/sci/ - Science & Math » Thread #11639281

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Anonymous Tue 05 May 13:06:46 2020 No.11639281 View Reply Original Report

Quoted By: >>11639284 >>11639289 >>11639293 >>11639300 >>11639355 >>11639402 >>11639483 >>11639503 >>11639537 >>11639578 >>11640723 >>11640850 >>11641624

I've got a new debate for you /sci/

Anonymous

Anonymous Tue 05 May 2020 13:08:05 No.11639284 Report

Quoted By: >>11639548

>>11639281
>e + -e = 0 /sci/ btfo

Anonymous

Anonymous Tue 05 May 2020 13:08:45 No.11639285 Report

Quoted By: >>11639317 >>11639548

Im unfamiliar with the question mark above the equals sign symbol.

Anonymous

Anonymous Tue 05 May 2020 13:10:47 No.11639289 Report

Quoted By: >>11639388 >>11639548

>>11639281
The FTC works only for the continuous function by definition

Anonymous

Anonymous Tue 05 May 2020 13:12:49 No.11639293 Report

Quoted By: >>11639548 >>11643391

>>11639281
>integral
>literally calculates the surface
>equal positive and negative surfaces that reduce each other to nil

anon, this is disappointing bait

Anonymous

Anonymous Tue 05 May 2020 13:14:38 No.11639300 Report

Quoted By: >>11639315 >>11639548

>>11639281
Anon, do you know what even and odd functions are?

Anonymous

Anonymous Tue 05 May 2020 13:20:27 No.11639315 Report

Quoted By: >>11639347 >>11639548

>>11639300
sure. what's your point ?

Anonymous

Anonymous Tue 05 May 2020 13:21:29 No.11639317 Report

Quoted By: >>11639548

>>11639285
read it like a question asking if the equality holds or not

Anonymous

Anonymous Tue 05 May 2020 13:25:18 No.11639323 Report

Quoted By: >>11639548

what is there to debate?

Anonymous

Anonymous Tue 05 May 2020 13:39:13 No.11639347 Report

Quoted By: >>11639388 >>11639548 >>11642453

>>11639315
im not sure what your point is, int of 1/x from a to -a = 0 holds for all real values of a since its an odd function

Anonymous

Anonymous Tue 05 May 2020 13:42:09 No.11639355 Report

Quoted By: >>11639385 >>11639388 >>11639548

>>11639281
The Riemann integral is not defined, since the integrand is not a function defined on the entire integration domain.

Anonymous

Anonymous Tue 05 May 2020 13:58:00 No.11639385 Report

Quoted By: >>11639396 >>11639397 >>11639548 >>11640050

>>11639355
We can ignore that since it is discontinous on a finite number of points

Anonymous

Anonymous Tue 05 May 2020 13:59:10 No.11639388 Report

Quoted By: >>11639548

>>11639347
>int of 1/x from a to -a = 0 holds for all real values of a
is it though? see>>11639289 and >>11639355

Anonymous

Anonymous Tue 05 May 2020 14:00:51 No.11639396 Report

Quoted By: >>11639466 >>11639548

>>11639385
It is not discontinuous in any points. It is however undefined in one point.

Anonymous

Anonymous Tue 05 May 2020 14:01:35 No.11639397 Report

Quoted By: >>11639548

See>>11639385
Also since f(x) is an odd function, f(x) = - f(-x), so the total integral is 0

Anonymous

Anonymous Tue 05 May 2020 14:05:48 No.11639402 Report

Quoted By: >>11639548

>>11639281
Something something principal value

Anonymous

Anonymous Tue 05 May 2020 14:18:37 No.11639425 Report

Quoted By: >>11639548

I think the real debate here is see if we can define a Riemann Integral in a formal and mathematical way (wich I think we can, because the only discontinuity is in x=0). If the we can define it, the result is obviously 0, we can not, then the integlar simply dosent exists.

Anonymous

Anonymous Tue 05 May 2020 14:36:27 No.11639466 Report

Quoted By: >>11639475 >>11639548

>>11639396
if we were anal about the notation the integral should be split in two:
we have
integral on [-1,0) + (0,1]. And since f is an odd function both of these integral cancel out. And in regards to continuity, as I think of it now since 0 is not in its domain, the function is continuous after all on all the points in its domain. So I guess you are right.

Anonymous

Anonymous Tue 05 May 2020 14:40:28 No.11639475 Report

Quoted By: >>11639482 >>11639495 >>11639548

the answer is that the integral is undefined.
you can make sense of the integral by treating it as two improper integrals like >>11639466 did. but the result is still undefined.

Anonymous

Anonymous Tue 05 May 2020 14:41:57 No.11639482 Report

Quoted By: >>11639548

>>11639475
actually, saying the integral doesn't exist is a bit better than saying it's undefined

Anonymous

Anonymous Tue 05 May 2020 14:42:02 No.11639483 Report

Quoted By: >>11639548

>>11639281
Yes, principal value

Anonymous

Anonymous Tue 05 May 2020 14:46:03 No.11639495 Report

Quoted By: >>11639548

>>11639475
You are right, I think it goes like this:
integral (0,1] of 1/x dx =
lim l -> 0 (integral (l,1] of 1/x dx)=
lim l->0 (ln(1) - ln(l))=
lim l->0 (- ln(l)) = infinity
Similarly on the other side.

Anonymous

Anonymous Tue 05 May 2020 14:47:26 No.11639503 Report

Quoted By: >>11639513 >>11639548

>>11639281
I'll do you one better OP
$\int^a_{-a}\frac{1}{x}dx=\ln|a|-\ln|-a|=0$

Anonymous

Anonymous Tue 05 May 2020 14:49:53 No.11639513 Report

Quoted By: >>11639523 >>11639548

>>11639503
see the discussion just above, 1/x is not defined on 0. 0 is not in its domain. You have two divergent integrals on [-1,0) and (0, 1] and you end up with infinity - infinity

Anonymous

Anonymous Tue 05 May 2020 14:54:03 No.11639523 Report

Quoted By: >>11639530 >>11639540 >>11639548

>>11639513
Finitely many point discontinuities are able to be ignored in Riemann integrals, the value at 0 doesn't matter
The upper and lower Riemann sums of the interval will always be some value $a$ plus its inverse, so even by not using the Fundamental Theorem of Calculus it's not too hard to prove the integral is equal to 0

Anonymous

Anonymous Tue 05 May 2020 14:55:53 No.11639530 Report

Quoted By: >>11639548

>>11639523
>Finitely many point discontinuities are able to be ignored in Riemann integrals, the value at 0 doesn't matter
only if the function is Riemann integrable on the interval. this is not the case.

Anonymous

Anonymous Tue 05 May 2020 14:57:59 No.11639537 Report

Quoted By:

>>11639281
both integrals are divergent you dweeb

Anonymous

Anonymous Tue 05 May 2020 14:58:36 No.11639540 Report

Quoted By: >>11639554

>>11639523
we talked about that above. I said look it up and you did not :l as one already replied, function is not defined on 0 so it is not continous or discontinous here either

Anonymous

Anonymous Tue 05 May 2020 15:01:02 No.11639548 Report

Quoted By: >>11640154

>>11639284
>>11639285
>>11639289
>>11639293
>>11639300
>>11639315
>>11639317
>>11639323
>>11639347
>>11639355
>>11639385
>>11639388
>>11639396
>>11639397
>>11639402
>>11639425
>>11639466
>>11639475
>>11639482
>>11639483
>>11639495
>>11639503
>>11639513
>>11639523
>>11639530
neeeerds

Anonymous

Anonymous Tue 05 May 2020 15:03:59 No.11639554 Report

Quoted By: >>11639566

>>11639540
It's continuous on $\mathbb{R}\setminus\{0\}$ (where the function is defined) and on a finite interval, so it's integrable

Anonymous

Anonymous Tue 05 May 2020 15:06:37 No.11639566 Report

Quoted By: >>11639569

>>11639554
Riemann integrability is defined only on compact intervals. you're wrong.

Anonymous

Anonymous Tue 05 May 2020 15:07:28 No.11639569 Report

Quoted By: >>11639573

>>11639566
[-a,a] is compact

Anonymous

Anonymous Tue 05 May 2020 15:09:18 No.11639573 Report

Quoted By:

>>11639569
1/x is not defined on this interval.
you can define it at 0 to be anything, but the extended function will not be continuous. therefore you cannot deduce integrability.

Anonymous

Anonymous Tue 05 May 2020 15:11:45 No.11639578 Report

Quoted By:

>>11639281
It is not uniformly continuous on the selected interval. We have to handle it with limits as an improper integral and in that case there isn't any debate about it. The symbol $\int_{-1}^1\frac{1}{x}dx$ is as of yet meaningless.

Anonymous

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Anonymous Tue 05 May 2020 16:18:36 No.11639739 Report

Quoted By: >>11639881 >>11640230 >>11643446

/thread

Anonymous

Anonymous Tue 05 May 2020 17:13:05 No.11639881 Report

Quoted By: >>11639981 >>11642236

>>11639739
Lmao Wolfram btfo by simple undergrad problem xD

Anonymous

Anonymous Tue 05 May 2020 17:44:22 No.11639981 Report

Quoted By: >>11640195

>>11639881
got it right tho

Anonymous

Anonymous Tue 05 May 2020 17:56:01 No.11640019 Report

Quoted By:

1.
$\int_{-1}^1 \frac{dx}{x}[\eqn] It's not well defined. 2. <script type="math/tex; mode=display">\int_{-1}^0 \frac{dx}{x} + \int_0^{1} \frac{dx}{x}$
Each summand doesn't converge.

3.
$\limit_{a \to 0+}\left(\int_{-1}^{-a} \frac{dx}{x} + \int_a^{1} \frac{dx}{x}\right) = 0$

Anonymous

Anonymous Tue 05 May 2020 18:00:11 No.11640031 Report

Quoted By:

1.

$\int_{-1}^1 \frac{dx}{x}[\eqn] Not well defined. 2. <script type="math/tex; mode=display">\int_{-1}^0 \frac{dx}{x}+\int_0^1 \frac{dx}{x}[\eqn] Each summand doesn't converge. 3. <script type="math/tex; mode=display">\lim_{a\to0^+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_a^1 \frac{dx}{x})[\eqn] Valid and equals 0.$

Anonymous

Anonymous Tue 05 May 2020 18:03:14 No.11640050 Report

Quoted By:

>>11639385
It isnt bounded faggot

Anonymous

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Anonymous Tue 05 May 2020 18:28:07 No.11640154 Report

Quoted By:

>>11639548
>neeeerds

Anonymous

Anonymous Tue 05 May 2020 18:29:14 No.11640157 Report

Quoted By:

This nigger doesn't know about improper integrals lmfao

Anonymous

Anonymous Tue 05 May 2020 18:39:29 No.11640195 Report

Quoted By: >>11642221 >>11642444

>>11639981
It converges to size of you're penis, faggot

Anonymous

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Anonymous Tue 05 May 2020 18:47:36 No.11640230 Report

Quoted By:

>>11639739
1/0 = undefined

Anonymous

Anonymous Tue 05 May 2020 21:10:44 No.11640723 Report

Quoted By:

>>11639281
It's an odd function's integral (just plug in values, if when you plug x you get y it's odd if when you plug -x you get -y) defined from -a to a, it's always going to be 0.

Anonymous

Anonymous Tue 05 May 2020 21:25:40 No.11640759 Report

Quoted By:

https://www.youtube.com/watch?v=dHwrzLDmdT8

Anonymous

Anonymous Tue 05 May 2020 21:54:41 No.11640850 Report

Quoted By:

>>11639281
engineer here, yes it equals zero

Anonymous

Anonymous Wed 06 May 2020 02:34:24 No.11641624 Report

Quoted By: >>11642201

>>11639281
one may define the cauchy principal value of this to be 0, but the function 1/x is not lebesgue integrable over the interval -1 to 1.

Anonymous

Anonymous Wed 06 May 2020 06:30:43 No.11642201 Report

Quoted By:

>>11641624
best answer

Anonymous

Anonymous Wed 06 May 2020 06:43:39 No.11642221 Report

Quoted By:

>>11640195
Impressive, did you quote that from your PhD?

Anonymous

Anonymous Wed 06 May 2020 06:57:48 No.11642236 Report

Quoted By:

>>11639881
>i can't cope
wa is #teamdivergent as it should be

Anonymous

Anonymous Wed 06 May 2020 10:33:11 No.11642444 Report

Quoted By:

>>11640195
my penis has size indeterminate. it depends, sometimes it grows beyond all bounds.

Anonymous

Anonymous Wed 06 May 2020 10:41:31 No.11642453 Report

Quoted By: >>11643194

>>11639347
>-inf = +inf
Thing is, 1/0 diverges. Does that integral exist?

Anonymous

Anonymous Wed 06 May 2020 16:24:20 No.11643194 Report

Quoted By: >>11643205

>>11642453
why wouldn't it?

Anonymous

Anonymous Wed 06 May 2020 16:28:57 No.11643205 Report

Quoted By:

>>11643194
the function is not bounded on [-a,a], therefore it's not Riemann integrable. one needs to handle this using improper integrals.

Anonymous

Anonymous Wed 06 May 2020 16:33:00 No.11643215 Report

Quoted By:

Anonymous

Anonymous Wed 06 May 2020 17:05:12 No.11643337 Report

Quoted By: >>11643568

What OP is asking is why should that improper integral be infinity instead of zero, as it is in common integration theory, you engenieering morons.

Anonymous

Anonymous Wed 06 May 2020 17:15:58 No.11643376 Report

Quoted By:

okay retards.
$ \lim_{\epsilon\to0^+} \left( \int_{-1}^{-\epsilon} + \int^{1}_{+\epsilon} \right) \frac{1}{x} dx = 0 \\ \int_{-1}^{1} \frac{1}{x} dx = \text{undefined} $

Anonymous

Anonymous Wed 06 May 2020 17:21:42 No.11643391 Report

Quoted By:

>>11639293
>negative areas
get back to high school

Anonymous

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Anonymous Wed 06 May 2020 17:36:31 No.11643446 Report

Quoted By:

>>11639739

Anonymous

Anonymous Wed 06 May 2020 18:10:39 No.11643568 Report

Quoted By:

>>11643337
OP here. I just wanted to see the opinions. it's very tempting to say it's 0 without giving it a second thought.

Anonymous

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Anonymous Wed 06 May 2020 18:13:08 No.11643575 Report

Quoted By:

https://en.wikipedia.org/wiki/Cauchy_principal_value

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