Difference between revisions of "Forward regressive"
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==Related definitions== | ==Related definitions== | ||
| − | * | + | *We say that a function $p \colon \mathbb{T} \rightarrow \mathbb{R}$ is $\nu$-regressive if |
$$1-\nu(t)p(t) \neq 0$$ | $$1-\nu(t)p(t) \neq 0$$ | ||
for all $t \in \mathbb{T}_{\kappa}$. We use notation $\mathcal{R}_{\nu}$ for the set of $\nu$-regressive functions. | for all $t \in \mathbb{T}_{\kappa}$. We use notation $\mathcal{R}_{\nu}$ for the set of $\nu$-regressive functions. | ||
Revision as of 16:38, 21 September 2014
Let $\mathbb{T}$ be a time scale. Let $p \colon \mathbb{T} \rightarrow \mathbb{R}$. We say that $p$ is regressive if for all $t \in \mathbb{T}^{\kappa}$ $$1+\mu(t)p(t)\neq 0.$$ We let $\mathcal{R}(X,Y)$ denote the set of regressive functions $p \colon X \rightarrow Y$. Let $p,q \in \mathcal{R}$ and define the "circle plus" operation $\oplus \colon \mathcal{R} \times \mathcal{R} \rightarrow \mathcal{R}$ by the formula, for $t \in \mathbb{T}^{\kappa}$, $$(p \oplus q)(t) = p(t)+q(t)+\mu(t)p(t)q(t).$$ We define the inverse operation of $\oplus$ by the formula $$(\ominus p)(t) = -\dfrac{p(t)}{1+\mu(t)p(t)}.$$ The ordered pair $(\mathcal{R},\oplus)$ is an Abelian group with subtraction $$(p \ominus q)(t) = (p \oplus (\ominus q))(t) = \dfrac{p(t)-q(t)}{1+\mu(t)q(t)}.$$
Related definitions
- We say that a function $p \colon \mathbb{T} \rightarrow \mathbb{R}$ is $\nu$-regressive if
$$1-\nu(t)p(t) \neq 0$$ for all $t \in \mathbb{T}_{\kappa}$. We use notation $\mathcal{R}_{\nu}$ for the set of $\nu$-regressive functions.
- The set of positively regressive functions is
$$\mathcal{R}^+(\mathbb{T},X)=\{p \in \mathcal{R} \colon \forall t \in \mathbb{T}, 1+\mu(t)p(t)>0 \}.$$
- The set of negatively regressive functions is
$$\mathcal{R}^-(\mathbb{T},X)=\{p \in \mathcal{R} \colon \forall t \in \mathbb{T}, 1+\mu(t)p(t)<0 \}.$$
- Consider the dynamic equation $y^{\Delta \Delta}(t)+p(t)y^{\Delta}(t)+q(t)y(t)=f(t)$. We say this equation is regressive if $,p,q,f \in C_{rd}$ and for all $t \in \mathbb{T}$
$$1-\mu(t)p(t)+\mu^2(t)q(t)\neq 0.$$
- If we define the "circle dot" multiplication for $p \colon \mathbb{T} \rightarrow \mathbb{R}$ and $f \colon \mathbb{T} \rightarrow \mathbb{R}$ by
$$(f\odot p)(t) = \left\{ \begin{array}{ll} \dfrac{(1+\mu(t)p(t))^{f(t)}-1}{\mu(t)} &; \mu(t) > 0 \\ f(t) p(t) &; \mu(t)=0 \end{array} \right.$$ then if we restrict $f$ to real constant functions then the triple $(\mathcal{R}^+,\oplus,\odot)$ is a real vector space.
