Calculus II, Spring 2016 (http://www.math.nthu.edu.tw/ wangwc/)

Solutions to selected problems in HW for Week 05

1. Section 10.9: Problem 41. By The Remainder Estimation Theorem, ∃ c between 0 and x such that

|error| ≤ e^{c} \frac{|x|^{3}}{3!} ≤ e^{0.1} \frac{(0.1)^{3}}{6} .

2. Section 10.9: Problem 42. By Alternating Series Estimation Theorem,

|error| ≤ \frac{|x|^{3}}{3!} ≤ \frac{(0.1)^{3}}{6} .

3. Section 10.9: Problem 50a. Let p = π + x, then |x| < 10^{−n}.

p + sin p = π + x + sin(π + x) = π + x − sin x = π + \frac{x^{3}}{3!} + O(x^{5})

⇒ |p + sin p − π| ≤ \frac{|x|^{3}}{6} + O(|x|^{5}) < \frac{1}{6}10^{−3n} + O(10^{−5n}) < \frac{1}{6}10^{−3n} + \frac{1}{2}10^{−3n} < 10^{−3n}.

4. Section 10.10: Problem 17. By Alternating Series Estimation Theorem,

|error| ≤ ∫_{0}^{0.1} | ( \begin{matrix}−\frac{1}{2} \\ n \end{matrix} ) x^{4n}| dx = | ( \begin{matrix}−\frac{1}{2} \\ n \end{matrix} ) | ∫_{0}^{0.1} x^{4n} dx = | ( \begin{matrix}−\frac{1}{2} \\ n \end{matrix} ) |\frac{(0.1)^{4n+1}}{4n + 1} < 10^{−3}

It follows that n = 1 is enough. Therefore, the approximation is

∫_{0}^{0.1} 1 dx = 0.1.

5. Section 10.10: Problem 19. By Alternating Series Estimation Theorem,

|error| ≤ ∫_{0}^{0.1} \frac{x^{2n}}{(2n + 1)!} dx = \frac{(0.1)^{2n+1}}{(2n + 1)!(2n + 1)} < 10^{−8}

. It follows that n = 3 is enough. Therefore, the approximation is

∫_{0}^{0.1} (1 − \frac{x^{2}}{3!} + \frac{x^{4}}{5!} ) dx = 0.099944461111111.

6. Section 10.10: Problem 27. Apply Alternating Series Estimation Theorem. (a)

|error| ≤ \frac{x^{2n+2}}{(2n + 1)(2n + 2)} = \frac{(0.5)^{2n+2}}{(2n + 1)(2n + 2)} < 10^{−3} ⇒ n = 2.

Therefore, the approximation is

\frac{x^{2}}{2} − \frac{x^{4}}{12}.

(a)

|error| ≤ \frac{x^{2n+2}}{(2n + 1)(2n + 2)} = \frac{1}{(2n + 1)(2n + 2)} < 10^{−3} ⇒ n = 16.

Therefore, the approximation is

∑_{n=0}^{15} \frac{(−1)^{n}x^{2n+2}}{(2n + 1)(2n + 2)}.

7. Section 10.10: Problem 64. (a)

(1 + x)f′(x) = (1 + x) ∑_{k≥1} ( \begin{matrix}m \\ k \end{matrix} ) kx^{k−1} = ( \begin{matrix}m \\ 1 \end{matrix} ) + ∑_{k≥1} [( \begin{matrix}m \\ k + 1 \end{matrix} ) (k + 1) + ( \begin{matrix}m \\ k \end{matrix} ) k] x^{k}

= ( \begin{matrix}m \\ 1 \end{matrix} ) + ∑_{k≥1} [k ( \begin{matrix}m + 1 \\ k + 1 \end{matrix} ) + ( \begin{matrix}m \\ k + 1 \end{matrix} )] x^{k}

= m + m ∑_{k≥1} ( \begin{matrix}m \\ k \end{matrix} ) x^{k} = mf(x).

(b) g′(x) = −m(1 = x)^{−m−1}f(x) + (1 + x)^{−m}f′(x) = 0.

(c) g is constant. g(x) = g(0) = f(0) = 1 ⇒ f(x) = (1 + x)^{m}.

8. Section 10.10: Problem 66. For |t| > 1,

\frac{1}{1 + t^{2}} = \frac{1}{t^{2}} \frac{1}{1 + \frac{1}{t^{2}}} = \frac{1}{t^{2}} (1 − \frac{1}{t^{2}} + \frac{1}{t^{4}} + …) = \frac{1}{t^{2}} − \frac{1}{t^{4}} + \frac{1}{t^{6}} + ….

For x > 1,

\frac{π}{2} − tan^{−1} x = ∫_{x}^{∞} \frac{1}{1 + t^{2}} dt = \frac{1}{x} − \frac{1}{3x^{3}} + \frac{1}{5x^{5}} − ….

For x < −1,

tan^{−1} x + \frac{π}{2} = ∫_{−∞}^{x} \frac{1}{1 + t^{2}} dt = −\frac{1}{x} + \frac{1}{3x^{3}} − \frac{1}{5x^{5}} + ….

9. Section 10.10: Problem 74.

∫ e^{ax}(cos bx + i sin bx) dx

= ∫ e^{(a+ib)x} dx
= \frac{a − ib}{a^{2} + b^{2}} e^{(a+ib)x} + C_{1} + iC_{2}
= \frac{e^{ax}}{a^{2} + b^{2}} (a cos bx + b sin bx) + C_{1}
+ i [\frac{e^{ax}}{a^{2} + b^{2}} (a sin bx − b cos bx) + C_{2}] .