MATH319 Slides

136 Partial fractions

where here q∈𝐂 is a constant since f∈𝒮 is proper, and all the λj have ℜ⁡λj<0. So we can take linear combinations of such f, and stay in 𝒮. Also f1⁢(s)+f2⁢(s)=(g1⁢(s)⁢h2⁢(s)+h1⁢(s)⁢g2⁢(s))/h1⁢(s)⁢h2⁢(s)∈𝒮.

(C) Commutativity of multiplication follows from the corresponding property for polynomials;

(ID2) likewise;

(Diff) Also, we can differentiate

f′⁢(s)=∑j=1N-nj⁢aj(s-λj)nj+1,

and the poles are at λj in open left half plane.

(ii) Whereas 1/(s+1) belongs to 𝒮, the inverse s+1 is not proper, hence not in 𝒮.