Question Number 59733 by otchereabdullai@gmail.com last updated on 14/May/19
$$\mathrm{If}\:\mathrm{y}=\sqrt{\mathrm{x}}\:\:,\mathrm{find}\:\mathrm{approximate}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\sqrt{\mathrm{101}} \\ $$
Commented by prakash jain last updated on 14/May/19
$$\left(\mathrm{101}\right)^{\mathrm{1}/\mathrm{2}} =\mathrm{100}^{\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$=\mathrm{10}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{200}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{100}^{\mathrm{2}} }+…\right) \\ $$$$\mathrm{ignore}\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }\:\mathrm{and}\:\mathrm{smaller}\:\mathrm{terms} \\ $$$$\approx\mathrm{10}.\mathrm{05} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/May/19
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 14/May/19
$${y}=\sqrt{{x}}\: \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\frac{{dy}}{{dx}}\approx\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}×\bigtriangleup{x} \\ $$$${now}\:{x}=\mathrm{100}\:\:\:\:\:{y}=\mathrm{10} \\ $$$${x}+\bigtriangleup{x}=\mathrm{101}\:\:\:\:{y}+\bigtriangleup{y}=? \\ $$$$\bigtriangleup{x}=\mathrm{1} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}×\bigtriangleup{x} \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}×\bigtriangleup{x} \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}×\sqrt{\mathrm{100}}}=\mathrm{0}.\mathrm{05} \\ $$$${so}\:{y}+\bigtriangleup{y}=\mathrm{10}+\mathrm{0}.\mathrm{05}=\mathrm{10}.\mathrm{05} \\ $$$${so}\:\sqrt{\mathrm{101}}\: \\ $$$$=\mathrm{10}.\mathrm{05} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/May/19
$$\mathrm{fantastic}\:\mathrm{solution}\:\mathrm{prof} \\ $$
Answered by MJS last updated on 14/May/19
$$\mathrm{Heron}'\mathrm{s}\:\mathrm{Algorithm}\:\mathrm{for}\:\sqrt[{{n}}]{{a}}: \\ $$$${x}_{{i}+\mathrm{1}} =\frac{\left({n}−\mathrm{1}\right){x}_{{i}} ^{{n}} +{a}}{{nx}_{{i}} ^{{n}−\mathrm{1}} } \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{any}\:{x}_{\mathrm{1}} >\mathrm{0} \\ $$$$\mathrm{for}\:{n}=\mathrm{2}: \\ $$$${x}_{{i}+\mathrm{1}} =\frac{{x}_{{i}} ^{\mathrm{2}} +{a}}{\mathrm{2}{x}_{{i}} } \\ $$$${a}=\mathrm{101};\:{x}_{\mathrm{1}} =\mathrm{10} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{101}}{\mathrm{2}×\mathrm{10}}=\frac{\mathrm{201}}{\mathrm{20}}=\mathrm{10}.\mathrm{050}\:\mathrm{000} \\ $$$${x}_{\mathrm{3}} =\frac{\left(\frac{\mathrm{201}}{\mathrm{20}}\right)^{\mathrm{2}} +\mathrm{101}}{\mathrm{2}×\frac{\mathrm{201}}{\mathrm{20}}}=\frac{\mathrm{80}\:\mathrm{801}}{\mathrm{8}\:\mathrm{040}}\approx\mathrm{10}.\mathrm{049}\:\mathrm{876} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/May/19
$$\mathrm{fantastic}\:\mathrm{solution}\:\mathrm{prof} \\ $$