Thursday, March 19, 2020

p+ , p , x(T) , f(x)


Form the \(x^2-xT-1=0\) where \(T\in\mathbb{R}^{+}\ ;\ T≠0\)
\(x-\frac1x=T\ →\ \frac xT=\frac{xT+1}{xT}=\frac{x+\frac1T}x\)
Def.: \(x≡p^+\) and \(\frac1x≡p^-\)
\(p^+=\frac{\sqrt{T^2+4}+T}2\ →\ p^-=\frac1{p^+}=\frac2{\sqrt{T^2+4}+T}=\frac{\sqrt{T^2+4}-T}2\)

Def.: \(x\left({T}\right)=T+p^-\left({\frac1x}\right)=T+\frac{\sqrt{T^2+4}-T}2=\frac{\sqrt{T^2+4}+T}2\) Def.: \(f\left({x}\right)≡f\left({x_T}\right)≡x\left({T}\right)\)

The properties of the \(f\left({x}\right)\) :
  1. \(f\left({-x}\right)=\frac1{f\left({x}\right)}=f\left({x}\right)-x\)
    \(\frac1{f\left({-x}\right)}=f\left({x}\right)=f\left({-x}\right)+x\)
  2. \(f\left({-x}\right)+x=\frac1{f\left({x}\right)-x}\)
  3. \(x·f\left({x}\right)-x·f\left({-x}\right)=x^2\)
    \(x·f\left({-x}\right)-x·f\left({x}\right)=-x^2\)


No comments: