Skip to main content

equation solving - Q: Problem - FullSimplify returns 0, FindRoot returns value & FindInstance requires system abend


The function at issue:


fa[a_] = 99999.99999999999` (-426.3417145941241` + 2.25` a - 
      2.25` a Erf[
        99999.99999999999` (0.4299932790728411` - 
           0.18257418583505533` Log[a])] + 
      23.714825526419478` Erf[
        99999.99999999999` (0.42999327934670234` - 

           0.18257418583505533` Log[a])]) + 
   9.999999999999998`*^9 a^3.1402269146507883`*^9 E^(
    9.999999999999998`*^9 (-0.36978844033114555` - 
       0.06666666666666667` Log[a]^2)) (a E^(
       1.8749999999999996`*^10 (0.3140226915650789` - 
          0.13333333333333333` Log[a])^2) (0.24259772920294995` - 
         0.10300645387285048` Log[a]) +
      E^(1.8749999999999996`*^10 (-0.3140226913650789` + 
          0.13333333333333333` Log[a])^2) (-2.556961252217486` + 
         1.085680036306589` Log[a]));


Build and plot the function - show it is not always zero and there is one root.


dataa = {#, fa[#]} & /@ Range[1, 1000, 10];
imagea = Plot[fa[a], {a, 0, 400}, Epilog -> {Red, PointSize[0.005], Point[dataa]}];
imagea

Illustration of non zero function


FindRoot is able to find the root - only if searching from above:


FindRoot[fa[a] == 0, {a, 10^10}]



{a -> 100.013}



Show that FullSimplify returns zero - with no warning:


Assuming[a > 0, FullSimplify[fa[a]]]


0.



The following consumes all memory, then thrashes the swap space. The only way to interrupt was: alt+ctl+sysrq+REISUB.



FindInstance[fa[a] == 0 && a > 0, {a}, Reals]

Does anyone observe the same behavior?
Is this expected or should be reported as a bug?


System Information:


SystemInformationData[{"Kernel" -> {
    "Version" -> "11.3.0 for Linux x86 (64-bit) (March 7, 2018)", 
    "ReleaseID" -> "11.3.0.0 (5944640, 2018030701)", 
    "PatchLevel" -> "0", 
    "MachineType" -> "PC", 

    "OperatingSystem" -> "Unix", 
    "ProcessorType" -> "x86-64", 
    "Language" -> "English", 
    "CharacterEncoding" -> "UTF-8", 
    "SystemCharacterEncoding" -> "UTF-8"

...

  "Machine" -> {"MemoryAvailable" -> 
     Quantity[11.852828979492188, "Gibibytes"], 

    "PhysicalUsed" -> Quantity[5.171413421630859, "Gibibytes"], 
    "PhysicalFree" -> Quantity[10.363525390625, "Gibibytes"], 
    "PhysicalTotal" -> Quantity[15.53493881225586, "Gibibytes"], 
    "VirtualUsed" -> Quantity[5.171413421630859, "Gibibytes"], 
    "VirtualFree" -> Quantity[14.234615325927734, "Gibibytes"], 
    "VirtualTotal" -> Quantity[19.406028747558594, "Gibibytes"], 
    "PageSize" -> Quantity[4., "Kibibytes"], 
    "PageUsed" -> Quantity[3.8710899353027344, "Gibibytes"], 
    "PageFree" -> Quantity[0, "Bytes"], 
    "PageTotal" -> Quantity[3.8710899353027344, "Gibibytes"], 

    "Active" -> Quantity[3.342662811279297, "Gibibytes"], 
    "Inactive" -> Quantity[1.4980888366699219, "Gibibytes"], 
    "Cached" -> Quantity[1.8926506042480469, "Gibibytes"], 
    "Buffers" -> Quantity[225.7890625, "Mebibytes"], 
    "SwapReclaimable" -> Quantity[96.015625, "Mebibytes"]}}]

Answer



It seems to me that fa[a] behaves relatively well with respect to FindRoot and plotting but not simplification. That a computation, whether or not it is FindInstance[], might exhaust system resources is unremarkable, but it's likely that an exact-symbolic computation with floating-point numbers will be worse. On things that would cancel or equal each other, round-off error sometimes makes them not do so, thereby making the problem more complicated. Also, Mathematica might rationalize the numbers and convert them to ratios of very large integers; while that cures the round-off problem, it can make the algebra seem very complicated.


The reason for the failure of Simplify[fa[a]] is the behavior of machine underflow. Somehow, Simplify suppresses the message; probably someone on the site can say how to get the message to be unsuppressed. A vestige of it can be summoned with $MessagePrePrint:


ClearSystemCache[];
Block[{$MessagePrePrint = (Print[FullForm@#]; #) &},

 Simplify[fa[a]]
 ]
(*
HoldForm[$MessageList]

HoldForm[Times[-224999.99999999997`,
 8.42822696534888596`6.386636439297712*^-1605970792]]

0.
*)


We can see that the expression culled results in underflow:


Times[-224999.99999999997`, 
 8.42822696534888596`6.386636439297712*^-1605970792]


General::munfl: -225000. 8.42823*10^-1605970792 is too small to represent as a normalized machine number; precision may be lost.



    (*  0.  *)


It's one of the pitfalls of using machine-precision in an exact-symbolic way. Arbitrary-precision is more robust, but it has its limitations, too.


Remark: You might notice the second factor of Times is an arbitrary precision number. This arose from machine-number overflow. (Underflow used to work in the same way, but Wolfram changed the behavior in V11.3.)


Update: I found the command, or at least another one, that leads to the problem, Factor.


Factor[fa[a]]


General::munfl: -225000. 8.42823*10^-1605970792 is too small to represent as a normalized machine number; precision may be lost.



(*  0.  *)

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more