Solutions to Problem Set 27

Math 211-03

11-13-2017

[Taylor's formula]

1. Suppose that

$$f(1) = 3, \quad f'(1) = -2, \quad f''(1) = 6, \quad f^{(3)}(1) = -3.$$

Use the $3^{\rm rd}$ degree Taylor polynomial $p_3(x; 1)$ to approximate $f(1.2)$ .

$$p_3(x; 1) = 3 - 2(x - 1) + 3(x - 1)^2 - \dfrac{1}{2} (x - 1)^3.$$

$$f(1.2) \approx 3 - 2(1.2 - 1) + 3(1.2 - 1)^2 - \dfrac{1}{2} (1.2 - 1)^3 = 2.716.\quad\halmos$$


2. Suppose that

$$f(2) = -5, \quad f'(2) = 1, \quad f''(2) = -4, \quad f^{(3)}(2) = -12.$$

Use the $3^{\rm rd}$ degree Taylor polynomial $p_3(x; 2)$ to approximate $f(1.9)$ .

$$p_3(x; 2) = -5 + (x - 2) - 2(x - 2)^2 - 2(x - 2)^3.$$

$$f(1.9) \approx -5 + (1.9 - 2) - 2(1.9 - 2)^2 - 2(1.9 - 2)^3 = -5.118.\quad\halmos$$


3. Suppose that $f(1) = 2$ , and for $n \ge 1$ ,

$$f^{(n)}(1) = \dfrac{\sqrt{n}}{(n + 1)^2}.$$

Write down the $3^{\rm rd}$ degree Taylor polynomial $p_3(x; 1)$ (with each of the coefficients simplified).

$$f'(1) = \dfrac{1}{4}, \quad f''(1) = \dfrac{\sqrt{2}}{9}, \quad f^{(3)}(1) = \dfrac{\sqrt{3}}{16}.$$

So

$$p_3(x; 1) = 2 + \dfrac{1}{4} (x - 1) + \dfrac{\sqrt{2}}{18} (x - 1)^2 + \dfrac{\sqrt{3}}{96} (x - 1)^3.\quad\halmos$$


4. The $6^{\rm th}$ degree term in the Taylor series for $f(x)$ at $x = 1$ is $\dfrac{7}{18} (x - 1)^6$ . Find $f^{(6)}(1)$ .

By Taylor's formula, the $6^{\rm
   th}$ degree term in the Taylor series for $f(x)$ at $x
   = 1$ is $\dfrac{f^{(6)}(1)}{6!} (x - 1)^6$ .

So

$$\eqalign{ \dfrac{f^{(6)}(1)}{6!} & = \dfrac{7}{18} \cr \noalign{\vskip2pt} f^{(6)}(1) & = \dfrac{7}{18} \cdot 6! = 280 \quad\halmos \cr}$$


5. Find the $4^{\rm th}$ degree Taylor polynomial for $f(x) = \ln (\sec x)$ expanded at $c = 0$ .

$$\eqalign{ f'(x) & = \dfrac{\sec x \tan x}{\sec x} = \tan x \cr f''(x) & = (\sec x)^2 \cr f^{(3)}(x) & = 2 (\sec x)^2 \tan x \cr f^{(4)}(x) & = 2 (\sec x)^4 + 4 (\sec x)^2 (\tan x)^2 \cr}$$

So

$$f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 1, \quad f^{(3)}(0) = 0, \quad f^{(4)}(0) = 2.$$

Therefore,

$$p_4(x; 0) = \dfrac{1}{2} x^2 + \dfrac{1}{12} x^4.\quad\halmos$$


6. Find the $3^{\rm rd}$ degree Taylor polynomial for $f(x) = \dfrac{1}{1 + e^x}$ expanded at $c = 0$ .

$$f'(x) = -\dfrac{e^x}{(1 + e^x)^2}.$$

$$f''(x) = -\dfrac{(1 + e^x)^2(e^x) - (e^x)(2)(1 + e^x)(e^x)} {(1 + e^x)^4} = -\dfrac{(1 + e^x)(e^x) - 2(e^x)(e^x)}{(1 + e^x)^3} = -\dfrac{e^x - e^{2 x}}{(1 + e^x)^3}.$$

$$f^{(3)}(x) = -\dfrac{(1 + e^x)^3(e^x - 2 e^{2 x}) - (e^x - e^{2 x})(3)(1 + e^x)^2(e^x)}{(1 + e^x)^6} =$$

$$-\dfrac{(1 + e^x)(e^x - 2 e^{2 x}) - (e^x - e^{2 x})(3)(e^x)} {(1 + e^x)^4} = -\dfrac{e^x - e^{2 x} - 2 e^{3 x} - 3 e^{2 x} + 3 e^{3 x}}{(1 + e^x)^4} = -\dfrac{e^x - 4 e^{2 x} + e^{3 x}}{(1 + e^x)^4}.$$

Then

$$f(0) = \dfrac{1}{2}, \quad f'(0) = -\dfrac{1}{4}, \quad f''(0) = 0, \quad f^{(3)}(0) = \dfrac{1}{8}.$$

Hence,

$$p_3(x; 0) = \dfrac{1}{2} - \dfrac{1}{4} x + \dfrac{1}{48} x^3.\quad\halmos$$


[Taylor series applications]

7. (a) Find the first 4 nonzero terms of the Taylor series at $c = 0$ for $\cos x$ .

(b) Use the series in (a) to obtain the value of the limit $\displaystyle \lim_{x \to 0}
   \dfrac{1 - \cos x}{x^2}$ .

(a)

$$\cos x = 1 - \dfrac{1}{2} x^2 + \dfrac{1}{24} x^4 - \dfrac{1}{720} x^6 + \cdots.\quad\halmos$$

(b) Using the series in (a),

$$\dfrac{1 - \cos x}{x^2} = \dfrac{1}{x^2} \left(1 - 1 + \dfrac{1}{2} x^2 - \dfrac{1}{24} x^4 + \dfrac{1}{720} x^6 - \cdots\right) = \dfrac{1}{x^2} \left(\dfrac{1}{2} x^2 - \dfrac{1}{24} x^4 + \dfrac{1}{720} x^6 - \cdots\right) =$$

$$\dfrac{1}{2} - \dfrac{1}{24} x^2 + \dfrac{1}{720} x^4 - \cdots.$$

Hence,

$$\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \lim_{x \to 0} \left(\dfrac{1}{2} - \dfrac{1}{24} x^2 + \dfrac{1}{720} x^4 - \cdots\right) = \dfrac{1}{2}.\quad\halmos$$


8. (a) Find the first 4 nonzero terms of the Taylor series at $c = -1$ for $\ln (x +
   2)$ .

(b) Use the series in (a) to obtain the value of the limit $\displaystyle \lim_{x \to -1}
   \dfrac{\ln (x + 2)}{x + 1}$ .

(a) Set $u = x + 1$ in the series for $\ln (1 + u)$ :

$$\ln (x + 2) = \ln [1 + (x + 1)] = (x + 1) - \dfrac{(x + 1)^2}{2} + \dfrac{(x + 1)^3}{3} - \dfrac{(x + 1)^4}{4} + \cdots.\quad\halmos$$

(b) Using the series in (a),

$$\dfrac{\ln (x + 2)}{x + 1} = \dfrac{1}{x + 1} \left((x + 1) - \dfrac{(x + 1)^2}{2} + \dfrac{(x + 1)^3}{3} - \dfrac{(x + 1)^4}{4} + \cdots\right) =$$

$$1 - \dfrac{x + 1}{2} + \dfrac{(x + 1)^2}{3} - \dfrac{(x + 1)^3}{4} + \cdots.$$

Hence,

$$\lim_{x \to -1} \dfrac{\ln (x + 2)}{x + 1} = \lim_{x \to -1} \left(1 - \dfrac{x + 1}{2} + \dfrac{(x + 1)^2}{3} - \dfrac{(x + 1)^3}{4} + \cdots\right) = 1.\quad\halmos$$


Too many people overvalue what they are not and undervalue what they are. - Malcolm Forbes


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