the range is given in this way:

y1=-1

y(n+1)=1/3y(n)+3

Find a f function y(n)=f(n)

thanks

Printable View

- Feb 8th 2009, 12:39 PMblerttarecurrence
the range is given in this way:

y1=-1

y(n+1)=1/3y(n)+3

Find a f function y(n)=f(n)

thanks - Feb 8th 2009, 03:46 PMJester
For $\displaystyle y_{n+1} = \frac{1}{3} y_n + 3$ for let $\displaystyle y_n = z_n + k $ so the difference equation becomes

$\displaystyle z_{n+1} + k = z_n + \frac{1}{3}\left( z_n + k \right) + 3$

and choosing $\displaystyle k = \frac{9}{2}$ gives

$\displaystyle z_{n+1} = \frac{1}{3} z_n$

Seeking a solution of this in the form

$\displaystyle z_n = c \rho^n$ gives $\displaystyle \rho = \frac{1}{3}$, thus giving the solution

$\displaystyle y_n = c \left(\frac{1}{3}\right)^n + \frac{9}{2}$

Now use your initial condition to find c.

Note. If the difference equation is $\displaystyle y_{n+1} = \frac{1}{3 y_n} + 3$, it's harder problem. (Giggle)