MHB First, second and third derivatives of a polynomial

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $x$.
 
Mathematics news on Phys.org
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 $x$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
5
Views
1K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
7
Views
2K
Replies
1
Views
1K
Back
Top