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:

\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 &gt;1. Is this a good way to start?

 

[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):

<br /> \frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)} \geq 0 --&gt; \frac{(a_n^2+1)}{a_n} \geq 2^{\lambda^{n-2}}<br />

So then:
\frac{(a_2^2+1)}{a_2} \geq 2 --&gt; \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}&gt;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.