5
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ Hint: What values does $P(x)$ give as $x$ runs through the real numbers? What values does $e^{P(x)}$ give? $\endgroup$ – mickep Dec 31 '14 at 21:19
8
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ I understand that is has least one real root, but how do you conclude that exist $x_0$ such that $e^P(x_0)=1$? $\endgroup$ – JaVaPG Dec 31 '14 at 21:25
  • $\begingroup$ @JaVaPG $x_0$ is a real root of $P(x)$. This means, by definition, $P(x_0)=0$. Since $e^0=1$, we have $e^{P(x_0)}=1$ $\endgroup$ – Zubin Mukerjee Dec 31 '14 at 21:26
  • $\begingroup$ 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. $\endgroup$ – JaVaPG Dec 31 '14 at 21:52
  • $\begingroup$ @JaVaPG I think you have the right idea, but you might want to specify that $n <x_0$. $\endgroup$ – Zubin Mukerjee Dec 31 '14 at 21:58
  • $\begingroup$ 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. $\endgroup$ – JaVaPG Dec 31 '14 at 22:09
0
$\begingroup$

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

sinx oscillates ,and equation has solution

$\endgroup$
2
  • 3
    $\begingroup$ No, it does not hold that $\sin x\to 0^+$. $\endgroup$ – mickep Dec 31 '14 at 21:31
  • $\begingroup$ $\sin x$ oscillates ... $\endgroup$ – MathMajor Dec 31 '14 at 21:38

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.