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/sci/ - Science & Math

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9312123 No.9312123 [Reply] [Original]

One might want to define a "derivative" on [math] \mathbb{N}, \mathbb{Z}, \mathbb{Q} [/math].

A candidate is the Lagarias arithmetic derivative (https://en.wikipedia.org/wiki/Arithmetic_derivative).).

Define [math] d(n): \mathbb{N} \cup \{0\} \to \mathbb{N}\cup \{0\} [/math] by the rules
[math] d(p)=1 [/math] when [math] p [/math] is prime.
[math] d(nm)=d(n)*m+n*d(m) [/math] for all [math] n,m\in \mathbb{N}\cup \{0\}[/math]

Then [math] d(0)=d(0*0)=d(0)*0+0*d(0)=0 [/math], and [math] d(1)=d(1*1)=d(1)*1+1*d(1)=2d(1) [/math] gives [math]d(1)=0[/math]. More results are in the image.

This may be extended to [math] d(n): \mathbb{Z} \to \mathbb{Z} [/math] by setting [math] d(-n) = -d(n) [/math] for all [math] n \geq 1 [/math].

A further extension of the derivative to [math] \mathbb{Q} [/math] is obtained by using a quotient rule [math] d(\frac{p}{q}) = \frac{d(p)*q-p*d(q)}{q^2} [/math].

>> No.9312139

Nice blog post, anon. Tell me more, please

>> No.9312143

>>9312123
Just look at kahler differentials

>> No.9312144

>>9312123
could anyone invent a more useless tool and even give it a name?

>> No.9312166

Where are going with this, anon?

I mean yes, there are a dozen operators that fulfill the Leibnitz rule, aka derivations.

>> No.9312177

>>9312123
bookstore.ams.org/surv-222/
it's happening guise

>> No.9312488
File: 240 KB, 320x320, 1502783753828.gif [View same] [iqdb] [saucenao] [google]
9312488

>>9312177

>the spectrum of the integers is “intrinsically curved”

>> No.9312499

>>9312488
>>the spectrum of the integers is “intrinsically curved”
He's not wrong.

>> No.9312504

>>9312123
>[math]\color{#b5bd68}{d(nm)=d(n)*m+n*d(m)}[/math]
have you really thought this through

>> No.9312505

>>9312504
>have you really thought this through
What's wrong with it?

>> No.9312540

I suppose this is equal to [math]\sum_p \frac{n}{p}[/math] for prime factors p of n (potentially repeated) or if you prefer [math]\sum_{p,k} \frac{nk}{p}[/math] for distinct prime factors p of n with exponents k.

>> No.9312557

>>9312499
>He's not wrong.
I'm not a "he".

>> No.9312565

>>9312557
are you are Alexandru Buium they clearly weren't talking about you retard

>> No.9312566

>>9312565
>are you are Alexandru Buium they clearly weren't talking about you retard
Can you rewrite that into something makes sense please?

>> No.9312568

> the ring of integers plays the role of a ring of functions on an infinite dimensional manifold

I WANT OUT.
>Before you compute 1+1=2 you must first invent infinite dimensional manifolds

>> No.9312579

>>9312568
what the fuck are you talking about you autist

>> No.9312581

>>9312566
not who you're quoting but since you're retarded i'll help you:
>are you are Alexandru Buium? they clearly weren't talking about you retard
hope that helped, retard

>> No.9312586

>>9312581
>hope that helped, retard
It doesn't, what does "are you are Alexandru Buium?" mean?

>> No.9312592

>>9312586
not who you're quoting but since you're retarded i'll help you:
>are you Alexandru Buium? they clearly weren't talking about you retard
hope that helped, retard

>> No.9312600

>>9312144
isn't math about simply making novel conjectures, not strictly useful ones?

>> No.9312604

>>9312144
it's not useless, what if you need to find velocity in some Z space?

>> No.9312615

>>9312592
>are you Alexandru Buium?
Yes.