# How would you take the limit of this?

1. May 21, 2012

### 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. :)

2. May 21, 2012

### 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.

3. May 21, 2012

### 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??

4. May 22, 2012

### haruspex

5. May 22, 2012

### D H

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

6. May 22, 2012

### micromass

Oops

7. May 22, 2012

### 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

8. May 22, 2012

### Mandlebra

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

9. May 23, 2012

### T.H. Caro

Something interseting useing the sum of powers rules mentioned earlier:

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

Rearrange to get:

xn = xn-1+xn-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."

Last edited: May 23, 2012
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