MHB First, second and third derivatives of a polynomial

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The discussion centers on proving that if the expression \( p(x) - p'(x) - p''(x) + p'''(x) \) is non-negative for all real \( x \), then the polynomial \( p(x) \) must also be non-negative for all real \( x \). Participants explore the implications of the derivatives of the polynomial and how they relate to the behavior of \( p(x) \). The proof likely involves analyzing the properties of polynomials and their derivatives, particularly focusing on the conditions under which the original polynomial remains non-negative. The conversation emphasizes the relationship between the derivatives and the original polynomial's values across the real number line. Ultimately, the conclusion drawn is that the non-negativity of the derivative expression ensures the non-negativity of the polynomial itself.
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Let $p(x)$ be a polynomial with real coefficients. Prove that if $p(x)-p'(x)-p''(x)+p'''(x)\ge 0$ for every real $x$, then $p(x)\ge 0$ for every real $x$.
 
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If $f(x)$ is a polynomial with real coefficients then $f(x)\to +\infty$ as $x\to\infty$ if the coefficient of its leading term is positive, and $f(x)\to -\infty$ as $x\to\infty$ if the coefficient of its leading term is negative. The degree of the derivative $f'(x)$ is lower than the degree of $f(x)$, so the polynomials $f(x)$ and $f(x) - f'(x)$ have the same leading term. Therefore either they both go to $+\infty$ or they both go to $-\infty$, as $x\to\infty$.

Now suppose that $f(x) - f'(x) \geqslant0$ for every real $x$. Then $f(x) - f'(x)\to +\infty$ as $x\to\infty$ and therefore $f(x)\to +\infty$ as $x\to\infty$. By the same argument, $f(x)\to +\infty$ as $x\to-\infty$. It follows that $f(x)$ has a finite minimum value, which it attains at some point $x_0$. Then $f'(x_0) = 0$. But $f(x_0) - f'(x_0) \geqslant0$. So $f(x_0)\geqslant0$. But if the minimum value of $f(x)$ is nonnegative then the function must be nonnegative everywhere. That proves
Fact 1: If $f(x)$ is a polynomial with real coefficients, and $f(x) - f'(x) \geqslant0$ for every real $x$, then $f(x)\geqslant0$ for every real $x$.
It follows that
Fact 2: If $f(x)$ is a polynomial with real coefficients, and $f(x) +f'(x) \geqslant0$ for every real $x$, then $f(x)\geqslant0$ for every real $x$.
To prove Fact 2, let $g(x) = f(-x)$. Then $g'(x) = -f'(-x)$ and so $g(x) - g'(x) = f(-x) + f'(-x) \geqslant0$ for every real $x$. It follows from Fact 1 that $g(x)\geqslant0$ for every real $x$. Therefore $f(x)\geqslant0$ for every real $x$.

Write $D= \frac d{dx}$ for the operator of differentiation, and $I$ for the identity operator. Then those two Facts can be written as
Fact 1: If $f(x)$ is a polynomial with real coefficients, and $(I-D)f(x) \geqslant0$ for every real $x$, then $f(x)\geqslant0$ for every real $x$;
Fact 2: If $f(x)$ is a polynomial with real coefficients, and $(I+D)f(x) \geqslant0$ for every real $x$, then $f(x)\geqslant0$ for every real $x$.

Now let $p(x)$ be a polynomial with real coefficients such that $p(x) - p'(x) - p''(x) + p'''(x) \geqslant0$ for every real $x$. That condition says that $(I-D - D^2 + D^3)p(x)\geqslant0$ for every real $x$. But $I-D-D^2+D^3 = (I-D)(I-D^2)$. Therefore $(I-D - D^2 + D^3)p(x) = (I-D)\bigl((I-D^2)p(x)\bigr) \geqslant0$ for every real $x$. It follows from Fact 1 that $(I-D^2)p(x) \geqslant0$ for every real $x$. But $I-D^2 = (I-D)(I+D)$. Therefore $(I-D)\bigl((I+D)p(x)\bigr) \geqslant0$ for every real $x$, and from Fact 1 again it follows that $(I+D)p(x) \geqslant0$ for every real $x$. Finally, by Fact 2 it then follows that $p(x) \geqslant0$ for every real $x$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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