Taylor's Theorem with Remainder

Taylor's theorem quantifies the error in approximating a function by its Taylor polynomial, providing the rigorous foundation for power series representations.

The Theorem

Theorem7.4Taylor's Theorem (Lagrange Form)

Let $f : [a, b] \to \mathbb{R}$ be $(n+1)$ times differentiable on $(a, b)$ with $f^{(n+1)}$ continuous on $[a, b]$ . Then for any $x \in [a, b]$ and any point $a_0 \in [a, b]$ , $f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a_0)}{k!}(x - a_0)^k + R_n(x)$ where the Lagrange remainder is $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a_0)^{n+1}$ for some $c$ strictly between $a_0$ and $x$ .

Theorem7.5Taylor's Theorem (Integral Form)

Under the same hypotheses, the remainder can be expressed as $R_n(x) = \frac{1}{n!}\int_{a_0}^{x} f^{(n+1)}(t)(x - t)^n \, dt$ This form is often more useful for obtaining sharp error bounds.

Error Estimation

ExampleApproximating $e$

Using the Maclaurin series for $e^x$ at $x = 1$ with $n$ terms: $e = \sum_{k=0}^{n} \frac{1}{k!} + R_n, \quad |R_n| = \frac{e^c}{(n+1)!} \leq \frac{3}{(n+1)!}$ since $e^c < e < 3$ for $c \in (0, 1)$ . With $n = 9$ : $|R_9| < \frac{3}{10!} \approx 8.3 \times 10^{-7}$ , giving $e \approx 2.718282$ accurate to six decimal places.

ExampleApproximating $\sin(0.1)$

$\sin(0.1) = 0.1 - \frac{(0.1)^3}{3!} + R_3$ The bound $|R_3| \leq \frac{(0.1)^5}{5!} = \frac{10^{-5}}{120} \approx 8.3 \times 10^{-8}$ shows that even two terms give accuracy to seven decimal places: $\sin(0.1) \approx 0.0998333$ .

Convergence Criteria

RemarkWhen does the Taylor series converge to $f$?

The Taylor series of $f$ converges to $f(x)$ if and only if $R_n(x) \to 0$ as $n \to \infty$ . A sufficient condition is the existence of a constant $M$ such that $|f^{(n)}(x)| \leq M^n$ for all $n$ in a neighborhood of $a_0$ . This holds for $e^x$ , $\sin x$ , $\cos x$ (where derivatives are bounded by $1$ or the function itself), but fails for functions like $f(x) = e^{-1/x^2}$ .