10.1, #18. P_{3}(x) = \sqrt{2} + \frac{(x − 1)}{2\sqrt{2}} − \frac{(x − 1)^{2}}{16\sqrt{2}} + \frac{(x − 1)^{3}}{64\sqrt{2}} since

\begin{aligned}f^{(0)}(x) & = (1 + x)^{1/2}, \\ f^{(1)}(x) & = (1/2)(1 + x)^{−1/2}, \\ f^{(2)}(x) & = −(1/2)(1/2)(1 + x)^{−3/2}, \\ f^{(3)}(x) & = (3/2)(1/2)(1/2)(1 + x)^{−5/2}, \end{aligned}  \begin{aligned}f^{(0)}(1) & = \sqrt{2}, \\ f^{(1)}(1) & = 1/2\sqrt{2}, \\ f^{(2)}(1) & = −1/8\sqrt{2}, \\ f^{(3)}(1) & = 3/32\sqrt{2}. \end{aligned}

10.1, #19. P_{4}(x) = \frac{1}{3} − \frac{(x − 2)}{9} + \frac{(x − 2)^{2}}{27} − \frac{(x − 2)^{3}}{81} + \frac{(x − 2)^{4}}{243} since

\begin{aligned}f^{(0)}(x) & = (1 + x)^{−1}, \\ f^{(1)}(x) & = (−1)(1 + x)^{−2}, \\ f^{(2)}(x) & = (−1)^{2}2!(1 + x)^{−3}, \\ f^{(3)}(x) & = (−1)^{3}3!(1 + x)^{−4}, \\ f^{(4)}(x) & = (−1)^{4}4!(1 + x)^{−5}, \end{aligned}  \begin{aligned}f^{(0)}(2) & = 1/3, \\ f^{(1)}(2) & = −1/9, \\ f^{(2)}(2) & = 2!/27, \\ f^{(3)}(2) & = −3!/81, \\ f^{(4)}(2) & = 4!/243. \end{aligned}

10.1, #23–26. a = f(0), b = f′(0), and c = f′′(0)/2. For each picture a is the y-intercept, b is positive or negative according as the graph is increasing or decreasing, and c is positive or negative according as the graph is concave-up or concave-down. There are 2^{3} cases, the four here and four more you should try to draw.

10.2, #1–3. #1 is a Taylor series. #2 is not a Taylor series since it contains negative powers of x. #3 is not a Taylor series since the center of each additional term changes.

10.2, #7. The first four derivatives of \sqrt{1 + x} were calculated in 10.1, #2:

\begin{aligned}f^{(0)}(x) & = (1 + x)^{1/2}, \\ f^{(1)}(x) & = (1/2)(1 + x)^{−1/2}, \\ f^{(2)}(x) & = −(1/2)(1/2)(1 + x)^{−3/2}, \\ f^{(3)}(x) & = (3/2)(1/2)(1/2)(1 + x)^{−5/2}, \\ f^{(4)}(x) & = −(5/2)(3/2)(1/2)(1/2)(1 + x)^{−7/2}. \end{aligned} \begin{aligned}f^{(0)}(0) & = 1, \\ f^{(1)}(0) & = 1/2, \\ f^{(2)}(0) & = −1/4, \\ f^{(3)}(0) & = 3/8, \\ f^{(4)}(0) & = −15/16 \end{aligned}

For an arbitrary n

f^{(n)}(x) = (−1)^{n−1} \frac{(2n − 3)(2n − 5) ⋯ 3 · 1}{2^{n}}(1 + x)^{−2n+1/2}
f^{(n)}(0) = (−1)^{n−1} \frac{(2n − 3)(2n − 5) ⋯ 3 · 1}{2^{n}}

The Taylor series for \sqrt{1 + x} is

\begin{matrix}1 + \frac{x}{2} − \frac{x^{2}}{8} + \frac{x^{3}}{16} − \frac{5x^{4}}{128} + ⋯ + (−1)^{n−1} \frac{(2n − 3)(2n − 5) ⋯ 3 · 1}{2^{n}n!}x^{n} + ⋯ \\ 1 + \frac{1}{2}x − \frac{1 · 1}{2 · 4}x^{2} + \frac{1 · 1 · 3}{2 · 4 · 6}x^{3} − \frac{1 · 1 · 3 · 5}{2 · 4 · 6 · 8}x^{4} + − · · · . \end{matrix}

10.2, #9. The first several derivatives of (1 − y)^{1/3} are

\begin{aligned}f^{(0)}(y) & = (1 − y)^{1/3}, \\ f^{(1)}(y) & = −(1/3)(1 − y)^{−2/3}, \\ f^{(2)}(y) & = −(2/3)(1/3)(1 − y)^{−5/3}, \\ f^{(3)}(y) & = −(5/3)(2/3)(1/3)(1 − y)^{−8/3} \end{aligned} \begin{aligned}f^{(0)}(0) & = 1 \\ f^{(1)}(0) & = −1/3 \\ f^{(2)}(0) & = −2/9 \\ f^{(3)}(0) & = −10/27 \end{aligned}

For arbitrary n ≥ 1

f^{(n)}(y) = −\frac{(3n − 4)(3n − 7) ⋯ 5 · 2}{3^{n}}(1 − y)^{−3n+1/3}
f^{(n)}(0) = −\frac{(3n − 4)(3n − 7) ⋯ 5 · 2}{3^{n}}

The Taylor series for (1 − y)^{1/3} is

1 − \frac{y}{3} − \frac{y^{2}}{9} − \frac{5y^{3}}{81} − ⋯ − \frac{(3n − 4)(3n − 7) ⋯ 2}{3^{n}n!}y^{n} − ⋯