3

For any $a>0$, define $$f(x):=\cosh(a\sin x)-\cos(a\cos x).$$

  1. It appears from plots to be true that $f(x)\ge 1-\cos a$ and $f(x)$ is convex over $\big[0,\frac\pi4\big]$. How would one prove this?
  2. Better yet, is there a lower bound simple function, such as an exponential function, which is also bounded from below by some positive number such as $1-\cos a$ and is convex on $\big[0,\frac\pi4\big]$?

We can of course examine the second derivative of $f$ which is \begin{align}&\frac{d^2 f(x)}{dx^2} \\ =& (a\cos x)^2\cosh(a\cos x)-a\sin x\sinh(a\sin x)+(a\sin x)^2\cos(a\cos x) \\ &-a\cos x\sin(a\cos x) \end{align} But how do we show the above is positive for $x\in \big[0,\frac\pi4\big]$?

Hans
  • 9,804

2 Answers2

1

Partial answer for convexity.

Let us prove that $f(x)$ is convex if $a \ge 3$.

We have \begin{align*} f''(x) &= (a\cos x)^2\cosh(a\sin x)-a\sin x\, \sinh(a\sin x)+(a\sin x)^2\cos(a\cos x) \\ &\qquad -a\cos x\,\sin(a\cos x)\\ &\ge (a\cos x)^2\cosh(a\sin x)-a\sin x\, \cosh(a\sin x) - (a\sin x)^2 - a\\ &= (a^2 \cos^2 x - a\sin x)\cosh(a\sin x) - a^2\sin^2 x - a\\ &\ge (a^2 \cos^2 x - a\sin x)\cdot \left(1 + \frac12a^2 \sin^2 x\right) - a^2\sin^2 x - a\\ &> 0 \tag{1} \end{align*} where we use $\cosh u \ge \sinh u$ for all $u \ge 0$, and $a^2 \cos^2 x - a\sin x \ge 0$, and $\cosh u \ge 1 + \frac12 u^2$ for all $u \ge 0$.

Proof of (1):

Letting $u = \sin x \in [0, \frac{1}{\sqrt 2}]$, we have \begin{align*} &(a^2 \cos^2 x - a\sin x)\cdot \left(1 + \frac12a^2 \sin^2 x\right) - a^2\sin^2 x - a\\ ={}& (u^2 - u^4)a^4/2 - u^3 a^3/2 + (1 - 2u^2)a^2 - (1 + u)a\\ \ge{}& (u^2 - u^4)a^3\cdot 3/2 - u^3 a^3/2 + (1 - 2u^2)a^2 - (1 + u)a\\ ={}& (-3u^4 - u^3 + 3u^2)a^3/2 + (1 - 2u^2)a^2 - (1 + u)a\\ \ge{}& (-3u^4 - u^3 + 3u^2)a^2\cdot 3/2 + (1 - 2u^2)a^2 - (1 + u)a\\ ={}& (-9u^4 - 3u^3 + 5u^2 + 2)a^2/2 - (1 + u)a\\ \ge{}& (-9u^4 - 3u^3 + 5u^2 + 2)a\cdot 3/2 - (1 + u)a\\ ={}& (-27u^4 - 9u^3 + 15u^2 - 2u + 4)a/2\\ \ge{}& \left(-27u^2\cdot \frac12 - 9u\cdot \frac12 + 15u^2 - 2u + 4\right)a/2\\ ={}& (3u^2 - 13u + 8)a/4\\ ={}& \frac{(13 - 6u)^2 - 73}{48}\cdot a\\ \ge{}& \frac{(13 - 6/\sqrt 2)^2 - 73}{48}\cdot a\\ >{}& 0. \end{align*}

We are done.

River Li
  • 37,323
  • Will you give a bit detail for the last step? 2) Is there a direct way to prove $f(x)\ge 1-\cos a$?
    • – Hans Aug 07 '23 at 15:31
  • @Hans Do you mean $-27u^4 - 9u^3 + 15u^2 - 2u + 4 > 0$? – River Li Aug 07 '23 at 15:34
  • Yes. I suppose you can divide the interval up into to 2 parts where $f$ is convex and concave on them respectively. But maybe you have a more elegant way to deal with it. – Hans Aug 07 '23 at 18:30
  • @Hans I edited it. – River Li Aug 07 '23 at 22:38
  • +1. Very impressive with the successive inequality reducing the order of $a$ in your proof of $(1)$. What prompted you to use this approach? – Hans Aug 08 '23 at 14:19
  • @Hans It is a simple trick. – River Li Aug 08 '23 at 14:37
  • Let us continue this discussion in chat. – Hans Aug 08 '23 at 15:20
  • @Hans It is fine to comment here. – River Li Aug 08 '23 at 15:32
  • The website prompts people to enter into a chat room when the conversation gets long. I was asking if you would list other examples on math.stackexchange of employing this successive bounding trick if you see any. – Hans Aug 08 '23 at 15:36
  • 1
    @Hans It is simple. I used many times. For example, here, we need to prove that $16,{z}^{9}-56,{z}^{6}+64,{z}^{5}+65,{z}^{3}-64,{z}^{2}+7 > 0$. First we deal with the term $16z^9$, we can use $\frac{1}{16}(16z^4 - 1)^2z \ge 0$ to get $16z^9 \ge 2z^5 - \frac{1}{16}z$. Then it suffices to prove that $2z^5 - \frac{1}{16}z -56,{z}^{6}+64,{z}^{5}+65,{z}^{3}-64,{z}^{2}+7 > 0$. Why $\frac{1}{16}(16z^4 - 1)^2z$? Because the minimizer is near $z = 1/2$ (so $16z^4 \approx 1$). – River Li Aug 08 '23 at 15:45
  • Are you still working on the case $a<3$? – Hans Aug 09 '23 at 15:26
  • @Hans No. However, since $a < 3$, I think we may use some bounds. – River Li Aug 09 '23 at 15:28