Applications of Taylor series: approximation, error, limits, sums

Last week you learned to write the Taylor series. This week we put it to work: computing a number approximately by hand, bounding the error of that approximation provably, solving in one line limits for which L’Hôpital would take long, matching a series with an unknown sum to a familiar function, and computing numerically integrals that have no antiderivative. The exam most often has “approximate with error less than …” and “use series to evaluate the limit” questions.

Golden rules:

1. Approximation $=$ Taylor polynomial. Error $=$ the part that was dropped.
2. If the series is alternating and decreasing, the error is smaller than the first dropped term. This is the easiest bound, try it first.
3. If it is not alternating, Taylor’s inequality (the Lagrange error bound) is used; for $M$, estimate the largest value of the derivative on the interval.
4. In a limit, expand the numerator and denominator into series and cancel the lowest-degree term.

Taylor’s theorem and the remainder term

If $f$ is $n+1$ times differentiable near $a$, there is a $c$ between $a$ and $x$ such that

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

$R_n$ is called the remainder. We do not know $c$, but we can bound the derivative.

Taylor’s inequality. If $\lvert f^{(n+1)}(t)\rvert\le M$ between $a$ and $x$, then

$\lvert R_n(x)\rvert\le\frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}.$

Intuition. The remainder is very much like the next term to be added to the polynomial; only the derivative is evaluated at an unknown point $c$ instead of $a$. So you can think of it as “the worst case of the next term”.

Example 1. Compute $e^{0.1}$ with $P_3$ and bound the error.

$P_3(0.1) = 1+0.1+\frac{0.01}{2}+\frac{0.001}{6} = 1.1051667.$

$f^{(4)}(t) = e^t$, whose largest value on $[0,0.1]$ is $e^{0.1}<3$ (crude but safe). $M=3$:

$\lvert R_3\rvert\le\frac{3}{4!}(0.1)^4 = \frac{0.0003}{24} = 1.25\times10^{-5}.$

The true value is $1.1051709$, the actual error is $4.3\times10^{-6}$, below the bound.

PITFALL · choosing M. $M$ must be the largest absolute value of the derivative between $a$ and $x$. Choosing $M=e^0=1$ for $e^t$ is wrong; the function is increasing, and the largest value is at the right end. Putting in place of an unknown value ($e^{0.1}$) a simple number that exceeds it (3) is allowed.

Alternating series estimate

If the value of $x$ makes the series alternating and decreasing, the job is much easier: $\lvert R_n\rvert<\lvert\text{first dropped term}\rvert$.

Example 2. Compute $\ln(1.2)$ with three terms of $\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$

$0.2-\frac{0.04}{2}+\frac{0.008}{3} = 0.1826667.$

The error is $<\dfrac{(0.2)^4}{4} = 0.0004$. The true value is $\ln1.2 = 0.1823216$, the error is $0.000345$.

“How many terms are needed?” questions

Find the smallest $n$ that brings the error bound below the desired tolerance. In the alternating case you compute the dropped terms in order; in Taylor’s inequality you solve the inequality $\dfrac{M\lvert x-a\rvert^{n+1}}{(n+1)!}<\varepsilon$ by trying $n=1,2,3,\dots$

Example 2b. How many terms are needed to compute $\sin(0.5)$ with an error smaller than $10^{-6}$? $\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$ is alternating and decreasing for $x=0.5$. Compute the terms in order:

x^3/3! = 0.125/6        = 0.0208333
x^5/5! = 0.03125/120    = 0.0002604
x^7/7! = 0.0078125/5040 = 0.0000015500   > 1e-6
x^9/9! = 0.00195/362880 = 0.0000000054   < 1e-6

The term $x^9/9!$ is the first to drop below $10^{-6}$. So you have to take the four terms before it ($x$, $x^3$, $x^5$, $x^7$): $0.5-0.0208333+0.0002604-0.0000016 = 0.4794255$. The true value is $\sin0.5 = 0.4794255$.

Example 2c (with Taylor’s inequality). If you use $\cos x$ with $P_2 = 1-\frac{x^2}{2}$ on the interval $\lvert x\rvert\le0.1$, what is the largest the error can be? The derivatives of $\cos$ are $\pm\sin$, $\pm\cos$, all bounded by $1$, $M=1$. But note: there is no $x^3$ term in the series of $\cos$, so $P_2 = P_3$. The bound is better with $n=3$: $\lvert R_3\rvert\le\dfrac{1}{4!}(0.1)^4\approx4.2\times10^{-6}$. Even if the exam says “use $P_2$”, if you notice that it equals $P_3$ you can write a sharper bound; but the bound given with $n=2$ ($\frac{(0.1)^3}{6}\approx1.7\times10^{-4}$) is also correct, just cruder.

Intuition: how does the polynomial hug the curve?

Imagine drawing $P_1 = 1+x$ (the tangent line), $P_2$, $P_3$ next to the graph of $e^x$. Near $x=0$ they all sit almost on the curve. As $x$ grows, first $P_1$ separates, then $P_2$, then $P_3$. The factor $\lvert x-a\rvert^{n+1}$ in the error formula says exactly this: as the distance from the center grows, the error grows quickly, while the $(n+1)!$ in the denominator crushes the error as the number of terms increases. The practical conclusion: always make the approximation at a point close to the center. For $\ln(1.2)$ the series of $\ln(1+x)$ is suitable; for $\ln5$ the same series does not work at all, because $x=4$ is outside the interval.

Limits with series

In $0/0$ limits that would need L’Hôpital several times, expanding with series is faster and less error-prone.

Example 3. $\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim\frac{x-\left(x-\frac{x^3}{6}+\dots\right)}{x^3} = \lim\left(\frac16-\frac{x^2}{120}+\dots\right) = \frac16.$

Example 4. $\displaystyle\lim_{x\to0}\frac{\sin(x^2)}{1-\cos x}$. Numerator: $x^2-\frac{x^6}{6}+\dots$ Denominator: $\frac{x^2}{2}-\frac{x^4}{24}+\dots$ Divide both by $x^2$: $\dfrac{1-\dots}{\frac12-\dots}\to2$.

Rule: in the numerator and denominator, expand up to the lowest-degree nonzero term, that is enough.

Finding sums with series

See a number series as a particular value of $x$ in a familiar Maclaurin series.

Tip: if there is $n!$ in the denominator think of $e^x$, if $(2n)!$ think of $\cos$, if $(2n+1)!$ think of $\sin$, if only $n$ think of $\ln$, if $2n+1$ think of $\arctan$.

Integrals with series

$\int_0^1\sin(x^2)dx$ cannot be computed with elementary functions. Expand into a series:

$\sin(x^2) = x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\dots\quad\Rightarrow\quad\int_0^1\sin(x^2)dx = \frac13-\frac{1}{42}+\frac{1}{1320}-\frac{1}{75600}+\dots$

Alternating and decreasing. The first three terms give $0.3102814$, with error $<\frac{1}{75600}\approx1.3\times10^{-5}$. Four terms give $0.3102682$.

The binomial series

For every real $k$ and $\lvert x\rvert<1$

$(1+x)^k = \sum_{n=0}^\infty\binom kn x^n = 1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3+\dots$

If $k$ is a positive integer the series terminates (the binomial expansion you know). Otherwise it is infinite.

Example 5. $\dfrac{1}{\sqrt{1.1}}$: $k = -\frac12$, $x = 0.1$. $1-\frac12(0.1)+\frac{(-\frac12)(-\frac32)}{2}(0.01) = 1-0.05+0.00375 = 0.95375$. The true value is $0.95346$.

Common mistakes