f(xy)=f(x)f(y)－ohe09nz81r的部落格

標題:

f(xy)=f(x)f(y)

發問:

prove f(x)>0 for all x ＞0

最佳解答:

資料不足，要加番ｄ先ｐｒｏｖｅ到的。 f(x)應該是non-constant function先得。做之前,可以先看看有咩function 符合這個條件: f(x)=x 就是了(f(x)=x, f(xy)=xy=f(x)f(y)) 先要check f(0)=f(0*0)=f(0)f(0) f(0)=0 or 1 咁你一看就知f(0)應該是=0的,要prove就要用contradiction了: if f(0)=1, there exist x belongs to set REAL such that f(0*x)=f(0)f(x)=1 兩邊除f(0) (可以咁做因為f(0)不等於0) f(x)=1 即是f是constant function!!! (contradiction) 好了,現在看下點樣做f(x)>0 通常做到d square term出來就可以>0 or =0 f(xx)=f(x)f(x)=[f(x)]^2 >0 or =0 let y =x^2 >0 or =0 f(y) >0 or =0 if y不等於０，即是： f(x)>0 or =0 for all x >0 or =0 f(x) 只會在＝０時先會＝０（因為f(x)=C^2 =0 only if f(x)=0, C是一ｄ野）如果ｘ不等於０，f(x)就不會等於０即是：f(x)>0 for all x ＞0 2007-01-29 22:25:36 補充：寫得ｃｌｕｍｓｙ左小小，因為想人們清楚每一個思考的步驟。

其他解答:

f(xy) ≡ f(x) f(y)for... all real numbers x, y? or all positive real numbers x, y? I suppose the question states that f(x) is a non-constant function for all x. Otherwise two obvious counter-examples are (1) f(x) = 0, and (2) f(x)=1, for any real number x (either of which contradicts with what we're going to prove). Also, I suppose the question states that f(x) is a continuous function for all x. Putting y = 1, f(x ? 1) = f(x) ? f(1) f(x) = f(x) ? f(1) f(x) [f(1) - 1] = 0 f(1) = 1( "f(x) = 0 for all x" is rejected as f(x) is non-constant. ) For all x＞0, f(x) ? f(1/x) = f(x ? 1/x) = f(1) = 1 Thus, f(x) ≠ 0 for any x＞0, as otherwise f(x) ? f(1/x) = 0 ≠ 1. As f(x) is continuous and f(x) ≠ 0 for any x＞0, we can conclude that either f(x) ＞0 for all x＞0, or f(x) ＜0 for all x＞0. Now, f(1) = 1 ＞0. Therefore, f(x) ＞0 for all x＞0.|||||這是AL pure maths 題目WO! 但是你是不是少了些資料啊......PASPAPER這類數會話條式是non-constant function才會證明到不是0阿 FROM f(xy)=f(x)f(y) PUT y=1 f(x X 1)=f(x)f(1) f(x)=f(x)f(1) 0=f(x)f(1)-f(x) 0=f(x)[f(1)-1] THEN f(1)=1 or f(x)=0 通常f(x)=0會REJECT因為題目話 f 是non-constant function 若無寫是會等於０的,你再CHECK CHECK題目吧