STEVIE-G™ asked in 科學及數學數學 · 1 decade ago

Year 1的Maths問題,但用的是pure的技巧

這是我中七pure mock的paper ii問題,如下:

1 Answer

  • 1 decade ago
    Best Answer

    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)




    =- 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


    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

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