5
$\begingroup$

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

 FullSimplify[Exp[(9 Zeta[3]^(1/3) n^(2/3))/(4 Log[8]^(2/3)) + (3 c Log[4] Zeta[3]^(1/3) n^(1/3))/(2 Log[8]^(2/3)) - (c^2 Log[4]^2 Zeta[3]^(1/3))/(4 Log[8]^(2/3))] /. c -> ((1 - Log[2]) (3 Zeta[3])^(1/3))/(4  Log[2]^(5/3)), n > 0]

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

but in version 11.3 I get an error message

 (* FullSimplify::infd: Expression 4 n^(1/3) (-3 n^(1/3) Log[2]^(4/3) (3 Zeta[3])^(1/3)-(Log[8] Zeta[3])^(2/3)+Log[2] (Log[8] Zeta[3])^(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).

$\endgroup$
3
  • $\begingroup$ I get no errors on 10.4.1 for Microsoft Windows (64-bit), nor on 11.0.1 for Linux x86 (64-bit). $\endgroup$ Jun 20, 2018 at 17:53
  • $\begingroup$ I get Indeterminate on MMA 11.2 Win7-64 $\endgroup$
    – MarcoB
    Jun 20, 2018 at 18:18
  • $\begingroup$ I also get Indeterminate on 11.1.1, Win10-64 $\endgroup$ Jun 20, 2018 at 21:59

1 Answer 1

2
$\begingroup$
$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[3]^(1/3) n^(2/3))/(4 Log[8]^(2/3)) + (3 c Log[
         4] Zeta[3]^(1/3) n^(1/3))/(2 Log[8]^(2/3)) - (c^2 Log[4]^2 Zeta[
          3]^(1/3))/(4 Log[8]^(2/3))] /. 
   c -> ((1 - Log[2]) (3 Zeta[3])^(1/3))/(4 Log[2]^(5/3))]


(* E^(-((Zeta[3]^(
  1/3) (-144 n^(2/3) Log[2]^(10/3) + 
    24 n^(1/3) (-1 + Log[2]) Log[2]^(5/3) Log[4] (3 Zeta[3])^(
     1/3) + (1 + Log[2]^2 - Log[4]) Log[4]^2 (3 Zeta[3])^(2/3)))/(
 64 Log[2]^(10/3) Log[8]^(2/3)))) *)

Some alternate forms are

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

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

or

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

(* E^(((-1 - Log[2]^2 + Log[4]) (3/Log[8])^(2/3) Zeta[3] + 
 12 n^(2/3) Log[2]^(2/3) (3 Zeta[3])^(1/3) - 
 4 n^(1/3) (Log[2] (3 Zeta[3])^(2/3) - 
    3 3^(1/3) ((Log[2] Zeta[3])/Log[8])^(2/3)))/(16 Log[2]^(4/3))) *)

or

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

(* Cosh[(1/(64 Log[2]^(10/3) Log[8]^(2/3)))
  Zeta[3]^(1/
    3) (-144 n^(2/3) Log[2]^(10/3) + 
     24 n^(1/3) (-1 + Log[2]) Log[2]^(5/3) Log[4] (3 Zeta[3])^(
      1/3) + (1 + Log[2]^2 - Log[4]) Log[4]^2 (3 Zeta[3])^(2/3))] - 
 Sinh[(1/(64 Log[2]^(10/3) Log[8]^(2/3)))
  Zeta[3]^(1/
    3) (-144 n^(2/3) Log[2]^(10/3) + 
     24 n^(1/3) (-1 + Log[2]) Log[2]^(5/3) Log[4] (3 Zeta[3])^(
      1/3) + (1 + Log[2]^2 - Log[4]) Log[4]^2 (3 Zeta[3])^(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[3]^(1/3) (-36 n^(2/3) Log[2]^(4/3) + 
           12 n^(1/3) (-1 + Log[2]) Log[
              2]^(2/3) (3 Zeta[3])^(1/3) + (-1 + Log[2])^2 (3 Zeta[3])^(2/
               3)))/(16 3^(2/3) Log[2]^2))) // FullSimplify

(* True *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.