Difference between revisions of "Unilateral Laplace transform of delta derivative"
From timescalewiki
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==Theorem== | ==Theorem== | ||
| − | If $\mathbb{T}$ is a time scale and $f \colon \mathbb{T} \rightarrow \mathbb{C}$, then | + | 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 | + | 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 $f$.
