0
$\begingroup$

The expression which I need to simplify in mathematica into simplest form

Cos[\[Theta]14]]^3 Sin[\[Theta]13]^2 Sin[\[Theta]14] Sin[\[Theta]24] Exp[-i (\[Delta]14 + \[Delta]24 + 2\[Delta]13)] - Sin[\[Theta]14]]^2 Cos[\[Theta]14]] Sin[\[Theta]13]] Exp[-i (\[Delta]13 + 2\[Delta]14)] (Cos[\[Theta]24] Sin[\[Theta]23] Cos[\[Theta]13] -Sin[\[Theta]13] Sin[\[Theta]24] Exp[i(\[Delta]14 - \[Delta]24 - \[Delta]13)] 

On Calculating manually it is simplified into

Cos[\[Theta]14] Sin[\[Theta]13]^2 Sin[\[Theta]14] Sin[\[Theta]24] Exp[-i (\[Delta]14 +\[Delta]24 + 2\[Delta]13)] - Sin[\[Theta]14]^2 Cos[\[Theta]14] Sin[\[Theta]13] Cos[\[Theta]13] Cos[\[Theta]24] Sin[\[Theta]23] Exp[-i (\[Delta]13 + 2\[Delta]14)]
$\endgroup$
9
  • $\begingroup$ Please post the Mathematica code instead of LaTeX code. $\endgroup$
    – cvgmt
    Jul 18 at 7:11
  • $\begingroup$ I've given the exact mathematica code using which I need to simplify the expression into the simplest form possible. $\endgroup$ Jul 18 at 7:58
  • 2
    $\begingroup$ it is a good idea to copy your own code back to your notebook to verify it is correct. If you do that, you will see it is not valid code. To paste code here, best to use the InputForm version of it. Your code as shown does not parse. Try it yourself and see. This is what it looks like !Mathematica graphics also, best to avoid using subscripted variables. $\endgroup$
    – Nasser
    Jul 18 at 8:29
  • 1
    $\begingroup$ In addition to what Nasser said, you are using invalid syntax. Cos^2[x] does not mean what you think it does (in fact it's meaningless). Use Cos[x]^2. Please edit the question, correct these mistakes, and show a complete minimal example of an attempt to simplify an expression. The example should include the original expression in correct syntax, the command you used for simplification, the output you obtain, and the output you want to obtain instead. $\endgroup$
    – Szabolcs
    Jul 18 at 9:07
  • 2
    $\begingroup$ The code is still wrong. $\endgroup$
    – yarchik
    Jul 18 at 12:30
2
$\begingroup$

It is not an answer. I tried to correct the errors in your code. After that the initial expression is

expr1 = Cos[\[Theta]14]^3 Sin[\[Theta]13]^2 Sin[\[Theta]14] Sin[\
\[Theta]24] Exp[-i (\[Delta]14 + \[Delta]24 + 2 \[Delta]13)] - 
  Sin[\[Theta]14]^2 Cos[\[Theta]14] Sin[\[Theta]13] Exp[-i \
(\[Delta]13 + 
       2 \[Delta]14)] (Cos[\[Theta]24] Sin[\[Theta]23] \
Cos[\[Theta]13] - 
     Sin[\[Theta]13] Sin[\[Theta]24] Exp[-i (\[Delta]14 - \[Delta]24 \
- \[Delta]13)]);

The final expression you have simplified manually is

expr2=Cos[\[Theta]14] Sin[\[Theta]13]^2 Sin[\[Theta]14] Sin[\[Theta]24] \
Exp[-i (\[Delta]14 + \[Delta]24 + 2 \[Delta]13)] - 
 Sin[\[Theta]14]^2 Cos[\[Theta]14] Sin[\[Theta]13] Cos[\[Theta]13] \
Cos[\[Theta]24] Sin[\[Theta]23] Exp[-i (\[Delta]13 + 2 \[Delta]14)];

If I corrected everything right it is not difficult to make sure that these two expressions are not equal to one another:

expr1 - expr2 // FullSimplify

(*  E^(-i (2 \[Delta]13 + 4 \[Delta]14 + \[Delta]24))
  Cos[\[Theta]14] Sin[\[Theta]13]^2 Sin[\[Theta]14]^2 (E^(
   i (2 \[Delta]13 + \[Delta]14 + 2 \[Delta]24)) - 
   E^(3 i \[Delta]14) Sin[\[Theta]14]) Sin[\[Theta]24]  *)

To be absolutely sure let us make an arbitrary substitution

(expr1 - expr2 // 
   FullSimplify) /. {\[Theta]14 -> \[Pi]/4, \[Theta]24 -> \[Pi]/
    4, \[Theta]13 -> \[Pi]/4, \[Theta]23 -> 0, \[Delta]13 -> 
   0, \[Delta]14 -> 0, \[Delta]24 -> 0}

(*  1/8 (1 - 1/Sqrt[2])  *)

Again, if I did not introduce something wrong during the correction of your code, my calculations demonstrate that your hand-made simplification contains an error.

On my side, I tried to simplify your expression (In my correction, of course). It is possible to rewrite it in several other forms, which find to be not much simpler than the initial one.

$\endgroup$

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