My buddy and I cannot figure out how to do this problem in preparation for an exam. It goes:
Let A(n) be a sequence defined by A(1) = 3.
A(n) = (2 * A(n-1))^0.5, for all integers n >= 2.
Prove that for all A(n), A(n) > 2.
How can I go about this?
My buddy and I cannot figure out how to do this problem in preparation for an exam. It goes:
Let A(n) be a sequence defined by A(1) = 3.
A(n) = (2 * A(n-1))^0.5, for all integers n >= 2.
Prove that for all A(n), A(n) > 2.
How can I go about this?
I would look at the first several terms in the sequence:
$\displaystyle A_1=3$
$\displaystyle A_2=2^{\Large\frac{1}{2}}\cdot3^{\Large\frac{1}{2} }$
$\displaystyle A_3=2^{\Large\frac{1}{2}}\left(2^{\Large\frac{1}{2 }}\cdot3^{\Large\frac{1}{2}}\right)^{\Large\frac{1 }{2}}=2^{\Large\frac{3}{4}}\cdot3^{\Large\frac{1}{ 4}}$
I think at this point, we may state our induction hypothesis $\displaystyle P_n$:
$\displaystyle A_n=2^{\Large\frac{2^{n-1}-1}{2^{n-1}}}\cdot3^{\Large\frac{1}{2^{n-1}}}$
I would use the recursive definition as the induction step...what do you get?
I have come to a similar conclusion, except my formula is:
A(n) = (2^{0.5})^{n-1} * ((3)^{0.5})^{n-1}
My question is more how can I use this to explain that it never goes below 2? I am lost in this part of the process.
Of course the problem, as initially stated, doesn't require that we find a formula, just that we show that every term is greater than 2. We are given that A(1)= 3> 2 so that is taken care of. Now, suppose that, for some k, A(k)> 2. Since A(k)> 2, 2A(k)> 4 and then immediately A(k+ 1)= (2A(k))^{1/2}> 2.