YLL討論網

由 **scsnake** 於星期三五月 21, 2003 9:45 pm

沒有啦～我表達錯了～我是指那裡的算式太少了...

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你的圖像好藝術！

由 **---** 於星期三五月 21, 2003 9:43 pm

You meant "q(x)-q(x-1)=0 ==> q(x)=c "不嚴謹?

由 **---** 於星期三五月 21, 2003 8:58 pm

siuhochung 寫到:how?
I just know p(0)=p(1)=0
So p(x)=0 is always an answer, but how to do then?

let p(x)=x(x-1)q(x)
then q(x)-q(x-1)=0
==>
q(x)=c
p(x)=cx(x-1)

由 **Raceleader** 於星期三五月 21, 2003 6:44 pm

You mean p(x)=Ax(x-1)

由 **scsnake** 於星期三五月 21, 2003 6:40 pm

由 **scsnake** 於星期三五月 21, 2003 6:14 pm

剛才看錯題目，sorry ^^

由 **----** 於星期三五月 21, 2003 6:04 pm

how?
I just know p(0)=p(1)=0
So p(x)=0 is always an answer, but how to do then?

由 **scsnake** 於星期三五月 21, 2003 6:00 pm

you can do this , i think !!

由 **----** 於星期三五月 21, 2003 5:49 pm

Find all polynomials p(X) satisfying
x p(x-1) = (x-2)p(x) for all real number of x.

(請將產生的程式碼，剪貼至上方文章區！)
[quote="---"][quote="siuhochung"]how? I just know p(0)=p(1)=0 So p(x)=0 is always an answer, but how to do then?[/quote] let p(x)=x(x-1)q(x) then q(x)-q(x-1)=0 ==> q(x)=c p(x)=cx(x-1)[/quote]