Solve $a_n>0$ Math Problem for $a_n$ with Given Sequence

[Albert1]

$a_n>0, \,\, for \,\, all \,\, n$

$a_1=1,a_2=3$

$a_{n+2}=\dfrac {(a_{n+1})^6}{(a_n)^9}$

$find:\,\, a_n$

 

[chisigma]

$a_n>0, \,\, for \,\, all \,\, n$

$a_1=1,a_2=3$

$a_{n+2}=\dfrac {(a_{n+1})^6}{(a_n)^9}$

$find:\,\, a_n$

Spoiler

Setting $\displaystyle \lambda_{n} = \ln a_{n}$ You arrive at the difference equation...

$\displaystyle \lambda_{n+2} - 6\ \lambda_{n+1} + 9\ \lambda_{n} = 0, \ \lambda_{1}=0,\ \lambda_{2}= \ln 3\ (1)$

The general solution of (1) is...

$\displaystyle \lambda_{n} = c_{1}\ 3^{n} + c_{2}\ n\ 3^{n}\ (2)$

... where the constant $c_{1}$ and $c_{2}$ can be found from the initial conditions. Once You have the $\lambda_{n}$ then is simply $\displaystyle a_{n} = e^{\lambda_{n}}$...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Spoiler

Setting $\displaystyle \lambda_{n} = \ln a_{n}$ You arrive at the difference equation...

$\displaystyle \lambda_{n+2} - 6\ \lambda_{n+1} + 9\ \lambda_{n} = 0, \ \lambda_{1}=0,\ \lambda_{2}= \ln 3\ (1)$

The general solution of (1) is...

$\displaystyle \lambda_{n} = c_{1}\ 3^{n} + c_{2}\ n\ 3^{n}\ (2)$

... where the constant $c_{1}$ and $c_{2}$ can be found from the initial conditions. Once You have the $\lambda_{n}$ then is simply $\displaystyle a_{n} = e^{\lambda_{n}}$...

Kind regards

$\chi$ $\sigma$

It's a challenge question...go ahead and finish it on up, my friend! (Tongueout)

 

[Saitama]

$a_n>0, \,\, for \,\, all \,\, n$

$a_1=1,a_2=3$

$a_{n+2}=\dfrac {(a_{n+1})^6}{(a_n)^9}$

$find:\,\, a_n$

I recently started with recurrence relations so expect the following to be wrong.

Spoiler

Take logarithm with base 3 on both the sides i.e
$$\log_3 a_{n+2}=6\log_3 a_{n+1}-9\log_3 a_n$$
Define $b_n=\log_3 a_n$ so the above relation is:
$$b_{n+2}=6b_{n+1}-9b_n$$
The characteristic polynomial of the above is $r^2-6r+9=0 \Rightarrow (r-3)^2=0$. Since the characteristic polynomial has a repeated root 3, the solution is of the form:
$$b_n=c_13^n+nc_2 3^n$$
From the initial conditions, we have: $b_1=0$ and $b_2=1$ so we have the following system of linear equation:
$$c_1+c_2=0$$
and
$$c_1+2c_2=\frac{1}{3^2}$$
Solving the above and plugging in $b_n$,
$$b_n=3^{n-2}(n-1)$$
Since $b_n=\log_3 a_n$, hence $a_n=3^{b_n}$
$$\Rightarrow b_n=3^{3^{n-2}(n-1)}$$
$\blacksquare$

 

[chisigma]

It's a challenge question...go ahead and finish it on up, my friend! (Tongueout)

I apologize for my slovenliness:(...

Spoiler

We are arrived to write... $\lambda_{n} = c_{1}\ 3^{n} + c_{2}\ n\ 3^{n}\ (1)$ ... and because is $\lambda_{1}=0$ and $\lambda_{2} = \ln 3$ we obtain... $\lambda_{n}= \ln 3\ (n-1)\ 3^{n-2}\ (2)$ ... so that is... $\displaystyle a_{n}= e^{\lambda_{n}} = 3\ e^{(n-1)\ 3^{n-2}}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[Albert1]

I recently started with recurrence relations so expect the following to be wrong.

Spoiler

Take logarithm with base 3 on both the sides i.e
$$\log_3 a_{n+2}=6\log_3 a_{n+1}-9\log_3 a_n$$
Define $b_n=\log_3 a_n$ so the above relation is:
$$b_{n+2}=6b_{n+1}-9b_n$$
The characteristic polynomial of the above is $r^2-6r+9=0 \Rightarrow (r-3)^2=0$. Since the characteristic polynomial has a repeated root 3, the solution is of the form:
$$b_n=c_13^n+nc_2 3^n$$
From the initial conditions, we have: $b_1=0$ and $b_2=1$ so we have the following system of linear equation:
$$c_1+c_2=0$$
and
$$c_1+2c_2=\frac{1}{3^2}$$
Solving the above and plugging in $b_n$,
$$b_n=3^{n-2}(n-1)$$
Since $b_n=\log_3 a_n$, hence $a_n=3^{b_n}$
$$\Rightarrow b_n=3^{3^{n-2}(n-1)}$$
$\blacksquare$

you expect the following to be wrong.but unfortunately it is correct
very nice (Yes)