Skip to main content
MathOverflow

Return to Answer

added 2 characters in body
Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_i')^k. $$$$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_{i'})^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_i')^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_{i'})^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

deleted 13 characters in body
Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} z_i^k \overline{z_{i'}^k}. $$$$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_i')^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} z_i^k \overline{z_{i'}^k}. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_i')^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} z_i^k \overline{z_{i'}^k}. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.