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

Let $$P(x)$$ be a polynomial that satisfies: $$P(0) = f(0)$$, $$P(2) = f(2)$$, $$P^{'} (2) = f^{'}(2)$$

(a) Using Hermite interpolation, find $$P(x)$$.

(b) Compute the expression $$\int^∞_0 e^{−x}P(x)dx$$ using the approximation from part (a).

Proof:

Using the Hermite formula, it seems that the minimum degree we can compute using the 3 given values for 2 points is $$2$$. However, the formula gives a 3rd degree polynomial and not a 2nd degree polynomial. I need the other missing derivative value. How do I get around that?

Concretely, Since the Hermite polynomial is $$H_n(x)=\sum_{k=1}^n f(x_k)h_k(x) +\sum_{k=1}^n f^{'}(x_k)\hat{h}_k(x)$$ where

$$h_k=[1-2L_k^{'}(x_k)(x-x_k)](L_k(x))^2$$ and $$\hat{h}_k=[1-2L_k^{'}(x_k)(x-x_k)](L_k(x))^2$$

where

$$L_k(x)=\prod_{j=1,j\neq k}^n \frac{x-x_j}{x_k-x_j}$$

It is easy to see for instance that $$L_1(x)=-\frac{x-2}{2}$$ and $$L^{'}_1(0)=-1/2$$ thus $$h_1(x)=\frac{1}{4}(1+x)(x-2)^2$$ Which is already a 3rd degree polynomial. Not a 2nd degree. Which means I would need to know also $$P^{'}(0)$$ which I do not have. Can I use the Hermite Interpolating Theorem to build a second degree polynomial? What am I missing here?

I read here about specialized Hermite interpolation but there are no details.

Thanks and regards,

• Can I assume $P^{'}(0)=f^{'}(0)$ ? Nov 30, 2021 at 1:34

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 $$f(x) = x^3$$.