- #1
tim_lou
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I haven't taken any abstract algebra course so I do not know if this is the right section to post this question in.
Anyway, I am learning differential equation right now and my prof. recently showed factorization of differential operator.
For example, let D be the differential operator, he then said that any thing in the form:
[tex](D^n+A_{n-1}D^{n-1}+...+A_1D+A_0I)f[/tex]
where I is the identity can be factorized into:
[tex](D-z_1I)(D-z_2I)...(D-z_nI)f[/tex]
where multiplication means composition of operator
I am curious to know what kind of algebraic structure has this property... and if I have a random operator, let's say F, what axioms does F have to satisfy in order for
[tex]F^n+A_{n-1}F^{n-1}+...+A_1F+A_0I[/tex]
to be factorizable like a polynomial?
more specifically, If I define an operator [tex]\Delta x(n)=x(n+1)-x(n)[/tex]
how can I show (what axioms does Delta have to satisfy) that polynomials in [tex]\Delta[/tex] can be factorized?
(I'm reading a book on differences equation... so I would like to know)
Anyway, I am learning differential equation right now and my prof. recently showed factorization of differential operator.
For example, let D be the differential operator, he then said that any thing in the form:
[tex](D^n+A_{n-1}D^{n-1}+...+A_1D+A_0I)f[/tex]
where I is the identity can be factorized into:
[tex](D-z_1I)(D-z_2I)...(D-z_nI)f[/tex]
where multiplication means composition of operator
I am curious to know what kind of algebraic structure has this property... and if I have a random operator, let's say F, what axioms does F have to satisfy in order for
[tex]F^n+A_{n-1}F^{n-1}+...+A_1F+A_0I[/tex]
to be factorizable like a polynomial?
more specifically, If I define an operator [tex]\Delta x(n)=x(n+1)-x(n)[/tex]
how can I show (what axioms does Delta have to satisfy) that polynomials in [tex]\Delta[/tex] can be factorized?
(I'm reading a book on differences equation... so I would like to know)
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