Difference between revisions of "Delta derivative of unilateral convolution"

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(Created page with "==Theorem== If $f$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$ If $g$ is $\Delta$-differentiable, then...")
 
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==Theorem==
 
==Theorem==
 
If $f$ is [[delta derivative|$\Delta$-differentiable]], then
 
If $f$ is [[delta derivative|$\Delta$-differentiable]], then
$$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$
+
$$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g,$$
If $g$ is [[delta derivative|$\Delta$-differentiable]], then
+
and if $g$ is [[delta derivative|$\Delta$-differentiable]], then
$$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0).$$
+
$$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0),$$
 +
where $(f*g)$ denotes [[unilateral convolution]].
  
 
==Proof==
 
==Proof==

Revision as of 13:41, 20 January 2023

Theorem

If $f$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g,$$ and if $g$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0),$$ where $(f*g)$ denotes unilateral convolution.

Proof

References