I managed to prove until $|\cos x|$ less than or equal to $1$, but not sure how to continue, please help
Asked
Active
Viewed 57 times
-1
-
Do you mean $|p(x)-cos(x)|$? This is usually know as Theorem of Stone-Weierstraß, see here https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem – bayes2021 Nov 30 '22 at 17:10
-
This is not true, as stated. You would need a bounded domain for this to be possible. In that case, a sufficiently high degree Taylor approximation would do the trick. – Steven Gubkin Nov 30 '22 at 17:10
-
1Another typo: I assume you mean $|p(x) -\cos(x)|$? – Steven Gubkin Nov 30 '22 at 17:11
-
Thanks! Edited the typo. How do I use the Taylor's Theorem to prove? – omelette Nov 30 '22 at 17:21
-
2@omlette This isn't true as stated. There must be some restriction on $\theta$ – Dionel Jaime Nov 30 '22 at 17:29
-
How do I prove that it doesn't exist then? – omelette Nov 30 '22 at 17:29
1 Answers
0
There is no polynommial $p$ such that, for all real numbers $\theta$, $|p(\theta) - \cos\theta|<10^{-6}$.
If the polynomial $p$ is not constant, then $\lim_{\theta \to \infty} |p(\theta)| = \infty$, so in particular $|p(\theta)|> 2$ for some theta, and thus $|p(\theta) - \cos\theta| > 1$.
If $p$ is constant, say $p(\theta) = c$ for all $\theta$, then either $c\ge 0$ or $c < 0$. In the first case, there exists $\theta$ so that $\cos\theta = -1$ and in the second case there exists $\theta$ so that $\cos\theta = 1$. In both cases, $|p(\theta)-\cos\theta| \ge 1$.
GEdgar
- 111,679