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Question Number 40173 by scientist last updated on 16/Jul/18

Commented by math khazana by abdo last updated on 17/Jul/18

$$letf\left(x\right)={\left(\frac{1+x}{1-x}\right)}^{\frac{1}{2}}\Rightarrow ln\left(f\left(x\right)\right)=\frac{1}{2}\{ln(1+x)-ln(1-x)\}$$$$but{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=1-x+x^2+o\left(x^3\right)\Rightarrow$$$$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o\left(x^4\right)alsowehave$$$$ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}+o\left(x^4\right)\Rightarrow$$$$\frac{1}{2}(ln(1+x)-ln(1-x))=\frac{1}{2}(2x+\frac{2x^3}{3}+o\left(x^4\right))$$$$=x+o\left(x^2\right)ifweneglige{x}^3\Rightarrow$$$$f\left(x\right)=e^{x+o\left(x^2\right)}=1+x+\frac{x^2}{2}ifweneglige{x}^3$$$$forx=\frac{1}{7}weget$$$${\left(\frac{1+\frac{1}{7}}{1-\frac{1}{7}}\right)}^{\frac{1}{2}}=1+\frac{1}{7}+\frac{1}{2.7^2}\Rightarrow$$$$\sqrt{\frac{8}{6}}=\frac{8}{7}+\frac{1}{98}\Rightarrow \sqrt{\frac{4}{3}}=\frac{8.98+7}{7.98}....$$

Answered by MJS last updated on 16/Jul/18

$${\left(\frac{1+x}{1-x}\right)}^{\frac{1}{2}}=1+x+\frac{1}{2}{x}^2$$$$1+x=(1-x){(1+x+\frac{1}{2}{x}^2)}^2$$$$1+x=1+x-(x^3+\frac{3}{4}{x}^4+\frac{1}{4}{x}^5)$$$$\mathrm{if}\mathrm{we}\mathrm{ignore}{x}^n\mathrm{with}n⩾3\mathrm{this}\mathrm{is}\mathrm{true}$$$$x=\frac{1}{7}$$$${\left(\frac{1+\frac{1}{7}}{1-\frac{1}{7}}\right)}^{\frac{1}{2}}=\frac{2\sqrt{3}}{3}\approx 1.15470$$$$1+\frac{1}{7}+\frac{1}{2}\times \frac{1}{7^2}=\frac{113}{98}\approx 1.15306$$$$\mathrm{the}\mathrm{mistake}\mathrm{is}\frac{1}{610}\mathrm{or}0.14\mathrm{\%}$$$$\left(\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{where}\frac{166}{113}\mathrm{comes}\mathrm{from}\right)$$

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