weidade37211

And the use of single quote in Maple 2019 and 2017 are giving different results. that is, the plot (note, not plot3d) results in the attched script by Maple 2017 and 2019 are different...

obtain the second-last result ...

Asked: weidade37211 110 Product: Maple

November 15 2019

1 3

Just a quick question, I know we can use "%" to obtain the las result in maple. what if I want to obtain the second last result? or in general the n-last result ? Thx

fsolve a equation set with 5 variables.....

Asked: weidade37211 110 Product: Maple

November 14 2019

0 1

I am trying to use Maple to solve a set of 5 equations, but cannot get a solution. Or there is no solution??

Any help? (Yes, the L function is a likelihood function and I am doing MLE for 5 variables..)

(1)

k := Matrix(3, 4, {(1, 1) = 11.0, (1, 2) = 7.0, (1, 3) = 7.0, (1, 4) = 11.0, (2, 1) = 5.0, (2, 2) = 7.0, (2, 3) = 12.0, (2, 4) = 12.0, (3, 1) = 1., (3, 2) = 9.0, (3, 3) = 7.0, (3, 4) = 19.0})

Matrix(%id = 18446746279852723246)

(2)

(3)

(4)

"l(N1,M1,lambda,phi,r):=product((phi*(N1-'B[i]'+r*'A[i]'))^('k[i][1]')*(1/(2)*lambda*(M1-'A[i]'))^('k[i][2]'+'k[i][3]')*(1-phi*(N1-'B[i]'+r*'A[i]')-lambda*(M1-'A[i]'))^('k[i][4]') ,i=1..(h-1))"

proc (N1, M1, lambda, phi, r) options operator, arrow, function_assign; product((phi*(N1-'B[i]'+r*'A[i]'))^'k[i][1]'*((1/2)*lambda*(M1-'A[i]'))^('k[i][2]'+'k[i][3]')*(1-phi*(N1-'B[i]'+r*'A[i]')-lambda*(M1-'A[i]'))^'k[i][4]', i = 1 .. h-1) end proc

(5)

fsolve({diff(ln(l(N1, M1, lambda, phi, r)), M1) = 0, diff(ln(l(N1, M1, lambda, phi, r)), N1) = 0, diff(ln(l(N1, M1, lambda, phi, r)), lambda) = 0, diff(ln(l(N1, M1, lambda, phi, r)), phi) = 0, diff(ln(l(N1, M1, lambda, phi, r)), r) = 0}, {M1, N1, lambda, phi, r}, N1 = 0 .. infinity, M1 = 0 .. infinity, lambda = 0 .. 1, phi = 0 .. 1, r = 0 .. 1)