# FullSimplify: Expression simplified to Indeterminate (in 11.3)

The following expression is correctly simplified in versions 7, 10.2, 11.0

 FullSimplify[Exp[(9 Zeta^(1/3) n^(2/3))/(4 Log^(2/3)) + (3 c Log Zeta^(1/3) n^(1/3))/(2 Log^(2/3)) - (c^2 Log^2 Zeta^(1/3))/(4 Log^(2/3))] /. c -> ((1 - Log) (3 Zeta)^(1/3))/(4  Log^(5/3)), n > 0]

(* E^(-((Zeta^(1/3) (-36 n^(2/3) Log^(4/3) + 12 n^(1/3) (-1 + Log) Log^(2/3) (3 Zeta)^(1/3) + (-1 + Log)^2 (3 Zeta)^(2/3)))/(16 3^(2/3) Log^2))) *)


but in version 11.3 I get an error message

 (* FullSimplify::infd: Expression 4 n^(1/3) (-3 n^(1/3) Log^(4/3) (3 Zeta)^(1/3)-(Log Zeta)^(2/3)+Log (Log Zeta)^(2/3)) simplified to Indeterminate. *)


Why ? There is no reason for such message.

With Simplify instead FullSimplify is the calculation (in 11.3) correct (with a little longer output).

• I get no errors on 10.4.1 for Microsoft Windows (64-bit), nor on 11.0.1 for Linux x86 (64-bit). – AccidentalFourierTransform Jun 20 '18 at 17:53
• I get Indeterminate on MMA 11.2 Win7-64 – MarcoB Jun 20 '18 at 18:18
• I also get Indeterminate on 11.1.1, Win10-64 – JungHwan Min Jun 20 '18 at 21:59

\$Version

(* "11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)" *)


Using Simplify rather than FullSimplify

f[n_] = Simplify[
Exp[(9 Zeta^(1/3) n^(2/3))/(4 Log^(2/3)) + (3 c Log[
4] Zeta^(1/3) n^(1/3))/(2 Log^(2/3)) - (c^2 Log^2 Zeta[
3]^(1/3))/(4 Log^(2/3))] /.
c -> ((1 - Log) (3 Zeta)^(1/3))/(4 Log^(5/3))]

(* E^(-((Zeta^(
1/3) (-144 n^(2/3) Log^(10/3) +
24 n^(1/3) (-1 + Log) Log^(5/3) Log (3 Zeta)^(
1/3) + (1 + Log^2 - Log) Log^2 (3 Zeta)^(2/3)))/(
64 Log^(10/3) Log^(2/3)))) *)


Some alternate forms are

f2[n_] = f[n] // ExpandAll // FullSimplify

(* 2^(-((n^(1/3) (-1 + Log) (3 Zeta)^(2/3))/(
4 Log^(7/3)))) E^(-(((-1 + Log)^2 Zeta)/(16 Log^2)) + (
3 n^(2/3) (3 Zeta)^(1/3))/(4 Log^(2/3))) *)


or

f3[n_] = f[n] // ExpandAll // PowerExpand // Simplify

(* E^(((-1 - Log^2 + Log) (3/Log)^(2/3) Zeta +
12 n^(2/3) Log^(2/3) (3 Zeta)^(1/3) -
4 n^(1/3) (Log (3 Zeta)^(2/3) -
3 3^(1/3) ((Log Zeta)/Log)^(2/3)))/(16 Log^(4/3))) *)


or

f4[n_] = f[n] // ExpToTrig // Simplify

(* Cosh[(1/(64 Log^(10/3) Log^(2/3)))
Zeta^(1/
3) (-144 n^(2/3) Log^(10/3) +
24 n^(1/3) (-1 + Log) Log^(5/3) Log (3 Zeta)^(
1/3) + (1 + Log^2 - Log) Log^2 (3 Zeta)^(2/3))] -
Sinh[(1/(64 Log^(10/3) Log^(2/3)))
Zeta^(1/
3) (-144 n^(2/3) Log^(10/3) +
24 n^(1/3) (-1 + Log) Log^(5/3) Log (3 Zeta)^(
1/3) + (1 + Log^2 - Log) Log^2 (3 Zeta)^(2/3))] *)


All of these expressions are equivalent to the result that you provided from earlier versions

f[n] == f2[n] == f3[n] == f4[n] ==
E^(-((Zeta^(1/3) (-36 n^(2/3) Log^(4/3) +
12 n^(1/3) (-1 + Log) Log[
2]^(2/3) (3 Zeta)^(1/3) + (-1 + Log)^2 (3 Zeta)^(2/
3)))/(16 3^(2/3) Log^2))) // FullSimplify

(* True *)