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Jan
7
comment xkcd-style graphs
Mathematica 9 users please see the follow-up post at mathematica.stackexchange.com/questions/17272/… concerning slower speeds.
Jan
7
comment Correct way to populate a DiagonalMatrix?
+1 for the edit. But have you honestly compared the timing by forcing an evaluation of both results, such as with an ArrayPlot? When I do that, I find the compiled solution is only 25% faster. That's a nice achievement, but because it is so little, I would in many cases prefer a clear simple native Mathematica solution to a compiled (and perhaps obscure) solution. As you hinted, speed isn't everything!
Jan
7
comment Correct way to populate a DiagonalMatrix?
+1 Given that the final matrix is not sparse, it is noteworthy that the SparseArray representation of the diagonal matrix gives a faster calculation.
Jan
7
comment Efficient code for the Ten True Sentences puzzle
@Silvia Thank you--it seems like almost any one-line function I can write has already been incorporated in the software somewhere :-).
Jan
7
answered Efficient code for the Ten True Sentences puzzle
Jan
7
reviewed Close How to search for initialization cells?
Jan
7
comment How to get exact roots of this polynomial?
+1 Example of another way: the output of MinimalPolynomial[ Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 1] - Cos[10 \[Pi]/11]] is #1 &.
Jan
6
comment How to get exact roots of this polynomial?
@Artes Thank you for the suggestion. I posted them as comments because I don't think they actually answer the question: they require one to anticipate that the root really is a trigonometric value and will not work in general.
Jan
6
answered Correct way to populate a DiagonalMatrix?
Jan
5
comment Can some one explain perplexing behavior of arbitrary precision arithmetic?
It may be of interest that MMA can recognize the equivalence of Root[1] and Root[1 + 6 #1 - ... &, 5]: applying MinimalPolynomial to both of these yields the same expression. Equivalently, applying Root[MinimalPolynomial[#], 1]& to their difference--rather than N--produces $0$ (and you can just as easily check that this is the only root).
Jan
5
comment How to obtain a smaller-sized output from Solve
Because these are linear equations, why aren't you using LinearSolve?
Jan
5
reviewed Close Sharing an axis between two plots
Jan
4
reviewed Leave Open Lorenz map for the Rössler system
Jan
4
reviewed Close How to use the same color bar for different DensityPlot
Jan
4
comment How to get exact roots of this polynomial?
@minthao: Those roots are exact. What you seem to mean, then, is that you wish to identify some of the roots with some of the roots of another (unspecified) polynomial (which is how those cosines are defined).
Jan
4
revised How to get exact roots of this polynomial?
deleted 26 characters in body; edited title
Jan
4
comment How to get exact roots of this polynomial?
@b.gates And the next two steps are to let $x\to z/2$ to clear out powers of $2$ and then to take the big factor, $p(z)=1+3 z-3 z^2-4 z^3+z^4+z^5$ and symmetrize it via $p(z+1/z)z^5$: the primitive eleventh roots of unity pop right out.
Jan
2
comment Turn list of edges into a polygon function
@Daniel The example output is an intersection of half-planes, which can describe only a convex figure.
Jan
2
comment Turn list of edges into a polygon function
The type of example you give will work only for convex polygons. Is that a fair assumption to make in your application? If so, may we also assume the vertices have already been sorted in the order they appear around the polygon's boundary, and that the sorting follows a conventional orientation (such as keeping the interior of the polygon always to the left)? A solution for non-convex polygons can be obtained but would require more work (equivalent to triangulating them). Is it possible you only need some procedure to solve the point-in-polygon problem?
Jan
2
reviewed Leave Open Domain Coloring