# Lower Bound of Sequence

1. Nov 13, 2012

### sunnybrooke

1. The problem statement, all variables and given/known data

2. Relevant equations
3. The attempt at a solution
This is what I have so far:
$$x_{n+1}=\frac{x^5_n + 1}{5x_n}=1$$

$$x_{n+2}=\frac{x^5_{n+1} + 1}{5x_{n+1}}=\frac{1^5+1}{5}$$

I think you have to do some sort of repeated substitution but I don't quite see it. Any help? Thanks.

2. Nov 14, 2012

### HallsofIvy

Staff Emeritus
Why would this be equal to 1? As far as you know to start with, only x1 is equal to 1 and then
$x_2= 2/5$, $x_3= (32/3125+ 1)/2= 1580/3125= 316/625'$ which is about 0.5056. x1 is the only term that is equal to 1.

Use induction. You already know that $x_1= 1> 3/11$. Show that if $x_n> 3/11$ then $x_{n+1}= (x_n^5+ 1)/5x_n> 3/11$.

3. Nov 14, 2012

### sunnybrooke

I meant to say $$x_{1+1}=\frac{x^5_1 + 1}{5x_1}=1$$
Since the first term of the sequence on the left is $$x_{((1)+1)}$$

won't it be $x_2=1$ and not $x_1=1$?

Thanks.

4. Nov 14, 2012

### haruspex

No, the sequence is defined from n=1, so the first term is x1. You are confusing that with the first opportunity to use the recurrence relation.

5. Nov 14, 2012

### sunnybrooke

The notation is confusing me. The problem is equating two different sequences, right? And the sequence on the left is defined as xn+1 from n=1.

So for n=1: xn+1 = x(1)+1 = x2

Isn't this the first term?

Last edited: Nov 14, 2012
6. Nov 14, 2012

### sunnybrooke

Never mind; silly me. It is the definition of a recursive sequence.

I don't understand how exactly to prove by induction. Can I do the following?

$$x_{n+1}> ((3/11)^5+ 1)/5(3/11)= 0.74> 3/11$$

7. Nov 14, 2012

### haruspex

No, that's not valid. What would happen to the bound on xn+1 if you made xn a bit larger? The numerator is ok, that would just boost xn+1 a bit more, but the denominator goes up too, and it's not obvious what the net effect is.
You need to show that x > 3/11 implies (x5+ 1)/5x > ((3/11)5+ 1)/5(3/11)

8. Nov 14, 2012

### sunnybrooke

Unfortunately, I don't understand why that is implied.

Is this a valid solution?

$$x_n = \frac{x_{n-1}^5 +1}{5x_{n-1}}\\ f(x) = \frac{x^5+1}{5x}$$

Find the minimum using the first derivative.

$$f '(x)=\frac{4x^5-1}{5x^2}\\ f '(x)=0\\ x=(\frac{1}{4})^\frac{1}{5}\\ f(x) = \frac{((\frac{1}{4})^\frac{1}{5})^5+1}{5((\frac{1}{4})^\frac{1}{5})} \approx 0.33 \\ \frac{3}{11} \approx 0.273$$

The lowest possible value is greater than 3/11.

9. Nov 14, 2012

### haruspex

That proof is valid, but I suspect it is not in the spirit of the question. I'll give it some thought.

10. Nov 14, 2012

### sunnybrooke

Thank you. That's what I thought too but at least now I've solved it. :) I'd love to find another way to solve it but I don't seem to understand your solution. :( I will look at it again.