Timeline for Linear combinations of roots of unity
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8, 2021 at 15:15 | vote | accept | Beni Bogosel | ||
| Jun 7, 2021 at 17:03 | answer | added | Emil Jeřábek | timeline score: 7 | |
| Jun 5, 2021 at 15:20 | comment | added | Beni Bogosel | Thank you very much for these computations. I will check them. | |
| Jun 5, 2021 at 11:36 | comment | added | Emil Jeřábek | I did the calculation after all. It turns out to reduce to just 4 cases. First, by symmtery considerations, max Re z equals either max |z| if $n\not\equiv0\pmod4$, or max Im z if $n\equiv0\pmod4$. The rest only depend on the parity of $n$. If $n$ is even, then max |z| is $2\csc\frac\pi n$ and max Im z is $2\cot\frac\pi n$. If $n$ is odd, then max |z| is $\csc\frac\pi{2n}$ and max Im z is $\csc\frac\pi n+\cot\frac\pi n$. This is all valid for $n\ge2$. I don’t have the time to write it up now, but may do so some time next week if needed. | |
| Jun 4, 2021 at 20:13 | comment | added | François Brunault | Another way of putting Emil Jeřábek's argument is that by rotational symmetry, the problems with $\mathrm{Re}(z)$, $\mathrm{Im}(z)$ and $|z|$ are all (roughly) equivalent. | |
| Jun 4, 2021 at 16:40 | comment | added | Emil Jeřábek | Let me put it this way. The arguments above serve to show that in all cases, the optimal bound is attained when all the $x_i$ are $\pm1$, with $+1$ in one half-plane and $-1$ in the other half. For Re z, take the right half-plane; for Im z, take the upper half-plane; for |z|, it makes no difference what half-plane you take, as the relevant sums will have the same absolute value. Then it is a matter of summing a geometric series and computing its Re/Im/|z|. The latter is kind of tedious as there will be annoying minor differences depending on $n\mod4$, hence I’m not volunteering to do it. | |
| Jun 4, 2021 at 15:49 | comment | added | Emil Jeřábek | No, the $x_i$ are real ($\pm1$). The bounds for |z|, Re z, and Im z will not be exactly the same for very small $n$, but they will all be roughly $\frac2\pi n$ for larger $n$. | |
| Jun 4, 2021 at 15:40 | comment | added | Beni Bogosel | @EmilJeřábek: Adding the parameter alpha you modify the $\omega^i$ using some complex numbers. Therefore, the new coefficients $x_i$ would be complex, not real, no? For example if $n=3$, the best bound for the imaginary part is smallest than the one for the real part ($2\sin 2\pi/3$ vs $2$). | |
| Jun 4, 2021 at 15:32 | comment | added | Emil Jeřábek | Actually, the same bound should hold for $|z|$ as well: if $\alpha=\overline z/|z|$, then $|z|=\operatorname{Re}\sum_ix_i\alpha\omega^i$, which is for a given $\alpha$ maximized for $x_i=\operatorname{sgn}\operatorname{Re}(\alpha\omega^i)$. This leads to the same calculation as before (only with a different starting angle). | |
| Jun 4, 2021 at 15:18 | comment | added | Beni Bogosel | I am well aware that $n$ is a valid upper bound for $|z|$, but it's not the best one. I agree that there is not a big improvement over $n$, but I was wondering if the best bounds are known. As you say, for the real and imaginary parts, the computation is straightforward: a sum of cosines or sines. | |
| Jun 4, 2021 at 15:16 | history | edited | Beni Bogosel | CC BY-SA 4.0 |
added 11 characters in body
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| Jun 4, 2021 at 15:11 | comment | added | Emil Jeřábek | It’s clear that the worst case for $\def\Im{\operatorname{Im}}\Im z$ is when $x_i=\operatorname{sgn}(\Im \omega^i)$, thus $\Im z=\sum_{i<n}|\Im\omega^i|$. By a back of the envelope calculation, this is about $4/\sin(2\pi/n)\sim\frac2\pi n$. Just about the same holds for $\operatorname{Re}z$. Thus, you can’t improve much over the trivial bound given by Leo Moos. | |
| Jun 4, 2021 at 15:05 | comment | added | Leo Moos | What is wrong with $\lvert z \rvert \leq n$? | |
| Jun 4, 2021 at 15:00 | history | asked | Beni Bogosel | CC BY-SA 4.0 |