Question Number 39837 by math khazana by abdo last updated on 12/Jul/18 $${let}\:{S}_{{n}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{e}^{−{n}\left[\frac{{x}}{{n}}\right]} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{S}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$ Terms of Service Privacy…
Question Number 39835 by math khazana by abdo last updated on 12/Jul/18 $${simplify}\:\left[\frac{\left[{nx}\right]}{{n}}\right]\:{with}\:{n}\:{natural}\:{integr}\:{not}\mathrm{0}\:\:{and}\:{x}\:{real} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 39703 by math khazana by abdo last updated on 10/Jul/18 $${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{{x}+\mathrm{2}−\left[{x}\right]}{dx} \\ $$$$\left.\mathrm{1}\right)\:\:{calculate}\:{A}_{{n}} \:\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} \: \\ $$$$\left.\mathrm{2}\right)\:{let}\:{S}_{{n}} \:=\sum_{{n}=\mathrm{0}}…
Question Number 39702 by math khazana by abdo last updated on 10/Jul/18 $${let}\:{f}\left({x}\right)=\mathrm{2}\sqrt{{x}−\sqrt{{x}−\mathrm{3}}\:+\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)…
Question Number 39699 by maxmathsup by imad last updated on 09/Jul/18 $${let}\:{f}\left({x}\right)=\:{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{\left({n}\right)} \left({n}\right) \\ $$$$\left.\mathrm{3}\right){developp}\:{f}\:\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}…
Question Number 39695 by maxmathsup by imad last updated on 09/Jul/18 $${let}\:{f}\left({x}\right)={ln}\left(\mathrm{2}{xarctan}\sqrt{\left.\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}\right)}\right. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{'} \left({x}\right)\:{and}\:\:{determine}\:{its}\:{sign}. \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{the}\:{equation}\:{of}\:{assymptote}\:{at}\:{pont}\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{a}\:{and}\:{b}\:{from}\:{R}\:/\:\:{f}\left({x}\right)\sim\:{a}\left({x}−\mathrm{1}\right)\:+{b}\:\:\left({x}\rightarrow\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 39671 by KMA last updated on 09/Jul/18 $${Three}\:{function}\:{are}\:{given}\:{f}\left({x}\right)=\sqrt{{x}} \\ $$$${g}\left({x}\right)={x}+\mathrm{5}\:{and}\:{h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{5}\:.{Express} \\ $$$${h}\left({x}\right)\:{interms}\:{of}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right). \\ $$ Answered by MrW3 last updated on 10/Jul/18 $${h}\left({x}\right)={x}^{\mathrm{2}}…
Question Number 39632 by abdo mathsup 649 cc last updated on 08/Jul/18 $${let}\:{S}_{{n}} \:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\sqrt{{k}}} \\ $$$${find}\:{a}\:{equivalent}\:\:{of}\:{S}_{{n}} \:\:\:\:\:\left({n}\rightarrow+\infty\right) \\ $$ Terms of Service Privacy…
Question Number 39631 by abdo mathsup 649 cc last updated on 08/Jul/18 $${let}\:{S}_{{n}} \:\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\:\sqrt{{k}}} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{S}_{{n}} \:{when}\:{n}\:\rightarrow+\infty \\ $$ Terms of Service Privacy…
Question Number 39519 by math khazana by abdo last updated on 07/Jul/18 $${simplify}\: \\ $$$$\left.\mathrm{1}\right)\:{A}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{{a}}}\left\{\:\left(\frac{\mathrm{1}+\sqrt{{a}}}{\mathrm{2}}\:\right)^{{n}} \:−\left(\frac{\mathrm{1}−\sqrt{{a}}}{\mathrm{2}}\right)^{{n}} \right\}\:{with}\:{n}\:{natural}\: \\ $$$${integr}\:{and}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}\left\{\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} \:−\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} \right\} \\…