Menu Close

Category: Relation and Functions

z-1-2-z-1-z-z-and-z-are-complex-numbers-show-that-z-2e-i-show-that-M-describes-a-conic-section-

Question Number 208533 by alcohol last updated on 18/Jun/24 $${z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}+\frac{\mathrm{1}}{{z}}\right) \\ $$$${z}\:{and}\:{z}'\:{are}\:{complex}\:{numbers} \\ $$$${show}\:{that}\:{z}\:=\:\mathrm{2}{e}^{{i}\theta} \\ $$$${show}\:{that}\:{M}'\:{describes}\:{a}\:{conic}\:{section} \\ $$ Answered by Berbere last updated on 18/Jun/24…

u-n-1-u-n-u-n-3-u-0-0-1-show-that-u-n-0-1-show-that-u-n-converges-to-0-v-n-1-u-n-1-2-1-u-n-2-express-v-n-interms-of-u-n-show-that-v-n-converges-to-2-f-

Question Number 208418 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} \:;\:{u}_{\mathrm{0}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\left..\:{show}\:{that}\:{u}_{{n}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$.\:{show}\:{that}\:{u}_{{n}} \:{converges}\:{to}\:\mathrm{0} \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}}…

Find-inf-m-n-m-n-N-m-lt-n-2-

Question Number 208103 by depressiveshrek last updated on 05/Jun/24 $$\mathrm{Find}\:\mathrm{inf}\left\{\frac{{m}}{{n}}\:\mid\:{m},\:{n}\:\in\:\mathbb{N},\:{m}<{n}−\mathrm{2}\right\} \\ $$ Answered by A5T last updated on 05/Jun/24 $${Since}\:{m},{n}\in\mathbb{N},\:\frac{{m}}{{n}}>\mathrm{0},{so}\:\mathrm{0}\:{is}\:{a}\:{lower}\:{bound}. \\ $$$$\frac{{m}}{{n}}\geqslant\frac{\mathrm{1}}{{n}}\:{and}\:\frac{\mathrm{1}}{{n}}\:{is}\:{in}\:{the}\:{set}. \\ $$$${For}\:{any}\:\epsilon>\mathrm{0},\:{one}\:{can}\:{always}\:{find}\:{n}\:{such}\:{that} \\…

let-f-R-R-be-a-continuous-function-then-show-that-1-if-f-x-f-x-2-x-R-then-f-is-a-constant-function-2-if-f-x-f-2x-1-x-R-then-f-is-a-constant-function-

Question Number 207314 by universe last updated on 11/May/24 $$\:\mathrm{let}\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{then} \\ $$$$\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:\forall\:\mathrm{x}\:\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\:\mathrm{function} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{constant}\:\mathrm{function}\: \\ $$ Answered by…

Let-f-be-a-function-with-the-following-properties-i-f-1-1-ii-f-2n-n-f-n-for-any-positive-integer-n-Find-the-value-of-f-2-10-a-1-b-2-10-c-2-35-d-2-45-

Question Number 207085 by necx122 last updated on 06/May/24 $${Let}\:{f}\:{be}\:{a}\:{function}\:{with}\:{the}\:{following} \\ $$$${properties}:\:\left({i}\right)\:{f}\left(\mathrm{1}\right)\:=\mathrm{1}\:\left({ii}\right)\:{f}\left(\mathrm{2}{n}\right)={n}.{f}\left({n}\right)\:{for} \\ $$$${any}\:{positive}\:{integer}\:{n}.\:{Find}\:{the}\:{value} \\ $$$${of}\:{f}\left(\mathrm{2}^{\mathrm{10}} \right) \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:{b}\right)\:\mathrm{2}^{\mathrm{10}\:} {c}\right)\:\mathrm{2}^{\mathrm{35}} \:{d}\right)\:\mathrm{2}^{\mathrm{45}} \\ $$ Answered by…

If-the-nth-term-of-a-sequence-is-given-by-n-2-2n-4-what-is-the-sum-of-n-terms-of-the-sequence-

Question Number 206993 by necx122 last updated on 03/May/24 $${If}\:{the}\:{nth}\:{term}\:{of}\:{a}\:{sequence}\:{is} \\ $$$${given}\:{by}\:\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}}{\mathrm{4}}\:,{what}\:{is}\:{the}\:{sum}\:{of}\:{n} \\ $$$${terms}\:{of}\:{the}\:{sequence}? \\ $$ Answered by Rasheed.Sindhi last updated on 03/May/24 $$\underset{{k}=\mathrm{1}}…

Question-206442

Question Number 206442 by universe last updated on 14/Apr/24 Answered by Berbere last updated on 14/Apr/24 $${let}\:{f}\left({x}\right)={e}^{{g}\left({x}\right)} ;\:{particular}\:{Solution}\:{just}\:{to}\:{simplifie} \\ $$$${the}\:{problems};{f}\left(\mathrm{0}\right)=\mathrm{1}={e}^{{g}\left(\mathrm{0}\right)} \Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}'\left({x}\right)={g}'{e}^{{g}\left({x}\right)} ;{f}''\left({x}\right)=\left({g}'^{\mathrm{2}} +{g}''\right){e}^{{g}\left({x}\right)}…

Question-206151

Question Number 206151 by cortano21 last updated on 08/Apr/24 Answered by dimentri last updated on 08/Apr/24 $$\:\:\mathrm{x}=\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\mathrm{4}\: \\ $$$$\:\:\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\:\:\mathrm{g}\left(\mathrm{4}\right)=\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{3}\right)\right)\:=\:\mathrm{g}\left(\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right)\right)\:=\:\mathrm{1} \\ $$…