Is this true?

[velvet falcon]

$$ \prod_{a}^{b} f(k) = \sum_{a}^{b} \ln(f(k)) \to \prod_{n=1}^{\infty} \frac{1}{p_{n}^s-1} = \sum_{n=1}^{\infty} \ln(\frac{1}{p_{n}^s-1}) $$

[calm berry]

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[wispy raptor]

don mastro

[undone pagoda]

It's not even really meaningful.

[velvet falcon]

what do you mean

sorry i forgot a log

on the first product

[undone pagoda]

...I mean, it's just a bunch of random letters. Even if I ignore that k is literally just some random value since you forgot to specify that it's the index value of the product, there's absolutely no relation between f and p, no indication of what the sequence of p even is, and you have p_n^s where s is similarly completely undefined.

[velvet falcon]

it is supposed to be a general rule duh

and yes, k=a

sorry sir

[undone pagoda]

A general rule about what?

[keen mantle]

6=2*3, but ln(2)+ln(3) isnt 6

so it's false in general

[velvet falcon]

@keen mantle

[velvet falcon]

why so aggressive you are on discord lol

calm down

[undone pagoda]

You picked literally the only thing that makes some degree of sense.

[velvet falcon]

already pointed out my typo no need to continue further sir!

[keen mantle]

yes cause the rest is just symbols looking like an attempt at trying to prove Riemann hyp

[velvet falcon]

hm?

[keen mantle]

so of course im going to disregard it lol

[undone pagoda]

I mean because that part is the antecedent of the conditional.

[velvet falcon]

sooo just because i'm studying at the moment number theory and was playing around with known identites im trying to prove riemann's hypothesis?

[undone pagoda]

It's the "if" of the "if/then".

[velvet falcon]

gotcha

[keen mantle]

cool, but didnt ask

you should focus on discussing with the person who made the channel, not me

[undone pagoda]

I'm saying that that part doesn't need to be true. It just needs to logically imply the consequent in the case that it is true.

[keen mantle]

k

you done?

[velvet falcon]

nah not really, this guy seems a douchebag

prefer not to

[keen mantle]

well anyway yes, the log of a product is the sum of the logs, the question of does that hold for infinite series is a different question

[undone pagoda]

Hey, guys, is it true that $abc = d + e^f \to g(h(i)) = k$?

[wispy raptor]

Techie Literate

[undone pagoda]

Not a man.

[keen mantle]

Ok

being's off their rocker

but anyway, since you're clearly getting upset by this channel you should probably just leave

[desert acorn]

what studying arithmetic functions does to a mf

snap out of it

[velvet falcon]

?

[velvet falcon]

do you know any source online to understand it better?

[keen mantle]

I'm sure stack exchange has something on it

[velvet falcon]

thanks man

[wraith girder]

@velvet falcon has given 1 rep to @keen mantle

[merry turtle]

product convergence is actually defined in terms of the sum of the ln's of the terms

so there is definitely a correspondence even down to definitionally

[velvet falcon]

could you explain better please?

[merry turtle]

when we say $\prod^\infty f(n)$ converges, what that means is $\sum^\infty \ln(f(n))$ converges

[wispy raptor]

cute rizzly bear (nom nom nom)

[velvet falcon]

oh, thanks

[merry turtle]

op: "does ln(ab) = ln(a) + ln(b)"
techie: "we don't even know what a and b are here. this is just a bunch of letters. send the problem statement"

[velvet falcon]

i have another question, as saw before ln(prod f(k)) = sum ln(f(k)) but is there a way to “convert” starting from prod f(k) and not ln(prod f(k))?

[merry turtle]

yeah, just raise each side from the e

$\prod f(n) = e^{\sum \ln(f(n))}$

[wispy raptor]

cute rizzly bear (nom nom nom)

[velvet falcon]

yea yea

thanks

[undone pagoda]

Not what I was saying at all.

[velvet falcon]

+close