Calculus I, Fall 2012 (http://www.math.nthu.edu.tw/˜ wangwc/)

Homework Assignment for Chap 06

1. Section 6.1: Problems 45, 47, 59, 65.

2. Section 6.2: Problems 28(c), 32, 45, 46.

3. Section 6.3: Problems 41, 50, 51, 57, 65, 77, 87, 89, 91, and as many as time permits in problems 31-40, 85-90.

4. Section 6.4: Problems: 9, 19, 25, 41, 47, 57, 71.

5. Evaluate lim_{n→∞} ∑_{k= \frac{n}{2}}^{n} \frac{1}{k}. Hint: try to express it in terms of \frac{k}{n}.

6. Section 6.6: Problems: 17, 21, 27, 33, 37-46, 51, 54, 55.

7. (s6.6-extra1) Cauchy’s Mean Value Theorem Prove the following variant of the Mean Value Theorem: Suppose f and g are continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that

| \begin{matrix}f(b) − f(a) & f′(c) \\ g(b) − g(a) & g′(c) \end{matrix} | = 0.  (1)

Note that, this c need not satisfy \frac{f(b) − f(a)}{b − a} = f′(c), nor \frac{g(b) − g(a)}{b − a} = g′(c), but only \frac{f(b) − f(a)}{g(b) − g(a)} = \frac{f′(c)}{g′(c)}, which is the same as (1) provided g(b) − g(a) ≠ 0. Hint: Apply standard Mean Value Theorem to

F (x) = | \begin{matrix}f(b) − f(a) & f(x) − f(a) \\ g(b) − g(a) & g(x) − g(a) \end{matrix} | on [a, b].

8. (s6.6-extra2) Use Cauchy’s Mean Value Theorem to prove the strong form of l’Hôpital’s rule.

9. Section 6.7: Problems: 5, 7, 8, 11, 12, 17(a), 18.

10. Section 6.9: Problems: 9, 13, 21, 23, 31, 33, 39, 43, 57, 61, 67, 71, 79.

11. (s6.9-extra1) Show that

\frac{d csc^{−1} y}{dy} = \frac{−1}{|y|\sqrt{y^{2} − 1}}, |y| > 1

Explain why the negative sign is chosen. You will need to start from the ‘restricted’ domain of csc to see this.

12. Section 6.10: Problems: 15, 31, 43, 67.

For Section 6.10, skip the inverse hyperbolic functions part and concentrate on hyperbolic functions and calculus related to them.

13. Section 6.11: Problems: 17, 21, 23, 27.

14. (s6.11-extra1) Verify that both sinh x are cosh x are solutions of y′′ = y. Then solve

for

y′′ = y, y(0) = 1, y′(0) = 2.

15. (s6.11-extra2) Verify that both e^{2x} and xe^{2x} are solutions of y′′ −4y′ +4y = 0, therefore so is the combination a_{1}e^{2x} + a_{2}xe^{2x}. This is an example of the multiple root case: (λ − 2)^{2} = 0. You can either verify by direct differentiation, or try to look for solutions of the form z(x)e^{2x} and find that this leads to z′′ = (y(x)e^{−2x})′′ = 0.

16. (s6.11-extra3) Verify by direct differentiation that

\frac{d}{dx}(cos(kx) + i sin(kx)) = ik(cos(kx) + i sin(kx))

This is a good explanation why one defines exp(ikx) to be cos(kx) + i sin(kx).