Calculus II, Spring 2013 (http://www.math.nthu.edu.tw/˜ wangwc/)

Homework Assignment for Chap 09

1. Section 9.1: Problems: 7, 41, 57, 59, 60, 65, 71, 72(abe).

2. Section 9.2: Problems: 14, 20, 33, 62, 63.

3. Section 9.3: Problems: 23, 24, 27, 39, 41, 47, 61, 62.

4. Section 9.4: Problems 3, 5, 7, 9, 11, 13, 15, 19, 23, 25, 28, 34, 35, 39, 40, 41, 44 (Does it converge? Why?).

5. Section 9.5: Problems 4, 8, 7, 9, 24, 25, 33, 35, 37, 43, 45, 47.

6. We know that if ∑_{n=1}^{∞}a_{n} converges, then lim_{n→∞}a_{n} = 0. How about the counter part for integration? Suppose that f(x) is non-negative, continuous on [0, ∞) and ∫_{0}^{∞}f(x) dx < ∞. Is lim_{x→∞}f(x) = 0 necessarily true?

7. Is ∑_{n=1}^{∞} sin \frac{1}{n} convergent? How about ∑_{n=1}^{∞}(1 − cos \frac{1}{n})?

8. We know that the p = 1/2 series ∑_{k=1}^{∞} k^{\frac{−1}{2}} diverges. The question here is how fast does

the partial sum grows with n, or ∑_{k=1}^{n} k^{\frac{−1}{2}} = O(n^{?})? In other words, can you evaluate

lim_{n→∞} \frac{log (∑_{k=1}^{n}k^{\frac{−1}{2}} )}{log n} if it exists?

9. Section 9.6: Problems 11, 15, 21, 23, 27, 28, 35, 39, 41, 44, 47.

10. Find a power series whose interval of convergence is [1, 3). Do the same for (1, 3), [1, 3] and (1, 3], respectively.

11. Section 9.7: Problems 1, 3, 7, 15, 19, 25, 29, 33 (show that equality holds), 35, 47(a), 50, 57, 58.

12. Continue on problem 50. Show that f′(0) = 0 and f′′(0) = 0.

Remark: In fact, it can be shown that f^{(n)}(0) = 0 for all n by induction. The calculation is lengthier, but not more difficult.

13. Prove the following version of the Taylor’s Theorem.

(a) If f, f′, ⋯, f^{(n+1)} are all continuous in (a − h, a + h), h > 0. Then for any x ∈ (a − h, a + h), we have

f(x) = f(a) + f′(a)(x − a) + \frac{f′′(a)}{2!}(x − a)^{2} + ⋯ \frac{f^{(n)}(a)}{n!}(x − a)^{2} + R_{n+1}(x),

where

R_{n+1}(x) = \frac{1}{n!} ∫_{a}^{x} f^{(n+1)}(t)(x − t)^{n}dt.  (1)

(b) Show that (1) leads to

R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n + 1)!}(x − a)^{n+1}  (2)

for some c between a and x.

14. Find the interval of convergence for the power series

1 + \frac{1}{2}x + \frac{1 · 3}{2 · 4}x^{2} + \frac{1 · 3 · 5}{2 · 4 · 6}x^{3} + ⋯ + \frac{1 ⋯ (2n − 1)}{2 ⋯ 2n}x^{n} + · · · .

Hint: Denote by a_{n} = \frac{1···(2n−1)}{2···2n} and define b_{2} = \frac{2}{3}, b_{3} = \frac{2·4}{3·5}, ⋯, b_{n} = \frac{2···(2n−2)}{3···(2n−1)}. For x = 1, the fact that b_{n} < 1 will help. For x = −1, compare a_{n} with b_{n} and 2b_{n} to conclude that a_{n} ∼ n^{−p}. What is p? Do the same for

1 + \frac{1}{3}x + \frac{1 · 4}{3 · 6}x^{2} + \frac{1 · 4 · 7}{3 · 6 · 9}x^{3} + ⋯ + \frac{1 ⋯ (3n − 2)}{3 ⋯ 3n}x^{n} + · · · .

15. Section 9.8: Problems 9, 13, 28, 29, 31, 32, 33, 34(a), 35, 43.

16. Chap 9: Problems 27, 28, 29, 30, 32, 37, 47(a), 49-54(important), 71, 72.

17. Show that the Taylor series generated by (1 + x)^{m}, m ∈ R, converges on |x| < 1. This does not mean that the series indeed converges to (1 + x)^{m}. Analyzing R_{n}(x) directly is not quite easy (try it and youll see why). An alternative approach is through the following steps:

(a) Verify that

(k + 1) ( \begin{matrix}m \\ k + 1 \end{matrix} ) + k ( \begin{matrix}m \\ k \end{matrix} ) = m ( \begin{matrix}m \\ k \end{matrix} )

(b) Define, for |x| < 1,

f(x) = ∑_{k=0}^{∞} ( \begin{matrix}m \\ k \end{matrix} ) x^{k} = 1 + mx + \frac{m(m − 1)}{2!}x^{2} \frac{m(m − 1)(m − 2)}{3!}x^{3} + ⋯

Show that f(0) = 1 and

(1 + x)f′(x) = mf(x), |x| < 1.

(c) Show that f(x) = (1 + x)^{m} on |x| < 1.

18. Solve the differential equation

dy/dx = 1 + y^{2}, y(0) = 0,

by power series expansion. That is, assume y(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯, then compare the coefficients on both sides to solve for a_{0}, a_{1}, ⋯, successively. The differential equation can also be integrated directly. Verify your answer by comparing the first 2 nonzero coefficients of the Taylor series expansion of the exact solution.