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Research

# Approximate controllability of fractional integro-differential equations involving nonlocal initial conditions

NI Mahmudov* and S Zorlu

Author Affiliations

Eastern Mediterranean University, via Mersin 10, Gazimagusa, T.R. North Cyprus, Turkey

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Boundary Value Problems 2013, 2013:118  doi:10.1186/1687-2770-2013-118

 Received: 12 December 2012 Accepted: 22 March 2013 Published: 8 May 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

We discuss the approximate controllability of nonlinear fractional integro-differential system under the assumptions that the corresponding linear system is approximately controllable. Using the fixed-point technique, fractional calculus and methods of controllability theory, a new set of sufficient conditions for approximate controllability of fractional integro-differential equations are formulated and proved. The results in this paper are generalization and continuation of the recent results on this issue. An example is provided to show the application of our result.

### 1 Introduction

Controllability is one of the fundamental concepts in mathematical control theory, which plays an important role in control systems. The controllability of nonlinear systems represented by evolution equations or inclusions in abstract spaces and qualitative theory of fractional differential equations has been extensively studied by several authors. An extensive list of these publications can be found in [1-44] and the references therein. Recently, the approximate controllability for various kinds of (fractional) differential equations has generated considerable interest. A pioneering work on the approximate controllability of deterministic and stochastic systems has been reported by Bashirov and Mahmudov [5], Dauer and Mahmudov [8] and Mahmudov [10]. Sakthivel et al.[28] studied the approximate controllability of nonlinear deterministic and stochastic evolution systems with unbounded delay in abstract spaces. On the other hand, the fractional differential equation has gained more attention due to its demonstrated applications in numerous seemingly diverse and widespread fields of science and engineering. Yan [45] derived a set of sufficient conditions for the controllability of fractional-order partial neutral functional integro-differential inclusions with infinite delay in Banach spaces. Debbouche and Baleanu [1] established the controllability result for a class of fractional evolution nonlocal impulsive quasi-linear delay integro-differential systems in a Banach space using the theory of fractional calculus and fixed point technique. However, there exists only a limited number of papers on the approximate controllability of the fractional nonlinear evolution systems. Sakthivel et al.[28] studied the approximate controllability of deterministic semilinear fractional differential equations in Hilbert spaces. Wang [40] investigated the nonlocal controllability of fractional evolution systems. Surendra Kumar and Sukavanam [33] obtained a new set of sufficient conditions for the approximate controllability of a class of semilinear delay control systems of fractional order using the contraction principle and the Schauder fixed-point theorem. More recently, Sakthivel et al.[27] derived a new set of sufficient conditions for approximate controllability of fractional stochastic differential equations.

In this paper, we discuss the approximate controllability of nonlinear fractional integro-differential system under the assumption that the corresponding linear system is approximately controllable. We consider the following fractional integro-differential control system involving nonlocal conditions,

(1)

in , where , , stands for the Caputo fractional derivative of order β, and , , , are given functions to be specified later. Here, is the infinitesimal generator of a compact analytic semigroup of bounded linear operators , , on a real Hilbert space X. B is a linear bounded operator from a real Hilbert space U to X.

The rest of this paper is organized as follows. In Section 2, we give some preliminary results on the fractional powers of the generator of an analytic compact semigroup and introduce the mild solution of system (1). In Section 3, we study the existence of mild solutions for system (1) under the feedback control defined in (5). We show that the control system (1) is approximately controllable on provided that the corresponding linear system is approximately controllable. Finally, an example is given to demonstrate the applicability of our result.

### 2 Preliminaries

In this section, we introduce some facts about the fractional powers of the generator of a compact analytic semigroup, the Caputo fractional derivative that are used throughout this paper.

We assume that X is a Hilbert space with norm . Let be the Banach space of continuous functions from into X with the norm , here . In this paper, we also assume that is the infinitesimal generator of a compact analytic semigroup , , of uniformly bounded linear operator in X, that is, there exists such that for all . Without loss of generality, let , where is the resolvent set of A. Then for any , we can define by

It follows that each is an injective continuous endomorphism of X. Hence we can define , which is a closed bijective linear operator in X. It can be shown that each has dense domain and that for . Moreover, for every and with , where , I is the identity in X. (For proofs of these facts, we refer to the literature [15,20,22].)

We denote by the Hilbert space of equipped with norm for , which is equivalent to the graph norm of . Then we have , for (with ) and the embedding is continuous. Moreover, has the following basic properties.

Lemma 1[42]

andhave the following properties.

(i) for eachand.

(ii) for eachand.

(iii) For every, is bounded inXand there existssuch that

(iv) is a bounded linear operator for.

Let us recall the following known definitions of fractional calculus. For more details, see [43,44].

Definition 2 The fractional integral of order with the lower limit 0 for a function f is defined as

provided the right-hand side is pointwise defined on , where Γ is the gamma function.

Definition 3 The Caputo derivative of order with the lower limit 0 for a function f can be written as

The Caputo derivative of a constant is equal to zero. If f is an abstract function with values in X then the integrals which appear in Definitions 2 and 3 are taken in Bochner’s sense.

According to Definitions 2 and 3, it is suitable to rewrite the problem (1) in the equivalent integral equation

(2)

provided that the integral in (2) exists. Applying the Laplace transform

to (2) and using the method similar to that used in [38] we get

where

Here, is a probability density function defined on , that is , and .

For , we define two families and of operators by

respectively.

The following lemma follows from the results given in [37-39].

Lemma 4The operatorsandhave the following properties.

(i) For any fixed, and any, we have the operatorsandare linear and bounded operators, i.e. for any,

(ii) The operatorsandare strongly continuous for all.

(iii) andare norm continuous inXfor.

(iv) andare compact operators inXfor.

(v) For every, the restriction oftoand the restriction oftoare norm continuous.

(vi) For every, the restriction oftoand the restriction oftoare compact operators in.

(vii) For alland,

In this paper, we adopt the following definition of mild solution of equation (1).

Definition 5 A function is said to be a mild solution of (1) if for any the integral equation

(3)

is satisfied.

It is clear that is bounded if . In what follows, we assume that .

### 3 Approximate controllability

In this section, we state and prove conditions for the approximate controllability of semilinear fractional control integro-differential systems. To do this, we first prove the existence of a fixed point of the operator defined below using Krasnoselskii’s fixed-point theorem. Secondly, in Theorem 11, we show that under the uniform boundedness of f and g the approximate controllability of fractional systems (1) is implied by the approximate controllability of the corresponding linear system (4).

Let be the state value of (1) at terminal time T corresponding to the control u and the initial value . Introduce the set , which is called the reachable set of system (1) at terminal time T, its closure in is denoted by .

Definition 6 The system (1) is said to be approximately controllable on if , that is, given an arbitrary it is possible to steer from the point to within a distance ε from all points in the state space at time T.

Consider the following linear fractional differential system:

(4)

The approximate controllability for linear fractional system (4) is a natural generalization of approximate controllability of linear first order control system [9,10,12]. It is convenient at this point to introduce the controllability and resolvent operators associated with (4) as

respectively, where denotes the adjoint of B and is the adjoint of . It is straightforward that the operator is a linear bounded operator.

Theorem 7[10]

LetZbe a separable reflexive Banach space and letstands for its dual space. Assume thatis symmetric. Then the following two conditions are equivalent:

1. is positive, that is, for all nonzero.

2. For allstrongly converges to zero as. Here, Jis the duality mapping ofZinto.

Lemma 8The linear fractional control system (4) is approximately controllable onif and only ifasin the strong operator topology.

Proof The lemma is a straightforward consequence of Theorem 7. Indeed, the system (4) is approximately controllable on if and only if for all nonzero , see [7]. By Theorem 7, as for all . □

Remark 9 Notice that positivity of is equivalent to . In other words, since , approximate controllability of the linear system (4) is equivalent to , .

Before proving the main results, let us first introduce our basic assumptions.

(H1) are continuous and for each , there exists a constant and functions , such that

(H2) is a Lipschitz function with Lipschitz constant .

(Hc) The linear system (4) is approximately controllable on .

Using the hypothesis (Hc), for an arbitrary function , we choose the feedback control function as follows:

(5)

Let , where r is a positive constant. Then is clearly a bounded closed and convex subset in . We will show that when using the above control the operator defined by

where

has a fixed point in .

Theorem 10Let the assumptions (H1) and (H2) be satisfied. Then for, the fractional Cauchy problem (1) withhas at least one mild solution provided that

(6)

where

Proof It is easy to see that for any the operator maps into itself.

Let and . Using assumption (H1) yield the following estimations,

and

(7)

From (6) and the assumption (H2), it follows that for any there exists such that

(8)

Therefore, from (7) and (8), it follows that for any there exists such that for every . Therefore, for any the fractional Cauchy problem (1) with the control (5) has a mild solution if and only if the operator has a fixed point in .

In what follows, we will show that and satisfy the conditions of Krasnoselskii’s fixed-point theorem. From (H2) and (6), we infer that is a contraction. Next, we show that is completely continuous on .

Step 1: We first prove that is continuous on . Let be a sequence such that as in . Therefore, it follows from the continuity of f, g and that for each ,

Also, by (H1), we see that

Since

using the Lebesgue dominated convergence theorem that for all , we conclude

implying that as . This proves that is continuous on .

Step 2. is compact on .

For the sake of brevity, we write

Let be fixed and be small enough. For , we define the map

Therefore, from Lemma 4, we see that for each , the set is relatively compact in . Since

approaches to zero as , using the total boundedness, we conclude that for each , the set is relatively compact in .

On the other hand, for and small enough, we have

where

Therefore, it follows from (H1) and Lemma 4 that

and

from which it is easy to see that all , , tend to zero independent of as and . Thus, we can conclude that

and the limit is independent of . The case is trivial. Consequently, the set is equicontinuous. Now applying the Arzela-Ascoli theorem, it results that is compact on .

Therefore, applying Krasnoselskii’s fixed-point theorem, we conclude that has a fixed point, which gives rise to a mild solution of Cauchy problem (1) with control given in (5). This completes the proof. □

Theorem 11Let the assumptions (H1), (H2) and (Hc) be satisfied. Moreover, assume the functionsandare bounded and. Then the semilinear fractional system (3) is approximately controllable on.

Proof It is clear that all assumptions of Theorem 10 are satisfied with . Let be a fixed point of in . Any fixed point of is a mild solution of (3) under the control

and satisfies the equality

(9)

where

Moreover, by the boundedness of the functions f and g and Dunford-Pettis theorem, we have that the sequences and are weakly compact in , so there are subsequences still denoted by and , that weakly converge to, say, f and g in . On the other hand, there exists such that converges to weakly in . Denote

It follows that

as because of compactness of the operator

Then from (9), we obtain

(10)

as . This proves the approximate controllability of (1). □

### 4 Applications

Example 1 As an application to Theorem 11, we study the following simple example. Consider a control system governed by the fractional partial differential equation of the form

(11)

where , , .

Let us take and define the operator A by with the domain . Then

where , ,  . Clearly −A generates a compact analytic semigroup , in X and it is given by

Clearly, the assumption (H1) is satisfied. On the other hand, it can be easily seen that the deterministic linear system corresponding to (11) is approximately controllable on ; see [12].

The operator is given by

where and .

Let , where for . Assume that satisfies the following conditions:

1. The functions , are continuous and uniformly bounded.

2. .

3. is continuously differentiable, and

Denote by , the Mittag-Leffler special function defined by

Therefore,

Define

Then, for each we have

and

It follows that is bounded and Lipschitz continuous. On the other hand, it is not difficult to verify that are continuous.

Next, we show that the linear system corresponding to (11) is approximately controllable on . It is clear that is defined as follows:

By Remark 9, the linear system corresponding to (11) is approximately controllable on if and only if , implies that . This follows from the representation of .

Now, we note that the problem (11) can be reformulated as the abstract problem. Thus, by Theorem 11, the system (11) is approximately controllable on , provided that

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to thank the reviewers for their valuable comments and helpful suggestions that improved the note’s quality.

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Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Stage: 3 and 4 Short Challenge Level:

Starting at the square containing the $2$, you are allowed to move from one square to the next either across a common edge, or diagonally through a common corner.

How many different routes are there passing through exactly two squares containing a $0$ and ending in one of the squares containing a $9$?

If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.

This problem is taken from the UKMT Mathematical Challenges.
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This is the archived website of SI 486H from the Spring 2016 semester. Feel free to browse around; you may also find more recent offerings at my teaching page.

# Unit 1: Discrete Math

After some introductory remarks, we'll go into some of the underlying math on probability and information theory that will get used over and over throughout the class. We'll just cover the basics here; expect more math as we go along and need to use it!

# 1 Introduction

In this class, we are going to learn about some of the ways random numbers get used in computing. To paraphrase the textbook, programs that make use of random numbers are often simpler or faster than their non-random counterparts (or both!). We will start by examining what random numbers are and how computer programs can obtain them. Then the bulk of the course will be spent looking at some well-known problems that can be solved more quickly or more simply using randomized algorithms. Finally, we'll look at things from a theoretical standpoint and ask whether there are some larger implications of all this.

But first, let's look at an example that illustrates the kind of way randomness can be used effectively.

## 1.1 Airplane terrorists

Consider the following scenario: An airplane carries 200 passengers. Due to strictly-enforced regulations, each passenger can carry on at most 3 ounces of liquid. But with 60 ounces of liquid, a bomb could be constructed to crash the plane. Therefore, if 20 or more passengers all manage to get 3 ounces of bomb-making liquid on the plane, there is a big problem. Fortunately, the vast majority of flights have no terrorists whatsoever: on at least 999 out of every 1000 flights, not a single passenger carries any bomb-making liquid.

To be clear, the input to any algorithm is going to be a list of 200 passengers, in some arbitrary order. All the algorithm can do is screen a given passenger for bomb-making liquid, and get a yes/no answer.

We want to develop effective ways of detecting and ultimately avoiding flights that are hijacked as described above. There is a way to test if any single passenger has bomb-making liquid, but it is somewhat disruptive and so we'd like to perform as few tests as possible.

Therefore we have two (possibly competing) goals:

1. Allow only non-hijacked planes to fly
2. Screen the fewest number of passengers for bomb-making liquid

The first goal can be classified as correctness, and the second as efficiency. In designing typical algorithms, we view the correctness requirement as absolute and work hard to optimize the efficiency. A theme we will repeatedly see with randomized algorithms is that they allow trade-offs between correctness and efficiency. Often, the loss of a correctness guarantee (a small chance that the algorithm gives the wrong answer) is more than offset by the gain in efficiency.

# 2 Discrete Probability

In order to understand randomness, we need to understand probability. We'll be covering some math concepts as needed throughout the class, but just to get started we need to cover some basics. You are strongly encouraged to read the entire chapter "An introduction to discrete probability", linked in the readings above.

## 2.1 Experiments and Outcomes

An experiment is anything for which the outcome may not be certain. This could be something like rolling a die, or flipping a coin, dealing a hand of cards in poker, drawing winning lottery numbers, or running a randomized algorithm.

More formally, an experiment is defined as a set of outcomes. (This set is usually written as $$\Omega$$ and is also called the sample space.) Very importantly, the set of outcomes must satisfy two properties:

1. They must be exhaustive, meaning that they contain every possibility.
2. They must be mutually exclusive, meaning that it's impossible to simultaneously have two different outcomes.

For example:

• Flipping a coin: There are two outcomes, heads or tails.
• Rolling a six-sided die: There are 6 outcomes, one for each side of the die.
• Dealing a hand in poker: There are $$\binom{52}{5} = 2598960$$ ways to choose 5 cards out of 52.

When we are talking about a complicated process (such as a sophisticated randomized algorithm), the outcomes are always simple; there just might be way too many of them to count. And the outcomes depend on the random choices only, not on whatever we're trying to accomplish. For example, in poker, you're trying to win (perhaps by getting a good hand). Or maybe you're trying to lose. Or maybe you're playing cribbage instead. None of that changes the set of outcomes for a five-card hand that is dealt randomly from a 52-card deck.

## 2.2 Probability

Every outcome in the sample space has some known probability that it will occur. This probability is a number between 0 and 1, where 0 means "definitely won't happen" and 1 means "definitely will happen." (Note that 1 = 100%, but in hardcore math mode we just stick with the decimals thank you very much.)

This set of probability assignments is called a probability distribution, and formally it's a function $$\Pr$$ that maps from the sample space $$\Omega$$ to real numbers in the range $$[0,1]$$.

For some outcome $$\omega$$, we write $$\Pr[\omega]$$ as the probability that $$\omega$$ will happen. Because the set of outcomes has to be exhaustive and mutually exclusive, it's guaranteed that the sum of probabilities of all outcomes must equal 1; that is:

$\sum_{\omega \in \Omega} \Pr[\omega] = 1$

The probability distribution of the outcomes is usually known in advance. And frequently, the probability of every outcome is the same. This is called the uniform distribution. For example:

• Flipping a coin: $$\Pr[\text{heads}] = \Pr[\text{tails}] = \tfrac{1}{2}$$
• Rolling a six-sided die: The probability of rolling a 2 is $$\tfrac{1}{6}$$.

## 2.3 Combining Experiments

Most complex situations, such as playing a complete game or running an algorithm, don't involve just a single randomized experiment but many of them. Fortunately, combining multiple independent experiments is really easy.

Say I have experiments $$Exp_1$$ and $$Exp_2$$ with sample spaces $$\Omega_1$$ and $$\Omega_2$$. If these experiments are conducted independently from each other (so that the outcome of one doesn't in any way influence the outcome of the other), then they are combined as follows:

• The combined sample space is the set of pairs of outcomes from the two experiments, written formally as a cross-product $$\Omega_1 \times \Omega_2$$.
• The probability of any pair of outcomes is the product of the original probabilities.

For example, say I roll a 6-sided die with one hand and flip a coin with the other hand. This combined experiment now has 12 outcomes:

$(1,H), (2,H), (3,H), (4,H), (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)$

Each the probability of rolling a 3 and getting tails is

$$\Pr[(3,T)] = \Pr[3] \cdot \Pr[T] = \tfrac{1}{6} \tfrac{1}{2} = \tfrac{1}{12}.$$

## 2.4 Events

So far, none of this should be really surprising or even interesting. That's because we usually don't care about the outcomes - their probabilities are already known to start out with, and are usually uniformly distributed so the probabilities are all the same. Boring.

What we really care about is some combination of outcomes that has some meaning relative to whatever game or algorithm or whatever else we're trying to accomplish. For example, in poker, you would really like to get a straight flush. This doesn't represent a single outcome, but multiple possible outcomes, because there are multiple different hands that would give you a straight flush.

An event in probability is a combination (subset) of outcomes, that has some meaning. Each outcome itself is called a simple event. The probability of an event is the sum of the probabilities of the outcomes that make it up.

Back to poker, there are 40 possible hands (outcomes) that result in a straight flush (the low card can be A up to 10, in any of 4 suits). So the straight flush "event" consists of these 40 hands. And the probability of getting a straight flush is

$\frac{40}{\binom{52}{5}} = \frac{40}{2598960} \approx 0.000015391.$

For another example, consider the rolling two six-sided dice (independently). We know there are $$6\times 6 = 36$$ equally-likely outcomes. What's the chance that the sum of the two dice equals 10? Well, there are 3 outcomes that make this happen: (4,6), (5,5), and (6,4). So the probability of getting a sum of 10 is $$\tfrac{3}{36} = \tfrac{1}{12}$$.

## 2.5 Random variables

We've already talked about one function on the outcomes of an experiment, which is the probability measure $$\Pr$$ that maps each outcome to some number between 0 and 1.

Sometimes we also want to associate some numerical value with each outcome, which has nothing to do with its probability. For example, each possible lottery ticket might be associated to the amount of money that ticket will win. The probabilities of every possible ticket are the same, but the winnings are definitely not!

For largely historical reasons, these functions are not called (or written as) functions, but instead are called . They are usually written with a capital letter like $$X$$ or $$Y$$.

For example, if we go back to the experiment of rolling two six-sided dice, define the random variable $$X$$ to be the sum of the two dice. We just figured out the probability that this equals 10, which can now be written conveniently as

$\Pr[X = 10] = \tfrac{1}{12}.$

The random variable notation is a little strange to get used to, but it also allows us to write down more complicated events easily. For example, consider the event that the two dice add up to 10 . There are actually 6 outcomes that make this happen: the three ways to make 10, plus the two ways (5,6) and (6,5) to make 11, and the single way (6,6) to make 12. Therefore

$\Pr[X \ge 10] = \tfrac{6}{36}.$

In this way, random variables are nice not only as a way of assigning a value to each outcome, but also as a way of describing more complicated events. You just have to remember one thing (repeat after me): A random variable is a function, not a variable.

Once we have a random variable, it's natural to ask about some statistical measures on that random variable, such as mean, mode, variance (standard deviation), and so on. The mean, or average, is the most important one, and it has a special name: the expected value.

Formally, the expected value of a random variable $$X$$, written as $${\mathop{\mathbb{E}}}[X]$$, is a weighted average formed by summing the value of $$X$$ for each outcome, times the possibility of each outcome. Here's a complete definition:

Definition: Expected value

Suppose an experiment has $$n$$ outcomes, each of which occurs with probability $$p_i$$. And suppose that a random variable $$X$$ associates with each outcome some value $$v_i$$. Then the expected value of $$X$$ is the sum of the value of each event, times the probability of that event, or

${\mathop{\mathbb{E}}}[X] = \sum_{i=1}^n p_i\ v_i$

Note we can also write this sum in another way as:

${\mathop{\mathbb{E}}}[X] = \sum_i i \cdot \Pr[X = i]$

## 2.6 More distributions

We already said that the probability distribution of the set of outcomes is usually (but not always) uniform - meaning every outcome is equally likely. A random variable also defines a probability distribution, if we consider the probability that the random variable takes on each possible value. And these distributions are usually not uniform.

For example, consider the experiment of flipping three coins, and the random variable $$X$$ defined as the number of heads among the three coin flips. For each of the $$2^3 = 8$$ outcomes, we can easily see the value of this random variable:

Outcome $$X$$
HHH 3
HHT 2
HTH 2
HTT 1
THH 2
THT 1
TTH 1
TTT 0

The distribution of the outcomes is uniform - each occurs with probability $$\tfrac{1}{8}$$. But the distribution of $$X$$ is seen by flipping this table around, adding together the rows with the same value:

$$X$$ Num. of Outcomes $$\Pr[X=i]$$
0 1 1/8
1 3 3/8
2 3 3/8
3 1 1/8

In other words, you're much more likely to get 1 head or 2 heads than you are to get 0 or 3.

This distribution actually has a name - it's called a binomial distribution with parameters $$p=.5$$ and $$n=3$$. You can find much more about the binomial distribution on Wikipedia.

Other distributions that might come up in this class, along with my pithy description, are:

• Uniform distribution (rolling a fair $$n$$-sided die)
• Bernoulli distribution (flipping a weighted coin)
• Binomial distribution (flipping a weighted coin $$n$$ times and counting how many heads)
• Geometric distribution (flipping a weighted coin until the first time it comes up heads, and counting how many flips that took)

## 2.7 Calculating probabilities and expectations

There are going to be many, many times when we want to compute the probability of some event, or the expected value of some random variable. And of course these skills are useful in a variety of real-like situations, not just in analyzing randomized algorithms.

There are basically three paths to take to perform one of these calculations:

1. List out every one of the outcomes, and then add up the probabilities (or the weighted sum, for an expectation). Since we always know the probability of the outcomes - they're usually all equally likely - this is mostly a matter of counting.
2. Estimate the probability using some approximations or bounds. We will see some of these as we go along in this class, and if you look up the distributions mentioned above you can find likes to some popular bounds or estimates for those distributions.
3. Approximate by actually running an experiment. This involves programming of course, and is a really excellent way to check your work from one of the first two methods.

Methods 1 and 3 are immediately available to you, and based on the definitions you should technically be able to calculate any probability or expectation that we could come up with. Of course the problem will often be that there are far too many possibilities to list them all out manually. Throughout the course, we will learn some counting techniques and math tricks to make method 1 easier, and we'll also see in many cases that just having an estimate or a bound (method 2) is good enough.

# 3 Back to algorithms

Now that we understand a little bit about probability, let's go back to the original example and see how probabilistic concepts show up in designing and analyzing algorithms.

## 3.1 Deterministic algorithm

A deterministic algorithm is the kind of algorithm you are used to: it takes some input, does some computation that depend only on that input, and then produces some output. The important thing here is that the actions taken by the algorithm depend only on whatever the input is. So if you run the algorithm twice on the same input, the algorithm will do exactly the same thing.

In the case of the "airplane terrorists", we came up with the following deterministic strategy:

1. Screen the first 181 passengers
2. If any of the first 181 has bomb liquid, screen all 200
3. If at least 20 passengers have bomb liquid, don't let the plane fly.

This algorithm is correct because it's guaranteed not to let a hijacked plane fly. If the plane has at least 20 terrorists, we will definitely screen at least one of them on step 1, and then the entire plane will be checked and ultimately grounded. Conversely, if the plane does not have at least 20 terrorists, there is no way that it will be grounded by our algorithm.

A standard, worst-case analysis of this algorithm is pretty straightforward: the worst case is that we have to screen all 200 passengers. So by this analysis our algorithm is no better than just screening every passenger, every time. That seems wrong through, because we know that most flights won't have any terrorists at all.

So instead, we will do a probabilistic analysis of the number of passengers screened. Rather than just focusing on the worst-case cost, we will instead focus on the "expected" cost, which corresponds roughly to the average cost over many, many flights.

Using our terminology from probability theory here, the experiment in this case will be the choice of which passengers (if any) have bomb-making liquid. And the random variable $$X$$ is defined as the number of passengers screened by the algorithm.

In our case, the two possible events are "there is a hijacker within the first 181 passengers" or "there isn't a hijacker in the first 181 passengers", as this is what determines the total number of passengers screened. From the problem definition, the probability that there is no hijacker on the plane at all - and so certainly not one within the first 181 passengers - is at least 999/1000. Therefore the expected cost is at most:

$E(\text{passengers screened}) \le \frac{999}{1000}\cdot 181 + \frac{1}{1000}\cdot 200 = 181.019$

In other words, the number of passengers screened on average is less than 182.

## 3.2 Randomized algorithm

The basic idea of a randomized algorithm is pretty obvious here: we will randomly select which passengers to screen. In class we looked at a variety of ways to do this random selection. As is often the case (in math, in life, in computer science, ...) the best approach is the simplest one:

1. Select $$n$$ passengers at random
2. Screen these $$n$$ passengers for bomb-making liquid
3. If any of the $$n$$ has bomb liquid, screen all 200 passengers
4. If at least 20 passengers have bomb liquid, don't let the plane fly.

Here $$n$$ is some constant that is selected beforehand. The big question is: how large should $$n$$ be? If it's small, then the algorithm will be rather efficient (not many passengers screened), but it will also not be very correct (some planes will blow up). On the other hand, if $$n$$ is very large, the algorithm will be more correct but less efficient.

One extreme is the equivalent of the deterministic algorithm: let $$n=181$$. Then there is no chance of failure, so the algorithm is perfectly correct, but not very efficient. The other extreme would be to set $$n=0$$ and not screen any passengers. Then the algorithm is quite efficient, but there is a one-in-a-thousand chance of failure.

What we see is something very typical in randomized algorithms: a trade-off between correctness and efficiency. It is easy to view this as a shortcoming of randomized algorithms, since the algorithm will fail with some (small) probability. But in fact it is a feature that we don't get with deterministic algorithms. In Problem 3, you showed that the deterministic algorithm is completely uncompromising: if we improve the efficiency even the smallest amount, the correctness (probability of failure) goes immediately to the max. On the other hand, Problem 2 showed that, by only scanning about one-fourth of the passengers, the probability of failure is less than one in a million.

The general theme here is called the principle of many witnesses and we will see it crop up repeatedly in this class. In this scenario, we are taking advantage of the fact that many hijackers must collude in order to blow up the plane: the 20 (or more) hijackers are the "many witnesses". Deterministically, it can be difficult to find a single witness, but with randomization, it becomes much easier.

The other principle at work here is called foiling an adversary. The consistency of deterministic algorithms is actually their weakness: they can be fooled by someone who knows the algorithm! On the other hand, if the strategy is random, then while it can be beaten some of the time, there is no way to beat it all of the time. This is a powerful idea behind many randomized algorithms that we will see.


text:
Bode Plots & High Pass, Low Pass, Band Stop and Pass Frequencies

Discussion in 'Homework Help' started by libnitz, Feb 20, 2016.

1. libnitz Thread Starter New Member

Jan 29, 2016
17
0
Hi,

I just need some confirmation on 3 questions

1. I was asked to calculate the poles and zeros. I believe

zeros: none
poles: 2 (or p1,2) = -1

2. I was also asked to calculate the slope of the magnitude response was w -> 0
Here is my sketched graph

From my pic, it shows that as w -> 0, H(s) -> 0.

3. What type of filter response does this network exhibit.
I believe it will be a Low Pass because as

s->0, H(s) -> 1 and as s-> infinity, H(s) -> 0?

Does this reasoning make it a low pass filter, high pass, band stop or band pass?

Any help is appreciated

2. shteii01 AAC Fanatic!

Feb 19, 2010
3,708
556
3.
Yes.
As w increases, s=jw will also increase. s is in the denominator so as s increases, the denominator also increases, so as s becomes bigger, $\frac{1}{s^2}$ will become smaller and smaller and smaller. So near 0 rad/s (or Hz) you will have some useful response, but as you move further away (toward infinity), the response will sharply drop.

libnitz likes this.
3. libnitz Thread Starter New Member

Jan 29, 2016
17
0
Thanks! I used the same reasoning as you wrote, however I was not too sure I got it right.

Could you also double check my calculations on this problem also?

MOD NOTE: You already have an active thread on this other problem.


text:
## Mathematical properties – wikiversity harry mileaf electricity 1 7 pdf

#########

1. Commutative Property of Addition/Multiplication – The word, commutative, from French, commuter (to switch), means move around. Such as in problem, a {\displaystyle a} • b {\displaystyle b} = b {\displaystyle b} • a {\displaystyle a} , the a {\displaystyle a} and the b {\displaystyle b} move around about the equal sign, but nonetheless, still equal to the same sum (in this situation, we can say gas house edwards co c {\displaystyle c} ). Now, let’s implement this on real numbers, such as:

2. Associative Property of Multiplication/Addition – Etymology here: The word, associative, is derived from the word associate, which electricity facts for 4th graders comes from the Latin word, associo, which means to unite together, associate. This property basically reflects the same rules as the Commutative property: No matter the order of the parentheses, you will get the same result. The parenthesis can go wherever you like it to be! So:

3. Identity Property – Specifically the Additive Identity Property and the Multiplicative a gas station near me Identity Property, is when the total (number) does not change throughout the equation. The word, identity, comes from the Latin word for idem, which means the same. So, you can think of it as the number is still the same after the math equation. In 5 {\displaystyle 5} + 0 {\displaystyle 0} = 5 {\displaystyle 5} : The total (number) stayed the same throughout the math e payment electricity bill up problem–It is still 5! For the Additive Identity Property, it is always zero (this works for subtraction as well)–For the Multiplicative Identity Property, it is always one (this works for division c gastronomie vitam as well).

4. Inverse Property, also the Additive Inverse Property and the Multiplicative Inverse Property, state that any number to added to its opposite counterpart ( 5 {\displaystyle 5} and − 5 {\displaystyle -5} ) will equal either zero or one. This zero or one depends on which Inverse Property you are using gas and water company. If you are using the Additive Inverse Property, you should be getting zero. If you are using the Multiplicative Inverse Property, you should be getting one. The Additive Inverse Property works with subtraction as well, and the Multiplicative Inverse Property works with division as well. In the Additive Inverse Property, the was electricity invented during the industrial revolution number must to be added to its negative counterpart, such as 29 {\displaystyle 29} and − 29 {\displaystyle -29} , 6 {\displaystyle 6} and − 6 {\displaystyle -6} , 91 {\displaystyle 91} and − 91 {\displaystyle -91} , 47 {\displaystyle 47} and − 47 {\displaystyle -47} , etc.. In the Multiplicative Inverse Property, the number must be multiplied to its reciprocal, which electricity dance moms choreography is the number found by flipping the numerator and the denomerator of a fraction ( 2 {\displaystyle 2} = 2 1 {\displaystyle {\tfrac {2}{1}}} ), such as 3 4 {\displaystyle {\tfrac {3}{4}}} and 4 3 {\displaystyle {\tfrac {4}{3}}} , 9 2 {\displaystyle {\tfrac {9}{2}}} and 2 9 {\displaystyle {\tfrac {2}{9}}} , 82 93 {\displaystyle {\tfrac {82}{93}}} and 93 82 {\displaystyle {\tfrac {93}{82}}} , 2 {\displaystyle 2} and 1 2 {\displaystyle {\tfrac {1}{2}}} , etc..

5. Property of Zero – Specifically the Multiplicative Property of Zero and the Additive Property of Zero, is when zero plays a role in a math equation. Specifically in the Multiplicative Property of Zero, anything multiplying zero gas works park fireworks is zero. This rule is true no matter what number it is. Big or small, this rule works in every and all cases of problems that follow the Multiplicative Property of Zero. For example:

The Additive Property of Zero is part of the property of zero group and the identity group, which will be explained later in this section. The Additive Property of Zero states that british gas jokes any number adding zero will remain the same (thus, identity): r {\displaystyle r} + 0 {\displaystyle electricity electricity lyrics 0} = r {\displaystyle r} . Here are a few problems to help explain:

6. Substitution property is the act of substituting factors with the answer. An example is − 5 x + 8 + x = − 4 x + 8 {\displaystyle -5x+8+x=-4x+8} . The substitution property simple takes − 5 x {\displaystyle -5x} , and adds it to x {\displaystyle x} , which equals − 4 x {\displaystyle -4x} and the constant: 8 {\displaystyle 8} (since there are no like terms).


text:
Average Potential Electricity and Magnetism

1. Sep 30, 2008

jrc5135

1. The problem statement, all variables and given/known data
Find the average potential over a spherical surface of radius R due to a point charge q located inside. Show that in general: (EQ 1 below), where Vcenter is the potential at the center due to all external charges and Qenc is the total enclosed charge

2. Relevant equations
EQ 1 Vave = Vcenter + (Qenc/4*pi*ε₀*R)

3. The attempt at a solution
I am just really confused on where to at least get started. I'll be at the computer for awhile so feel free to ask questions I just want to get it started.

2. Sep 30, 2008

gabbagabbahey

Try placing a point charge on the z-axis a distance z'<R from the origin. What is the potential $V(\vec{r})$ at a general point $\vec{r}$ in spherical coordinates? Average this potential over the surface of a sphere of radius R. What do you get?

3. Sep 30, 2008

jrc5135

1/(4*pi*epsilon nought) q/r

4. Sep 30, 2008

gabbagabbahey

Is it really?! I thought the potential due to a point charge located at $\vec{r'}$ was:

$$\frac{1}{4\pi \epsilon_0} \frac{q}{|\vec{r}-\vec{r'}|}$$

Of course, when the charge is at the origin, $\vec{r'}=0$ and the potential reduces to the one you gave. But(!) what about when the charge is located along the z-axis a distance z' from the origin (i.e.$\vec{r'}=z'\hat{z} \neq 0$) ??

5. Sep 30, 2008

jrc5135

not too sure?

6. Sep 30, 2008

gabbagabbahey

Have you learned about vectors yet? If so, what is $|\vec{r}-z'\hat{z}|$ in spherical coordinates?

7. Sep 30, 2008

jrc5135

i really dont know?

8. Sep 30, 2008

gabbagabbahey

you don't know if you've studied vectors yet?

9. Oct 1, 2008

jrc5135

I have and I know the potential outside the sphere has to only be in the +z direction because of symmetry, but I dont know the second part of your question.

10. Oct 1, 2008

gabbagabbahey

Well,

$$|\vec{r}-z'\hat{z}|=\sqrt{(r\hat{r}-z'\hat{z}) \cdot (r\hat{r}-z'\hat{z})}=\sqrt{r^2-2rz' (\hat{r} \cdot \hat{z})+z'^2}=\sqrt{r^2-2rz'cos(\theta)+z'^2}$$

$$\Rightarrow V(\vec{r})= \frac{1}{4 \pi \epsilon _0} \frac{q}{\sqrt{r^2-2rz'cos(\theta)+z'^2}}$$

And so on the spherical surface $r=R$,

$$\Rightarrow V(R,\theta,\phi)= \frac{1}{4 \pi \epsilon _0} \frac{q}{\sqrt{R^2-2Rdcos(\theta)+d^2}}$$

(where I have set z'=d the distance of the point charge from the origin)

Do you know how to average a function over a surface?

11. Oct 1, 2008

jrc5135

no i dont

12. Oct 1, 2008

gabbagabbahey

The average of a function $f(\vec{r})$ over any surface $\mathcal{S}$ is defined as

$$f_{ave}=\frac{\int_{\mathcal{S}} f(\vec{r})da }{\int_\mathcal{S} da}=\frac{1}{A} \int_{\mathcal{S}} f(\vec{r})da$$

where $A$ is the area of the surface, and $da$ is the infitesimal area element for said surface.

What is $da$ for a spherical surface of radius R (in spherical coordinates)?

Can you apply this to $V(R,\theta,\phi)$?

13. Oct 1, 2008

jrc5135

da would be r^2sin(theta)drdtheta
so then
Vave = 1/a(integral (k q/r)*r^2sin(theta)drdtheta

14. Oct 1, 2008

gabbagabbahey

Does $r$ really vary on a spherical surface of fixed radius $r=R$ ? If not, why is there a $dr$ in your $da$ ? Shouldn't there be a $d\phi$ term instead?

And why are you using kq/r for $V(R,\theta,\phi)$?

What are the limits of integration for $\theta$ and $\phi$?


text:
5

# Consider the reaction:B = â‚¬ + JDsolution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I...

## Question

###### Consider the reaction:B = â‚¬ + JDsolution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I6 MD. At equilibrium; the concentration of D was measured to be 6,0 x 10-M: Calculate the equilibrium concentrations of each speciesb) Calculate the equilibrium constant;

Consider the reaction: B = â‚¬ + JD solution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I6 MD. At equilibrium; the concentration of D was measured to be 6,0 x 10-M: Calculate the equilibrium concentrations of each species b) Calculate the equilibrium constant;

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text:
# Trouble with meeting design specifications of a second order system using Matlab

I am trying to solve the following question.

Consider the transfer function $$G(s) = \frac{1.247}{s^2+9.76s+23.8}$$ is in the forward path of a unity feedback control loop. Assume that it is compensated using a static gain K in the forward path. Now I have to plot the root locus of closed-loop system and determine if it is possible to find a value for K such that the settling time is 0.7963 s and the rise time is 0.4445 s.

I know how to do it on paper. The overall closed loop transfer function become $$G(s) = \frac{K*1.247}{s^2+9.76s+23.8+(K*1.247)}$$

If I compare this with standard form of second order system, I can see that $$2*\zeta*\omega_n = 9.76$$ and $$\zeta\omega_n = 4.88$$ From the formula for settling time, $$t_s = \frac{4}{\zeta\omega_n}$$ $$\zeta\omega_n = \frac{4}{0.7963} = 5.02$$

Therefore, it is not possible to design the system by just varying the value of K. But my doubt is, is there any way to directly find it from the root-locus plot in Matlab. How do I plot the root locus in the first place without knowing the value of K?

• The formula for settling time is approximate. In this case, $t_s \approx \frac{4}{4.88}=0.82$, and you cannot change this value. You can, however, select K to achieve the required 10%-90% rise time (once again, the formula for this is approximate). So, yes, this can all be done via the root locus, but don't expect the system to produce the exact values given in the design specification.
– Chu
Apr 12, 2021 at 12:02
• Please add a plot of the root locus. I think the answer will be obvious from the root locus plot itself. Mark the area on the complex s-plane where settling time is 0.79 and rise time is 0.44 (formulae for both exist for second order systems). See if the root locus passes through that area.
– AJN
Apr 12, 2021 at 12:21
• "How do I plot the root locus in the first place without knowing the value of K?" Isn't root locus plotted by varying K from zero to infinity ? So, there is no need to know the "final" value of K while plotting the locus.
– AJN
Apr 12, 2021 at 12:22
• Why do you evaluate the closed loop, since you said it is in feed forward path, therefore open loop? make a depiction of the block diagram of the entire loop. Apr 12, 2021 at 12:48

I think that the mistake you are making is assuming $$\\omega_n\$$ is unaffected by $$\K\$$. In your closed-loop $$\G(s)\$$ formula, $$\\omega_n\$$ is in fact this: -
$$\sqrt{23.8 + 1.247\cdot K}$$
Hence, $$\\zeta = \dfrac{5.02}{\sqrt{23.8 + 1.247\cdot K}}\$$


text:
How many boiling points does a mixture of liquids have?

Let's say I have a mixture of two miscible liquids. I want to get the liquids by fractional/pure distillation. My textbook says a mixture boils over a range of temperatures. On the contrary, another book I read says a mixture has only one boiling point and that the liquid with a lower boiling point adds more of itself to the mixture. I am wondering which of those is correct and will work with distillation.

• The mixture boils over a range of temperatures. The lowest point of that range is called the boiling point. – Ivan Neretin May 14 at 11:58
• There is too few information to anticipate if a separation of the two liquids by distillation may be achieved, or not. Depending on their mutual interaction, perhaps they form an azeotrope and you only reach a maximal concentration of A in B (c /= 100%), e.g., HCl/water. – Buttonwood May 14 at 12:12
• The actual mixture (at a given x) has one defined boiling point (at given P) but as boiling change composition the boiling occurs over a range of T. – Alchimista May 14 at 13:03
• There is no unique answer to this question. A mixture water-ethanol may boil at a temperature between $78$°C and $100$°C, depending on the relative proportion of the constituants. A mixture of two non miscible liquids like oil + water boil at a given temperature, whatever the proportions of the constituants. – Maurice May 14 at 20:37
• It would be nice to know which books you are reading. – Martin - マーチン May 15 at 9:03


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HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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# Solution - Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0. - HSC Science (Computer Science) 12th Board Exam - Mathematics and Statistics

ConceptGeneral and Particular Solutions of a Differential Equation

#### Question

Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0.

#### Solution

dy/dx=1/cos(x+y)

Substitute x + y = v...(1)
Differentiating w.r.t. 'x', we get

1+dy/dx=(dv)/dx

dy/dx=(dv)/dx-1

The original differential equation becomes

(dv)/dx-1=1/cosv

(dv)/dx=(1+cosv)/cosv

int(cosvdv)/(1+cosv)=intdx

int((cosv+1-1)dv)/(1+cosv)=x+C

int(1+cosv)/(1+cosv)dv-int1/(1+cosv)dv=x+C

intdv-int1/(2cos^2(v/2))dv=x+C

v-1/2intsec^2(v/2)dv=x+C

v-1/2tan(v/2)dv=x+C

Resubstituting the value of 'v' from (1), we get

x+y-1/2tan((x+y)/2)=x+C

y=1/2tan((x+y)/2)=x+C

For x = 0 and y= 0, we get C=0

Hence, particular solution is y=1/2tan((x+y)/2)

Is there an error in this question or solution?

#### APPEARS IN

2015-2016 (March) (with solutions)
Question 6.2.1 | 4 marks

#### Reference Material

Solution for question: Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0. concept: General and Particular Solutions of a Differential Equation. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Arts, HSC Science (General)
S


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# What is the origin of the names P-1 and P-2 of nickel boride catalysts?

From Chapter 2 of Catalytic Hydrogenation in Organic Syntheses:

Highly stereospecific hydrogenations of acetylenes to cis-olefins can be achieved with a nickel (P2) catalyst, prepared by sodium borohydride reduction of nickel acetate in ethanol, and ethylenediamine as a promoter.

The Wikipedia article on nickel boride describes two forms, P-1 and P-2, differing in their ratio of $\ce{Ni2B}$ to $\ce{NaBO2}$.

What is the origin of the names "P1" and "P2"? Does it have to do with whether the catalysts add hydrogen in a syn or anti fashion?

• Have a look at Wikipedia, it's right there: en.wikipedia.org/wiki/Nickel_boride#Preparation – Pritt says Reinstate Monica May 20 '17 at 12:22
• The meaning of ''p -2" , the reason behind this terminology is what I wanted to know. – Iype Eldho May 21 '17 at 18:12
• I'm speculating things: the only google books reference I'm able to find is this (it's not letting me preview the entire book). Notice closely, Raney Nickel has been called W2, and a footnote at this wikipedia article tells me that's a grade of nickel. So, perhaps P2 is also a grade? – Berry Holmes May 22 '17 at 7:38


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## danielcb114 Group Title How to factor y^4+1/2+1/(16y^4)? I know that it factors into (y^2+1/(4y^2))^2, but I can't figure out why. I need to simplify this for an arc length integral. Any help is greatly appreciated. 11 months ago 11 months ago

1. danielcb114

It is copied correctly; a 3 term sum.

Have you copies it correctly or is Y^4 + (1/2) (y^2) + (1/16)(6^4)?

Does the middle term have the variable y^2 in it?

4. danielcb114

Unfortunately, no. I have checked my algebra up to that point and I'm also looking at a solution manual; which skips the factoring step. I haven't seen anything like it before. I'm thinking something similar to completing the square is the way to go, but I'm lost at this point. Thanks for your help, by the way.:)

O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.

6. satellite73

there is a typo in your question somewhere for sure

7. satellite73

$y^2+\frac{1}{4}y^2=\frac{5}{4}y^2$ once you combine like terms

8. danielcb114

Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.

9. danielcb114

I have since corrected it; thank you.

10. satellite73

still a typo, since $$(\frac{5}{4}y^2)^2=\frac{25}{16}y^4$$

11. danielcb114

Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)

12. satellite73

oooh now i get it $\left(y^2+\frac{1}{4y^2}\right)^2$

13. satellite73

happy holidaze! you done?

14. myininaya

$y^4+\frac{1}{2}+\frac{1}{16y^4}$ $(y^2)^2+2 y^2 \frac{1}{4y^2}+(\frac{1}{4y^2})^2$ This is in the form $a^2+2ab+b^2$ which can be factored into the form: $(a+b)^2$

15. satellite73

probably easiest to see if you multiply out and see that it works

16. danielcb114

Ahh, yes; I don't know why I didn't see that before. I wanted to express it that way, but for some reason my mind turned off the possibility. Thank you, myininaya, satellite73 and radar. You've made me very happy!

17. satellite73

yw (from all of us)

18. myininaya

I hope you all have a great happy holiday! :)


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# Is direction of angular velocity just a definition or has a physical significance?

I am a high schooler so I don't know a lot of fancy maths but I do know some of calculus and multiplication of vectors as dot or cross product. I am learning about Angular velocity. And I am confused that is the direction of angular velocity just a definition or has a physical significance. I looked and searched for this on the internet and several other places and of course, I found the answer but its too diverse as someone says that it is defined and others say that it has some significance. I was amazed and confused much more when I saw gyroscopes in action.

Here are some of the search work I did:

And there are several other pages on the internet that I tried but this remains the same all over. So what I want is not only the answer but also its validity. Thanks and appreciation to anyone who answers or puts his/her effort into this question.

# Edit

Many people were getting confused by what I mean by Physical Significance. Here's what I mean If a thing has physical significance then its effects will be real and you will be able to see them. As a force, although the force itself is not visible its effects are and that too in the same direction in which a force is said to be in. So a direction is real but a quantity assigned in that direction can be just to help us solve some problems or fix some glitches and it could very well be a mathematical trick like a pseudo force in an accelerated frame. Hence for this question, has the direction which is told to be the direction of angular velocity something physical that is happening in that direction? Like a motion, you can not say that a car is moving in $$-X$$ direction if it is moving in $$+X$$ direction if the coordinate system is already defined of course.

# Edit 2

Everyone confused due to a lot of ambiguity in the question. Here's the final Edit and this is the actual question whose answer would be indirectly the answer to this entire title- Could we have defined the direction of Angular Velocity to any other direction if we had more options or let's say we had 4-dimensions reality?

• The stackexchange answer doesn't say that the direction has no physical significance. The direction is perpendicular to the plane. But there are two directions perpendicular to the plane. The convention is which direction we equate to clockwise and anticlockwise rotation. The fact that it's perpendicular to the plane is not a convention. Oct 3 '20 at 6:35
• @BrainStrokePatient Please refer to this link, it also states thats its just a mere defenition Oct 3 '20 at 6:39
• @BrainStrokePatient Where can I get a perfect answer which is true, any ideas Oct 3 '20 at 6:40
• @BrainStrokePatient It indirectly says the same that it doesn't have physical significance, we just did it to remove confusion, we could have also found any other method to do so it means Oct 3 '20 at 6:41
• I don't really understand what you mean by physical significance now. The answer you linked does connect the direction of angular velocity to something physical: which direction the rotation is in. Oct 3 '20 at 6:45

There is in fact a way of expressing angular velocity in such a way that there is no ambiguity of what-part-of-this-is-convention.

Angular velocity occurs in a plane, and it has direction and magnitude. To specify a plane two vectors that lie in that plane are specified, with the order of the two vectors giving the direction of the angular velocity. The magnitude of the angular velocity can for example be specified with a separate number.

If you would be working with a space with four spatial dimensions the above way to specify would be the only possible way; with four spatial dimensions each plane has two vectors that are perpendicular to it. (And with a higher number of spatial dimensions there are more perpendicular vectors.)

Our space has three spatial dimensions, and with three spatial dimensions a shorthand notation of angular velocity is available.

In a space with three spatial dimensions every plane has a single vector that is perpendicular to it. So: to specify a particular plane in a space with three spatial dimensions it is sufficient to specify the vector perpendicular to that plane. And then you can make the magnitude of that single vector represent the magnitude of the angular velocity.

That notation is so much shorter, so much more compact, using that notation is a no-brainer. (It should be kept in mind thoug though that it is something of a fluke; it works only with a space with three spatial dimensions.)

Direction of the rotation
There is one thing, of course. The notation is so compact that there is no room to specify the direction of the rotation. It's literally one bit of information: this-way-around or the-other-way-around. But the shorthand notation has no room to spare; it cannot express that bit.

That is why the shorthand notation is supplemented with the right-hand-rule, the right hand rule fills in that one necessary bit of information.

• So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? Oct 3 '20 at 11:29
• I just want to know that is it just a mathematical fitting to make things work? Oct 3 '20 at 11:32
• I got my query solved by combining the answer from J Thomas and Cleonis. So thanks for helping me and I respect your effort to help people like us. Oct 12 '20 at 7:42

Since the question mentioned higher-dimensional space, I wanted to give an answer that works in any-dimensional space, not just 3. I'll start with formal, mathematical definitions and then connect them to physical intuition.

Rotations in $$n$$-dimensional space form a group. Specifically, they form a group called the special orthogonal group, which is denoted by $$\mathrm{SO}(n)$$. $$\mathrm{SO}(n)$$ is also a smooth manifold, so we call it a Lie group.

Every point on a manifold has a tangent space. The elements of this tangent space are called tangent vectors. Intuitively, a tangent vector tells us which direction to move in and how fast to move in that direction. That is, it gives us a velocity, as illustrated below:

The Lie algebra of a Lie group is simply the tangent space at the identity element of the group. For $$\mathrm{SO}(n)$$, the identity element is the rotation that does nothing, i.e. no rotation.

Therefore, an angular velocity is an element of the Lie algebra of $$\mathrm{SO}(n)$$, which is denoted by $$\mathfrak{so}(n)$$.

Side note: In terms of matrices, $$\mathrm{SO}(n)$$ can be represented as the set of $$n \times n$$ orthogonal matrices with determinant 1, while $$\mathfrak{so}(n)$$ can be represented as the set of $$n \times n$$ antisymmetric matrices. The matrix exponential gives us the exponential map from the latter to the former.

So what is $$\mathfrak{so}(n)$$ like? Intuitively, we can specify any angular velocity $$\omega$$ as follows:

• Rotate this fast ($$a_1$$) on this plane ($$p_1$$) through the origin.
• Rotate this fast ($$a_2$$) on this plane ($$p_2$$) through the origin.
• etc.

Each plane $$p_i$$ also carries an orientation which tells us which way we intend to rotate.

In short, we can think of $$\omega$$ as a weighted sum $$a_1 p_1 + a_2 p_2 + \dots$$. But what is $$p_i$$, mathematically? To specify a plane, we only need 2 unit vectors (say $$\mathbf{u}$$ and $$\mathbf{v}$$), as illustrated below:

The resulting plane is the wedge product of $$\mathbf{u}$$ and $$\mathbf{v}$$, which is denoted by $$\mathbf{u} \wedge \mathbf{v}$$. Switching the order of $$\mathbf{u}$$ and $$\mathbf{v}$$ switches the orientation of the plane. When combined, they cancel out: \begin{align} \mathbf{u} \wedge \mathbf{v} + \mathbf{v} \wedge \mathbf{u} = 0 \end{align}

This corresponds to the fact that if we rotate this fast in one direction and equally fast in the opposite direction, we get nothing. Scaling either vector by a scalar $$a$$ simply scales the resulting angular velocity:

\begin{align} a \mathbf{u} \wedge \mathbf{v} = \mathbf{u} \wedge a \mathbf{v} = a (\mathbf{u} \wedge \mathbf{v}) \end{align}

Thus each summand $$a_i p_i$$ of our angular velocity is the wedge product of 2 vectors, i.e. a blade. Thus our angular velocity $$\omega$$ is a sum of blades, i.e. a bivector. The set of bivectors is denoted by $$\wedge^2 \mathbb{R}^n$$.

In 2 and 3 dimensions, something special happens: Any sum of blades is a blade. Thus we only need a single blade to specify an angular velocity. Consequently, every rotation is a simple rotation.

Furthermore, in 3 dimensions, the dual of a bivector is a vector, i.e. $$\star \left( \wedge^2 \mathbb{R}^3 \right) = \mathbb{R}^3$$. This is why in 3D we typically describe planes using "normal vectors": \begin{align} \mathbf{u} \times \mathbf{v} &\stackrel{\text{def}}{=} \star (\mathbf{u} \wedge \mathbf{v}) \end{align}

and rotations using "rotation axes" (see Euler's rotation theorem).

This trick doesn't work in other dimensions. For example, in 2 dimensions, the dual of a bivector is a scalar, which is why we typically describe 2D rotations using scalars.

In 4-dimensional space, something even stranger happens: Not only is the dual of a bivector not a vector, but there are bivectors that are not blades. Consequently, there are rotations in 4-dimensional space that cannot be described as rotations on a single plane. These are called double rotations. An example is the rotation given by

\begin{align} \mathbf{u} \wedge \mathbf{v} + \mathbf{w} \wedge \mathbf{x} \end{align}

where $$\mathbf{u},\mathbf{v},\mathbf{w},\mathbf{x}$$ are all mutually orthogonal. The animation below shows a double rotation acting on the 4D unit cube (stereographically projected to 3D, of course):

• While a bit dense for a high schooler, this one is the right answer. The wedge product is the thing that is "meaningful," including in higher dimensions. It just happens that in 2 or 3 dimensions, the math of these wedge products is identical to the math required for cross products. So the direction of the cross product itself isn't all that meaningful, but it's really close to something which is. Oct 12 '20 at 0:21

I am learning about Angular velocity. And I am confused that is the direction of angular velocity just a definition or has a physical significance.

You are going to get confusing answers, because your question as stated doesn't mean much. But it means something....

There are things in mathematical notation that are basicly arbitrary. Somebody chose to write them that way, and they worked, and now everybody does it that way. Like multiplication distributes over addition, and we write $$a(b+c)$$. We could have used any other symbol in place of (). $$a:b,c:$$ would have worked as well. For $$a(b-c)$$ we could do $$a:b,-c:$$.

We could have a convention that each time you have a newline.

$$a(b(d+e-f)+c)$$ becomes

a:
b:
d,e,-f
,c

That would some ways work better though it would take more space on the page. It's basicly arbitrary which way we use.

But the fact that $$a(b+c)=ab+ac$$ is not arbitrary. It's important.

It looks to me like you're asking what's the important part, and what's just convention.

Could we have defined the direction of Angular Velocity to any other direction if we had more options or let's say we had 4-dimensions reality?

It would have to amount to the same thing -- if it gave a different answer then it would be a wrong answer. Unless we changed the concepts around somehow so they combined differently to get the same end result.

But yes, instead of defining a vector axis as the defining direction, we could have two vectors to define the plane the rotation is in. And then at any one moment the velocity would be something in that plane. That wouldn't make any practical difference in 3D but it might be clearer.

A rotation is in some particular plane. If you use polar notation, rotation changes the angle but not the length. Polar coordinates (or for that matter cartesian coordinates) gives you an arbitrary zero point, and whatever point you rotate around, you arbitrarily subtract its displacement from all the locations so it will be at zero to do the rotation. You can add the displacement back later with no loss.

Using the normal vector is only one possible way to describe which plane the rotation is in. That's arbitrary notation.

• I got my query solved by combining the answer from J Thomas and Cleonis. So thanks for helping me and I respect your effort to help people like us. Oct 12 '20 at 7:42

## First lets be on the same page regarding what is angular velocity ?

Angular velocity often, denoted as $$\omega$$ is the rate of angular displacement, denoted as $$\theta$$ with respect to time i.e. you might have seen this equation a lot $$\displaystyle{\vec{\omega} = \frac{\vec{\theta}}{t}}$$ and if we are talking about instantaneous angular velocity then : $$\displaystyle{d\vec{\omega} = \frac{d\vec{\theta}} {dt}}$$ The angular displacement is the change in the plane angle, subtended by the body which is performing the motion, at some reference point.

## Direction of $$\vec\omega$$

The direction of angular velocity, basically tells you in which direction is a body rotating or performing circular motion with respect to a reference point i.e. it tells you the direction of the angular displacement.

Just like you can assume what direction you want to take for positive $$x$$ axis and which one for positive $$y$$ axis, you can also assume which direction to take positive and which one to take negative. Say, you took counterclockwise motion as positive then you have to take clockwise motion as negative.

And of course you can use the Right Hand thumb rule, also referred to as Maxwell's Corkscrew Rule to find the direction of angular velocity. Some teachers in Highschools, while teaching mechanics to students, say that angular velocity is a vector but they treat it like a scalar, which is wrong.

• So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? Oct 3 '20 at 11:29
• I just want to know that is it just a mathematical fitting to make things work? Oct 3 '20 at 11:32
• I couldn't clearly understand your question properly. If the property you are talking about in your first question is angular velocity, then it has 1 value just different signs. Kinda like how we teach counting to children, if you go to the left of zero, you have negative numbers, we told them that if they count to the left of 0 it is considered negative. That is a general convention. But, If you want to you can take going left as positive. Oct 3 '20 at 14:16
• I don't know if you have been taught about/ learned about torque or not ? But, after seeing your edit, I would like to make an analogy to how you said force has a physical significance. If a body has angular velocity in say +x direction and a torque is being applied to it in that direction that it will start spinning faster in that direction. But if the direction of both is opposite then the body will slow down, eventually. Oct 3 '20 at 14:22
• Or for example, there exists a sorta definition of angular velocity that the cross product of the position vector and angular velocity of a body is its linear velocity with respect to a reference point. So you can see that if the direction of angular velocity changes then so does linear velocity Oct 3 '20 at 14:24

Defining angular velocity as a vector that is perpendicular to the plane of rotation is useful in three dimensional scenarios because it allows angular velocities to be added using the rules of vector addition (the parallelogram rule). If an object is rotating with angular velocity vector $$\vec \omega_1$$ relative to a reference frame $$F_1$$, and $$F_1$$ is rotating about the same centre with angular velocity vector $$\vec \omega_2$$ relative to reference frame $$F_2$$ then the angular velocity of the object relative to $$F_2$$ is the vector sum $$\vec \omega_1 + \vec \omega_2$$. So, yes, the angular velocity vector does have physical significance.

However, the additional of angular velocity vectors would still work if we replaced $$\vec \omega_1$$ with $$-\vec \omega_1$$ and $$\vec \omega_2$$ with $$-\vec \omega_2$$ i.e. if we used a left hand rule instead of a right hand rule to find the direction of the angular velocity vector. So the use of a right hand rule to determine the direction of the angular velocity vector is the part that is convention.

• So let's say that if there was a property having two types of values A and B of which one is applied when a body is rotating clockwise and the other applies the other way around then would we have assigned directions in direction of A and B? Oct 3 '20 at 11:28
• I just want to know that is it just a mathematical fitting to make things work? Oct 3 '20 at 11:32
• @user921307 I don't understand what you mean by "two types of values A and B". If two objects are rotating in the same plane, one clockwise and the other anti-clockwise, then their angular velocity vectors point in opposite directions. Oct 3 '20 at 11:33
• I know, but in short, I want to ask that is it just a mathematical trick to help resolve ambiguities and make things work because it seems the same for the definition of the cross product too Oct 3 '20 at 11:34

The vectors representing rotation are chosen along the axis if rotation because that is the only direction in the system which is usually not continuously changing direction. That said, such vectors can accurately represent the direction and magnitude of rotational quantities.

• Thanks, I got the point Oct 3 '20 at 17:18

The magnitude of $$\vec{\omega}$$ gives you the speed of rotation. But the direction of rotation gives you the orientation of the rotation axis.

The general motion of a 3D solid body is a rotation about an instantaneous axis (line in space) and a parallel translation along the axis. This is the so called Chasles's theorem.

To describe a line in space you need the direction of the line, as well as any point along the line.

The motion of a solid body is described by the following properties, derived from the rotation vector $$\vec{\omega}$$, as well as the velocity vector $$\vec{v}$$ of some point on the body (the reference point).

• Magnitude of rotation $$\omega = \| \vec{\omega} \| \tag{1}$$
• Direction of rotation $$\hat{z} = \frac{ \vec{\omega}}{\omega} \tag{2}$$
• Point on rotation axis closest to reference point $$\vec{r} = \frac{ \vec{\omega} \times \vec{v} }{\omega^2} \tag{3}$$ this position is measured from the reference point.
• Parallel motion along the rotation axis $$\vec{v}_\parallel = \left( \frac{\vec{\omega} \cdot \vec{v}}{\omega^2} \right) \vec{\omega} \tag{4}$$

So the direction of $$\vec{\omega}$$ not only gives you (2), but contributes to the location of the rotation axis via (3).

In reverse, you can transform the rotation about the axis plus the parallel motion to the velocity of the reference point with the following expression

$$\vec{v} = \vec{v}_\parallel + \vec{\omega} \times (-\vec{r}) \tag{5}$$

PS> Something similar happens with force vector $$\vec{F}$$ given the torque of the force at some reference point $$\vec{\tau}$$. You can find similarly the position $$\vec{r}$$ and the parallel torque of the line of action of the force.


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How to avoid dyed liquids staining reflective/transparent surfaces

I'm dealing with some hydrocarbons (diesel, gas oil, kerosene, ethanol) where I'm putting them in a chamber and shining light through them to hit a sensor on the other side in order to measure their colour. Unfortunately what I'm finding is that the lenses and reflective surfaces in the chamber that I'm using for these optical measurements are being stained by the products and subsequently affecting the measurements. When the chamber is emptied, small amounts of product remain clinging to the internal surfaces. As this dries a residue is left which remains on the surface.

I've tried a couple of different materials: stainless steel reflectors, transparent Grilamid plastic lenses, a stain-resistant transparent potting and regular glass. All of them end up with a residue that clings to the surface that needs to be wiped clean. Unfortunately, for my application I'm not going to be able to get access to the internal surface to clean it so I need to find a way to prevent this residue from clinging to the internal surfaces.

The main restriction for me is that the system needs to basically be 100% maintenance free, and unfortunately I can't control the fluid coming in and out and therefore can't rinse the system out.

I was wondering if there are any products (both transparent and reflective) that are basically completely "hydrophobic" but for hydrocarbons and will prevent the product clinging at all to the surface and staining the lenses and reflectors?

A super-hydrophobic coating would work well for you because super-hydrophobic materials are self-cleaning. Sandia National Labs has a nice report on super-hydrophobic coatings which can be found here (associated patent here). In it they report the invention of a coating who's hydrophobicity can be varied by application of UV light. The contact angle of their coating is greater than 150$^\circ$, and they reference two other patents which reach contact angles of 169$^\circ$ and 172$^\circ$ which are both very high, the maximum is 180$^\circ$.

The problem is, all of this research is fairly recent and doesn't seem to have found its way to industry yet. I can't find many coating houses who sell super-hydrophobic optical coatings. Aculon's website says that they sell one, but they don't publish the contact angle, and the picture they show makes it appear to only be ~90$^\circ$.

Aculon also claims to sell an oleophobic (oil repelling) coating. This may work for you, but you would need to test a sample with the materials you plan to use. The video on their website shows someone trying to write on it with a Sharpie and it just won't take to the surface.

The problem with all of these coatings is longevity. Rain-X is an excellent example of a hydrophobic coating, but it must be reapplied once a month or so. My eyeglasses came with a hydrophobic coating which was very impressive when they were new. Now that they are 10 months old, everything sticks to them again. So, whatever you decide to build, you should be able to service it regularly.

• Yeah, coatings seem to be the best answer except that they don't last forever. There doesn't seem to be any materials that are oleophobic or hydrophobic by design that don't require any costing. Apr 22 '15 at 21:58

You will want to rinse the chamber between uses with something that dries clear and is a non-polar solvent. A typical solvent like pentane or diethyl-ether should do the trick though keep in mind the fire hazard.

• Thanks for the suggestion. Unfortunately for me I don't control the fluid coming in and out of the chamber and in our application unfortunately it just isn't feasible to clean the system out like that. Apr 22 '15 at 22:17


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# Probability for $3$ siblings

What is the probability for $3$ siblings' birthdays to fall on the same day of the week every year since the second sibling was born? This has occurred over 100 years before the first sibling was born and is still re-occurring $70$ years after the first sibling was born. Leap years make no difference.

The dates are April $6$, June $15$, and July $27$.

Thank you!

Once they are born, if the birthdays are all on the same side of February $29$, the pattern (whatever it is) will recur every year. In your example, June $15$ is $70$ days after April $6$. Because $70$ is divisible by $7$, they will fall on the same day. Similarly, July $27$ is $42$ days after June $15$, so they will fall on the same day.


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LocalReferenceClasses {methods} R Documentation

## Localized Objects based on Reference Classes

### Description

Local reference classes are modified ReferenceClasses that isolate the objects to the local frame. Therefore, they do not propagate changes back to the calling environment. At the same time, they use the reference field semantics locally, avoiding the automatic duplication applied to standard R objects.

The current implementation has no special construction. To create a local reference class, call setRefClass() with a contains= argument that includes "localRefClass". See the example below.

Local reference classes operate essentially as do regular, functional classes in R; that is, changes are made by assignment and take place in the local frame. The essential difference is that replacement operations (like the change to the twiddle field in the example) do not cause duplication of the entire object, as would be the case for a formal class or for data with attributes or in a named list. The purpose is to allow large objects in some fields that are not changed along with potentially frequent changes to other fields, but without copying the large fields.

### Usage

setRefClass(Class, fields = , contains = c("localRefClass",....),
methods =, where =, ...)

Localization of objects is only partially automated in the current implementation. Replacement expressions using the $<- operator are safe. However, if reference methods for the class themselves modify fields, using <<-, for example, then one must ensure that the object is local to the relevant frame before any such method is called. Otherwise, standard reference class behavior still prevails. There are two ways to ensure locality. The direct way is to invoke the special method x$ensureLocal() on the object. The other way is to modify a field explicitly by x$field <- ... It's only necessary that one or the other of these happens once for each object, in order to trigger the shallow copy that provides locality for the references. In the example below, we show both mechanisms. However it's done, localization must occur before any methods make changes. (Eventually, some use of code tools should at least largely automate this process, although it may be difficult to guarantee success under arbitrary circumstances.) ### Author(s) John Chambers ### Examples ## class "myIter" has a BigData field for the real (big) data ## and a "twiddle" field for some parameters that it twiddles ## ( for some reason) myIter <- setRefClass("myIter", contains = "localRefClass", fields = list(BigData = "numeric", twiddle = "numeric")) tw <- rnorm(3) x1 <- myIter(BigData = rnorm(1000), twiddle = tw) # OK, not REALLY big twiddler <- function(x, n) { x$ensureLocal() # see the Details.  Not really needed in this example
for(i in seq_len(n)) {
x$twiddle <- x$twiddle + rnorm(length(x$twiddle)) ## then do something .... ## Snooping in gdb, etc will show that x$BigData is not copied
}
return(x)
}

x2 <- twiddler(x1, 10)

stopifnot(identical(x1$twiddle, tw), !identical(x1$twiddle, x2\$twiddle))

[Package methods version 4.2.1 Index]