Find the polynomial p ( x ) = x 2 + p x + q for which max { | p ( x ) | : x ∈ [ − 1 , 1 ] } is minimal.

Question

Find the polynomial   p  (  x  )  =      x    2    +  p  x  +  q for which   max  {        |    p  (  x  )      |      :    x  ∈  [  −  1  ,  1  ]    } is minimal.

vyhlodatis · Accepted Answer

Note that   p  (  0  )  =  q  ,  p  (  1  )  =  1  +  p  +  q and   p  (  −  1  )  =  1  −  p  +  qThus  p  (  −  1  )  +  p  (  1  )  −  2  p  (  0  )  =  2    .Thus  2  =  p  (  −  1  )  +  p  (  1  )  −  2  p  (  0  )  ≤  4  max  {  p  (  −  1  )  ,  p  (  1  )  ,  −  p  (  0  )  }    .It follows that   max  {  p  (  −  1  )  ,  p  (  1  )  ,  −  p  (  0  )  }  ≤      1    2    (  ∗  )This proves that  max  {        |    p  (  x  )      |      :    x  ∈  [  −  1  ,  1  ]    }  ≥      1    2      .Moreover, we can only get equality in   (  ∗  ) if and only if   p  (  −  1  )  =      1    2    ,  p  (  1  )  =      1    2    ,  p  (  0  )  =  −      1    2  , if and only if   q  =  −      1    2    ,  p  =  0It is now easy to check that   f  (  x  )  =      x    2    −      1    2   satisfies  max  {        |    p  (  x  )      |      :    x  ∈  [  −  1  ,  1  ]    }  =      1    2      .This proves that  max  {        |    p  (  x  )      |      :    x  ∈  [  −  1  ,  1  ]    }  ≥      1    2      ,with equality if and only if   p  =  0  ,  q  =  −      1    2

Find the polynomial p(x)=x^2+px+q for which max{\:|p(x)|\::\:x\in[-1,1]\:} is minimal.

Answered question

Answer & Explanation

New Questions in Precalculus