How to create an XML sitemap without XML sitemap generators

Hi everyone. I am currently trying to work on my website.

How does the XML language work? Is it basically HTML but different?

I have searched … | Read the rest of https://www.webhostingtalk.com/showthread.php?t=1796424&goto=newpost

bip 32 hd wallets: why the keys generated in Bitcoin Core do not match those of online generators despite using the same seed?

I have the following problem in Bitcoin Core. With this command:

echo carpet begin bacon master draft fortune food cherry cage axis vault clown | 
bx mnemonic-to-seed | 
bx hd-new -v 70615956 | 
bx hd-to-ec -c bx-testnet.cfg | 
sed 's/$/01/' | 
bx base58check-encode -v 239

What I am doing is the mnemonic carpet begin bacon master draft fortune food cherry cage axis vault clown, I am generating a hdseed for bitcoin core in regtest. Results:

cNQQqyR81GZXU3Pa1pK1gQQp6jgni5HyLYM99nL2JpzVRmdCyrFE

Then I use the command:

bitcoin-cli -regtest sethdseed true cNQQqyR81GZXU3Pa1pK1gQQp6jgni5HyLYM99nL2JpzVRmdCyrFE

Where I change the hdseed for which I just bought you and Bitcoin Core generates a group of keys. When I put that same mnemonic on the BIP39 page, it gives me as the root key the same one that gives me the command of this part:

echo carpet begin bacon master draft fortune food cherry cage axis vault clown | 
bx mnemonic-to-seed | 
bx hd-new -v 70615956

which: tprv8ZgxMBicQKsPd5cY6ZZzsRzkcfDANp2y3jFtYNQZsp953sXiFNP9ZFod1fAhkgtCJZGvTcLJVdhFCj7VDNu68nuP4m2QGLcd4JpkteG8Ccc

However, the derivations of private keys generated by Bitcoin Core do not match those of BIP39, nor does the BIP32 root key of BIP39 match the Bitcoin Core private master key, while hdseed if it matches the one that happens in the command. Both derivations, hdseed and the Bitcoin Core master key are obtained with the command bitcoin-cli -regtest dumpwallet test_wallet

I have two big unknowns:

  1. Is the way I am generating hdseed from the mnemonic correct?
  2. Bitcoin Core generates its private master key from hdseed? And if so, is that operation reversible?

But in general I want to know why they are giving different!

Thanks in advance.

nt.number theory – Generators for the cyclic group $ ( mathbb {Z} / p mathbb {Z}) ^ * $

For any prime number $ p $, leave $ ( mathbb {Z} / p mathbb {Z}) ^ * $ denotes the multiplicative group of the field $ ( mathbb {Z} / p mathbb {Z}) $. The group $ ( mathbb {Z} / p mathbb {Z}) ^ * $ it's cyclic

Yes $ A subseteq mathbb {N} $, we let the higher density of $ A $ be defined by $$ mu ^ + (A) = frac {| A cap {1, ldots, n } |} {n}. $$

For any $ n in mathbb {N} $ we left $ p_n $ be the $ n $Prime number For any integer $ k geq 2 $ we consider the whole $ G_k $ from $ n in mathbb {N} $ such that $ p_n> k $ Y $ k $ is a generator of $ ( mathbb {Z} / p_n mathbb {Z}) ^ * $.

What rank do the values ​​have? $ mu ^ + (G_k) $ where $ k geq 2 $ drink?

Theory of the CT category: generators and colimit-closures in higher categories

In ordinary categories $ mathcal {C} $, there are good conditions under which a generator it is also a colimit generator for $ mathcal {C} $. In other words, under appropriate conditions, if $ mathcal {C} $ contains a family of objects $ Xα such that for all $ C in mathcal {C} $ there is an effective epimorphism $ coprod X { alpha} a C $, then the smallest cocomplete subcategory of $ mathcal {C} $ which contains all the $ Xαis $ mathcal {C} $ itself. (See the notes https://home.sandiego.edu/~shulman/papers/generators.pdf by Mike Shulman)

I wonder if something similar applies in the higher categories, where "higher category" could mean anything from $ 2 $-category to a $ infty $-category. Since we have an notion of effective epimorphism in higher categories, we can also say what it means for a collection of objects to generate the category. Under what conditions will these objects generate the category under colimits (iterated)?

characteristic classes – On the proof of Milnor-Novikov's theorem on multiplicative generators of the bordismo ring (complex)

I am trying to understand part of Milnor-Novikov's theorem about multiplicative generators of $ MU_ * cong mathbb Z (x_1, x_2, dots) $ using the "Spectral sequences of Bordsim, Stable Homotopy and Adams of S. Kochman".

Namely define
begin {equation}
gamma_n =
begin {cases}
p & {text {if $ n = p ^ t -1 $ for some cousins ​​$ p $ and positive integers $ t $};
\
1 & {text {otherwise}.
end {cases}
end {equation}

Theorem. (Proposition 4.3.16 in the book)
Element $ x_n in MU _ * $ is an iff polynomial generator
begin {equation}
h (x_n) = pm gamma_n A_n + mathrm {decomposable} in H _ * (MU; mathbb Z).
end {equation}

I have understood the case $ n neq p ^ t-1 $.
Can anyone explain to me what happens next?
Why $ h (x_ {p ^ t-1}) $ Divisible by $ p $ and not divisible by and another cousin $ p ’?

It seems that this is due to the bidegree of $ h (x_ {p ^ t-1}) $ in ASS, but I can't understand why.

Can you suggest a readable source (different from Stong or Ravenel's books) to find proof?

Questions about generators

I have a problem understanding the meaning of a group generated by two elements. For example, yes $ G =$ that is to say $ G $ it is a group generated by two elements, does that mean for any $ g in G $, $ g $ can be represented as $ (ab) i $ where $ i in mathbb {Z} $?

I know for a group $ G =$ means, medium $ G = {b ^ n: n in mathbb {Z} }, $ so yes $ G =$This implies that $ G = {(abc) ^ n: n in mathbb {Z} } $.

Thank you.

abstract algebra: a discrete set of generators

I used this fact in my solution of a problem, but I don't know how to prove it.

Leave $ G $ denote a group of connected lies. If the whole $ X subset G $ is discrete, then the subgroup generated by $ X $ It is also discreet.

It seems very intuitive but I have no idea … Can anyone help me (or give me a counterexample)?

Does the multiplicative group of integers module $ p $ have $ p-1 $ generators?

This is probably a simple question, but I came across this video that said order group $ p $ to have $ p-1 $ generators

But then I thought about the multiplicative module of integer module $ p $, set $ G = {0,1,2, … p-1 } $. Shouldn't the item $ 0 $ Y $ 1 $ both are not considered generators, since $ 0 ^ k = 0, 1 ^ k = 1 $, where $ k en mathbb Z $? Therefore, you will not find all the elements of the group.

abstract algebra – Generators for the positive general linear group

Anyone knows a set of generators for the positive general linear group (it's really just a monoid!):
$$ mathrm {GL} _n ^ + ( mathbb {Z}) = left {M in mathrm {GL} _n ( mathbb {Z}): M_ {ij}> 0, forall i, j right } $$
I think maybe the set of matrices of the form.
$$ M = mI + kE_ {ij} $$
where $ I $ is the identity matrix, $ k geq 0, m geq 1 $ Y $ E_ {ij} $ is the matrix with $ 1 $ at $ ij $ Position and zero elsewhere.

Why does this algebra have an adequate set of generators? (Proving the existence of Hilbert polynomial)

Leave $ A = bigoplus ^ n_ {i = 0} A_i $ be a commutative graduate algebra over the field $ k $.
The Hilbert series of $ A $ it is a series of formal power
$$ H_A (x) = sum_ {i = 0} ^ infty H ^ i_A x ^ i: , = sum ^ infty_ {i = 0} ( dim A_i) x ^ i. $$

I want to show that if $ A $ is generated by $ n $ homogeneous elements $ (a_i) _ {i = 1} ^ n $, there is a polynomial $ P_A in mathbb {Z}[x]$ such that
$$ tag {1} H_A (x) = frac {P_A (x)} { prod ^ n_ {i = 1} (1 -x ^ { deg a_i})}. $$

The standard proof of this fact uses induction in $ n $. Of course if $ n = 0 $, so $ H_A (x) in {1,0 } $ and we are finished. On the other hand, the induction step of the test that I have seen uses algebras. $ B = frac {A} {a_nA} $ Y $ C = ker lambda_ {a_n} = {c in A: ac = 0 } $. I do not see any problem with $ C $ not being unital, since the Hilbert series works for graduate modules as well. However, the test then uses additive decomposition.
$$
H_A (x) = frac {H_B (x) – x ^ { deg a_n} H_C (x)} {1 – x ^ { deg a_n}}
$$

and the hypothesis of induction states that both $ H_B (x) $ Y $ H_C (x) $ have a form $ (1) $ for $ n-1 $ generators $ a_1 $. I understand why this works for $ B $, but I do not understand why this works $ C $. It seems that $ C $ They could have had generators of a different set of degrees.

why $ C = {c in A | ca = 0 } $ It can be generated by elements. $ (c_i) ^ {n-1} i = 1} $ with $ deg c_i = deg a_i $? Where there must be a theorem in a commutative algebra but I do not remember its name. Maybe you remember?

For example, yes $ A = langle x, y | x ^ 2y = 0 rangle $, so
$ B $ is generated by $[x]$ with $ deg [x] = 1 $ in fact, but $ C $ is generated by $ x ^ 2 $ with $ deg x ^ 2 = 2 $, While $ deg x = 1 $ Y $ deg y = 1 $. What's wrong here?

Thanks in advance.