Cauchy recursive sequense

Question

I am wondering if this sequence is a Cauchy one or not?

$X_{n+1}=\beta+\underbrace{k\cdot\sqrt{d}}_{\alpha}\cdot X_n$.

Here, $\beta=1$, $\alpha\in (0,+\infty)$, $X_0=1$, $k\in (0,1)$ and $d\in(0,+\infty)$.

I understand that it is defined over $\mathbb{R}$ as a Banach space. Hence, my ultimate goal is to see if the sequence is convergent.

@Atbey Thanks for the hint. The question is now updated to have a suitable form of the sequence in question. Thanks for sharing the point of view in advance. — user571752, Jun 23 '18 at 13:01
Hint: Can you identify the limit, assuming it exists? (Oh, and proving Cauchy property is not required at all, actually.) — Did, Jun 23 '18 at 13:08
@Did I could implement it in Matlab, as part of my modelling project. I see, it converges but theoritically I dont see it. — user571752, Jun 23 '18 at 13:12
@Did , Note: The sequence is a bit expanded to its original form. — user571752, Jun 23 '18 at 13:19
Because of the very precise and specific approach I outlined in my first comment. Did you read it (the comment)? Did you try it (the method)? I mean, writing a proof, not "implementing in matlab" or whatever you could do to avoid doing the maths... — Did, Jun 23 '18 at 20:35
@Did , It is a bit hazzy for me. Please read me more in this direction. — user571752, Jun 25 '18 at 07:36
OK, here is "more in this direction": Hint: Can you identify the limit, assuming it exists? — Did, Jun 25 '18 at 17:24
@Did ،Hmmmm.... If I assume the limi exist then my sequence should be non-decreasing and bounded. Then I need to prove them. — user571752, Jun 26 '18 at 07:53
@Did , Identifying the limit in an analytical way could be done by approach followed by the user parsiad , namely converting the iteration scheme to a form that has a known solution. I am not sure if I understood your question properly. — user571752, Jun 26 '18 at 13:55
Perhaps you did not even read it? Eight words and a comma... — Did, Jun 26 '18 at 17:59
OK. FYI, if you think the "answer" you posted below proves anything, by recurrence or otherwise, and if you give it to your TA as such, then be aware that you might be in for a violent reaction from them... — Did, Jun 27 '18 at 04:46
@Did , Among those 8 words I do not understand what do you meand by saying identify the limit. Do you mean solving analytically ti find the limit? Could you elaborate your question more precisely? — user571752, Jun 27 '18 at 04:56
Sorry but you explicitely stated you were not interested. One cannot slap people in the face (why delete your comment?) and then expect this attitude to have no consequences. The thing is in the hands of your TA now... — Did, Jun 27 '18 at 05:00

score 0 · Answer 1 · answered Jun 23 '18 at 13:25

0

You should always try writing a few terms of the sequence out: \begin{align*} X_{0} & =1\\ X_{1} & =\beta+\alpha X_{0}=\alpha^{\phantom{1}}+\beta(1)\\ X_{2} & =\beta+\alpha X_{1}=\alpha^{2}+\beta(1+\alpha)\\ X_{3} & =\beta+\alpha X_{2}=\alpha^{3}+\beta(1+\alpha+\alpha^{2})\\ \text{etc.} \end{align*} The above suggests $$ X_{n}=\alpha^{n}+\beta\sum_{k=0}^{n-1}\alpha^{k} $$ (you can prove this by induction).

Now, the summation (from $k=0$ to $n-1$) is a geometric series, which has a closed form solution. If you plug that in, then you will have an expression for $X_{n}$ whose limit you can take.

answered Jun 23 '18 at 13:25

parsiad

25,154

I deeply appreciate the elabirated answer. It is a lot helpful. With this, I conclude that no Banach space as the background is needed to prove the convergence of sequence. – user571752 Jun 23 '18 at 13:34
Feel free to accept if this answered your question. Were you able to get the answer from the hint? – parsiad Jun 23 '18 at 13:43
Is it accepted now? – user571752 Jun 23 '18 at 14:09
No, but it's not a huge deal. There's a checkmark you have to click to accept an answer. – parsiad Jun 23 '18 at 14:39
Right but the problem is I do not see the check mark! – user571752 Jun 23 '18 at 14:44
Might have to do with you being a new user. You might have to wait a certain amount of time before you can accept. Anyways, it's not a big deal. (More important that you understood the answer) – parsiad Jun 23 '18 at 14:46
,Exactly it is the first post. Thanks indeed. I'll do it. – user571752 Jun 23 '18 at 15:24
Now I reach to the point that, if \alpha\in(0,1) then the sum as a geometric series converges to \frac{1}{1-\alpha}. So, I need to work on my \alpha and find the condition that makes the whole scheme convergent. – user571752 Jun 27 '18 at 12:16
That's correct. You need $|\alpha| < 1$ to ensure convergence. – parsiad Jun 27 '18 at 22:20
great support. I deeply appreciate. – user571752 Jun 28 '18 at 06:40

score -1 · Answer 2 · answered Jun 25 '18 at 07:30

Let me please try to prove the raised induction part of the problem as$\colon$

Prove $X_n=\alpha^n+\beta\sum_{k=0}^{n-1}{\alpha^k}$ for all $n\ge 0$.

We start by reformulating the problem so that we have$\colon$

Prove $\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

We proceed by proof as$\colon$

Induction basis$\colon$ In case $n=1$,

$\sum_{k=0}^{n-1}{\alpha^k}=\sum_{k=0}^{0}{\alpha^k}=\alpha^0=1$.

Induction hypothesis: we assume in case $n\ge 0$,

$\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

holds true.

Induction step: we need to show:

$\sum_{k=0}^{n}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

holds also true. In this direction, we take the left hand side of the latter equation and expand it as:

$\sum_{k=0}^{n}{\alpha^k}=\sum_{k=0}^{n-1}{\alpha^k}+\sum_{k=n}^{n}{\alpha^k}$

or further as:

$\sum_{k=0}^{n}{\alpha^k}=\underbrace{\frac{X_n-\alpha^n}{\beta}}_{\sum_{k=0}^{n-1}{\alpha^k}}+\sum_{k=n}^{n}{\alpha^k}$

or,

$\sum_{k=0}^{n}{\alpha^k}-\sum_{k=n}^{n}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

that could be simplified as:

$\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

A final re-arrangement leads us to:

$X_n=\alpha^n+\beta\sum_{k=0}^{n-1}{\alpha^k}$

that is true based on our induction hypothesis.

Cauchy recursive sequense

2 Answers2