6.4 Working with Taylor Series

Big idea

Having a series for a function is only worth the trouble if the series does work that the closed form cannot. This section is that payoff. Three kinds of problem that are hard or impossible in closed form become routine once the function is a series: integrals with no elementary antiderivative, indeterminate limits, and numerical approximation with a guaranteed accuracy.

The common mechanism is that a series turns analysis into algebra. An integral of an infinite sum becomes an infinite sum of integrals of powers, and integrating a power is trivial. A limit of a quotient becomes a quotient of series, where the lowest surviving power in the numerator and denominator decides the answer immediately. Approximation becomes a matter of truncating, and truncating is exact bookkeeping as long as you can say how large the discarded part is.

Before any of that, you need the series, and the practical rule is never to compute derivatives if you can avoid it. The exponential, sine and cosine series are exact on the whole line, the geometric and logarithmic series are exact on their intervals, and from those five you reach a very large catalogue by substituting an expression for the variable, multiplying by a power, differentiating, or integrating. Each of those moves is legal inside the interval of convergence and takes one line.

The last piece is honesty about error. An approximation with no error estimate is a guess. Two tools cover almost every case: for an alternating series with decreasing terms, the error after truncating is smaller than the first discarded term, which is as convenient a statement as analysis ever offers. For everything else there is the Lagrange remainder, which converts a bound on one higher derivative into a bound on the error.

Decoder

Truncating a Taylor series after the degree $n$ term leaves an error equal to the next derivative evaluated at some unknown interior point, times the next power, divided by the next factorial.

Three of the four ingredients are known to you: the power $(x-a)^{n+1}$, the factorial $(n+1)!$, and the interval on which the unknown point $c$ must lie. Only $c$ itself is out of reach, and the whole technique consists of not needing it. Replace $f^{(n+1)}(c)$ by any number $M$ that is at least as large as $|f^{(n+1)}|$ everywhere on the interval between $a$ and $x$, and the equality becomes an inequality you can actually evaluate.

The word that gets misread is some. Taylor’s theorem asserts that such a point exists, not that you can find it, and not that it is the midpoint or the endpoint. Any argument that pins $c$ down to a specific value is doing something the theorem does not license. Any argument that bounds $f^{(n+1)}$ over the whole interval is doing exactly what the theorem was designed for.

The classic mistake is choosing $M$ by evaluating the derivative at the centre. The bound has to hold across the entire interval, so you want the maximum of the derivative’s absolute value there, not its value at one convenient point. When in doubt, choose an $M$ that is obviously too generous. A loose bound that is correct still proves the accuracy you need; a tight bound that is wrong proves nothing.

Definitions and results

The five series to know. For $|x| < 1$, $\frac{1}{1-x} = \sum x^n$ and $\ln(1+x) = \sum_{n \ge 1} (-1)^{n-1}x^n/n$. For all real $x$,

$$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \qquad \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}, \qquad \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} $$

Substitution. Replacing $x$ by an expression in a known series gives the series for the composed function, valid wherever the substituted expression lies in the original interval. This is the single most useful move.

Multiplication by a power. Multiplying a series by $x^m$ raises every exponent by $m$ and leaves the interval of convergence alone.

Differentiation and integration. Term-by-term differentiation and integration are legal inside the interval of convergence and preserve the radius, so a new series can be produced from a known one by calculus rather than by the coefficient formula.

Integrating a nonelementary function. If $f(x) = \sum c_n x^n$ on an interval containing $[0,b]$, then

$$ \int_0^b f(x)\,dx = \sum_{n=0}^{\infty} \frac{c_n b^{n+1}}{n+1} $$

This is how you compute definite integrals of functions like $e^{-x^2}$ and $\sin(x^2)$, which have no antiderivative expressible in elementary terms.

Limits from lowest-order terms. In an indeterminate quotient, replace numerator and denominator by their series, cancel the common power of $x$, and take the limit. The leading surviving coefficients decide the answer, and no repeated differentiation is required.

The alternating series error bound. If a convergent series alternates in sign and its terms decrease in absolute value to zero, then the error made by stopping after any term is smaller in absolute value than the first term you dropped, and has that term’s sign.

The Lagrange error bound. If $|f^{(n+1)}(t)| \le M$ for all $t$ between $a$ and $x$, then

$$ |R_n(x)| \le \frac{M\,|x-a|^{n+1}}{(n+1)!} $$

The binomial series. For any real exponent $r$ and $|x| < 1$,

$$ (1+x)^r = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \cdots $$

which terminates into the ordinary binomial theorem when $r$ is a nonnegative integer.

Worked examples

New series without differentiating

Start from the exponential series and substitute $-x^2$ for $x$:

$$ e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots $$

valid for every real $x$, since the original series was. Computing the fourth derivative of $e^{-x^2}$ by hand would take a page; the substitution takes a line.

Multiplication is just as cheap. Multiplying the sine series by $x$ gives

$$ x\sin x = x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \cdots $$

and differentiating the sine series term by term gives $1 - x^2/2! + x^4/4! - \cdots$, which is the cosine series, as it should be.

An integral with no elementary antiderivative

Approximate $\displaystyle \int_0^1 e^{-x^2}\,dx$.

No elementary function differentiates to $e^{-x^2}$, so the fundamental theorem is unavailable in closed form. Integrate the series instead, term by term, from $0$ to $1$:

$$ \int_0^1 e^{-x^2}\,dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!\,(2n+1)} = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \frac{1}{1320} + \cdots $$

The terms alternate and decrease, so stopping anywhere gives an error smaller than the next term. Adding the six terms shown:

$$ 1 - 0.333333 + 0.100000 - 0.023810 + 0.004630 - 0.000758 = 0.746729 $$

The first discarded term is $1/(720 \cdot 13) = 0.000107$, so the true value lies within $0.000107$ of $0.746729$. It is in fact $0.746824$, and the gap is $0.000095$, comfortably inside the bound.

A limit decided by the lowest power

Evaluate $\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x^3}$.

Both parts vanish at $0$, so the quotient is indeterminate. Substitute the sine series into the numerator and watch the cancellation:

$$ x - \sin x = x - \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \right) = \frac{x^3}{6} - \frac{x^5}{120} + \cdots $$

The linear terms cancel exactly, which is why this limit is delicate. Dividing by $x^3$:

$$ \frac{x - \sin x}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots $$

Every remaining term after the first carries a positive power of $x$, so the limit is $\frac{1}{6}$. The series also tells you how the quotient approaches its limit: from below, and quadratically.

A numeric error bound

Approximate $e^{0.5}$ with the degree three Maclaurin polynomial and bound the error.

$$ p_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}, \qquad p_3(0.5) = 1 + 0.5 + 0.125 + 0.0208333 = 1.6458333 $$

For the bound, the fourth derivative of $e^x$ is $e^x$ again, and on $[0, 0.5]$ its largest value is $e^{0.5}$. You need a number you are certain about rather than the answer itself, so use $e < 4$, which gives $e^{0.5} < 2$. Take $M = 2$:

$$ |R_3(0.5)| \le \frac{2 \cdot (0.5)^4}{4!} = \frac{2 \cdot 0.0625}{24} = 0.0052083 $$

So the approximation is correct to within about $0.0053$, which guarantees two decimal places. The true value is $1.6487213$, and the actual error is $0.0028879$, about half the bound. That gap is normal: the bound replaced the derivative at an unknown interior point with its worst case over the whole interval, so it is designed to be conservative.

If you needed three decimal places instead, keep one more term. The degree four bound is $2(0.5)^5/120 = 0.00052$, which delivers them.

Practice

Start with the error question, since it is the one that has a definite right answer and a definite method. Identify the degree, bound the relevant derivative on the interval, and evaluate the Lagrange expression, or use the alternating bound when the series alternates.

Then the constructions. These ask you to produce a series from a known one and use it, whether the goal is an integral, a limit, or a numerical value.

Quiz

Five items on deriving series from known ones, series-based integrals and limits, and error bounds.