Question Number 67974 by mathmax by abdo last updated on 02/Sep/19 $${let}\:{F}\left({x}\right)\:=\int_{{x}} ^{{x}^{\mathrm{2}} } \:\:\:\frac{{arctan}\left({xt}\right)}{{x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt}\:\:{calculate}\:{F}\:^{'} \left({x}\right). \\ $$ Commented by mathmax by abdo…
Question Number 67972 by mathmax by abdo last updated on 02/Sep/19 $${if}\:{F}\left({x}\right)=\int_{{u}\left({x}\right)} ^{{v}\left({x}\right)} {g}\left({x},{t}\right){dt}\:\:\:\:\:{determine}\:{a}\:{expression}\:{for}\:{F}\:^{'} \left({x}\right). \\ $$ Answered by Tanmay chaudhury last updated on 03/Sep/19…
Question Number 133431 by Algoritm last updated on 22/Feb/21 Answered by benjo_mathlover last updated on 22/Feb/21 $$\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{5xy}−\mathrm{y}=\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{1} \\ $$$$\left(\mathrm{y}−\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} −\left(\mathrm{5y}−\mathrm{3}\right)\mathrm{x}+\mathrm{1}−\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{5y}−\mathrm{3}\pm\sqrt{\left(\mathrm{5y}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{y}\right)}}{\mathrm{2}\left(\mathrm{y}−\mathrm{1}\right)}…
Question Number 2344 by 123456 last updated on 17/Nov/15 $${f}\left({z}\right){e}^{\mathrm{1}−{z}} ={f}\left(\mathrm{1}−{z}\right)\pi^{{z}} \mathrm{sin}\:\left(\pi{z}\right) \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\mathrm{1}/\mathrm{2} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=?? \\ $$ Commented by Yozzi last updated on…
let-A-p-0-pi-x-p-cos-nx-dx-1-calculate-A-0-A-1-A-2-2-determine-a-relation-of-recurrence-between-A-p-
Question Number 67795 by mathmax by abdo last updated on 31/Aug/19 $${let}\:\:{A}_{{p}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{{p}} \:{cos}\left({nx}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{\mathrm{0}} ,{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){determine}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:\:{A}_{{p}} \\ $$ Commented…
Question Number 2253 by 123456 last updated on 11/Nov/15 $${f}_{{n}} :\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{1}\right],{g}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}_{{n}+\mathrm{1}} \left({x}\right)={g}\left[{f}_{{n}} \left({x}\right)\right]+{f}_{{n}} \left[{g}\left({x}\right)\right] \\ $$$${f}_{\mathrm{0}} \left({x}\right)={x} \\ $$$${f}_{\mathrm{4}} \left({x}\right)=? \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} ,{f}_{\mathrm{2}}…
Question Number 2205 by prakash jain last updated on 08/Nov/15 $${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\mathrm{1}+{x} \\ $$$${f}\left({x}\right)=? \\ $$ Commented by Yozzi last updated on 08/Nov/15 $${f}\left({x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}+{x} \\ $$$$\Rightarrow{f}^{'}…
Question Number 2187 by 123456 last updated on 07/Nov/15 $${f}:\mathbb{N}\rightarrow\mathbb{N} \\ $$$${f}\left({n}+{m}\right)={f}\left({n}\right)+\alpha{m} \\ $$$${f}\left(\mathrm{0}\right)={k} \\ $$$${f}\left({n}\right)=? \\ $$ Answered by prakash jain last updated on…
Question Number 67672 by Abdo msup. last updated on 30/Aug/19 $${decompose}\:{the}\:{folowing}\:\:{fraction}\:{at}\:{R}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right)\:{G}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by…
Question Number 133168 by bemath last updated on 19/Feb/21 Answered by floor(10²Eta[1]) last updated on 19/Feb/21 $$\mathrm{x}\rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{I}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{but}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right),\:\mathrm{so} \\ $$$$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{II}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{−\mathrm{x}^{\mathrm{2}}…