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# Applications of Trig Functions (Grades 11-12)

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## Applications of Trig Functions

1.
Sound waves can be represented as trigonometric functions. Which of the following functions correctly describes the note A above middle C, which has a frequency of 440 Hz? 1 Hz (Hertz) is equal to one cycle per second. Assume that the note played has an amplitude of 40 dB. Also assume that if the function is graphed, the midline is the x-axis.
1. $f(t) = 40 sin(880 pi t)$
2. $f(t) = 20 sin(880 pi t) + 20$
3. $f(t) = 20 sin(440 t)$
4. $f(t) = 40 sin(440 t)$
2.
A bicycle, which is stopped at the moment, has one wheel with a slight imperfection at the top of it. The wheel is 0.6 m in diameter. If the bicycle starts moving at a steady speed of 4 m/s, which of the following functions would best model the vertical height from the ground to the imperfection on the wheel, depending on how long the bike has been in motion (in seconds)? Assume that that $t>0$.
1. $h(t) = 3/5 cos(40/3 t) + 3/10$
2. $h(t) = 3/10 cos((3pi)/5 t) + 3/10$
3. $h(t) = 3/10 cos(40/3 t) + 3/10$
4. $h(t) = 3/10 sin((3pi)/5 t) + 3/10$
3.
The following table gives the average maximum temperature each month for a small town based on historical averages. Which of the following functions best models this data, where $T$ is measured in degrees Celsius and $m$ in months? Let $m=0$ be January.

 $\ \ \ \mathbf{Month} \ \ \$ $"Jan"$ $"Feb"$ $"Mar"$ $"Apr"$ $"May"$ $"June"$ $"July"$ $"Aug"$ $"Sept"$ $"Oct"$ $"Nov"$ $"Dec"$ $\mathbf{Temp \ (°C)}$ $-15$ $-10$ $-2$ $7$ $20$ $28$ $31$ $29$ $22$ $9$ $-3$ $-10$
1. $T(m) = -23cos(pi/12 m) + 8$
2. $T(m) = 23sin(pi/6 m) + 8$
3. $T(m) = 23cos(pi/12 m) + 8$
4. $T(m) = -23cos(pi/6 m) + 8$
4.
The electricity supplied to people's homes in North America is alternating current, meaning that the current switches direction. In terms of the voltage, it oscillates between positive and negative. If the frequency of the voltage is 60 cycles per second, and the maximum voltage is 170 V (and the minimum is the negative of the maximum), which of the following functions best describes the voltage, $V$, of a typical North American outlet depending on the time, $t$, in seconds?
1. $V(t) = 170 sin(120 pi t)$
2. $V(t) = 170 sin(60 t)$
3. $V(t) = 85 sin(60 t)$
4. $V(t) = 85 sin(120 pi t) + 105$
5.
A metronome helps keep time for a musician and usually has a solid wooden base with an upside down pendulum. The way a particular metronome is currently set, the tip of of pendulum rod reaches a maximum horizontal distance of 5 cm from the center of the metronome base on each side. It takes 0.4 seconds for the pendulum to go from the center of the metronome to one of the extreme positions. If the metronome was started in a position of 2 cm to the left of the metronome's center (measured horizontally), and began moving towards the right, which of the following functions would best model the pendulum tip's horizontal distance (in centimeters) from the center of the metronome, depending on the time (in seconds) since it was started? Assume that left of the metronome base is negative.
1. $D(t) = 5 sin((5pi)/4 (t-0.105))$
2. $D(t) = 5 sin(1.6(t-0.105))$
3. $D(t) = 5 sin((5pi)/4 t)$
4. $D(t) = 5 sin((5pi)/4 (t+0.105))$
6.
A certain island has a population of deer and wolves. The deer have no other predator and the wolves mainly eat the deer. These two species exhibit a cyclical predator-prey relationship, where the deer population will grow larger and therefore provide more food for the wolves. As the wolf population grows, it begins to reduce the deer population. As their food source diminishes, the wolf population also decreases. As the wolf population decreases, the deer population will grow, and the cycle will repeat. The full cycle repeats itself once every 130 weeks. The wolf population grows to a maximum of 35 wolves and a minimum of 11 wolves. If there are currently about 23 wolves and the population is declining, approximately how many wolves will there be in 83 weeks?
1. 13 wolves
2. 32 wolves
3. 35 wolves
4. 11 wolves
7.
A circular clock is positioned on a wall such that the lowest point on the clock is 2.8 meters above the floor. The clock has a diameter of 30 cm. The minute hand of the clock reaches to the very circumference of the clock's face. The vertical distance from the floor to the tip of the minute hand can be modeled by a trigonometric function. Create a trigonometric function to model this situation, and determine about how far the minute hand is above the floor at 1:17 pm.
1. 3.10 m
2. 3.08 m
3. 2.92 m
4. 2.87 m
8.
A certain retail store has noticed that their sales increases immediately after they do a large advertising campaign on social media, but then eventually their sales start to decrease. After running 5 advertising campaigns equally spaced during the year for a few years, they notice that their sales can be modeled by a trigonometric function. They earn about $550 in sales (the maximum they earn in a single day) on the following dates: February 14, April 28, July 10, September 21, and December 3. The lowest amount they earn on a given day is$100. How much does the store typically earn on August 1? Assume all years have 365 days.
1. $429 2.$344
3. $310 4.$254
9.
The number of daylight hours varies throughout the year and can be approximately modeled by trigonometric functions. For a certain city, the maximum number of daylight hours occurs on June 21 and is 16 hours and 7 minutes. The minimum of 8 hours and 19 minutes occurs on December 21. Using this information, create a trigonometric function which models this data and determine approximately how many hours of daylight there will be on April 5. Assume that the year has 365 days.
1. 8 hours and 18 minutes
2. 11 hours and 48 minutes
3. 13 hours and 8 minutes
4. 16 hours and 8 minutes
10.
The following table gives the percentage of the moon illuminated (P.I.) on dates in August and early September for a certain city. Given that the phases of the moon are periodic and can be approximately modeled by a trigonometric function, create a function which describes the data presented in the table. Then, use this function to find what approximate percentage of the moon will be illuminated on December 1. Round the answer to the nearest percent.

 $\ \ \ \ \ \ \ \ mathbf{"Date"} \ \ \ \ \ \ \ \$ $\ \ \ \ \ \ \ \mathbf{"P.I."} \ \ \ \ \ \ \$ $\ \$ $\ \ \ \ \ \ \ \ mathbf{"Date"} \ \ \ \ \ \ \ \$ $\ \ \ \ \ \ \ \mathbf{"P.I."} \ \ \ \ \ \ \$ $\ \$ $\ \ \ \ \ \ \ \ mathbf{"Date"} \ \ \ \ \ \ \ \$ $\ \ \ \ \ \ \ \mathbf{"P.I."} \ \ \ \ \ \ \$ $"Aug 1"$ $96$  $"Aug 13"$ $31$  $"Aug 25"$ $40$ $"Aug 2"$ $99$  $"Aug 14"$ $22$  $"Aug 26"$ $50$ $"Aug 3"$ $100$  $"Aug 15"$ $14$  $"Aug 27"$ $60$ $"Aug 4"$ $99$  $"Aug 16"$ $8$  $"Aug 28"$ $71$ $"Aug 5"$ $96$  $"Aug 17"$ $3$  $"Aug 29"$ $80$ $"Aug 6"$ $91$  $"Aug 18"$ $0$  $"Aug 30"$ $88$ $"Aug 7"$ $85$  $"Aug 19"$ $0$  $"Aug 31"$ $94$ $"Aug 8"$ $77$  $"Aug 20"$ $3$  $"Sept 1"$ $96$ $"Aug 9"$ $69$  $"Aug 21"$ $9$  $"Sept 2"$ $99$ $"Aug 10"$ $60$  $"Aug 22"$ $16$  $"Sept 3"$ $100$ $"Aug 11"$ $50$  $"Aug 23"$ $23$  $"Sept 4"$ $99$ $"Aug 12"$ $41$  $"Aug 24"$ $29$  $"Sept 5"$ $96$
1. 3%
2. 55%
3. 84%
4. 96%
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## Abstract

While most proposed methods for solving classification problems focus on minimization of the classification error rate, we are interested in the receiver operating characteristic (ROC) curve, which provides more information about classification performance than the error rate does. The area under the ROC curve (AUC) is a natural measure for overall assessment of a classifier based on the ROC curve. We discuss a class of concave functions for AUC maximization in which a boosting-type algorithm including RankBoost is considered, and the Bayesian risk consistency and the lower bound of the optimum function are discussed. A procedure derived by maximizing a specific optimum function has high robustness, based on gross error sensitivity. Additionally, we focus on the partial AUC, which is the partial area under the ROC curve. For example, in medical screening, a high true-positive rate to the fixed lower false-positive rate is preferable and thus the partial AUC corresponding to lower false-positive rates is much more important than the remaining AUC. We extend the class of concave optimum functions for partial AUC optimality with the boosting algorithm. We investigated the validity of the proposed method through several experiments with data sets in the UCI repository.

## 1.  Introduction

The receiver operating characteristic (ROC) curve and the area under the ROC curve (AUC) are widely employed performance measures of classifiers (Pepe & Thompson, 2000), especially in medical diagnosis and psychological studies. An important property of the AUC is that it can be used when the prior probability of each class is biased, whereas the error rate counts only the number of wrong classifications regardless of the bias of samples of each class. Also, when the goal of a problem is to find a discriminant function with a high AUC value, it is natural to use an algorithm that directly maximizes the AUC; however, in general, direct construction of a discriminant function that maximizes the AUC value is difficult because of the nondifferentiability of the definition of the AUC. Brefeld and Scheffer (2005) proposed a support vector machine (SVM) type of algorithm (Vapnik, 1998) maximizing the AUC. The optimization problem for this algorithm can be solved by standard QP solvers; however, its computational cost grows quadratically with the number of examples, and optimization is hard for large-scale problems. Boosting is a learning method that builds up a stronger classification machine from a set of base learning machines that can be easily learned, and AdaBoost (Freund & Schapire, 1997) is a typical implementation of boosting. Murata, Takenouchi, Kanamori, and Eguchi (2004) extended the algorithm to general U-Boost and discussed its statistical properties such as consistency and robustness. In this letter, we extend the concept of the AUC and propose a new performance measure along the lines of Murata et al. (2004) and a boosting-type of algorithm that maximizes the measure. We also propose an algorithm maximizing the partial AUC (pAUC), a measure of performance whose false positive rate is controlled, and investigate the statistical properties and robustness of the algorithms. In other words, the algorithm is designed to attain a high true positive rate within “a specific range of the false positive rate.” This idea is common in medical screening where the false-positive rate (the probability that healthy subjects are incorrectly judged to be positive) should be kept as low as possible or in preference-ranking problems, where only the top of the ranking list is the user's interest. Recently Rudin and Schapire (2009) proposed an algorithm that focuses on the left-most portion of the ROC curve based on RankBoost (Freund, Iyer, Schapire, & Singer, 2003). The method controls the weights on the exponential loss so that it puts more importance on low values of a false-positive rate, in contrast with the direct restriction of the range of the false positive in our approach.

This letter is organized as follows. In section 2, we describe the basic settings and briefly review the conventional AUC. We formulate a new measure, U-AUC, and propose a boosting-type algorithm, U-AUCBoost. In section 3, we propose an algorithm maximizing pAUC and investigate some of its statistical properties. The performances of the proposed methods are evaluated with some experiments in section 4.

## 2.  U-AUC

In this section, we define a new performance measure, U-AUC, which is a lower bound of the AUC, and propose an algorithm, U-AUCBoost, that can maximize the AUC. Adopting the framework of boosting makes it possible to deal with large-scale data sets. We also investigate a relationship with a statistical model and statistical properties of the algorithm such as consistency or robustness.

### 2.1.  Settings of the Problem.

Let x be a feature vector and y ∈ {+1, − 1} be a binary label of the feature vector x. A data set is given. Function I(·) is an indicator function:
For a discriminant function F(x), the conventional true-positive rate (TPR) and false-positive rate (FPR) at a threshold value c are defined as follows:
2.1
2.2
where H(z) is the Heaviside step function and py(x) is a conditional probability p(x|y) of x given the label y. The AUC is a performance measure of the discriminant function F(x) over all threshold value c and is defined by
2.3
Its empirical version is written as
2.4
where ny is the number of samples having label y in the data set.

A direct construction of F that maximizes equation 2.4 is difficult, because this equation is not continuously differentiable. Replacement of the Heaviside function with a sigmoid function solves this nondifferentiability (Herschtal & Raskutti, 2004; Ghosh & Chaudhuri, 2005) or the normal distribution function (Komori & Eguchi, 2010); however, it may not be necessary to find the global maximum because of the nonconvexity of the sigmoid function. To overcome the difficulty, in this letter we consider a true positive utility (TPU) rather than the conventional TPR and propose a new performance measure, U-AUC, that approximately maximizes equation 2.4. A similar approach, RankBoost, has been investigated by Freund et al. (2003) and Rudin and Schapire (2009), who considered combining preferences to solve ranking problems and pointed out that RankBoost is equivalent to maximization of the AUC value (Cortes & Mohri, 2004). Our proposed method includes RankBoost as a special case and has such statistical properties as consistency and robustness.

### 2.2.  U-AUC.

In this section, we define the new performance measure U-AUC and derive the algorithm for maximizing it in the framework of boosting. First, we define a quantity TPU as follows:
2.5
where U is a concave and monotonically increasing function, satisfying
2.6
Based on this quantity, we propose a new performance measure, U-AUC,
2.7
and denote its empirical version as ,
2.8
Note that the AU(F) is a lower bound of AUC,
2.9
because of equation 2.6. Then maximization of AU(F) implies maximization of the lower bound of the AUC, which is our purpose. Also note that AU(F) is concave with respect to F, i.e.,
because of the concavity of U. The equality holds if and only if F1(x) = F2(x) + c, where c is an arbitrary constant. This property also gives the uniqueness (except for c) of the optimal function .
Remark 1.

U-AUC is threshold free because AU(F + κ) = AU(F) holds for an arbitrary threshold value κ. In a situation where sample numbers of classes are biased, this threshold-free property is a virtue.

Remark 2.
We can extend the FPR to FPU as
where U2 is a concave and monotonically increasing function and an associated performance measure as
where U2 is the derivative of U2.

The maximization of AU(F) is a concave optimization problem that can be efficiently solved when a problem is not too large (Boyd & Vandenberghe, 2004). However, recently, there has been a great need for analyzing large-scale data sets having numerous examples with high dimension, and the direct optimization of AU(F) is difficult for such large-scale problems. To deal with such problems, a boosting method has been proposed: this is a learning method that builds up a stronger classification machine from a set of base learning machines that can be easily learned. A lot of boosting-type algorithms have been proposed, and theoretical properties such as consistency and robustness have been intensively investigated. In this letter, to maximize with respect to F for large-scale data sets, we derive a boosting-type algorithm, U-AUCBoost, as follows. Let be a set of base classifiers. We then construct F by ∑Tt=1αtft where ft is a component of and αt is a coefficient of ft, as follows:

1. Initialize a weight for a pair of examples as and discriminant function as F0(x) = 0.

2. For t = 1, …, T,

1. Find a base classifier as
where
2. Calculate a coefficient of the selected classifier:
3. Update the weight as
where Ft(x) = Ft−1(x) + αtft(x) and is a normalization term.

3. Output the discriminant function FT(x).

In this algorithm, the weight is defined for a pair of examples while the weight is defined for an example in the usual boosting algorithm such as AdaBoost.

A derivation of the algorithm is the same as the usual boosting-type algorithm (Murata et al., 2004). When the discriminant function Ft−1 is given, we consider an update from Ft−1 to Ft−1 + εf where ε is a small constant. Then we observe
2.11
and maximization of this quantity corresponds to the selection of the base classifier. For the selected base classifier, we determine the coefficient as in step b. In the maximization in equation 2.10, the solution often diverges. Then in our experiments, for the purpose of regularization, we threshold α at 1 and optimize in the range of [0, 1]. This kind of implementation for numerical problems is also found in LogitBoost (Friedman, Hastie, & Tibshirani, 2000).

In the following sections, we discuss statistical properties of the U-AUCBoost such as the empirical AUC attained by the algorithm (see section 2.3), a related statistical model and consistency (see section 2.4), robustness (see section 2.5), and relation with the conventional methods such as the RankBoost or AdaBoost (see section 2.6).

### 2.3.  Evaluation of Empirical AUC.

Using the boosting-type algorithm just described, we can construct the discriminant function FT by linear combination of base classifiers. We investigate how the algorithm behaves by evaluating a lower bound of the empirical U-AUC for FT as follows:

Theorem 1.
For the discriminant function FT(x) = ∑Tt=1αft(x), we observe that
where γt is a constant satisfying
and is on [0, 1].1 Note that acct(ft) − errt(ft) ⩾ 0 holds for all t.
Proof.

See appendix  A.

If acct(ft) − errt(ft)>0 and γt>0 hold for all t, the lower bound of and monotonically increases against T.

### 2.4.  Consistency of U-AUCBoost.

We consider a population maximizer of (2.7), and investigate its property. Then we observe the following theorem:

Theorem 2.
The maximizer F* satisfies
2.12
where
2.13
Proof.
Let η(x) be an arbitrary function of x and ε be a small constant, and we consider the functional derivative of AU(F). From
we observe that the maximizer F* of the concave function AU(F) satisfies
which is rewritten as
2.14
The bracketed term in equation 2.14 becomes 0 for any x because η(x) is arbitrary.

Note that F* + c where c is an arbitrary constant, also satisfies theorem 2.4 because of the threshold-free property of U-AUC. In the following, we ignore the arbitrariness of the constant.

With the concavity of U and a simple calculation, we observe
2.15
Then is a monotonically increasing function. Then we know the following:
Remark 3.
The optimal function F* is connected to the ratio p+1/p−1 or in one-to-one correspondence, and we observe that
2.16
where πy is a marginal distribution of y.
The statistical properties of equation 2.16 depend on the function U. In a context of U-Boost (Murata et al., 2004), for example, a boosting algorithm with a function U(z) = (1 − η)exp(z) + ηz where η is a tuning parameter, is closely related to a probabilistic model of mislabeling as
where δ(x) is a probability of mislabeling, which can depend on the input x. Takenouchi, Eguchi, Murata, and Kanamori (2008) investigated a model of δ(x) where the input has a higher mislabeling probability when the decision for predicting the label of x is more difficult.
We focus on a class of function U satisfying
2.17
Then we observe the following theorem:
Theorem 3.
If the function U satisfies equation 2.17, the maximizer F* of AU(F) is equal to
2.18
except for the arbitrariness of the constant.
Proof.
From relationship 2.17, it follows that
which implies that the bracketed term in equation 2.14 becomes 0 with the F‡ for any x. Then from the concavity of AU(F), we conclude that F* = F‡ except for the arbitrariness of the constant.

We next focus on a special type of function U associated with the logistic model.

#### 2.4.1.  Class of Function Associated with the Logistic Model.

Here we consider a family {F(x; α)} of discriminant functions parameterized by a vector α of parameters and the logistic model as
2.19
where c is a constant. We assume that the true conditional distribution is represented by the logistic model with parameter α0. Under these conditions, we focus on estimating the parameter α. Let be an estimator maximizing equation 2.7 where a true joint distribution of (X, Y) is (=πypy(x)).
Definition 1:

Fisher consistency. When the estimator of the parameter vector maximizing equation 2.7 associated with the distribution is consistent with the true parameter α0, that is, , then the estimator is Fisher consistent.

Note that Fisher consistency implies
2.20
and a class of function U associated with the logistic model is specified by the following theorem:
Theorem 4.

Assume that F(·; α) is a one-to-one mapping with respect to the parameter α. If the function U(z) satisfies equation 2.17 with λ = 2, the estimator maximizing equation 2.7 is Fisher consistent.

Proof.
A necessary and sufficient condition is that F(·; α0) satisfies equation 2.12. Under assumptions 2.19, we observe that
where c0 is a constant associated with α0. From the relationship 2.17, we know that
2.21
2.22
Then we observe equation 2.12 and conclude the theorem.
Remark 4.

These properties are derived from the concavity of the function U and do not hold for the sigmoid function or the normal distribution function.

We consider some examples of the function U that satisfy equation 2.17 with λ = 2.

1. Exponential type: Urank(z) = 1 − exp(−z). We discussed some properties in section 2.6.

2. Logistic type: .

The statistical properties of a boosting algorithm with this third function, called MadaBoost, is discussed in Kanamori, Takenouchi, Eguchi, and Murata (2007). MadaBoost is the most B-robust boosting algorithm for outliers in the sense of gross error sensitivity (Hampel, Rousseeuw, Ronchetti, & Stahel, 1986) when the true distribution is modeled by the logistic model. We next investigate the robustness of the loss function in the framework of U-AUC maximization. These functions are plotted in Figure 1.

Figure 1:

Examples of U-functions.

Figure 1:

Examples of U-functions.

### 2.5.  Robustness of MadaBoost Type Loss.

We consider the class of fun-ction U satisfying equation 2.17 and assume that the true distribution is written as
where F† is a discriminant function. Let us consider an update of the discriminant function with a base classifier f ∈ {+1, − 1} as
and a statistical model of the conditional distribution p(y|x) as
2.23
Note that p(y|x) = p0(y|x) and from theorem 4, the estimator maximizing equation 2.7 is Fisher consistent,
2.24
which implies .
The robustness of the estimator is measured by the gross error sensitivity defined by
2.25
where is the probability distribution with a point mass at .
Lemma 1.
The gross error sensitivity of the estimator is written as
2.26
where V(z) = 1 − U(−z).
Proof.

See appendix  B.

Then we observe the following theorem:

Theorem 5.
We assume that the base classifier f satisfies the following conditions,2
2.27
Then the estimator associated with the MadaBoost-type function Umada is the most B-robust estimator, which minimizes the gross error sensitivity (see equation 2.26).
Proof.
If the function V′ is not upper-bounded, the numerator of equation 2.26 diverges. Therefore we focus on the case where V′ is upper-bounded. Without loss of generality, we can assume
2.28
because multiplication by a positive constant for V′ does not change the estimator. From assumptions 2.27, we observe
2.29
2.30
The gross error sensitivity is minimized when the denominator of equation 2.26 is maximized. The term is calculated over the region {x|F†(x′) ⩾ F†(x)}, and then we find that V′(z) = 2 should hold for z ⩾ 0. From equation B.8, we observe that the MadaBoost-type function minimizes the gross error sensitivity.

### 2.6.  Special Example of U: Exponential Type.

The algorithm-maximizing U-AUC with a special function
is equivalent to RankBoost, whose properties were investigated in Freund et al. (2003). A relationship between the algorithm and AUC maximization was discussed in Cortes and Mohri (2004). While the selection of base classifier and calculation of αt cannot be explicitly solved in the algorithm with a general function U, the maximization of can be explicitly solved with corresponding steps of the algorithm modified as follows:
• •
Selection of a base classifier:
2.31
• •
Calculation of a coefficient for the selected classifier:
2.32

Also, as with theorem 1, theoretical evaluation of the empirical AUC value by the algorithm with the function Urank was analyzed in Freund et al. (2003).

In the following, we investigate the relationship between RankBoost associated with Urank and AdaBoost. AdaBoost is the most popular boosting algorithm for classification problems and is derived from sequential minimization of the following exponential loss function:
2.33
We observe the following:
Proposition 1.

RankBoost is equivalent to AdaBoost with an optimized threshold value.

Proof.
Here we consider a modified version of equation 2.33, which optimizes a threshold value of the discriminant function as
2.34
where
By plugging into equation 2.34, we find
2.35
and then minimization of equation 2.34, that is, AdaBoost is equivalent to the maximization of associated with the function Urank, or RankBoost.

### 2.7.  Related Work.

In this section, we briefly review related work to clarify our own contributions in this letter.

The AdaBoost procedure is derived from a sequential minimization of the loss function,
2.36
and its statistical properties have been investigated in detail (Friedman et al., 2000). Lebanon and Lafferty (2001) pointed out a relationship between AdaBoost and the maximum likelihood estimation, that is, the minimization of the extended Kullback-Leibler divergence. Also, the loss function (see equation 2.36) can be extended with any arbitrary convex function (Mason, Baxter, Bartlett, & Frean, 1999). Based on these insights, Murata et al. (2004) extended the framework of AdaBoost with a statistical version of the Bregman divergence and proposed a U-Boost algorithm. Statistical properties of the U-Boost such as the consistency or robustness were investigated, and the information geometrical structure (Amari & Nagaoka, 2000) of U-Boost algorithm was discussed, which clearly shows an intuitive interpretation of how U-Boost works. Kanamori et al. (2007) considered the robustness of the boosting method in the class of U-Boost and revealed that the MadaBoost-type loss function is the most robust loss function for the coefficient estimation.

RankBoost is a boosting method to learn accurate rank or preference by combining multiple preferences such as the results of different search engines (Freund et al., 2003), and it was pointed out that the minimization of the loss function of RankBoost is equivalent to the maximization of the surrogate of the empirical AUC (Cortes & Mohri, 2004). While RankBoost optimizes the surrogate of the empirical AUC, measure of performance of classifier, another important performance measure is the margin of classifier. However, RankBoost does not always converge to a maximum margin solution because its objective function is the empirical AUC rather than the margin. Rudin and Schapire (2009) proposed the smoothed margin ranking algorithm, a modified version of RankBoost that enables us to obtain a maximum ranking-margin solution. While the RankBoost and its variant, including the proposed method, are based on the maximization of concave objective functions, which are surrogates of the empirical AUC value, there are methods that approximate the Heaviside function by the sigmoid function (Herschtal & Raskutti, 2004; Ghosh & Chaudhuri, 2005) or the cumulative distribution function of the normal distribution (Komori & Eguchi, 2010). Algorithms are derived with the usual coordinate ascent method; however, the objective function is not a concave function, implying that the algorithm can be trapped in local maxima, and this kind of algorithm requires regularization to avoid the problem of local maxima.

Our proposed method is a generalization of the RankBoost with any arbitrary concave function for maximizing the surrogate of the empirical AUC, which is motivated by the extension of AdaBoost to U-Boost (Murata et al., 2004). Also we investigated the relationship between the class of concave functions and the associated statistical model, and proposed a robust algorithm for AUC maximization (in section 2.5) based on the framework of Kanamori et al. (2007).

## 3.  Partial U-AUC

The AUC provides an overall assessment of classification based on FPR and TPR as described in equation 2.3, each of which expresses a different accuracy for discrimination performance. In some situations, especially in disease screening, a high TPR in a range of a low FPR is important (Baker, 2003; Qi, Joseph, & Seetharaman, 2006), and hence the partial area under the ROC curve (pAUC) has been gaining in popularity. Moreover, a specific value of TPR given a fixed FPR is also used as a summary measure of discrimination in medical screening (Pepe, Longton, Anderson, & Schummer, 2003; Pepe, 2003). Pepe and Thompson (2000) proposed a statistical method that maximizes the pAUC by linearly combining two predictors, and Komori and Eguchi (2010) made it possible to nonlinearly combine more than two predictors using the approximate pAUC. In this context, we consider the partial U-AUC as an extension of the U-AUC.

### 3.1.  Definition of Partial U-AUC.

We focus on a range of FPR between τ1 and τ2 that corresponds to the threshold values c1,F and c2,F:
3.1
Note that c2,F < c1,F if τ1 < τ2. For the range of FPR, partial AUC pA(F) is defined as
3.2
To approximate the partial AUC, we define the partial U-AUC pAU(F) as
3.3
The empirical version of pAU is given as
3.4
where and are corresponding threshold values. In the empirical versions, a threshold value c associated with a specific value τ is chosen from the middle points of sorted values F(xj)’s having class label yj = −1. Note that from the definition of the function U, we observe that
3.5
and the following theorem:
Theorem 6.

pAU(F, τ1, τ2) is concave with respect to F.

Proof.

See appendix  C.

Note that the concavity for is easily shown by the same argument as that in the proof of theorem 6.

### 3.2.  Algorithm for Maximizing Partial U-AUC.

In this section, we consider a boosting-type algorithm to maximize ; however, its derivation is not straightforward because the threshold values c1,F and c2,F in definition 3.3 depend on the function F. If we update the discriminant function Ft−1 to Ft−1 + αf with a fixed f, the threshold values also change depending on the value of α, and then we cannot consider a derivation like equation 2.11. This implies that f and α cannot be independently determined, and then it is not computationally tractable. To overcome the difficulty, we assume that the coefficient α is in a range [0, 1] and consider the following lower bound of the difference of the partial U-AUC with fixed threshold values:
3.6
3.7
3.8
3.9
where the inequality of equation 3.8 is derived from the concavity of . The approximate lower bound is defined with thresholds for the function Ft−1 + f and does not depend on α, which enables us to select f independent of αt and leads to a drastic reduction in computational cost. Then the algorithm maximizing the partial U-AUC, pU-AUCBoost is written as follows:
1. Initialize the discriminant function to F0(x).

1. Find a base classifier,
where
and
2. Calculate the coefficient of the selected classifier:
3. Update the discriminant function:
2. Output the discriminant function FT(x) = F0(x) + ∑Tt=1αtft(x).

Note that in the optimization, equation 3.15, the threshold values depend on the value of α; however, we can easily detect when the threshold values change, and then the optimization is not difficult. During the optimization procedure, the value of Zt−1(f) in the denominator of equation 3.13 often becomes exactly 0 because of the restriction of the number of samples by and ; the result is that one base classifier is chosen repeatedly, and the algorithm does not move ahead. To avoid this problem, the base classifier that is chosen is discarded in later steps.

### 3.3.  Consistency of the Algorithm for pAUC.

Similar to the consistency of AU, we have the following remark:

Remark 5.
The maximizer F* of equation 3.3 is connected with the ratio p+1/p−1 or in a one-to-one correspondence, and we observe
3.17
where πy is a marginal distribution of y, and .
Proof.

See appendix  D.

Note that this one-to-one correspondence between F* and p+1/p−1 holds in the restricted range of . For the complement set that consists of and , we have
3.18
3.19
where . Hence, we have a weak Bayes consistency in the sense that the consistency holds only for .

### 3.4.  Evaluation of the Empirical pAU.

Regarding the bounds of the empirical pAU, we have the following theorem:

Theorem 7.
For the discriminant function FT(x) = ∑Tt=1αtft(x), is bounded as
3.20
where
Proof.

See appendix  E.

From relationship 3.5, we immediately observe the following corollary:

Corollary 1.
For the discriminant function FT(x) constructed by the pAU boosting algorithm, is lower-bounded as
3.21
Note that holds for all t(t = 1, …, T).

## 4.  Experiments

In this section, we clarify the validity of proposed method by comparing it with the existing binary classification method using benchmark data sets from the UCI repository (Blake & Merz, 1998). Through experiments, we investigated the performance of the proposed methods (U-AUCBoost and pU-AUCBoost) with three functions (Urank, Ulogit, Umada) and employed “stumps” (Friedman et al., 2000) as the base classifier of boosting-type algorithms. The proposed methods were compared with several boosting-type algorithms (AdaBoost, LogitBoost, and MadaBoost defined by functions Urank, Ulogit, and Umada, respectively) and SVM with an RBF kernel and a first-order polynomial (i.e., linear) kernel.

The coefficient α of the base classifiers was optimized in a range of [0, 1] for the purpose of regularization. The initial function of the pU-AUCBoost was set to a function constructed by RankBoost (the U-AUCBoost with Urank) whose step number T was determined by the AUC value or pAUC value for the validation data set, because pU-AUCBoost can use only part of all examples associated with a focused range of FPR, so it had difficulty in working smoothly in these data sets. (See Komori & Eguchi, 2010, for the preprocessing of the data by a pAUC-based method in high dimensions.) We employed the SVM in a package program e1071, which is an LIBSVM (Chang & Lin, 2011) implementation in R language (R Development Core Team, 2011), and used a first-order polynomial (i.e., linear) kernel and a radial basis function (RBF) kernel for their kernel functions (Schölkopf & Smola, 2001).

We divided the original data set into a training data set and a test data set with the ratios of 80% and 20%, respectively, and evaluated the generalization performance by using the test data set. We repeated this procedure 100 times by changing the data set division and observed the average performance. For the tuning of hyperparameters of each method (i.e., the step number T of the boosting-type algorithms or regularization parameter C controlling the number of support vectors and parameter of kernel γ of the SVM),3 we subdivided the training data set into a training subset and a validation subset with the ratios of 80% and 20%, respectively, and set hyperparameters so as to maximize the AUC value or pAUC value for the validation subset. Details of the data set are summarized in Table 1. We first removed examples with unobserved variables.

Table 1:
Data Set Information.
Data SetTotal NumberNumber of Attributes
Australian 690 14 −0.221
Breast Cancer 683 −0.619
Diabetes 768 0.624
German 1000 24 −0.847
Heart 270 13 −0.223
Ionosphere 351 34 0.580
Liver Disorders 345 0.322
Pima 768 −0.624
Sonar 208 60 −0.135
Data SetTotal NumberNumber of Attributes
Australian 690 14 −0.221
Breast Cancer 683 −0.619
Diabetes 768 0.624
German 1000 24 −0.847
Heart 270 13 −0.223
Ionosphere 351 34 0.580
Liver Disorders 345 0.322
Pima 768 −0.624
Sonar 208 60 −0.135

Figures 2 and 3 show box-whiskers plots of AUC and partial AUC (τ1 = 0, τ2 = 0.1) of each method over 100 trials, respectively. We observe that the proposed methods (especially U-AUCBoost) worked well and had improved better performance then the existing methods except for a few data sets; even for those few data sets, the existing method did not significantly outperform U-AUCBoost. Although improvement of the AUC value by the proposed method seemed to be small, this was not negligible because the AUC criteria were insensitive to the improvement of classifier (Pencina, D'Agostino, D'Agostino, & Vasan, 2008). On the other hand, pU-AUCBoost sometimes had degraded performance compared with that of U-AUCBoost. This may be because pU-AUCBoost focuses on a specific range of FPR, and its target function (see equation 3.4) of the pU-AUCBoost is defined with fewer negative examples compared with U-AUCBoost. This restriction of examples made its optimization processes less stable than that of U-AUCBoost. In summary, the algorithm pU-AUCBoost for maximizing the partial AUC can be formally defined, but we could not find a special situation or data set where it consistently works well. We recommend employing U-AUCBoost to maximize partial AUC rather than pU-AUCBoost, because U-AUCBoost performed well in the sense of the partial AUC.

Figure 2:

Box plots of AUC values for each data set. (A) SVM with RBF kernel. (B) SVM with linear kernel. (C) Boosting with Urank. (D) Boosting with Ulogit. (E) Boosting with Umada. (F) U-AUCBoost with Urank. (G) U-AUCBoost with Ulogit. (H) U-AUCBoost with Umada. (I) pU-AUCBoost with Urank. (J) pU-AUCBoost with Ulogit. (K) pU-AUCBoost with Umada.

Figure 2:

Box plots of AUC values for each data set. (A) SVM with RBF kernel. (B) SVM with linear kernel. (C) Boosting with Urank. (D) Boosting with Ulogit. (E) Boosting with Umada. (F) U-AUCBoost with Urank. (G) U-AUCBoost with Ulogit. (H) U-AUCBoost with Umada. (I) pU-AUCBoost with Urank. (J) pU-AUCBoost with Ulogit. (K) pU-AUCBoost with Umada.

Figure 3:

Box plots of partial AUC values for each data set. (A) SVM with RBF kernel. (B) SVM with linear kernel. (C) Boosting with Urank. (D) Boosting with Ulogit. (E) Boosting with Umada. (F) U-AUCBoost with Urank. (G) U-AUCBoost with Ulogit. (H) U-AUCBoost with Umada. (I) pU-AUCBoost with Urank. (J) pU-AUCBoost with Ulogit. (K) pU-AUCBoost with Umada.

Figure 3:

Box plots of partial AUC values for each data set. (A) SVM with RBF kernel. (B) SVM with linear kernel. (C) Boosting with Urank. (D) Boosting with Ulogit. (E) Boosting with Umada. (F) U-AUCBoost with Urank. (G) U-AUCBoost with Ulogit. (H) U-AUCBoost with Umada. (I) pU-AUCBoost with Urank. (J) pU-AUCBoost with Ulogit. (K) pU-AUCBoost with Umada.

## 5.  Conclusion

In this letter, we formulated a new performance measure U-AUC for AUC-optimal classification and proposed a boosting-type algorithm maximizing U-AUC, U-AUCBoost, which was formulated as sequential maximization of a concave function. The statistical properties of the maximizer of U-AUC were discussed, and we evaluated the robustness of the proposed algorithm. We also extended U-AUCBoost into pU-AUCBoost for maximizing U-AUC at a specific range of FPR. We experimentally investigated the performance of the proposed algorithms with real data sets and clarified the validity of the U-AUCBoost. pU-AUCBoost did not work so well because of the shortage of examples caused by the restriction of FPR. Nevertheless, numerical experiments showed that U-AUCBoost can maximize partial AUC.

## Appendix A:  Proof of Theorem 1

For convenience, we introduce a function V(z) that satisfies U(z) = 1 − V(−z). The function V is nonnegative, convex, and monotonically increasing. With this function, maximization of equation 2.8 is equivalent to minimization of
A.1
Note that . From the definition of γt, we observe that
Then we observe that
where
A.2
By recursively considering this inequality, we obtain
because F0(x) = 0 and V(0) = 1. From an inequality 1 + z ⩽ exp(z), we obtain
A.3
From , we conclude the theorem.

## Appendix B:  Proof of Lemma 1

By simple calculation, we observe that the conditional distribution of the contaminated distribution is written as
B.1
where . The estimator with the contaminated distribution satisfies
B.2
Note that because of the Fisher consistency, when ε = 0, and by considering the Taylor expansion, we observe that
Here, we can assume that without loss of generality. Then we observe that
B.3
Note that the first term of the above equation is zero. Then we observe that
B.4
The same holds for a case . Then the gross error sensitivity of the estimator is written as
B.5
Because F(x) is in {+1, − 1}, the numerator of equation B.5 is rewritten as
B.6
The denominator is rewritten as
B.7
From the definition of V, we observe
B.8
and then the above equation is rewritten as
B.9
Equation B.9 comes from the following relationship:
B.10
Then we conclude the lemma.

## Appendix C:  Proof of Theorem 6

For an arbitrary function η(x) and a scalar ε, we have
Note that and because of equations 3.1. In the same way, the second derivative is given as
C.1
because of the concavity of U(z). This indicates that pAU(F, τ1, τ2) is concave with respect to F.

## Appendix D:  Proof of Remark 5

The maximizer satisfies
D.1
for an arbitrary function η(x). From the same calculation as in the proof of theorem 6, we have
D.2
Note that equation D.2 can attain 0 for any η only if x is in a region , and hence F* satisfies
D.3
D.4
where Ψ(z) is defined in equation 2.13. From equation 2.15, F*(x) has a one-to-one correspondence with p+1(x)/p−1(x), which directly leads to remark 5.

## Appendix E:  Proof of Theorem 7

E.1
E.2
where 0 ⩽ α* ⩽ αT. From the same argument as in the proof of theorem 6, we have
E.3
Hence, we have
E.4
E.5
where
E.6
E.7
Using notation such as
E.8
E.9
we have
E.10
E.11
E.12
where F0(x) = 0. In the same way, we have
E.13
E.14
which concludes theorem 7.

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## Notes

1

From the concavity of with respect to α, we observe that , which implies 0 ⩽ γt ⩽ 1.

2

The function F† is associated with the true distribution rather than that constructed by the boosting algorithm, and the condition is not too strong; F† can converge when p(+1|x) (or p(−1|x)) is nearly equal to 0.

3

We optimize C and γ in a range {2−2, …, 25} and {2−3, …, 23}, respectively.


text:
Function to conduct multivariate regression analyses of survey data with the randomized response technique using Bayesian MCMC.

rrreg.bayes(formula, p, p0, p1, design, data, group.mixed,
formula.mixed = ~1, verbose = FALSE, n.draws = 10000, burnin = 5000, thin =
1, beta.start, beta.mu0, beta.A0, beta.tune, Psi.start, Psi.df, Psi.scale,
Psi.tune)

## Arguments

formula An object of class "formula": a symbolic description of the model to be fitted. The probability of receiving the sensitive question (Mirrored Question Design, Unrelated Question Design); the probability of answering truthfully (Forced Response Design); the probability of selecting a red card from the 'yes' stack (Disguised Response Design). The probability of forced 'no' (Forced Response Design). The probability of forced 'yes' (Forced Response Design). Character indicating the design. Currently only "forced-known" is supported. A data frame containing the variables in the model. A string indicating the variable name of a numerical group indicator specifying which group each individual belongs to for a mixed effects model. To specify a mixed effects model, include this formula object for the group-level fit. ~1 allows intercepts to vary, and including covariates in the formula allows the slopes to vary also. A logical value indicating whether model diagnostics are printed out during fitting. Number of MCMC iterations. The number of initial MCMC iterations that are discarded. The interval of thinning between consecutive retained iterations (1 for no thinning). Optional starting values for the sensitive item fit. This should be a vector of length the number of covariates. Optional vector of prior means for the sensitive item fit parameters, a vector of length the number of covariates. Optional matrix of prior precisions for the sensitive item fit parameters, a matrix of dimension the number of covariates. A required vector of tuning parameters for the Metropolis algorithm for the sensitive item fit. This must be set and refined by the user until the acceptance ratios are approximately .4 (reported in the output). Optional starting values for the variance of the random effects in the mixed effects models. This should be a scalar. Optional prior degrees of freedom parameter for the variance of the random effects in the mixed effects models. Optional prior scale parameter for the variance of the random effects in the mixed effects models. A required vector of tuning parameters for the Metropolis algorithm for variance of the random effects in the mixed effects models. This must be set and refined by the user until the acceptance ratios are approximately .4 (reported in the output).

## Value

rrreg.bayes returns an object of class "rrreg.bayes". The function summary is used to obtain a table of the results.

beta

The coefficients for the sensitive item fit. An object of class "mcmc" that can be analyzed using the coda package.

data

The data argument.

coef.names

Variable names as defined in the data frame.

x

The model matrix of covariates.

y

The randomized response vector.

design

Call of standard design used: "forced-known", "mirrored", "disguised", or "unrelated-known".

p

The p argument.

p0

The p0 argument.

p1

The p1 argument.

beta.tune

The beta.tune argument.

mixed

Indicator for whether a mixed effects model was run.

call

the matched call.

If a mixed-effects model is used, then several additional objects are included:
Psi

The coefficients for the group-level fit. An object of class "mcmc" that can be analyzed using the coda package.

gamma

The random effects estimates. An object of class "mcmc" that can be analyzed using the coda package.

coef.names.mixed

Variable names for the predictors for the second-level model

z

The predictors for the second-level model.

groups

A vector of group indicators.

Psi.tune

The Psi.tune argument.

## Details

This function allows the user to perform regression analysis on data from the randomized response technique using a Bayesian MCMC algorithm.

The Metropolis algorithm for the Bayesian MCMC estimators in this function must be tuned to work correctly. The beta.tune and, for the mixed effects model Psi.tune, are required, and the values, one for each estimated parameter, will need to be manipulated. The output of the rrreg.bayes function displays the acceptance ratios from the Metropolis algorithm. If these values are far from 0.4, the tuning parameters should be changed until the ratios approach 0.4.

Convergence is at times difficult to achieve, so we recommend running multiple chains from overdispersed starting values by, for example, running an MLE using the rrreg() function, and then generating a set of overdispersed starting values using those estimates and their estimated variance-covariance matrix. An example is provided below for each of the possible designs. Running summary() after such a procedure will output the Gelman-Rubin convergence statistics in addition to the estimates. If the G-R statistics are all below 1.1, the model is said to have converged.

## References

Blair, Graeme, Kosuke Imai and Yang-Yang Zhou. (2014) "Design and Analysis of the Randomized Response Technique." Working Paper. Available at http://imai.princeton.edu/research/randresp.html.

## Examples

data(nigeria)

## Define design parameters
p <- 2/3  # probability of answering honestly in Forced Response Design
p1 <- 1/6 # probability of forced 'yes'
p0 <- 1/6 # probability of forced 'no'

## run three chains with overdispersed starting values

set.seed(1)

## starting values constructed from MLE model
mle.estimates <- rrreg(rr.q1 ~ cov.asset.index + cov.married +
I(cov.age/10) + I((cov.age/10)^2) + cov.education + cov.female,
data = nigeria,
p = p, p1 = p1, p0 = p0,
design = "forced-known")

library(MASS)
draws <- mvrnorm(n = 3, mu = coef(mle.estimates),
Sigma = vcov(mle.estimates) * 9)# NOT RUN {
## run three chains
bayes.1 <- rrreg.bayes(rr.q1 ~ cov.asset.index + cov.married +
I(cov.age/10) + I((cov.age/10)^2) + cov.education + cov.female,
data = nigeria, p = p, p1 = p1, p0 = p0,
beta.tune = .0001, beta.start = draws[1,],
design = "forced-known")
bayes.2 <- rrreg.bayes(rr.q1 ~ cov.asset.index + cov.married +
I(cov.age/10) + I((cov.age/10)^2) + cov.education + cov.female,
data = nigeria, p = p, p1 = p1, p0 = p0,
beta.tune = .0001, beta.start = draws[2,],
design = "forced-known")

bayes.3 <- rrreg.bayes(rr.q1 ~ cov.asset.index + cov.married +
I(cov.age/10) + I((cov.age/10)^2) + cov.education + cov.female,
data = nigeria, p = p, p1 = p1, p0 = p0,
beta.tune = .0001, beta.start = draws[3,],
design = "forced-known")

bayes <- as.list(bayes.1, bayes.2, bayes.3)

summary(bayes)
# }


text:
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Geology
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Question:

According to Bowen's Reaction Series, quartz is the mineral that forms at the very last stage of crystallization, provided that there is excess silica. But we clearly see an abundance of quartz around us. Look at the buildings: the main constituent is sand, which is mostly of quartz. Look at the beaches, most of them are covered with sand. Even dirt, which comes from mudstones, contains quartz as framework grain. Explain - despite always forming lately, why quartz is so abundant in nature.

If we analyze Bowen's Reaction Series, we can see that at the very beginning of the crystallization, when the temperature is around 1200°C, olivine starts to crystallize first. As the temperature falls slowly, minerals of medium silication like albite, pyroxene, amphibole, biotite, anorthite crystallize. These minerals occupy most of the space in magma. At around 600°C, if there's an excess of silica $$(SiO_2)$$ in the magma, quartz begins to crystallize very slowly. This is why quartz is a mineral of high silication $$(> 60\%)$$. Now, as other minerals have occupied most of the space, there's little space for quartz to crystallize. So, due to $$\text{lack of free space}$$ and $$\text{extremely slow crystallization}$$, quartz grains become tightly bonded within the crystal lattice. That's why quartz has a very well-ordered atomic arrangement. This makes quartz one of the hardest $$(7 - \text{Mohs Hardness Scale})$$ minerals of the earth. Consequently all of these minerals undergo weathering & erosion by geomorphic agents (wind, water). Quartz being so well-crystallized, doesn't weather & erode as much as mica, feldspars do. So even when the other minerals have entirely been transformed to clay minerals like kaolinite, quartz sustains its composition and lattice structure. And quartz doesn't react with most of the chemical entities. So even after undergoing geologically extensive transportation, quartz is not much affected by weathering & erosion. And that's why there's an abundance of quartz around us. Quartz has a protection from weathering & erosion - a shield of well-crystallized crystal structure.

C Programming
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Question:

$( S_n = 1 + 2 - 3 \times 4\div 5 + 6 - 7 \times 8 \div 9 +10- \cdots n$) Consider the above sum where the first $$n$$ positive integers are listed in ascending order, and the mathematical operations $$(+,-,\times,\div)$$ are repeated in that order. Find the value of $$S_{31417}$$. Give your answer to 3 decimal places. (Little More) Details: - BODMAS (Order of Operations) is obeyed. - A couple of explicit examples: (\begin{align} S_4 &= 1 + 2 - 3\times4 = (1+2) - (3\times4) = 3 - 12=-9 \\ S_5 &= 1+2-3\times4\div5 = (1+2) - (3\times4\div5) = 3 - 12\div5 = 0.6. \end{align})

Answer: $$-3.598$$ Explanation: You can identify the sequence of the operators. $(s=1+(2-3\times \frac{4}{5})+(6-7\times \frac{8}{9})+(10-11\times \frac{12}{13})+\dots \dots$) Now, get ready for the loop. As you see, that $$1$$ is kept out of the braces, you have to set the initial value of Sum, $$s=1;$$. Then you take four variables: a,b,c,d. You see, always, $(b=a+1$)$(c=b+1$)$(d=c+1$) And it occurs again and again. In the 1st loop, $$a=2$$. You can declare it inside the $$for$$ loop. To get the numbers of the 2nd loop, $$a+=4$$. Because then you get $$a=6$$ in the 2nd, and $$a=10$$ in the 3rd loop and blah blah blah! So, the for loop is $$for(a=2; ; a+=4)$$ . Look carefully, we didn't specify the condition for the loop, because using the $$break;$$ statement, we can do that job. Then, inside the loop, first produce the other numbers, $$b,c,d$$. Now comes the most important. If $$a=n$$ , $$s+=a;$$ then $$Break$$ the loop. But if $$b=n$$, then $$s+=a-b;$$ , and $$Break$$ ! Similar for $$c=n$$ and $$d=n$$ . Notice the corresponding lines of these statements in the sample code below. If none of them is equal to $$n$$, that means that they haven't reached the value of $$n$$ yet, which will happen from the beginning. For that, the statement is, $$s+=a-((b*c)/d);$$ . Remember, no $$break;$$ this time, as it needs to continue until any of the variables reach the $$n$$ . So, that's it. Here's the sample code in C++ after all this detailed explanation... Compile and run it, enter $$31417$$ and get the answer! #include<stdio.h> #include<math.h> int main() { double a, b, c, d, s, n; printf("What's the n: "); scanf("%lf", &n); s = 1; for (a = 2;; a += 4) { b = a + 1; c = b + 1; d = c + 1; if (a == n) { s += a; break; } else if (b == n) { s += (a - b); break; } else if (c == n) { s += a - (b * c); break; } else if (d == n) { s += a - ((b * c) / d); break; } else { s += a - (b * c / d); } } printf("The Summation, s= %lf", s); printf("\n"); return 0; }

Physics
TutorMe
Question:

An ice cube of side length $$a_0$$ is placed in a cylindrical container with a base area of $$A$$. After some time, ice starts to melt uniformly. Find the expression for the length of the ice cube's edge $$a$$, right at the moment when it leaves the contact with the base of the container. Given, Density of water = $$\rho_w$$ Density of ice = $$\rho_i$$

Answer: $(\text{Edge Length, a}={\frac{ {a_0}^3}{A}}$) Its one of the many possible solutions. Explanation: Assume, Edge length (before)=$$a_0$$ Exactly when the cube starts floating, Edge length=$$a$$ Height of the submerged portion=$$h$$ Density of ice=$$\rho_i$$ Density of water=$$\rho_w$$ So, Mass of previous cube = Mass of current cube + Mass of water Water volume $( =\text{A}h-a^2h$) So, $(\rho_i{a_0}^3=\rho_ia^3+\rho_w h (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}} h (\text{A}-a^2)$) Then, Mass of the current cube = Mass of $$a^2 h$$ Volume of water. $(\rho_i a^3=\rho_w a^2 h$) $(\rightarrow h={\frac{\rho_i}{\rho_w}} a$) Plug the value to previous equation: $({a_0}^3-a^3={\frac{\rho_w}{\rho_i}}h (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}}{\frac{\rho_i}{\rho_w}} a (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3=a(\text{A}-a^2)$) See, the density is gone from the expression! So let's do the final job: $({a_0}^3-a^3=\text{A}a-a^3$) $(\rightarrow a=\boxed{{\frac{ {a_0}^3}{A}}}$) The problem seems so complex at first glance. But its simple, and excellently interesting, isn't it?

Send a message explaining your


text:
## Elementary Algebra

The conjugate of an expression $(a+b)$ must be $(a-b)$. Therefore, according to this rule, the conjugate of $5+\sqrt 3$ must be $5-\sqrt 3$. Therefore, the question statement is false.


text:
# Floor of ratio of Fermat numbers

1. Apr 1, 2015

### fourier jr

my turn to have something answered

1. The problem statement, all variables and given/known data

The problem is to show that Fermat numbers are relatively prime, which I could just look up somewhere but I want to use the mod operator that is used in Concrete Math, which says $n \bmod m = n - m\lfloor{\frac{n}{m}}\rfloor$ So if m<n the problem is to find

$$(2^{2^n} + 1)\bmod(2^{2^m} + 1) = (2^{2^n} + 1) - (2^{2^m} + 1)\Big\lfloor{\frac{2^{2^n} + 1}{2^{2^m} + 1}}\Big\rfloor$$

which the book says is 2. It's the part about simplifying the floor part that I'm hung up on.

2. Relevant equations
$n \bmod m = n - m\lfloor{\frac{n}{m}}\rfloor$

fractional part of x: $\{x\} = x - \lfloor x \rfloor$

$\lfloor x+y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$ or $\lfloor x \rfloor + \lfloor y \rfloor + 1$ depending on the size of $\lfloor \{x\} + \{y\} \rfloor$

$\lfloor x + n \rfloor = \lfloor x \rfloor + n$ (n an integer)

3. The attempt at a solution
I have a feeling I'm just missing something obvious about exponents or floors. Maybe it's the exponent in the exponent that's throwing me off. Anyway a couple things I've tried:

$$\Big\lfloor \frac{2^{2^n}}{2^{2^m}+1} + \frac{1}{2^{2^m}+1} \Big\rfloor = \Big\lfloor \frac{2^{2^n}}{2^{2^m}+1} \Big\rfloor + \Big\lfloor \frac{1}{2^{2^m}+1} \Big\rfloor + \Big\lfloor \Big\{ \frac{2^{2^n}}{2^{2^m}+1} \Big\} + \Big\{ \frac{1}{2^{2^m}+1}\Big\}\Big\rfloor$$

$$\Big\lfloor \frac{2^{2^n}+1}{2^{2^m}+1}\Big\rfloor = \Big\lfloor \frac{2^{2^n - 2^m}+ 2^{-2^m}}{1 + 2^{-2^m}} \Big\rfloor$$ (thinking that the denominator on the right-hand side would be close enough to 1 that it would simplify easily)

2. Apr 1, 2015

### Zerodivisor

I do not get this, Fy = 2 * prod(k=0 to y, Fk), where Fk = 2^(2^k))-1 ,
so Fy is relatively prime to all Fermat numbers less than it. qed

3. Apr 2, 2015

### fourier jr

What about the question I actually asked, about the floor, etc? I know I could look up another proof of that just about anywhere but it seems not this one. I doubt it's even very hard since it's in the 'basics' section of the problems in this chapter.

4. Apr 2, 2015

### Dick

Just because it's in the 'basics' section doesn't mean that every method of approaching the problem is equally tractable. I have no idea how you would approach simplifying that floor function. I don't think this problem even belongs in the 'basics' section unless the text has already given you some hints about properties of the Fermat numbers. They must have said some stuff beyond the basic definition.

5. Apr 3, 2015

### Zerodivisor

OK, my error.
The correct equation is (Concrete Mathematics, page 501):
fn = 2 + f0*f1*...*fm*...*f(n-1) = 2 + k * fm
so that fn % fm = 2.

This is in fact the simple way to prove the fractional expression is 2.

6. Apr 3, 2015

### fourier jr

...yeah a lot simpler. I even saw that identity in the answers but didn't make the connection. I thought I was just missing something simple about floors or exponents but I wasn't even right about that but I see it now...

7. Apr 4, 2015

### Zerodivisor

You probably figured out to show f0*f1*...*fn = 2^2^n - 1 by expressing the terms in binary.
So f0 = 3 = 11 binary, f1 = 101 binary, f0*f1 = 1100 + 11 = 1111 binary = 2^2^2 - 1,
f0*f1*f2 = 1111 * 10001 = 11110000 + 1111 = 11111111 bin = 2^2^3 - 1,
It's pretty how there are just enough zeros to avoid carries.

8. Apr 4, 2015

### fourier jr

I didn't try the arithmetic but I remember reading in the text that the reason a lot of a biggest prime numbers known are Fermat primes is that they have the form (1111....11)2 which makes it easier for a computer to work with, or that it avoids unnecessary calculations converting it to base 10.

9. Apr 4, 2015

### fourier jr

I meant Mersenne primes. I was wrong about that too.


text:
# Time dilation, and curvature of space caused by two black holes of unequal masses

If you were to try to find the time dilation in a region of space near a black hole you would use the equation

$$t_r=\sqrt{1-\frac{2GM}{rc^2}}$$

Would the time dilation from two black holes be this?

$$t_r=\sqrt{1-\frac{2G}{c^2}\left(\frac{M_1}{r_1}+\frac{M_2}{r_2}\right)}=\sqrt{1+\frac{2P_C}{c^2}}$$

Where $$P_C$$ is the classical gravitational potential.

• That equation makes no sense. The left side is a time and the right side is dimensionless. – G. Smith Dec 25 '18 at 23:44
• The left side is the ratio of the rate at which time passes at a point to the rate time passes at infinity. Both sides of the equation are unitless. The top formula is the formula for gravitation time dilation around a black hole. – Laff70 Dec 26 '18 at 0:40
• OK, thank you for explaining what you meant. I think using the symbol $t_r$ to mean the ratio of two time intervals is confusing, and I have never seen this usage anywhere. – G. Smith Dec 26 '18 at 0:52
• I think your formula is correct when $M_1/r_1<<1$ and $M_2/r_2<<1$. When this is not the case, I don’t think the formula is known, because there is no analytic solution for two black holes. The nonlinearity of the Einstein equations means that you can’t superpose them. – G. Smith Dec 26 '18 at 1:00
• Well I just tested out the second equation for a distribution of mass that is a hollow spherical shell. It gave a constant gravitational time dilation inside the shell and the correct gravitation time dilation outside the shell. So it seems that my equation is correct at least for spherically symmetric distributions. – Laff70 Dec 26 '18 at 1:25

You need to distinguish two different regions. As long as you are not too close to either black hole their gravitational potentials will just add. This is the linear region. Close to either black hole this linearity breaks down and we enter the non-linear region.

In the linear region the time dilation (relative to an observer at infinity) is well approximated by the weak field expression:

$$t_r = \sqrt{1 +\frac{2\Phi}{c^2}} \tag{1}$$

where $$\Phi$$ is the Newtonian gravitational potential. In fact for a single black hole the Newtonian potential is:

$$\Phi = -\frac{GM}{r}$$

and substituting this in equation (1) gives us the relativistic expression that you cite:

$$t_r = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1}$$

However be cautious about assigning too much significance to this as it's just a coincidence. The meaning of the variable $$r$$ is different in the Newtonian and relativistic theories.

Anyhow, provided we are in the linear region to calculate the time dilation for two (or more) black holes we need only calculate the total gravitational potential, and this is just the sum of the two separate potentials:

$$\Phi_t = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2}$$

and substituting this in equation (1) is going to give the expression that you suggest in your question:

$$t_r = \sqrt{1 +\frac{2\Phi_t}{c^2}} = \sqrt{1 -\frac{2G}{c^2}\left( \frac{M_1}{r_1} + \frac{M_2}{r_2} \right)} \tag{2}$$

This works for the linear region, but there is still the non-linear region to consider i.e. close to one or both of the black holes. The trouble is that there is no analytic solution for the metric that describes two orbiting black holes. All we can say is that equation (2) will become increasingly inaccurate the closer we approach. We would need to do a numeric calculation to get an accurate result.

• Actually if we are in the linear (weak field) region square roots are useless. You may simply write $t_r = 1+\Phi_t/c^2$. – Elio Fabri Dec 26 '18 at 10:33
• @Laff70 In hypothetical scenarios anything can happen. You can't imagine one contradicting basic physical laws and then demand to draw physical conclusions in it. – Elio Fabri Dec 26 '18 at 10:33
• You have made a few sign errors in your answer. – A.V.S. Dec 26 '18 at 14:43
• @Laff70 you could ask a new question about the numerical calculation, but it's formidably hard. Basically if you need to ask you can't do it. But then the linear approximation works surprisingly well as long as you're more than a handful of Schwarzschild radii away from both horizons. If your mass distribution isn't dense enough to contain a horizon then the linear approximation is effectively perfect. – John Rennie Dec 26 '18 at 16:32
• – John Rennie Dec 26 '18 at 16:37

No, it would not. Time dilation is a relative effect from one frame of reference and location to another. First define those two points, then the total effect will be that due to the cumulative difference in gravity at those points (assuming they are stationary).

• When one is talking about gravitational time dilation, they usually have the reference frame be at infinity. This can be assumed for this question. Also the system is static. It would be helpful if you were to give the correct equation for the time dilation. – Laff70 Dec 25 '18 at 23:26
• @Laff70 If 2 black holes are close enough to each other for there to be a region between them with non-negligible combined gravitational time dilation, then the system won't be static. And binary black hole systems are notoriously nonlinear. – PM 2Ring Dec 25 '18 at 23:44
• This is a hypothetical scenario for crying out loud!!! Imagine these are black holes(which don't evaporate in this scenario) with the masses of pebbles spaced meters apart from each other. Let's also say all contributions are non-negligible. Also, for good measure, lets say we have some super scifi tech that lets you anchor black holes in place. – Laff70 Dec 26 '18 at 0:37


text:
# In Faster R-CNN, how can I get the predicted bounding box given the neural network's output?

The RPN loss in Faster RCNN paper is

$$L({p_i}, {t_i}) = \frac{1}{N_{cls}} \sum_{i} L_{cls}(p_i,p_i^*) + \lambda \frac{1}{N_{reg}} \sum_i p_i^* L_{reg}(t_i, t_i^*)$$

For regression problems, we have the following parametrization

$$t_x=\frac{x - x_a}{w_a}, \\ t_y=\frac{y−y_a}{h_a}, \\ t_w= \log \left( \frac{w}{w_a} \right),\\ t_h= \log \left(\frac{h}{h_a} \right)$$

and the ground-truth labels are

$$t_x^*=\frac{x^* - x_a}{w_a},\\ t_y^*=\frac{y^*−y_a}{h_a}, \\ t_w^*= \log \left( \frac{w^*}{w_a} \right), \\ t_h^*= \log \left(\frac{h^*}{h_a} \right)$$

where

• $$x$$ and $$y$$ are the two coordinates of the center, $$w$$ the width, and $$h$$ the height of the predicted box.

• $$x$$ and $$y$$ are the two coordinates of the center, $$w$$ the width, and $$h$$ the height of the anchor box.

• $$L_{reg}(t_i, t_i^*) = R(t_i − t_i^*)$$, where $$R$$ is a robust loss function (smooth $$L_1$$)

These equations are unclear to me, so here are my two questions.

1. How can I get the predicted bounding box given the neural network's output?

2. What exactly is the smooth $$L_1$$ here? How is it defined?

## 1 Answer

If I understood well you have 2 questions.

• How to get the bounding box given the network output
• What Smooth L1 loss is

The answer to your first question lies in the equation (2) in the section 3.2.1 from the Faster R-CNN paper. As all anchor based object detector (Faster RCNN, YOLOv3, EfficientNets, FPN...) the regression output from the network are not the bounding box coordinates. The regression output predicts the shift of the predicted bounding box with respect to the selected anchor (all of these networks, use more than 1 anchor per location, check section 3.1.1 from the paper).

So basically what your network predict is $$t_x, t_y, t_w, t_h$$:

And the bounding box coordinates are $$x, y, w, h$$, and the anchor coordinates are $$x_a, y_a, w_a, h_a$$. So in order to compute $$x, y, w, h$$ from $$t_x, t_y, t_w, t_h$$, you just have to invert the equations above. However I think you could gain more intuition about it if you take your time and read the whole section 3.1 from the paper. I know sometimes is a pain, but you will grasp the high level concept.

With regard to your second question. Yes the loss is computed with the output from the network and the "coded" ground truth, meaning you compute the loss with the paramters $$t$$ (predicted) against $$t^*$$ (coded ground truth) instead for computing loss with the real coordinates of the bounding boxes (decoded output from the network). For the equation on Smooth L1 loss check this wonderful documentation.


text:
# Polarisation directions¶

The polarisation directions of the electric and magnetic fields are given by:

\begin{eqnarray} \alpha_E & = & \arctan{\frac{2 Re({E_x E_y^*})}{[E_x E_x^*] - [E_y E_y^*]}} \\ \alpha_B & = & \arctan{\frac{2 Re({B_x B_y^*})}{[B_x B_x^*] - [B_y B_y^*]}} \end{eqnarray}

The equations describe how two orthogonal field components with constant phasing vary.

Important

The resistics name for the polarisation direction statistic is: polarisationDirection.

The components of polarisation direction statistic are:

• pol Ex Ey describing the time-harmonic variation between the two orthogonal components of the electric field, corresponding to $$\alpha_E$$.

• pol Hx Hy describing the time-harmonic variation between the two orthogonal components of the magnetic field, corresponding to $$\alpha_B$$.

For magnetotelluric data, which has numerous sources of electromagnetic energy (solar activity, atmospheric currents etc.), it is not expected that there is a coherent polarisation direction i.e. the polarisation direction should vary.

When the polarisation direction holds a consistent value, this implies that there is an unexpected source of electromagnetic energy. This is particularly true for the magnetic field, which is the input to our system.

Below is an example of polarisation direction over time. The polarisation directions here appear to be quite consistent, suggesting an unexpected source of electromagnetic energy.

Polarisation directions plotted over time for evaluation frequency 1024 Hz

A histogram of the polarisation direction confirms that they are consistent over time.

Polarisation direction plotted as a histogram for evaluation frequency 1024 Hz

By contrast, the below two examples demonstrate large variations in polarisation directions, in concordance with natural sources of magnetotelluric energy.

Polarisation directions plotted over time for evaluation frequency 32 Hz

Polarisation direction plotted as a histogram for evaluation frequency 32 Hz


text:
[1] In practice, we provide these functions in a library and make sure that the library is loaded whenever we need them. One way of defining these functions is to use arrays, one of JavaScript's primitive data structures.
[2] Another way to define the selectors and constructor is
const make_rat = pair;


text:
# International Workshop on: Simulating gravitation and cosmology in condensed matter and optical systems

Jul 22 – 25, 2019
ECT* - Villa Tambosi
Europe/Rome timezone
Abstracts submission open. Posters do not require abstract submission.

## Towards exotic gravitational analogue in quantum fluids of light

Jul 23, 2019, 12:30 PM
15m
Meeting room (ECT* - Villa Tambosi)

### Meeting room

#### ECT* - Villa Tambosi

Strada delle Tabarelle, 286 38123 - Villazzano (TN) Italy

### Speaker

Dr Maxime Richard (CNRS - Institut Néel)

### Description

Exciton-polaritons in semiconductor microcavities can be understood essentially as interacting photons that are confined into a bidimensional degree of freedom. Collectively, polaritons behave as a quantum fluid, with its typical features such as a superfluid state with sonic excitations. As such, and considering their photonic character, polaritons offer an exciting platform to simulate the physics of gravitational structures, such as black holes.

In this context, the main peculiarity of polariton fluids, i.e. their its driven-dissipative character, offers original physical situations and resources that are not readily available in equilibrium system. I will illustrate this idea with a few recent experimental investigations. I will show for instance that a polaritonic black hole, featuring a maximal surface gravity, can be engineered by exploiting their hysteretic dynamics. I will also present preliminary results towards the realization of a quantum fluid of polaritons in the analogue of the relativistic regime, i.e. in which the free particle effective mass $m_{\rm eff}c^2$ in the condensate is comparable or smaller than the interaction energy $gn$.

### Primary author

Dr Maxime Richard (CNRS - Institut Néel)


text:
bug-grub
[Top][All Lists]

## Re: Cannot pass a single backslash in multiboot cmdline

 From: Jakub Jermář Subject: Re: Cannot pass a single backslash in multiboot cmdline Date: Mon, 26 Dec 2016 19:56:44 +0100 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:45.0) Gecko/20100101 Thunderbird/45.4.0

On 12/26/2016 07:24 PM, Andrei Borzenkov wrote:
> 26.12.2016 21:12, Jakub Jermář пишет:
>>>> I am observing a strange behavior when passing boot arguments with a
>>>> backslash to the kernel (the multiboot cmd_line via the multiboot
>>>> command in grub.cfg). I would like to pass foo\bar to the kernel, but to
>>>> no avail. I tried:
>>>>
>>>
>>> Which kernel? What do you load?
>>
>> The kernel is a modified version of HelenOS, but IMHO this issue is
>> kernel agnostic.
>
> Not really. It depends on target program being loaded which is why I

Ok, I submitted my grub.cfg and stated that HelenOS is
multiboot-compliant, gets loaded by the multiboot command. We boot from
an ISO image, which was created like this:

"The binary files of GRUB boot loader in this directory have been
created by compiling GRUB for the 'i386-pc' target and then using the
grub-mkrescue script to create the El Torito boot image."

Is there anything else I should include in order to answer your question?

>> The cmd_line is already wrong when picked up from the
>> multiboot info and printed out (so there is no processing on it from the
>> HelenOS side).
>
> Sure, but /if/ your kernel interpreted \' as quoting character, the
> result would be correct.

But instead of the plain actual data there would be escaped data, which
is a hassle. And also a possible workaround.

>> I think I understand why it is eating the single backslash (expected
>> behavior),
>
> You have two levels here. First level is grub command line processing.
> It is loosely compatible with (Bourne) shell so yes, backslash is
> treated as escape character. You can still pass backslash using usual
> shell quoting, e.g. foo\\bar or 'foo\bar'. Second level is
> grub_create_loader_cmdline() function. This one escapes \' with second
> \'. So in the above case this function receives foo\bar' as adds
> second \'.

This explains a lot, but does not make much sense for my usecase.

Thank you,
Jakub

`


text:
# Algorithm: Permutations & Signature

Theoretical side

Simple example: If I have two sets $$A_1=\{1,3\} ,A_2=\{2,3\}$$

and Permutations[{1, 2, 3}]= $$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \\ 3 & 2 & 1 \\ \end{array} \right)$$

Solution steps

1- Choose the rows whose sets $$A_1=\{1,3\} ,A_2=\{2,3\}$$ are partial and then count the number of rows $$D_2=\{1,3,2\},\{3,1,2\},\{2,3,1\},\{3,2,1\}$$

$$S_2=Count[D_2]/Count[Permutations[\{1, 2, 3\}]]=4/6$$

2- Delete $$D_2$$ from $$Permutations[\{1, 2, 3\}]$$ and choose from the triple order whose sets $$A_1=\{1,3\} ,A_2=\{2,3\}$$ are partial and then count the number of rows

$$D_3=\{1,2,3\},\{2,1,3\}$$

$$S_3=Count[D_3]/Count[Permutations[\{1, 2, 3\}]]=2/6$$

How can you write code that achieves this algorithm and maintains globality?

###########

#Edit

sets are partial:The group is contained within the second order

$$D_2$$:Such that

$$A_1=\{1,3\}\subset\{1,3,-\}$$ take $$\{1,3,-\}$$ $$A_1=\{1,3\}\subset\{3,1,-\}$$ take $$\{3,1,-\}$$ $$A_2=\{2,3\}\subset\{2,3,-\}$$ take $$\{1,3,-\}$$ $$A_2=\{2,3\}\subset\{3,2,-\}$$ take $$\{3,1,-\}$$

Then; We obtain $$D_2=\{1,3,2\},\{3,1,2\},\{2,3,1\},\{3,2,1\}$$

$$D_3$$:Such that

$$A_1=\{1,3\}\subset\{1,2,3\}$$ take $$\{1,2,3\}$$ $$A_1=\{2,3\}\subset\{2,1,3\}$$ take $$\{2,1,3\}$$

Then; We obtain $$D_3=\{1,2,3\},\{2,1,3\}$$

• Please clarify what you mean by "sets are partial". May 10 at 0:34
• @CarlWoll sets are partial: The set is contained within the second order $D_2$; $A_1=\{1,3\}\subset\{1,3,-\}$ or $A_1=\{1,3\}\subset\{3,1,-\}$ And, The set is contained within the thirds order $D_3$; $A_1=\{1,3\}\subset\{1,2,3\}$ May 10 at 0:46


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# Convergence of a Trigonometric Series

After working with a Fourier series for a while, I realized that it would be of great help to me if I could prove that the following limit is zero: $$\lim_{N\to\infty}\frac{1}{N}\sum_{m=1}^\infty\frac{(-1)^m\cos(m(N+1))\sin(mN)}{(4m^2-1)\sin(m)}$$

I tried to do several things and I can show that with a trivial upper bound that if $\sum_{m=1}^\infty\frac{1}{m^2|\sin(m)|}$ converges then the limit will clearly be zero, but this only seems to complicate things, as I think I would have to introduce notions of Diophantine approximation to show that the denominator doesn't get too crazy.

• The numerator is missing a ")". If it's supposed to be $\cos(m(N+1)) \sin(mN)$ then it converges because $|\sin(mN) / \sin(m)| \leq N \ll_N 1$ and the other factors are $O(1/m^2)$. Jan 27, 2014 at 0:07
• Cross post: math.stackexchange.com/questions/652609/… Jan 27, 2014 at 0:09
• @rghthndsd Those two questions don't look the same to me. Jan 27, 2014 at 0:17
• @NoamD.Elkies Let $N$ get arbitrarily large and it is no longer bounded. Jan 27, 2014 at 0:19
• Is your question whether the sum converges (in which case Noam's answer is right), or whether it converges to a value bounded independently of $N$? Jan 27, 2014 at 0:26

As pointed out earlier, $$a_{N,m}:=\frac{(-1)^m\cos(m(N+1))\sin(mN)}{N(4m^2-1)\sin(m)}$$ satisfies $$\left|a_{N,m}\right|\leq \frac{1}{(4m^2-1)}, \mbox{ and }\sum_{n=1}^\infty \frac{1}{(4m^2-1)} < \infty.$$ On the other hand, for every $m$, there holds $$\lim_{N\to\infty} a_{N,m} = 0.$$ Thus the Dominated Convergence Theorem does it for you, $$\lim_{N\to\infty} \sum_{m} a_{N,m} =0.$$ (If you have never seen this Theorem, it says, for series: Given ${a_{N,m}}$,, if
• $|a_{N,m}| < b_m$ for all $N$ and $\sum_{m} b_m$ converges
• for $m$ fixed, $a_{N,m}\to c_m$
• $\sum_{m} |a_{N,m} -c_m| \to 0$
• $\sum_m c_m$ converges
and consequently $\lim_{N\to \infty} \sum_m a_{N,m} = \sum_m c_m$).


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# How to show this equation have exactly single solution?

Consider $a=1,\:b\in \mathbb{R}$.
Show that there is single solution for the equation:$$x-a\sin x\:=\:b$$
So far I defined funtion $f\left(x\right)=x-a\sin \left(x\right)-b\:=\:x-\sin \left(x\right)-b$ and learned that $\lim _{x\to \infty }\left(f\left(x\right)\right)\:=\:\infty \:,\:\lim _{x\to -\infty }\left(f\left(x\right)\right)=-\infty \:$.
Now, the derivative of this function not always increasing because in $x_0=0$, $f'\left(x\right)\:=\:0$, so I can't use the Intermediate value theorem..
Let $f(x)=x-\sin x-b$. Then $$f(1+b)=1-\sin(1+b)>0$$ and $$f(-1+b)=-1-\sin(-1+b)<0$$ but then since $f$ is continuous, it follows that $f$ has a root in $(-1+b,1+b)$. For the uniqueness of the root, suppose that $f$ has two roots $x_1$ and $x_2$. Then $$x_1 -x_2=\sin x_1 -\sin x_2.$$ Take absolute value of both sides and use $$\sin p -\sin q= 2\sin \frac{p-q}{2}\cos\frac{p+q}{2}$$ to get a contradiction!
Edit: since for any nonzero value of $x$ (in radian) one has $|\sin x|\lt |x|$ so
$|x_1 -x_2|=| \sin x_1 -\sin x_2|=2|\sin \frac{x_1-x_2}{2}||\cos\frac{x_1+x_2}{2}|<2\times |\frac{x_1-x_2}{2}|\times 1=|x_1 -x_2|$
Hints: use that $f_0(x)=x-\sin x$ is strictly increasing in $(-\infty,0)$ and $(0,+\infty)$. Also: $f_0<0$ in $(-\infty,0)$ and $f_0>0$ in $(0,+\infty)$


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# How to calculate cycles per byte [closed]

Sometimes, websites and scientific papers that introduce and/or handle cryptographic algorithms also provide speed-analysis and tables comparing the individual performance of cryptographic implementations. I would like to do the same, but I'm not sure how to calculate the “cycles”.

I have this data:

• processor clock frequency: 2,1 ghz
• message length: 16 byte
• Speed: 4,3 Mbytes/s

How can I calculate “cycles” and “cycles per byte” from that data?

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## closed as off topic by Hendrik Brummermann♦May 19 '13 at 8:32

Questions on Cryptography Stack Exchange are expected to relate to cryptography within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

Welcome to Crypto.SE! Please note that this question is off-topic, and in general the community expects someone to do some research before asking a question. –  rath May 18 '13 at 21:21
I'm voting to leave closed since whilst this is now a clear and well-worded question I think it's still off-topic for this site: its about measuring computational speed, albeit the speed of a cryptographic algorithm. –  figlesquidge Mar 25 '14 at 10:59

## 1 Answer

$\displaystyle \text{cycles per byte} = \frac{\text{cycles per second}}{\text{bytes per second}} = \frac{2.1 ~ \text{GHz}}{4.3 ~ \text{MiB}} = \frac{2.1 \times 10^9}{4.3 \times 1024^2} \approx 466 ~ \text{cpb}$

Of course this may be way off because processors are complex beasts these days, and may not work at their full potential all the time, and the calculations may very well be reordered, optimized, or vectorized.

So it's a little better than raw speed in that it is not dependent of processor clock speed but it is still quite dependent on the underlying hardware details. It is a decent measure of cryptographic performance.

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thanks for the reply, I do not understand why data i see are very different ... for example this report fastcrypto.org/umac/2004/perf04.html cycles per byte are much less, why? –  Luca Del Giudice May 18 '13 at 17:56
@LucaDelGiudice Maybe the algorithm you are using has not been optimized during compilation, which would explain the slow speeds you see. But without additional details it's hard to tell. –  Thomas May 18 '13 at 18:40
@Thomas captain to the resque! –  Smit Johnth May 18 '13 at 18:49
@LucaDelGiudice Just noticed the "hash" tag. If you are measuring the performance of a hash function, you want to use messages longer than 16 bytes, otherwise most of the time will be spend doing initialization/finalization work (which do take up a lot of clock cycles for most hash functions). Try a 32MB buffer, you should see better results. –  Thomas May 19 '13 at 7:34


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# Double orthogonal complement of a finite module

Crossposted to mathoverflow

Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise. Let $V$ be a submodule of $W$. Let $V^{\perp} = \{w \in W \, : \, \forall v \in V \quad v.w = 0 \}$ where "." is the dot product. Is it true that ${(V^{\perp})}^{\perp} = V$?

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I would use the notation $V^{\perp}$. – Qiaochu Yuan Sep 12 '11 at 22:26
I don't think you waited long enough before crossposting. I would give it at least a day. – Qiaochu Yuan Sep 13 '11 at 2:04

The answer is yes. The easiest way for me is to appeal to character theory. If $\zeta$ is a complex $q$-th root of 1 then the map from $\mathbb{Z}_q^n$ to $\mathbb{C}$ given by $$x \mapsto \zeta^{a^Tx},\qquad (a\in\mathbb{Z}_q^n)$$ is a character of the abelian group $W=\mathbb{Z}_q^n$. The set of characters obtained as $a$ varies over the elements of $W$ is the character group $W^*$ of $W$. If $V$ is a subgroup of $W$, then $V^\perp=V^*$ is isomorphic to the subgroup $(W/V)^*$ of $W^*$.
A convenient source for the relevant character theory is from our own KConrad: http://www.math.uconn.edu/~kconrad/blurbs/ (under characters of finite abelian groups).


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Randomized Methods for Linear Constraints: Convergence Rates and Conditioning

Published Online:https://doi.org/10.1287/moor.1100.0456

We study randomized variants of two classical algorithms: coordinate descent for systems of linear equations and iterated projections for systems of linear inequalities. Expanding on a recent randomized iterated projection algorithm of Strohmer and Vershynin (Strohmer, T., R. Vershynin. 2009. A randomized Kaczmarz algorithm with exponential convergence. J. Fourier Anal. Appl.15 262–278) for systems of linear equations, we show that, under appropriate probability distributions, the linear rates of convergence (in expectation) can be bounded in terms of natural linear-algebraic condition numbers for the problems. We relate these condition measures to distances to ill-posedness and discuss generalizations to convex systems under metric regularity assumptions.