Radius of Convergence of Fibonacci sequence

[Susanne217]

Radius of Convergence of Fibonacci sequence :)

Homework Statement

Given the Fibonacci sequence where

\frac{1}{1-x-x^2} = \sum_{n=0}^{\infty} F_{n} x^n

find the radius of convergence around zero.

Homework Equations

Ratio test

The Attempt at a Solution

By the radio test \lim_{n \to \infty} \left|\frac{F_{n+1}x^{n+1}}{F_{n}x^n}\right| =|x| = 0 Thus the radius of Convergence since R = 0.

Am I correct?

 

[jbunniii]

\lim_{n \to \infty} \left|\frac{F_{n+1}x^{n+1}}{F_{n}x^n}\right| = 0

How did you get that? It's not true.

\frac{F_{n+1}}{F_n}

converges to a certain positive number (the so-called "golden section") as n \rightarrow \infty. What does that say about

\lim_{n \rightarrow \infty} \left|\frac{F_{n+1}x^{n+1}}{F_n x^n}\right|?

 

[Susanne217]

How did you get that? It's not true.

\frac{F_{n+1}}{F_n}

converges to a certain positive number (the so-called "golden section") as n \rightarrow \infty. What does that say about

\lim_{n \rightarrow \infty} \left|\frac{F_{n+1}x^{n+1}}{F_n x^n}\right|?

I will look at it again, sorry :)

 

[Dickfore]

You can also find the zeros of the generating function on the left. The one with the smalles absolute value (smallest distance from the origin around which you perform your Taylor expansion) should give you the radius of convergence.

 

[Susanne217]

I can see I was way of and I'm sorry for that. Golden ratio and all that. I didn't learn about that mathwise in High School here in my part of the world. So I will read up on it :)

However I'm now told that the Fibonacci sequence convergeces towards

the socalled golden ratio \frac{1+\sqrt{5}}{2}

 

[HallsofIvy]

Well, not the Fibonacci sequence itself- that increases without bound. It is the ratio
\frac{F_{n+1}}{F_n}
that converges to the "golden ratio"
\frac{1+ \sqrt{5}}{2}

 

[Dickfore]

But your radius of convergence is not that.

 

[Susanne217]

Well, not the Fibonacci sequence itself- that increases without bound. It is the ratio
\frac{F_{n+1}}{F_n}
that converges to the "golden ratio"
\frac{1+ \sqrt{5}}{2}

HallsofIvy we all know you are like a Jedi then it comes to Math and Science compared to rest of us.

So to show the radius of convergence of the fibunacci sequence do I need to show first that whole thing convergences to the golden ratio?

 

[Susanne217]

HallsofIvy we all know you are like a Jedi then it comes to Math and Science compared to rest of us.

So to show the radius of convergence of the fibunacci sequence do I need to show first that whole thing convergences to the golden ratio?

so for the ratio I get

that \lim_{n \to \infty} |\frac{F_{n+1}}{F_n}| = \frac{1+\sqrt{5}}{2}

thus the radius of convergence R = \frac{1+\sqrt{5}}{2}

how is that?

 

[Dickfore]

so for the ratio I get

that \lim_{n \to \infty} |\frac{F_{n+1}}{F_n}| = \frac{1+\sqrt{5}}{2}

thus the radius of convergence R = \frac{1+\sqrt{5}}{2}

how is that?

It is not correct. The general element of the series is not F_{n}, but F_{n} x^{n}.

 

[Susanne217]

It is not correct. The general element of the series is not F_{n}, but F_{n} x^{n}.

this maybe be a stupid question but if as a previous post surgests I do Taylor series expansion

F_0 + F_1 z + F_2 z^2 + ...+ F_(n+1) z^(n+1)

isn't the ratio which I need to find

lim = (F_(n+1) z^(n+1)/F_nx^n) = ?

for n - > infty

 

[Dickfore]

You need to find the constraints on z, and you have not done so.

 

[Susanne217]

You need to find the constraints on z, and you have not done so.

Its suppose to be the radius of convergence around zero but if I set z to zero then the whole fraction turns zero. That can't the right can't it?

 

[Dickfore]

Please use the ratio test correctly as you did in your original post, although you made a mistake evaluating the relevant limit there.

EDIT:

On further inspection, I don't know why you equate the limit to zero?

 

[Susanne217]

Please use the ratio test correctly as you did in your original post, although you made a mistake evaluating the relevant limit there.

EDIT:

On further inspection, I don't know why you equate the limit to zero?

so my original post was right?

 

[Dickfore]

No. Please refer to the proper use of the ratio test for any power series.

 

[wisvuze]

we have the power series:

\sum_{n=1} a_{n} x^{n-1}

from 1 to infinity, where a_{n} is the nth term of the fibonacci sequence.

From the ratio test, we have the form

a_{n+1} / a_{n} * x and we need this to be < 1 to determine our radius of convergence.. So, we must decide how a_{n+1} / a_{n} behaves.

We have a0 = 1, a1 = 1 , a2 = 2, a3 = 3

a1 / a0 = 1 , a2 / a1 = 2, a3 / a2 = 3/2 .. do this enough and convince yourself that the ratio between two consecutive an+1 / an is bounded above by something. Prove by induction.