MHB Forward Unit Push Operator Equation

[topsquark]

Hopefully the symbols I am using are standard. I will define them upon request.

I have a theorem that says, given a difference equation [math]\left ( \sum_{j = 0}^m a_j E^j \right ) y_n = \alpha ^n F(n)[/math], we can define a polynomial function [math]\phi (E) = \sum_{j = 0}^m a_j E^j [/math] such that [math]\phi (E) y_n = \alpha ^n F(n)[/math]. I can follow a proof to the following result:
(1) [math]\phi (E) \left ( \alpha ^n F(n) \right ) = \alpha ^n \phi ( \alpha E ) F(n)[/math]

The notes then go on to say, "therefore"
(2) [math]\dfrac{1}{ \phi (E) } \left ( \alpha ^n F(n) \right ) = \alpha ^n \dfrac{1}{ \phi ( \alpha E )} F(n)[/math]

Now, the particular solution to the difference equation is written as [math]y_p = \dfrac{1}{ \phi (E) } \left ( \alpha ^n F(n) \right )[/math] and I can use (2) to evaluate this and get the correct result. So I know that (2) is right, without any typos. But how do I get from (1) to (2)?

More details upon request.

-Dan

 

[topsquark]

I might have what I would call a "formal" solution.

[math]\dfrac{1}{ \phi (E) } \alpha ^n F(n) = \dfrac{1}{ \phi (E) } \dfrac{1}{ \dfrac{1}{ \alpha ^n F(n) } } = \dfrac{1}{ \phi (E) } \dfrac{1}{ \alpha ^{-n} F^{-1}(n)}[/math]

Now, I see no reason why we can't use a negative exponent in the equation (1) in the OP and [math]F^{-1}(n)[/math] is just a function of n. So we can apply (1):
[math]\dfrac{1}{ \phi (E) } \dfrac{1}{ \alpha ^{-n} F^{-1}(n)} = \dfrac{1}{ \alpha ^{-n} \phi ( \alpha E ) F^{-1}(n)} = \alpha ^n \dfrac{1}{ \phi ( \alpha E) } F(n)[/math]

So we get
[math]\dfrac{1}{ \phi (E) } \alpha ^n F(n) = \alpha ^n \dfrac{1}{ \phi ( \alpha E) } F(n)[/math]

as required.

Now, I call this formal because I haven't really "gotten into the gears" of the equation. I've used it but haven't fully explored it so there may be some surprises I'm not aware of. The other reason is that the derivation requires F(n) to never be 0, which wasn't in the notes that I downloaded. Granted, the notes weren't expected to be a complete introduction and they were shot through with typos. So F(n) not zero may well be a requirement on this equation. I'll have to mess with it.

Any thoughts?

-Dan

 

[GJA]

Hi Dan,

I like the symbolic argument above. Here is something I scratched out in an attempt to avoid the $F^{-1}(n)$ term:

By definition of the inverse,
$$\frac{1}{\phi(\alpha E)}\phi(\alpha E) F(n) = F(n)$$
From (1) we get
$$\left[\frac{1}{\phi(\alpha E)}\alpha^{-n}\phi(E)\alpha^{n}\right] F(n) = F(n)$$
This implies
$$\left[\alpha^{-n}\phi(E)\alpha^{n}\right]F(n) = \phi(\alpha E)F(n)$$
Since $F(n)$ is an arbitrary function of $n$, it follows that
$$\alpha^{-n}\phi(E)\alpha^{n} = \phi(\alpha E)$$
Thus,
$$1 = \alpha^{-n}\phi(E)\alpha^{n}\frac{1}{\phi(\alpha E)},$$
from which it follows
$$\frac{1}{\phi(E)}\alpha^{n} = \alpha^{n}\frac{1}{\phi(\alpha E)}$$
The above, when applied to $F(n)$, is the desired result in (2).

 

[topsquark]

Hi Dan,

I like the symbolic argument above. Here is something I scratched out in an attempt to avoid the $F^{-1}(n)$ term:

By definition of the inverse,
$$\frac{1}{\phi(\alpha E)}\phi(\alpha E) F(n) = F(n)$$
From (1) we get
$$\left[\frac{1}{\phi(\alpha E)}\alpha^{-n}\phi(E)\alpha^{n}\right] F(n) = F(n)$$
This implies
$$\left[\alpha^{-n}\phi(E)\alpha^{n}\right]F(n) = \phi(\alpha E)F(n)$$
Since $F(n)$ is an arbitrary function of $n$, it follows that
$$\alpha^{-n}\phi(E)\alpha^{n} = \phi(\alpha E)$$
Thus,
$$1 = \alpha^{-n}\phi(E)\alpha^{n}\frac{1}{\phi(\alpha E)},$$
from which it follows
$$\frac{1}{\phi(E)}\alpha^{n} = \alpha^{n}\frac{1}{\phi(\alpha E)}$$
The above, when applied to $F(n)$, is the desired result in (2).

I like it! Yes, I have solved a couple of systems and the [math]F(n) \neq 0[/math] condition does not seem to be a problem.

Thanks!

-Dan

 

[topsquark]

Well, I just found about half the derivation in an (old) online text that was missing some pages. It was far more direct.

[math]\dfrac{1}{ \phi (E)} \alpha ^n F(n) = \dfrac{1}{ \phi (E) \alpha ^{-n} } F(n) = \dfrac{1}{ \alpha ^{-n} \phi ( \alpha E)} F(n) = \alpha ^n \dfrac{1}{ \phi ( \alpha E)} F(n)[/math]

I hadn't thought about separating the factors but [math]E^j ( \alpha ^n F(n) ) = \alpha ^{n + j} E^j (F(n)) = \alpha ^{n + j} F(n + j)[/math] so it makes sense.

-Dan