Chapter 3 Problem16 from rudin

[Poopsilon]

I can't really type it up on an iPod touch but the title says it all, I can't prove the sequence decreases monotonically I've tried literally EVERYTHING I am so frustrated please help me solve this I will be infinitely grateful, thanks.

 

[jbunniii]

In the future, please include the problem statement in your post, even if it takes more work. There are people who could answer your question but do not necessarily have the book at hand.

I do have the book, so this time I will summarize the question for the sake of others.

16. Fix a positive number \alpha. Choose x_1 > \sqrt{\alpha}, and define x_2, x_3, x_4, \ldots by the recursion formula

x_{n+1} = \frac{1}{2}\left(x_n + \frac{\alpha}{x_n}\right)

(a) Prove that \{x_n\} decreases monotonically and that \lim x_n = \sqrt{\alpha}

OK, I'll give you two suggestions.

1. To prove that the sequence decreases monotonically, observe that if x_n > \sqrt{\alpha}, then you can write it as x_n = p \sqrt{\alpha} for some p > 1. (p depends on n).

2. *If* the limit exists, it's very easy to see that it must be \sqrt{\alpha}. Just take limits of both sides of the recursion formula. So you just need to justify why the limit must exist. Use part 1 for this.

 

[╔(σ_σ)╝]

Use AM-GM on the formula and do the following a_{n} - a_{n-1}. Combine both facts to show the sequence is decreasing.

 

[Poopsilon]

Thanks for putting up the problem, so I basically proved the biconditional: x_n > x_n+1 <=> sqrt(a) < x_n and then was able to show the right side which basically followed the path of AM-GM for n=2. So now I have my monotonically decreasing sequence bounded below by sqrt(a). But I'm having trouble with showing that this is a g.l.b for the set of all x_n in order to prove that lim of {x_n} is equal to sqrt(a). How do I know there isn't some bound greater then that? I've tried a couple different epsilon N methods both directly and by contradiction but I can't seem to find the correct formulation.

 

[╔(σ_σ)╝]

Thanks for putting up the problem, so I basically proved the biconditional: x_n > x_n+1 <=> sqrt(a) < x_n and then was able to show the right side which basically followed the path of AM-GM for n=2. So now I have my monotonically decreasing sequence bounded below by sqrt(a). But I'm having trouble with showing that this is a g.l.b for the set of all x_n in order to prove that lim of {x_n} is equal to sqrt(a). How do I know there isn't some bound greater then that? I've tried a couple different epsilon N methods both directly and by contradiction but I can't seem to find the correct formulation.

x_{n+1} - \sqrt{ \alpha} = \frac{ \left(x_{n} - \sqrt{\alpha}\right)^{2}}{2x_{n}}

That could have been used to show x_n &gt; \sqrt{\alpha} since the right hand side is strictly greater than zero.
Look over the following and verify that the argument works.
Okay so I don’t know how much stuff you are willing to use but pulling out the quadratic equation shouldn’t be too much of a stretch.
x_{n+1} - \beta =\frac{ \left(x_{n} \right)^{2} -2\beta x_{n} +\alpha}{2x_{n}}
Using the quadratic equation on the numerator it is clear that we have real roots only when \beta &gt; \sqrt{ \alpha}. In this case we have two roots and it is clear that x_{n+1} - \beta &lt; 0 for some n. This would mean that \beta cannot be the g.l.b.
Any number ( beta) less than \sqrt{\alpha} results complex roots. Hence, the quadratic does not cross the “x-axis” and x_{n+1} - \beta &gt; 0 thus \beta is simply a lower bound which less than \alpha
So our only choice for the g.l.b is \sqrt{ \alpha}.

 

[jbunniii]

Thanks for putting up the problem, so I basically proved the biconditional: x_n > x_n+1 <=> sqrt(a) < x_n and then was able to show the right side which basically followed the path of AM-GM for n=2. So now I have my monotonically decreasing sequence bounded below by sqrt(a). But I'm having trouble with showing that this is a g.l.b for the set of all x_n in order to prove that lim of {x_n} is equal to sqrt(a). How do I know there isn't some bound greater then that? I've tried a couple different epsilon N methods both directly and by contradiction but I can't seem to find the correct formulation.

If it's monotonically decreasing and bounded below, then it converges. Call the limit L.

Then take limits of both sides of the recursion formula:

L = \frac{1}{2}\left(L + \frac{\alpha}{L}\right)

Now solve for L.

 

[╔(σ_σ)╝]

If it's monotonically decreasing and bounded below, then it converges. Call the limit L.

Then take limits of both sides of the recursion formula:

L = \frac{1}{2}\left(L + \frac{\alpha}{L}\right)

Now solve for L.

I think his/her difficulty lies in the proof that, that is actually the limit.

 

[jbunniii]

I think his/her difficulty lies in the proof that, that is actually the limit.

I guess I don't understand the difficulty, because this proves that if the limit exists, then it must be \sqrt{\alpha}.

Existence of the limit is guaranteed because the sequence is monotonically decreasing and bounded below (e.g., by zero). This is Theorem 3.14 in Rudin.

The only part that takes much work is showing that the sequence is decreasing.

But maybe there's a subtlety I'm missing?

 

[╔(σ_σ)╝]

I guess I don't understand the difficulty, because this proves that if the limit exists, then it must be \sqrt{\alpha}.

Existence of the limit is guaranteed because the sequence is monotonically decreasing and bounded below (e.g., by zero).

The only part that takes much work is showing that the sequence is decreasing.

But maybe there's a subtlety I'm missing?

You are right XD i took a slightly different apporoach to showing the what the limit is.

OP was probably thinking about a theorem that states that a bounded monotonic sequence either converges to the infima or suprema. So he was trying to show the limit is the infima in this case but first he wanted to establish the infima using the definition of a limit.

Your approach works just fine. :-)