Prove that there exists a function $f$ such that
$$\forall x \in \mathbb{R}: (f(x))^5 + 5 f(x) + \sin(f(x)) = x$$
I've been trying to use inverse theorem or using the derivative and can't get anything out of it.
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8$\begingroup$ Consider the function $g(t)=t^5+5t+\sin t$, then $g'(t)=5t^4+5+\cos t\ge 0+5-1=4>0$ for all $t\in\mathbb{R}$, so $g$ is strictly increasing and therefore 1-1 on $\mathbb{R}$. $\endgroup$– Alexander BursteinCommented Apr 10, 2018 at 5:03
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$\begingroup$ I was able to get there myself but I might be missing some knowledge for understanding the implication. What if f(t) /= t, how can you be sure g'(t) > 0 for all t. If a function is one to one then it's enough of an argument to prove that such function exists to describe said above equality? $\endgroup$– FatFeynmanCommented Apr 10, 2018 at 15:32
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1 Answer
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The most elegant and quickest way is that based on the comment by @Alexander Burstein.
Here are two other ways based on your ideas:
Using implicit function theorem (which boils down to above mentioned comment): $$F(x,y) = y^5+5y+\sin y - x = 0 $$ satisfies the assumtions of the implicit function theorem: $$\partial_yF(x,y) = 5y^4+5+\cos y > 0$$ For any $(x_0,y_0) \in \mathbb{R}$ with $F(x_0,y_0)$ the equation is invertible, which means it defines a function $y=f(x)$ in an open neighbourhood of $(x_0,y_0)$.
Using derivatives: Just assuming $y$ is differentiable, we get $$5y^4y' + 5y'+\cos{(y)}y' - 1 = 0 \Rightarrow y' =\frac{1}{5y^4+5+\cos y}$$ For any $(x_0,y_0) \in \mathbb{R}$ with $F(x_0,y_0)$ we have an initial value problem $$y'(x)=g(x,y(x))= \frac{1}{5y^4+5+\cos y}, \; y(x_0) = y_0$$ Function $g$ satisfies the assumptions of the Picard-Lindelöf theorem, as $g$ is uniformly Lipschitz continuous in $y$, because $g'$ is bounded. So, we get the existence of the function looked for.
In both cases you get the unique function $f(x)$ locally. You may reason for yourself why these local functions represent "snippets" of a global $f(x)$.
answered Apr 10, 2018 at 7:20