Hermite interpolation for 2 data points and one only one datapoint derivative available

Solution 1:

Let's denote by $p=f(0)$, $q = f(2)$, $r=f'(2)$. We have the Hermite interpolation polynomial of degree ($\le$) $2$ $$P(x)=a x^2 + b x + c$$ satisfying $$P(0)=c = p \\ P(2) = 4 a + b + c = q \\ P'(2) = 4 a + b = r $$ so we get $$P(x) = \frac{p-q+2 r}{4} x^2 + (-p+q-r)x + p$$

Now we calculate $$\int_{0}^{\infty} P(x) e^{-x} d x= \frac{p+q}{2} = \frac{P(0)+P(2)}{2}$$

The surprise is that this formula is valid for any $P$ polynomial of degree $\le 2$, but it involves only $2$ nodes ($r$ does not appear --- if we added another condition $ P(a_3)= f(a_3)$ ( Lagrange) or $P'(0) = f(0)$ (another Hermite), still we would get the same result). The reason behind it is the equality $$\int_{0}^{\infty} x(x-2) e^{-x} d x = 0$$

Now the approximation formula is $$\int_{0}^{\infty}f(x) e^{-x} dx \simeq \frac{f(0)+f(2)}{2}$$

Since the formula is an equality for polynomials of degree $\le 2$ ( i.e. those functions satisfying $f^{(3)} =0$), the error will be $$R \le C \cdot \sup_{x\in [0, \infty)} |f^{(3)}(x)| $$

You can find $C$ by calculating the error in the formula for $f(x) = x^3$.