Proving Cauchy Sequences with Cosine Function

[Dollydaggerxo]

Homework Statement

Well, my problem is proving that sequences are in fact Cauchy sequences. I know all the conditions that need to be satisfied yet I cannot seem to apply it to questions. (Well, only the easy ones!)

My question is, prove that $$X_{n}$$ is a Cauchy sequence, given that $$X_{n+1}= f(X_{n})$$ where f(x)=cos(x).

I have been told that $$|X_{n+1}-X_{n}| \leq 2^{(1-n)}$$

The attempt at a solution

Well basically, I tried using m,n>N and N>0 and m>n>N to say that

$$|X_m-X_n| = cos(X_{m-1}) - cos(X_{n-1})$$

because this is the way i was doing it for simpler sequences.

However, I haven't got a clue where I would go from here which leads me to thinking I have gone the wrong way about it!
Any help would be greatly appreciated.
Thanks

 

[hgfalling]

OK, how do we prove that something is a Cauchy sequence?

First, a very mean man who hates undergrad analysis students is going to choose an $\epsilon > 0$. What we have to do is, for whatever $\epsilon$ he chooses, be able provide a number N such that for any two numbers m and n that are both greater than N,

$$|{X_n - X_m}| < \epsilon$$.

If we can do this, then it's a Cauchy sequence.

So now it would be nice if we were able to do cool things with $cos X_{m-1}$, but I don't really see how that's going to help us. However, the inequality they told you about looks promising. Try to figure out a way to provide a number N in response to any $\epsilon$ using that inequality.

 

[Dollydaggerxo]

Hello, thanks for the reply.

Here is what I have done:

Take m,n>N
N>0
$$\epsilon > 0$$

Take m = n+1 then

$$|X_m-X_n| \leq 2^{1-n} \leq 2^{-n}$$

But then i don't know where to fit epsilon into it?

I actually have the answer in front of me, but it doesn't use epsilon at all, or N, which is what I have been used to using. So I wanted to see if there was a way to do it with them included.

 

[hgfalling]

What definition of Cauchy sequences are you using that doesn't contain an epsilon? Maybe there is some other variable used, like a lot of calculus textbooks now use "h" instead of $\DELTA x$ in the definition of a derivative?

In any event, you're not allowed to choose m and n; the inequality has to be true for ANY m and n which are both greater than N. Otherwise, you'd have the following (FLAWED) proof:

Let $\{ x_n \} = \ln x$. Let $m = n+1$. Then $|X_m - X_n| = | \ln (n+1) - \ln n | = \ln \frac{n+1}{n}$.

Now $\lim_{n \rightarrow \infty} \ln \frac{n+1}{n} = 0.$ Hence for any $\epsilon$, there is an N such that for n > N $\ln \frac{n+1}{n}$ < $\epsilon$. So $\{ x_n \}$ is Cauchy.

Of course, this isn't true; ln x doesn't converge at all.

 

[Dollydaggerxo]

Yes I thought that was odd. But it is a follow on question so perhaps they had proved it before, I'm not sure.
But I want to be able to do it just based on this.

So I can't take m = n+1, that makes sense. I was just clutching at straws really, trying to use the inequality provided.

So could we take $$\epsilon > 0$$ and then we have $$|X_m-X_n| \leq 2^{1-n}$$
but m,n>N, so if we take N > 1

$$|X_m-X_n| \leq 2^{1-n} \leq 2^{1-N} \leq 2^{-N}$$

Is that wrong to just assume that?
Sorry, I am not very good at these at all! Thanks for your patience :)

 

[hgfalling]

You might try doing this.

Let $\epsilon = .1$.

Now give me N such that that no matter what m and n I choose greater than N, $| X_m - X_n | < \epsilon$.

Then try it for $\epsilon = .001$. Then .00001, etc, until you get the method down.

Then generalize so that you can give me an N for any $\epsilon$ I choose.

Then in your proof, just write down the method of generating Ns in response to an $\epsilon$ and show that it makes $| X_m - X_n | < \epsilon$.

There will be a DIFFERENT N for each $\epsilon$, by the way. As $\epsilon$ gets smaller, N gets bigger.

 

[wisvuze]

you don't have to go back to the cauchy definition, perhaps it was proven somewhere that either

$$| a_{n} - a_{n+1} | \leq c^{n}$$ where |c| < 1
or
$$| a_{n} - a_{n+1} | \leq c^{-n}$$
implies that the sequence is cauchy. In both cases, you can obviously see that as n gets large, c^n (where 0 < c < 1) or c^(-n) can be made arbitrarily small.

even if it was not proven formally anywhere, you can still use this

 

[Dollydaggerxo]

Okay that sort of makes sense, i have C defined in the set [-1,1] in a previous question. (if it is the same C !)
but what about the $$2^{n-1}$$
where does that come in in terms of C ?

 

[wisvuze]

$$2^{1-n}$$ becomes small as n becomes large