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发表于 18-11-2004 10:59 AM
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[size=-1]because sinhBK = (1/2)*(e^{BK} - e^{-BK}), and
Z{e^{-BK}} = z/(z - e^{-B}),
Z{e^{BK}} = z/(z - e^{B}).
Therefore,
Z{sinhBK} = (1/2)*Z{e^{BK} - e^{-BK}}
= (1/2)*[z/(z - e^{B}) - z/(z - e^{-B})]
= (1/2)*[(z^2 - z^2 + z*(e^{B}-e^{-B})) / (z^2 + 1 - z*(e^{B}+e^{-B}))]
= z*sinhB / (z^2 + 1 - z*2coshB)
[ Last edited by fadeev_popov on 18-11-2004 at 11:11 AM ] |
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