/sci/ - Science & Math

What if G(r) = min(|r - sqrt n|) for integer n? Then obviously the limit would not exist. Why should the limit exist when the argument inside the sqrt has even fewer options than integers?

>>5649051
You can choose between any m and n, so there are more points (n,m) which give a length d(n,m):=sqrt(n^2+2*m^2), which could be close to r.
I guess one way to think of it is in terms of circles (elipses really) on a grid with points (n,m). There is an elipse for all (n,m) and if you consider bigger n,m (bigger values of d(n,m) really), the circles get closer.
Naively, I'd expect them to get closer and closer so that for bigger r, you always find a closer and closer ellipse making G(r) going against 0

I was thinking of this
http://mathworld.wolfram.com/GausssCircleProblem.html
and how the points between the r's get more and more. But since the quation says n^2+2*m^2, this is all just fuzzy talk.

>>5649067
I'm thinking about using either m or n to approximate r roughly, then use the other integer to get a more subtle approximation.

>>5649108
like <span class="math"> m=\floor{r} [/spoiler], then choose <span class="math"> n[/spoiler] small enough.

I'm currently testing something

>>5649144
that was supposed to say m=floor(r)

Pointing out that G(r) = 0 if r is an integer, so the limit is zero if it's anything.

>>5649219

Every integer r divisible by 3 makes G(r) 0.
If 2m=n, then m^2 + 2n^2 = m^2 + n^2 + n^2,
which equals m^2 + 2mn + n^2 = (m+n)^2=(3m)^2
Take the square root and |r - 3m| should be 0.

>>5649258
you know that every integer works right?

>>5649262
Oh wait duh, I was forgetting about the trivial case..

Another idea I have is to pull m out and use

<span class="math">\sqrt{1 + x} = \sum_{n=0}^\infty \left(\frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)}\right)\ x^n = 1 + \textstyle \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 + ...[/spoiler]

<span class="math">x:=2(\frac{n}{m})^2[/spoiler]

<span class="math">r \ge m \ge 0[/spoiler]

might be that for larger n, this series is surjective on some relevant codomain

>>5649051
Since >>5649067 doesn't seem to understand your question ("You can choose between any m and n, so there are more points (n,m)" => no), here's my answer:

You're wrong. The limit does exist. Take floor(n)=r^2. When r goes to infinity, using the Taylor development <span class="math">\sqrt{r^2-a}= r-\frac{a}{2r}\dots[/spoiler], we have <span class="math">0 \leq r-\sqrt{ \lfloor r^2 \rfloor < r - \sqrt{ r^2-1} = r - r + o(1) = o(1)[/spoiler], therefore <span class="math">r-\sqrt{ \lfloor r^2[/spoiler] goes to 0 when r goes to infinity.

>>5649351
Fixing LaTeX:

we have <span class="math">0 \leq r-\sqrt{ \lfloor r^2 \rfloor} < r - \sqrt{ r^2-1} = r - r + o(1) = o(1)[/spoiler], therefore <span class="math">r-\sqrt{ \lfloor r^2 \rfloor}[/spoiler] goes to 0 when r goes to infinity.

>>5649005

>e^i*theta = sin(theta) + i*cos(theta)

Sin is the imaginary component though...

>>5649416
no OP, but i originally typed that up in a hurry in word for a thread and made that mistake. the fixed one is also flouting around on sci.

>>5649698
You assume that <span class="math">\lfloor \sqrt{x}\rfloor[/spoiler] "approaches" <span class="math">\sqrt{x}[/spoiler]. I'm not sure what your definition of "approaches" is, but what you actually want to use is that the difference between the two goes to 0, which isn't true (it covers the whole [0,1) interval regardless of how high x is).

On an unrelated topic, you should use \lim and \min instead of lim and min in LaTeX, it looks much nicer (especially in \displaymode).

>>5649746
The last limit still goes to zero though.

>>5649824
I'm pretty sure that it does, indeed, but I think proving it requires careful bounding of <span class="math">\lfloor \sqrt{d\lfloor r\rfloor} \rfloor[/spoiler], for instance by bounding it between <span class="math">\sqrt{d\lfloor r\rfloor} \pm 1[/spoiler], then writing that as <span class="math">\sqrt{d}\left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right)[/spoiler]. The whole expression becomes bounded by:
<div class="math">d- \frac{\left( \sqrt{d}\left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right) \right)^2}{\lfloor r\rfloor} = d\left( 1 - \frac{ \left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right)^2 }{\lfloor r\rfloor} \right) = d\left( 1 - \frac{ \lfloor r\rfloor +\frac{1}{d} \pm \frac{2}{\sqrt{d}} }{\lfloor r\rfloor} \right) = \left( \frac{ -1 \pm 2\sqrt{d} }{\lfloor r\rfloor} \right) \xrightarrow{ r\to\infty} 0</div>

>>5649824
I'm pretty sure that it does, indeed, but I think proving it requires careful bounding of <span class="math">\lfloor \sqrt{d\lfloor r\rfloor} \rfloor[/spoiler], for instance by bounding it between <span class="math">\sqrt{d\lfloor r\rfloor} \pm 1[/spoiler], then writing that as <span class="math">\sqrt{d}\left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right)[/spoiler]. The whole expression becomes bounded by:
<div class="math">d- \frac{\left( \sqrt{d}\left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right) \right)^2}{\lfloor r\rfloor} = d\left( 1 - \frac{ \left( \sqrt{\lfloor r\rfloor} \pm \frac{1}{\sqrt{d}} \right)^2 }{\lfloor r\rfloor} \right) = d\left( 1 - \frac{ \lfloor r\rfloor +\frac{1}{d} \pm \frac{2}{\sqrt{d}} }{\lfloor r\rfloor} \right) = \left( \frac{ -1 \pm 2\sqrt{d} }{\lfloor r\rfloor} \right) \to 0</div>

Bump.

>>5649698
You know you could have set n to $\lfloor\sqrt{d\lfloorr\rfloor}\rfloor$ from the very beginning before the inequality and the limit.

>>5649966
I think >>5649698 with the addition of >>5649932 solve it.

>>5649698
You know you could have set n to <span class="math">$\lfloor\sqrt{d\lfloorr\rfloor}\rfloor$[/spoiler] in the very beginning before the inequality and the limit.

>>5649698
You know you could have set n to <span class="math">\lfloor \sqrt{d\lfloor r\rfloor} \rfloor[/spoiler] in the very beginning before the inequality and the limit.

sage

>>5650557
AAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHH stop using "lim" instead of "\lim"!

<span class="math">lim=l\times i\times m[/spoiler].
<span class="math">\lim[/spoiler] is a mathematical operator.

NERD OCD RAAAAAAAAGE!

>>5649067
It's not completely fuzzy talk. All you're doing in this problem is to consider circles centered at the origin and passing through points of <span class="math">\mathbb{Z} \times \frac{1}{\sqrt{2}}\mathbb{Z}[/spoiler], and to prove that the distance between successive radii become smaller (appropriately understood) as the radii increases to infinity.

>>5650669
I like that.

>>5650669
How does that happen?

>>5650704
I'd like to explain but I don't understand the question. Are you asking why the radii become closer and closer to each other as they grow, or are you asking why it is equivalent to prove that or directly OP's question?

>>5650623
HAPPY?