Question Number 64574 by ajfour last updated on 19/Jul/19
Commented by ajfour last updated on 19/Jul/19
$${Find}\:{polynomial}\:{function}\:{p}\left({x}\right) \\ $$$${if}\:{coloured}\:{area}\:{is}\:{a}\:{maximum}. \\ $$
Commented by mr W last updated on 19/Jul/19
$${nice}\:{question}\:{sir}! \\ $$
Commented by behi83417@gmail.com last updated on 19/Jul/19
$$. \\ $$
Commented by ajfour last updated on 19/Jul/19
$${very}\:{brief}\:{solution}\:{behi}\:{Sir}. \\ $$
Commented by behi83417@gmail.com last updated on 19/Jul/19
$$\mathrm{sir}\:\mathrm{Ajfour}!\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{for}\:\mathrm{shearing} \\ $$$$\mathrm{this}\:\:\mathrm{nice}\:\mathrm{question}.\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{for} \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{now},\mathrm{but}\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{follow} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{and}\:\mathrm{to}\:\mathrm{recive}\:\mathrm{notifications} \\ $$$$\mathrm{of}\:\mathrm{Q},\mathrm{I}\:\mathrm{put}\:\mathrm{a}\:'.'.\mathrm{please}\:\mathrm{excuse}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 19/Jul/19
$$\mathcal{H}{appy}\:{is}\:{he}\:{who}\:{knows}\:{the} \\ $$$${reason}\:{of}\:{things},\:{thanks}\:{for} \\ $$$${the}\:{reason}.\:{It}\:{dint}\:{occur}\:{me}. \\ $$
Commented by MJS last updated on 19/Jul/19
$$\mathrm{if}\:\mathrm{any}\:\mathrm{function}\:\mathrm{was}\:\mathrm{allowed}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{an} \\ $$$$\mathrm{ellipse}\:\mathrm{I}\:\mathrm{guess}\:\left(\mathrm{please}\:\mathrm{check}\:\mathrm{this}\:\mathrm{idea}\right) \\ $$$$\mathrm{a}\:\mathrm{circle}\:\mathrm{only}\:\mathrm{if}\:{s}=\frac{\pi}{\mathrm{2}}{a} \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{then}\:\mathrm{would}\:\mathrm{be},\:\mathrm{the}\:\mathrm{perimeter} \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{found}\:\mathrm{exactly} \\ $$$$\mathrm{but}\:\mathrm{maybe}\:\mathrm{we}\:\mathrm{could}\:\mathrm{find}\:\mathrm{a}\:\mathrm{taylor}\:\mathrm{polynome} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{half}\:\mathrm{circle} \\ $$$${y}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{taylor}\:\mathrm{polynomes}\:\mathrm{are} \\ $$$$\mathrm{1} \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}} \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}−\frac{{x}^{\mathrm{6}} }{\mathrm{16}} \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}−\frac{{x}^{\mathrm{6}} }{\mathrm{16}}−\frac{\mathrm{5}{x}^{\mathrm{8}} }{\mathrm{128}} \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}−\frac{{x}^{\mathrm{6}} }{\mathrm{16}}−\frac{\mathrm{5}{x}^{\mathrm{8}} }{\mathrm{128}}−\frac{\mathrm{7}{x}^{\mathrm{10}} }{\mathrm{256}} \\ $$$$… \\ $$