/sci/ - Science & Math

>>10671742
If you're so smart you should know it's not worth the hassle

>>10619719
Sex with a 3D woman is overrated.

>>10349789
be slightly less beta than all the retards around you

>>10344657
Because I'm back faggot

Yes, should I publish it?

hi

>>9805129
do you even lift bro

>>9803563

>>8163245
Sorry I might not have been clear.
This is not a mathematical conclusion, this is a postulate:
the first universe (1) is analyzed through a discrete fourier space (2). Just like you can sample a continuous signal and analyze it, even if the signal itself is continuous.


the universe that we observe (3) is the inverse dft of what is in (2).
(3) is different from (1) because of aliasing.

>>8162965
there are three "worlds".

A base universe (1) where all the particles and waves live, continuous.
A fourier domain (2) associated to that world, discrete. This causes aliasing in the sampling.
The world we experience, where the objects are inverse discrete fourier transform of the objects in (2). The aliasing in (2) leads to uncertainty in (3).

Notice that the finer the discretization in (2), the smaller the distortion, and in other words the uncertainty. Which is exactly the same as the uncertainty principle.

>>6642261
I'd like to suggest the name "bias"

the longer the displacement is not equal to 0, and the further it gets from 0, the bigger bias is.
Likewise, if positive and negative displacement compensate each other, there is no bias. You can have positive or negative bias. And just like for a politician, the current position of the point is given by the variation in their bias. Where is my Nobel Prize ?

(idea came to me from robotics and stats courses)

I...I'm here guize

>>6166349
>implying I didn't provide real help and contribution to scientific posts
I'm roaming /sci/ without a tripcode nowadays though

hi

hi guys

>>5888161
which leads simply to <span class="math">S=2.(1 + \frac{1}{2})=3 [/spoiler] (S is a telesoping series)

>>5888139
forget u_n.

anyway the sum is also equal to
<span class="math">\displaystyle S= \sum_{p=1}^{\infty} \sum_{n=p(p-1)+1}^{p(p+1)} \frac{2^p + 2^{-p}}{2^n}[/spoiler].

modulo my mistakes, I find <span class="math"> \displaystyle S=2. \sum_{p=1}^{\infty} \frac{1}{2^{(p-1)^2}} - \frac{1}{2^{(p+1)^2}}[/spoiler].

>>5888127
sorry the end should be: <span class="math">p(p-1)<n\leq p(p+1)[/spoiler] as I replaced the left side with the first smaller integer.
now we can calculate the sum <span class="math">\displaystyle \sum_{n=1}^{\infty} \frac{2^{<n>}+2^{-<n>}}{2^n}= \sum_{n=1}^{\infty} u_n[/spoiler].

>>5888127
sorry the end should be: [/math]p(p-1)<n\leq p(p+1)[/math] as I replaced the left side with the first smaller integer.
now we can calculate the sum
<span class="math"> \displaystyle \sum_{n=1}^{infty} \frac{2^{<n>}+2^{-<n>}}{2^n}= \sum_{n=1}^{infty} u_n [/spoiler].

>>5888099
I'll show later that the square root of n is a half integer if and only if it is an integer.
Assuming that for the moment, we have <span class="math"><n>=p [/spoiler] if and only if:
<span class="math">p-1/2 < \sqrt{n} < p+1/2 [/spoiler], ie <span class="math">p^2-p+1/4 < n <p^2+p+1/4 [/spoiler].
The left side and right side can never be integers, so if n is a half integer, it is necessarily an integer anyway.
The left side can be replaced by the first bigger integer, and the right side can be replaced by the first smaller integer.
We obtain: <span class="math"> p^2-p=p(p-1) < n < p^2 + p=p(p+1) [/spoiler], which is what we wanted to show.

alright:
I want to show that <span class="math"><n>=p [/spoiler] if and only if <span class="math">p(p-1)<n \leq p(p+1)[/spoiler].

let <span class="math">x \in \mathbb{R}^n[/spoiler], <span class="math">x=(a_1,...,a_n)^T[/spoiler].

In <span class="math">mathbb{R}^n[/spoiler], all norms are equivalent, and in particular:
<span class="math">||x||_{\infty} \leq ||x||_2 \leq \sqrt{n} ||x||_{\infty}[/spoiler].
Here, <span class="math">(||x||_2)^2 = \sum_{k=1}^{n} (a_k)^2 \geq n^2 [/spoiler], so <span class="math">max(|a_1|,...,|a_n|)=||x||_{\infty} \geq \sqrt{n} \geq 2 [/spoiler].

Now I assume we can remove the absolute values by using <span class="math">||x||_1[/spoiler] somehow?

>>5861895
>>5861886
come on man!

substitute <span class="math">u=(2n+1)x, dx=du/(2n+1)[/spoiler].
<span class="math"> K_n= \int{0}^{(2n+1)\pi/2} \frac{sin(u)}{u/(2n+1)} du/(2n+1) [/spoiler] so <span class="math">K_n = \int_0^{(2n+1)\pi/2} \frac{sin(u)}{u} du [/spoiler]
so when n goes to infinity...

let <span class="math">J_n = \int_0^{\pi/2} \frac{sin((2n+1)x)}{sin(x)} dx[/spoiler]

and <span class="math">K_n = \int_0^{\pi/2} \frac{sin((2n+1)x)}{x} dx [/spoiler]

show that <span class="math">J_n - K_n \rightarrow 0 [/spoiler], that <span class="math">\forall n, J_n = \frac{\pi}{2}[/spoiler], and that <span class="math">K_n \rightarrow \int_0^{+\infty} \frac{sin(x)}{x}dx[/spoiler].