Taylor's theorem with remainder

Theorem

Taylor's theorem with remainder

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Statement

Let \( n \in \mathbb{N} \cup \{0\} \), let \( I \subseteq \mathbb{R} \) be an interval, and let \( a, x \in I \) with \( a \ne x \). Suppose \( f : I \to \mathbb{R} \) is such that the derivatives \( f, f', \dots, f^{(n)} \) all exist and are continuous on the closed interval with endpoints \( a \) and \( x \), and \( f^{(n+1)} \) exists at every point of the open interval between \( a \) and \( x \). Then there exists a point \( \xi \) strictly between \( a \) and \( x \) such that \[ f(x) \;=\; \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\,(x-a)^{k} \;+\; \frac{f^{(n+1)}(\xi)}{(n+1)!}\,(x-a)^{n+1}. \] The polynomial \( T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \) is the degree-\(n\) Taylor polynomial of \( f \) at \( a \), and \( R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} \) is the Lagrange form of the remainder.

Why it matters

Taylor's theorem is the bridge between the local and the global: it converts knowledge of finitely many derivatives at a single point into a quantitative statement about \( f \) on a whole interval, with an error term that is not merely "small" but explicitly controlled by one higher derivative at one unknown point. It is the engine behind every truncation-error estimate in numerical analysis, every big-\(\mathcal{O}\) asymptotic expansion, the second-derivative test, and the standard proofs that concrete series (for \( e^x \), \( \sin \), \( \cos \), \( \ln \)) actually converge to their functions.

Conceptually, it generalises the Mean Value Theorem (the case \( n = 0 \)) up the ladder of derivatives, and it isolates precisely what smoothness buys you: polynomial approximation of controllable quality — while its famous failure cases (such as \( e^{-1/x^2} \)) mark the exact boundary between smoothness and analyticity.

Hypotheses

\( f \) is real-valued.For \( f(t) = e^{it} : [0, 2\pi] \to \mathbb{C} \) with \( n = 0 \), the conclusion would demand \( \xi \) with \( i e^{i\xi} (2\pi - 0) = e^{2\pi i} - e^{0} = 0 \); but \( |i e^{i\xi}| = 1 \) everywhere. The Lagrange form is genuinely a theorem about \( \mathbb{R} \)-valued functions (it rests on Rolle, i.e. on order).

\( f^{(n)} \) is continuous on the closed interval, endpoints included.Take \( n = 0 \), \( f(t) = 0 \) for \( 0 \le t \lt 1 \) and \( f(1) = 1 \), on \( [0,1] \). Then \( f'(t) = 0 \) on \( (0,1) \), yet \( f(1) - f(0) = 1 \ne f'(\xi)(1-0) \) for every \( \xi \). One bad endpoint kills the conclusion.

\( f^{(n+1)} \) exists at every point of the open interval.Take \( n = 0 \), \( f(t) = |t| \) on \( [-1, 1] \): \( f \) is continuous, differentiable except at the single point \( 0 \), and \( f(1) - f(-1) = 0 \), but \( f'(\xi) = \pm 1 \) wherever it exists — never \( 0 \). One missing interior point of differentiability is fatal.

\( a \) and \( x \) lie in one interval of the domain (connectedness).For \( f(t) = 1/t \) on \( \mathbb{R} \setminus \{0\} \), \( a = -1 \), \( x = 1 \), \( n = 0 \): \( f(1) - f(-1) = 2 \) but \( f'(\xi) = -1/\xi^2 \lt 0 \) everywhere. All derivatives exist on the domain, yet the conclusion fails because \( [-1,1] \not\subseteq \operatorname{dom} f \). The theorem is a statement about intervals.

Nothing more than \( f^{(n+1)} \) pointwise: no continuity of \( f^{(n+1)} \) is assumed.This is a genuine strength, not a gap — but beware the flip side: the integral form of the remainder, \( R_n(x) = \frac{1}{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt \), needs \( f^{(n+1)} \) integrable (e.g. continuous), which the Lagrange form does not. The two forms have different hypotheses and are not interchangeable.

Proof

Assume \( a \lt x \); the case \( x \lt a \) is identical with the roles of the endpoints exchanged (run the whole argument on \( [x,a] \)); in both cases \( \xi \) lands strictly between \( a \) and \( x \).

Symmetry of the hypotheses in \( a \) and \( x \); no step below uses which endpoint is larger. A

Since \( x \ne a \), \( (x-a)^{n+1} \ne 0 \), so there is a unique real number \( M \) with \[ f(x) \;=\; \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \;+\; M\,(x-a)^{n+1}. \] It now suffices to produce \( \xi \in (a,x) \) with \( M = \dfrac{f^{(n+1)}(\xi)}{(n+1)!} \).

Definition of \( M \) by solving a linear equation with nonzero coefficient; this reduces the theorem to identifying \( M \). A

Define the auxiliary function \( g : [a,x] \to \mathbb{R} \) by \[ g(t) \;=\; f(x) \;-\; \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x-t)^k \;-\; M\,(x-t)^{n+1}. \] Here the expansion point of the Taylor polynomial is the variable \( t \), while the evaluation point \( x \) is frozen.

This is the key idea of the proof: measure the error of the degree-\(n\) expansion as the basepoint slides from \( a \) to \( x \), then apply Rolle. B

\( g \) is continuous on \( [a,x] \) and differentiable on \( (a,x) \).

For \( 0 \le k \le n \), \( f^{(k)} \) is continuous on \( [a,x] \) by hypothesis and \( t \mapsto (x-t)^k \) is a polynomial, so each summand is continuous on the closed interval (algebra of continuous functions). On \( (a,x) \), \( f^{(n+1)} \) exists, so each \( f^{(k)} \) with \( k \le n \) is differentiable there; each summand is then differentiable by the product rule, and finite sums of differentiable functions are differentiable. C

\[ g(x) = f(x) - f(x) - 0 = 0. \]

At \( t = x \), every term of the sum with \( k \ge 1 \) carries a factor \( (x-x)^k = 0 \); the \( k = 0 \) term is \( f(x) \); and \( M(x-x)^{n+1} = 0 \). A

\[ g(a) = f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k - M(x-a)^{n+1} = 0. \]

This is exactly the defining equation of \( M \) from Step 2, rearranged. A

There exists \( \xi \in (a,x) \) with \( g'(\xi) = 0 \).

Rolle's theorem: \( g \) is continuous on \( [a,x] \), differentiable on \( (a,x) \) (Step 4), and \( g(a) = g(x) = 0 \) (Steps 5–6). Rolle's theorem is itself proved from the extreme value theorem plus the interior-extremum (Fermat) lemma. B

For \( t \in (a,x) \), the sum in \( g \) telescopes under differentiation: \[ \frac{d}{dt}\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x-t)^k \;=\; \sum_{k=0}^{n} \frac{f^{(k+1)}(t)}{k!}(x-t)^k \;-\; \sum_{k=1}^{n} \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}. \] Reindexing the second sum by \( j = k-1 \) turns it into \( \sum_{j=0}^{n-1} \frac{f^{(j+1)}(t)}{j!}(x-t)^{j} \), which cancels every term of the first sum except \( k = n \). Hence \[ g'(t) \;=\; -\,\frac{f^{(n+1)}(t)}{n!}(x-t)^n \;+\; (n+1)\,M\,(x-t)^{n}. \]

Product rule on each summand \( \frac{f^{(k)}(t)}{k!}(x-t)^k \) (legal by Step 4), noting the derivative of \( (x-t)^k \) is \( -k(x-t)^{k-1} \) and the \( k=0 \) term contributes only \( f'(t) \); then an index shift and exact pairwise cancellation. This telescoping is the entire computational content of the proof. C

Evaluating at \( t = \xi \): \[ 0 = g'(\xi) = (x-\xi)^n\left[ (n+1)M - \frac{f^{(n+1)}(\xi)}{n!} \right]. \] Since \( \xi \lt x \), \( (x-\xi)^n \ne 0 \), so \( M = \dfrac{f^{(n+1)}(\xi)}{(n+1)!} \). Substituting into Step 2 gives the theorem. \( \blacksquare \)

A product of reals vanishes only if a factor does (field axioms); \( (n+1)\cdot n! = (n+1)! \). A

Result

\[ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}, \qquad \xi \text{ strictly between } a \text{ and } x. \]

Reading. A function with \( n+1 \) derivatives is its degree-\(n\) Taylor polynomial exactly, up to an error that looks like the "next term" of the series — except that the \((n+1)\)-st derivative is evaluated not at \( a \) but at some unknown intermediate point \( \xi \). You never learn where \( \xi \) is; you only learn it exists, which is enough whenever you can bound \( f^{(n+1)} \) on the whole interval.

Scope. Real-valued functions of one real variable, on an interval containing both the expansion point \( a \) and the evaluation point \( x \), with \( f^{(n)} \) continuous up to and including both endpoints and \( f^{(n+1)} \) existing (no continuity needed) at every interior point between them. The point \( \xi \) depends on \( x \), on \( a \), and on \( n \). For complex- or vector-valued \( f \) the Lagrange form fails; only the weaker bound \( |R_n(x)| \le \frac{\sup |f^{(n+1)}|}{(n+1)!}|x-a|^{n+1} \) and the integral form survive.

Corollaries & converses

Mean Value Theorem. The case \( n = 0 \) reads \( f(x) = f(a) + f'(\xi)(x-a) \): Taylor's theorem is the MVT's higher-order generalisation, and the proof above degenerates to the standard MVT proof when \( n = 0 \).
The workhorse estimate. If \( |f^{(n+1)}| \le M_{n+1} \) on the interval, then \( |f(x) - T_n(x)| \le \dfrac{M_{n+1}}{(n+1)!}|x-a|^{n+1} \). If moreover \( \dfrac{M_{n+1}\,|x-a|^{n+1}}{(n+1)!} \to 0 \) as \( n \to \infty \) (as for \( e^x, \sin, \cos \) on any bounded set), the Taylor series converges to \( f \).
Polynomial rigidity. If \( f^{(n+1)} \equiv 0 \) on an interval, then \( f \) is a polynomial of degree at most \( n \) there (the remainder vanishes identically; see Problem 4).
Second-derivative test, quantitatively. With \( n = 1 \): if \( f'(a) = 0 \) and \( f'' \ge m \gt 0 \) near \( a \), then \( f(x) \ge f(a) + \frac{m}{2}(x-a)^2 \) near \( a \) — a strict quantitative local minimum.
Peano's form is a different theorem. If merely \( f^{(n)}(a) \) exists (no hypotheses away from \( a \)), then \( f(x) = T_n(x) + o\!\left((x-a)^n\right) \); weaker conclusion, weaker hypotheses. Neither subsumes the other.
Converse fails. Being approximable to order \( o((x-a)^n) \) by a polynomial does not imply \( n \)-fold differentiability: \( f(x) = x^3 \sin(1/x) \), \( f(0)=0 \), satisfies \( |f(x)| \le |x|^3 = o(x^2) \), so the zero polynomial is a degree-2 approximation of Peano quality at \( 0 \); but \( f''(0) \) does not exist, since \( \frac{f'(h)-f'(0)}{h} = 3h\sin(1/h) - \cos(1/h) \) has no limit as \( h \to 0 \).

Fails without

Differentiability missing at one interior point. \( f(t) = |t| \) on \( [-1,1] \), \( n = 0 \), \( a = -1 \), \( x = 1 \): the conclusion demands \( \xi \) with \( f'(\xi)\cdot 2 = f(1) - f(-1) = 0 \), i.e. \( f'(\xi) = 0 \); but \( f'(t) = 1 \) for \( t \gt 0 \), \( f'(t) = -1 \) for \( t \lt 0 \), and \( f'(0) \) does not exist. No such \( \xi \). One point of non-differentiability, strictly inside the interval, destroys the theorem.
Real-valuedness dropped. \( f(t) = e^{it} \) on \( [0, 2\pi] \), \( n = 0 \): \( f(2\pi) - f(0) = 0 \), so the Lagrange form would force \( f'(\xi) = i e^{i\xi} = 0 \) for some \( \xi \) — impossible, since \( |ie^{i\xi}| = 1 \). The same failure occurs for the vector-valued \( t \mapsto (\cos t, \sin t) \). Rolle's theorem, and with it the pointwise-\( \xi \) form of the remainder, is an artefact of the total order on \( \mathbb{R} \).
Domain not an interval. \( f(t) = 1/t \), \( a = -1 \), \( x = 1 \), \( n = 0 \): every derivative of \( f \) exists at every point of its domain, yet \( f(1) - f(-1) = 2 \) while \( f'(\xi) = -1/\xi^2 \lt 0 \) for all \( \xi \). The hypotheses require the entire closed segment from \( a \) to \( x \) to lie in the domain; \( 0 \) is missing.
Endpoint continuity dropped. \( n = 0 \), \( f(t) = 0 \) on \( [0,1) \), \( f(1) = 1 \): \( f' \equiv 0 \) on \( (0,1) \) but \( f(1) - f(0) = 1 \). Continuity on the closed interval is where the extreme value theorem (inside Rolle) earns its keep.

Common errors

Treating \( \xi \) as a constant. \( \xi \) depends on \( x \), on \( a \), and on \( n \). Writing \( \xi \) outside a quantifier, differentiating an identity in \( x \) that contains \( \xi(x) \), or reusing "the same \( \xi \)" for two different \( n \) are all invalid moves.
"Smooth implies the series converges to \( f \)". False: \( f(x) = e^{-1/x^2} \) (with \( f(0) = 0 \)) is \( C^\infty \) with every \( f^{(k)}(0) = 0 \), so its Taylor series at \( 0 \) is identically zero and converges everywhere — to the wrong function. Taylor's theorem never asserts \( R_n \to 0 \); that must be proved separately, and can be false.
Off-by-one in the remainder. Writing \( \frac{f^{(n+1)}(\xi)}{n!} \) instead of \( \frac{f^{(n+1)}(\xi)}{(n+1)!} \), or evaluating the \((n+1)\)-st derivative at \( a \) rather than at \( \xi \). The remainder is the "\( (n+1) \)-st term with a displaced evaluation point".
Using the Lagrange form under Peano hypotheses. If you only know \( f^{(n)}(a) \) exists (nothing about a neighbourhood), the Lagrange form is unavailable; you get only \( T_n(x) + o((x-a)^n) \). Conversely, quoting "\( R_n = o((x-a)^n) \)" and then bounding it numerically at a fixed \( x \) is meaningless — little-\( o \) says nothing at any fixed point.
Applying the pointwise form to vector-valued functions. For \( f : \mathbb{R} \to \mathbb{R}^m \), there is in general no single \( \xi \) that works for all coordinates simultaneously (see the circle example above). Use the integral remainder or a norm estimate instead.
Bounding \( f^{(n+1)}(\xi) \) by its value at \( a \) or at \( x \). You must bound \( \sup |f^{(n+1)}| \) over the whole interval between \( a \) and \( x \), since \( \xi \) is unknown.

Discussion

Brook Taylor published the expansion without any remainder in his Methodus Incrementorum (1715), deriving it formally from finite differences; as stated it was a heuristic about infinite series, silent on convergence. The theorem as a tool of rigour is really Lagrange's: in the Théorie des fonctions analytiques (1797) he introduced the remainder term that now bears his name, precisely because he wanted to found calculus on power series and needed to control truncation. Cauchy later gave his own form of the remainder and the integral form, and — decisively — the counterexample \( e^{-1/x^2} \), which shattered Lagrange's programme by showing that a \( C^\infty \) function is not determined by its derivatives at a point.

The right way to read the theorem is as a family of statements, one for each \( n \), about a fixed function on a fixed interval — not as a statement about infinite series. For a single \( n \) it is a finite, exact identity, as concrete as the MVT. Questions about the Taylor series (does \( T_n \to f \)?) are questions about the behaviour of \( R_n \) as \( n \to \infty \), and the theorem's role there is to hand you \( R_n \) in a form you can estimate. This division of labour — exact identity first, limit analysis second — is the pattern for most of hard analysis.

The theorem also marks the boundary between real and complex analysis. On \( \mathbb{R} \), infinite smoothness and analyticity are far apart: by Borel's theorem, every sequence of reals \( (c_k) \) is the derivative sequence \( f^{(k)}(0) \) of some \( C^\infty \) function, so Taylor series at a point carry no convergence information whatsoever. On \( \mathbb{C} \), a single complex derivative on an open set forces analyticity and the Taylor series converges automatically on discs. Taylor's theorem with remainder is exactly the tool real analysis has instead of that miracle.

There is a whole family of remainders beyond Lagrange's. Applying the Cauchy MVT to the auxiliary function \( g \) of the proof against \( \varphi(t) = (x-t)^p \) for any \( p \gt 0 \) yields the Schlömilch–Roche remainder \( R_n(x) = \frac{f^{(n+1)}(\xi)}{p \cdot n!}(x-\xi)^{n+1-p}(x-a)^{p} \); the choice \( p = n+1 \) recovers Lagrange, and \( p = 1 \) gives the Cauchy form \( R_n(x) = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^{n}(x-a) \), which is sharper for functions like \( (1+x)^{\alpha} \) near the edge of the interval of convergence. If \( f^{(n+1)} \) is additionally continuous, repeated integration by parts gives the integral form \( R_n(x) = \frac{1}{n!}\int_a^x (x-t)^n f^{(n+1)}(t)\,dt \), which — unlike the Lagrange form — survives verbatim for maps into any Banach space and is the correct statement in \( \mathbb{R}^m \), on manifolds, and in the calculus of variations. Finally, under the weakest hypothesis (only \( f^{(n)}(a) \) exists), Peano's form \( R_n(x) = o((x-a)^n) \) holds, proved by \( n-1 \) applications of L'Hôpital plus the definition of the derivative.

Common misconceptions. The theorem does not say the Taylor series converges, nor that it converges to \( f \) when it does converge (\( e^{-1/x^2} \)); it does not tell you where \( \xi \) is, only that it exists; and "the remainder is the next term" is a mnemonic, not a fact — the next term has \( f^{(n+1)}(a) \), the remainder has \( f^{(n+1)}(\xi) \), and the gap between those is the whole content.

Worked examples

Example 1 — computing \( e \) to within \( 10^{-3} \), with certainty.

Apply the theorem to \( f = \exp \) on \( [0,1] \) with \( a = 0 \), \( x = 1 \). Every derivative of \( \exp \) is \( \exp \), which is continuous on \( [0,1] \) and differentiable on \( (0,1) \), so for each \( n \) there is \( \xi \in (0,1) \) with \[ e \;=\; \sum_{k=0}^{n} \frac{1}{k!} \;+\; \frac{e^{\xi}}{(n+1)!}. \]

Hypotheses verified directly; \( f^{(k)}(0) = e^0 = 1 \) for all \( k \). A

Bound the remainder: \( 0 \lt \xi \lt 1 \) gives \( 1 \lt e^{\xi} \lt e \lt 3 \), hence \[ 0 \;\lt\; e - \sum_{k=0}^{n}\frac{1}{k!} \;\lt\; \frac{3}{(n+1)!}. \]

Monotonicity of \( \exp \), plus the crude a priori bound \( e \lt 3 \) (e.g. from \( e \lt 1 + 1 + \frac12 + \frac14 + \frac18 + \dots = 3 \) by comparing \( k! \ge 2^{k-1} \)). B

Choose \( n = 6 \): \( \dfrac{3}{7!} = \dfrac{3}{5040} \lt 6 \times 10^{-4} \lt 10^{-3} \).

Direct computation; smallest \( n \) for which \( 3/(n+1)! \lt 10^{-3} \). A

\[ \sum_{k=0}^{6}\frac{1}{k!} = 1 + 1 + \frac12 + \frac16 + \frac1{24} + \frac1{120} + \frac1{720} = \frac{1957}{720} = 2.718055\ldots \]

Exact rational arithmetic over the common denominator \( 720 \): \( 720 + 720 + 360 + 120 + 30 + 6 + 1 = 1957 \). A

\[ 2.71805 \;\lt\; \frac{1957}{720} \;\lt\; e \;\lt\; \frac{1957}{720} + \frac{3}{5040} \;\lt\; 2.71866. \]

Reading. Seven terms of the exponential series pin \( e \) to within \( 6 \times 10^{-4} \), and — because the remainder is manifestly positive — we even know the approximation is an underestimate. (True value: \( e = 2.71828\ldots \))

Scope. The same template bounds \( e^x \) for any fixed \( x \), with \( 3 \) replaced by \( e^{|x|} \le 3^{\lceil |x| \rceil} \).

Example 2 — the inequality \( \sin x \gt x - \dfrac{x^3}{6} \) for all \( x \gt 0 \).

Apply the theorem to \( f = \sin \) with \( a = 0 \), \( n = 3 \), at any \( x \gt 0 \). Since \( \sin' = \cos \), \( \sin'' = -\sin \), \( \sin''' = -\cos \), \( \sin^{(4)} = \sin \), the values at \( 0 \) are \( 0, 1, 0, -1 \), so there is \( \xi \in (0,x) \) with \[ \sin x \;=\; x - \frac{x^3}{6} + \frac{\sin \xi}{24}\,x^4. \]

All derivatives of \( \sin \) are continuous on all of \( \mathbb{R} \), so the hypotheses hold on \( [0,x] \) for every \( x \); \( T_3(x) = x - x^3/6 \). A

Case \( 0 \lt x \le \pi \): then \( \xi \in (0,x) \subseteq (0,\pi) \), so \( \sin \xi \gt 0 \), hence \[ \sin x - \left(x - \frac{x^3}{6}\right) = \frac{\sin \xi}{24}x^4 \gt 0. \]

Positivity of \( \sin \) on \( (0,\pi) \) and of \( x^4 \); this is where knowing \( \xi \) lies strictly inside \( (0,x) \) matters. B

Case \( x \gt \pi \): the cubic \( h(x) = x - \frac{x^3}{6} \) has \( h'(x) = 1 - \frac{x^2}{2} \lt 0 \) for \( x \gt \sqrt{2} \), so \( h \) is strictly decreasing on \( [\sqrt2, \infty) \), giving \[ h(x) \lt h(\pi) = \pi - \frac{\pi^3}{6} \lt 3.15 - \frac{31}{6} \lt -2 \lt -1 \le \sin x. \]

Monotonicity from the sign of \( h' \) (a corollary of the MVT, i.e. the \( n=0 \) case of this very theorem), plus \( \pi^3 \gt 31 \) and the universal bound \( \sin x \ge -1 \). B

\[ \forall\, x \gt 0: \qquad \sin x \;\gt\; x - \frac{x^3}{6}. \]

Reading. The cubic Taylor polynomial of \( \sin \) is a genuine global lower bound on the positive half-line — the remainder's sign, not just its size, is readable from the Lagrange form.

Scope. The same technique gives the interlacing \( T_{4m+3}(x) \lt \sin x \lt T_{4m+1}(x) \) for \( x \gt 0 \), and by oddness the reversed inequalities for \( x \lt 0 \).

Problems

Write down the Taylor polynomial \( T_3 \) of \( f(x) = \ln(1+x) \) at \( a = 0 \) and its Lagrange remainder, and bound the error of \( T_3(0.5) \) as an approximation to \( \ln 1.5 \). Is \( T_3(0.5) \) an over- or under-estimate?
Solution
By induction, \( f^{(k)}(x) = \dfrac{(-1)^{k-1}(k-1)!}{(1+x)^k} \) for \( k \ge 1 \), so \( f^{(k)}(0) = (-1)^{k-1}(k-1)! \) and \[ T_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}, \qquad R_3(x) = \frac{f^{(4)}(\xi)}{4!}x^4 = -\frac{x^4}{4(1+\xi)^4}, \quad \xi \in (0,x). \] All hypotheses hold on \( [0, 0.5] \) since \( f \) is \( C^\infty \) on \( (-1, \infty) \). At \( x = 0.5 \): since \( 0 \lt \xi \lt 0.5 \) we have \( 1 \lt (1+\xi)^4 \), so \[ |R_3(0.5)| = \frac{(0.5)^4}{4(1+\xi)^4} \lt \frac{1}{64} = 0.015625, \] and \( R_3(0.5) \lt 0 \), so \( T_3(0.5) = 0.5 - 0.125 + \frac{1}{24} = \frac{5}{12} = 0.41\overline{6} \) is an overestimate: \[ \frac{5}{12} - \frac{1}{64} \lt \ln 1.5 \lt \frac{5}{12}. \] (Indeed \( \ln 1.5 = 0.405465\ldots \))
Let \( f \) be twice differentiable on an interval \( I \) with \( f''(t) \ge 0 \) for all \( t \in I \). Prove that the graph of \( f \) lies above every tangent line: \( f(x) \ge f(a) + f'(a)(x-a) \) for all \( a, x \in I \).
Solution
If \( x = a \) the claim is an equality. Otherwise apply Taylor's theorem with \( n = 1 \) on the interval between \( a \) and \( x \): the hypotheses hold because \( f'' \) exists everywhere on \( I \), which forces \( f' \) to be differentiable, hence continuous, on all of \( I \) (and \( f \) likewise). So there is \( \xi \) strictly between \( a \) and \( x \) with \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(\xi)}{2}(x-a)^2. \] Since \( f''(\xi) \ge 0 \) and \( (x-a)^2 \gt 0 \), the last term is \( \ge 0 \), which is the claim. (If \( f'' \gt 0 \) throughout, the inequality is strict for \( x \ne a \): tangent lines touch the graph only at the point of tangency.)
How many terms of the Taylor expansion of \( \cos \) at \( 0 \) suffice to compute \( \cos(0.2) \) with error at most \( 10^{-8} \)? Justify with the Lagrange remainder, and exploit the vanishing odd-degree terms.
Solution
Every derivative of \( \cos \) is \( \pm\sin \) or \( \pm\cos \), so \( |\cos^{(m)}(t)| \le 1 \) for all \( m, t \), and the hypotheses of the theorem hold on \( [0, 0.2] \) for every \( n \). Thus \[ |\cos(0.2) - T_n(0.2)| \le \frac{(0.2)^{n+1}}{(n+1)!}. \] Take \( n = 7 \). Because all odd Taylor coefficients of \( \cos \) vanish, \( T_7 = T_6 = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} \), and \[ |\cos(0.2) - T_6(0.2)| = |\cos(0.2) - T_7(0.2)| \le \frac{(0.2)^8}{8!} = \frac{2.56 \times 10^{-6}}{40320} \lt 6.4 \times 10^{-11} \lt 10^{-8}. \] So the four terms through degree \( 6 \) suffice (comfortably: even the cruder bound \( (0.2)^7/7! \lt 2.6 \times 10^{-9} \) applied to \( T_6 \) directly already beats \( 10^{-8} \)). Numerically \( T_6(0.2) = 1 - 0.02 + \frac{0.0016}{24} - \frac{0.000064}{720} = 0.98006657\ldots \)
Prove: if \( f^{(n+1)}(t) = 0 \) for every \( t \) in an interval \( I \), then \( f \) restricted to \( I \) is a polynomial of degree at most \( n \).
Solution
Fix \( a \in I \) and let \( x \in I \), \( x \ne a \). Since \( f^{(n+1)} \) exists on all of \( I \), the derivative \( f^{(n)} \) is differentiable, hence continuous, on \( I \), and likewise all lower derivatives are continuous; so the hypotheses of Taylor's theorem hold on the interval between \( a \) and \( x \). Therefore there is \( \xi \) between \( a \) and \( x \) with \[ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k, \] because \( f^{(n+1)}(\xi) = 0 \) by hypothesis. As \( x \in I \) was arbitrary (and the identity is trivial at \( x = a \)), \( f \) coincides on \( I \) with the fixed polynomial \( T_n \), which has degree at most \( n \). Note the converse is immediate, so \( f^{(n+1)} \equiv 0 \) on an interval characterises polynomials of degree \( \le n \) — and note also that connectedness of \( I \) is essential: on \( (0,1) \cup (2,3) \) a function can be two different polynomials.
(Harder.) Use Taylor's theorem for \( e^x \) to prove that \( e \) is irrational.
Solution
As in Worked Example 1, for every \( n \in \mathbb{N} \) there is \( \xi_n \in (0,1) \) with \[ e = \sum_{k=0}^{n}\frac{1}{k!} + \frac{e^{\xi_n}}{(n+1)!}, \qquad\text{hence}\qquad 0 \lt e - \sum_{k=0}^{n}\frac{1}{k!} \lt \frac{3}{(n+1)!}. \] Suppose for contradiction \( e = p/q \) with \( p, q \in \mathbb{N} \). Choose \( n \ge \max(q, 3) \) and multiply the displayed inequalities by \( n! \): \[ 0 \;\lt\; n!\,e - \sum_{k=0}^{n}\frac{n!}{k!} \;\lt\; \frac{3\,n!}{(n+1)!} = \frac{3}{n+1} \;\le\; \frac{3}{4} \;\lt\; 1. \] Now \( n!\,e = n!\,p/q \) is an integer because \( q \mid n! \) (as \( q \le n \)), and each \( n!/k! = n(n-1)\cdots(k+1) \) is an integer for \( k \le n \). So the middle quantity is an integer lying strictly between \( 0 \) and \( 1 \) — a contradiction. Hence \( e \) is irrational. (The strict positivity of the remainder, which the Lagrange form hands us for free since \( e^{\xi_n} \gt 0 \), is exactly what rules out the integer \( 0 \).)