/sci/ - Science & Math » Thread #12186340

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Thread about derivatives

El Arcón Fri 02 Oct 17:46:54 2020 No.12186340 View Reply Original Report

Quoted By: >>12186346 >>12186387 >>12187091 >>12187249 >>12187261 >>12187310 >>12188184 >>12188199 >>12188342

Look at this derivative. It works, unsurprisingly, just like
$\dfrac{d}{dx}\sqrt{x}=\sqrt{ \dfrac{1}{4x} }$

Anonymous

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Anonymous Fri 02 Oct 2020 17:48:15 No.12186346 Report

Quoted By: >>12186352

>>12186340

El Arcón

El Arcón Fri 02 Oct 2020 17:50:25 No.12186352 Report

Quoted By:

>>12186346
pretty much

Anonymous

Anonymous Fri 02 Oct 2020 18:03:14 No.12186387 Report

Quoted By: >>12186400

>>12186340
i hate when people choose weird letters for the sake of being fancy. man you 28 latin letters why do you have to use the unreadables? just make it a capital X if you want them look related.

El Arcón

El Arcón Fri 02 Oct 2020 18:07:45 No.12186400 Report

Quoted By:

>>12186387
I chose aleph because it was already associated with infinity.

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

Anonymous

Anonymous Fri 02 Oct 2020 21:00:30 No.12187091 Report

Quoted By: >>12187224 >>12187235 >>12187238

>>12186340
are you developing calculus in the neighborhood of infinity?

El Arcón

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El Arcón Fri 02 Oct 2020 21:41:37 No.12187224 Report

Quoted By:

>>12187091
I don't know if "developing calculus" is what I'm doing. I was examining the integral
$I_4=\int_{-\infty}^\infty \!dx\,e^{-x^4}=?$
because this book I'm reading says it's important for QFT. I thought if I could use some trick in the neighborhood of infinity to solve it, that would be good. If there is some trick, then it would equally for
$I_4=\int_{-\infty}^\infty \!dx\,e^{-x^2}=\sqrt{\pi}$
but I didn't have any ideas about how I might get $\pi$ in there without converting to polar.

El Arcón

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El Arcón Fri 02 Oct 2020 21:42:59 No.12187235 Report

Quoted By:

>>12187091
I don't know if "developing calculus" is the right word for it. I was examining the integral
$I_4=\int_{-\infty}^\infty \!dx\,e^{-x^4}=\,\,?$
because this book I'm reading says it's important for QFT. I thought if I could use some trick in the neighborhood of infinity to solve it, that would be good. If there is some trick, then it would equally for
$I_2=\int_{-\infty}^\infty \!dx\,e^{-x^2}=\sqrt{\pi}$
but I didn't have any ideas about I migth get $\pi$ in there without converting to polar.

El Arcón

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El Arcón Fri 02 Oct 2020 21:44:07 No.12187238 Report

Quoted By: >>12188167

>>12187091
I don't know if "developing calculus" is the right word for it. I was examining the integral
$I_4=\int_{-\infty}^\infty \!dx\,e^{-x^4}=\,\,?$
because this book I'm reading says it's important for QFT. I thought if I could use some trick in the neighborhood of infinity to solve it, that would be good. If there is some trick, then it would equally for
$I_2=\int_{-\infty}^\infty \!dx\,e^{-x^2}=\sqrt{\pi}$
but I didn't have any ideas about how I might get $\pi$ in there without converting to polar.

Anonymous

Anonymous Fri 02 Oct 2020 21:49:01 No.12187249 Report

Quoted By: >>12187286 >>12187296

>>12186340
You left out the part where you prove that $\sqrt{\frac{\aleph_x}{4x^2}}=\sqrt{\aleph_\frac{1}{4x}}$

Anonymous

Anonymous Fri 02 Oct 2020 21:52:28 No.12187261 Report

Quoted By: >>12187292

>>12186340
>works
Holy shit this is cringe.

El Arcón

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El Arcón Fri 02 Oct 2020 22:00:29 No.12187286 Report

Quoted By:

>>12187249
It is true that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and the axioms give
$a\aleph_b=\aleph_{ab}\,\,.$

El Arcón

El Arcón Fri 02 Oct 2020 22:01:43 No.12187292 Report

Quoted By: >>12188246

>>12187261
Actually, if you write $\aleph_x=x\cdot\widehat\infty$ , then it is trivial that it works.

El Arcón

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El Arcón Fri 02 Oct 2020 22:02:29 No.12187296 Report

Quoted By: >>12187336

>>12187249
It is true that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and the axioms give
$a\aleph_b=\aleph_{ab}.$

Anonymous

Anonymous Fri 02 Oct 2020 22:05:58 No.12187310 Report

Quoted By: >>12187436

>>12186340
Welcome back "the acorn" stop being such a nut and gtfo

Anonymous

Anonymous Fri 02 Oct 2020 22:14:20 No.12187336 Report

Quoted By:

>>12187296
Axioms of what? I thought we were talking about ordinals or cardinals, but I'm not familiar with whatever this structure is.

El Arcón

El Arcón Fri 02 Oct 2020 22:58:10 No.12187436 Report

Quoted By: >>12187444

>>12187310
>"the acorn"
one day I'm going to make you know what a fool you have been

Anonymous

Anonymous Fri 02 Oct 2020 23:03:17 No.12187444 Report

Quoted By:

>>12187436
Oh tough guy, are we? I'd like to see you try, buddy.

El Arcón

El Arcón Sat 03 Oct 2020 02:13:00 No.12188167 Report

Quoted By: >>12188207

>>12187238
Actually, I think I found an exact solution to this integral. I must be fooling myself thinking I solved something new. Is $I_4$ known?

Anonymous

Anonymous Sat 03 Oct 2020 02:18:47 No.12188184 Report

Quoted By: >>12188207

>>12186340
What the actual fuck is $\aleph_x$ ???

Anonymous

Anonymous Sat 03 Oct 2020 02:23:42 No.12188199 Report

Quoted By:

>>12186340
TOOKER I TOLD YOU TO FUCKING INVEST YOUR TIME INTO REWRITING MOCHI'S PROOF OF ABC AND THIS IS WHAT YOU DO? JERKING OFF WITH FUCKING MADE UP INTEGRALS? JESUS CHRIST TOOKER YOU REALLY ARE A FUCKING ACORN.

Anonymous

Anonymous Sat 03 Oct 2020 02:29:02 No.12188207 Report

Quoted By:

>>12188167
You can just put it on wolfram alpha and it gives you a closed-form solution, you fucking acorn.
>>12188184
Autism mixed with schizophrenia.

Anonymous

Anonymous Sat 03 Oct 2020 02:48:36 No.12188246 Report

Quoted By:

>>12187292
unlike your meds

Anonymous

Anonymous Sat 03 Oct 2020 03:47:30 No.12188342 Report

Quoted By: >>12188348

>>12186340
In that limit, how do you get from X(x-dx) to Xx -Xdx??

El Arcón

El Arcón Sat 03 Oct 2020 03:49:50 No.12188348 Report

Quoted By: >>12188351

>>12188342
X(x-dx) = Xx -Xdx

Anonymous

Anonymous Sat 03 Oct 2020 03:52:15 No.12188351 Report

Quoted By:

>>12188348
Can you prove that? What does that mean?

Anonymous

Anonymous Sat 03 Oct 2020 04:12:29 No.12188387 Report

Quoted By:

I find this thread derivative

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