📂Numerical Analysis

Hermite Interpolation

Definition ¹

For data $(x_{1}, y_{1} , y’_{1}) , \cdots , (x_{n} , y_{n}, y’_{n})$ of different $x_{1} , \cdots , x_{n}$ that satisfies $\begin{cases} p (x_{i} ) = y_{i} \\ p '(x_{i} ) = y’_{i} \end{cases}$ and $\deg H \le 2n-1$, the polynomial function $H$ is referred to as Hermite Interpolation.

Theorem

Existence and Uniqueness

• [1]: For the given data, $H$ uniquely exists.

Lagrange Form

• [2]: $$H_{n} (x) = \sum_{i=1}^{n} y_{i} h_{i} (x) + \sum_{i=1}^{n} y’_{i} \tilde{h}_{i} (x)$$

Newton Divided Differences Form

• [3]: $$\begin{align*}H_{n} (x) =& f(x_{1}) + (x - x_{1}) f [ x_{1} , x_{1} ] + (x - x_{1})^2 f [ x_{1} , x_{1} , x ] + \cdots \\ & + (x - x_{1})^2 \cdots (x - x_{n-1})^2 (x - x_{n}) f [ x_{1} , x_{1} , \cdots, x_{n} , x_{n} ]\end{align*}$$

Error Analysis

• [4]: Error with the actual function: For a $2n$ times differentiable $f : \mathbb{R} \to \mathbb{R}$ and some $\xi_{x} \in \mathscr{H} \left\{ x_{1} , \cdots , x_{n} , x \right\}$, the Hermite Interpolation $H_{n}$ satisfies the following. $$f(x) - H_{n} (x) = {{ [ \Psi_{n} (x) ]^2 } \over { (2n)! }} f^{(2n)} ( \xi_{x} )$$

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• $\mathscr{H} \left\{ a,b,c, \cdots \right\}$ represents the smallest interval that includes $a,b,c, \cdots$.

Explanation

Hermite Interpolation is interested in the polynomial that interpolates not only the function values but also the derivative values in Polynomial Interpolation. Since the differentiation is reflected, it is more accurate than Polynomial Interpolation.

However, it can also be applied to more than one differentiation, but its form is extremely complicated.

Proof

[1]

Strategy: Assume that multiple Hermite interpolations exist and show that it is a contradiction.

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Assuming $H$ and a different Hermite Polynomial $G$ exist.

If $R := H - G$, then for $x = x_{1}, \cdots , x_{n}$ $\begin{cases} H (x_{i} ) = G (x_{i} ) = y_{i} \\ H’ (x_{i} ) = G' (x_{i} ) = y’_{i} \end{cases}$, meaning $$R(x_{i}) = R’(x_{i}) = 0$$ This means $x_{1}, \cdots , x_{n}$ are all $R(x) = 0$th multiple roots of $2$, so for some polynomial $q(x)$, $$R = q(x) (x - x_{1} )^2 \cdots (x - x_{n} )^2$$ If $q(x) \ne 0$, it contradicts because $\deg R \ge 2n$, and if $q(x) = 0$, $R(x) = 0$ must be $H(x) = G(x)$.

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[2]

Strategy: Similar to Lagrange’s formula, specifically find $h_{i}$ and $\tilde{h}_{i}$.

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Part 1.

$$\Psi_{n} (x) := (x-x_{1}) \cdots (x - x_{n})$$ Differentiating $\Psi_{n}$ with respect to $x$ and substituting $x = x_{j}$ yields $$\Psi’_{n} (x_{j}) := (x-x_{1}) \cdots (x - x_{j-1})(x - x_{j+1}) \cdots (x - x_{n})$$

$$\implies {{ \Psi (x)_{n} } \over { (x - x_{j}) \Psi’_{n}(x_{j}) }} \equiv {{ (x - x_{1} ) \cdots (x - x_{j-1} ) (x - x_{j+1} ) \cdots (x - x_{n} ) } \over { (x_{j} - x_{1} ) \cdots (x_{j} - x_{j-1} ) (x_{j} - x_{j+1} ) \cdots (x_{j} - x_{n} ) }} = l_{j} (x)$$ Here, $l_{i}$ indicates the Kronecker delta function with respect to the index.

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Part 2.

$$h_{i} (x) := \left[ 1 - 2 l’_{i} (x_{i}) (x -x_{i} ) \right] \left[ l_{i}(x) \right]^2$$ Defining $h_{i}$ as such, since $l_{i} (x_{j}) = \delta_{ij}$, $$\begin{cases} h_{i} ( x_{j} ) = \delta_{ij} \\ h’_{i} ( x_{j} ) = 0 \end{cases}$$

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Part 3.

$$\tilde{h}_{i} (x) := (x -x_{i} ) \left[ l_{i}(x) \right]^2$$ Defining $\tilde{h}_{i}$ as such, since $l_{i} (x_{j}) = \delta_{ij}$, $$\begin{cases} \tilde{h}_{i} ( x_{j} ) = 0 \\ \tilde{h}’_{i} ( x_{j} ) = \delta_{ij} \end{cases}$$

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Part 4.

$$H_{n} (x) := \sum_{i=1}^{n} y_{i} h_{i} (x) + \sum_{i=1}^{n} y’_{i} \tilde{h}_{i} (x)$$ Defining $H_{n}$ as such, by Part 2 and Part 3, $$\begin{cases} H_{n}(x_{i}) = y_{i} \\ H'_{n}(x_{i}) = y’_{i} \end{cases}$$

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[3]

Strategy: Start from the Newton’s Divided Differences Formula and directly show that the formula holds.

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According to the Newton’s Divided Differences Formula for $z_{1}, \cdots , z_{2n}$, $$f(x) - p_{2n-1} (x) = (x - z_{1}) \cdots (x - z_{2n}) f [ z_{1} , \cdots , z_{n} , x ]$$ If $z_{2i-1} , z_{2i} \to x_{i}$, then $$\begin{equation} f(x) - p_{2n-1} (x) = (x - x_{1})^2 \cdots (x - x_{n})^2 f [ x_{1} , x_{1} , \cdots , x_{n} , x_{n} , x ] \end{equation}$$ Directly substituting $x = x_{1}, \cdots , x_{n}$ yields $$f(x_{i} ) - p_{2n-1} (x_{i}) = 0$$ Meanwhile, differentiating both sides of $(1)$ yields $$\begin{align*} A(x) :=& (x - x_{1})^2 \cdots (x - x_{n})^2 {{ d } \over { dx }} f [ x_{1} , x_{1} , \cdots , x_{n} , x_{n} , x ] \\ B(x) :=& 2 f [ x_{1} , x_{1} , \cdots , x_{n} , x_{n} , x ] \sum_{i=1}^{n} \left[ (x -x_{i} ) \prod_{i \ne j} (x - x_{j})^2 \right] \end{align*}$$ For which, $$f ' (x) - p '_{2n-1} (x) = A(x) + B(x)$$ Direct substitution of $x = x_{1}, \cdots , x_{n}$ gives $A(x_{i}) = B(x_{i}) = 0$, so $$f ' (x_{i} ) - p '_{2n-1} (x_{i}) = 0$$ Therefore, we obtain $H_{n} (x) = p_{2n-1} (x)$.

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[4]

According to [3], $$f(x) - H_{n} (x) = f [ x_{1} , x_{1} , \cdots , x_{n} , x_{n} , x ]$$

Corollary of the Hermite-Genocchi Formula:

• [4]’: $$f [ \underbrace{ x_{i} , \cdots , x_{i} }_{ n+1 } ] = {{1} \over {n!}} f^{(n)} ( x_{i} )$$

According to the Corollary of the Hermite-Genocchi Formula, for some $\xi_{x} \in \mathscr{H} \left\{ x_{1} , \cdots , x_{n} , x \right\}$, $$f(x) - H_{n} (x) = {{ [ \Psi_{n} (x) ]^2 } \over { (2n)! }} f^{(2n)} ( \xi_{x} )$$

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