mbras

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8 years, 202 days

MaplePrimes Activity


These are questions asked by mbras

I've read the spec online for LinearSolve but it's not clear on the function of inplace on nonsquare matrices. For example, consider the following:

A:=<0;1>;
B:=<0:2>;
LinearSolve(A,B,inplace=true)

This last line outputs <2;0>, which is also the value stored in B after the operation, whereas [2] would be the desired result. In the case of A being a row vector an error is encountered due to lack of storage. Does what happened above generalise for any matrix with more rows than columns: storing the result in B, but adding zeroes to the bottom unused parts of B, due to B being larger than the solution?

Also, does anyone have any advice on an efficient method for solving A.x=b, in the case where b is a vector, and A is a large but tall (varying size, but often 5x as many rows as columns, e.g. 10000x2000) integer matrix, where most of the entries in any particular column are zero (more than 90%)? I've found the option method='modular' helps quite a lot, but not enough, any ideas for quick fixes?

I'd like to have an algorithm for computing the ring of invariants of a reductive group (i.e. G acts on V, compute C[V]^G). I know that there are algorithms in existance for this, e.g. see http://www.math.lsa.umich.edu/~hderksen/Publications/invar1999.pdf. Does anyone know if this has been implemented in Maple, or where to find code in another language? Any ideas on what sort of data structure would be appropriate

Hi,

I'm relatively new to using Maple.

I'm looking for information on how the "factor" function works. I printed its definition, and it refers to "factor/factor" and I can't find any more information on this. I'd like to know so that I could have more trust that it works correctly. Specifically, I'd like to be able to believe that if it does not factor a cubic, then the cubic is irreducible.

I'm specifically looking for rational roots of cubic polynomials. The "solve" function seems to work, and gives me the roots in terms of square roots, cubic roots and rationals. I have no idea why I should believe that "type(x,rational)" would work when the description of "x" is quite complicated.

Does anyone know anything about how "factor" works, or how "type" works when testing whether an expression evaluates to a rational number? Any information would be much appreciated.

 

Thanks,
Matt

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