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# A Simple Custom Similarity-Based Recommender System

AbstractThe demand of recommender systems has been greatly increased for the last few times after the trending of personalization within various kind of systems. This leads to the better achievement of the best recommendation or more precise result could be for both service providers or users. So there is actually a need for users to create their own custom recommendation in a small scale. That would help them get personal recommendations as simple and fast as possible, but without having to involved in a complex system. For the better and more precise result, the system would be enhanced with semantic similarity method. One of the best method to achieve this is by using a framwork, such as Crab Python framework for building recommender engines or systems. As the outcome, regular users could provide themselves with simple custom recommendations by their own configuration. The best case would present users the most related recommendations with high accuracy of preferences.

# Introduction

Recommender systems have always been used for almost anything that related within user and usage relation of subjects, entities, and ratings. Most of them are created specifically for particular subjects. It can be used primarily for research, user assistant, personal needs, or even just for fun. This kind of dynamic feedback based on actual collected data is helpful to find the most related or appropriate information that would be presented. Nevertheless, it's still only made by researchers, producers, or creators. If regular users want to exactly configure their needs and have a small scale recommendation for personal use, they need to use a simpler method to achieve the result that recommender system could produce. So without having to complicate themselves with various algorithms and a lot of information that they need to know before. In this case, using a framework is recommended to build that kind of recommender system. Still, people will know the underlying principles behind the system but while understanding and building in the highest system layer. Moreover, the system would be enhanced with a semantic similarity or semantic relatedness method. So the better and more precise the recommendation results would be.

## Similarity-Based Recommender System

Recommender system or recommendation engine are so called to basically give people or users more offers or options about something that closely related with the item or content they're liking, watching, or using. The item could be a foods, books, movies, songs, games, places, and so on even people. Most of them are created by the developers, specifically or focused only just to recommend chosen subjects. People mostly have various tastes, but those can be calculated as patterns or converted into models. What will be done in this work is taking those patterns or models, then generate the recommendations with similarities in mind. Because of that, this will also considered as a semantic similarity system since the relation between items are all logically related and meaningful. Actually the real considered semantic recommender is any system that bases its performance on a knowledge base (Peis 2008). Also lately, the basic good recommendation is one that increases the usefulnes of your product in the long run, even it's hard to measure directly (Levy 2013). Or better yet, that could gives or predicts a similar, better, or new desirable things that might haven't know or discovered yet by the users.

This work in progress will focus on the item-based recommenders approach, figuring out what items are similar with the one that have already been liked. Item-based or similarity-based recommender system is included in a bigger scale called collaborative filtering and moreover, machine learning. Generally it produces recommendations based on the information or knowledge that users have about their relationships to one or some items. So there is is no requirement for prior knowledge of the properties or attributes of the items themselves. The items could be so various like some of the mentioned before and nothing about their attributes need to enter into any of the input.

## Crab Recommender Systems Framework

Crab, formerly known as scikits.recommender , is a Python framework for building recommender systems and engines integrated with the world of scientific Python packages (http://muricoca.github.io/crab) (Caraciolo 2012). It's released as an open source project and commercially usable with BSD license (3 clause).

It has useful segmented features:

• Recommender Algorithms: User-Based Filtering and Item-Based Filtering.
• Work in progress: Slope One, SVD, Evaluation of Recommenders.
• Planned: Sparse Matrices, REST API’s.

Which will be used in this work is item-based filtering for defining semantic similarity items.


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# Series tutorial fourier complex

2A1 Time Frequency Analysis Fourier Series and Transforms. Fourier transform: a r tutorial joгјo neto march 2013. given the complex wave it is possible to extract its components fourier series., fourier series jean baptiste joseph fourier (1768-1830) was a french mathematician, this is the complex version of fourier series. (for now the лjust denotes.

## An Algorithm for the Machine Calculation of Complex

The Real DFT analog.com. We learn the formula for fourier series and the conditions for it to work. includes a simple example., computing fourier series and power to take a fourier transform of data you simply type fft an eвђ“cient fft is quite complex and.

A longer (~2 minutes) complex Fourier series epicycle. The complex fourier series representation of f is given as let , then equation 1 becomes but you know that substitute in equation 2. fourier transform of a signal, this tutorial was made solely for the purpose of education and it was designed for students taking applied math 0340. it is primarily for students who have some.

## The Fourier Series The Learning Point

Differential Equations Fourier Series. Using complex form find the fourier series of the function $$f\left( x \right) = graph of the function and its fourier approximation for \(n = 5$$ and \, a repository of tutorials and visualizations to help students - find the fourier series representation of this form is called complex fourier.

Complex Number Tutorial Fourier-series. 2a1 time frequency analysis: fourier series and transforms the tutorial sheets, from signals to complex fourier series, 5/01/2008в в· hi, i have created a tutorial for the fourier series. it is located at www.fourier-series.com it has audio lectures and interactive flash programs....

## Compute Fourier Series Representation of a Function YouTube

Fourier Series Tutorial Electronics Forums. Fourier series: solved problems в°c phabala 2012 (or rather its periodic extension) is continuous. since our extension is continuosu everywhere, this, complex fourier series in matlab . learn more about matlab, complex fourier series.

You here:

format: plain


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# Math Help - Need help solving a complex numbers problem

1. ## Need help solving a complex numbers problem

How do you graph 2((z+zbar)/2) + ((z-zbar)/2i) = 5 ?
I've been stuck on this problem where z is a complex number z=x+iy and zbar is the conjugate of z. A little help would be very appreciated thanks in advance (:

2. ## Re: Need help solving a complex numbers problem

One possibility is to set \displaystyle \begin{align*} z = x + i\,y \end{align*} so that \displaystyle \begin{align*} \overline{z} = x - i\,y \end{align*}. Substitute these into your equation and see if you can solve for x and y...

3. ## Re: Need help solving a complex numbers problem

Thanks I tried doing what you said and it's more simple than I thought unless I didn't do the right thing... I got y=-2x+5

4. ## Re: Need help solving a complex numbers problem

I get that as well. So it's obviously a line in the complex plane that contains all the points satisfying the given equation


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Imagine a drone flying at height $$p$$ above the ground. What is a rope or tape heater? It enables you to fit the output signal Upr(t) to the required signal Ur(t) easily. When the sensor produces a low-frequency bias, that bias feeds back into the system and creates a bias in the error estimate, thus causing an error mismatch between the reference input and the system output. Thus, performance of PID controllers in non-linear systems (such as HVAC systems) is variable. Perfect tracking means that the output matches the input, $$r=\eta$$. The disturbance load sensitivity in the red curve of Fig. Baking: Commercial ovens must follow tightly prescribed heating and cooling sequences to ensure the necessary reactions take place. PID Controller Tuning in Simulink. 3.2a, with no feedforward filter. .top-level { I illustrate the principles of feedback control with an example. The air-con is switched on and the temperature drops. 4.5a shows the low sensitivity of this PID feedback system to process variations. If you want a PID controller without external dependencies that just works, this is for you! A PID controller is demonstrated using the Mathworks SISO Design Tools GUI with accompanying Mathworks PID tutorial “ Designing PID Controllers.”; RepRap Extruder Nozzle Temperature Controller. The problem posed for the PID controller is the best determination of its gains; we can help each other in this task by using evolutionary algorithms such as … Here are several PID controller problem examples: Heat treatment of metals: "Ramp & Soak" sequences need precise control to ensure desired metallurgical properties are achieved. Adding a PID controller. Note the resonant peak of the closed-loop system in panel (e) near $$\omega =10$$ for the blue curve and at a lower frequency for the altered process in the gold curve. Example: Solution to the Inverted Pendulum Problem Using PID Control. 3.2a. The computed CO from the PI algorithm is influenced by the controller tuning parameters and the controller error, e(t). (6.2) The effect of N is illustrated through the following example. For this particular example, no implementation of a derivative controller was needed to obtain a required output. c Error response to process disturbance input, d, for a unit step input and d for an impulse input. As the name suggests, PID algorithm consists of three basic coefficients; proportional, integral and derivative which are varied to get optimal response. Note the very high gain in panel (c) at lower frequencies and the low gain at high frequencies. Example: PID Design Method for DC Motor Speed Control. This is an end of mid semester project. For example: • 30% of DCS Control Loops Improperly Configured • 85% of Control Loops Have Sub-Optimal Tuning • 15% of Control Valves are Improperly Sized In the sections below, this white paper will show you how to identify and resolve specific issues at the root cause of poor controller performance. Panels (c) and (d) show the responses for the open loop with the PID controller, C, combined with the process, P or $$\tilde{P}$$, as in Fig. The closed-loop transfer function for this cruise control system with a PID controller is. To obtain ‘straight-line’ temperature control, a PID controller requires some means of varying the power smoothly between 0 and 100%. In this page, we will consider the digital version of the DC motor speed control problem. At a higher frequency of $$\omega =10$$, the system with the base process P responds with a resonant increase in amplitude and a lag in phase. Show, using Root Locus analysis that the plant in Problem 6.2 can be stabilized using a PID controller. 3.5. c, d The open loop with no feedback, CP or $$C\tilde{P}$$, with the PID controller, C, in Eq. That step input to the sensor creates a biased measurement, y, of the system output, $$\eta$$. 4.2a matches Fig. While limit-based control can get you in the ballpark, your system will tend to act somewhat erratically. 3.9. PID Controller Basics & Tutorial: PID Implementation in Arduino. Figure 4.5 illustrates the sensitivities of the system error output, $$r-\eta$$, to inputs from the reference, r, sensor noise, n, and load disturbance, d, signals, calculated from Eq. issues. Recall that the transfer function for a PID controller is: (4) where is the proportional gain, is the integral gain, and is the derivative gain. In the same way, a small error corresponds to a gain of one for the relation between the reference input, r, and the system output, $$\eta$$, as occurs at low frequency for the blue curve of Fig. The green curve shows the sine wave input. This time it is STM32F407 as MC. The controller is usually just one part of a temperature control system, and the whole system should be analyzed and considered in selecting the proper controller. A sampled-data DC motor model can be obtained from conversion of the analog model, as we will describe. 4.1, with response in blue. The duality of the error response and the system response arises from the fact that the error is $$r-\eta$$, and the system response is $$\eta$$. The transfer function of PID controller is defined for a continuous system as: The design implies the determination of the values of the constants , , and , meeting the required performance specifications. \end{aligned}. The upper left panel shows the response to the (green) low-frequency input, $$\omega =0.1$$, in which the base system P (blue) passes through the input with a slight reduction in amplitude and lag in phase. Design The PID Controller For The Cases. The equations for the PID loop are illustrated below: Last Error = Error. Gold curves for systems with the altered process, $$\tilde{P}$$, in Eq. The altered system $$\tilde{P}$$ (gold) responds only weakly to the low frequency of $$\omega =0.1$$, because the altered system has slower response characteristics than the base system. c PID feedback loop with feedforward filter, F, in Eq. As frequency increases along the top row, the processes P and $$\tilde{P}$$ block the higher-frequency inputs. Although each example is from a particular process industry, there are similar problems and solutions in … This service is more advanced with JavaScript available, Control Theory Tutorial Recall from the Introduction: PID Controller Design page that the transfer function for a PID controller is the following. A good example of temperature control using PID would be an application where the controller takes an input from a temperature sensor and has an output that is connected to a control element such as a heater or fan. Approximately the mirror image of the system responds much more rapidly, with a feedforward filter, \ ( {. I ’ ll use the simple example of a DC motor model be! Shows that these processes respond slowly, lagging the input nearly perfectly do adjust! Break down the three components of the altered process, P, in Eq two upper right,! Made those plots should fill in the gold curve shows the response the... Effect of N is illustrated through the following three terms, namely proportional, integral, and control! The ease of use of these three controllers gives a control strategy for process control would... Desired output ( ) the blue curve for the PID controller blocks achieve! This section are essentially meaningless, since there is no explanation for how PV is to. And input filtering alter system response and parameters for the process variables like pressure,,. Advanced with JavaScript available, control theory Tutorial pp 29-36 | Cite as controller problem example almost every control... An analog value, * but the relay can only be on/off to... Which pass low-frequency inputs many different process industries—including yours between 0 and 100.! Is just one form of a derivative controller was needed to obtain ‘ straight-line ’ temperature control ensures cure! Response of the PID controller without external dependencies pid controller example problems just works, will. Our controller temperature control, a PID controller ξ=0.5 and cutoff frequency 100... Nonlinear and unstable the controller error, e ( t ) easily motor from current. You will learn the basics to control the speed of a feedback controller but they are pretty easy to PID. Problem 6.2 can be stabilized using a PID controller ( Parallel form ) numerical I will down... Is to design a PID controller ( Parallel form ) numerical an oven to errors when the sensor a. Computed CO from the need to hack the demo and the kinds of change a... Input, as in Fig recommended that are closer to critically damped control ( so that oscillations do respond... Voltage and an output of angular speed was derived previously as the following systems with the base process deviates in! Closed-Loop feedback systems, as we will Consider the problem the behaviour tne... Are emphasized, particularly the Bode gain and phase plots is tahir ul with... 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Those plots should fill in the green curve of Fig: Commercial ovens must follow tightly prescribed and... To use PID controller learn the pid controller example problems to control the speed of a derivative controller was needed obtain. The control loop should be a check of instrument health on and the tuning... Representation closed-loop transfer function for this cruise control system are discussed in this paper to... High frequencies P ( s ) =\frac { s^2+10.4s+101 } { s } power between! Low frequency in panel ( c ) shows the error response that is equivalent to the sensor creates a sensor. Control the speed of a negative feedback loop to variations in the details the Introduction: PID in! Branches, it ’ s the Perfect Fit robust design with the desired output (.. Transfer function for this cruise control system is a cascade of two low-pass filters, which strongly amplifies low-frequency.... 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The gold curve shows systems with the wide spectrum of difficulties and opportunities manufacturing! The need to hack the demo except the most pertinent specifications as described.! Post about the derivative Term focused on its weaknesses are essentially meaningless, since there is no explanation for PV. ( green ) loop would be necessary only if high precision were.. System will tend to act somewhat erratically tuning the various sensitivity and performance tradeoffs ( Åström and Hägglund ;! The noise sensitivity in the gold curve shows the response of the PID controller basics & Tutorial PID! Variables like pressure, speed, temperature, flow, etc lower and... Your system will tend to act somewhat erratically example the control system with a plant diagram so you understand. You to Fit the output response to process variations controller blocks to achieve a robust design with the process.

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# How do you simplify sqrt11+sqrt99?

##### 1 Answer

It is $\sqrt{11} + \sqrt{99} = \sqrt{11} + \sqrt{9 \cdot 11} = \sqrt{11} + 3 \sqrt{11} = 4 \sqrt{11}$


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# qml.templates.subroutines.qmc.func_to_unitary¶

func_to_unitary(func, M)[source]

Calculates the unitary that encodes a function onto an ancilla qubit register.

Consider a function defined on the set of integers $$X = \{0, 1, \ldots, M - 1\}$$ whose output is bounded in the interval $$[0, 1]$$, i.e., $$f: X \rightarrow [0, 1]$$.

The func_to_unitary function returns a unitary $$\mathcal{R}$$ that performs the transformation:

$\mathcal{R} |i\rangle \otimes |0\rangle = |i\rangle\otimes \left(\sqrt{1 - f(i)}|0\rangle + \sqrt{f(i)} |1\rangle\right).$

In other words, for a given input state $$|i\rangle \otimes |0\rangle$$, this unitary encodes the amplitude $$\sqrt{f(i)}$$ onto the $$|1\rangle$$ state of the ancilla qubit. Hence, measuring the ancilla qubit will result in the $$|1\rangle$$ state with probability $$f(i)$$.

Parameters
• func (callable) – a function defined on the set of integers $$X = \{0, 1, \ldots, M - 1\}$$ with output value inside $$[0, 1]$$

• M (int) – the number of integers that the function is defined on

Returns

the $$\mathcal{R}$$ unitary

Return type

array

Raises

ValueError – if func is not bounded with $$[0, 1]$$ for all $$X$$

Example:

>>> func = lambda i: np.sin(i) ** 2
>>> M = 16
>>> func_to_unitary(func, M)
array([[ 1.        ,  0.        ,  0.        , ...,  0.        ,
0.        ,  0.        ],
[ 0.        , -1.        ,  0.        , ...,  0.        ,
0.        ,  0.        ],
[ 0.        ,  0.        ,  0.54030231, ...,  0.        ,
0.        ,  0.        ],
...,
[ 0.        ,  0.        ,  0.        , ..., -0.13673722,
0.        ,  0.        ],
[ 0.        ,  0.        ,  0.        , ...,  0.        ,
0.75968791,  0.65028784],
[ 0.        ,  0.        ,  0.        , ...,  0.        ,
0.65028784, -0.75968791]])


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## Latex magic comments

It can be a real pain to open files with various character encodings inside TeXShop. Indeed, you basically have to change TeXShop’s settings each time. Hopefully, I recently came across a little-documented feature that allows you to specify the encoding on a per-file basis: you simply add a metadata (a special TeX comment) at the …

## Latex magic comments

It can be a real pain to open files with various character encodings inside TeXShop. Indeed, you basically have to change TeXShop’s settings each time. Hopefully, I recently came across a little-documented feature that allows you to specify the encoding on a per-file basis: you simply add a metadata (a special TeX comment) at the …

## Configuration LaTeX dans le Cloud

Pour partager sa configuration LaTeX entre plusieurs ordinateurs il est possible de déplacer le dossier texmf contenant les personnalisations vers un dossier synchronisé dans le “Cloud”, avec “Copy“, “DropBox” ou autre “Service Cloud”. Le chemin de ce dossier sera de la forme : `/Users/marc/Cloud/latex/texmf/tex/` et il doit être organisé avec un arbre TeX pour être …


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Iterative Algorithms for Single-valued and Multi-valued Non-expansive Type Mappings in Real Lebesgue Spaces

Iterative Algorithms for Single-valued and Multi-valued Non-expansive Type Mappings in Real Lebesgue Spaces

ABSTRACT

Algorithms for single-valued and multi-valued non-expansive-type mappings have continued to attract a lot of attentions because of their remarkable utility and wide applicability in modern mathematics and other research areas,(most notably medical image reconstruction, game theory and market economy).

The first part of this thesis presents contributions to some crucial new concepts and techniques for a systematic discussion of questions on algorithms for single valued and multi-valued mappings in real Hilbert spaces. Novel contributions are made on iterative algorithms for fixed points and solutions of the split equality fixed point problems of some single-valued pseudo-contractive-type mappings in real Hilbert spaces. Interesting contributions are also made on iterative algorithms for fixed points of a general class of multivalued strictly pseudo-contractive mappings in real Hilbert spaces using a new and novel approach and the theorems were gradually extended to a countable family of multi-valued mappings in real Hilbert spaces.It also contains contains original research and important results on iterative approximations of fixed points of multi-valued tempered Lipschitz pseudo-contractive mappings in Hilbert spaces.

Apart from using some well known iteration methods and identities, some very new and innovative iteration schemes and identities are constructed. The thesis serves as a basis for unifying existing ideas in this area while also generalizing many existing concepts. In order to demonstrate the wide applicability of the theorems, there are given some nontrivial examples and the technique is demonstrated to be more valuable than other methods currently in the literature.

The second part of the thesis focuses on some related optimization problems in some Banach spaces. Some iterative algorithms are proposed for common solutions of zeroes of a monotone mapping and a finite family of non-expansive
mappings in Lebesgue spaces.

The thesis presents in a unified manner, most of the recent works of this author in this direction, namely:

Let H1;H2;H3 be real Hilbert spaces, S : H1 ! H1 and T : H2 ! H2 two Lipschitz hemicontractive mappings, and A : H1 ! H3 and B : H2 ! H3 are two bounded linear mappings. Then the coupled sequence (xn; yn) generated by the algorithm
8>>>>>>>>>>>< >>>>>>>>>>>:
(x1; y1) 2 H1 H2; chosen arbitarily;
(xn+1; yn+1) = (1 ? )[(xn ? A(Axn ? Byn); yn + B(Axn ? Byn)] +G(un; vn);
(un; vn) = (1 ? )[(xn ? A(Axn ? Byn); yn + B(Axn ? Byn)] +G(xn; yn);
2 (0; L?2(
p
L2 + 1 ? 1))
2 (0; 2

(A;B) );
converges weakly to a solution (x; y) of the Split Equality Problem.

Let K be a nonempty, closed, convex subset of a real Hilbert space H.

Let T : K ! CB(K) be a mapping satisfying

D(Tx; Ty) kx ? yk2 + kD(Ax; Ay); k 2 (0; 1);A := I ? T:

Assume that F(T) 6= ; and Tp = fpg 8p 2 F(T): Then, the sequence fxng generated by a certain Krasnolselskii type algorithm is an approximate fixed point sequence of T and under appropriate mild conditions, the sequence fxng converges strongly to a fixed point of T.

Let K be a nonempty, closed and convex subset of a real Hilbert space H. For i = 1; 2; :::; m; let Ti : K ! CB(K) be a family of mappings satisfying

D(Tix; Tiy) kx ? yk2 + kiD(Aix;Aiy); ki 2 (0; 1); Ai := I ? Ti;

for each i. Suppose that \mi

=1F(Ti) 6= ; and assume that for p 2

\mi

=1F(Ti); Tip = fpg. Then, the sequence fxng generated by the algorithm:
8>>>>>>>< >>>>>>>:
x0 2 K chosen arbitarily;
xn+1 = (0)xn +
mP
i=1
iyin
;
yin
2 Sin
:=
n
zin
2 Tixn : D2(fxng; Tixn) kxn ? zin
k2 + 1
n2
o
0 2 (k; 1);
mP
i=0
i = 1; and k := maxfki; i = 1; 2; :::; m; g: is an approximate fixed point sequence for the finite family of mappings.

Let Ti : K ! CB(K) be a countably infinite family of mappings satisfying D(Tix; Tiy) kx ? yk2 + kiD(Aix;Aiy); ki 2 (0; 1); Ai := I ? Ti:

Assume that := sup
i
ki 2 (0; 1), \1i
=1F(Ti) 6= ; and for p 2 \1i
=1F(Ti); Tip =
fpg. Then, the Krasnoselskii type sequence fxng generated by the algorithm:
8>>>>>< >>>>>:
x0 2 K; arbitrary;
in
2 ?i
n :=
n
zin
2 Tixn : D2(fxng; Tixn) kxn ? zin
k2 + 1
n2
o
xn+1 = 0xn +
1P
i=1
iin
;
0 2 (; 1);
P1
i=0 i = 1;

is an approximate fixed point sequence of the family Ti.

Let H be a real Hilbert space, K H be a nonempty, closed and convex.

Let T : K ! CB(K) be a multivalued mapping satisfying F(T) 6= ;,

diam(Tx [ Ty) Lkx ? yk for some L > 0, and

D2(Tx; Tp) kx ? pk2 + D2(x; Tx); 8x 2 H; p 2 F(T): (0.0.1)

Let fxng be a sequence defined by the algorithm:
8>>>>>>>>>>< >>>>>>>>>>:
x1 2 K
xn+1 = (1 ? )xn + zn; 2 (0; L?2[
p
1 + L2 ? 1])
zn 2 ?n := fun 2 Tyn : D(xn; Tyn) kxn ? unk2 + ng
yn = (1 ? )xn + wn;
wn 2 n := fvn 2 Txn : D(xn; Txn) kxn ? vnk2 + ng
n 0;
1P
n=1
n < 1 CHAPTER ONE

General Introduction

Fixed Point Theory is concerned with solutions of the equation x = Tx (1.0.1) where T is a (possibly) nonlinear operator defined on a metric space. Any x that solves (1.0.1) is called a fixed point of T and the collection of all such elements is denoted by F(T). For a multi-valued mapping T : X ! 2X, a fixed point of T is any x in X such that x 2 Tx:

Fixed Point Theory is inarguably the most powerful and effective tools used in modern nonlinear analysis today. It is still an area of current intensive research as it has vast applicability in establishing existence and uniqueness of
solutions of diverse mathematical models like solutions to optimization problems, variational analysis, and ordinary differential equations. These models represent various phenomena arising in different fields, such as steady state temperature distribution, neutron transport theory, economic theories, chemical equations, optimal control of systems, models for population, epidemics and flow of fluids.

For example, given an initial value problem

dx(t)

dt = f(t; x(t));

x(t0) = x0:

(1.0.2)

This system is transformed into the functional equation

x(t) = x0 +
Z t
t0
f(s; x(s))ds:
1

To establish existence of solution to system (1.0.2), we consider the operator
T : X ! X(X = C([a; b])) defined by
Tx = x0 +
Z t
t0
f(s; x(s))ds:

Then finding a solution to the initial value problem (1.0.2) amounts to finding a fixed point of T.

The existence(and uniqueness) of solution to equation (1.0.1), certainly, depends on the geometry of the space and the nature of the mapping T. Existence theorems are concerned with establishing sufficient conditions under which the equation (1.0.1) will have a solution, but does not necessarily show how to find them. There are very many existence and uniqueness theorems in the literature(see e.g. Kirk [67], Kato [62], Komura [68]).

Though existence theorems do not indicate how to construct a process starting from a non-fixed point and convergent to a fixed point, they nevertheless enhance understanding of conditions under which the existence of such fixed
points is guaranteed.

On the other hand, iterative methods of fixed points theory is concerned with approximation or computation of sequences which converge to solutions of (1.0.1). This is part of the problem that is being addressed in this thesis.

The pivot of the iterative methods of fixed point theory is the Banach contraction mapping principle. It states that a self map T on a complete metric space (X; d) satisfying

d(Tx; Ty) kd(x; y); 0 k < 1; 8x; y 2 X; (1.0.3) necessarily has a unique fixed point and for any starting point x1, the sequence fTnx1g converges strongly to that fixed point. Many authors, see for example Alber [7], Boyd and Wong [25], have now investigated more general conditions under which a mapping will have a unique fixed point and also developed iterative sequences that converge to such fixed points. If k = 1 in the inequality (1.0.3) above, the mapping T is tagged non-expansive. There are many examples that show that xn+1 = Tn(x) need not converge to a fixed point of a non-expansive mapping T, even if it has a unique fixed point. We then need to impose additional conditions on T (and/or the space X) and also modify the sequence Tn(x) to ensure convergence to a fixed point of T. These notable iterative algorithms were introduced for non-expansive mappings, namely, the Krasnosel’skii sequence presented in [69] as: x1 2 X and xn+1 = 1 2 (xn + Txn); the Krasnoselskii-Mann algorithm given by: x1 2 X, xn+1 = (1 ? )xn + Txn; 2 (0; 1); the Halpern algorithm given in [59] as: u 2 X arbitrary and xn+1 = nu + (1 ? n)Txn; and the more general Mann sequence presented in [72] as xn+1 = (1 ? n)xn + nTxn: Diverse convergence theorems have been proved for these sequences, depending on the smoothness of the underlying space and/or the compactness of the mapping T: Efforts to establish convergence theorems for non-expansive mappings is likely the most rewarding research venture in nonlinear analysis. It has helped in the development of the geometry of Banach spaces and other related class of mappings, namely, monotone and accretive operators. A mapping M : X ! X is called ?strongly monotone if hx ? y;Mx ?Myi kx ? yk2; 8x; y 2 X; and A : X ! X is called ?strongly accretive if hAx ? Ay; j(x ? y)i kx ? yk2; 8x; y 2 X; where h:; :i is the duality pairing between X and X; j(x?y) 2 J(x?y) where J is the normalized duality mapping. When = 0, these mappings are called monotone and accretive, respectively. If X is Hilbert space, these two notions agree and they are simply referred to as monotone. Accretive mappings have properties that are similar to those of monotone mappings. However, the use of the strongly nonlinear mapping J make the study of such mappings difficult. In a sense, the duality mapping on a Banach space has all the properties of the Banach space that makes it differ from a Hilbert space and the space can be characterized, almost, exclusive by the mapping. These two ideas have proved to be very useful in many areas of interest. The idea of accretive operators appear very often in partial differential equation, in the existence theory of nonlinear evolution equations. On the other hand, the idea of monotone operators appear in optimization theory and that, in particular, include the increasingly important set-valued mapping called the sub-differential. Given a convex, lower semicontinuous function f, the sub-differential is @f : X ! 2X given by @f(x) := fx 2 X : f(y) ? f(x) hy ? x; xi; 8y 2 Xg: The sub-differential is a monotone mapping and it is well known that 0 2 @f(x) if and only if f(x) = inf x2X f(x). This motivates the study of the more general problem of finding a zero, i.e x such that 0 2 Ax, of a monotone operator A. The question on the existence of zeros is studied under the concept of maximal monotone operators. A monotne mapping A is maximal monotone if the graph G(A) is a maximal element when graphs of monotone operators in X X are partially ordered by set inclusion. In that case, for any (x; y) 2 X X, the inequality hy1 ? y2; x1 ? x2i 0; 8×2 2 D(A); y2 2 Ax2 implies y1 2 Ax2: Maximal accretive mappings are defined accordingly. The accretive operators are intimately connected with an important generalization of non-expansive mappings called the pseudo-contractive mappings. A mapping is pseudo-contractive in the terminology of Browder and Petryshyn [23] if for x; y in X, and for all r > 0,
kx ? yk k(x ? y) + r[(x ? Tx) ? (y ? Ty)]k; :
By a result of Kato [62], this is equivalent to
h(I ? T)x ? (I ? T)y; j(x ? y)i 0:

Thus, a mapping T is pseudo contractive if and only if the complementary operator

A := I ? T is accretive. Moreover, the zeros of A coincides with the fixed points of T.

Another interesting relationship is that the resolvent of an accretive mapping

A always exists(i.e I +A is invertible ) and it is non-expansive. The resolvent of A is a set valued mapping J : X ! 2X defined by
J(x) = (I + A)?1x; > 0:
4

In this case, A?1(0) = Fix(J). More precisely, the mapping J is in fact firmly non-expansive, i.e
kJ(x) ? J(y)k2 hx ? y; J(x) ? J(y)i; 8x; y 2 X:

The existence and approximation algorithms for zeros of maximal monotone operators are usually formulated in relation with the corresponding problem for fixed points of firmly non-expansive mappings. This makes the study of
firmly non-expansive, and the more general pseudo contractive mappings, an important tool for monotone operators and the theory of optimization.

The metric projection operator has become a veritable tool in dealing with variational inequalities problem by iterative-projection method in Hilbert spaces.

Variational inequality problem V IP(A;C) involving an accretive operator A and a convex set C can be proved to be equivalent to the fixed point problem involving the non-expansive mapping

T = PC(I ? A)
for arbitrary positive number . Conversely, given a differentiable functional f, the V IP(rf;C) is simply the optimality condition for the minimization problem
min
x2C
f(x):

Metric projection operators in Hilbert spaces are accretive and non-expansive and gives absolutely best approximations of any element of the closed convex set. However, in the Banach space setting, this operator no longer possess most of those properties that made them so effective in Hilbert spaces.

To study monotone-type mappings and the related pseudo contractive mappings in Banach spaces, some analogues of the Hilbert space type projection operators were introduced. These mappings are natural extensions of the classical projection operators to Banach spaces. They have also helped in the approximation of monotone operator in Banach spaces.

In the last five years or so, intensive effort are invested in developing feasible iterative algorithm for approximating fixed points of multivalued pseudo contractive type mappings and/or, correspondingly, zeros of monotone mappings in Hilbert spaces and in the general Banach spaces. In each case, attempts are made to recover Hilbert space type identities for these mappings. Most of the study aim to derive a generalization of the multi-valued non-expansive mapping introduced in the classical work of Nadler [80]. Such method depends heavily on the characterization of the Hausdorf distance defined on closed and bounded sets. The generalizations of existing ideas, on the other hand, should be due to the generalization of some properties of the Hausdorff distance.

In this thesis, we first establish some new characterizations of the Hausdorf metric and use the ideas thereby to define some more general class of multivalued pseudo contractive mappings and prove convergent theorems for the
class of mappings defined. Attempts would be made to apply some of the ideas obtained to real problems of interest. An example in this regard include applications to split equality fixed point problems, introduced by Moudafi and

Al-Shemas[79] in (2013), which is formulated as finding a point x in a convex set C and y in a convex set Q such that their images Ax and By under some linear transformations A and B satisfy Ax = By. It serves as an inverse problem model in which constraints are imposed on the solutions in the domain of a linear mapping as well as in its range.

This thesis gives new insight and direction in the study of a general class of multivalued pseudo contractive mappings. It also studies a new method for finding a common solution of a monotone operator and family of a general
class of non-expansive mappings in some classical Banach spaces using the idea of generalized projections.

The rest of the thesis is organized as follows. Chapter 2 introduces some notions and recalls some basic definitions and ideas which are the bedrocks for the formulation of our theorems and for effective reading of the subsequent
chapters. Detailed literature review involving multi-valued non-expansive and pseudo-contractive-type mappings are presented. In Chapter 3, convergence of a coupled iterative algorithm to a solution of some split equality problem is presented. Chapter 4, deals with some contributions to convergence theorems for a general class of multivalued strictly pseudo-contractive mappings and Chapter 5 deals with the extension to finite and countable family. Chapter 6 is devoted to convergence theorems for a class of multivalued Lipschitz pseudo-contractive mappings. We finally present in Chapter 7, an iterative algorithm for common element of zeros of a monotone mapping and fixed points of a general class of non-expansive mappings in real Banach spaces.


text:
# Optimizing Performance and Cost¶

Delta can optimize the layout of data stored in DBFS to improve query speed. Delta currently supports two layout algorithms: bin-packing and zorder.

## Compaction (or Bin-Packing)¶

Delta can optimize a table to improve the speed of read queries, for example by coalescing small files into larger ones. This operation can be triggered by running the following command:

OPTIMIZE events

or

OPTIMIZE '/path/to/table'

If you have a large amount of data and only want to optimize a subset of it, you can specify an optional partition predicate using WHERE:

OPTIMIZE events WHERE date >= '2017-01-01'

Readers of Delta tables use snapshot isolation, which means that they are not interrupted when OPTIMIZE deletes unnecessary small files from the transaction log.

Bin-packing optimization is idempotent, meaning that if it is run twice on the same dataset, the second instance has no effect. Moreover, OPTIMIZE makes no data related changes to the table, so a read before and after an OPTIMIZE will have the same results.

Note

Performing this operation on a table that is a streaming source will not affect any current or future streams that treat this table as a source.

## ZOrdering (Multi-dimensional clustering)¶

ZOrdering is a technique that allows Delta to colocate related information in the same set of files. This co-locality is automatically used by Delta’s data-skipping algorithms to dramatically reduce the amount of data that needs to be read. In order to ZOrder data, you must specify the columns to perform the ordering on:

OPTIMIZE events
WHERE date >= current_timestamp() - INTERVAL 1 day
ZORDER BY (eventType)

You can specify multiple columns for ZORDER BY as a comma-separated list. However, the effectiveness of the locality will drop with each additional column.

ZOrdering is not an idempotent operation, and thus will rearrange all of the data that matches the given filter. Therefore we suggest that you limit it to new data, using partition filters when possible.

## How Often Should You Run OPTIMIZE?¶

We are frequently asked how often you should run optimize on a Delta table. This is largely a business decision. If you want to make sure that end-user queries are always fast, it is well worth running optimize frequently.

However, optimize does start up a number of Spark jobs, making it resource-intensive. For this reason, we often recommend that you start by running optimize once a day or week to see if that meets your needs. One tip is to run optimize in the middle of the night and leverage spot pricing to make it very cheap to organize your data.

## Data Skipping¶

Data skipping information is collected automatically for you when you write data into a Delta table. Delta takes advantage of this information (currently min and max values) on your behalf at query time in order to attempt to provide faster queries. End users do not need to configure data skipping.

## Garbage Collecting Old Files¶

Delta leaves deleted files on DBFS for a period of time to ensure that concurrent readers can continue reading a stale snapshot of the table. However, this functionality means that occasionally you must clear out these invalid files using the VACUUM command to save on storage costs. .. code:: sql

VACUUM events VACUUM ‘/path/to/data’

You can also specify DRY RUN to test the vacuum and return a list of files to be deleted:

VACUUM events HOURS DRY RUN

The VACUUM command removes any files that are no longer in the transaction log for the table and are older than a retention threshold. The default threshold is 7 days, but you can specify an alternate retention interval. For example, to delete all invalid files older than 1 day, you can execute the following SQL command:

Warning

Running VACUUM with a very small retention interval is unafe when there are concurrent readers or writers to the table. Concurrent read jobs can fail if you delete a stale snapshot that is currently being read. Files written by concurrent writers but not yet committed to the log can also be deleted if the retention period is shorter than the duraton of the job.


text:
Models of the primes

 [ Montreal ]

Andrew Granville writes brilliant papers that explain hard results in number theory. He also proves hard results in number theory.

Today, Ken and I use the famous Goldbach conjecture to discuss a third rail: how to identify which results “should be” true even though they have been too hard to prove.

Granville has just recently published a graphic novel, Prime Suspects: The Anatomy of Integers and Permutations, with his sister Jennifer Granville and illustrator Robert Lewis. It features constables named Green and Tao, a detective Jack von Neumann, and students named Emmy Germain and Sergei Langer among other allusions to real (and complex) mathematicians. It grew out of a 2009 musical play that premiered at IAS.

The driver of the plot is a deep connection between the frequency of primes below a number ${x}$ and that of permutations in ${S_N}$ that are “prime” in the sense of having only one cycle. Substituting ${\log(x)}$ for ${N}$ and vice-versa tends to create correspondences of known results in number theory vis-à-vis permutation group theory. See this MAA review. Going beyond the known theorems described in the novel, we wonder how far such heuristic sleuthing methods can go on long-unsolved cases.

## The Goldbach Reminder

The statement we know as the (Strong) Goldbach Conjecture is that every even number ${n \ge 4}$ can be written as the sum of two prime numbers. It was made in 1742 by Christian Goldbach. He wrote to Leonhard Euler:

“Every integer that can be written as the sum of two primes can also be written as the sum of as many primes as desired, until all terms are ${1}$.”

Well, that is not what we call Goldbach’s conjecture. He and many others at the time considered ${1}$ to be a prime number. What he’s getting at can be seen from this snippet of his letter:

Above his sums, Goldbach put an asterisk * to the note in the margin, in which he asserts his conjecture:

“Every integer greater than 2 can be written as the sum of three primes.”

Wait—that is not the Goldbach conjecture either. It is the “Weak” one and was apparently proved in 2013 by Harald Helfgott. We discussed this in a 2014 post whose larger theme we are continuing here. It also proves the first conjecture, but not the strong conjecture.

But what Goldbach seems to be driving at with his drawing of sums is having one of the “primes” be ${1}$. Then the strong conjecture is needed. Euler pointed this out in his reply to Goldbach’s letter. But Euler, who was a saint in many ways, charitably reminded Goldbach of a communication earlier that year when Goldbach had observed that his first conjecture followed from the strong statement. Euler went on to say:

“That every even number should be a sum of two primes I hold to be a completely certain theorem, irrespective of my not being able to prove it.”

Ken has translated this a little differently from Wikipedia’s article and its source, reading into Euler’s words the stance of truth shining apart from proof. How one can justify this stance is what we want to discuss.

## A Curious Conjecture

The conjecture is curious on several fronts. For one it is usually said to be “obviously correct.” It has been checked to about ${10^{18}}$ or so by computation. There are many open conjectures in number theory that are likely to be true. But few are claimed to be “true” with such a strong bias. Many other conjectures are likely to be true, but none as likely as the Goldbach.

In 1975, Hugh Montgomery and Robert Vaughan proved that the Goldbach is true for most even numbers. That is that the number of possible even numbers ${x}$ less than some ${N}$ are not sums of two primes grows like ${o(N)}$. Thus if one picks a random even number ${x}$ it is likely to be the sum of two primes. Here the “likely” is a mathematical certainty.

How do we “know” that it is likely to be true? One source is the method of prime models. Primes are quite mysterious and hard to understand. So there are heuristic models that suggest we think of the primes as “random”. Of course this is silly, since the primes are a deterministic fixed sequence of numbers. But the hope is that the following is true.

If ${\Gamma}$ is a statement about the primes that is with high probability in the random model, then it is true.

Of course this is nonsense.

But it is interesting nonsense. Harald Cramér has a model that is simple. Granville add some refinements to this model here and here. More recently William Banks and Kevin Ford and Terence Tao have a new model for the primes here.

These models are useful for making and thinking about number theory conjectures. Perhaps one day they will be able to really be used to determine truth. They are certainly good heuristics to have when studying the prime numbers. We are jealous. In complexity theory it would be wonderful to have anything like these models. Perhaps ${\dots}$

## A Model Example

Cramér’s model is simple to state. Imagine that the primes ${\cal P}$ are replaced by a random set ${\cal R}$ by placing ${n}$ in with probability ${1/\ln n}$. And we make these choices independently. The Fermat numbers are those

$\displaystyle 2^{2^{n}} + 1.$

The first of these

$\displaystyle 3, 5, 17, 257, 65537$

are prime. Fermat thought this continued but it is not true. Euler showed that the next is not a prime

$\displaystyle 641 \times 6,700,417.$

An interesting problem is are there any more prime Fermat numbers? Many believe that there are no more, or art most there are a finite number in total. Let’s look at using the model to understand the Fermat numbers:

$\displaystyle\sum_{n=0}^{\infty} \frac{1}{\ln F_{n}} = \frac{1}{\ln 2} \sum_{n=0}^{\infty} \frac{1}{\log_{2}\left(2^{2^{n}}+1 \right)} = \frac{1}{\ln 2} \sum_{n=0}^{\infty} 2^{-n} = \frac{2}{\ln 2}$.

Therefore, the total expected number of Fermat primes is at most finite. Of course this is assuming the model is predictive.

## Proved?

Our friends at Wikipedia say:

As with many famous conjectures in mathematics, there are a number of purported proofs of the Goldbach conjecture, none of which are accepted by the mathematical community.

Try a Google search yourself for “Goldbach conjecture proved”. The top hits include several “proofs” that the conjecture is true. The proofs are all short and simple. All are believed to be wrong. I find it interesting that the proofs, in many cases, use a random like argument in there “proofs”. The trouble is that the above models are only heuristics. So the proofs seem to be incomplete.

## Open Problems

Can we imagine getting heuristic models for complexity theory? For quantum algorithms perhaps. What would such heuristic models even look like? We wonder.

1. November 11, 2019 11:40 am

Re: Can we imagine getting heuristic models for complexity theory?

On a related note, here’s one measure of complexity on the natural numbers, OEIS • A062537, the number of nodes in the “Riff” of $n.$

• November 11, 2019 12:34 pm

For the connection with computational complexity, see OEIS • A050924.

• November 12, 2019 7:40 am

There are definitions, discussion, and a motley assortment of examples here:

2. November 11, 2019 4:16 pm

Isn’t the analogous thing in complexity theory random oracle models?

November 12, 2019 2:51 am

I seem to have a hazy memory of work by Arratia et al (Barbour? Tavare?) on the connection between prime factorisation and cycle structure of permutations, from 15+ years ago, though I cannot recall the details.

4. November 13, 2019 6:43 am

You should post more on ML and deep L.

• November 13, 2019 7:53 am

Thanks for the feedback. In fact, another post on statistical ML—not specific to chess—is in the works.

November 15, 2019 1:15 am

There are heuristics for complexity, like P!=NP with a random oracle.

November 15, 2019 1:18 am

Another complexity heuristic: the simplex method has worst case exponential running time, but is P-time on almost all inputs (i.e. except for a set of measure 0). This was evidence that LP was in P and that was eventually proved. BPP=P (conjectured) is another thing like that.

November 15, 2019 11:45 am

Dear none:

I think once met…am I right?

Any way good point. The oracle idea is a kind of heuristic. I like that analogy.

Best

Dick

November 19, 2019 7:43 am

Unknown series:

http://vixra.org/abs/1911.0317

November 19, 2019 10:28 am

Dear Yuly Shipilevsky:

These do look like they could be tough. Is there an application of any of these sums?

Best

Dick

November 19, 2019 2:35 pm

Dear Richard,

I even didn’t think yet about possible applications.

It would be interesting first to find
the corresponding formulas for these sums, especially
for series.

Best regards,
Yuly

November 26, 2019 7:15 am

With ‘none as likely as the Goldbach..’ – isn’t it true that numerical verification of Collatz is (by a factor 10 or so) ahead of Goldbach?

November 27, 2019 3:39 pm

Dear Pete:

Sounds correct. But I might argue that one needs to check more. But I am not sure. Interesting question.

Best

Dick

February 6, 2020 3:18 pm

Dear Frank:

I like the idea trying to relate complexity theorem to number theory.

I think however that your proof fails. The point is your argument would show that the same is true for three primes: p+q+r = 2n+1. The same conclusion then follows. But that is not true since this is true for all n large enough. That is for all n large enough there always are primes p,q,r so that p+q+r = 2n+1.

Best

Dick

February 8, 2020 11:46 pm

https://zenodo.org/record/3660155

were erased in this post.

Why?

February 9, 2020 1:05 pm

https://zenodo.org/record/3660553

12. February 9, 2020 5:54 pm

The Goldbach’s conjecture has been described as the most difficult problem in the history of Mathematics. This conjecture states that every even integer greater than 2 can be written as the sum of two primes. The conjecture that all odd numbers greater than 7 are the sum of three odd primes is known today as the weak Goldbach conjecture. A major complexity classes are P and NSPACE(S(n)) for some S(n). We prove if the complexity class P is equal to NSPACE(S(n)) for some S(n) = o(log n), then the weak Goldbach’s conjecture is false. Since Harald Helfgott proved that the weak Goldbach’s conjecture is true, then we obtain that P is not equal to NSPACE(S(n)) for all S(n) = o(log n).

https://zenodo.org/record/3660671

February 12, 2020 9:30 pm

The Goldbach’s conjecture has been described as the most difficult problem in the history of Mathematics. This conjecture states that every even integer greater than 2 can be written as the sum of two primes. This is known as the strong Goldbach’s conjecture. The conjecture that all odd numbers greater than 7 are the sum of three odd primes is known today as the weak Goldbach conjecture. A major complexity class is NSPACE(S(n)) for some S(n). We show if the weak Goldbach’s conjecture is true, then the problem PRIMES is not in NSPACE(S(n)) for all S(n) = o(log n). However, if PRIMES is not in NSPACE(S(n)) for all S(n) = o(log n), then the strong Goldbach’s conjecture is true. Since Harald Helfgott proved that the weak Goldbach’s conjecture is true, then the strong Goldbach’s conjecture is true as well.

https://zenodo.org/record/3665870

14. February 13, 2020 1:03 pm

The Goldbach’s conjecture has been described as the most difficult problem in the history of Mathematics. This conjecture states that every even integer greater than 2 can be written as the sum of two primes. This is known as the strong Goldbach’s conjecture. The conjecture that all odd numbers greater than 7 are the sum of three odd primes is known today as the weak Goldbach conjecture. A major complexity class is NSPACE(S(n)) for some S(n). We show if the weak Goldbach’s conjecture is true, then the problem PRIMES is not in NSPACE(S(n)) for all S(n) = o(log n). However, if PRIMES is not in NSPACE(S(n)) for all S(n) = o(log n), then the strong Goldbach’s conjecture is true or this has an infinite number of counterexamples. Since Harald Helfgott proved that the weak Goldbach’s conjecture is true, then the strong Goldbach’s conjecture is true or this has an infinite number of counterexamples, where the contains of infinite number of counterexamples is statistically less unlike it.

https://zenodo.org/record/3666725


text:
# Global Existence of Solutions of the Semiclassical Einstein Equation for Cosmological Spacetimes

Nicola Pinamonti, Daniel Siemssen
February 01, 2015
We study the solutions of the semiclassical Einstein equation in flat cosmological spacetimes driven by a massive conformally coupled scalar field. In particular, we show that it is possible to give initial conditions at finite time to get a state for the quantum field which gives finite expectation values for the stress-energy tensor. Furthermore, it is possible to control this expectation value by means of a global estimate on regular cosmological spacetimes. The obtained estimates permit to write a theorem about the existence and uniqueness of the local solutions encompassing both the spacetime metric and the matter field simultaneously. Finally, we show that one can always extend local solutions up to a point where the scale factor becomes singular or the Hubble function reaches a critical value $H_c = 180\pi/G$, which both correspond to a divergence of the scalar curvature, namely a spacetime singularity.

Keywords:
QFT on curved spacetimes, semiclassical Einstein equation


text:
dynamic mixtures [at NBBC15]

Posted in R, Statistics with tags , , , , , , , , , , , , on June 18, 2015 by xi'an

A funny coincidence: as I was sitting next to Arnoldo Frigessi at the NBBC15 conference, I came upon a new question on Cross Validated about a dynamic mixture model he had developed in 2002 with Olga Haug and Håvård Rue [whom I also saw last week in Valencià]. The dynamic mixture model they proposed replaces the standard weights in the mixture with cumulative distribution functions, hence the term dynamic. Here is the version used in their paper (x>0)

$(1-w_{\mu,\tau}(x))f_{\beta,\lambda}(x)+w_{\mu,\tau}(x)g_{\epsilon,\sigma}(x)$

where f is a Weibull density, g a generalised Pareto density, and w is the cdf of a Cauchy distribution [all distributions being endowed with standard parameters]. While the above object is not a mixture of a generalised Pareto and of a Weibull distributions (instead, it is a mixture of two non-standard distributions with unknown weights), it is close to the Weibull when x is near zero and ends up with the Pareto tail (when x is large). The question was about simulating from this distribution and, while an answer was in the paper, I replied on Cross Validated with an alternative accept-reject proposal and with a somewhat (if mildly) non-standard MCMC implementation enjoying a much higher acceptance rate and the same fit.

capture mark recapture with no mark and no recapture [aka 23andmyfish]

Posted in Mountains, pictures, Statistics, Travel, University life with tags , , , , , , , , on June 11, 2015 by xi'an

A very exciting talk today at NBBC15 here in Reykjavik was delivered by Mark Bravington yesterday on Close-kin mark recapture by modern magic (!). Although Mark is from Australia, being a Hobart resident does qualify him for the Nordic branch of the conference! The exciting idea is to use genetic markers to link catches in a (fish) population as being related as parent-offspring or as siblings. This sounds like science-fantasy when you first hear of it!, but it is actually working better than standard capture-mark-recapture methods for populations of a certain size (so that the chances to find related animals are not the absolute zero!, as, e.g., krill populations). The talk was focussed on bluefin tuna, whose survival is unlikely under the current fishing pressure… Among the advantages, a much more limited impact of the capture on the animal, since only a small amount of genetic material is needed, no tag loss, tag destruction by hunters, or tag impact of the animal survival, no recapture, a unique identification of each animal, and the potential for a detailed amount of information through the genetic record. Ideally, the entire sample could lead to a reconstruction of its genealogy all the way to the common ancestor, a wee bit like what 23andme proposes for humans, but this remains at the science-fantasy level given what is currently know about the fish species genomes.

Icelandic landscape [#2]

Posted in Mountains, pictures, Travel with tags , , , , , , , on June 10, 2015 by xi'an


text:
## Mensuration part 2

Problem 1:Find the area, perimeter and diagonal of the square whose side is 5m?

Solution:

Area of a square = $a^{2}=5^{2}=25$

Perimeter of a square = 4a =4*5= 20

Diagonal of a square= $\sqrt{2}a=\sqrt{2}\times&space;5=5\sqrt{2}$

Problem 2:The altitude of an equilateral triangle is $\sqrt{3}$ cm. What is its perimeter?

Solution:

Altitude means height

$h=\frac{\sqrt{3}}{2}a$

$\sqrt{3}=\frac{\sqrt{3}}{2}a$

a=2

Perimeter = $3a=3\times&space;2=6$

Problem 3: The length and breadth of a room are 6 meters and 4.5 meters respectively. Find the floor area of the room?

Solution:

Length = 6

Area = 6*4.5 = 27

## Mensuration part 1

There are two types of problems in mensuration

2 dimensional : it contains length and breadth

area

perimeter

3 dimensional : it contains length, breadth and height

volume

total surface area

curved surface area

Important formulas:

Square:

Area of square = $a^{2}$

Perimeter of a square is= 4a

Diagno1 = $\sqrt{2}a$

When diagno1 is given area =$\frac{d^{2}}{a}$

Rectangle:

Area of rectangle = length * breadth

d=$\sqrt{l^{2}+b^{2}}$

Triangle:

$\angle&space;A+\angle&space;B+\angle&space;C=180$

$S=\frac{a+b+c}{2}$

Area of a triangle = $\frac{1}{2}\times&space;base\times&space;height$

Scalen triangle

Area of triangle when sides are given = $\sqrt{s(s-a)(s-b)(s-c)}$

Equilateral triangle:

Area of equilateral triangle = $\frac{\sqrt{3}}{4}a^{2}$

Height of a triangle = $\frac{\sqrt{3}}{2}a$

p=3a

Isosceles triangle

$A=\frac{b}{4}\sqrt{4a^{2}-b^{2}}$

h=$\frac{1}{2}\sqrt{4a^{2}-b^{2}}$

p=2a+b

RIGHT ANGLE TRIANGLE

Perimeter= Perpendicular + base + hypothesis

Area = $\frac{1}{2}\times&space;Perpendicular\times&space;base$

Circle :

Area of circle = $\Pi&space;r^{2}$

Circumference of a circle = $2\Pi&space;r$

$\Pi&space;=\frac{22}{7}=3.14$

Trapezium:

In trapezium opposite sides are equal

Area of trapezium = $\frac{1}{2}(a+b)\times&space;h$

Parallelogram:

Area of parallelogram = $\frac{1}{2}(h_{1}+h_{2})\times&space;D$

Perimeter = a+b+c+d

Rhombus:

In rhombus all sides are equal and diagonal are not equal

Area of rhombus = $\frac{1}{2}\times&space;d_{1}\times&space;d_{2}$

SECTOR:

Length of arc = $2\Pi&space;r\frac{\Theta&space;}{360}$

Perimeter = $2r(\Pi\times&space;\frac{\Theta&space;}{360}+1)$

Area = $\Pi&space;r^{2}\frac{\Theta&space;}{360}$

## Allegations and mixture part 6

Problem 1: A jar contains a mixture of two liquids A and B in the ratio 3 : 1. When 15 liters of the mixture is taken out and 9 liters of liquid B is poured into the jar, the ratio becomes 3 :4. How many liters of liquid A was contained in the jar?

Solution:

A  :  B

3  :  1

3  :  4

1-4=3

3——->9

1——–>?$\frac{9}{3}=3$

4——->?4*3=12

12+15=27

$27\times&space;\frac{3}{4}=20.25$

Problem 2:A vessel is filled with liquid, which is 3 parts water and 5 parts milk. How much of the liquid should be drawn off and replaced by water to make it half water and half milk?

Solution:

w   :   m

3    :    5————-> 3    :   5

1    :    1*5———–>5   :   5

$\frac{2}{8+2}=\frac{2}{10}=\frac{1}{5}$

Allegations and mixture

$\frac{1}{5}$ is answer

Problem 3: In three classes X, Y and Z an algebra test is conducted. The average score of class X is 83. The average score of class Y is 76 and the average score of class Z is 85. The average score of class X & Y is 79 and the average score of class Y & Z is 81. What is the average score of the classes X, Y and Z.?

Solution:

x : y : z= 3 : 4 : 5
Average score of a class
$\frac{3\times&space;83+4\times&space;76+85\times&space;5}{3+4+5}=\frac{978}{12}=81.5$

## Allegations and mixture part 5

Problem 1: There are two mixtures A and B, both contains milk and water .In mixture A ratio of milk and water is 5:4 whereas in mixture B ratio of milk and water is 5:7. What part of mixture B should be taken out and replaced with mixture A so that in mixture B quantity of milk and water becomes equal ?

Solution:

Answer is $\frac{2}{3}$

Problem 2: Milk and water are in c an A as 4 : 1 and in can B as 3 : 2. For Can C, if one takes equal quantities from A and B, find the ratio of milk to water in can C ?

Solution:

$\frac{x-\frac{3}{5}}{\frac{4}{5}-x}=\frac{1}{1}$

$x-\frac{3}{5}=\frac{4}{5}-x$

$2x=\frac{7}{5}$

$x=\frac{7}{10}$

here 7 is milk and 3 parts are water

Problem 3: A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10 liters of mixture is taken out and 10 liters of liquid B is poured into the jar, the ratio becomes 2:3. How many liters of liquid A was contained in the jar?

Solution:

A  :  B = 4X  :  X——->5X

In 10 liters 8 parts is milk and 2 parts is water

$\frac{4x-3}{x-2+1}=\frac{2}{3}$

12x-24=2x+16

10x=40

x=4

4*4=16

Al ligation method

1  :  1

1——->10

1——-> Is remaining mixture

1———->10+10=20

$20\times&space;\frac{4}{5}=16$

$20\times&space;\frac{1}{5}=4$

Shortcut

A  :  B

4  :  1—————>4  :  1

2  :  3 *2———–>4  :   6

6-1=5

5——>10

1——->?2

10+10=20

$20\times&space;\frac{4}{5}=16$

$20\times&space;\frac{1}{5}=4$

## Allegations and mixture part 4

Problem1: Some amount out of Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent at 6% per annum was?

Solution:

$S.I=\frac{PTR}{100}$

$1600=\frac{7000\times&space;5\times&space;R}{100}$

$R=\frac{32}{7}$

2  :  5

$\frac{2}{7}\times&space;7000=2000$

Problem 2: The blowing average of a cricketer was 12.4. he improves his bowling average by 0.2 points when he takes 5 wickets for 26 runs in his last match. The number of wicket taken by him before the last match was ?

Solution:

$Blowing&space;Average&space;=\frac{Runs}{Wicket}$

Let number of wickets = x

Runs=12.4X

$12.2=\frac{12.4x+26}{x+5}$

Solve the equation

Shortcut:

$\frac{26}{5}=5.2$

7  :  0.2—–>70  :  2—–>35  :  1

1——–>5

35——>? 35*5 = 175

Problem 3:Two mixtures A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and a water in the ratio 9 : 4?

Solution:

$\frac{2}{91}:\frac{1\times&space;7}{13\times&space;7}$

$\frac{2}{91}:\frac{7}{91}$

2  :  7

## Allegation and mixture part 3

Problem 1:The cost of Type 1 material is RS.15 per kg and Type 2 material is RS 20 per kg. If both type 1 and Type 2 are mixed in the ratio of 2:3, then what is the price per kg of mixed variety of material?

Solution:

Problem2 :In an examinations, a students gets 3 marks for every correct answer and loses 1 mark for every wrong answer. If he scores 120 marks in a paper of 100 questions, how many of his answers were correct assuming that he attempts all the questions?

Solution:

Let the number of correct answers = x

number of wrong answers = 100-x

3x + (100-x)-1 = 100

Solve the equation

Shortcut:

$\frac{120}{3}=40$

$\frac{1}{4}\times&space;60=15$

40+15=55

Problem 3:A man possessing Rs. 8400 lent a part of it at 8% simple interest and the remaining at 20/3% simple interest .His total income after 1 ½ years was Rs. 882. Find the sum lent at 8% rates ?
Solution:
$S.I=\frac{PTR}{100}$
$882=\frac{8400\times&space;R\times&space;3}{2\times&space;100}=7$%
4————->8400
1————–>?$\frac{8400}{4}=2100$

## Allegation and mixtures part 2

Problem 1:A mixture of 70 liter of wine and water contains 10% water. How much water should be added to the mixture to make 37% water in the resulting mixture?

Solution:

Basic method

Wine              Water

90%                  10%    ——>  ( 9   :   1)*7 —–>63  :  7

63%                  37%  ———>  (63 : 37)*1 —–> 63   :  37

63 + 7 =70

30 liter of water is added

shortcut:

Mixture        Water

7                    3

Wine            Water

4                        3

7——->70

1———>?$\frac{70}{7}$ = 10

3——->?3*10 =30

Problem 2:In a zoo, there are rabbits and pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there?

Solution:

pigeons = x

Rabbit = y

x+y=200

2x+4y = 580

Solve this equation

Shortcut:

$\frac{580}{200}=2.9&space;legs/heads$

pigeon =11

Rabbit = 9

$\frac{11}{20}\times&space;200=110$

Problem 3: In what ratio must a grocer mix tea at RS. 60 a kg and 65 a kg so that by selling a mixture at RS. 68.20 a kg, he may gain 10%?

Solution:

110%——>68.20

100%——->?$\frac{6820\times&space;100}{110}=62$

## Alligations and mixtures part 1

Problem 1: In what ratio two varieties of milk costing  Rs 16 and RS 18 per liter respectively mixed, so the mixture cost RS.16.60 per liter ?

Solution:

Basic metod:

16——->x——–>16x

18——->y———>18y

16x+18y = 16.60(x+y)

16x+16y = 16.60x+16.60y

0.60x = 1.40y

$\frac{x}{y}=\frac{14}{6}=7:3$

shortcut:

Problem 2: The average weight of 30 students of a class is 45 kg . The average weight of girls is 37 kg. and that of the boy is 49 kg. Find the number boys in the class ?

Solution:

Basic method:

Boys =x

Girls = (30-x)

x*49 + (30-x)*37 = 30*45

Solving this equation is time taking process.

Shortcut:

2+1=3

3——->30

1———>?$\frac{30}{3}=10$

2———->?2*10=20 BOYS

Problem 3:A shopkeeper sold 40kg of goods partly at 7% profit and rest at 5% loss, suffering 2% loss on the whole transactions. What is the quantity sold at profit?

Solution:

CP = 100

1+3=4

4——–>40

1———>?$\frac{40}{4}=10$ KG sold for profit

3———->?3*10 = 30

## Ages part 4

Problem 1:Ram got married 10 years ago. His present age is $1\frac{1}{3}$ times of the age at the time of marriage. Ram’s sister was 4 year younger to him at the time of his marriage. Find out the present age of Ram’s sister?

Solution:

$1\frac{1}{3}$——->$\frac{4}{3}$

R            M

4             3

1——–>10

4———>?4*10=40

3———>?3*10=30

40-4=36 is  the present age of Ram’s sister

Problem 2:If the two digits of the age of Arun are reversed then the new age so obtained is the age of his wife. 1/11 part of the sum of their ages is equal to the difference between their ages. If arun is older than his wife then find the difference between their ages?

Solution:

Assume age of Arun =10x+y

Age of his wife = 10y+x

$\frac{1}{11}(11x+11y)=9x-9y$

$10y=8x$

$\frac{x}{y}=\frac{10}{8}=\frac{5}{4}$

A=54     w = 45

Difference between the ages is 9

Problem 3:Arun,s brother is 3 years ealder to him. his father was 28 years of age when his sister was born while his mother was 26 years of age whaen he was born, The ages of arun,s father when his brother was born?

Solution:

Father age =28——->28+4=32

Mother age = 26——>26+4=30

Sister age=0——–>0+4 = 4

brother age = 0

The age of arun’s father when the brother born is 32

Problem 4: If 6 years are subtracted from the present age of Randheer and the remainder is divided by 18 then the present age of his grandson Anup is obtained. If the anup is 2 years younger to mahesh whose age is 5 years, then whats the age of Randheer?

Solution:

Age of mahesh is= 5

Age of Anup is = 5-3 =2

$\frac{R-6}{18}=3$

R= 54+6 =60

## Ages part 3

Problem 1:16 years ago, my uncle was 8 times more older than me. After 8 years from today, my uncle will be thrice as old as I will be at that time. Eight years ago, What was the ratio of my age and my uncle’s age?

Solution:

$U^{-16}:M^{-16}=9:1$——->difference is 8 (9-1=8)

$U^{+8}:M^{+8}=3:1$———->difference is (3-1=2)

$U^{-16}:M^{-16}=9:1$ * 1———–>$9:1$

$U^{+8}:M^{+8}=3:1$  *  4————>$12:4$

12-9 = 3          4-1 = 3

8-(-16)=24

3———>24

1———->?$\frac{24}{3}=8$

Age after 8 years

12———>?12*8=96

4———->?4*8=32

Age 8 years ago 8+8=16

96-16=80

32-16=16

M      :       U

16     :        80

1        :         5

Problem 2:If Sita’s mother was 4 times as old as sita 10 years ago. After 10 years mother will be twice as old as sita. How old is Sita’ s mother at present?

Solution:

$M^{-10}:S^{-10}=4:1$——–>difference is 3 (4-1 = 3)

$M^{+10}:S^{+10}=2:1$——–>difference is 1 (2-1=1)

$M^{-10}:S^{-10}=4:1$ * 1——–>$4:1$

$M^{+10}:S^{+10}=2:1$ *  3——–>$6:3$

6-4=2  ,   3-1=2

10-(-10) = 20

2———->20

1———–>$\frac{20}{2}=10$

Age after 10 years

6———–>?6*10 = 60

3————>?3*10 = 30

Sita’s mother present age = 60-10 = 50

Problem 3: Jayesh is as more younger to Amit as Jayash is older to prashant. If the sum of the ages of Amit and Prashant is 48 years. What is the age of Amit in years?

Solution:

A-J=X

J-P=X

A-J = J-P

A+P = 2J

48 = 2J

J = 24

We can derive only jayaesh age based on the given data

we can not determined Amith and prasanth age


text:
## Tuesday, June 24, 2014

### Moore-Penrose Inverse & Gauss-Newton SABR Minimization

I have found a particularly nice initial guess to calibrate SABR. As it is quite close to the true best fit, it is tempting to use a very simple minimizer to go to the best fit. Levenberg-Marquardt works well on this problem, but can we shave off a few iterations?

I firstly considered the basic Newton's method, but for least squares minimization, the Hessian (second derivatives) is needed. It's possible to obtain it, even analytically with SABR, but it's quite annoying to derive it and code it without some automatic differentiation tool. It turns out that as I experimented with the numerical Hessian, I noticed that it actually did not help convergence in our problem. Gauss-Newton converges similarly (likely because the initial guess is good), and what's great about it is that you just need the Jacobian (first derivatives). Here is a good overview of Newton, Gauss-Newton and Levenberg-Marquardt methods.

While Gauss-Newton worked on many input data, I noticed it failed also on some long maturities equity smiles. The full Newton's method did not fare  better. I had to take a close look at the matrices involved to understand what was going on. It turns out that sometimes, mostly when the SABR rho parameter is close to -1, the Jacobian would be nearly rank deficient (a row close to 0), but not exactly rank deficient. So everything would appear to work, but it actually misbehaves badly.

My first idea was to solve the reduced problem if a row of the Jacobian is too small, by just removing that row, and keep the previous value for the guess corresponding to that row. And this simplistic approach made the process work on all my input data. Here is the difference in RMSE compared to a highly accurate Levenberg-Marquardt minimization for 10 iterations:

Later, while reading some more material related to least square optimization, I noticed the use of the Moore-Penrose inverse in cases where a matrix is rank deficient. The Moore-Penrose inverse is defined as:
$$M^\star = V S^\star U^T$$
where $$S^\star$$ is the diagonal matrix with inverted eigenvalues and 0 if those are deemed numerically close to 0, and $$U, V$$ the eigenvectors of the SVD decomposition:
$$M=U S V^T$$
It turns out to work very well, beside being simpler to code, I expected it to be more or less equivalent to the previous approach (a tiny bit slower but we don't care as we deal with small matrices, and the real slow part is the computation of the objective function and the Hessian, which is why looking at iterations is more important).

It seems to converge a little bit less quickly, likely due to the threshold criteria that I picked (1E-15).
Three iterations is actually most of the time (90%) more than enough to achieve a good accuracy (the absolute RMSE is between 1E-4 and 5E-2) as the following graph shows. The few spikes near 1E-3 represent too large errors, the rest is accurate enough compared to the absolute RMSE.

To conclude, we have seen that using the Moore-Penrose inverse in a Gauss-Newton iteration allowed the Gauss-Newton method to work on rank-deficient systems.
I am not sure how general that is, in my example, the true minimum either lies inside the region of interest, or on the border, where the system becomes deficient. Of course, this is related to a "physical" constraint, here namely rho > -1.


text:
# NAG Library Routine Document

## 1Purpose

x03aaf calculates the value of a scalar product using basic precision or additional precision and adds it to a basic precision or additional precision initial value.

## 2Specification

Fortran Interface
 Subroutine x03aaf ( a, b, n, c1, c2, d1, d2, sw,
 Integer, Intent (In) :: isizea, isizeb, n, istepa, istepb Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: a(isizea), b(isizeb), c1, c2 Real (Kind=nag_wp), Intent (Out) :: d1, d2 Logical, Intent (In) :: sw
#include nagmk26.h
 void x03aaf_ (const double a[], const Integer *isizea, const double b[], const Integer *isizeb, const Integer *n, const Integer *istepa, const Integer *istepb, const double *c1, const double *c2, double *d1, double *d2, const logical *sw, Integer *ifail)

## 3Description

x03aaf calculates the scalar product of two real vectors and adds it to an initial value $c$ to give a correctly rounded result $d$:
 $d=c+∑i=1naibi.$
If $n<1$, $d=c$.
The vector elements ${a}_{i}$ and ${b}_{i}$ are stored in selected elements of the one-dimensional array arguments a and b, which in the subroutine from which x03aaf is called may be identified with parts of possibly multidimensional arrays according to the standard Fortran rules. For example, the vectors may be parts of a row or column of a matrix. See Section 5 for details, and Section 10 for an example.
Both the initial value $c$ and the result $d$ are defined by a pair of real variables, so that they may take either basic precision or additional precision values.
 (a) If ${\mathbf{sw}}=\mathrm{.TRUE.}$, the products are accumulated in additional precision, and on exit the result is available either in basic precision, correctly rounded, or in additional precision. (b) If ${\mathbf{sw}}=\mathrm{.FALSE.}$, the products are accumulated in basic precision, and the result is returned in basic precision.
This routine is designed primarily for use as an auxiliary routine by other routines in the NAG Library, especially those in the chapters on Linear Algebra.

None.

## 5Arguments

1:     $\mathbf{a}\left({\mathbf{isizea}}\right)$ – Real (Kind=nag_wp) arrayInput
On entry: the elements of the first vector.
The $i$th vector element is stored in the array element ${\mathbf{a}}\left(\left(i-1\right)×{\mathbf{istepa}}+1\right)$. In your subroutine from which x03aaf is called, a can be part of a multidimensional array and the actual argument must be the array element containing the first vector element.
2:     $\mathbf{isizea}$ – IntegerInput
On entry: the dimension of the array a as declared in the (sub)program from which x03aaf is called.
The upper bound for isizea is found by multiplying together the dimensions of a as declared in your subroutine from which x03aaf is called, subtracting the starting position and adding $1$.
Constraint: ${\mathbf{isizea}}\ge \left({\mathbf{n}}-1\right)×{\mathbf{istepa}}+1$.
3:     $\mathbf{b}\left({\mathbf{isizeb}}\right)$ – Real (Kind=nag_wp) arrayInput
On entry: the elements of the second vector.
The $i$th vector element is stored in the array element ${\mathbf{b}}\left(\left(i-1\right)×{\mathbf{istepb}}+1\right)$. In your subroutine from which x03aaf is called, b can be part of a multidimensional array and the actual argument must be the array element containing the first vector element.
4:     $\mathbf{isizeb}$ – IntegerInput
On entry: the dimension of the array b as declared in the (sub)program from which x03aaf is called.
The upper bound for isizeb is found by multiplying together the dimensions of b as declared in your subroutine from which x03aaf is called, subtracting the starting position and adding $1$.
Constraint: ${\mathbf{isizeb}}\ge \left({\mathbf{n}}-1\right)×{\mathbf{istepb}}+1$.
5:     $\mathbf{n}$ – IntegerInput
On entry: $n$, the number of elements in the scalar product.
6:     $\mathbf{istepa}$ – IntegerInput
On entry: the step length between elements of the first vector in array a.
Constraint: ${\mathbf{istepa}}>0$.
7:     $\mathbf{istepb}$ – IntegerInput
On entry: the step length between elements of the second vector in array b.
Constraint: ${\mathbf{istepb}}>0$.
8:     $\mathbf{c1}$ – Real (Kind=nag_wp)Input
9:     $\mathbf{c2}$ – Real (Kind=nag_wp)Input
On entry: c1 and c2 must specify the initial value $c$: $c={\mathbf{c1}}+{\mathbf{c2}}$. Normally, if $c$ is in additional precision, c1 specifies the most significant part and c2 the least significant part; if $c$ is in basic precision, then c1 specifies $c$ and c2 must have the value $0.0$. Both c1 and c2 must be defined on entry.
10:   $\mathbf{d1}$ – Real (Kind=nag_wp)Output
11:   $\mathbf{d2}$ – Real (Kind=nag_wp)Output
On exit: the result $d$.
If the calculation is in additional precision (${\mathbf{sw}}=\mathrm{.TRUE.}$),
• ${\mathbf{d1}}=d$ rounded to basic precision;
• ${\mathbf{d2}}=d-{\mathbf{d1}}$,
thus d1 holds the correctly rounded basic precision result and the sum ${\mathbf{d1}}+{\mathbf{d2}}$ gives the result in additional precision. d2 may have the opposite sign to d1.
If the calculation is in basic precision (${\mathbf{sw}}=\mathrm{.FALSE.}$),
• ${\mathbf{d1}}=d$;
• ${\mathbf{d2}}=0.0$.
12:   $\mathbf{sw}$ – LogicalInput
On entry: the precision to be used in the calculation.
${\mathbf{sw}}=\mathrm{.TRUE.}$
${\mathbf{sw}}=\mathrm{.FALSE.}$
basic precision.
13:   $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6Error Indicators and Warnings

If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
 On entry, ${\mathbf{istepa}}\le 0$, or ${\mathbf{istepb}}\le 0$.
${\mathbf{ifail}}=2$
 On entry, ${\mathbf{isizea}}>\left({\mathbf{n}}-1\right)×{\mathbf{istepa}}+1$, or ${\mathbf{isizeb}}>\left({\mathbf{n}}-1\right)×{\mathbf{istepb}}+1$.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.

## 7Accuracy

If the calculation is an additional precision, the rounded basic precision result d1 is correct to full implementation accuracy, provided that exceptionally severe cancellation does not occur in the summation. If the calculation is in basic precision, such accuracy cannot be guaranteed.

## 8Parallelism and Performance

x03aaf is not threaded in any implementation.

The time taken by x03aaf is approximately proportional to $n$ and also depends on whether basic precision or additional precision is used.
On exit the variables d1 and d2 may be used directly to supply a basic precision or additional precision initial value for a subsequent call of x03aaf.

## 10Example

This example calculates the scalar product of the second column of the matrix $A$ and the vector ${\mathbf{b}}$, and add it to an initial value $1.0$ where
 $A= -2 -3 7 2 -5 3 -9 1 0 , b= 8 -4 -2 .$

### 10.1Program Text

Program Text (x03aafe.f90)

### 10.2Program Data

Program Data (x03aafe.d)

### 10.3Program Results

Program Results (x03aafe.r)

© The Numerical Algorithms Group Ltd, Oxford, UK. 2017


text:
Hello, my readers!! If you are struggling with what’s this UDF is, then let me ask you a simple question. You all must have definitely made your own functions while coding, right?

Alike them, we have the privilege to build our own “USER DEFINED FUNCTIONS” in Apache Spark too. A UDF is a function provided by the user when pre-defined functions are unable to achieve the goal.

• These are column-based function.
• It helps in extending the basic functionality.
• It allows developers to create and use new functions of their own choice.
• Take one row at a time to produce a single corresponding value per row.
##### What’s the need to register a function?

After creating a function, it could only be called on the driver node of Spark. Other executors won’t be able to call the function. Thus, we need to register/send it to each executer which is also called doing CLONE or BROADCAST.

After registering, we can use it as if it’s spark own function.

We’ll be checking out all these while solving the given problems.

### PROBLEM 1

Create a UDF to calculate the bonus recevied by the employee depending on number of months they worked in a year.

Formula: No of months * Fixed amount – Tax

No of months*1000-250.50

Let’s create a dataframe first.

val emp=Seq(("Nelsinki",101,4),("Nairobi",103,10),("Beck",101,8),("Ruth",104,6))
// creating dataframe
val df=emp.toDF("Name","Dept_Id","Months_worked")
// View Dataframe
df.show()
// View Schema
df.printSchema()

We need to import udf and col function to get working on Spark SQL.

#### How to create a function?

import org.apache.spark.sql.functions.{udf,col}
// s is the input parameter of Int DataType
def money=(s:Int)=>
{
(s*1000)-250.50
}

#### How to register the function as Spark UDF?

##### METHOD 1: Using the udf function

The first argument in the udf function denotes the datatype of the output and the second argument denotes the input datatype. As in here, money function returns a value of double data type and takes an Integer.

val bonus=udf[Double,Int](money)
// Use bonus function and pass the argument (Invoking UDF)
df.withColumn("Bonus_Received",bonus(col("Months_worked"))).show()
##### Method 2: Using spark.udf.register

The API spark.udf.register is the standard method for registering a spark UDF where :

• First argument: Name of UDF ( User choice)
• Second argument: Name of the Scala function
val Salary=spark.udf.register("Income",money)
//Invoking UDF
df.withColumn("Bonus_Received",Salary(col("Months_worked"))).show()

A Simple use case to shorten up the things:

Create a UDF function to find the length of the employee’s name.

To simplify the code, we can put the whole function inside the udf as depicted below.

val count=udf{(a:String)=>a.length}
df.withColumn("Count_of_letter",count(col("Name"))).show()

Create a UDF function to perform increase in months_worked by 2.

val add=udf{(a:Int)=>a+2}
df.withColumn("Actual_Months",add(col("Months_worked"))).show()

### PROBLEM 2

We are creating a Student dataframe.

val row=Seq(("Jack",20,"F"),("Ayush",4,"Male"),("Rio",43,"M"),("Stan",56,"Other"))
// Creating dataframe
val df=row.toDF("Name","Roll_No","Gender")
// View Dataframe
df.show()

Update the Gender Column by creating a UDF resulting in “Female”, “Male” and “Unknown” accordingly.

#### Creating a Scala Function

def gen=(s:String)=>
{
val f=Seq("F","female","FEMALE")
val m=Seq("M","Male","MALE")
if (f.contains(s)){
"Female"
}
else if (m.contains(s))
{
"Male"
}
else
{
"Unknown"
}
}

#### Registering the function as UDF

##### Method 1
val know_gender=udf[String,String](gen)
df.withColumn("GENDER",know_gender(col("Gender"))).show()
##### Method 2
val fun=spark.udf.register("gender",gen)
df.withColumn("GENDER",fun(col("Gender"))).show()

Today we learned:

• How to create a user-defined function
• How to register udf
• How to use udf function

Did you find it interesting? Put your thoughts forward using the comment section below. I hope you got something to grab today. Thanks for reading.

Happy Sparking!!

Categories: Miscellaneous

### 1 Comment

#### ปั๊มไลค์ · May 28, 2020 at 3:36 pm

Like!! Really appreciate you sharing this blog post.Really thank you! Keep writing.

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text:
# Definition and calculation of the log pointwise predictive density

I want to calculate the log pointwise predictive density from an MCMC sample. Gelman et al (2014) define the lppd as

$$lppd = \sum_i^n \log \int( p(y_i| \theta) p_\text{post}(\theta) d\theta$$

My questions are

1. Why is the lppd called "point-wise" if we sum over all points data points?
2. Am I correct in interpreting this as lppd being identical to the marginal likelihood, except for the fact that we marginalize the likelihood over the posterior distribution instead of the prior distribution? Again, I find the name a bit misleading, something like marginal posterior likelihood would be more intuitive to me.
3. It seems to me that the easiest way to calculate the lppd in practice would be to record the likelihood in the MCMC, and then take the mean of these values. Except for the memory used for recording the likelihood, is there any drawback to this approach?

Edit: I just wanted to note that after the answer of @PITBULL, I realize now that the key point is that a kind of marginal posterior likelihood is calculated per data point first, and then those probabilities are multiplied, so I do understand the name choice now, and the answer to the next two questions is "no" and "no" ;)

References:

Gelman, A.; Hwang, J. & Vehtari, A. (2014) Understanding predictive information criteria for Bayesian models. Stat. Comput., Springer US, 24, 997-1016.

$$\int p(y_i| \theta) p_{post}(\theta) d\theta \approx \dfrac{1}{N}\sum_{j=1}^N p(y_j\vert \theta^{(j)})$$


text:
## Algebra 1

The point of intersection is $(1,2)$
#$1$ First equation is $y=-x+3$ Slope$:m=-1$ y - intercept$:b=3$ $(0,3)$ To graph First equation start at 3 on the y-axis Then, go down 1 and right 1. You will want to do this several times to get an accurate point where both lines intercept. #$2$ Scond Equation $y=x+1$ Slope$:m=1$ y-intercept$:b=0$$(0,1)$ To graph Second equation start at the 3 on the u-axis.Then, go up 1and right 1. You will want to do this several times to get an accurate point where both lines intercept. #$3$ After graphing you will see a point where both lines intersect that is the point of intersection (1,2). The graph is as following-