Difference between revisions of "Diamond alpha derivative"

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The $\Diamond_{\alpha}$-derivative of a function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is formally defined to be the number $f^{\Diamond_{\alpha}}(t)$ such that for all $\epsilon>0$ there is a neighborhood $U$ of $t$ such that for all $s \in U$,
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Let $\mathbb{T}$ be a [[time scale]] and let $0\leq \alpha \leq 1$. The $\Diamond_{\alpha}$-derivative of a function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is formally defined to be the number $f^{\Diamond_{\alpha}}(t)$ such that for all $\epsilon>0$ there is a neighborhood $U$ of $t$ such that for all $s \in U$,
 
$$\left| \alpha[f^{\sigma}(t)-f(s)]\eta_{ts} + (1-\alpha)[f^{\rho}(t)-f(s)]\mu_{ts}-f^{\Diamond_{\alpha}}\mu_{ts}\eta_{ts} \right| < \epsilon |\mu_{ts}\eta_{ts}|,$$
 
$$\left| \alpha[f^{\sigma}(t)-f(s)]\eta_{ts} + (1-\alpha)[f^{\rho}(t)-f(s)]\mu_{ts}-f^{\Diamond_{\alpha}}\mu_{ts}\eta_{ts} \right| < \epsilon |\mu_{ts}\eta_{ts}|,$$
 
where $\mu_{ts}=\sigma(t)-s$ and $\eta_{ts}=\rho(t)-s$.
 
where $\mu_{ts}=\sigma(t)-s$ and $\eta_{ts}=\rho(t)-s$.
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<strong>Theorem:</strong> Let $0 \leq \alpha \leq 1$. If $f$ is both $\Delta$ and $\nabla$ differentiable at $t \in \mathbb{T}$ then $f$ is $\Diamond_{\alpha}$-differentiable at t and  
 
<strong>Theorem:</strong> Let $0 \leq \alpha \leq 1$. If $f$ is both $\Delta$ and $\nabla$ differentiable at $t \in \mathbb{T}$ then $f$ is $\Diamond_{\alpha}$-differentiable at t and  
 
$$f^{\Diamond_{\alpha}}(t)=\alpha f^{\Delta}(t) + (1-\alpha)f^{\nabla}(t).$$
 
$$f^{\Diamond_{\alpha}}(t)=\alpha f^{\Delta}(t) + (1-\alpha)f^{\nabla}(t).$$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> Let $f \colon \mathbb{T} \rightarrow \mathbb{C}$. If $f$ is $\Diamond_{\alpha}$-differentiable at $t$, then $f$ is [[continuous]] at $t$.
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> Let $f \colon \mathbb{T} \rightarrow \mathbb{C}$. If $f$ is $\Diamond_{\alpha}$-differentiable at $t$, then $f$ is both [[Delta derivative|$\Delta$-differentiable]] and [[Nabla derivative|$\nabla$-differentiable]] at $t$.
 
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<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  

Latest revision as of 08:32, 12 April 2015

Let $\mathbb{T}$ be a time scale and let $0\leq \alpha \leq 1$. The $\Diamond_{\alpha}$-derivative of a function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is formally defined to be the number $f^{\Diamond_{\alpha}}(t)$ such that for all $\epsilon>0$ there is a neighborhood $U$ of $t$ such that for all $s \in U$, $$\left| \alpha[f^{\sigma}(t)-f(s)]\eta_{ts} + (1-\alpha)[f^{\rho}(t)-f(s)]\mu_{ts}-f^{\Diamond_{\alpha}}\mu_{ts}\eta_{ts} \right| < \epsilon |\mu_{ts}\eta_{ts}|,$$ where $\mu_{ts}=\sigma(t)-s$ and $\eta_{ts}=\rho(t)-s$.

Properties

Theorem: Let $0 \leq \alpha \leq 1$. If $f$ is both $\Delta$ and $\nabla$ differentiable at $t \in \mathbb{T}$ then $f$ is $\Diamond_{\alpha}$-differentiable at t and $$f^{\Diamond_{\alpha}}(t)=\alpha f^{\Delta}(t) + (1-\alpha)f^{\nabla}(t).$$

Proof:

Theorem: Let $f \colon \mathbb{T} \rightarrow \mathbb{C}$. If $f$ is $\Diamond_{\alpha}$-differentiable at $t$, then $f$ is continuous at $t$.

Proof:

Theorem: Let $f \colon \mathbb{T} \rightarrow \mathbb{C}$. If $f$ is $\Diamond_{\alpha}$-differentiable at $t$, then $f$ is both $\Delta$-differentiable and $\nabla$-differentiable at $t$.

Proof:

Problem with the $\Diamond_{\alpha}$-integral

References

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[2]