{5}}{2} \right) + B^{2n}\left(\frac{5 – \sqrt{5}}{2} \right)}{5}\\
&= \frac{A^{2n} \times \sqrt{5} \left( \frac{\sqrt{5} + 1}{2} \right) – \sqrt{5} B^{2n}\left(\frac{1-\sqrt{5}}{2} \right)}{5}\\
&= \frac{A^{2n} \times \left( \frac{\sqrt{5} + 1}{2} \right) – B^{2n}\left(\frac{1-\sqrt{5}}{2} \right)}{ \sqrt{5}}
\end{align*}

Plugging $A$ and $B$ back in gives:
\begin{align*}
u_n^2 + u_{n+1}^2 &= \frac{\left( \frac{\sqrt{5} + 1}{2} \right)^{2n} \times \left( \frac{\sqrt{5} + 1}{2} \right) – \left(\frac{1-\sqrt{5}}{2} \right)^{2n}\left(\frac{1-\sqrt{5}}{2} \right)}{ \sqrt{5}}\\&= \frac{\left( \frac{\sqrt{5} + 1}{2} \right)^{2n+1} – \left(\frac{1-\sqrt{5}}{2} \right)^{2n+1}}{ \sqrt{5}}\\
u_{2n+1} &= \frac{\left( \frac{\sqrt{5} + 1}{2} \right)^{2n+1} – \left(\frac{1-\sqrt{5}}{2} \right)^{2n+1}}{ \sqrt{5}}
\end{align*}

Therefore, $\displaystyle u_n^2 + u_{n+1}^2 = u_{2n+1}$.

Let’s verify if this equation is right by substituting some numbers.

$n=1 \Rightarrow \left(u_1\right)^2 + \left(u_2\right)^2 = u_3 \Rightarrow (1)^2 + (1)^2 = 2$

$n=2 \Rightarrow \left(u_2\right)^2 + \left(u_3\right)^2 = u_5 \Rightarrow (1)^2 + (2)^2 = 5$

$n=3 \Rightarrow \left(u_3\right)^2 + \left(u_4\right)^2 = u_7 \Rightarrow (2)^2 + (3)^2 = 2$

Another pattern can be found when adding up the first few Fibonacci numbers.

$ \left(u_1\right)^2 = 1$

$ \left(u_1\right)^2 + \left(u_2\right)^2 = 1 + 1 = 2$

$ \left(u_1\right)^2 + \left(u_2\right)^2 + \left(u_3\right)^2 = 1 + 1 +4 = 6$

$ \left(u_1\right)^2 + \left(u_2\right)^2 + \left(u_3\right)^2 + \left(u_4\right)^2 = 1 + 1 +4 + 9 = 15$

$ \left(u_1\right)^2 + \left(u_2\right)^2 + \left(u_3\right)^2 + \left(u_4\right)^2 + \left(u_5\right)^2= 1 + 1 +4 + 9 + 25 = 40$

There seems to be no pattern in the numbers we got from adding up the squares of Fibonacci numbers. However, take a look at the factors of these numbers:

$1 = 1 \times 1$

$2 = 1 \times 2$

$6 = 2 \times 3$

$15 = 3 \times 5$

$40 = 5 \times 8$

The factors are two consecutive Fibonacci numbers. An equation can be derived from this pattern.

$ \left(u_1\right)^2 + \left(u_2\right)^2 + \left(u_3\right)^2 +\ldots + \left(u_{n-2}\right)^2 + \left(u_{n-1}\right)^2 + \left(u_n\right)^2 = u_n \times u_{n+1}$

Applications of the Fibonacci sequence

The Fibonacci sequence can be found in mathematical situations as well as non-mathematical situations. As for mathematical situations, the Fibonacci sequence can be detected in Pascal’s triangle. Pascal’s triangle is usually used in probability, combinatorics or algebra. However, it is possible to find Fibonacci numbers in this triangle, although it is quite challenging to see them.

Pascal’s triangle. Image: Wikimedia Commons user drini, CC BY-SA 3.0

The Fibonacci numbers can be seen when adding up the numbers in the “shallow” diagonals in Pascal’s triangle.

Sum of the numbers in the first shallow diagonal: $1$
Sum of the numbers in the second shallow diagonal: $1$
Sum of the numbers in the third shallow diagonal: $1+1=2$
Sum of the numbers in the fourth shallow diagonal: $1+2=3$
Sum of the numbers in the fifth shallow diagonal:$1+3+1=5$
Sum of the numbers in the sixth shallow diagonal: $1+4+3=8$

1, 1, 2, 3, 5, and 8 are all consecutive Fibonacci numbers.

In fact, the Fibonacci sequence can be commonly found in nature.

For example, this pineapple has 13 spirals clockwise and 8 spirals anticlockwise. 13 and 8 are adjacent Fibonacci numbers.

The number of flower petals are often Fibonacci numbers.

Number of petals Type of flowers
3 Lilies, amaryllis, iris, tulips
5 Wild rose, columbine, larkspur, buttercup
8 Clematis, delphiniums
13 Cineraria, ragwort, corn marigold
21 Aster, black-eyed susan, chicory

The above table shows the type of flowers depending on the number of petals.

Other than pineapples and flower petals, there are many more situations in which you can encounter a Fibonacci sequence. They might be in your town, in your house, or even inside your room! Look carefully around you. You might see Fibonacci numbers hiding somewhere. Somewhere, maybe right next to you.

Doyeon Jang is a grade 10 student studying at Branksome Hall Asia, an international school located in Korea. She is interested in discovering mathematics in everyday life and thinking about various applications of maths.
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