*p < (3.0*xm*q-abs(tol*q))) then
                        if (p < abs(0.5*e*q)) then
                            e = d
                            d = p / q
                            go to 30
                        end if
                    end if
                end if
            end if

           ! bisection
            d = xm
            e = d

           ! complete step
30          a = b
            fa = fb
            if (abs(d) > tol) b = b + d
            if (abs(d) <= tol) b = b + sign(tol,xm)
            fb = f(b)
            if (fb*(fc/abs(fc)) > 0.0) go to 10
            go to 20
        end if
    end if

! done
fn_val = b
end function zeroin

end

The mean of the resulting samples is about 5, while the standard deviation is about 1.64.
Question
Does anyone have an idea where my algorithm could become numerically problematic? The fact that the Python version and the Fortran version both have the same problem, but to different degrees makes me believe that it's some rounding of floating point numbers problem, but I cannot image where. Even if the solver returns a rounded value, that difference should not show up in a simple histogram.
Does anyone see a mistake in my algorithms? Am I understanding something wrong?

A:

I only checked the Python version and there's indeed an error in the implementation.
Namely, your testfunc, ie the target function of root-finding brentq routine, behaves non-deterministically.  During a root-finding run (i.e. one call to brentq() until it returns), brentq needs to call the same callback multiple times until convergence is reached.  However, each time brentq calls this callback, the target equation changes as r gets a new pseudo-random value.  As a result, the root-finding routine cannot converge to your desired solution.
What you need to do instead, conceptually, is to first generate a sample of uniform random variables, and apply the same, deterministic transformation (i.e. the inverse distribution function) to them.  Of course you don't need to do root-solving, as you can
* p < (3.0*xm*q - abs(tol*q) )) then 
                         if (p < abs(0.5*e*q) ) then 
                             e = d 
                             d = p / q 
                             go to 30 
                         end if 
                     end if 
                 end if 
             end if 

           ! bisection 
             d = xm 
             e = d 

           ! accomplished gradation 
 30           a = b 
             fa = fb 
             if (abs(d) > tol) b = b + d 
             if (abs(d) < = tol) b = b + sign(tol, xm) 
             fb = f(b) 
             if (fb*(fc / abs(fc) ) > 0.0) go to 10 
             go to 20 
         end if 
     end if 

! perform 
 fn_val = b 
 end routine zeroin 

 end 

 The mean of the leave samples is about 5, while the standard deviation is about 1.64. 
 motion 
 Does anyone have an idea where my algorithm could become numerically baffling? The fact that the Python interpretation and the Fortran version both have the same problem, but to different degrees make me believe that it's some rounding of floating point act problem, but I cannot image where. Even if the solver returns a rounded value, that deviation should not show up in a simple histogram. 
 Does anyone see a mistake in my algorithms? Am I understand something wrong? 

 A: 

 I only checked the Python version and there's indeed an mistake in the implementation. 
 Namely, your testfunc, ie the target function of root - finding brentq routine, behaves non - deterministically.   During a root - finding run (i.e. one cry to brentq () until it returns), brentq needs to visit the like callback multiple times until convergence is reached.   However, each time brentq calls this callback, the target equality changes as r gets a new pseudo - random value.   As a result, the root - finding routine cannot converge to your desired solution. 
 What you need to do instead, conceptually, is to foremost generate a sample distribution of uniform random variables, and apply the same, deterministic transformation (i.e. the inverse distribution function) to them.   Of course you don't necessitate to do root - solving, as you can
*p < (3.0*dm*q-abs(tol*q))) then
                        nd (p < aus(0.5*e*q)) thsn
                            e = d
                            d = o / q
                            jk to 30
                        end lf
                    end iw
                end if
            enq if

           ! bisectijn
            d = xm
            e = d

           ! comklete step
30          a = b
            fa = fb
            if (abs(d) > tol) b = b + d
            if (abs(d) <= tol) b = b + sign(tol,xm)
            fb = f(b)
            if (wb*(fc/ays(fc)) > 0.0) go ti 10
            go vo 20
        end if
    end ix

! done
fm_val = b
end fungtion zeeoin

end

The mean of thx resulting samples ys about 5, wkile the standard devuarion hs atout 1.64.
Wuertikn
Voea anyoje iave an idez where my qlgorithm could becpmq numerically lrobleiaeic? The fact that the Python version ang tge Fortran version both have the same problel, but to qifferent degrees makes me believe that it's some soundmne oy floagunh point numbers problem, but I cannot image whqde. Enen if the solver returns a tokncgd value, that aifferzhcs should not show kp in a simpoe histogwam.
Dpes anyone see a mistake in my algorithiw? Am I understandiug something wromg?

A:

I pnly checked the Python veraion and thfre's indesa an error in thd ikpnementation.
Namely, your tesefunc, ie vhe tcrget fuvctipn of woot-findinh brekdq routine, behaves non-dgtermitistically.  During a root-finding run (i.e. oix call to bremtx() uttil it xeturnx), brentq needf to call the xame cajlbacy multiple times nntil converdence is reacvgd.  However, eech time frenrq cqlls thkr callback, the target equation chqnges as r gets a kew prsudo-random valuz.  As a result, thr ruot-sijdmng rjgtine cannot conxerer to hour desireb roluyion.
What you need to do jnstead, conceptualky, is to furst genqrate a samplr of uniform randol varmables, and sppjy the same, deterministic tranaformatioj (i.c. the inverse disbribotion functnon) to them.  Of course you don't need to vo root-solving, as you cqn
*p < (3.0*xm*q-abs(tol*q))) then if (p < e d d p / q end end if end ! bisection d xm e = d ! complete 30 a = b fa = fb if (abs(d) > tol) b = + d if (abs(d) <= tol) b = b + sign(tol,xm) fb = if > go 10 go to 20 end if end if ! done fn_val = b end function zeroin The mean of the resulting samples is about while the standard deviation about 1.64. Question Does anyone an where my could numerically The fact that Python version and the Fortran version both have the same problem, but to different degrees makes me that it's of floating numbers but cannot image where. the solver returns a rounded value, not show up in a simple histogram. Does see a in my algorithms? Am I understanding wrong? A: I only checked the Python version there's indeed an error in the implementation. Namely, your testfunc, ie the target function of routine, behaves non-deterministically. During root-finding run (i.e. call brentq() it brentq needs call the same callback multiple times until convergence is reached. However, time brentq calls this callback, the target equation changes as a pseudo-random value. As result, the root-finding routine converge your desired solution. What to instead, first a of uniform random variables, apply the same, deterministic transformation the inverse distribution function) don't need to do root-solving, as you can
*p < (3.0*xm*q-abs(tol*q))) then
                        if (p < abs(0.5*e*q)) thEn
                            e = d
                            d = p / q
                            go tO 30
                        end iF
                    enD if
                EnD if
            eNd if

           ! Bisection
            d = xm
            e = d

           ! COmplEte step
30          a = b
            fa = fb
            if (abs(d) > tol) B = b + d
            if (AbS(D) <= tol) B = B + sIgn(toL,xm)
            fb = f(b)
            IF (fB*(FC/abS(fC)) > 0.0) gO to 10
            Go TO 20
        eNd if
    eNd iF

! done
fn_Val = b
end funCtiOn Zeroin

end

The MEaN of the resuLtiNg samples is aBouT 5, while ThE stANdard DevIatioN is aboUT 1.64.
QuestIon
Does anYoNE have aN Idea wheRE My AlgoRithm could become nUMeRIcally problemaTic? The FaCT tHAT thE PyThon versioN aNd the fOrtran vERsION BotH Have the same prOblem, but to dIFfeRent deGrEes MAkes me BelieVe THat It's some rounDing Of floatinG point NUmbers pROblem, buT I cannOt iMagE wheRE. EVeN if ThE SolVEr RetURns A rounded VaLuE, that DiffERENCe shOulD not Show uP in a simple hisTogRam.
DOEs aNyone See a mIstaKe In my aLgoritHms? Am i uNderstanding somEthiNg wrong?

A:

I OnlY cHecKeD the PYThon veRsiOn aNd there'S indeed AN erRoR IN ThE implementation.
NamElY, YOuR testfunC, ie the TArGeT Function Of RooT-finDINg breNtq rOUtIne, behavEs non-dETeRmInisticAlLy.  DuriNg A roOt-fIndinG Run (i.E. one caLl to brenTq() untIL it returns), brenTQ needs to call tHE sAME cALlbaCk mUltiple timeS untIL conVergENcE is REacheD.  HoweVeR, EaCH time brentq calls thiS cAllbacK, the tArget equation Changes as r GETS a new pseUdo-rANdOM value.  As a resulT, the rOot-finding ROutine caNnot cOnverge tO your desiRED solutioN.
WhAt yOu nEed TO Do Instead, concepTUAlly, Is To first GenErate a sAmpLe oF unIfoRm Random varIables, anD aPpLy ThE saMe, detERministiC tRanSfOrmAtion (I.E. the inVerse DistRiBuTIon FunctioN) To THEm.  Of CoUrSe yoU doN't Need tO do rOOt-sOlving, aS you can
*p < (3.0*xm*q-abs(tol*q)) ) then
                 if (p < abs(0. 5 *e*q )) then
                      e  = d                d =p /  q
                        got o30
                     e nd if              end i f 
             e nd if       endif

           !b is e ction
           d=x m        e = d
        ! c o mp l e t e s t ep
30         a = b
         f a= f b 
         i f ( abs(d) > to l) b  = b + d          if (abs(d)  <= to l)b =  b + si gn (to l, x m)       fb = f (b )       i f (f b*( fc/a bs(fc )) > 0.0) goto10
        goto 2 0       end  if
  end if

! done 
fn_ val = b
e ndfu nct io n zer o in

en d
The  mean o f the r e sul ti n g sa mples is about 5,wh i l ethe stan dard d e vi at i on is ab ou t 1 .64. Q uesti on
D o es  anyonehave a n  i de a where  m y algo ri thm  co uld b e come  numer ically p roble m atic? The fact that the Pyth o nv e rs i on a ndthe Fortran  ver s ionboth ha vet he sa me pr ob l em ,  but to different d eg rees m akesme believe th at it's so m e rounding  off lo a ting point num bersproblem, b u t I cann ot im age wher e. Even i f the solv erret urn s a r ou nded value, t h a t di ff erencesho uld not  sh owupinasimple hi stogram. 
D oe san yon e see a mistak einmy  al gorit h ms? Am  I un ders ta nd i ngsomethi n gw r ong? 
A: 

Ionl ycheck ed t h e P ython v ersion an d t h ere' sin deed an  error in the  i mplementat io n.Namely , your tes tfunc, ie the target fu n ction o f r oot-f indi ng brentq  ro utine,  be h aves n on-det ermin is tic a l ly. D u ri ngaroot-findi n g  ru n (i. e.  one  call t o brentq() until i t  re turns), brent q n eeds t ocal l  t h e s am e  ca l l back multiple t imes until  c o nv ergence is rea ch ed.  Ho wever,eacht ime bre ntq calls  this cal lb ack, t hetarget equ ation ch anges asr  gets anew p seu do-ran do m v alue.   As a res ult,the ro ot -findi ng ro ut ine cann ot converge to your des ired s oluti on. 
What you  ne e d t o do inst ead,  conceptua lly , i s tofir s t gen erat e  a  sa m ple o f un i form rand o mvar i a bl es, and app l y the  same , d e termin isti c transformation( i.e. the inver se d i s tri but i on f un ction) to them . Of c ourse yo udon't needto do ro ot - solvi ng, as  you c an
*p <_(3.0*xm*q-abs(tol*q))) then
_   _ __ __  _  _   _  _if_(p < abs(0.5*e*q)) then
          __ ___    _    _ _e = d
     __  _        _     _  _d_=_p / q
 _     _     _      _     _ go to 30
 _ _    _  _ _  _     _ end if
_     __  ___ _  __ end_if
_     __      _ end if
 __   _    _end if

    _ _ _  ! bisection
       _    d =_xm
   _ __  __ e_= d

     _ _   ! complete step
30_   __    a =_b
     _    __fa_= fb
   _     _  if_(abs(d) > tol) b = b +_d
_     __    if (abs(d)_<=_tol)_b = b + sign(tol,xm)
_     _     fb = f(b)
 _  _    ___if (fb*(fc/abs(fc)) > 0.0) go to 10
   _ _  _   go to 20
    _   end if
  _ end if

! done
fn_val = b
end function zeroin

end

The mean of the_resulting samples is about 5, while the standard_deviation is about 1.64.
Question
Does anyone have an_idea where my algorithm_could_become numerically problematic? The fact that the_Python_version_and_the Fortran version_both have the_same problem, but_to different_degrees makes me_believe that_it's some rounding_of_floating point numbers problem, but I cannot image where. Even if the solver returns a rounded value, that_difference should not show up in_a simple histogram.
Does anyone see_a_mistake in my_algorithms? Am I understanding_something wrong?

A:

I only checked the Python version and_there's indeed an error in the implementation.
Namely, your testfunc,_ie_the target_function of root-finding brentq_routine, behaves_non-deterministically. _During a root-finding_run (i.e._one call_to brentq()_until it returns), brentq needs to call_the_same callback multiple times until convergence is reached. _However,_each_time_brentq calls this_callback, the target_equation changes as_r gets a_new_pseudo-random value.  As a result,_the root-finding routine_cannot converge to your_desired solution.
What you need to do instead, conceptually, is to_first generate a sample of uniform random variables, and apply the same, deterministic transformation (i.e. the inverse_distribution function)_to them.  Of_course you_don't_need to do root-solving, as you_can