0
$\begingroup$

I have read lots of questions on square root simplification on MMA, but I haven't really seen my problem posted (or understood a clear way to deal with this issue).

I am simplifying some expressions "manually" as the results from MMA seems too complex. I find out that MMA do not check correctly the third and fourth following simplification.

To explain things in order, look at the following simplifications, which have increasing complexity. The left hand side is the original form, the right one is the one I propose. The third and fourth one do not returns the value True.

Am I missing something or it is a MMA issue? Note: in my original code I put only the assumptions -1<=p<=1 && 0<=t<=2\[Pi] in $Assumptions, but I still have the problem. Also, the assumption 1-2 Q+Sqrt[1-p^2] Cos[t]>0 should be implicit since MMA should recognize from -1<=p<=1 && 0<=t<=2\[Pi] that 1-Sqrt[1-p^2] Cos[t]>=0 and then 1-2 Q+Sqrt[1-p^2] Cos[t]>0.

Assuming[a > 0 && b >= 0, Simplify[Sqrt[a b]/a == Sqrt[b/a]]]
(*returns True*)

Assuming[1-2 Q+Sqrt[1-p^2] Cos[t]>0 &&b>=0,Sqrt[(1-2 Q+Sqrt[1-p^2] Cos[t])(b)]/(1-2 Q+Sqrt[1-p^2] Cos[t]) == Sqrt[b/(1-2 Q+Sqrt[1-p^2] Cos[t])] // Simplify]
(*returns True*)

Assuming[1-2 Q+Sqrt[1-p^2] Cos[t]>0 && -1<=p<=1 && 0<=t<=2\[Pi],Sqrt[(1-2 Q+Sqrt[1-p^2] Cos[t])(1-Sqrt[1-p^2] Cos[t])]/(1-2 Q+Sqrt[1-p^2] Cos[t]) == Sqrt[(1-Sqrt[1-p^2] Cos[t])/(1-2 Q+Sqrt[1-p^2] Cos[t])]// Simplify]
(*returns the expression*)

Assuming[1-2 Q+Sqrt[1-p^2] Cos[t]>0 &&1-Sqrt[1-p^2] Cos[t]>=0,Sqrt[(1-2 Q+Sqrt[1-p^2] Cos[t])(1-Sqrt[1-p^2] Cos[t])]/(1-2 Q+Sqrt[1-p^2] Cos[t]) == Sqrt[(1-Sqrt[1-p^2] Cos[t])/(1-2 Q+Sqrt[1-p^2] Cos[t])]// Simplify]
(*returns the expression*)
$\endgroup$
1
$\begingroup$

For your third and fourth cases you need to use FullSimplify rather than Simplify. As stated in the documentation for Simplify:

FullSimplify does more extensive simplification than Simplify.

Whenever, Simplify "fails" you should try using FullSimplify before concluding that there is an issue.

Assuming[1 - 2 Q + Sqrt[1 - p^2] Cos[t] > 0 && -1 <= p <= 1 && 0 <= t <= 2 Pi,
  Sqrt[(1 - 2 Q + Sqrt[1 - p^2] Cos[t]) (1 - Sqrt[1 - p^2] Cos[t])]/(1 - 2 Q +
       Sqrt[1 - p^2] Cos[t]) == 
   Sqrt[(1 - Sqrt[1 - p^2] Cos[t])/(1 - 2 Q + Sqrt[1 - p^2] Cos[t])] // 
  FullSimplify]

True

Assuming[1 - 2 Q + Sqrt[1 - p^2] Cos[t] > 0 && 1 - Sqrt[1 - p^2] Cos[t] >= 0, 
 Sqrt[(1 - 2 Q + Sqrt[1 - p^2] Cos[t]) (1 - Sqrt[1 - p^2] Cos[t])]/(1 - 2 Q + 
      Sqrt[1 - p^2] Cos[t]) == 
   Sqrt[(1 - Sqrt[1 - p^2] Cos[t])/(1 - 2 Q + Sqrt[1 - p^2] Cos[t])] // 
  FullSimplify]

True

$\endgroup$
  • $\begingroup$ Thank you. Do you agree that the assumption 1 - Sqrt[1 - p^2] Cos[t] >= 0 is not necessary? If I omit that however, even FullSimplify fails. EDIT the original assumptions are 0 <= Q <= 1 && -1 <= p <= 1 && 0 <= t <= 2 [Pi] $\endgroup$ – Nicola Jan 30 '15 at 3:43
  • $\begingroup$ I rephrase. Do you agree that the assumption 1 - Sqrt[1 - p^2] Cos[t] >= 0 should not be necessary? $\endgroup$ – Nicola Jan 30 '15 at 3:50
  • $\begingroup$ No. The assumption that you use is 1-2Q+Sqrt[1-p^2] Cos[t]>0. This is not implied by 1-Sqrt[1-p^2] Cos[t]>=0 since nothing is known about Q. $\endgroup$ – Bob Hanlon Jan 30 '15 at 3:59
  • $\begingroup$ But if you study the sign of the term in the square root on the LHS (remember the LHS is the original one given by MMA), i.e. Sqrt[(1 - 2 Q + Sqrt[1 - p^2] Cos[t]) (1 - Sqrt[1 - p^2] Cos[t])], you should conclude that 1 - 2 Q + Sqrt[1 - p^2] Cos[t]>=0 since 1 - Sqrt[1 - p^2] Cos[t]>=0 (by assumption on p,t or on the whole term). And with a not null denominator you have 1 - 2 Q + Sqrt[1 - p^2] Cos[t]>0. In true, Q do not really matter. Do you agree? $\endgroup$ – Nicola Jan 30 '15 at 4:11
  • 2
    $\begingroup$ Use Reduce to find the conditions for the inequality to be valid. $\endgroup$ – Bob Hanlon Jan 30 '15 at 4:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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