Notes - Numerical Analysis HT24, Hermite interpolation
Flashcards
Suppose:
- Have $x _ 0 < x _ 1 < \cdots < x _ n$ and $f _ i$, $g _ i$ for each $0 \le i \le n$
- Want to find some degree $2n+1$ polynomial $p$ such that $p(x _ i) = f _ i$ and $p’(x _ i) = g _ i$
Describe the construction used to do this and briefly justify why it works.
This is Hermite interpolation. Let
\[\begin{aligned} &L_ {n,k}(x) := \prod^n_ {i = 0, i \ne k} \frac{(x -x_ i)}{(x_ k - x_ i)} \\\\ &F_ {n,k}(x) := (L_ {n, k}(x))^2 (1 - 2(x - x_ k)L'_ {n, k}(x_ k)) \\\\ &G_ {n,k}(x) := (L_ {n, k}(x))^2 (x - x_ k) \end{aligned}\]then
\[p_{2n+1}(x) := \sum^n_{k = 0} \left( f_k F_{n, k}(x) + g_k G_{n, k}(x) \right)\]is a unique degree $2n+1$ polynomial that satisfies the required properties.
Quick justification:
- $F _ {n, k}(x _ i) = \delta _ {ik}$ and $F’ _ {n,k}(x _ i) = 0$
- $G _ {n,k}(x _ i) = 0$ and $G’ _ {n, k}(x _ i) = \delta _ {ik}$
Suppose:
- Have $x _ 0 < x _ 1 < \cdots < x _ n$ and $f _ i$, $g _ i$ for each $0 \le i \le n$
- Want to find some degree $2n+1$ polynomial $p$ such that $p(x _ i) = f _ i$ and $p’(x _ i) = g _ i$
To do this, let
\[\begin{aligned}
&L_
{n,k}(x) := \prod^n_
{i = 0, i \ne k} \frac{(x -x_
i)}{(x_
k - x_
i)} \\\\
&F_
{n,k}(x) := (L_
{n, k}(x))^2 (1 - 2(x - x_
k)L'_
{n, k}(x_
k)) \\\\
&G_
{n,k}(x) := (L_
{n, k}(x))^2 (x - x_
k)
\end{aligned}\]
and then the Hermite interpolating polynomial is defined by
\[p_{2n+1}(x) := \sum^n_{k = 0} \left( f_k F_{n, k}(x) + g_k G_{n, k}(x) \right)\]
Can you state a theorem which gives you a bound on the error, assuming $f$ has at least $2n+2$ smooth derivatives?
For every $x \in [x _ 0, x _ n]$, there exists $\xi _ x \in (x _ 0, x _ n)$ such that
\[e(x) = \frac{f^{(2n+2)}(\xi_x)}{(2n+2)!}\left(\prod^n_{k = 0}(x-x_k)\right)^2\]where $e(x) = f(x) - p _ {2n+1}(x)$. This is almost the same as the Lagrange interpolating case, but now the “product term” has been squared, and we are taking the $2n+2$th derivative instead of the $n+1$st derivative.