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Question Number 2032 by 123456 last updated on 31/Oct/15

$$f:[0,+1]\to \mathbb{R}$$$$\eta \left(f\right):=\underset{0}{\stackrel{1}{\int }}{f}^{2}dx$$$$\mathrm{and}\mathrm{if}\mathrm{\forall }x\in [0,1],f\in [m,\mathrm{M}]\mathrm{for}\mathrm{some}(m,\mathrm{M})\in {\mathbb{R}}^2$$$$f\uparrow :=\max \left(f\right)-f$$$$f\downarrow :=f-\min \left(f\right)$$$$\mu \left(f\right):=\underset{0}{\stackrel{1}{\int }}f\downarrow f\uparrow dx$$$$\mathrm{then}$$$$f=e^x,\eta \left(f\right)\stackrel{?}{=}\mu \left(f\right)$$$$f=x,\eta \left(f\right)\stackrel{?}{=}\mu \left(f\right)$$$$\mathrm{\exists }f,\eta \left(f\right)=0???$$$$\mathrm{\exists }f,\mu \left(f\right)=0???$$$$\mu \left(f\right)=0,\mathrm{does}\eta \left(f\right)=0???$$$$\eta \left(f\right)=0,\mathrm{does}\mu \left(f\right)=0???$$

Commented by Yozzi last updated on 31/Oct/15

$$Anattempt$$$$max\left(f\right)=M,min\left(f\right)=m$$$$\therefore f\uparrow =M-f,f\downarrow =f-m$$$$\mu \left(f\right)={\stackrel{1}{\int }}_{0}f\uparrow f\downarrow dx={\int }_0^1(M-f)(f-m)dx$$$$\mu \left(f\right)={\int }_0^1(M+m)fdx-{\int }_0^1{f}^{2}dx-Mm{\int }_0^{1}dx$$$$\mu \left(f\right)=(M+m){\int }_0^{1}fdx-\eta \left(f\right)-Mm$$$$\mu \left(f\right)+\eta \left(f\right)=(M+m){\int }_0^{1}fdx-Mm$$$$f\left(x\right)=e^x$$$$\eta \left(e^x\right)={\int }_0^1{e}^{2x}dx=\frac{1}{2}(e^2-1)=\frac{1}{2}(e+1)(e-1)$$$$\mathrm{\forall }x\in [0,1],e^x\in [1,e]\Rightarrow m=1,M=e$$$$\therefore \mu \left(e^x\right)=(1+e){\int }_0^1{e}^{x}dx-\frac{1}{2}(e+1)(e-1)-e$$$$\mu \left(e^x\right)=(1+e)(e-1)-0.5(e+1)(e-1)-e$$$$\mu \left(e^x\right)=0.5(e+1)(e-1)-e\ne 0.5(e-1)(e+1)$$$$\mu \left(e^x\right)\ne \eta \left(e^x\right)$$$$$$$$f\left(x\right)=x$$$$\eta \left(x\right)={\int }_0^1{x}^{2}dx=\frac{1}{3}$$$$\mathrm{\forall }x\in [0,1],f\in [0,1]\Rightarrow m=0,M=1.$$$$\therefore \mu \left(x\right)=(0+1){\int }_0^{1}xdx-\frac{1}{3}-0$$$$\mu \left(x\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\ne \frac{1}{3}$$$$\therefore \mu \left(x\right)\ne \eta \left(x\right)$$$$$$$$Forf\left(x\right)\ne 0\mathrm{\forall }x\in [0,1],if\eta \left(f\right)=0\Rightarrow f^{2}isoddforx\in [0,1]$$$$Thus,if\varphi \left(x\right)=f^2\left(x\right)\Rightarrow \varphi (-x)=-\varphi \left(x\right)=-f^2\left(x\right)$$$$But,\varphi (-x)=f^2(-x)and,\mathrm{\forall }f\left(x\right)\in \mathbb{R},f^2(-x)⩾0.$$$$\Rightarrow \varphi (-x)⩾0\Rightarrow \varphi (-x)\nless 0asimpliedby$$$$\varphi (-x)=-f^2\left(x\right).\therefore f^{2}isnotoddforany$$$$x\in \mathbb{R}.\Rightarrow \nexists f\in \mathbb{R}\mid \eta \left(f\right)=0iff\left(x\right)\ne 0.$$$$Ifweonlyconsiderthestatement$$$$\eta \left(f\right)=0,wecanhavef\left(x\right)=0beinga$$$$function.$$$$$$$$If\mu \left(f\right)=0\Rightarrow \eta \left(f\right)=0,0+0=(M+m){\int }_0^{1}fdx-Mm$$$$\Rightarrow {\int }_0^{1}f\left(x\right)dx=\frac{Mm}{m+M}=\frac{max\left(f\right)min\left(f\right)}{max\left(f\right)+min\left(f\right)}$$$$Isthereafunctionf\in [m,M]suchthat$$$${\int }_0^{1}f\left(x\right)dx=\frac{mM}{m+M}?$$$$Bythefundamentaltheoremofcalculus$$$$F\left(1\right)-F\left(0\right)=\frac{max\left(F^{{}^{\prime }}\left(x\right)\right)min\left(F^{{}^{\prime }}\left(x\right)\right)}{max\left(F^{{}^{\prime }}\left(x\right)\right)+min\left(F^{{}^{\prime }}\left(x\right)\right)}$$$$If\eta \left(f\right)=0\Rightarrow \mu \left(f\right)=(m+M){\int }_0^{1}fdx-Mm$$$$$$$$\therefore {\int }_0^1{f}^{2}dx=(m+M){\int }_0^{1}fdx-mM$$$${\int }_0^1(f^2\left(x\right)-(m+M)f\left(x\right))dx=-mM$$$${\int }_0^{1}f\left(x\right)(f\left(x\right)-m+M)dx=-mM$$$$Thesimplerequationtotheaboveone$$$$is{\int }_0^1(M-f\left(x\right))(f\left(x\right)-m)dx=0.$$$$\therefore Let\varphi \left(x\right)=(M-f\left(x\right))(f\left(x\right)-m)$$$$where\varphi \left(x\right)isoddforx\in [0,1].$$$$$$$$$$$$$$$$$$

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