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=√3/16AQ = \sqrt{3/16 A} one would get Y=Z=3AY = 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 QQ where Y=ZY = Z? (..I found this one value special value of QQ by just staring at the equation for sometime…)

Rewriting the functions again.

Y=12Q√A√2−3(21/3)a(√b2−4c3+b)1/3−(√b2−4c3+b)1/33(21/3)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=1A(4+(√b2−4c3+b)1/33(21/3)+3(21/3)a(√b2−4c3+b)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=√316AQ = \sqrt{\frac{3}{16A} } one has a=14a = \frac{1}{4} and b=−274b = – \frac{27}{4} and b2−4c3=0b^2 – 4 c^3 = 0. Also whenever I encounter b1/3b^{1/3} I am writing that as −3×2−2/3-3 \times 2^{-2/3}. Substituting these into the above one gets Y=Z=3AY = Z = \frac{3}{A}

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You can try $Assumptions = {A > 0}; Y – Z // FullSimplify but it seems there is a mistake somewhere.

– Kuba

Jun 26 ’14 at 10:28

Getting 4√3√1A√A−92A\frac{4 \sqrt{3} \sqrt{\frac{1}{A}} \sqrt{A}-9}{2 A} and with assuming A > 0 4√3−92A\frac{4 \sqrt{3}-9}{2 A}. No Zero.

– Phab

Jun 26 ’14 at 10:36

@Phab But did you do the calculation by hand? Isn’t it zero doing so?

– Anirbit

Jun 26 ’14 at 11:35

@Anirbit Not on the first try. No time for a second so far.

– Phab

Jun 26 ’14 at 12:36

@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.

– Anirbit

Jun 26 ’14 at 13:09

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1 Answer

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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]

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