Timeline for Integer multiples of a irrational dense in R/Z ?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2013 at 21:18 | comment | added | Barry Cipra | I just noticed, the question about the divergence of $\sin n$ was asked last year at math.stackexchange.com/questions/238997/… | |
| Apr 16, 2013 at 21:04 | comment | added | Barry Cipra | Another trig-identity-based approach starts with $\sin 2n = 2\sin n\cos n$ and $\cos 2n = 2\cos^2n - 1$ to argue that the limit $S$, if it existed, would have to be 0: If $S\ne0$, then $S\approx2S\cos n$ implies $\cos n$ tends to the limit $C=1/2$, which does not satisfy $C=2C^2-1$. But now $\sin(n+1) = \sin n\cos1+\cos n\sin1$ implies $\cos n$ tends to $0$ (since $\sin1\ne0$), and this contradicts the identity $\sin^2n+\cos^2n=1$. | |
| Apr 16, 2013 at 19:04 | comment | added | Douglas Zare | You don't have to prove anything like this to show that $\sin n$ diverges. $\sin (n+1) - \sin (n-1) = 2 \sin 1 \cos n$. If this is not close to $0$, then $\sin n$ can't be close to both. If it is close to $0$, then $\cos n$ must be close to $0$, which implies that $\sin n$ is close to $1$ or $-1$, so $\sin (n+1) = \sin n \cos 1 + \sin 1 \cos n$ is close to $\pm \cos 1$, hence far from $\sin n$. You can also find the minimum of $(\sin (x-1)-\sin x)^2 + (\sin x - \sin(x+1))^2$ (at $x=\pi/2 + k \pi$) and check that it is above $0$. | |
| Apr 16, 2013 at 18:22 | history | edited | Richard Rast | CC BY-SA 3.0 |
Improved question (transc. -> irrat.)
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| Apr 16, 2013 at 18:09 | vote | accept | Richard Rast | ||
| Apr 16, 2013 at 18:08 | answer | added | Tom Lovering | timeline score: 9 | |
| Apr 16, 2013 at 17:59 | history | asked | Richard Rast | CC BY-SA 3.0 |