If`D=d i ag[d_1, d_2, d_n]` , then prove that `f(D)=d i ag[f(d_1),f(d_2), ,f(d_n)],w h e r ef(x)` is a polynomial with scalar coefficient.
If`D=d i ag[d_1, d_2, d_n]` , then prove that `f(D)=d i ag[f(d_1),f(d_2), ,f(d_n)],w h e r ef(x)` is a polynomial with scalar coefficient.
4 views
1 Answers
Let
`f(x)=a_(0)+a_(1)x+a_(2)x^(2)+...+a_(n)x^(n)`
`implies f(D)=a_(0)I+a_(1)D+a_(2)D^(2)+...+a_(n)D^(n)`
`=a_(0)xx"diag. "(1, 1, ... , 1)+a_(1)xx"diag. "(d_(1), d_(2), ..., d_(n))+a_(2)xx"diag. "(d_(1)^(2), d_(2)^(2), ..., d_(n)^(2))`
...
...
...
`="diag."(a_(0)+a_(1)d_(1)+a_(2)d_(1)^(2)+...+a_(n)d_(1)^(n), a_(0)+a_(1)d_(2)+a_(2)d_(2)^(2)+...+a_(n)d_(2)^(n), a_(0)+a_(1)d_(3)+a_(2)d_(3)^(2)+...+ a_(n)d_(3)^(n)`,
...
...
`a_(0)+a_(1)d_(n)+a_(2)d_(n)^(2)+...+a_(n)d_(n)^(2))`
`="diag."(f(d_(1)), f(d_(2)), ..., f(d_(n)))`
4 views
Answered