CODEWITHSUNDEEP
wolfram-mathematica
rprospero
rprospero

Reputation: 961

Simplify Absolute Value in Mathematica

I currently have a large expression with many terms of the form 

Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]

I know, from the geometry of my problem, that

-2 b + 2 d1 m + l Tan[\[Theta]] > 0

However, when I try to simplify my expression,

Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + l Tan[\[Theta]] > 0]

I just get back

Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]

How can I make Mathematica simplify out the unnecessary absolute value?

EDIT 1

The full expression which I'm trying to simplify is 

-(1/(2 (m - Tan[\[Theta]])))
 Sqrt[1 + m^2] (B2 Sqrt[(-2 b + 2 d1 m + l Tan[\[Theta]])^2] + 
    B4 Sqrt[(-2 b + 2 d2 m + l Tan[\[Theta]])^2] + 
    B5 Sqrt[(2 b + 2 d3 m + l Tan[\[Theta]])^2] + 
    B7 Sqrt[(2 b + 2 d4 m + l Tan[\[Theta]])^2] + 
    B1 Sqrt[(2 b - 2 (d1 + l) m + l Tan[\[Theta]])^2] + 
    B3 Sqrt[(2 b - 2 (d2 + l) m + l Tan[\[Theta]])^2] + 
    B6 Sqrt[(-2 (b + (d3 + l) m) + l Tan[\[Theta]])^2] + 
    B8 Sqrt[(-2 (b + (d4 + l) m) + l Tan[\[Theta]])^2])

The terms being squared under each of the radicals is known to be a positive real number.

Upvotes: 6

Views: 6594

Answers (4)

Travis Bemrose
Travis Bemrose

Reputation: 442

I'm constantly stimied by things like Abs[a]^2, and stuff like using Assuming with a\[Element]Reals doesn't help.

I found some help here WolframMathWorld - Absolute Square with
ComplexExpand[Abs[a]^2, TargetFunctions -> {Conjugate}], but sometimes it still returns stuff like Conjugate[Sqrt[a^2 + b^2]] and I've found wrapping a second ComplexExpand (without parameters) around that helps.

Upvotes: 1

Mike Bailey
Mike Bailey

Reputation: 12817

If you only wish to remove specific instances of absolute value, you could do something along these lines:

Clear[removeAbs]
removeAbs[expr_, r_] := expr /. {Sqrt[r^2] :> r, Abs[r] :> r}

That way it only removes the absolute value from any expressions you specify:

In: removeAbs[Abs[x] + Abs[y], x]
Out: x + Abs[y]

I'll see if I can find a nicer looking solution than this.

Upvotes: 1

abcd
abcd

Reputation: 42225

Since the terms are all known to be real and positive, squaring and taking the square-root will only give you the same number. Hence, you could do something like 

expr /. Sqrt[(x___)^2] :> x

where expr is your giant expression above.

Upvotes: 6

user1054186
user1054186

Reputation:

Here are two ideas:

1) 

Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], 
 0 < \[Theta] < \[Pi]/2 && l > 0 && 2 d1 m > 0 && -2 b > 0]

2)

f[e_] := 100 Count[e, _Abs, {0, Infinity}] + LeafCount[e]
Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + 
   l Tan[\[Theta]] > 0, ComplexityFunction -> f]

Th complexity function f makes Abs more expensive than Times. See docu for Simplify. Does that help?

Upvotes: 4

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