Simbolic Toolbox: how to avoid conj() in matrix *

Question

Igor 2011-12-28

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/24846-simbolic-toolbox-how-to-avoid-conj-in-matrix

回答： moh ab 2019-6-26

N=10;x=0:(N-1);
syms real r;
a=r.^(x),
b=r.^(-x),
M=b'*a

Result is

a =
[ 1, r, r^2, r^3, r^4, r^5, r^6, r^7, r^8, r^9]
b =
[ 1, 1/r, 1/r^2, 1/r^3, 1/r^4, 1/r^5, 1/r^6, 1/r^7, 1/r^8, 1/r^9]
M =
[           1,           r,           r^2,           r^3,           r^4,           r^5,           r^6,           r^7,           r^8,           r^9]
[   1/conj(r),   r/conj(r),   r^2/conj(r),   r^3/conj(r),   r^4/conj(r),   r^5/conj(r),   r^6/conj(r),   r^7/conj(r),   r^8/conj(r),   r^9/conj(r)]
[ 1/conj(r)^2, r/conj(r)^2, r^2/conj(r)^2, r^3/conj(r)^2, r^4/conj(r)^2, r^5/conj(r)^2, r^6/conj(r)^2, r^7/conj(r)^2, r^8/conj(r)^2, r^9/conj(r)^2]
[ 1/conj(r)^3, r/conj(r)^3, r^2/conj(r)^3, r^3/conj(r)^3, r^4/conj(r)^3, r^5/conj(r)^3, r^6/conj(r)^3, r^7/conj(r)^3, r^8/conj(r)^3, r^9/conj(r)^3]
[ 1/conj(r)^4, r/conj(r)^4, r^2/conj(r)^4, r^3/conj(r)^4, r^4/conj(r)^4, r^5/conj(r)^4, r^6/conj(r)^4, r^7/conj(r)^4, r^8/conj(r)^4, r^9/conj(r)^4]
[ 1/conj(r)^5, r/conj(r)^5, r^2/conj(r)^5, r^3/conj(r)^5, r^4/conj(r)^5, r^5/conj(r)^5, r^6/conj(r)^5, r^7/conj(r)^5, r^8/conj(r)^5, r^9/conj(r)^5]
[ 1/conj(r)^6, r/conj(r)^6, r^2/conj(r)^6, r^3/conj(r)^6, r^4/conj(r)^6, r^5/conj(r)^6, r^6/conj(r)^6, r^7/conj(r)^6, r^8/conj(r)^6, r^9/conj(r)^6]
[ 1/conj(r)^7, r/conj(r)^7, r^2/conj(r)^7, r^3/conj(r)^7, r^4/conj(r)^7, r^5/conj(r)^7, r^6/conj(r)^7, r^7/conj(r)^7, r^8/conj(r)^7, r^9/conj(r)^7]
[ 1/conj(r)^8, r/conj(r)^8, r^2/conj(r)^8, r^3/conj(r)^8, r^4/conj(r)^8, r^5/conj(r)^8, r^6/conj(r)^8, r^7/conj(r)^8, r^8/conj(r)^8, r^9/conj(r)^8]
[ 1/conj(r)^9, r/conj(r)^9, r^2/conj(r)^9, r^3/conj(r)^9, r^4/conj(r)^9, r^5/conj(r)^9, r^6/conj(r)^9, r^7/conj(r)^9, r^8/conj(r)^9, r^9/conj(r)^9]