- #1

- 682

- 1

## Main Question or Discussion Point

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)

Last edited: