Difference between revisions of "Derivative of delta cosine"

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==Theorem==
<strong>[[Derivative of delta cosine|Proposition]]:</strong> The following formula holds:
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The following formula holds:
 
$$\cos_p^{\Delta}(t,t_0)=-p(t)\sin_p(t,t_0),$$
 
$$\cos_p^{\Delta}(t,t_0)=-p(t)\sin_p(t,t_0),$$
 
where $\cos_p$ denotes the [[Delta cosine|$\Delta$-$\cos_p$]] function and $\sin_p$ denotes the [[Delta sine|$\Delta$-$\sin_p$]] function.
 
where $\cos_p$ denotes the [[Delta cosine|$\Delta$-$\cos_p$]] function and $\sin_p$ denotes the [[Delta sine|$\Delta$-$\sin_p$]] 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}
 
\cos_p^{\Delta}(t,t_0) &= \dfrac{1}{2} \dfrac{\Delta}{\Delta t} (e_{ip}(t,t_0) + e_{-ip}(t,t_0) \\
 
\cos_p^{\Delta}(t,t_0) &= \dfrac{1}{2} \dfrac{\Delta}{\Delta t} (e_{ip}(t,t_0) + e_{-ip}(t,t_0) \\
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\end{array}$$
 
\end{array}$$
 
as was to be shown. █  
 
as was to be shown. █  
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==References==

Revision as of 21:26, 9 June 2016

Theorem

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

Proof

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

References