Test 1

Linear Methods of Applied Mathematics Evans M. Harrell II and James V. Herod*

SAMPLE TEST SOLUTIONS

At Georgia Tech, students were allowed to take one problem home to work out overnight. The rest of the test took 50 minutes (calculators allowed but no notes). Most students did reasonably well, but few finished early.

1.

a) Construct an orthonormal set of three functions on the interval -1 < x < 1, the span of which is the same as the span of {f1(x) := 1, f2(x) := sin( \pi x), f3(x) := x4}. Call the resulting set {g1(x), g2(x), g3(x)}.

The thing which makes this relatively easy is that f2(x) is odd, while f1(x) and f3(x) are even. Therefore we merely need to normalize f1(x) and f2(x), and then project away the part of f3(x) which correlates with f1(x) (and then normalize. We find

g1(x) = 1/Sqrt(2)
g2(x) = f2 = sin( \pi x) (already normalized)
f3(x) - projection onto f1 = f3(x) - average
&160;&160;&160;&160;&160;&160; = x4 - 1/5
g3(x) = (x4 - 1/5) *15/(4 Sqrt(2))

b) Find the function f(x) in the span of these three functions, which is closest in the r.m.s. sense to the function |x|.

Calculate this by projecting |x| onto g1 and g3. Forget g2, since it is odd and |x| is even. Also, remember to calculate the integrals involving |x| times other even functions as 2* the same integrals from 0 to 1. Then you can drop the absolute value signs.

&160;&160;&160;The best approximation to |x| is 1/2 + 0 sin( \pi x) + (15/16) (x4 - 1/5

&160;&160;&160;I.e., 5/16 + 0 sin( \pi x) + (15/16)x4

2. Find the (full) Fourier series for the function f(x) = |sin([[pi]]x)| on the interval -1 < x < 1. Sketch the sum of this series for all x, -[[infinity]] < x < [[infinity]]. Find the formula for all coefficients.

The absolute value makes this function even, so there are no sine terms. The entire series involves cosines, with coefficients

a0 = 2/ \pi (the average!) am = (2/2) integral from -1 to +1 of |sin( \pi x)| cos(2 m \pi x/2) dx
&160;&160;&160; = 2 integral from 0 to +1 of sin( \pi x) cos(m \pi x) dx

This works out to be 0 for m odd (can you interpret this differently?), while for m even we get

am = 2/(1 - m2 \pi2).

3. Let D f = f'(x), and consider the differential operator

A = D2 + 3 D + 2 Id

a) The dimension of the null space of A is 2 (it is second order

b) A basis for the null space of A is {exp(-x). exp(-2x)}