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]}];

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]]]


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" -> " (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"]}}]
  • $\begingroup$ What is the definition of fa? $\endgroup$ – Rohit Namjoshi Jan 14 '20 at 4:36
  • $\begingroup$ Fixed function name. $\endgroup$ – Hedgehog Jan 14 '20 at 5:28
  • $\begingroup$ Adding the WorkingPrecision->20 option, I immediately obtain "General::ovfl: Overflow occurred in computation." $\endgroup$ – user64494 Jan 14 '20 at 7:06
  • $\begingroup$ @user64494 which function are you referring to? Do this mean that, otherwise, you observe the same behavior? Particularly interested if FindInstance cripples your system. $\endgroup$ – Hedgehog Jan 14 '20 at 7:51
  • 1
    $\begingroup$ FindRoot works for lower values of the starting point for me, greater than around 10.7 below which the derivative suffers from underflow. FindRoot[fa[a] == 0, {a, 1}, WorkingPrecision -> $MachinePrecision] fixes that. I take it the main question is why does Simplify return an unreliable answer? $\endgroup$ – Michael E2 Jan 14 '20 at 17:07

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:

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



We can see that the expression culled results in underflow:


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.


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

(*  0.  *)

I think that mixing of huge numbers and machine precision is what's making Mathematica go crazy (so this is not a bug, but a lack of feature). In general, running functions best suited for analytic expression transformations like FullSimplify and FindInstance on expressions with floating point numbers is a bad idea except in simple, well-understood cases.

Let's find a good approximation to your function.

E^(9.999999999999998*^9 (-0.36978844033114555 - 0.06666666666666667` Log[a]^2)) is almost zero to millions of digits of precision when a>11, so let's replace it exactly with 0.

We're left with a much simpler expression:

100000. (-426.342 + 2.25 a - 2.25 a Erf[100000. (0.429993 - 0.182574 Log[a])] + 23.7148 Erf[100000. (0.429993 - 0.182574 Log[a])])

When a>11, both Erf functions here become almost equal to -1, so the final approximation is

fb[a_] = 100000. (-450.057 + 4.5 a)

It's easy to check that with machine precision and a>11 the approximation is equivalent to the original, and finding its zero becomes trivial.

  • 1
    $\begingroup$ FindRoot seems to do a better job finding the root of the original expression, though. And it's fast and simple. $\endgroup$ – Michael E2 Jan 14 '20 at 17:04

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