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.