/sci/ - Science & Math

>>9234961
Calm down, cutie. No need to worry about those baka engineers.

Study for a while, go for a walk in cold air, and study more.

>>8416744
It seems to be false, like these guys are saying here. Try this one on your own first:

[math]1^2+2^2 \dots n^2 = \frac{n(n+1)(2n+1)}{6}[/math]

>>8408416
>>8408459
Reformulating the post now:
[math]\textbf{Cat}[/math] is the category of categories and functors. Taking any categories C, D, define [math]H=Hom_{\textbf{Cat} } (C, D)[/math]. For any [math]F \in H[/math], define a decomposition of [math]F[/math] to be an ordered n-tuple [math](F_i )_{1 \le i \le n }[/math], where [math]F_i \colon C_{i-1} \to C_i[/math] is a functor for all [math]i[/math], [math]C_0=C[/math] and [math]C_n=D[/math], and [math]F=F_n F_{n-1} \cdots F_1[/math]. Call a decomposition proper if [math]F_{i+1} F_i[/math] is not naturally equivalent to the identity functor [math]1_{C_{i-1} }[math] for any [math]1 \le i < n[/math]. If [math]n \ge 1[/math] is such that [math](F_i)_{1 \le i \le n}[/math] can be a proper decomposition, call the functor [math]n-1[/math] times differentiable.

Let [math]\partial _n[/math] be the category with its objects all [math]n[/math] times differentiable functors, and as morphisms ordered [math]n[/math]-tuples [math]\tau = (\tau _i)_{1 \le i \le n} \colon (F_i)_{1 \le i \le n} \to G_{1 \le i \le n}[/math], where [math]\tau _i \colon F_i \to G_i [/math] is a natural transformation for each [math]i[/math]. Clearly, [math]m < n[/math] implies that [math]\partial _n[/math] is a subcategory of [math]\partial _m[/math].

>>8383325
i too am your board buddy

>>8277293
Thanks.