If your function has nice derivatives, you may want to look at the Newton implementation in<div><br></div><div><a href="http://hackage.haskell.org/packages/archive/ad/0.44.4/doc/html/Numeric-AD-Newton.html#v:findZero">http://hackage.haskell.org/packages/archive/ad/0.44.4/doc/html/Numeric-AD-Newton.html#v:findZero</a></div>
<div><br></div><div><a href="http://hackage.haskell.org/packages/archive/ad/0.44.4/doc/html/Numeric-AD-Newton.html#v:findZero"></a>or if you have enough derivatives, you can even move up to the next Householder method at Numeric.AD.Halley.findZero</div>
<div><br></div><div>These have the benefit of using exact derivatives, and returning a stream of successively better approximations.</div><div><br></div><div>-Edward</div><div><br></div><div><br><div class="gmail_quote">On Fri, Mar 18, 2011 at 7:39 PM, Artyom Kazak <span dir="ltr">&lt;<a href="mailto:artyom.kazak@gmail.com">artyom.kazak@gmail.com</a>&gt;</span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><br>
Hi Café!<br>
<br>
roots (<a href="http://hackage.haskell.org/package/roots" target="_blank">http://hackage.haskell.org/package/roots</a>) is a package to solve equations like &quot;f(x)==0&quot;.<br>
<br>
In RootFinder class there is an &#39;defaultNSteps&#39; value, which is used as maximal count of iterations functions like findRoot and traceRoot can make. By default it is 250, but sometimes it&#39;s not enough. How can I use another value instead of 250? Should I write my own RootFinder instance, or findRoot function?<br>

<br>
Thanks in advance.<br>
— Artyom.<br>
<br>
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</blockquote></div><br></div>