Mathematics, Simon YAU

先回答第1題
(第2題請至台灣Yahoo知識家提問,讓我好收藏)
2步驟解第1題: (1)binomial thm. differentiation &linear combination整理結論
(2)L'Hospital's rule

(1)binomial thm. differentiation & linear combination
(1-x)^n=Σ[j=0~n] C(n,j) (-x)^j (以下Σ均為Σ[j=0~n] )
Putting x=0, then 0=Σ C(n, j)(-1)^j -----(A)
(d/dx) (1-x)^n = -n(1-x)^(n-1) = Σ C(n,j) (-1)^j *j x^(j-1)
Putting x=1, then 0=Σ j C(n,j) (-1)^j ------(B)
(d/dx)^2 (1-x)^n = n(n-1)(1-x)^(n-2)= Σ j(j-1)C(n,j) (-1)^j x^(j-2)
Putting x=1, then 0=Σ j(j-1)C(n, j) (-1)^j C(n, j)
Taking suitable linear combination with (B), then
0= Σ j^2 C(n, j) (-1)^j -----(C)
Similarly, 0=Σ j^k C(n, j) (-1)^j , k=0, 1, 2, ..., n-1
(d/dx)^n (1-x)^n =Σ j(j-1)..(j-n) C(n, j) (-1)^j x^(j-n)
thus, (-1)^n n!= Σ j(j-1)...(j-n) C(n, j) (-1)^j
hence, (-1)^n n!= Σ j^n C(n, j) (-1)^j
so, Σ j^k C(n, j) (-1)^j = 0 for k=0,1,2,3,..., n-1 -----(D)
and Σ j^n C(n,j) (-1)^j = (-1)^n n! ----(E)

(2)L'Hospital's rule
(Set h=△x)
Let F(h)=Σ C(n, j) f(x+jh) (-1)^j, then
F(0)=Σ C(n, j) f(x) (-1)^j = 0 (By (D))
F'(0)= Σ j C(n, j) f'(x) (-1)^j =0 (By (D))
F"(0)= Σ j^2 C(n,j) f"(x) (-1)^j =0 (By (D))
...
F^(n-1) (0)=Σ j^(n-1) C(n,j) f^(n-1)(x) (-1)^j =0 (by (D))
F^(n) (0) = Σ j^n C(n, j) f^n (x) (-1)^j = (-1)^n f^(n) (x) *n! (by (E))
Hence, appliny L' rule n times, then

Σ C(n, j) f(x+jh) (-1)^jF(h)
------------------------------ = ----------
h^n h^n

approaches to (-1)^n f(x).

註:若有打錯字,請包含. (高手級,看到開始指明的兩個步驟應該就懂了,同意?)

2010-12-29 10:42:28 補充：
最後排列錯亂,更正如下:
Σ C(n, j) f(x+jh) (-1)^j
------------------------------
h^n

=

F(h)
------
h^n

approaches to (-1)^n f(x)

2010-12-29 10:44:54 補充：
更正: Equation (A)處, 應是 putting x=1.