/sci/ - Science & Math » Thread #11153000

132KiB, 823x903, 25fcbc380085cff11327d3285f29f6e4.jpg

View Same Google iqdb SauceNAO

Anonymous Sat 16 Nov 23:34:48 2019 No.11153000 View Reply Original Report

Quoted By: >>11153125 >>11154044 >>11154085 >>11154089 >>11154363

Alright this is what I'm talking about why are math people so bad at explaining things. So divergent means that it doesn't have a set value right? So I just read that 1 + 1/2 + 1/3 ... etc is divergent, but since each number is getting smaller as you go wouldn't that make it convergent instead? I don't get this crap.

Anonymous

Anonymous Sat 16 Nov 2019 23:36:35 No.11153012 Report

Quoted By: >>11154089

And why is all the reciprocals of the squares equal to pi square over 6? Does that mean that all the squares equal 6 over pi squared? What ???

Anonymous

Anonymous Sat 16 Nov 2019 23:45:53 No.11153038 Report

Quoted By:

For fuck sake somebody answer already, I could learn about pokemon easier because pokemon autists actually can explain themselves but mathematicians try to make everything impossible on purpose because they don't realize that other people are able to comprehend their shit if its actually explained correctly.

Anonymous

Anonymous Sat 16 Nov 2019 23:54:49 No.11153059 Report

Quoted By:

I swear I'm going to have to rewrite all this math correctly.

Anonymous

Anonymous Sun 17 Nov 2019 00:03:34 No.11153077 Report

Quoted By:

draw a graph of the series and you'll see that it gets longer faster than it gets shallower

Anonymous

Anonymous Sun 17 Nov 2019 00:19:24 No.11153125 Report

Quoted By:

>>11153000
>but since each number is getting smaller as you go wouldn't that make it convergent instead?
no
I could name an arbitrarily large number and that series would surpass that value at some point.
That's what it means for it to be divergent.

I'm not sure what you're asking.. Do you want a proof for why it's divergent? If so:
https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Comparison_test

Anonymous

Anonymous Sun 17 Nov 2019 06:50:52 No.11154044 Report

Quoted By:

>>11153000
It's very simple. It is always getting bigger, right? Can you find a number it is never greater than?

Anonymous

View Same Google iqdb SauceNAO bored.jpg, 24KiB, 310x303

Anonymous Sun 17 Nov 2019 06:56:29 No.11154056 Report

Quoted By:

$\sum\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ ...\ =S\leq1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\ ...\ =\infty$
therefore the harmonic series has no bound. it does not converge to a finite number.

Anonymous

View Same Google iqdb SauceNAO bored.jpg, 24KiB, 310x303

Anonymous Sun 17 Nov 2019 06:59:52 No.11154064 Report

Quoted By: >>11154092

$\sum\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ ...\ \leq1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\ ...\ =1+\frac{1}{2}+\frac{1}{2}+\ ...\ =\infty$
therefore the harmonic series has no bound. it does not converge to a finite number.

Anonymous

Anonymous Sun 17 Nov 2019 07:14:58 No.11154085 Report

Quoted By:

>>11153000
$\displaystyle \begin{align*} \sum_{n=1}^{ \infty} \frac{1}{n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\ &> 1 + \frac{1}{2} + \left (\frac{1}{4} + \frac{1}{4} \right ) + \left ( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right ) + \cdots \\ &= 1 + \lim_{n \rightarrow \infty} n \cdot \frac{1}{2} = \infty \end{align*} $

Anonymous

View Same Google iqdb SauceNAO e3ty1pzq7e721.jpg, 222KiB, 1080x1313

Anonymous Sun 17 Nov 2019 07:16:12 No.11154089 Report

Quoted By:

>>11153000
The sum grows logarithmically as u add terms. Log doesnt converge. Just grows more slow as input grows
>>11153012
Fuck no division doesnt distribute like that you nut

Anonymous

Anonymous Sun 17 Nov 2019 07:17:05 No.11154092 Report

Quoted By: >>11154122 >>11154124

>>11154064
"<"
nope

Anonymous

Anonymous Sun 17 Nov 2019 07:28:16 No.11154122 Report

Quoted By:

>>11154092
why

Anonymous

Anonymous Sun 17 Nov 2019 07:29:26 No.11154124 Report

Quoted By:

>>11154092
>*

Anonymous

Anonymous Sun 17 Nov 2019 10:21:03 No.11154363 Report

Quoted By: >>11154372

>>11153000
Any series of the form 1/k^q diverges if q <= 1 and converges for q > 1. A simple integral test will prove this. Let f(x)=1/k^q. We then evaluate the integral from 1 to infinity of f(x). With a bit of manipulation you will see that the only case where this is convergent is where q > 1.

Anonymous

Anonymous Sun 17 Nov 2019 10:24:58 No.11154372 Report

Quoted By:

>>11154363
this right here is the basis for infinitesimal calculus, which I think is what OP is probably learning right now. I learned about divergent and convergent sets in calc.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches