【Study Notes】Supercomplex Numbers

This is a study note that will be continuously supplemented and improved.

hypercomplex number

Hypercomplex numbers are a generalized complex number . Hamilton generalized complex numbers and introduced quaternions . This generalized complex number is a hypercomplex number .

Combined with ordinary plurals, it is easy to understand the following.

1 definition

A = a 0 + a 1 i + a 2 j + a 3 k A=a_0+a_1i+a_2j+a_3k A=a0+a1i+a2j+a3k is a multiple number.
a = ∣ A ∣ = ( a 0 2 + a 1 2 + a 2 2 + a 3 2 ) 1 / 2 a=|A|=(a_0^2+a_1^2+a_2^2+a_3^2)^ {1/2}a=A=(a02+a12+a22+a32)1/2 is called hyperparameterAAThe modulus of A.
i , j , ki,j,ki,j,k is called hyperparameterAAThe imaginary units,second, and third dimensions of A.

2 Related properties

2.1 Hypercomplex equality

Two supermultiples A = a 0 + a 1 i + a 2 j + a 3 k A=a_0+a_1i+a_2j+a_3kA=a0+a1i+a2j+a3k sumB = b 0 + b 1 i + b 2 j + b 3 k B=b_0+b_1i+b_2j+b_3kB=b0+b1i+b2j+b3k
(1) A = B ⇔ a n = b n   ( n = 0 , 1 , 2 , 3 ) A=B\Leftrightarrow a_n=b_n \ (n=0,1,2,3) A=Ban=bn (n=0,1,2,3 )
(2)m ( a 0 + a 1 i + a 2 j + a 3 k ) = at 0 + at 1 i + at 2 j + at 3 km(a_0+a_1i+a_2j+a_3k)=ma_0+ma_1i +ma_2j+ma_3km(a0+a1i+a2j+a3k)=ma0+ma1i+ma2j+ma3k m m m is a real number.

2.2 Conjugate hypercomplex numbers

Supermultiple A = a 0 + a 1 i + a 2 j + a 3 k A=a_0+a_1i+a_2j+a_3kA=a0+a1i+a2j+a3k A ˉ = a 0 − a 1 i − a 2 j − a 3 k \bar{A}=a_0-a_1i-a_2j-a_3k Aˉ=a0a1ia2ja3k isAATheconjugate hypercomplex numbersof A.

∣ A ∣ = ∣ A ˉ ∣ , A ˉ = B ˉ ⇔ a n = b n   ( n = 0 , 1 , 2 , 3 ) |A|=|\bar{A}|, \bar{A}=\bar{B}\Leftrightarrow a_n=b_n \ (n=0,1,2,3) A=Aˉ,Aˉ=Bˉan=bn (n=0,1,2,3)

2.3 Multiplication rule

Two supermultiples A = a 0 + a 1 i + a 2 j + a 3 k A=a_0+a_1i+a_2j+a_3kA=a0+a1i+a2j+a3k sumB = b 0 + b 1 i + b 2 j + b 3 k B=b_0+b_1i+b_2j+b_3kB=b0+b1i+b2j+b3product of k

A B = [ a 0 + a 1 i + a 2 j + a 3 k ] [ b 0 + b 1 i + b 2 j + b 3 k ] = ⋯ = a 0 b 0 − a 1 b 1 + ( a 0 b 1 + b 0 a 1 ) i \begin{aligned} AB&=[a_0+a_1i+a_2j+a_3k][b_0+b_1i+b_2j+b_3k] \\ &=\cdots \\ &=a_0b_0-a_1b_1+(a_0b_1+b_0a_1)i \end{aligned} AB=[a0+a1i+a2j+a3k][b0+b1i+b2j+b3k]==a0b0a1b1+(a0b1+b0a1)i
That is, the regular product of complex numbers.

3 Trigonometric expressions

Supermultiple A = a 0 + a 1 i + a 2 j + a 3 k A=a_0+a_1i+a_2j+a_3kA=a0+a1i+a2j+a3k namesu , v , wu,v,wu,v,w forAAThe first, second and third corners of A.
a 0 = a cos ⁡ u cos ⁡ v cos ⁡ w a_0=a\cos u\cos v\cos wa0=acosucosvcosw a 1 = a cos ⁡ u cos ⁡ v sin ⁡ w a_1=a\cos u\cos v\sin w a1=acosucosvsinw a 2 = a cos ⁡ u sin ⁡ v a_2=a\cos u\sin v a2=acosusinv a 3 = a sin ⁡ u a_3=a\sin u a3=asinu

A = a 0 + a 1 i + a 2 j + a 3 k = ( a , u , v , w ) = a ( cos ⁡ u cos ⁡ v cos ⁡ w + i cos ⁡ u cos ⁡ v sin ⁡ w + j cos ⁡ u sin ⁡ v + k sin ⁡ u ) \begin{aligned} A&=a_0+a_1i+a_2j+a_3k \\ &=(a,u,v,w) \\ &=a(\cos u\cos v\cos w+i\cos u\cos v\sin w+j\cos u\sin v+k\sin u) \end{aligned} A=a0+a1i+a2j+a3k=(a,u,v,w)=a(cosucosvcosw+icosucosvsinw+jcosusinv+ksinu)

some conclusions

Set-up supermultiple A = ( a , u 1 , v 1 , w 1 ) , B = ( b , u 2 , v 2 , w 2 ) A=(a,u_1,v_1,w_1),B=(b, u_2,v_2,w_2)A=(a,u1,v1,w1),B=(b,u2,v2,w2) , real numbermmm,则
(1)m ( a , u 1 , v 1 , w 1 ) = ( ma , mu 1 , mv 1 , mw 1 ) m(a,u_1,v_1,w_1)=(ma,mu_1,mv_1, mw_1)m(a,u1,v1,w1)=(ma,m u1,mv1,mw1)
(2)( a , u 1 , v 1 , w 1 ) ( b , u 2 , v 2 , w 2 ) = ( ab , u 1 + u 2 , v 1 + v 2 , w 1 + w 2 ) (a,u_1,v_1,w_1)(b,u_2,v_2,w_2)=(ab,u_1+u_2,v_1+v_2,w_1+w_2)(a,u1,v1,w1)(b,u2,v2,w2)=(ab,u1+u2,v1+v2,w1+w2)
(3) ( a , u 1 , v 1 , w 1 ) / ( b , u 2 , v 2 , w 2 ) = ( a / b , u 1 − u 2 , v 1 − v 2 , w 1 − w 2 ) (a,u_1,v_1,w_1)/(b,u_2,v_2,w_2)=(a/b,u_1-u_2,v_1-v_2,w_1-w_2) (a,u1,v1,w1)/(b,u2,v2,w2)=(a/b,u1u2,v1v2,w1w2)
(4)( a , u 1 , v 1 , w 1 ) n = ( an , nu 1 , nv 1 , nw 1 ) (a,u_1,v_1,w_1)^n=(a^n,nu_1,nv_1 ,nw_1)(a,u1,v1,w1)n=(an,n u1,nv1,nw1)
(5) SupermultipleA = ( a , u , v , w ) A=(a,u,v,w)A=(a,u,v,w ) conjugate hypercomplex numberA ˉ = ( a , − u , − v , − w ) \bar{A}=(a,-u,-v,-w)Aˉ=(a,u,v,w ) , 则
( A ˉ ) n (\bar{A})^n(Aˉ)n isA n A^nAThe conjugate hypercomplex numbersof n .
(6) Let hypercomplex numberA = ( a , u , v , w ) A=(a,u,v,w)A=(a,u,v,w) f ( A ) = c 0 + c 1 A + ⋯ + c n A n f(A)=c_0+c_1A+\cdots+c_nA^n f(A)=c0+c1A++cnAn c j ( j = 0 , 1 , ⋯ n ) c_j (j=0,1,\cdots n) cj(j=0,1,n ) is a real number, thenf ( A ˉ ) f(\bar{A})f(Aˉ )isf ( A ) f(A)Conjugate hypercomplex numbers of f ( A ) .

For (6), there is a simple way to understand: we can put A ˉ \bar{A}AˉBring intof ( A ) f(A)f(A)中,可以得到:
f ( A ) = c 0 + c 1 A + ⋯ + c n A n = c 0 + c 1 ( a , u , v , w ) + ⋯ + c n ( a n , n u , n v , n w ) \begin{aligned} f(A)&=c_0+c_1A+\cdots+c_nA^n \\ &=c_0+ c_1(a,u,v,w)+\cdots+c_n(a^n,nu,nv,nw) \\ \end{aligned} f(A)=c0+c1A++cnAn=c0+c1(a,u,v,w)++cn(an,no , _nv,nw)
f ( A ˉ ) = c 0 + c 1 A ˉ + ⋯ + c n ( A ˉ ) n = c 0 + c 1 ( a , − u , − v , − w ) + ⋯ + c n ( a n , − n u , − n v , − n w ) \begin{aligned} f(\bar{A})&=c_0+c_1\bar{A}+\cdots+c_n(\bar{A})^n \\ &=c_0+ c_1(a,-u,-v,-w)+\cdots+c_n(a^n,-nu,-nv,-nw) \\ \end{aligned} f(Aˉ)=c0+c1Aˉ++cn(Aˉ)n=c0+c1(a,u,v,w)++cn(an,n u ,nv,nw)

It is not difficult to see that the absolute value of the quaternion is completely symmetrical, and the three angles are mutually opposite numbers. After spatial addition, f ( A ) f(A)f(A) f ( A ˉ ) f(\bar{A}) f(Aˉ )andAAA A ˉ \bar{A} AˉSimilarity is the conjugate relationship.

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Origin blog.csdn.net/xhyu61/article/details/130022589