/sci/ - Science & Math

[Anonymous No.8274750]

Well /sci/, what is it?

[Anonymous No.8274757]

A bait thread.

[Anonymous No.8274760]

Why are people shocked that it equals 1?

[Anonymous No.8274766]

>>8274750
3 * 1/3

[Anonymous No.8274770]

Because it can be treated like an infinite geometric series where a1/1-r =1. Where r is the common ratio.

[Anonymous No.8274776]

[math] \displaystyle
1 = \frac {3}{3} = 3 \cdot \frac {1}{3} = 3 \cdot 0.333... = 0.999...
[/math]

[Anonymous No.8274785]

>>8274776

[Anonymous No.8274820]

>>8274760
Because it doesn't look like what they think "1" is. They picture 1 as something that has concrete, almost physical bounds and that can't be "infinite." I also don't think they really understand what geometric series are.

[Anonymous No.8274825]

>>8274750
It's George lucas way of writing 1

[Anonymous No.8274828]

>>8274760
Because it doesn't.

[Anonymous No.8275000]

>>8274750

x=0.999999……
->1000x-x =999.99999…...-0.999999……=999
->1000x-x =999x

999x=999
x=1

[Anonymous No.8275010]

It does not equal 1, it tends to 1. Is this highschool all over again ?

[Anonymous No.8275016]

>>8274750
[eqn]0.\overline{9} = \sum_{k=1}^{\infty}(\frac{9}{10})^k= \lim_{n\to \infty}(1-(\frac{1}{10})^n) =1[/eqn]

[Anonymous No.8275020]

>>8275016
fix those brackets with \left( ... \right) pls

[Anonymous No.8275036]

>>8275020
[eqn]\left\text{You}\right[/eqn]

[Anonymous No.8275058]

You don't need to separate out the terms where k=l.
Here's how to solve.

First square the absolute value term as the other anon suggested. The square of the aboslute value is monotonically increasing preserving order, so [math]\alpha_n[/math] is also just squared.
[math]\displaystyle \alpha_n^2 = \underset{(\epsilon_1,\dots,\epsilon_n) \in \{-1,1\}^n }{\inf} \underset{ \theta \in [0,2 \pi] }{\sup} \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2 [/math]

Then some algebra. Don't know why I bothered to write all this out.
[math]\displaystyle \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2[/math]
[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k e^{i k \theta} \right) \left( \sum_{k=1}^n \epsilon_k e^{-i k \theta} \right)[/math]
[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) + i \sum_{k=1}^n \epsilon_k \sin(k \theta) \right) \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) - i \sum_{k=1}^n \epsilon_k \sin(k \theta) \right) [/math]
[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) \right)^2 + \left( \sum_{k=1}^n \epsilon_k \sin(k \theta) \right)^2 [/math]
[math]\displaystyle \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \cos(k \theta) \cos(l \theta) + \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \sin(k \theta) \sin(l \theta) [/math]
[math]\displaystyle \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \cos((k-l) \theta) [/math]

Now we can establish an inequality by noting the supremum is greater than the average.
The average of [math]\cos(n \theta)[/math] over [math]\theta \in [0,2 \pi][/math] is [math]\mathrm{sinc}(2 \pi n)[/math] so all but the [math]l=k[/math] term of inner sum vanish. All the epsilons are just -1 or 1 so
[math]\displaystyle \underset{ \theta \in [0,2 \pi] }{\sup} \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2 >= \sum_{k=1}^n \epsilon_k^2 = n[/math]
The infimum of a finite number of a constant is obviously just the constant so [math]\alpha_n >= \sqrt{n}[/math]

[Anonymous No.8275115]

>>8275010
No. It's equal to 1. The symbol 0.999... *represents* the limit of the series S_n = 9/10 + ... + (9/10)^n, which is 1.

[Anonymous No.8275127]

>>8275036
[math] \left ( 4U \right ) [/math]

[Anonymous No.8275191]

>>8274785
is it wrong though?!

[Anonymous No.8275256]

>>8275191
only if you assume 1/3 isn't 0.333...

[Anonymous No.8275266]

>>8275256
is it?

[Anonymous No.8275332]