Difference between revisions of "Derivation of delta sin sub p for T=Z"

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$$\begin{array}{ll}
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Using the properties of [[Delta exponential|$e_p$]], it is clear that $\sin_p(t,s) = \dfrac{1-1}{2i} = 0$. Furthermore if $t>s$, then
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\sin_p(t,s) &= \dfrac{e_{ip}(t,s)-e_{-ip}(t,s)}{2i} \\
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$$\sin_p(t,s) = \dfrac{e_{ip}(t,s)-e_{-ip}(t,s)}{2i} = \dfrac{\displaystyle\prod_{k=s}^{t-1}1+ip(k) - \displaystyle\prod_{k=s}^{t-1}1-ip(k)}{2i}.$$
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&= \dfrac{\displaystyle\prod_{k=s}^{t-1}1+ip(k) - \displaystyle\prod_{k=s}^{t-1}1-ip(k)}{2i}
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If $t<s$ then
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\end{array}$$
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$\sin_p(t,s) = \dfrac{e_{ip}(t,s)-e_{-ip}(t,s)}{2i} = \dfrac{\displaystyle\prod_{k=t}^{s-1} \frac{1}{1+ip(k)} - \displaystyle\prod_{k=t}^{s-1} \frac{1}{1-ip(k)}}{2i}.$$

Revision as of 20:43, 29 April 2015

Using the properties of $e_p$, it is clear that $\sin_p(t,s) = \dfrac{1-1}{2i} = 0$. Furthermore if $t>s$, then $$\sin_p(t,s) = \dfrac{e_{ip}(t,s)-e_{-ip}(t,s)}{2i} = \dfrac{\displaystyle\prod_{k=s}^{t-1}1+ip(k) - \displaystyle\prod_{k=s}^{t-1}1-ip(k)}{2i}.$$ If $t<s$ then $\sin_p(t,s) = \dfrac{e_{ip}(t,s)-e_{-ip}(t,s)}{2i} = \dfrac{\displaystyle\prod_{k=t}^{s-1} \frac{1}{1+ip(k)} - \displaystyle\prod_{k=t}^{s-1} \frac{1}{1-ip(k)}}{2i}.$$

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