Thread: solve equation

and

What do you mean "solve equation"? There isn't anything to solve. You have a recurrence relation. What is the exact wording of the question?

IN this case we don't want recurrence relationship, we need determine using only.

Hello, greg1987!

Have you been taught anything about recurrence relations?

It's impossible to help you if I don't know what you have learned so far
. . and where you are having difficulty.

. .

By the way, the answer is:.

Originally Posted by greg1987

and

Is the term of an arithmetic progression?

Originally Posted by sbhatnagar

Is the term of an arithmetic progression?

Clearly not, since the recurrence relation would then take the form where d is a constant, and obviously this is not the case. Not to mention the fact that post #4 obviously answers your question in the negative.

For "linear recurrences with constant coefficients", things such as where b and c are numbers, there is a very general method: first drop the number that is not multiplying and "a": and "try" . Then so the equation becomes and, dividing both sides by , r= b. That is satisfies . In fact it is easy to see that any constant times that, , is a solution. To find a solution to the entire equation, "try" a constant: if the equation becomes so and (if b is not 1), . The solution to the entire equation is the sum of those: for any constant p. (If b= 1, the equation is and we have an arithmetic sequence.)


This same idea can be used for higher order recurrences: If we have , setting gives [tex]r^{n+2}= 3r^{n+1}- 2r^n[tex] and dividing through by and moving everything to the left, . That is, both and satisfy the recurrence and the general solution is for any constants, A and B.