@Thorgott cheat proof that a^n/n! converges to 0: by the quotient criterion (for large enough n), the sum of a^n/n! converges, so the terms must go to zero
it's not much of a cheat though, since once you unwind the proof of the quotient criterion and forget all the summation parts, it just goes like: Pick $N$ large enough such that $N>a$. Then $\frac{a^n}{n!}=\frac{a^N}{N!}\frac{a^{n-N}}{(N+1)\cdot\dots\cdot n}\le\frac{a^N}{N!}\frac{a^{n-N}}{N^{n-N}}=\frac{a^N}{N!}\left(\frac{a}{N}\right)^{n-N}$ for $n>N$ and the RHS goes to $0$ as the bracketed term is $<1$.
Ah, well, okay I should therefore qualify my statement. "You understand the natural numbers as a monoid, which is a nontrivial part of their structure."
I can agree with that. I was about to argue that polynomial rings might take that title, but then I realized they're just monoid rings where the monoid is the natural numbers
Oh, you were definitely right that epsilons and deltas can be wholely circumvented in that proof
a^n is monotonely decreasing so has a limit which must naturally be 0, and if one goes by the geometric series proof, one can argue that it's monotonely increasing and bounded above by 1/(1-a), without proving that a^n goes to 0 first
@Astyx: I lied. The answer to that geometry problem was $\arctan(1/2)$. I made a dumb, dumb mistake. Nevertheless, I contend that a math multiple choice question should say "which is these is approximately the correct answer?" ...
Let X be a non-empty set and n ∈ N. Then X n is the Cartesian product of n copies of X. A relation can be defined on Xn by (a1,a2,...,an) ∼ (b1,b2,...,bn) if and only if every x ∈ X appears the same the number of times in the first list as it does in the second.
Question:Let X=R^3. For the relation above, list all elements which are related to (0,1,2).
@geocalc33 You have a bijective map $(0,1)^2 \to \Bbb R^2$, so you can use the vector space structure on $R^2$. Of course the operations are not going to be the standard real operations on $(0,1)$
A vector space is no just a set, it's a set and its operations
I don't remember if you listen to them (though you should if you like post rock), but god is an astronaut released a new single a few days ago and there will be a new album in February @Balarka
1. Questions: Find the lowest rank differential equation that has the following solution; $y=a\cos(\ln{x})+b\sin(\ln{x})$, a and b are arbitrary fixeds
Yeah, it is a trend that a good few of us come around to just hang out. The more commitment a problem looks likely to require, the less likely people are to engage with it. That being said, never say "never."
a peer of mine during my bachelor tried to get my dissertation supervisor to critique his water-tight proof of the Collatz conjecture and when he refused he just submitted his paper to some journal with the profs name on it
he also tried to get me to read his proof and the first line was like.. some incorrect modular arithmetic
Okay, so the upper triangular matrices can be written as a direct sum of their columns, and these columns will form left submodules/ideals. How does one show they are indecomposable at modules over the upper triangular matrices (everything is over a field $k$, btw).
Let $T_n(k)$ be the algebra of upper triangular matrices over the field $k$, and define $$S_j = \{A \in T_n(k)$ \mid \text{ every column of A, except possibly the j-th column, is 0 } \}$$. Then it's clear that $T_n(k) = S_1 \oplus ... \oplus S_n$. I am trying to show that the $S_j$ are indecomposable ($T_n(k)$-modules?). Is it possible that they are actually simple modules?
one way to show that the $T_n(k)$ are indecomposable is to compute $\mathrm{soc}(T_n(k))$. If an Artinian module has simple socle, then it is indecomposable
basically if you can show that $P_j$ has a unique simple submodule, then it follows that it is indecomposable. That's because if we had $P_j=A \oplus B$ with $A$ and $B$ non-zero, both $A$ and $B$ have a simple submodule, so there's more than one simple submodule
You can also (more elementarily) show that all submodules are of the form I gave above, and it's clear that the intersection of these submodules is never 0
So, for $n=3$, we have $$S_3 = \{\begin{pmatrix} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & c \\ \end{pmatrix} \mid a,b,c \in k \}$$. Would one submodule of this be $$\{\begin{pmatrix} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & 0 \\ \end{pmatrix} \mid a,b \in k \}$$?
Let $C$ be a smooth manifold and $c \in \Bbb N$, then we have $C^c(C) \supset C^c_c(C) \subset C_c(C)$. Now write that sloppily on a blackboard so that $($, $\subset$ and $C$ look the same
How do I show that any vector field on $S^n$ can be extended to a vector field on $\Bbb{R}^{n+1} \setminus \{0\}$? My intuition tells me that I should extend the vector field on $S^n$ radially throughout $\Bbb{R}^{n+1} \setminus \{0\}$, but how do I write this down?
compose a deformation retraction $\mathbb{R}^{n+1}\setminus\{0\}\rightarrow S^n$ with the vector field $S^n\rightarrow TS^n$ and then with the embedding $TS^n\rightarrow T\mathbb{R}^{n+1}\setminus\{0\}$ induced by the inclusion $S^n\rightarrow\mathbb{R}^{n+1}\setminus\{0\}$?
@MadSpaces I am trying to figure out what you are asking. Aside from the $x$ inside the integral and outside the integral conflicting, is $n$ a constant, or a function of $x$?
I do not know. apparently the notation provided by my prof is plane crap and hides the true mathematical discrebtion of the matter... i guess i cant do anything about it.. dumb physicts right?
So, if that is the case, then $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^bn(t)\,\mathrm{d}t=\frac{\mathrm{d}b}{\mathrm{d}x}\cdot n(b(x))-\frac{\mathrm{d}a}{\mathrm{d}x}\cdot n(a(x))$$
Can somebody explain to me the concept of codistribution in differential geometry, or link me to a good source where it is explained clearly? I am a real beginner in differential geometry. Thanks in advance.
Hello everyone. I have a very simple question about the simple proof that is the uniqueness of a limit of a real function $f\colon X\to \mathbb{R}$, with $X\subseteq\mathbb{R}$.
I have proved it before, but this book I'm utilizing gives us the following step with an additional information concerning $a\in X'$.
Ok, we have $\lim_{x\to a}f(x) = L_1$ and $\lim_{x\to a}f(x) = L_2$, and then for every $\varepsilon > 0 $ there exists $\delta_1,\delta_2 > 0$ such that $x\in X, 0 < \left|x-a\right| <\delta_1 \implies \left|f(x) - L_1 \right| < \varepsilon$ and $x\in X, 0 < \left|x-a\right| <\delt…
@J.D. I have been a differential geometer for 40+ years, and this is the first time I've heard that. This is a distribution as a subbundle of the cotangent bundle rather than as a subbundle of the tangent bundle?
@Attractor The key is to choose $\epsilon$ small enough so that you cannot have both inequalities.
My days of scheming are long gone. I know what reduced and irreducible mean, and integral is about integral ring extensions, but I don't remember much.
Reduced means that intuitively that the implicit function theorem applies: you're looking at $x=0$ rather than $x^2=0$, which is the same underlying set but has nilpotents in the coordinate ring.
Blah @ dense point ...
Doesn't a reducible one have that too?
You just want to say you cannot decompose into pieces.
I am fairly certain irreducible does not in general imply that there is a dense point. It might hold fro schemes in general, and I at least recall showing that it holds for affine schemes
In fact, there's a soberification functor, if you apply that to the maximal spectrum of an affine algebra A over an algebraically closed field (which is "drunk" in general), you get Spec(A) back.
So, the point is that the implicit function theorem may fail to prove something is a manifold even though it is ... just because you're giving the set a non-reduced scheme structure.
I already mentioned this recently in chat, but my favourite paper shows that the Scott topology on a poset is not necessarily Sober and it is called "Scott is not always sober". The first letter in each paragraph spell out the sentence "I'm not being personal Dana"
@user2103480 can you be more specific? i have had a course in statmech but never heard of the terminology "discretization of phase space", i can only guess what you mean
user486313
anyone here know much about partial differential equations?
