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)

I thank you in advance.

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- February 11th 2008, 05:32 AMvalalRecurrence 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)

I thank you in advance. - February 11th 2008, 06:18 AMSoroban
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 wouldthat it is: .*guess*

**(*)**.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 tothe first four terms.*list*

. . . .But they are expected tous the first term,*give*

- February 11th 2008, 06:34 AMvalal
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?

In advance thank you. - February 12th 2008, 06:25 AMvalal
Nobody who is able to help? (Worried) It's just the basics (Speechless)...