I'm trying to solve the following equation: $$ f(n+1)=\sqrt{\frac{3}{2}f(n)}$$
with the condition:
$$ f(1)=1$$
Now I tried to put: $$f(n)=Az^n$$
in the equation where $z$ is a parameter and $A$ is a constant.Using that and the boundary condition I found: $$ f(n) = (3/2)^{\frac{1-n}{1+n}}$$
but that is not the answer since it doesn't give correct values of: $f(2)$,$f(3)$,... I know the answer is: $$f(n)=(3/2)^{1-2^{1-n}}$$
I don't know why I can't get to the answer.Please Help!
Take the logarithm,
$$\log f(n+1)=\log\sqrt{a f(x)},$$ or
$$g(n+1)=\frac{g(n)}2+\frac{\log a}2.$$
The general solution of this linear recurrence is
$$g(n)=\frac C{2^n}+\log a$$
and with the given initial condition $g(1)=0$,
$$g(n)=\log a\left(1-\frac2{2^n}\right),$$ or
$$f(n)=a^{1-2/2^n}.$$
As you can verify,
$$\sqrt{af(n)}=a^{(1+1-2/2^n)/2}=a^{1-2/2^{n+1}}=f(n+1).$$