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在解题中,我们往往不自觉地应用了下面关于多项式函数奇偶性的定理: 定理多项式函数f(x)为奇函数(或偶函数)的充要条件是f(x)只含奇次项(或偶次项)。这个定理由于教材上未作介绍,而在解决这方面的问题时又经常用到,为此,笔者将此定理的证明写出,供参考。证明充分性是显然的。下证必要性。若f(x)为奇函数,即有f(x)=-f(-x)。我们写出多项式函数的一般形式,就有a_n(-x)~n+a_(n-1)(-x)~(n-1)+…+a_1(-x)+a。=a_nx~n-a_(n-1)x~(n-1)-…-a_1x-a (1) 若n为偶数,则有 2a_nx~n+2a_(n-2)a(n-2)+…+2a_2x~2+2a_o=0从而 a_n=0,a_(m-2)=0,…,a_2=0,a_0=0。
In the problem solving, we often unconsciously apply the following theorems about the parity of polynomial functions: Theorem The polynomial function f(x) is an odd function (or even function) if and only if f(x) contains only odd-order terms. (or even item). This theorem was often used in the solution of this problem because it was not introduced in the textbook. For this reason, the author wrote the proof of this theorem for reference. Proof of adequacy is obvious. The necessity of certification. If f(x) is an odd function, there is f(x)=-f(-x). We write the general form of a polynomial function with a_n(-x)~n+a_(n-1)(-x)~(n-1)+...+a_1(-x)+a. =a_nx~n-a_(n-1)x~(n-1)-...-a_1x-a (1) If n is an even number, then there are 2a_nx~n+2a_(n-2)a(n-2) +...+2a_2x~2+2a_o=0 so that a_n=0, a_(m-2)=0,..., a_2=0, a_0=0.