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| Binet returns (Posted on 2007-05-04) |
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Given that f(x)= 7f(x-1) - 14f(x-2) + 8f(x-3) and that f(0)=0, f(½)=1, and f(1)=3, find f(x).
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Submitted by Kurious
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Rating: 4.0000 (2 votes)
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Solution:
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We begin by defining:
f(x) = cx
where c is constant.
Then:
f(x) = 7 f(x-1) - 14 f(x-2) + 8 f(x-3)
cx = 7 cx-1 - 14 cx-2 + 8 cx-3
c0 = 7 c-1 - 14 c-2 + 8 c-3
c3 = 7 c2 - 14 c1 + 8 c0
c3 = 7 c2 - 14 c + 8
c3 - 7 c2 + 14 c - 8 = 0
The last term suggests we try c = 1, -1, 2, -2, 4, -4, 8, -8
Trying c = 1 we find that it is a solution.
Thus, "c - 1" is a factor.
Dividing "c3 - 7 c2 + 14 c - 8" by "c - 1":
c3 - 7 c2 + 14 c - 8 = 0
(c2 - 6c + 8)(c - 1) = 0
(c-4)(c-2)(c-1) = 0
c = 4, 2, or 1
Because there are multiple values of c, we need to use linear independence for completeness.
f(x) = cx
f(x) = p 4x + q 2x + r 1x
f(x) = p 4x + q 2x + r
where p, q and r are arbitrary constants.
To recap, the general solution to:
f(x) = 7 f(x-1) - 14 f(x-2) + 8 f(x-3)
is:
f(x) = p 4x + q 2x + r
where p, q and r are arbitrary constants.
However, we have three initial conditions to care about.
Substituting, we obtain:
0 = p + q + r
1 = 2p + sqrt(2) q + r
3 = 4p + 2q + r
Solving, we obtain:
p = 1, q = 0, r = -1
Therefore, the final solution is:
f(x) = 4x - 1
Which can be easily verified to comply with all conditions.
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