当k≤n时,比较上式两端x的系数,因为deg(x)<m,得到 k) 01Sk-1+02Sk-2+ 故有 Sk-01sk-1+a25k-2+…+(-1)-1ak-151+(-1)kok=0. 当k>n时,上式左端xn的系数为0,故有 Sk-01Sk-1+O2Sk-2+ 例3n=3时, S2=a2+2+n3=(x1+x2+x3)2-2(x1x2+x1x3+x23)=012-202 S3=x2+m2+x3=(x1+x2+x23)3-3(x12x2)-6x1x2x3 而s(x2x2)=x2+1x3+x1m2+x1x32+2x3+x2n3=0102-303,所以 求s3第二种方法:由 Newton公式,S3-8201+102-303=0.所以s3=s201 s102+33=(-22)1-a102+303=03-30102+303.同理,s4-8301+s202=0 故S4=8901-8202=(01-30102)01-(0-202)02=a1-30102-0102+272 例4解下列方程组: r1+x2+x3+4=4 r2+n2+n3+2=4 x1+a2+r3+x3 x4+x4+x4+4=4 解设x1,x2,x3,x4是4次复系数多项式∫(x)=x4-01x3+02x2-03x+σ4的根 由已知s1=52=83=84=4,利用 Newton公式可得a1=4,02=6,03=4,4=1, 即x1,x2,x3,x4是∫(x)=x4-4x3+6x2-4x+1=0的4个根,但f(x)=(x-1) 故x1=1,1≤i≤4为方程组的解 例5求一个n次方程使81=s2 Sn-1=0. k ≤ n n￾IlqS& x n  x￾ degg(x) < n,  (−1)k (n − k)σk = sk − σ1sk−1 + σ2sk−2 + · · · + (−1)kσks0, 8 sk − σ1sk−1 + σ2sk−2 + · · · + (−1)k−1σk−1s1 + (−1)k kσk = 0.  k > n n￾lq:& x n  x 0, 8 sk − σ1sk−1 + σ2sk−2 + · · · + (−1)nσnsk−n = 0. 2 H 3 n = 3 n￾ s1 = x1 + x2 + x3 = σ1, s2 = x 2 1 + x 2 2 + x 2 3 = (x1 + x2 + x3) 2 − 2(x1x2 + x1x3 + x2x3) = σ 2 1 − 2σ2, s3 = x 3 1 + x 3 2 + x 3 3 = (x1 + x2 + x3) 3 − 3s(x 2 1x2) − 6x1x2x3. * s(x 2 1x2) = x 2 1x2 + x 2 1x3 + x1x 2 2 + x1x 2 3 + x 2 2x3 + x2x 2 3 = σ1σ2 − 3σ3, } s3 = σ 3 1 − 3(σ1σ2 − 3σ3) − 6σ3 = σ 3 1 − 3σ1σ2 + 3σ3. d s3 "+4., Newton 7q￾s3−s2σ1+s1σ2−3σ3 = 0. } s3 = s2σ1− s1σ2+3σ3 = (σ 2 1 −2σ2)σ1−σ1σ2+3σ3 = σ 3 1 −3σ1σ2+3σ3. P￾s4−s3σ1+s2σ2 = 0. 8 s4 = s3σ1 − s2σ2 = (σ 3 1 − 3σ1σ2)σ1 − (σ 2 1 − 2σ2)σ2 = σ 3 1 − 3σ 2 1σ2 − σ 2 1σ2 + 2σ 2 2 . H 4 JT.8    x1 + x2 + x3 + x4 = 4 x 2 1 + x 2 2 + x 2 3 + x 2 4 = 4 x 3 1 + x 3 2 + x 3 3 + x 3 4 = 4 x 4 1 + x 4 2 + x 4 3 + x 4 4 = 4 F m x1, x2, x3, x4 t 4 2 x)q f(x) = x 4−σ1x 3+σ2x 2−σ3x+σ4 6
- s1 = s2 = s3 = s4 = 4, R Newton 7qN σ1 = 4, σ2 = 6, σ3 = 4, σ4 = 1, C x1, x2, x3, x4 t f(x) = x 4 −4x 3 + 6x 2 −4x+ 1 = 0  4 46￾ f(x) = (x−1)4 , 8 xi = 1, 1 ≤ i ≤ 4 .8J
H 5 d4 n .p s1 = s2 = · · · = sn−1 = 0. 5