9.2 Properties of Power Series
Inside its interval of convergence a power series behaves like a polynomial — add it, multiply it, differentiate and integrate it term by term — giving series for $\ln(1+x)$, $\arctan x$ and even $\pi$.
In Section 9.1 the geometric series gave us power series for functions such as $\dfrac{1}{1+x^2}$ and $\dfrac{x^3}{2-x}$. But what about $\ln(1+x)$ or $\arctan x$? No amount of substituting into $\dfrac{1}{1-x}$ will produce them. Notice, however, that their derivatives are $\dfrac{1}{1+x}$ and $\dfrac{1}{1+x^2}$, which we already know how to expand. If we could integrate a power series one term at a time — just as we integrate a polynomial — we would be done.
This section shows that inside its interval of convergence a power series really does behave like a polynomial: we can add, multiply, differentiate and integrate term by term, and the radius of convergence does not change. These rules open the door to series for logarithms and inverse trigonometric functions, a famous formula for $\pi$, exact sums such as $\sum n/2^n=2$, and the fact that a function has only one power series about a given centre.
Combining Power Series
📐 Theorem 9.3 — Algebra of power series
Suppose $\displaystyle\sum_{n=0}^{\infty}c_nx^n$ and $\displaystyle\sum_{n=0}^{\infty}d_nx^n$ converge to $f(x)$ and $g(x)$ respectively for every $x$ in an interval $I$.
- Sum and constant multiple: $\displaystyle\sum_{n=0}^{\infty}(c_n\pm d_n)x^n=f(x)\pm g(x)$ and $\displaystyle\sum_{n=0}^{\infty}k\,c_nx^n=k\,f(x)$ for every $x$ in $I$ (any constant $k$).
- Multiplying by a power: for any integer $m\ge0$, $\displaystyle\sum_{n=0}^{\infty}c_nx^{n+m}=x^mf(x)$ for every $x$ in $I$.
- Substitution: for any constant $b\ne0$ and integer $k\ge1$, $\displaystyle\sum_{n=0}^{\infty}c_nb^nx^{kn}=f\left(bx^k\right)$ for every $x$ with $bx^k$ in $I$.
- Product: if both series converge absolutely for $|x|\lt R$, then for $|x|\lt R$ $$f(x)g(x)=\sum_{n=0}^{\infty}e_nx^n,\qquad e_n=c_0d_n+c_1d_{n-1}+\cdots+c_nd_0=\sum_{k=0}^{n}c_kd_{n-k}.$$
Parts 1–3 are just the algebra of convergent series of numbers applied at each fixed $x$; we used parts 2 and 3 throughout Section 9.1. Part 4, called the Cauchy product, says we may multiply two power series like long polynomials and collect equal powers: the coefficient of $x^n$ collects every product $c_kx^k\cdot d_{n-k}x^{n-k}$.
✏️ Example 1 — Adding two series
Use $\dfrac1{1-x}$ and $\dfrac1{1+x}$ to find a power series for $f(x)=\dfrac{2}{1-x^2}$, and state where it is valid. Confirm the answer by substituting into the geometric series directly.
💡 Show solution▶
Split $\dfrac1{1-x}+\dfrac1{1+x}=\dfrac{(1+x)+(1-x)}{1-x^2}=\dfrac{2}{1-x^2}$.
Add For $|x|\lt 1$ both $\sum x^n$ and $\sum(-1)^nx^n$ converge, so by Theorem 9.3 $$\frac{2}{1-x^2}=\sum_{n=0}^{\infty}\left(1+(-1)^n\right)x^n=2+0x+2x^2+0x^3+2x^4+\cdots=\sum_{n=0}^{\infty}2x^{2n}.$$ The odd powers cancel and the even ones double.
Interval Valid on the common interval $(-1,1)$; at $x=\pm1$ the series is $2+2+2+\cdots$, which diverges.
Check Directly: $\dfrac{2}{1-x^2}=2\sum\left(x^2\right)^n=\sum 2x^{2n}$ for $x^2\lt 1$ — the same series. (Theorem 9.5 below says this is no coincidence.)
Differentiating and Integrating Power Series
📐 Theorem 9.4 — Term-by-term differentiation and integration
Suppose the power series $\displaystyle\sum_{n=0}^{\infty}c_n(x-a)^n$ has radius of convergence $R\gt 0$, and let $$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots\qquad\text{for }|x-a|\lt R.$$ Then $f$ is differentiable (and hence continuous) on $(a-R,\,a+R)$, and on that interval
$$f^\prime(x)=\sum_{n=1}^{\infty}n\,c_n(x-a)^{n-1}=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots$$
$$\int f(x)\,dx=C+\sum_{n=0}^{\infty}c_n\frac{(x-a)^{n+1}}{n+1}=C+c_0(x-a)+c_1\frac{(x-a)^2}{2}+c_2\frac{(x-a)^3}{3}+\cdots$$
Both new series have the same radius of convergence $R$. Their behaviour at the endpoints $x=a\pm R$ may be different from that of the original series.
Applying the theorem again to $f^\prime$ shows that a power series can be differentiated as many times as we like inside its interval: a function given by a power series is infinitely differentiable there. The proof of Theorem 9.4 is beyond this course, but the radius claim is easy to believe: multiplying $c_n$ by $n$ or dividing by $n+1$ does not change $\lim|c_{n+1}/c_n|$, because $\frac{n+1}{n}\to1$.
⚠️ Endpoints can change — an illustration
The series $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$ has interval of convergence $[-1,1)$ (Section 9.1). Differentiating and integrating term by term:
| series | radius | $x=-1$ | $x=1$ | interval |
|---|---|---|---|---|
| $\sum\dfrac{x^n}{n}$ | 1 | converges | diverges | $[-1,1)$ |
| derivative $\sum x^{n-1}$ | 1 | diverges | diverges | $(-1,1)$ |
| integral $\sum\dfrac{x^{n+1}}{n(n+1)}$ | 1 | converges | converges | $[-1,1]$ |
The radius never changes, but differentiation can lose an endpoint and integration can gain one. Always re-check the endpoints after differentiating or integrating.
✏️ Example 2 — $\dfrac{1}{(1-x)^2}$ two ways
Find a power series for $\dfrac{1}{(1-x)^2}$ (a) by differentiating the geometric series and (b) by multiplying $\dfrac1{1-x}$ by itself. Then find a series for $\dfrac{2}{(1-x)^3}$.
💡 Show solution▶
(a) Differentiate For $|x|\lt 1$, $\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots$. Since $\dfrac{d}{dx}\dfrac{1}{1-x}=\dfrac{1}{(1-x)^2}$, Theorem 9.4 gives $$\frac{1}{(1-x)^2}=0+1+2x+3x^2+4x^3+\cdots=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n,\qquad|x|\lt 1.$$
(b) Multiply With $c_k=d_k=1$, the Cauchy product has $e_n=\displaystyle\sum_{k=0}^{n}1\cdot1=n+1$, so again $\dfrac{1}{(1-x)^2}=\sum(n+1)x^n$ for $|x|\lt 1$. ✓
Again Differentiate once more: $$\frac{2}{(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)x^n=2+6x+12x^2+\cdots,\qquad|x|\lt 1.$$
Endpoints At $x=\pm1$ the terms $(n+1)(\pm1)^n$ do not tend to $0$, so both series have interval $(-1,1)$.
✏️ Example 3 — A power series for $\ln(1+x)$
Find a power series for $\ln(1+x)$ and its interval of convergence. Use four terms to estimate $\ln1.2$ and bound the error.
💡 Show solution▶
Start For $|t|\lt 1$, $\dfrac{1}{1+t}=\displaystyle\sum_{n=0}^{\infty}(-1)^nt^n=1-t+t^2-t^3+\cdots$
Integrate For $|x|\lt 1$, integrate from $0$ to $x$ term by term (Theorem 9.4): $$\ln(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$ Using the definite integral from $0$ fixes the constant: both sides are $0$ at $x=0$.
Interval The radius is still $1$. At $x=1$: $1-\tfrac12+\tfrac13-\cdots$ converges (alternating series test). At $x=-1$: $-\left(1+\tfrac12+\tfrac13+\cdots\right)$ diverges. Interval $(-1,1]$. (A result called Abel's theorem guarantees that at $x=1$ the sum equals $\ln2$, because $\ln(1+x)$ is continuous there: $\ln2=1-\tfrac12+\tfrac13-\tfrac14+\cdots$)
Estimate With $x=0.2$: $$\ln1.2\approx0.2-\frac{0.04}{2}+\frac{0.008}{3}-\frac{0.0016}{4}=0.182267.$$ The series is alternating with decreasing terms, so the error is less than the next term, $\dfrac{0.2^5}{5}=0.000064$.
Check A calculator gives $\ln1.2=0.182322$; the actual error is $0.000055\lt 0.000064$. ✓
✏️ Example 4 — $\arctan x$, and a series for $\pi$
Find a power series for $\arctan x$ and its interval of convergence. Deduce a series for $\pi$ and discuss how quickly it converges.
💡 Show solution▶
Start From Section 9.1, $\dfrac{1}{1+t^2}=\displaystyle\sum_{n=0}^{\infty}(-1)^nt^{2n}$ for $|t|\lt 1$.
Integrate Since $\arctan x=\displaystyle\int_0^x\frac{dt}{1+t^2}$, $$\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots,\qquad|x|\lt 1.$$
Interval $R=1$. At $x=\pm1$ the series is $\pm\left(1-\tfrac13+\tfrac15-\cdots\right)$, which converges by the alternating series test. Interval $[-1,1]$, and (Abel's theorem again) the sum at $x=1$ is $\arctan1=\tfrac{\pi}{4}$: $$\frac{\pi}{4}=1-\frac13+\frac15-\frac17+\cdots,\qquad\pi=4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}.$$ This formula was known to the Kerala mathematician Madhava in the 14th century and rediscovered in Europe by Gregory and Leibniz; it is often called the Madhava–Leibniz series.
Speed By the alternating series bound, $N$ terms give $\pi$ with error less than $\dfrac{4}{2N+1}$, and in fact the error is about $\dfrac1N$. For $1000$ terms we get $3.1406$ — only three correct digits. Guaranteeing an error below $10^{-4}$ needs $\dfrac{4}{2N+1}\lt 10^{-4}$, i.e. about $20\,000$ terms. The trouble is that $x=1$ sits on the edge of the interval of convergence, where the terms shrink only like $\tfrac1n$.
Faster Machin's identity $\dfrac{\pi}{4}=4\arctan\dfrac15-\arctan\dfrac1{239}$ uses $x$-values deep inside the interval. With just $7$ terms of each arctan series it gives $\pi$ to $10$ decimal places. Interactive 2 compares the two.
✏️ Example 5 — Summing $\sum n x^n$ by differentiation
Find a closed form for $\displaystyle\sum_{n=1}^{\infty}nx^n$ when $|x|\lt 1$, and use it to evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac12+\frac24+\frac38+\cdots$
💡 Show solution▶
Idea The factor $n$ is what differentiation produces: $\dfrac{d}{dx}x^n=nx^{n-1}$. So write $nx^n=x\cdot nx^{n-1}$.
Differentiate From Example 2, $\displaystyle\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$ for $|x|\lt 1$. Multiply by $x$ (Theorem 9.3): $$\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2},\qquad|x|\lt 1.$$
Evaluate $x=\tfrac12$ lies inside the interval, so $$\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1/2}{(1-1/2)^2}=\frac{1/2}{1/4}=2.$$
Check Partial sums: $0.5,\ 1,\ 1.375,\ 1.625,\ 1.78,\ 1.875,\dots$; after $30$ terms the sum is $1.99999997$. ✓
Uniqueness of Power Series
Example 1 found the same series for $\dfrac{2}{1-x^2}$ by two different routes, and Example 2 did the same for $\dfrac{1}{(1-x)^2}$. This always happens.
📐 Theorem 9.5 — Uniqueness of power-series representations
Suppose that $$\sum_{n=0}^{\infty}c_n(x-a)^n=\sum_{n=0}^{\infty}d_n(x-a)^n$$ for every $x$ in some open interval containing $a$. Then $c_n=d_n$ for every $n$. In fact, if $f(x)=\sum c_n(x-a)^n$ on such an interval, then necessarily $$c_n=\frac{f^{(n)}(a)}{n!},\qquad n=0,1,2,\dots$$
Why. Put $x=a$: every term except the first vanishes, so $f(a)=c_0$. Differentiate (Theorem 9.4): $f^\prime(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots$, and putting $x=a$ gives $f^\prime(a)=c_1$. Differentiating again, $f^{\prime\prime}(a)=2c_2$; then $f^{\prime\prime\prime}(a)=3\cdot2\,c_3$, and in general $f^{(n)}(a)=n!\,c_n$. The coefficients are fixed by $f$ — so any correct method gives the same series. This formula is the starting point of Section 9.3 (Taylor series).
✏️ Example 6 — Using uniqueness to solve a differential equation
Find a power series $y=\displaystyle\sum_{n=0}^{\infty}c_nx^n$ (with positive radius of convergence) that satisfies $y^\prime=y$ and $y(0)=1$.
💡 Show solution▶
Differentiate $y^\prime=\displaystyle\sum_{n=1}^{\infty}nc_nx^{n-1}=\sum_{n=0}^{\infty}(n+1)c_{n+1}x^n$ (shift the index so both sides show $x^n$).
Compare coefficients $y^\prime=y$ means $\displaystyle\sum_{n=0}^{\infty}(n+1)c_{n+1}x^n=\sum_{n=0}^{\infty}c_nx^n$ on an interval, so by uniqueness $(n+1)c_{n+1}=c_n$, i.e. $c_{n+1}=\dfrac{c_n}{n+1}$.
Solve the recurrence $y(0)=c_0=1$, so $c_1=1$, $c_2=\tfrac12$, $c_3=\tfrac1{2\cdot3}$, $c_4=\tfrac{1}{2\cdot3\cdot4}$, …, $c_n=\dfrac1{n!}$. Hence $$y=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
Interpret This series converges for all $x$ (Example 1 of 9.1). From Chapter 7 we know the solution of $y^\prime=y$, $y(0)=1$ is $y=e^x$, so $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\quad\text{for all }x.$$ Check at $x=1$: $1+1+0.5+0.1667+0.0417+0.0083+\cdots\approx2.718=e$. ✓
Exercises
I
Use differentiation, integration, substitution or algebra with known series to find a power series centred at $0$ for each function. State the radius of convergence (and, where asked, the interval).
- 1 $\dfrac{1}{(1+x)^2}$
- 2 $\dfrac{1}{(1-x)^3}$
- 3 $\ln(1-x)$ (give the interval)
- 4 $\ln(1+2x)$ (give the interval)
- 5 $x\ln(1+x)$
- 6 $\arctan\left(x^2\right)$
- 7 $\arctan\dfrac x2$
- 8 $\ln\dfrac{1+x}{1-x}$ (give the interval)
- 9 $\dfrac{x^2}{(1-x)^2}$
- 10 $\displaystyle F(x)=\int_0^x\frac{dt}{1+t^4}$
- 11 $\ln(3+x)$ (hint: $3+x=3\left(1+\tfrac x3\right)$; give the interval)
💡 Show solutions▶
1 Differentiate $\dfrac1{1+x}=\sum(-1)^nx^n$: $-\dfrac{1}{(1+x)^2}=\sum_{n\ge1}(-1)^nnx^{n-1}$. So $\dfrac{1}{(1+x)^2}=\displaystyle\sum_{n=0}^{\infty}(-1)^n(n+1)x^n=1-2x+3x^2-\cdots$, $R=1$.
2 Halve Example 2's second derivative: $\dfrac{1}{(1-x)^3}=\displaystyle\sum_{n=0}^{\infty}\frac{(n+2)(n+1)}{2}x^n=1+3x+6x^2+10x^3+\cdots$, $R=1$.
3 Replace $x$ by $-x$ in the $\ln(1+x)$ series: $\ln(1-x)=\displaystyle-\sum_{n=1}^{\infty}\frac{x^n}{n}$. $R=1$; at $x=-1$ it is $\sum\frac{(-1)^{n+1}}{n}$ (converges), at $x=1$ it is $-\sum\frac1n$ (diverges). Interval $[-1,1)$.
4 Replace $x$ by $2x$: $\ln(1+2x)=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}2^nx^n}{n}$, valid for $-1\lt 2x\le1$. $R=\tfrac12$, interval $\left(-\tfrac12,\tfrac12\right]$.
5 Multiply by $x$: $x\ln(1+x)=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n+1}}{n}=x^2-\frac{x^3}{2}+\frac{x^4}{3}-\cdots$, $R=1$.
6 Substitute $x^2$: $\arctan\left(x^2\right)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+2}}{2n+1}=x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\cdots$, valid for $x^2\le1$; $R=1$.
7 Substitute $\tfrac x2$: $\arctan\dfrac x2=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)\,2^{2n+1}}=\frac x2-\frac{x^3}{24}+\frac{x^5}{160}-\cdots$, valid for $\left|\tfrac x2\right|\le1$; $R=2$.
8 $\ln\dfrac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=\displaystyle\sum\frac{(-1)^{n+1}x^n}{n}+\sum\frac{x^n}{n}$. Even powers cancel, odd powers double: $$\ln\frac{1+x}{1-x}=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right).$$ $R=1$; at $x=\pm1$ it is $\pm2\sum\frac{1}{2n+1}$, which diverges. Interval $(-1,1)$.
9 $\dfrac{x^2}{(1-x)^2}=x^2\displaystyle\sum_{n=0}^{\infty}(n+1)x^n=\sum_{n=0}^{\infty}(n+1)x^{n+2}=\sum_{n=2}^{\infty}(n-1)x^n=x^2+2x^3+3x^4+\cdots$, $R=1$.
10 $\dfrac{1}{1+t^4}=\sum(-1)^nt^{4n}$ for $|t|\lt 1$; integrate: $F(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+1}}{4n+1}=x-\frac{x^5}{5}+\frac{x^9}{9}-\cdots$, $R=1$.
11 $\ln(3+x)=\ln3+\ln\left(1+\tfrac x3\right)=\ln3+\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n\,3^n}$, valid for $-1\lt\tfrac x3\le1$. $R=3$, interval $(-3,3]$.
II
Find the exact sum, or answer the estimation question. Name the known series you use.
- 1 $\displaystyle\sum_{n=1}^{\infty}\frac{n}{3^n}$
- 2 $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ (hint: differentiate $\sum nx^n$ and multiply by $x$)
- 3 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n\,2^n}$
- 4 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\,3^n}$
- 5 $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\,3^n}$
- 6 $\displaystyle\sum_{n=0}^{\infty}\frac{n+1}{4^n}$
- 7 How many terms of the $\ln(1+x)$ series are needed to compute $\ln1.1$ with error less than $10^{-6}$? Give the estimate.
- 8 How many terms of $4\left(1-\tfrac13+\tfrac15-\cdots\right)$ does the alternating series bound require to guarantee $\pi$ within $0.01$?
- 9 Estimate $\displaystyle\int_0^{0.5}\frac{dx}{1+x^4}$ to four decimal places using the series of Exercise I.10.
Check: watch the partial sums of parts 1–6 settle on the exact value
5
partial sum—
exact value—
difference—
💡 Show solutions▶
1 $\sum nx^n=\dfrac{x}{(1-x)^2}$ at $x=\tfrac13$: $\dfrac{1/3}{(2/3)^2}=\dfrac{1/3}{4/9}=\dfrac34$.
2 Differentiate $\sum nx^n=\dfrac{x}{(1-x)^2}$: $\sum n^2x^{n-1}=\dfrac{1+x}{(1-x)^3}$, so $\sum n^2x^n=\dfrac{x(1+x)}{(1-x)^3}$ for $|x|\lt 1$. At $x=\tfrac12$: $\dfrac{\frac12\cdot\frac32}{\frac18}=6$.
3 $-\ln(1-x)=\sum\dfrac{x^n}{n}$ at $x=\tfrac12$: the sum is $-\ln\tfrac12=\ln2\approx0.6931$.
4 $\ln(1+x)=\sum\dfrac{(-1)^{n+1}x^n}{n}$ at $x=\tfrac13$: $\ln\tfrac43\approx0.2877$.
5 $\arctan x=\sum\dfrac{(-1)^nx^{2n+1}}{2n+1}$ with $x=\tfrac1{\sqrt3}$ gives $x^{2n+1}=\tfrac{1}{\sqrt3}\cdot\tfrac{1}{3^n}$. So the sum is $\sqrt3\arctan\dfrac1{\sqrt3}=\sqrt3\cdot\dfrac{\pi}{6}=\dfrac{\pi}{2\sqrt3}\approx0.9069$.
6 $\sum(n+1)x^n=\dfrac1{(1-x)^2}$ at $x=\tfrac14$: $\dfrac{1}{(3/4)^2}=\dfrac{16}{9}\approx1.7778$.
7 With $x=0.1$ the terms are $\dfrac{0.1^n}{n}$: the 6th is $\dfrac{10^{-6}}{6}\approx1.7\times10^{-7}\lt 10^{-6}$ while the 5th is $2\times10^{-6}$. So $5$ terms suffice: $\ln1.1\approx0.1-0.005+0.000333-0.000025+0.000002=0.0953103$ (true value $0.0953102$).
8 After $N$ terms the error is less than the next term, $\dfrac{4}{2N+1}$. We need $\dfrac{4}{2N+1}\lt 0.01$, i.e. $2N+1\gt 400$, so $N=200$ terms.
9 $F(0.5)=0.5-\dfrac{0.5^5}{5}+\dfrac{0.5^9}{9}-\dfrac{0.5^{13}}{13}+\cdots=0.5-0.00625+0.000217-0.0000094+\cdots$ Three terms give $0.493967$ with error less than $0.0000094$; so $\displaystyle\int_0^{0.5}\frac{dx}{1+x^4}\approx0.4940$ (more precisely $0.493958$).
III
Concepts, uniqueness and applications.
- 1 Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n^2}$, of its derivative series and of its term-by-term integral (from $0$). Which endpoints change?
- 2 Use uniqueness to find the power series $y=\sum c_nx^n$ with $y^\prime=2xy$ and $y(0)=1$. Which function is it?
- 3 Find the power series $y=\sum c_nx^n$ with $y^{\prime\prime}=-y$, $y(0)=0$, $y^\prime(0)=1$. Which function is it?
- 4 Let $f(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(n+1)x^n}{3^n}$. Find $f^\prime(0)$ and $f^{\prime\prime}(0)$ without differentiating a closed form, then find the closed form of $f$ and its interval of validity.
- 5 Applied — quality control. Each chip on a production line is defective with probability $p=0.04$, independently. The expected number of chips tested up to and including the first defective one is $\displaystyle\sum_{n=1}^{\infty}n\,p(1-p)^{n-1}$. Find a closed form in terms of $p$ and evaluate it.
- 6 Applied — acoustics. When a sound's power increases by a small fraction $x$, its level rises by $\Delta L=10\log_{10}(1+x)=\dfrac{10}{\ln10}\ln(1+x)$ decibels. Use two terms of the $\ln(1+x)$ series to estimate $\Delta L$ for $x=0.05$, and bound the error.
- 7 The series $\sum\frac{(-1)^nx^{2n+1}}{2n+1}$ is a formula for $\arctan x$. Explain why it cannot be used to compute $\arctan2$, even though $\arctan2$ exists.
- 8 Show that $\displaystyle\int_0^1\frac{\ln(1+x)}{x}\,dx=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$, and use $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6}$ to evaluate it.
💡 Show solutions▶
1 $\sum\frac{x^n}{n^2}$: $R=1$, both endpoints converge, $[-1,1]$. Derivative $\sum\frac{x^{n-1}}{n}$: $R=1$, converges at $-1$, diverges at $1$, so $[-1,1)$ — the right endpoint is lost. Integral $\sum\frac{x^{n+1}}{n^2(n+1)}$: both endpoints converge (compare with $\sum\frac1{n^3}$), $[-1,1]$ — no change.
2 $\sum(n+1)c_{n+1}x^n=\sum2c_{n-1}x^n$ (right side starting at $n=1$). Coefficient of $x^0$: $c_1=0$; for $n\ge1$: $c_{n+1}=\dfrac{2c_{n-1}}{n+1}$. With $c_0=1$: $c_2=1$, $c_4=\tfrac12$, $c_6=\tfrac16$, … and all odd $c_n=0$; in general $c_{2k}=\dfrac1{k!}$. So $y=\displaystyle\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}=e^{x^2}$.
3 $y^{\prime\prime}=\sum(n+2)(n+1)c_{n+2}x^n=-\sum c_nx^n$, so $c_{n+2}=-\dfrac{c_n}{(n+2)(n+1)}$. $c_0=0$ makes all even coefficients $0$; $c_1=1$ gives $c_3=-\tfrac1{3!}$, $c_5=\tfrac1{5!}$, … So $y=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$, which is $\sin x$ (it satisfies the same initial-value problem).
4 By uniqueness $c_n=\dfrac{f^{(n)}(0)}{n!}$. $c_1=\tfrac23$ so $f^\prime(0)=\tfrac23$; $c_2=\tfrac39=\tfrac13$ so $f^{\prime\prime}(0)=2!\cdot\tfrac13=\tfrac23$. Closed form: $\sum(n+1)u^n=\dfrac1{(1-u)^2}$ with $u=\tfrac x3$, so $f(x)=\dfrac{1}{(1-x/3)^2}=\dfrac{9}{(3-x)^2}$ for $|x|\lt 3$.
5 With $q=1-p$: $\displaystyle\sum_{n=1}^{\infty}n\,p\,q^{n-1}=p\cdot\frac{1}{(1-q)^2}=p\cdot\frac{1}{p^2}=\frac1p$, valid since $0\lt q\lt 1$. For $p=0.04$ the expected number is $25$ chips.
6 $\ln1.05\approx0.05-\dfrac{0.05^2}{2}=0.04875$, so $\Delta L\approx\dfrac{10}{\ln10}(0.04875)\approx4.3429\times0.04875\approx0.2117$ dB. Alternating series bound: error $\lt\dfrac{10}{\ln10}\cdot\dfrac{0.05^3}{3}\approx0.00018$ dB. (Exact: $10\log_{10}1.05=0.2119$ dB.)
7 The series has radius $1$. At $x=2$ the terms $\dfrac{2^{2n+1}}{2n+1}$ grow without bound, so the series diverges; the formula $\arctan x=\sum\cdots$ holds only on $[-1,1]$. (One can instead use $\arctan2=\tfrac\pi2-\arctan\tfrac12$.)
8 For $0\lt x\lt 1$, $\dfrac{\ln(1+x)}{x}=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n-1}}{n}$. Integrating term by term from $0$ to $1$ (the integrated series converges at $x=1$, and Abel's theorem justifies the endpoint) gives $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$. Splitting into even and odd terms, this equals $\displaystyle\sum\frac1{n^2}-2\sum\frac{1}{(2n)^2}=\frac{\pi^2}{6}-\frac12\cdot\frac{\pi^2}{6}=\frac{\pi^2}{12}\approx0.8225$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Why did the series $\pi=4\left(1-\tfrac13+\tfrac15-\cdots\right)$, known since Madhava of Kerala in the 14th century, never become the practical way to compute $\pi$ — and what replaced it?
Why it matters. Every scientific library must evaluate $\pi$, logarithms and inverse trigonometric functions to full machine precision in microseconds. Choosing where to evaluate a power series — at the edge of its interval or deep inside it — is the difference between a usable algorithm and a useless one.
What to do.
- In a spreadsheet, compute $4S_N$ for $N=10,100,1000$ terms of the arctan series at $x=1$. Record the error and check that it is close to $\tfrac1N$.
- Estimate how many terms this series would need for $10$ correct decimal places of $\pi$.
- Verify numerically that $4\arctan\tfrac15-\arctan\tfrac1{239}=\tfrac\pi4$ (John Machin, England, 1706). Then use the arctan series at $x=\tfrac15$ and $x=\tfrac1{239}$: how many terms of each give $10$ correct decimals?
- Explain the difference using the size of the terms: compare $\tfrac{1}{2n+1}$ with $\tfrac{(1/5)^{2n+1}}{2n+1}$.
Tools: Spreadsheet or Python, calculator
Hand in: A one-page note with your table of errors, the two term counts, and a short explanation linking speed of convergence to the distance from the centre relative to the radius of convergence.
Going further. Use Exercise I.8 with $x=\tfrac13$ (so that $\tfrac{1+x}{1-x}=2$) to compute $\ln2$, and compare with the slow series $1-\tfrac12+\tfrac13-\cdots$.
Instructor note: $N=1000$ gives $3.14059$ (error $\approx0.0010$); about $10^{10}$ terms would be needed for 10 decimals. Machin's formula reaches error $\approx3\times10^{-11}$ with $7$ terms of each series. For $\ln2$, $2\sum\frac{(1/3)^{2n+1}}{2n+1}$ is within $10^{-6}$ after $6$ terms.
🧪 Self-check quiz
0/5 correct
Q1. A power series $\sum c_nx^n$ has radius of convergence $4$. What is the radius of convergence of its term-by-term derivative?
Q2. Which series equals $\ln(1+x)$ for $-1\lt x\le1$?
Q3. For $|x|\lt 1$, $\displaystyle\sum_{n=1}^{\infty}nx^n=$
Q4. The interval of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$ (the series for $\arctan x$) is
Q5. Why is $\pi=4\left(1-\tfrac13+\tfrac15-\cdots\right)$ a poor way to compute $\pi$?