Taylor Series

Summary

The Taylor series of $f$ about $a$ is the power series built from derivatives of $f$ at $a$ . Truncations give polynomial approximations; Lagrange’s form bounds the remainder.

Prerequisites

Higher-Order Derivatives, Power Series, Maclaurin Series

Formula

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} (x - a)^{n},

when the series represents $f$ . The degree- $n$ Taylor polynomial is

P_{n} (x) = \sum_{k = 0}^{n} \frac{f^{(k)} (a)}{k!} (x - a)^{k} .

Lagrange remainder

If $f^{(n + 1)}$ exists on an interval containing $a$ and $x$ , then

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

for some $c$ between $a$ and $x$ . Hence if $| f^{(n + 1)} | \leq M$ on that interval,

| R_{n} (x) | \leq \frac{M}{(n + 1)!} | x - a |^{n + 1} .

Radius via coefficients

If $a_{n} = f^{(n)} (a) / n!$ and $lim | a_{n} / a_{n + 1} |$ exists, that limit is the radius $R$ .

Worked Example

About $a = 0$ : $e^{x}$ , $\sin x$ , and $\cos x$ have the standard series with $R = \infty$ .

For $\cos x$ , $| f^{(n + 1)} | \leq 1$ , so the remainder of $P_{n}$ satisfies $| R_{n} (x) | \leq | x |^{n + 1} / (n + 1)!$ .

Common Mistakes

Confusing the formal Taylor series with equality $f (x) = series$ without remainder control.
Applying the ratio formula to $a_{n} (x - a)^{n}$ including the $(x - a)^{n}$ factor instead of pure coefficients.

Connections

Maclaurin Series, Power Series, Estimating Series Sums

References

Taylor series and remainders are in OpenStax Calculus Volume 2.^[1]

OpenStax, Calculus Volume 2, Section 6.3, https://openstax.org/details/books/calculus-volume-2 ↩︎