1
$\begingroup$

Let $g_1 = x_1+x_2+x_3+x_4$, $g_2 = x_1^2+x_2^2+x_3^2+x_4^2$, $g_3= x_1x_2+x_3x_4$, $g_4 = x_1x_3+x_2x_4$, $g_5 = x_1^3+x_2^3+x_3^3+x_4^3$. I want to construct a polynomial $s(y_1,\cdots,y_5)$ such that $s(g_1,\cdots,g_5) = x_1 \cdots x_4$. Does somebody have an idea on how to do this?

What I have done so far: $s_1 = y_1$, $s_2=1/2(y_1^2-y_2)$, $s_3 = 1/6 y_1^3-1/2y_1 y_2+1/3y_5$ This gives the three elementary symmetric polynomials when pluging in $g_i$ for $y_i$ in $s_j$. Now how to proceed?

$\endgroup$
  • 1
    $\begingroup$ Here is an idea. From $g_1,g_2,g_5$, you can determine $\sum_{1\leq i<j<k\leq 4}\,x_ix_jx_k$. Now, you know the first three elementary symmetric sums of $x_1,x_2,x_3,x_4$. Employ $g_3=x_1x_2+x_3x_4$, $g_4=x_1x_3+x_2x_4$, and $\gamma:=x_1x_4+x_2x_3$ (note that $\gamma$ can be found as a polynomial in $g_1,g_2,g_3,g_4$). I think you can get $x_1x_2x_3x_4$ easily now. $\endgroup$ – Batominovski Jan 9 '17 at 19:11
  • $\begingroup$ Thanks for your suggestion. Would you mind telling in an answer how to proceed? $\endgroup$ – orgesleka Jan 9 '17 at 19:20
  • $\begingroup$ Sorry, I am a bit sick and don't want to type long mathematical expressions. I suggest two more things: use Newton's Identities and play around with $g_3^2+g_4^2+\gamma^2$. (WARNING: I haven't completely worked this out, so I can't promise it will come out nicely.) $\endgroup$ – Batominovski Jan 9 '17 at 19:27
  • $\begingroup$ ok thanks. und gute besserung ;) $\endgroup$ – orgesleka Jan 9 '17 at 19:29
1
$\begingroup$

I managed to find a solution, it is: $s = 1/24*y_1^4 - 1/8*y_1^2*y_2 - 1/8*y_1^2*y_3 - 1/8*y_1^2*y_4 + 1/8*y_2*y_3 + 1/4*y_3^2 + 1/8*y_2*y_4 + 1/4*y_3*y_4 + 1/4*y_4^2 + 1/12*y_1*y_5$

$\endgroup$

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.