fields and extended arithmetic operators: videos

3D-animations

Representation of the respective operator function of two variables x and y (since the function value for negative x and y can be complex, the absolute value of the complex function value has been plotted).

Method 1, addition:

\text{(a$\oplus $b):= }\left(a^{\frac{1}{n}}+b^{\frac{1}{n}}-1\right)^n
neutral element: 1
inverse element: \left(2-a^{1/n}\right)^n
\left(\underset{n\to \infty}{\text{lim}}a\oplus b\right)=a b (normal multiplication)
\left(\underset{n\to 1}{\text{lim}}a\oplus b\right)=a+b-1 (normal addition)

n=10, viewpoint is changing:

n varies from 1 up to 20, viewpoint is changing:

 

n varies from 1 up to 20, fixed viewpoint:

 

Method 1, multiplication:

\text{(a$\otimes $b):= }\left(\left(\left(a^{\frac{1}{n}}-1\right) \left(b^{\frac{1}{n}}-1\right)\right) n+1\right)^n
neutral element: \left(\frac{1}{n}+1\right)^n
inverse element: (1+1/((-1+a^(1/n)) n^2))^n
\left(\underset{n\to \infty }{\text{lim}}a\otimes b\right)=a^{\log (b)} (logarithmic exponentiation)
\left(\underset{n\to 1}{\text{lim}}a\otimes b\right)=(a-1) (b-1)+1 (multiplication)

n varies from 0 up to 10, fixed viewpoint:

 

n varies from 0 bis 10, viewpoint is changing:

 

n=3, viewpoint is changing:

 

n=5, viewpoint is changing:

 

Method 2, addition:

\text{(a$\oplus $b):= }\left(\left(\left(\frac{a}{n}+1\right)^n+\left(\frac{b}{n}+1\right)^n\right)^{\frac{1}{n}}-1\right) n
neutral element: -n
inverse element: n \left(\left(-\left(\frac{a+n}{n}\right)^n\right)^{1/n}-1\right)
\left(\underset{n\to \infty }{\text{lim}}a\oplus b\right)=\log \left(e^a+e^b\right) (Jacobi-addition)
\left(\underset{n\to 1}{\text{lim}}a\oplus b\right)=a+b+1 (normal addition)

 

n=5, viewpoint is changing:

 

n=13, viewpoint is changing:

 

n=2, viewpoint is changing:

 

n=3, viewpoint is changing:

 

Method 2, multiplication:

\text{(a$\otimes $b):= }\left(\left(\left(\frac{a}{n}+1\right)^n \left(\frac{b}{n}+1\right)^n\right)^{\frac{1}{n}}-1\right) n
neutral element: 0
inverse element: n (-1+(((a+n)/n)^-n)^(1/n))
\left(\underset{n\to \infty }{\text{lim}}a\otimes b\right)=\log \left(e^{a+b}\right)=a+b (normal addition)
\underset{n\to 1}{\text{lim}}a\otimes b=a b+a+b (normal addition and multiplication)

n=3, viewpoint is changing:

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