Here are solutions to the reset of the 20Apr1998 Exam C, in MAS3114.
tortoise: Mon Apr 27 14:59:52 EDT 1998
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C1(c)
# You can compute the desired polynomial just as you did for Project Interpolate:
> read(`polyinter.mp`);
tuple2poly := proc(low2highcoefs,quotedvar) ... end
polyinter := proc(ytuple,xtuple) ... end
> poly_for_C1_c := polyinter([1,1,13],[-1,1,3]);
2
poly_for_C1_c := 3/2 x - 1/2
# Or, you can use the method of basis vectors, as I showed in class, which
# derives the Lagrange Interpolation formula.
================================================================
C1(d)
[ a b c ]
[ ]
C := [ d e f ]
[ ]
[ g h i ]
> det(C);
a e i - a f h - d b i + d c h + g b f - g c e
Notice that we have 6=3! monomials. The sign of the coefficient for each
monomial is that of the permutation of the corresponding generalized-diagonal.
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C1(e)
As explained in class, since the two occurrences of "y" appear in the same row,
the mapping
y -> E(y)
is of the form
(*) y -> a + b*y,
for numbers {a,b} that we could solve for. In this case, we don't even need
to solve for them. The function (*) is a first-order map (an affine map), so
E(3)-E(2)
must equal
E(2)-E(1) which is 3-1 or 2.
Thus E(3) = 2 + E(2) = 2+3 = 5.
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