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

Homework Assignment for Week 04

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

2. (s9.7-extra1) 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.

3. (s9.7-extra2) 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}(x),

where

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

(b) Show that (1) leads to

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

for some c between a and x.

4. Read Taylor’s Theorem and proof on p609-610. Note the difference on the assumption on f^{(n+1)} between the two versions. The version in problem 3 has stronger assumption and leads to stronger result (1). On the other hand, to get (2), the weaker assumption in Theorem 15 on p609 is enough.

5. (s9.8-extra1) 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.