Skip to main content
added 72 characters in body
Source Link
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$$|A+A|+|A\cdot A|\gg |A|^{4/3-o(1)}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-pageshort proof for $4/3$$4/3-o(1)$ by Solymosi.

The conjecture of Erdős is that the exponent approaches 2.

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

The conjecture of Erdős is that the exponent approaches 2.

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3-o(1)}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the short proof for $4/3-o(1)$ by Solymosi.

The conjecture of Erdős is that the exponent approaches 2.

Post Made Community Wiki by S. Carnahan
added 62 characters in body
Source Link
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

The conjecture of Erdős is that the exponent approaches 2.

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

The conjecture of Erdős is that the exponent approaches 2.

Recently Konyagin and Shkredov improved (arxiv:1602.03473)Konyagin and Shkredov improved the exponent of $4/3$ exponent in the sum-products estimate in $|A+A|+|A\cdot A|\gg |A|^{4/3}, A\subset \mathbb{R}$$\mathbb{R}$, bynamely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

Recently Konyagin and Shkredov improved (arxiv:1602.03473) the $4/3$ exponent in the sum-products estimate $|A+A|+|A\cdot A|\gg |A|^{4/3}, A\subset \mathbb{R}$, by $4/3+5/9813$. This appears to be much harder than one-page proof for $4/3$.

Recently Konyagin and Shkredov improved the exponent of $4/3$ in the sum-products estimate in $\mathbb{R}$, namely that $|A+A|+|A\cdot A|\gg |A|^{4/3}$ for every $A\subset \mathbb{R}$, to $4/3+5/9813$. This appears to be much harder than the one-page proof for $4/3$.

Source Link
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459
Loading