Question Number 192481 by Tomal last updated on 19/May/23
Answered by aleks041103 last updated on 19/May/23
$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+{x}}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{4}} +…\right) \\ $$$${if}\:\:\mid\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid<\mathrm{1},\:{i}.{e}.\:\mid\mathrm{1}+{x}\mid>\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{1}=\mathrm{2}+\mathrm{2}{x} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}\Rightarrow{x}=\pm\sqrt{\mathrm{2}} \\ $$$$\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid=\mathrm{1}+\sqrt{\mathrm{2}}>\mathrm{1} \\ $$$$\mid\mathrm{1}−\sqrt{\mathrm{2}}\mid\approx\mathrm{0}.\mathrm{4142}<\mathrm{1} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}} \\ $$
Commented by Tomal last updated on 19/May/23
$${Thanks}\:\lambda\underline{\underbrace{\:}} \\ $$