I'm such a goddamn genus

[Anonymous No.5941060]

1 + 2 + 3 ... + n = n(n+1)/2

therefore


1 * 2 * 3 ... *n = n^(n+1)/2

[Anonymous No.5941062]

[Anonymous No.5941063]

The second one doesn't work with n = 2.

[Anonymous No.5941066]

Nope not working.

Resort to the Gamma Function

[Anonymous No.5941074]

>>5941060
OP, did you just incorrectly derive?
did you, punk?
did you?

[Anonymous No.5941138]

>>5941060
http://en.wikipedia.org/wiki/Sterling%27s_approximation

~(n/e)^n*√2πn

[Anonymous No.5941157]

>>5941063
& what's the formula for what else?

[Anonymous No.5941194]

>>5941066
top lel

[Anonymous No.5941204]

this better be a troll thread.
never heard the word "factorial"?

[Anonymous No.5941235]

>>5941060
>such a goddamn genus
>genus
Mmmmno.

[Anonymous No.5941248]

>>5941235
I saw that too. X3

[Anonymous No.5942566]

Why does it not work? How is multiplication different from addition?

[Anonymous No.5942574]

>since something is true, something entirely unrelated is true

cool jump to conclusions, bra

[Anonymous No.5942603]

>>5942574
and you should at least fail properly :

>1 * 2 * 3 ... *n = n^(n+1)/2
what about the n+1 ? and the /2 ? they don't change ?

-->you fail at failing

[Anonymous No.5944400]

I wish I was a genus like OP.

[Anonymous No.5946529]

Can you prove it with induction?

[Anonymous No.5947754]

>>5946529
Induction is not logically sound.

[Anonymous No.5947784]

>>5947754
>axioms are not logically sound

[Anonymous No.5949183]

>>5947784
Induction is not an axiom.

[Anonymous No.5949239]

>>5949183
Peano Axioms:
1. <span class="math">0\in\mathbb{N}[/spoiler]
2. <span class="math">n\in\mathbb{N}\Longrightarrow[/spoiler]<span class="math">S(n)\in\mathbb{N}[/spoiler]
3. <span class="math">\nexists n\in\mathbb{N}:S(n)=0[/spoiler]
4. <span class="math">\forall x,y\in\mathbb{N}[/spoiler] <span class="math">S(x)=S(y)\Longrightarrow x=y[/spoiler]
5. <span class="math">(0\in U)(\forall n\in\mathbb{N}[/spoiler] <span class="math">n\in U\Longrightarrow S(n)\in U)[/spoiler]<span class="math">\Longrightarrow (\forall n\in\mathbb{N}~n\in U)[/spoiler] (Induction axiom)

http://mathworld.wolfram.com/PeanosAxioms.html

[Anonymous No.5949242]

>>5949183
Peano Axioms:
1. <span class="math">0\in\mathbb{N}[/spoiler]
2. <span class="math">n\in\mathbb{N}\Longrightarrow[/spoiler]<span class="math">S(n)\in\mathbb{N}[/spoiler]
3. <span class="math">\not\exists n\in\mathbb{N}:S(n)=0[/spoiler]
4. <span class="math">\forall x,y\in\mathbb{N}[/spoiler] <span class="math">S(x)=S(y)\Longrightarrow x=y[/spoiler]
5. <span class="math">(0\in U)(\forall n\in\mathbb{N}[/spoiler] <span class="math">n\in U\Longrightarrow S(n)\in U)[/spoiler]<span class="math">\Longrightarrow (\forall n\in\mathbb{N}~n\in U)[/spoiler] (Induction axiom)

http://mathworld.wolfram.com/PeanosAxioms.html

[Anonymous No.5949258]

>>5941066
Why do you need it to be continuous?

[Anonymous No.5949275]

>>5946529

Can you prove it with conduction?

[Anonymous No.5949294]

>>5941204

What do fractions have to do with this?

[Anonymous No.5949340]

>>5949242
That's awesome!...
I had never seen that. Thank you master for this I'll go to sleep happier than before. I love you.

[Anonymous No.5949354]

>>5949340
By the way, if you're the same guy. If you have the other four Peano axioms, you can either take the well ordering principle (i.e. every set of natural numbers has a minimum element) or induction because they both imply each other (i.e. equivalent). So just assume whichever you personally find intuitive and prove the other to yourself.

[Anonymous No.5949452]

>>5949354
As soon as I saw this I just started deducing basic algebra out of it. I've done a similar thing a long time ago, I should have looked for this sooner.
I'll certainly look at that. Thanks.

[Anonymous No.5949455]

>>5947754
never change /sci/

[Anonymous No.5950991]

>>5949242
>piano axioms

Pianos belong on >>>/mu/