Question Number 17600 by mondodotto@gmail.com last updated on 08/Jul/17 Answered by alex041103 last updated on 08/Jul/17 $$\mathrm{We}\:\mathrm{use}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{sub}: \\ $$$${x}={tan}\left(\theta\right)\Rightarrow{dx}={sec}^{\mathrm{2}} \left(\theta\right){d}\theta \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{1}+{tan}^{\mathrm{2}} \theta={sec}^{\mathrm{2}} \theta \\…
Question Number 148645 by metamorfose last updated on 29/Jul/21 $${u}_{{n}+\mathrm{3}} =\frac{{u}_{{n}+\mathrm{2}} +{u}_{{n}+\mathrm{1}} +{u}_{{n}} }{\mathrm{3}}\:,\:\forall{n}\in{IN} \\ $$$${find}\:{u}_{{n}} \: \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 148636 by tabata last updated on 29/Jul/21 $${find}\:{tylor}\:{series}\:{of}\:{f}\left({z}\right)={logz}\: \\ $$$${about}\:{z}_{{o}} =−\mathrm{1}+{i} \\ $$ Commented by tabata last updated on 29/Jul/21 $$???? \\ $$…
Question Number 148638 by tabata last updated on 29/Jul/21 $${find}\:{laurent}\:{series}\:{f}\left({z}\right)=\frac{{cos}\left({iz}\right)}{{z}^{{n}} }\:\:,\mid{z}−{i}\mid>\mathrm{2} \\ $$ Commented by tabata last updated on 29/Jul/21 $$???? \\ $$ Terms of…
Question Number 148639 by tabata last updated on 29/Jul/21 $${find}\:{singular}\:{point}\:{of}\:{this}\:{following}\:{and} \\ $$$${whats}\:{the}\:{type}\:{of}\:{singular}\:{point}\:? \\ $$$$ \\ $$$$\left(\mathrm{1}\right){f}\left({z}\right)=\frac{\mathrm{1}}{{lnz}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right){f}\left({z}\right)=\frac{\mathrm{1}−{cos}\left({z}+{i}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\…
Question Number 148635 by learner001 last updated on 29/Jul/21 $$\mathrm{show}\:\mathrm{that}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}:=\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}…+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}!}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cauchy} \\ $$$$\mathrm{sequence}. \\ $$$$\mathrm{my}\:\mathrm{attempt}: \\ $$$$\mathrm{let}\:\epsilon>\mathrm{0}\:\mathrm{we}\:\mathrm{have}\:\mid\mathrm{a}_{\mathrm{m}} −\mathrm{a}_{\mathrm{n}} \mid=\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{1}\right)!}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{3}} }{\left(\mathrm{n}+\mathrm{2}\right)!}+…+\frac{\left(−\mathrm{1}\right)^{\mathrm{m}+\mathrm{1}} }{\mathrm{m}!}\mid \\ $$$$\leqslant\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}}…
Question Number 148631 by tabata last updated on 29/Jul/21 $${find}\:{the}\:{region}\:{converge}\:{of}\:{the}\:{series}\:{and}\: \\ $$$${find}\:{the}\:{sum}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{n}} \left({z}−\mathrm{1}\right)^{{n}} }\:? \\ $$ Commented by tabata last updated on 29/Jul/21…
Question Number 148609 by learner001 last updated on 29/Jul/21 $$\mathrm{prove}\:\mathrm{that}\:\left(\mathrm{a}_{\mathrm{n}} \right)_{\mathrm{n}\geqslant\mathrm{1}\:} \mathrm{defined}\:\mathrm{by}\:\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:\mathrm{is}\: \\ $$$$\mathrm{cauchy}\:\mathrm{sequence}. \\ $$ Commented by learner001 last updated on 29/Jul/21 $$\mathrm{This}\:\mathrm{is}\:\mathrm{what}\:\mathrm{i}\:\mathrm{tried}.…
Question Number 83037 by otchereabdullai@gmail.com last updated on 27/Feb/20 $$\mathrm{If}\:\mathrm{m}=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{sin}\theta}\:,\:\:\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{m}}=\:\frac{\mathrm{1}+\mathrm{sin}\theta}{\mathrm{sin}\theta} \\ $$ Commented by Tony Lin last updated on 27/Feb/20 $$\frac{\mathrm{1}}{{m}}=\frac{{sin}\theta}{\mathrm{1}−{cos}\theta}=\:\frac{{sin}\theta\left(\mathrm{1}+{cos}\theta\right)}{\left(\mathrm{1}−{cos}\theta\right)\left(\mathrm{1}+{cos}\theta\right)} \\ $$$$=\frac{\mathrm{1}+{cos}\theta}{{sin}\theta} \\ $$…
Question Number 83036 by otchereabdullai@gmail.com last updated on 27/Feb/20 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{cos}^{\mathrm{4}} \theta−\mathrm{sin}^{\mathrm{4}} \theta=\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta \\ $$ Commented by Tony Lin last updated on 27/Feb/20 $${cos}^{\mathrm{4}}…