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# 2 ht tu eNW clncs paint: atan acd be frctation unich afw flldcng faroe ntt AS (alpa ? mals c8 aeld mdes_&Cbei MaUa Mb Vp Tk Concenftafian chOH ( flutneus_1M Ik ...

## Question

###### 2 ht tu eNW clncs paint: atan acd be frctation unich afw flldcng faroe ntt AS (alpa ? mals c8 aeld mdes_&Cbei MaUa Mb Vp Tk Concenftafian chOH ( flutneus_1M Ik 4d and)bale Vawf huutcallu d_each atne â‚¬

2 ht tu eNW clncs paint: atan acd be frctation unich afw flldcng faroe ntt AS (alpa ? mals c8 aeld mdes_&Cbei MaUa Mb Vp Tk Concenftafian chOH ( flutneus_1M Ik 4d and)bale Vawf huutcallu d_each atne â‚¬

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Good reduction of isogenous abelian varieties over finitely generated fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:56:07Z http://mathoverflow.net/feeds/question/81884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81884/good-reduction-of-isogenous-abelian-varieties-over-finitely-generated-fields Good reduction of isogenous abelian varieties over finitely generated fields Martin Orr 2011-11-25T13:58:54Z 2011-11-25T13:58:54Z
Let $K$ be a finitely generated field over $\mathbb{Q}$. Let $A$ and $B$ be abelian varieties over a field $K$, isogenous over some finite extension $L$ of $K$. I want to ask if they have the same points of good reduction.

More precisely, suppose that $S$ is a normal affine variety over $\mathbb{Q}$ with function field $K$, such that there is an abelian scheme $\mathcal{A}$ over $S$ with generic fibre $A$.

Does there exist an abelian scheme $\mathcal{B}$ over $S$ with generic fibre isomorphic to $B$, and such that the isogeny $A_L \to B_L$ extends to an isogeny $\mathcal{A} \times_S T \to \mathcal{B} \times_S T$ where $T$ is the normalisation of $S$ in $L$?

I believe that this is true if I suppose that $L=K$, as then we can extend the kernel of the isogeny to a finite subgroup scheme of $\mathcal{A}$ and form the quotient.

I ask this because the analogous statement is true if we replace $S$ by the spectrum of the ring of integers of a number field, by the Néron-Ogg-Shafarevich criterion. But I am not sure if this extends to $\dim S > 1$.

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# The Function class in JavaScript

Most programmers know how to define and use a function in JavaScript. For example the following function displays an alert containing the string passed in as a parameter:

}
// Then you call it this way

How many developers, however, know that JavaScript functions are actually objects? Indeed, you can define the previous function using the Function class:

//...and you call it the same way

For those who didn’t figure it out, the syntax of the Function class is the following:

var functionName = new Function(arg_1, arg_2, ..., arg_n, functionBody);

where arg_i (i = 1,…,n) is the i-th argument and functionBody is (you guessed it) the body of the function. Although you can define functions using the Function class it’s best to avoid it because it’s less performant than the traditional way. Anyway, knowing that even a function is an object is useful because you know you can call on it the valueOf() and toString() methods shared by all objects. However, when called on a function, these two methods return the source code for the function as a string. You can also call the one property defined in ECMAScript which is length. This property returns the number of arguments as defined in the function. For example:


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# Lazy Programmer

## Your source for the latest in deep learning, big data, data science, and artificial intelligence. Sign up now

Featured partners

# Linear Regression in the Wild – AV1: Next Generation Video Codec

April 10, 2018

A lot of students come up to me and ask about when they’re going to learn the latest and greatest new deep learning algorithm.

Sometimes, it’s easy to forget how applicable even the most basic of tools are.

As you know, I consider Linear Regression to be the best starting point for deep learning and machine learning in general.

And wouldn’t you know, here it is being used in the most advanced, state-of-the-art video codec we have today:

next generation video: Introducing AV1

Check it out!

This new state-of-the-art video codec is based on research done by multiple big companies, such as Google, Cisco, and Mozilla.

As you can see, the final equation is just a line ($$y = mx + b$$).

$$CfL(\alpha) = \alpha L^{AC} + DC$$

# $10 Deep Learning Courses (+more) July Special part 2! July 24, 2017 Since I am still busy hacking away at my next course, we are going to do another HUGE sale. ALL courses on Udemy are now$10. Take this opportunity to grab as many courses as you can because you never know when the next sale is going to be!

As usual, I’m providing $10 coupons for all my courses in the links below. Please use these links and share them with your friends! You can also just type in the coupon code “JUL456”. The$10 promo doesn’t come around often, so make sure you pick up everything you are interested in, or could become interested in later this year. The promo goes until July 31. Don’t wait!

At the end of this post, I’m going to provide you with some additional links to get machine learning prerequisites (calculus, linear algebra, Python, etc…) for $10 too! If you don’t know what order to take the courses in, please check here: https://deeplearningcourses.com/course_order Here are the links for my courses: Deep Learning Prerequisites: Linear Regression in Python https://www.udemy.com/data-science-linear-regression-in-python/?couponCode=JUL456 Deep Learning Prerequisites: Logistic Regression in Python https://www.udemy.com/data-science-logistic-regression-in-python/?couponCode=JUL456 Deep Learning in Python https://www.udemy.com/data-science-deep-learning-in-python/?couponCode=JUL456 Practical Deep Learning in Theano and TensorFlow https://www.udemy.com/data-science-deep-learning-in-theano-tensorflow/?couponCode=JUL456 Deep Learning: Convolutional Neural Networks in Python https://www.udemy.com/deep-learning-convolutional-neural-networks-theano-tensorflow/?couponCode=JUL456 Unsupervised Deep Learning in Python https://www.udemy.com/unsupervised-deep-learning-in-python/?couponCode=JUL456 Deep Learning: Recurrent Neural Networks in Python https://www.udemy.com/deep-learning-recurrent-neural-networks-in-python/?couponCode=JUL456 Advanced Natural Language Processing: Deep Learning in Python https://www.udemy.com/natural-language-processing-with-deep-learning-in-python/?couponCode=JUL456 Advanced AI: Deep Reinforcement Learning in Python https://www.udemy.com/deep-reinforcement-learning-in-python/?couponCode=JUL456 Easy Natural Language Processing in Python https://www.udemy.com/data-science-natural-language-processing-in-python/?couponCode=JUL456 Cluster Analysis and Unsupervised Machine Learning in Python https://www.udemy.com/cluster-analysis-unsupervised-machine-learning-python/?couponCode=JUL456 Unsupervised Machine Learning: Hidden Markov Models in Python https://www.udemy.com/unsupervised-machine-learning-hidden-markov-models-in-python/?couponCode=JUL456 Data Science: Supervised Machine Learning in Python https://www.udemy.com/data-science-supervised-machine-learning-in-python/?couponCode=JUL456 Bayesian Machine Learning in Python: A/B Testing https://www.udemy.com/bayesian-machine-learning-in-python-ab-testing/?couponCode=JUL456 Ensemble Machine Learning in Python: Random Forest and AdaBoost https://www.udemy.com/machine-learning-in-python-random-forest-adaboost/?couponCode=JUL456 Artificial Intelligence: Reinforcement Learning in Python https://www.udemy.com/artificial-intelligence-reinforcement-learning-in-python/?couponCode=JUL456 SQL for Newbs and Marketers https://www.udemy.com/sql-for-marketers-data-analytics-data-science-big-data/?couponCode=JUL456 PREREQUISITE COURSE COUPONS And last but not least,$10 coupons for some helpful prerequisite courses. You NEED to know this stuff before you study machine learning:

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Remember, these links will self-destruct on July 31 (7 days). Act NOW!

# AI news of the week: Tensorflow Object Detection API and Free Online Data Science Books

July 11, 2017

Tensorflow Object Detection API

Summary: Creating accurate machine learning models capable of localizing and identifying multiple objects in a single image remains a core challenge in computer vision. The TensorFlow Object Detection API is an open source framework built on top of TensorFlow that makes it easy to construct, train and deploy object detection models. At Google we’ve certainly found this codebase to be useful for our computer vision needs, and we hope that you will as well.

TechCrunch Article

Free Online Data Science Books

The title says they are “mathematics” textbooks but there’s tons of material for data scientists (who do lots of math) too.

Particularly relevant: probability, statistics, linear algebra, graph theory, inference, learning algorithms, convex optimization, statistical signal processing, and more!

# New machine learning course! Cluster Analysis and Unsupervised Machine Learning in Python

April 20, 2016

[Scroll to the bottom if you want to jump straight to the coupon]

Cluster analysis is a staple of unsupervised machine learning and data science.

It is very useful for data mining and big data because it automatically finds patterns in the data, without the need for labels, unlike supervised machine learning.

In a real-world environment, you can imagine that a robot or an artificial intelligence won’t always have access to the optimal answer, or maybe there isn’t an optimal correct answer. You’d want that robot to be able to explore the world on its own, and learn things just by looking for patterns.

Do you ever wonder how we get the data that we use in our supervised machine learning algorithms?

We always seem to have a nice CSV or a table, complete with Xs and corresponding Ys.

If you haven’t been involved in acquiring data yourself, you might not have thought about this, but someone has to make this data!

Those “Y”s have to come from somewhere, and a lot of the time that involves manual labor.

Sometimes, you don’t have access to this kind of information or it is infeasible or costly to acquire.

But you still want to have some idea of the structure of the data. If you’re doing data analytics automating pattern recognition in your data would be invaluable.

This is where unsupervised machine learning comes into play.

In this course we are first going to talk about clustering. This is where instead of training on labels, we try to create our own labels! We’ll do this by grouping together data that looks alike.

There are 2 methods of clustering we’ll talk about: k-means clustering and hierarchical clustering.

Next, because in machine learning we like to talk about probability distributions, we’ll go into Gaussian mixture models and kernel density estimation, where we talk about how to “learn” the probability distribution of a set of data.

One interesting fact is that under certain conditions, Gaussian mixture models and k-means clustering are exactly the same! You can think of GMMs as a “souped up” version of k-means. We’ll prove how this is the case.

All the algorithms we’ll talk about in this course are staples in machine learning and data science, so if you want to know how to automatically find patterns in your data with data mining and pattern extraction, without needing someone to put in manual work to label that data, then this course is for you.

All the materials for this course are FREE. You can download and install Python, Numpy, and Scipy with simple commands on Windows, Linux, or Mac.

#agglomerative clustering #cluster analysis #data mining #data science #expectation-maximization #gaussian mixture model #hierarchical clustering #k-means clustering #kernel density estimation #pattern recognition #udemy #unsupervised machine learning

# New Deep Learning course on Udemy

February 26, 2016

This course continues where my first course, Deep Learning in Python, left off. You already know how to build an artificial neural network in Python, and you have a plug-and-play script that you can use for TensorFlow.

You learned about backpropagation (and because of that, this course contains basically NO MATH), but there were a lot of unanswered questions. How can you modify it to improve training speed? In this course you will learn about batch and stochastic gradient descent, two commonly used techniques that allow you to train on just a small sample of the data at each iteration, greatly speeding up training time.

You will also learn about momentum, which can be helpful for carrying you through local minima and prevent you from having to be too conservative with your learning rate. You will also learn aboutadaptive learning rate techniques like AdaGrad and RMSprop which can also help speed up your training.

In my last course, I just wanted to give you a little sneak peak at TensorFlow. In this course we are going to start from the basics so you understand exactly what’s going on – what are TensorFlow variables and expressions and how can you use these building blocks to create a neural network? We are also going to look at a library that’s been around much longer and is very popular for deep learning – Theano. With this library we will also examine the basic building blocks – variables, expressions, and functions – so that you can build neural networks in Theano with confidence.

Because one of the main advantages of TensorFlow and Theano is the ability to use the GPU to speed up training, I will show you how to set up a GPU-instance on AWS and compare the speed of CPU vs GPU for training a deep neural network.

With all this extra speed, we are going to look at a real dataset – the famous MNIST dataset (images of handwritten digits) and compare against various known benchmarks.

# Covariance Matrix: Divide by N or N-1?

February 14, 2016

This is a statistics post. It’s probably very boring. I am posting it for my own reference, because I seem to forget how this is derived every time I need it.

Sometimes you see the sample variance defined as:

$$\hat{\sigma}^2 = \frac{1}{N} \sum_{n=1}^{N} (X_n – \mu)^2$$

But you might also see it defined as:

$$\hat{\sigma}^2 = \frac{1}{N-1} \sum_{n=1}^{N} (X_n – \hat{\mu})^2$$

Where as usual the “hat” symbol means that is our prediction.

Why do statisticians sometimes divide by N, and sometimes divide by N-1?

The same question arises for the calculation of the sample covariance matrix, and this is what we will work with in this post.

This has to do with whether you want your estimate to be a biased estimate or an unbiased estimate.

For any parameter $$\theta$$, our estimate $$\hat{ \theta }$$ is unbiased if:

$$E\{ \hat{ \theta } \} = 0$$

In this tutorial we will calculate the bias of the sample covariance on the multivariate Gaussian, which is defined as:

$$p(x | \mu, \sigma) = \frac{1}{\sqrt{(2\pi)^D |\Sigma|}} exp( -\frac{1}{2} (x – \mu)^T \Sigma^{-1} (x – \mu) )$$

The maximum likelihood estimates of $$\mu$$ and $$\Sigma$$ can be found by taking the derivative of the log-likelihood and setting it to 0.

The likelihood is:

$$p(X | \mu, \Sigma) = \prod_{n=1}^{N} p(x_n | \mu, \Sigma)$$

So the log-likelihood (expanded) is:

$$L(X | \mu, \Sigma) = -\frac{ND}{2} log(2\pi) -\frac{N}{2} log(|\Sigma|) -\sum_{n=1}^{N} \frac{1}{2}(x_n – \mu)^T \Sigma^{-1} (x_n – \mu)$$

To take “vector derivatives” and “matrix derivatives” you’ll want to consult the Matrix Cookbook.

If you made it this far it is almost trivial to calculate $$\frac{ \partial L }{ \partial \mu } = 0$$ to get the usual result:

$$\hat{ \mu } = \frac{1}{N} \sum_{n=1}^{N} x_n$$

To get the sample covariance, we calculate:

$$\frac{ \partial L }{ \partial \Sigma } = -\frac{N}{2} (\Sigma^{-1})^T – \frac{1}{2} \sum_{n=1}^{N} – \Sigma^{-T} (x_n – \mu) (x_n – \mu)^T \Sigma^{-T}$$

Set that t0 0 and solve for $$\Sigma$$ to get:

$$\hat{ \Sigma } = \frac{1}{N} \sum_{n=1}^{N} (x_n – \hat{\mu}) (x_n – \hat{\mu})^T$$

Note that we are assuming we don’t have the “true mean”, so we are estimating the mean using the maximum likelihood estimate before calculating the maximum likelihood estimate for the covariance.

Now we will show that $$E\{ \hat{ \Sigma } \} \neq 0$$. By definition:

$$E\{ \hat{ \Sigma } \} = \frac{1}{N} E\{ \sum_{n} (x_n – \hat{\mu})(x_n – \hat{\mu})^T \}$$

Expand:

$$E\{ \hat{ \Sigma } \} = \frac{1}{N} E\{ \sum_{n} ((x_n – \mu) – (\hat{\mu} – \mu))((x_n – \mu) – (\hat{\mu} – \mu)))^T \}$$

Expand that:

$$E\{ \hat{ \Sigma } \} = \Sigma + \frac{1}{N} E\{ \sum_{n} (\hat{\mu} – \mu)) (\hat{\mu} – \mu))^T – (x_n – \mu)(\hat{\mu} – \mu))^T – (\hat{\mu} – \mu))(x_n – \mu)^T \}$$

Multiply out the terms:

$$E\{ \hat{ \Sigma } \} = \Sigma + E\{ \hat{\mu}\hat{\mu}^T \} – \frac{1}{N} \sum_{n} E\{ x_n\hat{\mu}^T \} – \frac{1}{N} \sum_{n} E\{ \hat{\mu} x_n^T \} + \mu\mu^T$$

We can combine the expected values to get:

$$E\{ \hat{ \Sigma } \} = \Sigma + \mu\mu^T – E\{ \hat{\mu}\hat{\mu}^T \}$$

Now the exercise becomes finding the expected value.

We need some identities.

First, for $$m \neq n$$:

$$E\{ x_m x_n^T \} = E\{ x_m \} E\{ x_n^T \} = \mu\mu^T$$

Because each sample is IID: independent and identically distributed.

Next, the definition of covariance:

$$\Sigma = E\{ (x – \mu)(x – \mu)^T \} = E\{ xx^T \} – \mu\mu^T$$

We can rearrange this to get:

$$E\{ xx^T \} = \Sigma + \mu\mu^T$$

The term $$E\{\hat{\mu}\hat{\mu}^T \}$$ can be expanded as:

$$E\{\hat{\mu}\hat{\mu}^T \} = \frac{1}{N^2} E\{ (x_1 + x_2 + … + x_N)(x_1 + x_2 + … + x_N)^T \}$$

When expanding the multiplication, there are $$N$$ terms that are the same, so that would be a $$N E\{ x_n x_n^T \}$$ contribution. There are $$N(N-1)$$ terms that are different, and since different terms are independent that is a $$N(N-1)\mu\mu^T$$ contribution.

So in total:

$$E\{\hat{\mu}\hat{\mu}^T \} =\frac{1}{N^2}(N(\Sigma + \mu\mu^T) + N(N-1)\mu\mu^T)$$

Or:

$$E\{\hat{\mu}\hat{\mu}^T \} = \frac{1}{N}\Sigma + \mu\mu^T$$

Plugging this back into the expression for the bias:

$$E\{ \hat{ \Sigma } \} = \Sigma + \mu\mu^T – \frac{1}{N}\Sigma – \mu\mu^T$$

Or:

$$E\{ \hat{ \Sigma } \} = \frac{N-1}{N} \Sigma \neq \Sigma$$

So, if you want the unbiased estimator, you can multiply the biased maximum likelihood estimator by $$\frac{N}{N-1}$$, which gives the expected unbiased formula.

#covariance #maximum likelihood #MLE #multivariate Gaussian #multivariate normal #sample variance #statistics #unbiased estimator

# The Simpsons: Tapped Out and Prisoner’s Dilemma

December 12, 2015

I have to admit that, despite all this science-y stuff I do, I still have my vices.

You know when you’re just sitting there waiting for the train to take you home, and you wish you had a game to play on your iPhone? Then you go home and download that game, fully intending to only play it when you’re bored or waiting? Then you end up sacrificing the time you should have spent doing important things playing that game instead?

That’s The Simpsons: Tapped Out for me (along with new contender Marvel Contest of Champions, but that’s another story).

Every holiday, and even sometimes when it’s not a holiday, The Simpsons has a special holiday update. This year’s Christmas special lets you play a classic game originally known as the Prisoner’s Dilemma.

It is the canonical example of a “game” in Game Theory. John Nash, the subject of the movie A Beautiful Mind, made fundamental contributions to the theory. It has applications in military strategy, politics, and even sports.

## So how does it work?

In Tapped Out, every player has their own town with their own characters and houses, etc. You can visit other towns to get points.

In the Christmas update, when you visit a friend’s town, you drop a present off, and you are asked if you want to make the present “nice” (Lisa), or “naughty” (Bart). Naturally.

When you receive a present, you also have to choose nice/Lisa, or naughty/Bart.

So there are 4 possibilities:

• You choose Bart, other player chooses Lisa
• You choose Bart, other player chooses Bart
• You choose Lisa, other player chooses Lisa
• You choose Lisa, other player chooses Bart

This is a picture of the game that explains how it is played.

So as you can see, the point system is as follows:

• You choose Bart, other player chooses Lisa – You get 25, other player gets 1
• You choose Bart, other player chooses Bart – You get 5, other player gets 5
• You choose Lisa, other player chooses Lisa – You get 15, other player gets 15
• You choose Lisa, other player chooses Bart – You get 1, other player gets 25

In the original prisoner’s dilemma, the problem is posed as follows:

2 prisoners are given a choice of whether to cooperate with or betray the other. If they both cooperate, they both serve 1 year. If they both betray each other, they both serve 2 years. But if one betrays and one cooperates, then the one who betrays goes free, while the one who cooperated gets 3 years.

The problem can be summarized in a table like this:

Where P, R, S, and T are called “payoffs”. To be a prisoner’s dilemma game, we must have T > R > P > S, which is true for The Simpsons: Tapped Out version also.

Once you understand the problem, the question becomes – how do I make a choice that will maximize my winnings?

## How do I maximize my winnings in prisoner’s dilemma?

The Prisoner’s Dilemma is a dilemma because it seems obvious that to maximize both our winnings (altruism), we should cooperate.

But if I know you’re rational and will cooperate with me, and I am selfish, then I will defect, so I win more and you get nothing.

The problem is, if we both think that way, then we will both defect and we will both end up with less.

I would also be hesitant to cooperate because if you are selfish, then you will defect and I will end up with nothing.

The only “rational” solution is to be selfish and betray. Why? Because it is the only choice for which choosing the other thing will result in a worse outcome.

Regardless of what you do – if I choose to betray, it is better than if I had chosen to cooperate.

If you choose cooperate – I choose betray, I get 25 points. I choose cooperate, I only get 15.

If you choose betray – I choose betray, I get 5 points, but if I choose cooperate, I only get 1.

So no matter what you choose, my choice of betray is better than my choice of cooperate.

Of course, since I can drop off multiple presents to my neighbors’ towns, there isn’t only one “game” to play. In fact, multiple or iterated games make the prisoner’s dilemma even more interesting.

## Repeated Games

There is a bigger social element to repeated games.

If you betray me, then now I know you are a betrayer, so I will not cooperate with you.

If you cooperate with me, then perhaps I can trust you, and we can cooperate with each other until the end of the holiday special. That will result in more points for both of us.

But not as many for me compared to what I’d get if I continually betrayed you.

Then again, if I kept betraying you, you would stop cooperating with me.

You see how this gets complex?

## What is the best strategy?

Research done by Robert Axelrod in his 1984 book The Evolution of Cooperation helps to shed light on this issue.

He studied competing strategies in a tournament and discovered that purely selfish strategies performed very poorly in the long run, and that more altruistic strategies performed better, even when measured from a selfish perspective (how many points did I win?).

In terms of natural selection, this may explain why we have evolved over time to be both altruistic and self-interested.

The winning strategy was called “tit for tat”, which in short, means you just play the move your opponent previously played. The tit for tat strategy can be improved upon slightly by adding a small random chance that you will forgive your opponent if they betray you, and cooperate.

Interestingly, Axelrod’s analysis of the top strategies have a very “human” component. The conditions for a successful strategy can be summarized as:

• Be nice. Cooperate unless your opponent betrays you.
• Have a spine. If your opponent betrays you, don’t keep cooperating with them.
• Forgive. Don’t fall into a cycle of repeated betrayals.
• Don’t be jealous. Don’t strive to get more points than the other player and don’t be outcome dependent.

## Limited number of games

In The Simpsons: Tapped Out, you don’t have an infinite number of games to learn from or an infinite number of gifts to give. You are limited to 5 per day. Thus your ultimate strategy will probably be adaptive as well – choose only the 5 players you want to mutually benefit from your gift giving and choose the ones who have a high probability of cooperating.

## Conclusion

Let’s be real though. I don’t think the people who are playing this game are really considering this stuff. For the first few hours I just choose Bart, because that was the naughty one.

So yeah.

#a beautiful mind #game theory #john nash #prisoner's dilemma #tapped out #the simpsons #tsto

# Big Mac Index

July 26, 2014

Source

Food for thought: Economists use the price of a Big Mac to determine if a country’s currency is undervalued or overvalued.

#big mac #economics #gdp #mcdonalds

# Linear Regression

July 25, 2014

Code for this tutorial is here:

http://bit.ly/2IV9BGW

Prerequisites for understanding this material:

• calculus (taking partial derivatives)

Linear regression is one of the simplest machine learning techniques you can use. It is often useful as a baseline relative to more powerful techniques.

To start, we will look at a simple 1-D case.

Like all regressions, we wish to map some input X to some input Y.

ie.

Y = f(X)

With linear regression:

Y = aX + b

Or we can say:

h(X) = aX + b

Where “h” is our “hypothesis”.

You may recall from your high school studies that this is just the equation for a straight line.

When X is 1-D, or when “Y has one explanatory variable”, we call this “simple linear regression”.

When we use linear regression, we are using it to model linear relationships, or what we think may be linear relationships.

As with all supervised machine learning problems, we are given labeled data points:

(X1, Y1), (X2, Y2), (X3, Y3), …, (Xn, Yn)

And we will try to fit the line (aX + b) as best we can to these data points.

This means we have to optimize the parameters “a” and “b”.

How do we do this?

We will define an error function and then find the “a” and “b” that will make the error as small as possible.

You will see that many regression problems work this way.

What is our error function?

We could use the difference between the predicted Y and the actual Y like so:

But if we had equal amounts of errors where Y was bigger than the prediction, and where Y was smaller than the prediction, then the errors would cancel out, even though the absolute difference in errors is large.

Typically in machine learning, the squared error is a good place to start.

Now, whether or not the difference in the actual and predicted output is positive or negative, its contribution to the total error is still positive.

We call this sum the “sum of squared errors”.

Recall that we want to minimize it.

Recall from calculus that to minimize something, you want to take its derivative.

Because there are two parameters, we have to take the derivatives both with respect to a and with respect to b, set them to 0, and solve for a and b.

Luckily, because the error function is a quadratic it increases as (a,b) get further and further away from the minimum.

As an exercise I will let you calculate the derivatives.

You will get 2 equations (the derivatives) and 2 unknowns (a, b). From high school math you should know how to solve this by rearranging the terms.

Note that these equations can be solved analytically. Meaning you can just plug and chug the values of your inputs and get the final value of a and b by blindly using a formula.

Note that this method is also called “ordinary least squares”.

Measuring the error (R-squared)

To determine how well our model fits the data, we need a measure called the “R-square”.

Note that in classification problems, we can simply use the “classification rate”, which is the number of correctly classified inputs divided by the total number of inputs. With the real-valued outputs we have in regression, this is not possible.

Here are the equations we use to predict the R-square.

SS(residual) is the sum of squared error between the actual and predicted output. This is the same as the error we were trying to minimize before!

SS(total) is the sum of squared error between each sample output and the mean of all the sample outputs, i.e. What the residual error would be if we just predicted the average output every time.

So the R-square then, is just how much better our model is compared to predicting the mean each time. If we just predicted the mean each time, the R-square would be 1-1=0. If our model is perfect, then the R-square would be 1-0=1.

Something to think about: If our model performs worse than predicting the mean each time, what would be the R-square value?

Limitations of Linear Regression

• It only models linear equations. You can model higher order polynomials (link to later post) but the model is still linear in its parameters.
• It is sensitive to outliers. Meaning if we have one data point very far away from all the others, it could “pull” the regression line in its direction, away from all the other data points, just to minimize the error.


text:
# Thread: u substitution help -- lots of e^x

1. ## u substitution help -- lots of e^x

hi there,

I need help solving the following problem using U-substitution:

any idea what I should set my u variable equal to?

any help would be greatly appreciated

thanks so much & all the best
- xecutioner

2. ## Re: u substitution help -- lots of e^x

Originally Posted by xecutioner
hi there,

I need help solving the following problem using U-substitution:

any idea what I should set my u variable equal to?

any help would be greatly appreciated

thanks so much & all the best
- xecutioner
Let \displaystyle \begin{align*} u = e^x + 4 \implies du = e^x\,dx \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{\frac{e^x}{\left(e^x + 4\right)^3}\,dx} &= \int{\frac{1}{u^3}\,du} \\ &= \int{u^{-3}\,du} \end{align*}

I'm sure you can go from here

3. ## Re: u substitution help -- lots of e^x

hi there,

your post has helped me significantly, thanks so much!

all the best,

xecutioner


text:
# 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.3

Students can Download Basic Maths Exercise 15.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.3

Part – A

2nd PUC Basic Maths Circles Ex 15.3 One Mark Questions and Answers

1. Show that the circle.

Question 1.
x2 + y2 – 8x + 2y + 16 = 0 touches x= axis.
Given g = -4, f = 1, & C = 16

Question 2.
x2 + y2 – 4x + 4y + 4 = 0 touches x-axis.
Here g = -2, C = 4
Here (-2)2 = 4 ⇒ g2 = C
∴ The circle touches x-axis.

Question 3.
x2 + y2 + 8x + 6y + 9 = 0 touches y = axis.
Here f = 3 & C = 9
& f2 = C ⇒ (3)2 = 9
⇒ Circle touches y-axis.

Question 4.
x2 + y2 – x + 4y + 4 = 0 touches y-axis.
Here f= 2 & C = 4
& f2 = C ⇒ (2)2 = 4
⇒ Circle0 touches y-axis

Question 5.
Show that the circle x2 + y2 + 4x – 3y + 4 = 0 touches x- axis.
Here g = 2 & C = 4
g2 = C ⇒ (2)2 = 4
⇒ Circle touches x-axis.

Question 6.
Show that the circle x2 + y2 – 3x + 8y + 16 = 0 touches y – axis.
Here f = 4, C = 16 & f2 = C
⇒ (4)2 = 16 .
∴ Circle touches y-axis.

Part – B

2nd PUC Basic Maths Circles Ex 15.3 Two Marks Questions and Answers

Question 1.
Find the length of the chord of the circle x2 + y2 – 6x + 4y + 5 = 0 intercepted by x-axis.
Centre of given circle = (3,-2), c = 5
Length of chord intercepted by x – axis = 2$$\sqrt{g^{2}-c}$$ = 2$$\sqrt{9-5}$$ = 2$$\sqrt{4}$$ = 2.2 = 4 units.

Question 2.
Find the length of the chord of the circle x2 + y2 – 6x + 15y – 16 = 0 intercepted by the y-axis.
centre = $$\left(3, \frac{-15}{2}\right)$$ & C = -16
Length of the chord intercepted by y – axis = 2$$\sqrt{f^{2}-c}$$ = 2$$\sqrt{\left(\frac{15}{2}\right)^{2}+16}$$

Question 3.
Find the length of the chord of the circle x2 + y2 + 3x – 2 = 0 intercepted by y-axis.
Centre = $$\left(-\frac{3}{2}, 0\right)$$ c = -2
Length of the chord by y – axis = 2$$\sqrt{f^{2}-c}$$ = 2$$\sqrt{0+2}$$ = 2$$\sqrt{2}$$ units.

Question 4.
Find the length of the chord of the circle x2 + y2 + 3x – y – 6 = 0 intercepted by the y-axis.
Centre = $$\left(-\frac{3}{2}, \frac{1}{2}\right)$$ c = -6
Length of the chord by axis = 2 $$\sqrt{f^{2}-c}$$ = 2$$\sqrt{\frac{1}{4}+6}$$ = 2$$\sqrt{\frac{25}{4}}$$ = 5 units

Question 5.
Find the length of the chord of the circle x2 + y2 + 4x + 6y – 12 = 0 and x + 4y – 6 = 0.

Question 6.
Find the length of the chord intercepted by the circle x2 + y2 – 8x – 6y – 6 = 0 and the line x – 7 y – 8 = 0.

Question 7.
Find the length of the chord intercepted by the circle x2 + y2 – 6x – 2y + 5 = 0 and the line x – y + 1 = 0.

Question 8.
Prove that the length of the chord x + 2y = 5 of the circle x2 + y2 = 9 is 4 units.
Length of the chord intercepted by x – axis = 2$$\sqrt{g^{2}-c}$$
Here g = 5, f = -3, c = 9
∴ Length = $$2 \cdot \sqrt{25-9}=2 \sqrt{16}$$
= 2(4) = 8 units
Hence proved.
Centre of the given circle = (0,0), r = 3 units line = x + 2y – 5 = 0
$$d=\left|\frac{0+2(0)-5}{\sqrt{1^{2}+2^{2}}}\right|=\frac{5}{\sqrt{5}}=\sqrt{5}$$
∴ Length of the chord = $$2 \cdot \sqrt{r^{2}-d^{2}}$$
$$=2 \cdot \sqrt{(3)^{2}-(\sqrt{5})^{2}}=2 \sqrt{9-5}=2.2=4 \text { units. }$$
Hence proved.

Part – C

2nd PUC Basic Maths Circles Ex 15.3 Five Marks Questions and Answers

Question 1.
Show that the line 3x + 4y + 7 = 0 touches the circle x2 + y2 – 4x – 6y – 12 = 0 and find the point of contact.
Given centre = (2, 3) and r = $$\sqrt{4+9+12}$$ = $$\sqrt{25}$$ = 5
If the length of the perpendicular is equal to radius of the circle than the line touches the circle

∴ The line 3x + 4y + 7 touches the circle
Any line perpendicular to 3x + 4y + 7 = 0 is of the form 4x – 3y + k = 0
It passes through (2,3) ∴ 8 – 9 + k = 0 ⇒ k = 1
∴ The line is 4x – 3y + 1 = 0
Point of contact is the intersection of 3x + 4y + 7 = 0 and 4x – 3y + 1 = 0

∴ The point of contact is (-1,-1)

Question 2.
Find k if the line 3x + y + k = 0 touches the circle x2 + y2 – 2x – 4y – 5 = 0.
Given C = (1, 2) C = -5
Radius = Length of the perpendicular from (1, 2) to 3x + y + k = 0.

Question 3.
Find k if the 2x + y + k = 0 touches the circle x2 + y2 + 6x + 2y + 5 = 0.
Hence centre = (-3,-1), c = 5
$$r=\sqrt{g^{2}+f^{2}+c}=\sqrt{9+1-5}=\sqrt{5}$$
r = length of the perpendicular from (-3,-1) to the line 2x + y + k = 0
$$r=\sqrt{g^{2}+f^{2}+c}=\sqrt{9+1-5}=\sqrt{5}$$

Question 4.
Find k if the line 4x – y + k = 0 touches the circle x2 + y2 + 4x – 8y + 3 = 0.
Here centre = (-2,4), r = $$\sqrt{4+16-3}$$ = $$\sqrt{17}$$
∴ $$\sqrt{17}$$ = $$\left|\frac{4(-2)-4+k}{\sqrt{16+1}}\right|$$
-12 + k = ±$$(\sqrt{17})^{2}$$ ⇒ -12 + K = ±17 ⇒ k = 28 or -5
Here centre $$\left(-\frac{3}{2}, 1\right)$$, $$r=\sqrt{\frac{9}{4}+1-2}=\sqrt{\frac{5}{4}}$$
r = length of perpendicular from $$\left(-\frac{3}{2},-1\right)$$ to the line x – 2y + k = 0


text:
# Sniffing Crazyflie’s radio with HackRF blue

While we where in the US we finally received our long-awaited HackRF Blue. Our plan was to use it to sniff the Crazyradio and Crazyflie communication in order to be able to better debug the communication.

HackRF Blue is a lower cost build of the open source HackRF One. It is a Software Defined Radio (SDR), you can think of it as a sound card for radio. It allows to observe and manipulate radio signals from ~1MHz up to 6GHz within a maximum bandwidth of 20MHz. We use it with GNU Radio on the PC which is a signal processing library that contains all we need to do using SDR. Gnuradio has a nice GUI, the Gnu Radio companion, that allows to start testing without having to write code (this GUI actually output a Python program). Getting into SDR is not easy, we have been looking at the Michael Ossmann’s SDR videos (I suggest you look at them if you want to learn about SDR!) and it helps a lot understanding what to do. In this post I will try to briefly explain the step to detect and decode the Crazyradio nRF24 signal. We wrote an howto in the wiki if you want to set up an nRF24 sniffer.

To test the HackRF I just created a very simple python script that sends 10 packets per seconds with Crazyradio:

from crazyradio import Crazyradio
import time

cr.set_channel(26)
cr.set_data_rate(cr.DR_1MPS)

while True:
cr.send_packet((0,1,2,3,4))
time.sleep(0.1)

Then we just tune HackRF to the Crazyradio frequency, and we can see the GFSK signal!

GFSK is a kind of Frequency Modulation. Which means that it should be a cosine wave of constant amplitude. So calulating the magnitude of the complex signal allows to locate data packets by setting the scope trigger:

Now that we can synchronize on a packet, we can add a filter and a quadrature demodulator to demodulate the fm signal and show the data packet (in green):

The preamble (series of 0101010101) is clearly visible followed by the radio address which is 0xe7e7e7e7e7. Now the ‘only’ things left would be to decode the packet. Hopefully for us Cyber Explorer already did the hard work and all we have to do is to send the demodulated data in a unix fifo and send the fifo in the decoder. This procedure is explained in the wiki. As a result we receives the packets:

As a conclusion we found that with the current setup we have a lot of packet lost. We also have a sniffer made out of an nRF51 evaluation kit and it gets much more packets so it is still preferred to analyse protocols. However we can still enhance the SDR algorithm and the 20MHz bandwidth of the HackRF will allow us to sniff on many channels at once, making it perfect to debug channel hopping when we implement it for the Crazyflie.

• Martin says:

Hey, very nice article. I’m trying to sniff data from an XN297 (nRF24 clone), but running into some issues. (It’s a drone).
When I use exactly the same grc setup like you, I just get a lot of random spikes all over the place (the first grc already)! And I didn’t even turn the drone controller on. How come you see the cosine wave so clearly and I don’t?

• arnaud says:

Hi Martin, Do you mean that you haven’t turned ON the transmitter? You need to turn on the controller to have a chance to see the signal, the random spikes are normal this is noise. You also need to make sure that you are close enough to the transmitting frequency (the hackrf has a 20MHz bandwidth but the nRF can transmit in a 125MHz range).

• Martin says:

I also have an nRF24 with Arduino, so I figured I’d first give that a try. I keep sending messages at channel 26 from the nRF, and I get a nice spike in the FFT Sink at 2.4257, but the Scope sink doesn’t show a nice cosine like yours. It does occasionally jump up and down, but that’s just for a millisecond or so. It DOES detect the nRF, but not consistently and with such a cosine :/

• Martin says:

Ah, it looks like I did manage to get a cosine now. Will try the follow-up steps now ;) Thanks so far!

• Martin says:

Alright so I managed to get it to work, but I’m not getting consistent output when running the decoder at all. I’m getting the idea that I’m not receiving the packages correctly for some reason (the cosine isn’t pretty and doesn’t show very often.. mainly noise). Also, how did you know the address is 0xe7e7e7e7e7 just from the cosine? The data the decoder is giving me is like this:
1469092232.277075 NRF24 Packet start sample 2697101, Threshold:4254, Address: 0x49994A299D length:8, pid:1, no_ack:0, CRC:0x9060 data:6D A8 42 AC 92 62 15 C1
1469092233.963550 NRF24 Packet start sample 6067541, Threshold:-2360, Address: 0x69554AAAAD length:22, pid:2, no_ack:0, CRC:0xCCAF data:D6 94 B5 55 5B 2B D6 AB 6C A9 5A B4 AD 6B DA 95 DA B7 55 57 77 FB
1469092235.203272 NRF24 Packet start sample 8550385, Threshold:-857, Address: 0x96D2EAAAB6 length:29, pid:2, no_ack:1, CRC:0x2ADB data:2C DA 38 B7 0D 2E 49 DC B9 AC BA 72 E7 29 69 26 AA BF 4D B9 4C C6 99 6E A7 B4 A6 75 4F
1469092235.669475 NRF24 Packet start sample 9479401, Threshold:37, Address: 0x7B9B595798 length:10, pid:1, no_ack:1, CRC:0xB254 data:11 68 A9 92 CE 18 6C DE 98 82
1469092235.786139 NRF24 Packet start sample 9722024, Threshold:-2432, Address: 0xB97B97BA7B length:22, pid:3, no_ack:1, CRC:0x5F2E data:77 53 A9 AA BA 57 7B 37 33 1D AE 4E 95 E3 F7 5D F7 55 EE D6 F9 77

• arnaud says:

This could be noise, you can see one noisy packet in my screenshot and the address is random.

From the 3rd plot in the article the E7 sequence is visible, E7 E7 is 1110 0111 1110 0111 in binary and this looks like the sequence just after the preample.

If you are sending packet with your Arduino using the 0xe7e7e7e7e7 address you should be able to look at similar packet being demodulated. For testing, it is easier to work with known packet like I did (you can see in my screenshot that I only sent packet containing 00 01 02 03 04).

Do you manage to synchronise on one packet and observe the demodulated signal?

• Martin says:

Just be sure:
– From time 305us to 315us (approx.) is the preamble?
– Package is the part from 315 to 350?

So how are you reading this graph? I find it hard to read 1110 or 0111 from those tiny bumps, but maybe that’s just me :P. Anyway, I changed my address to 0xe7e7e7e7e7 as well and the message to ‘123’. If I stop the plot quickly, I sometimes get a cosine that looks quite similar to yours (like in your 3rd plot). So that’s good. The blue line however (channel 1) stays flat. Not sure if that matters. My Arduino program looks like this: https://codeshare.io/tUIpE
For some reason I get only noise, and not even 1 successful package :/

• Martin says:

Ah I found out how to read the binary. Stupid of me that it took so long >.>. However, I’m still experiencing issues with retrieving the data from the decoder :/

• arnaud says:

Hi, this is a serial communication and so the time is important: the bits are sampled (ie. read) at regular interval. The preamble allows to synchronize the receiver. I made a drawing in inkscape to show the decoding: https://goo.gl/XnHfsC. If you want to play with the file this is the original you can edit: https://goo.gl/lkVYK4.

• Martin says:

Ah, very nice! Thank you so much :). I applied it on my own cosine and it matches up. Another question: Why can we just see the address and not also the message following in the plot?
Also, is there a way to decrease noise so I can better catch the actual signal? Because I still can’t seem to demodulate it and retrieve the message. Could it have to do with the fact that I have an nRF24L01+ instead of a regular one?

• Martin says:

Ah, scratch the question about the ‘message’ – because it IS there haha.

• Martin says:

Alright so just out of curiousity, I’d like to decode the package message manually (also because I don’t really understand the nRF24 decoder script yet). The message after the address is as followed (https://drive.google.com/open?id=0B9CJ42CGPiF2YnhuX0hWMG1fVUk):
CE 26 80 30 80 39 00 3..

The ‘message’ I sent was “Martijn”, which is indeed 7 bytes long. (last 2 bytes are CRC I guess). How would I go about decoding this message?

• Martin says:

Ahhh, apparently the address is followed by one extra bit. I’ve now succesfully identified ‘martijn’ as the message, yay! https://drive.google.com/open?id=0B9CJ42CGPiF2TWoyelRmWldZcU0
My remaining questions are:
– What are the ‘CC’ and the highlighted ‘0’ after the address?
– I still can’t get the live decoder to work >.< Do you have any suggestions left for me to try out?

• arnaud says:

Hi, Great job so far decoding in the dark :-). I assume you have not looked at the Nordic nRF24L01 product specification? On page 28 the packet format is documented and there is a 9 bit control field (the CC plus one bit). This contains the payload length (6 bits) the packet ID (2 bits, is incremented for each packet) and the no_ack flag (1 bit) that is set to 1 if no acknowledge is requested.

One reason for the decoder not to work might be that you do not seems to be exactly at the right frequency: the signal is never crossing 0 while on my graph it was right around 0. For this graph the value of the signal actually corresponds to a frequency difference from the channel frequency. Since your signal is always positive it seems that you are not tuned exactly on the right channel but one up or one down.

• Martin says:

Thanks! I now understand the packet format :).
About the decoder: I’ve tuned the frequency to better line up with the 0, but I’m still not even receiving one package :/
I set the address to 0xe7e7e7e7e7 on my nRF but as you can see, it doesn’t pop up at all (while I DO see packages coming by in the plot!):

For the decoder I’m using the GRC setup from the wiki (https://wiki.bitcraze.io/_media/misc:hacks:grc_nrf24_demod.png), but I’ve also tried playing with the Quadrature Demod gain, sample rates, etc. – no result.

Could it be that I should tweak the decoder in some way? For example, I think I’m not using any CRC.
My arduino code atm: https://codeshare.io/ptMHb

• Martin says:

Could you maybe show me how I could binary slice the data and put that into a file, instead of the fifo way?
I can’t find any good documentation on how to use the binary slicer. The file gets huge once I try..

• arnaud says:

I am not sure what you mean by binary slicer? If you want to send the data in a file instead of the fifo you can just change the filename in the file sink block. I have not written the nrf24 decoder so I am not sure how it works and why it is not working for you.

• ddvv says:

HI,arnaud
Could you explain how the ‘quadrature_demod’ can be used to demodulate the fsk signal?
I search a long time in google, but there is any useful information about it.
Can you give me a hand?

• Alexandre says:

I’m having the same issues as mentioned by Martin – completely random addresses in the decoded NRF24 data.

• Juan M. Castro says:

Hi, I have read your article and it is very interesting. I want write a python script to detect a drone syma X5SC-1 (NRF24L01) with HackRF. I don’t need decode information, only produce a warning on screen like “drone detected”. Can you help me with de python script? If it is possible send me information about this. Thanks.


text:
# Thread: Is derived set closed in any topological space?

1. Maybe.
Hope I can find a counterexample that A' is not closed in the future. Thank you Plato, for your patient replies.

2. ## A subset of a T0 space whose derived set is not closed

Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.

3. A beautiful counterexample! Thank you for your brilliant first-post.

4. Originally Posted by austinmohr
Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.
Ack - a typo. The set $\displaystyle (0, \infty)$ is neither open nor closed in this topology. The counterexample still works, however.

5. Originally Posted by austinmohr
Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.
Originally Posted by austinmohr
Ack - a typo. The set $\displaystyle (0, \infty)$ is neither open nor closed in this topology. The counterexample still works, however.
If this is Lynn Steen’s Left Order Topology, then isn’t it true that $\displaystyle 0$ is a limit point of $\displaystyle \{0\}?$
If so, then is $\displaystyle 0$ in the derived set of $\displaystyle \{0\}?$
If that is so, what is its complement?

6. Originally Posted by Plato
If this is Lynn Steen’s Left Order Topology, then isn’t it true that $\displaystyle 0$ is a limit point of $\displaystyle \{0\}?$
If so, then is $\displaystyle 0$ in the derived set of $\displaystyle \{0\}?$
If that is so, what is its complement?
We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set $\displaystyle A$. Hence, if $\displaystyle A$ is singleton, it cannot be contained in its own derived set by definition.

7. Originally Posted by austinmohr
We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set $\displaystyle A$. Hence, if $\displaystyle A$ is singleton, it cannot be contained in its own derived set by definition.
You are correct. I was not thinking of a singleton set.
In fact, I don't know what I had in mind.
Maybe $\displaystyle ( - \infty ,0]$ I really don't know.

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# Conductors and Insulators Definition and Examples

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Conductors carry electric current. Insulators protect conductors and protect people from conductors. Whether a material is a conductor or an insulator depends on its atoms and on the relationship of each atom to its surrounding atoms.

Insulators may break down if subjected to excessive voltages. Similarly, conductors may be destroyed if too much current is passed through them. The best conductor materials have the lowest resistivity, that is, resistance per cubic meter. The best insulator materials have the highest breakdown voltage for a given thickness.

The resistance of any conductor can be calculated from a knowledge of its cross-sectional area, its length, and the resistivity of the material. Once the resistance is known, the voltage drop along a conductor and the power dissipated in it can be determined for any given current.

## Insulators

Figure 1 shows some typical arrangements of conductors and insulators. Electric cable usually consists of conducting copper wire surrounded by an insulating sheath of rubber or plastic. Sometimes there is more than one conductor, and these are, Of course, individually insulated.

Fig. 1: Conductors and Insulators

Conductors employed for industrial and domestic purposes normally have stranded copper wires with rubber or plastic insulation. In electronic equipment, flat cables of fine wires and thin printed circuit conductors are widely used.

Even in the best of insulating materials, there are some free electrons drifting about between the atoms. Therefore, a very small electric current can flow through an insulator. In normal circumstances, this current is so small that it is absolutely negligible. A current can also flow along the surface of an insulator, especially if the insulator is dirty or wet. This is particularly evident with the porcelain insulators on high-voltage overhead cables during a rainstorm, when arcing occurs over the insulator surface.

If a sufficiently high potential difference is applied across an insulator, electrons can be pulled out of the atoms, and substantial current flow may occur. This is known as an insulator breakdown, and it normally results in the destruction of the insulator. To help avoid breakdown, all insulating materials are rated according to the maximum voltage that may be safely applied. Table 1 lists typical breakdown voltages (or dielectric strengths) for various insulating materials. It is seen that the air has the lowest breakdown voltage, at 30 kV/cm, and mica has the highest, at 2000 kV/cm.

The electric field strength is a measure of the electric stress set up in an insulator. The field strength should always be lower than that which might cause insulator breakdown. The field strength is simply the applied voltage divided by the insulator thickness.

$\begin{matrix} \text{electricfield strength,}\xi \text{=}\frac{E}{d} & {} & \left( 1 \right) \\\end{matrix}$

Where E is in volts and d is in meters, the unit of field strength is volts/meter (V/m). The situation is illustrated in figure 2 and further explored in the following example.

Table.1: Typical breakdown voltages of some insulating materials

 Material Breakdown voltage (kV/cm) Air 30 Porcelain 70 Rubber 270 Glass 1200 Mica 2000

Fig.2 (a): Two Conductor Cable

Fig.2 (b): Cable Cross-Section

Example

Two electric conductors encased in rubber are exactly 0.25 cm apart.

(a). Calculate the electric field strength in the insulation when the potential difference between the conductors is 25 V.

(b). Determine the minimum potential difference between conductors that may cause insulator breakdown.

Solution

From equation (1)

$\text{electricfield strength,}\xi \text{=}\frac{E}{d}=\frac{25}{0.25}=100V/cm$

From table 1, insulation breakdown for rubber occurs at 270 kV/cm.

$E\text{=}d\times \text{breakdown voltage=0}\text{.25*270=67}\text{.5kV}$

## Conductors

The function of a conductor is to conduct current from one point to another in an electric circuit. Electric cables usually consist of copper conductors sheathed with rubber or plastic insulating material. Cables that have to carry large currents must have relatively thick conductors. Where very small currents are involved, the conductors may be a thin strip of copper or even an aluminum foil. Between these two extremes, a wide range of conductors exists for various applications. Three different types of cables used in electronics equipment are illustrated in Figure 3. The one identiﬁed as a coaxial cable has a central insulated conductor and a circular plaited conducting screen, as well as an outer insulating sheath. The other two are multi-conductor cables, one circular, and one ﬂat.

Fig.3: Different types of cables

Many different types of cables are used with electronics equipment.

Because each conductor has a ﬁnite resistance, a current passing through it causes a voltage drop from one end of the conductor to the other (Figure 4). When conductors are long and/or carry large currents, the conductor voltage drop may cause unsatisfactory performance of the equipment supplied. Power (I2R) is also dissipated in every current-carrying conductor, and this is, of course, wasted power.

Fig.4 (a): Current Flow through a conductor produces a voltage drop

Fig.4 (b): Conductor resistance causes voltage drop

Conductor resistance (R) is determined by applying the voltage drop and current level to Ohm’s law. The resistance per unit length (R/l) is then used to select a suitable wire gauge.

In extreme cases, the power dissipated in the conductors may generate sufﬁcient heat to destroy the insulation or even melt the conductor. Where the resistance per unit length of a conductor is known, the conductor voltage drop and power dissipation are easily calculated.

To select a conductor for a given application, its resistance per unit length is first determined from the current level and the acceptable voltage drop along the cable. The table is then used to find a wire gauge with the calculated resistance per kilometer (R/km) or with the next lower R/km value.

## Conductor Resistivity

The resistance per unit length of a given conductor is not always available. Therefore, to allow for all possible combinations of cable length and cross-sectional area. The specific resistance of the conducting material is employed.

The specific resistance (or resistivity) of a material is the resistance of 1-meter cube of the material.

Obviously, as some metals are better conductors than others, each has its own specific resistance. Table 2 lists the specific resistances of venous metals.

The symbol used for specific resistance is ρ (Greek lowercase letter rho), and the units of p are ohm meters (Ωm).

Consider the meter cube of copper illustrated in Figure 5 (a). From Table 2, the speciﬁc resistance of copper is p =1.72 X 10-8 Ωm.

Fig.5: Block Resistances

If a 1-meter cube of copper has a resistance of ρ Ω, three cubes in series will have a resistance of 3 ρ Ω, and the resistance of two in parallel will be ρ/2 Ω.

If the length of copper is doubled or tripled by placing blocks side by side as shown in Figure 5(b), the resistance is increased. (Blocks or other components, connected together in this way are said to be connected in series. The resistance of two series-connected blocks is 2p Ω and that of three is 3p Ω, and so on. From this, it is derived that the total resistance of any length of copper with a cross-sectional area of 1 m2 is,

\begin{align} & R=\rho \times (\text{length in meters}) \\ & =1.72*{{10}^{-8}}\times (\text{length in meters) }\Omega \\\end{align}

Now consider the effect of doubling the cross-sectional area of the copper by placing two blocks as illustrated in Figure 5 (c).  (Blocks or other components, connected together in this way are said to be connected in parallel. Because the cross-sectional area is doubled, the resistance is halved. If four blocks are parallel-connected [Figure 5 (d)] the resistance is quartered. In the case of blocks connected in parallel, the resistance can be written as,

$R=\frac{\rho }{\text{cross-sectional area in }{{\text{m}}^{\text{2}}}}$

Table.2: Specific resistance for electrical conducting metals

 Metal Specific resistance at 20 oC (Ωm) Silver 1.64*10-8 Copper 1.72*10-8 Gold 2.45*10-8 Aluminum 2.83*10-8 Tungsten 5.5*10-8 Nickel 7.8*10-8

When blocks are put in series and parallel [Figure 5(e)], the two previous equations are combined to give

$\begin{matrix} R=\rho \frac{l}{a} & {} & \left( 2 \right) \\\end{matrix}$

Where R is the resistance in ohms, p is the specific resistance of the material, l is the length in meters, and a is the cross-sectional area in m2.

Cylindrical shaped conductors with different dimensions are illustrated in Figure 6. It is seen that the resistance of a conductor with length l1 and cross-section a1 is calculated as

${{R}_{1}}=\rho \frac{{{l}_{1}}}{{{a}_{1}}}$

Fig.6: Conductors Resistance for different lengths

Conductor resistance is determined by the length, cross-sectional area, and the specific resistance of the conductor material.

A similar conductor with the same cross-sectional area and twice the length has a resistance of 2R1. Also, a conductor with the same length as the ﬁrst but a cross-sectional area of 2a has a resistance of R1/2.

The resistance of any length of a conductor of any cross-sectional area can be calculated from equation (2). Where R, l, and a are known, the equation can be rewritten to determine the specific resistance:

Example

Determine the resistance of 20 m of annealed copper wire 2 mm in diameter.

Solution

Cross sectional area:

$a=\frac{\pi {{D}^{2}}}{4}=\frac{\pi {{\left( 2\times {{10}^{-3}} \right)}^{2}}}{4}=\pi \times {{10}^{-6}}{{m}^{2}}$

From equation (2)

$R=\rho \frac{l}{a}$

From table 2, for Copper

$\rho =1.72*{{10}^{-8}}$

Therefore,

$R=1.72\times {{10}^{-8}}\times \frac{20}{\pi \times {{10}^{-6}}}=0.109\Omega$

The resistance of any conductor can be calculated from a knowledge of its cross-sectional area, its length, and the resistivity of the material. Once the resistance is known, the voltage drop along a conductor and the power dissipated in it can be determined for any given current.

## Temperature Effects on Conductors

The values of specific resistance stated in Table 2 refer only to conducting materials at a temperature of 20oC. Thus, in all resistance calculations using the specific resistance, the conductor temperature is assumed to remain constant at 20oC. Because the resistance of metals changes with temperature, it is necessary to be able to calculate the resistance values at higher or lower temperature levels.

The resistance of all pure metals tends to increase as the temperature of the metal rises. This may be explained in terms of the atoms actually vibrating at elevated temperatures and, thus, becoming greater obstructions in the path of moving electrons. Because the resistance increases with increasing temperatures, metals are said to have a positive temperature coefficient (PTC). Some materials, notably semiconductors, exhibit a decrease in resistance as their temperature rises. These have a negative temperature coefficient (NTC). Over the normal range of operating temperatures, all metals exhibit a nearly linear relationship between resistance and temperature.

In Figure 7(a) the variation in resistance of copper is plotted versus temperature. Note that the resistance of copper appears to go to zero at a temperature of -234.5oC. The actual graph is not as completely linear as illustrated, however, the straight line passing through -234.5oC is a projection of the linear portion of the graph. If the resistance R1 at a temperature of 20oC is known, the new value of resistance at any other temperature can be calculated.

Fig.7 (a): Resistance of Copper plotted versus Temperature

Suppose that R1 is 1 Ω at 20oC. If the temperature is reduced to -234.5oC, the temperature change is

$\Delta T={{20}^{o}}C-\left( -{{234.5}^{o}}C \right)={{254.5}^{o}}C$

And the resistance change per degree of temperature change is

$\frac{\Delta {{R}_{1}}}{\Delta T}=\frac{1}{254.5}=0.00393{\Omega }/{^{o}C}\;$

If R1 were 10 Ω (instead of 1 Ω) at 20oC, the resistance change with temperature would be

$\frac{\Delta {{R}_{1}}}{\Delta T}=\frac{10}{254.5}=0.0393{\Omega }/{^{o}C}\;$

The quantity 0.00393 Ω/oC is the temperature coefficient for copper at 20oC.

The symbol α or (Greek lowercase letter alpha) is used for temperature coefficient. Table 3 lists the temperature coefficients for various metals at 20oC. Note that the alloy constantan has an extremely small temperature coefficient. This characteristic makes constantan useful in applications where resistance is required to remain as constant as possible when the temperature changes. Temperature coefficients can also be determined at temperatures other than 20oC. Column 3 in table 3 lists temperature coefficients at 0oC.

Table.3: Temperature coefficients for metals

 Material α at 20oC α at 0oC Silver 0.003 8 0.004 12 Copper 0.003 93 0.004 26 Gold 0.003 4 0.004 65 Aluminum 0.003 9 0.004 24 Tungsten 0.004 5 0.004 95 Nickel 0.006 Constantan 0.000 008

Fig. 7 (b): The resistance (R2) at a new temperature (T2) can be determined from a knowledge of R1 and T1 and the temperature coefficient α

As explained, for any conducting material having a resistance R1,

$\frac{\Delta {{R}_{1}}}{\Delta T}=\alpha {{R}_{1}}$

Or

$\begin{matrix} \Delta {{R}_{1}}=\alpha {{R}_{1}}\Delta T\text{ } & \cdots & [\text{ see Figure 7 (a) and (b) }] \\\end{matrix}$

And resistance R2 at a new temperature T2 is

$\begin{matrix} {{R}_{2}}={{R}_{1}}+\Delta {{R}_{1}}={{R}_{1}}+\alpha {{R}_{1}}\Delta T\text{ } & {} & [\text{ see Figure 7 (b) }] \\\end{matrix}$

Therefore,

$\begin{matrix} {{R}_{2}}={{R}_{1}}\left( 1+\alpha \Delta T \right) & {} & \left( 3 \right) \\\end{matrix}$

Where R1 is the resistance at 20oC, α is the temperature coefficient of the material, and ∆T is the temperature change from 20oC.

If R1 and R2 are known, equation (3) can be rearranged to determine ∆T or α:

$\Delta T=\frac{1}{\alpha }\left( \frac{{{R}_{2}}}{{{R}_{1}}}-1 \right)$

$\alpha =\frac{1}{\Delta T}\left( \frac{{{R}_{2}}}{{{R}_{1}}}-1 \right)$

The temperature coefficient at 0oC may also be used in equation (3) if the resistance R1 is measured at 0oC and ∆T is the temperature increase from 0oC.

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# Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?

Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?

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What have you tried in order to prove this? –  Mariano Suárez-Alvarez Nov 2 '10 at 3:58
the derivative operation commutes over the integral (of convolution) but i get two different scenarios when i change the order of convolution, i.e fmoll or mollf ? –  Rajesh D Nov 2 '10 at 4:04
Yes, this is actually a very important property of mollifiers (maybe the most important?). Do you know what an approximate identity is? You can approximate your function with the mollified version which often has much nicer properties. –  Jonas Teuwen Nov 2 '10 at 13:29

Yes.

The key is that when you (as you say in the comments) get the two scenarios:

$$f \star (D g) = D(f \star g) = (D f) \star g$$

then you get to choose which!

So if $D f$ doesn't make sense, then you can ignore it and choose to use the identity $D(f \star g) = f \star (D g)$.

Taking this to the extreme, you get the - bizarre, in my opinion - result that if $p$ is a polynomial, then $f \star p$ is always a polynomial.

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Why do we get to choose ? It isnt part of a proof. Looks like its some sort of an axiom or something ! BTW convolution is not always commutative according to statements in your answer. –  Rajesh D Nov 2 '10 at 9:49
@Andrew a polynomial is neither compact nor in L^2.I guess you mean a polynomial multiplied with some bump function. –  Rajesh D Nov 2 '10 at 10:18
@Rajesh: I don't understand that comment. There is no notation in the question (other than "f") and there's not a lot of notation in my answer either, so I'm not sure what you mean by "not in sync". –  Loop Space Nov 2 '10 at 20:35
@Rajesh: Ah, I see. When I write "If $f$ ..." then I am implicitly making $f$ a local variable whose properties and so forth are only set for the scope of that statement. So the $f$s can all be different functions with different properties. I suggest that at this point you look in an introductory book on Fourier Theory. –  Loop Space Nov 3 '10 at 9:48
@Rajesh, all Andrew is saying is that for any f,g in certain spaces, $D(f\star g)$ is equal to $Df\star g$ OR $f\star Dg$, provided either of the latter expressions are well-defined, i.e. the derivatives exist and the integrals defining the convolution are convergent. For instance, in your scenario, since f is continuous and g is smooth with compact support, we have $D(f\star g)$ = $f\star Dg$. However, if f was $C^1$, we also would have $D(f\star g) = Df\star g$. This is proved by using the Lebesgue Dominated Convergence Theorem to move the derivative inside the integral. –  Greg O. Nov 7 '10 at 16:50