It involved a progression of states, so similar to this in outline. Also, Shannon's work introducing entropy to information theory talks about a similar problem (to better compress texts).
I study the something similar in game theory 2 years ago, I don't remember it but I remember it being something something on progression(we didn't use entropy), but it wasn't exactly the same
@Holo and @KarlKronenfeld here's what I meant previously
sequence is 'abca?' probability that the last letter is 'a' is given by: $P_a = b_a*c_b*a_c*a_a$ probability that the last letter is 'b' is given by: $P_a = b_a*c_b*a_c*b_a$ probability that the last letter is 'c' is given by: $P_a = b_a*c_b*a_c*c_a$
if we consider conditional probabilities ^^^ will be the results
then all we are doing is considering the last term
You can strike some sort of balance and consider previous probabilities based on coming after duos or trios (such as $b_{ca}$ or $b_{bca}$). I do not think using probabilities involving arbitrary-length predecessors is even possible w/ or w/o weights since you cannot sample for them, and if you knew them why are you doing this at all.
The reason I thought of Shannon + entropy is that the techniques should resemble what techniques you need here. The overview article only references probabilities of involving a single symbol, but strings of symbols should work equally well.
Oh, you need to put it in a db? I tried to develop what I did 2 years ago and from the little I remember it will need to keep track of $bn^m$ numbers where $b$ is a large number $n$ is the numbers of possible letters and $m$ is the current length$+1$, so it is no good I guess
I'm specifically interested as to whether there are substantial applications for these isometric embeddings on the riemann sphere so i might do a reference request
What people thought you meant was just whether the equivalence relation $A = P^{-1}DP$ for some $P$ was the same as that but with the inverse swapped, which is the case
no problem? I can tell a you a number of problems the first and most damaging one in Australian society at present day is there is no where that is licensed to sell alcohol yet also allows you to bring your computer for the periods where people are annoying
that way you never get called an asshole because whenever you feel like saying something inappropriate you just go back and do math. its the perfect form of social interaction and it doesn't happen
@Holo sure social interaction with the idiots is well overrated because it sucks, and then all the interesting people never bother with it and the infinite cycle continues. But conversation with interesting people is actually pretty worthwhile I think you'd agree
In all of my heart I hate small talk face to face, but conversation with meaning and with someone that is not an idiot can be worthwhile(I would still prefer it to be not face to face)
anyway I am going thru my notes that were done at some point where I must have believed copying things of Wikipedia aimlessly was effective, and need help with the questions for things I encounter
sure if that's what you prefer there is nothing wrong with that
"A pole of a meromorphic function is a certain type of singularity that behaves like the singularity of $\frac{1}{z^n}$ at $z=0$"
can someone please tell what is used to characterize the degree or "exactness" of this behaviour for me please
ok well definitely stop doing everything after the 48 hr mark if you believe you can keep going at that point, you absolutely can but it will result in a psychosis
possibly anyway. you can go anything up to 10 days without any real consequence but I am just informing you of the risks there
@Adam for $f : B_a(r) \setminus \{a\} \to \Bbb C$, equivalent: 1. $f$ has a pole of degree $n$ at $a$ 2. there is $g : B_a(r) \to \Bbb C$ such that $g(a) \ne 0$ and $g(z) = (z-a)^n f(z)$ 3. the smallest non-zero term in the laurent series of $f$ is the $-n$ term
This is for $N = 7$, and N is always of the form $N = 2^n - 1$ for some $n$.
The product terms (inside the square brackets) increase recursively. I can post the series for $N=15$, if someone is interested in having a go at it. I have given the above example for $N=7$ for clarity.
$$a_n=f([2^{n-1}],[2^{n}]\setminus[2^{n-1}])H(n-1,2^{n-1}),a_0=g(0)$$Where $[n]=\{0,1,...,n-1\}$ and $H(n,m)$ is the sum of $a_0$ to $a_n$ with the inputs "shifted" by $m$(instead of, for example, $g(0)+f(\{0\},\{1\})g(1)...$ you have $g(m+0)+f(\{m+0\},\{m+1\})g(m+1)...$)
It is not exactly what you searched for but maybe this will help @BrownNinja
Hey guys, it's been a long time since I've exercised anything math intensive. I have a problem where I want to merge 2 range of numbers that given a condition will equal [0,1]
I could specify if that isn't anywhere clear enough to understand what I need, but what's an approach I could use for this?
To be more specific: Y is a range from [-2, 2] Z is a range from [-2, 2] I need a function that will return 1 if Y < 0 and Z > 0.5, otherwise gradually down to 0
On the right side there is a link($\LaTeX$ in chat: ...) and they explain there what to do
The first case you gave me(Y<0,Z>0.5) for the second case I took what left from $Y$, the set $[0,2]$ and "squeezed" it to $[0,1]$ by dividing by $2$, and then I for $Z$ I took what is left, the set $[-2,0.5]$, shifted it to $[-2.5,0]$ by subtracting $0.5$ and then "squeezed" it to $[0,1]$ by dividing by $-2.5$. This gives a linear map that goes from $0$ to $1$, and to make it go down(from $1$ to $0$) I took $1$ minus the map
A terminology question, in the characterisitc property of free abelian groups it's stated that given any abelian group $H$ and any map $\varphi : S \to H$ there exists a unique homomorphism $\Phi : \mathbb{Z}S \to H$ extending $\varphi$
But strictly speaking I don't see how $\varphi$ is extended in any literal way, because its domain is the same, its codomain is the same and it's still a map nothing more
Does the author mean that the inclusion homomorphism $i : S \to \mathbb{Z}S$ can be composed with $\Phi$ to yield a homomorphism $\Phi \circ i : S \to H$?
Okay yeah whoops, it's just a set theoretic inclusion
What do authors mean then by "extending" $\varphi$? Do they just mean set theoretic inclusion of $S$ into $\mathbb{Z}S$ 'induces' a homomorphism $\Phi : \mathbb{Z}S \to H$?
Because $\Phi \circ i = \varphi$, but $\varphi$ is just a map and $\Phi$ is a homomorphism (so it's extending it in this sort of way)
hmm... while I cannot find anything with K (other than the K function), it seems number theory did use a lot of sums, products and integrals in their formulae
So in my textbook it says that for some stuff in algebraic topology "we need to extend the notion of rank to finitely generated abelian groups that are not necessarily free abelian". But finitely generated abelian groups have a basis and so are thus free abelian...
Why can't we just define rank for free abelian groups and be done with it?
Ahh okay thanks for those examples, so $\mathbb{Z} /2\mathbb{Z} = \{0, 1\}$ is finitely generated but has no basis, because linear independance fails for the only possible basis sets $\{0\}$, $\{1\}$ and $\{0, 1\}$
But yeah, in general, I found negative abstractions hard to grasp especially when proving something, because for example you can have some set A with some members I, and another set with some members J where J is opposite to I (thus you basically have two objects, A and anti A). However not A is not necessary anti A, it can be B, C, D etc.
Behrend's theorem regarding AP states that
There exists an absolute constant $C$ such that for all sufficiently large integers $N$ there exists a subset $A$ of $\{1, 2, \cdots, N \}$ with atleast $Ne^{-c \sqrt{\ln N}}$ such that no three elements of $A$ forms an arithmetic progression.
Th...
Hello. I am stuck on an easy problem. Given a bin with 50 balls, 20 red and 30 blue. A ball is drawn at random from the bin and then placed back in it. Another draw is made, and the ball drawn is blue, find the probability that this is the first time blue ball has been drawn. So it implies the first ball must be red wit h probability 2/5 and now blue with probabilty 3/5 so the answer is 6/25. Is this correct?
So if i say let $A$ : the first ball is red and $B$: the second ball is blue. Now i need to calculate the conditional probability $P(A|B) = P (A \cap B) / P(B)$ which gives me $(2/5 * 3/5 )/(3/5) = 2/5$
actually, the conditioning on $B$ doesn't make a difference when you phrase it like "given $B$, what's the probability of the 1st ball being red"
this conditioning was important when the question was still in its original form: "given $B$, what's the probability that the second ball was the first blue ball"
in this original form, your first attempt was to calculate the probability without the conditioning
intuitively, you just need to read the inference and ask yourself: would I be willing to believe the conclusion solely from the fact that I believe the assumptions?
or "is it logically impossible for the assumption to be true and yet the conclusion be false?"
So $\frac{\partial}{\partial x} u(x,-y)$ is the derivative in respect to $x$ and compare it to $-\frac{\partial}{\partial y} v(x,-y)$ and then compare $\frac{\partial}{\partial y} u(x,-y)$ to $-(-\frac{\partial}{\partial x} v(x,-y))$
Behrend's theorem regarding AP states that
There exists an absolute constant $C$ such that for all sufficiently large integers $N$ there exists a subset $A$ of $\{1, 2, \cdots, N \}$ with atleast $Ne^{-c \sqrt{\ln N}}$ such that no three elements of $A$ forms an arithmetic progression.
Th...
