第14章§5 - 1

$\log (x+1)$のマクローリン展開と剰余項の評価は $$\begin{array}{c}\log (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots +\frac{(-1{)}^{n-1}}{n}{x}^n+R_{n+1}\\ |R_{n+1}|\leqq \{\begin{array}{ll}\frac{a^{n+1}}{n+1}& (x>0;x\leqq a\leqq 1)\\ \frac{|a{|}^{n+1}}{(n+1)(1+a)}& (x<0;-1<a\leqq x)\end{array}\end{array}$$ である．

(a)

$$\begin{array}{c}\log 1.2=\log (1+0.2)=\sum _{k=1}^n\frac{(-1{)}^{k-1}}{k}{0.2}^k+R_{n+1}\\ |R_{n+1}|\leqq \frac{{0.2}^{n+1}}{n+1}=\frac{2^{n+1}}{n+1}\cdot {10}^{-(n+1)}\end{array}$$ となる．$n=3$のとき$|R_4|\leqq 4.0\times {10}^{-4}$だから誤差を${10}^{-3}$以内とでき， $$\begin{array}{rl}\log 1.2& \approx \sum _{k=1}^3\frac{(-1{)}^{k-1}}{k}{0.2}^k=0.2-\frac{{0.2}^2}{2}+\frac{{0.2}^3}{3}\\ & \approx 0.182\end{array}$$

(b)

$$\begin{array}{c}\log 0.9=\log (1-0.1)=\sum _{k=1}^n\frac{(-1{)}^{k-1}}{k}(-0.1{)}^k+R_{n+1}\\ |R_{n+1}|\leqq \frac{{0.1}^{n+1}}{(n+1)(1-0.1)}=\frac{1}{0.9(n+1)}\cdot {10}^{-(n+1)}\end{array}$$ となる．$n=2$のとき$|R_3|\leqq 3.7\times {10}^{-4}$だから誤差を${10}^{-3}$以内とでき，  $$\begin{array}{rl}\log 0.9& \approx \sum _{k=1}^2\frac{(-1{)}^{k-1}}{k}(-0.1{)}^k=-0.1-\frac{(-0.1{)}^2}{2}\\ & \approx -0.105\end{array}$$

※ 巻末解答の$1/27\times {10}^{-4}$は$1/27\times {10}^{-4}$の誤りと思われる．

(c)

$$\begin{array}{c}\log 1.05=\log (1+0.05)=\sum _{k=1}^n\frac{(-1{)}^{k-1}}{k}{0.05}^k+R_{n+1}\\ |R_{n+1}|\leqq \frac{{0.05}^{n+1}}{n+1}=\frac{1}{2^{n+1}(n+1)}\cdot {10}^{-(n+1)}\end{array}$$ となる．$n=2$のとき$|R_3|\leqq 4.2\times {10}^{-5}$だから誤差を${10}^{-3}$以内とでき， $$\begin{array}{rl}\log 1.05& \approx \sum _{k=1}^3\frac{(-1{)}^{k-1}}{k}{0.05}^k=0.05-\frac{{0.05}^2}{2}\\ & \approx 0.048\end{array}$$

※ 巻末解答の$1/16\times {10}^{-4}$は$1/24\times {10}^{-4}$の誤りではないだろうか．

(d)

$\log {\displaystyle \frac{9}{10}}=\log 0.9$だから(b)と同じである．

(e)

$$\begin{array}{c}\begin{array}{rl}\log \frac{24}{25}& =\log 0.96=\log (1-0.04)\\ & =\sum _{k=1}^n\frac{(-1{)}^{k-1}}{k}(-0.04{)}^k+R_{n+1}\end{array}\\ |R_{n+1}|\leqq \frac{{0.04}^{n+1}}{(n+1)(1-0.04)}=\frac{2^{2n+2}}{0.96(n+1)}\cdot {10}^{-2n-2}\end{array}$$ となる．$n=1$のとき$|R_2|\leqq 8.3\times {10}^{-4}$だから誤差を${10}^{-3}$以内とでき， $$\begin{array}{rl}\log \frac{24}{25}& \approx \sum _{k=1}^1\frac{(-1{)}^{k-1}}{k}(-0.04{)}^k\\ & \approx -0.04\end{array}$$

(f)

$$\begin{array}{c}\begin{array}{rl}\log \frac{26}{25}& =\log 1.04=\log (1+0.04)\\ & =\sum _{k=1}^n\frac{(-1{)}^{k-1}}{k}{0.04}^k+R_{n+1}\end{array}\\ |R_{n+1}|\leqq \frac{{0.04}^{n+1}}{n+1}=\frac{2^{2n+2}}{n+1}\cdot {10}^{-2n-2}\end{array}$$ となる．$n=1$のとき$|R_2|\leqq 8.0\times {10}^{-4}$だから誤差を${10}^{-3}$以内とでき， $$\begin{array}{rl}\log \frac{26}{25}& \approx \sum _{k=1}^1\frac{(-1{)}^{k-1}}{k}{0.04}^k\\ & \approx 0.04\end{array}$$