Martin Maxa's profile - activity

I need to solve this third-order linear partial differential equation:

d^2/dx^2 d/dy f(x,y) = f(x,y) - x*y Could you please help me to do this?

Thank you very much for your advise!

Hello, I would like to compute for example this double integral: exp(ixcos(y))

0 < x < pi

0 < y < 2

Could you please advice how to do that?

Thank you very much for your answer.

Martin

sorry for my answer...I didnt understand when reading first

Hello all! I would like to know if there is some list of algorithms that are not implemented in Sage and that are important (for ex. because of its speed). I think that it would be great for contributors. I believe the greatest things are mostly free and that Sage is very good example of this. Lot of algorithms in Sage are the fastest (when I compare it with 'ma' softwares) and that is really amazing that something free and open source for everyone can do it! For this reason I would like to contribute to this project and think that the list of important algorithms that arent still implemented in Sage would be great 'to do list' for contributors. Thanks, Martin

Thank you DSM for your comment, I edited my question with example. Thanks, Martin

Just write M.LLL() instead of M.lll()

Hello all! I have following question. I have given matrix with numbers and one variable a. I need compute LLL reduction of the matrix, during the algorithm there are steps like this: if a< 'something' then do...for this reason I need substitution for a before the algorithm. Now here is my question: I want to solve the algorithm symbolically with variable a, but with the substitution for a that I define before- the algorithm should solve it with substitution but during its steps reminder symbolic value a and the result should be with a as variable.

Example:

I have matrix A=[[a, 23],[a^2, 12]] now the algorithm compute for ex. this: if a<10 then B=A-[[a^2, a+5],[a-1,2]] if a>10 then B= A-[[a^2+23, a],[a-1,2*a^2]] now I want to solve A symbolically, I want obtain result as for example B=A-[[a^2, a+5],[a-1,2]], but here is the problem, if I dont define the a before the algorithm, it would have exponential complexity / because in LLL algo. there are many steps like if a<10. For this reason I need substitution for a before the alogrithm, for example a=6, then the result in example above should be B=A-[[a^2, a+5],[a-1,2]] It is unexact result for a>10 but I want try just this in my analysis of LLL. Hope now it is clear.

Thanks you very much for any advice! Martin