## 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).

$\text{(a\oplus b):= }\left(a^{\frac{1}{n}}+b^{\frac{1}{n}}-1\right)^n$
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:

$\text{(a\oplus b):= }\left(\left(\left(\frac{a}{n}+1\right)^n+\left(\frac{b}{n}+1\right)^n\right)^{\frac{1}{n}}-1\right) n$
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:

[/vc_column_text][/vc_column][/vc_row]