$W(0)=2$ and $W(x_1+x_2)=W(x_1)+W(x_2)+2x_1x_2-2$
The first condition I know (theorem Bezout). What does the second condition say to us? How to solve this math problem?
$W(0)=2$ and $W(x_1+x_2)=W(x_1)+W(x_2)+2x_1x_2-2$
The first condition I know (theorem Bezout). What does the second condition say to us? How to solve this math problem?
Since we have $W(0)=2$ we can write $W(x) = xQ(x)+2$ so
\begin{eqnarray}W'(x)&=&\lim _{h\to 0}{W(x+h)-W(x)\over h} \\&=&\lim _{h\to 0}{W(h) +2xh-2\over h}\\ &=&\lim _{h\to 0}{hQ(h)+2xh\over h} \\ &=& Q(0)+2x \end{eqnarray}
So $W(x) = 2+ax+x^2$ where $a=Q(0)$. Pluging this in to original equation we get: $$ 2+a(x_1+x_2)+(x_1+x_2)^2 = 2+ax_1+x_1^2+2+ax_2+x_2^2 +2x_1x_2-2$$ which means that $W(x)$ works for any $a$.