Error in polynomial interpolation: Question bank

Intuition How to use this bank

These are the conceptual landmines of the interpolation error topic — the ones that trip people even after they can plug numbers into the formula. Each line is a question ::: answer reveal. Cover the answer, commit to a position and a reason, then check. If your reason was wrong even when your answer was right, the item still caught you.

Notation used everywhere below. We work on a closed interval $[a, b]$ — this is the domain: it contains the evaluation point $x$ and all the nodes $x_{0}, x_{1}, \dots, x_{n}$ . The master formula is $E (x) = f (x) - p_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} ω (x), ω (x) = \prod_{i = 0}^{n} (x - x_{i}),$ where $f$ is the true function, $p_{n}$ the degree- $\leq n$ polynomial through the $n + 1$ nodes, $ω$ the node polynomial that is zero at every node, and $ξ$ some point in $(a, b)$ that merely exists — we are never handed its value. The pictures below fix these ideas visually before the traps begin.

Figure — Error in polynomial interpolation

The two panels above show the two independent knobs. On the left, the node polynomial $ω (x)$ : it is pinned to zero at each node and swings widest between nodes — especially near the endpoints for equal spacing. On the right, why moving nodes toward the ends (Chebyshev nodes) flattens those swings.

And the classic failure to keep in mind: raising the degree with equally spaced nodes makes the true error grow near the endpoints — the Runge phenomenon — even though the function is perfectly smooth.

True or false — justify

TRUE / FALSE — At every node

x_{i}

, the interpolation error is exactly zero.

True. By construction

p_{n} (x_{i}) = f (x_{i})

, so

E (x_{i}) = f (x_{i}) - p_{n} (x_{i}) = 0

; equivalently

ω (x_{i}) = 0

because the factor

(x - x_{i})

vanishes there.

TRUE / FALSE — Between two adjacent nodes the error must also be zero somewhere.

False in general. The error is guaranteed zero at the nodes but is typically nonzero throughout the open interval between them — that "guessing gap" is the whole reason the topic exists.

TRUE / FALSE — Adding one more node always shrinks the maximum error.

False. Adding a node raises

n

n + 1

, so there are now

n + 2

nodes and the error term uses one order higher derivative,

f^{(n + 2)} / (n + 2)!

times the new

ω

(each extra node contributes an extra factor and an extra required derivative). If

max ∣ f^{(n + 2)} ∣

grows faster than the factorial (and

max ∣ ω ∣

blows up at the ends, as for equally spaced nodes), the bound and the true error can grow — this is the Runge phenomenon.

TRUE / FALSE — The factor

\frac{1}{( n + 1 )!}

guarantees the error goes to zero as

n \to \infty

False. The factorial alone does not decide the limit; it fights against

max ∣ f^{(n + 1)} ∣

, which can grow super-factorially, and against

max ∣ ω ∣

. Convergence depends on all three factors together, not on the factorial in isolation.

TRUE / FALSE — For a polynomial

f

of degree

\leq n

, the interpolation error is identically zero.

True. Then

f^{(n + 1)} \equiv 0

, so the whole error term vanishes for every

x

; equivalently

p_{n}

f

because the interpolating polynomial of degree

\leq n

through

n + 1

points is unique.

TRUE / FALSE — The single point

ξ

in the formula is the same for every

x

False.

ξ = ξ (x)

moves as

x

moves; the theorem only promises that for each

x

some

ξ \in (a, b)

works. Treating it as one fixed number is a classic error.

TRUE / FALSE — The error formula requires

f

to have

n + 1

continuous derivatives.

True. The hypothesis is

f \in C^{n + 1} [a, b]

; the derivation differentiates the helper function

n + 1

times via Rolle, so

f^{(n + 1)}

must exist and be continuous for the argument to land.

TRUE / FALSE — Choosing better nodes can reduce the

f^{(n + 1)}

factor.

False. The derivative

f^{(n + 1)}

belongs to the function itself and is untouched by node placement; you only control

ω (x)

through where you sample. Node choice fixes the "distance from data" knob, never the "twistiness" knob.

TRUE / FALSE — Equally spaced nodes minimise

max ∣ ω (x) ∣

over

[a, b]

False. They actually make

∣ ω ∣

blow up near the endpoints; the placement that minimises

max ∣ ω ∣

clusters points toward the ends — the Chebyshev nodes.

TRUE / FALSE — Interpolation error and Taylor's remainder are unrelated results.

False. They share the identical

\frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!}

factor; Taylor uses one node repeated

n + 1

times

(x - x_{0})^{n + 1}

, interpolation spreads the same degree across distinct nodes

\prod (x - x_{i})

. Interpolation is "Taylor with derivatives traded for data."

Spot the error

WRONG: "

E (x) = f^{(n + 1)} (ξ) ω (x)

." What is missing?

The

\frac{1}{( n + 1 )!}

denominator. The

n + 1

Rolle differentiations turn

ω

into

ω^{(n + 1)} = (n + 1)!

, and that constant lands in the denominator — dropping it overstates the error by a factor of

(n + 1)!

WRONG: "Since

E (x_{i}) = 0

at nodes, the polynomial equals

f

everywhere, so error is always zero." Where's the flaw?

Matching at finitely many points does not force agreement between them.

E

is only pinned to zero at the

n + 1

nodes; the free "wiggle"

ω (x)

is exactly what measures the disagreement off the nodes.

WRONG: "I'll compute

ξ

by solving

g^{(n + 1)} (ξ) = 0

and plug it in for an exact error." What's wrong operationally?

The theorem is an existence statement;

ξ

is generally not computable in closed form and depends on

x

. In practice you replace

f^{(n + 1)} (ξ)

M_{n + 1} = max_{t \in [a, b]} ∣ f^{(n + 1)} (t) ∣

, the largest magnitude of the

(n + 1)

-th derivative over the domain, to get a usable bound, not an exact value.

WRONG: "For linear interpolation the worst error sits at a node." Correct it.

For

n = 1

ω (x) = (x - x_{0}) (x - x_{1})

is a downward parabola with roots at the nodes, so

∣ ω ∣

is largest at the midpoint, giving

∣ ω ∣_{m a x} = h^{2} /4

and

max ∣ E ∣ \leq \frac{h ^{2}}{8} max ∣ f^{''} ∣

. The nodes are where error is smallest (zero), not largest.

WRONG: "The formula needs

ξ

inside

[x_{0}, x_{n}]

specifically." Fix the interval.

ξ

is only guaranteed to lie between

min {x, x_{0}, \dots, x_{n}}

and

max {x, x_{0}, \dots, x_{n}}

— the smallest interval containing the evaluation point and every node. If

x

sits outside the outermost nodes (extrapolation),

ξ

can lie beyond

[x_{0}, x_{n}]

too.

WRONG: "

ω^{(n + 1)} (t)

depends on

t

, so it changes across the interval." Correct it.

ω (t) = t^{n + 1} + (lower degree)

, and the

(n + 1)

-th derivative of a degree-

(n + 1)

polynomial is the constant

(n + 1)!

. All the lower-degree terms differentiate away, so

ω^{(n + 1)}

is a pure constant.

Why questions

WHY does the derivation invent the auxiliary

g (t) = f (t) - p_{n} (t) - K ω (t)

, and what is the constant

K

K

is chosen to make

g

vanish at one extra chosen point

\overset{x}{ˉ}

(the fixed evaluation point where we want the error): set

K = \frac{f ( x ˉ ) - p _{n} ( x ˉ )}{ω ( x ˉ )}

, i.e.

K

is the unknown error-per-unit-

ω

we are trying to identify. With that choice

g

has

n + 2

known zeros (the

n + 1

nodes plus

\overset{x}{ˉ}

), exactly what Rolle needs to be applied

n + 1

times — a lever

E

alone doesn't give you.

WHY do we use a new symbol

\overset{x}{ˉ}

(bar-x) rather than the page's

x

\overset{x}{ˉ}

is one frozen evaluation point held fixed while the auxiliary function

g (t)

runs over a free variable

t

; keeping the names separate stops us confusing "the place we are measuring the error" (

\overset{x}{ˉ}

) with "the running variable Rolle chases zeros in" (

t

). Once

K

is pinned we rename

\overset{x}{ˉ}

back to

x

WHY does applying Rolle

n + 1

times produce precisely

f^{(n + 1)}

and not some other derivative order?

Each Rolle step reduces the guaranteed zero count by one, so starting from

n + 2

zeros you can differentiate exactly

n + 1

times before you are down to a single guaranteed zero

ξ

; that final order matches the number of nodes-plus-one and pins the factor.

WHY does

p_{n}^{(n + 1)} \equiv 0

appear in the derivation, and what does it accomplish?

A polynomial of degree

\leq n

has zero

(n + 1)

-th derivative, so

p_{n}

silently drops out of

g^{(n + 1)}

, leaving only

f^{(n + 1)} (ξ) - K (n + 1)!

— which is what lets us solve cleanly for

K

WHY do "wiggly" functions interpolate badly even with many nodes?

Wiggliness means large high-order derivatives

∣ f^{(n + 1)} ∣

, and that factor multiplies the whole error; no amount of clever node placement touches it, so the roller-coaster function fights every polynomial fit.

WHY do Chebyshev nodes beat equally spaced nodes for the worst-case error?

They are chosen to minimise

max_{[a, b]} ∣ ω (x) ∣

, flattening the node polynomial's peaks (which for equal spacing spike near the endpoints), so the one factor you can control is made as small as possible.

WHY is the error

O (h^{n + 1})

for

n + 1

equally spaced nodes with spacing

h

The node polynomial

ω (x) = \prod (x - x_{i})

is a product of

n + 1

factors each of size

O (h)

, so

∣ ω ∣ = O (h^{n + 1})

; multiplied by the bounded derivative factor, the error scales like

h^{n + 1}

WHY can't we simply keep raising the degree to reach any accuracy we like?

Because increasing

n

can inflate both

max ∣ f^{(n + 1)} ∣

and (for equal spacing)

max ∣ ω ∣

faster than

(n + 1)!

suppresses them, producing divergence near the endpoints — the Runge warning that piecewise/spline methods sidestep.

Sign of the error

WHY is the sign of the error at a given

x

, not just its size, worth thinking about?

Because

E (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} ω (x)

is a product, so its sign is (sign of

f^{(n + 1)} (ξ)

)

\times

(sign of

ω (x)

); knowing whether

p_{n}

over- or under-shoots

f

often matters as much as the magnitude.

SIGN: For linear interpolation (

n = 1

) of a convex function (

f^{''} > 0

), does the line lie above or below

f

between the nodes?

Below. Between the two nodes

ω (x) = (x - x_{0}) (x - x_{1}) < 0

and

f^{''} (ξ) > 0

, so

E = f - p_{1} = \frac{f ^{''} ( ξ )}{2} ω (x) < 0

, meaning

f < p_{1}

: the chord sits above the convex curve, exactly the picture of a chord over a bowl.

SIGN: How does

sign ω (x)

behave as

x

crosses a node?

Each factor

(x - x_{i})

flips sign as

x

passes

x_{i}

, so

ω

alternates sign in successive between-node intervals; the error therefore tends to swing over-then-under-then-over as you sweep across the nodes.

SIGN: If

f^{(n + 1)}

keeps a constant sign on

[a, b]

, what can you say about where

p_{n}

over- and under-estimates

f

The over/under pattern is then dictated entirely by

ω (x)

's alternating sign, so

p_{n}

crosses

f

near each node in a predictable ripple — useful for guessing the shape of the error curve without computing it.

Edge cases

EDGE: What is the error when the evaluation point

x

coincides with a node?

It is exactly zero, because

ω (x) = 0

there; the derivation's construction of

K

explicitly sets this trivial case aside and handles only non-nodes.

EDGE: What happens to the formula if two nodes are allowed to merge,

x_{1} \to x_{0}

The distinct-node hypothesis breaks; in the limit the repeated node forces

p_{n}

to match a derivative too (Hermite interpolation), and

ω

gains a squared factor

(x - x_{0})^{2}

— the interpolation picture bends toward Taylor.

EDGE: Is the error formula valid when

x

lies outside the node interval (extrapolation)?

The formula still holds with

ξ

between

min {x, x_{0}, \dots, x_{n}}

and

max {x, x_{0}, \dots, x_{n}}

, but

∣ ω (x) ∣

grows very fast beyond the outermost nodes, so the bound explodes — extrapolation is where interpolation error is least trustworthy.

EDGE: What does the error term say when

f

is a constant function?

Every derivative

f^{(n + 1)} = 0

, so

E (x) = 0

for all

x

and any node set — a constant is fit perfectly by even a single node, matching intuition.

EDGE: With only one node (

n = 0

, so

p_{0}

is a constant), what is the error term?

It becomes

E (x) = f^{'} (ξ) (x - x_{0})

, which is exactly the Mean Value Theorem statement — the

n = 0

interpolation error is the MVT, the base case the whole formula generalises.

EDGE: If

f^{(n + 1)}

changes sign on

[a, b]

, does the bound

∣ E ∣ \leq \frac{M _{n + 1}}{( n + 1 )!} ∣ ω ∣

still hold?

Yes; the bound uses

M_{n + 1} = max_{[a, b]} ∣ f^{(n + 1)} ∣

, the largest magnitude regardless of sign, so sign changes only make the true error smaller than the bound, never larger.

EDGE: Does the error formula apply to the connected topic of numerical integration error?

Yes — integrating the interpolation error term over the interval is precisely how quadrature error formulas (e.g. trapezoid's

O (h^{2})

, Simpson's

O (h^{4})

) are derived, inheriting the same

f^{(n + 1)}

-and-

ω

structure.

Recall One-line survival summary

Error = (twistiness $f^{(n + 1)}$ ) $\times$ (distance-from-data $ω$ ) $\div$ (softening factorial). You control only $ω$ ; $ξ$ merely exists and moves with $x$ ; the sign comes from $f^{(n + 1)}$ times $ω$ ; more nodes is not always better.