litun

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These are questions asked by litun

I am trying to extract the coefficients of z from its series expansion. In two cases I succeed in finding the coefficients, but in the last one I fail to get the correct coefficients. Some garbage value is obtained. What is the reason behind this? I have attached my maple program.

 

 

test.mw

I have the following function

where A,B,Ψ, K1,K2,K3,α,β are all constants.

How to find the value of m for which the above expression is 0 or approximate to 0 for different values fo the constants.

e.g., Fixing all the parameters except A, I want to find the values of m for different values of A. How to do that in maple?

 

I have the following transition rate matrix. Upto m the matrix follows a patern and after that the patern changes but repetative structure is there upto n. In this problem m=4, n=9. How to write these entries using for loop in maple?

Matrix.pdf

I have the following mathematical expression

where a=0.2, b=0.09, c=0.57, p=0.3 and q=1-p=0.7, Now i want to find the value of n for which the value of the expression will be 1, i.e., for what value of n , A_n will be 1?

test.mw

restart; with(LinearAlgebra)

``

dF := -.525*exp(-7*t)+2.625*exp(-3*t)+.8*exp(-4*t);

-.525*exp(-7*t)+2.625*exp(-3*t)+.8*exp(-4*t)

(1)

``

e3 := `<,>`(1, 1, 1); E := proc (m) options operator, arrow; IdentityMatrix(m) end proc; beta := `<|>`(.1, .6, .3); S := `<|>`(`<,>`(-3, 1, 1), `<,>`(1, -5, 2), `<,>`(0, 2, -4)); S0 := -S.e3

beta := Vector[row](3, {(1) = .1, (2) = .6, (3) = .3})

 

S := Matrix(3, 3, {(1, 1) = -3, (1, 2) = 1, (1, 3) = 0, (2, 1) = 1, (2, 2) = -5, (2, 3) = 2, (3, 1) = 1, (3, 2) = 2, (3, 3) = -4})

 

S0 := Vector(3, {(1) = 2, (2) = 2, (3) = 1})

(2)

Z := `<|>`(x, y, z)

Z := Vector[row](3, {(1) = x, (2) = y, (3) = z})

(3)

ME := MatrixExponential(S+Typesetting:-delayDotProduct(S0, Z), t);

`[Length of output exceeds limit of 1000000]`

(4)

MEint := map(int, ME.dF, t = 0 .. infinity)

Error, (in int) wrong number (or type) of arguments: wrong type of integrand passed to definite integration.

 

`&beta;plus&Assign;solve`(Z = beta.MEint, Z)

"(RTABLE(18446744074195006390,VECTOR([x, y, z]),Vector[row])=RTABLE(18446744074193876574,VECTOR([.1, .6, .3]),Vector[row]).MEint) betaplus:=solve (RTABLE(18446744074195006390,VECTOR([x, y, z]),Vector[row]))"

(5)

``

1step- I want to integrate the (ME*dF) from t=0 to ∞ .

2step- Evaluate Z=<x,y,z> by solving Z=β*MEint.

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