casperyc

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These are questions asked by casperyc

 

restart:
tmp:=Vector(
[
1+(-s[2]-s[4]+2*w[1]/(1+1/exp(mu[p]))^2+(2*(-w[1]+1))/(1+1/(exp(mu[p])*exp(eta[p2])))^2)*s[1]^3+(-s[2]+s[3])*s[1]^2-s[2]*s[1],

(s[2]+s[4]-2*w[1]/(1+1/exp(mu[p]))^2-(2*(-w[1]+1))/(1+1/(exp(mu[p])*exp(eta[p2])))^2)*s[1]^3
]
);

tmp := Vector(2, {(1) = 1+(-s[2]-s[4]+2*w[1]/(1+1/exp(mu[p]))^2+(-2*w[1]+2)/(1+1/(exp(mu[p])*exp(eta[p2])))^2)*s[1]^3+(-s[2]+s[3])*s[1]^2-s[2]*s[1], (2) = (s[2]+s[4]-2*w[1]/(1+1/exp(mu[p]))^2-(-2*w[1]+2)/(1+1/(exp(mu[p])*exp(eta[p2])))^2)*s[1]^3})

(1)

rule3:=w[1]/(1+1/exp(mu[p]))^2+(-w[1]+1)/(1+1/(exp(mu[p])*exp(eta[p2])))^2 = s[3];

w[1]/(1+1/exp(mu[p]))^2+(-w[1]+1)/(1+1/(exp(mu[p])*exp(eta[p2])))^2 = s[3]

(2)

applyrule(rule3,tmp[1]);

1+(-s[2]-s[4]+2*w[1]/(1+1/exp(mu[p]))^2+2*(-w[1]+1)/(1+1/(exp(mu[p])*exp(eta[p2])))^2)*s[1]^3+(-s[2]+s[3])*s[1]^2-s[2]*s[1]

(3)

 

``

 

Download problem.mw

 

This is part of a large simplifcation where lots of terms are being substituted. In two of those terms, it did not simplify as we would expect.

I think the main thing is trying to find a way to factor out the "2".

 

I could do this

> rule3:=w[1]/(1+1/exp(mu[p]))^2+(-w[1]+1)/(1+1/(exp(mu[p])*exp(eta[p2])))^2 = s[3];
>rule3:=2*rule3;

> rule3ne:=-(w[1]/(1+1/exp(mu[p]))^2+(-w[1]+1)/(1+1/(exp(mu[p])*exp(eta[p2])))^2) = -s[3];
> rule3ne:=2*rule3ne;

> applyrule(rule3,tmp[1]);
> applyrule(rule3ne,tmp[2]);

For this example, this works.

But I hope for a more generic approach.

 

Thanks,

 

casper

 

 

 

M:=Array([[a+2*b,b],[x1+y1,z1+z2]]);
pars:=indets(M);
npars:=numelems(pars);

myRand:=RandomTools[Generate](float(range=-3..3),makeproc=true):

set1:={seq(pars[i]=myRand(),i=1..npars)};
numM:=subs(set1,M);
LinearAlgebra:-Rank(numM);

is there a way to make sure that the random values generated from myRand(), subs into M, to make sure that each cell of M is between 0 and 1?

Basically, what I need is being able to simulate some values, subject to each cell of M is between 0 and 1, and find the rank of M. I want to do that efficiently 20 times (or more possibly). Those values that make the cells out of 0 and 1 do not count as 1 of the 20 simulations.

 

Thanks,

 

casper

 

 

Say for a complicated function f(x), there are 10 solutions. I want to check for any of the solutions, the matrix R is defined or not.

 

Is there a way to force execution like the following ones?

seq(1/i,i=0..3);

for i from 0 to 3 do
    1/i;
end do;

 

I think the problem is that, Maple stops immediately when it's undefined. But the following terms could be defined.

Or is there a way to put this check into "solve"? That's

> solve(f(x),x, "some way of making sure that the solution has to be defined for the matrix R" )

 

 

Is there a way to achieve this?

Thanks

I am a research student and I am writing up my thesis. I was reading one of the paper written earlier by authors in 2003.

The authors calculated a symbolic rank of a matrix A, and got 9, using Maple 6.

 

Then new methods developed in 2012 and proved that the "true" rank should be 8. The later paper that "Most  recent versions of Maple have this simplification built in and are able to return a rank of 8.".

 

I just noticed that in Maple 18.01, this matrix was still evaluted for a symbolic rank 9, if no simplifcaiton was done before using Rank(), see attached.

 

I didnt explore a lot, but just as a notice. I am a bit concerned as most of my research was trying to deal with exponentials.

 

Is that something to be fixed in future versions?

 

Thanks!

 

 Bug.mw

Hi,

I have coded up a vector that is of my interest. The code runs witout any problem and gives me exactly what I want.

newtest.mw

All I want to know, is that if there are more efficient way to do so?

Any tricks, or better use of a particular function that I wasn't aware?

The only tiny bit of unsatisfactory is that, the (1-w) term is at the first term of the addition, is that possible to move it to the last term? Which is more conventional to read.

 

Thanks,

 

casper

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