By Todorova G., Vitillaro E.

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**Example text**

Finally we use the Newton formulas to construct the resultant R(w). 5. Let there be given a system f1 = a0 x + a1 y + a2 f2 = b0 x2 + b1 xy + b2 y2 + b3 x + b4 y + b5 f3 = c0 x2 + c1xy + c2y2 + c3x + c4y + c5: We shall assume that the coe cients of these polynomials are polynomials in w. Also let the system f1 = 0 f2 = 0 be0non-degenerate 1 for some w. (k21 + k22 )! k22 ! ksm 0 k11 +k12 +k21 +k22 2j j =1 2 " # j k +k k +k Aak1111 : : : ak2222 : M f3 Q111 x2(21kQ112+12k12)+122 Jyf2(det k21 +k22 )+1 In order to nd Sj we must substitute the monomials xt ym, t + m 4, into this formula (the result is denoted St m).

By the Hilbert Nullstellensatz (see 142, Sec. 130]) there exists a matrix A = jjajkjjnj k=1 CHAPTER 2. 38 consisting of homogeneous polynomials ajk , such that zjNj +1 = n X k=1 ajk (z)Pk (z) j = 1 : : : n: Moreover, we can choose the number Nj less than the number k1 + : : : + kn ; n (thanks to the Macaulay theorem 115, 136]). 1 (Kytmanov). For each polynomial R(z) of degree M we have X R(a) = a2Ef X = k Pn kssjj where sj 0 M ! 2 3 n P Qn P n (;1) ks j ks j ! P Q det AJf Q aksjs j s=1 j =1 M 664 Qn j Nj + sjj+=1Nj 775 Qn (k )!

In exactly the same way we nd S2 S3 S4 . 10) has four roots. e. 10), then we arrive at the same answer. The general case can be handled in the same way. 1 with R(z) equal to z1k . 1) we nd the coe cients j of the polynomial R(z1). 2). 2) all variables except z1 . A MODIFIED ELIMINATION METHOD 43 The principal di culty of this method is the determination of the matrix A. We now discuss some ways of calculating A. 1. The general method for nding A is the method of undetermined coe cients. Since we have an estimate of the degree Nj , we can write down ajk (z) with unknown coe cients, then perform arithmetic operations, equating the coe cients at equal degrees of z, and hence get a system (in general under determined) of linear equations.