Taylor's Theorem: The Remainder Envelope

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

T_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 = 3

: over the

T_3

of

\sin

, the same remainder argument gives

e^x

the bound

\frac{e}{24} \approx 0.1133

.

Show proof

Every derivative of $e^x$ is $e^x$ , so $f^{(k)}(0) = 1$ . Hence $T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ and $T_3(1) = 1 + 1 + \frac12 + \frac16 = \frac83 \approx 2.6667$ .
By Taylor's theorem (Lagrange remainder), $R_3(1) = \frac{e^c}{4!} \cdot 1^4$ for some $c \in (0, 1)$ . Since $e^c \le e$ , $|R_3(1)| \le \frac{e}{24} \approx 0.1133$ .
The true error is $e - \frac83 \approx 0.0516 < 0.1133$ , inside the bound — a remainder bound is a safe overestimate.

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

|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \xrightarrow{n \to \infty} 0 \quad \text{for every fixed } x

n = 9

: on

[-\pi, \pi]

the envelope's half-width narrows to

\pi^{10}/10! \approx 0.0258

.

Show proof

Every derivative of $\sin$ is $\pm\sin$ or $\pm\cos$ , so $|f^{(n+1)}(c)| \le 1$ . The Lagrange remainder is $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ , hence $|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!}$ .
For fixed $x$ the numerator is a power and the denominator a factorial, so $\frac{|x|^{n+1}}{(n+1)!} \to 0$ as $n \to \infty$ — the factorial beats the power.
Hence $|T_n(x) - \sin x| \to 0$ : the Maclaurin series converges to $\sin x$ for every $x$ .

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

f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k \;\Rightarrow\; c_j = \frac{f^{(j)}(a)}{j!}

n = 5

: the coefficients of

T_5(x) = x - x^3/6 + x^5/120

match

\frac{f^{(k)}(0)}{k!}

.

Show proof

Assume $f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k$ converges near $a$ and can be differentiated term by term.
Differentiating $j$ times gives $f^{(j)}(x) = \sum_{k=j}^{\infty} k(k-1)\cdots(k-j+1)\,c_k\,(x-a)^{k-j}$ (terms with $k < j$ vanish in the differentiation).
Setting $x = a$ kills every term with $k > j$ via $(x-a)^{k-j}$ ; only $k = j$ survives, giving $f^{(j)}(a) = j!\,c_j$ , i.e. $c_j = \frac{f^{(j)}(a)}{j!}$ — the Taylor coefficients are unique.

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