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Category: Relation and Functions

let-x-arctan-2x-1-x-2-1-calculate-n-x-2-calculate-n-0-anddevelpp-at-integr-serie-

Question Number 53961 by maxmathsup by imad last updated on 27/Jan/19 $${let}\:\varphi\left({x}\right)\:=\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left({x}\right)\: \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left(\mathrm{0}\right)\:{anddevelpp}\:\varphi\:{at}\:{integr}\:{serie} \\ $$ Commented by maxmathsup by…

let-f-x-arctan-1-2x-1-calculate-f-n-x-then-f-n-0-2-developp-f-at-integr-serie-we-have-f-x-2-1-1-2x-2-f-n-x-2-1-2x-1-2-1-n-1-with-n-gt-0-let-W-x-

Question Number 53957 by maxmathsup by imad last updated on 05/Feb/19 $${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=\frac{\mathrm{2}}{\mathrm{1}+\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\:\Rightarrow\:{f}^{\left({n}\right)} \left({x}\right)\:=\mathrm{2}\:\left\{\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right\}^{\left({n}−\mathrm{1}\right)} \:\:{with}\:{n}>\mathrm{0}…

Given-that-f-x-3-x-2-12x-41-find-an-explicit-expression-for-f-x-please-I-need-the-procedure-

Question Number 119391 by Don08q last updated on 24/Oct/20 $$\:\mathrm{Given}\:\mathrm{that}\:{f}\left({x}−\mathrm{3}\right)\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{12}{x}\:+\:\mathrm{41} \\ $$$$\:\mathrm{find}\:\mathrm{an}\:\mathrm{explicit}\:\mathrm{expression}\:\mathrm{for}\:{f}\left({x}\right) \\ $$$$ \\ $$$$\:{please}\:{I}\:\:{need}\:{the}\:{procedure} \\ $$ Answered by bemath last updated on…

u-n-is-a-convergent-serie-u-n-gt-0-find-nature-of-the-serie-1-u-n-n-2-u-n-1-u-n-

Question Number 53779 by maxmathsup by imad last updated on 25/Jan/19 $$\Sigma\:{u}_{{n}} {is}\:{a}\:{convergent}\:{serie}\:\left({u}_{{n}} >\mathrm{0}\right)\:\:{find}\:{nature}\:{of}\:{the}\:{serie} \\ $$$$\left.\mathrm{1}\right)\:\Sigma\:\frac{\sqrt{{u}_{{n}} }}{{n}} \\ $$$$\left.\mathrm{2}\right)\Sigma\:\:\frac{{u}_{{n}} }{\mathrm{1}+{u}_{{n}} } \\ $$ Terms of…

let-U-n-1-nH-n-with-H-n-k-1-n-1-k-study-the-convergence-of-n-1-U-n-2-study-the-convergence-of-n-1-U-n-2-

Question Number 53778 by maxmathsup by imad last updated on 25/Jan/19 $${let}\:{U}_{{n}} =\frac{\mathrm{1}}{{nH}_{{n}} }\:\:\:\:{with}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$${study}\:{the}\:{convergence}\:{of}\:\sum_{{n}\geqslant\mathrm{1}} \:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\sum_{{n}\geqslant\mathrm{1}} {U}_{{n}} ^{\mathrm{2}} \\…