Real analysis problem with set convergence

In summary, the sequence converges for all positive real numbers but to find out what any given choice will converge to, you need to find the limit.
  • #1
geoman
8
0
Define the sequence A where Asub1=a and Asubn=sqrt(a+Asub(n-1)) for n greater than or equal to 2
I need to determine what positive choices of a will make the sequence converge and to what limit. I also need to prove it.


I plugged some values into the sequence and found that it seems to converge for all positive real numbers but I don't know how find out what any given choice of a will converge to
 
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  • #2
geoman said:
Define the sequence A where Asub1=a and Asubn=sqrt(a+Asub(n-1)) for n greater than or equal to 2
I need to determine what positive choices of a will make the sequence converge and to what limit. I also need to prove it.


I plugged some values into the sequence and found that it seems to converge for all positive real numbers but I don't know how find out what any given choice of a will converge to
If that sequence converges, then taking the limit of both sides of [itex]A_n= \sqrt{a+ A_{n-1}}[/itex], we have, because square root is continuous, [itex]\lim_{n\rightarrow\infty}A_n= \sqrt{a+ \lim_{n\rightarrow\infty}A_{n-1}}[/itex]. Of course, those two limits are of the same sequence so, calling that limit L, [itex]L= \sqrt{a+ L}[/itex]. Squaring both sides, [itex]L^2= a+ L[/itex] or [itex]L^2- L- a= 0[/itex]. By the quadratic formula, [itex]L= (1\pm\sqrt{1+ 4a})/2[/itex]. Which of the two roots (assuming 1+4a> 0) is the limit will depend on a.

(What does this have to do with "set convergence"? I see no sets here.)
 
  • #3
geoman said:
Define the sequence A where Asub1=a and Asubn=sqrt(a+Asub(n-1)) for n greater than or equal to 2
I need to determine what positive choices of a will make the sequence converge and to what limit. I also need to prove it.


I plugged some values into the sequence and found that it seems to converge for all positive real numbers but I don't know how find out what any given choice of a will converge to

Is the sequence you are considering the following??


1) [tex]x_{1} =a[/tex] ...

2) [tex] x_{n} = \sqrt(a + x_{n-1})[/tex]...for [tex]n\geq 2[/tex]
 
  • #4
yes that is the sequence. And sorry I did not meant to say set convergence. I meant convergence of sequences.
 
  • #5
geoman

This is a complete study of the convergence of the sequence:


PART A
Assume the sequence converges,then [tex]\lim _{ n\rightarrow\infty}\ x_n=x[/tex] and thus;

[tex] x=\sqrt{a + x}[/tex]

And the solutions of the equation are:

[tex] y_{1} = \frac{1+\sqrt{1+4a}}{2}[/tex] ..........1



[tex] y_{2}=\frac{1-\sqrt{1+4a}}{2}[/tex]............2


Now we have the following cases:


a) For a< -1/4 one of the subsequences of the sequence might have a limit on complex Nos .This is a point for research.


b) For [tex] -\frac{1}{4}\leq a<0[/tex] ,the sequence has no limit


c) For a=0 the sequence obviously has 0 as a limit ,although the equation of the sequence give us two limits 1 and 0.


d)For a>0 the equation has as its limit the positive root (1) for the following reasons:


We can prove by induction that [tex] x_{n}>0 \forall n\in N[/tex]...................3

Also a>0 ===> 4a+1>1 =====> root (2) is negative.Hence if the above sequence converges then its limit is the positive root (1)

So we have proved:

(the sequence{[tex]x_{n}[/tex]} converges)=====>(converges to root (1))
 
  • #6
PART B

Now we must prove that the sequence converges

For that we must form two differences:


The difference [tex] x_{n+1}-x_{n}[/tex] and the difference [tex] x_{n+1}-y_{1}[/tex]

Thus:

[tex] x_{n+1}-x_{n}=\sqrt{a+x_{n}}-x_{n} =\frac{a+x_{n}-x_{n}^2}{\sqrt{a+x_{n}}+x_{n}} =-\frac{(x_{n}-y_{1})(x_{n}-y_{2})}{\sqrt{a+x_{n}}+x_{n}}[/tex]...................4

Note [tex](x_{n}-y_{2})>0\forall n\in N[/tex] since root (2) is negative .


That difference will show us whether the sequence increases or decreases and that will depend on the factor [tex] x_{n}-y_{1}[/tex] since all the other factors in (4) are positive.

For that we form the difference:

[tex] x_{n+1}-y_{1}= \sqrt{a+x_{n}}-y_{1} =\frac{a+x_{n}-y_{1}^2}{\sqrt{a+x_{n}}+y_{1}}[/tex],and if we substitute the value of [tex]y_{1}[/tex] we get:


[tex]x_{n+1}-y_{1}=\frac{2x_{n}-1-\sqrt{1+4a}}{2(\sqrt{a+x_{n}}+y_{1})}[/tex]..................5

That difference will help us prove by induction whether [tex]x_{n}-y_{1}[/tex] is positive or negative and hence establish thru (4) whether the sequence is increasing or decreasing.

Indeed:

For [tex] x_{1}=a>y_{1}[/tex] we have if [tex] x_{n}>y_{1}[/tex] then [tex]x_{n}-y_{1}= 2x_{n}-1-\sqrt{1+4a}>0[/tex] and using (5) [tex]x_{n+1}-y_{1}>0 [/tex],hence

[tex]x_{n}-y{1}>0 \forall n\in N[/tex]............6

And

For [tex] x_{1}=a< y_{1}[/tex] we have if [tex] x_{n}< y_{1}[/tex] then [tex]x_{n}-y_{1}= 2x_{n}-1-\sqrt{1+4a}<0[/tex] and using (5) [tex]x_{n+1}-y_{1}<0[/tex],hence

[tex]x_{n}-y{1}<0 \forall n\in N[/tex]............7

So in either case by substituting (6) or (7) into (4) we see that the sequence is increasing or decreasing and also either bounded from above or below ,since by (6) or (7):

[tex] x_{n}\leq y_{1}[/tex] or [tex] x_{n}\geq y_{1}[/tex] for all nεN

THUS the sequence converges and since in part A WE proved if it converges it converges to root (1), we conclude:

For a>0 the sequence converges to the positive root (1)
 
  • #7
In part A OF the convergence of the sequence the equation [tex] x=\sqrt{a+x}[/tex] give us two solutions,and for a>0 we have two solutions ,one -ve and one +ve.

So one may wonder why the sequence converges to the +ve root ,due to the fact that,[tex]x_{n}>0 \forall n\in N[/tex] and not to the -ve root.The following proof shows why:

Suppose the sequence tends to the -ve root [tex] y_{2}[/tex].

From the definition of the limit we have :

For all ε>0 ,there exists a kεN SUCH that for all [tex] n\geq k[/tex] ,then [tex]|x_{n}-y_{2}|[/tex]<ε

Ιn the above definition put ε= [tex] -y_{2}[/tex]... n=k,then [tex]|x_{k}-y_{2}|< -y_{2}[/tex] which implies [tex] x_{k}<0[/tex]

But [tex] x_{n}>0 \forall n\in N[/tex] ,and for n=k we get :[tex]x_{k}>0[/tex].

therefor we end with the contradiction [tex] x_{k}>0[/tex] and [tex] x_{k}<0[/tex].


THUS we deduce that the sequence converges to the +ve root
 

1. What is the difference between pointwise and uniform convergence?

Pointwise convergence refers to the behavior of a sequence of functions at individual points. A sequence of functions is pointwise convergent if, for every fixed point in the domain, the value of the function at that point converges to a limit as the sequence progresses. On the other hand, uniform convergence refers to the behavior of a sequence of functions across the entire domain. A sequence of functions is uniformly convergent if, for any given epsilon, the values of the function at all points in the domain eventually fall within an epsilon neighborhood of the limit as the sequence progresses.

2. How is convergence related to continuity?

In general, a function is said to be continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. If a sequence of functions converges to a limit function, and each of the functions in the sequence is continuous at a point, then the limit function will also be continuous at that point.

3. What is the Cauchy criterion for convergence of a sequence?

The Cauchy criterion states that a sequence of real numbers is convergent if and only if it satisfies the Cauchy condition. The Cauchy condition requires that for any given epsilon, there exists a positive integer N such that for all n and m greater than N, the difference between the nth and mth terms of the sequence is less than epsilon.

4. How does the Bolzano-Weierstrass theorem relate to convergence?

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. This means that if a sequence of real numbers is bounded, then it must have at least one convergent subsequence. This theorem is often used in real analysis to prove the convergence of certain sequences.

5. What is the difference between absolute and conditional convergence?

A series is said to be absolutely convergent if the series of absolute values of its terms is convergent. On the other hand, a series is conditionally convergent if it is convergent but not absolutely convergent. In other words, a conditionally convergent series may converge, but rearranging the terms of the series can result in a different sum.

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