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**Extra resources for Seminaire Bourbaki vol 1975 76 Exposes 471-488**

**Example text**

4c2 - έ· The above computations lead to the conclusion that υ = 1 X ΧΔ 1 4 64 is an approximate solution to the given equation. More terms can be obtained by further iterations of this procedure. y As is the case for equations with a single unknown, solutions of algebraic equations need not be unique. 3 Suppose F(x, y) = 3z 2 + 2xy + y2 + x - y - 2 = 0. (6) This gives F0(x,t0): 3x2 + 2xt0 + t% +x-t0-2 = 0. Iteration # 0 Step 1: Ignoring all the nonconstant powers of x in FQ(X, to), we get THE SOLUTION OF ALGEBRAIC EQUATIONS 45 c2, - CQ - 2 = 0, x° level: which has solutions Co = — 1, 2.

Step 2: ci = - . Step 3: In Fi(x,t\) replace ¿i with x / 3 4- t 2 to obtain 3 (Ι + ί 2 ) + 3 (Ι + ί 2 ) 2 + (Ι + ί 2 ) 3 -^= 0 ' which simplifies to F2(x, t 2 ): 3i 2 + 3*1 + i | + 2t2x + t\x + y + ^ - + | ^ = 0. Iteration # 2 Step 1: Since t2 = c 2 x 2 + C3X3 + C4X4 + ■ ■ ■ and since the x2-level equation ignores all terms of degree >2 in F2(x, t2), it follows that x? level: 3c2 + \ = 0. Step 2: c 2 = — - . Step 3: Stop. Thus, xz 2: y 3 9 is an approximate solution to the given equation. The Fractional Binomial Theorem affords us with another way to solve this example's equation.

Accordingly, # 1 + x == ( l + x ) 1/3 1/3 1+ X — +m. = 1 + 1/3- - 1 2—x ( 2 ■ 1 M K 2 1+ - ■ f X2 9 " We now restate and comment on the three steps in each iteration of Newton's procedure. THE SOL UTION OF AL GEBRAIC EQUATIONS 43 Step 1: Extract the xfc-level equation out of Fk(x,tk) = 0. If the term tk in the equation Fk(x, tk) =0 were replaced by its power series expansion of Equation (5), each summand would have xk as a factor. The reason for this is that the Step 3 substitutions of iterations 0 , 1 , 2 , .