標題:
show that f(x)=f(1)x
發問:
http://www.hkedcity.net/sch_files/a/kst/kst-06185/public_html/fx=f1x.jpg using the above result, show that f(x)=f(1)x after reading the suggested answer, i know what it said, but i don't know how a human being to think of it! plz tell me what you think in each IMPORTANT steps. 更新: http://www2.hkedcity.net/sch_files/a/kst/kst-06185/public_html/f2.jpg here also! as same as part a.
(i) 只係change of variable "Let u = t+b", 應該冇間題吧? (ii) 的第一part 是用 f(t+1) = f(t)+f(1), 於是 Int(0->x)f(t+1)dt = Int(0->x)f(t)dt + Int(0->x)f(1)dt = Int(0->x)f(t)dt + f(1)x, 因為f(1) 係常數. ------------------------------------------------------------- 最後, 佢要你用(i), 所以就compare 兩條formula 左手邊係 Int(0->x)f(t+1)dt, 所以你想要 a =x, b=1, Part (i) 變左 Int(0->x)f(t+1)dt = Int(0->x+1)f(t)dt - Int(0->1) f(t)dt 而你想要右手邊係f(1)x, 所以你要掉(ii)的 Int(0>x)f(t)dt 去左手邊. 整條式變成: f(1)x = Int(0->x)f(t+1)dt - Int(0->x)f(t) dt = Int(0->x+1)f(t)dt - Int(0->1) f(t)dt - Int(0->x)f(t) dt = Int (x->x+1)f(t)dt - Int(0->1)f(t)dt 想將佢地combine, 就要change variable: t-> t+x = Int(0->1) f(t+x)dt - Int(0->1) f(t)dt = Int(0->1) f(t+x)-f(t) dt 最後再用given: f(t+x) = f(t) + f(x) = Int(0->1) f(t)+f(x)-f(t) dt = Int(0->1) f(x)dt = f(x). ----------------------------------------------- 所以成個題目的結論係: f(x+y)=f(x)+f(y) ---> f(x)=f(1)x 最後, part b 叫你用f(t)=g(e^t), 好容易就check 到 f(x+y) = f(x)+f(y), 於是 g(e^t) = f(t) = f(1)t 代 t = logx, 就得 g(x) = f(1) logx = log_a x, 因為佢地只差一個constant, 而f(1) =/= 0, 因為g 話左係 non-constant function. 2008-01-07 01:35:10 補充： you need to check f(1)=/=0 because, you don't want g(x) to be a constant.and you also notice that the expression in the answer: log_a(x), can never be constant zero. So you don't want f(1) to be zero, otherwise you can never get to the answer.
其他解答:
show that f(x)=f(1)x
發問:
http://www.hkedcity.net/sch_files/a/kst/kst-06185/public_html/fx=f1x.jpg using the above result, show that f(x)=f(1)x after reading the suggested answer, i know what it said, but i don't know how a human being to think of it! plz tell me what you think in each IMPORTANT steps. 更新: http://www2.hkedcity.net/sch_files/a/kst/kst-06185/public_html/f2.jpg here also! as same as part a.
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最佳解答:(i) 只係change of variable "Let u = t+b", 應該冇間題吧? (ii) 的第一part 是用 f(t+1) = f(t)+f(1), 於是 Int(0->x)f(t+1)dt = Int(0->x)f(t)dt + Int(0->x)f(1)dt = Int(0->x)f(t)dt + f(1)x, 因為f(1) 係常數. ------------------------------------------------------------- 最後, 佢要你用(i), 所以就compare 兩條formula 左手邊係 Int(0->x)f(t+1)dt, 所以你想要 a =x, b=1, Part (i) 變左 Int(0->x)f(t+1)dt = Int(0->x+1)f(t)dt - Int(0->1) f(t)dt 而你想要右手邊係f(1)x, 所以你要掉(ii)的 Int(0>x)f(t)dt 去左手邊. 整條式變成: f(1)x = Int(0->x)f(t+1)dt - Int(0->x)f(t) dt = Int(0->x+1)f(t)dt - Int(0->1) f(t)dt - Int(0->x)f(t) dt = Int (x->x+1)f(t)dt - Int(0->1)f(t)dt 想將佢地combine, 就要change variable: t-> t+x = Int(0->1) f(t+x)dt - Int(0->1) f(t)dt = Int(0->1) f(t+x)-f(t) dt 最後再用given: f(t+x) = f(t) + f(x) = Int(0->1) f(t)+f(x)-f(t) dt = Int(0->1) f(x)dt = f(x). ----------------------------------------------- 所以成個題目的結論係: f(x+y)=f(x)+f(y) ---> f(x)=f(1)x 最後, part b 叫你用f(t)=g(e^t), 好容易就check 到 f(x+y) = f(x)+f(y), 於是 g(e^t) = f(t) = f(1)t 代 t = logx, 就得 g(x) = f(1) logx = log_a x, 因為佢地只差一個constant, 而f(1) =/= 0, 因為g 話左係 non-constant function. 2008-01-07 01:35:10 補充： you need to check f(1)=/=0 because, you don't want g(x) to be a constant.and you also notice that the expression in the answer: log_a(x), can never be constant zero. So you don't want f(1) to be zero, otherwise you can never get to the answer.
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