Question Number 71776 by TawaTawa last updated on 19/Oct/19 Answered by tw000001 last updated on 22/Oct/19 $$\mathrm{Use}\:\mathrm{Harmonic}\:\mathrm{series}\:\mathrm{to}\:\mathrm{solve}. \\ $$$${A}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2015}}−\frac{\mathrm{1}}{\mathrm{2016}} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{1008}}\right) \\ $$$$={H}_{\mathrm{2016}}…
Question Number 137311 by bramlexs22 last updated on 01/Apr/21 Answered by liberty last updated on 01/Apr/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 71761 by TawaTawa last updated on 19/Oct/19 $$\mathrm{Find}\:\mathrm{at}\:\mathrm{least}\:\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{non}\:\mathrm{zero}\:\mathrm{term}\:\mathrm{in}\:\mathrm{a}\:\mathrm{power} \\ $$$$\mathrm{series}\:\mathrm{expansion}\:\mathrm{about}\:\:\mathrm{x}\:\:=\:\:\mathrm{0}\:\:\mathrm{for}\:\mathrm{a}\:\mathrm{general}\:\mathrm{solution} \\ $$$$\mathrm{to}\:\:\:\:\mathrm{z}''\:\:−\:\:\mathrm{x}^{\mathrm{2}} \mathrm{z}\:\:\:=\:\:\mathrm{0} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 71756 by aliesam last updated on 19/Oct/19 $${A}=\sqrt[{\mathrm{3}}]{\mathrm{8}+\mathrm{3}\sqrt{\mathrm{21}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{8}−\mathrm{3}\sqrt{\mathrm{21}}} \\ $$$$ \\ $$$${find}\:{A} \\ $$ Answered by MJS last updated on 19/Oct/19 $${A}={p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}}…
Question Number 137290 by JulioCesar last updated on 31/Mar/21 Answered by EDWIN88 last updated on 01/Apr/21 $$\mathrm{Ostrogradski}\:\mathrm{Integral}\: \\ $$ Answered by MJS_new last updated on…
Question Number 6216 by sanusihammed last updated on 18/Jun/16 $${If}\:\:\:\:\:{x}^{\mathrm{2}} \:\:=\:\:\mathrm{2}^{{x}} \:\:\:\:\:\:{find}\:\:{x}\: \\ $$$$ \\ $$$${please}\:{i}\:{need}\:{workings} \\ $$ Answered by prakash jain last updated on…
Question Number 71750 by TawaTawa last updated on 19/Oct/19 $$\int\:\mathrm{cos}^{\mathrm{3}} \theta\:\left(\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{3}} \theta\right) \\ $$$$\mathrm{Using}\:\mathrm{beta}\:\mathrm{function} \\ $$ Commented by mathmax by abdo last updated on 20/Oct/19…
Question Number 71729 by peter frank last updated on 19/Oct/19 $${find}\:{dU}\:\:\:{if}\:\:\:{U}={x}^{\mathrm{2}} {e}^{\frac{{x}}{{y}}} \\ $$$$ \\ $$ Commented by peter frank last updated on 19/Oct/19 $${thank}\:{you}…
Question Number 137269 by mohssinee last updated on 31/Mar/21 Commented by mohssinee last updated on 31/Mar/21 $${n}\in\mathbb{N}^{\ast} \\ $$ Answered by aleks041103 last updated on…
Question Number 71696 by TawaTawa last updated on 18/Oct/19 $$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$ Answered by MJS last updated…