# Finding zero linear combinations of polynomials (numerically)

I have two functions of real variables defined on two compact spaces

f1[k_,x_,y_,z_,w_]=(9 (144 - 528 k + 532 k^2 - 200 k^3 + 25 k^4) w^2 - 60 x^2 + 10 (8 - 12 k + 3 k^2) y^2 + 900 z^2 - 1872 k z^2 + 1188 k^2 z^2 - 360 k^3 z^2 + 45 k^4 z^2)
f2[k_,x_,y_,z_,w_]=(12 (17 - 24 k + 6 k^2) w^2 - 15 x^2 + 5 (8 - 12 k + 3 k^2) y^2 + 9 (17 - 24 k + 6 k^2) z^2)


where $$x,y,z,w \in S_3$$, namely $$x^2+y^2+z^2+w^2=1$$ and $$k\in [0,4]$$.

I have checked that these functions admit a zero somewhere.

My first question is: How can I check if these functions admit the same zero?

My secondon question is: Let us suppose the answer to the 1st question is NO. How can I found (efficiently) a set of points $$x_n,y_n,z_n,w_n$$ and strictly-positive coefficients $$a_n>0$$ such that the following equations are true?

$$\sum_n a_n f_1(k_n,x_n,y_n,z_n,w_n) = 0\\ \sum_n a_n f_2(k_n,x_n,y_n,z_n,w_n) = 0$$

In particular, I can traslate this problem into a geometrical one. Given the 2D vector $$\vec{v}(k,x,y,z,w) = (f_1,f_2)$$, how can I find (efficiently) the solution to the problem

$$\sum_n a_n \vec{v}_n(k_n,x_n,y_n,z_n,w_n) =0 \,\,\qquad (\text{Eq}.1)$$

For example, If for a fixed $$\vec{v}_n$$ it does not exist a $$\vec{v}_m = -\vec{v}_n$$, then Eq.1 is equivalent of finding a set of vectors $$a_n\vec{v}_n$$ which form a closed path in 2D.

• @KraZug yes, there was a typo. Now, even if I clear the kernel, the results are the one in the OP. – apt45 Nov 8 '18 at 9:58
• Basically it looks like the NMaximize has found a local maximum, not the global one. If you use a different method (SimulatedAnnealing or DifferentialEvolution) then it finds the correct maximum – KraZug Nov 8 '18 at 10:01
• Alternatively you could use FindInstance with $f_i==0$ to try and find if a zero exists. I think the interesting question here is the second and third part, and you might want to rewrite the question to focus on those – KraZug Nov 8 '18 at 10:32
• Thanks, I edited the question – apt45 Nov 8 '18 at 11:12