IMC 2020 · Day 1 · P4

Denote $h(z) = p(1 - \bar{z}) - p(z)$ for complex $z$. For $t \in \mathbb{R}$ we have $h(it) = p(1 + it) - p(it) = t^{100}$, $h(1/2 + it) = 0$.

If $p(z) = c_n z^n + \dots + c_0$, $c_n \ne 0$, we have $$h(a + it) = p((1 - a) + it) - p(a + it) = (1 - 2a) \bigl( n c_n i^{n-1} t^{n-1} + Q(t, a) \bigr)$$ for some polynomial $Q$ having degree at most $n - 2$ with respect to the variable $t$. Substituting $a = 0$ we get $n = 101$, $c_n = 1/101$.

Next, for large $|t|$ we see that $\Re(h(a + it)) > 0$ for $0 \leqslant a < 1/2$.

Therefore by Maximum Principle for the harmonic function $\Re h$ and the rectangle $[0, 1/2] \times [-N, N]$ for large enough $N$ we conclude that $\Re h$ is non-negative in this rectangle, in particular on $[0, 1/2]$, as we need.

Let $p(x) = \sum_{j=0}^{m} a_j x^j$. Then $$p(x+1) - p(x) = \sum_{j=0}^{m} a_j \bigl( (x+1)^j - x^j \bigr) = a_1 + a_2 (2x + 1) + \dots + a_m \left( m x^{m-1} + \binom{m}{2} x^{m-2} + \dots + 1 \right).$$ This implies that $m = 101$, $m a_m = 1$ so $a_{101} = \frac{1}{101}$, $(m-1) a_{m-1} + a_m \binom{m}{2} = 0$ so $a_{100} = -\frac12$ etc. For $j \geqslant 1$ $a_j$ is uniquely defined, $a_0$ may be chosen arbitrarily.

The equality $p_{2n}(\frac12) = 0$ holds because $0 = p_{2n}(\frac12) + p_{2n}(1 - \frac12) = 2 p_{2n}(\frac12)$. Let $n \geqslant 1$ be an integer and let $p_n$ be a polynomial such that $p_n(x+1) - p_n(x) = x^n$ for all $x$ and $p_n(0) = 0 = p_n(1)$. The above considerations prove the uniqueness of $p_n$. We have $p_1(x) = \frac12 x^2 - \frac12 x$. Also $p'_n(x+1) - p'_n(x) = n x^{n-1} = n (p_{n-1}(x+1) - p_{n-1}(x))$. Therefore $p'_n(x) = n p_{n-1}(x) + c_{n-1}$ for a properly chosen constant $c_{n-1}$. We shall prove that $$\tag{1} p_{2n-1}(x) - p_{2n-1}(1-x) = 0, \quad p_{2n}(x) + p_{2n}(1-x) = 0, \quad c_{2n} = 0, \quad p''_{2n}(x) = 2n (2n-1) p_{2n-2}(x)$$ for $n = 1, 2, \dots$ and for all $x$. Simple computation shows that $p_1(x) - p_1(1-x) = 0$. We have $(p_2(x) + p_2(1-x))' = 2 p_1(x) + c_1 - (2 p_1(1-x) + c_1) = 0$ so the map $x \mapsto p_2(x) + p_2(1-x)$ is constant thus $p_2(x) + p_2(1-x) = p_2(0) + p_2(1-0) = 0$. If the first two equalities hold for some $n$ then $$(p_{2n+1}(x) - p_{2n+1}(1-x))' = (2n+1) p_{2n}(x) + c_{2n} + (p_{2n}(1-x) + c_{2n})$$

= 2 c_{2n} so there exists $b \in \mathbb{R}$ such that $p_{2n+1}(x) - p_{2n+1}(1-x) = 2 c_{2n} x + b$ for all $x$. $p_{2n+1}(0) - p_{2n+1}(1-0) = 0$ and $p_{2n+1}(1) - p_{2n+1}(1-1) = 0$ so $2 c_{2n} = 0 = b$. This proves that $p_{2n+1}(x) - p_{2n+1}(1-x) = 0$ for all $x$. In a similar way we shall prove the second equality: $$(p_{2n+2}(x) + p_{2n+2}(1-x))' = (2n+2) p'_{2n+1}(x) + c_{2n+1} - (2n+2)(p_{2n+1}(1-x) + c_{2n+1}) = 0$$ so the map $x \mapsto p_{2n+2}(x) + p_{2n+2}(1-x)$ is constant hence $p_{2n+2}(x) + p_{2n+2}(1-x) = p_{2n+2}(0) + p_{2n+2}(1-0) = 0$ for all $x$. Now $p''_{2n+2}(x) = ((2n+2) p_{2n+1}(x) + c_{2n+1})' = (2n+2) p'_{2n+1}(x) = (2n+2)((2n+1) p_{2n}(x) + c_{2n}) = (2n+2)(2n+1) p_{2n}(x)$. Since $p'_2(x) = 2 p_1(x) + c_1 = x^2 - x + c_1$ we obtain $p''_2(x) = 2x - 1 < 0$ for $x < \frac12$. The function $p_2$ is strictly concave on $[0, \frac12]$ and $p_2(0) = 0 = p_2(\frac12)$. Therefore $p_2(x) > 0$ for $x \in (0, \frac12)$. This together with the equality $p''_4(x) = 12 p_2(x)$ implies that $p_4$ is strictly convex on $[0, \frac12]$ so in view of $p_4(0) = 0 = p_4(\frac12)$ we conclude that $p_4(x) < 0$ for $x \in (0, \frac12)$. Easy induction shows that for $x \in (0, \frac12)$ one has $p_{2n}(x) > 0$ for an odd $n$ and $p_{2n}(x) < 0$ for an even $n$. If $t \in (0, \frac12)$ then by (1) we get $p_{100}(1-t) - p_{100}(t) = -2 p_{100}(t) > 0$ as required.