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Am I doing this right? What's a good way to think about the different f(n)'s. f(n + 1) is the "next" element in the series? f(n) is the current element? The previous element?

Find $f(1), f(3), f(4),$ and $f(5)$ if $f(n)$ is defined recursively by $f(0) = 3$ and for $n = 0,1,2...$

[a.] $f(n+1) = -2f(n)$

\begin{align*} \text{Basis Step: } f(0) = 3& \\ \\ f(0 + 1) = -2(3) = -6 \\ f(1 + 1) = -2(-6) = 12 \\ f(2 + 1) = -2(12) = -24 \\ f(3 + 1) = -2(-24) = 48 \\ f(4 + 1) = -2(48) = -96 \end{align*}$

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2 Answers

up vote 1 down vote accepted

The computations are correct. Yes, $f(n)$ can be thought of as the current element of the sequence, and $f(n+1)$ as the next element. The recurrence (in this case) tells you how to compute the next element, if you know the current one.

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A clearer way for me to see the recurrence relation is by rewriting it as

$f(0) = 0$

$f(n) = -2f(n-1)$

Looking at it this way says whatever term in the sequence you are interested in is twice the previous term.

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I agree, this is a much more intuitive way to look at it. Does this mean if I want to change from f(n+1) to f(n) I should subtract 1 from each argument inside f()? –  papercuts Feb 10 '13 at 5:40
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You can just do a simple change of variable: let $n=m-1$: hence $f(n+1)=f(m)=-2*f(m-1)$ –  jesterII Feb 10 '13 at 5:50
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