Year 1的Maths問題,但用的是pure的技巧

Hey,the question is interesting and I think you can do it in your mock exam
(a) (i) F(t)=f(x)-Σf^(k) (t)/k! (x-t)^k
F(x)=f(x)-Σf^(k) (x)/k! (x-x)^k = f(x)-f(x)=0 (since 0^0=1)
G(x)=F(x)=0
(ii)
F'(t)
=- Σ(d/dt) [f^(k) (t)/k! (x-t)^k]
=-Σ[ f^(k+1) (t)/k! (x-t)^k+f^(k) (t)/k! k(x-t)^(k-1)(-1)]
=-{Σ[ f^(k+1) (t)/k! (x-t)^k-Σ f^(k) (t)/(k-1)! (x-t)^(k-1)]
After cancelling the equal term, the remaining term is just
- f^(n+1) (t)/n! (x-t)^n
Now
F(x)-F(a)=F'(c)(x-a)=- f^(n+1) (c)/n! (x-c)^n (x-a)
But since x<c<a, and a,x are arbitrary. We can choose c such that
(x-c)^n=(x-a)^n/(n+1) please verfied by yourself.
So 0-F(a)=- f^(n+1) (c)/(n+1)! (x-a)^n+1
Or F(a)=f^(n+1) (c)/(n+1)! (x-a)^n+1
(iii) F(t)=f(x)-Σf^(k) (t)/k! (x-t)^k
Sub. t=a, F(a)=f(x)-Σf^(k) (a)/k! (x-a)^k
or f(x)=Σf^(k) (a)/k! (x-a)^k+f^(n+1) (c)/(n+1)! (x-a)^n+1
(b) Sub. f^(k) (a)=0 where k=1,2,...2k+1
f(x)=f(a)+f^(2n+2) (c)/(n+1)! (x-a)^(2n+2)
Since f^(2n+2) (x) > 0 and (x-a)^(2n+2) is positive. f(x) will get min. at x=a
(c) Again | f(x)-Σf^(k) (a)/k! (x-a)^k|= f^(n+1) (c)/(n+1)! (x-a)^n+1
Since |f^(n) (x)|<M
| f(x)-Σf^(k) (a)/k! (x-a)^k| < M /(n+1)! (x-a)^n+1
Let n-> infinity, | f(x)-Σf^(k) (a)/k! (x-a)^k| = 0 by the hint
and so f(x) =Σf^(k) (a)/k! (x-a)^k