Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 191960 by Rupesh123 last updated on 04/May/23

Answered by a.lgnaoui last updated on 05/May/23

$$\mathbf{f}(\mathbf{f}\left(\mathbf{x}\right)-\mathbf{f}\left(\mathbf{y}\right))=\mathbf{f}(\mathbf{f}\left(\mathbf{x}\right)-2{\mathbf{x}}^2\mathbf{f}\left(\mathbf{y}\right)+\mathbf{f}\left({\mathbf{y}}^2\right)$$$$\Rightarrow {\mathbf{f}}^{-1}(\mathbf{f}\left(\mathbf{x}\right)-\mathbf{f}\left(\mathbf{y}\right))={\mathbf{f}}^{-1}\left(\mathbf{f}\left(\mathbf{x}\right)\right)-2{\mathbf{x}}^2{\mathbf{f}}^{-1}\left(\mathbf{y}\right)+{\mathbf{f}}^{-1}\left({\mathbf{y}}^2\right)$$$$\Rightarrow \mathbf{f}\left(\mathbf{x}\right)-\mathbf{f}\left(\mathbf{y}\right)=\mathbf{f}\left(\mathbf{x}\right)-2{\mathbf{x}}^2\mathbf{y}+{\mathbf{y}}^2$$$$\Rightarrow \mathbf{f}\left(\mathbf{y}\right)=\mathbf{y}(2{\mathbf{x}}^2-\mathbf{y})$$$$$$

Commented by mehdee42 last updated on 05/May/23

$$alloperationhavenomathemticalbasis.??!!$$

Answered by deleteduser1 last updated on 25/May/23

$$P(x,y):f(f\left(x\right)-f\left(y\right))=f\left(f\left(x\right)\right)-2x^{2}f\left(y\right)+f\left(y^2\right)$$$$f\left(f\left(0\right)\right)=0;f(f\left(0\right)-f\left(1\right))=f\left(1\right)$$$$f(f\left(1\right)-f\left(0\right))=f\left(f\left(1\right)\right)-f\left(0\right);f\left(0\right)=f\left(f\left(1\right)\right)-f\left(1\right)$$$$f(f\left(0\right)-f\left(y\right))=f\left(y^2\right)$$$$f\left(0\right)=f\left(f\left(x\right)\right)-2x^{2}f\left(x\right)+f\left(x^2\right)$$$$f\left(f{\left(0\right)}^2\right)=0\Rightarrow f{\left(0\right)}^2=f\left(0\right)\Rightarrow f\left(0\right)=0or1$$$$CaseI:f\left(0\right)=0\Rightarrow f(-f\left(y\right))=f\left(y^2\right)$$$$f(-f(-y))=f\left(y^2\right)$$$$\Rightarrow -f\left(y\right)=-f(-y)\Rightarrow f\left(y\right)=f(-y)$$$$f\left(f\left(1\right)\right)=f\left(1\right);f\left(0\right)-f\left(1\right)=f\left(1\right)\Rightarrow f\left(1\right)=0$$$$\Rightarrow f(-f\left(y\right))=-2f\left(y\right)+f\left(y^2\right);f(-f\left(y\right))=f\left(y^2\right)$$$$\Rightarrow f\left(y^2\right)=-2f\left(y\right)+f\left(y^2\right)$$$$\Rightarrow f\left(y\right)=0\Rightarrow f\equiv 0$$$$fisidenticallyzerowhenf\left(0\right)=0$$$$CaseII:f\left(0\right)=1\Rightarrow f\left(f\left(0\right)\right)=f\left(1\right)=0.....\left(a\right)$$$$Butalsoobservethat$$$$f(f\left(0\right)-f\left(1\right))=f\left(1\right)\Rightarrow f(1-f\left(1\right))=f\left(1\right)...\left(i\right)$$$$andthatf(f\left(1\right)-f\left(0\right))=f\left(f\left(1\right)\right)-f\left(0\right)$$$$\Rightarrow f(f\left(1\right)-1)=f\left(f\left(1\right)\right)-1...\left(ii\right)$$$$Wealsohavethat$$$$f\left(0\right)+f\left(1\right)=1+f\left(1\right)=f\left(f\left(1\right)\right)...\left(iii\right)$$$$Combiningi,iiandiii$$$$\Rightarrow f(f\left(1\right)-1)=f\left(1\right)=f(1-f\left(1\right))$$$$\Rightarrow f\left(1\right)-1=1-f\left(1\right)\Rightarrow f\left(1\right)=1$$$$Thiscontradicts\left(a\right)$$$$Hence,f\left(0\right)cannotbeequalto1$$

Commented by witcher3 last updated on 06/May/23

$$5\mathrm{ligne}\mathrm{this}\mathrm{is}\mathrm{true}\iff \mathrm{f}\mathrm{injective}$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2\mathrm{worcks}\mathrm{also}$$$$$$

Commented by deleteduser1 last updated on 06/May/23

$$f\left(x\right)=x^{2}isnotinjective.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com