Interpolation Error & Cauchy's Theorem

We know how to build an interpolating polynomial through a set of nodes. But between the nodes, how far can the polynomial stray from the true function? Cauchy’s theorem — a direct consequence of the Taylor remainder — gives a precise upper bound on this error without requiring us to know the exact answer first.

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The Error Formula

Theorem (Cauchy — Interpolation Error). Let $f$ be $n+1$ times differentiable on $[a,b]$, and let $P_n(x)$ be the polynomial interpolating $f$ at the $n+1$ distinct nodes $x_0, x_1, \ldots, x_n \in [a,b]$. Then for any $x \in [a,b]$, there exists $\xi \in (a,b)$ such that:

$f(x) - P_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}\,(x-x_0)(x-x_1)\cdots(x-x_n)$

Taking absolute values and bounding $|f^{(n+1)}(\xi)|$ by its maximum over the interval:

$|f(x) - P_n(x)| \le \frac{\max_{\xi \in [a,b]}|f^{(n+1)}(\xi)|}{(n+1)!} \cdot |W(x)|$

where $W(x) = (x-x_0)(x-x_1)\cdots(x-x_n)$ is the node product polynomial.

Note

The error has two independent components: the derivative bound (how wildly the function behaves beyond degree $n$) and the node product $|W(x)|$ (which depends on where the nodes sit). Both must be small for the error to be small — and the node product can be controlled by choosing nodes wisely.

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Worked Example — Bounding the Error for $\cos(x)$

Problem: Find the upper bound on the interpolation error for $f(x) = \cos(x)$ on $[-1, 1]$, using three equally spaced nodes.

Node	Position
$x_0$	$-\pi/4$
$x_1$	$0$
$x_2$	$\pi/4$

Step 1 — Identify the Error Derivative

With $n = 2$ (three nodes, degree-2 polynomial), we need the $(n+1) = 3$rd derivative of $\cos(x)$:

$f'(x) = -\sin(x), \qquad f''(x) = -\cos(x), \qquad f'''(x) = \sin(x)$

Step 2 — Bound $|f'''(\xi)|$ on $[-1, 1]$

$\max_{\xi \in [-1,1]} |f'''(\xi)| = \max_{\xi \in [-1,1]} |\sin(\xi)| = \sin(1) \approx 0.8415$

Step 3 — Construct the Node Product $W(x)$

$W(x) = \left(x + \frac{\pi}{4}\right)(x - 0)\left(x - \frac{\pi}{4}\right) = x\!\left(x^2 - \frac{\pi^2}{16}\right) = x^3 - \frac{\pi^2}{16}\,x$

Step 4 — Find the Maximum of $|W(x)|$ on $[-1, 1]$

To find the absolute maximum of $|W(x)|$ on a closed interval, we must evaluate the function at its critical points and its endpoints, then compare the magnitudes.

First, set $W'(x) = 0$ to locate the critical points inside the interval:

$W'(x) = 3x^2 - \frac{\pi^2}{16} = 0 \implies x = \pm\frac{\pi}{4\sqrt{3}} \approx \pm 0.4534$

Next, we evaluate $W(x) = x^3 - \frac{\pi^2}{16}x$ at these critical points and at the endpoints ($x = -1$ and $x = 1$):

$x$	$W(x)$	$\lvert W(x) \rvert$
$-1$ (Endpoint)	$-0.3831$	$0.3831$
$-\dfrac{\pi}{4\sqrt{3}}$ (Critical Point)	$0.1865$	$0.1865$
$\dfrac{\pi}{4\sqrt{3}}$ (Critical Point)	$-0.1865$	$0.1865$
$1$ (Endpoint)	$0.3831$	$0.3831$

Comparing the magnitudes, the maximum value of $|W(x)|$ does not occur at the critical points where the curve turns, but rather at the extreme edges of the interval.

Therefore:

$\max_{x \in [-1,1]} |W(x)| \approx 0.3831$

Step 5 — Compute the Upper Bound

$|f(x) - P_2(x)| \le \frac{|f'''(\xi)|_{\max}}{3!} \cdot |W(x)|_{\max} = \frac{0.8415}{6} \times 0.3831 \approx \mathbf{0.0537}$

Note

This bound of $0.0537$ holds for all $x \in [-1,1]$ — no matter where between the nodes you evaluate the interpolant, the true value of $\cos(x)$ and the polynomial $P_2(x)$ cannot differ by more than about $0.054$.

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Key Takeaways

Quantity	Role in the Error
$\max\lvert f^{(n+1)}(\xi) \rvert$	How rapidly the function varies beyond degree $n$
$(n+1)!$	Grows quickly, dampens error as degree increases
$\max\lvert W(x) \rvert$	Controlled by node placement (the motivation for Chebyshev nodes)

To reduce interpolation error you can:

1. Increase the degree $n$ : Adds more $(n+1)!$ damping in the denominator and reduces $\lvert W(x) \rvert$
2. Choose better nodes : Chebyshev nodes minimise $\max\lvert W(x) \rvert$ among all possible placements of $n+1$ nodes in $[a,b]$, and are covered in Section 8