10.4, #15.

3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + ⋯ + \frac{3}{2^{10}}

= 3(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ⋯ + \frac{1}{2^{10}} )

= \frac{3(1 − 1/2^{11})}{1 − 1/2}

= \frac{3(2^{11} − 1)}{2^{10}}.

10.4, #23. (a) The height to which the ball rises after it hits the floor for the n-th time is 10(3/4)^{n} feet.

(b) The ball has traveled a total vertical distance of 10 feet when it hits the floor the first time, 10 + 2(10)(3/4) = 25 feet the second time, 10 + 2(10)(3/4) + 2(10)(3/4)^{2} = 36.25 feet the third time, and 10 + 2(10)(3/4) + 2(10)(3/4)^{2} + 2(10)(3/4)^{3} = 44.69 feet the fourth time.

(c) When the ball hits the floor the n-th time, it has traveled a total vertical distance

10 + 2(10)(3/4) + 2(10)(3/4)^{2} + 2(10)(3/4)^{3} + ⋯ + 2(10)(3/4)^{n−1} = 10 + 15[1 + (3/4) + (3/4)^{2} + ⋯ + (3/4)^{n−2}]

= 10 + 15 (\frac{1 − (3/4)^{n−1}}{1 − (3/4)} )

= 10 + 60(1 − (3/4)^{n−1}) feet.

10.4, #30. The present value of a $1000 bond is

1000(1.06)^{−10} + 50[(1.06)^{−1} + (1.06)^{−2} + + ⋯ + (1.06)^{−10}]

The term 1000(1.06)^{−10} appears since the principal will have been held by the issuer of the bond for 10 years. The term 50(1.06)^{−j} is the present value of the coupon paid at the end of the j-th year. Thus present value of the bond is

1000(1.06)^{−10} + \frac{50}{1.06} (\frac{1 − (1.06)^{−10}}{1 − (1.06)^{−1}} ) = 926.40 dollars.

10.5, #1–2. The derivatives of f(t) = e^{t} are e^{t}. Thus |f^{n}(t)| ≤ e^{0.5} on [0, 0.5]. In particular, E_{0} = f(t) − P_{0}(t) and E_{2} = f(t) − P_{2}(t) satisfy the inequalities

|E_{0}| ≤ e^{0.5}(0.5) = 0.8243, |E_{2}| ≤ e^{0.5}(0.5)^{3}/3! = 0.0343

by the second boxed remark on page 631. P_{0}(t) and P_{2}(2) are underestimates for e^{t} since they are truncations of the series 1+t+t^{2}/2!+t^{3}/3!+⋯, which contains only non-negative numbers when t is in [0, .05].

10.5, #9. The (n + 1)-st derivative of f(x) = cos x is − sin x, − cos x, sin x, or cos x according as n = 4t, 4t + 1, 4t + 2, or 4t + 3 for some non-negative integer t In particular,

|f^{(n+1)}(x)| = \{ \begin{matrix}| sin x| & if n is even \\ | cos x| & if n is odd \end{matrix}

If 0 ≤ x ≤ 1, then |f^{(n+1)}(x)| ≤ sin 1 = 0.841 for even n, and |f^{(n+1)}(x)| ≤ cos 0 = 1 for odd n. The error term is thus bounded above by (0.841/(n + 1)!)1^{n+1} = 0.841/(n + 1)! for even n and by 1/(n + 1)!)1^{n+1} = 1/(n + 1)! for odd n. A similar analysis works for sin x if the words “even” and “odd” are interchanged.