/sci/ - Science & Math

>>11770228
Yeah, it's a bit hard to guess what the time is in East Mongolia or some other country around that like Australia maybe. However, I recommend you do these tomorrow:
>prove the sum formula (f+g)' = f'+g'
>prove the product formula (fg)' = f'g + fg'
>use the product formula to get the formula (af)' = af'
(where a is a constant)
>use the previous formula and the sum formula to get the formula for difference (f-g)' = f' - g'
>prove the division formula (f/g)' = (f'g - fg')/g^2
>check how the chain rule is derived
Those will help you when you need to differentiate stuff like [math]e^{x^2 + x}[/math]. Then you just have to go look for ways to obtain the derivatives of some concrete functions like x^n, e^x, sin(x), etc. The first one is easy to get directly from the definition, but the rest will require some trickery, so maybe check the details and try to understand what is going on.

Did you know Feynman was a fan of the Tuvan republic? Well now you know. Also, reindeer are cute https://www.youtube.com/watch?v=O5DCxaGy3wY

>>11722386
Another nice thing about it is that you can combine it with the fundamental theorem of finitely generated abelian groups. If [math]A[/math] is f.g., then [math]A \cong \mathbb{Z}^n \oplus \sum\limits_{i_1}^m \mathbb{Z}/k_i\mathbb{Z}[/math], and so it suffices to find one of those spaces for each summand and take the wedge of them, as [math]H^n(X \vee Y) \cong H^n(X) \oplus H^n(Y)[/math]. As mentioned before, [math]S^n[/math] can be used as [math]P^n(\mathbb{Z})[/math], so we need to get the [math]P^n(\mathbb{Z}/k_i\mathbb{Z})[/math]s somehow. This happens by taking the degree [math]k_i[/math] map [math]f\colon S^{n-1}\to S^{n-1}[/math], and then its mapping cone gives you [math]H^(C_f) \cong \mathbb{Z}/k_i \mathbb{Z}[/math] and trivial reduced cohomology otherwise. Taking the wedge of these gives you [math]P^n(A)[/math]. Easy to construct.

>>11722871
When [math]n=2[/math], you get just a set, but if [math]n\ge 3[/math], then it is a group, abelian when [math]n\ge 4[/math]. You get the group structure using the comultiplication of co-H spaces, as [math]\Sigma P^n \simeq P^{n+1}[/math] and suspensions are co-H. Then you apply the contravariant functor [math][-, X]_*[/math].

>>8508445
That's like the total opposite to what I've been doing. Nevertheless, good luck!

>an iranian man discovers how to make a car go with water
>oil companies get mad
>iran suddenly has some experimental nukes that require the us to invade the country and make this guy collateral damage

>>8472689
You have triangulations for polyhedra, and a hypercube would be a polyhedron. These are topology.

>>8472695
Oops, I missed that part completely. Then you can discard my proof of impossibility, too. It holds only for the whole surface, not for a surface with edges and vertices removed.

>>8451685
It fails in the part that it should be an open map, but the embedded disk is not open in R^3. As a subspace with its subspace topology, that way they are homeomorphic, but it makes no difference here, since it is not an open set in R^3.

>>8409050
Dunno. Nevertheless, I feel happier than for ages. This is still irrelevant, and I won't take part in derailing this thread any more than these few posts have done. Your fanmail has been read, though.

>>8409060
Cheers, m8!

>>8409097
Ah, yes ofcourse. My bad.