20221014 Complex numbers, hyperbolic complex numbers, dual numbers

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This blog post introduces three plural forms.

1. Hyperbolic sine function and hyperbolic cosine function

- hyperbolic sine function

$$\mathrm{born}x=\frac{e^x-e^{-x}}{2}$$
The Taylor expansion of the hyperbolic sine function is:
$$\mathrm{born}x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots \cdots$$
即:$\mathrm{born}x={\sum }_{n=0}^{\infty }\frac{x^{2n+}}{(2n+1)!}$

- hyperbolic cosine function

$$\cosh x=\frac{e^x+e^{-x}}{2}$$

The Taylor expansion of the hyperbolic sine function is: $$\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots$$

- hyperbolic tangent function

$$\mathrm{fishy}x=\frac{\mathrm{born}x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$

2. Angle definition

Rounded corners (from trigonometric and circular functions): $\alpha$ corresponds to twice the area of ​​the graph.

Hyperbolic angle (derived from hyperbolic functions): $\alpha$ corresponds to twice the area of ​​the graph.

Note: It does not say which angle is $\alpha$ , which can be understood as a monotonically increasing function about the included angle.

The specific area of ​​the right figure is equal to $\frac{1}{2}ln(x+y)$ , according to the definition of hyperbolic angle$a=\frac{2}{2}ln(x+y)$，即$x+y=\exp \left(\alpha \right)$ ;且$x^2-y^2=1$ , then$x-y=\exp (-\alpha )$，则有$x=\frac{exp\left(\alpha \right)+exp(-\alpha }{2}$和$y=\frac{exp\left(\alpha \right)-exp(-\alpha }{2}$。

The very ingenious simple form here comes from a virtual angle $\alpha$ , or defines the angle in terms of area, really opens up new ideas.

3. Complex numbers, hyperbolic complex numbers, even numbers (extension of real numbers)

- unit circle $x^2+y^2=1$

Define complex units on the unit circle $i$ , the corresponding area is$\frac{}{2}$, the corresponding complex number is $\cos a+i\sin \alpha$ ,in the case of$\exp (i\alpha )=\cos a+i\sin \alpha$，展开为$$\exp \left(i\alpha \right)=1+i\alpha +\frac{1}{2!}{i}^2{a}^2+\frac{1}{3!}{i}^3{a}^3+\frac{1}{4!}{i}^4{a}^4+\cdots$$$$\cos a=\sum _{\mathrm{k}=0}^{\infty }\frac{(-1{)}^{\mathrm{k}}}{(2k)!}{a}^{2k}=1-\frac{1}{2!}{a}^2+\frac{1}{4!}{a}^4-\frac{1}{6!}{a}^6+\cdots$$$$\sin a=\sum _{\mathrm{k}=0}^{\infty }\frac{(-1{)}^{\mathrm{k}}}{(2k+1)!}{a}^{2k+1\_}=a-\frac{1}{3!}{a}^3+\frac{1}{5!}{a}^5-\frac{1}{7!}{a}^7+\cdots$$ Therefore, the property is$i^2=-1$ .$\_$
At the same time, it can be seen that$a=\sqrt{a^2+b^2}\cos \alpha$，$b=\sqrt{a^2+b^2}\sin \alpha$ ,if$a+ib=\sqrt{a^2+b^2}\cos a+i\sqrt{a^2+b^2}\sin \alpha$ , so the scaling is$a+ib=\sqrt{a^2+b^2}\exp (i\alpha )$。

- Unit hyperbola $x^2-y^2=1$

Define the hyperbolic complex unit jj on the unit hyperbola$j$ , the corresponding area is$\frac{}{2}$, the corresponding complex number is $\cosh a+j\mathrm{born}\alpha$ , exponential form$\exp (j\alpha )=\cosh a+j\mathrm{born}\alpha$ , according to cosh ⁡ \coshin Section 1$\cosh$和$\sinh$ 的展开形式，$$\exp \left(j\alpha \right)=1+ja+\frac{1}{2!}{j}^2{a}^2+\frac{1}{3!}{j}^3{a}^3+\frac{1}{4!}{j}^4{a}^4+\cdots$$$$\cosh a=1+\frac{a^2}{2!}+\frac{a^4}{4!}+\frac{a^6}{6!}+\cdots$$$$\mathrm{born}a=a+\frac{a^3}{3!}+\frac{a^5}{5!}+\frac{a^7}{7!}+...$$ Therefore, the property is$j^2=1$。

同时可知，$$\frac{a}{\cosh a}=\frac{b}{\mathrm{born}a}=\frac{\sqrt{a^2-b^2}}{\sqrt{{\cosh }^{2}a-{\mathrm{born}}^{2}a}}=\sqrt{a^2-b^2},\text{if}a>b,a>0$$ Considering the other three directions, there is$$a=\sqrt{|a^2-b^2|}\cosh a，b=\sqrt{|a^2-b^2|}\mathrm{born}\alpha$$ is scaled to$a+jb=\sqrt{|a^2-b^2|}\exp (j\alpha )$。

- unit double vertical bar $x^2=1$

A point on the right branch is defined as $1+ϵ\alpha$ , the corresponding area is$\frac{1}{2}\alpha$ ,$ϵ$ is called a dual even number

To make it also have a corresponding version in exponential form, $\exp \left(ϵ\alpha \right)=1+ϵ\alpha$ , then the corresponding Taylor expansion shows that once${ϵ}^2=0$ , the above formula holds.

Then the even number is defined as $ϵ\ne 0$ and${ϵ}^2=0$。

$ϵ(a+ϵb)=aϵ$ , projected to the abscissa and then rotated 90 degrees counterclockwise

The value $a+\epsilon b=|a|\exp (j\alpha )$

- Modulo: the root sign times itself by its conjugate

Why are the specific definitions of moduli different in the three figures? This is because the defined energy functions are different on the three same graphs.

Note:

Reference: Bilibili Fengzhuyunmo
https://www.bilibili.com/video/BV1xp4y1v7cw/?vd_source=5138fcb56aada2b6f1c51dfff686251a

https://baike.baidu.com/item/%E5%8F%8C%E6%9B%B2%E6%AD%A3%E5%BC%A6%E5%87%BD%E6%95%B0/4395524?fromtitle=sinh&fromid=1965202&fr=aladdin

https://zhuanlan.zhihu.com/p/444445404