MHB Proving $2P(x)>P'(x)$ with Continuous Third Derivative of $P(x)$

AI Thread Summary
The discussion centers on proving the inequality $2P(x) > P'(x)$ given that $P(x)$ has a continuous third derivative and that $P(x)$, $P'(x)$, $P''(x)$, and $P'''(x)$ are all positive for all $x$. It is established that if $P(x) > P'''(x)$ for all $x$, then this condition can be leveraged to demonstrate the desired inequality. The proposed solution involves analyzing the behavior of the derivatives and applying properties of continuous functions. The thread emphasizes the importance of the assumptions about the positivity and continuity of the derivatives in reaching the conclusion. Overall, the discussion aims to provide a rigorous proof of the stated inequality under the given conditions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $P$ be a real function with a continuous third derivative such that $P(x),\,P'(x),\,P''(x),\,P'''(x)$ are greater than zero for all $x$.

Suppose that $P(x)>P'''(x)$ for all $x$, prove that $2P(x)>P'(x)$ for all $x$.
 
Mathematics news on Phys.org
Solution proposed by other:

For simplicity, we will show $2P(0)>P'(0)$. Applying this result to $P(x+c)$ shows that $2P(c)>P'(c)$ for all $c$.

Since $P'(x)$ is positive, $P$ is an increasing function. Thus for $x\le 0$, $P'''(x)< P(x)\le P(0)$. Integrating $P'''(x)\le P(0)$ from $x$ to 0 gives $P''(x)\le P''(0)+P(0)x$ for $x\le 0$, and integrating again gives the second inequality in $0<P'(x)\le P'(0)+P''(0)x+P(0)\dfrac{x^2}{2}$. Thus the polynomial $P'(0)+P''(0)x+P(0)\dfrac{x^2}{2}$ has no negative zeros. It also has no nonnegative zeros (because all its coefficients are positive). Therefore its discriminant $P''(0)^2-2P(0)P'(0)$ must be negative.

In a similar vein, since $P'''$ is positive, $P''$ is increasing. Thus for $x\le 0$, $P''(x)\le P''(0)$, so $P'(x)\le P'(0)+P''(0)x$ and $0<P(x)\le P(0)+P'(0)x+P''(0)\dfrac{x^2}{2}$.

Again, the discriminant of the quadratic must be negative: $P'(0)^2-2P(0)P''(0)<0$.

Combining these conclusion, we obtain $P'(0)^4<4P(0)^2P''(0)^2<8P(0)^3P'(0)$, which implies $2P(x)>P'(x)$ for all $x$..
 
Last edited:
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top