Carl Love

Carl Love

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13 years, 122 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

@brian bovril Are you still restricted to 3 groups of 3, although now the products of the groups may possibly be unequal?

@candy898 Try this.

 

restart:

A:= Matrix([[.7, .3, .3], [.2, .6, .1], [.1, .1, .6]]);

A := Matrix(3, 3, {(1, 1) = .7, (1, 2) = .3, (1, 3) = .3, (2, 1) = .2, (2, 2) = .6, (2, 3) = .1, (3, 1) = .1, (3, 2) = .1, (3, 3) = .6})

macro(LA= LinearAlgebra):

The matrix A is singular, so it doesn't make sense to refer to 1/(I-A). Rather, we solve the system (I-A).X = 0.

Sol:= LA:-LinearSolve(convert~(LA:-IdentityMatrix(3)-A, rational), <0,0,0>);

Sol := Vector(3, {(1) = (5/3)*_t[2], (2) = _t[2], (3) = (2/3)*_t[2]})

solve({Sol[3]=100});

{_t[2] = 150}

eval(Sol, %);

Vector(3, {(1) = 250, (2) = 150, (3) = 100})

 




Download Leontief2.mw

Why are you setting the datalimit to such a small value? Of course it will run out of memory. I'd like to see the output of kernelopts(datalimit) before you make any changes to it. You obscure this information by ending the command with a colon.

I don't understand the question. Is it that you have a surface in a plot structure and you want to find eight points on that surface which are the vertices of a cube?

@candy898 

Replace LA with LinearAlgebra.

@candy898 What version of Maple are you using?

 

@saysics 

In the command

F := eval(-.2*s*Eta*r*u(t)*exp(-r^2/omega^2-t^2*1.177^2/tau^2)*cos(k*z-omega*t), params);

replace u(t) with U(t).

Do you have some reason to believe that Maple is not recognizing the amount of memory that you have?

What number do you get from the command kernelopts(datalimit)?

@saysics There was a mistake in my code in the specification of Eta. That did not affect the dsolve answer because Eta does not occur in the ODEs.

I had

Eta= 1.05457173*e-34

but it should be

Eta= 1.05457173e-34

@fluff I take it that you want the option of doing 0, 1, or 2 switches, right? You can't get two switches by including the switch argument twice.You make switches take a value in {0, 1, 2} by specifying the parameters like this

MontyHall:= proc(N::posint, {switches::identical(0,1,2):= 0})

and then making the calls like this

MontyHall(100000);  MontyHall(100000, switches=1);

etc.

@Markiyan Hirnyk Note that the prime notation for derivatives, A'(t), only works with 2D input.

@Kitonum It's O(n^4) because inside the loop for choose(S,3) there is a seq that scans over the remaining vertices.

I agree with you on points (1) and (3).

Do you mean that you want to do a simulation of the situation, with random selection?

Do you want the contestant to choose to switch always?

@rwooduk The green arrow for file upload is the last (rightmost) item in the second row of the toolbar.

(But I was able to download your file from zippyshare.)

@jelmew None of the basic simplifying commands work with the non-commutative multiplication operator `.`. If you are working with tensors, you should probably load the Physics package, which has its own simplifiers. Outside of that, the expression at hand can be further simplified via

eval(%, `.`= `*`);

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