Mathematics

Howdy, I have a minor question. Let $\alpha \in [0,1]$. $\forall n, \vert \alpha - \frac{p}{q} \vert <\frac{1}{q^n}$ has $\alpha$ many solutions. How is this the set of all Liouville numbers in [0,1]?

Peter Tamaroff

12:29 AM

@LinearMan $\alpha$ many? What if $\alpha = 1/\sqrt 3$?

@robjohn $$s_n=\sum_{k=1}^n \lambda_{n,k} \chi_{\lambda(n,k)}$$

Where $\lambda_{n,k}=\inf \{f(x):x\in \lambda(n,k)\}$

and $\lambda(n,k)=\{x\in[a,b]:t_{k-1}<f(x)<t_k\}$

and $t_k=m_f+k \Delta_n $

And $\Delta_n=(M_f-m_f)/n$, with $m_f=\inf f$ ,$M_f=\sup f$ over $[a,b]$

Lebsgue, stahp!

I think $\Lambda_{n,k}=\sup \{\cdots\}$ would have worked too.

Or any inbetween values

a Dangerous Idea

0.75 Epic

Peter Tamaroff

Now, I'm thinking that $f(x)-s(x)$ can be at most $$\max_{1\leq k \leq n}\{\lambda_{n,k}-\lambda_{n,k-1}\}$$

and I can make that as small as I please, right?

@robjohn

Peter Tamaroff

12:55 AM

rooooooooooooooooooooooooob

Peter Tamaroff

1:05 AM

$${s_n} = \sum\limits_{k = 1}^n {{\lambda _{n,k}}} {{\bf{1}}_{\lambda \left( {n,k} \right)}}$$

@robjohn Am I left or right?

@BrianMScott

Peter Tamaroff

1:33 AM

@amWhy

amWhy

@PeterTamaroff hey, there! I just "saw" you (and your post)!

Peter Tamaroff

@amWhy Hahaha yes. It is a funny example.

@amWhy COuld you look up here what I wrote?

amWhy

@PeterTamaroff here?

Peter Tamaroff

@amWhy Yes, more or less.

amWhy

Yes, I meant the whole "chunk"...I'll take a look, but first, will upvote your answer!

Peter Tamaroff

1:36 AM

@amWhy Oh, thanks for the former! (I don't really mind about the later =P)

amWhy

@PeterTamaroff What you wrote looks good!

Peter Tamaroff

@amWhy GAWD, it does get messy though, right?

amWhy

@PeterTamaroff Yes, it gets messy. I'm not very quick with reading unrendered formatting. I can write it fairly quickly, but I wish mathjax were enabled in chat!

Peter Tamaroff

@amWhy We have it!!!

50

A: Should chat have TeX support?

This bookmark processes the current page with MathJax. It has been modified from the bookmark on this page to handle $$...$$ and to handle \\[...\\] properly, to include AMS additions, and to update automatically. The COPY TO CLIPBOARD link on pastebin.com should copy the bookmark to your clipbo...

amWhy

@PeterTamaroff Ahhhh! Thanks for pointing that out!

Peter Tamaroff

1:52 AM

My work today! =)

leo

2:19 AM

@PeterTamaroff nice!

Peter Tamaroff

@leo Ei! Que bueno que estas por aca

robjohn

@PeterTamaroff what is $1_{\lambda(n,k)}$?

Peter Tamaroff

@robjohn indicator function?

leo

@PeterTamaroff ya no vengo muy seguido :-P

Peter Tamaroff

@robjohn There was an overloa d of greek letters

robjohn

2:21 AM

@PeterTamaroff yes, of what?

Peter Tamaroff

@robjohn $\lambda(n,k)=\{x \in [a,b]:t_{k-1}<f(x)\leq t_k\}$

robjohn

@PeterTamaroff then $\lambda_{n,k}$ is different from $\lambda(n,k)$

Peter Tamaroff

@robjohn yep

Bad choice of notation

robjohn

that is confusing

Peter Tamaroff

$\lambda_{n,k}$ is the infimum of $f$ over $\lambda(n,k)$

robjohn

2:23 AM

I'm off to pick up dinner... bbl

Peter Tamaroff

@robjohn I made a mess unnecessarily

I'll try and unwind all this tomorrow

I get the idea, now I have tco write it down

leo

3:12 AM

On the other hand, physicists like to say physics is to math as sex is to masturbation.

2

KaliMa

anyone know wolfram?

peoplepower

conocer? no.

leo

@KaliMa why you ask?

2

don't know that there were Gauss Facts

KaliMa

3:28 AM

@leo I am trying to add a condition in my inequality that says the variables involved must be integers

like if i write a<=X, I want to make it clear that a is integral

leo

@KaliMa can you provide a link so we can see what are you requesting to WolframAlpha

or just paste here the code

peoplepower

Obviously, an ad hoc way is to add the condition $a=[a]$.

But I do not use wolfram alpha, so I would not know whether even that is feasible.

KaliMa

-X<=a+b<=X, -X<=-ab<=X, X>0, a,b,X integers

leo

@KaliMa Try Reduce[-X<=a+b<=X&&-X<=-ab<=X&&X>0,{a,b,X},Integers]

KaliMa

trying now

hm i am not sure

one of the resulting solutions is 0<a<1

which shouldn't be considered since a is an integer

leo

3:37 AM

@KaliMa are you trying in WolframAlpha or Mathematica?

KaliMa

i have both

should i be doing this in mathematica?

and if so what should i be typing in?

leo

@KaliMa In Mathematica it should work

KaliMa

i just tried the above line and it gave me some weird results

leo

@KaliMa Like what?

KaliMa

C[1] | C[2] | C[3] | C[4] | C[5] | C[6] | C[7] | C[8] | C[9] | C[10] |
C[11] | C[12] | C[13] | C[14] | C[15] | C[16] |
C[17]) \[Element] Integers && C[1] >= 0 && C[2] >= 0 &&
C[3] >= 0 && C[4] >= 0 && C[5] >= 0 && C[6] >= 0 && C[7] >= 0 &&
C[8] >= 0 && C[9] >= 0 && C[10] >= 0 && C[11] >= 0 && C[12] >= 0 &&
C[13] >= 0 && C[14] >= 0 && C[15] >= 0 && C[16] >= 0 &&
C[17] >= 0 && ((X ==
1 + C[1] + C[2] + C[3] + C[4] + C[5] + C[6] + C[7] + ...

etc

leo

3:42 AM

@KaliMa I see. I don't know how to interpret that

KaliMa

yeah it's completely different output from wolfram

robjohn

4:07 AM

@KaliMa looks like at least a missing paren

KaliMa

I just posted a subsection

just the sort of output it provides

leo

4:23 AM

@KaliMa can you post the full output

KaliMa

(C[1] | C[2] | C[3] | C[4] | C[5] | C[6] | C[7] | C[8] | C[9] |
C[10] | C[11] | C[12] | C[13] | C[14] | C[15] | C[16] |
C[17]) \[Element] Integers && C[1] >= 0 && C[2] >= 0 &&
C[3] >= 0 && C[4] >= 0 && C[5] >= 0 && C[6] >= 0 && C[7] >= 0 &&
C[8] >= 0 && C[9] >= 0 && C[10] >= 0 && C[11] >= 0 && C[12] >= 0 &&
C[13] >= 0 && C[14] >= 0 && C[15] >= 0 && C[16] >= 0 &&
C[17] >= 0 && ((X ==
1 + C[1] + C[2] + C[3] + C[4] + C[5] + C[6] + C[7] + C[8] +
C[9] + C[10] + C[11] + C[12] + C[13] + C[14] + C[15] &&

i've added more conditions Reduce[-X <= a + b <= X && -X <= -ab <= X && X > 0 && (a + b)^2 - 4 ab = (b - a)^2 && b >= a, {a, b, X}, Integers]

still errors out though

"(-a + b) && b >= a is not a quantified system of equations and inequalities"

leo

@KaliMa you must write equations with ==

KaliMa

ah ok

leo

specifically (a + b)^2 - 4 ab == (b - a)^2

KaliMa

yeah

it's still not working properly