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Category: Relation and Functions

f-n-4f-n-1-4f-n-2-n-2-and-f-0-2-f-1-5-

Question Number 129403 by bemath last updated on 15/Jan/21 $$\:\mathrm{f}\left(\mathrm{n}\right)\:=\:\mathrm{4f}\left(\mathrm{n}−\mathrm{1}\right)\:−\mathrm{4f}\left(\mathrm{n}−\mathrm{2}\right)\:+\:\mathrm{n}^{\mathrm{2}} \: \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{2}\:,\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}\: \\ $$ Commented by talminator2856791 last updated on 15/Jan/21 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}? \\ $$…

find-the-sequence-u-n-wish-verify-u-n-p-u-n-p-1-p-n-p-for-all-n-p-

Question Number 129383 by mathmax by abdo last updated on 15/Jan/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \:\mathrm{wish}\:\mathrm{verify}\:\mathrm{u}_{\mathrm{n}−\mathrm{p}} +\mathrm{u}_{\mathrm{n}+\mathrm{p}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{n}+\mathrm{p}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{n}\geqslant\mathrm{p} \\ $$ Terms of Service Privacy Policy…

If-a-b-and-c-are-the-roots-of-x-3-21x-35-0-what-is-the-value-of-a-b-c-a-b-c-

Question Number 129336 by bramlexs22 last updated on 15/Jan/21 $$\mathrm{If}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{21}{x}−\mathrm{35}\:=\:\mathrm{0}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\left({a}−{b}\right)\left({c}−{a}\right)\left({b}−{c}\right)\:? \\ $$ Commented by MJS_new last updated on 22/Jan/21 $${x}^{\mathrm{3}}…

Given-f-x-2x-1-2x-1-and-f-1-x-2-1-f-1-x-2-1-1-x-2-then-x-

Question Number 129213 by bramlexs22 last updated on 13/Jan/21 $$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\:\mathrm{and}\:\frac{\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{1}}{\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{then}\:\mathrm{x}\:=? \\ $$ Answered by liberty last updated on 13/Jan/21 $$\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…

let-S-n-x-k-0-n-e-k-sin-k-2-x-1-determine-2-sequence-U-n-x-and-V-n-x-wich-verify-U-n-S-n-V-n-2-let-S-lim-n-S-x-study-the-convergence-of-S-

Question Number 63651 by mathmax by abdo last updated on 06/Jul/19 $${let}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{−{k}} {sin}\left({k}^{\mathrm{2}} {x}\right) \\ $$$$\left.\mathrm{1}\right)\:{determine}\:\mathrm{2}\:{sequence}\:\:{U}_{{n}} \left({x}\right)\:{and}\:{V}_{{n}} \left({x}\right)\:{wich}\:{verify}\:{U}_{{n}} \leqslant\:{S}_{{n}} \leqslant\:{V}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:\:{S}\:={lim}_{{n}\rightarrow+\infty}…

developp-at-laurent-series-1-f-z-1-z-2-2-g-z-3-z-2-3z-2-3-h-z-1-z-2-4-

Question Number 63560 by mathmax by abdo last updated on 05/Jul/19 $${developp}\:{at}\:{laurent}\:{series} \\ $$$$\left.\mathrm{1}\right)\:{f}\left({z}\right)\:=\frac{\mathrm{1}}{{z}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){g}\left({z}\right)\:=\frac{\mathrm{3}}{{z}^{\mathrm{2}} −\mathrm{3}{z}\:+\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){h}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{4}} \\ $$ Commented by mathmax…

Let-p-and-Q-be-points-on-the-curve-y-x-2-2x-while-x-2-and-x-2-h-respectively-Epress-the-gradient-of-PQ-in-terms-of-h-

Question Number 129052 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${Let}\:\boldsymbol{\mathrm{p}}\:{and}\:\boldsymbol{\mathrm{Q}}\:{be}\:{points}\:{on}\:{the}\:{curve} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}\:{while}\:\boldsymbol{{x}}=\mathrm{2}\:{and}\:\boldsymbol{{x}}=\mathrm{2}+\boldsymbol{{h}} \\ $$$$\boldsymbol{{respectively}}.\:\boldsymbol{{E}}{press}\:{the}\:{gradient} \\ $$$${of}\:\boldsymbol{\mathrm{P}}{Q}\:{in}\:{terms}\:{of}\:\boldsymbol{{h}}. \\ $$ Commented by benjo_mathlover last updated on…

solve-y-2y-3y-xe-x-sin-2x-with-y-0-0-and-y-0-1-

Question Number 128951 by mathmax by abdo last updated on 11/Jan/21 $$\mathrm{solve}\:\mathrm{y}^{,,} −\mathrm{2y}^{'} \:+\mathrm{3y}\:=\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right)\:\:\mathrm{with}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$ Answered by mnjuly1970 last updated on 11/Jan/21…