10.10 Taylor Series and Lagrange Error Bound

I. Concept

1. The Taylor Series Formula

The Taylor series for a function $f(x)$ centered at $x = a$ is defined as:

\[f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\]

Maclaurin Series is a Taylor series centered at $a=0$:

\[f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]

Why the separate name?

While it might seem redundant to have two names for the same concept, the distinction exists for historical and practical reasons:

The "Special Case" Logic: A Maclaurin Series is simply a Taylor Series centered at $a = 0$. In mathematics, the most frequently used "special case" often earns its own name to simplify communication.
Historical Influence: Although Brook Taylor published the general version in 1715, Colin Maclaurin wrote the most influential calculus textbook of the 18th century (Treatise of Fluxions). He used the $a=0$ case so effectively that the mathematical community began associating that specific form with him.
The "Standard" Form: Most "Common Series" you memorize ($\sin x, e^x, \cos x$) are Maclaurin series. Having a specific name helps students and engineers immediately know the expansion is centered at the origin without needing to specify "$a=0$".

In short: Taylor gave us the tool, but Maclaurin gave us the manual that everyone actually read.

2. Essential Maclaurin Series to Memorize

In the BC exam, you must know these by heart:

Function	Maclaurin Series Expansion	General Term	Radius of Convergence
$\frac{1}{1-x}$	$1 + x + x^2 + x^3 + \dots$	$\sum_{n=0}^{\infty} x^n$	$R=1$
$e^x$	$1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$	$\sum_{n=0}^{\infty} \frac{x^n}{n!}$	$R=\infty$
$\sin x$	$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$	$R=\infty$
$\cos x$	$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$	$R=\infty$
$\arctan x$	$x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$	$R=1$
$\ln(1+x)$	$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$	$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$	$R=1$
$(1+x)^k$	$1 + kx + \frac{k(k-1)}{2!}x^2 + \dots$	$\sum_{n=0}^{\infty} \binom{k}{n} x^n$	$R=1$

3. Lagrange Error Bound (The Hard Part)

When we use a Taylor polynomial of degree $n$, $P_n(x)$, to approximate $f(x)$, the remainder (error) $R_n(x)$ is:

\[|R_n(x)| = |f(x) - P_n(x)| \leq \left| \frac{M}{(n+1)!} (x-c)^{n+1} \right|\]

Where $M$ is: The maximum value of the absolute value of the $(n+1)^{th}$ derivative of $f$ on the interval between $c$ and $x$: $$|f^{(n+1)}(z)| \leq M$$

4. Problem Solving Strategy

Find $n$: Identify the degree of the polynomial used.
Find $f^{(n+1)}(x)$: Calculate the next derivative.
Find $M$: Look for the maximum value of that derivative on the given interval.
Plug into the formula: Calculate the upper bound of the error.

II. Practice

🟢 例题 1：基础计算题 (Basic Calculation)

Question: The function $f$ has a Taylor polynomial of degree $n=3$ about $x=0$. If the fourth derivative satisfies $|f^{(4)}(x)| \leq 3$ for all $x$ in the interval $[0, 0.2]$, find the Lagrange error bound for the approximation of $f(0.2)$ using $P_3(0.2)$.

查看解析 (Click to Expand)

Identify $n$: $n=3$, so the formula uses $(n+1) = 4$.
Find $M$: Given $|f^{(4)}(x)| \leq 3$, so $M = 3$.
Find $(x-c)$: $x=0.2$ and $c=0$, so $(x-c) = 0.2$.
Plug into formula: $$|R_3(0.2)| \leq \frac{3}{4!}(0.2)^4$$ $$|R_3(0.2)| \leq \frac{3}{24}(0.0016) = \frac{1}{8}(0.0016) = 0.0002$$ Answer: The maximum error is $0.0002$.

🔴 例题 2：AP 考试真题风格 (FRQ Style)

Question: A function $f$ is approximated by $P_4(x)$ about $x=2$. The graph of $|f^{(5)}(x)|$ is shown to be decreasing on $[2, 2.5]$ with $f^{(5)}(2) = 10$ and $f^{(5)}(2.5) = 2$. Find the Lagrange error bound for $|f(2.5) - P_4(2.5)|$.

查看解析 (Click to Expand)

Analyze $M$: On the interval $[2, 2.5]$, $|f^{(5)}(x)|$ is decreasing. Therefore, its maximum value $M$ occurs at the left endpoint, $x=2$.
$M = 10$.
Apply Formula ($n=4$): $$|R_4(2.5)| \leq \frac{10}{(4+1)!}(2.5-2)^{4+1}$$ $$|R_4(2.5)| \leq \frac{10}{5!}(0.5)^5$$ $$|R_4(2.5)| \leq \frac{10}{120}(0.03125) \approx 0.0026$$

注意 (Common Mistake)

在计算 $M$ 时，一定要看清题目给的是第几阶导数。如果是 $P_n$，公式里永远用的是 $n+1$ 阶。

https://math.stackexchange.com/questions/1974159/algebraic-expansion-of-1x1-x-is

Function	Maclaurin Series Expansion	General Term	Radius of Convergence
\(\frac{1}{1-x}\)	\(1 + x + x^2 + x^3 + \dots\)	\(\sum_{n=0}^{\infty} x^n\)	\(R=1\)
\(e^x\)	\(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\)	\(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)	\(R=\infty\)
\(\sin x\)	\(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots\)	\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)	\(R=\infty\)
\(\cos x\)	\(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots\)	\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)	\(R=\infty\)
\(\arctan x\)	\(x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots\)	\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\)	\(R=1\)
\(\ln(1+x)\)	\(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots\)	\(\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}\)	\(R=1\)
\((1+x)^k\)	\(1 + kx + \frac{k(k-1)}{2!}x^2 + \dots\)	\(\sum_{n=0}^{\infty} \binom{k}{n} x^n\)	\(R=1\)