Timeline for Limit of the real part of a geometric sequence
Current License: CC BY-SA 4.0
8 events
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| Jun 18, 2019 at 17:08 | comment | added | Renaud Detcherry | @Anthony Quas: No it was fine with denominator $2$ and not $2^n.$ As $z=\frac{1+i\sqrt{7}}{2}$ is an algebraic integer, $z^n$ is a $\mathbb{Z}$-linear combination of $1$ and $z$. Thus $a_n, b_n \in \mathbb{Z}.$ PoundSterling: Thanks for the answer. Judging by where I found this problem, I thought there would be an elementary solution, but I guess it can't be helped. | |
| Jun 18, 2019 at 3:11 | comment | added | fedja | Looks somewhat similar to mathoverflow.net/questions/273112/… | |
| Jun 17, 2019 at 23:20 | comment | added | Pound Sterling | These type of problems are easy consequences of the finiteness of solutions (in number fields) to the $S$-unit equation, but I don't expect more elementary arguments (for the general version of the problem). You can also use Baker's theorem to get effective bounds. | |
| Jun 17, 2019 at 18:35 | comment | added | Somos | @user64494 I would not trust Mathematica on this particular kind of problem. Also, the output gives no hint of any kind of proof of the result. | |
| Jun 17, 2019 at 16:59 | comment | added | user64494 | The command of Mathematica DiscreteMaxLimit[RealAbs[ComplexExpand[Re[((1 + I*Sqrt[7])/2)^n]]], n -> Infinity] outputs $\infty$. | |
| Jun 17, 2019 at 16:05 | comment | added | Anthony Quas | I added a power of $n$ in the denominator of $z^n$. I presume this is what you intended. | |
| Jun 17, 2019 at 16:04 | history | edited | Anthony Quas | CC BY-SA 4.0 |
added 2 characters in body
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| Jun 17, 2019 at 14:56 | history | asked | Renaud Detcherry | CC BY-SA 4.0 |