Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

restart:with(plots):
eq:=(diff(f(eta),eta$2))-a*f(eta)+b*(1+diff(f(eta),eta)^2)^(-1/2)=0;
bc:=f(1)=0,D(f)(0)=0;
ans := dsolve(eq);

hi.how i can determind  eignvalue of matrix in the form parametric?

thanks1.mw

T := Matrix(5, 5, {(1, 1) = -b*beta*k/(c*u)-d, (1, 2) = 0, (1, 3) = -beta*lambda*c*u/(b*beta*k+c*d*u), (1, 4) = 0, (1, 5) = 0, (2, 1) = b*beta*k/(c*u), (2, 2) = -s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P-a, (2, 3) = beta*lambda*c*u/(b*beta*k+c*d*u), (2, 4) = -s*b/c, (2, 5) = r*(s-p)/s, (3, 1) = 0, (3, 2) = k, (3, 3) = -u, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = -s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P, (4, 3) = 0, (4, 4) = -s*b/c-b, (4, 5) = r*(s+c)/s, (5, 1) = 0, (5, 2) = s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P, (5, 3) = 0, (5, 4) = s*b/c, (5, 5) = -r})

Matrix(5, 5, {(1, 1) = -b*beta*k/(c*u)-d, (1, 2) = 0, (1, 3) = -beta*lambda*c*u/(b*beta*k+c*d*u), (1, 4) = 0, (1, 5) = 0, (2, 1) = b*beta*k/(c*u), (2, 2) = -s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P-a, (2, 3) = beta*lambda*c*u/(b*beta*k+c*d*u), (2, 4) = -s*b/c, (2, 5) = r*(s-p)/s, (3, 1) = 0, (3, 2) = k, (3, 3) = -u, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = -s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P, (4, 3) = 0, (4, 4) = -s*b/c-b, (4, 5) = r*(s+c)/s, (5, 1) = 0, (5, 2) = s*(lambda*c*k*beta/(b*beta*k+c*d*u)-a)/P, (5, 3) = 0, (5, 4) = s*b/c, (5, 5) = -r})

(1)

``

 

Download 1.mw

 

Hi All

Assume that we have a stochastic model with following density function

and our goal is to estimate unknown parameters namely, alpha, beta, landa, mu and sigma by any available method especially maximum likelihood estimation method.
How can we do it with maple software?

Does the "MaximumLikelihoodEstimate" command can help?

or should i define Maximum Likelihood function first and then differentiate it according to unknown parameters?

 

thanks in advance

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hi everybody,

how about a little arithmetical challenge?

 

known quantities : Z, t, epsilon : reals

let W = X+Y*t    X, Y integers, t real (t with unlimited precision)

let W = Z + epsilon (epsilon < 1/W), Z real (Z with unlimited precision)

W and Z have same decimal expansion ( W - Integerpart[W], Z - IntegerPart[Z] ), up to a certain "rank" (determined by the precision : epsilon)

Y*t and Z have same decimal expansion, up to a certain "rank".

The problem : build an algorithm to find the integer Y, using the "usable" decimal expansion of Z.

 

Thanks in advance for your contributions!

how to convert system of differential equations to differential form for evalDG?

 

[a(t)*(diff(c(t), t))+b(t), a(t)*(diff(b(t), t))+c(t)*(diff(b(t), t)), a(t)*(diff(c(t), t))+a(t)*(diff(b(t), t))+b(t)];

when i try eliminate dt which is the denominator

eliminate([a(t)*dc(t) + b(t)*dt,a(t)*db(t)+dt*c(t)*db(t),a(t)*dc(t)+a(t)*db(t)+b(t)*dt],dt);

[{dt = -a(t)/c(t)}, {a(t)*(c(t)*dc(t)-b(t)), a(t)*(db(t)*c(t)+c(t)*dc(t)-b(t))}]

 

i got two solutions, which one is correct?

a(t)*(c(t)*dc(t)-b(t)), a(t)*(db(t)*c(t)+c(t)*dc(t)-b(t))

does it mean that two have to use together to form a differential form?

 

update1

with(DifferentialGeometry):
DGsetup([a,b,c], M);
X := evalDG({a*(c*D_c-b), a*(D_b*c+c*D_c-b(t))});
Flow(X,t);
Flow(X, t, ode = true);

got error when run with above result

 

Dear All

I have downloaded second version of DGApplications to work with abstract Lie algebra. The file is actually .mla file and it is executale(as when we open it, a prompt ask, "do you want to execute this file"), but when I press ok for execution, a file open with command like as

"march('open',"C:\\Users\\Manjit\\Downloads\\DGApplications.mla");",

what should I do after this, is it a some sort installation procedure. I keep all my Maple file in E drive with following path:

E:\Maple work\General Maple Workout

Please guide me in simple way, as I failed to install Maple package many times.

Regards

I am currently working on FDM ,i have 2 coupled nonlinear pde ,i need help in solving these equation using maple code.

> restart:

> alias(f=f(tau,eta), theta=theta(tau,eta));

 

>

 

> PDE1:=S*diff(f,tau,eta)=eta^2*diff(f,eta)^2+(6*eta^2-2*f*eta)*diff(f,eta)+(6*eta^3-f*eta)*diff(f,eta,eta)-eta^4*diff(f,eta,eta,eta);

 

> PDE2:=eta^4*diff(theta,eta,eta)+2*eta^3*diff(theta,eta)-Pr*(f*eta^2*diff(theta,eta)+S*diff(theta,tau))=0;

 

i need this equation solve by maple

eql := alap*[(la(a)^4+6*la(a)+1+alap*(la(a)^4-1)/difl1)*la(a)*j(a)-((la(a)^2+1)*(1+2/difl)+la(a)^2-1)*intdj/a^4+[(4*(la(a)^4-1))/difl-8*la(a)^2]*intdj/a^2]-(4*(la(a)^2+1))*intdj/a^4+(4*(la(a)^4-1))*intdj1/a^2 = 0;

Hello all,

I have a hash structure with arrays in a YAML file that I would like to read in Maple.

Does there exist a module, so I can read it?

 

I'm trying to implement the QR algorithm to find the Eigenvalues of the input matrix which will be forwarded to another implementation (of the SVD alg.) to find the singular values. My implementation goes as follows:

1. feeding input: A::Matrix(datatype=float) # a bidiagonal matrix
2. construct input matrix for the QR alg. of matrix A and Z (zeros of size A): C := Matrix([[Z,Transpose(A)],[A,Z]], datatype=float); # therefore C should be symmetric
3. find the eigenvalues of matrix C with an implementation of the QR alg.:

for k from 1 to 400 do
Q, R := QRDecomposition(C);
C:=R.Q;
end do:

At this point, the eigenvalues of C should be placed in the diagonal of the matrix, but they're randomly placed around the diagonal, with only ~0 elements (like 2,xxx * 10^(-13)) in the diagonal.

If anyone knows how to resolve this, let the knowledge flow through. Any help will be appriciated, thanks in advance.

Please check why Maple is not returning location of Minima in following case:

 

-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x)

-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x)

(1)

plot(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = -3.2 .. 3.2)

 

readlib(extrema):

{-0.6447467154e-1, 0.6447467152e-1}

(2)

Minima := op(1, {-0.6447467154e-1, 0.6447467152e-1}); 1; Maxima := op(2, {-0.6447467154e-1, 0.6447467152e-1})

-0.6447467154e-1

 

0.6447467152e-1

(3)

minimize(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = 0 .. 3.5, location)

minimize(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = 0 .. 3.5, location), {}

(4)

Why Maple is not returning location of minima?

 

Download Location_for_Max_Min.mw

Regards

 

ANALYSIS AND DESIGN OF MACHINE FOUNDATION

 


restart

Loading Optimization

 

Loading LinearAlgebra  

 

Loading plots  

with(ScientificConstants):

Loading DynamicSystems  

with(Units:-Standard)

with(Units:-Natural)

with(StringTools)

FormatTime("%m-%d-%Y, %H:%M")

FormatTime("%m-%d-%Y, %H:%M")

(1)

NULL

Introduction

 

This document deals with vibration analysis and design of machine foundations subjected to dynamic load.

NULL

NULL

NULL

NULL

GetConstant(g);

standard_acceleration_of_gravity, symbol = g, value = 9.80665, uncertainty = 0, units = m/s^2

(2)

g__SI := evalf(Constant(g, system = SI, units))

9.80665*Units:-Unit(('m')/('s')^2)

(3)

NULL

Richart and Lysmer's Model

 

Richart et al. (1970) idealised the foundation as a lumped mass supported on soil which is idealised as frequency independent springs which he described in term of soil parameter

dynamic shear modulus or shear wave velocity of the soil for circular footing when footings having equivalent circular radius. The Tables below shows the different values of spring and damping vlaues as per Richart and Lysmer.

NULL

In which, G = dynamic shar modulus of he soil and is given G = `&rho;__s`*V__s^2 ; ν = Piosson's ratio of the soil; ρs = mass density of the soil; Vs = shear wave velocity of the soil obtained

from soil testing; g = acceleration due to gravity; m = mass of the machine and foundation; J = mass moment of inertia of the machine and foundation about the appropriate axes; K = equivalent spring stiffness of the soil; C = damping value of the soil; B = interia factor contributing to the damping factor; D = damping ratio of the soil; r = equivalent radius of a circular foundation; L = length of foundation, and B = width of the foundation.

NULL

NULL

NULL

Sketch

 

NULL

NULL

NULL

NULL

nu := .25

Table : Values of soil springs as per Richart and Lysmer (1970) model

 

NULL

NULL

SI No.

Direction

Spring value

Equivalent radius

Remarks

1

Vertical

K__z = 4*G*r__z/(1-nu)"(->)"

r__z = sqrt(L*B/Pi)"(->)"

This is in vertical Z direction

2

Horizontal

K__x = (32*(1-nu))*G*r__x/(7-8*nu)
"(->)"

r__x = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal X

2.1

Horizontal

K__y = (32*(1-nu))*G*r__y/(7-8*nu)
"(->)"

r__y = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal Y

3

Rocking

`K__&phi;x` = 8*G*`r__&phi;x`^3/(3*(1-nu))"(->)"

`r__&phi;x` = (L*B^3/(3*Pi))^(1/4)"(->)"

This produces roxking about Y axis

3.1

Rocking

`K__&phi;y` = 8*G*`r__&phi;y`^3/(3*(1-nu))"(->)"

`r__&phi;y` = (L^3*B/(3*Pi))^(1/4)"(->)"

This produces roxking about X axis

4

Twisting

`K__&psi;` = 16*G*`r__&psi;`^3*(1/3)"(->)"

`r__&psi;` = ((B^3*L+B*L^3)/(6*Pi))^(1/4)
"(->)"

This produces twisting about vertical Z axis

 

NULL

NULL

NULL

NULL

NULL

Table : Values of soil damping as per Richart and Lysmer (1970) model

 

NULL

SI No.

Direction

Mass ratio (B)

Damping ratio and Damping values

Remarks

1

Vertical

B__z = .25*m__U*(1-nu)*g__SI/(`&rho;__s`*r__z^3)
"(->)"

`&zeta;__z` = .425/sqrt(B__z)"(->)"C__z = 2*`&zeta;__z`*sqrt(K__z*m__U)"(->)"

This damping value is in vertical Z direction

2

Horizontal

B__x = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`&rho;__s`*r__x^3)
"(->)"

`&zeta;__x` = .288/sqrt(B__x)"(->)"

C__x = 2*`&zeta;__x`*sqrt(K__x*m__U)"(->)"

This damping value is in lateral X direction

2.1

Horizontal

B__y = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`&rho;__s`*r__y^3)
"(->)"

`&zeta;__y` = .288/sqrt(B__y)"(->)"

`&zeta;__y` = .288/((2.145204688-2.451662500*nu)*m__U*Units:-Unit(('m')/('s')^2)/((1-nu)*`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)

(5.1)

NULLError, invalid left hand side in assignmentError, invalid left hand side in assignment

`&zeta;__&psi;` = .5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.2)

C__y = 2*`&zeta;__y`*sqrt(K__y*m__U)"(->)"

This damping value is in lateral Y direction

3

Rocking

`B__&phi;x` = (.375*(1-nu))*`J__&phi;x`*g__SI/(`&rho;__s`*`r__&phi;x`^5)
"(->)"

`&zeta;__&phi;x` = .15/((1+`B__&phi;x`)*sqrt(`B__&phi;x`))
"(->)"

`&zeta;__&phi;x` = .15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.3)

Error, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2)) = .15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.4)

`C__&phi;x` = 2*`&zeta;__&phi;x`*sqrt(`K__&phi;x`*`J__&phi;x`)"(->)"

NULL

This damping value is for rocking about Y direction

3.1

Rocking

`B__&phi;y` = (.375*(1-nu))*`J__&phi;y`*g__SI/(`&rho;__s`*`r__&phi;y`^5)
"(->)"

NULL

`&zeta;__&phi;y` = .15/((1+`B__&phi;y`)*sqrt(`B__&phi;y`))
"(->)"

.15/((1+`B__&phi;y`)*`B__&phi;y`^(1/2)) = .15/((1+`B__&phi;y`)*`B__&phi;y`^(1/2))

(5.5)

`C__&phi;y` = 2*`&zeta;__&phi;y`*sqrt(`K__&phi;y`*`J__&phi;y`)"(->)"

NULL

This damping value is for rocking about X direction

4

Twisting

`B__&psi;` = `J__&psi;`*g__SI/(`&rho;__s`*`r__&psi;`^5)"(->)"

`&zeta;__&psi;` = .5/(1+2*`B__&psi;`)"(->)"NULLError, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4))) = .5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.6)

`C__&psi;` = 2*`&zeta;__&psi;`*sqrt(`K__&psi;`*`J__&psi;`)"(->)"NULL

``

NULL

This damping value is valid for twisting about vertical Z axis

 

NULL

NULL

NULLNULL

(4)

``

NULL

Vertical Motion Considering damping of the Soil

 

For vertical direction the equation becomes that of a lumped mass having single degree of freedom when

deq := m__U*(diff(z(t), t, t))+'C__z'*(diff(z(t), t))+'K__z'*`#mi("z")` = P__0*sin(`&omega;__m`*t)

m__U*(diff(diff(z(t), t), t))+C__z*(diff(z(t), t))+K__z*`#mi("z")` = P__0*sin(`&omega;__m`*t)

(6.1)

NULL

t1 := subs(P__0*sin(`&omega;__m`*t)/m__U = F, expand(deq/m__U));

diff(diff(z(t), t), t)+1.085721853*(G*(L*B/Pi)^(1/2)*m__U/(1-nu))^(1/2)*(diff(z(t), t))/(m__U*(m__U*Units:-Unit(('m')/('s')^2)/(`&rho;__s`*(L*B/Pi)^(3/2))-m__U*Units:-Unit(('m')/('s')^2)*nu/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))+4*G*(L*B/Pi)^(1/2)*`#mi("z")`/(m__U*(1-nu)) = F

(6.2)

NULL

(6.3)

By algebraically manipulating the expression, the form traditionally used by engineers is derived:

t2 := algsubs('C__z'/m__U = 2*zeta*omega, t1)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.4)

NULL

t3 := algsubs('K__z'/m__U = omega^2, t2)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.5)

This form includes the damping ratio , the natural frequency , and the external forcing term .  Consider only free vibration by setting

gen3 := subs(F = 0, t3)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = 0

(6.6)

NULL

sol1 := dsolve({gen3, z(0) = P, (D(z))(0) = V}, z(t))

z(t) = -(1000000000/1178791942081753609)*Pi*m__U*exp(-(1085721853/1000000000)*Units:-Unit(('s')/('m')^(1/2))*t/(Pi*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*m__U))*(1085721853*V*Units:-Unit(('s')/('m')^(1/2))*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*`&rho;__s`*L^2*B^2-4000000000*(L*B/Pi)^(1/2)*Pi*nu*`#mi("z")`+4000000000*`#mi("z")`*Pi*(L*B/Pi)^(1/2))/(Units:-Unit(('s')/('m')^(1/2))^2*B^2*L^2*`&rho;__s`)+(4000000000/1085721853)*`#mi("z")`*G*Pi*(L*B/Pi)^(1/2)*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*t/((nu-1)*Units:-Unit(('s')/('m')^(1/2)))+(1/1178791942081753609)*(1085721853000000000*V*Units:-Unit(('s')/('m')^(1/2))*Pi*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*m__U*`&rho;__s`*L^2*B^2+1178791942081753609*Units:-Unit(('s')/('m')^(1/2))^2*P*`&rho;__s`*L^2*B^2-4000000000000000000*(L*B/Pi)^(1/2)*Pi^2*m__U*nu*`#mi("z")`+4000000000000000000*`#mi("z")`*Pi^2*(L*B/Pi)^(1/2)*m__U)/(B^2*L^2*`&rho;__s`*Units:-Unit(('s')/('m')^(1/2))^2)

(6.7)

NULL

NULL

 

Download Analysis_and_Design_of_Machine_Foundations_1.mw

Good Morning Mapleprime Community,

Would anybody please help in the attached worksheet. I'm trying to use the new function in Maple that is the clicable method, but I was having problem in some of my output such as zeta_y and zeta_phi as this two equations are generating an error message.

 

Regards,

Moses

I would like to add threads to this code that compiles cleanly:

RECTmatrixSEQ :=  (FD,m, n, xw, yw, tr, k, rho, dmax) ->
 
   [seq
     (seq
        # A return of 0 means there is no choice - no digit possible
        # result of a 4 means there is a choice of (-1 or 0)
        (add(RECTpred(m, n, i, j, iter1, k, rho)*(2*abs(iter1)+1), iter1 = -dmax .. dmax)
        + (2*dmax+1)*(RECTpredR(m, n, i, j, dmax, k, rho)+
        RECTpredL(m, n, i, j, -dmax, k, rho))+
        .5*RECTpredOR(m, n, i, j, dmax, k, rho)+
        .5*RECTpredOL(m, n, i, j, -dmax, k, rho),
      i = -ceil(2*dmax+2*rho)*2^m .. ceil(2*dmax+2*rho)*2^m-1),
    j = 0 .. 2^(n-yw)-1)]:

I've added thead calls like this, still no syntax errors:

RECTmatrixSEQ :=  (FD,m, n, xw, yw, tr, k, rho, dmax) ->

   CodeTools:-Usage( Threads:-Wait(
   [seq (Threads:-Create
     (seq
        # A return of 0 means there is no choice - no digit possible
        # result of a 4 means there is a choice of (-1 or 0)
        (add(RECTpred(m, n, i, j, iter1, k, rho)*(2*abs(iter1)+1), iter1 = -dmax .. dmax)
        + (2*dmax+1)*(RECTpredR(m, n, i, j, dmax, k, rho) +
        RECTpredL(m, n, i, j, -dmax, k, rho))+
        .5*RECTpredOR(m, n, i, j, dmax, k, rho)+
        .5*RECTpredOL(m, n, i, j, -dmax, k, rho),
      i = -ceil(2*dmax+2*rho)*2^m .. ceil(2*dmax+2*rho)*2^m-1),
    j = 0 .. 2^(n-yw)-1)) ] )
    ):

 

I do get a runtime error:

                 "PROC:feasibilitycheckproONCE

                   "
Error, (in CodeTools:-Usage) invalid input: too many and/or wrong type of arguments passed to Threads:-Create; first unused argument is j = 0 .. 3

Any help spotting this problem would be appreciated. Thanks in advance.

Bonnie

 

The following code is part of my attempt to answer the recent Question about the bifurcation of the map f:= x-> exp(x^2*(a-x)). Two very weird things are happening. They can be seen by applying trace to f. The first is that the input argument to f seems to be changed to a very large integer. The second is that for some real values of a and x, I get imaginary results from this obviously real-valued function. Why are these things happening?

restart:

f:= x-> exp(x^2*(a-x)):

trace(f):

Iterate:= proc(a, x0:= 1., n:= 2000)
local A:= hfarray(1..n, [x0]), f:= subs(:-a= a, eval(:-f));          
     #evalhf(
          proc(f, A, n)
          local k;
               for k from 2 to n do A[k]:= f(A[k-1]) end do
          end proc
          (f, A, n);
     #);
     evalf[4]~(convert(A[1000..], set))
end proc:

Iterate(1.05);

{--> enter f, args = 4607182418800017408

 

HFloat(1.0512710963760241)

 

<-- exit f (now in unknown) = 4607413323290551347}
{--> enter f, args = 4607413323290551347

 

HFloat(0.9985962074909431)

 

<-- exit f (now in unknown) = 4607169774561176020}
{--> enter f, args = 4607169774561176020

 

HFloat(1.0525960836530153)

 

Warning,  computation interrupted

 

Iterate(.75);

{--> enter f, args = 4607182418800017408

 

.754589752755861+.192678397202388*I

 

<-- exit f (now in unknown) = HFloat(0.7545897527558614)+HFloat(0.19267839720238844)*I}

Error, (in unknown) unable to store 'HFloat(0.7545897527558614)+HFloat(0.19267839720238844)*I' when datatype=float[8]

 

 

 

Download bifurcation.mw

hi.i am a problem for solving this non linear algebric equation.

please help me...thanks

FSOLVE.mw

FSOLVE.mw

 

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