Solution We will need the fact that for any integer n ≥
0, we have
∫0∞ xne-
xdx = n!.
Using this and the Gram-Schmidt process we get the
polynomials {1,x-1,½(x2 - 4x + 2)}.
t | 0 | 1 | 2 | 3 | 4 |
---|---|---|---|---|---|
ln(C) | -0.1 | -0.4 | -0.8 | -1.1 | -1.5 |
Solution Let e = [1 1 1 1 1]T, t = [0 1 2 3 4]T, and y = [-0.1 -0.4 -0.8 -1.1 -1.5]T. To do a least squares straight-line fit y = a + bt to the data, we need to find a and b such that ||y - ae - bt|| is a minimum. The norm and inner product are the standard ones for R5. This is equivalent to finding u in the span of e and t that minimizes ||y - u||. Recall that we showed that there is a unique solution to this problem,
u* = < y,u1> u1 + < y,u2> u2
where u1, u2 form an orthonormal basis for the span of e and t. To obtain a and b, we use the Gram-Schmisdt process to find an orthonormal set. First, we letu* = - 1.7441×5-½ e - 1.1068×10-½ (t -2e) = - 0.78e -0.35(t -2e) = - 0.08e - 0.35t,
which yields a = -0.08 and b = -0.35. The line we want is thus y = - 0.08 - 0.35t.
Solution This problem is really a straightforward application of the minimization results we got earlier. The inner product is the one found in Example 11 on p. 166 of the text. The normalized Legendre polynomials of degree 0, 1, and 2 are found at the bottom of p. 166. To simplify notation here, we will let
p0(x) = 2-1/2, p1(x) = (3/2)1/2x, and p2(x)= (5/8)1/2(3x2-1).
The polynomial p(x) that minimizes ||f-q|| among all q in P2 is
p(x) = < f, p0> p0(x) + < f, p1> p1(x) + < f, p2> p2(x).
The inner products are all integrals of products of polynomials and exponentials; doing them results in these values:
< f, p0> = 8-1/2(e2 -
e-2)
< f, p1> = (3/32)1/2(e2 +
3e-2)
< f, p2> = (5/128)1/2(e2 -
13e-2)
The final polynomial we get is
p(x) = (1/4)(e2 - e-2) + (3/8)(e2 +
3e-2)x + (5/32)(e2 -
13e-2)(3x2-1).
Here is the required plot.