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Friday, October 3, 2014

Mathematical Reflections 2014, Issue 4 - Problem U307

Problem:
Prove that any polynomial f \in \mathbb{R}[X] can be written as a difference of increasing polynomials.

Proposed by Jishnu Bose.

Solution:
Let f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0. We construct the following algorithm. Set g_0(X)=a_0, h_0(X)=0 and for k=1,\ldots,n, if k is odd, then g_k(X)=\begin{cases} a_kX^k & \textrm{if } a_k \geq 0 \\ 0 & \textrm{if } a_k<0, \end{cases} h_k(X)=\begin{cases} 0 & \textrm{if } a_k \geq 0 \\ -a_kX^k & \textrm{if } a_k<0, \end{cases} and if k is even, then g_k(X)=\begin{cases}a_k(X^{k+1}+X^k+X^{k-1}) & \textrm{if } a_k \geq 0 \\ -a_k(X^{k+1}+X^{k-1}) & \textrm{if } a_k < 0,  \end{cases} h_k(X)=\begin{cases}a_k(X^{k+1}+X^{k-1}) & \textrm{if } a_k \geq 0 \\ -a_k(X^{k+1}+X^k+X^{k-1}) & \textrm{if } a_k < 0. \end{cases}
Observe that for all k=0,\ldots,n, we have g_k(X)-h_k(X)=a_kX^k and g'_k(X) \geq 0, h'_k(X) \geq 0 for all x \in \mathbb{R}. Setting g(X)=\sum_{k=0}^n g_k(X), \qquad h(X)=\sum_{k=0}^n h_k(X), we have that g(X) and h(X) are increasing polynomials and f(X)=g(X)-h(X).

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