November 3

Is there anywhere in Wikipedia that discusses techniques for estimating the uncertainty on ${\displaystyle \mathbf {x} }$ where it is the solution to ${\displaystyle A\mathbf {x} =\mathbf {y} }$. I am specifically interested in the case where A is a square matrix of full rank (so the solution to the linear equation is exact), but that both the elements of A and of ${\displaystyle \mathbf {y} }$ have previously estimated uncertainties. Dragons flight (talk) 14:50, 3 November 2011 (UTC)[reply]

The uncertainty in x will depend on the magnitude of det(A). If det(A) is close to zero, even small uncertainties in A can become large uncertainties in A^-1 and hence large uncertainties in x = A^-1y . For example, solving

${\displaystyle x-y=1}$

${\displaystyle (1.001)x-y=2}$

gives (x, y) = (1000, 999), but with just a small change in the coefficient of x in the second equation we have

${\displaystyle x-y=1}$

${\displaystyle (0.999)x-y=2}$

and now (x, y) = (-1000, -1001). Gandalf61 (talk) 15:11, 3 November 2011 (UTC)[reply]

Condition number will be a good place to start, though it seems to focus on errors in y rather than A. -- Meni Rosenfeld (talk) 15:13, 3 November 2011 (UTC)[reply]

Your A must be a positive definite matrix presumably, so a Wishart distribution might be a good place to start. Then its inverse will have an inverse Wishart distribution. HTH, Robinh (talk) 20:47, 3 November 2011 (UTC)[reply]

In engineering, we sometimes use sensitivity analysis and root locus analysis and graphs. These are mathematically equivalent to computing the condition number of the system description matrix, or taking the partial derivative with respect to the input variables (in some cases, this means constructing the matrix of "Fréchet derivative"s or calculating the Jacobian matrix). Such techniques are convenient for handling linear algebra engineering problems in "standard form," because you can apply some shortcuts to determine system stability - often without explicitly calculations (saving time, and turning intractable problems into ... tractable problems). For "impractically large" linear systems, this allows us to use heuristics to approximately analyze stability. Nimur (talk) 21:14, 3 November 2011 (UTC)[reply]

http://www.bmoc.maths.org/home/bmo1-2011.pdf Can someone have look at this paper from the British Maths Olympiad in December 2010 for me please? I'm looking for some hints to solve question 6. Basically me and my teacher have been agonising for the last 2 days trying different methods from expanding and simplyfing to area=1/2absinC to sine and cosine rule to logical reasoning, and still it remains unsolved. So any help is welcome, preferably not the full solution but hints that will direct us on the right path.

In case the link doesnt work here's the question in short

"If a, b and c are the lengths of the sides of a triangle and ab+bc+ac=1, prove that (a+1)(b+1)(c+1)<4." 81.174.172.79 (talk) 19:44, 3 November 2011 (UTC)[reply]

The worst case scenario is

${\displaystyle a=b=c={\frac {\sqrt {3)){3))}$

where

${\displaystyle (a+1)(b+1)(c+1)=\left({\frac {\sqrt {3)){3))+1\right)^{3}\approx 3.9245<4}$.

Bo Jacoby (talk) 22:02, 3 November 2011 (UTC).[reply]

Another nice way to do it is to consider replacing 4 with 4(ab + bc + ac) and comparing the left-hand side with 0. You're going to need (a − 1)(b − 1)(c − 1) < 0; reason why that should be true from the inequalities you can derive. —Anonymous Dissident^Talk 22:16, 3 November 2011 (UTC)[reply]

Thank you very much I think I've got it now. But I still don't understand where Bo Jacoby gets his ${\displaystyle a=b=c={\frac {\sqrt {3)){3))}$ from? 81.174.172.79 (talk) 22:33, 3 November 2011 (UTC)[reply]

Sorry, I was too fast. (if a=b=c and ab+bc+ca=1 then ${\displaystyle a={\frac {\sqrt {3)){3))}$, but it is not worst case).

The limiting case

${\displaystyle a=b=1,\qquad c=0}$

gives

${\displaystyle (a+1)(b+1)(c+1)=4}$.

The other limiting case

${\displaystyle a=b={\frac {1}{\sqrt {5))},\qquad c={\frac {2}{\sqrt {5))))$

gives

${\displaystyle (a+1)(b+1)(c+1)=\left({\frac {1}{\sqrt {5))}+1\right)^{2}\left({\frac {2}{\sqrt {5))}+1\right)\approx 3.96774<4}$.

I suppose the remaining can be done using a lagrange multiplier to find maximum under constraint. Bo Jacoby (talk) 22:44, 3 November 2011 (UTC).[reply]

Note that the question is equivalent to "If a, b and c are the lengths of the sides of a triangle and ${\displaystyle d={\sqrt {ab+bc+ac))}$, prove that ${\displaystyle ({\frac {a}{d))+1)({\frac {b}{d))+1)({\frac {c}{d))+1)<4}$." Bo Jacoby (talk) 19:50, 6 November 2011 (UTC).[reply]