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Aug
20
comment Solving very large matrices in “pieces”
Ah, now I have seen where I obtained the formula from: check out "Computational Methods of Linear Algebra" by Faddeev and Faddeeva; the treatment is a bit dated but can still be useful.
Aug
20
comment How to tell if a line segment intersects with a circle?
Two tiny notes: you did not have to square Norm[], and why did the multiplier for Q become (t-1)?
Aug
19
comment How to tell if a line segment intersects with a circle?
I got momentarily confused with the last formula until I realized you meant $\|tP+(1-t)Q-C\|_2=r$ ... :D
Aug
19
comment How to tell if a line segment intersects with a circle?
Now I'm confused... try this: how would a line segment that would have intersected the circle if extended in the right direction be treated?
Aug
19
comment How to tell if a line segment intersects with a circle?
So you do mean the line determined by two given points... :) Then the formulae in the link I gave should apply.
Aug
19
comment How to tell if a line segment intersects with a circle?
If you want the intersection of a line (and not a segment) with a circle: mathworld.wolfram.com/Circle-LineIntersection.html
Aug
19
comment How to tell if a line segment intersects with a circle?
"Line segment" or "line"? How would you want the case of a line segment being entirely inside the circle handled?
Aug
19
comment Blow up of a solution
You can't always blame the numerical method; if you use the wrong set of variables (i.e. not figuring out a proper variable substitution), such numerical disasters are likely to happen.
Aug
19
comment Solving very large matrices in “pieces”
Aha: web.cs.wpi.edu/~matt/courses/cs563/talks/radiosity.html and particularly "it is diagonally dominant ... and the upper right of the matrix is computable from the lower left." Maybe your proposal would be a better use of computing power than mine!
Aug
19
comment Solving very large matrices in “pieces”
Indeed, he would do well to have a look at the condition number of his matrix; if its logarithm is even just half of the number of digits of precision available he's short on luck.
Aug
19
comment Solving very large matrices in “pieces”
Otherwise, the only other observation I can make is that if all the eigenvalues of $\mathbf{P}\mathbf{F}$ are less than 1 in magnitude, you can expand the inverse of your matrix $\mathbf{I}-\mathbf{P}\mathbf{F}$ as a geometric series in $\mathbf{P}\mathbf{F}$: $\mathbf{I}+\mathbf{P}\mathbf{F}+\mathbf{P}^2\mathbf{F}^2+\dots$; this might prove to be useful.
Aug
19
comment Solving very large matrices in “pieces”
Looking at your link to your other question, the only discernible structure your matrix has from that presentation is that it can be expressed in the form $\mathbf{I}-\mathbf{P}\mathbf{F}$, $\mathbf{I}$ an identity matrix, $\mathbf{P}$ a diagonal matrix of the reflectivities $\mathrm{diag}(\rho_1\dots\rho_n)$, and $\mathbf{F}$ the matrix of form factors. I would hope that there is something special about $\mathbf{F}$'s structure that can be exploited.
Aug
19
comment Solving very large matrices in “pieces”
It's supposed to be a comment, hence "community wiki".
Aug
19
comment Solving very large matrices in “pieces”
Dense, possibly unsymmetric and unstructured... I don't have high hopes for iterative approaches working well in his application.
Aug
19
answered Solving very large matrices in “pieces”
Aug
19
comment Solving very large matrices in “pieces”
bobobobo: "Recursively"... that would sound like setting yourself up for a administrative nightmare when you start writing the program. Two things: 1. you still haven't said why you have a dense matrix in the first place or where it came from. I would hope that it would have some sort of exploitable structure that you won't have to employ the baroque formula in my answer. 2. Again, it is a bad idea to maintain explicit inverses; store the matrices indicated as inverses in the formula above as an appropriate decomposition.
Aug
19
comment Help in getting the Quadratic Equation
SB, your dividing of the depression term $-\frac{b}{2a}$ again with $2a$ was what threw me off. :) At least WWright has already pointed you in the proper direction.
Aug
19
comment Help in getting the Quadratic Equation
Looks good now, you've done it right! :)
Aug
19
comment Help in getting the Quadratic Equation
WWright, this may be a useful crutch: codecogs.com/latex/eqneditor.php
Aug
19
comment Help in getting the Quadratic Equation
A note on the TeX subscripts: try "x_{1,2}". As for the derivation, are you already familiar with "completing the square"? Otherwise, one thing you can try is to make the substitution $x=u-\frac{b}{2a}$, solve for u, and then reexpress the whole mess in terms of x. Good luck!