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Category: Relation and Functions

f-n-0-1-0-1-g-0-1-0-1-f-n-1-x-g-f-n-x-f-n-g-x-f-0-x-x-f-4-x-g-x-x-2-f-2-2-

Question Number 2253 by 123456 last updated on 11/Nov/15 $${f}_{{n}} :\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{1}\right],{g}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}_{{n}+\mathrm{1}} \left({x}\right)={g}\left[{f}_{{n}} \left({x}\right)\right]+{f}_{{n}} \left[{g}\left({x}\right)\right] \\ $$$${f}_{\mathrm{0}} \left({x}\right)={x} \\ $$$${f}_{\mathrm{4}} \left({x}\right)=? \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} ,{f}_{\mathrm{2}}…

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Question Number 67672 by Abdo msup. last updated on 30/Aug/19 $${decompose}\:{the}\:{folowing}\:\:{fraction}\:{at}\:{R}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right)\:{G}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by…

Question-133168

Question Number 133168 by bemath last updated on 19/Feb/21 Answered by floor(10²Eta[1]) last updated on 19/Feb/21 $$\mathrm{x}\rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{I}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{but}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right),\:\mathrm{so} \\ $$$$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{II}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{−\mathrm{x}^{\mathrm{2}}…

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Question Number 67540 by mathmax by abdo last updated on 28/Aug/19 $${prove}\:{that}\:\:\:\mid\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{it}\right)\mid\:=\sqrt{\frac{\mathrm{2}\pi}{{e}^{\pi{t}} \:+{e}^{−\pi{t}} }} \\ $$$${and}\:\mid\Gamma\left(\mathrm{1}+{it}\right)\mid\:=\sqrt{\frac{\mathrm{2}\pi{t}}{{e}^{\pi{t}} −{e}^{−\pi{t}} }} \\ $$ Commented by ~ À ®…