9.4 Working with Taylor Series
Use a small library of known series as building blocks: the binomial series, substitution, multiplication, differentiation and integration — then put them to work on impossible integrals, tricky limits, physics approximations and differential equations.
Finding a Taylor series straight from Definition 9.4 means computing $f^{(n)}(a)$ for every $n$ — fine for $e^x$ or $\sin x$, hopeless for $e^{-x^2}$ or $\tan^{-1}x$, whose derivatives get messier at every step. Fortunately we rarely have to. The uniqueness theorem (Theorem 9.7) says that any power series that equals $f$ on an interval is its Taylor series, however we found it. So we can start from a handful of known series and build new ones by algebra and calculus, exactly as we build new functions from old.
Once we have series, they become a powerful tool. They let us compute integrals like $\displaystyle\int_0^1e^{-x^2}dx$ that have no elementary antiderivative (and which appear everywhere in probability and statistics), evaluate awkward limits in one line, explain the approximations physicists use every day — $\sin\theta\approx\theta$ for a pendulum, $\tfrac12mv^2$ as the low-speed limit of Einstein's kinetic energy — and even solve differential equations.
The Binomial Series
For a positive integer $k$, the binomial theorem from school expands $(1+x)^k$ as a finite sum. Newton discovered that the same pattern works for any real exponent — $k=\tfrac12$, $k=-1$, $k=-\tfrac32$ — provided we allow the sum to be infinite. We need a generalised binomial coefficient.
📘 Definition 9.5 — Binomial coefficients for any real $k$
For a real number $k$ and an integer $n\ge0$,
$$\binom{k}{0}=1,\qquad\binom{k}{n}=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}\quad(n\ge1).$$
There are $n$ factors on top. Example: $\dbinom{1/2}{3}=\dfrac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{3!}=\dfrac{3/8}{6}=\dfrac1{16}$. When $k$ is a non-negative integer and $n\gt k$, one factor on top is $0$, so $\binom kn=0$.
📐 Theorem 9.10 — The binomial series
Let $k$ be any real number. Then
$$(1+x)^k=\sum_{n=0}^\infty\binom{k}{n}x^n=1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3+\cdots\qquad\text{for }-1\lt x\lt1.$$
If $k$ is a non-negative integer the series stops after the $x^k$ term and the formula holds for all $x$ (the ordinary binomial theorem). Otherwise the radius of convergence is exactly $1$; whether the series also converges at $x=\pm1$ depends on $k$.
💡 Show proof▶
Radius If $k$ is not a non-negative integer, no coefficient is $0$ and $\left|\dfrac{\binom{k}{n+1}x^{n+1}}{\binom{k}{n}x^n}\right|=\dfrac{|k-n|}{n+1}|x|\to|x|$. By the ratio test the series converges for $|x|\lt1$ and diverges for $|x|\gt1$: $R=1$.
A differential equation Let $g(x)=\sum\binom knx^n$ for $|x|\lt1$. From the definition, $(n+1)\binom{k}{n+1}=(k-n)\binom kn$. Differentiating term by term (Section 9.2) and collecting the coefficient of $x^n$ in $(1+x)g^\prime(x)$ gives $(n+1)\binom k{n+1}+n\binom kn=k\binom kn$. Hence $$(1+x)\,g^\prime(x)=k\,g(x).$$
Identify $g$ Let $h(x)=g(x)(1+x)^{-k}$. Then $h^\prime(x)=(1+x)^{-k-1}\left[(1+x)g^\prime(x)-kg(x)\right]=0$, so $h$ is constant on $(-1,1)$, and $h(0)=g(0)=1$. Therefore $g(x)=(1+x)^k$. $\blacksquare$
✏️ Example 1 — Square roots from the binomial series
Write the first five terms of the Maclaurin series of $\sqrt{1+x}$. Use it to approximate $\sqrt{17}$ and estimate the error.
💡 Show solution▶
Coefficients $k=\tfrac12$: $\binom{1/2}{1}=\tfrac12$, $\binom{1/2}{2}=\dfrac{\frac12\cdot\left(-\frac12\right)}{2}=-\dfrac18$, $\binom{1/2}{3}=\dfrac1{16}$, $\binom{1/2}{4}=\dfrac{\frac12\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{24}=-\dfrac{5}{128}$. So for $|x|\lt1$ $$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots$$
Rewrite We need $|x|\lt1$, and the smaller the better: $\sqrt{17}=\sqrt{16\left(1+\tfrac1{16}\right)}=4\sqrt{1+\tfrac1{16}}$.
Approximate Using four terms with $x=\tfrac1{16}$: $$\sqrt{17}\approx4\left(1+\frac1{32}-\frac1{2048}+\frac1{65\,536}\right)=4(1.030\,777\,6)=4.123\,108.$$
Error From the $x$ term on, the terms alternate in sign and decrease in size, so by the alternating series estimation theorem (Section 8.5) the error is less than the first omitted term: $4\cdot\dfrac{5}{128}\left(\dfrac1{16}\right)^4\approx2.4\times10^{-6}$. Check: $\sqrt{17}=4.123\,105\,6$, error $2.3\times10^{-6}$. ✓
Building New Series from Old
🧭 Problem-solving strategy — four ways to make a new series
Start from a known series (Theorem 9.9, Theorem 9.10, $\ln(1+x)$, $\tan^{-1}x$ below) with radius $R$.
1. SubstituteReplace $x$ by $cx^m$, e.g. $e^{-x^2}$ from $e^u$ with $u=-x^2$. The series converges where $|cx^m|\lt R$.
2. MultiplyBy a polynomial (just shift powers), or by another series: multiply like long polynomials and collect each power of $x$ in turn.
3. DifferentiateTerm by term, inside $|x-a|\lt R$. The radius does not change (Section 9.2).
4. IntegrateTerm by term, inside $|x-a|\lt R$; remember the constant (fix it by evaluating at the centre). The radius does not change.
By uniqueness (Theorem 9.7), whatever series you obtain is the Taylor series of the new function.
✏️ Example 2 — Substitution, integration, multiplication and differentiation
Find Maclaurin series (with radius of convergence) for (a) $e^{-x^2}$; (b) $\ln(1+x)$ and $\tan^{-1}x$; (c) the first four non-zero terms of $e^x\sin x$; (d) $\dfrac{1}{(1-x)^2}$.
💡 Show solution▶
(a) Substitute Put $u=-x^2$ in $e^u=\sum\frac{u^n}{n!}$ (valid for all $u$): $$e^{-x^2}=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{n!}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots,\qquad R=\infty.$$ (Compare: finding $f^{(6)}(0)$ directly would take six product-rule differentiations.)
(b) Integrate Substituting $-t$ and $-t^2$ in the geometric series: $\dfrac1{1+t}=\sum(-1)^nt^n$ and $\dfrac1{1+t^2}=\sum(-1)^nt^{2n}$ for $|t|\lt1$. Integrating from $0$ to $x$ term by term: $$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}=x-\frac{x^2}2+\frac{x^3}3-\cdots,\qquad \tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}3+\frac{x^5}5-\cdots$$ both with $R=1$. (No constant appears because $\ln1=0$ and $\tan^{-1}0=0$.) One can show both also hold at $x=1$, giving $\ln2=1-\tfrac12+\tfrac13-\cdots$ and $\tfrac\pi4=1-\tfrac13+\tfrac15-\cdots$.
(c) Multiply $$\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots\right)\left(x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots\right).$$ Collect powers: $x$: $1$. $x^2$: $1$. $x^3$: $\tfrac12-\tfrac16=\tfrac13$. $x^4$: $\tfrac16-\tfrac16=0$. $x^5$: $\tfrac1{24}-\tfrac1{12}+\tfrac1{120}=-\tfrac1{30}$. So $$e^x\sin x=x+x^2+\frac{x^3}{3}-\frac{x^5}{30}+\cdots,\qquad R=\infty.$$
(d) Differentiate $\dfrac{d}{dx}\dfrac1{1-x}=\dfrac1{(1-x)^2}$, so differentiating $\sum x^n$ term by term: $$\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty(n+1)x^n=1+2x+3x^2+\cdots,\qquad R=1.$$ Check with Theorem 9.10, $k=-2$, $x\to-x$: $\binom{-2}{n}(-1)^n=n+1$. ✓
Integrals You Cannot Do Any Other Way
The bell-curve function $e^{-x^2}$ and the "sinc" function $\dfrac{\sin x}{x}$ are perfectly continuous, so they have antiderivatives (by the Fundamental Theorem of Calculus) — but those antiderivatives cannot be written using finitely many elementary functions. Series solve the problem: integrate term by term, then use an error estimate to decide how many terms to keep.
✏️ Example 3 — Non-elementary integrals to a stated accuracy
(a) Evaluate $\displaystyle\int_0^1e^{-x^2}dx$ with error less than $0.001$. (b) Evaluate $\displaystyle\int_0^1\frac{\sin x}{x}dx$ with error less than $10^{-5}$ (the integrand is taken to be $1$ at $x=0$).
💡 Show solution▶
(a) Series From Example 2(a), integrating term by term (allowed: $R=\infty$): $$\int_0^1e^{-x^2}dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1x^{2n}dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!\,(2n+1)}=1-\frac13+\frac1{10}-\frac1{42}+\frac1{216}-\frac1{1320}+\cdots$$
(a) How many terms The series alternates and its terms decrease to $0$, so the error is less than the first term left out. $\tfrac1{216}\approx0.0046$ is too big but $\tfrac1{1320}\approx0.000\,76\lt0.001$, so we keep the five terms up to $\tfrac1{216}$: $$\int_0^1e^{-x^2}dx\approx1-\frac13+\frac1{10}-\frac1{42}+\frac1{216}\approx0.7475.$$ (Accurate value: $0.746\,824$; error $0.000\,66$. ✓)
(b) Series $\dfrac{\sin x}{x}=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots$ (dividing the sine series by $x$; at $x=0$ this gives $1$, matching the limit). Integrating: $$\int_0^1\frac{\sin x}{x}dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\,(2n+1)!}=1-\frac1{18}+\frac1{600}-\frac1{35\,280}+\frac1{3\,265\,920}-\cdots$$
(b) How many terms $\tfrac1{35\,280}\approx2.8\times10^{-5}$ is still too big; $\tfrac1{3\,265\,920}\approx3.1\times10^{-7}\lt10^{-5}$. Keep four terms: $1-\tfrac1{18}+\tfrac1{600}-\tfrac1{35\,280}\approx0.946\,083$. (Accurate value $0.946\,083\,07$. ✓)
Limits and Physical Approximations
Near $x=0$ a Maclaurin series tells us how fast each piece of an expression goes to zero, which makes many limits immediate — often faster and safer than repeated L'Hôpital.
✏️ Example 4 — Limits by series
Evaluate (a) $\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^3}$; (b) $\displaystyle\lim_{x\to0}\frac{e^{x^2}-1-x^2}{x^4}$; (c) $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$.
💡 Show solution▶
(a) $x-\sin x=x-\left(x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots\right)=\frac{x^3}{6}-\frac{x^5}{120}+\cdots$, so $$\frac{x-\sin x}{x^3}=\frac16-\frac{x^2}{120}+\cdots\ \longrightarrow\ \frac16.$$
(b) $e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots$ (substitute $u=x^2$), so $\dfrac{e^{x^2}-1-x^2}{x^4}=\dfrac12+\dfrac{x^2}{6}+\cdots\to\dfrac12$. (L'Hôpital would need four rounds here.)
(c) $\ln(1+x)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$, so the limit is $-\dfrac12$.
Check Numerically at $x=0.01$: (a) $0.166\,666$, (b) $0.500\,017$, (c) $-0.496\,7$ — each close to the limit, approaching at the rate the next series term predicts.
Small-angle approximation. A pendulum of length $L$ obeys $\theta^{\prime\prime}+\frac gL\sin\theta=0$, a non-linear equation with no elementary solution. Since $\sin\theta=\theta-\frac{\theta^3}{6}+\cdots$, for small swings we replace $\sin\theta$ by $\theta$, get $\theta^{\prime\prime}+\frac gL\theta=0$, whose solution $\theta=\theta_0\cos\left(\sqrt{g/L}\,t\right)$ has period $T_0=2\pi\sqrt{L/g}$ — independent of the amplitude. The alternating series tells us the price: $|\sin\theta-\theta|\le\frac{|\theta|^3}{6}$. A more careful analysis (using the binomial series inside an integral — see the Research–Teaching Nexus) gives the true period $T\approx T_0\left(1+\frac{\theta_0^2}{16}\right)$, so big swings are slower. Interactive 2 lets you see both effects.
✏️ Example 5 — Einstein's kinetic energy at everyday speeds
In special relativity a mass $m$ moving at speed $v$ has kinetic energy $K=mc^2\left(\dfrac{1}{\sqrt{1-v^2/c^2}}-1\right)$, where $c\approx3.00\times10^8$ m/s. Show that $K\approx\tfrac12mv^2$ when $v$ is small compared with $c$, find the first correction, and estimate its relative size for a car at $30$ m/s and for an electron at $v=0.1c$.
💡 Show solution▶
Binomial series With $k=-\tfrac12$ and $x=-\dfrac{v^2}{c^2}$ (allowed since $v\lt c$): $\binom{-1/2}{1}=-\tfrac12$, $\binom{-1/2}{2}=\tfrac38$, $\binom{-1/2}{3}=-\tfrac5{16}$, so $$\left(1-\frac{v^2}{c^2}\right)^{-1/2}=1+\frac12\frac{v^2}{c^2}+\frac38\frac{v^4}{c^4}+\frac5{16}\frac{v^6}{c^6}+\cdots$$
Kinetic energy $$K=mc^2\left(\frac12\frac{v^2}{c^2}+\frac38\frac{v^4}{c^4}+\cdots\right)=\frac12mv^2+\frac38\frac{mv^4}{c^2}+\cdots$$ The leading term is Newton's $\tfrac12mv^2$. Relative to it, the first correction is $\dfrac{3mv^4/(8c^2)}{mv^2/2}=\dfrac34\dfrac{v^2}{c^2}$.
Car $v/c=30/(3\times10^8)=10^{-7}$: correction $\tfrac34\times10^{-14}=7.5\times10^{-15}$ — far below anything measurable. Newtonian mechanics is perfectly adequate for cars.
Electron $v/c=0.1$: correction $\tfrac34(0.01)=0.0075$, i.e. $\tfrac12mv^2$ underestimates $K$ by about $0.75\%$. Check with the exact formula: $\dfrac{K}{\frac12mv^2}=\dfrac{2\left(1/\sqrt{0.99}-1\right)}{0.01}=1.007\,56$. ✓ Particle accelerators, where $v$ is close to $c$, must use the full formula.
Solving Differential Equations with Power Series
Many differential equations in physics and engineering have no solution in terms of elementary functions. A powerful alternative is to assume the solution is a power series $y=\sum c_nx^n$, substitute, and use uniqueness to match the coefficient of each power of $x$. This turns the differential equation into a recurrence relation for the coefficients.
✏️ Example 6 — A power-series solution
Solve $y^\prime=2xy$, $y(0)=1$ by assuming $y=\displaystyle\sum_{n=0}^\infty c_nx^n$. Identify the function you obtain and check it.
💡 Show solution▶
Substitute $y^\prime=\displaystyle\sum_{n=1}^\infty nc_nx^{n-1}=c_1+2c_2x+3c_3x^2+\cdots$ and $2xy=\displaystyle\sum_{n=0}^\infty2c_nx^{n+1}=2c_0x+2c_1x^2+\cdots$
Match powers Constant term: $c_1=0$. Coefficient of $x^m$ ($m\ge1$): $(m+1)c_{m+1}=2c_{m-1}$, i.e. $$c_{m+1}=\frac{2}{m+1}\,c_{m-1}.$$ The initial condition gives $c_0=y(0)=1$.
Solve the recurrence Odd coefficients: $c_1=0\Rightarrow c_3=c_5=\dots=0$. Even ones: $c_2=\tfrac22c_0=1$, $c_4=\tfrac24c_2=\tfrac12$, $c_6=\tfrac26c_4=\tfrac16$, and in general $c_{2k}=\dfrac{1}{k!}$. So $$y=\sum_{k=0}^\infty\frac{x^{2k}}{k!}=\sum_{k=0}^\infty\frac{\left(x^2\right)^k}{k!}=e^{x^2}.$$
Check $y=e^{x^2}$ gives $y^\prime=2xe^{x^2}=2xy$ and $y(0)=1$. ✓ (The equation is also separable — Section 7.3 — and gives the same answer.)
⚠️ Common mistakes when working with series
1. Forgetting the new radius after substitution. $\dfrac1{1-x}$ needs $|x|\lt1$, so $\dfrac1{1+4x^2}=\sum(-4x^2)^n$ needs $|4x^2|\lt1$, i.e. $|x|\lt\tfrac12$.
2. Dropping the constant after integrating. $\int\frac{dx}{1-x}=-\ln(1-x)+C$; evaluate at the centre to find $C$.
3. Using the alternating-series bound too early. The bound "error $\lt$ first omitted term" needs the terms to be decreasing. In $\int_0^3e^{-x^2}dx$ the terms grow up to $n=7$ before shrinking (try it in Interactive 1).
4. Truncating too soon in a limit. In $\lim\frac{x-\sin x}{x^3}$, using only $\sin x\approx x$ gives $\frac00$; you must keep terms up to the power in the denominator.
Exercises
I
The binomial series.
- 1 Find the first four terms of the Maclaurin series of $(1+x)^{-2}$ and state the radius of convergence.
- 2 Find the first four terms of the Maclaurin series of $\sqrt[3]{1+x}$.
- 3 Find the first four terms of the Maclaurin series of $\dfrac{1}{\sqrt{1-x}}$.
- 4 Find the first four terms of the Maclaurin series of $\dfrac{1}{\sqrt{1+2x}}$ and its radius of convergence.
- 5 Use the binomial series to compute $\sqrt{26}$ to four decimal places.
- 6 Find the Maclaurin series of $(1-x)^{-3}$ (general term) and its radius.
💡 Show solutions▶
1 $k=-2$: $\binom{-2}{n}=(-1)^n(n+1)$. $(1+x)^{-2}=1-2x+3x^2-4x^3+\cdots$, $R=1$.
2 $k=\tfrac13$: $\binom{1/3}{2}=\dfrac{\frac13\left(-\frac23\right)}{2}=-\dfrac19$, $\binom{1/3}{3}=\dfrac{\frac13\left(-\frac23\right)\left(-\frac53\right)}{6}=\dfrac{5}{81}$: $\sqrt[3]{1+x}=1+\dfrac x3-\dfrac{x^2}{9}+\dfrac{5x^3}{81}-\cdots$ ($|x|\lt1$).
3 $k=-\tfrac12$, $x\to-x$: $\binom{-1/2}{n}(-1)^n=\dfrac{1\cdot3\cdot5\cdots(2n-1)}{2^nn!}$: $\dfrac1{\sqrt{1-x}}=1+\dfrac x2+\dfrac{3x^2}{8}+\dfrac{5x^3}{16}+\cdots$ ($|x|\lt1$).
4 $k=-\tfrac12$ with $2x$ in place of $x$: $1-\tfrac12(2x)+\tfrac38(2x)^2-\tfrac5{16}(2x)^3+\cdots=1-x+\dfrac32x^2-\dfrac52x^3+\cdots$. Need $|2x|\lt1$: $R=\tfrac12$.
5 $\sqrt{26}=5\left(1+\tfrac1{25}\right)^{1/2}=5\left(1+\tfrac1{50}-\tfrac1{5000}+\tfrac1{250\,000}-\cdots\right)\approx5.099\,02$. The next term is $5\cdot\tfrac5{128}\cdot25^{-4}=5\times10^{-7}$, so $\sqrt{26}\approx5.0990$. (Calculator: $5.099\,019\,5$.)
6 Differentiate $\dfrac1{1-x}=\sum x^n$ twice: $\dfrac{2}{(1-x)^3}=\sum_{n\ge2}n(n-1)x^{n-2}$, so $(1-x)^{-3}=\displaystyle\sum_{n=0}^\infty\frac{(n+1)(n+2)}{2}x^n=1+3x+6x^2+10x^3+\cdots$, $R=1$.
II
Build each series from known ones (substitution, multiplication, differentiation, integration). Give the radius of convergence where asked.
- 1 $\dfrac{x}{1+x^2}$ (general term and radius).
- 2 $x^2e^{-3x}$ (general term and radius).
- 3 $\cos\sqrt x$ for $x\ge0$ (general term).
- 4 $\ln\left(1-x^2\right)$ (general term and radius).
- 5 Show $\ln\dfrac{1+x}{1-x}=2\displaystyle\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$ for $|x|\lt1$, and use four terms with $x=\tfrac13$ to approximate $\ln2$.
- 6 The first four non-zero terms of $e^x\cos x$.
- 7 The first three non-zero terms of $\tan x=\dfrac{\sin x}{\cos x}$, by long division of series.
- 8 Find the sum $\displaystyle\sum_{n=1}^\infty\frac{n}{3^n}$. (Hint: differentiate the geometric series, then multiply by $x$.)
- 9 Find the sum $\displaystyle\sum_{n=1}^\infty\frac{1}{n\,2^n}$.
💡 Show solutions▶
1 $x\cdot\dfrac1{1+x^2}=x\sum(-x^2)^n=\displaystyle\sum_{n=0}^\infty(-1)^nx^{2n+1}=x-x^3+x^5-\cdots$, $R=1$.
2 $x^2\sum\dfrac{(-3x)^n}{n!}=\displaystyle\sum_{n=0}^\infty\frac{(-3)^nx^{n+2}}{n!}=x^2-3x^3+\frac92x^4-\cdots$, $R=\infty$.
3 Put $u=\sqrt x$ in $\cos u=\sum\frac{(-1)^nu^{2n}}{(2n)!}$: $\cos\sqrt x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^n}{(2n)!}=1-\frac x2+\frac{x^2}{24}-\cdots$
4 $\ln(1+u)$ with $u=-x^2$: $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}(-x^2)^n}{n}=-\sum_{n=1}^\infty\frac{x^{2n}}{n}=-x^2-\frac{x^4}2-\cdots$; need $|x^2|\lt1$: $R=1$.
5 $\ln(1+x)-\ln(1-x)$: the even powers cancel and the odd ones double, giving $2\left(x+\frac{x^3}3+\frac{x^5}5+\cdots\right)$. $\frac{1+x}{1-x}=2$ when $x=\frac13$: $\ln2\approx2\left(\tfrac13+\tfrac1{81}+\tfrac1{1215}+\tfrac1{15\,309}\right)\approx0.693\,135$ (true $0.693\,147$; error $1.2\times10^{-5}$ — far faster than $1-\frac12+\frac13-\cdots$).
6 $\left(1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}\right)\left(1-\frac{x^2}2+\frac{x^4}{24}\right)$: $x^2$: $\tfrac12-\tfrac12=0$; $x^3$: $\tfrac16-\tfrac12=-\tfrac13$; $x^4$: $\tfrac1{24}-\tfrac14+\tfrac1{24}=-\tfrac16$. So $e^x\cos x=1+x-\dfrac{x^3}3-\dfrac{x^4}6+\cdots$
7 Write $\tan x=a_1x+a_3x^3+a_5x^5+\cdots$ (odd function) and require $\left(1-\frac{x^2}2+\frac{x^4}{24}\right)(a_1x+a_3x^3+a_5x^5)=x-\frac{x^3}6+\frac{x^5}{120}$: $a_1=1$; $a_3-\tfrac12=-\tfrac16\Rightarrow a_3=\tfrac13$; $a_5-\tfrac16+\tfrac1{24}=\tfrac1{120}\Rightarrow a_5=\tfrac2{15}$. So $\tan x=x+\dfrac{x^3}3+\dfrac{2x^5}{15}+\cdots$
8 $\sum_{n\ge1}nx^{n-1}=\dfrac1{(1-x)^2}$, so $\sum nx^n=\dfrac{x}{(1-x)^2}$ for $|x|\lt1$. At $x=\tfrac13$: $\dfrac{1/3}{4/9}=\dfrac34$.
9 $-\ln(1-x)=\displaystyle\sum_{n=1}^\infty\frac{x^n}{n}$ for $|x|\lt1$. At $x=\tfrac12$: $\displaystyle\sum\frac1{n2^n}=-\ln\tfrac12=\ln2\approx0.6931$.
III
Integrals and limits with series.
- 1 $\displaystyle\int_0^{0.5}e^{-x^2}dx$ with error less than $10^{-4}$.
- 2 $\displaystyle\int_0^1\sin\left(x^2\right)dx$ with error less than $0.001$.
- 3 $\displaystyle\int_0^{0.5}\frac{dx}{1+x^4}$ with error less than $10^{-4}$.
- 4 $\displaystyle\int_0^1\frac{e^x-1}{x}dx$ with error less than $10^{-4}$. (The terms are positive here — bound the tail yourself.)
- 5 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}$
- 6 $\displaystyle\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$
- 7 $\displaystyle\lim_{x\to0}\frac{\sin x-x+\frac{x^3}{6}}{x^5}$
- 8 $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-x+\frac{x^2}{2}}{x^3}$
- 9 $\displaystyle\lim_{x\to\infty}x^2\left(1-\cos\frac1x\right)$
Check: partial sums of your series against an accurate numerical integral
number of terms—
partial sum—
accurate integral—
actual error—
💡 Show solutions▶
1 $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n(0.5)^{2n+1}}{n!(2n+1)}=0.5-0.041\,667+0.003\,125-0.000\,186+0.000\,009-\cdots$ Alternating and decreasing; $0.000\,009\lt10^{-4}$, so four terms: $\approx0.461\,272$. (Accurate: $0.461\,281$.)
2 $\sin(x^2)=\sum\frac{(-1)^nx^{4n+2}}{(2n+1)!}$, so $\displaystyle\int_0^1=\sum_{n=0}^\infty\frac{(-1)^n}{(4n+3)(2n+1)!}=\frac13-\frac1{42}+\frac1{1320}-\frac1{75\,600}+\cdots$ $\tfrac1{1320}\approx0.000\,76\lt0.001$ already, so two terms suffice: $\approx0.3095$; with three, $0.310\,281$. (Accurate: $0.310\,268$.)
3 $\dfrac1{1+x^4}=\sum(-1)^nx^{4n}$ for $|x|\lt1$, so $\displaystyle\int_0^{0.5}=\sum_{n=0}^\infty\frac{(-1)^n(0.5)^{4n+1}}{4n+1}=0.5-0.006\,25+0.000\,217-0.000\,009\,4+\cdots$ Three terms: $\approx0.493\,967$ (error $\lt9.4\times10^{-6}$; accurate $0.493\,958$).
4 $\dfrac{e^x-1}{x}=\displaystyle\sum_{n=1}^\infty\frac{x^{n-1}}{n!}$, so the integral is $\displaystyle\sum_{n=1}^\infty\frac1{n\cdot n!}=1+\frac14+\frac1{18}+\frac1{96}+\frac1{600}+\frac1{4320}+\cdots$ The tail after $n=6$ is less than $\dfrac{1}{7\cdot7!}\left(1+\tfrac18+\tfrac1{64}+\cdots\right)=\dfrac{1}{35\,280}\cdot\dfrac87\approx3.2\times10^{-5}$ (each term is less than $\tfrac18$ of the previous one). So six terms: $\approx1.317\,870$, i.e. $1.3179$. (Accurate: $1.317\,902$.)
5 $e^x-1-x=\frac{x^2}2+\frac{x^3}6+\cdots$: limit $\dfrac12$.
6 $\tan^{-1}x-x=-\frac{x^3}3+\frac{x^5}5-\cdots$: limit $-\dfrac13$.
7 $\sin x-x+\frac{x^3}6=\frac{x^5}{120}-\frac{x^7}{5040}+\cdots$: limit $\dfrac1{120}$.
8 $\ln(1+x)-x+\frac{x^2}2=\frac{x^3}3-\frac{x^4}4+\cdots$: limit $\dfrac13$.
9 Put $u=\tfrac1x\to0^+$: $\dfrac{1-\cos u}{u^2}=\dfrac{\frac{u^2}2-\frac{u^4}{24}+\cdots}{u^2}\to\dfrac12$.
IV
Applications and differential equations.
- 1 Pendulum. For a swing of $\theta=10^\circ$, compute the error and relative error of $\sin\theta\approx\theta$, and compare with the bound $\frac{|\theta|^3}{6}$.
- 2 Particle physics. An electron moves at $v=0.1c$. By what percentage does $\tfrac12mv^2$ underestimate the relativistic kinetic energy? Compare the exact answer with the series prediction $\tfrac34(v/c)^2$.
- 3 Space station. Gravity at height $h$ above the Earth is $g(h)=\dfrac{g_0}{(1+h/R)^2}$ with $R=6371$ km. Use the binomial series to give linear and quadratic approximations in $h/R$, and compare them with the exact value of $g/g_0$ at the International Space Station's height $h=400$ km.
- 4 Solve $y^{\prime\prime}+y=0$, $y(0)=1$, $y^\prime(0)=0$ with a power series and identify the solution.
- 5 Solve $y^\prime=xy$, $y(0)=1$ with a power series and identify the solution.
- 6 Challenge (Airy's equation, used in optics and quantum mechanics). Find the first four non-zero terms of the power-series solution of $y^{\prime\prime}=xy$, $y(0)=1$, $y^\prime(0)=0$.
💡 Show solutions▶
1 $\theta=\tfrac{\pi}{18}=0.174\,533$ rad, $\sin\theta=0.173\,648$. Error $0.000\,885$ rad; relative error $0.000\,885/0.173\,648\approx0.51\%$. Bound: $\theta^3/6=0.000\,886$ rad — essentially exact, since the next series term $\theta^5/120\approx1.3\times10^{-6}$ is tiny.
2 Exact: $\dfrac{K}{\frac12mv^2}=\dfrac{2(\gamma-1)}{(v/c)^2}$ with $\gamma=\dfrac1{\sqrt{0.99}}=1.005\,038$: ratio $1.007\,56$, so $\tfrac12mv^2$ is about $0.76\%$ too small. Series: $\tfrac34(0.1)^2=0.75\%$ — the next term, $\tfrac58(v/c)^4\approx0.006\%$, accounts for the difference.
3 With $u=h/R$: $(1+u)^{-2}=1-2u+3u^2-\cdots$ ($|u|\lt1$). For $h=400$ km, $u=0.062\,78$: linear $1-2u=0.8744$; quadratic $1-2u+3u^2=0.8863$; exact $(1.062\,78)^{-2}=0.8853$. So astronauts on the ISS still feel about $89\%$ of surface gravity (they float because they are in free fall). The linear model is off by about $1.2\%$, the quadratic by about $0.1\%$.
4 $y^{\prime\prime}=\sum(n+2)(n+1)c_{n+2}x^n$, so $(n+2)(n+1)c_{n+2}=-c_n$. With $c_0=1$, $c_1=0$: odd $c$'s vanish, $c_2=-\tfrac12$, $c_4=\tfrac1{24}$, $c_{2k}=\dfrac{(-1)^k}{(2k)!}$. So $y=\sum\dfrac{(-1)^kx^{2k}}{(2k)!}=\cos x$. (Check: $-\cos x+\cos x=0$, $y(0)=1$, $y^\prime(0)=0$.)
5 $(m+1)c_{m+1}=c_{m-1}$, $c_1=0$, $c_0=1$: $c_{2k}=\dfrac{c_{2k-2}}{2k}=\dfrac{1}{2^kk!}$. So $y=\sum\dfrac{(x^2/2)^k}{k!}=e^{x^2/2}$. (Check: $y^\prime=xe^{x^2/2}=xy$.)
6 $\sum(n+2)(n+1)c_{n+2}x^n=\sum c_nx^{n+1}$: constant term $2c_2=0$, and $(m+2)(m+1)c_{m+2}=c_{m-1}$ for $m\ge1$. With $c_0=1$, $c_1=0$: $c_3=\tfrac{c_0}{6}=\tfrac16$, $c_6=\tfrac{c_3}{30}=\tfrac1{180}$, $c_9=\tfrac{c_6}{72}=\tfrac1{12\,960}$, all others $0$. So $$y=1+\frac{x^3}{6}+\frac{x^6}{180}+\frac{x^9}{12\,960}+\cdots$$
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes
Research question. Does a pendulum's period really depend on how far it swings — and does the series $T\approx T_0\left(1+\frac{\theta_0^2}{16}\right)$ predict the effect you can measure?
Why it matters. Pendulum clocks kept the world's time for nearly three centuries, and their makers fought "circular error" — the slowing of large swings. The same small-angle question arises today in building sway under wind, ship roll, seismometers and the swinging loads of construction cranes.
What to do.
- Theory: energy conservation gives the exact period $T=\dfrac{2T_0}{\pi}\displaystyle\int_0^{\pi/2}\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}}$ with $k=\sin\frac{\theta_0}{2}$. Expand $(1-k^2\sin^2\varphi)^{-1/2}$ with the binomial series and integrate term by term (use $\int_0^{\pi/2}\sin^2\varphi\,d\varphi=\frac\pi4$) to get $T\approx T_0\left(1+\frac{k^2}{4}\right)\approx T_0\left(1+\frac{\theta_0^2}{16}\right)$.
- Experiment: hang a small heavy object (keys, a nut) on a string about $1$ m long. Time $10$ swings with a phone stopwatch for $\theta_0\approx10^\circ,20^\circ,40^\circ,60^\circ$ (measure the angle with a protractor or a phone app). Repeat each three times.
- Plot measured $T/T_0$ against $\theta_0$ together with $1+\theta_0^2/16$ and the small-angle prediction $T/T_0=1$. Use Interactive 2 for the exact curve.
Tools: String, small mass, phone stopwatch, protractor, spreadsheet
Hand in: A one-page report: the derivation, your table and graph, the size of your timing uncertainty, and whether you could detect the amplitude effect at each angle.
Going further. Keep the next binomial term to get $T\approx T_0\left(1+\frac{\theta_0^2}{16}+\frac{11\theta_0^4}{3072}\right)$ and compare all three models at $60^\circ$ and $90^\circ$.
Instructor note: the effect is about $0.19\%$ at $10^\circ$, $0.77\%$ at $20^\circ$, $3.1\%$ at $40^\circ$ and $7.3\%$ at $60^\circ$; with ten-swing timing (reaction error about $0.1$ s over about $20$ s) students can usually detect it from roughly $30^\circ$ upwards. At $60^\circ$ the two-term series gives $6.9\%$, slightly low.
🧪 Self-check quiz
0/5 correct
Q1. The binomial coefficient $\dbinom{1/2}{2}$ equals
Q2. In the Maclaurin series of $e^{-x^2}$, the coefficient of $x^6$ is
Q3. $\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^3}=$
Q4. For $v\ll c$, the relativistic kinetic energy is $K=\tfrac12mv^2+(\text{first correction})+\cdots$. The first correction is
Q5. The Maclaurin series of $\dfrac{1}{1+4x^2}$ converges for