Calculus I, Fall 2014 (http://www.math.nthu.edu.tw/~wangwc/)

Brief answer to selected problems in HW05

1. Section 3.8: Problem 98. Let

f(x) = ln x ⟹ f′(x) = \frac{1}{x}

By definition, for fixed x > 0

f′(1) = lim_{\frac{x}{n} →0} \frac{f(1+ \frac{x}{n})−f(1)}{\frac{x}{n}} = lim_{n→∞} \frac{n}{x}(ln(1 + \frac{x}{n}) − ln 1)
= \frac{1}{x} lim_{n→∞} ln(1 + \frac{x}{n})^{n}

which implies

x = ln( lim_{n→∞}(1 + \frac{x}{n})^{n}) ⟹ lim_{n→∞}(1 + \frac{x}{n})^{n} = e^{x}

2. Section 3.9: Problem 55. After differentiation, we find out

f′(x) = g′(x), x > 0

which means

f(x) = g(x) + c, x ≥ 0

where c is some constant. And x = 0 gives us

−\frac{π}{2} = 0 + c, x ≥ 0

hence

f(x) = g(x) − \frac{π}{2} , x ≥ 0

On the other hand, we also can derive same result by following process. For x ≥ 0, let

θ = sin^{−1} \frac{x − 1}{x + 1} ⟹ sin θ = \frac{x − 1}{x + 1}

then

tan(\frac{θ + \frac{π}{2}}{2} ) = \sqrt{\frac{1 − cos(θ + \frac{π}{2})}{1 + cos(θ + \frac{π}{2})}} = \sqrt{\frac{1 + sin θ}{1 − sin θ}} = \sqrt{x}

hence

tan^{−1} \sqrt{x} = \frac{θ + \frac{π}{2}}{2} ⟹ sin^{−1} \frac{x − 1}{x + 1} = 2 tan^{−1} \sqrt{x} − \frac{π}{2}

3. Hw05: Problem 3. Domain and image for csc and csc^{−1}

csc x : {x|x ∈ R, x ≠ kπ, ∀k ∈ Z} → (−∞, −1] ∪ [1, ∞)
csc^{−1} x : (−∞, −1] ∪ [1, ∞) → \{y| − \frac{π}{2} ≤ y ≤ \frac{π}{2} , y ≠ 0\}

For |x| > 1, let

y = csc^{−1} x ⟹ csc y = x
⟹ − csc y cot y \frac{dy}{dx} = 1
⟹ \frac{dy}{dx} = \frac{−1}{csc y cot y}

where

cot y = \{ \begin{matrix}\sqrt{x^{2} − 1} & , x > 1 \\ −\sqrt{x^{2} − 1} & , x < −1 \end{matrix}

hence

\frac{d csc^{−1} x}{dx} = \frac{−1}{|x|\sqrt{x^{2} − 1}}, |x| > 1

4. Section 3.11: Problem 64.

(a)b_{0} = f(a), b_{1} = f′(a), and b_{2} = \frac{f′′(a)}{2} (b)Quadratic approximation of \frac{1}{1−x} at x = 0 is 1 + x + x^{2} (d)Quadratic approximation of \frac{1}{x} at x = 1 is 1 − (x − 1) + (x − 1)^{2} (e)Quadratic approximation of \sqrt{1 + x} at x = 0 is 1 + \frac{1}{2}x − \frac{1}{8}x^{2} Graphs f, g, and h appear to be identical with their own quadratic approximation around the given points.

5. Hw05: Problem 5. For ^{3} \sqrt{1.009}, let

f(x) = (1 + x)^{\frac{1}{3}} ⟹ f′′(x) = −\frac{2}{9}(1 + x)^{− \frac{5}{3}}

Then ^{3} \sqrt{1.009} = f(0.009). Hence we use linear approximation at x = 0. By the error formula, we can get error bound by following process.

|f(0.009) − L(0.009, 0)| ≤ \frac{1}{2}( max_{0≤c≤0.009} |f′′(c)|)(0.009 − 0)^{2}
= \frac{1}{2} × \frac{2}{9} × 0.009^{2} = 9 × 10^{−6}