6.3 Taylor and maclaurin series  (Page 7/13)

$f\left(\mathrm{-1}\right)=1;f^{\prime }\left(\mathrm{-1}\right)=\mathrm{-1};f\text{″}\left(\mathrm{-1}\right)=2;f\left(x\right)=1-\left(x+1\right)+{\left(x+1\right)}^2$

$f^{\prime }\left(x\right)=2\text{cos}\left(2x\right);f\text{″}\left(x\right)=\mathrm{-4}\text{sin}\left(2x\right);p_2\left(x\right)=\mathrm{-2}\left(x-\frac{\pi }{2}\right)$

$f^{\prime }\left(x\right)=\frac{1}{x};f\text{″}\left(x\right)=-\frac{1}{x^2};p_2\left(x\right)=0+\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^2$

$p_2\left(x\right)=e+e\left(x-1\right)+\frac{e}{2}{\left(x-1\right)}^2$

$\frac{d^2}{d{x}^2}{x}^{1\text{/}3}=-\frac{2}{9x^{5\text{/}3}}\ge \mathrm{-0.00092}\text{…}$ when $x\ge 28$ so the remainder estimate applies to the linear approximation $x^{1\text{/}3}\approx p_1\left(27\right)=3+\frac{x-27}{27},$ which gives ${\left(28\right)}^{1\text{/}3}\approx 3+\frac{1}{27}=3.\overline{037},$ while ${\left(28\right)}^{1\text{/}3}\approx 3.03658.$

Using the estimate $\frac{2^{10}}{10\text{!}}<0.000283$ we can use the Taylor expansion of order 9 to estimate e ^x at $x=2.$ as $e^2\approx p_9\left(2\right)=1+2+\frac{2^2}{2}+\frac{2^3}{6}+\text{⋯}+\frac{2^9}{9\text{!}}=7.3887\text{…}$ whereas $e^2\approx 7.3891.$

Since $\frac{d^n}{d{x}^n}\left(\text{ln}x\right)={\left(\mathrm{-1}\right)}^{n-1}\frac{\left(n-1\right)\text{!}}{x^n},R_{1000}\approx \frac{1}{1001}.$ One has $p_{1000}\left(1\right)={\displaystyle \sum _{n=1}^{1000}\frac{{\left(\mathrm{-1}\right)}^{n-1}}{n}}\approx 0.6936$ whereas $\text{ln}\left(2\right)\approx 0.6931\text{⋯}.$

${\displaystyle {\int }_0^1\left(1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\frac{x^8}{24}-\frac{x^{10}}{120}+\frac{x^{12}}{720}\right)dx}$

$=1-\frac{1^3}{3}+\frac{1^5}{10}-\frac{1^7}{42}+\frac{1^9}{9·24}-\frac{1^{11}}{120·11}+\frac{1^{13}}{720·13}\approx 0.74683$ whereas ${\displaystyle {\int }_0^1{e}^{\text{−}{x}^2}dx}\approx 0.74682.$

Since $f^{\left(n+1\right)}\left(z\right)$ is $\text{sin}z$ or $\text{cos}z,$ we have $M=1.$ Since $\left|x-0\right|\le \frac{\pi }{2},$ we seek the smallest n such that $\frac{{\pi }^{n+1}}{2^{n+1}\left(n+1\right)\text{!}}\le 0.001.$ The smallest such value is $n=7.$ The remainder estimate is $R_7\le 0.00092.$

Since $f^{\left(n+1\right)}\left(z\right)=\text{±}{e}^{\text{−}z}$ one has $M=e^3.$ Since $\left|x-0\right|\le 3,$ one seeks the smallest n such that $\frac{3^{n+1}{e}^3}{\left(n+1\right)\text{!}}\le 0.001.$ The smallest such value is $n=14.$ The remainder estimate is $R_{14}\le 0.000220.$