另明由引理,存等∫(x),1≤i≤m,使f(x)=l(x)n(x)+1,f(x) hy(x)p/(x),i≠j令∫(x)=∑m1Jf(x)a.根据带余除法,有f(x)=t(x)II1P1(x)+ g(x),这里degg(x)<degI1p(x)=∑1degp(x).我们指出这里g(x)≠0.否 则,对于固定的j,大为p(x)∑1f(x)a2和p|f(x),r≠j,所以plf(x)a,与 f(x)=l(x)n(x)+1矛盾 9(x)=f(x)-t(x)I=1n2(x) ∑=1f(x)1(x)+f(x)a)-t(x)I=12(x) =()(∑出1()1+()-(x)(2)+2(a) h(x)p(x)a+(l(x)p2(x)+1)a1-t(x) 的证明唯一性:若9(x)≠91(x),g(x)=p(x)(x)+a1,1≤i≤m,g1(x)= n(x)t(x)+a1,1≤i≤m,gi(x)≠t(x),1≤i≤m.则0≠g(x)-91(x) n(x)(q(x)-t(x).故p(x)g(x)-g1(x),1≤i≤m,而p(x),i≠j两两互素, 所以p1(x)p2(x)-pm(x)9(x)-91(x),∑1degp(x)≤max{deg(x), peggy()}< degp(x),此为矛盾.所以g(x)=91(x),即g(x)是唯一的.进一步,根据带 余除法的商的唯一性,二a(x)是唯一的.口 例3( Lagrange插值或式)设a1,a2,…,am∈K为m个不同数,则对任意 b1,b2,…,bmn∈K,存等唯一次数必于m的多项式 L(x)=∑h (x-a3) j≠i 满足L(a)=b,1≤i≤m 另明设p(x)=x-a,1≤i≤m,则p2(x)两两互素,degp(x)=1,显然 L(a1)=b,degL(x)=m-1<∑m1degf(x),L(x)=(x-a1)q(x)+b,根据中国 剩余定理,满足条件的L(x)唯一确定的.口 作充:P1942(1),3,4;P271,2. 补充作充1:设p1(x),P2(x),…,pm(x)∈K]是两两互素的多项式,s1(x),s2(x), sm(x)∈K回]则存等唯一g(x),qa(x),1≤i≤m,使得deg(x)<∑degp(x) 且g(x)=9(x)p(x)+s(x),1≤i≤mNB H￾" fi(x), 1 ≤ i ≤ m, l fi(x) = li(x)pi(x) + 1, fi(x) = hij (x)pj (x), i 6= j. O f(x) = Pm i=1 fi(x)ai . 2E+￾ f(x) = t(x) Qm i=1 pi(x)+ g(x), 'I degg(x) < deg Qm i=1 pi(x) = Pm i=1 degpi(x). T,'I g(x) 6= 0. . #￾&5%! j, ~ pj (x)| Pm i=1 fi(x)ai 8 pj |fr(x), r 6= j, w pj |fj (x)aj ,  fi(x) = li(x)pi(x) + 1 R'
g(x) = f(x) − t(x) Qm i=1 pi(x) = (Pm j6=i,j=1 fj (x)aj (x) + fi(x)ai) − t(x) Qm i=1 pi(x) = P j6=i hj (x)pi(x)aj + (li(x)pi(x) + 1)ai − t(x) Qm i=1 pi(x) = pi(x)(P j6=i hj (x)aj + li(x) − t(x) Q j6=i pj (x)) + ai . !)V} e g(x) 6= g1(x), g(x) = pi(x)qi(x) + ai , 1 ≤ i ≤ m, g1(x) = pi(x)ti(x) + ai , 1 ≤ i ≤ m, qi(x) 6= ti(x), 1 ≤ i ≤ m. # 0 6= g(x) − g1(x) = pi(x)(qi(x) − ti(x)). 4 pi(x)|g(x) − g1(x), 1 ≤ i ≤ m, ) pi(x), i 6= j LL:u￾ w p1(x)p2(x)...pm(x)|g(x) − g1(x), Pm i=1 degpi(x) ≤ max{degg(x), degg1(x)} < Pm i=1 degpi(x), ~R'
w g(x) = g1(x), < g(x) n}!
D ￾2E +!g!} ￾* qi(x) n}!
✷  3 (Lagrange 6O;F) i a1, a2, ..., am ∈ K ~ m 1{r￾#&a b1, b2, ..., bm ∈ K, "}r  m !(m L(x) = Xm i=1 bi( Y j6=i (x − aj ) (ai − aj ) ) Q2 L(ai) = bi , 1 ≤ i ≤ m. NB i pi(x) = x − ai , 1 ≤ i ≤ m, # pi(x) LL:u￾ degpi(x) = 1, ` L(ai) = bi , degL(x) = m − 1 < Pm i=1 degfi(x), L(x) = (x − ai)qi(x) + bi . 2E.6 j%H￾Q2z?! L(x) }_%!
✷ 5 P194 2(1), 3, 4; P227 1, 2. 5 1: i p1(x), p2(x), ...pm(x) ∈ K[x] nLL:u!(m￾s1 (x), s2(x), · · · , sm(x) ∈ K[x] #"} g(x), qi(x), 1 ≤ i ≤ m, l degg(x) < Pdegpi(x) \ g(x) = qi(x)pi(x) + si(x), 1 ≤ i ≤ m. 6