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Category: Relation and Functions

If-f-2-x-3-2-4x-and-f-1-p-2-then-p-

Question Number 124444 by benjo_mathlover last updated on 03/Dec/20 $$\:{If}\:{f}\left(\frac{\mathrm{2}}{{x}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}+\mathrm{4}{x}}\:{and}\:{f}^{−\mathrm{1}} \left({p}\right)=\mathrm{2}\: \\ $$$${then}\:{p}\:=? \\ $$ Answered by bemath last updated on 03/Dec/20 $$\:\Leftrightarrow\:{f}^{−\mathrm{1}} \left({p}\right)=\mathrm{2}\:{then}\:{f}\left(\mathrm{2}\right)={p} \\…

advanced-calculus-p-q-are-positive-integers-and-p-q-let-p-q-0-1-x-p-1-x-q-dx-prove-that-n-n-1-1-n-1-k-1-n-k-1-

Question Number 124102 by mnjuly1970 last updated on 30/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:\:{calculus}… \\ $$$$\:\:{p}\:,\:{q}\:{are}\:{positive}\:{integers}\:{and} \\ $$$${p}\geqslant{q}\:\::\:{let}\::\phi\left({p},{q}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{p}} \left\{\frac{\mathrm{1}}{{x}}\right\}^{{q}} {dx} \\ $$$$\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\phi\left({n},{n}\right)\overset{?} {=}\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\zeta\left({k}+\mathrm{1}\right)…

let-f-x-arctan-2-x-calculate-f-n-x-and-f-n-1-find-f-7-1-7-

Question Number 124066 by Bird last updated on 30/Nov/20 $${let}\:{f}\left({x}\right)={arctan}\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$${calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$${find}\:{f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$ Answered by Olaf last updated on…

let-U-n-1-2-2-2-3-2-n-2-1-4-2-4-3-4-n-4-1-find-lim-n-U-n-2-calculate-n-1-U-n-

Question Number 58354 by maxmathsup by imad last updated on 21/Apr/19 $${let}\:{U}_{{n}} =\frac{\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+….+{n}^{\mathrm{2}} }{\mathrm{1}^{\mathrm{4}} \:+\mathrm{2}^{\mathrm{4}} \:+\mathrm{3}^{\mathrm{4}} \:+….+{n}^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\…