BigBugBuzz

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2 years, 118 days

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These are questions asked by BigBugBuzz

I am attempting to make the transition from Mathematica to Maple. I that regard, I would like to know how I would implement something like Mathematica's "Conditioned" in Maple. For example, how would I implement the example given in the diagram shown below involving the Poisson Distribution?

I am trying to solve an equation with respect to the variable w. However, although there seems to be a solution (see plot indicating a root), Maple produces the wron solution (0, when P = 1/2 and d = 1/10):


 

equations := S-P*S*((ln(-2*d*w+P+d)-ln(P-d))*(P-d)/(-1+w)+(P+d)*(ln(-P-d)-ln(d*(2*w-1)-P))/w)/(2*d^2) = 0

S-(1/2)*P*S*((ln(-2*d*w+P+d)-ln(P-d))*(P-d)/(-1+w)+(P+d)*(ln(-P-d)-ln(d*(2*w-1)-P))/w)/d^2 = 0

(1)

variables := w

w

(2)

solutions := solve(equations, variables)

(1/2)*(P+d-exp(RootOf(-P^2*exp(_Z)*ln(1/(P+d))-P*d*exp(_Z)*ln(1/(P+d))+P^3*ln(1/(P+d))-P*d^2*ln(1/(P+d))-ln(P-d)*P^2*exp(_Z)+ln(P-d)*P*d*exp(_Z)-2*P*d*exp(_Z)*_Z+ln(P-d)*P^3-ln(P-d)*P*d^2+d*(exp(_Z))^2-2*P*d*exp(_Z)+P^2*d-d^3)))/d

(3)

eval({solutions}, [P = 1/2, d = 1/10])

{3-5*exp(RootOf(-25*(exp(_Z))^2+75*exp(_Z)*ln(5/3)+50*ln(2/5)*exp(_Z)+25*exp(_Z)*_Z+25*exp(_Z)-30*ln(5/3)-30*ln(2/5)-6))}

(4)

evalf({%})

{{0.}}

(5)

plot(eval(S*(-P*(-2*d*w+P+d)*ln(-2*d*w+P+d)+P*w*(P-d)*ln(P-d)-(-1+w)*(P*(P+d)*ln(P+d)-2*d^2*w))/(2*d^2*(-1+w)*w), {P = 1/2, S = 1, d = 1/10}), w = 0 .. 1)

 

``


 

Download MaplePrimes_03072018.mw

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