Taylor's remainder theorem — error estimation: Question bank

The core objects, so nothing here is a mystery symbol:

$f$ is the true function, $a$ the centre, $x$ the target point.
$P_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ is the degree- $n$ Taylor polynomial (see Taylor & Maclaurin Series).
$R_{n} (x) = f (x) - P_{n} (x)$ is the exact error.
Lagrange form: $R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1}$ for some unknown $c$ strictly between $a$ and $x$ .
Bound: $∣ R_{n} (x) ∣ \leq \frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}$ where $M \geq ∣ f^{(n + 1)} ∣$ on the interval.

True or false — justify

The Lagrange remainder gives the exact error, not just a bound.

True —

R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1}

is an exact equality; it becomes a bound only when we replace the unknown

f^{(n + 1)} (c)

by a max

M

The bound

\frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}

always shrinks as

n

grows.

False — only if the numerator

M ∣ x - a ∣^{n + 1}

eventually loses the race to

(n + 1)!

; for large

∣ x - a ∣

or functions whose derivatives blow up (like

\frac{1}{1 - x}

), the bound can grow before it shrinks or never shrink at all.

R_{n} (x) \to 0

n \to \infty

, the Taylor series converges to

f (x)

at that

x

True — this is precisely the definition: the series equals

f

exactly when the remainder vanishes, which is the analytic link to Radius of Convergence.

A Taylor series that converges must converge to the function it came from.

False — convergence of the series and

R_{n} \to 0

are different statements; the classic

e^{- 1/ x^{2}}

has a convergent (all-zero) Maclaurin series that equals

0

, not the function, so

R_{n} \neq \to 0

The point

c

in the Lagrange form is the same for every

n

False —

c

depends on

n

, on

x

, and on the function; it is just "some point in between" guaranteed to exist, re-chosen for each remainder.

The bound requires

f^{(n + 1)}

to exist on the whole interval between

a

and

x

True — the derivation applies Cauchy Mean Value Theorem on

[a, x]

; if

f^{(n + 1)}

fails to exist anywhere inside, the theorem's hypotheses break and no bound is guaranteed.

Making

M

larger than necessary makes your error claim invalid.

False — a larger

M

gives a valid but looser (more conservative) bound; the claim "

∣ R_{n} ∣ \leq

bound" still holds, you just promise less accuracy than you actually have.

For

sin x

and

cos x

, you can always take

M = 1

regardless of

n

a

, or

x

True — every derivative of

sin

cos

\pm sin

\pm cos

, all bounded by

1

everywhere, so

M = 1

is always valid (though often loose).

Spot the error

"The error uses the

n

-th derivative and

n!

, matching the polynomial's degree."

Wrong — the remainder reaches one step beyond: the

(n + 1)

-th derivative and

(n + 1)!

, because

P_{n}

already matched everything up to the

n

-th derivative, so the first mismatch lives at order

n + 1

c

is between

a

and

x

, so I'll just set

c = x

to evaluate."

Wrong —

c

is unknown; picking an endpoint may under-estimate

∣ f^{(n + 1)} ∣

and produce a bound smaller than the true error. You must use

M = max ∣ f^{(n + 1)} ∣

over the interval.

"Since

x - a = 0.3

is small,

(x - a)^{n + 1}

is negligible, so I can drop it."

Wrong — you must keep the factor; it is one of the three parts of the bound. Dropping it changes the number entirely, and if

∣ x - a ∣ > 1

that factor actually grows and can dominate.

P_{2}

has three terms, so I use

M = max ∣ f^{(2)} ∣

Wrong —

n = 2

means the remainder needs

f^{(3)}

and

3!

, not

f^{(2)}

; count derivatives from the remainder, never from the highest term of the polynomial.

e^{0.1}

is about

1.105

, so I'll take

M = 1.105

for the

e^{x}

bound."

Wrong idea, though it happens to work —

M

must bound

f^{(n + 1)} = e^{t}

over the whole

[0, 0.1]

, and its max there is

e^{0.1} \approx 1.105

, so this is fine for that interval; the error is thinking

M

tracks

f (x)

rather than

max ∣ f^{(n + 1)} ∣

"The alternating-series bound and Taylor bound must give the same number."

Wrong — they are different tools; Alternating Series Estimation Theorem bounds by the first omitted term (only when terms alternate and shrink), while Lagrange bounds by

\frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}

. They can disagree, and one may be sharper.

Why questions

Why is the factorial

(n + 1)!

rather than

(n + 1)

2^{n + 1}

It comes directly from the Cauchy-MVT derivation: differentiating the Taylor structure produces

\frac{f ^{(n + 1)} ( c )}{n !} (x - c)^{n}

, and dividing by

h^{'} = - (n + 1) (x - c)^{n}

contributes the extra factor, giving

(n + 1)!

— it is earned, not chosen.

Why do we bound

∣ f^{(n + 1)} ∣

instead of computing

R_{n}

directly?

If we could compute

R_{n} = f (x) - P_{n} (x)

exactly we'd know

f (x)

and wouldn't need the approximation at all; bounding the next derivative sidesteps knowing

f (x)

Why does the Mean Value Theorem (or its Cauchy version) even appear here?

MVT is the tool that converts "

g

starts at

R_{n}

and ends at

0

" into "there exists a point

c

where the rate equals the average rate," letting an existence claim about an unknown

c

replace the impossible exact computation. See Mean Value Theorem and Cauchy Mean Value Theorem.

Why does the error get small so fast for a short walk (small

∣ x - a ∣

The factor

∣ x - a ∣^{n + 1}

is a high power of a small number — squaring or cubing

0.1

crashes it toward zero — and the

(n + 1)!

denominator adds even faster growth, so short walks are guaranteed almost perfect.

Why can the remainder theorem guarantee accuracy even though

c

is never found?

We never need

c

's value — only that some valid

c

exists in the interval; bounding

∣ f^{(n + 1)} ∣

over the whole interval covers every possible location of that

c

Why is

R_{n} = O (∣ x - a ∣^{n + 1})

near

a

a fair summary of the bound?

x \to a

the factor

∣ x - a ∣^{n + 1}

controls the size while

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

is just a constant, so the error's leading behaviour is that power — the language of Big-O and asymptotic error.

Edge cases

x = a

, what is

R_{n} (x)

Exactly

0

— the polynomial equals the function at the centre, and the bound confirms it since

∣ x - a ∣^{n + 1} = 0

What happens to the bound when

f^{(n + 1)}

is unbounded on the interval (no finite

M

exists)?

The theorem gives no usable bound; you must shrink the interval so a finite

M

exists, or the guarantee simply fails.

Does the theorem work when

x < a

(target left of centre)?

Yes —

c

lies strictly between

a

and

x

in either direction, and the bound uses

∣ x - a ∣^{n + 1}

, so orientation doesn't matter.

For a polynomial

f

of degree

d

, what is

R_{n}

when

n \geq d

Exactly

0

— the

(n + 1)

-th derivative of a degree-

d

polynomial is

0

once

n + 1 > d

, so both the Lagrange term and the bound vanish.

At the boundary

∣ x - a ∣ = 1

, which factor decides whether the bound shrinks with

n

The factor

∣ x - a ∣^{n + 1} = 1

is inert, so the race is purely

M

versus

(n + 1)!

; the factorial wins as long as

M

doesn't grow faster than

(n + 1)!

f^{(n + 1)}

is continuous but changes sign on the interval, is the bound still valid?

Yes —

M

bounds the absolute value

∣ f^{(n + 1)} ∣

, so sign changes are irrelevant; you only need the largest magnitude.

Connections

Taylor & Maclaurin Series — the polynomials these traps quiz you on.
Mean Value Theorem / Cauchy Mean Value Theorem — why "some $c$ exists".
Radius of Convergence — the $R_{n} \to 0$ true/false items.
Alternating Series Estimation Theorem — the "same bound?" trap.
Big-O and asymptotic error — the asymptotic "why" item.