Difference between revisions of "Derivative of delta cosine"

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(Proof)
 
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The following formula holds:
 
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 $\cos^{\Delta}_p$ denotes the [[Delta derivative|delta derivative]] of the [[Delta cosine|delta cosine]] function and $\sin_p$ denotes the [[Delta sine|delta sine]] function.
  
 
==Proof==
 
==Proof==
 
Compute
 
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} \Big(e_{ip}(t,t_0) + e_{-ip}(t,t_0) \Big) \\
 
&= \dfrac{ip}{2} (e_{ip}-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)) \\
 
&= -\dfrac{p}{2i} (e_{ip}-e_{-ip}(t,t_0)) \\

Latest revision as of 01:51, 6 February 2023

Theorem

The following formula holds: $$\cos_p^{\Delta}(t,t_0)=-p(t)\sin_p(t,t_0),$$ where $\cos^{\Delta}_p$ denotes the delta derivative of the delta cosine function and $\sin_p$ denotes the delta sine function.

Proof

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

References