# $P(x)=x^3+ax^2+bx+c$, Proof $e^{P(x)}=\sin x$ has a solution.

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!

• Hint: What values does $P(x)$ give as $x$ runs through the real numbers? What values does $e^{P(x)}$ give? – mickep Dec 31 '14 at 21:19

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?

• I understand that is has least one real root, but how do you conclude that exist $x_0$ such that $e^P(x_0)=1$? – JaVaPG Dec 31 '14 at 21:25
• @JaVaPG $x_0$ is a real root of $P(x)$. This means, by definition, $P(x_0)=0$. Since $e^0=1$, we have $e^{P(x_0)}=1$ – Zubin Mukerjee Dec 31 '14 at 21:26
• I thought of something i'll be glad to receive feedback, for all $\epsilon>0$ exist $N>0$ so for all $x<-N$ implies $|e^{P(x)}|<\epsilon$. Let choose natural $n$ such $n<-N$ for $\implies sin(0)=0$ And for $sin(n\pi/2)=1$, we also know that exist $x_0 \implies e^{P(x_0)}=1$ and $\lim_{x\to-\infty} e^{P(x)}=0$ Then we can use the IVT. – JaVaPG Dec 31 '14 at 21:52
• @JaVaPG I think you have the right idea, but you might want to specify that $n <x_0$. – Zubin Mukerjee Dec 31 '14 at 21:58
• Thank you for your help, I understand your proof, its just hard to write it on a well construct form If I choose $n<x_0 \implies e^P(x_0) \geq \sin(x_0)$, Can you explain that that establish $e^{P(x)}=\sin x$, Like how to write this part of the prove. – JaVaPG Dec 31 '14 at 22:09

$$x \rightarrow -\infty \rightarrow P(x)=x^3+ax^2+bx+c\rightarrow -\infty \\e^{p(x)} \rightarrow 0^+ \\ \sin x \rightarrow 0^+$$ so $$e^{p(x)} =\sin x$$ flag

sinx oscillates ,and equation has solution

• No, it does not hold that $\sin x\to 0^+$. – mickep Dec 31 '14 at 21:31
• $\sin x$ oscillates ... – MathMajor Dec 31 '14 at 21:38