I think of it as hard proof by induction

[rock.freak667]

[SOLVED] I think of it as hard proof by induction

Homework Statement

given for the sequence $a_1,a_2,a_3,a_4,...$,$a_1=1$ and that

$$a_{n+1}=(a_n+\frac{1}{a_n})^{\lambda}$$

where$\lambda>1$.Prove by mathematical induction that,for $n \geq 2$

$$a_n \geq 2^{g(n)}$$

where $g(n)=\lambda^{n-1}$

Prove also,for $n \geq 2$

$$\frac{a_{n+1}}{a_n}>2^{(\lambda -1)g(n)}$$

Homework Equations

The Attempt at a Solution

well what I tried to do was to consider $a_{n+1}-2^{g(n)}$

which after simplification gives:

[tex]\frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)}[/itex]<br /> <br /> which I then said should be greater than or equal to zero since the fraction on the left is +ve and 2^g(n) is positive since lambda is >1. Is this a good way to start?[/tex]

 

[Dick]

Not really. I think you are supposed to use induction. Prove the statement is true for n=1. Then prove is the statement is true for n, then it's true for n+1. Isn't that how induction is supposed to go? What you've done so far doesn't look very 'inductive'.

 

[rock.freak667]

Not really. I think you are supposed to use induction. Prove the statement is true for n=1. Then prove is the statement is true for n, then it's true for n+1. Isn't that how induction is supposed to go? What you've done so far doesn't look very 'inductive'.

Well what I was essentially trying to do was to find an expression for a_n+1-2^g(n) and then add,subtract,multiply and divide various things to $a_n \geq 2^{g(n)}$ and then eventually say that from my expression a_n+1 >= 2^g(n)

 

[Dick]

Yeah, I know. But the problem says 'use induction'. And that might be the best way to go. I'll warn you, I haven't actually tried to do it.

 

[exk]

As Dick mentioned induction may be the way to go:

$$a_2=2^{\lambda}$$

Take your expression now and show that the statement is true for n=2 (I guess this is also your inductive hypothesis):

$$\frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)} \geq 0 --> \frac{(a_n^2+1)}{a_n} \geq 2^{\lambda^{n-2}}$$

So then:
$$\frac{(a_2^2+1)}{a_2} \geq 2 --> \frac{(2^{2\lambda}+1)}{2^{\lambda}} \geq 2$$

So now you need to find a way to express the relation for n=k+1. I guess expanding a couple of terms may be helpful, but I always came back to the general form that you proposed.

 

[rock.freak667]

Then just say:

Assume true for n=N

$a_N \geq 2^{g(n)}$

+ $\frac{1}{a_N}$

$$a_N +\frac{1}{a_N} \geq 2^{g(n)} +\frac{1}{a_N}$$


Raise both sides to the power of $\lambda$


$$(a_N +\frac{1}{a_N})^{\lambda} \geq (2^{g(n)} +\frac{1}{a_N})^{\lambda}$$

and then say that

$$(2^{g(n)} +\frac{1}{a_N})^{\lambda} \geq 2^{g(N+1)}$$

and so

$$a_{N+1}>2^{g(N+1)}$$

and hence true by PMI ?

 

[Vid]

We need to show that

(an + 1/an)^t >= 2^(t^(n))
Observation:
t*ln(an + 1/an) >=t^(n)ln(2)
ln(an + 1/an) >= t^(n-1)(ln(2))
an + 1/an >= 2^(t^(n-1))
Since t >= 1, it's an increasing sequence. Thus,
an + 1/an > an >= 2^(t^(n-1)) by the inductive hypothesis. QED.

 

[Vid]

I see no problems with your proof.

 

[rock.freak667]

We need to show that

(an + 1/an)^t >= 2^(t^(n))
Observation:
t*ln(an + 1/an) >=t^(n)ln(2)
ln(an + 1/an) >= t^(n-1)(ln(2))
an + 1/an >= 2^(t^(n-1))
Since t >= 1, it's an increasing sequence. Thus,
an + 1/an > an >= 2^(t^(n-1)) by the inductive hypothesis. QED.

hm.didn't think about taking ln...interesting

I see no problems with your proof.

But I might need to justify why

$$(2^{g(n)} +\frac{1}{a_N})^{\lambda} \geq 2^{g(N+1)}$$

so I would just say that the 2^g(n)+(anything) must be bigger than 2^(g(n+1)) since a_n is greater than zero

 

[Vid]

(2^(t^(n-1) + 1/an)^t >= 2^(t^(n-1)*t) = 2^(t^n) = 2^(g(n+1)). The inequality holds since t> 1 and an is positive.