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polinomialsp=x^2+mx+n `max|p(x)|_(-1<=x<=1) - min|p(x)|_(-1<=x<=1) >=...

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drifterkay | Student, Undergraduate | eNoter

Posted January 5, 2013 at 8:00 AM via web

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polinomials

p=x^2+mx+n

`max|p(x)|_(-1<=x<=1) - min|p(x)|_(-1<=x<=1) >= 1/2`

is true maxim absolute value p(-1equal <x equal< 1)-minim abs val p(1equal <x equal< 1) > equal 1/2?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 5, 2013 at 3:04 PM (Answer #1)

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Considering `p(-1)*p(1) =< 0` , hence the polynomial `p(x)`  cancels over the interval `[-1,1],`  thus `min|p(x)|_(-1<=x<=1) = 0`  and the given inequality that you need to prove is `max|p(x)|>=1/2`

`|p(1)| + |p(-1)| + 2|p(0)| = |1^2+m*1+n| + |(-1)^2+m*(-1)+n| + 2|n| > |1+m+n|+|1-m+n|+2|n| >= 2`

You need to study the case `p(-1)*p(1)> 0` , hence, considering `min|p(x)|_(-1<=x<=1) = a`  and `max|p(x)|_(-1<=x<=1) = b`  yields:

`b-a >= ||p(1)| - |p(-1)||`

Since `p(-1)*p(1) > 0`  yields that `p(-1)`  and `p(1)`  are either both positive, or both negative, hence `||p(1)| - |p(-1)|| = |p(1) - p(-1)| = |1+m+n-1+m-n| = 2|m|`

If 2`|m|>=1/2 => |m| >= 1/4, hence b-a >= 1/2.`

If `2|m|<1/2 => |m|< 1/4 and p(-1)*p(1) > 0, hence -m/2 in (-1,1).`

`b-a >= ||p(1)| - |p(-m/2)|| = |p(1) - p(-m/2)| = |1 + m + n + (m^2-4n^2)/4| = |1 + m + m^2/4| = (m+2)^2/4 >= (2 - 1/4)^2/4 = 49/64 >= 1/2`

Hence, considering both cases `p(-1)*p(1) <= 0`  and `p(-1)*p(1)> 0`  yields that the inequality `max|p(x)|_(-1<=x<=1) - min|p(x)|_(-1<=x<=1) >= 1/2`  holds.

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