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發問:
這是我中七pure mock的paper ii問題,如下: http://i707.photobucket.com/albums/ww74/stevieg90/01-50.gif
最佳解答:
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 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)| 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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Year 1的Maths問題,但用的是pure的技巧發問:
這是我中七pure mock的paper ii問題,如下: http://i707.photobucket.com/albums/ww74/stevieg90/01-50.gif
最佳解答:
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
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