# Math Help - A topic on series subjects. The Fibonacci Sequence.

1. ## A topic on series subjects. The Fibonacci Sequence.

Suppose the Fibonacci sequence is a geometric sequence (since the differences of the first few terms repeats the sequence), how to show the proof that it has a common ratio that would give an approximate solution to find the nth term of the sequence?

I want to have an idea how the proof goes.

Thank you!!

2. The Fibonacci sequence is NOT a geometric sequence.

3. You can use the theory of recurrence relations (difference equations) to find the nth term. It should have the square root of 5 and the golden ratio in it.

4. The Fibonacci sequence obeys to the difference equation...

$x_{n}= x_{n-1} + x_{n-2}$ (1)

... with the 'initial conditions' $x_{0}=0$ and $x_{1}=1$. The general solution of (1) is...

$x_{n} = c_{1}\ (\frac{1-\sqrt{5}}{2})^{n} + c_{2}\ (\frac{1+\sqrt{5}}{2})^{n}$ (2)

... and the constants $c_{1}$ and $c_{2}$ are found from the 'initial constants'. Properly speacking the Fibonacci sequence is the sum of two geometric sequences...

Kind regards

$\chi$ $\sigma$

5. Hello, meow91006!

I can't find a way to use Geometric Sequences on this problem.
So I solved the old-fashioned way.

Suppose the Fibonacci sequence is a geometric sequence.
How to prove that it has a common ratio that would give an approximate
. . solution to find the $n^{th}$ term of the sequence?

We have this sequence:

. . $\begin{array}{|c||c|c|c|c|c|c|c|c|c|c}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \hdots\\ \hline
F_n & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & \hdots \end{array}$

. . where: . $F_n \;=\;F_{n-1} + F_{n-2}$

Let $X = F_n$

Then: . $X^n \;=\;X^{n-1} + X^{n-2} \quad\Rightarrow\quad X^n - X^{n-1} - X^{n-2} \:=\:0$

. . Divide by $X^{n-2}\!:\;\;X^2 - X - 1 \;=\;0$

. . Quadratic Formula: . $X \:=\:\dfrac{1\pm\sqrt{5}}{2}$

$\text{Hence: }\;F(n) \;=\;A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$

We know that: . $F(1) = 1,\;F(2) = 1$

. . $\text{We have: }\;\begin{array}{ccccc}
A\left(\frac{1+\sqrt{5}}{2}\right) + B\left(\frac{1-\sqrt{5}}{2}\right) &=& 1 \\
A\left(\frac{1+\sqrt{5}}{2}\right)^2 + B\left(\frac{1-\sqrt{5}}{2}\right)^2 &=& 1 \end{array}$

. . Solve the system: . $A \,=\,\frac{1}{\sqrt{5}},\;\;B \,=\,-\frac{1}{\sqrt{5}}$

$\text{Hence: }\;F(n) \;=\;\dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2 }\right)^n - \dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

. . . . . . $F(n) \;=\;\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right]$

Note that: . $\frac{1+\sqrt{5}}{2} \:=\:\phi\,\text{ and }\;\frac{1-\sqrt{5}}{2} \:=\:\text{-}\frac{1}{\phi}$

. . . . . . . . .where $\phi$ = the Golden Ratio.

$\text{So we have: }\;F(n) \;=\;\dfrac{1}{\sqrt{5}}\bigg[\phi^n - \left(\text{-}\dfrac{1}{\phi}\right)^n\bigg]$

$\displaystyle \text{Note that: }\;\lim_{n\to\infty}\left(\text{-}\frac{1}{\phi}\right)^n \;=\;0$

$\text{Therefore, for large }n\!:\;\;f(n) \;\approx\;\dfrac{\phi^n}{\sqrt{5}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The approximations are quite close!

. . $\begin{array}{c|c|c}
n & \phi^n/\sqrt{5} & F(n) \\ \hline \\[-4mm]
1 & 0.723606798 & 1 \\
2 & 1.170820394 & 1 \\
3 & 1.894427192 & 2 \\
4 & 3.065247586 & 5 \\
5 & 4.959674779 & 5 \\
6 & 8.024922367 & 8 \\
7 & 12.98459715 & 13 \\
8 & 21.00951952 & 21 \\
9 & 33.99411668 & 34 \\
10 & 55.00363621 & 55
\end{array}$

6. ## Merged posts.

Thank you! My paper has shown me a bit of this but i don't really get it because it stopped at the middle and assumed that i know the conclusion. Thank you very much, this will help me understand my paper.

Thank you! Do you mean i have to read a book on differential equations? What is difference equation anyway? Is that different from differential equations? Well, i believe, the theory of recurrence relations will help.

Thank you very much! God bless!!

7. Difference equations are different from differential equations, but a lot of the solution methods are analogous. You even have an analogy of the Laplace transform in the Z transform.

There's even another, tighter, relationship between the two: if you discretize a differential equation, you get a difference equation.

8. Originally Posted by chisigma
The thread is duplicated and, because is not elementary, the answer will be done here. The Fibonacci sequence obeys to the difference equation...

$x_{n}= x_{n-1} + x_{n-2}$ (1)

... with the 'initial conditions' $x_{0}=0$ and $x_{1}=1$. The general solution of (1) is...

$x_{n} = c_{1}\ (\frac{1-\sqrt{5}}{2})^{n} + c_{2}\ (\frac{1+\sqrt{5}}{2})^{n}$ (2)

... and the constants $c_{1}$ and $c_{2}$ are found from the 'initial constants'. Properly speacking the Fibonacci sequence is the sum of two geometric sequences...

Kind regards

$\chi$ $\sigma$
In a hand waving sense we may say that a Fibonacci sequence is asymptotically a geometric sequence since the ratio of consecutive terms tends to the golden number.

CB

9. Originally Posted by meow91006
Thank you! What's a difference equation? Is that different from differential equations?
For an accurate enough deswcription see...

Recurrence relation - Wikipedia, the free encyclopedia

Effectively the precise general term is recurrence relation , even if the term difference equation is more 'traditional' and widely used again...

Kind regards

$\chi$ $\sigma$

10. Originally Posted by meow91006
Suppose the Fibonacci sequence is a geometric sequence (since the differences of the first few terms repeats the sequence), how to show the proof that it has a common ratio that would give an approximate solution to find the nth term of the sequence?

I want to have an idea how the proof goes.

Thank you!!
I have a book at home on Fibonacci numbers. I'll review it tonight for geometric sequences.

11. Reviewed my book, but no geometric sequences. The author is excellent along with the book so I'm assuming there is no geometric sequence you can derive from Fibonacci numbers.