barybash

by v_Enhance, Jan 12, 2012, 3:38 AM

+1
circumcenters suck
cramer's rule is imba

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6 Comments

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cramer's rule is awesome. It solves system of equations in $O(n!)$ with surprisingly large constants.
which is clearly very good, cause obviously $O(n^{2.6})$ is worse.
This post has been edited 1 time. Last edited by dinoboy, Jan 12, 2012, 4:01 AM

by dinoboy, Jan 12, 2012, 4:00 AM

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You're kidding. O(n!) is definitely worse than O(n^2.6).

by phiReKaLk6781, Jan 12, 2012, 6:54 AM

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it appears someone doesn't understand sarcasm.

by dinoboy, Jan 12, 2012, 7:41 AM

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Well for $n=3$, if the coefficients are either terrible algebraic expressions or $\pm 1$, Cramer's rule definitely helps. The other day I was solving for $y_0/z_0$ in the system:

\[ \begin{align}x_{0}+y_{0}+z_{0}&= 1\\ x_{0}-y_{0}+\frac{a^{2}-b^{2}}{c^{2}}z_{0}&=\frac{c^{2}}{b^{2}+c^{2}}\\ x_{0}+\frac{a^{2}-c^{2}}{b^{2}}y_{0}-z_{0}&=\frac{b^{2}}{b^{2}+c^{2}}\end{align*} \]
This post has been edited 1 time. Last edited by v_Enhance, Jan 14, 2012, 6:23 PM

by v_Enhance, Jan 14, 2012, 6:23 PM

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hello, the solution is:

$x_0 = \frac{a^4 b^2 - 2 a^2 b^4 + b^6 + a^4 c^2 - 2 a^2 b^2 c^2 - b^4 c^2 - 2 a^2 c^4 - b^2 c^4 + c^6}{(b^2 + c^2) (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 c^2 - 2 b^2 c^2 + c^4)}$
$y_0 = \frac{-a^2b^2 c^2 -b^4c^2 + b^2c^4}{(b^2 + c^2) (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 c^2 - 2 b^2 c^2 + c^4))}$
$z_0 = \frac{-a^2 b^2 c^2 + b^4 c^2 - b^2 c^4}{(b^2 + c^2) (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 c^2 - 2 b^2 c^2 + c^4)}$

dinoboy
This post has been edited 1 time. Last edited by dinoboy, Jan 16, 2012, 11:41 PM

by dinoboy, Jan 16, 2012, 11:37 PM

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Hello,

what is a "barybash?"

yugrey does not know what a "barybash" is, and neither do I.

sonnhard

by yugrey, Jan 18, 2012, 6:32 PM

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