Difference between revisions of "Delta derivative of squaring function"

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(Created page with "==Theorem== Let $\mathbb{T}$ be a time scale. The following formula holds for $f \colon \mathbb{T} \rightarrow \mathbb{R}$ defined by $f(t)=t^2$: $$f^{\Delta}(t)=t \sigma(...")
 
 
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==Theorem==
 
==Theorem==
 
Let $\mathbb{T}$ be a [[time scale]]. The following formula holds for $f \colon \mathbb{T} \rightarrow \mathbb{R}$ defined by $f(t)=t^2$:
 
Let $\mathbb{T}$ be a [[time scale]]. The following formula holds for $f \colon \mathbb{T} \rightarrow \mathbb{R}$ defined by $f(t)=t^2$:
$$f^{\Delta}(t)=t \sigma(t),$$
+
$$f^{\Delta}(t)=t+\sigma(t),$$
 
where $f^{\Delta}$ denotes the [[delta derivative]] and $\sigma$ denotes the [[forward jump]].
 
where $f^{\Delta}$ denotes the [[delta derivative]] and $\sigma$ denotes the [[forward jump]].
  

Latest revision as of 03:02, 19 December 2016

Theorem

Let $\mathbb{T}$ be a time scale. The following formula holds for $f \colon \mathbb{T} \rightarrow \mathbb{R}$ defined by $f(t)=t^2$: $$f^{\Delta}(t)=t+\sigma(t),$$ where $f^{\Delta}$ denotes the delta derivative and $\sigma$ denotes the forward jump.

Proof

References