### Differential Equations to Difference Equations

#### Answers for selected problems

Chapter 5

Section 5.1

1. $\displaystyle{\lim_{h \to 0}\frac{\tan^2(h)}{h} = 0}$

3. $\displaystyle{\lim_{h \to 0}\frac{h^2\sin(h)}{h^2} = 0}$ and $\displaystyle{\lim_{h \to 0}\frac{h^2\sin(h)}{h^3} = 1}$

5. $\displaystyle{\lim_{h \to 0}\frac{\sin^2(3h)}{h^2} = 9}$

7. (a) $\displaystyle{P_4(x) = 2x - \frac{4x^3}{3}}$

(c) $\displaystyle{P_4(z) = 2 + \frac{1}{4}(z - 4) - \frac{1}{64}(z - 4)^2 + \frac{1}{512}(z - 4)^3 - \frac{5}{16384}(z - 4)^4}$

(e) $\displaystyle{P_4(x) = -(x - \pi) + \frac{1}{6}(x - \pi)^3}$

(g) $\displaystyle{P_4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4}$

(h) $\displaystyle{P_4(x) = -9 + 2x + 3x^2}$

(i) $\displaystyle{P_4(x) = 1 - x^2 + x^4}$

(k) $\displaystyle{P_4(t) = 1 - 2(t - 1) + 3(t - 1)^2 - 4(t - 1)^3 + 5 (t - 1)^4}$

(m) $\displaystyle{P_4(t) = 1 + \frac{1}{4}t^2 + \frac{5}{24}t^4}$

(n) $\displaystyle{P_4(x) = -5 - (x - 1) + 7(x - 1)^2 + 8(x - 1)^3 +3(x - 1)^4}$

Section 5.2

1. (a) $\displaystyle{P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}}$

$\sin(0.8) \approx P_5(0.8) = 0.717397$, rounded to 6 decimal places

$\displaystyle{|\sin(0.8) - P_5(0.8)| \le \frac{0.8^7}{7!} = 0.000041610}$, rounded to 9 decimal places

(c) $\displaystyle{P_5(x) = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5}$

$\sin(-1) \approx P_5(-0.5) = -0.841667$, rounded to 6 decimal places

$\displaystyle{|\sin(-1) - P_5(-0.5)| \le \frac{2^7(0.5)^7}{7!} = 0.000198413}$, rounded to 9 decimal places

(e) $\displaystyle{P_5(x) = 3 + \frac{1}{6}(x - 9) - \frac{1}{216}(x - 9)^2 + \frac{1}{3888}(x - 9)^3}$

$\displaystyle{- \frac{5}{279,936}(x - 9)^4 + \frac{7}{5,038,848}(x - 9)^5}$

$\sqrt{10} \approx P_5(10) = 3.1622778$, rounded to 7 decimal places

$\displaystyle{|\sqrt{10} - P_5(10)| \le \frac{945}{(64)(3^{11})(6!)} = 0.000000115767}$, rounded to 12 decimal places

(g) $\displaystyle{P_5(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4 - (x - 1)^5}$

$\displaystyle{\frac{1}{0.8} \approx P_5(0.8) = 1.24992}$, rounded to 5 decimal places

$\displaystyle{\left|\frac{1}{0.8} - P_5(0.8)\right| \le \frac{720(0.2)^6}{(0.8)^7(61)} = 0.000305176}$, rounded to 9 decimal places

3. $\sin(-1.3) \approx P_9(-1.3) = -0.963559$, rounded to 6 decimal places

5. $\displaystyle{P_9(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}}$

7. (a) $\displaystyle{P_{10}(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \frac{x^9}{9!} + \frac{x^{10}}{10!}}$

(b) $E(1) \approx P_{10}(1) = 2.7182818011$, rounded to 10 decimal places

(c) $\displaystyle{|E(1) - P_{10}(1)| \le \frac{3}{11!} = 0.00000007515633}$, rounded to 14 decimal places

(d) $\displaystyle{P_{11}(x) = 1 + x + \frac{x^2}{2} \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \frac{x^9}{9!} + \frac{x^{10}}{10!} + \frac{x^{11}}{11!}}$

8. $\displaystyle{\int_0^3 \frac{\sin(x)}{x} \ dx \approx \int_0^3 \frac{P_9(x)}{x} \ dx = \int_0^3 \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}\right)dx = 1.849037}$, rounded to 6 decimal places

9. (a) $\displaystyle{P_6(x) = x^2 - \frac{x^6}{6}}$

(b) $\displaystyle{P_7(x) = \frac{x^3}{3} - \frac{x^7}{42}}$

Section 5.3

1. (a) The series converges because it is a multiple of a geometric series with ratio $r = \dfrac{1}{5}$. Moreover, $\displaystyle{\sum_{n=0}^\infty \frac{3}{5^n} = \frac{15}{4}}$

(c) The series converges because it is the sum of two geometric series with ratios $r =\dfrac{1}{2}$ for the first and $r = \dfrac{1}{3}$ for the second. Moreover, $\displaystyle{\sum_{n=1}^\infty \left(\frac{5}{2^n} + \frac{2}{3^n}\right) = 6}$

(e) Since $\displaystyle{\lim_{n \to \infty}\left(1 - \frac{1}{n}\right) = 1}$, the series diverges by the $n$th term test for divergence.

(g) Since $\dfrac{(-1)^n}{1000}$ does not have a limit as $n \rightarrow \infty$, the series diverges by the $n$th term test for divergence.

2. (a) The series converges because it is a multiple of a $p$-series with $p = 3$.

(c) The series converges because it is the sum of multiples of two $p$-series with $p = 2$ for the first and $p = 4$ for the second.

(e) The series diverges because it is a $p$-series with $p = \dfrac{1}{3}$.

(g) The series diverges because it is a multiple of a $p$-series with $p = \dfrac{1}{2}.$

5. (a) The series converges.

(c) The integral diverges.

Section 5.4

1. (a) The series converges because it is a multiple of a geometric series with ratio $r = \dfrac{1}{2}$.

(c) The series diverges by comparion with a $p$-series with $p = \dfrac{1}{2}$.

(e) The series converges because it is a geometric series with ratio $r = \dfrac{1}{\pi}$.

(g) The series converges because is the sum of a convergent geometric series (with ratio $r = \dfrac{1}{3}$) and a convergent $p$-series (with $p = \dfrac{3}{2}$).

2. (a) The series diverges by comparison with the harmonic series.

(c) The series converges by the limit comparison test with a $p$-series with $p = 3$.

(e) The series converges by the limit comparison test with a geometric series with ratio $r = \dfrac{1}{3}$.

(g) The series diverges by the limit comparison test with a $p$-series with $p = 2$.

4. The series $\displaystyle{\sum_{n=1}^\infty \frac{1}{4n^5 - 2}}$ converges by the limit comparison test with a $p$-series with $p = 5$. Hence $\displaystyle{\int_1^\infty \frac{1}{4x^5 - 2} \ dx}$ convergess by the integral test.

Section 5.5

1. (a) The series converges by the ratio test: $\rho = \dfrac{1}{2}$.

(c) The series converges by the ratio test: $\rho = 0$.

(e) The series converges by the ratio test: $\rho = 0$.

(g) The series converges by the ratio test: $\rho = 0$.

(h) The series diverges by the ratio test: $\rho = \pi$.

2. (a) The series diverges by the limit comparison test with a $p$-series with $p = \dfrac{1}{2}$.

(c) The series converges by the ratio test: $\rho = 0$.

(e) The series converges by the ratio test: $\rho = \dfrac{1}{4}$.

(g) The series diverges by the $n$th term test for divergence since $\displaystyle{\lim_{n \to \infty}\frac{3n + 1}{2n - 1} = \frac{3}{2}}$.

3. (a) $(-1, 1)$

4. (a) $(-\infty, \infty)$

5. The series converges by the ratio test since $\displaystyle{\rho = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \frac{1}{2}}$.

Section 5.6

1. (a) The series converges absolutely ($p$-series with $p = 3$), and so converges.

(c) The series diverges by the $n$th term for divergence since <table align=“center”> <tr><td align=“center”>$\displaystyle{\lim_{n \to \infty}\frac{3n^2 - 1}{4n^2 + 2} = \frac{3}{4}}$.</td></tr> </table>

(e) The series does not converge absolutely (limit comparison test with a $p$-series with $p = \dfrac{1}{2}$), but does converge conditionally by Leibniz's theorem. Hence the series converges.

(g) Since $\displaystyle{\lim_{n \to \infty}\frac{n!}{2^n} = \infty}$, the terms $\dfrac{(-1)^n n!}{2^n}$ do not have a limit as $n \rightarrow \infty$. Hence the series diverges by the $n$th term test for divergence.

2. (a) The series converges absolutely (limit comparison with a $p$-series with $p = 3$), and so converges.

(c) The series converges absolutely (ratio test, $\rho = 0$), and so converges.

(e) The series converges absolutely (ratio test, $\rho = 0$), and so converges.

(g) Since $\displaystyle{\lim_{n \to \infty}\frac{n + 1}{2n - 1} = \frac{1}{2}}$, the terms $\dfrac{(-1)^n (n + 1)}{2n - 1}$ do not have a limit as $n \rightarrow \infty$. Hence the series diverges by the the $n$th term test for divergence.

3. (a) $\displaystyle{s_{15} = \frac{34,361,893,981}{93,405,312,000} = 0.367879441171397}$, to 15 decimal places

(b) $|s - s_{15}| \le \dfrac{1}{16!} = 4.779 \times 10^{-14}$, rounded to 17 decimal places

(c) $s_9 = \dfrac{16,687}{45,360} = 0.367879$ to 6 decimal places

Section 5.7

1. (a) The interval of convergence is $(-1, 1)$.

The series converges at $x = -1$ and diverges at $x = 1$.

The series converges absolutely on $(-1, 1)$, converges conditionally at $x = -1$, and diverges for $x < -1$ and $x \ge 1$.

$\displaystyle{\sum_{n = 1}^\infty \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \cdots}$

(c) The interval of convergence is $(-1, 1)$.

The series diverges at both $x = -1$ and $x = 1$.

The series converges absolutely on $(-1, 1)$ and diverges for $x \le -1$ and $x \ge 1$.

$\displaystyle{\sum_{n = 1}^\infty nx^n = x + 2x^2 + 3x^3 + 4x^4 + 5x^5 + \cdots}$

(e) The interval of convergence is $(1, 3)$.

The series converges at $x = 1$ and diverges at $x = 3$.

The series converges absolutely on $(1, 3)$, converges conditionally at $x = 1$, and diverges for $x < 1$ and $x \ge 3$.

$\displaystyle{\sum_{n = 0}^\infty \frac{(x-2)^{n+1}}{n+1} = (x - 2) + \frac{(x-2)^2}{2} + \frac{(x-2)^3}{3} + \frac{(x-2)^4}{4} + \frac{(x-2)^5}{5} + \cdots}$

(g) The interval of convergence is $(0, 2)$.

The series converges at both $x = 0$ and $x = 2$.

The series converges absolutely on $[0, 2]$ and diverges for $x < 0$ and $x > 2$.

$\displaystyle{\sum_{n = 1}^\infty \frac{(x-1)^n}{n^2} = (x-1) + \frac{(x-1)^2}{4} + \frac{(x-1)^3}{9} + \frac{(x-1)^4}{16} + \frac{(x-1)^5}{25} + \cdots}$

2. (a) The interval of convergence is $(-\infty, \infty)$.

The series converges absolutely on $(-\infty, \infty)$.

$\displaystyle{\sum_{n = 0}^\infty \frac{(x-3)^n}{n!} = 1 + (x - 3) + \frac{(x-3)^2}{2} + \frac{(x-3)^3}{6} + \frac{(x-3)^4}{24} + \cdots}$

(c) The interval of convergence is $(-\sqrt{2}, \sqrt{2})$.

The series diverges at both $x = -\sqrt{2}$ and $x = \sqrt{2}$.

The series converges absolutely on $(-\sqrt{2}, \sqrt{2})$ and diverges for $x \le -\sqrt{2}$ and $x \ge \sqrt{2}$.

$\displaystyle{\sum_{n = 0}^\infty \frac{x^{2n}}{2^n} = 1 + \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{8} + \frac{x^8}{16}+ \cdots}$

(e) The interval of convergence is $(-1, 1)$.

The series diverges at both $x = -1$ and $x = 1$.

The series converges absolutely on $(-1, 1)$ and diverges for $x \le -1$ and $x \ge 1$.

$\displaystyle{\sum_{n = 1}^\infty \frac{x^{2n+1}}{n} = x^3 + \frac{x^5}{2} + \frac{x^7}{3} + \frac{x^9}{4} + \frac{x^{11}}{5} + \cdots}$

(g) The interval of convergence is $\displaystyle{\left(-\frac{1}{3}, \frac{1}{3}\right)}$.

The series diverges at both $x = -\dfrac{1}{3}$ and $x = \dfrac{1}{3}$.

The series converges absolutely on $\displaystyle{\left(-\frac{1}{3}, \frac{1}{3}\right)}$ and diverges for $x \le -\dfrac{1}{3}$ and $x \ge \dfrac{1}{3}$.

$\displaystyle{\sum_{n = 1}^\infty 3^nx^n = 3x + 9x^2 + 27x^3 + 81x^4 + 243x^5 + \cdots}$

3. (a) For $-1 < x < 1$, $\frac{1}{1 + x} = \sum_{n=0}^\infty (-1)^nx^n = 1 - x + x^2 - x^3 + x^4 - \cdots.$

(b) $f^{(35)}(0) = -35!$

(c) $\displaystyle{\int_0^x \frac{1}{1 + t} \ dt = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n + 1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots}$

The series converges absolutely on $(-1, 1)$, converges conditionally at $x = 1$, and diverges for $x \le -1$ and $x > 1$.

(d) $\displaystyle{\int_0^1 \frac{1}{1 + t} \ dt = \sum_{n=0}^\infty \frac{(-1)^n}{n + 1} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots}$

$\displaystyle{\int_0^1 \frac{1}{1 + t} \ dt \approx \sum_{n=0}^{998} \frac{(-1)^n}{n + 1} = 0.6936}$, rounded to 4 decimal places

5. $\displaystyle{\cos(1) = \sum_{n=0}^\infty \frac{(-1)^{2n}}{(2n)!} = 1 - \frac{1}{2} + \frac{1}{4!} - \frac{1}{6!} + \frac{1}{8!} - \cdots}$

$\displaystyle{\cos(1) \approx 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \frac{1}{40,320} = 0.5403026}$, rounded to 7 decimal places

7. $\displaystyle{\sum_{n=1}^\infty \frac{n}{2^{n-1}} = \frac{1}{\left(1 - \frac{1}{2}\right)^2} = 4}$

8. (a) $\displaystyle{A = \sum_{n=1}^\infty \frac{n}{2^n} = \frac{1}{2}\sum_{n=1}^\infty \frac{n}{2^{n-1}} = 2}$

(b) $\displaystyle{A = q\sum_{n=1}^\infty np^{n-1} = \frac{q}{(1 - p)^2} = \frac{1}{q}}$

Section 5.8

2. (a) For $-\infty < x < \infty$, $\cos(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n}}{(2n)!} = 1 - \frac{x^4}{2} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \frac{x^{16}}{8!} - \cdots .$

(c) For $-1 < t < 1$, $\frac{1}{1 - t^2} = \sum_{n=0}^\infty t^{2n} = 1 + t^2 + t^4 + t^6 + t^8 + \cdots .$

(d) For $-\dfrac{1}{2} < x < \dfrac{1}{2}$, $\frac{1}{2x - 1} = -\sum_{n=1}^\infty 2^nx^n = -1 - 2x - 4x^2 - 8x^3 - 16x^4 - \cdots .$

(e) For $-1 < t < 1$, $\frac{1}{(1 + t)^2} = \sum_{n=1}^\infty (-1)^{n+1}nt^{n-1} = 1 - 2t + 3t^2 - 4t^3 + 5t^4 - \cdots .$

(g) For $-\infty < x < \infty$, $f(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-1}}{(2n)!} = \frac{x}{2} - \frac{x^3}{4!} + \frac{x^5}{6!} - \frac{x^7}{8!} + \frac{x^9}{10!} - \cdots .$

3. (a) For $-\infty < x < \infty$, $\cos^2(x) = 1 + \sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}x^{2n}}{(2n)!} = 1 - x^2 + \frac{x^4}{3} - \frac{2x^6}{45} + \frac{x^8}{315} - \cdots .$

(b) $\displaystyle{P_8(x) = 1 - x^2 + \frac{x^4}{3} - \frac{2x^6}{45} + \frac{x^8}{315}}$

5. (a) For $-\infty < x < \infty$, $\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!} = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \frac{x^{18}}{9!} - \cdots .$

(b) $\displaystyle{P_{10}(x) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}}$

(c) For $-\infty < x < \infty$, $S(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!} = \frac{x^3}{3} - \frac{x^7}{7(3!)} + \frac{x^{11}}{11(5!)} - \frac{x^{15}}{15(7!)} + \frac{x^{19}}{19(9!)} - \cdots .$

(d) $\displaystyle{P_{11}(x) = \frac{x^3}{3} - \frac{x^7}{7(3!)} + \frac{x^{11}}{11(5!)}}$

(e) $S(1) \approx P_{15}(1) = 0.310268$, rounded to 6 deciimal places

7. $\displaystyle{\frac{d^9}{dx^9}\mathrm{Si}(x)\Bigg|_{x=0} = \frac{1}{9}}$

Section 5.9

1. (a) $\displaystyle{\lim_{x \to 0} \frac{\sin(3x)}{x} = 3}$

(c) $\displaystyle{\lim_{x \to 0} \frac{\cos(x) - 1 + \dfrac{x^2}{2}}{x^4} = \frac{1}{24}}$

(e) $\displaystyle{\lim_{u \to 0} \frac{\tan(u)}{\sin(u)} = 1}$

(g) $\displaystyle{\lim_{y \to 0} \frac{\tan(3y)}{\tan(5y)} = \frac{3}{5}}$

(i) $\displaystyle{\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{3x} = \frac{1}{6}}$

2. (a) $\displaystyle{\lim_{x \to 0} \frac{\sin(5x)}{3x} = \frac{5}{3}}$

(c) $\displaystyle{\lim_{x \to 0} \frac{1 - \sec(x)}{x} = 0}$

(e) $\displaystyle{\lim_{x \to 0^+} \frac{\sin(2x)}{\sqrt{x}} = 0}$

(g) $\displaystyle{\lim_{x \to \infty}x\sin\left(\frac{1}{x^2}\right) = 0}$

3.(a) $\displaystyle{\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{16x^2 + 2} = \frac{3}{16}}$

(c) $\displaystyle{\lim_{x \to 0}\frac{1 - \dfrac{\sin(x)}{x}}{3x^2} = \frac{1}{18}}$

(e) $\displaystyle{\lim_{x \to 0} \frac{(1+x)^{\frac{1}{3}} - 1 - \dfrac{x}{3}}{x^2} = -\frac{1}{9}}$

(g) $\displaystyle{\lim_{x \to 1}\frac{\sqrt{x} - 1}{x - 1} = \frac{1}{2}}$

(i) $\displaystyle{\lim_{t \to 0}\frac{\cos(t) - 1 + \dfrac{t^2}{2}}{t^2\sin(t^2)} = \frac{1}{24}}$