# 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 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"]}}]

• What is the definition of fa? – Rohit Namjoshi Jan 14 at 4:36
• Fixed function name. – Hedgehog Jan 14 at 5:28
• Adding the WorkingPrecision->20 option, I immediately obtain "General::ovfl: Overflow occurred in computation." – user64494 Jan 14 at 7:06
• @user64494 which function are you referring to? Do this mean that, otherwise, you observe the same behavior? Particularly interested if FindInstance cripples your system. – Hedgehog Jan 14 at 7:51
• 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? – Michael E2 Jan 14 at 17:07 ## 2 Answers 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,
8.428226965348885966.386636439297712*^-1605970792]]

0.
*)


We can see that the expression culled results in underflow:

Times[-224999.99999999997,
8.428226965348885966.386636439297712*^-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.  *)


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.

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