от nevrodermit » 21 Май 2016, 09:12
Too many signs there since [tex]|P(x)|=|ax^3+bx^2+cx+d|=\pm|ax^3|\pm|bx^2|\pm |cx|\pm d[/tex].
First, we will show that the problem doesn't change if we require [tex]a\ge 0, b\ge 0[/tex].
We have 3 cases (when at least one of them is negative):
1) If [tex]a\le 0, b\le 0[/tex] then choose [tex]Q(x)=-P(x)=(-a)x^3+(-b)x^2-cx-d[/tex]. Then [tex]Q(x)[/tex] satisfies the conditions in the problem because [tex]|Q(x)|=|-P(x)|=|P(x)|\le 1 \forall |x|\le 1[/tex] and have positive coefficients [tex]a,b[/tex].
2) If [tex]a\le 0, b\ge 0[/tex] then choose [tex]Q(x)=P(-x)=(-a)x^3+(b)x^2-cx+d[/tex]. Then [tex]Q(x)[/tex] satisfies the conditions in the problem because [tex]|Q(x)|=|P(-x)|\le 1\forall |x|\le 1[/tex] and have positive coefficients [tex]a,b[/tex].
3) If [tex]a\ge 0, b\le 0[/tex] then choose [tex]Q(x)=-P(-x)=(a)x^3 + (-b)x^2+cx+d[/tex]. Then [tex]Q(x)[/tex] satisfies the conditions in the problem because [tex]|Q(x)|=|-P(-x)|\le 1 \forall |x| \le 1[/tex] and have positive coefficients [tex]a,b[/tex].
So, WLOG, we assume [tex]a\ge0, b\ge 0[/tex].
Now, we again we have several cases:
1) [tex]c\ge 0, d\ge0[/tex]. Then [tex]|a|+|b|+|c|+|d|=a+b+c+d=P(1)\le 1[/tex]
2) [tex]c\ge 0, d<0[/tex]. Then [tex]|a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0)\le 3[/tex]
3) [tex]c<0, d\ge 0[/tex]. Then [tex]|a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}P(1)-\frac{1}{3}P(-1)-\frac{8}{3}P(\frac{1}{2})+\frac{8}{3}P(-\frac{1}{2})\le 7[/tex]
4) [tex]c<0, d<0[/tex]. Then [tex]|a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}P(1)-4P(\frac{1}{2})+\frac{4}{3}P(-\frac{1}{2})\le 7[/tex]
So the maximum value is [tex]7[/tex].