MATH319 Slides

154 Example

Let

G⁢(s)=s2+5s2-s+1.

Then with s=(1-λ)/λ, we have

G⁢(s)=(1-λ)2+5⁢λ2(1-λ)2-λ⁢(1-λ)+λ2=6⁢λ2-2⁢λ+13⁢λ2-3⁢λ+1.

Then by the Euclidean algorithm

(24⁢λ-27)⁢(3⁢λ2-3⁢λ+1)-(12⁢λ-97)⁢(6⁢λ2-2⁢λ+1)=1

so letting λ=1/(1+s), we have P⁢X+Q⁢Y=1 with P,Q,X,Y∈𝒮

(22-2⁢s7⁢(1+s))⁢(3(1+s)2-31+s+1)-(3-9⁢s7⁢(1+s))⁢(6(1+s)2-21+s+1)=1.