A topic on series subjects. The Fibonacci Sequence.

**meow91006** · August 20th 2010, 05:34 AM

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!!

**Prove It** · August 20th 2010, 05:37 AM

The Fibonacci sequence is NOT a geometric sequence.

**Ackbeet** · August 20th 2010, 05:46 AM

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.

**chisigma** · August 20th 2010, 07:28 AM

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

$\chi$ $\sigma$

**Soroban** · August 20th 2010, 09:56 AM

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!

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

**meow91006** · August 20th 2010, 02:33 PM

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!!

**Ackbeet** · August 20th 2010, 04:40 PM

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.

**CaptainBlack** · August 20th 2010, 09:15 PM

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

**chisigma** · August 20th 2010, 10:49 PM

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

$\chi$ $\sigma$

**wonderboy1953** · August 21st 2010, 08:58 AM

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.

**wonderboy1953** · August 23rd 2010, 02:16 PM

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.

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