Mathematics

well assuming the $a_i$ are positive, any of those expressions diverges iff any other diverges too

These manipulations wouldn't be justified, for instance, if $\sum_i a_i$ or $\sum_i ia_i$ aren't convergent.

N3buchadnezzar

Everything is zero or positive so no problems with reordering yeah.

@SteamyRoot i thought some of them are also called physicists

I call them lower species

6:01 PM

The distinction there is probably how much you get paid to ignore said limits :P

yes, probably

N3buchadnezzar

sub-humans

Sub-humans with good job satisfaction and retirement plans :>

N3buchadnezzar

That is why they are sub-humans

Suffering is exsistence :p

exsistence is a good word, i like it

i have always been curious about how some people can write things like 2^{sqrt(2)} and when you ask them "what does that mean?", they throw their hands up and say "idk"

6:09 PM

I was like that when i was very young o.o

If, given enough time, I can compute approximations to that, then I'm fine with just saying "it is what it is."

playing with volumes of spheres and thinking "hey wouldn't it be really cool if $(-1/2)! = \sqrt \pi$ it would make the formula very simple"

@mercio do you mean sqrt(pi)/2

I'm not actually sure

I forget which one has the factor of 1/2.

I think it's (1/2)! that's sqrt(\pi)/2.

6:10 PM

yeah

and then i asked my calculator about $(-1/2)!$ and it gave the result I wanted

my calculator was pretty good it seems

ya, okay, i just can't do u-substitution

yes, very good calculator

i never had a calculator like that, but i have wolfram alpha on my phone

so i suppose that's fine

Secret

[Abstract algebra]
Let $Z$
associative, distributive (+ over *), commutative, existence of unique identities 0 and 1, additive inverses, and induction 1+1=2. Then:

$\forall n\in Z, n*(n+1)=(n+0)*(n+1)=n+(0*1)=n+0=n$
$\forall n\in Z, n+n=n+(n*1)=(n+n)*(n+1)$
TBC

SoumyoB

well well well what have we here... Oh roasting physicists?

@Secret perhaps you should phrase this as a coherent question, and post it somewhere on the website

then link back here

Secret

that one is not really a question. more like progress so far

6:16 PM

Guys i have a problem

what's the problem?

the derivative of this function looks like the gamma function but kind of shifted in both axis

Sophie

@SteamyRoot relevant youtube.com/watch?v=rp8hvyjZWHs

excuse me

Mahmoud

Hi.

6:17 PM

hmm

@PearlSek the term in the parantheses looks very much like a form for the sum of a geometric series

@Starfall yes i know

I guess it can be extended to a function on $\Bbb R$

with the integral formula

oh, yes, sure

well what would happen if you could differentiate under the integral

Mahmoud

6:18 PM

@Sophie Lol

Balarka Sen

@Starfall that's precisely the reason it's the sum of $k!$'s isn't it

it comes from there :

@BalarkaSen indeed

Balarka Sen

write down the geometric series, integrate termwise

6:19 PM

well, it's more likely it was obtained the other way around, as seen in this picture

oh wait a sec i'll write something

well, if you can differentiate under the integral, you pick up a factor of 1/(1-t) in addition to the usual derivative of the gamma function

inside the integral

You can also differentiate it when it's still in geometric sum form, i.e. $\sum_{k=1}^n t^k\mapsto \sum_{k=1}^n k t^{k-1}$.

But then you've gotta resum it, etc.

Though if you return to the gamma function at that point, you have $\sum_{k=1}^n k \Gamma(k) = \sum_{k=1}^n k(k-1)!=\sum_{k=1}^n k!$...somehow, I feel like I've made an algebra error.

but you're not differenting with respect to $t$

Oh, derp. I did $t^k\mapsto kt^{k-1}$, which is differentiating w/r/t $t$ not $k$.

Yeah.

Should be $t^k\mapsto (\ln t)t^k$.

6:28 PM

ln(k) still sounds wrong

isn't it ln(t)