Reference Request – Is there logic with three values ​​whose game semantics matches potentially infinite games?

Consider game trees with the following features:

  • Each node in the tree is one of the following:
    • Verifier Choice: Has one or more children
    • Choice of counterfeiters: Has one or more children
    • No choice: Has exactly one child always selected (could be represented by a verifier or a fake selection, but is considered a separate node type for clarity)
    • Verifier Victory: Is a leaf
    • Counterfeit victory: is a hand
  • The branches of the tree can be infinitely long (ie the tree is not necessarily justified).

These game trees are the perfect piece of information for two players. It is said that a game is a true statement if Verifier has a winning strategy, a false statement if Falsifier has a winning strategy, or an indefinite statement if none of the winning strategies have. It is clear that only an unfounded game tree can make an indefinite statement.

Is there a logic whose game semantics corresponds to the games considered above?


If there is such a logic, it would contain Kleene logic. Kleene logic, however, has no quantifiers, so we would need something more complex.

An interesting example of an indefinite statement would be likely $ {x: x notin x } in {x: x notin x } $ in the naive set theory. In classical logic, this would lead to Russell's paradox. If there is a logic that has been discussed above, this would not be the case. Rather, the statement would consist of an infinite row without selection nodes and thus would have an indeterminate truth value. The same applies to $ {x: x in x } in {x: x in x } $, The statement "The set of all sets contains the set of all sets" would be true, but would consist of a single Verifier victory knot. Other interesting examples arise when you have defined sets of quantifiers (this includes verifier selections and falsifier selectors).

Sum of the (almost) infinite geometric series

I've recently stumbled upon this old problem to prove it $ sum_ {k = 0} ^ { inf} [[dfrac{n+2^k}{2^{k+1}}]]= n $, from where [[x]]denotes the largest integer function of $ x $, One possible solution was that $[[2x]]= [[x]]+ [[x+dfrac{1}{2}]]$but that's not the solution I'm looking for. I remember it had something to do with it $ dfrac {n} {2} + dfrac {n} {4} + dfrac {n} {8} + cdots = n $,

Reference request – for certain infinite products

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I'm interested in knowing how to calculate (or read a reference to) infinite products like the one below:
Dear Sirs, I am interested in knowing how to calculate an infinite number of products of the form below. Inserted into a Mathematica Worksheet (Wolfram Reasearch) I get the nice result
Product = product_ {j = 1} ^ {} j = infty {(1- frac {x} {a + j pi}) ^ 2}. When inserting into a Mathematica worksheet (tungsten research), the following nice formula is returned:

Product = frac { pi ^ 2 Gamma ^ 2 ( Gamma ^ 2 ( frac { pi + a} { pi})} {a ^ 2 Gamma ( frac {ax {} { pi} )) Gamma ( frac {a + x} { pi})},

where Γ (𝑥)
is the gamma function of Euler. The parameters x and a are positive real numbers.

Thank you in advance,

Gustavo.

Linear Algebra – Infinite products with complex numbers or matrices resulting from rank-in-rank embedding

I wonder what kinds of infinite products of matrices, elements of Banach algebras, and complex numbers emerge from rankings in rank.

Suppose that $ lambda $ is a cardinal and $ j_ {1}, dots, j_ {k}: V _ { lambda} rightarrow V _ { lambda} $ are not trivial elementary embeddings. To let $ mathrm {crit} _ {n} (j_ {1}, dots, j_ {k}) $ be that $ n $-th element in the sentence $ { mathrm {crit} (j) midj in langle j_ {1}, dots, j_ {k} rangle } $,

To let $ p_ {n, j_ {1}, dots, j_ {k}} (x_ {1}, dots, x_ {k}) $ denote the defined non-commutative polynomial
$$ 1 + sum {x_ {a_ {1}} dots x_ {a_ {s}} mid mathrm {crit} {j_ {a_ {1}} * dots * j_ {a_ {s}} )
= mathrm {crit} _ {n} (j_ {1}, dots, j_ {k}), $$

$$ mathrm {critical} (j_ {a_ {1}} * dots * j_ {a_ {r}}) < mathrm {critical} _ {n} (j_ {1}, dots, j_ {k} )) , text {for all} , 1 leq r <s }. $$

If the elementary embeddings $ j_ {1}, dots, j_ {k} $ are unique, then we will write $ p_ {n} (x_ {1}, dots, x_ {k}) $ to the $ p_ {n, j_ {1}, dots, j_ {k}} (x_ {1}, dots, x_ {k}) $,

The variables $ x_ {1}, dots, x_ {k} $ do not commute with each other (like that $ x_ {1}, dots, x_ {k} $ should be considered as matrices of elements of a Banach algebra.

The polynomials $ p_ {n} (x_ {1}, dots, x_ {k}) $ fulfill the infinite product formula
$$ lim_ {n rightarrow infty} p_ {n} (x_ {1}, dots, x_ {k}) cdot dots cdot p_ {0} (x_ {1}, dots, x_ { k}) = frac {1} {1- (x_ {1} + dots + x_ {k})}. $$

For example when $ j_ {1} = dots = j_ {k} $, then $ p_ {n} (x_ {1}, dots, x_ {k}) = 1+ (x_ {1} + dots + x_ {k}) ^ {2 ^ {n}} $ for all $ n in omega. $

Can anyone give a nontrivial example of a sequence of different nontrivial elementary embeddings $ j_ {1}, dots, j_ {k} $ along with $ r times r $ matrices $ A_ {1}, dots, A_ {k} $ where if $ B_ {n} = p_ {n} (A_ {1}, dots, A_ {k}), $ then

  1. $ 1- (A_ {1} + dots + A_ {k}) $ is not singular,

  2. There is a closed expression for the sequence $ (B_ {n}) _ {n in omega} $ (especially if every entry in each $ A_ {i} $ is algebraically over $ mathbb {Q} $, then the coefficients in $ B_ {n} $ should be calculable at least in polynomial time),

  3. $$ lim_ {n rightarrow infty} B_ {n} cdot dots cdot B_ {0} = frac {1} {1- (A_ {1} + dots + A_ {k})}, $$

  4. If $ alpha = mathrm {crit} (j_ {i}) $ for some $ i $, then $$ sum {A_ {i} mid 1 leq i leqk, mathrm {crit} (j_ {i}) = alpha } $$ is not singular and

  5. The sequence $ (B_ {n}) _ {n in omega} $ is not identical after all $ 1. $

I hope that conditions 4 and 5 rule out all trivial cases.

I am also interested in some generalizations of this question and am pleased about answers to the general questions. One way to generalize polynomials, for example, is to use an infinite number of variables $ x_ {r} $ corresponding to infinitely many elementary embeddings. Another generalization would be the choice $ A_ {1}, dots, A_ {k} $ from some banach algebra (or another room) or for $ A_ {1}, dots, A_ {k} $ simply be complex numbers. Another way to generalize this question is to use algebraic structures that resemble rank-in-rank embedding in terms of critical points, a composition operation, and so forth, but that actually arise in the algebras of elementary embedding.

Some of my computer calculations indicate that there are these non-trivial infinite products, including the following candidate to answer this question.

Suppose that $ k = 2 $ and $ j: V _ { lambda} rightarrow V _ { lambda} $, To let $ j_ {1} = j * j, j_ {2} = j * j * j $, Then the following sequence is the sequence $ (p_ {0} (x, ix), dots, p_ {15} (x, ix)) $:

$ (i cdot x + 1, x + 1, 1, i cdot x ^ 2 + 1, -x ^ 4-x ^ 3 + 1, i cdot x ^ 3 + 1, x ^ 7 + 1, (-i) cdot x ^ 6 + 1, 1, 1, 1, 1, 1, 1, 1) $,

The value of the polynomial $ p_ {16} (x, ix) $ is unknown, and I do not know if the order of polynomials $ (p_ {n} (x, ix)) _ {n} $ is in some closed form or can be calculated in polynomial from $ n $ Time. The expression for $ (p_ {0} (x, y), dots, p_ {15} (x, y)) $ is 33982 characters long, hence the friendliness of the polynomials $ (p_ {0} (x, ix), dots, p_ {15} (x, ix)) $ is pretty unusual.

We also have $ (p_ {0, j, j * j} (i, 1), points, p_ {11, j, j * j} (i, 1)) = (1 + i, 1, 1 + i, 1 ) 0, 1, 1, 1, 1, 1, 1), $ and

$ (p_ {0, j * j, j * (j * j) * j} (1, i), dots, p_ {15, j * j, j * (j * j) * j} (1, i)) =
(1 + i, 1, 1, 1 + i, 1 + i, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) $

One should be careful, and Laver tables should never be assumed to continue a pattern, as the Laver tables are filled with temporary patterns and even some long-lived patterns that must end up under great cardinal hypotheses.

Fix: "ValueError: Input contains NaN, infinite, or too large a value for dtype (& # 39; float32 & # 39;)."?

import numpy as np
import matplotlib.pyplot as plt
Import pandas as pd
from Pandas import Series, DataFrame
from sklearn.model_selection import train_test_split
From sklearn.metrics import r2_score

train = pd.read_csv (r> C: Users Parth Desktop Data Science Materials Regression Project-5 (Regression Practice – Sales Forecasting) Record Train_UWu5bXk (1) .csv & # 39;)

test = pd.read_csv (r> C: Users Parth Desktop Data Science Materials Regression Project-5 (Regression Practice – Sales Forecasting) Record Train_UWu5bXk (1) .csv & # 39;)

From sklearn.linear_model import LinearRegression
lreg = linear regression ()

X = train[:,[‘Outlet_Establishment_Year’,’Item_MRP’,’Item_Weight’]]

x_train, x_cv, y_train, y_cv = train_test_split (X, train.Item_Outlet_Sales)
lreg.fit (x_train, y_train)

train[‘Item_Weight’].fillna ((train[‘Item_Weight’].mean ()), inplace = true)

numerical integration – infinite sum + integral

If we use Fubini's theorem, or whatever its discrete equivalent, unfounded, we exchange the sum and the integral. First, do the integration:

Accepted,[Element[a | b | c, Reals] && element[m | n, Integers],
Integrate[t^((m - 3)/2) Exp[-t (a^2 + b^2)] Exp[-t (c^2 n^2 + 2 b c n)],
{t, 0, ∞}]// FullSimplify]

ConditionalExpression[(a^2 + (b + c n)^2)^((1 – m)/2) Gamma[1/2 (-1 + m)]m> 1]

Then we make the sum for a certain value of $ m ge2 $:

With[{M}=2Total[{2}m=Sum[{m=2}Summe[{m=2}Sum[(a^2 + (b + c n)^2)^((1 - m)/2) Gamma[1/2 (-1 + m)], {n, -∞, ∞}]]

Sum does not converge.

With[{3}m=sum[{3}m=Sum[{m=3}Summe[{m=3}Sum[(a^2 + (b + c n)^2)^((1 - m)/2) Gamma[1/2 (-1 + m)], {n, -∞, ∞}]]

$ frac { pi left ( left lfloor frac {2 arg (a-i b) -2 arg (c) + pi} {4 pi} right rfloor
+ left lfloor frac {-2 arg (a-i b) + 2 arg (c) + pi} {4 pi} right rfloor right)} {a
c} + frac { pi coth left ( frac { pi a -i pi b} {c} right)} {2 a c} + frac { pi coth
left ( frac { pi a + i pi b} {c} right)} {2 a c} $

With[{M}=4Total[{4}m=Sum[{m=4}Summe[{m=4}Sum[(a^2 + (b + c n)^2)^((1 - m)/2) Gamma[1/2 (-1 + m)], {n, -∞, ∞}]]

(no closed form)

With[{M}=5Total[{M=5}Sum[{m=5}Summe[{m=5}Sum[(a^2 + (b + c n)^2)^((1 - m)/2) Gamma[1/2 (-1 + m)], {n, -∞, ∞}]]

$ frac { pi left lfloor frac {2 arg (a-i b) -2 arg (c) + pi} {4 pi} right rfloor} {2
a ^ 3 c} + frac { pi left lfloor frac {-2 arg (a-i b) +2 arg (c) + pi} {4 pi
} right rfloor} {2 a ^ 3 c} + frac { pi coth left ( frac { pi a -i pi b} {c} right)} {4
a ^ 3 c} + frac { pi coth left ( frac { pi a + i pi b} {c} right)} {4 a ^ 3 c} + frac { pi ^ 2
text {c} ^ 2 left ( frac { pi a -i pi b} {c} right)} {4 a ^ 2 c ^ 2} + frac { pi ^ 2
text {c} ^ 2 left ( frac { pi a + i pi b} {c} right)} {4 a ^ 2 c ^ 2} $

etc.

Machine Learning – For Linear Equation Ax = b: Why is there not more than an infinite number of solutions for a particular b?

I'm new to deep learning. I read the book by Ian Goodfellow entitled Deep Learning. I am in Chapter 1 about linear algebra. There in the section Linear dependence and span they say that for a linear equation ON,x=b There can only be one solution, none or infinitely many. However, there can be no more than one solution and less than an infinite number of solutions. How come?

Here, ON is a matrix b and x are vectors.

GR group theory – Breuer-Guralnick-Kantor conjecture and infinite 3/2 generated groups

A group $ G $ is called $ frac {3} {2} $generated when each non-trivial element is contained in a generation pair, i. H. $$ forall g in G setminus {e }, g exists in G text {so} langle g, g & # 39; Rank = G. $$

See this beautiful poster by Scott Harper.
suggestion: If $ G $ is $ frac {3} {2} $-generated then every correct quotient of $ G $ is cyclic (proof).
guess (B.G.K.): A finite group is $ frac {3} {2} $generated when every correct quotient is cyclic.
sentence (G.K.): Every finite simple group is $ frac {3} {2} $-Generates.

questionCan the above assumption be extended to finitely generated groups?
In other words, is there a (known) counter-example for such groups?

Of course, any simple group that is not finally generated (like the infinite alternating group) $ A _ { infty} $) has any correct quotient cyclically, but it is not $ frac {3} {2} $-Generates.

Formal Grammars – Can I still have infinite recursion if I remove all remaining recursions?

Look at the following rule in the original grammar.
$ quad S to aSb mid bAS $
When you start generating $ S $You can not get rid of it $ S $,
$ quad A to AaA mid bAA mid AAa mid b $ $
If you continue to apply production rules from a form that contains $ A $You can not get rid of it $ A $, either.

These rules are called unproductive rules. Unproductive rules and unattainable rules are called useless rules. You can look at this question.

If there are meaningless rules in the original grammar, there are useless rules after being transformed by a pure method of removing the left recursion. Therefore, you see "there is still an infinite recursion because $ A $ keeps calling. "There is nothing wrong with your conversion to this non-recursive grammar.

These useless production rules can be removed by some algorithms.
We can see that the original grammar becomes empty in the question of such a purification. In fact, the original grammar generates the empty language, the language that does not contain words.