Interpolation Error

An interpolating polynomial matches the data at the nodes. How far is it from the function in between? The answer looks just like the Taylor remainder, with one important new piece, and it uses the polynomial from Newton's Divided Differences or any other form, since they are all the same polynomial.

Idea clave

If $f$ has $n + 1$ continuous derivatives and $p$ (degree at most $n$) interpolates $f$ at $x_0, \ldots, x_n$, then for each $x$ there is a $\xi$ in the smallest interval containing $x$ and the nodes with $$f(x) - p(x) = \frac{f^{(n+1)}(\xi)}{(n + 1)!}\,\omega(x),\qquad \omega(x) = (x - x_0)(x - x_1)\cdots(x - x_n).$$

Por qué funciona

Fix an $x$ that is not a node and let $g(t) = f(t) - p(t) - \big(f(x) - p(x)\big)\frac{\omega(t)}{\omega(x)}$. Then $g$ is $0$ at the $n + 1$ nodes and at $t = x$: $n + 2$ zeros. By Rolle's theorem $g'$ has $n + 1$ zeros, $g''$ has $n$, and so on, until $g^{(n+1)}$ has at least one zero $\xi$. But $p^{(n+1)} = 0$ and $\omega^{(n+1)} = (n + 1)!$, so $0 = f^{(n+1)}(\xi) - \big(f(x) - p(x)\big)\frac{(n + 1)!}{\omega(x)}$. Solve for $f(x) - p(x)$.

Two factors control the error. The derivative factor depends on $f$; you cannot change it. The node product $\omega(x)$ depends only on where the nodes are: it is small among the nodes and grows quickly outside them, and for equally spaced nodes it is much larger near the ends of the interval than in the middle.

Ejemplo resuelto: An exact error

Let $p$ interpolate $f(x) = x^{3}$ at $0, 1, 2$. Find $f(3) - p(3)$ and then $p(3)$, without finding $p$.

Here $f'''(x) = 6$ is constant, so the formula is exact: $f(x) - p(x) = \frac{6}{3!}x(x - 1)(x - 2) = x(x - 1)(x - 2)$.
At $x = 3$: $3 \cdot 2 \cdot 1 = 6$, so $p(3) = 27 - 6 = 21$.
Check: $p(x) = 3x^{2} - 2x$ fits $(0, 0), (1, 1), (2, 8)$, and $p(3) = 21$.

Respuesta: $6$, and $p(3) = 21$

Ejemplo resuelto: A bound

Bound the error at $x = 0.25$ when $p$ interpolates $\sin x$ at $0, 0.5, 1$.

Every derivative of $\sin$ is at most 1 in size, so $|f'''(\xi)| \le 1$.
$|\omega(0.25)| = |0.25| \cdot |{-0.25}| \cdot |{-0.75}| = 0.046875$.
Bound: $\frac{1}{3!}(0.046875) = 0.0078125$. The actual error is about $0.0070$, close to the bound.

Respuesta: at most $0.0078$

Ejemplo resuelto: Linear interpolation in a table

A table of $e^{x}$ on $[0, 1]$ has spacing $h$. How small must $h$ be for linear interpolation to be good to $10^{-6}$?

Between neighbors $x_0$ and $x_1 = x_0 + h$, $|\omega(x)| = |(x - x_0)(x - x_1)|$ is largest at the midpoint, where it is $\frac{h^{2}}{4}$.
With $|f''| \le e$: error $\le \frac{e}{2} \cdot \frac{h^{2}}{4} = \frac{eh^{2}}{8}$.
$\frac{eh^{2}}{8} \le 10^{-6}$ gives $h \le \sqrt{8 \times 10^{-6}/e} \approx 0.0017$.

Respuesta: $h \lesssim 0.0017$

Below the graph: the error f − p (red) and the node product ω(x), scaled (dashed). For a function like eˣ, whose derivatives barely change, the error has almost exactly the shape of ω.

Error común

Using the $n$th derivative instead of the $(n + 1)$st, or dropping the product $\omega(x)$. With $n + 1$ nodes the formula has $(n + 1)!$, $f^{(n+1)}$ and $n + 1$ factors in $\omega$; the three counts match. And $\omega$ matters: it is largest near the ends of the interval, which is where interpolation errors concentrate.

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