"For the particular case of a simplicial complex structure for a torus, David Eppstein is right: the minimal triangulation has 7 vertices, 21 edges, and 14 triangles."
I had a quick quesiton about extensions. If $S$ is a collection of elements in the closure of $F$, can a general element $\alpha\in F(S)$ be represented as an $F-$linear combination of elements in $S$?
@gian: Do you know nullity-rank in linear algebra or the fundamental homomorphism theorem in group theory? Surely you do, or you shouldn't be playing with homology.
So what's the dimension of $\text{im}\,\partial_1$?
Last night dream: (forgot except there's an n-cube lattice where a graph consists of two connected components attached at a single vertex is traced and the total number of some unspecified type of component is given by the expression: $$\sum_{r=1}^{n}\binom{n}{r}(n-r)+1$$. In particular, I asked the tutor about the infinite dimensional case $n\to \infty$
So for any $\alpha\in F(S)$, we need only consider $\alpha^{-1}\in F(S)$, run the argument you just made to get that $\alpha$ is an $F-$linear combination.
If I have a collection of proper submodules of some module $M$, and my book tells me their sum is also a submodule of $M$, then surely they don't mean direct sum right? Since say $\Bbb R\oplus \Bbb R\not\subset \Bbb R$. Is this normal terminology, and is it equivalent to just taking a union of their basis (say they are free modules)
@PVAL-inactive I guess this is just $F[S]\subset F(S)$ is obvious, and then we show that any element in $F[S]$ is invertible by your argument, which makes $F[S]$ a field. The proof is done by minimality?
@Semiclassical $$\sum_{r=0}^{n}\binom{n}{r}rx^{r-1}=n(1+x)^{n-1}$$. Then set x=1 in the GF to get the required sum $$\sum_{r=1}^{n-1}\binom{n}{r}r+1=(n-1)2^{n-1}$$ which diverges as $n\to \infty$ as suspected
In Did's answer here: math.stackexchange.com/questions/2436627/… I'm trying to show that $G^{\prime} = -g$ like he said, but I don't know how to use the fundamental theorem on an improper integral.
I know the equation $ f^{\prime} (x) = f(x) f^{\prime} (x-1) $ is solved by $f(x) = C$ or by tetration ( $ f(x+1) = \exp(f(x)) $).
So I wonder What are the solutions to
$$f ^{\prime \prime}(x) = f(x) f^{\prime \prime} (x-1) ?$$
What does it mean the logarithmic derivative of $R_{x}(t) = \frac{G(x+t)}{G(t)}$? The regular derivative of $R_{x}(t)$ is $\frac{ G(t)\cdot G^{\prime}(x+t) - G(x+t)G^{\prime}(t)} {[G(t)]^{2}}$ And since $G(x) = -g(x)$, this derivative comes out to $\frac{g(t)G(x+t) - g(x+t)G(t)}{[G(t)]^{2}}$
I need to prove that the age behavior of the Gamma Distribution with probability density function $$ f(x)=\frac{\lambda^{\alpha}x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)}$$
For $x\geq 0$, $\lambda, \alpha >0$; I.e., the conditional pprobability $P(X>x+t|X>t)$, increases in $t$ whenever $\alpha >...
I don't know why he said logarithmic derivative — you're right — he's doing the quotient rule. But the denominator is positive, so he's looking just at the sign of the numerator of the quotient rule.
I know that if $f$ is surjective then $f(\mathfrak{a})$ is an ideal for every ideal $\mathfrak{a}$ in $A$. I want to first prove that if $\mathfrak{m}$ is maximal in $A$ then $f(\mathfrak{m}) = \mathfrak{m'}$ is maximal in $B$.
$A/\mathfrak{m}$ is a field by maximality of $\mathfrak{m}$ and $f$...
let $f(x)$ in $ f(\mathfrak{a}) \cap f(\mathfrak{b})$ but not in $f(\mathfrak{a} \cap \mathfrak{b})$ . I think you can say $x$ is in the kernel of the map A to B.
or it differs from something in $\mathfrak{a} \cap \mathfrak{b}$ by an element of the kernel
but since you know what the kernel is contained in $\mathfrak{a} \cap \mathfrak{b}$ you are done.
I worry because this implies, as far as I can see, something that isn't true: every normal extension $E/F$ is galois
Let $K$ be the separable closure of $F$ in $E$. So we have $E/K/F$. By some lemma, $E/K$ is separable. Since $K$ is just $F$ adjoin some separable elements, $K/F$ is separable also. By some tower theorem, $E/F$ is separable.
holy crap forgive me
I found the mistake
:( gosh it sucks when you spend so much time on something... trudging through a mistake
@Prototank One standard example of a non-separable extension is F_p(x^(1/p)) over F_p(x). In this case the separable closure is just the base field, so your "some lemma" isn't true.