Taylor's Theorem: The Remainder Envelope

The error Rn(x)R_n(x) is trapped in the band Tn(x)±∣x∣n+1/(n+1)!T_n(x) \pm |x|^{n+1}/(n+1)!. Raise nn and the band shrinks to zero.

f(x)=Tn(x)+Rn(x),∣Rn(x)∣≤∣x∣n+1(n+1)!f(x) = T_n(x) + R_n(x), \quad |R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!}

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Examples & Exercises

Approximate ee with the degree-3 Taylor polynomial of exe^x, and check the Lagrange bound e24≈0.1133\frac{e}{24} \approx 0.1133 covers the true error.

T3(1)=83≈2.6667,∣R3(1)∣≤e24≈0.1133,e−83≈0.0516T_3(1) = \frac{8}{3} \approx 2.6667, \quad |R_3(1)| \le \frac{e}{24} \approx 0.1133, \quad e - \frac{8}{3} \approx 0.0516
n=3n = 3: over the T3T_3 of sin⁡\sin, the same remainder argument gives exe^x the bound e24≈0.1133\frac{e}{24} \approx 0.1133.
Show proof
  1. Every derivative of exe^x is exe^x, so f(k)(0)=1f^{(k)}(0) = 1. Hence T3(x)=1+x+x22+x36T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} and T3(1)=1+1+12+16=83≈2.6667T_3(1) = 1 + 1 + \frac12 + \frac16 = \frac83 \approx 2.6667.
  2. By Taylor's theorem (Lagrange remainder), R3(1)=ec4!⋅14R_3(1) = \frac{e^c}{4!} \cdot 1^4 for some c∈(0,1)c \in (0, 1). Since ec≤ee^c \le e, ∣R3(1)∣≤e24≈0.1133|R_3(1)| \le \frac{e}{24} \approx 0.1133.
  3. The true error is e−83≈0.0516<0.1133e - \frac83 \approx 0.0516 < 0.1133, inside the bound — a remainder bound is a safe overestimate.

Use the remainder bound ∣x∣n+1(n+1)!→0\frac{|x|^{n+1}}{(n+1)!} \to 0 to prove the Maclaurin series of sin⁡x\sin x converges to sin⁡x\sin x for every xx.

∣Rn(x)∣≤∣x∣n+1(n+1)!→n→∞0for every fixed x|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \xrightarrow{n \to \infty} 0 \quad \text{for every fixed } x
n=9n = 9: on [−π,π][-\pi, \pi] the envelope's half-width narrows to π10/10!≈0.0258\pi^{10}/10! \approx 0.0258.
Show proof
  1. Every derivative of sin⁡\sin is ±sin⁡\pm\sin or ±cos⁡\pm\cos, so ∣f(n+1)(c)∣≤1|f^{(n+1)}(c)| \le 1. The Lagrange remainder is Rn(x)=f(n+1)(c)(n+1)!xn+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}, hence ∣Rn(x)∣≤∣x∣n+1(n+1)!|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!}.
  2. Put an=∣x∣n+1(n+1)!a_n = \frac{|x|^{n+1}}{(n+1)!}; then an+1an=∣x∣n+2\frac{a_{n+1}}{a_n} = \frac{|x|}{n+2}. Once n+2>2∣x∣n + 2 > 2|x| we get an+1≤12ana_{n+1} \le \frac{1}{2}a_n, so for n≥N:=⌈2∣x∣⌉n \ge N := \lceil 2|x| \rceil, an≤aN(12)n−N→0a_n \le a_N\left(\frac{1}{2}\right)^{n-N} \to 0 — the factorial beats the power geometrically.
  3. Hence ∣Tn(x)−sin⁡x∣→0|T_n(x) - \sin x| \to 0: the Maclaurin series converges to sin⁡x\sin x for every xx.

If f(x)=∑k=0∞ck(x−a)kf(x) = \sum_{k=0}^{\infty} c_k (x-a)^k can be differentiated termwise, prove ck=f(k)(a)k!c_k = \frac{f^{(k)}(a)}{k!} (uniqueness of Taylor coefficients).

f(x)=∑k=0∞ck(x−a)k  ⇒  cj=f(j)(a)j!f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k \;\Rightarrow\; c_j = \frac{f^{(j)}(a)}{j!}
n=5n = 5: the coefficients of T5(x)=x−x3/6+x5/120T_5(x) = x - x^3/6 + x^5/120 match f(k)(0)k!\frac{f^{(k)}(0)}{k!}.
Show proof
  1. Assume f(x)=∑k=0∞ck(x−a)kf(x) = \sum_{k=0}^{\infty} c_k (x-a)^k converges near aa and can be differentiated term by term.
  2. Differentiating jj times gives f(j)(x)=∑k=j∞k(k−1)⋯(k−j+1) ck (x−a)k−jf^{(j)}(x) = \sum_{k=j}^{\infty} k(k-1)\cdots(k-j+1)\,c_k\,(x-a)^{k-j} (terms with k<jk < j vanish in the differentiation).
  3. Setting x=ax = a kills every term with k>jk > j via (x−a)k−j(x-a)^{k-j}; only k=jk = j survives, giving f(j)(a)=j! cjf^{(j)}(a) = j!\,c_j, i.e. cj=f(j)(a)j!c_j = \frac{f^{(j)}(a)}{j!} — the Taylor coefficients are unique.

10. Differentiation