Prove that $e^{\sin x}-\sin e^x\neq0$ $\forall x\in [-1, \pi]$, an equation in the form $f(g(x))=g(f(x))$

Question

As the title say, we need to prove that:

$e^{\sin x}-\sin e^x\neq0$ $\forall x\in [-1, \pi]$

Now, I am not sure if this can lead somewhere but we can notice that $e^{\sin x}-\sin e^x=0$ is equivalent to $f(g(x))=g(f(x))$ for $f(x)=e^x$ and $g(x)=\sin x$.

Is there a general approach for equations of this form?

Well, I am more than a little embarrassed as I included this question in a quiz and I am having difficulty solving it myself. (Not a quiz addressed to students but part of a little project..).

Anyway, this is a transcendental equation and seeking a closed form solution that would inform us also about the presence of roots in $[-1, \pi]$ seems unlikely to lead anywhere.

But we can make some observations.

We know that $-1\le\sin e^x\le1$ holds so we can show that $e^{\sin x}\ge1$ $\forall x\in [0, \pi]$.

Indeed, let $r(x)=e^{\sin x}$.

We have $r'(x)=(\cos x)e^{\sin x}\Rightarrow r'(x_0)=0\Rightarrow x_0=\pi/2$ the only root in $ [0, \pi]$

So, we verify that $\forall x\in [0,\pi/2], r'(x)\ge0\Rightarrow f$ is increasing.

and similarly

$\forall x\in (\pi/2,\pi], r'(x)\le0\Rightarrow f$ is decreasing.

Since $r(0)=1, r(\pi/2)=e, r(\pi)=1$ in both cases it holds that $1\le r(x)\le e$.

Now we know that $r(x)=1$ only for $x=0$ or $x=\pi$

but we check (numerically) that, say $t(x)=\sin e^x$ for these two values $t(0)=\sin1\neq1$ and $t(\pi)=\sin e^{\pi}\neq1$.

So we have showed that $e^{\sin x}-\sin e^x\neq0$ $\forall x\in [0, \pi]$

But the case $[-1, \pi]$ is not suited for a similar treatment so I am having difficulty proceeding. Perhaps I need to resort to Lambert W? (a technic that sadly is not very familiar to me.)

Any suggestions welcome..I also include a graph of $h(x)=e^{\sin x}-\sin e^x$

Answer 1

If $x<0$, then $x<\sin(x)$ and if $x>0$, then $x>\sin(x)$.

Let $x\in [-1,0)$. Then, $e^{\sin(x)} - \sin(e^x)>e^x - \sin(e^x) > 0$. (The last inequality holds because $e^x>0$)

Also, note that $e^x>1+x$ for $x>0$.

Let $x\in (0,\pi)$.

Then, $e^{\sin(x)} - \sin(e^x)> 1+ \sin x - \sin(e^x)> \sin x>0$.

Now, it remains to just substitute $x=0$ and $x=\pi$ to show that the equality does not hold.