7
$\begingroup$

I would like to prove that :

$$\forall x \in [0,1], e^x\cos(\sqrt{x^2+1}) \leq 1$$

When plotting the graph this inequality is not sharp at all and we even have : $$\forall x \in [0,1], e^x \cos(\sqrt{x^2+1}) \leq 0.8$$

I tried several things such has :

Calculating the derivative and try to apply the mean value theorem to get an upper bound, but the derivative is hard to manipulate and it doesn’t seem I am getting something.

Moreover trying something on convexity but once again this is difficult due to the horrible looking of the derivative.

I am very interested in sharper upper bound, even if I can’t manage to prove the inequality for $1$...

$\endgroup$
  • $\begingroup$ “sharp”? ${}{}{}$ $\endgroup$ – let's have a breakdown Nov 7 '18 at 22:41
  • 3
    $\begingroup$ @ChaseRyanTaylor It means that the inequality achieves equality somewhere, creating (somewhat incorrectly) the impression that the inequality cannot be "improved". $\endgroup$ – Theo Bendit Nov 7 '18 at 22:45
  • $\begingroup$ The maximum value is about $0.722$ $\endgroup$ – Yuriy S Nov 8 '18 at 1:07
5
$\begingroup$

We can rewrite the inequality to be proved as

$$e^{-x}-\cos(\sqrt{1+x^2})\ge0$$

for $0\le x\le1$.

By truncating Taylor series and using a crude estimate to keep things quadratic, we have

$$e^{-x}\ge1-x+{x^2\over2}-{x^3\over6}=1-x+{x^2\over2}\left(1-{x\over3}\right)\ge1-x+{x^2\over2}\left(1-{1\over3}\right)=1-x+{x^2\over3}$$

and (since $\sqrt{1+x^2}\le\sqrt2$ for $0\le x\le1$ and the alternating terms in the Taylor series for $\cos\sqrt2$ are monotonically decreasing)

$$\cos(\sqrt{1+x^2})\le1-{1+x^2\over2}+{(1+x^2)^2\over24}={13-10x^2+x^4\over24}\le{13-10x^2+1\over24}={7-5x^2\over12}$$

It follows that

$$e^{-x}-\cos(\sqrt{1+x^2})\ge1-x+{x^2\over3}-{7-5x^2\over12}={9x^2-12x+5\over12}={(3x-2)^2+1\over12}\gt0$$

Remark: Part of what makes this work is that the inequality, as the OP observed, is not sharp, so there is a fair amount of room for crude estimates to keep things simple.

$\endgroup$
3
$\begingroup$

$e^x$ is increasing on $[0,1]$, and $\cos\sqrt{x^2+1}$ is decreasing. Therefore:

  • if $x\in[0,\frac12]$, then $e^x\le e^\frac12$ and $\cos\sqrt{x^2+1}\le\cos 1$, so $$e^x\cos\sqrt{x^2+1}\le e^\frac12\cos 1 < 0.9$$

  • if $x\in[\frac12,\frac34]$, then $e^x\le e^\frac34$ and $\cos\sqrt{x^2+1}\le\cos\sqrt\frac54$, so $$e^x\cos\sqrt{x^2+1}\le e^\frac34\cos \sqrt\frac54 < 0.93$$

  • if $x\in[\frac34,1]$, then $e^x\le e$ and $\cos\sqrt{x^2+1}\le\cos\sqrt{(\frac34)^2+1}=\cos\frac54$, so $$e^x\cos\sqrt{x^2+1}\le e\cos\frac54 < 0.86$$

This proves that your function is less than $0.93$ on $[0,1]$. And you can get as close to the true maximum as you like using this method, by dividing $[0,1]$ into ever more intervals.

$\endgroup$
  • $\begingroup$ That’s clever. I didn’t think about dividing $[0,1]$ into smaller part. I’ve seen that $e^1 \cdot \cos(1) >1$ unfortunately... The only problem with this kind of proofs is that we obvisouly need some calculator to continue the proof. $\endgroup$ – Interesting problems Nov 8 '18 at 7:42
3
$\begingroup$

We can reduce to a polynomial inequality using that

  • $e^x\le 1+x+x^2 \quad 0\le x \le 1\quad $ proved here

  • $\cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2 \quad $ to be proved

therefore

$$e^x \cos (\sqrt{x^2+1})< (1+x+x^2)\left(1-\frac25 (x^2+1)\right)\stackrel{?}<0.90<1$$

which can be easily checked by derivatives and IVT.

Let define

  • $f(x)= (1+x+x^2)\left(1-\frac25 (x^2+1)\right)-\frac9{10}$
  • $f(0)<0$
  • $f(1)<0$

and

  • $g(x)=f'(x)= -\frac85x^3-\frac65x^2+\frac25 x+3$
  • $g(0)>0$
  • $g(1)<0$

and

  • $h(x)=g'(x)=-\frac{24}5x^2-\frac{12}5x+\frac25$
  • $h(0)>0$
  • $h(1)<0$
  • $h'(x)=-\frac{48}5x-\frac{12}5<0$

then

  • $h(x)$ is strictly decreasing and has exactly one root on that interval
  • $g(x)$ has a local maximum at that point and exactly one root on that interval that is $x_0 \approx 0.622$
  • $f(x_0)<0$ is a maximum and therefore $f(x)$ is always negative on the interval that is

$$f(x)= (1+x+x^2)\left(1-\frac25 (x^2+1)\right)-\frac9{10}<0 \\\implies e^x \cos (\sqrt{x^2+1})< (1+x+x^2)\left(1-\frac25 (x^2+1)\right)<\frac9{10}<1$$


To prove $\cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2 \quad$ let consider

  • $f(x)=\cos x-1+\frac25 x^2$
  • $f(1)<0$
  • $f(\sqrt 2)<0$

and

  • $g(x)=f'(x)=-\sin x+\frac45 x$
  • $g(1)<0$
  • $g(\sqrt 2)>0$
  • $g'(x)=-\cos x+\frac45>0$

therefore

  • $g(x)$ is strictly increasing and has exactly one root on that interval
  • $f(x)$ has a local minimum and is negative on that interval that is

$$f(x)=\cos x-1+\frac25 x^2< 0 \implies \cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2 $$

$\endgroup$
  • $\begingroup$ Thank you for the links about the two inequality you use. It can be helpful to have them in mind in the future :) $\endgroup$ – Interesting problems Nov 8 '18 at 7:40
  • 1
    $\begingroup$ You are welcome, I'm also expanding the answer with the full solution. $\endgroup$ – gimusi Nov 8 '18 at 7:41
  • $\begingroup$ Great explanation thank you ! $\endgroup$ – Interesting problems Nov 8 '18 at 8:21
  • $\begingroup$ That’s a simple but powerful way to prove inequalities by a chain of derivatives. You are welcome! Bye $\endgroup$ – gimusi Nov 8 '18 at 8:22
1
$\begingroup$

Not a proof, but some considerations which could be developed into a proof.

Let's change the variable to $$x=\sinh t$$

Now the function becomes:

$$f(t)=e^{\sinh t} \cos (\cosh t), \quad t \in [0, \ln (1+\sqrt{2})]$$

Note that $$\ln (1+\sqrt{2}) < \ln 2.5 < 1$$

This function is easy to differentiate:

$$f'(t)=e^{\sinh t} (\cosh t \cos (\cosh t) - \sinh t \sin (\cosh t))$$

The extrema obey the equation:

$$\tanh t \tan (\cosh t)=1$$

Remember that: $$\cosh t \geq 1 > \frac{\pi}{4}$$

Considering the ranges of both tangents, it's not hard to guess that there should only be a single solution in our interval.

To approximate the value, we can use bisection method:

$$g(t)=\tanh t \tan (\cosh t)-1$$

$$g(0) = -1 <0 \\ g(\ln (1+\sqrt{2})=\frac{\tan \sqrt{2}}{\sqrt{2}}-1 >0$$

Because for the range $a \in [0, \pi/2]$ we have $\tan a > a$.

Now we check the middle of the interval:

$$g(1/2)=\tanh \frac{1}{2} \tan \left(\cosh \frac{1}{2} \right)-1=\frac{e-1}{e+1} \tan \frac{e+1}{2 \sqrt{e}}-1=-0.0264\dots$$

We had to use calculators here, but with some care and knowledge of the first few digits of $e$ we can at least prove the the value is negative.

Which means that the extremum point $t_0 \in (1/2,1)$. However, we don't need to check the middle of this new interval, because $g(1/2)$ is much smaller than $g(0)$ and it's quite clear that $t_0$ is close to $1/2$. Moreover, we know* that the function $f(t)$ is growing on the interval $(0,1/2)$. So we can write:

$$f(t)< f(1/2) = e^{\sinh \frac{1}{2}} \cos \left( \cosh \frac{1}{2} \right)=0.722 \dots$$

This is again done with a calculator, however we can fix that by picking some number close to $1/2$ which gives us a more simple expression.

$^*$ It remains to be shown that $t_0$ is a maximum, but it can be done in principle by checking a few suitable values of $f(t)$ around it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.