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.

================================================================

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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