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Question Number 199405 by mnjuly1970 last updated on 03/Nov/23

$$$$$$\mathrm{calculate}...$$$$\mathrm{Q}:\mathrm{I}f,f\left(x\right)=2e^x-1+\lfloor e^x+\frac{3}{2}+\lfloor e^x\rfloor \rfloor$$$$\Rightarrow f{}^{-1}\left(\frac{\pi }{4}\right)=?$$$$$$

Answered by Frix last updated on 03/Nov/23

$${\mathrm{e}}^x=i+f;i\in \mathbb{Z}\wedge 0⩽f<1$$$${\mathrm{e}}^{\pi }=23+\zeta$$$$2(i+f)-1+\lfloor i+f+1+\frac{1}{2}+\lfloor i+f\rfloor \rfloor =23+\zeta$$$$2i-1+2f+\lfloor 2i+1+f+\frac{1}{2}\rfloor =23+\zeta$$$$4i+2f+\lfloor f+\frac{1}{2}\rfloor =23+\zeta$$$$0⩽f<\frac{1}{2}$$$$4i+2f=23+\zeta \Rightarrow i=\frac{23}{4}\notin \mathbb{Z}$$$$\frac{1}{2}⩽f<1\Rightarrow 0⩽2f-1<1$$$$4i+2+(2f-1)=23+\zeta \Rightarrow i=\frac{21}{4}\notin \mathbb{Z}$$$$\Rightarrow f\left(x\right)\ne {\mathrm{e}}^{\pi }$$$$\Rightarrow f^{-1}\left({\mathrm{e}}^{\pi }\right)\mathrm{does}\mathrm{not}\mathrm{exist}$$

Commented by Frix last updated on 03/Nov/23

$$\mathrm{Please}\mathrm{correct}\mathrm{me}\mathrm{if}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{wrong}.$$

Commented by mnjuly1970 last updated on 03/Nov/23

$$thankyousomuchsir$$$$foryoureffort.$$

Answered by mnjuly1970 last updated on 03/Nov/23

$$mysolution$$$$f\left(x\right)=2e^x-1+\lfloor e^x\rfloor +\lfloor e^x+\frac{1}{2}\rfloor +1$$$$=2e^x+\lfloor 2e^x\rfloor$$$$-----$$$$y=2e^x+\lfloor 2e^x\rfloor \Rightarrow x=2e^y+\lfloor 2e^y\rfloor$$$$\lfloor x\rfloor =2\lfloor 2e^y\rfloor \Rightarrow \lfloor 2e^y\rfloor =\frac{\lfloor x\rfloor }{2}$$$$----$$$$x=2e^y+\frac{\lfloor x\rfloor }{2}\Rightarrow y=ln(\frac{x}{2}-\frac{\lfloor x\rfloor }{4})$$$$$$$$f^{-1}\left(x\right)=ln(\frac{x}{2}-\frac{\lfloor x\rfloor }{4})$$$$f{}^{-1}\left(\frac{\pi }{4}\right)=ln\left(\frac{\pi }{8}\right)\iff f\left(ln\left(\frac{\pi }{8}\right)\right)=\frac{\pi }{4}$$$$$$

Commented by Frix last updated on 03/Nov/23

$$f\left(x\right)={2\mathrm{e}}^x+\lfloor {2\mathrm{e}}^x\rfloor$$$$\underset{x\to {\left(\ln 6\right)}^{-}}{\lim }f\left(x\right)=23$$$$\underset{x\to {\left(\ln 6\right)}^{+}}{\lim }f\left(x\right)=24$$$${\mathrm{e}}^{\pi }\approx 23.1407$$$$\Rightarrow \mathrm{\forall }x\in \mathbb{R}:f\left(x\right)\ne {\mathrm{e}}^{\pi }$$$$======================$$$$f\left(x\right)={2\mathrm{e}}^x+\lfloor {2\mathrm{e}}^x\rfloor$$$$f^{-1}\left(x\right)=\ln \frac{2x-\lfloor x\rfloor }{4}$$$$f^{-1}\left(f\left(x\right)\right)=x\mathrm{but}f\left(f^{-1}\left(x\right)\right)\ne x$$$$\mathrm{This}\mathrm{is}\mathrm{strange}...$$$$f^{-1}\left({\mathrm{e}}^{\pi }\right)=\ln \frac{{2\mathrm{e}}^{\pi }-23}{4}$$$$\mathrm{But}$$$$f\left(\ln \frac{{2\mathrm{e}}^{\pi }-23}{4}\right)={\mathrm{e}}^{\pi }-\frac{1}{2}$$$$\mathrm{So}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{something}\mathrm{wrong}.$$

Commented by Frix last updated on 03/Nov/23

$$v=u+\lfloor u\rfloor$$$$\mathrm{The}\mathrm{range}\mathrm{of}\mathrm{this}\mathrm{function}\mathrm{is}$$$$\mathbb{R}\mathrm{\setminus }\{v\in \mathbb{R}\wedge n\in \mathbb{Z}\mid 2n-1⩽v<2n\}$$$$\Rightarrow v^{-1}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{in}\mathrm{these}\mathrm{intervals}$$$$\Rightarrow$$$$f^{-1}\left(x\right)=\{\begin{array}{l}\ln \frac{2x-\lfloor x\rfloor }{4},2n⩽x<2n+1\\ \mathrm{undefined},2n-1⩽x<2n\end{array}\mathrm{\forall }n\in \mathbb{Z}$$$$23<{\mathrm{e}}^{\pi }<24$$$$2\times 12-1<{\mathrm{e}}^{\pi }<2\times 12$$$$\Rightarrow$$$$f^{-1}\left({\mathrm{e}}^{\pi }\right){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by mnjuly1970 last updated on 03/Nov/23

$$thanksalotsir.youareright\underset{―}{\underbrace{⋚}}$$

Commented by Frix last updated on 03/Nov/23

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{welcome},\mathrm{working}\mathrm{on}\mathrm{these}\mathrm{things}$$$$\mathrm{keeps}\mathrm{my}\mathrm{brain}\mathrm{young}.$$

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