Calculus I, Fall 2014 (http://www.math.nthu.edu.tw/^{∼}wangwc/)

Brief answer to selected problems in Homework 09

1. Section 4.7: Problem 18: Solve, for example, f(x) = tan(\frac{x}{4}) − 1 = 0.

Problem 29: The solution: x_{∗} = 1. Newton’s method: x_{n+1} = x_{n} − \frac{x_{n} − 1}{40}. Therefore the error between nth and (n + 1)th iteration is given by x_{n+1} − x_{∗} = \frac{39}{40}(x_{n} − x_{∗}). It takes about \frac{3}{log_{10}(\frac{40}{39})} ≈ 118 iterations.

Problem 30: Combine rθ = 3 and r sin(\frac{θ}{2}) = 1 to get f(r) = r sin(\frac{3}{2r}) − 1 = 0 and solve for r.

2. Section 4.8: Problem 84: (b) and (c) are right. Problem 128: Yes. Since F − G is a constant and F (y_{0}) − G(y_{0}) = 0.

3. Section 5.3: Problem 87: We take for granted from the problem that a continuous function on a closed interval [a, b] must be uniformly continuous (ie. Assume this statement is correct. This is an advanced calculus Theorem). Therefore given ϵ > 0, one can find δ > 0 such that |x_{1} − x_{2}| < δ implies |f(x_{1}) − f(x_{2})| < ϵ. It is not difficult to see that for this δ, ∥P ∥ < δ implies U − L < ϵ(∑_{k} Δx_{k}) = ϵ(b − a).