MaplePrimes - answers and comments on Question, What's wrong with fsolve? Crash test.

tolerances

pagan — Mon, 09 Jan 2012 02:53:59 Z

fsolve may not have been satisfied that at the working precision it was not able to attain some tolerance. It's a flaw, that precision and tolerance are not separate for fsolve.

Using operators instead of expressions seems to use a weaker tolerance

ans:=fsolve({(lu,lv)->FDerivative2YYMain(lu,lv, 12.5)/(lv+12.5),
             (lu,lv)->FDerivative3YYYMain(lu,lv, 12.5)/(lv+12.5)},
            [0.05435690547,0.54502809258264223447]);

  [0.054356905499263310199322866010579288988739481809993666893229496592579505151375,
   0.54502809258264223446637739872970465118604413199322232443888293557355194978678]

evalf(FDerivative2YYMain(ans[],12.5));

                                   -62
                          -7.287 10  
 
evalf(FDerivative3YYYMain(ans[],12.5));

                                   -60
                         5.34464 10

Found

icegood — Tue, 10 Jan 2012 07:44:18 Z

After understatnding that fsolve/sysnewton called and checking it

i see such peace of code:

      for it2 to 10 do
        lsub2:=seq(lvars[k]=x[k] - C[k],k=1..n);
        if Bproc then
          valsub2:=op(map(x->rhs(x),[lsub2]))
        end if;
        if not `fsolve/checkrange`(lsub2) then
          if 5<=it2 then
            for k to n do
              if not `fsolve/checkrange`(lvars[k]=x[k] - C[k]) then
                C[k]:=0
              end if
            end do;
            if C=[`$`(0,n)] then
              it2:=11;
              break
            end if;
            C:=C*factorial(it2);
            it2:=0;
            next
          else
            C:=C/(it2+1);
            next
          end if
        end if;
        if Bproc then
          B2:=traperror(evalf(leqns(valsub2),Digits+n));
          if not type(B2,'list'('complex'('numeric'))) then
            error "procedures don't evaluate to numeric types."
          end if
        else
          B2:=traperror(evalf(subs(lsub2,leqns),Digits+n))
        end if;
        if B2=lasterror or isreal and hastype(B2,nonreal) then
          C:=C/(it2+1);
          next
        end if;
        aB2:=add(abs(Re(z))+abs(Im(z)),z=B2);
        userinfo(2,fsolve,`new norm:`,aB2);
        if aB2<aB then
          break
        end if;
        C:=C/(it2+1)
      end do;
      if 10<it2 then
        break
      end if;

----------

So, 10 subiterations done with hope that norm simply will reduce (aB2<aB)! And if not (it2>10) then just go to new initial guess. That's why on practice some functions don't work. Because you have just two limit points instead of one with relatevely big norm of them difference due round off. It's not shame to have such errors, in my case i saw such situation under debugger only. But it rather shame not to repair them since 1996 in code that so "most wanted".

operators

icegood — Mon, 09 Jan 2012 04:14:05 Z

"using operators instead of expressions seems to use a weaker tolerance".

Yeah, it looks natural. All of that functions was created via codegen stuff so they are optimized => less operations made => precision of result is higher. But problem not in tolerance i.e. smth that checked and smth else under check fails. There could be another stuff that i had with newton under c on double type.Namely, because of round-off error newton algorythm could be divergent i.e. stuck on pairs (or even triples and generally on arbitrary number, i suspect) of points x1, x2 such that

x2= newton_step(x1) and x1= newton_step(x2).

To overcome such situation after "critical times of guesses" i fixed delta=|x2-x1| and

after "other critical times of guesses" i calculate new delta and if i see that new delta wasn't twice less than old one then i changed guess to random perturbated:

x2+=unit_rand()*(x1-x2). Sequence of guesses still inside bounds and more probably covergent to only one point.

I just wonder, wheather maple developers did smth like that, because such behavior doesn't depend on neither tolerance cryterium nor precision of calculations as well. It depends only on precision in that sense if precision is higher then such situation is less probable to occur but still probable.