Timeline for Sharpenings of Liouville's inequality
Current License: CC BY-SA 3.0
9 events
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| May 31, 2012 at 9:25 | comment | added | Jyrki Lahtonen | Isn't the power of X supposed to be in the denominator in Liouville bound (or have a negative exponent)? Also isn't the logic sorta flowing backwards here? The Liouville bound in a way is a result of the fact that the norm of an algebraic integer is an integer. Say, if $|m−n\sqrt2|$ is small, say much smaller than 1, then $|m+n\sqrt2|$ is approximately $2|n|\sqrt2$. As the norm $(m−n\sqrt2)(m+n\sqrt2)$ has absolute value $\ge1$, then we get the Liouville bound $$|m−n\sqrt2|>\frac{C}n.$$ Probably I misunderstood something about what you want to show | |
| May 31, 2012 at 7:17 | comment | added | Gerry Myerson | If $\gamma\gt1$, then $|x-\alpha y|\gt cX^{n-\gamma}$ is a weaker hypothesis than $|x-\alpha y|\gt cX^{n-1}$, while $|N(x-\alpha y|\lt CX^{n-\gamma}|x-\alpha y|$ is a stronger conclusion than $|N(x-\alpha y|\lt CX^{n-1}|x-\alpha y|$. Also, your comment is unreadable for TeXnical reasons. | |
| May 30, 2012 at 22:44 | comment | added | Kale | Gerry Myserson: I'm not sure what is stronger/weaker here. I would like to use a stronger Liouiville's inequality to imply an upper bound of a certain form on the norm of the algebraic number $x-\alpha*y$ (in absolute value). Basically I want $\prod (x-\alpha_i*y) (where i ranges over all conjugates of $\alpha$ other than itself to be $ \leq C*X^(\gamma=1)$ given that the stronger Liouville inequality $|x-\alpha*y| \leq c *X^{\gamma-1}$ does hold. | |
| May 30, 2012 at 22:37 | comment | added | Gerry Myerson | So, you're asking whether a weaker hypothesis can imply a stronger conclusion? | |
| May 30, 2012 at 22:29 | history | edited | Kale | CC BY-SA 3.0 |
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| May 30, 2012 at 22:23 | comment | added | Kale | To make my questions a bit more clear: In the case of $\gamma=1$, the inequality follows immediately, I would like a proof that works for any $\gamma$, given the lower bound on the absolute value of $x-\alpha*y|$ I think there should be some way to use a lower bound on the absolute value of an algebraic number and transform it into an upper bound on the norm as is the case for $\gamma=1$. | |
| May 30, 2012 at 22:14 | history | edited | Kale | CC BY-SA 3.0 |
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| May 30, 2012 at 22:14 | history | edited | Charles Matthews | CC BY-SA 3.0 |
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| May 30, 2012 at 22:12 | history | asked | Kale | CC BY-SA 3.0 |