Calculus II, Spring 2014

Brief solutions to quiz 02

1. Suppose that f is infinitely many times differentiable on R. State Taylor’s Theorem for f and write down the formula for R_{n}(x) (in any form you choose). Ans: See page 607 of the textbook. Both formula are correct.

2. Find the Taylor series generated by sin x at x = 0. For what values of x does sin x equal this series? Explain. Ans:

(a) sin x = ∑_{n=0}^{∞} \frac{(−1)^{n} x^{2n+1}}{(2n + 1)!} .

(b) The series converges for all x ∈ (−∞, ∞).

(c) Since for all x ∈ (−∞, ∞),

lim_{n→∞} \begin{matrix}| & \frac{(−1)^{n+1}x^{2n+3}}{(2n+3)!} & | \\ | & \frac{(−1)^{n}x^{2n+1}}{(2n+1)!} & | \end{matrix} = lim_{n→∞} \begin{matrix}|x| & 2 \\ (2n + 2) & (2n + 3) \end{matrix} = 0 < 1,

by the Ratio Test, the series converges for all x ∈ (−∞, ∞).

3. Find the sum of the series

\frac{1}{2} − \frac{1}{8} + \frac{1}{24} − ⋯ + (−1)^{n−1} \frac{1}{n2^{n}} + ⋯

Ans: Replace \frac{1}{2^{n}} by x^{n}. Then take differentiate it to eliminate \frac{1}{n} term. This will result in a series that you recognize. Then you can integrate once to get the original series and then evaluate at x = 1/2. Answer: ln(1 + \frac{1}{2}).

4. Let f(x) = exp(\frac{−1}{x^{2}}) for x ≠ 0 and f(0) = 0. Is f differentiable everywhere? Explain. Ans:

(a) Yes, f is differentiable everywhere.

(b) For c ≠ 0, since g(y) = e^{y} is differentiable at −\frac{1}{c^{2}} and h(x) = −\frac{1}{x^{2}} is differentiable at c, by the Chain Rule, e^{−1/x^{2}} is differentiable at c, and

f′(c) = \frac{2 e^{− \frac{1}{c^{2}}}}{c^{3}} .

(c) Since

lim_{h→0^{+}} \frac{f(0 + h) − f(0)}{h} = lim_{h→0^{+}} \frac{e^{− \frac{1}{h^{2}}} − 0}{h} = lim_{y→+∞} \frac{e^{−y^{2}}}{\frac{1}{y}} = lim_{y→+∞} \frac{y}{e^{y^{2}}} = 0,
lim_{h→0^{−}} \frac{f(0 + h) − f(0)}{h} = lim_{h→0^{−}} \frac{e^{− \frac{1}{h^{2}}} − 0}{h} = lim_{y→−∞} \frac{e^{−y^{2}}}{\frac{1}{y}} = lim_{y→−∞} \frac{y}{e^{y^{2}}} = 0,

f is differentiable at x = 0, and f′(0) = 0.

5. Find the area of the surface generated by evolving x = cos t, y = 2 + sin t, 0 ≤ t ≤ 2π

about the x-axis. Ans:

∫_{0}^{2π} 2π(2 + sin t)\sqrt{x′(t)^{2} + y′(t)^{2}}dt = ∫_{0}^{2π} 2π(2 + sin t)dt = 8π^{2}  (1)