Difference between revisions of "Unilateral Laplace transform of delta derivative"

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==Theorem==
 
==Theorem==
If $\mathbb{T}$ is a time scale, then
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If $\mathbb{T}$ is a time scale and $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is [[delta derivative|delta differentiable]], then
 
$$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$
 
$$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$
where $\mathscr{L}_{\mathbb{T}}$ denotes the [[unilateral Laplace transform]] and $f^{\Delta}$ denotes the [[delta derivative]] of $f$.
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where $\mathscr{L}_{\mathbb{T}}$ denotes the [[unilateral Laplace transform]] and $f^{\Delta}$ denotes the delta derivative of $f$.
  
 
==Proof==
 
==Proof==

Latest revision as of 15:11, 21 January 2023

Theorem

If $\mathbb{T}$ is a time scale and $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is delta differentiable, then $$\mathscr{L}_{\mathbb{T}}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z),$$ where $\mathscr{L}_{\mathbb{T}}$ denotes the unilateral Laplace transform and $f^{\Delta}$ denotes the delta derivative of $f$.

Proof

References