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I would like to calculate the following two expressions using Wolfram Alpha:

$$z = (x (d^2 + d (4 y - 6) - 8 y^2 + 12 y - 3) + 6 (d - 1) (y - 1) y)/(d^2 - 2 d y + 4 y^2 - 6 y + 3)$$ and

$$w = -(\sqrt3 (d^2 (x + 2 y - 1) - 2 d (x (2 y - 1) + y^2) + x (4 y - 3) - 2 y^2 + 6 y - 3))/(d^2 - 2 d y + 4 y^2 - 6 y + 3)$$

where $$x = 1/2 (-d + \sqrt3 \sqrt{d - 1)^2} - 3)$$ and $$y = 1/2 ((\sqrt3 d^2)/\sqrt{(d - 1)^2} - d - \sqrt3/\sqrt{(d - 1)^2} + 3)$$

so I write in the bar

if x = 1/2 (-d + sqrt(3) sqrt((d - 1)^2) - 3) and y = 1/2 ((sqrt(3) d^2)/sqrt((d - 1)^2) - d - sqrt(3)/sqrt((d - 1)^2) + 3) find z = (x (d^2 + d (4 y - 6) - 8 y^2 + 12 y - 3) + 6 (d - 1) (y - 1) y)/(d^2 - 2 d y + 4 y^2 - 6 y + 3) and w = -(sqrt(3) (d^2 (x + 2 y - 1) - 2 d (x (2 y - 1) + y^2) + x (4 y - 3) - 2 y^2 + 6 y - 3))/(d^2 - 2 d y + 4 y^2 - 6 y + 3)

but I get all the time a message that Wolfram Alpha cannot understand my query...

Any help is really appreciated.

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    $\begingroup$ This seems like much too complicated an expression for Wolfram Alpha to understand. You could write up code that solved this Mathematica, but there is no guarantee that Wolfram Alpha is going to be able to translate that long of a sentence $\endgroup$
    – Tyberius
    Aug 30, 2019 at 21:37
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    $\begingroup$ I'm voting to close this question as off-topic because it is better suited for mathematica.stackexchange.com $\endgroup$
    – Mauro Vanzetto
    Sep 6, 2019 at 10:28

1 Answer 1

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Are you asking for a plot of $z(d)$ and $w(d)$? Wolfram Alpha might not like the long query, but Mathematica simplifies your number/letter soup just fine:

$$w(d)=-\sqrt{3}\frac{ d^2 (x+2 y-1)-2 d \left(x (2 y-1)+y^2\right)+x (4 y-3)-2 y^2+6 y-3}{d^2-2 d y+4 y^2-6 y+3} = \frac{1}{2} \left(\sqrt{3} d-3 \sqrt{(d-1)^2}+\sqrt{3}\right)$$

$$z(d)=\frac{x \left(d^2+d (4 y-6)-8 y^2+12 y-3\right)+6 (d-1) (y-1) y}{d^2-2 d y+4 y^2-6 y+3} = \frac{1}{2} \left(5 d-\sqrt{3} \sqrt{(d-1)^2}+3\right)$$

Which are just piece-wise linear functions. In the future, it's best not to feed computer algebra software equations you don't understand yourself.

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  • $\begingroup$ thanks for the answer but wolframalpha itself produced such expression of $x$ and $y$...and I preffered not to change it. Actually, why mathematica prefers $\sqrt{d^2}$ than $\vert d \vert$? $\endgroup$
    – dmtri
    Aug 31, 2019 at 19:04
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    $\begingroup$ @dmtri Those aren't necessarily equal whenever $d$ is a complex number, and Mathematica wants to be able to handle complex numbers too. You have to simplify like this: FullSimplify[Sqrt[x^2], x \[Element] Reals] == Abs[x] with explicit assumptions about real variables. $\endgroup$
    – Kirill
    Aug 31, 2019 at 19:48
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    $\begingroup$ Yes, I left complex $d$ a possibility. $\endgroup$
    – user20857
    Aug 31, 2019 at 19:49

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