text:
TheInfoList

In
physics Physics (from grc, φυσική (ἐπιστήμη), physikḗ (epistḗmē), knowledge of nature, from ''phýsis'' 'nature'), , is the natural science that studies matter, its Motion (physics), motion and behavior through Spacetime, spa ...
, a sign convention is a choice of the physical significance of signs (plus or minus) for a set of quantities, in a case where the choice of sign is arbitrary. "Arbitrary" here means that the same physical system can be correctly described using different choices for the signs, as long as one set of definitions is used consistently. The choices made may differ between authors. Disagreement about sign conventions is a frequent source of confusion, frustration, misunderstandings, and even outright errors in scientific work. In general, a sign convention is a special case of a choice of
coordinate system In geometry Geometry (from the grc, γεωμετρία; ''wikt:γῆ, geo-'' "earth", ''wikt:μέτρον, -metron'' "measurement") is, with arithmetic, one of the oldest branches of mathematics. It is concerned with properties of space t ...
for the case of one dimension. Sometimes, the term "sign convention" is used more broadly to include factors of '' i'' and 2 π, rather than just choices of sign.

Relativity

Metric signature

In relativity, the
metric signatureIn mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ha ...
can be either (+,−,−,−) or (−,+,+,+). (Note that throughout this article we are displaying the signs of the eigenvalues of the metric in the order that presents the timelike component first, followed by the spacelike components). A similar convention is used in higher-dimensional relativistic theories; that is, (+,−,−,−,...) or (−,+,+,+,...). A choice of signature is associated with a variety of names: + − − −: *'' Timelike convention'' *''
Particle physics Particle physics (also known as high energy physics) is a branch of physics Physics (from grc, φυσική (ἐπιστήμη), physikḗ (epistḗmē), knowledge of nature, from ''phýsis'' 'nature'), , is the natural science that ...
convention'' *'' West coast convention'' *''Mostly minuses'' *''
Landau Landau, officially Landau in der Pfalz, is an autonomous (''kreisfrei'') town surrounded by the Südliche Weinstraße Südliche Weinstraße ( en, Southwest Wine Route) is a district (''Kreis'') in the south of Rhineland-Palatinate, Germany. Neigh ...
Lifshitz sign convention''. − + + +: *'' Spacelike convention'' *'' Relativity convention'' *'' East coast convention'' *''Mostly pluses'' *''Pauli convention'' We catalog the choices of various authors of some graduate textbooks: (+,−,−,−): *'' Landau & Lifshitz'' *
Gravitation: an introduction to current research
' (Louis Witten, L. Witten) *
Ray D'Inverno, Introducting Einstein's relativity
'' (−,+,+,+): *''Gravitation (book), Misner, Thorne and Wheeler'' *
Spacetime and Geometry: An Introduction to General Relativity
' *''General Relativity (book), General Relativity (Wald)'' (Note that Wald changes signature to the timelike convention for Chapter 13 only.) The signature + − − − corresponds to the metric tensor: :$\begin 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end$ whereas the signature − + + + corresponds to: :$\begin -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end$

Curvature

The Ricci tensor is defined as the contraction of the Riemann tensor. Some authors use the contraction $R_ \, = R^c_$, whereas others use the alternative $R_ \, = R^c_$. Due to the Riemann tensor#Symmetries and identities, symmetries of the Riemann tensor, these two definitions differ by a minus sign. In fact, the second definition of the Ricci tensor is $R_ \, = ^c$. The sign of the Ricci tensor does not change, because the two sign conventions concern the sign of the Riemann tensor. The second definition just compensates the sign, and it works together with the second definition of the Riemann tensor (see e.g. Barrett O'Neill's Semi-riemannian geometry).

Other sign conventions

* The sign choice for arrow of time, time in frames of reference and proper time: + for future and − for past is universally accepted. * The choice of $\pm$ in the Dirac equation. * The sign of the electric charge, electromagnetic tensor, field strength tensor $\, F_$ in Gauge theory, gauge theories and Maxwell's equations, classical electrodynamics. * Time dependence of a positive-frequency wave (see, e.g., the electromagnetic wave equation): ** $\, e^$ (mainly used by physicists) ** $\, e^$ (mainly used by engineers) * The sign for the imaginary part of Permittivity#Complex permittivity, permittivity (in fact dictated by the choice of sign for time-dependence) * The signs of distances and radius of curvature (optics), radii of curvature of optical surfaces in optics * The sign of work in the first law of thermodynamics. * The sign of the weight of the determinant of the metric tensor when dealing with tensor density. It is often considered good form to state explicitly which sign convention is to be used at the beginning of each book or article. The sign of spherical mirrors are also represented by sign convention


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# Definition:First Order Ordinary Differential Equation

(Redirected from Definition:First Order ODE)

## Definition

A first order ordinary differential equation is an ordinary differential equation in which any derivatives with respect to the independent variable have order no greater than $1$.

The general first order ODE can be written as:

$\map F {x, y, \dfrac {\d y} {\d x} }$

or, using prime notation:

$\map F {x, y, y'}$

If it is possible to do so, then it is often convenient to present such an equation in the form:

$\dfrac {\d y} {\d x} = \map f {x, y}$

that is:

$y' = \map f {x, y}$

It can also be seen presented in the form:

$\map \phi {x, y, y'} = 0$

format: plain


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# Binary heap

Binary Heap
Type Tree
Time complexity
in big O notation
Average Worst case
Space O(n) O(n)
Search Not supported Not supported
Insert O(1) O(log n)
Delete O(log n) O(log n)
Example of a complete binary max heap
Example of a complete binary min heap

A binary heap is a heap data structure created using a binary tree. It can be seen as a binary tree with two additional constraints:

Shape property
The tree is a complete binary tree; that is, all levels of the tree, except possibly the last one (deepest) are fully filled, and, if the last level of the tree is not complete, the nodes of that level are filled from left to right.
Heap property
All nodes are either [greater than or equal to] or [less than or equal to] each of its children, according to a comparison predicate defined for the heap.

Heaps with a mathematical "greater than or equal to" (≥) comparison predicate are called max-heaps; those with a mathematical "less than or equal to" (≤) comparison predicate are called min-heaps. Min-heaps are often used to implement priority queues.[1][2]

Since the ordering of siblings in a heap is not specified by the heap property, a single node's two children can be freely interchanged unless doing so violates the shape property (compare with treap).

The binary heap is a special case of the d-ary heap in which d = 2.

## Heap operations

Both the insert and remove operations modify the heap to conform to the shape property first, by adding or removing from the end of the heap. Then the heap property is restored by traversing up or down the heap. Both operations take O(log n) time.

### Insert

To add an element to a heap we must perform an up-heap operation (also known as bubble-up, percolate-up, shift-up, trickle up, heapify-up, or cascade-up), by following this algorithm:

1. Add the element to the bottom level of the heap.
2. Compare the added element with its parent; if they are in the correct order, stop.
3. If not, swap the element with its parent and return to the previous step.

The number of operations required is dependent on the number of levels the new element must rise to satisfy the heap property, thus the insertion operation has a time complexity of O(log n). However, in 1974, Thomas Porter and Istvan Simon proved that the function for the average number of levels an inserted node moves up is upper bounded by the constant 1.6067.[3] The average number of operations required for an insertion into a binary heap is 2.6067 since one additional comparison is made that does not result in the inserted node moving up a level. Thus, on average, binary heap insertion has a constant, O(1), time complexity. Intuitively, this makes sense since approximately 50% of the elements are leaves and approximately 75% of the elements are in the bottom two levels, it is likely that the new element to be inserted will only move a few levels upwards to maintain the heap.

As an example of binary heap insertion, say we have a max-heap

and we want to add the number 15 to the heap. We first place the 15 in the position marked by the X. However, the heap property is violated since 15 is greater than 8, so we need to swap the 15 and the 8. So, we have the heap looking as follows after the first swap:

However the heap property is still violated since 15 is greater than 11, so we need to swap again:

which is a valid max-heap. There is no need to check the children after this. Before we placed 15 on X, the heap was valid, meaning 11 is greater than 5. If 15 is greater than 11, and 11 is greater than 5, then 15 must be greater than 5, because of the transitive relation.

### Delete

The procedure for deleting the root from the heap (effectively extracting the maximum element in a max-heap or the minimum element in a min-heap) and restoring the properties is called down-heap (also known as bubble-down, percolate-down, sift-down, trickle down, heapify-down, cascade-down and extract-min/max).

1. Replace the root of the heap with the last element on the last level.
2. Compare the new root with its children; if they are in the correct order, stop.
3. If not, swap the element with one of its children and return to the previous step. (Swap with its smaller child in a min-heap and its larger child in a max-heap.)

So, if we have the same max-heap as before

We remove the 11 and replace it with the 4.

Now the heap property is violated since 8 is greater than 4. In this case, swapping the two elements, 4 and 8, is enough to restore the heap property and we need not swap elements further:

The downward-moving node is swapped with the larger of its children in a max-heap (in a min-heap it would be swapped with its smaller child), until it satisfies the heap property in its new position. This functionality is achieved by the Max-Heapify function as defined below in pseudocode for an array-backed heap A of length heap_length[A]. Note that "A" is indexed starting at 1, not 0 as is common in many real programming languages.

Max-Heapify (A, i):
left ← 2i
right ← 2i + 1
largesti
if leftheap_length[A] and A[left] > A[largest] then:
largestleft
if rightheap_length[A] and A[right] > A[largest] then:
largestright
if largesti then:
swap A[i] ↔ A[largest]
Max-Heapify(A, largest)

For the above algorithm to correctly re-heapify the array, the node at index i and its two direct children must violate the heap property. If they do not, the algorithm will fall through with no change to the array. The down-heap operation (without the preceding swap) can also be used to modify the value of the root, even when an element is not being deleted.

In the worst case, the new root has to be swapped with its child on each level until it reaches the bottom level of the heap, meaning that the delete operation has a time complexity relative to the height of the tree, or O(log n).

## Building a heap

A heap could be built by successive insertions. This approach requires $O(n \log n)$ time because each insertion takes $O(\log n)$ time and there are $n$ elements. However this is not the optimal method. The optimal method starts by arbitrarily putting the elements on a binary tree, respecting the shape property (the tree could be represented by an array, see below). Then starting from the lowest level and moving upwards, shift the root of each subtree downward as in the deletion algorithm until the heap property is restored. More specifically if all the subtrees starting at some height $h$ (measured from the bottom) have already been "heapified", the trees at height $h+1$ can be heapified by sending their root down along the path of maximum valued children when building a max-heap, or minimum valued children when building a min-heap. This process takes $O(h)$ operations (swaps) per node. In this method most of the heapification takes place in the lower levels. Since the height of the heap is $\left\lfloor \lg (n) \right\rfloor$, the number of nodes at height $h$ is $\le \left\lceil 2^{\left(\lg n - h\right) - 1} \right\rceil = \left\lceil \frac{2^{\lg n}}{2^{h+1}}\right\rceil = \left\lceil\frac{n}{2^{h+1}}\right\rceil$. Therefore, the cost of heapifying all subtrees is:

\begin{align} \sum_{h=0}^{\lceil \lg n \rceil} \frac{n}{2^{h+1}}O(h) & = O\left(n\sum_{h=0}^{\lceil \lg n \rceil} \frac{h}{2^{h + 1}}\right) \\ & \le O\left(n\sum_{h=0}^{\infty} \frac{h}{2^h}\right) \\ & = O(n) \end{align}

This uses the fact that the given infinite series h / 2h converges to 2.

The exact value of the above (the worst-case number of comparisons during the heap construction) is known to be equal to:

$2 n - 2 s_2 (n) - e_2 (n)$,[4]

where s2(n) is the sum of all digits of the binary representation of n and e2(n) is the exponent of 2 in the prime factorization of n.

The Build-Max-Heap function that follows, converts an array A which stores a complete binary tree with n nodes to a max-heap by repeatedly using Max-Heapify in a bottom up manner. It is based on the observation that the array elements indexed by floor(n/2) + 1, floor(n/2) + 2, ..., n are all leaves for the tree, thus each is a one-element heap. Build-Max-Heap runs Max-Heapify on each of the remaining tree nodes.

Build-Max-Heap (A):
heap_length[A] ← length[A]
for ifloor(length[A]/2) downto 1 do
Max-Heapify(A, i)

## Heap implementation

A small complete binary tree stored in an array
Comparison between a binary heap and an array implementation.

Heaps are commonly implemented with an array. Any binary tree can be stored in an array, but because a heap is always an almost complete binary tree, it can be stored compactly. No space is required for pointers; instead, the parent and children of each node can be found by arithmetic on array indices. These properties make this heap implementation a simple example of an implicit data structure or Ahnentafel list. Details depend on the root position, which in turn may depend on constraints of a programming language used for implementation, or programmer preference. Specifically, sometimes the root is placed at index 1, sacrificing space in order to simplify arithmetic. The peek operation (find-min or find-max) simply returns the value of the root, and is thus O(1).

Let n be the number of elements in the heap and i be an arbitrary valid index of the array storing the heap. If the tree root is at index 0, with valid indices 0 through n − 1, then each element a at index i has

• children at indices 2i + 1 and 2i + 2
• its parent ⌊(i − 1) ∕ 2⌋ where ⌊…⌋ is the floor function.

Alternatively, if the tree root is at index 1, with valid indices 1 through n, then each element a at index i has

• children at indices 2i and 2i +1
• its parent at index ⌊i ∕ 2⌋.

This implementation is used in the heapsort algorithm, where it allows the space in the input array to be reused to store the heap (i.e. the algorithm is done in-place). The implementation is also useful for use as a Priority queue where use of a dynamic array allows insertion of an unbounded number of items.

The upheap/downheap operations can then be stated in terms of an array as follows: suppose that the heap property holds for the indices b, b+1, ..., e. The sift-down function extends the heap property to b−1, b, b+1, ..., e. Only index i = b−1 can violate the heap property. Let j be the index of the largest child of a[i] (for a max-heap, or the smallest child for a min-heap) within the range b, ..., e. (If no such index exists because 2i > e then the heap property holds for the newly extended range and nothing needs to be done.) By swapping the values a[i] and a[j] the heap property for position i is established. At this point, the only problem is that the heap property might not hold for index j. The sift-down function is applied tail-recursively to index j until the heap property is established for all elements.

The sift-down function is fast. In each step it only needs two comparisons and one swap. The index value where it is working doubles in each iteration, so that at most log2 e steps are required.

For big heaps and using virtual memory, storing elements in an array according to the above scheme is inefficient: (almost) every level is in a different page. B-heaps are binary heaps that keep subtrees in a single page, reducing the number of pages accessed by up to a factor of ten.[5]

The operation of merging two binary heaps takes Θ(n) for equal-sized heaps. The best you can do is (in case of array implementation) simply concatenating the two heap arrays and build a heap of the result.[6] A heap on n elements can be merged with a heap on k elements using O(log n log k) key comparisons, or, in case of a pointer-based implementation, in O(log n log k) time.[7] An algorithm for splitting a heap on n elements into two heaps on k and n-k elements, respectively, based on a new view of heaps as an ordered collections of subheaps was presented in.[8] The algorithm requires O(log n * log n) comparisons. The view also presents a new and conceptually simple algorithm for merging heaps. When merging is a common task, a different heap implementation is recommended, such as binomial heaps, which can be merged in O(log n).

Additionally, a binary heap can be implemented with a traditional binary tree data structure, but there is an issue with finding the adjacent element on the last level on the binary heap when adding an element. This element can be determined algorithmically or by adding extra data to the nodes, called "threading" the tree—instead of merely storing references to the children, we store the inorder successor of the node as well.

It is possible to modify the heap structure to allow extraction of both the smallest and largest element in $O$$(\log n)$ time.[9] To do this, the rows alternate between min heap and max heap. The algorithms are roughly the same, but, in each step, one must consider the alternating rows with alternating comparisons. The performance is roughly the same as a normal single direction heap. This idea can be generalised to a min-max-median heap.

## Derivation of index equations

In an array-based heap, the children and parent of a node can be located via simple arithmetic on the node's index. This section derives the relevant equations for heaps with their root at index 0, with additional notes on heaps with their root at index 1.

To avoid confusion, we'll define the level of a node as its distance from the root, such that the root itself occupies level 0.

### Child nodes

For a general node located at index $i$ (beginning from 0), we will first derive the index of its right child, $\text{right} = 2i + 2$.

Let node $i$ be located in level $L$, and note that any level $l$ contains exactly $2^l$ nodes. Furthermore, there are exactly $2^{l + 1} - 1$ nodes contained in the layers up to and including layer $l$ (think of binary arithmetic; 0111...111 = 1000...000 - 1). Because the root is stored at 0, the $k$th node will be stored at index $(k - 1)$. Putting these observations together yields the following expression for the index of the last node in layer l.

$\text{last}(l) = (2^{l + 1} - 1) - 1 = 2^{l + 1} - 2$

Let there be $j$ nodes after node $i$ in layer L, such that

\begin{alignat}{2} i = & \quad \text{last}(L) - j\\ = & \quad (2^{L + 1} -2) - j\\ \end{alignat}

Each of these $j$ nodes must have exactly 2 children, so there must be $2j$ nodes separating $i$'s right child from the end of its layer ($L + 1$).

\begin{alignat}{2} \text{right} = & \quad \text{last(L + 1)} -2j\\ = & \quad (2^{L + 2} -2) -2j\\ = & \quad 2(2^{L + 1} -2 -j) + 2\\ = & \quad 2i + 2 \end{alignat}

As required.

Noting that the left child of any node is always 1 place before its right child, we get $\text{left} = 2i + 1$.

If the root is located at index 1 instead of 0, the last node in each level is instead at index $2^{l + 1} - 1$. Using this throughout yields $\text{left} = 2i$ and $\text{right} = 2i + 1$ for heaps with their root at 1.

### Parent node

Every node is either the left or right child of its parent, so we know that either of the following is true.

1. $i = 2 \times (\text{parent}) + 1$
2. $i = 2 \times (\text{parent}) + 2$

Hence,

$\text{parent} = \frac{i - 1}{2} \textbf{ or } \frac{i - 2}{2}$

Now consider the expression $\left\lfloor \dfrac{i - 1}{2} \right\rfloor$.

If node $i$ is a left child, this gives the result immediately, however, it also gives the correct result if node $i$ is a right child. In this case, $(i - 2)$ must be even, and hence $(i - 1)$ must be odd.

\begin{alignat}{2} \left\lfloor \dfrac{i - 1}{2} \right\rfloor = & \quad \left\lfloor \dfrac{i - 2}{2} + \dfrac{1}{2} \right\rfloor\\ = & \quad \frac{i - 2}{2}\\ = & \quad \text{parent} \end{alignat}

Therefore, irrespective of whether a node is a left or right child, its parent can be found by the expression:

$\text{parent} = \left\lfloor \dfrac{i - 1}{2} \right\rfloor$

## References

1. ^ "heapq – Heap queue algorithm". Python Standard Library.
2. ^ "Class PriorityQueue". Java™ Platform Standard Ed. 6.
3. ^ Thomas Porter; Istvan Simon (1974). "Random Insertion into a Priority Queue Structure". Stanford University Reports. Stanford University. p. 13. Retrieved 31 January 2014.
4. ^ Suchenek, Marek A. (2012), "Elementary Yet Precise Worst-Case Analysis of Floyd's Heap-Construction Program", Fundamenta Informaticae (IOS Press) 120 (1): 75–92, doi:10.3233/FI-2012-751.
5. ^ Poul-Henning Kamp. "You're Doing It Wrong". ACM Queue. June 11, 2010.
6. ^ Chris L. Kuszmaul. "binary heap". Dictionary of Algorithms and Data Structures, Paul E. Black, ed., U.S. National Institute of Standards and Technology. 16 November 2009.
7. ^ J.-R. Sack and T. Strothotte "An Algorithm for Merging Heaps", Acta Informatica 22, 171-186 (1985).
8. ^ . J.-R. Sack and T. Strothotte "A characterization of heaps and its applications" Information and Computation Volume 86, Issue 1, May 1990, Pages 69–86.
9. ^ Atkinson, M.D., J.-R. Sack, N. Santoro, and T. Strothotte (1 October 1986). "Min-max heaps and generalized priority queues.". Programming techniques and Data structures. Comm. ACM, 29(10): 996–1000.


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# Finding the Roots of Unity

I have the following equation, $$z^5 = -16 + (16\sqrt 3)i$$I am asked to write down the 5th roots of unity and find all the roots for the above equation expressing each root in the form $re^{i\theta}$. I am just wondering if my solutions are correct. Here are my solutions,

5th roots of unity, $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$

and for all the roots for the above equation, $$z = e^{\frac{2\pi k} {5}i}, 2e^{\frac{(6k+2)\pi}{15}i}$$ where k = 0, 1, 2, 3, 4.

Please correct me if there are any mistakes. I will leave my working at the answer section for future reference.

• Do you mean 2k$\pi$/n ? (where n is the power of z) – Lily L Oct 9 '15 at 1:52
For the fifth roots of unity, the roots are $\frac{2\pi}{5}$ radian apart from one another. Since the principal value for arg(z) is (-$\pi$,$\pi$], hence, the fifth roots of unity are $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$ For the all the roots of the above equation, one of the roots will always be $$z = e^{\frac{2k\pi}{n}i}$$ where n = the power of z and k = 0, 1, 2, ..., n-1
Hence, $$z = e^{\frac{2k\pi}{5}i}$$ where k = 0, 1, 2, 3, 4
For the other root, $$|z^5| = \sqrt {(-16)^2 + (16\sqrt3)^2}$$ $$|z| = \sqrt[5] {32} = 2$$ and since the coordinate lies in the 2nd quadrant, $$arg(z) = \pi - tan^{-1} (\frac{16\sqrt3}{16}) = \frac{2\pi}{3}$$ Therefore, \begin{align} z & = re^{i\theta}\\ & = 2e^{(\frac{2k\pi}{5}+\frac{1}{5}\cdot\frac{2\pi}{3})i}\\ & = 2e^{\frac{(6k+2)\pi}{15}i}\\ \end{align}


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# How do you find the vertex for y=x^2-6x+1?

Write the equation as $y = {x}^{2} - 6 x + 9 - 8$
Or $\left(y + 8\right) = {\left(x - 3\right)}^{2}$.


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Two weeks ago I used STAN to create predictions after just throwing in all independent variables. This week I aim to refine the STAN model. For this it is convenient to use the loo package (Efficient Leave-One-Out Cross-Validation and WAIC for Bayesian Models). See also the paper by Aki Vehtari, Andrew Gelman and Jonah Gabry.
Since the package does the heavy lifting, it only remains to wrap it a function so I can quickly compare some models. A potential next step is to automate some more, but I did not do that pending current results.

### Data

Same as last time.

### Model

To keep the model similar as last time, I need to get a full design matrix for each independent variable in the model. So I made a mechanism which takes a model formulation and creates both the design matrix and a bunch of indices to keep track which column corresponds to which part of the model. To be specific, terms contains 1 to nterm if the corresponding column in xmat is from variable 1 (intercept) to the last variable. This sits in the function des.matrix.
The generated quantities block is purely for the LOO statistic.
It is preferred to compile the model only once, hence fit1 is calculated beforehand. Having done that preparation, MySmodel  is a function which does model fitting, LOO statistic and output it all in one step. In this function I can just drop in the formula and get something usable as output, so I can easily examine a bunch of models. It seemed to me that forward selection was a suitable way to examine the model space. I know it is not ideal, but at this point I mainly want to know if this actually will function.

### Results

To my surprise, Title was the parameter which gave the best predictions. I had expected sex to play that role.
Survived ~ Title -445.4972 16.46314
Computed from 4000 by 891 log-likelihood matrix

Estimate   SE
elpd_loo   -445.5 16.5
p_loo         4.1  0.2
looic       891.0 32.9

All Pareto k estimates OK (k < 0.5)
The next variable was passenger class

Survived ~ Title + Pclass -395.5926 17.42705
Unfortunately after adding a few independent variables things gave only minor improvements. This os not because of anything faulty, I made a classical mechanism to leave 10% out and predict the remainder. Those results were similar, but took more time and showed more run to run variation in the results. The only true advantage was that it gave results on the same scale as previous cross validations.
I expanded the model formula to about 10 terms. At that point, the expected prediction error decreased so slow that I decided on an eight term model. (Title + Pclass + sibsp + Title:Pclass + Embarked + oe + Title:sibsp + parch). The functions myPmodel and mySpred refit the model and perform the actual predictions. The result was a disappointing 0.78 on Kaggle. A minor improvement on the previous STAN result, but boosting is still better.

### Code

library(rstan)
rstan_options(auto_write = TRUE)
options(mc.cores = parallel::detectCores())
library(loo)

train$status <- 'train' test <- read.csv('test.csv') test$status <- 'test'
test$Survived <- NA tt <- rbind(test,train) # generate variables tt$Embarked[tt$Embarked==”] <- 'S' tt$Embarked <- factor(tt$Embarked) tt$Pclass <- factor(tt$Pclass) tt$Survived <- factor(tt$Survived) tt$age <- tt$Age tt$age[is.na(tt$age)] <- 999 tt$age <- cut(tt$age,c(0,2,5,9,12,15,21,55,65,100,1000)) tt$Title <- sapply(tt$Name,function(x) strsplit(as.character(x),'[.,]')[[1]][2]) tt$Title <- gsub(' ','',tt$Title) tt$Title[tt$Title==’Dr’ & tt$Sex==’female’] <- 'Miss'
tt$Title[tt$Title %in% c(‘Capt’,’Col’,’Don’,’Sir’,’Jonkheer’,’Major’,’Rev’,’Dr’)] <- 'Mr'
tt$Title[tt$Title %in% c(‘Lady’,’Ms’,’theCountess’,’Mlle’,’Mme’,’Ms’,’Dona’)] <- 'Miss'
tt$Title <- factor(tt$Title)
# changed cabin character
tt$cabchar <- substr(tt$Cabin,1,1)
tt$cabchar[tt$cabchar %in% c(‘F’,’G’,’T’)] <- 'X';
tt$cabchar <- factor(tt$cabchar)
tt$ncabin <- nchar(as.character(tt$Cabin))
tt$cn <- as.numeric(gsub('[[:space:][:alpha:]]','',tt$Cabin))
tt$oe <- factor(ifelse(!is.na(tt$cn),tt$cn%%2,-1)) tt$Fare[is.na(tt$Fare)]<- median(tt$Fare,na.rm=TRUE)
tt$ticket <- sub('[[:digit:]]+$','',tt$Ticket) tt$ticket <- toupper(gsub('(\\.)|( )|(/)','',tt$ticket)) tt$ticket[tt$ticket %in% c(‘A2′,’A4′,’AQ3′,’AQ4′,’AS’)] <- 'An' tt$ticket[tt$ticket %in% c(‘SCA3′,’SCA4′,’SCAH’,’SC’,’SCAHBASLE’,’SCOW’)] <- 'SC' tt$ticket[tt$ticket %in% c(‘CASOTON’,’SOTONO2′,’SOTONOQ’)] <- 'SOTON' tt$ticket[tt$ticket %in% c(‘STONO2′,’STONOQ’)] <- 'STON' tt$ticket[tt$ticket %in% c(‘C’)] <- 'CA' tt$ticket[tt$ticket %in% c(‘SOC’,’SOP’,’SOPP’)] <- 'SOP' tt$ticket[tt$ticket %in% c(‘SWPP’,’WC’,’WEP’)] <- 'W' tt$ticket[tt$ticket %in% c(‘FA’,’FC’,’FCC’)] <- 'F' tt$ticket[tt$ticket %in% c(‘PP’,’PPP’,’LINE’,’LP’,’SP’)] <- 'PPPP' tt$ticket <- factor(tt$ticket) tt$fare <- cut(tt$Fare,breaks=c(min(tt$Fare)-1,quantile(tt$Fare,seq(.2,.8,.2)),max(tt$Fare)+1))
tt$sibsp=factor(c(1:4,rep(4,6))[tt$SibSp+1])
tt$parch=factor(c(1:4,rep(4,6))[tt$Parch+1])

train <- tt[tt$status=='train',] test <- tt[tt$status=='test',]

#end of preparation and data reading

options(width=90)

#################4444444444444444444##############################

des.matrix <- function(formula,data) {
form2 <- strsplit(as.character(formula),'~',fixed=TRUE)
resp <- form2[[length(form2)]]
form3 <- strsplit(resp,'+',fixed=TRUE)[[1]]
la <- lapply(form3,function(x)
model.matrix(as.formula(paste(‘~’ , x, ‘-1’ )),data) )
nterm <- c(1,sapply(la,ncol))
terms <- rep(1:length(nterm),nterm)
ntrain <- nrow(data)
mat <- do.call(cbind,la)
mat <- cbind(rep(1,ntrain),mat)
np <- ncol(mat)
list(
survived = c(0,1)[data$Survived], np=np, ntrain=nrow(data), terms=terms, nterm=max(terms), tx=mat) } datain <- des.matrix(~ Sex+Pclass,data=train) parameters=c(‘std’,’f’,’log_lik’) my_code <- ' data { int ntrain; int survived[ntrain]; int np; int nterm; int terms[np]; matrix [ntrain,np] tx; } parameters { vector[np] f; real std[nterm]; real stdhyp; } model { stdhyp ~ normal(0,2); std ~ normal(0,stdhyp); for (i in 1:np) { f[i] ~ normal(0,std[terms[i]]); } survived ~ bernoulli_logit(tx*f); } generated quantities { vector [ntrain] log_lik; for (i in 1:ntrain) { log_lik[i] <- bernoulli_logit_log(survived[i], tx[i]*f); } } ‘ fit1 <- stan(model_code = my_code, data = datain, pars=parameters, iter = 1000, chains = 4, open_progress=FALSE) #fit1 #log_lik1 <- extract_log_lik(fit1) #loo1 <- loo(log_lik1) #print(loo1, digits = 3) print.mySmodel <- function(x) { print(x$loo1)
cat(‘\n’)
invisible(x)
}

mySmodel <- function(formula,data) {
datain <- des.matrix(formula,data)

fitx <-
stan(model_code = my_code,
data = datain,
pars=parameters,
fit=fit1,
iter = 2000,
chains = parallel::detectCores(),
open_progress=FALSE)
log_lik1 <- extract_log_lik(fitx)
loo1 <- loo(log_lik1)
ll <-  list(myform=formula,fitx=fitx,loo1=loo1)
class(ll) <- 'mySmodel'
cat(format(formula),loo1$elpd_loo,loo1$se_elpd_loo,’\n’)
ll
}
mySmodel(Survived ~
Title ,
data=train)

################# prediction functions

myPmodel <- function(formula,data) {
datain <- des.matrix(formula,data)

fitx <-
stan(model_code = my_code,
data = datain,
pars=parameters,
fit=fit1,
iter = 2000,
chains = parallel::detectCores(),
open_progress=FALSE)
list(myform=formula,fitx=fitx)
}

PredM <- myPmodel(~ Title + Pclass + sibsp + Title:Pclass + Embarked + oe + Title:sibsp + parch
,data=train)

mySpred <- function(mymodel,newdata) {
pfit <- as.matrix(mymodel$fitx) fmat <- pfit[,grep('^f\\[',colnames(pfit))] px <- des.matrix(mymodel$myform,data=newdata)$tx mylpred <- tcrossprod(fmat,px) mpred <- apply(mylpred,2,function(x) mean(x)) pred <- as.numeric(gtools::inv.logit(mpred)>.5) factor(pred) } preds <- mySpred(PredM,test) out <- data.frame( PassengerId=test$PassengerId,
Survived=as.numeric(as.character(preds)),
row.names=NULL)
write.csv(x=out,
file=’stanqua.csv’,
row.names=FALSE,
quote=FALSE)


text:
Article Contents
Article Contents

# Quiescent phases with distributed exit times

• Diffusive coupling of a dynamical system to a quiescent (zero) phase, with the same rates for all variables, stabilizes against oscillations. When the coupling rates are increased then, at a stationary point, the eigenvalues of the Jacobian matrix with positive real parts and large imaginary parts may move towards the imaginary axis of the complex plane and eventually enter the left half-plane. Diffusive coupling means that holding times in the active and in the quiescent phase are exponentially distributed. Here, we ask whether similar phenomena occur if the exponential distributions are replaced by other distributions. A general stability result can be shown for arbitrary distributions, and several more specific results for Gamma distributions and delta peaks (leading to delay equations). Some of the results apply to traveling fronts in reaction diffusion equations with quiescent phase.
Mathematics Subject Classification: Primary: 34K08, 34K20, 45D05; Secondary: 92B05.

 Citation:


text:
Location of Repository

## Probing Non-leptonic Two-body Decays of $B_c$ meson

### Abstract

Rates and CP asymmetries of the non-leptonic two-body decay of $B_c$ are calculated based on the low energy effective Hamiltonian. We concentrate on such $b$ quark decays of the processes with $0^-$ and $1^-$ S-wave particles and/or $0^+$ and $1^+$ P-wave particles in the final states. The Salpeter method, which is the relativistic instantaneous approximation of the original Bethe-Salpeter equation, is used to derive hadron transition matrix elements. Based on the calculation, it is found that the best decay channels to observe CP violation are $B_c^-\rightarrow \eta_c +D^-(D^{*-}_0)$, which need about $\sim10^7$ $B_c$ events in experiment. Decays to $\eta_c +D^{*-}(D_s^-), h_c+D^-(D^{*-})(D^{*-}_0), J/\Psi+D^{*-}$ are also hopeful channels.Comment: 17 pages, 2 figures, 7 table

Topics: High Energy Physics - Phenomenology, High Energy Physics - Experiment
Year: 2011
DOI identifier: 10.1007/JHEP06(2011)015
OAI identifier: oai:arXiv.org:1102.5399


text:
## Cryptology ePrint Archive: Report 2021/308

Threshold Garbled Circuits and Ad Hoc Secure Computation

Michele Ciampi and Vipul Goyal and Rafail Ostrovsky

Abstract: Garbled Circuits (GCs) represent fundamental and powerful tools in cryptography, and many variants of GCs have been considered since their introduction. An important property of the garbled circuits is that they can be evaluated securely if and only if exactly 1 key for each input wire is obtained: no less and no more.

In this work we study the case when: 1) some of the wire-keys are missing, but we are still interested in computing the output of the garbled circuit and 2) the evaluator of the GC might have both keys for a constant number of wires. We start to study this question in terms of non-interactive multi-party computation (NIMPC) which is strongly connected with GCs. In this notion there is a fixed number of parties ($n$) that can get correlated information from a trusted setup. Then these parties can send an encoding of their input to an evaluator, which can compute the output of the function. Similarly to the notion of ad hoc secure computation proposed by Beimel et al. [ITCS 2016], we consider the case when less than $n$ parties participate in the online phase, and in addition we let these parties colluding with the evaluator. We refer to this notion as Threshold NIMPC. In addition, we show that when the number of parties participating in the online phase is a fixed threshold $l\leq n$ then it is possible to securely evaluate any $l$-input function. We build our result on top of a new secret-sharing scheme (which can be of independent interest) and on the results proposed by Benhamouda, Krawczyk and Rabin [Crypto 2017]. Our protocol can be used to compute any function in NC1 in the information-theoretic setting and any function in $P$ assuming one-way functions. As a second (and main) contribution, we consider a slightly different notion of security in which the number of parties that can participate in the online phase is not specified, and can be any number $c$ above the threshold $l$ (in this case the evaluator cannot collude with the other parties). We solve an open question left open by  Beimel, Ishai and Kushilevitz [Eurocrypt 2017] showing how to build a secure protocol for the case when $c$ is constant, under the Learning with Errors assumption.

Category / Keywords: cryptographic protocols / non-interactive multi-party computation, ad hoc private simultaneous messages, garbled circuits

Original Publication (with major differences): IACR-EUROCRYPT-2021


text:
basic value co to jest
ABAP definicja basic value. Co znaczy that displays the past level of the forecast model, during.
Znaczenie basic value definicja.

Czy przydatne?

# Definicja basic value

Co znaczy:

A part of the forecast model that displays the past level of the forecast model, during forecasting.

Depending on the forecast model used, the system adds a trend value to basic value or it multiplies the basic value by the seasonal index.

The basic value is calculated first by the system by means of historical data and is then updated after every period with the help of the alpha factor.

Słownik i definicje SAPa na .


text:
### Boris Veytsman

[completed 2011-01-19]

Boris Veytsman has written and maintains many LaTeX and font-related packages and recently joined the TUG Board of Directors.

Boris Veytsman, interviewee:     I was born in Odessa, then USSR, and later Ukraine. I got my undergraduate degree, and then PhD from the Odessa National University.

I am a theoretical physicist by training, and I owe a lot to my physics teachers, from grade school to graduate school. I did many things in my life, but they are more or less related to the general idea of understanding the world in mathematical terms — and then teaching others this understanding.

I worked in engineering companies, research institutes, universities and schools. Now I am employed by a high tech company, involved in air traffic modeling and management. I also do some part time research and occasional teaching in the local university — plus some TeX consulting. My research interests are really scattered around: from complex liquids to mathematical genetics to neurology.

I think I am incredibly lucky because I always have been enjoying what I did for a living. You ask about hobbies; sometimes I think that everything I do is a hobby. If you want a hobby in the strict sense of the word — I play Russian style trivia tournaments and support their web pages.

I am married; our son Max is finishing his undergraduate course in the University of Toronto. My wife Maria is a mathematician by training; she works as an engineer, and is an avid ballroom dancer. She plays trivia tournaments with me. Max chose computer science and mathematics.

DW:     How did you first become involved with TeX?

BV:     I started to write (and publish) scientific texts in the pre-computer era. My first paper (and my thesis) was typed on a typewriter with the equations inked in by my wife (my handwriting was never up to the task). Then I discovered Chiwriter and used it a lot — but the quality of output and the ease of use was not good. The same can be said about MS Word Equation Editor, which I also used for some time. In 1994 my colleague at Penn State, where I was a visiting professor, showed me TeX — and I immediately fell in love with its elegance and quality. I think it was OzTeX for Macintosh — a great distribution.

I found out that I can write TeX expressions very fast, almost with the same speed I write them in long hand. Thus I solved a problem that bugged me for a long time: my notes were unreadable even for me because of my awful handwriting. I started to store my calculations and notes in TeX, and now most of my work hours are spent in editing TeX files in an Emacs window.

I started to customize my environment and this led me to TeX programming. My first TeX package (envlab) was written because I wanted to address my envelopes automatically. Actually all my first packages were caused by some need — for example, a journal requirement to create an index of notation used in equations led to nomencl package, etc.

Our son took this love for TeX from me: at the age of eight he wrote his first book (a short novel) completely in LaTeX. He still sometimes calls me with TeX questions.

DW:     What brought you to the United States?

BV:     In the end of 1992, after I got my degree and published a couple of papers, I got an invitation from Penn State for a position of Visiting Professor — basically a glorified postdoc. We (my wife, our then four-year-old son and I) thought it would be a relatively short stay — one or two years. However, after the breakup of the Soviet Union the new states found that they did not have resources or will to support science and technology on the same level as in the USSR. Many people changed their careers: for example, my Quantum Mechanics teacher became a very successful banker. Actually he told me he always wanted to be a banker, but under the Socialist regime there was much more fun in science than in banking. Unfortunately, I never wanted to become a banker or a full time businessman. Anyway, my position in Odessa was eliminated, and there was not much for me to do back there. We decided to stay.

By the way, to keep this interview somewhat TeX-related, I should mention that our applications for green cards were written in LaTeX, and contained some macrocode — these were among my first TeX macros. They were very simple: we ought to supply some evidence to support the applications, so I decided to automatically number the “exhibits”:

   \newcounter{Ex}
\newcommand{\Exhibit}[1]{(Exhibit~\refstepcounter{Ex}\theEx\label{#1})}

and then I could refer to the evidence in the main text using the usual \label-\ref mechanism.

Our petitions were granted (I do not want to claim that the beauty of TeX typesetting helped to convince the immigration officials, but it could be the case), and we have been here since then.

DW:     My memory is that you have an affiliation with George Mason University, but I also know you have some sort of consultancy? How does this all fit together?

BV:     After we came to the United States, I worked at Penn State for six years. Well, full-time university work is fun, but at some time I decided to try to do something completely different and moved to industry. I found that this work is also interesting and fun. However there is one big advantage of working in academia, which I really missed: an easy access to a good library. Someone said that a university is nothing more than a community of scholars around a collection of books and journals. I found that I needed this a lot, so I asked my friends in a university whether they could help. They told me that I could become affiliated with the university, but I must teach. Well, I love teaching, so it was not a problem — and the university was very accommodating, putting my class hours in the evenings. I have been teaching Introduction to Nanoscience — a mix of Quantum Mechanics, Statistics and Solid State customized for our Nanotechnology program. We now are redesigning this program, so I am not sure what will be next for me — but I am sure it will be interesting.

Besides access to the library, this affiliation allowed me to apply for grants, do collaborative research, etc. I still do some research, both in the theory of hydrogen bonded systems (this was the topic of my PhD thesis) and in many other fields. Usually it works this way: I talk to my friends, and they describe some interesting problem. If I can solve it, we publish it — and if this is both interesting and somewhat important, we try to get a grant for mode detailed study.

Consultancy started somewhat differently. Once I opened TUGboat and found that the advertisement fee there was ridiculously low — \$35 for a year. I thought it might be interesting to try and put an ad there. Soon I got my first engagement: LaTeX lectures for helpdesk staff. I really love teaching, so it looked like a good idea. I did this, and then I started to get offers to do other things: writing LaTeX styles for publishers, helping to set up typesetting workflows, and, of course, more lectures and seminars. I learned a lot this way, and it has been a lot of fun.

Of course both university and consultancy require a lot of time, but I enjoy both.

DW:     Do you have any interesting stories to tell about package writing. You have written a bunch of them: http://az.ctan.org/author/id/veytsman

BV:     Well, each package has its own story. Let me tell you just a couple of them. When I wrote several packages for US Army Corps of Engineers, I found out that the tales of military exactitude and discipline are actually true. This package provides some decorative lines for the cover page, and the customer wanted them to exactly correspond to the specifications. I got e-mails like “Please move this line 0.1pt to the left and make it 0.1pt thinner.” I was impressed by this.

Another story is about a package I did not write. I got a request from an astrologer to create a style for customized horoscopes. This presented a dilemma for me: I do not believe in astrology and would not like to participate in such activity. On the other hand, I think a businessman has an implied obligation to try to serve all legal customers — you cannot be thrown out of a restaurant because the owner does not like your face. Fortunately, this customer did not like my rates.

Answering your question about the code not on CTAN. I believe in Free Software, and am a card-carrying member of FSF. Fortunately, most TeX code written for publishers is designed to be distributed: the publishers want the authors to use their styles. However, there are special cases: for example, a customized package to produce journal covers is not really useful for anybody but the people who actually print this journal. I usually put in my contracts the clause that the code is going to be released under LPPL or other free license, but I realize that in some cases it is not practical. Thus I wrote a small number of private packages. I guess RMS might consider this position too opportunistic.

DW:     Do you have a sense for whether usage of TeX is increasing or decreasing in the worlds in which you work?

BV:     I do not have statistics, and the plural of anecdote is not data. What I can say is that I have been feeling recently a lot of interest in TeX from quite different people and organizations, sometimes those that you would not normally mention in the same sentence with TeX. I had quite surprising customers, like the US Census Bureau or a dermatopathologist (he used TeX to format lab reports). I sometimes look at the properties of PDF documents I receive; in many cases the Creator field is set to TeX. So I would say (with a lot of caution) that there are plenty of TeX users around there.

DW:     You've also done a lot with fonts. Has that been a natural outgrowth of job need or was it an inherent interest?

BV:     Well, both. Many customers wanted some font effects, so I got to do this. On the other hand, I love book design and fonts: I have read some books on the subject and spent time studying letter forms. I am by no means an artist, and never designed fonts myself — but I appreciate the beauty of the work of others.

DW:     Tell me about the diversity of your TeX format and distribution use. And how is the evolution of TeX engines and distributions working out for you?

BV:     Almost all my work is in LaTeX; of course, serious LaTeX programming requires some knowledge of plain TeX. I would like to have a customer who needs ConTeXt work: this would provide me the incentive to learn this format. It seems that most publishers, at least here in the USA, work with LaTeX. Maybe the situation in Europe is different.

I started with OzTeX for Macintosh. I still fondly remember its features — even the ugly DVI previewer. Then I switched to Linux and teTeX, and later to TeX Live. I use more or less the usual setup for Linux: Emacs with AUCTeX as a front end, pdfetex as the engine. I use PSTricks for graphics and Gnuplot for most of my plots (I need to learn Asymptote, and I plan to do this at some point). Many of my plots and figures are automatically generated, so I developed over the years a complex set of Makefiles.

DW:     At at least three TUG conferences I remember, you have reported on work done jointly with Leila Akhmadeeva. How did that collaboration come about and where is it going?

BV:     I have known Leila for more than a decade now. Our common friend introduced us since she had an interesting mathematical problem, where I could help. She studied Myotonic Dystrophy, a genetic disease with an unusual mechanism of inheritance. We developed a theory for it and published it in biology and medical journals. Then I was asked to co-advise her thesis. Well, it was a medical thesis, so I had some misgivings — I know nothing about medicine. But I was told that the main advisor was a prominent physician, and I ought to consult with her about “just a couple of chapters with the equations”. I thought this might be interesting, so I took this offer, and everything went fine. I liked this experience, so we started to look for whether there were some other areas where we could do something interesting together.

Now Leila is a great practicing physician and a Neurology Professor with the Bashkir State Medical University (she was among the youngest full professors in Russia when she got this position). We continue to cooperate. We published some stuff about lower back pains, genetics, etc. The pedigree work is actually a spinoff of another project: we plan to write a textbook for medical students with problems about hereditary disorders and diagnosis, and I wanted the illustrations to look good. So I wrote an algorithm to draw pedigrees. It seems that we created a nice tool, and some people — including students and professionals — seem to find it useful. We have now got some interesting ideas about stabilometry and posture studies, so we plan to continue and extend this collaboration.

DW:     Many people gripe about the difficulty of writing packages in TeX. What is your take on this?

BV:     Macro programming is indeed quite hard. It is very different from the other styles of programming, and sometimes looks quite alien. Especially for people that got used to the other languages. When I code in this language, I am painfully reminded of the fact that DEK is a genius — and I am not. The LaTeX3 project tries to change the situation, making TeX programming easy for the rest of us. It will be interesting to see the results of this effort.

Having said this, I think that there is a good side in this strangeness. Computer languages are meant to express thoughts, much like natural languages are. However, it is virtually impossible to separate the thought from its expression. Thus when we are exposed to different languages — computer or natural ones — we extend the ways we can express our thoughts, and by this we expand our thoughts themselves. TeX programming is hard — but it is also rewarding.

DW:     Thank you, Boris, for taking the time to participate in this interview. I look forward to seeing you again in person at a future TUG conference.

Interview pages regenerated January 26, 2017;


text:
# Nisnevich covers of algebraic spaces

Does every algebraic space have a Nisnevich cover by a scheme?

(Assume that the algebraic space is quasi-separated, quasi-compact and over a scheme $S$.)

Background:

Every algebraic space has an etale cover by a scheme by definition: an algebraic space is a sheaf $$F : (Sch/S)_{fppf} \to (Set)$$ with representable diagonal $\Delta : F \to F \times F$ and an etale, surjective morphism $U \to F$ from a sheaf associated to an $S$-scheme $U$.

A Nisnevich cover in this context is a surjective, etale morphism $f : U \to X$ such that for all $x \in X$, there exists a $u \in U$ with $f(u) = x$ such that the induced morphism $k(x) \to k(u)$ is an isomorphism.

-
Yes, if you restrict to quasi-separated algebraic spaces (to avoid pathologies). Donald Knutson proved that for each $x \in X$ there is an etale scheme neighborhood $(U,u) \rightarrow (X,x)$ with $k(u) = k(x)$. (This underlies how $k(x)$ is defined.) So with a gigantic disjoint union you get a huge Nisnevich scheme cover. If $X$ is also quasi-compact then it has a dense open subscheme $V$ (also proved by Knutson) and by noetherian induction you get a quasi-compact Nisnevich scheme cover applying the previous Knutson theorem at the finitely many generic points of $X-V$. – user30180 Jun 8 '13 at 3:29
In the preceding comment, to carry out noetherian induction it is of course necessary to assume $X$ is noetherian. But the general qcqs case can be reduced to the noetherian case because any qcqs algebraic space is affine over a noetherian algebraic space (by a variant of the result of Thomason & Trobaugh according to which any qcqs scheme is affine over a noetherian scheme). – user30180 Jun 8 '13 at 5:14
A reference for what ayanta explains above is [Knutson, Algebraic spaces, Chapter 2, Theorem 6.4]. – Olivier Benoist Jun 8 '13 at 13:30
Ayanta, thanks for the thorough explanation. If you post it as an answer, I can accept it. I should have specified quasi-separated in the definition, and quasi-compact is a reasonable hypothesis. I should have also clarified what you assumed about etale meaning locally of finite presentation. A definition of $k(x)$ in terms of a universal factorization of $Spec K \to X$ (for $X$ locally noetherian) can be found in Laumon,Moret-Bailly, Definition 11.2. Maybe it uses the result you cite...I need Knuston's book. Olivier, I don't have the book in front of me, but thanks for the reference. – expz Jun 10 '13 at 10:55


text:
# An algebra problem by Department 8

Algebra Level 5

$\large{ \left\lfloor \dfrac{3x-1}{16}\right\rfloor+\left\lceil \dfrac{16}{3x-1} \right\rceil =3x }$

Find the value of $$x$$ that satisfies the above equation.

×


text:
# naginterfaces.library.stat.prob_hypergeom_vector¶

naginterfaces.library.stat.prob_hypergeom_vector(n, l, m, k)[source]

prob_hypergeom_vector returns a number of the lower tail, upper tail and point probabilities for the hypergeometric distribution.

For full information please refer to the NAG Library document for g01sl

https://www.nag.com/numeric/nl/nagdoc_29/flhtml/g01/g01slf.html

Parameters
nint, array-like, shape

, the parameter of the hypergeometric distribution.

lint, array-like, shape

, the parameter of the hypergeometric distribution.

mint, array-like, shape

, the parameter of the hypergeometric distribution.

kint, array-like, shape

, the integer which defines the required probabilities.

Returns
plekfloat, ndarray, shape

, the lower tail probabilities.

pgtkfloat, ndarray, shape

, the upper tail probabilities.

peqkfloat, ndarray, shape

, the point probabilities.

ivalidint, ndarray, shape

indicates any errors with the input arguments, with

No error.

On entry, .

On entry, , or, .

On entry, , or, .

On entry, , or, , or, , or, .

On entry, is too large to be represented exactly as a real number.

On entry, the variance (see Notes) exceeds .

Raises
NagValueError
(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

Warns
NagAlgorithmicWarning
(errno )

On entry, at least one value of , , or was invalid, or the variance was too large.

Notes

Let denote a vector of random variables having a hypergeometric distribution with parameters , and . Then

where , and .

The hypergeometric distribution may arise if in a population of size a number are marked. From this population a sample of size is drawn and of these are observed to be marked.

The mean of the distribution , and the variance .

prob_hypergeom_vector computes for given , , and the probabilities: , and using an algorithm similar to that described in Knüsel (1986) for the Poisson distribution.

The input arrays to this function are designed to allow maximum flexibility in the supply of vector arguments by re-using elements of any arrays that are shorter than the total number of evaluations required. See the G01 Introduction for further information.

References

Knüsel, L, 1986, Computation of the chi-square and Poisson distribution, SIAM J. Sci. Statist. Comput. (7), 1022–1036


text:
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collecte

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### Question Asked by a Student from EXXAMM.com Team

Q 3073112946.     A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Which method did you use for finding the mean, and why?

#### HINT

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text:
# Category: Processing

## Using Processing 2.0 with Notepad++ or SublimeText as external editor

I have used Processing with an external editor for years now, and I relied heavily on the “use external editor” option in the IDE. That option is now gone, and has been replaced by a command line option instead.

At first the removal of that option annoyed me, but it has grown on me after I managed to set it up properly.

To set it up in Notepad++ you have to install a plug-in called NppExec. That can be done with the plugin manager.

When that is done, open up the source file for a Processing sketch, and press F6.

As command enter:

C:\path\to\processing\processing-java.exe --run --force --sketch="$(CURRENT_DIRECTORY)" --output=c:\temp\p5 Just change it to point to the correct path, and for output have some temporary folder set up. Works like a charm As I was also tipped about Sublime Text on Twitter after posting this, I decided to try it out. I sat up a new build system with the following lines: { "cmd": ["C:\\path\\to\\processing\\processing-java", "--sketch=$file_path", "--run", "--output=c:\\temp\\p5", "--force"] }

Which also seems to work very well. As for how that editor is, I guess I will find out.

## Processing.js plugin for WordPress

I just installed the Processing JS WordPress plugin, and this is just a quick test to see if, and how, it works.

## My first commissioned work

At the last half of 2010, I was contacted by one of the professors at the department of music technology at NTNU, to do a work for them, and I gladly accepted.

They wanted some sort of permanent installation that would make their hallways a bit more interesting, and they had previously seen one of my prints from the “Wasted time” project, and so I used that as a basis when I started thinking.

"Wasted time, 2009-03-27 11:49:01"

I also wanted to do something that was tightly connected with the department, and what they do, so sound would have to be, in some way, an element in the work.

As the work was to be permanent, and is to be there for a long time, I wanted to make something that would need little to no maintenance, and not have the risk of stop working in some way. It should also not be to obtrusive, since people will need to walk past it every day, and I don’t want it to end up be an annoyance to the people who use the premises. With that in mind, I decided early on that I wanted to make some sort of generative prints, and started checking out possibilities at a prints shop. The choice I made was to make prints on acrylic plates.

I went to the location, and after deciding where I wanted the plates to hang, when they were done, I recorded the ambient sound in the hallways, with microphones placed at the spots where the pictures would be. The sound was then cut to find some interesting segments and then normalized. I then used the sound as data for drawing curves, circles and lines. This was done with utilizing the language Processing and the library minim.

## Contour tracing in Processing

I recently read the book Aarons code, and with that the thought of writing some contour tracing functions came up. When I also, a few days later got an idea for a project, I got to it. Do searches on Google pointed me in the direction of the blob extraction algorithm that OpenCV uses, and so I found “A linear-time component labeling algorithm using contour tracing technique”.

Lately I have also started looking into using NetBeans as an IDE for programming with the Processing library, so I decided that I would implement the algorithm in the paper in java, using Processing.

The code is provided below, and is provided as is. It can probably be optimized a bit, and/or rewritten to be better. But for now it works for me.

 /*
* An implementation of
* «A linear-time component labeling algorithm using contour tracing technique»
* by Fu Chang, Chun-Jen Chen, and Chi-Jen Lu
*
* Written in Java for use with the Processing (www.processing.org) library.
*
* Code written by Greger Stolt Nilsen
* (gregersn@gmail.com / http://gregerstoltnilsen.net)
* Copyright 2010, Greger Stolt Nilsen
*
* Code is provided «as is», and no warrenty is given. Use at your own risk.
*
* This program is free software: you can redistribute it and/or modify
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.

This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
GNU General Public License for more details.

You should have received a copy of the GNU General Public License
along with this program.  If not, see <http://www.gnu.org/licenses/>.

*/
/*
* Create a new instance with:
* ComponentFinder finder = new ComponentFinder(this, image);
* where "image" is the PImage (or PGraphics) to be traced.
* use the find()-function, to get all the contours in the contours arraylist.
* Each contour is by itself an ArrayList of PVectors.
*
* render_blobs() will return a PImage object that is a rendering of all the
* blobs in the traced image.
*
* The Image to be traced need to be black (0x000000) objects on white background.
* So use the threshold-filter to get to that.
*
* Also, unless the outer border of pixles is white, the finder will hang.
*/

package ComponentFinder;

import processing.core.*;
import java.util.ArrayList;

public class ComponentFinder
{
PApplet parent;
public ArrayList contours;
int[] inout;
int[] L;
PImage I;
int C;

// An array with the 8 direction coordinates (x1, y1, x2, y2... x8, y8)
int cd[] = {1, 0, 1, 1, 0, 1, -1, 1, -1, 0, -1, -1, 0, -1, 1, -1 };

public ComponentFinder(PApplet p, PImage img)
{
parent = p;
I = img;
C = 1;
L = new int[I.width*I.height];

}

int getNumContours()
{
if(contours != null)
return contours.size();
else return 0;
}

public int find()
{
// Initialize the blob map to zero.
for(int i = 0; i < I.width*I.height; i++)
{
L[i] = 0;
}

// Make a new array list to store each contour.
contours = new ArrayList();

C = 1;

int pos = I.width; // Start analyzing the image at the second line

// Label the first line as blank pixels.
for(int x = 0; x < I.width; x++)
L[x] = -1;

for(int y = 1; y < I.height; y++)
{
for(int x = 0; x < I.width; x++)
{
if(L[pos] == 0) // Only process unlabeled pixels
{
if(((I.pixels[pos]&0xffffff) != 0x000000))
{
// White pixel
L[pos] = -1;
}
else if(((I.pixels[pos]&0xffffff) == 0x000000))
{
if((I.pixels[pos-I.width]&0xffffff) != 0x000000)
{
// If the pixel above is a white pixel,
// then this is the start of an outer contour
L[pos] = C;
// Start tracing an outer contour
C++;
}
if(((I.pixels[pos+I.width]&0xffffff) != 0x000000) && (L[pos+I.width] == 0))
{
// If the pixel below is a white pixel, then it is the start of an
// inner contour, and we get the id for the outer contour
// then start tracing the inner contour
int l;
if(L[pos] != 0 ) l = L[pos];
else l  = L[pos-1];

if(L[pos] != 0) L[pos] = l;
}
if(L[pos] == 0) if(L[pos-1] > 0) L[pos] = L[pos-1];
}
}
pos++;
}
}
I.updatePixels();
return getNumContours();
}

// Traces a countour at the given pixel S,
// r is set to 0 for outer contours, and 1 for inner contours.
// c is the label to give the contour
ArrayList contourTrace(PVector S, int r, int c)
{
int next = 8;
ArrayList cont = new ArrayList(32);
PVector current = new PVector();
if(r == 0) // Outer contour
next = Trace((PVector)cont.get(cont.size()-1), 7, c);
else if(r == 1) // Inner contour
next = Trace((PVector)cont.get(cont.size()-1), 3, c);

if(next == 8) return cont; // The pixel is isolated.

PVector T = new PVector(S.x + cd[next<<1], S.y + cd[(next<<1)+1]);

next = Trace((PVector)cont.get(cont.size()-1), (next+6)%8, c);

if(next == 8) return cont;

while(true)
{

current.x = (float)((PVector)cont.get(cont.size()-1)).x + cd[(next<<1)%16];
current.y = (float)((PVector)cont.get(cont.size()-1)).y + cd[((next<<1)+1)%16];
next = Trace(current, (next+6)%8, c);

// Check if we are back at start position
if((current.x == S.x) && (current.y == S.y) && (current.x + cd[next<<1] == T.x) && (current.y + cd[(next<<1)+1] == T.y))
{
return cont;
}
else
{
}
}
}

int Trace(PVector P, int n, int c)
{
for(int i = 0; i < 8; i++)
{
// if pixel at position n+i is black, return n+i
PVector ni = new PVector(P.x+cd[((i+n)%8)*2],
P.y+cd[((i+n)%8)*2+1]);

if((I.pixels[(int)(ni.y)*I.width+(int)(ni.x)]&0xffffff) == 0x000000)
{

L[(int)(ni.y)*I.width+(int)(ni.x)] = c;
return (i+n)%8;
}
// if not, mark with a negative number
else
{
if(L[(int)(ni.y)*I.width+(int)(ni.x)] == 0)
L[(int)(ni.y)*I.width+(int)(ni.x)] = -1;
}
}
return 8; // Should only happen for isolated pixels.
}

public PImage render_blobs()
{
PImage blobs = parent.createImage(I.width, I.height, PApplet.RGB);
parent.colorMode(PApplet.HSB);
for(int i = 0; i < L.length; i++)
{
if(L[i] > 0)
blobs.pixels[i] = parent.color(PApplet.map(L[i], 1, C, 0, 255), 255, 255);//|0xff000000;
if(L[i] == -1)
blobs.pixels[i] = parent.color(0, 0, 255);
if(L[i] == -2)
blobs.pixels[i] = parent.color(0, 0, 0);
if(L[i] == 0)
blobs.pixels[i] = parent.color(0, 0, 128);
}
blobs.updatePixels();
return blobs;

}
}


text:
# Spin, orbital angular momentum and total angular momentum

If I understand correctly, spin is an intrinsic property of particles, which follows the algebra of angular momentum, but has nothing to do with an "orbital angular momentum" in that the particle is not like a small sphere that rotates on itself, to which we could attribute an angular momentum as usual in classical mechanics, cf. this and this Phys.SE posts.

My question is: why does spin combine with orbital angular momentum to give total angular momentum? To me this is surprising, as spin, in fact, has nothing to do with a classical angular momentum. Any insight on that?

• Check out the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Also see the electron magnetic moment, see what Amey said below, and see Gouldsmit on the discovery of electron spin: "It means that the electron has a spin, that it rotates". – John Duffield May 17 '15 at 13:35
• The theoretical explanation of ACuriousMind is correct, but you have to put the "faith" part into the right physical perspective. "Faith" always means "experimental observation" and it is also linked to a scale argument: an orbital angular momentum, when seen from a "large enough" scale, is indistinguishable from an intrinsic spin. This is already the case in classical mechanics. An object with "zero radius" is physically the same as an object with "a small enough radius". In that sense spin could indeed be an internal current of a compound object that we can't resolve, yet. – CuriousOne May 17 '15 at 15:46
• John - What is Gouldsmit talking about? The electron "rotates?" For starters, I've heard it is a "point particle" which doesn't even a radius we can measure? – Frank May 17 '15 at 17:57

The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to take it "on faith" or rather, as an experimental fact that spin is indeed (part of) the Noether charge associated to spatial rotations and not some other $\mathfrak{so}(3)$. But, when you enter QFT, you will find that every quantum field should transform in some representation of the spatial rotation group (or rather in relativistic QFT, in some representation of the Lorentz group, of which the rotations are a subgroup), and that is exactly what spin then is - the "label" of the representation the quantum field transforms in.
• @Frank: For a single non-relativistic particle in 3D, you would have $L^2(\mathbb{R}^3)\otimes\mathcal{H}_s$ as its space of states, where $\mathcal{H}_s$ is the $\mathfrak{so}(3)$-rep associated to the spin value $s$. The orbital angular momentum now comes from decomposing the "wavefunction space" $L^2(\mathbb{R}^3) = \oplus_l\mathcal{H}_l$, where $l$ is orbital angular momentum. Then you have $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s$, and you can now decompose this again as $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s = \oplus_j \mathcal{H}_j$ with $j$ total angular momentum. – ACuriousMind May 16 '15 at 22:53


text:
Poster
Algorithmic Regularization in Learning Deep Homogeneous Models: Layers are Automatically Balanced
Simon Du · Wei Hu · Jason Lee

Wed Dec 5th 05:00 -- 07:00 PM @ Room 210 #53
We study the implicit regularization imposed by gradient descent for learning multi-layer homogeneous functions including feed-forward fully connected and convolutional deep neural networks with linear, ReLU or Leaky ReLU activation. We rigorously prove that gradient flow (i.e. gradient descent with infinitesimal step size) effectively enforces the differences between squared norms across different layers to remain invariant without any explicit regularization. This result implies that if the weights are initially small, gradient flow automatically balances the magnitudes of all layers. Using a discretization argument, we analyze gradient descent with positive step size for the non-convex low-rank asymmetric matrix factorization problem without any regularization. Inspired by our findings for gradient flow, we prove that gradient descent with step sizes $\eta_t=O(t^{−(1/2+\delta)}) (0<\delta\le1/2)$ automatically balances two low-rank factors and converges to a bounded global optimum. Furthermore, for rank-1 asymmetric matrix factorization we give a finer analysis showing gradient descent with constant step size converges to the global minimum at a globally linear rate. We believe that the idea of examining the invariance imposed by first order algorithms in learning homogeneous models could serve as a fundamental building block for studying optimization for learning deep models.


text:
# Generalized stone residuated lattices

Document Type : Research Paper

Author

Department of Mathematics, College of science, Persian Gulf University, Bushehr, 7516913817, Iran

10.29252/as.2020.1885

Abstract

This paper introduces and investigates the notion of a generalized Stone residuated lattice. It is observed that a residuated lattice is generalized Stone if and only if it is quasicomplemented and normal. Also, it is proved that a finite residuated lattice is generalized Stone if and only if it is normal. A characterization for generalized Stone residuated lattices is given by means of the new notion of $\alpha$-filters. Finally, it is shown that each non-unit element of a directly indecomposable generalized Stone residuated lattice is a dense element.

Keywords

#### References

[1] G. Birkho , Lattice theory, Vol. 25, American Mathematical Soc. (1940).
[2] C. C. Chang, Algebraic analysis of many valued logics, Transactions of the American Mathematical society Vol. 88 No. 2 (1958), pp. 467-490.
[3] R. Cignoli and A. T. Torrell, Boolean products of MV-algebras: hypernormal MV-algebras, Journal of mathematical analysis and applications Vol. 199 No. 3 (1996), pp. 637-653.
[4] L. C. Ciungu, Classes of residuated lattices, Annals of the University of Craiova Mathematics and Computer Science Series Vol. 33 (2006), pp. 189-207.
[5] W. H. Cornish, Normal lattices, Journal of the Australian Mathematical Society Vol. 14 No. 2 (1972), pp.200-215.
[6] W. H. Cornish, Annulets and -ideals in a distributive lattice, Journal of the Australian Mathematical Society Vol. 15 No. 1 (1973), pp. 70-77.
[7] A. Di Nola and S. Sessa, On MV-algebras of continuous functions, in Non-classical Logics and Their Applications to fuzzy Subsets, Springer, (1995), pp. 23-32.
[8] N. Galatos, P. Jipsen, T. Kowalski and H. Ono, Residuated lattices: an algebraic glimpse at substructural logics, Vol. 151, Elsevier (2007).
[9] L. Gillman and M. Jerison, Rings of continuous functions, Courier Dover Publications (2017).
[10] G. Gratzer, Lattice theory: Foundation, Springer Science & Business Media (2011).
[11] G. Gratzer and E. T. Schmidt, On a problem of mh stone, Acta Mathematica Hungarica Vol. 8 No. 3-4 (1957), pp. 455-460.
[12] T. Katrinak, Remarks on stone lattices I,(russian) math, Fyz. Casopis Vol. 16 (1966), pp. 128.
[13] S. Rasouli, Generalized co-annihilators in residuated lattices, Annals of the University of Craiova Mathematics and Computer Science Series Vol. 45 No. 2 (2018a), pp. 190-207.
[14] S. Rasouli, Heyting, Boolean and Pseudo-MV lters in residuated lattices, J. Multiple Valued Log. Soft Comput. Vol 31 No 4 (2018b), pp. 287-322.
[15] S. Rasouli, The going-up and going-down theorems in residuated lattices, Soft Computing Vol. 23 No. 17 (2019), pp. 7621-7635.
[16] S. Rasouli, Quasicomplemented residuated lattices, Soft Computing Vol. 24 (2020), pp. 6591-6602.
[17] S. Rasouli and B. Davvaz, An investigation on boolean prime lters in BL-algebras, Soft Computing Vol. 19 No. 10 (2015), pp. 2743-2750.
[18] S. Rasouli and A. Dehghani, Topological residuated lattices, Soft Computing Vol. 24 No. 5 (2020), pp. 3179-3192.
[19] S. Rasouli and M. Kondo, n-normal residuated lattices, Soft Computing Vol. 24 No. 1 (2020), pp. 247-258.
[20] S. Rasouli and A. Radfar, PMTL lters, Rl lters and PBL lters in residuated lattices, Journal of Multiple Valued Logic and Soft Computing Vol. 29 No. 6 (2017), pp. 551-576.
[21] H. P. Sankappanavar and S. Burris, A course in universal algebra, Graduate Texts Math 78 (1981).
[22] T. Speed, Some remarks on a class of distributive lattices, Journal of the Australian Mathematical Society Vol. 9 No. 3-4 (1969a), pp. 289-296.
[23] T. Speed, Two congruences on distributive lattices, Bulletin de la Societe Royale des Sciences de Liege (1969b).
[24] J. Varlet, Contribution a l'etude des treillis pseudocomplementes et des treillis de Stone, Siege de la societe:Universite (1963).
[25] J. Varlet, A generalization of the notion of pseudo-complementedness, Bull. Soc. Roy. Sci. Liege Vol. 37 (1968), pp. 149-158.


text:
# Relativistic Force and Momentum

1. Oct 5, 2005

### Jibobo

I've been having a lot of trouble with this problem. There's definitely something I'm missing and it most likely has to do with the force.

"A mass $$m$$ is thrown from the origin at $$t = 0$$ with initial three-momentum $$p_0$$ in the y direction. If it is subject to a constant force $$F_0$$ in the x direction, find its velocity $$v$$ as a function of $$t$$ and by integrating $$v$$, find its trajectory. You will need to integrate functions such as $$t/\sqrt{a + bt^2}$$ and $$1/\sqrt{a + bt^2}$$.

In addition, check that in the non-relativistic limit, ($$c \rightarrow \infty$$), $$x(t)$$ is what you expect for motion in a constant field and check that $$y(t)$$ is what you expect for motion in a constant field when the force is orthogonal to the y direction.

HINT: You will need the Taylor expansion of the functions $$\sqrt{1 + x}$$ and $$\ln(1 + x)$$."

The 2nd part seems easy, but I'm simply not sure how to find $$x(t)$$ or $$y(t)$$ in the first place.

My work so far:
$$\gamma = 1/\sqrt(1 - v^2/c^2)\\ \gamma_v_0 = 1/\sqrt(1 - {v_0}^2/c^2)$$
$$p_0 = (0, p_0, 0) = m*\gamma_v0*(0, v_0, 0)$$
$$F_0 = (F_0, 0, 0)$$
$$F = dP/dt, \mbox{so } P - p_0 = F*t$$
$$P = F_0*t + p_0 = m*\gamma*v$$
$$v = (v_x, v_y, v_z), v_z = 0$$
$$m*\gamma*v_x = F_0*t$$
$$m*\gamma*v_y = p_0 = m*\gamma_v0*v_0$$

I'm not exactly sure how to proceed from here since I can't really isolate $$v_x$$ or $$v_y$$ because the gamma term contains only the magnitude of $$v$$. Should I use $$\|v\| = \sqrt{v_x^2 + v_y^2}$$ and then work through some really terrible algebra? Or is this even the right way to approach this problem?

Edit: I've actually done the terrible alegbra using $$\|v\| = \sqrt{v_x^2 + v_y^2}$$, but the equations I end up with are ridiculous. Can anyone suggest a different method?

Last edited: Oct 6, 2005
2. Oct 12, 2005

### pervect

Staff Emeritus
The 4-momentum of the particle, which I'll call P, is supposed to be a 4-vector.

Write down the four components of P:

E (the energy)
Px (the momentum in the x direction)
Py (the momentum in the y direction)
Pz (the momentum in the z direction, which is very easy, it's zero)

You have something that you call P, but it doesn't appear to be a 4-vector.


text:
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text:
Features: Faculty Insights

## James Maynard, an alumnus of the Maths Faculty, has won one of this year's Fields Medals at the International Congress of Mathematicians (ICM). The Fields Medal is one of the most prestigious prizes in mathematics, awarded every four years "to recognise outstanding mathematical achievement for existing work and for the promise of future achievement".

Up to four mathematicians under the age of 40 are awarded a Fields Medal at each ICM. Cambridge mathematician Caucher Birkar won a Fields Medal at the 2018 ICM. This year Maynard has been recognised for his "spectacular contributions in analytic number theory".

Maynard, now professor of number theory at the Mathematical Institute in Oxford, completed his undergraduate degree in the University of Cambridge Maths Faculty as a student at Queens College, and completed Part III in 2009. Part III, the final, optional, year of the Cambridge Mathematical Tripos, is one of the oldest taught masters courses in the world. Attracting outstanding students from across the globe, it is renowned for teaching content based on cutting-edge research that may not even have made it into the textbooks yet.

Maynard says: "I feel very grateful for the mathematical education I had at Cambridge, it was an amazing experience for me."

### The twin prime conjecture

"Number theory always had a slightly special status in my mind, even before I formally learned about it," says Maynard. Number theory – the study of whole numbers and how they can be combined with each other – has provided Maynard with an academic playground to build on his childhood fascination.

Maynard's first significant contribution in number theory came when he was a postdoctoral student at the University of Montreal when he was working on the twin prime conjecture. We've known for thousands of years that there are infinitely many primes but there is no discernible pattern in how they are sprinkled along the number line. "Typically as you go down the number line the gaps between prime numbers get bigger," says Maynard. "But the [twin prime conjecture] is saying that even if the gaps get typically bigger, occasionally you get these gaps where the primes are very close together. Understanding the gaps between prime numbers is fundamental to understanding the distribution of the primes."

Apart from the number 2, all the other primes are odd numbers, so the closest two prime numbers could ever be (once you get past the number 2) is separated by a difference of two. Mathematicians believe there are infinitely many pairs of twin primes which is known as the twin prime conjecture.

The twin prime conjecture is one of those famous problems in number theory that are simple to state and have fascinated mathematicians for hundreds of years and yet a proof still remains out of reach. After centuries of effort there was a big breakthrough in 2013, when Yitan Zhang proved that there were infinitely many pairs of primes that are separated by 70 million. "For mathematicians this was absolutely a huge breakthrough as this was the first time we had [a proof of gaps of] any finite number," says Maynard. "70 million is much bigger than 2 but it's a lot smaller than infinity."

Zhang's breakthrough involved sieve methods, which are ways of filtering numbers in your proof. An elementary example is the sieve of Eratosthenes where you filter out non-prime numbers up to, say 100, by crossing out all the multiples of 2, then all the multiples of the next remaining number, 3, then the multiples of the next remaining number, 5, and so on. Sundaram's sieve also filters out the non-prime numbers, but is based on arithmetic progressions – sequences of numbers that are separated from their neighbours by a fixed difference.

"Sieve methods are a mathematical tool for translating some information you understand about numbers to create some information you'd like to know," says Maynard. "[For example] if you understand certain technical questions about primes in arithmetic progressions, then you can translate that to information about small gaps between primes." Sieve methods might seem easy at first, but sometimes very strong mathematical results are needed to make them work. Zhang's breakthrough involved proving strong results about primes that could provide the input for the sieving method.

Maynard's approach was different: "Rather than improving the input to the method I changed the method itself. It became more efficient in turning one type of information into another and it meant that we needed much weaker inputs to get a result about boundary gaps between primes." With this new method he dramatically reduced the gap from 70 million to just 600. And after a flurry of collaborative work with a group of mathematicians, we now know there are infinitely many pairs of primes separated by a gap of just 246.

Even after such dramatic progress a proof of the twin prime conjecture still remains elusive. Work continues, but often involves taking new approaches. An example is Maynard's work proving there are infinitely many primes without certain digits. It's hard to know at this stage when the twin prime conjecture will finally be proved in full, but we came away from our discussion with Maynard feeling optimistic. "We're still one big idea away from proving the twin prime conjecture, but maybe we're only one big idea away."

### All or nothing

Number theory has a wealth of long-standing conjectures and open problems. A proof of the twin prime conjecture might still be a way off, but Maynard recently proved another significant conjecture with his colleague Dimitris Koukoulopoulos.

The Duffin-Schaeffer conjecture, first posed in 1941, concerns how well you can approximate real numbers with rational numbers. The real numbers are made up of rational numbers – numbers that can be written as a fraction $p/q$ for some integers $p$ and $q$ – and irrational numbers that can't be written in this way. For example the famous number $\pi$, 3.14151... is irrational, so it can't be written down as a fraction. What we can do, however, is approximate it by a fraction. . For example, when we only use the first two decimal places of $\pi$, 3.14, to approximate $\pi$, then what we are doing is approximating it by the fraction 314/100. But that approximation involves pretty big numbers, and actually the fraction 22/7 is a more accurate approximation of $\pi$.

"This means that 22/7 is somehow a more efficient way of capturing some of the arithmetic information about the number $\pi$," says Maynard. Understanding efficient approximations of real numbers (also known as diophantine approximations) and how they are distributed for the real numbers can lead to very important information for number theorists. The Duffin-Schaeffer conjecture makes it very straightforward to tell when these efficient approximations do and don't exist.

Suppose you want to guarantee your approximations will have a certain accuracy, and this accuracy can vary with the denominator $q$ used in your approximations, $p/q$. Then the Duffin-Schaeffer conjecture says there is an easy calculation that tells you you'll either have your specified type of efficient approximations for almost all the numbers, or for almost none of them.

"The Duffin-Schaeffer conjecture says either you only have a rare set of exceptions to this statement about efficient approximations, or it is basically never ever true," says Maynard. "And [the conjecture] gives you a very simple step to know whether you are ok or not."

This might not sound so useful at first, but it gives mathematicians a powerful tool. "There's lots of mathematical statements that you might wish were true for absolutely everything but it turns out there's a few annoying exceptions," says Maynard. "But provided those exceptions are suitably rare, the result tells us they don't actually matter."

### Prizes for playing

Maynard's work is described in the citation for his prize as "highly ingenious, often leading to surprising breakthroughs on important problems that seemed to be inaccessible by current techniques." But despite his many successes, the Fields Medal still comes as a surprise. "You don't think of yourself as getting a huge mathematical honour when you're sitting at your desk playing around with mathematical bobbins!"

And while receiving one of the biggest prizes in mathematics is a huge honour, Maynard also finds it daunting, and slightly surreal. "In some ways it's intimidating thinking of my name on this list of legends of mathematics from the past. People I looked up to, when I was a kid and thinking about mathematicians," he says. "It's certainly quite surreal in that way."

This year five Cambridge mathematicians have been invited to speak at the International Congress of Mathematicians – you can find out more about them here.

The photo of James Maynard is by Ryan Cowan and used by permission.


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# Sixfold fun in 2015, Part 37

$S=\sum_{d|12090}\frac{1}{\phi(d)}$

Find the smallest positive integer $$a$$ such that $$a\times S$$ is an integer.

Clarifications

• The sum $$S$$ is taken over all 32 positive integer divisors $$d$$ of $$12090=6\times 2015$$

• $$\phi(d)$$ denotes the Euler's totient function.

×


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# Why do we use only 1/2 for continuity correction in case of approximating binomial random varable to a standard normal random variable?

I have read about continuity correction in case of approximating a binomial random variable to a standard normal variable. But in all the examples , they only use 1/2 as a continuity correction factor.Let's take an example:

Let X be the number of times that a fair coin, flipped 40 times,lands heads. Find the probability that X=20. X is a binomial random variable with mean 20 and variance 10. The actual answer is 12.54%

Use the normal approximation for this question.

\begin{align} Ans: P(X=20) = P(19.5 <= X <= 20.5) = P( \frac{19.5-20}{\sqrt(10)} < \frac{X-20}{\sqrt(10)} < \frac{20.5-20}{\sqrt(10)} ) = P(-0.16 < \frac{X-20}{\sqrt(10)}< 0.16) = fi(-0.16) - fi(0.16) = 12.72 \% \end{align}

         ; where fi is the distribution function of a normal random variable

This is pretty close to the actual answer..

Now if I could have written

\begin{align} P(X=20) = P(19.99 <= X <= 20.01) = P( \frac{19.99-20}{\sqrt(10)} < \frac{X-20}{\sqrt(10)} < \frac{20.01-20}{\sqrt(10)} ) = P(-0.031 < \frac{X-20}{\sqrt(10)}< 0.031) = fi(0.031) - fi(-0.031) = 2.4 \% \end{align} I have brought the range very close to 20 instead of just using (19.5,20.5) .

Why is the approximation in second case worst than that in first case ??

First of all, you did the second calculation incorrectly. You should have had $F(0.0032)-F(-0.0032)$. If you do this, you will see that the result of the second calculation is approximately 50 times smaller than the result of the first. (This comes from continuity of the normal PDF.) This is to be expected because you're using an interval of length $0.02=1/50$ instead of an interval of length $1$. That is, a normal approximation of $50 (F(0.0032)-F(-0.0032))$ would give you a very similar result to the first calculation.

The rest of this is a somewhat subtle discussion about this issue, which may not be quite suitable for a beginning statistics student.

The meaningful thing is that the binomial CDF $F_b$ and the appropriate normal CDF $F_n$ are close to one another. The continuity correction is used to relate this to the binomial PMF $f_b$. The problem is that $f_b$ can be represented in any number of different ways in terms of $F_b$.

Specifically, for any two real numbers $x,y \in [0,1)$, $f_b(m)=F_b(m+x)-F_b(m-1+y)$. To approximate $f_b(m)$ using $F_n$, we choose $x$ and $y$ and then approximate these values of $F_b$ using the corresponding values of $F_n$.

But it is not good enough to just get an acceptable approximation of these values of $F_b$, because when $n$ (the number of binomial trials) is large, these two numbers are very close. As a result small relative errors in our approximations of $F_b(m+x)$ and $F_b(m-1+y)$ can cause large relative errors in the difference. (As an example, consider $x=1.001$, $y=1$, and attempting to approximate $x-y$ when replacing $x$ with $z=1.0015$. Here $z$ and $x$ are relatively quite close together but $x-y$ and $z-y$ are relatively quite far apart.)

It turns out that we can only make this error even moderately small if $x=y$. More precisely, if $x \neq y$ then our value will be off by a factor of about $x-y+1$, even as $n \to \infty$. This is essentially because the values of the binomial variable are separated by intervals of length $1$; the precise details involve analytical tricks that are beyond the scope of both calculus and elementary statistics.

Moreover the error is minimized in an appropriate sense when $x=y=1/2$. Nevertheless, if $X$ is $Bin(n,p)$ and $Y$ is $N(np,np(1-p))$, then $P(X=m)$ is still well-approximated by, for example, $P(m \leq Y \leq m+1)$. It is just that $P(m-1/2 \leq Y \leq m+1/2)$ is a better approximation, especially if $m$ is close to $np$.

The continuity correction takes into account that the number of occurrences is actually discrete and assumes values that are separated by unit intervals. So it is necessary to consider $\pm \frac{1}{2}$ intervals or adjustments when using the normal approximation. If you narrow the range in the approximated normal distribution you can obtain progressively decreasing probabilities (tending to zero), but these have no meaning.

If you are using a continuous variable to model (or approximate) a discrete (integer) quantity, you have to decide what to do when the continuous variable gives you a noninteger value. In order to get your continuous value to represent an integer value, you simply round the noninteger continuous quantity to the nearest integer. Hence the 'continuity correction factor'.