# Recurrence relations - Iterative Method

• Feb 11th 2008, 04:32 AM
valal
Recurrence relations - Iterative Method
Could somebody please solve y_t+1=0.5y_t+3 for me by using the iterative method (t=0,1,2 & 3 - please write down every step)? I don't understand it! (Worried)

• Feb 11th 2008, 05:18 AM
Soroban
Hello, valal!

If no one is responding, it's because we don't understand the problem.
Could you state the original problem?

Quote:

Could somebody please solve y_t+1=0.5y_t+3 for me (?)

by using the iterative method (*)

t=0,1,2 & 3 (#)

(?) .What do you mean by "solve"?
. . . .Do they want a "closed form"?

Also, a few spaces and parentheses would help us read what you wrote.
. . I would guess that it is: .$\displaystyle y_{t+1} \:=\:\frac{1}{2}y_t + 3$

(*) .We need to know what your teacher/course/textbook
. . . .means by the "iterative method".

(#) .This is strange.
. . . .We're involved with only the first four terms?

I would conjecture that they wants us to list the first four terms.
. . . .But they are expected to give us the first term, $\displaystyle t_o$

• Feb 11th 2008, 05:34 AM
valal
I want to solve the recurrence relation/equation - find the "common" solution. (contra the "exact" solution. I'm not sure if these terms are the right names in English)

And yes, it is y_{t+1} \:=\:\frac{1}{2}y_t + 3 which I mean. (Nod)

Example of iterative (= repeated) method:

y_{t+1} = y_t + 2

t=0: y_1 = y_0 + 2

t=1: y_2 = y_1 + 2 = (y_0 + 2) + 2 = y_0 + 2*(2)

t=2: y_3 = y_2 + 2 = [y_0 + 2*(2)] + 2 = y_0 + 3*(2)

t=3: y_4 = y_3 + 2 = [y_0 + 3*(2)] + 2 = y_0 + 4*(2)

From this you can see that:

y_t = y_0 + t*(2) = y_0 + 2t <--- (solution)

If you use my system, what is then the solution for y_{t+1} = 0.5y_t + 3?