Hi Café! roots (http://hackage.haskell.org/package/roots) is a package to solve equations like "f(x)==0". In RootFinder class there is an 'defaultNSteps' value, which is used as maximal count of iterations functions like findRoot and traceRoot can make. By default it is 250, but sometimes it's not enough. How can I use another value instead of 250? Should I write my own RootFinder instance, or findRoot function? Thanks in advance. — Artyom.
On Fri, Mar 18, 2011 at 6:39 PM, Artyom Kazak
Hi Café!
roots (http://hackage.haskell.org/package/roots) is a package to solve equations like "f(x)==0".
In RootFinder class there is an 'defaultNSteps' value, which is used as maximal count of iterations functions like findRoot and traceRoot can make. By default it is 250, but sometimes it's not enough. How can I use another value instead of 250? Should I write my own RootFinder instance, or findRoot function?
Either choice looks like it would work fine - however using a newtype wrapper around an existing instance of FindRoot would have the least amount of code duplication, I think. I recommend contacting the maintainer of the package about getting their input, and about solving the problem in the package itself. Antoine
Thanks in advance. — Artyom.
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
On Mar 18, 2011, at 7:39 PM, Artyom Kazak wrote:
Hi Café!
roots (http://hackage.haskell.org/package/roots) is a package to solve equations like "f(x)==0".
In RootFinder class there is an 'defaultNSteps' value, which is used as maximal count of iterations functions like findRoot and traceRoot can make. By default it is 250, but sometimes it's not enough. How can I use another value instead of 250? Should I write my own RootFinder instance, or findRoot function?
Those are both options, as is to simply restart findRoot if it returns a 'Left' vaule. I personally would incline toward a custom driver function (findRoot). I should probably add one to the library that accepts a step limit and/or one that just iterates until convergence. I'm curious, though - what sort of functions are you using, and what root finding algorithm, that it takes that long to converge? And are you sure that it ever does, if it were allowed to run longer? Typically, for functions on Double, if an algorithm fails to converge within around 50 steps it's fairly likely that it never will - especially with an algorithm like bisection or Brent's method. If you're using a higher precision type, then I probably need to change the default iteration limit to something more suited to the function type. -- James
James Cook schrieb:
Those are both options, as is to simply restart findRoot if it returns a 'Left' vaule. I personally would incline toward a custom driver function (findRoot). I should probably add one to the library that accepts a step limit and/or one that just iterates until convergence.
I thought that the mosts Haskellish way to do numerical iterations is to generate a list of successive approximations (by List.iterate, of course) and then let the user choose where and why he wants to abort the list, and thus the iteration.
On Mar 23, 2011, at 6:57 PM, Henning Thielemann wrote:
James Cook schrieb:
Those are both options, as is to simply restart findRoot if it returns a 'Left' vaule. I personally would incline toward a custom driver function (findRoot). I should probably add one to the library that accepts a step limit and/or one that just iterates until convergence.
I thought that the mosts Haskellish way to do numerical iterations is to generate a list of successive approximations (by List.iterate, of course) and then let the user choose where and why he wants to abort the list, and thus the iteration.
That's probably true, and the same module exports a function that does so. But sometimes you just want to ask "what is the root?", and get an answer without taking (possibly literally) forever. Overall, the code is relatively un-Haskellish to begin with though. It was assembled in a short period of time in order to simply have code that performs the task, and hasn't really been touched since. -- James
If your function has nice derivatives, you may want to look at the Newton
implementation in
http://hackage.haskell.org/packages/archive/ad/0.44.4/doc/html/Numeric-AD-Ne...
http://hackage.haskell.org/packages/archive/ad/0.44.4/doc/html/Numeric-AD-Ne...or
if you have enough derivatives, you can even move up to the next Householder
method at Numeric.AD.Halley.findZero
These have the benefit of using exact derivatives, and returning a stream of
successively better approximations.
-Edward
On Fri, Mar 18, 2011 at 7:39 PM, Artyom Kazak
Hi Café!
roots (http://hackage.haskell.org/package/roots) is a package to solve equations like "f(x)==0".
In RootFinder class there is an 'defaultNSteps' value, which is used as maximal count of iterations functions like findRoot and traceRoot can make. By default it is 250, but sometimes it's not enough. How can I use another value instead of 250? Should I write my own RootFinder instance, or findRoot function?
Thanks in advance. — Artyom.
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
participants (5)
-
Antoine Latter -
Artyom Kazak -
Edward Kmett -
Henning Thielemann -
James Cook