1
$\begingroup$

In the following simplification why doesn't Mathematica get a zero when asked to compute "Y-Z" eventually?

 a = 1 - 4 A Q^2; 
 b = (-972 + 648) A Q^2 + 54;
 c = 9 - 36 A Q^2; 

 Y = 12 Q/( 
 Sqrt[A] (  
   2 - (3 2^(1/3) a )/(Sqrt[b^2 - 4 c^3] + b  )^(1/3)  - ( 
       Sqrt[b^2 - 4 c^3] + b  )^(1/3)/( 3 2^(1/3) )     )   );

 Z = (4/A) + (3 2^(1/3) a )/(A (Sqrt[b^2 - 4 c^3] + b  )^(1/3) )  +  ( 
  Sqrt[b^2 - 4 c^3] + b  )^(1/3)/( 3 2^(1/3) A);

Q = Sqrt [ 3/(16 A)];
Y - Z // FullSimplify
(-9 + 4 Sqrt[3] Sqrt[1/A] Sqrt[A])/(2 A)

One can do the above calculation by hand and one would see that for the specific chosen value of $Q = \sqrt{3/16 A}$ one would get $Y = Z = \frac{3}{A}$. Why doesn't Mathematica see this?


  • And is there a way to get Mathematica to detect if there are other values of $Q$ where $Y = Z$? (..I found this one value special value of $Q$ by just staring at the equation for sometime...)

Rewriting the functions again.

$Y = \frac{12Q}{\sqrt{A}\sqrt{ 2 - \frac{3 (2^{1/3})a }{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} } - \frac{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }{3 (2^{1/3})} } }$

$Z = \frac{1}{A} \left (4 + \frac{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }{3 (2^{1/3})} + \frac{3 (2^{1/3})a }{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }\right ) $

at the chosen values of $Q = \sqrt{\frac{3}{16A} }$ one has $a = \frac{1}{4}$ and $b = - \frac{27}{4}$ and $b^2 - 4 c^3 = 0$. Also whenever I encounter $b^{1/3}$ I am writing that as $-3 \times 2^{-2/3}$. Substituting these into the above one gets $Y = Z = \frac{3}{A}$

$\endgroup$
16
  • $\begingroup$ You can try $Assumptions = {A > 0}; Y - Z // FullSimplify but it seems there is a mistake somewhere. $\endgroup$
    – Kuba
    Jun 26 '14 at 10:28
  • $\begingroup$ Getting $\frac{4 \sqrt{3} \sqrt{\frac{1}{A}} \sqrt{A}-9}{2 A}$ and with assuming A > 0 $\frac{4 \sqrt{3}-9}{2 A}$. No Zero. $\endgroup$
    – Phab
    Jun 26 '14 at 10:36
  • $\begingroup$ @Phab But did you do the calculation by hand? Isn't it zero doing so? $\endgroup$
    – Anirbit
    Jun 26 '14 at 11:35
  • $\begingroup$ @Anirbit Not on the first try. No time for a second so far. $\endgroup$
    – Phab
    Jun 26 '14 at 12:36
  • $\begingroup$ @Phab I am quite confident that it is zero. (there are other cross-checks which it passes given the larger context fro where it comes) It would be great if you could kindly check again. $\endgroup$
    – Anirbit
    Jun 26 '14 at 13:09
6
$\begingroup$

The problem is that b^(1/3) has three roots. You are choosing the real root, -(3/2^(2/3)), while Mathematica is choosing, (3 (-1)^(1/3))/2^(2/3) which is complex. To get the result you want, rewrite your expressions for Y and Z like this:

Y =
  12 Q/
    Sqrt[A (2 - 
            (3 2^(1/3) a)/Surd[Sqrt[b^2 - 4 c^3] + b, 3] - 
            Surd[Sqrt[b^2 - 4 c^3] + b, 3]/(3 2^(1/3)))];

Z = (4/A) + (3 2^(1/3) a)/A /Surd[Sqrt[b^2 - 4 c^3] + b, 3] + 
   Surd[Sqrt[b^2 - 4 c^3] + b, 3]/(3 2^(1/3) A);

With the above you will get

FullSimplify[Y - Z, Assumptions -> A > 0]
0
$\endgroup$

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.