✔ 最佳答案
(a) f(x)=(x-a_1)(x-a_2)...(x-a_n)
f'(x)=f(x)[1/(x-a_1)+1/(x-a_2)+...+1/(x-a_n)]
f'(a_i)=(a_i-a_1)(a_i-a_2)...(a_i-a_(i-1))(a_i-a_(i 1))...(a_i-a_n)
(b)(i) Observe that
g(x)=f(x)[A_1/(x-a_1)+A_2/(x-a_2)+...+A_n/(x-a_n)]
g(x)/f(x)=A_1/(x-a_1)+A_2/(x-a_2)+...+A_n/(x-a_n)
[Remark. It is in fact the partial fraction expansion of g(x)/f(x)]
Thus
A_i=g(a_i)/[(a_i-a_1)...(a_i-a_(i-1))(a_i-a_(i 1))...(a_i-a_n)]=g(a_i)/f'(a_i)
This prove the existence.
To prove the uniqueness, let there is another set of B_i satisfied he relation.
Expand it and by comparing the coefficients, we will get A_i=B_i.
Or we may simply expand the RHS as a degree-(n-1) polynomial, and by comparing the coefficients,
we get a system of linear equations with n equations n unknowns.
If we can show that the coefficient determinant is non-zero,
then the existence and uniqueness follows.
(ii) In (i), we have shown that A_i=g(a_i)/f'(a_i)
Note that the RHS of (*) is a polynomial of degree-(n-1),
and the coefficients of x^(n-1)=A_1+A_2+...+A_n.
Hence if g(x) is of degree less than (n-1), then the coefficient of x^(n-1)=0.
i.e. A_1+A_2+...+A_n=0, the result follows.
(iii) Given a_i=i and define g(x)=x^m where m<=n-2 is an integer.
Using (a), we have f'(i)=(i-1)(i-2)...(1)(-1)...(i-n)=(-1)^(n-i) (i-1)! (n-i)!
Since degree of g(x)<n-1, therefore by (ii),
Σ(i=1 to n) g(i)/f'(i)=Σ(i=1 to n) (-1)^(n-i) i^m/[(i-1)!(n-i)!]=0
(c) Let h(x)=Σ(i=1 to n) A_i (x-a_1)...(x-a_(i-1))(x-a_(i 1))...(x-a_n)
We want to find such an A_i which satisfied h(a_i)=b_i
Now h(a_i)=A_i (a_i-a_1)...(a_i-a_(i-1))(a_i-a_(i 1))...(a_i-a_n)=b_i
so take A_i=b_i/(a_i-a_1)...(a_i-a_(i-1))(a_i-a_(i 1))...(a_i-a_n)
then such h(x) is the polynomial we want.
[Remark. It is the well known Lagrange Interpolating Polynomial.]
