Let P ( z ) = a 0 + a 1 z + ⋯ + a n z n be a polynomial whose coefficients satisfy 0 &lt; a 0 &lt; a 1 &lt; ⋯ &lt; a n .

Question

Let   P  (  z  )  =      a    0    +      a    1    z  +  ⋯  +      a    n        z    n   be a polynomial whose coefficients satisfy  0  &amp;lt;      a    0    &amp;lt;      a    1    &amp;lt;  ⋯  &amp;lt;      a    n    .

Brenton Gay · Accepted Answer

The thing to do is to look instead at the polynomial  Q  (  z  )  =  (  1  −  z  )  P  (  z  )  =  (  1  −  z  )      (          ∑              i        =        0            n              a      i              z      i        )    =      a    0    −      a    n        z          n      +      1        +      ∑          i      =      1        n    (      a    i    −      a          i      −      1        )      z    i  Now, let       |    z      |    &amp;gt;  1 be a root of   P  (  z  ), and hence a root of   Q  (  z  ). Therefore, we have       a    0    +      ∑          i      =      1        n    (      a    i    −      a          i      −      1        )      z    i    =      a    n        z          n      +      1       Then, we have                              |                          a          n                          z                      n            +            1                                    |                                    =                  |                      a            0                    +                      ∑                          i              =              1                        n                    (                      a            i                    −                      a                          i              −              1                                )                      z            i                    |                                                  ≤                  a          0                +                  ∑                      i            =            1                    n                (                  a          i                −                  a                      i            −            1                          )                  |                          z          i                          |                                                  &amp;lt;                  a          0                          |                          z          n                          |                +                  ∑                      i            =            1                    n                (                  a          i                −                  a                      i            −            1                          )                  |                          z          n                          |                                                  =                  |                          a          n                          z          n                          |                    a contradiction.

Let P(z)=a_0+a_1z+⋯+a_nz^n be a polynomial whose coefficients satisfy 0<a_0<a_1<⋯<a_n.

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