# Linear Recurrence Relations

I'm having trouble understanding the process of solving simple linear recurrence relation problems. The problem in the book is this: $$0=a_{n+1}-1.5a_n,\ n \ge 0$$ What is the general process, and purpose, of solving this? Unfortunately there is a very large language barrier between my professor and myself, which is quite a problem.

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Check your equation again; it makes no sense as written. – Ron Gordon Apr 10 '13 at 16:45
Welcome to math.SE. Wrap your equations in \$s to make them look shiny. – Sammy Black Apr 10 '13 at 16:47 Thanks! The parts in the parentheses are supposed to be subscript, but I'm not quite sure how to do that. – Taylor Hill Apr 10 '13 at 16:51 You use an underline to get subscripts, and braces around multiple characters. So a_{n+1} in dollar signs gives$a_{n+1}$– Ross Millikan Apr 10 '13 at 16:52 See my answer below and copy the latex by right-clicking or editing. – Ron Gordon Apr 10 '13 at 16:52 ## 3 Answers The general solution to the equation $$a_{n+1} = k a_n$$ is $$a_n = B \cdot k^n$$ for some constant$B$, which is related to an initial condition. - What about the unique solution? – Taylor Hill Apr 10 '13 at 16:54 What of it? I gave you what is possible in terms of what you gave me. You need to specify some value at, say,$n=0$. That is,$a_0 = B$. – Ron Gordon Apr 10 '13 at 16:59 Hmm. That is exactly what it says in the book, with the instruction "find the unique solution for each of the following recurrence relations." I'm wondering if the first term is supposed to be 0 because otherwise it didn't specify. Or perhaps it's supposed to be unsolvable. – Taylor Hill Apr 10 '13 at 17:03 I hope the initial value is not zero. They may have meant what I said about$a_0$. But there is no unique solution to any difference or differential equation without an initial or boundary condition specified. For example, how can you tell what time it is on a clock without setting it first? – Ron Gordon Apr 10 '13 at 17:04 For this one, you can rewrite it as$a_{n+1}=1.5a_n$and then continue substituting to get$a_{n+1}=1.5a_n=(1.5)^2a_{n-1}$and see$a_n=1.5^na_0$-$a(n)=1,5^{n}a(0)$when$a(0)\$ should be given.

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Does the n>=0 imply that a_0 is 0? – Taylor Hill Apr 10 '13 at 16:56
No, it must be given arbitrarily. – Boris Novikov Apr 10 '13 at 17:00