International Mathematics Competition
for University Students
2020

IMC2020: Day 1, Problem 4

Problem 4. A polynomial \(\displaystyle p\) with real coefficients satisfies the equation \(\displaystyle p(x+1)-p(x)=x^{100}\) for all \(\displaystyle x\in\mathbb{R}\). Prove that \(\displaystyle p(1-t)\geqslant p(t)\) for \(\displaystyle 0\leqslant t\leqslant 1/2\).

Daniil Klyuev, St. Petersburg State University

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

If \(\displaystyle p(z)=c_nz^n+\ldots+c_0,c_n\ne 0\), we have

\(\displaystyle h(a+it)=p((1-a)+it)-p(a+it)=(1-2a)\left(nc_ni^{n-1}t^{n-1}+Q(t,a)\right) \)

for some polynomial \(\displaystyle Q\) having degree at most \(\displaystyle n-2\) with respect to the variable \(\displaystyle t\). Substituting \(\displaystyle a=0\) we get \(\displaystyle n=101\), \(\displaystyle c_n=1/101\).

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

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

Solution 2. Let \(\displaystyle p(x)=\sum_{j=0}^m a_jx^j\). Then

\(\displaystyle p(x+1)-p(x)=\sum_{j=0}^m a_j\left((x+1)^j-x^j\right)= a_1+a_2(2x+1)+\dots+a_m\left(mx^{m-1}+\binom{m}{2}x^{m-2}+\dots+1\right).\)

This implies that \(\displaystyle m=101\), \(\displaystyle ma_m=1\) so \(\displaystyle a_{101}=\frac1{101}\), \(\displaystyle (m-1)a_{m-1}+a_m\binom{m}{2}=0\) so \(\displaystyle a_{100}=-\frac12\) etc. For \(\displaystyle j\geqslant 1\) \(\displaystyle a_j\) is uniquely defined, \(\displaystyle a_0\) may be chosen arbitrarily.

The equality \(\displaystyle p_{2n}(\frac12)=0\) holds because \(\displaystyle 0=p_{2n}(\frac12)+p_{2n}(1-\frac12)=2p_{2n}(\frac12)\). Let \(\displaystyle n\geqslant1\) be an integer and let \(\displaystyle p_n\) be a polynomial such that \(\displaystyle p_n(x+1)-p_n(x)=x^n\) for all \(\displaystyle x\) and \(\displaystyle p_n(0)=0=p_n(1)\). The above considerations prove the uniqueness of \(\displaystyle p_n\). We have \(\displaystyle p_1(x)=\frac12x^2-\frac12x\). Also \(\displaystyle p_n'(x+1)-p_n'(x)=nx^{n-1}=n\left(p_{n-1}(x+1)-p_{n-1}(x)\right)\). Therefore \(\displaystyle p_n'(x)=np_{n-1}(x)+c_{n-1}\) for a properly chosen constant \(\displaystyle c_{n-1}\). We shall prove that

\(\displaystyle {(1)}\)

\(\displaystyle p_{2n-1}(x)-p_{2n-1}(1-x)=0,\ p_{2n}(x)+p_{2n}(1-x)=0, \ c_{2n}=0, \ p_{2n}''(x)=2n(2n-1)p_{2n-2}(x) \)

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