How would you take the limit of this?

[1MileCrash]

I was just thinking about the golden ratio, and how it is a solution to the equation x^2 - x - 1 = 0.

So I wondered what it would be like to make a similar higher degree equation.

x^3 - x^2 - x - 1 = 0

To which the positive solution is 1.84.. is there any significance to this ratio?

I kept adding more terms.

x^4 - x^3 - x^2 - x - 1 = 0
x^5 - x^4 - x^3 - x^2 - x - 1 = 0
Etc.

These solutions, as I add more terms, grow more and more slowly and approach 2. I haven't found it to go past 2.

How can I show that the solution to the equation approaches 2 as the number of terms in the format I've shown grows without bound?

Another weird question from me, but I feel compelled to know the answers to these questions I have. :)

 

[coolul007]

rewriting the equation as x^5 - 1 = x^4 + x^3 + x^2 + x is always a difference of 1 when x = 2, because X^5 - 1 = x^4 + x^3 + x^2 + x + 1. this is true for binary.

 

[micromass]

One can make the problem easier by remembering the partial sum of a geometric series:

1+x+x^2+...+x^{n-1}=\frac{1-x^{n+1}}{1-x}

Can you do it now??

 

[haruspex]

You might find this relevant: http://mathworld.wolfram.com/Fibonaccin-StepNumber.html?affilliate=1

 

[D H]

One can make the problem easier by remembering the partial sum of a geometric series:

1+x+x^2+...+x^{n-1}=\frac{1-x^{n+1}}{1-x}

Can you do it now??

Ahem.
1+x+x^2+...+x^{n-1}=\frac{1-x^n}{1-x}

 

[micromass]

Ahem.
1+x+x^2+...+x^{n-1}=\frac{1-x^n}{1-x}

Oops

 

[T.H. Caro]

To formalize what you're getting at:

{x∈R^+ : lim_(n→∞)⁡ (x^n-∑_(k=0)^(n-1) (x^k) )=0}={2} forgive the lack of "real" notation

 

[Mandlebra]

Taking the infinite sum on the left, x= inf sum, you get phi. Did this the other week actually

 

[T.H. Caro]

Something interseting useing the sum of powers rules mentioned earlier:

We began with:
xⁿ-x^n-1-x^n-2-...-x-1 = 0

Rearrange to get:

xⁿ = x^n-1+x^n-2+...+x+1

Then using sum of consecutive powers:

xⁿ = (1-xⁿ) / (1-x)

xⁿ(x-1) = xⁿ-1

xⁿ⁺¹-xⁿ = xⁿ-1

∴ xⁿ⁺¹-2xⁿ+1 = 0

However this sort of question (to me) sounds a little before it's time as it calls for the ability to solve very high degree polynomials. But what do I know? Answer: "All I know is I know nothing."