10.1, #2. P_{n}(x) = C_{0} + C_{1}x + C_{2}x^{2} + ⋯ + C_{n}x^{n}, where C_{j} = f^{(j)}(0)/j!. Now

\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}

Thus

P_{2}(x) = 1 + (1/2)x − (1/8)x^{2},
P_{3}(x) = 1 + (1/2)x − (1/8)x^{2} + (1/16)x^{3},
P_{4}(x) = 1 + (1/2)x − (1/8)x^{2} + (1/16)x^{3} − (5/128)x^{4}.

10.1, #4.

\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^{(n)}(x) & = (−1)^{n}n!(1 + x)^{−n−1}, \end{aligned}  \begin{aligned}f^{(0)}(0) & = 1, \\ f^{(1)}(0) & = −1, \\ f^{(2)}(0) & = 2!, \\ f^{(3)}(0) & = −3!, \\ & ⋯ \\ f^{(n)}(0) & = (−1)^{n}n!. \end{aligned}

Thus

P_{4}(x) = 1 − x + x^{2} − x^{3} + x^{4},
P_{6}(x) = 1 − x + x^{2} − x^{3} + x^{4} − x^{5} + x^{6},
P_{8}(x) = 1 − x + x^{2} − x^{3} + x^{4} − x^{5} + x^{6} − x^{7} + x^{8}.

10.1, #5. P_{3}(x) = P_{4}(x) = x + x^{3}/3, for

f^{(0)}(x) = tan x,
f^{(1)}(x) = sec^{2} x,
f^{(2)}(x) = 2 sec^{2} x tan x
f^{(3)}(x) = 4 sec^{2} x tan^{2} x + 2 sec^{4} x
f^{(4)}(x) = 8 sec^{2} x tan^{3} x + 16 sec^{4} x tan x

f^{(0)}(0) = 0,_{f^{(1)}(0) = 1,}_{f^{(2)}(0) = 0,}_{f^{(3)}(0) = 2,}_{f^{(4)}(0) = 0.}

10.1, #6. P_{3}(x) = P_{4}(x) = x − x^{3}/3, for

\begin{aligned}f^{(0)}(x) & = arctan x, \\ f^{(1)}(x) & = (1 + x^{2})^{−1}, \\ f^{(2)}(x) & = −2x(1 + x^{2})^{−2} \\ f^{(3)}(x) & = −2(1 + x^{2})^{−2} + 8x^{2}(1 + x^{2})^{−3} \\ f^{(4)}(x) & = 24x(1 + x^{2})^{−3} − 48x^{3}(1 + x^{2})^{−4} \end{aligned}  \begin{aligned}f^{(0)}(0) & = 0, \\ f^{(1)}(0) & = 1, \\ f^{(2)}(0) & = 0, \\ f^{(3)}(0) & = −2, \\ f^{(4)}(0) & = 0. \end{aligned}

10.1, #7.

\begin{aligned}f^{(0)}(x) & = ln(1 + x), \\ f^{(1)}(x) & = (1 + x)^{−1}, \\ f^{(2)}(x) & = (−1)(1 + x)^{−2}, \\ f^{(3)}(x) & = (−1)^{2}2!(1 + x)^{−3}, \\ & ⋯ \\ f^{(n)}(x) & = (−1)^{n−1}(n − 1)!(1 + x)^{−n}, \end{aligned}  \begin{aligned}f^{(0)}(0) & = 0, \\ f^{(1)}(0) & = 1, \\ f^{(2)}(0) & = −1, \\ f^{(3)}(0) & = 2!, \\ & ⋯ \\ f^{(n)}(0) & = (−1)^{n−1}(n − 1)!. \end{aligned}

Thus

P_{5}(x) = x − x^{2}/2 + x^{3}/3 − x^{4}/4 + x^{5}/5,
P_{7}(x) = x − x^{2}/2 + x^{3}/3 − x^{4}/4 + x^{5}/5 − x^{6}/6 + x^{7}/7,
P_{9}(x) = x − x^{2}/2 + x^{3}/3 − x^{4}/4 + x^{5}/5 − x^{6}/6 + x^{7}/7 − x^{8}/8 + x^{9}/9.

10.1, #10.

\begin{aligned}f^{(0)}(x) & = (1 + x)^{p}, \\ f^{(1)}(x) & = p(1 + x)^{p−1}, \\ f^{(2)}(x) & = p(p − 1)(1 + x)^{p−2}, \\ f^{(3)}(x) & = p(p − 1)(p − 2)(1 + x)^{p−3}, \\ f^{(4)}(x) & = p(p − 1)(p − 2)(p − 3)(1 + x)^{p−4}, \end{aligned}  \begin{aligned}f^{(0)}(0) & = 1, \\ f^{(1)}(0) & = p, \\ f^{(2)}(0) & = p(p − 1), \\ f^{(3)}(0) & = p(p − 1)(p − 2), \\ f^{(4)}(0) & = p(p − 1)(p − 2)(p − 3). \end{aligned}

Thus

P_{2}(x) = 1 + px − p(p − 1)x^{2}/2!,
P_{3}(x) = 1 + px − p(p − 1)x^{2}/2! + p(p − 1)(p − 2)x^{3}/3!,
P_{4}(x) = 1 + px − p(p − 1)x^{2}/2! + p(p − 1)(p − 2)x^{3}/3! − p(p − 1)(p − 2)(p − 3)x^{4}/4!.