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Answer:
solve for f(n) in terms of f(n-1)
Explanation:
In general, you solve for f(n) in terms of f(n-k) for k = 1, 2, 3, ....
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Usually, such questions arise in the context of arithmetic or geometric sequences.
Arithmetic sequence
The explicit formula for an arithmetic sequence has the general form ...
a(n) = a(1) +d(n -1) . . . . . . . first term a(1); common difference d
The recursive formula for the same arithmetic sequence will look like ...
a(1) = a(1) . . . . . . . the first term is the first term
a(n) = a(n-1) +d . . . the successive terms are found by adding the common difference to the term before
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Note: The explicit formula may be given as the linear equation a(n) = dn +b. Then the first term is a(1) = d+b.
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Geometric sequence
The explicit formula of a geometric sequence has the general form ...
a(n) = a(1)·r^(n -1) . . . . . . first term a(1); common ratio r
The recursive formula for the same geometric sequence will be ...
a(1) = a(1) . . . . . . the first term is the first term
a(n) = a(n-1)·r . . . the successive terms are found by multiplying the term before by the common ratio
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Note: The explicit formula may be given as the exponential equation a(n) = k·r^n. Then the first term is a(1) = kr.
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Other sequences
Suppose you're given the quadratic sequence ...
a(n) = pn^2 +qn +r
Since the sequence is known to be quadratic (polynomial degree 2), we expect that we will only need the two previous terms a(n-1) and a(n-2). Effectively, we want to solve ...
a(n) = c·a(n-1) +d·a(n-2) +e
for the values c, d, and e. Doing that, we find ...
(c, d, e) = (2, -1, 2p)
So, the recursive relation is ...
a(1) = p +q +r
a(2) = 4p +2q +r
a(n) = 2a(n-1) -a(n-2) +2p
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Additional comment
The basic idea is to write the expression for a(n) in terms of terms a(n-1), a(n-2) and so on. That will be easier for polynomial sequences than for sequences of arbitrary form.
There are some known translations between explicit and recursive formulas for different kinds of sequences, as we have shown above. If you recognize the sequence you have as being of a form with a known translation, then you would make use of that known translation. (For example, Fibonacci-like sequences are originally defined as recursive, but have explicit formulas of a somewhat complicated nature. If you recognize the form, translation from the explicit formula may be easy. If you must derive the recursive relation from the explicit formula, you may be in for a lot of work.)