Is this proof valid to prove that $F(x)-G(x)$ is always non zero?

Question

I have two functions:

$F(x)=16 \left(4 \pi x^2+\pi \right)^2 x^4+2$

$G(x)=4 x \left(x \left(96 x^4+104 x^2+\left(4 \pi x^2+\pi \right)^2+2\right)-2 \pi \left(4 x^2+1\right)^3 \coth (\pi x)\right)+\left(384 x^6-96 x^4+8 x^2-2\right) \cosh (2 \pi x)$

I want to show that $F(x)-G(x)$ can never be zero for any value of $0<x<0.5$. I assume that it can be zero, i.e. $F(x)=G(x)$ for all values of $x$, and try to find a contradiction. Then, we have:

$F(0)=2\quad$ and, $\quad\lim_{x\to 0} \, G(x)=-10$

Therefore, as I had supposed that $F(x)=G(x)$ should be held for all values of $x$ and now I have found a counterexample, so, the assumption is false and $F(x)-G(x)$ can never be zero for any value of $x$. Is this claim true? If not, does anyone have an idea or hint?

Actually $F$ is always positive, and $G$ is always negative, but it is not easy to prove the latter one.

I thank anyone in advance for their help.

Answer 1

If you have indeed shown that $F(0) - G(0) \neq 0$, this still does not rule out the possibility that $F\left(\frac14\right) - G\left(\frac14\right) = 0$ or $F(0.15) - G(1.5) = 0$ or even $F\left(\frac1\pi\right) - G\left(\frac1\pi\right) = 0,$ any one of which would be a counterexample to the claim that "$F(x) - G(x)$ can never be zero for any value of $0<x<0.5$."

The problem with your "proof" is that you're acting as if "always zero" is the opposite of "never zero." The opposite of "never zero" is "sometimes zero".

More generally, a claim like "$F(x) - G(x)$ can never be zero" is disproved if you find even one counterexample, that is, for a disproof all you need is one value of $x$ where $F(x) - G(x) = 0.$ If something can be disproved by a single counterexample, you're usually not going to be able to prove it by a single counterexample.

But if you know that $F(x) > 0$ for every $0 < x < 0.5$, and you also know that $G(x) < 0$ for every $0 < x < 0.5$, then you know $-G(x) > 0$ for every $0 < x < 0.5$ (because the negation of a negative number is a positive number) and therefore $F(x) - G(x) > 0$ for every $0 < x < 0.5$ (because $F(x) - G(x) = F(x) + (-G(x))$ and the sum of two positive numbers is positive).

Answer 2

This started out as an attempt to simplify the problem, but it turned out that it actually solves the problem with very little additinal effort in the Addendum.

If we replace $G(x)$ with a $G_1(x)$ that fulfills $G_1(x) \ge G(x)$ in $[0,0.5]$ and then can prove $F(x) > G_1(x)$ in $[0,0.5]$ we are done.

Choose

$$G_1(x)=4 x \left(x \left(96 x^4+104 x^2+\left(4 \pi x^2+\pi \right)^2+2\right)\right) -8(4x^2+1)^3.$$

If you compare it to your $G(x)$, you'll see 2 terms missing,

1. $-8x\pi(4x^2+1)^3\coth(\pi x)$, which I have replaced with $-8(4x^2+1)^3$, and
2. $\left(384 x^6-96 x^4+8 x^2-2\right) \cosh (2 \pi x)$, which I have replaced with $0$.

To justify 1., you need to apply that $x\coth(x) \ge 1$ for all $x$, see for example in the german Wikipedia under "Reihenentwicklungen" (the English version of the Wikepdia page does not seem to contain that expansion that is usefull here). So $\pi x \coth(\pi x) \ge 1$, and multiplying both sides with the negative $-8(4x^2+1)^3$ shows that I replace a term from $G(x)$ with one that is at least as big in $G_1(x)$.

To justify 2., note that $\left(384 x^6-96 x^4+8 x^2-2\right)$ is non-positive in $[0,0.5]$ (it has a root at $0.5$), while $\cosh (2 \pi x)$ is always positive. So again a I replace a term in $G(x)$ with one in $G_1(x)$ that's at least as big.

This proves $G_1(x) \ge G(x)$ in $[0,0.5]$.

Of course, all of this is for nought if not $F(x) > G_1(x)$, but Wolrfam Alpha shows otherwise:

The gain we have reached in my opinion is that $F(x)$ and $G_1(x)$ are "just" polynomials, no more unwieldy coth and cosh. Furthermore, both $F(x)$ and $G_1(x)$ are really functions of $x^2$, which means you can subsitute $t=x^2$ and $F$ becomes a 4th-degree polynomial and while $G_1$ becomes a 3rd degree polynomial.

The $-8$ constant term in $G_1$ is "dominating" that function for some time after $x=0$, so it seems substantial work may need to be done only near $x=0.5$.

In addition, even if this proves still too difficult, better inequalities that I used above (especially for 2.) may give more breathing room.

Addendum:

It turned out, the remaining way to go is not that hard. Let's start with writing $F(x)$ and $G_1(x)$ as polynomials in expanded form:

$$F(x)=256\pi^2x^8+128\pi^2x^6+16\pi^2x^4+2$$ $$G_1(x)=64(\pi^2-2)x^6+32(1+\pi^2)x^4+4(\pi^2-22)x^2-8$$

First, as already said, both polynomials contain only even powers of $x$, so a higher coefficient on the same power of $x$ means a higher value of that term.

Comparing them term by term on the power of $x$, we see that for $x^8$ and $x^6$, $F$ has the higher coefficient. For $x^4$ that's not the case, but we can amend that with taking the terms with $x^2$ into account as well:

$$ \begin{eqnarray} 16\pi^2x^4 & \ge & 32(1+\pi^2)x^4+4(\pi^2-22)x^2 & \Longleftrightarrow\\ (-32-16\pi^2)x^4 + 4(22-\pi^2)x^2 & \ge & 0 & \Longleftrightarrow\\ 4x^2((22-\pi^2) - (8+4\pi^2)x^2) & \ge & 0 & \Longleftrightarrow\\ \end{eqnarray} $$

Now, since we are only interested in $x \in [0,0.5]$, we find that $x^2 \le \frac14$ and hence $(8+4\pi^2)x^2 \le 2+\pi^2$, so we get

$$(22-\pi^2) - (8+4\pi^2)x^2 \ge (22-\pi^2) - (2+\pi^2)=20-2\pi^2 \approx 0.26 > 0.$$

Since also $4x^2 \ge 0$, we find that indeed the the quartic and quadratic terms of $F$ are larger than the quartic and quadratic terms of $G_1(x)$. The only remaining terms are the constant terms, where again $F$ is higher than $G_1$ with $2 > -8$.

So we see that actually $F(x)-G_1(x) \ge 10$ on the interval $[0,0.5]$, which proves the conclusion!

Answer 3

Let $p(x)=384x^6-96x^4+8x^2-2$, $q(x)=(4x^2+1)^3$ and $r(x)=96x^4+104x^2+(4\pi x^2+\pi)^2+2$. So we have $$G(x)=4x(x r(x)-2\pi q(x) \coth(\pi x))+p(x)\cosh(2\pi x).$$

Note that $p(x)=2(2x-1)(2x+1)(48x^4+1)$, so the only real roots of $p(x)$ are $\frac{1}{2}$ and $-\frac{1}{2}$. As $p(0)=-2<0$, we know that $p(x)<0$ for all $x\in (-\frac{1}{2}, \frac{1}{2})$.

As $\cosh(2\pi x)=\frac{1}{2} (e^{2\pi x}+e^{-2\pi x})>0$ for all $x\in\mathbb{R}$, we have $$p(x)\cosh{(2\pi x)}<0 $$ for all $x\in (0, \frac{1}{2})$.

Now for all $x>0$, we have $$\coth(\pi x)=\frac{e^{2\pi x}+1}{e^{2\pi x}-1}>1,$$ so $$2\pi q(x)\coth(\pi x) - x r(x) > 2\pi q(x) - x r(x).$$ We will prove that $2\pi q(x) - x r(x)>0$ using Sturm's theorem.

Expanding $2\pi q(x) - x r(x)$ gives $$128\pi x^6 - (16\pi^2+96) x^5 + 96\pi x^4 - (8\pi^2-104) x^3 + 24\pi x^2 - (\pi^2+2)x + 2\pi.$$ Using $\pi>3$ and $-\pi^2>-10$, we get $$2\pi q(x) - x r(x) > 2(64x^6-128x^5+144x^4-92x^3+36x^2-6x+3).$$ Denote the non-constant factor in the RHS by $s(x)$. Note that $s(0)=3>0$, so if $s(x)$ has no real root, then we are done. We construct the Sturm chain: \begin{align*} s(x) &= 64x^6 - 128x^5 + 144x^4 - 92x^3 + 36x^2 - 6x + 3 \\ s'(x) &= 384x^5 - 640 x^4 + 576x^3 - 276x^2+72x-6 \\ s_2(x) &= -\frac{1}{9} (112x^4 - 126 x^3 + 78 x^2 - 9x + 24) \\ s_3(x) &= -\frac{9}{7} (58x^3 - 78x^2 - 21x + 30) \\ s_4(x) &= \frac{2}{2523} (21258 x^2 - 8132 x + 1579) \\ s_5(x) &= \frac{1}{12552849} (- 753041492x + 551089639) \\ s_6(x) &= -\frac{937093362962103}{168570597108676} \end{align*} Signs at $-\infty$ are $(+,-,-,+,+,+,-)$. Number of sign changes is $3$.

Signs at $+\infty$ are $(+,+,-,-,+,-,-)$. Number of sign changes is $3$.

Number of real roots is $3-3=0$.