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: $$\sin_p^{\Del...") |
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− | + | ==Theorem== | |
− | + | The following formula holds: | |
$$\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),$$ | ||
− | where $\sin_p$ denotes the [[Delta sine|$\Delta \sin_p$]] function and $\cos_p$ denotes the [[Delta cosine|$\Delta\cos_p$]] function. | + | where $\sin_p$ denotes the [[Delta sine|$\Delta$-$\sin_p$]] function and $\cos_p$ denotes the [[Delta cosine|$\Delta$-$\cos_p$]] function. |
− | + | ||
− | + | ==Proof== | |
+ | Compute | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
\sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\ | \sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\ | ||
Line 12: | Line 13: | ||
\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
− | + | ||
− | + | ==References== | |
+ | |||
+ | [[Category:Theorem]] | ||
+ | [[Category:Proven]] |
Latest revision as of 21:28, 9 June 2016
Theorem
The following formula holds: $$\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. █