Solution of the recurrence relation

[mathmari]

Hey!

How can we solve the following recurrence relation?

$$f_n=\left (\frac{2}{a^2}+b\right )f_{n-1}-\frac{1}{a^4}f_{n-2} \\ f_0=1, f_{-1}=0$$

I calculated some values to see if there is a general pattern, but it doesn't seems so...

$$f_1=\left (\frac{2}{a^2}+b\right ) \\ f_2 =\left (\frac{2}{a^2}+b\right )^2-\frac{1}{a^4} \\ f_3 =\left (\frac{2}{a^2}+b\right )^3-\frac{2}{a^4}\left (\frac{2}{a^2}+b\right ) \\ f_4=\left (\frac{2}{a^2}+b\right )^4-\frac{3}{a^4}\left (\frac{2}{a^2}+b\right ) ^2+\frac{1}{a^8} \\ f_5=\left (\frac{2}{a^2}+b\right )^5-\frac{4}{a^4}\left (\frac{2}{a^2}+b\right ) ^3+\frac{3}{a^8}\left (\frac{2}{a^2}+b\right )$$

Have calculated these values right?

Is there an other way to find the solution of the recurrence relation? (Wondering)

 

[topsquark]

Hey!

How can we solve the following recurrence relation?

$$f_n=\left (\frac{2}{a^2}+b\right )f_{n-1}-\frac{1}{a^4}f_{n-2} \\ f_0=1, f_{-1}=0$$

I calculated some values to see if there is a general pattern, but it doesn't seems so...

$$f_1=\left (\frac{2}{a^2}+b\right ) \\ f_2 =\left (\frac{2}{a^2}+b\right )^2-\frac{1}{a^4} \\ f_3 =\left (\frac{2}{a^2}+b\right )^3-\frac{2}{a^4}\left (\frac{2}{a^2}+b\right ) \\ f_4=\left (\frac{2}{a^2}+b\right )^4-\frac{3}{a^4}\left (\frac{2}{a^2}+b\right ) ^2+\frac{1}{a^8} \\ f_5=\left (\frac{2}{a^2}+b\right )^5-\frac{4}{a^4}\left (\frac{2}{a^2}+b\right ) ^3+\frac{3}{a^8}\left (\frac{2}{a^2}+b\right )$$

Have calculated these values right?

Is there an other way to find the solution of the recurrence relation? (Wondering)

Start with:
\(\displaystyle f_n = c f_{n-1} + d f_{n-2}\)

\(\displaystyle f_{n + 2} - c f_{n + 1} - d f_n = 0\)

The characteristic equation is \(\displaystyle m^2 - cm - d = 0\)

Can you take it from here?

-Dan

 

[mathmari]

I did it! Thanks for the hint! (flower)