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对于二次函y_1(x)=a_1x~2+b_1x+c_1与y_2(x)=a_2x~2+b_2x+c_2,(a_1.a_2(/)0),能否找到常数λ,使叠加得到的y_0(x)=y_1(x)+λy_2(x)的函数值不改变符号(定正或定负)? 下面用纯粹初等的方法进行探索: 因y_0(x)=a_1[x~2+b_1/a_1x+c_1/a_1+λa_2/a_1(x~2+b_2/a_2x+c_2/a_2)],若记b_/a_1=b、c_/a_1=c、λa_2/a_1=μ、 b_2/a_2=b_0、c_2/a_2=c_0,即考查y(x)=x~2+bx+c+μ(x~2+b_0x+c_0) 仍记为y(x)=y_1(x)+μy_2(x)〕在哪些情况下可以选取到实数μ使其定号。
For the second-order functions y_1(x)=a_1x~2+b_1x+c_1 and y_2(x)=a_2x~2+b_2x+c_2, (a_1.a_2(/)0), whether or not the constant λ can be found is obtained by the superposition. The function value of y_0(x) = y_1(x) + λy_2(x) does not change the sign (fixed or negative)? The following is a purely elementary method to explore: Because y_0(x) = a_1[x~2+b_1 /a_1x+c_1/a_1+λa_2/a_1(x~2+b_2/a_2x+c_2/a_2)] if b_/a_1=b, c_/a_1=c, λa_2/a_1=μ, b_2/a_2=b_0 ,c_2/a_2=c_0, that is, examining y(x)=x~2+bx+c+μ(x~2+b_0x+c_0) still denoted as y(x)=y_1(x)+μy_2(x)] In some cases, the real number μ can be selected to determine the number.