IB AA: Maclaurin and Taylor Series

Powerful functions like $e^x$, $\sin x$, and $\ln (1+x)$ can be represented as infinite polynomials, giving mathematicians and scientists a versatile tool for approximation, analysis, and computation. This study of Maclaurin and Taylor series transforms how we handle complex functions by expressing them as sums of simpler terms, a concept central to calculus and its countless applications in physics and engineering.

From Polynomials to Power Series: The Maclaurin Foundation

A Maclaurin series is a specific type of power series that represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point—specifically, at $x=0$. It is the function's "infinite-degree polynomial" approximation centered at zero. The general formula is derived from the idea that if a function can be represented by such a series, its coefficients must be related to its derivatives.

For a function $f(x)$ that is infinitely differentiable at $x=0$, its Maclaurin series is given by:

$$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\cdots =\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$$

Here, $f^{(n)}(0)$ denotes the $n$-th derivative evaluated at zero, and $n!$ is factorial notation. You derive the series for standard functions by repeatedly differentiating, evaluating at zero, and identifying the pattern. For instance:

• $e^x$: All derivatives are $e^x$, and $e^0=1$, so the series is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$.
• $\sin x$: The derivatives cycle through $\sin x,\cos x,-\sin x,-\cos x$. Evaluating at zero gives the pattern $0,1,0,-1,...$, leading to the series $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$.

It is crucial to remember that this equality, $f(x)=\text{Series}$, only holds where the series converges to the function. A series is useless if it doesn't converge, which leads us to the next core concept.

Generalization: The Taylor Series About Any Point

The Taylor series generalizes the Maclaurin concept. It allows us to expand a function about any point $x=a$, not just zero. This is incredibly useful when the behavior of a function near a specific point $a$ is of interest, or when the function is not defined at zero (like $\ln x$).

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

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\frac{f^{\prime \prime \prime }(a)}{3!}(x-a{)}^3+\cdots =\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$

The expression $(x-a)$ is the key difference. The series is now in powers of $(x-a)$, meaning it is most accurate near $x=a$. A Maclaurin series is simply a Taylor series with $a=0$. For example, the Taylor series for $\ln x$ centered at $a=1$ is $(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\cdots$, which converges for $0<x\le 2$.

Convergence: Radius, Interval, and the Ratio Test

A series representation is only valid where it converges. The radius of convergence, $R$, is a non-negative number such that the power series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$. It tells you the "radius" of the interval around the center $a$ where the series works.

The interval of convergence is the actual set of $x$-values for which the series converges. You find it by first determining $R$ and then separately testing the endpoints $x=a\pm R$. The interval could be $(a-R,a+R)$, $[a-R,a+R)$, $(a-R,a+R]$, or $[a-R,a+R]$.

The primary tool for finding $R$ is the ratio test for convergence. For a series $\sum a_n$, you examine the limit $L={\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|$. For a power series $\sum c_n(x-a{)}^n$, this becomes:

$$L=\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}(x-a{)}^{n+1}}{c_n(x-a{)}^n}\right|=|x-a|\cdot \underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|$$

By the ratio test, the series converges absolutely if $L<1$. Setting $L<1$ and solving for $|x-a|$ gives you the radius: $R=1/{\lim }_{n\to \infty }|c_{n+1}/c_n|$, provided the limit exists.

Application: Using Series for Approximation

One of the most practical uses of these series is using series for approximation. Since calculating an infinite sum is impossible, we truncate the series after a finite number of terms to create a polynomial approximation. For example, using the first three non-zero terms of the Maclaurin series for $\sin x$ gives $\sin x\approx x-\frac{x^3}{6}$. For small values of $x$, this is remarkably accurate and far simpler to compute than the actual sine function.

The more terms you include, the better the approximation and the larger the range of $x$-values for which it is valid. This technique is fundamental in calculators, physics (like the small-angle approximation in pendulum motion), and engineering for simplifying complex models.

Quantifying Accuracy: Error Estimation Techniques

Whenever you use a truncated series, you introduce an error. Error estimation techniques allow you to bound or approximate this error, telling you how accurate your approximation is. The primary method in the IB syllabus involves the Lagrange form of the remainder.

If you approximate $f(x)$ using the first $(n+1)$ terms of its Taylor series (i.e., up to the term with the $n$-th derivative), the error $R_n(x)$ is given by:

$$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$$

for some value $c$ that lies between $a$ and $x$. You don't know $c$, but you can often find a maximum possible value for $|f^{(n+1)}(c)|$ on the interval between $a$ and $x$. This allows you to state that the absolute error is at most a certain value. For instance, to show that approximating $e^{0.1}$ with the first four terms of its Maclaurin series has an error less than $10^{-6}$, you would bound the size of the fifth derivative, $e^c$, on the interval $[0,0.1]$.

Common Pitfalls

1. Assuming Convergence Equals Representation: The biggest mistake is assuming the series equals the function for all $x$ once you've found the formula. You must determine the interval of convergence. The series for $\ln (1+x)$ is valid only for $-1<x\le 1$. Plugging in $x=2$ gives a divergent series, not the value of $\ln (3)$.
2. Misapplying the Ratio Test: When using the ratio test on $\sum c_n(x-a{)}^n$, students often forget to take the absolute value or mishandle the limit. Remember, you are solving ${\lim }_{n\to \infty }|a_{n+1}/a_n|<1$ for $x$. The limit should simplify to the form $|x-a|\cdot K$, where $K$ is a constant from the coefficients $c_n$.
3. Confusing Series Type and Center: Using a Maclaurin series $(a=0)$ for a function like $\ln x$ is impossible because $\ln 0$ is undefined. You must choose an appropriate center $a$ (like $a=1$) and use the general Taylor series form with $(x-a{)}^n$.
4. Error Estimation Oversight: When using the Lagrange remainder, a common error is using the wrong derivative degree. If your polynomial is of degree $n$ (last term is the $n$-th derivative), the error term uses the $(n+1)$-th derivative. Also, failing to find a maximum for $|f^{(n+1)}(c)|$ on the interval renders the bound useless.

Summary

• A Maclaurin series is a power series expansion of a function about $x=0$, with coefficients derived from the function's derivatives at zero: $f(x)=\sum \frac{f^{(n)}(0)}{n!}{x}^n$.
• The Taylor series generalizes this to expansions about any point $x=a$, using powers of $(x-a)$: $f(x)=\sum \frac{f^{(n)}(a)}{n!}(x-a{)}^n$.
• The radius of convergence $R$ defines the interval around $a$ where the series converges, found using the ratio test. The interval of convergence requires checking the endpoints $x=a\pm R$ separately.
• Truncated series provide powerful approximations for functions, with accuracy improving as more terms are included.
• The Lagrange remainder $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$ provides a method for error estimation, allowing you to bound the maximum error in a polynomial approximation.