Find $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for $f(x) = a e^x + b \cos(\pi x /2)$?

Question

I am reading through some exercises of the chapter I am reading about numerical integration. I stumbled upon the following question:

Find the formula $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for functions of the form $f(x) = a e^x + b \cos(\pi x/2)$

I know that for any polynomial of degree $\leq n$ its integral can be exactly estimated (that is, is equal to) by the formula $$ \int_a^b f(x) \,dx = \sum_{i=0}^{n} A_i f(x_i) $$ where $A_i = \int_a^b \ell (x)\, dx$ (Lagrange fundamental polynomial) for $n+1$ distinct nodes $x_i \quad 0 \leq i \leq n$.

However, the $f$ in the exercise is clearly not a polynomial. So no idea how to find such a formula that is exact as with numerical integration for polynomials. Can you help?

Answer 1

You can just set for instante $x_0=0$ and $x_1=1$ and compute $A_0,A_1$ such that $$ \int_0^1 e^x dx = A_0 e^0 + A_1 e^1, \quad \int_0^1 \cos \frac{\pi x}{2}\, dx = A_0 \cos \frac{\pi \cdot 0}{2} + A_1 \cos \frac{\pi \cdot 1}{2}. $$

This way you obtain the quadrature $$ Q(f)=\frac{2}{\pi} f(0) + \frac{e \pi -2-\pi }{e \pi } f(1). $$

If you require the formula to be exact to an extended set of functions, you'll be able to remove the initial arbitrarity in the choice of $x_0,x_1$. For instance, if you take $x_0 \approx 0.221611$, $x_1 \approx 0.79518$, $A_0 \approx 0.513634$, $A_1 \approx 0.486366$, the formula will be exact for functions of the form $$ f(x)= a e^x + b \cos \frac{\pi x}{2} + c \sin \frac{\pi x}{2} + d. $$

Answer 2

Integrating explicitly and equating to 0 the coefficients multiplying $a$ and $b$ individually one gets the linear system:

$e-1=A_0e^{x_0}+A_1e^{x_1}$ [1]

$\frac{2}{\pi}=A_0cos(\pi x_0/2)+A_1cos(\pi x_1/2)$ [2]

If we have $x_0$, $x_1$ and $A_0,A_1$ that solve [1] and [2] we are done.

Let's take $x_0$ and $x_1$ arbitrary. Since this system is linear in $A_0,A_1$ one can simply invert the matrix of coefficients and solve for these:

$\begin{pmatrix} A_0 \\ A_1 \end{pmatrix}$ =$\begin{pmatrix} e^{x_0} & e^{x_1}\\ cos(\pi x_0/2) & cos(\pi x_1/2) \end{pmatrix}^{-1}$ $\begin{pmatrix} e-1 \\ 2/\pi \end{pmatrix}$

It is simple to find $x_0$ and $x_1$ such that the system is invertible checking that the determinant is different from zero.