3.4 Taylor and Maclaurin Series with Error Bounds

3.4 Taylor and Maclaurin Series with Error Bounds

Taylor and Maclaurin expansions are among the highest-yield problem-solving tools on the GRE Mathematics Subject Test. The examination rewards candidates who recognize that Taylor expansions convert complex transcendental functions into elementary polynomial algebra to evaluate high-order derivatives, resolve indeterminate limits, and compute rigorous error bounds.

Taylor and Maclaurin Series Definitions and Analyticity

Let $f$ be infinitely differentiable ($C^\infty$) on an open interval containing $a$.

• The Taylor series of $f$ centered at $a$ is: $T(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$
• When centered at $a = 0$, the series is the Maclaurin series: $M(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$

Smoothness vs. Analyticity

A function is smooth if it has derivatives of all orders ($C^\infty$). It is real analytic at $a$ if its Taylor series converges to $f(x)$ on an open neighborhood of $a$.

• Counterexample: $f(x) = e^{-1/x^2}$ for $x \ne 0$ with $f(0) = 0$ is $C^\infty$ on $\mathbb{R}$ with $f^{(n)}(0) = 0$ for all $n$. Its Maclaurin series is identically zero, converging everywhere but equaling $f(x)$ only at $x = 0$.

Essential Maclaurin Expansions Reference Table

The following fundamental expansions must be memorized alongside their intervals of convergence:

Fast Extraction of High-Order Derivatives

Because power series representations are unique, if $f(x) = \sum_{n=0}^\infty c_n (x - a)^n$, the coefficients are directly linked to the derivatives: $c_n = \frac{f^{(n)}(a)}{n!} \iff f^{(n)}(a) = n! \cdot c_n$

• GRE Shortcut: To find $f^{(k)}(0)$ for large $k$, never take repeated derivatives. Expand $f(x)$ into its Maclaurin series by substitution, identify the coefficient $c_k$ of $x^k$, and multiply by $k!$: $f^{(k)}(0) = k! \cdot c_k$.

Evaluating Indeterminate Limits via Series Expansions

For indeterminate forms $\lim_{x \to 0} \frac{f(x)}{g(x)}$ of type $\frac{0}{0}$ of order $\ge 2$, repeated L'Hôpital applications invite algebraic sign errors. Substituting low-order series terms resolves the limit immediately: $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{x - (x - x^3/6 + O(x^5))}{x^3} = \frac{1}{6}$

Operations on Power Series

Power series can be combined within their shared convergence domains:

• Substitution: For $f(x) = \cos(x^2)$, substitute $u = x^2$ into $\cos u = 1 - u^2/2 + u^4/24 - \cdots$ to obtain $1 - x^4/2 + x^8/24 - \cdots$.
• Cauchy Product: $\left(\sum a_n x^n\right)\left(\sum b_n x^n\right) = \sum c_n x^n$ where $c_n = \sum_{k=0}^n a_k b_{n-k}$.

Taylor's Theorem and the Lagrange Form of the Remainder

Let $f$ have $n+1$ continuous derivatives on an interval containing $a$ and $x$. The $n$-th degree Taylor polynomial is $P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k$, with remainder $R_n(x) = f(x) - P_n(x)$.

Lagrange Remainder Formula

There exists a point $c$ strictly between $a$ and $x$ such that: $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1}$ To bound error $|R_n(x)| \le \epsilon$, maximize $|f^{(n+1)}(c)|$ over $t \in [a, x]$: $|R_n(x)| \le \frac{\max_{t \in [a, x]} |f^{(n+1)}(t)|}{(n+1)!} |x - a|^{n+1}$

Summary Comparison Table: Taylor Remainder Forms

Worked Problem: Extracting High-Order Derivatives

Problem: Let $f(x) = \frac{x^2}{1 + 2x^3}$. Find the exact value of $f^{(11)}(0)$.

Solution:

1. Expand $f(x)$ using geometric series $\frac{1}{1-u} = \sum_{k=0}^\infty u^k$ with $u = -2x^3$: $f(x) = x^2 \sum_{k=0}^\infty (-2x^3)^k = \sum_{k=0}^\infty (-1)^k 2^k x^{3k+2}$
2. Set exponent to 11: $3k + 2 = 11 \implies 3k = 9 \implies k = 3$.
3. The coefficient of $x^{11}$ is $c_{11} = (-1)^3 2^3 = -8$.
4. Since $c_{11} = \frac{f^{(11)}(0)}{11!}$, we obtain $f^{(11)}(0) = 11! \cdot c_{11} = -8 \cdot 11!$.

GRE Exam Traps & Fast Test-Taking Strategies

• Trap 1: Evaluating high derivatives by brute force. If an exam problem asks for $f^{(12)}(0)$, taking derivatives directly is a trap. Expand into a Maclaurin series and extract $c_{12} \cdot 12!$.
• Trap 2: Forgetting $(n+1)!$ in the Lagrange denominator. The remainder after degree $n$ uses $(n+1)!$ and the $(n+1)$-th derivative, not $n!$.
• Trap 3: Overlooking even/odd function remainder improvements. For $\cos x$, $P_2(x) = P_3(x)$ because the $x^3$ coefficient is 0. You can bound the remainder using $n = 3$ (with 4th derivative and $4!$) to achieve a much tighter error bound.