How to prove that $\frac{\cosh (2 \pi x)}{x^2}$ is convex?

Question

I want to prove that this function $$f(x)=\frac{\cosh (2 \pi x)}{x^2}$$ is convex $(0<x<10)$, and find its minimum. The first and second derivatives are a follows $$f'(x)=\frac{2 (\pi x \sinh (2 \pi x)-\cosh (2 \pi x))}{x^3}$$

$$f''(x)=\frac{\left(4 \pi ^2 x^2+6\right) \cosh (2 \pi x)-8 \pi x \sinh (2 \pi x)}{x^4}$$ How to prove that $f''$ is positive?

Answer 1

I think my hint might've been a little too vague so here's a direct method.

Since $x^4$ is always positive then we just have to show that the numerator is positive. Using the definitions $\mbox{cosh}(x) = \frac{e^x + e^{-x}}{2}$ and $\mbox{sinh}(x) = \frac{e^x - e^{-x}}{2}$ then we have

$$(4 \pi ^2 x^2+6) \cosh (2 \pi x)-8 \pi x \sinh (2 \pi x)=(4 \pi ^2 x^2+6)\frac{e^{2\pi x} + e^{-2\pi x}}{2} - 8\pi x\frac{e^{2\pi x} - e^{-2\pi x}}{2}$$

now simplifying and ignoring the $1/2$ factor because it's not important then we just have to show that

$$(4\pi^2 x^2 - 8\pi x + 6)e^{2\pi x} + (4\pi^2 x^2 + 8\pi x + 6)e^{-2\pi x}$$

is positive. Now, the quadratic $4\pi^2 x^2 - 8\pi x + 6$ is positive at $x=0$, and its determinant is negative. Therefore it has no real roots, and so it is positive for all values of $x$. And since the coefficient of $e^{-2\pi x}$ is positive in the range $0<x<10$ then the entire expression is positive in the range $0<x<10$. And so $f^{\prime\prime}$ is positive for $0<x<10$.

Answer 2

For starting the problem, let us make life simpler $$f(x)=\frac{\cosh (2 \pi x)}{x^2}$$ Let $2\pi x=t$ and consider the problem of $$g(t)=\frac{\cosh (t)}{t^2}$$ for which $$g'(t)=\frac{t \sinh (t)-2 \cosh (t)}{t^3}\qquad \text{and} \qquad g''(t)=\frac{\left(t^2+6\right) \cosh (t)-4 t \sinh (t)}{t^4}$$ Considering the numerator of the first derivative, we have $$y_1=t \sinh (t)-2 \cosh (t)=-2+2\sum_{n=2}^\infty \frac{ (n-1)}{\Gamma (2 n+1)}t^{2n}$$ In the summation, all coefficienst are positive and then $y_1$ cancels only once. Considering the first terms as a truncated Taylor expansion, using series reversion we have $$t=z-\frac{z^3}{60}+\frac{9 z^5}{11200}-\frac{z^7}{19200}+\frac{96463 z^9}{24837120000}+O\left(z^{11}\right) \qquad \text{where} \qquad z=\sqrt[4]{12 (y_1+2)}$$ Making $y_1=0$, that is to say $z=\sqrt{2 \sqrt{6}}$, this gives for the solution the approximate solution $$t_*=\frac{44048063 }{21560000}\sqrt[4]{\frac{3}{2}}-\frac{43}{100\, \sqrt[4]{24}}\approx 2.06673$$ Since we face an alternating series, this gives a slightly overestimate of the solution (effectively, using Newton method, the "exact" solution is $t=2.06534$.

Doing the same with the numerator of the second derivative gives $$y_2=\left(t^2+6\right) \cosh (t)-4 t \sinh (t)=6+\sum_{n=2}^\infty \frac{2 n^2-5n+3}{n\,\Gamma (2 n)}t^{2n}$$ All coefficients are positive; then $g''(t)$ is always positive.

Back to $x$ and $f(x)$, the minimum occurs at $x_*=\frac {t_*}{2 \pi} \approx 0.328930\sim \frac 1 \pi$. Expanding as a Taylor series, we then have $$f(x)=\pi ^2 \cosh (2)-\frac{2 \pi ^3 }{e^2}\left(x-\frac{1}{\pi }\right)+\pi ^4 \left(\frac{4}{e^2}+\cosh (2)\right)\left(x-\frac{1}{\pi }\right)^2 +O\left(\left(x-\frac{1}{\pi }\right)^3\right)$$ which gives as a refinement $$x_*=\frac 1 \pi \left(1+\frac{2}{9+e^4} \right)\approx 0.328320$$ $$f(x_*)=\frac{\left(5+10 e^4+e^8\right) \pi ^2}{2 e^2 \left(9+e^4\right)}\approx 37.0894$$ while a full numerical optimization would give $x_*=0.328709$ and $f(x_*)=37.0883$.

Answer 3

Convexity of $f(x)$ is easy to prove from the expansion:

$$ f(x) = \frac{1}{x^2} \sum_{n=0}^{\infty} \frac{(2\pi x)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(2\pi)^{2n}}{(2n)!} x^{2n-2}. $$

Since $x \mapsto x^{2n-2}$ is convex on $(0, \infty)$ for each $n \in \{0,1,2,\dots\}$ and the coefficients are positive, $f(x)$ must be also convex.