Even Degree Polynomial, show ##p(x) \geq 0 ## for all real x

[QuietMind]

Homework Statement

Prove that if p(x) has even degree with positive leading coefficient, and $p(x) - p''(x) \geq 0$ for all real x, then $$ p(x) \geq 0$$ for all real x

Homework Equations

N/A
Problem is from Art and Craft of Problem Solving, as an exercise left to the reader following a partial solution in the text. The text goes over a somewhat similar example: when we have a polynomial q(x) such that $q(x) - q'(x) \geq 0$. The text showed us that we can determine that q(x) must be of even degree with leading coefficient positive.

The Attempt at a Solution

I have thought about a few different possible strategies:

1) Show that p''(x) is always nonnegative. I'm not sure how to go about this strategy, or if this conjecture is even true.
2) Show that p(x) either has no zeroes, or only has zeroes that barely touch the x-axis (therefore it lies on the upper half plane). If p(x) does have a zero at some point, say $x_0$ then $p(x_0) - p''(x_0) \geq 0$. If $p(x_0) = 0$ then $p''(x_0) = 0$. This is a necessary but not sufficient condition for an inflection point at $x_0$. This seems significant but how does this help me?

 

[member 587159]

Homework Statement

Prove that if p(x) has even degree with positive leading coefficient, and $p(x) - p''(x) \geq 0$ for all real x, then $$ p(x) \geq 0$$ for all real x

Homework Equations

N/A
Problem is from Art and Craft of Problem Solving, as an exercise left to the reader following a partial solution in the text. The text goes over a somewhat similar example: when we have a polynomial q(x) such that $q(x) - q'(x) \geq 0$. The text showed us that we can determine that q(x) must be of even degree with leading coefficient positive.

The Attempt at a Solution

I have thought about a few different possible strategies:

1) Show that p''(x) is always nonnegative. I'm not sure how to go about this strategy, or if this conjecture is even true.
2) Show that p(x) either has no zeroes, or only has zeroes that barely touch the x-axis (therefore it lies on the upper half plane). If p(x) does have a zero at some point, say $x_0$ then $p(x_0) - p''(x_0) \geq 0$. If $p(x_0) = 0$ then $p''(x_0) = 0$. This is a necessary but not sufficient condition for an inflection point at $x_0$. This seems significant but how does this help me?

And what did you achieve trying those strategies?

 

[QuietMind]

I'm really not sure how to go about proving the first one, but the second derivative being zero should mean that the first derivative is not changing at $x_0$. This tells me that the first derivative is not changing at this point, but does this also tell me that the function never crosses the x-axis (only barely touching it)?

Or are you telling me that I should be trying something else?

 

[PeroK]

I'm really not sure how to go about proving the first one, but the second derivative being zero should mean that the first derivative is not changing at $x_0$. This tells me that the first derivative is not changing at this point, but does this also tell me that the function never crosses the x-axis (only barely touching it)?

Or are you telling me that I should be trying something else?

What can the second derivative be used to determine? Think elementary calculus.

 

[QuietMind]

The second derivative determines the concavity of a function. At a zero of f(x), $x_0$ , the function has $f''(x_0) \leq 0$ . The problem is I don't know anything about the first derivative, so I can't use the second derivative test to say if a point is a local max/min. If the second derivative is zero, it may be an inflection point. Were you referring to concavity?

For any zero $x_0$ , either the second derivative must be 0 (function is linear at that point, might be inflection point) or negative (concave down). Also, I also know that the second derivative is positive as x tends to plus or minus infinity because it also is a even degree polynomial with positive leading coefficient. I'm wondering if it's impossible for these statements to be concurrently true. If all the zeroes of f(x) are concave down or linear, does this prevent the function from having the correct concavity at very large and very negative x?

 

[Ray Vickson]

The second derivative determines the concavity of a function. At a zero of f(x), $x_0$ , the function has $f''(x_0) \leq 0$ . The problem is I don't know anything about the first derivative, so I can't use the second derivative test to say if a point is a local max/min. If the second derivative is zero, it may be an inflection point. Were you referring to concavity?

For any zero $x_0$ , either the second derivative must be 0 (function is linear at that point, might be inflection point) or negative (concave down). Also, I also know that the second derivative is positive as x tends to plus or minus infinity because it also is a even degree polynomial with positive leading coefficient. I'm wondering if it's impossible for these statements to be concurrently true. If all the zeroes of f(x) are concave down or linear, does this prevent the function from having the correct concavity at very large and very negative x?

For large $|x|$ we must have $p(x) > 0$ because the large-$|x|$ behavior of $p(x)$ is governed by the leading term $c_{2n} x^{2n}$, with $c_{2n} > 0$ (stated in the problem, but also provable on its own from the condition $p'' \leq p$). If $p(x_0)< 0$ for some $x_0$, we will have $p(x) < 0$ for all $x$ in some interval $I$ that contains $x_0$. Since $p'' \leq p$ we have that $p''(x) < 0$ on the interval $I$; that is, $p$ is strictly concave on the interval $I$.

If we have $p'(x_1) < 0$ for some $x_1 \in I$, the function will be going down as $x$ increases from $x_1$, so will become increasingly negative. There is no way that we could have $p(x) > 0$ for some $x > x_1$---which we would need because we have $p(x) > 0$ for large enough $x$---because in order for that to happen the graph of $p(x)$ would need to "turn around" and start increasing again, and that would mean that $p(x)$ would need to pass through a minimum $x_2$ for which $p''(x_2) > 0$. In other words, $p$ would have to switch from being concave to being convex at some point where it was still < 0. But that is impossible because $p(x_2) < 0$ implies $p''(x_2) < 0$. So, our assumption that $p'(x_1) < 0$ for some $x_1 \in I$ leads to a contradiction, hence it cannot occur.

You can look at similar arguments if $p'(x_1) > 0$ for some $x_1 \in I$.

 

[QuietMind]

For large $|x|$ we must have $p(x) > 0$ because the large-$|x|$ behavior of $p(x)$ is governed by the leading term $c_{2n} x^{2n}$, with $c_{2n} > 0$ (stated in the problem, but also provable on its own from the condition $p'' \leq p$). If $p(x_0)< 0$ for some $x_0$, we will have $p(x) < 0$ for all $x$ in some interval $I$ that contains $x_0$. Since $p'' \leq p$ we have that $p''(x) < 0$ on the interval $I$; that is, $p$ is strictly concave on the interval $I$.

If we have $p'(x_1) < 0$ for some $x_1 \in I$, the function will be going down as $x$ increases from $x_1$, so will become increasingly negative. There is no way that we could have $p(x) > 0$ for some $x > x_1$---which we would need because we have $p(x) > 0$ for large enough $x$---because in order for that to happen the graph of $p(x)$ would need to "turn around" and start increasing again, and that would mean that $p(x)$ would need to pass through a minimum $x_2$ for which $p''(x_2) > 0$. In other words, $p$ would have to switch from being concave to being convex at some point where it was still < 0. But that is impossible because $p(x_2) < 0$ implies $p''(x_2) < 0$. So, our assumption that $p'(x_1) < 0$ for some $x_1 \in I$ leads to a contradiction, hence it cannot occur.

You can look at similar arguments if $p'(x_1) > 0$ for some $x_1 \in I$.

I did out the argument for $p'(x_1) > 0$ for some $x_1 \in I$ by considering decreasing x from $x_1$. The derivative must be even more positive for these points, and because the second derivative is negative, this means that p(x) must approach negative infinity as x decreases from $x_1$. However, does one even have to consider this argument? I would expect that for the interval $I$ that you described, there must necessarily be a point $x_1$ with a negative derivative, because the function p(x) goes from zero, to some negative point $x_0$ and then back to zero. That would imply there must be a point with negative derivative, because polynomials are nice and smooth. So it technically shouldn't be necessary to consider $p'(x_1) > 0$ for some $x_1 \in I$, should it?

 

[Delta2]

Ehm sorry for being slow (lol) but all we prove is that $p'(x)=0$ for x in some interval I, so doesn't that mean that $p(x)=c$ for all x in I? And since our function is polynomial doesn't that mean that $p(x)=c$ for all x in real line?

is $p(x)=c>0$ the only polynomials that satisfy the conditions of this problem?

 

[PeroK]

I did out the argument for $p'(x_1) > 0$ for some $x_1 \in I$ by considering decreasing x from $x_1$. The derivative must be even more positive for these points, and because the second derivative is negative, this means that p(x) must approach negative infinity as x decreases from $x_1$. However, does one even have to consider this argument? I would expect that for the interval $I$ that you described, there must necessarily be a point $x_1$ with a negative derivative, because the function p(x) goes from zero, to some negative point $x_0$ and then back to zero. That would imply there must be a point with negative derivative, because polynomials are nice and smooth. So it technically shouldn't be necessary to consider $p'(x_1) > 0$ for some $x_1 \in I$, should it?

I think you're making this too complicated. Just consider any local mimimum for the polynomial.

 

[Ray Vickson]

I did out the argument for $p'(x_1) > 0$ for some $x_1 \in I$ by considering decreasing x from $x_1$. The derivative must be even more positive for these points, and because the second derivative is negative, this means that p(x) must approach negative infinity as x decreases from $x_1$. However, does one even have to consider this argument? I would expect that for the interval $I$ that you described, there must necessarily be a point $x_1$ with a negative derivative, because the function p(x) goes from zero, to some negative point $x_0$ and then back to zero. That would imply there must be a point with negative derivative, because polynomials are nice and smooth. So it technically shouldn't be necessary to consider $p'(x_1) > 0$ for some $x_1 \in I$, should it?

Well, you can show that having $p(x) < 0$ on an interval $I$ leads to a contradiction (because $x_1 \in I$ and $p'(x_1) < 0$ or $x_1 \in I$ and $p'(x_1) > 0$ both lead to contradictions, and $p'(x_1) = 0$ implies that $p'(x_{1a}) > 0, p'(x_{1b}) < 0$ for some $x_{1a} < x_1 < x_{1b}$ in $I$. In other words, you cannot have $p(x) < 0$ on an interval $I$, so cannot have $p(x_0)< 0$ at any point $x_0$.

Basically we cannot have $p(x) < 0$ along with $p'(x) < 0$, $p'(x) = 0$ or $p'(x) > 0$ anywhere---in other words, we cannot have $p(x) < 0$ anywhere.

 

[QuietMind]

I think you're making this too complicated. Just consider any local mimimum for the polynomial.

So in order to have the correct behavior at large positive or negative x, if the function crosses below the x axis, it must come back up and cross the x-axis again. This guarantees that a local minimum must exist somewhere on this interval I where this function goes below the x axis, but if the second derivative is negative whenever p(x) < 0, then we cannot have a local minimum, only a maximum. Hence this is a contradiction, and we cannot have the function p(x) < 0.

Ehm sorry for being slow (lol) but all we prove is that $p'(x)=0$ for x in some interval I, so doesn't that mean that $p(x)=c$ for all x in I? And since our function is polynomial doesn't that mean that $p(x)=c$ for all x in real line?

is $p(x)=c>0$ the only polynomials that satisfy the conditions of this problem?

I don't think p(x) has to be a constant, some function like $p(x) = x^4 + 12x^2 + 1000$ with $p''(x) = 12x^2 + 12$ should have $p(x)-p''(x) \geq 0$ for all real x. I made a mistake in an earlier post where I said that $p''(x_0)$ had to be 0.

 

[PeroK]

So in order to have the correct behavior at large positive or negative x, if the function crosses below the x axis, it must come back up and cross the x-axis again. This guarantees that a local minimum must exist somewhere on this interval I where this function goes below the x axis, but if the second derivative is negative whenever p(x) < 0, then we cannot have a local minimum, only a maximum. Hence this is a contradiction, and we cannot have the function p(x) < 0.

Yes, that's the core of the argument. You can also argue simply that at any local minimum, $x_0$, we have $p(x_0) \ge p''(x_0) \ge 0$.

Note that the proposition holds for any function $f$ which is non-negative as $x \rightarrow \pm \infty$ and has $\forall \ x \ f(x) \ge f''(x)$.

 

[QuietMind]

Yes, that's the core of the argument. You can also argue simply that at any local minimum, $x_0$, we have $p(x_0) \ge p''(x_0) \ge 0$.

Note that the proposition holds for any function $f$ which is non-negative as $x \rightarrow \pm \infty$ and has $\forall \ x \ f(x) \ge f''(x)$.

I think I see what you are saying with this holding for any function f, but I'm wondering if you'd need any additional constraint of it being "nice" and/or smooth. The argument hinges on if p(x) goes negative, there must necessarily be a local minimum at a negative value of p(x) to guarantee that the function curves back up. However, if a function goes negative and comes back up but doesn't have a local minimum (say a dirac delta valley or something very sharp like that) so the argument in my prior post using a local minimum wouldn't hold. This wouldn't happen for a polynomial, which tend to be "nice"/smooth, but I'm wondering what needs to be said about an arbitrary function f(x). Perhaps the existence of the second derivative of f(x) (or p(x) )for all x guarantee the function is "nice"/ smooth enough? I have no Real Analysis background yet, so this is an area that I have but a hazy understanding of.

 

[PeroK]

The fact that $f$ is twice differentiable is enough. The Dirac Delta isn't a function.