36#
大 中
小 發表於 2014-5-1 11:34 PM (第 3642 天)
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For 11c, the more "classical" approach would be completing the square. This is what we normally do when there is no part c(i).
To simplify notation, let z=e^-t.
y=240/(2+z-2z^2)
=-120/(z^2-0.5z-1)
=-120/[(z-0.25)^2-(17/16)], where z=e^-t
Note that z=e^(-alpha) and z=e^(alpha-3) gives the same value of y.
The y-z graph has an axis of symmetry z=0.25
Hence,
[e^(-alpha)+e^(alpha-3)]/2=0.25
e^(-alpha)+e^(alpha-3)=0.5
(e^-3)(e^alpha)^2-0.5e^alpha+1=0
This is a QE in e^alpha.
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Now, go back to the approach in this question.
To simplify notation, let z=e^-t.
y=240/(2+z-2z^2)
(2+z-2z^2)=240/y
2z^2-z+(240/y)-2=0
z^2-0.5z+(120/y)-1=0, where z=e^-t ------- (E)
Choose y to be the value when z=e^-alpha.
Note that this is also the value when z=e^(alpha-3).
Hence, z=e^-alpha and z=e^(alpha-3) are the roots of equation (E).
By sum of roots,
e^(-alpha)+e^(alpha-3)=0.5
This is the same equation we get above.
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Note: To address the concern of D.Y., I have chosen a particular value of y, which becomes a constant.
