Let $P(x)=x^3+ax^2+bx+c$
Proof : $e^{P(x)}=\sin x$ has a solution.
I thought about it, and still cannot find where to start.
Any ideas?, Thanks!
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Sign up to join this communityLet $P(x)=x^3+ax^2+bx+c$
Proof : $e^{P(x)}=\sin x$ has a solution.
I thought about it, and still cannot find where to start.
Any ideas?, Thanks!
Hint: $P(x)$ is a cubic, so it must have at least one real root. There exists an $x_0$ such that $e^{P(x_0)}=1$. This means $$e^{P(x_0)}\geq \sin x_0$$
The leading coefficient of $P(x)$ is $1$, so $$\lim_{x\to-\infty} P(x)=-\infty$$
This means that $$\lim_{x\to-\infty} e^{P(x)}=0$$
Now, why must there be a solution to $e^{P(x)}=\sin x$ in the interval $(-\infty,x_0]$? More strongly, why must there be infinitely many solutions in this interval?
$$ x \rightarrow -\infty \rightarrow P(x)=x^3+ax^2+bx+c\rightarrow -\infty \\e^{p(x)} \rightarrow 0^+ \\ \sin x \rightarrow 0^+ $$ so $$e^{p(x)} =\sin x $$
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sinx oscillates ,and equation has solution