Convergence of Sequence x_n = (1/2)(x_n + 2/x_n)

[linuxux]

Hello, I have a question i can figure out.

THE QUESTION:

Show that

$$\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...$$

converges and find the limit.From what I see, the first term is root 2, the second term will be the root of 2 times the first term making it larger than the first term, the third term will be the root of 2 times the second term making it larger than the second term.

How can this converge to a real number?

 

[Gib Z]

A formal proof would involve showing that the sequence $$x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2}$$ is strictly increasing and is bounded by a certain value, but really that's boring isn't it?

Heres a less rigorous but more fun way :)

Lets take some terms in the sequence and solve for 2 (Not that hard, just square both sides and divide both sides by 2 until the RHS is prepared :) :

$$x_1^2 = 2$$
$$\frac{ x_2^4}{4} = 2$$
$$\frac{ x_3^8}{64} = 2$$

Ooo i Think i see a pattern :) Write the Denominators on the LHS as a power of 2, and take them over to the RHS :)

$$x_3^8 = 2^7$$

We can see in general :
$$x_n^{2^n} = 2^{2^n -1}$$

So using this for the 5th term,
$$x_5^{32} = 2^{31}$$, which I've checked is true.

Now, Take the log base 2 of the general form and use the following log rule, where b can be any base: $\log_b y^k = k \log_b y$.

So we have: $$2^n \log_2 x_n = 2^n -1$$, dividing both sides by [itex]2^n[/tex] and then making both sides the exponents of 2 finally gives us:

$$x_n = 2^{ \frac{2^n -1}{2^n}}$$.

So now, take the limit as n approaches infinity, the exponent on the RHS becomes 1. ie
$$\lim_{n\to \infty} x_n = 2$$.

Very nice?

 

[linuxux]

A formal proof would involve showing that the sequence $$x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2}$$ is strictly increasing and is bounded by a certain value, but really that's boring isn't it?

Heres a less rigorous but more fun way :)

Lets take some terms in the sequence and solve for 2 (Not that hard, just square both sides and divide both sides by 2 until the RHS is prepared :) :

$$x_1^2 = 2$$
$$\frac{ x_2^4}{4} = 2$$
$$\frac{ x_3^8}{64} = 2$$

Ooo i Think i see a pattern :) Write the Denominators on the LHS as a power of 2, and take them over to the RHS :)

$$x_3^8 = 2^7$$

We can see in general :
$$x_n^{2^n} = 2^{2^n -1}$$

So using this for the 5th term,
$$x_5^{32} = 2^{31}$$, which I've checked is true.

Now, Take the log base 2 of the general form and use the following log rule, where b can be any base: $\log_b y^k = k \log_b y$.

So we have: $$2^n \log_2 x_n = 2^n -1$$, dividing both sides by [itex]2^n[/tex] and then making both sides the exponents of 2 finally gives us:

$$x_n = 2^{ \frac{2^n -1}{2^n}}$$.

So now, take the limit as n approaches infinity, the exponent on the RHS becomes 1. ie
$$\lim_{n\to \infty} x_n = 2$$.

Very nice?

Sir, thank you.

 

[Gib Z]

No, thank YOU, I think that's the first time I've ever been called sir :D

 

[Gib Z]

The monotone convergence theorem is the formal proof I mentioned you could do in my original post, Wikipedia states: If a_k is a monotone sequence of real numbers (e.g., if a_k ≤ a_{k+1},) then this sequence has a limit (if we admit plus and minus infinity as possible limits.) The limit is finite if and only if the sequence is bounded. I don't know why or how they could implement summation into it. Please post the proofs if the book included them.

Edit: It seems you deleted your last post, making me look like a crackpot talking to myself lol

 

[linuxux]

Please post the proofs if the book included them.

No, no proof. the previous ones i did all utilized summation and that was the only technique they showed, but I'll tinker with it until tomorrow if i can't get it I'll surely post.

Edit: It seems you deleted your last post, making me look like a crackpot talking to myself lol

Just trying to keep you on your toes.

;)

 

[linuxux]

Another Question

I have another question for you.

I have a sequence defined by:

$$x_1=2$$

$$x_{n+1}=\frac{1}{2}(x_n + \frac{2}{x_n})$$

and i have to show that

$$x^2_n$$

is always greater than 2.The only thing i can come up is this; I end up using two cases (its convoluted, its confusing, i don't even know if it qualifies as a proof, i don't know if its the best proof if it is a proof, but anyway) :

i have to show that:

$$x^2_n > 2$$ which is equivalent to $$x_n > \sqrt{2}$$we assume

$$x_n > \sqrt{2}$$.

we know

2 > $$\sqrt{2}$$.the expression

$$x_n + 2$$

has two numbers > $$\sqrt{2}$$.when we add 2 to $$x_n$$ the result will either be a number twice as great as $$\sqrt{2}$$ or not.case 1:
suppose adding 2 to $$x_n$$ does produce a number twice as great as $$\sqrt{2}$$.

we know

$$x^2_n > x_n > \sqrt{2}$$

so

$$\frac{x^2_n}{x_n} > \sqrt{2}$$.

but if

$$\frac{x^2_n}{2x_n} < \sqrt{2}$$

then

$$\frac{x^2_n + 2}{2x_n} > \sqrt{2}$$case 2:
suppose adding 2 to $$x_n$$ does not produce a number twice as great as $$\sqrt{2}$$.

then we know $$x^2_n$$ is a number at least 4 times greater than $$\sqrt{2}$$.

so

$$\frac{x^2_n}{2x_n}$$ is still greater than $$\sqrt{2}$$.does that work as a proof?

(by the way, I've still only been given Monotone Convergence and Cauchy Condensation to use to prove.)