btw to supplement, $|\mathcal O_L/\mathfrak P^n| = |\mathcal O_L/\mathfrak P|^n$ because $\mathcal O_L/\mathfrak P^n = \left( \mathcal O_L \right)_\mathfrak P /\mathfrak P^n \left( \mathcal O_L \right)_\mathfrak P$ and $\left( \mathcal O_L \right)_\mathfrak P$ is a DVR because $\mathcal O_L$ is a Dedekind domain
I conjecture that for any nonzero ideal $I$ there is an order $C \subseteq \mathcal O_L$ such that $C$ is free over $\mathcal O_K$ and $C/IC = \mathcal O_L/I\mathcal O_L$
I mean this sounds like the opposite of cheap but it is what it is
> Definition. Let X satisfy(*). Two divisors D and D' are said to be linearly equivalent, written D ~ D', if D - D' is a principal divisor. The group Div X of all divisors divided by the subgroup of principal divisors is called the divisor class group of X, and is denoted by Cl X.
So I'm given that the Partial Sum $S_N$ for $\sum_{n \geq 2} a_n$ is given by $S_n=\frac{n-2}{n+6}$. Then the terms $a_n = S_N - S_{N-1} = ... = \frac{8}{(n+5)(n+6)}$. And $a_2 = \frac{8}{(2+5)(2+6)} = \frac 17$. Since the series starts at 2, we should have that $S_2=a_2$. But $S_2 = \frac{2-2}{2+6}=0$. What am I doing wrong?
The question says summation from n=2 $\sum_{n=2}^{\infty} a_n$. It says $S_n = (n-2)/(n+6)$. Then it says a) Find the sum of $\sum_{n=1}^{\infty} a_n$ (but I think that's just a typo). It says "What is $a_1$?", which is also probably a typo. It says "Find $a_n$ for $n \geq 2$. So the question thinks that the formula for $a_n$ should match $S_n$ for $n=2$, right?
Oh, the conditionally convergent integral. Your teacher wasn't wrong. I think you misinterpreted or misquoted her.
Search is such a wonderful thing. ? not ?
Anyhow, lucky for you, I haven't read the problem you and Leaky were discussing. So I'm innocent this time.
By the way, the very first equation you wrote down, @Jeff, makes no sense with the letters here. What is the partial sum, explicitly? $S_n = \sum_{k=2}^n a_k$ ? Nowhere do I see this correctly written.
@Ted We are given that we are summing from 2, $\sum_{n=2}^{\infty}$ and that $S_n=(n-2)/(n+6)$. I've calculated $a_n = 8/((x+5)(x+6))$ (and checked it several times, including on wolfram). Since we're starting at 2, it seems $a_2$ should equal $S_2$. However, $S_2= (2-2)/(2+6)=0$, but $a_2 = 8/(7 \cdot 8) = 1/7$.
@TedShifrin Yes, we agree. I'm using the letters the prof. gave. I agree we shouldn't use the same letter (which was why I originally, and also incorrectly, used capital $N$).
@TedShifrin For $n \geq 2$. But I don't see why it wouldn't work for $a_2$, too. Every sequence/series I've seen till now always did.
Hello, I have a problem regarding first hitting time for irreducible Markov chains. Someone has given a solution, but I'd like to know other approaches. Thank you.
I am doing some exercises in E-Li-Vanden-Eijnden's Applied Stochastic Analysis and I meet this problem:
(Exercise 3.19) Consider an irreducible Markov chain $\{X_n\}$ on a
finite state space $S$. Let $H\subset S$ and define the first passage
time $T_H=\inf\{n:X_n\in H\}$, $h_i=P_i(T_H<\infty)$. ...
My confusion may solely be in the line "and since I is finitely generated, one can localize at an element h such that $I_h = 0$", but I'd have to see after someone explains that line
the point should be that localization at $\mathfrak{p}$ being $0$ is equivalent to every element being $R\setminus\mathfrak{p}$-torsion, but that + f.g. implies there's a single element of $R\setminus\mathfrak{p}$ annihilating $I$ and then localizing at that gives $0$ too, I think
Let me think about that, although it's not clear to me why $I$ is finitely generated
Wouldn't every element need to be torsion for 'some' element of $R\backslash \mathfrak{p}$ rather than being $R\backslash\mathfrak{p}$-torsion (which atleast by name seems to suggest that every element in $R\backslash\mathfrak{p}$-kills each element)
(Just by the definition of localisation I mean)
Namely having m/1 = 0/1 just means that there exists some s\in S such that sm = 0, rather than needing all elements of S to kill m
Hm? I assume the "standard" orientation on the torus but that is probably not what you mean? Also, since I honestly don't know this, what orientation (that is not dependent on the ambient space) does the outward pointing normal field correspond to?
OK, let's say you orient all the copies of S^1 in S^1 x S^1 x S^1 in the standard way; that gives a product orientation on all these various subtorii. Are we happy with that?
Then the intersection of S^1 x S^1 x pt and pt x pt x S^1 is a positively-oriented point.
If you negate the orientation of the last S^1 factor you get a negatively-oriented point
Locally the picture is the xy-plane intersection z-axis in $\Bbb R^3$.
If you have right hand rule, all those axes are positively oriented and $\Bbb R^3$ gets the product orientation, intersection is +. If you flip the orientation of the z-axis you are down to left-hand rule, so -
Right hand rule just means to me that x, y and z-axes are oriented as $-\infty \to \infty$
As in the title I found critical points to be $(1,0)$-saddle point, $(-1,0)$- local minimum and $(0,0)$ is the one I have problem with.
Second derivative test is inconclusive in this case. If it was a saddle point I should find some curve in the graph that has inflection point at $(0,0)$ to prove...
Let me explain, in my course on differential geometry we define a 2-d manifold to be a normal surface.
We usually take the zero function of that equation to look for the Jacobian to have the maximum rank and said that equation was an (m-p) manifold.
But we consider a triangle on $ \mathbb{R}^n ...
Let me explain, in my course on differential geometry we define a 2-d manifold to be a normal surface.
We usually take the zero function of that equation to look for the Jacobian to have the maximum rank and said that equation was an (m-p) manifold.
But we consider a triangle on $ \mathbb{R}^n ...
