# Tag Info

## New answers tagged mathematica

0

Mathematica doesn't just applies the Buchberger algorithm and returns a Groebner basis. It returns a reduced Groebner basis. So, if your polynomial set contains any non-zero constant, the result is {1}. For example GroebnerBasis[{ 1/y, z}, {x}] returns {1}. However, if your set consists of constants that can be eventually zero, Mathematica returns a reduced ...

2

In Mathematica, the bound variable in Sum starts at 1 when the second argument is in the form {x,n}. It is also defined locally, so setting its value outside changes nothing. To make it start from 2, use {x,2,4} as the second argument. But this is really a question for Mathematica Stack Exchange.

-1

I would recommend SyMAT for starting out with. Scripts are written in your choice of Python and JavaScript. (Full disclosure: I am the lead developer of SyMAT.)

0

Let $C_5 = C_2 i$ and $C_6 = C_4 i$. This will also catch wholly real solutions for wholly imaginary $C_2$ and $C_4$

0

It is the same answer, once in real and once in complex form. Note that the 4th roots of $-4λ^4$ are $\pmλ\pm i·λ$, with independent sign selection for 4 roots. One could guess that Mathematica understands $\sqrt[4]{-1}=(-1)^{1/4}=\frac{1+i}{\sqrt{2}}$ and $(-1)^{3/4}=\frac{-1+i}{\sqrt{2}}$. The combination of complex conjugate roots gives the ...

2

It is quite unsafe to use underscores for function names. Try instead: lemnx[t_]:=(t + t^3)/(1+ t^4) lemny[t_]:=( t-t^3)/( 1+ t^4) ParametricPlot[{lemnx[t], lemny[t]},{t, -10., 10.}]

0

In a trial version of Mathematica I used the Eliminate function like this: Eliminate[ {(x^2 + y^2) ^2 == A*(x^2-y^2) * z^2, u== 2*x*(x^2 + y^2) - A*x * z^2, v == 2*y*(x^2 + y^2) + A*y * z^2, w == A*z *(y^2 - x^2)}, {x,y,z}]. And the result is: $$F^d = A^3(u^2 - v^2)^3 + A^2(15u^4 +78u^2v^2 +15v^4)w^2 + A(48u^2 - 48 v^2) w^4 - 64w^6 = 0$$ The lemniscate ...

Top 50 recent answers are included