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Stock Speculation: In a classic paper on the theory of conflict, L.F. Richardson claimed that the proportion p of a population a...Asked by Wade Wilson
Stock Speculation: In a classic paper on the theory of conflict, L.F. Richardson claimed that the proportion p of a population advocating war or other aggressive action at a time t satisfies
p(t) = (Ce^kt) / (1 + Ce^kt)
where k and C are positive constants. Speculative day-trading in the stock market can be regarded as “aggressive action.” Suppose that initially, (1/200) of total daily market volume is attributed to day-trading and that 4 weeks later, the proportion is (1/100). When will the proportion be increasing most rapidly? What will the proportion be at that time?
p(t) = (Ce^kt) / (1 + Ce^kt)
where k and C are positive constants. Speculative day-trading in the stock market can be regarded as “aggressive action.” Suppose that initially, (1/200) of total daily market volume is attributed to day-trading and that 4 weeks later, the proportion is (1/100). When will the proportion be increasing most rapidly? What will the proportion be at that time?
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Answered by
Luisa
El valor del impuesto de un producto equivale a los 10/11 del costo de importación. Si el producto en cuestión vale 22 entonce los impuestos valen:
(de su respuesta en forma decimal)
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Answered by
Steve
surely as a calculus student, you can determine C and k:
at t=0,
C/(1+C) = 1/200
C = 1/199
Since you say daily volume, I assume t is in days, so
at t=28,
(1/199)e^28k/(1+(1/199)e^28k) = 1/100
k ≈ 0.025 = 1/40
p(t) = 1/199 e^(t/40)/(1+1/199 e^(t/40))
increasing most rapidly when dp/dt has a max. That is, when p" = 0
p' = 199/40 e^(t/40) / (1+1/199 e^(t/40))^2
p" = c(e^(t/40)-199)/(1+1/199 e^(t/40))^3
where c is a bunch of nonzero junk
p' has a max at 40 log199 ≈ 212 days
now just find p(212)
for the derivatives, wolframalpha is your friend:
http://www.wolframalpha.com/input/?i=p%28t%29+%3D+1%2F199+e^%28t%2F40%29%2F%281%2B1%2F199+e^%28t%2F40%29%29+for+0%3C%3Dt%3C%3D400
http://www.wolframalpha.com/input/?i=199%2F40+e^%28t%2F40%29+%2F+%281%2B1%2F199+e^%28t%2F40%29%29^2
at t=0,
C/(1+C) = 1/200
C = 1/199
Since you say daily volume, I assume t is in days, so
at t=28,
(1/199)e^28k/(1+(1/199)e^28k) = 1/100
k ≈ 0.025 = 1/40
p(t) = 1/199 e^(t/40)/(1+1/199 e^(t/40))
increasing most rapidly when dp/dt has a max. That is, when p" = 0
p' = 199/40 e^(t/40) / (1+1/199 e^(t/40))^2
p" = c(e^(t/40)-199)/(1+1/199 e^(t/40))^3
where c is a bunch of nonzero junk
p' has a max at 40 log199 ≈ 212 days
now just find p(212)
for the derivatives, wolframalpha is your friend:
http://www.wolframalpha.com/input/?i=p%28t%29+%3D+1%2F199+e^%28t%2F40%29%2F%281%2B1%2F199+e^%28t%2F40%29%29+for+0%3C%3Dt%3C%3D400
http://www.wolframalpha.com/input/?i=199%2F40+e^%28t%2F40%29+%2F+%281%2B1%2F199+e^%28t%2F40%29%29^2
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