CALCULUS III FALL 1999 HOMEWORK 10 – ANSWERS

§10.11 Question 6,8,12,16,20; §10.12 Questions 10,12

6.

\frac{1}{\sqrt{2 + x}} = \frac{1}{\sqrt{2}}(1 + \frac{x}{2})^{−1/2} = \frac{1}{\sqrt{2}} ∑_{n=0}^{∞} \frac{(−\frac{1}{2})(−\frac{3}{2}) ⋯ (−\frac{1}{2} − n + 1)}{n!}(\frac{x}{2})^{n}

= \frac{1}{\sqrt{2}} ∑_{n=0}^{∞}(−1)^{n} \frac{1.3 … (2n − 1)}{2^{2n}n!}x^{n} = \frac{1}{\sqrt{2}} ∑_{n=0}^{∞} \frac{(−1)^{n}(2n)!}{2^{3n}(n!)^{2}}x^{n}.

We applied the binomial theorem to the variable x/2 so the possible range of values is given by |x/2| < 1 which gives |x| < 2 and so the radius of convergence is 2.

8.

\frac{x^{2}}{\sqrt{1 − x^{3}}} = x^{2}(1 − x^{3})^{−1/2} = x^{2} ∑_{n=0}^{∞} \frac{(−1)^{n}(2n)!}{2^{2n}(n!)^{2}}(−x^{3})^{n}

= ∑_{n=0}^{∞} \frac{(2n)!}{2^{2n}(n!)^{2}} x^{3n+2}.

We applied the binomial theorem to the variable −x^{3} so the possible range of values is given by | − x^{3}| < 1 which is the same as |x| < 1 and so the radius of convergence is 1.

CALCULUS III  HOMEWORK 10 – ANSWERS  FALL 1999 2.

12.

(4 + x)^{3/2} = 8(1 + \frac{x}{4})^{3/2} = 8 ∑_{n=0}^{∞} \frac{(\frac{3}{2})(\frac{1}{2})(−\frac{1}{2}) ⋯ (\frac{3}{2} − n + 1)}{n!}(\frac{x}{4})^{n}

= 24 ∑_{n=0}^{∞}(−1)^{n} \frac{1.3 … (2n − 5)}{2^{n}n!}(\frac{x}{4})^{n} = 24 ∑_{n=0}^{∞} \frac{(2n − 4)!}{2^{4n−2}n!(n − 2)!}x^{n}.

It follows that the Taylor polynomials are

T_{1}(x) = 8(1 + \frac{3x}{8}) = 8 + 3x,
T_{2}(x) = 8(1 + \frac{3x}{8} + \frac{(\frac{3}{2})(\frac{1}{2})}{2})\frac{x^{2}}{16} = 8 + 3x + \frac{3x^{2}}{16} ,
T_{3}(x) = 8(1 + \frac{3x}{8} + \frac{(\frac{3}{2})(\frac{1}{2})}{2} + \frac{(\frac{3}{2})(\frac{1}{2})(−\frac{1}{2})}{3!} \frac{x^{3}}{64} )\frac{x^{2}}{16} = 8 + 3x + \frac{3x^{2}}{16}

= 8 + 3x + \frac{3x^{2}}{16} − \frac{x^{3}}{128}.

We applied the binomial theorem to the variable x/4 so the possible range of values is given by |x/4| < 1 which gives |x| < 4 and so the radius of convergence is 4. The graphs are given below 16.

(8 + x)^{1/3} = 2(1 + \frac{x}{8})^{1/3} = 2 ∑_{n=0}^{∞} \frac{(\frac{1}{3})(−\frac{2}{3}) ⋯ (\frac{1}{3} − n + 1)}{n!} \frac{x^{n}}{8^{n}}

= 2 ∑_{n=0}^{∞}(−1)^{n−1} \frac{2.5 … (3n − 4)}{2^{3n}3^{n}n!}x^{n}.

We applied the binomial theorem to the variable x/8 so the possible range of values is given by |x/8| < 1 which gives |x| < 8 and so the radius of convergence is 8.

For part b). we put x = 0.2 to get

2 ∑_{n=0}^{∞}(−1)^{n−1} \frac{2.5 … (3n − 4)}{2^{3n}3^{n}n!} \frac{1}{5^{n}} .

The alternating nature of this series suggests that the required answer is given by the partial sum s_{n} where

\frac{2.5 … (3n − 1)}{2^{3n+3}3^{n+1}n!} \frac{1}{5^{n+1}} < 0.00005.

We find that this is satisfied when n = 2 and so the required estimate is

\sqrt[3]{8.2} ≃ 2(1 + \frac{1}{120} − \frac{1}{14400}) ≃ 2.0165.

20.

(1 + x^{3})^{−1/2} = ∑_{n=0}^{∞} \frac{(−\frac{1}{2})(−\frac{3}{2}) ⋯ (−\frac{1}{2} − n + 1)}{n!}(x^{3})^{n}

= ∑_{n=0}^{∞}(−1)^{n} \frac{1.3 … (2n − 1)}{2^{n}n!}x^{3n} = ∑_{n=0}^{∞}(−1)^{n} \frac{(2n)!}{2^{2n}(n!)^{2}} x^{3n}.

It follows that

\frac{f^{(3n)}(0)}{(3n)!} = (−1)^{n} \frac{(2n)!}{2^{2n}(n!)^{2}}  and so  f^{(9)}(0) = −9!\frac{6!}{2^{6}r} = −113, 400.

10. We have f′(x) = −1/x^{2}, f′′(x) = 2/x^{3} and, in general, f^{(n)}(x) = (−1)^{n}n!/x^{n+1}, which gives f^{(n)}(1) = (−1)^{n}n!. It follows that the required Taylor series is

∑_{n=0}^{∞}(−1)^{n}(x − 1)^{n}.

Notice that there is another way to reach this same conclusion,

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

It follows that T_{1}(x) = 1−(x−1) = 2−x, T_{2}(x) = 1−(x−1)+(x−1)^{2} = 3−3x+x^{2} and T_{3}(x) = 1 − (x − 1) + (x − 1)^{2} − (x − 1)^{3} = 4 − 6x + 4x^{2} − x^{3}. The graphs are shown below 22. We have f^{(n)}(x) = (−1)^{n−1}(n − 1)!/x^{n} if n ⩾ 1. It follows that

ln x = ln 4 + ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{n4^{n}} (x − 4)^{n}.

The third Taylor polynomial is

T_{3}(x) = ln 4 + \frac{x − 4}{4} − \frac{(x − 4)^{2}}{32} + \frac{(x − 4)^{3}}{192} .

The remainder term is R_{3}(x) = −(x − 4)^{4}/(4z^{4}) where z is between x and 4. We have 3 ⩽ x ⩽ 5 and so 3 < z < 5, hence |R_{3}(x)| < 1/(4(3^{4})) = 1/324 ≃ 0.0031. The graph of f and T_{3} are given below