Talk:Biconjugate gradient stabilized method

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Step 2

Step 2 is confusing, there seems to be two different r0? —Preceding unsigned comment added by 130.37.28.128 (talk • contribs) 09:38, April 13, 2010 (UTC)

Could be a problem with your browser. The second r-zero is in fact "r hat"-zero. Kxx (talk | contribs) 14:07, 13 April 2010 (UTC)[reply]

I was confused too, as the spanish page says:

Elige un vector arbitrario

{\boldsymbol {\hat {r}}}_{0}

tal que

({\boldsymbol {\hat {r}}}_{0},{\boldsymbol {r}}_{0})\neq 0

, por ejemplo,

{\boldsymbol {\hat {r}}}_{0}={\boldsymbol {r}}_{0}

while the english one says:

Choose an arbitrary vector

r ̂ 0

such that

(r ̂ 0, r 0) \neq 0

, e.g.,

r ̂ 0 = r 0

Which one is right? Should the english page be changed? Nicolas Bigaouette (talk) 21:40, 26 April 2010 (UTC)[reply]

This is how your reply looks on my machine: http://img153.imageshack.us/img153/1053/61070581.png. I don't see any problems. Kxx (talk | contribs) 08:14, 27 April 2010 (UTC)[reply]

Fails if Residual is Eigenvector

I'm wondering if anyone else has noticed the problem that if $r_{0}$ is any eigenvector of $A$ and ${\hat {r_{0}}}=r_{0}$ , step 5.8 will evaluate as $\omega _{i}=0/0$ , thus wreaking havoc. Mind you, if this case does crop up, there's an exact solution of $x=x_{0}+r_{0}/\lambda$ , where $\lambda =(r_{0}^{T}Ar_{0})/(r_{0}^{T}r_{0})$ . Another instant failure case is if $r_{0}^{T}Ar_{0}=0$ and ${\hat {r_{0}}}=r_{0}$ , in which case step 5.5 divides by zero, though it might be that this only happens when either the above is true or the initial guess is orthogonal to the solution. Are there are any other instant failure cases to worry about? --Ndickson (talk) 05:36, 22 February 2012 (UTC)[reply]