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Pointwise definable and Leibnizian extensions of models of arithmetic and set theory, MOPA seminar CUNY, November 2022

This will be an online talk for the MOPA Seminar at CUNY on 22 November 2022 1pm. Contact organizers for Zoom access.

Abstract. I shall introduce a flexible new method showing that every countable model of PA admits a pointwise definable end-extension, one in which every individual is definable without parameters. And similarly for models of set theory, in which one may also achieve the Barwise extension result—every countable model of ZF admits a pointwise definable end-extension to a model of ZFC+V=L, or indeed any theory arising in a suitable inner model. A generalization of the method shows that every model of arithmetic of size at most continuum admits a Leibnizian extension, and similarly in set theory.

Pointwise definable and Leibnizian models of arithmetic and set theory, realized in end extensions of a given model, Notre Dame Logic Seminar, October 2022

This will be a talk for the Notre Dame logic seminar, 11 October 2022, 2pm in Hales-Healey Hall.

Abstract.  I shall present very new results on pointwise definable and Leibnizian end-extensions of models of arithmetic and set theory. Using the universal algorithm, I shall present a new flexible method showing that every countable model of PA admits a pointwise definable $\Sigma_n$-elementary end-extension. Also, any model of PA of size at most continuum admits an extension that is Leibnizian, meaning that any two distinct points are separated by some expressible property. Similar results hold in set theory, where one can also achieve V=L in the extension, or indeed any suitable theory holding in an inner model of the original model.

Every countable model of arithmetic or set theory has a pointwise definable end extension

• <NAME>, “Every countable model of arithmetic or set theory has a pointwise definable end extension,” mathematics arXiv, 2022.
[Bibtex]
@ARTICLE{Hamkins:Every-countable-model-of-arithmetic-or-set-theory-has-a-pointwise-definable-end-extension,
author = {<NAME>},
title = {Every countable model of arithmetic or set theory has a pointwise definable end extension},
journal = {mathematics arXiv},
year = {2022},
volume = {},
number = {},
pages = {},
month = {},
note = {manuscript under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {10.48550/ARXIV.2209.12578},
eprint = {2209.12578},
archivePrefix={arXiv},
primaryClass={math.LO},
url = {http://jdh.hamkins.org/pointwise-definable-end-extensions},
}

arXiv:2209.12578

Abstract. According to the math tea argument, there must be real numbers that we cannot describe or define, because there are uncountably many real numbers, but only countably many definitions. And yet, the existence of pointwise definable models of set theory, in which every individual is definable without parameters, challenges this conclusion. In this article, I introduce a flexible new method for constructing pointwise definable models of arithmetic and set theory, showing furthermore that every countable model of Zermelo-Fraenkel ZF set theory and of Peano arithmetic PA has a pointwise-definable end extension. In the arithmetic case, I use the universal algorithm and its $\Sigma_n$ generalizations to build a progressively elementary tower making any desired individual $a_n$ definable at each stage $n$, while preserving these definitions through to the limit model, which can thus be arranged to be pointwise definable. A similar method works in set theory, and one can moreover achieve $V=L$ in the extension or indeed any other suitable theory holding in an inner model of the original model, thereby fulfilling the resurrection phenomenon. For example, every countable model of ZF with an inner model with a measurable cardinal has an end extension to a pointwise-definable model of $\text{ZFC}+V=L[\mu]$.

The math tea argument—must there be numbers we cannot describe or define? Pavia Logic Seminar

This will be a talk for the Philosophy Seminar at the IUSS, Scuola Universitaria Superiore Pavia, 28 September 2022.

(Note: This seminar will be held the day before the related conference Philosophy of Mathematics: Foundations, Definitions and Axioms, Italian Network for the Philosophy of Mathematics, 29 September to 1 October 2022. I shall be speaking at that conference on the topic, Fregean abstraction in set theory, a deflationary account.)

Abstract. According to the math tea argument, perhaps heard at a good afternoon tea, there must be some real numbers that we can neither describe nor define, since there are uncountably many real numbers, but only countably many definitions. Is it correct? In this talk, I shall discuss the phenomenon of pointwise definable structures in mathematics, structures in which every object has a property that only it exhibits. A mathematical structure is Leibnizian, in contrast, if any pair of distinct objects in it exhibit different properties. Is there a Leibnizian structure with no definable elements? We shall discuss many interesting elementary examples, eventually working up to the proof that every countable model of set theory has a pointwise definable extension, in which every mathematical object is definable.

Pointwise definable end-extensions of the universe, Sophia 2022, Salzburg

This will be an online talk for the Salzburg Conference for Young Analytical Philosophy, the SOPhiA 2022 Salzburgiense Concilium Omnibus Philosophis Analyticis, with a special workshop session Reflecting on ten years of the set-theoretic multiverse. The workshop will meet Thursday 8 September 2022 4:00pm – 7:30pm.

The name of the workshop (“Reflecting on ten years…”), I was amazed to learn, refers to the period since my 2012 paper, The set-theoretic multiverse, in the Review of Symbolic Logic, in which I had first introduced my arguments and views concerning set-theoretic pluralism. I am deeply honored by this workshop highlighting my work in this way and focussing on the developments growing out of it.

In this talk, I shall engage in that discussion by presenting some very new work connecting several topics that have been prominent in discussions of the set-theoretic multiverse, namely, set-theoretic potentialism and pointwise definability.

Abstract. Using the universal algorithm and its generalizations, I shall present new work on the possibility of end-extending any given countable model of arithmetic or set theory to a pointwise definable model, one in which every object is definable without parameters. Every countable model of Peano arithmetic, for example, admits an end-extension to a pointwise definable model. And similarly, every countable model of ZF set theory admits an end-extension to a pointwise definable model of ZFC+V=L, as well as to pointwise definable models of other sufficient theories, accommodating large cardinals. I shall discuss the philosophical significance of these results in the philosophy of set theory with a view to potentialism and the set-theoretic multiverse.

Definability and the Math Tea argument: must there be numbers we cannot describe or define? University of Warsaw, 22 January 2021

This will be a talk for a new mathematical logic seminar at the University of Warsaw in the Department of Hhilosophy, entitled Epistemic and Semantic Commitments of Foundational Theories, devoted to formal truth theories and implicit commitments of foundational theories as well as their conceptual surroundings.

My talk will be held 22 January 2021, 8 pm CET (7 pm UK), online via Zoom https://us02web.zoom.us/j/83366049995.

Abstract. According to the math tea argument, perhaps heard at a good afternoon tea, there must be some real numbers that we can neither describe nor define, since there are uncountably many real numbers, but only countably many definitions. Is it correct? In this talk, I shall discuss the phenomenon of pointwise definable structures in mathematics, structures in which every object has a property that only it exhibits. A mathematical structure is Leibnizian, in contrast, if any pair of distinct objects in it exhibit different properties. Is there a Leibnizian structure with no definable elements? We shall discuss many interesting elementary examples, eventually working up to the proof that every countable model of set theory has a pointwise definable extension, in which every mathematical object is definable.

Pointwise definable models of set theory

• <NAME>, <NAME>, and <NAME>, “Pointwise definable models of set theory,” Journal of Symbolic Logic, vol. 78, iss. 1, p. 139–156, 2013.
[Bibtex]
@article {HamkinsLinetskyReitz2013:PointwiseDefinableModelsOfSetTheory,
AUTHOR = {Hamkins, <NAME> Linetsky, <NAME>, Jonas},
TITLE = {Pointwise definable models of set theory},
JOURNAL = {Journal of Symbolic Logic},
FJOURNAL = {Journal of Symbolic Logic},
VOLUME = {78},
YEAR = {2013},
NUMBER = {1},
PAGES = {139--156},
ISSN = {0022-4812},
MRCLASS = {03E55},
MRNUMBER = {3087066},
MRREVIEWER = {<NAME>},
DOI = {10.2178/jsl.7801090},
URL = {http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/},
eprint = "1105.4597",
archivePrefix = {arXiv},
primaryClass = {math.LO},
}

When does every definable set have a definable member? CUNY Set Theory Seminar, October 2014

This will be a talk for the CUNY set theory seminar, October 10, 2014, 12pm  GC 6417.

Abstract. Although the concept of being definable’ is not generally expressible in the language of set theory, it turns out that the models of ZF in which every definable nonempty set has a definable element are precisely the models of V=HOD.  Indeed, V=HOD is equivalent to the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, this is not true in the case of $\Sigma_2$-definability, because every model of ZFC has a forcing extension satisfying $V\neq\text{HOD}$ in which every $\Sigma_2$-definable set has an ordinal-definable element.

This is joint work with <NAME> and <NAME>, growing out of some questions and answers on MathOverflow, namely,

Definable collections without definable members
A question asked by Ashutosh five years ago, in which François and I gradually came upon the answer together.
Is it consistent that every definable set has a definable member?
A similar question asked last week by (anonymous) user38200
Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable member?
A question I had regarding the limits of an issue in my answer to the previous question.

In this talk, I shall present the answers to all these questions and place the results in the context of classical results on definability, including a review of basic concepts for graduate students.

Algebraicity and implicit definability in set theory

• <NAME> and <NAME>, “Algebraicity and Implicit Definability in Set Theory,” Notre Dame Journal of Formal Logic, vol. 57, iss. 3, p. 431–439, 2016.
[Bibtex]
@article{HamkinsLeahy2016:AlgebraicityAndImplicitDefinabilityInSetTheory,
author = "<NAME> <NAME>",
doi = "10.1215/00294527-3542326",
fjournal = "Notre Dame Journal of Formal Logic",
journal = "Notre Dame Journal of Formal Logic",
number = "3",
pages = "431--439",
publisher = "Duke University Press",
title = "Algebraicity and Implicit Definability in Set Theory",
volume = "57",
year = "2016",
url = {http://jdh.hamkins.org/algebraicity-and-implicit-definability},
eprint = {1305.5953},
archivePrefix = {arXiv},
primaryClass = {math.LO},
ISSN = {0029-4527},
MRCLASS = {03E47 (03C55)},
MRNUMBER = {3521491},
}`

We aim in this article to analyze the effect of replacing several natural uses of definability in set theory by the weaker model-theoretic notion of algebraicity and its companion concept of implicit definability. In place of the class HOD of hereditarily ordinal definable sets, for example, we consider the class HOA of hereditarily ordinal-algebraic sets. In place of the pointwise definable models of set theory, we examine its (pointwise) algebraic models. And in place of Gödel’s constructible universe L, obtained by iterating the definable power set operation, we introduce the implicitly constructible universe Imp, obtained by iterating the algebraic or implicitly definable power set operation. In each case we investigate how the change from definability to algebraicity affects the nature of the resulting concept. We are especially intrigued by Imp, for it is a new canonical inner model of ZF whose subtler properties are just now coming to light. Open questions about Imp abound.

Before proceeding further, let us review the basic definability definitions. In the model theory of first-order logic, an element $a$ is definable in a structure $M$ if it is the unique object in $M$ satisfying some first-order property $\varphi$ there, that is, if $M\models\varphi[b]$ just in case $b=a$. More generally, an element $a$ is algebraic in $M$ if it has a property $\varphi$ exhibited by only finitely many objects in $M$, so that $\{b\in M \mid M\models\varphi[b]\}$ is a finite set containing $a$. For each class $P\subset M$ we can similarly define what it means for an element to be $P$-definable or $P$-algebraic by allowing the formula $\varphi$ to have parameters from $P$.

In the second-order context, a subset or class $A\subset M^n$ is said to be definable in $M$, if $A=\{\vec a\in M\mid M\models\varphi[\vec a]\}$ for some first-order formula $\varphi$. In particular, $A$ is the unique class in $M^n$ with $\langle M,A\rangle\models\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, in the language where we have added a predicate symbol for $A$. Generalizing this condition, we say that a class $A\subset M^n$ is implicitly definable in $M$ if there is a first-order formula $\psi(A)$ in the expanded language, not necessarily of the form $\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, such that $A$ is unique such that $\langle M,A\rangle\models\psi(A)$. Thus, every (explicitly) definable class is also implicitly definable, but the converse can fail. Even more generally, we say that a class $A\subset M^n$ is algebraic in $M$ if there is a first-order formula $\psi(A)$ in the expanded language such that $\langle M,A\rangle\models\psi(A)$ and there are only finitely many $B\subset M^n$ for which $\langle M,B\rangle\models\psi(B)$. Allowing parameters from a fixed class $P\subset M$ to appear in $\psi$ yields the notions of $P$-definability, implicit $P$-definability, and $P$-algebraicity in $M$. Simplifying the terminology, we say that $A$ is definable, implicitly definable, or algebraic over (rather than in) $M$ if it is $M$-definable, implicitly $M$-definable, or $M$-algebraic in $M$, respectively. A natural generalization of these concepts arises by allowing second-order quantifiers to appear in $\psi$. Thus we may speak of a class $A$ as second-order definable, implicitly second-order definable, or second-order algebraic. Further generalizations are of course possible by allowing $\psi$ to use resources from other strong logics.

The main theorems of the paper are:

Theorem. The class of hereditarily ordinal algebraic sets is the same as the class of hereditarily ordinal definable sets: $$\text{HOA}=\text{HOD}.$$

Theorem. Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+V=HOD.

In the latter part of the paper, we introduce what we view as the natural algebraic analogue of the constructible universe, namely, the implicitly constructible universe, denoted Imp, and built as follows:

$$\text{Imp}_0 = \emptyset$$

$$\text{Imp}_{\alpha + 1} = P_{imp}(\text{Imp}_\alpha)$$

$$\text{Imp}_\lambda = \bigcup_{\alpha < \lambda} \text{Imp}_\alpha, \text{ for limit }\lambda$$

$$\text{Imp} = \bigcup_\alpha \text{Imp}_\alpha.$$

Theorem.  Imp is an inner model of ZF with $L\subset\text{Imp}\subset\text{HOD}$.

Theorem.  It is relatively consistent with ZFC that $\text{Imp}\neq L$.

Theorem. In any set-forcing extension $L[G]$ of $L$, there is a further extension $L[G][H]$ with $\text{gImp}^{L[G][H]}=\text{Imp}^{L[G][H]}=L$.

Open questions about Imp abound. Can $\text{Imp}^{\text{Imp}}$ differ from $\text{Imp}$? Does $\text{Imp}$ satisfy the axiom of choice? Can $\text{Imp}$ have measurable cardinals? Must $0^\sharp$ be in $\text{Imp}$ when it exists? (An affirmative answer arose in conversation with <NAME> and <NAME>, and we hope that $\text{Imp}$ will subsume further large cardinal features. We anticipate a future article on the implicitly constructible universe.)  Which large cardinals are absolute to $\text{Imp}$? Does $\text{Imp}$ have fine structure? Should we hope for any condensation-like principle? Can CH or GCH fail in $\text{Imp}$? Can reals be added at uncountable construction stages of $\text{Imp}$? Can we separate $\text{Imp}$ from HOD? How much can we control $\text{Imp}$ by forcing? Can we put arbitrary sets into the $\text{Imp}$ of a suitable forcing extension? What can be said about the universe $\text{Imp}(\mathbb{R})$ of sets implicitly constructible relative to $\mathbb{R}$ and, more generally, about $\text{Imp}(X)$ for other sets $X$? Here we hope at least to have aroused interest in these questions.

This article arose from a question posed on MathOverflow by my co-author <NAME> and our subsequent engagement with it.

Algebraicity and implicit definability in set theory, CUNY, May 2013

This is a talk May 10, 2013 for the CUNY Set Theory Seminar.

Abstract.  An element a is definable in a model M if it is the unique object in M satisfying some first-order property. It is algebraic, in contrast, if it is amongst at most finitely many objects satisfying some first-order property φ, that is, if { b | M satisfies φ[b] } is a finite set containing a. In this talk, I aim to consider the situation that arises when one replaces the use of definability in several parts of set theory with the weaker concept of algebraicity. For example, in place of the class HOD of all hereditarily ordinal-definable sets, I should like to consider the class HOA of all hereditarily ordinal algebraic sets. How do these two classes relate? In place of the study of pointwise definable models of set theory, I should like to consider the pointwise algebraic models of set theory. Are these the same? In place of the constructible universe L, I should like to consider the inner model arising from iterating the algebraic (or implicit) power set operation rather than the definable power set operation. The result is a highly interesting new inner model of ZFC, denoted Imp, whose properties are only now coming to light. Is Imp the same as L? Is it absolute? I shall answer all these questions at the talk, but many others remain open.

This is joint work with <NAME> (MIT).


text:
# Do all generalised eigenvalue problems with Hermitian $N \times N$ matrices $\mathbf{K}$, $\mathbf{M}$ have $N$ linearly independent eigenvectors?

Consider the generalised eigenvalue problem

$$$$\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}$$$$

where both $$N \times N$$ matrices are Hermitian.

I am trying to prove that one can form an $$\mathbf{M}$$-orthonormal basis using $$N$$ eigenvectors, regardless of whether there exists any repeated eigenvalues. ($$\mathbf{M}$$-orthogonality refers to $$\mathbf{a}^*\mathbf{M}\mathbf{b} = 0$$, and normalisation of eigenvectors is according to $$\mathbf{a}^*\mathbf{M}\mathbf{a} = 1$$)

However, in the process of doing so, I realised that I had been assuming that solving the generalised eigenvalue problem will always spits out $$N$$ linearly independent eigenvectors. Indeed, this is not always true in the general case, as seen for solving the ordinary eigenvalue problem for the non-Hermitian matrix,

$$$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$$$

which has eigenvectors that span only one dimension.

So, for generalised eigenvalue problems for Hermitian matrices $$\mathbf{K}, \mathbf{M}$$, is it possible to prove that it's always possible to obtain $$N$$ linearly independent eigenvectors? If not, under what additional conditions to Hermiticity are required?

With a bit extra poking around, I've found on the Wikipedia page Definite system matrix that, for the case where

• $$\mathbf{K}$$ is an $$N \times N$$ Hermitian matrix,
• $$\mathbf{M}$$ is an $$N\times N$$ Hermitian matrix and positive definite,

both matrices can be diagonalised simultaneously. From this, it can then be shown that $$N$$ linearly independent eigenvectors can be obtained. (Furthermore, it is possible to select the eigenvectors such that they are all $$\mathbf{M}$$-orthogonal.

If $$\mathbf{M}$$ is positive definite, it is invertible, and its inverse $$\mathbf{M}^{-1}$$ is also positive definite. Furthermore, since $$\mathbf{M}$$ is Hermitian, so is its inverse. Every positive definite Hermitian matrix has a Cholesky decomposition, so we can express a Cholesky decomposition for $$\mathbf{M}^{-1}$$:

$$$$\mathbf{M}^{-1} = \mathbf{L}\mathbf{L}^*$$$$

where $$\mathbf{L}$$ is a lower triangular matrix with real positive entries on the diagonal. Therefore, $$\mathbf{L}$$ is also positive definite, and can be inverted. Therefore, we can see that

$$\mathbf{M} = \left(\mathbf{L^{-1}}\right)^* \mathbf{L}^{-1}$$

or, equivalently

$$\mathbf{L}^* \mathbf{M}\mathbf{L} = \mathbf{I} \tag{1}\label{1}$$

This looks reminiscent of a similarity transformation from $$\mathbf{M}$$ to the identity $$\mathbf{I}$$. However, it is not quite, as $$\mathbf{L}$$ is generally not a unitary matrix. We can define a new matrix $$\mathbf{A}$$ by doing the same operation to $$\mathbf{K}$$:

$$\mathbf{L}^* \mathbf{K}\mathbf{L} = \mathbf{A} \tag{2}\label{2}$$

$$\mathbf{A}$$ is generally not diagonal. However, it can be readily shown that since $$\mathbf{K}$$ is Hermitian, then $$\mathbf{A}$$ is Hermitian. (In fact, if we let $$\mathbf{u} = \mathbf{L} \mathbf{v}$$, we can use Equations \eqref{1} & \eqref{2} to convert our generalised eigenvalue problem $$\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}$$ to the standard eigenvalue problem $$\mathbf{A}\mathbf{v} = \lambda \mathbf{v}$$)

Any complex matrix has a Schur decomposition $$\mathbf{A} = \mathbf{Q}\mathbf{T}\mathbf{Q}^*$$, where $$\mathbf{T}$$ is a triangular matrix, and $$\mathbf{Q}$$ is a unitary matrix. However, since in our case $$\mathbf{A}$$ is Hermitian, the triangular matrix is also Hermitian. A Hermitian triangular matrix must be a diagonal matrix! We will represent this matrix with $$\mathbf{D}$$:

$$\mathbf{A} = \mathbf{Q} \mathbf{D} \mathbf{Q}^*$$

or, equivalently,

$$\mathbf{Q}^* \mathbf{A} \mathbf{Q} = \mathbf{D}$$

Trivally, we may also express $$\mathbf{I}$$ as

$$\mathbf{Q}^* \mathbf{I} \mathbf{Q} = \mathbf{I}$$

Substituting Equations \eqref{1} & \eqref{2} into the above, and letting $$\mathbf{U} = \mathbf{L}\mathbf{Q}$$, we can now see that the operations that simulataneously diagonalise $$\mathbf{M}$$ and $$\mathbf{K}$$ are

$$\mathbf{U}^* \mathbf{M}\mathbf{U} = \mathbf{I} \tag{3}\label{3}$$

and

$$\mathbf{U}^* \mathbf{K}\mathbf{U} = \mathbf{D} \tag{4}\label{4}$$

By letting $$\mathbf{u} = \mathbf{U}\mathbf{q}$$, and substituting this and Equations \eqref{3} & \eqref{4}, we obtain the straightforward eigenvalue problem

$$\mathbf{D}\mathbf{q} = \lambda \mathbf{q}$$

whose $$n^\text{th}$$ solution is readily obtained: $$\lambda_n = D_{nn}$$ and $$\mathbf{q}_n = \mathbf{e}_n$$, where $$\mathbf{e}_n$$ is the unit column vector whose $$n^\text{th}$$ element is 1, with all other elements equal to 0.

Then, since $$\mathbf{u} = \mathbf{U}\mathbf{q}$$, we can see that the $$n^\text{th}$$ eigenvector of our original problem has the form

$$\mathbf{u}_n = \mathbf{U}\mathbf{e}_n$$

In other words, the columns of $$\mathbf{U}$$ are the eigenvectors of $$\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}$$.

Since $$\mathbf{U} = \mathbf{L}\mathbf{Q}$$, and $$\mathbf{L}$$ and $$\mathbf{Q}$$ are both invertible, it follows that $$\mathbf{U}$$ must also be invertible. As a result, this means that $$\mathbf{U}$$ must be full rank, which implies that the columns comprising the matrix must all be linearly independent.

Therefore, it is indeed possible to express $$N$$ linearly independent eigenvectors.

(In fact, since $$\mathbf{U}^* \mathbf{M}\mathbf{U} = \mathbf{I}$$, it is also clear that the linearly independent eigenvectors are also $$\mathbf{M}$$-orthogonal. Note that this $$\mathbf{M}$$-orthogonality still applies even between eigenvectors corresponding to repeating eigenvalues.)

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Knuth's up-arrow notation

In mathematics, Knuth's up-arrow notation is a method of notation for very large integers, introduced by <NAME> in 1976.[1] It is closely related to the Ackermann function and especially to the hyperoperation sequence. The idea is based on the fact that multiplication can be viewed as iterated addition and exponentiation as iterated multiplication. Continuing in this manner leads to iterated exponentiation (tetration) and to the remainder of the hyperoperation sequence, which is commonly denoted using Knuth arrow notation.

Introduction

The ordinary arithmetical operations of addition, multiplication and exponentiation are naturally extended into a sequence of hyperoperations as follows.

Multiplication by a natural number is defined as iterated addition:

$\begin{matrix} a\times b & = & \underbrace{a+a+\dots+a} \\ & & b\mbox{ copies of }a \end{matrix}$

For example,

$\begin{matrix} 4\times 3 & = & \underbrace{4+4+4} & = & 12\\ & & 3\mbox{ copies of }4 \end{matrix}$

Exponentiation for a natural power $b$ is defined as iterated multiplication, which Knuth denoted by a single up-arrow:

$\begin{matrix} a\uparrow b= a^b = & \underbrace{a\times a\times\dots\times a}\\ & b\mbox{ copies of }a \end{matrix}$

For example,

$\begin{matrix} 4\uparrow 3= 4^3 = & \underbrace{4\times 4\times 4} & = & 64\\ & 3\mbox{ copies of }4 \end{matrix}$

To extend the sequence of operations beyond exponentiation, Knuth defined a “double arrow” operator to denote iterated exponentiation (tetration):

$\begin{matrix} a\uparrow\uparrow b & = {\ ^{b}a} = & \underbrace{a^{a^{{}^{.\,^{.\,^{.\,^a}}}}}} & = & \underbrace{a\uparrow (a\uparrow(\dots\uparrow a))} \\ & & b\mbox{ copies of }a\uparrow & & b\mbox{ copies of }a\uparrow \end{matrix}$

For example,

$\begin{matrix} 4\uparrow\uparrow 3 & = {\ ^{3}4} = & \underbrace{4^{4^4}} & = & \underbrace{4\uparrow (4\uparrow 4)} & = & 4^{256} & \approx & 1.34078079\times 10^{154}& \\ & & 3\mbox{ copies of }4\uparrow & & 3\mbox{ copies of }4\uparrow \end{matrix}$

Here and below evaluation is to take place from right to left, as Knuth's arrow operators (just like exponentiation) are defined to be right-associative.

According to this definition,

$3\uparrow\uparrow 2=3^3=27$
$3\uparrow\uparrow 3=3^{3^3}=3^{27}=7625597484987$
$3\uparrow\uparrow 4=3^{3^{3^3}}=3^{3^{27}}=3^{7625597484987}$
$3\uparrow\uparrow 5=3^{3^{3^{3^3}}}=3^{3^{3^{27}}}=3^{3^{7625597484987}}$
etc.

This already leads to some fairly large numbers, but Knuth extended the notation. He went on to define a “triple arrow” operator for iterated application of the “double arrow” operator (also known as pentation):

$\begin{matrix} a\uparrow\uparrow\uparrow b= & \underbrace{a_{}\uparrow\uparrow (a\uparrow\uparrow(\dots\uparrow\uparrow a))}\\ & b\mbox{ copies of }a \end{matrix}$

followed by a 'quadruple arrow' operator (also known as hexation):

$\begin{matrix} a\uparrow\uparrow\uparrow\uparrow b= & \underbrace{a_{}\uparrow\uparrow\uparrow (a\uparrow\uparrow\uparrow(\dots\uparrow\uparrow\uparrow a))}\\ & b\mbox{ copies of }a \end{matrix}$

and so on. The general rule is that an $n$-arrow operator expands into a right-associative series of ($n - 1$)-arrow operators. Symbolically,

$\begin{matrix} a\ \underbrace{\uparrow_{}\uparrow\!\!\dots\!\!\uparrow}_{n}\ b= \underbrace{a\ \underbrace{\uparrow\!\!\dots\!\!\uparrow}_{n-1} \ (a\ \underbrace{\uparrow_{}\!\!\dots\!\!\uparrow}_{n-1} \ (\dots \ \underbrace{\uparrow_{}\!\!\dots\!\!\uparrow}_{n-1} \ a))}_{b\text{ copies of }a} \end{matrix}$

Examples:

$3\uparrow\uparrow\uparrow2 = 3\uparrow\uparrow3 = 3^{3^3} = 3^{27}=7,625,597,484,987$
$\begin{matrix} 3\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow3\uparrow\uparrow3 = 3\uparrow\uparrow(3\uparrow3\uparrow3) = & \underbrace{3_{}\uparrow 3\uparrow\dots\uparrow 3} \\ & 3\uparrow3\uparrow3\mbox{ copies of }3 \end{matrix} \begin{matrix} = & \underbrace{3_{}\uparrow 3\uparrow\dots\uparrow 3} \\ & \mbox{7,625,597,484,987 copies of 3} \end{matrix}=\underbrace{3^{3^{3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}}}}_{7,625,597,484,987}$

The notation $a \uparrow^n b$ is commonly used to denote $a \uparrow\uparrow \dots \uparrow b$ with n arrows.

Notation

In expressions such as $a^b$, the notation for exponentiation is usually to write the exponent $b$ as a superscript to the base number $a$. But many environments — such as programming languages and plain-text e-mail — do not support superscript typesetting. People have adopted the linear notation $a \uparrow b$ for such environments; the up-arrow suggests 'raising to the power of'. If the character set doesn't contain an up arrow, the caret ^ is used instead.

The superscript notation $a^b$ doesn't lend itself well to generalization, which explains why Knuth chose to work from the inline notation $a \uparrow b$ instead.

$a \uparrow^n b$ is a shorter alternative notation for n uparrows. Thus $a \uparrow^4 b = a \uparrow \uparrow \uparrow \uparrow b$.

Writing out up-arrow notation in terms of powers

Attempting to write $a \uparrow \uparrow b$ using the familiar superscript notation gives a power tower.

For example: $a \uparrow \uparrow 4 = a \uparrow (a \uparrow (a \uparrow a)) = a^{a^{a^a}}$

If b is a variable (or is too large), the power tower might be written using dots and a note indicating the height of the tower.

$a \uparrow \uparrow b = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{b}$

Continuing with this notation, $a \uparrow \uparrow \uparrow b$ could be written with a stack of such power towers, each describing the size of the one above it.

$a \uparrow \uparrow \uparrow 4 = a \uparrow \uparrow (a \uparrow \uparrow (a \uparrow \uparrow a)) = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{a} }}$

Again, if b is a variable or is too large, the stack might be written using dots and a note indicating its height.

$a \uparrow \uparrow \uparrow b = \left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} b$

Furthermore, $a \uparrow \uparrow \uparrow \uparrow b$ might be written using several columns of such stacks of power towers, each column describing the number of power towers in the stack to its left:

$a \uparrow \uparrow \uparrow \uparrow 4 = a \uparrow \uparrow \uparrow (a \uparrow \uparrow \uparrow (a \uparrow \uparrow \uparrow a)) = \left.\left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} a$

And more generally:

$a \uparrow \uparrow \uparrow \uparrow b = \underbrace{ \left.\left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \cdots \right\} a }_{b}$

This might be carried out indefinitely to represent $a \uparrow^n b$ as iterated exponentiation of iterated exponentiation for any a, n and b (although it clearly becomes rather cumbersome).

Using tetration

The tetration notation $^{b}a$ allows us to make these diagrams slightly simpler while still employing a geometric representation (we could call these tetration towers).

$a \uparrow \uparrow b = { }^{b}a$
$a \uparrow \uparrow \uparrow b = \underbrace{^{^{^{^{^{a}.}.}.}a}a}_{b}$
$a \uparrow \uparrow \uparrow \uparrow b = \left. \underbrace{^{^{^{^{^{a}.}.}.}a}a}_{ \underbrace{^{^{^{^{^{a}.}.}.}a}a}_{ \underbrace{\vdots}_{a} }} \right\} b$

Finally, as an example, the fourth Ackermann number $4 \uparrow^4 4$ could be represented as:

$\underbrace{^{^{^{^{^{4}.}.}.}4}4}_{ \underbrace{^{^{^{^{^{4}.}.}.}4}4}_{ \underbrace{^{^{^{^{^{4}.}.}.}4}4}_{4} }} = \underbrace{^{^{^{^{^{4}.}.}.}4}4}_{ \underbrace{^{^{^{^{^{4}.}.}.}4}4}_{ ^{^{^{4}4}4}4 }}$

Generalizations

Some numbers are so large that multiple arrows of Knuth's up-arrow notation become too cumbersome; then an n-arrow operator $\uparrow^n$ is useful (and also for descriptions with a variable number of arrows), or equivalently, hyper operators.

Some numbers are so large that even that notation is not sufficient. Graham's number is an example. The Conway chained arrow notation can then be used: a chain of three elements is equivalent with the other notations, but a chain of four or more is even more powerful.

$\begin{matrix} a\uparrow^n b & = & \mbox{hyper}(a,n+2,b) & = & a\to b\to n \\ \mbox{(Knuth)} & & & & \mbox{(Conway)} \end{matrix}$

It is generally suggested that Knuth's arrow should be used for smaller magnitude numbers, and the chained arrow or hyper operators for larger ones.

Definition

The up-arrow notation is formally defined by

$a\uparrow^n b= \left\{ \begin{matrix} a^b, & \mbox{if }n=1; \\ 1, & \mbox{if }b=0; \\ a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise} \end{matrix} \right.$

for all integers $a, b, n$ with $b \ge 0, n \ge 1$.

All up-arrow operators (including normal exponentiation, $a \uparrow b$) are right associative, i.e. evaluation is to take place from right to left in an expression that contains two or more such operators. For example, $a \uparrow b \uparrow c = a \uparrow (b \uparrow c)$, not $(a \uparrow b) \uparrow c$; for example
$3\uparrow\uparrow 3=3^{3^3}$ is $3^{(3^3)}=3^{27}=7625597484987$ not $\left(3^3\right)^3=27^3=19683.$

There is good reason for the choice of this right-to-left order of evaluation. If we used left-to-right evaluation, then $a \uparrow\uparrow b$ would equal $a \uparrow (a \uparrow (b - 1))$, so that $\uparrow\uparrow$ would not be an essentially new operation. Right associativity is also natural because we can rewrite the iterated arrow expression $a\uparrow^n\cdots\uparrow^na$ that appears in the expansion of $a \uparrow^{n + 1}b$ as $a\uparrow^n\cdots\uparrow^na\uparrow^n1$, so that all the $a$s appear as left operands of arrow operators. This is significant since the arrow operators are not commutative.

Writing $(a\uparrow ^m)^b$ for the bth functional power of the function $f(x)=a\uparrow ^m x$ we have $a\uparrow ^n b = (a\uparrow ^{n-1})^b 1$.

The definition could be extrapolated one step, starting with $a\uparrow^n b= ab$ if n = 0, because exponentiation is repeated multiplication starting with 1. Extrapolating one step more, writing multiplication as repeated addition, is not as straightforward because multiplication is repeated addition starting with 0 instead of 1. "Extrapolating" again one step more, writing addition of n as repeated addition of 1, requires starting with the number a. Compare the definition of the hyper operator, where the starting values for addition and multiplication are also separately specified.

Tables of values

Computing 2↑mn

Computing $2\uparrow^m n$ can be restated in terms of an infinite table. We place the numbers $2^n$ in the top row, and fill the left column with values 2. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $2\uparrow^m n$ = hyper(2, m + 2, n) = 2 → n → m
m\n 1 2 3 4 5 6 formula
1 2 4 8 16 32 64 $2^n$
2 2 4 16 65536 $2^{65536}\approx 2.0 \times 10^{19,729}$ $2^{2^{65536}}\approx 10^{6.0 \times 10^{19,728}}$ $2\uparrow\uparrow n$
3 2 4 65536 $\begin{matrix} \underbrace{2_{}^{2^{{}^{.\,^{.\,^{.\,^2}}}}}} \\ 65536\mbox{ copies of }2 \end{matrix}$     $2\uparrow\uparrow\uparrow n$
4 2 4 $\begin{matrix} \underbrace{2_{}^{2^{{}^{.\,^{.\,^{.\,^2}}}}}}\\ 65536\mbox{ copies of }2 \end{matrix}$       $2\uparrow\uparrow\uparrow\uparrow n$

The table is the same as that of the Ackermann function, except for a shift in $m$ and $n$, and an addition of 3 to all values.

Computing 3↑mn

We place the numbers $3^n$ in the top row, and fill the left column with values 3. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $3\uparrow^m n$ = hyper(3, m + 2, n) = 3 → n → m
m\n 1 2 3 4 5 formula
1 3 9 27 81 243 $3^n$
2 3 27 7,625,597,484,987 $3^{7{,}625{,}597{,}484{,}987}$   $3\uparrow\uparrow n$
3 3 7,625,597,484,987 $\begin{matrix} \underbrace{3_{}^{3^{{}^{.\,^{.\,^{.\,^3}}}}}}\\ 7{,}625{,}597{,}484{,}987\mbox{ copies of }3 \end{matrix}$     $3\uparrow\uparrow\uparrow n$
4 3 $\begin{matrix} \underbrace{3_{}^{3^{{}^{.\,^{.\,^{.\,^3}}}}}}\\ 7{,}625{,}597{,}484{,}987\mbox{ copies of }3 \end{matrix}$       $3\uparrow\uparrow\uparrow\uparrow n$

Computing 10↑mn

We place the numbers $10^n$ in the top row, and fill the left column with values 10. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $10\uparrow^m n$ = hyper(10, m + 2, n) = 10 → n → m
m\n 1 2 3 4 5 formula
1 10 100 1,000 10,000 100,000 $10^n$
2 10 10,000,000,000 $10^{10,000,000,000}$ $10^{10^{10,000,000,000}}$ $10^{10^{10^{10,000,000,000}}}$ $10\uparrow\uparrow n$
3 10 $\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10\mbox{ copies of }10 \end{matrix}$ $\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10\mbox{ copies of }10 \end{matrix}$ $\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10\mbox{ copies of }10 \end{matrix}$   $10\uparrow\uparrow\uparrow n$
4 10 $\begin{matrix} \underbrace{^{^{^{^{^{10}.}.}.}10}10}\\ 10\mbox{ copies of }10 \end{matrix}$ $\begin{matrix} \underbrace{^{^{^{^{^{10}.}.}.}10}10}\\ \underbrace{^{^{^{^{^{10}.}.}.}10}10}\\ 10\mbox{ copies of }10 \end{matrix}$     $10\uparrow\uparrow\uparrow\uparrow n$

Note that for 2 ≤ n ≤ 9 the numerical order of the numbers $10\uparrow^m n$ is the lexicographical order with m as the most significant number, so for the numbers of these 8 columns the numerical order is simply line-by-line. The same applies for the numbers in the 97 columns with 3 ≤ n ≤ 99, and if we start from m = 1 even for 3 ≤ n ≤ 9,999,999,999.

Numeration systems based on the hyperoperation sequence

<NAME>,[2] with a system of notation different from Knuth arrows, used the sequence of hyperoperators here denoted by $( +, \ \times, \ \uparrow, \ \uparrow\uparrow, \ \dots)\,\!$ to create systems of numeration for the nonnegative integers. Letting superscripts $\quad ( ^{(1)}, \ ^{(2)}, \ ^{(3)}, \ ^{(4)}, \ \dots )\,\!$ denote the respective hyperoperators $( +, \ \times, \ \uparrow, \ \uparrow\uparrow, \ \dots)\,\!$, the so-called complete hereditary representation of integer n, at level k and base b, can be expressed as follows using only the first k hyperoperators and using as digits only 0, 1, ..., b-1:

• For 0 ≤ nb-1, n is represented simply by the corresponding digit.
• For n > b-1, the representation of n is found recursively, first representing n in the form
$b^{(k)}{x_k}^{(k-1)}{x_{k-1}}^{(k-2)} \dots {x_2}^{(1)}x_1$
where xk, ..., x1 are the largest integers satisfying (in turn)
$b^{(k)}x_k \le n$
$b^{(k)}{x_k}^{(k-1)}x_{k-1} \le n$
...
$b^{(k)}{x_k}^{(k-1)}{x_{k-1}}^{(k-2)} \dots {x_2}^{(1)}x_1 \le n$.
Any xi exceeding b-1 is then re-expressed in the same manner, and so on, repeating this procedure until the resulting form contains only the digits 0, 1, ..., b-1.

The remainder of this section will use $( +, \ \times, \ \uparrow, \ \uparrow\uparrow, \ \uparrow\uparrow\uparrow, \ \dots)$, rather than superscripts, to denote the hyperoperators.

Unnecessary parentheses can be avoided by giving higher-level operators higher precedence in the order of evaluation; thus,

level-1 representations have the form $b + X$, with X also of this form;

level-2 representations have the form $b \times X + Y$, with X,Y also of this form;

level-3 representations have the form $b \uparrow X \times Y + Z$, with X,Y,Z also of this form;

level-4 representations have the form $b \uparrow\uparrow X \uparrow Y \times Z + T$, with X,Y,Z,T also of this form;

and so on.

The representations can be abbreviated by omitting any instances of $+0, \ \times1, \uparrow1, \ \uparrow\uparrow1,$ etc.; for example, the level-3 base-2 representation of the number 6 is $2\uparrow(2\uparrow1\times1+0)\times1+(2\uparrow1\times1+0)$, which abbreviates to $2 \uparrow 2 + 2$.

Examples: The unique base-2 representations of the number 266, at levels 1, 2, 3, 4, and 5 are as follows:

$\text{Level 1:} \ \ 266 = 2 + 2 + \dots + 2 \ \ \text{(with 133 2s)}$
$\text{Level 2:} \ \ 266 = 2 \times (2 \times (2 \times (2 \times 2 \times 2 \times 2 \times 2 + 1)) + 1)$
$\text{Level 3:} \ \ 266 = 2 \uparrow 2 \uparrow (2 + 1) + 2 \uparrow (2 + 1) + 2$
$\text{Level 4:} \ \ 266 = 2 \uparrow\uparrow (2 + 1) \uparrow 2 + 2 \uparrow\uparrow 2 \times 2 + 2$
$\text{Level 5:} \ \ 266 = 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 2 + 2 \uparrow\uparrow\uparrow 2 \times 2 + 2$.


text:
Radian Standard: SI derived unit Quantity: Angle Symbol: rad Symbol3: r Extralabel: In units Extradata: Dimensionless with an arc length equal to the radius, i.e. 1 Units1: milliradians Inunits1: 1000 mrad Units2: turns Inunits2: turn Units3: degrees Inunits3: ≈ 57.296° Units4: gradians Inunits4: ≈ 63.662g

The radian is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends; one radian is just under 57.3 degrees (expansion at). The unit was formerly an SI supplementary unit, but this category was abolished in 1995 and the radian is now considered an SI derived unit.[1]

The symbol rad can be used to represent the radian (for example, an angle of radians would be written as), but this is often omitted, especially in mathematical writing.

Definition

Radian describes the plane angle subtended by a circular arc as the length of the arc divided by the radius of the arc. One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle. More generally, the magnitude in radians of such a subtended angle is equal to the ratio of the arc length to the radius of the circle; that is,, where θ is the subtended angle in radians, s is arc length, and r is radius. Conversely, the length of the enclosed arc is equal to the radius multiplied by the magnitude of the angle in radians; that is, .

As the ratio of two lengths, the radian is a "pure number" that needs no unit symbol, and in mathematical writing the symbol "rad" is almost always omitted. When quantifying an angle in the absence of any symbol, radians are assumed, and when degrees are meant the symbol ° is used. The radian is defined as 1.[2]

It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or, or 2. Thus 2 radians is equal to 360 degrees, meaning that one radian is equal to 180/ degrees.

The relation

can be derived using the formula for arc length. Taking the formula for arc length, or

\ellarc=2\pir\left(

 \theta 360\circ

\right)

. Assuming a unit circle; the radius is therefore one. Knowing that the definition of radian is the measure of an angle that subtends an arc of a length equal to the radius of the circle, we know that

\right)

. This can be further simplified to
. Multiplying both sides by

360\circ

gives

.

History

The concept of radian measure, as opposed to the degree of an angle, is normally credited to <NAME> in 1714.[3] [4] He described the radian in everything but name, and he recognized its naturalness as a unit of angular measure. Prior to the term radian becoming widespread, the unit was commonly called circular measure of an angle.[5]

The idea of measuring angles by the length of the arc was already in use by other mathematicians. For example, al-Kashi (c. 1400) used so-called diameter parts as units where one diameter part was radian and they also used sexagesimal subunits of the diameter part.[6]

The term radian first appeared in print on 5 June 1873, in examination questions set by <NAME> (brother of <NAME>) at Queen's College, Belfast. He had used the term as early as 1871, while in 1869, <NAME>, then of the University of St Andrews, vacillated between the terms rad, radial, and radian. In 1874, after a consultation with <NAME>, Muir adopted radian.[7] [8] [9] The name radian was not universally adopted for some time after this. Longmans' School Trigonometry still called the radian circular measure when published in 1890.[10]

Unit symbol

The International Bureau of Weights and Measures[11] and International Organization for Standardization[12] specify rad as the symbol for the radian. Alternative symbols used 100 years ago are c (the superscript letter c, for "circular measure"), the letter r, or a superscript, but these variants are infrequently used as they may be mistaken for a degree symbol (°) or a radius (r). So, for example, a value of 1.2 radians would most commonly be written as 1.2 rad; other notations include 1.2 r, 1.2, 1.2, or 1.2.

Conversions

Conversion between radians and degrees

As stated, one radian is equal to 180/ degrees. Thus, to convert from radians to degrees, multiply by 180/.

 180\circ \pi

For example:

 180\circ \pi

57.2958\circ

 180\circ \pi

143.2394\circ

 \pi 3

 \pi 3

 180\circ \pi

=60\circ

Conversely, to convert from degrees to radians, multiply by /180.

 \pi 180\circ

For example:

1\circ=1

 \pi 180\circ

23\circ=23

 \pi 180\circ

Radians can be converted to turns (complete revolutions) by dividing the number of radians by 2.

Radian to degree conversion derivation

The length of circumference of a circle is given by

2\pir

, where

r

is the radius of the circle.

So the following equivalent relation is true:

360\circ\iff2\pir

[Since a $360^\circ$ sweep is needed to draw a full circle]

By the definition of radian, a full circle represents:

 2\pir r

Combining both the above relations:

 360\circ 2\pi

 180\circ \pi

2\pi

radians equals one turn, which is by definition 400 gradians (400 gons or 400g). So, to convert from radians to gradians multiply by

200/\pi

, and to convert from gradians to radians multiply by

\pi/200

. For example,

 200g \pi

76.3944g

50g=50

 \pi 200g

In calculus and most other branches of mathematics beyond practical geometry, angles are universally measured in radians. This is because radians have a mathematical "naturalness" that leads to a more elegant formulation of a number of important results.

Most notably, results in analysis involving trigonometric functions are simple and elegant when the functions' arguments are expressed in radians. For example, the use of radians leads to the simple limit formula

\limh

 \sinh h

=1,

which is the basis of many other identities in mathematics, including

 d dx

\sinx=\cosx

 d2 dx2

\sinx=-\sinx.

Because of these and other properties, the trigonometric functions appear in solutions to mathematical problems that are not obviously related to the functions' geometrical meanings (for example, the solutions to the differential equation

 d2y dx2

=-y

, the evaluation of the integral

\int

 dx 1+x2

, and so on). In all such cases it is found that the arguments to the functions are most naturally written in the form that corresponds, in geometrical contexts, to the radian measurement of angles.

The trigonometric functions also have simple and elegant series expansions when radians are used; for example, the following Taylor series for sin x :

\sinx=x-

 x3 3!

+

 x5 5!

-

 x7 7!

+.

If x were expressed in degrees then the series would contain messy factors involving powers of /180: if x is the number of degrees, the number of radians is, so

 \pi 180

x-\left(

 \pi 180

\right

 3 x3 3!
)

+\left(

 \pi 180

\right

 5 x5 5!
)

-\left(

 \pi 180

\right

 7 x7 7!
)

+.

Mathematically important relationships between the sine and cosine functions and the exponential function (see, for example, Euler's formula) are, again, elegant when the functions' arguments are in radians and messy otherwise.

Dimensional analysis

Although the radian is a unit of measure, it is a dimensionless quantity. This can be seen from the definition given earlier: the angle subtended at the centre of a circle, measured in radians, is equal to the ratio of the length of the enclosed arc to the length of the circle's radius. Since the units of measurement cancel, this ratio is dimensionless.

Although polar and spherical coordinates use radians to describe coordinates in two and three dimensions, the unit is derived from the radius coordinate, so the angle measure is still dimensionless.[13]

Use in physics

The radian is widely used in physics when angular measurements are required. For example, angular velocity is typically measured in radians per second (rad/s). One revolution per second is equal to 2 radians per second.

Similarly, angular acceleration is often measured in radians per second per second (rad/s2).

For the purpose of dimensional analysis, the units of angular velocity and angular acceleration are s−1 and s−2 respectively.

Likewise, the phase difference of two waves can also be measured in radians. For example, if the phase difference of two waves is (k⋅2) radians, where k is an integer, they are considered in phase, whilst if the phase difference of two waves is, where k is an integer, they are considered in antiphase.

SI multiples

Metric prefixes have limited use with radians, and none in mathematics. A milliradian (mrad) is a thousandth of a radian and a microradian (μrad) is a millionth of a radian, i.e. .

There are 2 × 1000 milliradians (≈ 6283.185 mrad) in a circle. So a trigonometric milliradian is just under of a circle. This “real” trigonometric unit of angular measurement of a circle is in use by telescopic sight manufacturers using (stadiametric) rangefinding in reticles. The divergence of laser beams is also usually measured in milliradians.

An approximation of the trigonometric milliradian (0.001 rad) is used by NATO and other military organizations in gunnery and targeting. Each angular mil represents of a circle and is % or 1.875% smaller than the trigonometric milliradian. For the small angles typically found in targeting work, the convenience of using the number 6400 in calculation outweighs the small mathematical errors it introduces. In the past, other gunnery systems have used different approximations to ; for example Sweden used the streck and the USSR used . Being based on the milliradian, the NATO mil subtends roughly 1 m at a range of 1000 m (at such small angles, the curvature is negligible).

Smaller units like microradians (μrad) and nanoradians (nrad) are used in astronomy, and can also be used to measure the beam quality of lasers with ultra-low divergence. More common is arc second, which is  rad (around 4.8481 microradians). Similarly, the prefixes smaller than milli- are potentially useful in measuring extremely small angles.

Notes and References

1. Web site: Resolution 8 of the CGPM at its 20th Meeting (1995) . 2014-09-23 . Bureau International des Poids et Mesures.
2. ISO 80000-3:2006
3. Web site: Biography of Roger Cotes . The MacTutor History of Mathematics . February 2005 . <NAME>. . <NAME>. . Robertson.
4. Roger Cotes died in 1716. By 1722, his cousin <NAME> had collected and published Cotes' mathematical writings in a book, Harmonia mensurarum … . In a chapter of editorial comments by Smith, he gives, for the first time, the value of one radian in degrees. See: Roger Cotes with <NAME>, ed., Harmonia mensurarum … (Cambridge, England: 1722), chapter: Editoris notæ ad Harmoniam mensurarum, top of page 95. From page 95: After stating that 180° corresponds to a length of (3.14159…) along a unit circle (i.e., radians), Smith writes: "Unde Modulus Canonis Trigonometrici prodibit 57.2957795130 &c. " (Whence the unit of trigonometric measure, 57.2957795130… [degrees per radian], will appear.)
5. <NAME>, Plane Trigonometry: For the Use of Colleges and Schools, p. 10, Cambridge and London: MacMillan, 1864
6. Book: Luckey, <NAME>. . Siggel. Berlin. Akademie Verlag. Translation of 1424 book. 1953. Der Lehrbrief über den kreisumfang von Gamshid b. Mas'ud al-Kasi. Treatise on the Circumference of al-Kashi. 6. 40.
7. Book: Cajori, Florian. <NAME>. 1929. History of Mathematical Notations. 2. 147–148. Dover Publications. 0-486-67766-4. registration.
8. Nature. 1910. 83. 156. 10.1038/083156a0. The Term "Radian" in Trigonometry. Muir. Thos.. 2110. 1910Natur..83..156M . Nature. 1910. 83. 217. 10.1038/083217c0. The Term "Radian" in Trigonometry. <NAME>. 2112. 1910Natur..83..217T . Nature. 1910. 83. 459–460. 10.1038/083459d0. The Term "Radian" in Trigonometry. Muir. Thos.. 2120. 1910Natur..83..459M .
9. Web site: Earliest Known Uses of Some of the Words of Mathematics. Nov 23, 2009. Sep 30, 2011. <NAME> .
10. <NAME>, Longmans' School Trigonometry, p. 6, London: Longmans, Green, and Co., 1890 (1891 edition)
11. https://www.bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9.pdf 2019 BIPM Brochure
12. https://www.iso.org/standard/31888.html ISO 80000-3:2006 Quantities and Units - Space and Time
13. For a debate on this meaning and use see:10.1119/1.18616. Angles—Let's treat them squarely. 1997. Brownstein. K. R.. American Journal of Physics. 65. 7. 605–614. 1997AmJPh..65..605B .,Angles as a fourth fundamental quantity. 1962. Romain. J.E.. Journal of Research of the National Bureau of Standards Section B. 66B. 3. 97. 10.6028/jres.066B.012. free.,10.1119/1.18964. Dimensional angles and universal constants. 1998. LéVy-Leblond. Jean-Marc. American Journal of Physics. 66. 9. 814–815. 1998AmJPh..66..814L ., and 10.1119/1.19185. Units—SI-Only, or Multicultural Diversity?. 1999. <NAME>.. American Journal of Physics. 67. 1. 13–16 . 1999AmJPh..67...13R .


text:
# Application of Vibrato Montecarlo methods

Ciao, I was studying Vibrato Montecarlo methods and I came up with a very simple question: what is an real application of this method? Let me explain.

In short the main idea of the method is the following: suppose you want to compute the sensitivities of a derivative with a discontinuous payoff then you should try the following:

1. Discretize time with pillars $t_1, \dots, t_N$.
2. Suppose you know the value of the stock at time $t_{N-1}$ and, by knowing it, analitically integrate the payoff function as $$V(S_{N-1}) = \mathbb{E}[f(S_N) | S_{N-1}]$$
3. $V(S_{N-1})$ is still a stochastic quantity depending on $S_{N-1}$ but know it is a differentiable function (essentially the integration works as molllification of the starting discontinuous payoff). At this point you can compute the sensitivity w.r.t. parameter $\lambda$ by using classical finite difference and Montecarlo methods applied to $V(S_{N-1})$ simulating trajectory $S_{N-1}$: $$\partial_\lambda V(\lambda) = \frac{V(\lambda + h, S_{N-1}) - V(\lambda, S_{N-1})}{h}$$

The procedure is quite smart and it works from a numerical point of view, but I do not understand how it could be usefull. In fact, usually, if you can integrate from $t_{N-1}$ to $t_N$, i.e. the step $2$, you can integrate starting from $t_0$ so that is does not make sense to use Montecarlo and finite difference.

This is evident for example for a Call option.

My question is then:

Can you present me a case in which the fully integrability is not possible (or too difficult) but the partial one (i.e. from $t_{N-1}$ to $t_N$ is easy to do (so that the Vibrato Montecarlo technique makes sense in such case)?

Thank you for help!!

Ciao ciao!

AM

To keep notations uncluttered assume zero funding costs and consider a European contingent claim priced as $$V = \Bbb{E}_0 \left[ h(S_N) \right]$$ Suppose you would like to make the dependence of your pricing on a specific model parameter $\theta$ appear. You then usually have the following representation choices \begin{align} V &= \int_0^\infty h(S) q(S,\theta) dS \tag{1} \\ &= \int_{-\infty}^\infty h(S(z,\theta)) p(z) dz \tag{2} \end{align} where $z$ is a random variable with known distribution. When expressing $\partial V/\partial \theta$ using the above expressions + Fubini, the first will give rise to the so-called Likelihood Ratio method the second to the Pathwise Sensitivities method \begin{align} \frac{\partial V}{\partial \theta} &= \int_0^\infty h(S) \frac{\partial q(S,\theta)}{\partial \theta} dS = \Bbb{E}_0 \left[ h(S_N) \frac{\partial \ln q}{\partial \theta} \right] \tag{1'} \\ &= \int_{-\infty}^\infty h(S(z,\theta)) p(z) dz = \Bbb{E}_0 \left[ h'(S_N) \frac{\partial S_N}{\partial \theta} \right] \tag{2'} \end{align}
The first is OK as long as the risk-neutral distribution of $S_N$ is tractable under the model at hand, which may not always be the case (think local volatility for instance). But when simulating the SDE the distribution of $S_N \vert S_{N-1}$ will be tractable so that the LR method is often rewritten as $$\Bbb{E}_0 \left[ h(S_N) \sum_{n=1}^N \frac{\partial \ln q(S_n \vert S_{n-1})}{\partial \theta}\right]$$ One of the issues with the LR method is it does not exhibit extraordinary convergence property. The Vibrato is presented by its author as a smart hybrid between the 2.
It is easily illustrated in the case of a digital option, which does not have a differentiable payout function so tranditional PS method cannot be applied. The idea is that conditionally on $S_{N-1}$ the distribution of $S_N$ is tractable. So we shall use the LR method on the last time step. The parameters of the latter distribution will emerge as functions of $\theta$. To obtain the sensitivities of these parameters to $\theta$, we use the PS method.
So Vibrato in that case boils down to applying PS method for $t \in [0, t_{N-1}[$ to obtain the sensitivity of the parmaeters of the conditional distribution $q(S_N \vert S_{N-1})$ to $\theta$. Then, for each generated path, use the LR method over the last time step. Note that this gives rise to some kind of "nested" Monte Carlo step (which can be done analytically for vanilla and digital payouts) but not at all to using 'Finite Differences' as you suggest?


text:
## July 10, 2003

### Not in Kyoto

Scheduling prevents me from being at the Strings Conference in Kyoto this week. But I can live vicariously over the web.

Here are some of the talks that struck me.

#### Strominger

<NAME>’s been thinking about spacelike branes. One version of the story is to introduce a boundary interaction on the string worldsheet,

(1)$S= \int_\Sigma (\partial x^i \partial x^i - \partial x^0 \partial x^0) + \lambda\int_{\partial \Sigma} \cosh (x^0)$

Following Sen, one notes that this is the analytic continuation of Boundary Sine-Gordon theory, which is exactly soluble. For $\lambda=\pm 1/2$, the theory is particularly interesting, as the boundary interaction is equivalent to setting Dirichlet boundary conditions for $x^0$ (an “SD-brane”).

The boundary state for $x^0$ is

(2)$| B\rangle = \left( \frac{1}{1+\sin(\pi\lambda)e^{x^0}} + \frac{1}{1+\sin(\pi\lambda)e^{-x^0}} -1\right) | 0\rangle + \text{oscillators}$

For $\lambda\to\pm 1/2$, this has the limit

(3)$| B\rangle = \left\{ \array{ \sum_n \delta(x^0- 2\pi i n) & \lambda\to -1/2 \\ \sum_n \delta(x^0- 2\pi i (n+1/2)) & \lambda\to 1/2 }\right.$

For $\lambda\to - 1/2$, this is indeed sets $x^0=0$ at the boundary. But the physics is a little awkward. It corresponds to a rolling tachyon which rolls up from $- \infty$ towards the local maximum of the tachyon potential and then rolls back down. This has no analogue in the superstring. $\lambda=+1/2$ seems to be even more boring. It looks like the boundary state vanishes (for real $x^0$), and Sen has interpreted this as the closed string vacuum.

“Not so!” says Strominger. The closed string field configuration which is sourced by the brane satisfies

(4)$(L_0+\overline{L}_0) |\phi\rangle = | B \rangle$

Clearly, $|\phi\rangle$ is a solution, but it is not, according to Strominger, the one dictated by the usual rules of string perturbation theory. Rather, one should take the analytic continuation from Euclidean signature. Equivalently, one should solve

(5)$\square \phi = g\sum_i \delta^4(\vec{x},t-t_i)$

using the Feynman propagator and analytically-continue to imaginary $t_i$. This gives

(6)$\phi = \frac{g}{r} \left( \tanh\frac{r+t}{2}+\tanh\frac{r-t}{2}\right)$

This, indeed, vanishes near the origin, but is nonetheless nontrivial.

This is a really interesting subtlety. For static branes, finding the closed string field configuration sourced by the brane involves solving an elliptic equation, and demanding that the solution die off at infinity gives a unique solution. In the present situation, one has a hyperbolic equation to solve, and there is some extra physical input that is required. The proposal for SD branes is very nice, but I wonder what the general prescription might be like.

#### Bousso

<NAME> talked about proving Bekenstein’s original Entropy Bound for weakly-gravitating systems. The Bound states that the entropy of a system of total mass, $M$, contained in as sphere of radius $R$ is bounded by

(7)$S \leq \frac{2\pi M R}{\hbar}$

Since $M\ll M_{bh} = R/2G$, the mass of a black hole whose Schwarzschild radius is $R$, this is much more stringent than the Holographic Bound

(8)$S \ll \frac{\pi R^2}{G \hbar} = \frac{A}{4 G \hbar}$

of 't Hooft and Susskind.

To prove Bekenstein’s bound, Raphael invokes a version of the Covariant Entropy Bound due to Flanagan, Marolf and Wald. Let $A_1$ and $A_2$ be spacelike hypersurfaces, and $L$ be a lightsheet whose boundary is $A_1-A_2$, which is non-expanding with respect to $A_1$ and expanding with respect to $A_2$. Then the entropy crossing $L$ is bounded by

(9)$S \lt \frac{A_1-A_2}{4 G \hbar}$

What’s cool about this version of the Covariant Entropy Bound is that the surfaces $A_i$ don’t need to be closed. Indeed, Raphael takes them to be disks, such that the lightsheet emanating from $A_1$ is just barely nonexpanding. The lightrays passing through our weakly-gravitating system are focussed, so $A_2$ is a slightly smaller disk than $A_1$. The radius of $A_1$ is roughly $\pi R^2$. Because of the focussing, $A_2$ is smaller by a fraction $2 M/R$. So, sloughing over all the technical details, one obtains something like Bekenstein’s bound.

Theoretically, the bound you obtain this way could actually be tighter than Bekenstein’s, but it’s not clear that, in practice, a self-gravitating system can give a tighter bound.

#### Aganagic

<NAME> gave a real tour de force talk about recent work on the open topological A-Model. In the A-Model, one has a Calabi-Yau manifold $X$, and a Riemann surface $\Sigma$ with boundaries. One is interested in counting holomorphic maps from $\Sigma$ into $X$ such that the boundaries of $\Sigma$ get mapped into Special Lagrangian submanifolds of $X$. In the case of $X$ a local Calabi-Yau, with special Lagrangian $T^2\times\mathbb{R}$ fibers over $\mathbb{R}^3$, these guys have an incredibly elegant solution which, alas, defies my humble attempts at a simple summary.

More next time …

Posted by distler at July 10, 2003 11:24 AM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/187


text:
 <NAME>/Examples/Example 6

Example of Problem from <NAME>

There is a pool $10$ feet square, with a reed growing vertically in the centre,
its roots at the bottom of the pool, which rises $1$ foot above the surface;
when drawn towards the shore it reaches exactly to the brink of the pool;
what is the depth of the water?

Solution

The water is $12$ feet deep.

Proof

Let the depth of the water be $d$.

The length of the reed is $d + 1$.

When drawn to the edge of the pool, the reed forms the hypotenuse of a right triangle.

One of the legs of that right triangle is the depth of the pool, which is $d$.

The other leg is the distance from the centre of the pool, which is $5$ feet.

Hence:

 $\ds \paren {d + 1}^2$ $=$ $\ds d^2 + 5^2$ Pythagoras's Theorem $\ds \leadsto \ \$ $\ds 2 d + 1$ $=$ $\ds 25$ simplification $\ds \leadsto \ \$ $\ds d$ $=$ $\ds 12$ simplification

The right triangle in question here is the $\text{5-12-13}$ triangle.

$\blacksquare$


text:
Preprint A266/2003
Curvature of pencils of foliations
<NAME>
Keywords: holomorphic foliations | pencil | curvature
\abstract{Let $\Cal{F}$ and $\Cal G$ be two distinct singular holomorphic foliations on a compact complex surface $M$, in the same class, that is $N_\Cal{F}=N_\Cal G$. In this case, we can define the {\it pencil $\Cal{P}=\Cal{P}(\Cal{F},\Cal G)$ of foliations generated by $\Cal{F}$ and} $\Cal G$. We can associate to a pencil $\Cal{P}$ a meromorphic 2-form $\Theta=\Theta(\Cal{P})$, the form of curvature of the pencil, which is in fact the Chern curvature (cf. [Ch]). When $\Theta(\Cal{P})\equiv 0$ we will say that the pencil is {\it flat}. In this paper we give some sufficient condictions for a pencil to be flat. (Theorem 2). We will see also how the flatness reflects in the pseudo-group of holonomy of the foliations of $\Cal{P}$. In particular, we will study the set $\lbrace\Cal{H}\in \Cal{P}|\,\Cal{H}$ has a first integral $\rbrace$ in some cases (Theorem 1).}


text:
# Calculate the mass of Mg needed to produce 5 kg of vanadium

I need to calculate the mass of magnesium needed to produce $\pu{5 kg}$ of vanadium from the following reaction:

$$\ce{3Mg + 2VCl3 -> 2V + 3MgCl2}$$

given relative atomic masses for $\ce{Mg}$ and $\ce{V}$ as $24$ and $51$, respectively.

I multiplied $24 × 3$ and $51 × 2$ to get

$72 : 102$

I changed $\pu{5 kg}$ into grams, which is $5000$.

I cross-multiplied to find the answer and then divided the answer by $1000$ to turn it back to kilograms because the question stated it. My answer is $\pu{3.5 kg}$. Is this correct?

It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.

First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:

$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$

Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:

$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$

Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:

$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$


text:
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### Renewable Integration

Renewable integration is the process of plugging renewable sources of energy into the electric grid.
FEBRUARY 15, 2022


text: CR Extension from Manifolds of Higher Type This paper deals with the extension of CR functions from a manifold $M\subset \mathbb C^n$ into directions produced by higher order commutators of holomorphic and antiholomorphic vector fields. It uses the theory of complex sectors'' attached to real submanifolds introduced in recent joint work of the authors with <NAME>. In addition, it develops a new technique of approximation of sectors by smooth discs. Categories:32V25, 32V35, 32C16, 32F18


text:
Randomizer Theory

Thread in 'Research & Development' started by colour_thief, 12 Jan 2007.

1. makalaka

I don't completely understand this

so explain to me how I got OOO at the beginning of one game

2. colour_thief

Digital's video and that CPU level 13 (human side) are definitely using 8-bag. The odds of any other randomizer matching up with that many pieces is astronomical.

I'd guess either you mis-remembered the OOO, or whatever you were playing used yet another randomizer.

3. caffeine

how does JIJJ say it's not another bag weaved between the bags? i'm lost.

4. colour_thief

Because no bag would have 3 Js in it? JIJJ is a sequence of 8th pieces.

Take a look at game #4, divide the pieces into groups of 8, and then delete one of every piece from each group of 8. What you're left with is the random 8th pieces. JIJJ were the 8th pieces for the first 4 consecutive bags of game 4, which is a sequence so hideous it can't be anything other than pure, memoryless piece selection.

5. caffeine

i must be missing something.
eight piece bags mean seven different pieces, plus one piece. so that would mean there will never be more than two of the same tetrominoes in any one bag. so JIJJ would mean you're ending one bag and going to the next. if that happened four times in a row, then it means no: every seven eight-bags, the eighth's are not a bag themselves. but is that what you meant?

6. colour_thief

I don't think I can make it clearer than this:

Sequence of pieces. From the start to the end of a match.

Code:
tojlijsztzlisjoilsztijjooisjjztljislozst
Bag-seperated sequence of pieces. By chance there is an 8-multiple number of pieces.
Code:
tojlijsz
tzlisjoi
lsztijjo
oisjjztl
jislozst
Making it really obvious which is the 8th piece in every bag...
Code:
--j--j--
---i---i
-----jj-
---jj---
--s---s-
Sequence of 8th pieces:
Code:
jijjs
If you don't now see how the sequence of 8th pieces is definitely not fed by it's own independent bag system, you're just not thinking about it enough.

7. Pineapple

No. J is the bonus piece in the first bag. I in the second bag, J in the 3rd bag, J in the 4th bag. That makes it somewhere between "very unlikely" and "impossible" that the bonus pieces are being generated from a bag.

8. makalaka

None of this makes sense. Why would they give the player an 8-bag randomizer, then give the CPU a different randomizer, and then have ANOTHER different one that allows OOO at the start? (luckily for you, I don't remember whether I got it when playing against the CPU, or in time attack mode.)

9. colour_thief

Well, you can look at it that way, but an overwhelming quantity of evidence supports 8-bag with 7-bag for the CPU. I'm not calling you a liar, and I strongly doubt they also included "9-bag" or something else in there somewhere. Maybe you witnessed a rare game bug? I don't know. For all intents and purposes your OOO never happened.

10. lee n

Did you get it right at the start, as in literally the first three pieces? .. or was it within the first 10 pieces or something?

11. tepplesLockjaw developer

It's likely that the first few pieces use a different randomizer to minimize unavoidable holds and soft drops without requiring the player to platform. TGM games are known to use a different randomizer for the first piece, with equal probability I, J, L, or T, for just this reason.

12. caffeine

oh okay, i got it.

i'll try to think harder next time.

13. Pineapple

We're not even sure if The New Tetris does use a memoryless randomizer. We suspect that it might do, but we don't know for sure...

14. Zaphod77Resident Misinformer

8 piece bag can give you the same piece 3 times in a row.

Maybe it starts with a random amount of the bag already dealt?

That could result in the OOO series.

this can be done on

piece 7,8,9

or

8.9.10

it's unlikely, but not quite as impossible as you'd think.

More calculations.

Odds of 2 of the same in a row.

For memoryless: 1/7

For 7 piece bag. 1/49 (1/7 on the first piece of the bag, if it's not bag 1)

Odds of getting a 4 snake series under 7 bag. 1/3087 (1/441 if you are at the right part in the bag, and 0 if not)

Here's a 7 piece bag variant that will be interesting to test.

Have a 1 in 7 chance to, instead of drawing the next peice from the bag, serve the same piece again. But at the start of a bag, if it matches the last piece of the previous bag, redraw.

That way, the odds of getting a repeat are exactly the same as memoryless, but otherwise it tends to an even distribution.

15. lee n

8 piece bag can give you the same piece 4 times in a row.

Also true.

17. Edoa.k.a. FSY

Okay, so tepples, you're assuming The New Tetris uses a memoryless randomizer, but am I right in saying that no one actually knows for sure? I mean, no one has ever investigated this, right? If that's the case, I wouldn't mind looking into it myself, it sounds quite interesting. I don't actually own The New Tetris, but I could play around with the square mode of Tetris Worlds for PS2.

18. colour_thief

The algorithm for Tetris Worlds square mode is known. It's:

I, O, J, L, T:

3/19 probability each.

S, Z:

2/19 probability each.

It's just random with skewed probabilities.

19. Pineapple

That doesn't tel the whole story, though. Tetris Worlds has 2 main ways of playing: Story mode and Arcade mode. Some of the 6 game modes differ between the two, and Square mode is one of them. (Note to anyone who has the game: the .tws files that start with M are the arcade mode versions)

In Story mode, it's 2/19 for S and Z and 3/19 for the others. In Arcade mode, it's 1/7 for all pieces.

As for TNT64, I have my suspicions that it's not traditional memoryless, but as I don't have an N64, I am unable to prove it...

Getting a few games transcribed would definitely help. Right now, we have just the one game transcribed, and that's Gilly's 868 video that's on YouTube.

Code:
sjozoojltlstszsliiizlojsojolszlittiititjozjtszzjjtillloozstzjisjizttstjjzliiltsojissloltisjlsjzlzoizotjoiltosiitzzjoolozsslztjss

===== Piece Counts =====
I: 18/128
Z: 18/128
S: 20/128
J: 18/128
L: 18/128
O: 18/128
T: 18/128

===== Interval Counts =====
Interval Size 24: Found 1/121, 0.008264
Interval Size 22: Found 1/121, 0.008264
Interval Size 21: Found 2/121, 0.016529
Interval Size 20: Found 1/121, 0.008264
Interval Size 16: Found 4/121, 0.033058
Interval Size 14: Found 4/121, 0.033058
Interval Size 13: Found 5/121, 0.041322
Interval Size 12: Found 6/121, 0.049587
Interval Size 11: Found 2/121, 0.016529
Interval Size 10: Found 10/121, 0.082645
Interval Size 9: Found 4/121, 0.033058
Interval Size 8: Found 4/121, 0.033058
Interval Size 7: Found 4/121, 0.033058
Interval Size 6: Found 10/121, 0.082645
Interval Size 5: Found 12/121, 0.099174
Interval Size 4: Found 5/121, 0.041322
Interval Size 3: Found 16/121, 0.132231
Interval Size 2: Found 11/121, 0.090909
Interval Size 1: Found 19/121, 0.157025
So far, the results are inconclusive...

20. colour_thief

Though I agree 128 pieces is inconclusive, I think this sample points towards it being true random.