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Category: Relation and Functions

given-f-R-R-f-x-a-n-x-n-a-0-with-a-n-0-a-i-R-i-0-1-n-n-N-supose-that-x-1-x-n-are-roots-the-n-roots-of-f-x-then-proof-or-give-a-counter-example-that-the-root-of-f-n-1-x-

Question Number 1120 by 123456 last updated on 17/Jun/15 $$\mathrm{given}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={a}_{{n}} {x}^{{n}} +\centerdot\centerdot\centerdot+{a}_{\mathrm{0}} \: \\ $$$$\mathrm{with}\:{a}_{{n}} \neq\mathrm{0},\:{a}_{{i}} \in\mathbb{R},{i}\in\left\{\mathrm{0},\mathrm{1},…,{n}\right\},{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{supose}\:\mathrm{that}\:{x}_{\mathrm{1}} ,…,{x}_{{n}} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\mathrm{the}\:{n}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right) \\…

Question-132135

Question Number 132135 by benjo_mathlover last updated on 11/Feb/21 Answered by Olaf last updated on 11/Feb/21 $${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{u}−\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\frac{{u}−\mathrm{1}}{{u}}−\mathrm{1}}{\frac{{u}−\mathrm{1}}{{u}}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}} \\ $$$${f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\::…

f-R-R-continuous-such-that-x-R-f-x-f-f-x-1-f-2004-2003-f-1999-

Question Number 950 by 123456 last updated on 06/May/15 $${f}:\mathbb{R}\rightarrow\mathbb{R},\mathrm{continuous}\:\mathrm{such}\:\mathrm{that}\:\forall{x}\in\mathbb{R},{f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1},{f}\left(\mathrm{2004}\right)=\mathrm{2003},{f}\left(\mathrm{1999}\right)=? \\ $$ Commented by 123456 last updated on 06/May/15 $${f}\left(\mathrm{2004}\right){f}\left[{f}\left(\mathrm{2004}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1}\Leftrightarrow{f}\left(\mathrm{2003}\right)=\frac{\mathrm{1}}{\mathrm{2003}} \\ $$$${f}\left(\mathrm{2003}\right){f}\left[{f}\left(\mathrm{2003}\right)\right]=\frac{\mathrm{1}}{\mathrm{2003}}{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{1}\Leftrightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{2003}={f}\left(\mathrm{2004}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right){f}\left[{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1} \\…

a-n-m-m-n-0-na-n-m-1-m-m-n-gt-0-m-0-na-n-m-m-1-m-m-2-n-gt-0-m-gt-0-a-5-5-a-10-10-

Question Number 926 by 123456 last updated on 26/Apr/15 $${a}\left({n},{m}\right)=\begin{cases}{{m}}&{{n}\leqslant\mathrm{0}}\\{{na}\left({n}+{m}−\mathrm{1},{m}\right)+{m}}&{{n}>\mathrm{0}\vee{m}\leqslant\mathrm{0}}\\{{na}\left({n}−{m},{m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} }&{{n}>\mathrm{0}\vee{m}>\mathrm{0}}\end{cases} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)=??? \\ $$$${a}\left(\mathrm{10},\mathrm{10}\right)=?? \\ $$ Commented by 123456 last updated on 27/Apr/15 $${a}\left(\mathrm{0},{m}\right)={m}…

1-simplify-S-n-x-k-0-n-C-n-k-cos-k-x-cos-kx-2-find-the-value-of-A-n-k-0-n-C-n-k-cos-k-pi-n-cos-kpi-n-

Question Number 66462 by mathmax by abdo last updated on 15/Aug/19 $$\left.\mathrm{1}\right){simplify}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}} \left({x}\right){cos}\left({kx}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}}…

f-R-R-g-R-R-f-x-g-x-f-x-g-x-f-x-g-x-f-x-g-x-f-x-g-x-f-x-f-x-g-x-g-x-f-x-g-x-f-x-g-x-

Question Number 874 by 123456 last updated on 08/Apr/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={g}\left({x}\right)−{f}\left(−{x}\right) \\ $$$${g}\left({x}\right)={f}\left({x}\right)+{g}\left(−{x}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$$${f}\left({x}\right)+{g}\left({x}\right)=? \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\overset{?} {=}\boldsymbol{{f}}\left(−\boldsymbol{{x}}\right)…