5.19 Maclaurin series

Polynomials are the easiest functions to work with: their values need only addition and multiplication, and they can be differentiated and integrated term by term. A Maclaurin series writes a function such as $e^x$ or $\sin x$ as a polynomial with infinitely many terms. Taking only the first few terms gives a polynomial that is very close to the function for values of $x$ near $0$.

Calculators and computer software evaluate functions such as $\sin x$ and $e^x$ with polynomial approximations of this kind (or with closely related methods), after first reducing $x$ to a small interval near $0$ where few terms are needed. The same series also give quick estimates of values such as $\sqrt{e}$

Approximating a function by a polynomial

Matching the derivatives at zero

Near $x = 0$, the simplest approximation to a function is its tangent line, $f(x) \approx f(0) + f'(0)x$. The tangent has the same value and the same gradient as $f$

Adding a term in $x^2$ allows the approximation to bend in the same way as the curve. The quadratic $1 + x + \frac{x^2}{2}$

The pattern continues. Suppose a polynomial

is to have the same value and the same first $n$ derivatives as $f$ at $x = 0$. Substituting $x = 0$ into $p$ and its derivatives gives:

and in general $p^{(k)}(0) = k!\,a_k$, because differentiating $x^k$

where $f^{(k)}$ is the $k$th derivative and $0! = 1$.

The Maclaurin series

Letting the number of terms grow without limit gives an infinite series.

A Maclaurin polynomial approximates the function near $0$, and for the functions in this course the approximation improves as more terms are taken, as long as $x$ is in the series' interval of convergence. The series is an example of a power series, a series in increasing whole-number powers of $x$.

The standard Maclaurin series

The series for e^x

Every derivative of $e^x$ is $e^x$, so $f^{(k)}(0) = e^0 = 1$

The factorials grow so quickly that the series converges for every real $x$. Substituting $x = 1$ gives a series for $e$ itself, $e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots$

The series for sin x and cos x

The derivatives of $\sin x$ repeat in a cycle of four: $\sin x$, $\cos x$, $-\sin x$, $-\cos x$, then $\sin x$

For $\cos x$, the values at $0$ are $1, 0, -1, 0, 1, \ldots$, so only the even powers survive:

Both series converge for every real $x$. They reflect the symmetry of the functions: $\sin x$ is odd and its series has only odd powers, while $\cos x$ is even and its series has only even powers.

The series for ln(1 + x)

The function $\ln x$ has no Maclaurin series, because $\ln 0$ is undefined. Shifting it to $\ln(1 + x)$ fixes this. Its derivatives are

and in general $f^{(k)}(0) = (-1)^{k - 1}(k - 1)!$. The coefficient of $x^k$

The denominators are $1, 2, 3, 4, \ldots$, not factorials: the factorials cancelled.

Unlike the series for $e^x$, $\sin x$ and $\cos x$, this series is valid only for $-1 < x \leq 1$. At $x = 1$

The series for (1 + x)^p

For $f(x) = (1 + x)^p$, the derivatives are $p(1 + x)^{p - 1}$

This is exactly the extended binomial theorem of 1.10. When $p \in \mathbb{N}$ the series stops after $p + 1$ terms and is valid for all $x$; otherwise, for $p \in \mathbb{Q}$, it is an infinite series valid for $|x| < 1$

Summary of the standard series

Finding a series by repeated differentiation

The general formula works for any function whose derivatives at $0$ can be found. The method is to differentiate several times, substitute $x = 0$ into each derivative, and then substitute the values into the formula. It is reliable but can involve heavy differentiation, so it suits the first three or four terms.

For a composite function, the chain rule and the product rule produce longer expressions at each stage. Simplifying each derivative before differentiating again keeps the working manageable, and a question often gives one derivative to show, which then serves as a check.

Building new series from known ones

Repeated differentiation quickly becomes long. Most series are found faster by adapting the standard series, in four ways: substitution, multiplication, differentiation and integration. These are valid because a power series behaves like a polynomial inside its interval of convergence.

Substitution

Replacing $x$ by an expression such as $x^2$, $-x$ or $3x$ throughout a known series gives the series of the new function. The interval of convergence changes with it.

Substitution also combines with the laws of logarithms. Since $\ln\left(\frac{1 + x}{1 - x}\right) = \ln(1 + x) - \ln(1 - x)$

for $-1 < x < 1$. The even powers cancel. Choosing $x = \frac{1}{3}$ makes $\frac{1 + x}{1 - x} = 2$

which is within $0.00015$ of $\ln 2 = 0.693147\ldots$ after only three terms. The series for $\ln(1 + x)$ with $x = 1$ would need thousands of terms for the same accuracy, which shows the benefit of choosing a small value of

Multiplication

The series of a product is the product of the two series, expanded and collected up to the power needed. Only the products of terms whose powers add to at most the highest power needed to be found.

Composition

A series can be substituted into another series. To expand $e^{\sin x}$, write $u = \sin x = x - \frac{x^3}{6} + \cdots$

Here $u^2 = x^2 + \ldots$ and $u^3 = x^3 + \ldots$

Differentiation and integration

Inside its interval of convergence, a power series can be differentiated or integrated term by term, like a polynomial. Differentiating the series for $\sin x$ gives

which is the series for $\cos x$, as it should be.

Integration gives the series for $\arctan x$, whose derivatives are awkward to find directly. Since $\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$

Integrate term by term:

Substituting $x = 0$ gives $\arctan 0 = 0 = C$, so

The series is valid for $-1 \leq x \leq 1$. At $x = 1$ it gives $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$

Integrating a series also gives approximate values of definite integrals that cannot be found exactly. The function $e^{-x^2}$, which underlies the normal distribution (4.9), has no antiderivative made of standard functions, but its series can be integrated term by term.

Maclaurin series from differential equations

Many differential equations (5.18) have no solution in terms of standard functions. If the value $y(0)$ is known, the differential equation itself gives $y'(0)$, and differentiating the equation gives $y''(0)$, $y'''(0)$

The method:

1. Substitute $x = 0$ and $y(0)$ into the differential equation to find $y'(0)$.

2. Differentiate the differential equation with respect to $x$, treating

When the differential equation contains products such as $xy$, each differentiation produces a pattern that can be used to find higher derivatives quickly.

Using Maclaurin series

Approximations

A Maclaurin polynomial gives good approximations for values of $x$ close to $0$, and the closer $x$ is to $0$, the fewer terms are needed. For the alternating series of $\sin x$, $\cos x$, $\ln(1 + x)$

To approximate a value, choose the function and the value of $x$ so that $x$ is small and inside the interval of convergence. To estimate $\ln 1.1$, use $\ln(1 + x)$ with $x = 0.1$

Evaluating limits

Limits of the indeterminate form $\frac{0}{0}$ as $x \to 0$ can be found by replacing each function by its series. After simplifying, the lowest power of $x$ cancels and the limit can be read off. This gives the same results as L'Hôpital's rule (5.13), and is often quicker when the rule would have to be applied several times.

The standard limit $\lim_{x \to 0}\frac{\sin x}{x} = 1$ follows in the same way: $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots \to 1$

Validity and the interval of convergence

A Maclaurin series is an infinite sum, and an infinite sum has a value only if it converges (1.8). For some functions this happens for every $x$; for others only on an interval.

The series for $\ln(1 + x)$ shows what goes wrong outside the interval. At $x = 2$ the partial sums are $2,\ 0,\ 2.667,\ -1.333,\ 5.067,\ -5.6,\ 12.7, \ldots$

The interval of convergence of a new series follows from the series it was built from:

• Substituting $kx$ for $x$ in a series valid for $|x| < 1$ gives a series valid for $|kx| < 1$, that is $|x| < \frac{1}{|k|}$

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Exam-style questions on Maclaurin series

Maclaurin series questions usually lead from finding a series, either by differentiation or from the standard series, to using it for an approximation, a limit or an integral. The first question below is for a paper without a GDC; the second allows one.

Self-check

• To do:

  I can explain why the coefficient of $x^n$ in a Maclaurin series is $\frac{f^{(n)}(0)}{n!}$