2007-10-07
给定函数给定函数F(x)=ax2+bx+c,以及G(x)=cx2x+a,其中|F(0)|≤1,|F(1)|≤1
|F(―1)|≤1
证明:对于|x|≤1
(1)|F(x)|≤5/4
(2)|G(X)|≤2
给定函数F(x)=ax2+bx+c,以及G(x)=cx2+a,其中|F(0)|≤1,|F(1)|≤1
|F(―1)|≤1
证明:对于|x|≤1 (1) |F(x)|≤5/4 (2)|G(X)|≤2
证明:(1)
c=f(0) c=f(0) c=f(0)
a+b+c=f(1) a+b=f(1)-f(0) 2a=f(1)+f(-1)-2f(0)
a-b+c=f(-1) a-b=f(-1)-f(0) 2b=f(1)-f(-1)
2|f(x)| = |2ax^2 + 2bx + 2c|
= |[f(1)+f(-1)-2f(0)]x^...全部
给定函数F(x)=ax2+bx+c,以及G(x)=cx2+a,其中|F(0)|≤1,|F(1)|≤1
|F(―1)|≤1
证明:对于|x|≤1 (1) |F(x)|≤5/4 (2)|G(X)|≤2
证明:(1)
c=f(0) c=f(0) c=f(0)
a+b+c=f(1) a+b=f(1)-f(0) 2a=f(1)+f(-1)-2f(0)
a-b+c=f(-1) a-b=f(-1)-f(0) 2b=f(1)-f(-1)
2|f(x)| = |2ax^2 + 2bx + 2c|
= |[f(1)+f(-1)-2f(0)]x^2 + [f(1)-f(-1)]x + 2f(0)|
= |f(1)(x^2+x) + f(-1)(x^2-x) + 2f(0)(1-x^2)|
≤ |f(1)||x^2+x| + |f(-1)||x^2-x| + 2|f(0)||1-x^2|
≤ |x^2+x| + |x^2-x| + 2|1-x^2|
当|x|≤1时必有 |x|≥x^2,从而:
若x=0,则 |x^2+x| + |x^2-x| + 2|1-x^2| = 2
若x>0,则 |x^2+x| + |x^2-x| + 2|1-x^2|
= x^2+x + x-x^2 + 2-2x^2
= 2 + 2x - 2x^2
≤ 5/2 (这里,开口向上,当x=1/2时取“等号”)
若x<0,则 |x^2+x| + |x^2-x| + 2|1-x^2|
= -x^2-x + x^2-x + 2-2x^2
= 2 - 2x - 2x^2
≤ 5/2(这里,开口向上,当x=-1/2时取“等号”)
即|x|≤1总有 2|f(x)| ≤ |x^2+x| + |x^2-x| + 2|1-x^2| ≤ 5/2
故 |f(x)| ≤ 5/4
(2)当|x|≤1时
2|G(x)| = |2cx^2 + 2a|
= |2f(0)x^2 + f(1) + f(-1) - 2f(0)|
= |2f(0)(x^2-1) + f(1) + f(-1)|
≤ 2|f(0)||x^2-1| + |f(1)| + |f(-1)|
= 2|f(0)|(1-x^2) + |f(1)| + |f(-1)|
≤ 2|f(0)| + |f(1)| + |f(-1)|
≤ 4
所以 |G(x)| ≤ 2
。
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