Continuous least squares. Use the orthonormal basis comprising the normalized Legendre polynomials to obtain the quadratic polynomial that gives the best continuous least-squares fit for the function f(x) = e^2x on the interval [-1,1]. Sketch both the function and the polynomial that fits it.

Solution This problem is really a straightforward application of the minimization results we got earlier. The inner product and norm are
< f, g > = ∫ ₋₁¹ f(x)g(x)dx and ||f|| = (∫ ₋₁¹ f(x)²dx)^½.
As we mentioned in class, the normalized Legendre polynomials of degree 0, 1, and 2 are

p₀(x) = 2^-1/2,   p₁(x) = (3/2)^1/2x,   and   p₂(x)= (5/8)^1/2(3x²-1).

The quadratic polynomial p that minimizes ||f-q|| among all quadratics q 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).

We now want to use Matlab to create the required plot. Plotting e^2x is easy. To plot the polynomials, we will do the following steps.

1. Put the coefficients of each Legendre polynomial into a row vector, in descending order.
   c0 = [0 0 1]; c1= [0 1 0]; c2 = [3 0 -1];
   
2. Calculate the coefficients for the Legendre polynomials in the polynomial p(x).
   a0 = (1/4)*(exp(2) - exp(-2));
   a1 = (3/8)*(exp(2) + 3*exp(-2));
   a2 = (5/32)*(exp(2) - 13*exp(-2));
3. Form the coefficient vector c for p(x). This contains the coefficients of p(x) relative to the basis {x²}
   c = a0*c0 + a1*c1 + a2*c2;
4. Select the values of x to be used in the plot. Find the corresponding values p(x) and e^2x. Plot these.
   x = linspace(-1,1,300);
   yp = polyval(c,x);
   yf = exp(2*x);
   plot(x,yf,x,yp,'r')