数学竞赛题,不多说,直接进入主题,Alice and Bob want to design a modern "anti-prime polynomial" f(x) that satisfies the conditions:(1).f(x) is degree 2010 and all of the coefficients of f(x) are either 0,1 or -1; and(2).f(-1)f(0)f(1)≠ 0;
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数学竞赛题,不多说,直接进入主题,Alice and Bob want to design a modern "anti-prime polynomial" f(x) that satisfies the conditions:(1).f(x) is degree 2010 and all of the coefficients of f(x) are either 0,1 or -1; and(2).f(-1)f(0)f(1)≠ 0;
数学竞赛题,
不多说,直接进入主题,
Alice and Bob want to design a modern "anti-prime polynomial" f(x) that satisfies the conditions:
(1).f(x) is degree 2010 and all of the coefficients of f(x) are either 0,1 or -1; and
(2).f(-1)f(0)f(1)≠ 0; and
(3).for every integer x,｜f(x)｜ is not prime.
Does such a polynomial f(x) exist?( 1 is not a prime)
就是这题目,有无数学达人懂的,希望可以屈尊赐教,
当 x=-1时，f(x)=x^2010+x^2009+x^2008+x^2+x+1= 2
但是题目中第三个条件是 ｜f(x)｜ is not prime
数学竞赛题,不多说,直接进入主题,Alice and Bob want to design a modern "anti-prime polynomial" f(x) that satisfies the conditions:(1).f(x) is degree 2010 and all of the coefficients of f(x) are either 0,1 or -1; and(2).f(-1)f(0)f(1)≠ 0;
其实很容易,这道题就是想考察因式分解,不需要想的太复杂
你就设计一个函数可以因式分解就行了
首先,f(0)≠ 0,说明x^0项系数是1或-1
然后,f(-1),f(1)都≠ 0,说明因式里面不能有(x+1)和(x-1)
那么,就考虑(x^2+x+1)这个因式就行了
你只要让f(x)里面系数是1的项数为3的倍数就行了,最简单的,比如f(x)=x^2010+x^2009+x^2008+x^2+x+1
那么f(x)=(x^2008+1)(x^2+x+1)
当然,这里还要注意除了(x^2+x+1)之外的其他因式不能有(x+1)和(x-1)
这个f(x)就满足条件了
其实答案有无数个
不知道我讲清楚没.
哦,我没注意哈,那就再加一项咯
f(x)=(x^2+x+1)(x^2008+x^2002+1)
这样就没问题了吧,哈哈
后悔没学好英语，半懂不懂，还是不懂。
函数可以因式分解 就满足第三个条件哦