Tutorial 01

1. Given $x_1$ = 1.333, $x_1^{\ast }$ = 1.334, $x_2$ = 0.001, $x_2^{\ast }$ = 0.002, which approximated value is better? State your reason.

$$\begin{array}{r}{e}_{abs}=|x-x^{\ast }|=|1.333-1.334|=0.001\\ e_1=\frac{|x-x^{\ast }|}{|x|}=\frac{0.001}{|1.333}=0.00075019\\ e_{abs}=|x-x^{\ast }|=|0.001-0.002|=0.001\\ e_1=\frac{|x-x^{\ast }|}{|x|}=\frac{0.001}{|1.333}=0.00075019\end{array}$$

$\therefore$ $x_1$ better because

2a)

$$lnx-1=lnx-lne=ln(\frac{x}{e})$$

b)

$$\begin{array}{r}logx-log(\frac{1}{x})=log(\frac{x}{\frac{1}{x}})\\ =log(x.\frac{x}{1})\\ =\frac{log{x}^2}{2logx}\end{array}$$

c)

$$\begin{array}{r}f(x)=\frac{sinx}{x-\sqrt{x^2-1}}\times \frac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\\ =\frac{sinx(x+\sqrt{x^2-1})}{x-\sqrt{x^2-1}}\\ =\frac{sinx(x+\sqrt{x^2-1})}{x^2-x^2+1}\\ sinx(x+\sqrt{x^2-1})\end{array}$$

d)

$$\begin{array}{r}\frac{x^2}{\sqrt{1+x^2}-1}\times \frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1}\\ =\frac{x^2(\sqrt{1+x^2}+1)}{(\sqrt{1+x^2}+1)-(1{)}^2}\\ =\frac{}{}\\ =\sqrt{1+x^2+1}\end{array}$$