Consider the following polynomial in x: p ( x ) = ∑ n = 0 2 k x n − 1 2 k + 1 ∑ n = 0 2 k ( − x ) n .I want to show that p ( x ) ≥ 0 for real x.

Question

Consider the following polynomial in x:  p  (  x  )  =      ∑          n      =      0              2      k            x    n    −      1          2      k      +      1            ∑          n      =      0              2      k        (  −  x      )    n    .I want to show that   p  (  x  )  ≥  0 for real x.

flonsHaroDolalv · Accepted Answer

You can do this by showing that the numerator of  (  2  k  +  1  )  p  (  x  )  =                    2        k                  x                      2            k            +            2                          +        (        2        k        +        2        )                  x                      2            k            +            1                          −        (        2        k        +        2        )        x        −        2        k                              x          2                −        1            has just two zeroes, at   x  =  ±  1 (so that elsewhere it has the same sign as the denominator, since it clearly tends to   +  ∞ as   x  →  ±  ∞ and is negative at   x  =  0). This can be done by looking closely at the numerator&#039;s first and second derivatives,  (  2  k  +  2  )  (  2  k  )      x          2      k      +      1        +  (  2  k  +  2  )  (  2  k  +  1  )      x          2      k        −  (  2  k  +  2  )and  (  2  k  +  2  )  (  2  k  +  1  )  (  2  k  )      x          2      k        +  (  2  k  +  2  )  (  2  k  +  1  )  (  2  k  )      x          2      k      −      1        =  (  2  k  +  2  )  (  2  k  +  1  )  (  2  k  )      x          2      k      −      1        (  x  +  1  )We see that the second derivative is positive to the left of −1 and to the right of 0, and negative in between. The implies the first derivative is increasing on   (  −  ∞  ,  −  1  ), decreasing on   (  −  1  ,  0  ), and increasing on   (  0  ,  ∞  ). Since the value of the first derivative at x=-1 is   (  2  k  +  2  )  (  −  2  k  +  (  2  k  +  1  )  −  1  )  =  0, the first derivative is non-positive on   (  −  ∞  ,  0  ), so the numerator is strictly decreasing on that interval, passing through 0 at   x  =  −  1. On   (  0  ,  1  ), the first derivative increases from a negative value at x=0 to a positive value at x=1 so the numerator decreases to a minimum (negative) value and then increases to pass through 0 at x=1. It continues increasing on   (  1  ,  ∞  ), and thus it has just the two zeroes, at   x  =  ±  1

Consider the following polynomial in x: p ( x ) = <munderover> ∑

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