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Read exp(x) as a power of e rather than as a product (#730) - #731
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`exp` is the ordinary name for the exponential function everywhere else
-- sympy, Mathematica, MATLAB, numpy -- and the grammar did not have it.
It did not fail. It fell through to implicit multiplication, the rule
that lets `a(b + c)` mean `a * (b + c)`, and came out as the product of
an undeclared variable named `exp` with the argument. Nothing was said
about it, so the misreading propagated into an answer that looked like
one: `exp(x) - 3x = 0` answered `{ 0 }`, which is a root of `exp * x -
3x` and of nothing else. The equation has two real roots, near 0.6191
and 1.5121.
That is the same shape as the arcsinh case -- a name that looks like a
function, is not one, and is quietly absorbed by implicit
multiplication. The difference is that arcsinh is a misnomer and was
right to refuse, whereas exp names something the library already has.
So it maps to `MathS.Pow(MathS.e, arg)` rather than to a new node. A
distinct Expf would have to be taught differentiation, integration and
simplification over again; a power of e already has all of it, which is
why `exp(x) * exp(y)` comes out `e^(x + y)` with nothing further added.
Only the exact name followed by a bracket is taken as the function, as
for every other function in the grammar: `expr(x)`, `expo(x)` and a bare
`exp` are the products they were.
Measured: `exp(x) - 3x = 0` from `{ 0 }` to both real roots. Parser
regenerated with antlr-4.13.1 and the post-processor; regenerating the
unmodified grammar first gave an empty diff, so the only change is the
rule. Full suite 4866 passed / 0 failed, F# 130/130, corpus 112/117 with
0 wrong, 0 error, 0 timeout.
Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
This was referenced Aug 5, 2026
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Closes #730.
expis the ordinary name for the exponential function everywhere else — sympy, Mathematica, MATLAB, numpy — and the grammar did not have it.What was wrong
It did not fail. It fell through to implicit multiplication, the rule that lets
a(b + c)meana * (b + c), and came out as the product of an undeclared variable namedexpwith the argument:Nothing was said about it, so the misreading propagated into an answer that looked like one:
0is a root ofexp * x - 3xand of nothing else. The equation has two real roots, near 0.6190612867 and 1.5121345517. A wrong answer to a misread question, rather than a refusal to answer an unsupported one.This is the same shape as the
arcsinhcase fixed earlier. The difference is thatarcsinhis a misnomer and was right to refuse, whereasexpnames something the library already has.The fix
One grammar rule, mapping to
MathS.Pow(MathS.e, arg)rather than to a new node. A distinctExpfwould have to be taught differentiation, integration and simplification over again; a power ofealready has all of it — which is whyexp(x) * exp(y)comes oute ^ (x + y)with nothing further added.Only the exact name followed by a bracket is the function, as for every other function in the grammar:
expr(x),expo(x),aexp(x)and a bareexpare the products they were, and a test pins that.Measured
exp(x)exp * xe ^ xexp(1)evaluatedexp2.71828…exp(x) - 3x = 0{ 0 }d/dx exp(2x)2 * exp2 * e ^ (2 * x)Parser regenerated with the committed
antlr-4.13.1-complete.jarand theAntlrPostProcessorReplacePublicWithInternalstep. Regenerating the unmodified grammar first produced an empty diff, so the toolchain here reproduces the committed files and the only change is the new rule.Full suite 4866 passed / 0 failed, F# 130/130, 117-problem corpus 112/117 with 0 wrong, 0 error, 0 timeout.
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