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Now I'm trying to use Mathematica to simplify an expression like this:

$P \sqrt{-P + \sqrt{M^2 + P^2}} + \sqrt{(M^2 + P^2) (-P + \sqrt{M^2 + P^2})} - M \sqrt{P + \sqrt{M^2 + P^2}}$

where $P,M>0$

It should be zero, obviously. But Mathematica cannot do this simplification. Is there any technique that will help Mathematica to deal with this kind of expression?

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  • $\begingroup$ There are three left brackets in your formulate but just two right brackets.Also "It should be zero" is not correct,for example P=1, M=-2. $\endgroup$
    – Apple
    Commented Jun 15, 2014 at 0:35

2 Answers 2

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I always find that I need to give Mathematica some help when manipulating this sort of expression, but the reason is (invariably?) that I have to make some assumptions in order to make progress through the simplification. For instance, you have to be careful with "simplifying" expressions such as Sqrt[M^2] to M (i.e. using PowerExpand), because this introduces the assumption M > 0.

Here is a sequence of steps that you could use to simplify your expression (evaluate it to see the intermediate outputs):

P Sqrt[\[Minus]P + Sqrt[M^2 + P^2]] + Sqrt[(M^2 + P^2) (\[Minus]P + Sqrt[M^2 + P^2])] \[Minus] M Sqrt[P + Sqrt[M^2 + P^2]]

% // PowerExpand

% // FullSimplify

% /. Sqrt[a_] b_ + a_ c_ :> Sqrt[a] (b + Sqrt[a] c)

% /. Sqrt[a_ + b_] Sqrt[a_ - b_] :> Sqrt[a^2 - b^2]

% // PowerExpand

(* 0 *)

M >= 0 is implicit in the final step of the derivation.

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    $\begingroup$ Great answer. So why isn't this stuff already in MMa functions??? $\endgroup$ Commented Sep 6, 2016 at 2:46
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I like such questions, and like very much the answer of Stephen Luttrell. Such questions are fun by themselves, and for this reason I am giving my version of the answer too. I propose to first multiply the result by Sqrt[p + Sqrt[m^2 + p^2]] and then to transform. This is the expression:

 expr1 = p*Sqrt[-p + Sqrt[m^2 + p^2]] + 
   Sqrt[(m^2 + p^2)*(Sqrt[m^2 + p^2] - p)] - 
   m*Sqrt[p + Sqrt[m^2 + p^2]];

now let us multiply and Expand:

 expr2 = expr1*Sqrt[p + Sqrt[m^2 + p^2]] // Expand

It gives this:

(*   -m p - m Sqrt[m^2 + p^2] + 
 p Sqrt[-p + Sqrt[m^2 + p^2]] Sqrt[p + Sqrt[m^2 + p^2]] + 
 Sqrt[(m^2 + p^2) (-p + Sqrt[m^2 + p^2])] Sqrt[p + Sqrt[m^2 + p^2]]  *)

Let us now teach Mma to fuse the square roots:

expr3 = Map[ReplaceAll[#, Sqrt[a_]*Sqrt[b_] -> Sqrt[a*b]] &, expr2]

yielding the following:

(*    -m p - m Sqrt[m^2 + p^2] + 
 p Sqrt[(-p + Sqrt[m^2 + p^2]) (p + Sqrt[m^2 + p^2])] + Sqrt[(m^2 + 
    p^2) (-p + Sqrt[m^2 + p^2]) (p + Sqrt[m^2 + p^2])]                *)

And as the final stroke let us open parentheses under the radicals:

Map[Expand, expr3, 4] // Simplify[#, {m > 0, p > 0}] &

giving

(* 0 *)

as you expected.

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