Errata in Farlow's Differential Equations — Chapter 4

Page 200: General comments on Section 4.1

The statement of Theorem 4.2 (Ratio Test) is correct but not directly applicable to solving some of the exercise problems. At times an appeal to the ratio test for general series (not necessarily power series) from calculus is more expedient.

Theorem (Ratio Test). Consider the series $\sum_{n=0}^\infty b_n$ where $b_n$ are nonzero when $n$ is large, and let $L = \lim_{n\to\infty} \Big| \frac{b_{n+1}}{b_n} \Big|$. Then

• if $L<1$, then the series converges absolutely;
• if $L>1$, then the series is divergent;
• if $L=1$, then the series may or may not converge.

You are advised to forget the book's Theorem 4.2 and use this version of the ratio test in all instances.

To illustrate, let's find the interval of convergence of the power series $\sum_{n=1}^\infty \frac{n^3 x^{2n}}{3^n}$ which involves only even powers of $x$. The odd powers of $x$ are missing, that is, the coefficients of the odd powers are zero, and consequently the book's Theorem 4.2 is not applicable since it requires the coefficients all powers to be nonzero. In contrast, the boxed version above, which makes no reference of powers of $x$, works just fine with $b_n = \frac{n^3 x^{2n}}{3^n}$. We have \[ L = \lim_{n\to\infty} \Big|\frac{b_{n+1}}{b_n}\Big| = \lim_{n\to\infty} \bigg| \frac{1}{3} \Big( \frac{n+1}{n} \Big)^3 x^2 \bigg| = \frac{1}{3} |x|^2. \] Then $L<1$ implies $\frac{1}{3} |x|^2 < 1$, and therefore $ -\sqrt{3} < x < \sqrt{3}$. Thus, the radius of convergence is $\sqrt{3}$. At the endpoints $x=\pm\sqrt{3}$ the series reduces to $\sum_{n=1}^\infty n^3$ which is divergent. We conclude that the interval of convergence is $(-\sqrt{3} , \sqrt{3})$.

See this section's Exercises 7 and 8 for similar problems.

To answer some of the questions in the exercises, you will also need the alternating series test also from calculus:

Theorem (Alternating Series Test). Consider the series $\sum_{n=0}^\infty (-1)^n b_n$, where $b_n$ are all positive. If $b_{n+1} \le b_n$ for all $n$, and if $\lim_{n\to\infty} b_n = 0$, then the series converges.

To illustrate, consider the power series $\sum_{n=0}^\infty \frac{x^n}{n+1}$. You should be able to verify that the series is absolutely convergent in $-1 < x < 1$. Let us examine what happens at the endpoints of that interval.

At $x=-1$ the series reduces to $\sum_{n=0}^\infty \frac{(-1)^n}{n+1}$, This alternating series satisfies the requirements of the Alternating Series Test, and therefore it converges. On the other hand, at $x=1$ the series reduces to $\sum_{n=0}^\infty \frac{1}{n+1}$, which is the well-known harmonic series which is divergent. We conclude that the interval of convergence of the original series is $[-1,1)$.

See this section's Exercises 3, 7, and 8 for similar problems.

Page 205, Exercise #11:

(Spotted by Paul Rydeen) The book's solution does not address the intervals of convergence question. The intervals of convergence for all cases are $(-1,1)$ except for part (c) where it is $[-1,0)$.

Page 206, Exercise #18–27:

(Spotted by Paul Rydeen) The book's solutions do not address the intervals of convergence question. For problems 18–22 the intervals of convergence are $(-\infty,\infty)$, for 23 it is $(-1,1)$, for 24 it is $(-1,3)$, for 25 it is $(0,2)$, for 26 it is $(-1,1)$, and for 27 it is $(-\infty,\infty)$.

Page 216, Exercise #4:

(Spotted by Paul Rydeen) The book's answer of $x=\pm i$ is not correct. The variable $x$ is real-valued, so $x=\pm i$ makes no sense. All points are ordinary points for this differential equation.

In the wider context of differential equations in the complex plane, then yes, $z=\pm i$ ($z$, not $x\,$!) would be singular points, but that's far from the subject of this book.

Page 216, Exercise #11:

(Spotted by Dr. Elliott Bertrand) The answer in the back of the book is not quite correct. The summation should begin at zero, and the numerator 2 is meant to be 1. Thus: $\ds y = a_0 \sum_{n=0}^\infty \frac{1}{n!}x^{2n}$.

Page 223, Exercise #8:

(Spotted by Paul Rydeen) The problem's statement is incomplete since initial conditions for $T_0(x)$ and $T_1(x)$ are missing. You will need: \begin{align*} \text{Initial conditions for $T_0$:}\qquad & T_0(0)=1, \quad T'_0(0)=0, \\ \text{Initial conditions for $T_1$:}\qquad & T_1(0)=0, \quad T'_1(0)=1. \end{align*} Then you will find that $T_0(x)=1$ and $T_1(x)=x$.

Page 227:

(Spotted by Paul Rydeen) Between equations (11) and (12) where it says “Setting the coefficient of $x^r$ to zero”, the $x^r$ should be $x^{r-1}$.

Page 231, Exercise #13:

(Spotted by Paul Rydeen) The book's answer is incorrect. The indicial equation is $r(r-1)=0$, which has roots $r_1=1$, $r_2=0$, and therefore, according to part 3(c) in the boxed prescription titled “The Method of Frobenius” on page 229, the correct answer is \[ y_1(x) = x\sum_{n=0}^\infty a_n x^n,\quad y_2(x) = \sum_{n=0}^\infty b_n x^n + c(\ln x) y_1(x). \]

Page 231, Exercise #22:

(Spotted by Paul Rydeen) The book’s answer for part (b) is labeled as part (c).

Page 238:

(Spotted by Paul Rydeen) In equation (16), the $J_{1/2}(x)$ should be $J_{-1/2}(x)$.

Page 241, Exercise #7:

(Spotted by Paul Rydeen) In parts (b) and (c), the variables of integration should be $x$, not $t$.