Difference between revisions of "Derivative of Delta sine"

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(Created page with "<div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> <strong>Proposition:</strong> The following formula holds: $$\sin_p^{\Del...")
 
 
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==Theorem==
<strong>[[Derivative of Delta sine|Proposition]]:</strong> The following formula holds:
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The following formula holds:
 
$$\sin_p^{\Delta}(t,t_0)=p(t)\cos_p(t,t_0),$$
 
$$\sin_p^{\Delta}(t,t_0)=p(t)\cos_p(t,t_0),$$
where $\sin_p$ denotes the [[Delta sine|$\Delta \sin_p$]] function and $\cos_p$ denotes the [[Delta cosine|$\Delta\cos_p$]] function.
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where $\sin_p$ denotes the [[Delta sine|$\Delta$-$\sin_p$]] function and $\cos_p$ denotes the [[Delta cosine|$\Delta$-$\cos_p$]] function.
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<strong>Proof:</strong> Compute
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==Proof==
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Compute
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
 
\sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\
 
\sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\
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\end{array}$$
 
\end{array}$$
 
as was to be shown. █   
 
as was to be shown. █   
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==References==
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[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 21:28, 9 June 2016

Theorem

The following formula holds: $$\sin_p^{\Delta}(t,t_0)=p(t)\cos_p(t,t_0),$$ where $\sin_p$ denotes the $\Delta$-$\sin_p$ function and $\cos_p$ denotes the $\Delta$-$\cos_p$ function.

Proof

Compute $$\begin{array}{ll} \sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\ &= \dfrac{ip}{2i} ( e_{ip}(t,t_0) + e_{-ip}(t,t_0) ) \\ &= \dfrac{1}{2} (e_{ip}(t,t_0)+e_{-ip}(t,t_0)) \\ &= p\cos_p(t,t_0), \end{array}$$ as was to be shown. █

References