 Solution

ID: past-exam-of-the-graph-theory-and-probabilistic-combinatorics-course-of-the-university-of-oxford/2009/a/2/c/solution

t = - x > 0

(- 1)^{n} p_{G} (x) = \sum_{i = 0}^{n - 1} a_{i} t^{n - i} .

Every summand is nonnegative and the leading term is

t^{n} > 0

, since

a_{0} = 1

. Therefore

(- 1)^{n} p_{G} (x) > 0 (x < 0) .

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