exponential Times
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Infinite interlude

ρz: z3z2z1z0z+1
z: z2z1z0z+1z+2
λz: z1z0z+1z+2z+3
Here we plan to look at the way characteristic solutions arise and perhaps glean a more solid understanding of what is going on with the characteristic polynomial and so on. Above I begin by suggesting an analogy of the infection data sequence with a sort of infinite vector, which we have designated z. At the same time we have also defined two linear shift ‘operators’: λ,ρ. The λ operator shifts the sequence entries left, and ρ, right. The thicker, red border indicates the zero component of the respective sequences. It is perhaps also useful to give a component definition for the shift operators:
(λz)n=zn+1,(ρz)n=zn1
I hope it is obvious that λρz=ρλz=ιz=z, where ι is the “identity”, do-nothing operator. Even more compactly: λρ=ρλ=ι.
We also define the hopefully “obvious” vector operations of addition and scalar multiplication: (w+z)n=wn+zn,(kz)n=kzn.
In this notation our primordial recurrence relation zn=azn1 can be expressed as z=aρz. Using λ as the inverse of ρ, we also derive λz=az. In (linear) operator theory, if an operation on a certain vector gives back the same vector multiplied by a scalar, the vector is called an eigenvector of the operator, and the scalar an eigenvalue. Older English texts often used “characteristic" in place of the German loan of “eigen” (meaning “own” as in “one’s own”, i.e. an adjective rather than a verb). We therefore know a whole range of eigenvectors of λ,ρ: αn=an, with a variable. We have λα=aα and α=aρα ρα=α/a. It also useful to put this in the form (λa)α=0.
Let us look at a product of two operators: (λa)(λb). If we apply this to the vector α, I hope it is not too difficult to believe that we get zero. Also the same result is obtained when the operator product is applied to βn=bn. Combining, (λa)(λb)(Aα+Bβ)=0 with A,B arbitrary constants. From our previous work, it seems likely that this is the general solution. Let us write out the equation for general z:
(λa)(λb)z=(λ2(a+b)λ+ab)z=0
or in components: zn+2(a+b)zn+1+abzn=0zn+2=(a+b)zn+1abzn. This is our recurrence relation form zn=g1zn1+g2zn2, shifting back two places. So long as we define two consecutive components of z, we can build up a forward solution. The backward solution can be built by the relation zn=((a+b)zn+1zn+2)/ab. The same two initial components can be used. These components can be used to determine a consistent pair A,B. Thinking back, we are just seeing our characteristic polynomial with powers of λ rather than powers of a number.
So we have a general solution z=Aα+Bβ. Applying one of the factors (λa)z=B(ba)β, remembering that λα=aα,λβ=bβ. Looking at the zero component, z1az0=B(ba)b0=B(ba)B=(z1az0)/(ba). Similarly, A=(z1bz0)/(ab). We can think of the factors of the recurrence relation characteristic polynomial as being projection operators onto the opposite eigenvector:
Pαz=(λb)zab=Aα
Pβz=(λa)zba=Bβ
If we have a longer product (λa1)(λa2)z=i(λai)z=0 with general solution z=iAiαi, the projections are:
Pαiz=ji(λαj)zαiαj=Aiαi
Each (λαj) factor acts as a filter nulling out the unwanted components.
The assumption in all the above is that the αi values are distinct — new problems arise when multiple zeros occur in the characteristic polynomial . . .
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