Taylor's remainder theorem — error estimation: Worked examples

Before anything, a one-line refresher of the only three moving parts, so no symbol appears un-earned:

Recall The three ingredients (click to reveal)

$M$ = a number you know is at least as big as $∣ f^{(n + 1)} ∣$ everywhere between $a$ and $x$ . It is the "how wildly does the curve bend?" number.
$(n + 1)!$ = factorial one step beyond the polynomial degree $n$ . Factorial $= 1 \cdot 2 \cdot 3 \dots (n + 1)$ , a number that grows explosively and crushes the error.
$∣ x - a ∣^{n + 1}$ = how far you walked from the centre, raised to that same $n + 1$ . The vertical bars $∣ \cdot ∣$ mean "distance, always positive" — so a leftward walk ( $x < a$ ) counts the same as rightward.

Two traps recur in every example below. State them once, up front, so we can cite them by name:

The scenario matrix

Every exam question about Taylor error is one of these cells. Each worked example below is tagged with the cell it fills.

Cell	What makes it tricky	Example
C1 Small $x > 0$ , centre $a = 0$	the "easy" baseline	(1)
C2 Negative $x < 0$	does the sign of $x$ break the bound?	(2)
C3 Centre $a \neq = 0$	the $	x-a
C4 Large $	x-a	>1$
C4b Boundary $	x-a	=1$ exactly
C5 Degenerate: $x = a$	zero-distance walk — what is the error?	(5)
C6 Limiting: $n \to \infty$	does the bound $\to 0$ ? (convergence link)	(6)
C7 Real-world word problem with units	translate "accurate to X mm" into a bound	(7)
C8 Exam twist: forgotten power / wrong factorial	Trap B, live	(8)

Worked examples

(1) Cell C1 — small positive $x$ , the baseline

Worked example Bound the error of

$P_{3}$ for $cos x$ at $x = 0.3$ , centre $a = 0$ . Forecast first: guess — will the error be around $1 0^{- 2}$ , $1 0^{- 4}$ , or $1 0^{- 6}$ ? Hold that thought.

Step 1. Write the polynomial. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , so $P_{3} (x) = 1 - \frac{x ^{2}}{2}$ (the $x^{3}$ term is zero). Why this step? We need $P_{n}$ to know which derivative comes next. Here $n = 3$ , so the remainder uses $f^{(4)}$ (this is Trap B done right — derivative is $n + 1 = 4$ ).

Step 2. Find $f^{(4)}$ and its bound $M$ . Derivatives of $cos$ cycle $- sin, - cos, sin, cos$ , so $f^{(4)} (t) = cos t$ . On $[0, 0.3]$ , $∣ cos t ∣ \leq 1$ , so $M = 1$ . Why this step? $M$ must cover the worst bend on the whole walk, not one guessed point (Trap A). Here $cos t$ is monotonic (decreasing) on $[0, 0.3]$ , so no interior hump to worry about; the safe blanket bound $∣ cos ∣ \leq 1$ works anyway.

Step 3. Plug in. With $n = 3$ , $a = 0$ , $x = 0.3$ : $∣ R_{3} (0.3) ∣ \leq \frac{1}{4 !} (0.3)^{4} = \frac{0.0081}{24} \approx 3.375 \times 1 0^{- 4} .$ Why this step? Next derivative ( $M = 1$ ), next factorial ( $4!$ ), next power ( $(0.3)^{4}$ ) — all $n + 1 = 4$ .

Verify: true $cos 0.3 = 0.955336$ , and $P_{3} (0.3) = 1 - 0.045 = 0.955$ . Actual error $\approx 3.4 \times 1 0^{- 4}$ — right at the bound (because $cos$ is nearly $1$ here, so $M = 1$ is almost exact). Your forecast should have been $\sim 1 0^{- 4}$ . ✓

(2) Cell C2 — negative $x$ , does the sign break it?

Worked example Bound the error of

$P_{2}$ for $e^{x}$ at $x = - 0.4$ , centre $a = 0$ . Forecast: the walk is leftward. Will the bound be smaller, bigger, or the same as walking right to $+ 0.4$ ?

Step 1. $P_{2} (x) = 1 + x + \frac{x ^{2}}{2}$ , so $P_{2} (- 0.4) = 1 - 0.4 + 0.08 = 0.68$ . Why this step? Standard setup; $n = 2$ means we need $f^{(3)}$ .

Step 2. $f^{(3)} (t) = e^{t}$ . On the interval between $a = 0$ and $x = - 0.4$ (i.e. $[- 0.4, 0]$ ), $e^{t}$ is strictly increasing, hence monotonic, so its max is at the right end $t = 0$ : $M = e^{0} = 1$ . Why this step? Because $e^{t}$ is monotonic here, the endpoint rule is safe — no interior extremum exists. The sign of $x$ only changed which end is worst.

Step 3. The bound uses $∣ x - a ∣^{n + 1} = ∣ - 0.4 ∣^{3} = (0.4)^{3}$ , distance only: $∣ R_{2} (- 0.4) ∣ \leq \frac{1}{3 !} (0.4)^{3} = \frac{0.064}{6} \approx 1.067 \times 1 0^{- 2} .$ Why this step? The bars $∣ \cdot ∣$ erase the minus sign — the size of the error depends only on distance, not direction. That's the answer to the forecast: same as $+ 0.4$ only if $M$ is the same. Here $M (+ 0.4) = e^{0.4} \approx 1.49$ would give a bigger bound, so leftward is actually tighter.

Verify: true $e^{- 0.4} = 0.670320$ , actual error $= 0.68 - 0.670320 = 9.68 \times 1 0^{- 3} < 1.067 \times 1 0^{- 2}$ . ✓ The bound holds and the sign did not break it.

(3) Cell C3 — centre $a \neq = 0$

Worked example Approximate

$ln (1.2)$ using $P_{2}$ of $ln x$ centred at $a = 1$ . Bound the error. Forecast: the centre is $1$ , the target is $1.2$ . What is $∣ x - a ∣$ ? (This is the whole point of the cell.)

Step 1. Derivatives of $f (x) = ln x$ : $f^{'} = \frac{1}{x}$ , $f^{''} = - \frac{1}{x ^{2}}$ , $f^{'''} = \frac{2}{x ^{3}}$ . At $a = 1$ : $f (1) = 0, f^{'} (1) = 1, f^{''} (1) = - 1$ . So $P_{2} (x) = 0 + 1 \cdot (x - 1) - \frac{1}{2} (x - 1)^{2} .$ At $x = 1.2$ : $P_{2} = 0.2 - \frac{1}{2} (0.04) = 0.2 - 0.02 = 0.18$ . Why this step? When $a \neq = 0$ every term uses $(x - a)$ , not $x$ . Forgetting this is the classic slip.

Step 2. $n = 2 \Rightarrow$ need $f^{(3)} (t) = \frac{2}{t ^{3}}$ . On $[1, 1.2]$ it is monotonic (decreasing, since $2/ t^{3}$ falls as $t$ grows), so its max is at the small end $t = 1$ : $M = \frac{2}{1 ^{3}} = 2$ . Why this step? We verified monotonicity before taking the endpoint — that is the correct order (Trap A avoided; interior-extremum check passed trivially).

Step 3. Here $∣ x - a ∣ = ∣1.2 - 1∣ = 0.2$ : $∣ R_{2} (1.2) ∣ \leq \frac{2}{3 !} (0.2)^{3} = \frac{2}{6} (0.008) \approx 2.667 \times 1 0^{- 3} .$ Why this step? The distance is $0.2$ , not $1.2$ — the centre shifted the walk. That is cell C3's entire lesson.

Verify: true $ln 1.2 = 0.182322$ , actual error $= 0.182322 - 0.18 = 2.32 \times 1 0^{- 3} < 2.667 \times 1 0^{- 3}$ . ✓

(4) Cell C4 — large distance, the power GROWS

Worked example You try to approximate

$e^{2}$ with $P_{3}$ centred at $a = 0$ . What does the error bound say? Forecast: with $∣ x - a ∣ = 2 > 1$ , will the power $2^{n + 1}$ help you or hurt you?

Step 1. $P_{3} (x) = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}$ , so $P_{3} (2) = 1 + 2 + 2 + \frac{8}{6} = 6.3333$ . Why this step? Baseline value; $n = 3$ needs $f^{(4)} = e^{t}$ .

Step 2. On $[0, 2]$ , $e^{t}$ is monotonic increasing, so the max is at the far end $t = 2$ : $M = e^{2} \approx 7.389$ . (We may round up to $M = 8$ for a clean safe bound.) Why this step? Monotonic ⇒ endpoint is safe. For a provable bound you may round $M$ up but never down.

Step 3. Now watch the power: $∣ R_{3} (2) ∣ \leq \frac{e ^{2}}{4 !} (2)^{4} = \frac{7.389}{24} \cdot 16 \approx 4.926.$ Why this step? $(2)^{4} = 16$ is big — the power did not shrink because $∣ x - a ∣ > 1$ . The bound is huge, warning us $P_{3}$ is a poor approximation far from the centre.

Verify: true $e^{2} = 7.389056$ , actual error $= 7.389056 - 6.3333 = 1.056$ . It is inside the bound $4.926$ (✓) but note both are large. Lesson of C4: far walks need far more terms. Your forecast should have been "hurt". ✓

(4b) Cell C4b — the boundary $∣ x - a ∣ = 1$ , where the danger is real

Worked example For

$f (x) = ln (1 + x)$ at $a = 0$ , the series is $x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ with radius of convergence $1$ . Examine the Lagrange bound at the boundary $x = 1$ and show it does not obviously shrink. Forecast: at $x = 1$ the power is frozen at $1$ . Guess: does the bound still die like $1/ (n + 1)!$ , or does $M$ sabotage it?

Step 1. Find the general derivative. For $g (x) = ln (1 + x)$ , $g^{(n + 1)} (t) = \frac{( - 1 ) ^{n} n !}{( 1 + t ) ^{n + 1}}$ , so $g^{(n + 1)} (t) = \frac{n !}{( 1 + t ) ^{n + 1}}$ . Why this step? The bound needs $M = max ∣ g^{(n + 1)} ∣$ on $[0, 1]$ . Notice the factor $n!$ inside the derivative — this is the sabotage.

Step 2. On $[0, 1]$ , $(1 + t)^{n + 1}$ is smallest at $t = 0$ , so $∣ g^{(n + 1)} ∣$ is largest at $t = 0$ : $M = \frac{n !}{1 ^{n + 1}} = n!$ . Why this step? Monotonic (decreasing in $t$ ) ⇒ endpoint $t = 0$ is the worst point (Trap A handled). But $M = n!$ grows with $n$ — a red flag.

Step 3. Assemble the bound with $∣ x - a ∣^{n + 1} = 1^{n + 1} = 1$ : $∣ R_{n} (1) ∣ \leq \frac{M}{( n + 1 )!} \cdot 1 = \frac{n !}{( n + 1 )!} = \frac{1}{n + 1} .$ Why this step? The $n!$ in $M$ cancels most of the $(n + 1)!$ , leaving only $\frac{1}{n + 1}$ . This does go to $0$ , but only like $1/ (n + 1)$ — painfully slow, and the bound barely proves convergence. Contrast $e^{x}$ where $M$ was a constant and the bound died like $1/ (n + 1)!$ .

Verify: at $n = 0$ , bound $= 1$ ; at $n = 9$ , bound $= 1/10 = 0.1$ . The true error of $ln 2$ 's partial sums (the alternating harmonic series) really does decay like $1/ (n + 1)$ — the tool did not give the fast factorial decay you get inside the disk. Forecast lesson: at $∣ x - a ∣ = 1$ , a growing $M$ can gut the bound. (Compare Radius of Convergence: $x = 1$ is exactly the boundary of the disk.) ✓

(5) Cell C5 — degenerate input $x = a$

Worked example What is the error bound for

any $P_{n}$ of any smooth $f$ when $x = a$ ? Forecast: you're standing exactly on the centre. Error = ?

Step 1. Distance $∣ x - a ∣ = ∣ a - a ∣ = 0$ . Why this step? The whole bound rides on this distance; make it explicit.

Step 2. Plug in: $∣ R_{n} (a) ∣ \leq \frac{M}{( n + 1 )!} \cdot 0^{n + 1} = 0.$ Why this step? $0$ raised to any positive power is $0$ , and $n + 1 \geq 1$ always. So the bound collapses to exactly zero.

Step 3. Cross-check with the definition. $P_{n} (a) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (a - a)^{k} = f (a)$ (only the $k = 0$ term survives). So $R_{n} (a) = f (a) - f (a) = 0$ . Why this step? Two independent routes agree — the polynomial is exact at its own centre. No approximation error at the anchor point.

Verify: for $f = e^{x}$ , $P_{2} (0) = 1 = e^{0}$ , error $= 0$ . ✓ The degenerate case is not a bug; it's the sanity anchor.

(6) Cell C6 — limiting behaviour $n \to \infty$

Worked example For

$f (x) = sin x$ at $a = 0$ , show the error bound $\to 0$ as $n \to \infty$ for every fixed $x$ . (This is the bridge to convergence.) Forecast: factorial in the denominator vs power in the numerator — who wins the race?

Step 1. Every derivative of $sin$ is $\pm sin$ or $\pm cos$ , all bounded by $1$ . So $M = 1$ works for all $n$ . Why this step? We need one $M$ that survives every degree; $sin$ gives it for free — and because $∣ sin ∣, ∣ cos ∣ \leq 1$ everywhere, we never even need the interior-extremum check.

Step 2. The bound becomes $∣ R_{n} (x) ∣ \leq \frac{1}{( n + 1 )!} ∣ x ∣^{n + 1} .$ Why this step? Now it's a pure race: $∣ x ∣^{n + 1}$ (grows) vs $(n + 1)!$ (grows much faster).

Step 3. Fix any $x$ . Figure 1 (below) plots exactly this bound, $∣ x ∣^{n + 1} / (n + 1)!$ , against the degree $n$ for the fixed choice $x = 3$ . Its pedagogical job is to show you the race visually: the coral curve first rises while $n + 1 < ∣ x ∣$ (each new factor $\frac{∣ x ∣}{n + 1} > 1$ makes terms grow), reaches a small hump near $n \approx 2$ , and then, once $n + 1$ passes $⌈ ∣ x ∣ ⌉ = 3$ , each new factor multiplies by $\frac{∣ x ∣}{n + 1} < 1$ , so the terms plunge geometrically toward $0$ . Formally, $\frac{∣ x ∣ ^{n + 1}}{( n + 1 )!} \to 0$ for every fixed $x$ . Why this step? Factorial growth always beats exponential growth eventually — that's why the Maclaurin series of $sin$ converges everywhere (link: Radius of Convergence is $\infty$ ).

Figure 1 — the factorial-vs-power race.

The horizontal axis is the degree $n$ ; the vertical axis is the error bound $∣ x ∣^{n + 1} / (n + 1)!$ . Watch the coral curve rise to the small hump near $n \approx 2$ (where $n + 1 \approx ∣ x ∣ = 3$ ) and then dive — that dive is convergence in a picture.

Verify: at $x = 3$ , the sequence $\frac{3 ^{n + 1}}{( n + 1 )!}$ for $n = 0, 1, \dots$ is $3, 4.5, 4.5, 3.375, \dots$ then plunges past $n = 3$ . The factorial wins. ✓ (See the coral curve diving in Figure 1.)

(7) Cell C7 — real-world word problem with units

Worked example A lens is ground so its surface height is

$h (r) = R - R^{2} - r^{2}$ for a sphere of radius $R = 50 mm$ . Engineers approximate it by the parabola $P_{2} (r) = \frac{r ^{2}}{2 R}$ (the $n = 2$ Maclaurin polynomial in $r$ ). For radii up to $r = 5 mm$ , is the approximation error below the $1 μ m = 1 0^{- 3} mm$ tolerance? Forecast: guess whether the parabola is "good enough" before computing.

Step 1. Set $a = 0$ , $n = 2$ . The exact $h$ has Maclaurin expansion $h (r) = \frac{r ^{2}}{2 R} + \frac{r ^{4}}{8 R ^{3}} + \dots$ , so indeed $P_{2} (r) = \frac{r ^{2}}{2 R}$ and the next live term is order $r^{4}$ . We need $M = max ∣ h^{(3)} (r) ∣$ on $[0, 5]$ . Why this step? $n = 2 \Rightarrow$ remainder uses $h^{(3)}$ . Units: $h$ is in mm and $r$ in mm, so $h^{(3)}$ has units $mm / mm^{3} = mm^{- 2}$ .

Step 2. Derive $h^{(3)}$ explicitly, then bound it. Differentiating $h (r) = R - (R^{2} - r^{2})^{1/2}$ : $h^{'} (r) = \frac{r}{R ^{2} - r ^{2}}, h^{''} (r) = \frac{R ^{2}}{( R ^{2} - r ^{2} ) ^{3/2}}, h^{'''} (r) = \frac{3 R ^{2} r}{( R ^{2} - r ^{2} ) ^{5/2}} .$ On $[0, 5]$ the numerator $3 R^{2} r$ rises with $r$ and the denominator $(R^{2} - r^{2})^{5/2}$ falls with $r$ , so $h^{'''}$ is monotonic increasing here and its max is at the far end $r = 5$ . Plug $R = 50, r = 5$ : $h^{'''} (5) = \frac{3 \cdot 2500 \cdot 5}{( 2500 - 25 ) ^{5/2}} = \frac{37500}{247 5 ^{5/2}} \approx 1.24 \times 1 0^{- 4} mm^{- 2} .$ Round up to a safe $M = 2 \times 1 0^{- 4} mm^{- 2}$ . Why this step? Because $h^{'''}$ is monotonic on $[0, 5]$ (checked, not assumed), the endpoint $r = 5$ is genuinely the worst; we round $M$ up for a guaranteed bound.

Step 3. Apply the box with $∣ x - a ∣ = 5 mm$ : $∣ R_{2} (5) ∣ \leq \frac{M}{3 !} (5)^{3} = \frac{2 \times 1 0 ^{- 4}}{6} \cdot 125 \approx 4.17 \times 1 0^{- 3} mm .$ Why this step? Units check: $mm^{- 2} \times mm^{3} = mm$ . Good — the answer is a length.

Verify: $4.17 \times 1 0^{- 3} mm = 4.17 μ m$ , which is above the $1 μ m$ tolerance — so the guaranteed bound does not certify the lens out to $5 mm$ . (Direct check: true $h (5) = 50 - 2500 - 25 = 0.250627 mm$ , $P_{2} (5) = \frac{25}{100} = 0.25 mm$ , actual error $= 6.27 \times 1 0^{- 4} mm = 0.63 μ m$ .) The true error (0.63 µm) actually passes, while the bound (4.17 µm) is conservative. Forecast lesson: the bound gives a guarantee; a loose $M$ can be pessimistic, so a failed bound does not always mean a failed part. ✓

(8) Cell C8 — the exam twist (Trap B, live)

Worked example "A student bounds the error of

$P_{4}$ for $f$ at $x = 3$ , $a = 1$ , using $f^{(4)} \leq M$ and writes $∣ R_{4} ∣ \leq \frac{M}{4 !} ∣3 - 1 ∣^{4}$ . Find and fix two errors, then evaluate the corrected bound for $M = 15$ ." Forecast: spot the two mistakes before reading on — one is the derivative order, one is the factorial-and-power.

Step 1. Error 1 — wrong derivative order (Trap B). $P_{4}$ has degree $n = 4$ , so the remainder uses $f^{(n + 1)} = f^{(5)}$ , not $f^{(4)}$ . The student bounded the wrong derivative. Why this step? "Next derivative" means $n + 1 = 5$ — the first half of Trap B.

Step 2. Error 2 — wrong factorial and power (Trap B, other half). It must be $(n + 1)! = 5!$ and $∣ x - a ∣^{n + 1} = ∣3 - 1 ∣^{5}$ , not $4!$ and the $4$ th power. Why this step? All three pieces — derivative, factorial, power — must carry the same $n + 1$ .

Step 3. Correct statement (with $M$ now the bound on $∣ f^{(5)} ∣$ ): $∣ R_{4} (3) ∣ \leq \frac{M}{5 !} ∣3 - 1 ∣^{5} = \frac{M}{120} \cdot 32 = \frac{32 M}{120} = \frac{4 M}{15} .$ With $M = 15$ : $\frac{4 \cdot 15}{15} = 4$ . Why this step? $5! = 120$ , $2^{5} = 32$ , and $\frac{32}{120} = \frac{4}{15}$ ; the clean $M = 15$ makes the arithmetic visible.

Verify: the corrected bound gives exactly $4$ . The student's wrong formula would have given $\frac{15}{24} \cdot 16 = 10$ — a different number, so the traps genuinely change the answer. ✓

Recall One-line recap of the whole matrix

Small/negative/far/off-centre only change which end (or interior point) gives $M$ and how big $∣ x - a ∣^{n + 1}$ is; $∣ x - a ∣ = 1$ freezes the power at $1$ so only the factorial-vs- $M$ race matters (and a growing $M$ , as for $ln (1 + x)$ , can slow the bound to $1/ (n + 1)$ ); $x = a$ gives exactly $0$ ; $n \to \infty$ lets the factorial win (convergence); word problems add units you must track; exam twists (Trap B) test that derivative, factorial, and power are all $n + 1$ ; and always check monotonicity before grabbing an endpoint for $M$ (Trap A).

Self-test:

Why did walking left to

x = - 0.4

give a tighter bound than right to

+ 0.4

for

e^{x}

Because

M = max e^{t}

is smaller on

[- 0.4, 0]

(max

= 1

) than on

[0, 0.4]

(max

= e^{0.4}

); the distance is identical.

In cell C4, why is

P_{3}

a bad approximation of

e^{2}

∣ x - a ∣ = 2 > 1

, so

2^{n + 1}

grows; the bound

\approx 4.9

is large — far walks need more terms.

At the boundary

x = 1

for

ln (1 + x)

, why does the bound only decay like

1/ (n + 1)

Because

M = n!

grows with

n

and cancels most of

(n + 1)!

, leaving

n! / (n + 1)! = 1/ (n + 1)

What makes the error bound exactly

0

x = a

∣ x - a ∣^{n + 1} = 0^{n + 1} = 0

; the polynomial equals

f

at its own centre.

When is it wrong to take

M

at an endpoint of

[a, x]

When

f^{(n + 1)}

is not monotonic there — an interior hump can be the true maximum, so check first.

Connections

Parent: remainder theorem — the bound these examples apply.
Taylor & Maclaurin Series — the polynomials $P_{n}$ used throughout.
Radius of Convergence — cells C6 and C4b are exactly the interior-vs-boundary story of the convergence disk.
Alternating Series Estimation Theorem — for $sin, cos$ and $ln (1 + x)$ an even tighter alternating bound is often available.
Big-O and asymptotic error — each bound is $O (∣ x - a ∣^{n + 1})$ , the near- $a$ behaviour.
Cauchy Mean Value Theorem — the engine behind the Lagrange form these examples use.

Worked examples — Taylor's remainder theorem — error estimation