Show that a_n --> 2: Using Calculus

[quasar987]

Homework Statement

Let $a_1=\sqrt{2}$, $a_2=(\sqrt{2})^{a_1}$,... $a_{n+1}=(\sqrt{2})^{a_n}$. Show that a_n-->2. you may use any relevant fact from calculus.

The Attempt at a Solution

I noticed that $\ln(a_1)=\ln(\sqrt{2})=\ln(2)/2$, $$\ln(a_2)=\ln((\sqrt{2})^{\ln(2)/2})=(\ln(2)/2)^2$$, ..., $\ln(a_n)=(\ln(2)/2)^n$. But ln(2) < 2. So ln(2)/2 < 1, and thus ln(a_n)-->0 <==> a_n-->1. What's wrong?

 

[Hurkyl]

Homework Statement

Let $a_1=\sqrt{2}$, $a_2=(\sqrt{2})^{a_1}$,... $a_{n+1}=(\sqrt{2})^{a_n}$. Show that a_n-->2. you may use any relevant fact from calculus.

The Attempt at a Solution

I noticed that $\ln(a_1)=\ln(\sqrt{2})=\ln(2)/2$, $$\ln(a_2)=\ln((\sqrt{2})^{\ln(2)/2})=(\ln(2)/2)^2$$, ..., $\ln(a_n)=(\ln(2)/2)^n$. But ln(2) < 2. So ln(2)/2 < 1, and thus ln(a_n)-->0 <==> a_n-->1. What's wrong?

$$\log a_2 = \log (\sqrt{2})^{a_1}$$

not

$$\log a_2 = \log (\sqrt{2})^{\log a_1}$$

 

[matt grime]

we know a_n= (sqrt(2))^{a_{n-1}}

eugh.

Suppose there is a limit. Let that limit be a. Can you show from this that a=2? I.e. what equality does a satisfy, and thus what is a (by inspection).

Now we sjust need to show that a_n converges to anything? What are the only things we know converge for sure? increasing bounded above sequences, or decreasing bounded below ones.

 

[Pseudo Statistic]

Did someone say... proof by induction!?

 

[NateTG]

Homework Statement

Let $a_1=\sqrt{2}$, $a_2=(\sqrt{2})^{a_1}$,... $a_{n+1}=(\sqrt{2})^{a_n}$. Show that a_n-->2. you may use any relevant fact from calculus.

One way to do this is to look at the sequence:
$$b_n=2-(a_n)$$
So if $b_n \rightarrow 0$ you're good. Maybe you can look at:
$$\frac{b_n}{b_{n+1}}$$.

 

[quasar987]

Suppose there is a limit. Let that limit be a. Can you show from this that a=2? I.e. what equality does a satisfy, and thus what is a (by inspection).

Nice. Suppoes a_n-->a. Then $a_{n+1}=\sqrt{2}^{a_n}\rightarrow a$. But also, $\sqrt{2}^{a_n}\rightarrow \sqrt{2}^a$. So $\sqrt{2}^a=a$. So a=2.

One way to do this is to look at the sequence:
$$b_n=2-(a_n)$$
So if $b_n \rightarrow 0$ you're good. Maybe you can look at:
$$\frac{b_n}{b_{n+1}}$$.

Did you follow through with that idea? Cuz it did cross my mind, but I rejected it quickly because that $$\frac{b_n}{b_{n+1}}$$ fraction is not too cooperative as far as i can see.

 

[quasar987]

How could we show the sequence is bounded though? :/

 

[NateTG]

Did you follow through with that idea? Cuz it did cross my mind, but I rejected it quickly because that $$\frac{b_n}{b_{n+1}}$$ fraction is not too cooperative as far as i can see.

$$b_{n+1}=2-a_{n+1}=2-\sqrt{2}^{2-b_n}=2-\frac{2}{\sqrt{2}^{b_n}}$$
I'm pretty sure it's possible to show that
$$0 < \frac{2-\frac{2}{\sqrt{2}^{b_n}}}{b_n} < 1$$
if [itex]b_n \in (0,1)[/tex]

 

[Hurkyl]

How could we show the sequence is bounded though? :/

One method is guessing what the bound is, and then proving that the sequence never exceeds the bound.

 

[matt grime]

Nice. Suppoes a_n-->a. Then $a_{n+1}=\sqrt{2}^{a_n}\rightarrow 2$. But also, $\sqrt{2}^{a_n}\rightarrow \sqrt{2}^a$. So $\sqrt{2}^a=a$. So a=2.

That doesn't follow at all. in fact you appear to be supposing the answer, unless it is just a typo, which seems most likely. If a_n converges to a, then a satisfies a=sqrt(2)^a.

 

[Pseudo Statistic]

How could we show the sequence is bounded though? :/

I'm thinking induction-- which would just be what Hurkyl hinted at.
$$a_1 = \sqrt{2}^{\sqrt{2}} < 2$$ => true for n=1.
Assume true for n, i.e. $$a_n = \sqrt{2}^{a_n -1} < 2$$... and show that this implies n+1 true, which would be something like:
$$a_{n+1} = \sqrt{2}^{a_n} < \sqrt{2}^{2} = 2$$
Or something.
And it's a monotone increasing sequence, so...

 

[quasar987]

Yeah I realized induction does the trick at work today (typo corrected in post #6)

 

[Office_Shredder]

That doesn't follow at all. in fact you appear to be supposing the answer, unless it is just a typo, which seems most likely. If a_n converges to a, then a satisfies a=sqrt(2)^a.

That's the point though... IF a_n is a convergent sequence, by algebra of limits, it must converge to 2 by the argument given. This is always a good way to start, because often to show a sequence converges, it's best to know what it converges to. For example, in this case if we didn't have the number 2, it would be difficult to pull the number out of mid-air and show it's an upper bound of the sequence (which it is regardless of whether we've verified it's going to be the limit, so there's no circular logic); so starting by showing it converges to something and then finding the limit would have been a much harder way to progress

 

[quasar987]

What guarantees us once we're at

$$a^{1/a}=2^{1/2}$$

that a=2 is the only solution?

 

[morphism]

What guarantees us once we're at

$$a^{1/a}=2^{1/2}$$

that a=2 is the only solution?

It's not. a=4 works too.

I guess you're going to need to use some bounds.

 

[quasar987]

How do we methodically find all the solutions to this?

 

[NateTG]

How do we methodically find all the solutions to this?

$$\frac{d}{da}a^{\frac{1}{a}}=\frac{1-\ln a}{a^2}a^{\frac{1}{a}}$$
Since the derivative only changes sign once for $a>0$, those are the only positive real solutions. I expect that there is an infinite number of complex solutions.