Hermite Interpolation

Standard polynomial interpolation matches function values at a set of nodes. But what if you also know the derivative at each node? Hermite interpolation incorporates both, effectively doubling the number of conditions — and by the Weierstrass principle, a higher-degree polynomial means lower error. Hermite interpolation achieves a significantly higher degree polynomial without requiring any additional data points.

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Motivation

Recall from the Weierstrass theorem: higher degree → smaller error. The bottleneck with standard interpolation is that each node contributes only one condition (the function value), capping the degree at $n$ for $n+1$ nodes. Hermite interpolation unlocks additional conditions by also matching the derivative at each node. With the same dataset, the polynomial’s degree more than doubles.

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Conditions and Degree

For $n+1$ nodes $x_0, x_1, \ldots, x_n$, Hermite interpolation imposes:

Condition Type	Count
Function values: $P(x_k) = f(x_k)$	$n+1$
Derivative values: $P'(x_k) = f'(x_k)$	$n+1$
Total conditions	$\mathbf{2n+2}$

A polynomial is uniquely determined by its coefficients. Matching $2n+2$ conditions requires exactly $2n+2$ free coefficients, which means a polynomial of degree $\mathbf{2n+1}$.

Nodes	Standard Degree	Hermite Degree
2 ($n=1$)	1	3 (cubic)
3 ($n=2$)	2	5 (quintic)
4 ($n=3$)	3	7
$n+1$	$n$	$2n+1$

Note

For the same set of nodes, Hermite interpolation achieves roughly twice the polynomial degree. Because the $(n+1)!$ factor in the error bound grows with degree, the interpolation error shrinks substantially — with no extra data points required.

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The Hermite Polynomial Formula

$P_{2n+1}(x) = \sum_{k=0}^{n} \Bigl[\,h_k(x)\,f(x_k) \;+\; \hat{h}_k(x)\,f'(x_k)\Bigr]$

where the two families of basis polynomials are built directly from the Lagrange basis $L_k(x)$:

$h_k(x) = \bigl[1 - 2(x - x_k)\,L_k'(x_k)\bigr]\,\bigl[L_k(x)\bigr]^2$

$\hat{h}_k(x) = (x - x_k)\,\bigl[L_k(x)\bigr]^2$

Why Two Families?

The two basis families are designed to never interfere with each other:

Basis	Property	Ensures
$h_k(x_j) = \delta_{kj}$	Equals 1 at own node, 0 elsewhere	Matches function value $f(x_k)$
$h_k'(x_j) = 0$ for all $j$	Zero derivative at every node	Does not disturb derivative conditions
$\hat{h}_k(x_j) = 0$ for all $j$	Zero value at every node	Does not disturb function value conditions
$\hat{h}_k'(x_j) = \delta_{kj}$	Derivative equals 1 at own node, 0 elsewhere	Matches derivative $f'(x_k)$

The $h_k$ terms carry the function values; the $\hat{h}_k$ terms carry the derivative values. Neither family disrupts the other’s conditions.

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Worked Example

Problem: Consider the following data points. Compute all Hermite basis elements and evaluate the Hermite interpolation polynomial.

$x$	$f(x)$	$f'(x)$
$0$	$0$	$1$
$\pi$	$1$	$0$

(i) Compute all Hermite basis elements

First, we find the standard Lagrange basis polynomials, $l_0(x)$ and $l_1(x)$, and their derivatives:

For $x_0 = 0$:

$l_0(x) = \frac{x - x_1}{x_0 - x_1} = \frac{x - \pi}{0 - \pi} = 1 - \frac{x}{\pi}$

$l_0'(x) = -\frac{1}{\pi}$

For $x_1 = \pi$:

$l_1(x) = \frac{x - x_0}{x_1 - x_0} = \frac{x - 0}{\pi - 0} = \frac{x}{\pi}$

$l_1'(x) = \frac{1}{\pi}$

Next, we use these to compute the Hermite basis elements $h_0, h_1, \hat{h}_0,$ and $\hat{h}_1$:

Computing $h_0(x)$:

$h_0(x) = l_0^2(x)[1 - 2(x - x_0)l_0'(x_0)] = \left(1 - \frac{x}{\pi}\right)^2 \left[1 - 2(x - 0)\left(-\frac{1}{\pi}\right)\right] = \left(1 - \frac{x}{\pi}\right)^2 \left[1 + \frac{2x}{\pi}\right]$

Computing $h_1(x)$:

$h_1(x) = l_1^2(x)[1 - 2(x - x_1)l_1'(x_1)]= \left(\frac{x}{\pi}\right)^2 \left[1 - 2(x - \pi)\left(\frac{1}{\pi}\right)\right] = \frac{x^2}{\pi^2} \left[1 - \frac{2(x - \pi)}{\pi}\right]$

$= \frac{x^2}{\pi^2} \left[1 - \frac{2x}{\pi} + 2\right] = \frac{x^2}{\pi^2} \left[3 - \frac{2x}{\pi}\right]$

Computing $\hat{h}_0(x)$:

$\hat{h}_0(x) = (x - x_0)(l_0(x))^2 = (x - 0)\left(1 - \frac{x}{\pi}\right)^2 = x\left(1 - \frac{x}{\pi}\right)^2$

Computing $\hat{h}_1(x)$:

$\hat{h}_1(x) = (x - x_1)(l_1(x))^2 = (x - \pi)\left(\frac{x}{\pi}\right)^2$

(ii) Evaluate the Hermite interpolation polynomial

Since there are 2 nodes ($n=1$), the polynomial will be of degree 3 ($2n+1=3$). Using the formula:

$P_3(x) = f(x_0)h_0(x) + f'(x_0)\hat{h}_0(x) + f(x_1)h_1(x) + f'(x_1)\hat{h}_1(x)$

Substitute the given data $f(0)=0$, $f'(0)=1$, $f(\pi)=1$, $f'(\pi)=0$:

$P_3(x) = (0)h_0(x) + (1)\hat{h}_0(x) + (1)h_1(x) + (0)\hat{h}_1(x)$

The first and last terms zero out, leaving:

$P_3(x) = \hat{h}_0(x) + h_1(x)$

Substituting our evaluated bases back in gives the final polynomial:

$P_3(x) = x\left(1 - \frac{x}{\pi}\right)^2 + \frac{x^2}{\pi^2}\left(3 - \frac{2x}{\pi}\right)$

Advantage Over Standard Interpolation

Method	Data Used	Degree Achieved	Error Bound
Standard interpolation	$n+1$ function values	$n$	Higher
Hermite interpolation	$n+1$ values $+$ $n+1$ derivatives	$2n+1$	Lower

For a fixed set of $n+1$ data points, Hermite interpolation achieves a polynomial of degree $2n+1$ rather than $n$. The error formula from the previous section shows that a higher-degree interpolant has a larger $(n+1)!$ damping factor, which leads directly to a tighter upper bound on the approximation error.

Caution

Hermite interpolation requires knowledge of $f'(x_k)$ at every node. If derivative values are unavailable — for example, in pure data-fitting scenarios where only measurements are given — standard interpolation or spline methods must be used instead.