Results 1 to 11 of 11

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

  1. #1
    Newbie
    Joined
    Aug 2010
    From
    Sibulan
    Posts
    12

    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!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,662
    Thanks
    1481
    The Fibonacci sequence is NOT a geometric sequence.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    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
    Last edited by mr fantastic; August 20th 2010 at 11:32 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,806
    Thanks
    697
    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}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \hdots\\ \hline<br />
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}<br />
A\left(\frac{1+\sqrt{5}}{2}\right) + B\left(\frac{1-\sqrt{5}}{2}\right) &=& 1 \\<br />
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}<br />
n & \phi^n/\sqrt{5} & F(n) \\ \hline \\[-4mm]<br />
1 & 0.723606798 & 1 \\<br />
2 & 1.170820394 & 1 \\<br />
3 & 1.894427192 & 2 \\<br />
4 & 3.065247586 & 5 \\<br />
5 & 4.959674779 & 5 \\<br />
6 & 8.024922367 & 8 \\<br />
7 & 12.98459715 & 13 \\<br />
8 & 21.00951952 & 21 \\<br />
9 & 33.99411668 & 34 \\<br />
10 & 55.00363621 & 55 <br />
\end{array}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2010
    From
    Sibulan
    Posts
    12

    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!!
    Last edited by mr fantastic; August 20th 2010 at 11:38 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by chisigma View Post
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by meow91006 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Oct 2009
    Posts
    769
    Quote Originally Posted by meow91006 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Oct 2009
    Posts
    769
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fibonacci sequence
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: April 6th 2010, 07:34 AM
  2. Fibonacci sequence
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 28th 2010, 03:54 AM
  3. Replies: 2
    Last Post: March 1st 2010, 11:57 AM
  4. Fibonacci sequence
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 26th 2009, 08:28 PM
  5. Fibonacci's sequence
    Posted in the Number Theory Forum
    Replies: 14
    Last Post: June 11th 2006, 08:04 AM

Search Tags


/mathhelpforum @mathhelpforum