Let f ( x ) be a polynomial function with non-negative coefficients such that f ( 1 ) = f ′ ( 1 ) = f ″ ( 1 ) = f ‴ ( 1 ) = 1. Find the minimum value of f ( 0 )

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Let   f  (  x  ) be a polynomial function with non-negative coefficients such that   f  (  1  )  =      f    ′    (  1  )  =      f    ″    (  1  )  =      f    ‴    (  1  )  =  1. Find the minimum value of   f  (  0  )

scoseBexgofvc · Accepted Answer

It is perhaps simpler to use Taylor&#039;s formula with fixed degree three and a Lagrange remainder:  f  (  x  )  =  1  +  (  x  −  1  )  +      1          2      !        (  x  −  1      )    2    +      1          3      !        (  x  −  1      )    3    +                    f                  (          4          )                    (      ξ      )              4      !        (  x  −  1      )    4    .where   ξ is between 1 and x. For   x  =  0 is   ξ  ≥  0 and the last term is non-negative, since       f          (      4      )       has non-negative coefficients as well. This gives  f  (  0  )  ≥  1  −  1  +      1    2    −      1    6    =      1    3      .The bound is sharp, equality holds for the function  f  (  x  )  =  1  +  (  x  −  1  )  +      1          2      !        (  x  −  1      )    2    +      1          3      !        (  x  −  1      )    3    =      1    3    +      1    2    x  +      1    6        x    3      .

Let f ( x ) be a polynomial function with non-negative coefficients such that f ( 1

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