Strategies for simplifying complicated expressions

Question

I have a very complicated expression involving trigonometric functions, complex numbers etc. You may find it here as it is too long to be pasted here. You may also find a screenshot of it here.

Assumptions: All variables are real and strictly positive. This expression should be real!

I'm dealing with this expression for days. Could someone tell me a strategy for simplifying it within Mathematica 8 ?

More generaly, are there good tutorials or textbooks or lessons on strategies for simplifications? Perhaps is simplification a subjective concept? In my case it means more readable, and why not something that could fit in a research paper.

Answer 1

I don't know why you would expect Mathematica to understand that this is a real expression when you have I, (-1)^(1/4) and (-1)^(3/4) at various places in the expression.

(-1)^(1/4) // N
(* Out[1]= 0.707107 + 0.707107 I *)

(-1)^(3/4) // N
(* Out[2]= -0.707107 + 0.707107 I *)

In this particular instance, the denominator is pretty obviously real.

Simplify[Sign@Denominator[finalnew], 
  Assumptions -> {ka > 0, kd > 0, L > 0, ω > 0, d > 0}]

(*
Out[3]= Sign[(ka^4 ω^2 + d^2 (kd^2 + ω^2)^2 - 4 d ka^2 ω (kd^2 - kd ω + ω^2)) 
    Cos[ Sqrt[2] L Sqrt[ω/d]] - (ka^4 ω^2 + d^2 (kd^2 + ω^2)^2 + 4 d ka^2 ω (kd^2 + kd ω + ω^2)) 
    Cosh[Sqrt[2] L Sqrt[ω/d]] - 2 Sqrt[2] ka ((d^(3/2) (kd - ω) ω^(5/2) - 
        kd^2 (d ω)^(3/2) + kd^3 Sqrt[d^3 ω] - ka^2 kd Sqrt[d ω^3] + ka^2 Sqrt[d ω^5]) 
    Sin[Sqrt[2] L Sqrt[ω/d]] + (kd^2 (d ω)^(3/2) + kd^3 Sqrt[d^3 ω] + ka^2 kd Sqrt[d ω^3] + 
        ka^2 Sqrt[d ω^5] + d^(3/2) ω^(5/2) (kd + ω)) 
    Sinh[ Sqrt[2] L Sqrt[ω/d]])
]
*)

Simplify[Im@Denominator[finalnew], Assumptions -> {ka > 0, kd > 0, L > 0, ω > 0, d > 0, u > 0}]
(* Out[4]= 0 *)

So concentrate on refining the numerator. The equivalent

Simplify[Im@Numerator[finalnew], 
 Assumptions -> {ka > 0, kd > 0, L > 0, ω > 0, d > 0, u > 0}]

does not simplify to zero. For example:

Expand@Im@Numerator[finalnew] /. {ka -> 1, kd -> 1, L -> 1, ω -> 1, d -> 1, u -> 1}

(*
Out[5]= 6 Im[(-24 - 168 I) Cos[(-1)^(1/4)] - (24 - 168 I) Cosh[(-1)^(1/4)] - 
    (78 - 102 I) (-1)^(1/4) Sin[(-1)^(1/4)] + (42 + 24 I) (-1)^(3/4) Sin[(-1)^(1/4)] +
    (108 + 144 I) (-1)^(1/4) Sinh[(-1)^(1/4)] + 6 (-1)^(3/4) Sinh[(-1)^(3/4)]]
*)

You can confirm the numerator is also real for specific values of the parameters:

FullSimplify@ Im[Numerator[finalnew] /. {ka -> 1, kd -> 1, L -> 1, ω -> 1,  d -> 1, u -> 1}] 
(* result is zero *)

But this takes an inordinate amount of time and the result is a long complex expression.

FullSimplify[Im[Numerator[finalnew]], Assumptions -> {ka > 0, kd > 0, L > 0, ω > 0, d > 0, u > 0}]

As for general strategies, FullSimplify with as many variables in the Assumptions option is a good bet in most cases, as is separately simplifying numerators and denominators. I don't know if there are best-practice strategies, though. I would expect it would depend on the kind of expression, for example, whether it is a polynomial or contains trigonometric expressions.

Answer 2

To show that your expression is real you could use ComplexExpand. By default, ComplexExpand will expand a complex expression into its real and imaginary part under the assumption that all undefined symbols occurring in the expression are real. The only problem here is that as far as I know there is no way to provide extra assumptions to ComplexExpand, especially that ω and d are positive which means that Sqrt[ω] and Sqrt[d] are real as well. The easiest way around this is to replace these symbols with the square of another symbol, e.g.

Simplify[ComplexExpand[Im[finalnew /. {ω -> om^2, d -> dd^2}]], dd > 0]

which returns 0.

Answer 3

Building on @Verbeia's observation that the denominator is real:

almostThere = ComplexExpand[1/2 (Numerator[finalnew] - Conjugate[Numerator[finalnew]]), 
    TargetFunctions -> {Re, Im}];

Then if you can make the assumptions ArcTan[x_, 0] == 0, ω > 0, d > 0 :

Simplify[almostThere /. (ArcTan[x_, 0] -> 0), Assumptions -> {ω > 0, d > 0}]
(* Out[1]= 0 *)