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Category: Relation and Functions

let-f-x-ln-tanx-developp-f-at-fourier-serie-

Question Number 121490 by mathmax by abdo last updated on 08/Nov/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{tanx}\right) \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$ Answered by mnjuly1970 last updated on 08/Nov/20 $$\:\:\:\mathrm{1}.{f}\left({x}\right)={ln}\left(\mathrm{2}{sin}\left({x}\right)\right)\overset{{fourier}\:{series}} {=}\underset{{n}=\mathrm{1}}…

known-a-lt-pi-2-If-M-lt-1-with-cos-x-cos-y-M-x-y-for-every-x-y-0-a-then-M-

Question Number 55908 by gunawan last updated on 06/Mar/19 $$\mathrm{known}\:{a}\:<\:\frac{\pi}{\mathrm{2}}\:. \\ $$$$\mathrm{If}\:\:\mathrm{M}<\mathrm{1}\:\mathrm{with}\:\mid\mathrm{cos}\:{x}−\mathrm{cos}\:{y}\mid\leqslant\mathrm{M}\:\mid{x}−{y}\mid \\ $$$$\mathrm{for}\:\mathrm{every}\:{x},\:{y}\:\in\:\left[\mathrm{0},{a}\right],\:\mathrm{then}\:\mathrm{M}=.. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 07/Mar/19 $${let}\:{y}>{x} \\…

let-f-n-m-1-n-1-m-1-n-2-i-0-m-f-n-1-i-n-gt-2-a-compute-f-3-4-b-find-a-function-tha-satisfies-above-recursion-

Question Number 55823 by 121194 last updated on 04/Mar/19 $${let} \\ $$$${f}\left({n},{m}\right)=\begin{cases}{\mathrm{1}}&{{n}\:=\:\mathrm{1}}\\{{m}+\mathrm{1}}&{{n}\:=\:\mathrm{2}}\\{\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{f}\left({n}−\mathrm{1},{i}\right)}&{{n}\:>\:\mathrm{2}}\end{cases} \\ $$$${a}.\mathrm{compute}\:{f}\left(\mathrm{3},\mathrm{4}\right) \\ $$$${b}.\mathrm{find}\:\mathrm{a}\:\mathrm{function}\:\mathrm{tha}\:\mathrm{satisfies}\:\mathrm{above}\:\mathrm{recursion}. \\ $$ Terms of Service Privacy Policy…

known-function-f-5-4-R-continues-then-E-x-5-4-f-x-then-closure-from-E-is-

Question Number 55636 by gunawan last updated on 01/Mar/19 $$\mathrm{known}\:\mathrm{function}\:{f}:\left[−\mathrm{5},\:\mathrm{4}\right]\rightarrow\mathbb{R}\:\mathrm{continues}, \\ $$$$\mathrm{then}\:{E}=\left\{{x}\:\in\:\left[−\mathrm{5},\:\mathrm{4}\right]\::\:{f}\left({x}\right)\right\}, \\ $$$$\mathrm{then}\:\mathrm{closure}\:\mathrm{from}\:{E}\:\mathrm{is}… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

The-function-pogof-x-x-4-2x-3-2x-2-is-divisible-by-the-half-of-the-function-of-p-Find-g-x-

Question Number 55621 by Tawa1 last updated on 28/Feb/19 $$\mathrm{The}\:\mathrm{function}\:\:\:\mathrm{pogof}\left(\mathrm{x}\right)\:\:=\:\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{2x}^{\mathrm{3}} \:+\:\mathrm{2x}^{\mathrm{2}} \:\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{the}\:\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{function}\:\mathrm{of}\:\:\mathrm{p}.\:\:\mathrm{Find}\:\:\mathrm{g}\left(\mathrm{x}\right). \\ $$ Answered by kaivan.ahmadi last updated on 28/Feb/19 $$\mathrm{2}{p}\left({gof}\left({x}\right)\right)=\mathrm{2}{x}^{\mathrm{2}}…

Question-55408

Question Number 55408 by Tinkutara last updated on 23/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19 $${a}_{{n}+\mathrm{1}} ={a}_{{n}} \left({a}_{{n}} +\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{{a}_{{n}} \left({a}_{{n}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{{a}_{{n}}…

If-f-x-2f-1-x-3f-x-x-1-x-what-is-f-x-

Question Number 120819 by bramlexs22 last updated on 03/Nov/20 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:+\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:+\mathrm{3f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)\:=\:\mathrm{x}\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{x}\right)\:? \\ $$ Commented by liberty last updated on 03/Nov/20 $$\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{3f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)\:=\:\mathrm{x} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{replace}\:\mathrm{x}\:\mathrm{with}\:\frac{\mathrm{1}}{\mathrm{x}},\:\mathrm{give} \\…