Lagrange Error Bound

AP Calculus BC · Infinite Sequences and Series

The Lagrange Error Bound

One more term of the polynomial, with the unknown derivative replaced by an upper bound.

key idea$\big\lvert R_n(x)\big\rvert\le\frac{M\,\lvert x-a\rvert^{\,n+1}}{(n+1)!},\quad M\ge\big\lvert f^{(n+1)}\big\rvert\ \text{on the interval}$

The rule

The Lagrange bound measures how far a Taylor polynomial can stray from the function it approximates. It looks exactly like one more term of the polynomial, with the unknown derivative replaced by an upper bound $M$ — which is the only genuinely new decision it asks you to make.

The remainder

\[R_n(x)=f(x)-P_n(x)\]

The exact error committed by stopping at degree $n$. It is unknown, which is why a bound rather than a value is what the theorem provides.

The bound

\[\big\lvert R_n(x)\big\rvert\le\frac{M\lvert x-a\rvert^{n+1}}{(n+1)!}\]

Here $M$ is any number with $\left\lvert f^{(n+1)}(t)\right\rvert\le M$ for all $t$ between $a$ and $x$. A larger $M$ still gives a valid — if weaker — bound, so it is safe to overestimate.

four ingredients

The bound uses the $(n+1)$st derivative, one past the last one in the polynomial.
$M$ bounds that derivative on the whole interval between $a$ and $x$ — not just at $a$.
$\lvert x-a\rvert^{n+1}$ is the distance from the centre, raised to the matching power.
Overestimating $M$ is safe; underestimating it invalidates the whole bound.

Piece	Where it comes from	Common error
$n+1$	one past the polynomial's degree	using $n$
$M$	a bound on $\left\lvert f^{(n+1)}\right\rvert$ over the interval	using $f^{(n+1)}(a)$ alone
$\lvert x-a\rvert$	distance from the centre to the evaluation point	using $x$ when $a\neq0$
$(n+1)!$	the factorial matching the power	using $n!$

Every one of these is off by one from something. Write $n+1$ in all three places at once — derivative order, exponent, factorial — and the bound assembles itself correctly.

Pattern recognition

Function	$\left\lvert f^{(n+1)}\right\rvert$	Usable $M$
$\sin x$ or $\cos x$	always a sine or cosine	$M=1$, on any interval
$e^{x}$ on $[0,b]$	$e^{t}$, increasing	$M=e^{b}$ — the right endpoint
$\ln x$ near $1$	$\dfrac{n!}{t^{\,n+1}}$	evaluate at the endpoint closest to zero
a polynomial of degree $\le n$	$0$	$M=0$ — the approximation is exact

Row 1 is why trigonometric estimates are the standard exam case: every derivative is bounded by $1$, so $M=1$ always works and the whole bound collapses to $\dfrac{\lvert x-a\rvert^{n+1}}{(n+1)!}$.

Worked examples

EX 1

A trigonometric bound

$M=1$

Bound the error when $P_3(x)$ for $\sin x$ about $0$ is used to estimate $\sin(0.5)$.

Identify $n$ and the derivative order

$n=3$, so the bound uses the fourth derivative.

Bound that derivative

$f^{(4)}(x)=\sin x$, and $\lvert\sin t\rvert\le1$ everywhere, so $M=1$.

Write the bound

$\left\lvert R_3(0.5)\right\rvert\le\dfrac{1\cdot(0.5)^{4}}{4!}$.

Evaluate

$\dfrac{0.0625}{24}\approx0.0026$.

Interpret

The estimate is correct to within about $0.003$ — roughly two decimal places.

EX 2

An exponential bound

$M$ at the far endpoint

Bound the error when $P_2(x)$ for $e^{x}$ about $0$ is used to estimate $e^{0.3}$.

Identify the derivative needed

$n=2$, so the third derivative, which is $e^{t}$.

Bound it on $[0,0.3]$

$e^{t}$ is increasing, so its maximum is $e^{0.3}\approx1.35$. Take $M=1.35$.

Write the bound

$\left\lvert R_2(0.3)\right\rvert\le\dfrac{1.35\,(0.3)^{3}}{3!}$.

Evaluate

$\dfrac{1.35\cdot0.027}{6}\approx0.0061$.

Note the safe simplification

Using $M=2$ instead gives $0.009$ — larger, but still a valid bound. Overestimating $M$ never breaks the inequality.

EX 3

How many terms are needed?

solve for $n$

What degree Maclaurin polynomial guarantees $\cos(1)$ to within $0.001$?

Set up the bound

All derivatives of $\cos$ are bounded by $1$, so $M=1$ and the bound is $\dfrac{1^{\,n+1}}{(n+1)!}=\dfrac{1}{(n+1)!}$.

Write the inequality

$\dfrac{1}{(n+1)!}\lt0.001$, that is $(n+1)!\gt1000$.

Test factorials

$6!=720$ is too small; $7!=5040$ works.

Solve for $n$

$n+1=7$, so $n=6$.

State the answer

$P_6$ suffices. Note that $P_5$ and $P_6$ are the same polynomial for $\cos$, since the $x^{5}$ coefficient vanishes — the bound is still stated with $n=6$.

EX 4

Why $M$ must cover the interval

not just the centre

A student bounding $P_2$ for $e^{x}$ on $[0,2]$ takes $M=f'''(0)=1$. Explain the error.

State what $M$ must do

It must satisfy $\left\lvert f'''(t)\right\rvert\le M$ for every $t$ between $0$ and $2$.

Test the claim

$f'''(2)=e^{2}\approx7.39$, which exceeds $1$.

Name the consequence

The hypothesis fails, so the resulting inequality is not a bound at all — the true error may be far larger.

Repair it

Take $M=e^{2}\approx7.39$, giving $\left\lvert R_2(2)\right\rvert\le\dfrac{7.39\cdot8}{6}\approx9.85$.

Read the result honestly

A useless-looking bound, but a true one. Far from the centre, low-degree Taylor polynomials genuinely are poor.

Watch out

Everything is $n+1$

Derivative order, exponent, and factorial all use $n+1$. Getting two of the three right and one wrong is the usual failure.

$M$ bounds the derivative over the whole interval

Evaluating $f^{(n+1)}$ at the centre is not enough. Find the maximum of its absolute value between $a$ and $x$.

Overestimate $M$ freely

A larger $M$ gives a weaker but still valid bound. If the resulting bound meets the required accuracy, the argument is complete.

Use the distance from the centre

The factor is $\lvert x-a\rvert^{n+1}$. For a Maclaurin polynomial $a=0$ and this is just $\lvert x\rvert^{n+1}$ — which is why the distinction is so easy to forget elsewhere.

Course map. Topic 10.12 · LO LIM-8.C — determine the error bound associated with a Taylor polynomial approximation. BC-only.