Solutions to Problem Set 29

Math 211-03

11-16-2017

[Remainder Term]

1. Write down the remainder term $R_3(x; 0)$ for $f(x) = \sin 2 x$ .

$$\eqalign{ f'(x) & = 2 \cos 2 x \cr f''(x) & = -4 \sin 2 x \cr f^{(3)}(x) & = -8 \cos 2 x \cr f^{(4)}(x) & = 16 \sin 2 x \cr}$$

$$R_3(x; 0) = \dfrac{16 \sin 2 z}{4!} x^4 = \dfrac{2}{3} (\sin 2 z) x^4.\quad\halmos$$


2. Write down the remainder term $R_{10}(x; 0)$ for $f(x) = e^{2 x}$ .

$$\eqalign{ f'(x) & = 2 e^{2 x} \cr f''(x) & = 2^2 e^{2 x} \cr f^{(3)}(x) & = 2^3 e^{2 x} \cr}$$

I can see that $f^{(n)}(x) = 2^n
   e^{2 x}$ . So

$$R_{10}(x; 0) = \dfrac{2^{11} e^{2 z}}{11!} x^{11} = \dfrac{8}{155925} e^{2 z} x^{11}.\quad\halmos$$


3. Write down the remainder term $R_n(x; 0)$ for $f(x) = \dfrac{1}{3 - x}$ .

$$\eqalign{ f'(x) & = \dfrac{1}{(3 - x)^2} \cr f''(x) & = \dfrac{2}{(3 - x)^3} \cr f^{(3)}(x) & = \dfrac{2 \cdot 3}{(3 - x)^4} \cr f^{(4)}(x) & = \dfrac{2 \cdot 3 \cdot 4}{(3 - x)^5} \cr}$$

I can see that $f^{(n)}(x) =
   \dfrac{n!}{(3 - x)^{n + 1}}$ . So

$$R_n(x; 0) = \dfrac{\dfrac{(n + 1)!}{(3 - z)^{n + 2}}}{(n + 1)!} x^{n + 1} = \dfrac{x^{n + 1}}{(3 - z)^{n + 2}}.\quad\halmos$$


4. Use the remainder term $R_3(x; 0)$ to estimate the largest possible error in using the $3^{\rm rd}$ degree Taylor polynomial $p_3(x; 0)$ to approximate $f(x) = e^{-2 x}$ for $0 \le x \le
   0.4$ .

$$\eqalign{ f'(x) & = -2 e^{-2 x} \cr f''(x) & = 4 e^{-2 x} \cr f^{(3)}(x) & = -8 e^{-2 x} \cr f^{(4)}(x) & = 16 e^{-2 x} \cr}$$

$$R_3(x; 0) = \dfrac{16 e^{-2 z}}{4!} x^4 = \dfrac{2}{3} e^{-2 z} x^4.$$

Since $0 \le x \le 0.4$ , I have $x^4 \le 0.4^4$ . Moreover, $0
   \le z \le x \le 0.4$ , so

$$\eqalign{ 0 \le & z \le 0.4 \cr 0 \ge & -2 z \ge -0.8 \cr 1 \ge & e^{-2 z} \ge e^{-0.8} \cr}$$

Thus,

$$|R_3(x; 0)| \le \dfrac{2}{3} \cdot 1 \cdot 0.4^4 = 0.01706 \ldots.\quad\halmos$$


5. Using the remainder term, find the minimum number n so that the $n^{\rm th}$ degree Taylor polynomial $p_n(x; 0)$ for $f(x) = e^{5
   x}$ will approximate $f(x)$ on the interval $0 \le x \le 0.2$ to an error of less than $0.0001 =
   10^{-4}$ .

$$\eqalign{ f'(x) & = 5 e^{5 x} \cr f''(x) & = 5^2 e^{5 x} \cr f^{(3)}(x) & = 5^3 e^{5 x} \cr}$$

I can see that $f^{(n)}(x) = 5^n
   e^{5 x}$ . So

$$R_n(x; 0) = \dfrac{5^{n + 1} e^{5 z}}{(n + 1)!} x^{n + 1}.$$

Since $0 \le x \le 0.2$ , I have $x^{n + 1} \le 0.2^{n + 1}$ . Also $0 \le z \le x \le 0.2$ , so $e^{5 z} \le e^{5(0.2)} = e$ . So

$$|R_n(x; 0)| \le \dfrac{5^{n + 1} e}{(n + 1)!} 0.2^{n + 1} = \dfrac{e}{(n + 1)!}.$$

(I used the fact that $5^{n + 1}
   \cdot 0.2^{n + 1} = (5 \cdot 0.2)^{n + 1} = 1^{n + 1} = 1$ .)

I want to find the smallest n for which this is less than $10^{-4}$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & n & & $\dfrac{e}{(n + 1)!}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 3 & & $0.11326 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 4 & & $0.02265 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 5 & & $0.00377 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 6 & & $5.39341 \ldots \cdot 10^{-4}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 7 & & $6.74177 \ldots \cdot 10^{-5}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first n for which $\dfrac{e}{(n + 1)!}$ is less than $10^{-4}$ is $n
   = 7$ .


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