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# simplify indices

Three Times Table Multiplication PowerPoint x3. Use the division law which tells you to subtract the indices. So 25 ^ 1/2 = √25 = 5. You can & download or print using the browser document reader options. Simplifying expressions using the laws of indices, Working with appreciation and depreciation, Religious, moral and philosophical studies. To download/print, click on pop-out icon or print icon to worksheet to print or download. So 25 ^ 1/2 = √25 = 5, Ha ha the guy above me put in an answer of 1 then went back and edited it - copied my answer. Sorry about the typo before, I am new to notation on the computer. An exponent of 1/2 means square root. Some of the worksheets for this concept are Indices and surds, More properties of exponents, Indices or powers, Laws of indices work, Mathematics linear 1ma0 algebra indices, 3 indices and standard form mep y9 practice book a, Surds and indices, Maths learning service revision index laws mathematics.

docx, 38 KB. A differentiated lesson on using the laws of indices.

Also, the fact that it explains how to use the rules. 25^1/2 … Using the Laws of Indices to Simplify Expressions. Get your answers by asking now. Still have questions? Use the multiplication law, add the numerator indices. And how do you do them? Unsere Kompetenz wird Ihr Erfolg. Ihr Bedarf, Ihre Anwender und Ihre Geschäftsprozesse sind die Basis unserer Lösungen. Preview and details Files included (3) docx, 26 KB. Use the division law, subtract the indices. what's the answer. Preview. Loading... Save for later.
I don't have much time, so please be as fast as possible, in half an hour I have to leave (it is due tomorrow morning!!! How to simplify your expression To simplify your expression using the Simplify Calculator, type in your expression like 2(5x+4)-3x. Join Yahoo Answers and get 100 points today. London WC1R 4HQ. Read about our approach to external linking. This website and its content is subject to our Terms and Try the Free Math Solver or Scroll down to Tutorials! Hierin haben wir Erfahrung seit 1999. 5 8 customer reviews. William Marks, OH, This version is 1000 times better then the last. Found worksheet you are looking for? Chemical Reactions Identify Reactants And Products. Square registered in England (Company No 02017289) with its registered office at 26 Red Lion

Conditions.

In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. Thanks. Answer Save. I love it! by simply. Favorite Answer. What is the radius of one kg platinum ball? Some of the worksheets for this concept are Indices and surds, More properties of exponents, Indices or powers, Laws of indices work, Mathematics linear 1ma0 algebra indices, 3 indices and standard form mep y9 practice book a, Surds and indices, Maths learning service revision index laws mathematics. So 25 ^ 1/2 = √25 = 5. We have a large amount of quality reference materials on subjects ranging from college mathematics to matrix algebra

True  False? Simplify Indices - Displaying top 8 worksheets found for this concept..

Remember these rules when you're working with exponents. I'm able to go into my class and follow along with the teacher; it's amazing! Simplify Indices - Displaying top 8 worksheets found for this concept. If ever you actually have service with algebra and in particular with online indices solver with step by step solution or basic concepts of mathematics come visit us at Pocketmath.net. An exponent of 1/2 means square root. Indices-Exit-Ticket . Simplify $${y^7} \times {y^3} \times {y^5}$$. Read more. Author: Created by l.orme. A negative exponent means do the reciprocal of what it would be with a positive exponent. Relevance? The average length of a smallmouth bass in Georgian Bay is 24.2 cm with a standard deviation of 2.3 cm. Some terms with indices can be simplified using the laws of indices. So 25 ^ -1/2 = 1/√25 = 1/5. I know coin flips are always 50/50 odds, but what is the scientific statistical term for being 'overdue' for a heads after 5 tails in a row? The "3" in the radical above is called the "index" of the radical (the plural being "indices", pronounced "INN-duh-seez"); ... To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square.

The calculator works for both numbers and expressions containing variables. 1/3 would be cube root, etc. Für eine benutzerfreundliche Bedienung von Geräten bieten graphische Benutzeroberflächen vielseitige Möglichkeiten. Diese Möglichkeiten für unsere Kunden einfach nutzbar zu machen, ist unser Ziel: - "simplify technologies" ! For instance: Simplify a 6 × a 5; The rules tell me to add the exponents. Our team of exam survivors will get you started and keep you going. Dazu bereichern technische Neuerungen ständig unseren Erfahrungsschatz. A differentiated lesson on using the laws of indices. Simplify Expression Calculator.

simplify x2 + 4x − 45 x2 + x − 30 simplify x2 + 14x + 49 49 − x2 simplify 6 x − 1 − 3 x + 1 simplify 5x 6 + 3x 2 Use the multiplication law. The simplify calculator will then show you the steps to help you learn how to simplify your algebraic expression on your own. It is often simpler to work directly from the definition and meaning of exponents. Show Instructions. The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. Created: Oct 20, 2015. Math Question? Enter the expression you want to simplify into the editor. S.D., Oregon. Partial Derivative help? How to Simplify indices ? Free. In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. Hi, what is 25^1/2 and 25 ^ -1/2? 1/3 would be cube root, etc.

Focus groups in Asian societies tend to have fewer number of respondents than those in Western societies This tells you to add the indices. Tes Global Ltd is Hi, I wanted to comment on your software, not only is it really easy to follow, but I can hide the step by step procedures, or choose to show them. Radio 4 podcast showing maths is the driving force behind modern science. 4 Answers. You are most welcome. Worksheet will open in a new window. Produkte und Dienstleistungen für graphische LCD- und Touch-Displays und für die Entwicklung von Embedded Systemen. Laws-of-Indices-Differentiated-Questions. Tabitha Wright, MN, I liked the detailed, clearly explained step by step process that Algebrator uses. notebook, 69 KB. An exponent of 1/2 means square root. Indices show where a number has been multiplied by itself, eg squared or cubed, or to show roots of numbers, eg square root. Is there a value for x when the given triangle is equilateral?


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Castle tower

The castle tower has a cone-shaped roof with a diameter of 10 meters and a height of 8 meters. Calculate how much m² of coverage is needed to cover it if we must add one-third for the overlap.

Correct result:

S =  197.585 m2

Solution:

$D=10 \ \text{m} \ \\ h=8 \ \text{m} \ \\ \ \\ r=D/2=10/2=5 \ \text{m} \ \\ \ \\ s=\sqrt{ h^2 + r^2 }=\sqrt{ 8^2 + 5^2 } \doteq \sqrt{ 89 } \ \text{m} \doteq 9.434 \ \text{m} \ \\ \ \\ S_{1}=\pi \cdot \ r \cdot \ s=3.1416 \cdot \ 5 \cdot \ 9.434 \doteq 148.1886 \ \text{m}^2 \ \\ \ \\ S=(1+\dfrac{ 1 }{ 3 } ) \cdot \ S_{1}=(1+\dfrac{ 1 }{ 3 } ) \cdot \ 148.1886=197.585 \ \text{m}^2$

We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!

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## Thursday, March 18, 2010

### Narrow Phase: Sweeping a Sphere Against a Polygon

Before sweeping a sphere against a polygon, you should determine if the sphere is already intersecting the polygon. I will discuss the case where the sphere is not initially intersecting the polygon.

We will use a 2D representation of the hypothetical setup, since it's easier to draw 2D pictures.

In this setup, there is a sphere with radius r located at point c, a relative velocity vector v that represents the difference of velocity between the sphere and the polygon. This relative velocity can be considered as the velocity of the sphere from the point of view of the polygon. The velocity is given in units that represent how much the sphere will move over the course of a given frame. Thus, at the end of the frame, the center of the sphere should be positioned at the tip of the velocity vector. The polygon lies entirely within some plane, and we are projecting the setup in such a way that the plane occupies a single line. The polygon itself is the bold section of the line. The normal n of the plane is pointing out of the front side of the polygon.

Before determining collision with the polygon, we will first determine collision with the plane. The first step is to find the point on the sphere that will first come into contact with the plane. This can be done scaling the normal of the plane by the radius of the sphere, and subtracting this from the center of the sphere.

$\mathbf{p}_1=\mathbf{c}-\mathbf{n}r$

We will sweep the point p1 forward. The sweep is accomplished by finding a line that passes through p1, and is parallel to v. We then find the point where the line intersects the plane. The point on the plane is the point where the sphere will first make contact with the plane.

The point p2 can be calculated using the following expression

$s=d-\mathbf{p}_1\cdot\mathbf{n}$
$\mathbf{p}_2=\mathbf{p}_1+\mathbf{v}s$

Here d is the plane parameter, and can be found by dotting the plane normal with any point that resides on the plane, such as one of the vertices of the polygon.

Before calculating p2, check if s is less than 0, or greater than 1. If it is then the collision is not going to happen during this frame, so you can return a false on your collision routine.

If s is between 0 and 1, then go ahead and calculate p2 and continue with the operation.

The next step is to find a point inside the polygon that is nearest to p2. The method for doing this is dependant on the polygon. If p2 already resides within the polygon, then the nearest point to p2 is just p2. Otherwise, the nearest point lies on the boundary of the polygon.

$\mathbf{p}_3={\rm nearestPoint}\left(\mathbf{p}_2\right)$

NOTE: Finding the nearest point on the polygon to p2 does not actually work in 3D, since it is possible to have components of velocity that are tangent to the polygon plane. To correct for this, you should try to find the point on the polygon that is closest to the ray that is cast from p1. This is done by finding which feature of the polygon (vertex, edge, or face) is closest to the line, and then finding the closest point on this feature to the line.

The point p3 is the point on the polygon that will first be touched by the sphere.

Now we will sweep the point p3 backward, and see if it touches the sphere. This sweep is performed by finding a line that passes throw p3 and is parallel with v. This line will intersect the sphere in 0, 1, or 2 places.

The formula for this sweep involves a square root. If the stuff we are trying to take the square root of is negative, then there is no intersection. So, we first check to see if this stuff is negative.

$\Delta =\left[\left(\mathbf{p}_3-\mathbf{c}\right)\cdot\mathbf{v}\right]^2-4\left(\mathbf{v}\cdot\mathbf{v}\right)\left[\left(\mathbf{p}_3+\mathbf{c}\right)\cdot\left(\mathbf{p}_3-\mathbf{c}\right)-r^2\right]$

If this is negative, then there is no collision and we can return false. Otherwise, we will continue to calculate the fraction of the time step t at which the collision takes place

$t=\frac{\sqrt{\Delta}-2\left(\mathbf{p}_3-\mathbf{c}\right)\cdot\mathbf{v}}{2\mathbf{v}\cdot\mathbf{v}}$

This quantity should not be less than 0. It may however be greater than 1. If it is, then there will be a collision, but not during this time interval, so return false. If however, t is less than 1, you can proceed to determine the point p4 that represents the point that the sphere will first come into contact with the polygon.

$\mathbf{p}_4=\mathbf{p}_3-\mathbf{v}t$

At this point, we can move the sphere forward so that it just touches the polygon. The task now is to determine how to resolve this collision, by adjusting the velocity of the sphere. Once we have determined the new, resolved, velocity we continue to sweep the sphere for the remainder of the time step - checking for collisions with other polygons along the way.

## Tuesday, March 9, 2010

### Collision Overview

Collision detection is one of the very key components of a physics engine. However, collision detection on its own is a much more broad topic. It is important to know what kinds of information your collision detection code should be providing, and the different means of accessing this information.

Generally, the primary thing we want the collision system to do is notify us if two objects are intersecting. As a secondary part of this computation, we often would like to know how to keep the objects from intersecting.

There are a couple of methods I have used, when writing collision systems, to deliver this information.

The first is a direct query method, which has the possible functional prototype

bool CollisionSystem::CollideObjects(const PhysObj& A, const PhysObj& B, ColResData* resolve = NULL);

Such a function would return true if the two objects were intersecting. We may optionally pass in a data structure to retrieve information about how to resolve the collision.

The second method is a bit more automatic. All collision information is passed to the user through callback functions. The collision system accumulates all motion of all objects, and automatically determines which objects need collision processing. If a collision event occurs, a callback function is called.

typedef void (*CollisionHandler)(PhysObj&, PhysObj&, ColResData*);
void collisionHandler(PhysObj& A, PhysObj& B, ColResData* resolve);

The physics objects are not passed in as const because the collision handler may change the state of these objects in order to resolve the collision.

We may want to have several different collision event handlers. For instance, we may want to handle an event which corresponds to objects touching, or an event which corresponds to the moment that they cease to touch. There are several different objects in the world, and we may want a different handler for different object types. Therefore, when we proved the collision system with collision handlers, we must supply the object types, and event types for which the collision handler applies

bool CollisionSystem::AddHandler(CollisionHandler handler, int AType, int BType, ColEventType event);

This function passes in a collision handler, and an int for the types of objects A and B. An int is used so that the user may map these types to any meaning desired. If the function returns false, it was not able to map the collision handler, probably because there is already a handler mapped to the given types and event. The ColEventType is an enum that may contains events like

enum ColEventType
{
ColEvt_OnNear,
ColEvt_Near,
ColEvt_OnIntersect,
ColEvt_Intersect,
ColEvt_Resolve,
ColEvt_OnSeperation,
ColEvt_OnFar
};

We may want to add more collision events, but these cover many of the bases. We want to know if two objects have just gotten near to each other, if they are near to each other, if they have just begun to intersect, if they are intersecting, if we want to resolve the collision, if they have just separated, and if they have just moved away from each other.

The way that we actually process collisions is broken up into three distinct portions
1. Broadphase - rough estimation if objects are even close to each other
2. Midphase - determines which parts of a complex object might be colliding
3. Narrowphase - performs collision on convex primitives, and calculates resolution information

In the following posts, I will discuss each of these topics.

## Thursday, March 4, 2010

### Quaternion Tricks

Trick #1 : Optimizing the Quaternion Rotation Formula

The quaternion rotation formula is given as

(0, v') = R(0, v)R

This formula can be expressed in terms of standard 3D vector products such as scaling, dot products, and cross products. This equation involves two quaternion products, each of which requires 16 multiplies and 12 additions. Thus there are 56 operations in all.

We can express the vector cross product in terms of the quaternion product. We know that if you switch the factors in a cross product, the product switches sign. Knowing this, you might accept without proof the following formula.

qV × pV = (1/2) (qp - pq)

Indeed if you flip the order of the arguments, the product will change sign.

We will use this relationship, in a rearranged form, on the first two factors in the quaternion rotation formula.

R(0, v) = (0, 2RV × v) + (0, v)R

Inserting the expression on the right hand side into the quaternion rotation formula yields

(0, v') = (0, 2RV × v)R + (0, v)RR

When examining the first term on the right hand side, it can be shown that the scalar portion is always zero. For the second term, we can use the fact that the rotation quaternion R obeys the identity RR = 1. Since all 3 terms have zero scalar component, we can now rewrite this expression in terms of vector operations.

v' = v + 2[ RS( RV × v ) + RV × ( RV × v ) ]

This expression employs 18 multiplications and 12 adds, totalling 30 unique operations in all. This optimisation basically cuts the expense of the quaternion rotation formula in half.

Here is a function that employs this omptimisation

Vector rotateVector(const Vector& v, const Quat& q){     Vector result;     float x1 = q.y*v.z - q.z*v.y;     float y1 = q.z*v.x - q.x*v.z;     float z1 = q.x*v.y - q.y*v.x;     float x2 = q.w*x1 + q.y*z1 - q.z*y1;     float y2 = q.w*y1 + q.z*x1 - q.x*z1;     float z2 = q.w*z1 + q.x*y1 - q.y*x1;     result.x = v.x + 2.0f*x2;     result.y = v.y + 2.0f*y2;     result.z = v.z + 2.0f*z2;     return result;}

Let's express the scalar and vector portions of the quaternion in terms of the axis and angle of the rotation

RS = cos( θ/2 )

RV = n sin( θ/2 )

Pluging these values into the optimized quaternion rotation formula, and using a little bit of trigonometric magic (double angle identity), the formula reduces to

v' = v + [ sin(θ) (n × v) + ( 1 - cos(θ) ) n × (n × v) ]

Using the vector triple product identity, and the fact the n is normalized, we can reduce it a bit further.

v' = cos(θ)v + sin(θ)(n × v) + (1 - cos(θ)) (nv) n

This operation has 19 mults and 12 adds, totalling 31 operations. We didn't do anything to optimize the formula, but we were able to transform it into a form involving axis and angle.

Trick #2: Determining a Rotation From an Initial and Final Vector

One of the questions which I am asked most frequently is how to find a rotation which takes some initial vector into some final vector. There are several variants of this problem, including "I am pointing my gun at Bob, and I would like to point my gun at Jim." I have seen this problem solved in many different ways, many times involving inverse trig functions. We are going to solve this problem with quaternions - and it's really easy!

We are going to call the initial vector A, and the final vector B. Both A and B are normalized 3D vectors.

We are going to multiply these two 3D vectors together, and if you remember, this is accomplished with a quaternion multiplication. To reitterate, our input vectors are just standard 3D vectors, but the output of the product is a quaternion, which represents the product of the inputs. Since the ordering matters, we are going to place A on the right, and B on the left.

BA = -AB - A × B

We can relate the dot and cross products to functions involving the angle between A and B. If A and B are normalized, then their product has the following form

BA = - cos(θ) - n sin(θ)

The angle θ is the angle between the initial and final vector, and the vector n is a normalized vector that is perpendicular to the input vectors. This is the exact form of a rotation quaternion. Don't worry about the minus sign, since we can flip the sign without changing the rotation we are representing. In fact, this almost represents the quaternion that we are looking for. We need a quaternion with a half angle instead of a full angle.

Instead of multiplying the initial vector A by the final vector B, let's instead find a vector that represents a halfway rotation between A and B. This is very easy to do - just find the average of A and B!

H = (1/2) (A + B)

Now, this H is not normalized, and it needs to be. If we are going to normalize it anyway, we don't need to worry about the factor of 1/2. The normalized version of H is

H = (A + B) / A + B

Knowing that the angle between A and H is half of the angle between A and B, we can use our previous result to show that the rotation which takes A into B is given by

R = HA

Here is a function that will give you the rotation quaternion that will rotate some initial vector into some final vector

Quat getRotationQuat(const Vector& from, const Vector& to){     Quat result;     Vector H = VecAdd(from, to);     H = VecNormalize(H);     result.w = VecDot(from, H);     result.x = from.y*H.z - from.z*H.y;     result.y = from.z*H.x - from.x*H.z;     result.z = from.x*H.y - from.y*H.x;     return result;}

It doesn't get much easier than that folks.

Trick #3: Tangent Space Compression / Quaternion Maps

Tangent space is a set of 3 vectors - normal, tangent, and bi-normal - which are used in lighting calculations such as bump mapping, or anisotropic shading techniques, like fur or hair shaders.

The tangent space is a representation of the surface of the mesh. The tangent space vectors are stored per-vertex, and interpolated across the polygon for per-pixel operations. If you use floats to represent the components of the tangent space vectors, then you are dedicating 3 x 3 x 4 bytes = 36 bytes per vertex to the storage of tangent space.

The tangent space vectors form an ortho-normal frame. Consider the rotation which will transform the ortho-normal frame of tangent space into the standard frame in world space. If this rotation is expressed as a matrix, then the 3 columns of this matrix represent the 3 tangent space vectors.

If you can express a rotation as a matrix, then you can also express it as a quaternion. This immediately reduces the tangent space representation to 4 components = 16 bytes. When decompressing the vertex, use the quat->matrix conversion, and pull out the 3 columns of the resulting matrix as your tangent space vectors. The quat->matrix conversion costs 13 multiplies and 15 subtractions. It may be worthwhile to determine if it is possible to optimize this conversion for shader operations.

If you want to save more space, you can go even further, if you have the time - computationally speaking. Since rotation quaternions are normalized, there are only 3 degrees of freedom instead of 4. Thus you only really need to store 3 components of the quaternion, and the 4th can be calculated using the formula
w = sqrt(1 - x2 - y2 - z2)

This brings the tangent space size down to 12 bytes.

Finally, if you are super hard pressed for space, you can take advantage of the fact that the components of a quaternion are within the range [-1,1], which may not require 32 bits to have a sufficient representation. You may possibly use 16 bit floats to represent these components. Doing so will reduce the size further - 6 bytes.

With a 6-byte tangent space representation, it seems conceivable to employ quaternion texture maps, which store the tangent space directly in the texture, rather than using bump map data to perturb the interpolated tangent vectors. You could then apply a texture compression, like 3DC which gets 2:1 compression. Now tangent space only consumes 3-bytes of texture, and 0-bytes on the vertex. If you were previously using bump maps, and you replace them with quaternion maps, then you aren't really increasing the amount of texture memory used, and you have successfully removed all 36 bytes of tangent space data from the vertex.

Of course, doing quaternion map lookups in the pixel shader would also mean doing the quat->matrix conversion in the pixel shader - or, alternatively recasting the pixel shader lighting equation in terms of quaternion algebra, rather than matrix algebra - whichever ends up being cheaper. To gain perspective on the cost of this conversion, you can perform 5 quat->matrix conversions in roughly the same amount of time as a single matrix multiply. A single quat->matrix conversion is about as costly as transforming 3 vectors with a matrix.

The quaternion map would take the place of bump maps altogether, and would provide the added feature of having fine tuned control over tangent space on a per-pixel level. For anisotropic shaders, like hair, fur, and velvet shaders, you could introduce swirls and wiggles in the middle of a polygon.

There are two rather large problems facing the idea of a quaternion map: 1) a lot of people get freaked out when you use the word quaternion. 2) I can't off the top of my head conceive of a way for an artist or content creator to generate such a map.

At any rate, it seems like a very intriguing idea.


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# Volume by integration

## Homework Statement

I have the attached problem ,and Show my answer and solution,

Can any one point to the mistake I made???

Related Calculus and Beyond Homework Help News on Phys.org
Homework Helper
Gold Member
Your calculation looks correct to me. I don't see any obvious errors.

Your work also looks good to me.
The one thing I noticed is the answer from the book would be right if you take the negative away from the 1.

I'm guessing the book has a typo.

Mark44
Mentor
If this is actually the book's answer, it makes no sense without something following ln.

Thanks
Absolutely right, but draw the function, you will see that volume is almost as his answer(using geometry not integration),and that why I am confused.

Ray Vickson
Homework Helper
Dearly Missed
Thanks
Absolutely right, but draw the function, you will see that volume is almost as his answer(using geometry not integration),and that why I am confused.
If you think that the book's answer ##\pi (15+8 \ln(4))## is close to the correct answer you are very much mistaken. The correct answer is yours, except that the approximation ##\pi = 3.14## is way too crude---just leave it as symbolic ##\pi## until it comes time to evaluate the final answer numerically. Look at the following:
$$\begin{array}{rccc} \pi (15 - 8 \ln\,4 )& \doteq & 12.28251236 & \doteq & 12.3 \\ \pi (15 + 8 \ln\,4 ) & \doteq & 81.96526726 & \doteq & 82.0 \\ \pi 15 & \doteq & 47.12388981 & \doteq & 47.1 \end{array}$$

$$\begin{array}{rccc} \pi (15 - 8 \ln\,4 )& \doteq & 12.28251236 & \doteq & 12.3 \\ \pi (15 + 8 \ln\,4 ) & \doteq & 81.96526726 & \doteq & 82.0 \\ \pi 15 & \doteq & 47.12388981 & \doteq & 47.1 \end{array}$$


text:
## Number of integral solutions to multi-variable polynomials

This question follows the article discussed here

## Problem

Suppose we're trying to bound the number of integral solutions to a system of multi-variable polynomials, say

$$\sum_{i=1}^n x_i^t = \sum_{i=1}^n y_i^t,$$ where each $x_i,y_i \in \mathbb N$ and for each $t < c$ for some constant $c$.

If we do not put any constrains on the solution, there are infinitely many possible solutions even when $n=C=1$. So if we put some constrains on {$x_i,y_i$} like $x_i,y_i \in$ {$0,1,\ldots,n$}, then how many possible solutions can we get? Naively there are $O(n^n)$ choices, but it seems highly unlikely that there are many solutions to the system of equations. Is there any exist bound on the number of solutions, say $O(n^k)$ for fixed $k$ or even better bounds? Are there some well-known approaches to bound the number of solutions of an equation?

## Motivation

This question arose when I'm trying to come up with some reasonable constrains with the equation in Prouhet-Tarry-Escott Problem. It seems like if we restrict the maximum value of variables, there aren't many solutions to the equation. I tried to add more constrains to get rid of the already few solutions, but it seems that there is no direct way making the solution set empty, that is, no possible solutions under such constrains.

So I turn to find some existing bounds for the equation, but sadly nothing occurred. Can it be still hard to find such results, or there are some theorems like the Fundamental Theorem of Algebra, concerning the number of solutions to a multi-variable equation? Any information is useful. Thank you all!

## Edited

According to Felipe Voloch (Thanks!), the general approach to the question is the Hardy-Littlewood method, which considers the number of solutions to an equal-power Diophantine equation. But it seems that the method gives a lower bound on the number of solutions (is this correct?), rather than an upper bound. Or there are some ways to give upper bounds by the same method?

One more question: How about further restricting the solutions to be prime numbers? Does this make any difference?

-
 In its most general form, this problem is undecidable; en.wikipedia.org/wiki/… . Any substantive answer to a question of this type therefore has to be very sensitive to the type of equations considered. – Qiaochu Yuan Apr 17 2010 at 18:58 @Qiaochu: He has a very specific equation in mind, sums of equal powers. I don't think undecidability is relevant here. – Felipe Voloch Apr 17 2010 at 21:26 The question is stated in quite some generality. I think information about what level of generality is appropriate is relevant. – Qiaochu Yuan Apr 17 2010 at 21:44 Thank you for all your comments! If I cannot provide any bounds on n (since the number of terms is generated by a combinatorial problem, which cannot guarentees the size of n), but I can restrict the solutions lie in prime numbers < n log n (since there are about n prime numbers in this range), and the solution must satisfies the equation above SIMULTANEOUSLY for every t

 The original problem itself is a part of Waring's problem: to estimate the number of solutions to $\sum_{i=1}^nx_i^t=N$ as $N\to\infty$. So, any monograph with emphasise on Waring's problem could be a good source. – Wadim Zudilin Apr 17 2010 at 23:48 Thank you for this wonderful method!! I'm trying to read it but since I'm not familiar with this topic, it may take me for a while to understand the context. But still thanks very much for the reference! @Wadim: Since I have to estimate the number of solutions when N is small (say O(n^t) for some constant t), is this method still works? – Hsien-Chih Chang Apr 18 2010 at 8:56 I found out that in Chapter 6 of the book you provide solved the problem for a single equation fixing the exponent t, and the work by Davenport can be extended to a system of equations. Thank you very much!!! – Hsien-Chih Chang Apr 20 2010 at 15:51


text:
### Interface conditions with normal unit vectors

78
views
-2
3 months ago by
Hi everyone!!!

I come to you for asking if there is anyway for implementing in FEniCs an interface condition of the form
$_{F^c\left(N^{^c}\right)n_{j^{^{\alpha}}}-F^{^{\alpha}}\left(N^{^{\alpha}}\right)n_{j^{^{\alpha}}}=f_{kl}n_j^{^{\alpha}}}$Fc(Nc)njαFα(Nα)njα=ƒ klnjα   on  $\Gamma$Γ when solving a PDE.

Where, N is the searched function, and F^c and F^a are known and different functions in two regions i.e. an sphere inside a cube. And   $n$n  is a normal unit vector

Thanks,
Jorge.
Community: FEniCS Project
1
what is  $f_{kl}$ƒ kl ? what is the PDE? Is the interface known? This looks like a nonlinear problem (I imagine F depends nonlinearly of N)
written 3 months ago by Miguel


text:
# In how many inequivalent ways can 8 people be seated at a round table?

How many different arrangements are there of eight people seated at a round table, where two arrangements are considered the same if one can be obtained from the other by a rotation?

-

Imagine that the $8$ people take their seats one at a time, from shortest to tallest. Since orders equivalent by rotation are considered the same, it doesn’t matter where the $A$, first person sits; the arrangement is determined by how the others are seated in relation to $A$. They can then be seated in any order clockwise around the table from $A$’s left. How many permutations of those $7$ people are there?
Number the seats from one to eight, and then we can try to count the number of ways to put the people around the table, starting by putting someone in seat one, then someone else in seat two, etc. There are eight different people that we could put in the first seat, and then only seven to put in the second, six for the third and so on. This gives us the total number of possible combinations of seatings if we didn't count seatings as the same if they differ only by a rotation. Call this number $n$. But here we have counted different seatings that can be obtained from each other by a rotation several times. In fact, since each seating can be rotated in eight different ways, we have over-counted by a factor of eight, so the answer we want should be $\frac{n}{8}$.


text:
### ko_osaga's blog

By ko_osaga, history, 3 years ago, ,

The traditional implementation of union find tree (aka disjoint set) utilizes rank compression — this is the code with it.

int parent[1000], rank[1000];

int find(int x){
return parent[x] = (parent[x] == x ? x : find(parent[x]));
}

void Union(int p, int q){
p = find(p);
q = find(q);
if(rank[p] < rank[q]) parent[p] = q;
else parent[q] = p;
if(rank[p] == rank[q]) rank[p]++;
}

But in competitive programming, I found out that the rank compression was useless. Without rank compression, the code was shorter, and even the program ran faster. This is my implementation without rank compression.

int parent[1000];

void Union(int p, int q){ parent[find(p)] = find(q); }

and for some cases, I found implementing rank compression is quite hard (maybe impossible).

I will give you a sample problem : "Given a tree T, you should process Q queries — "paint the edge's value into x (>0) from path s to e".

I reversed the query, broke down the path s — e to s — LCA(s,e) — e, and used this algorithm to paint the tree. (root_val[s] : value of edge s — parent(s), dep[s] = depth of s. nxt[s] = parent(s))

int paint(int s, int e, int x){
if(dep[s] <= dep[e]) return s;
if(!root_val[s]) root_val[s] = x;
return nxt[s] = paint(nxt[s],e,x);
}

All of those program ran very fast in practice, but I was curious about the worst-case runtime of such path compression without ranks, so I googled for a while but didn't find any answer. Here's my question:

1. Does the path compression runs in (amortized) sublinear time per query?
2. If it doesn't, is there any counterexamples for that?

•
• +14
•

 » 3 years ago, # | ← Rev. 4 →   0 Any Disjoint-set Structure requires amortize Ω(α(m, n)) for each operation, which means it cannot be ran at amortized linear time. You can see more information about time complexity of disjoint-set structure here
•  » » 3 years ago, # ^ |   0 Are you talking about the upper bound? I was asking about the time complexity of path compression in amortized sublinear time (per query). I will fix that part in the article.If you are talking about the upper bound, can I ask for the exact page of proof in that paper? I also read that paper shortly, but I couldn't find the part for disjoint set without rank compression.
 » 3 years ago, # |   0 Auto comment: topic has been updated by koosaga (previous revision, new revision, compare).
 » 3 years ago, # |   0 Found an article which claims it runs in amortized O(lgn) time. Thanks anyway! http://stackoverflow.com/questions/2323351/union-find-algorithm-without-union-by-rank-for-disjoint-set-forests-data-structu
 » 3 years ago, # |   0 I can't remember the problem, but once I wrote it without ranks, and I got TLE. When I added ranks to it, it passed. That's why I don't think it's always a good idea.
 » 3 years ago, # | ← Rev. 2 →   +11 Ignore me, I'm misreading things.
•  » » 3 years ago, # ^ |   +19 What do you mean by "got rid of path compression". It still seems to be in find, isn't it? And if you have path compression, you are guaranteed to be at most O(log N) on average per query.
•  » » » 3 years ago, # ^ |   +19 Thanks, that just reminds me that I should not post on internet before having a morning cup of coffee.
 » 3 years ago, # |   +8 Interesting post. I also never use rank (then I have to implement the whole thing) as it seemed not to be very useful in practice. Instead of it, several years ago, I started joining parents at random, that is: void set_union(int p, int q){ p = find(p), q = find(q); if(rand() % 2) parent[p] = q; else parent[q] = p; } After reading your post I tried to search for some complexity proofs for this type of linking and found this gem Disjoint Set Union with Randomized Linking. In article authors review most common types of linking (disclaimer: still haven't read the whole thing): by size/rank/randomly shuffled input without heuristic/by index Here n is number of items and m number of operations.So it seems that randomized union is also a viable simplification :) (unless calculating random becomes your bottleneck :D)


text:
Unexpected result calculating the determinant of a singular matrix (42S)
10-28-2019, 08:29 AM
Post: #29
 Werner Senior Member Posts: 648 Joined: Dec 2013
RE: Unexpected result calculating the determinant of a singular matrix (42S)
Ah yes indeed, you are right. So it is not impossible to have a 7x7 with integer elements < 100 and a condition number > 1e15, so that the cheat would not work.
Here's a 2x2 for which it fails:

Code:
[[ 49'999'999 49'999'998 ]  [ 50'000'000 49'999'999 ]]

Cheers, Werner
 « Next Oldest | Next Newest »

 Messages In This Thread Unexpected result calculating the determinant of a singular matrix (42S) - Dave Britten - 10-21-2019, 01:54 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-21-2019, 02:24 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Dave Britten - 10-21-2019, 02:30 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-21-2019, 02:36 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Thomas Okken - 10-21-2019, 03:11 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-21-2019, 04:23 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - pinkman - 10-21-2019, 07:39 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Moggul - 10-21-2019, 08:41 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Thomas Okken - 10-21-2019, 11:37 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-21-2019, 10:33 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - John Keith - 10-21-2019, 11:58 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - ttw - 10-21-2019, 12:48 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - John Keith - 10-21-2019, 06:12 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Dave Britten - 10-21-2019, 01:49 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Thomas Okken - 10-21-2019, 03:06 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Claudio L. - 10-21-2019, 04:46 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Thomas Okken - 10-21-2019, 05:54 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - toml_12953 - 10-21-2019, 06:00 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - ttw - 10-21-2019, 07:10 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Werner - 10-22-2019, 06:12 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Thomas Okken - 10-22-2019, 11:11 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-23-2019, 04:50 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Werner - 10-23-2019, 07:01 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-23-2019, 11:06 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Werner - 10-24-2019, 06:34 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Valentin Albillo - 10-25-2019, 11:00 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Paul Dale - 10-22-2019, 09:23 AM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Werner - 10-22-2019, 12:13 PM RE: Unexpected result calculating the determinant of a singular matrix (42S) - Werner - 10-28-2019 08:29 AM

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text:
# Minimum System of ATmega128

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Not open for further replies.

#### satria02

##### Junior Member level 2
Hi......
I'm new for using ATmega128 and I have a problem with the minimum system of ATmega128.

I was design the schematic of minimum system of ATmega128 from some literature (data sheet and from internet).....I think the minimum system that i designed was right, but when i try to access the I/O port (using C --> AVR GCC), i was failed....And i don't know why this happen....

when i test the I/O port using multimeter, the multimeter show less than 1 Volt...It's mean that I/O port which i accessed was failed....

This is the schematic that i designed...
MCU.pdf
MCU.SchDoc

can you help me why this problem happen.....??
thanks before....

best regards...

#### alexan_e

I can see the mosi, miso and rst connected but I can't see the sck connection, there is only the SCK which goes to the SDcard but no sck.
Check if it is connected, everything else seems to be fine

Alex

---------- Post added at 14:06 ---------- Previous post was at 13:54 ----------

Another obvious note would be to check if the chip is soldered correctly and there aren't any bridges between pins.

Alex

#### satria02

##### Junior Member level 2
Thanks for replay...

Beside connect to SD Card, sck also connect to ISP (programming)....
I have been tried to load program from PC to this minimum system and it doesn't matter.....the load process was success, but the program doesn't work (I/O port access was failed)....

#### alexan_e

So you program it correctly and you also verify the written checksum and it is correct?

Alex

#### satria02

##### Junior Member level 2
Actually i don't know what is the problem..... the circuit or the programs

#include <avr/io.h>
#include <avr/iom128.h>

int main(void)
{
PORTA = 0xFF;
DDRA = 0xFF;

PORTC = 0xFF;
DDRC = 0xFF;

PORTD = 0xFF;
DDRD = 0xFF;

PORTE = 0xFF;
DDRE = 0xFF;

PORTA= 0xFF;
PORTC= 0xFF;
PORTD= 0xFF;
PORTE= 0xFF;

while(1);
}

Anything wrong with my program....?

#### alexan_e

I assume you are using avrgcc,
#include <avr/iom128.h> is not needed, #include <avr/io.h> will point to the right library.

Do you use AVRstudio with winavr?

Alex

#### satria02

##### Junior Member level 2
yup.... i using AVRstudio 4 and winavr....

#### alexan_e

Have you set the correct device in the project options (atmega128) and also the frequency (only makes a difference when delays are involved but fill it anyway).
Your code has no problem, if you want instead of turning everything on try to blink a pin or port

Code C - [expand]1
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#include <avr/io.h>
#include <util/delay.h>

int main(void)
{
PORTA = 0xFF;
DDRA = 0xFF;

while(1){
_delay_ms(1000); // delay 1 sec
PORTA^=0xff;  // invert PORTA outputs
// PORTA^=0x01;  // or invert PORTA.0 output only

}
}

Alex

satria02

### satria02

points: 2

#### satria02

##### Junior Member level 2
can you tell me how to setting frequency clock in this compiler?

#### alexan_e

The frequency is set in the project menu -> configuration options
that is where you set the mcu model too.
The frequency should match the core speed of maga128 device, it can't change it just calculates correct the delays.
Have you changed the armega128 fuses to work with the external crystal at 16MHz?
If so then use 16000000 as frequency of the project

In m128 I think there is also a clock divider that can change the clock frequency , that one is controlled with a register value but you don't need to use it for now.

Alex

satria02

### satria02

points: 2

#### satria02

##### Junior Member level 2
How to changed the atmega128 fuse?

external crystal that i used was 16MHz

#### alexan_e

Your programmer application should have a screen to change the fuses,
I use USBasp ( fischl.de - USBasp - USB programmer for Atmel AVR controllers ) programmer (can't work from AVRstudio interface)
with AVRdude and AVR BURN-O-Mat front end ( AVR8 Burn-O-Mat avrdude GUI ) , it is very easy to change fuse with this application.

The default setting for m128 is internal RC clock with a frequency of 1 MHz and your code should work fine with that,
this is not the cause of your problem so maybe you should see why it doesn't work first before changing fuses.

What application do you use to program the device?

Alex

---------- Post added at 21:08 ---------- Previous post was at 20:32 ----------

Actually according to the datasheet , when the internal RC is used the XTAL pins should be left unconnected,
it is not the same as mega8 that I use in which the xtal pins turn into I/O when the internal RC is used.
In mega 8 you can have internal RC clock and also a crystal connected in XTAL and it doesn't matter,
in m128 the XTAL has dedicated pins and maybe things are different.

On the other hand if the clock wasn't working you wouldn't be able to program the device and you say that you did.

Alex

satria02

### satria02

points: 2

#### satria02

##### Junior Member level 2

thanks for your assist, i will try it again....

#### alexan_e

I haven't heard of Khazama but a google search has some pictures and I see the check boxes for the fuses http://www.khazama.com/project/programmer/
Check the drop down lists Hfuse, maybe there is a selector for clock source.

Alex

#### satria02

##### Junior Member level 2
still doesn't work.......

#### alexan_e

When you upload the program to the mcu there should be a function to verify and compare your hex file with the content of the chip, is the content of the chip verified?
Have you changed the fuses?

Alex

#### satria02

##### Junior Member level 2
I can read fuses and lock from mcu, but i haven't change anything....

Is it possible that mcu that i used was damage?

#### alexan_e

Your code was correct, if the mcu is also set correctly in the avrstudio properties then this part is fine.
You say that you can read the chip fuses and you can transfer the hex to the mcu and verify that it was transferred so the chip clock works (internal RC).
If the pin output of the mcu is not correct (you said 1v instead of 5) then it has nothing to do with the program, also the power supply must be correct and the mcu works because it can be programmed.

It is very strange, have you tried my blinking code to see if portA (or use any port) blinks?

Alex

---------- Post added at 11:58 ---------- Previous post was at 11:54 ----------

in your first post you said
but when i try to access the I/O port (using C --> AVR GCC), i was failed....And i don't know why this happen....
so you where not able to connect at all but now you say that you can transfer and verify the program.
What have you changed and what is the pin voltage now?

Alex

---------- Post added at 12:11 ---------- Previous post was at 11:58 ----------

Is anything connected to the pins of the mcu , for example LCD, or MMC etc?
If you have connected peripherals to the pins and you have turned every pin of the mcu to an output with high hevel (5V) then maybe this is the cause of your problem and some pin creates a "shortcut" or even worse something has been damaged.
It should probably be visible in the 5v power supply too, in the measured voltage.

Alex

#### satria02

##### Junior Member level 2
not only in port A, but also in the other port.....

hmm....may be the problem is fuse bits.....but i don't know how to setting the fuse bits....
when i read the fuse bits from mcu.....there are Lock Bits, H Fuse, and checklist box on H-Fuse, L-Fuse,E-Fuse, and Lock (like link that you shown),
and i don't know which box should be on the check list....

#### alexan_e

My suggestion would be to download AVR BURN-O-Mat front end ( AVR8 Burn-O-Mat avrdude GUI )

It will be very easy to set the fuses from there but the fusebits can't stop your code from working.

You haven't answered to my question if anything is connected to the mcu pins.

Alex

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text:
# Introduction

A lot has happened since my last post (and really, I mean A LOT). However, in the spirit of that last post, one good thing that has happened is that in the upcoming 1.3-0 release of the oce package on CRAN we have changed the default colormap (or “palette” as it’s often referred in the R world) to something not nearly sucky as the classic “jet” colormap (originally made popular by Matlab).

The purpose of this post is just to highlight the difference between the old and new defaults, and also to show many other great colormaps for plotting oceanographic data.

# Jet vs viridis

When the imagep() function was first added to oce, there was a limited selection of palettes that could be used. For a long time, the default of the imagep() function itself has been a palette called oceColorsPalette() (which is a blue-through-white-to-red palette most useful for diverging colormaps), however many of the internal functions that use colors defaulted to the oceColorsJet() palette, including: the plot,section-method, and more importantly, the colormap() function itself (which I wrote about here).

library(oce)
## Loading required package: gsw
## Loading required package: testthat
data(volcano)
par(mfrow=c(1, 3))
imagep(volcano, col=oceColorsPalette, main='oceColorsPalette()')
imagep(volcano, col=oceColorsJet, main='oceColorsJet')
imagep(volcano, col=oceColorsViridis, main='oceColorsViridis()')

The palette on the left, oceColorsPalette() has been retained as the default for the plot,adp-method (is velocities are typically diverging). The palette in the middle, Jet, is still available but is no longer the default for any plots. The viridis palette on the right is now the default for all palette-related functions.

I won’t go into the details of why viridis is so much better – a quick google will turn up lots of articles (though this new paper from Kristen Thyng has lots of great info.

# Other, even better palettes

Kristen Thyng is author the author of the cmocean series of oceanographic colormaps, originally created for python/matplotlib. There is now a cmocean package on CRAN, which makes all the python colormaps available in R.

library(ocedata)
library(cmocean)
data(levitus)
par(mfrow=c(1, 2))
with(levitus, imagep(longitude, latitude, SST, col=cmocean('thermal'), main="sea surface temperature"))
with(levitus, imagep(longitude, latitude, SST, col=cmocean('haline'), main="sea surface salinity"))

There is also the viridis package itself, which makes the series of palettes developed for matplotlib available in R.

library(viridis)
## Loading required package: viridisLite
imagep(volcano, col=inferno)


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[–] 0 points1 point  (6 children)

sorry, this has been archived and can no longer be voted on

I understand what you said but I can't see how you can change an eigenvalue without changing the whole transformation itself. Bear with me please, I'm just an undergrad.

Let's say we take the transformation represented by the identity matrix in GL(5,R) with the first three 1's changed to -1. This matrix is already diagonalized, so how could we possibly do this flipping only one vector of another base and turning the others?

[–]Mathematical Physics 4 points5 points  (5 children)

sorry, this has been archived and can no longer be voted on

I'm not talking about continuously transforming the eigenvectors themselves, I'm talking about continuously the matrix. To simplify your example, I'll explain how to continuously transform

-1   0   0
0  -1   0
0   0  -1

into

-1   0   0
0   1   0
0   0   1

First, by a continuous transformation from A to B, I mean a continuous map f:[0,1] -> M such that f(0)=A and f(1)=B. So in this case we're looking for a continuous map f:[0,1]->GL(3,R) such that

       -1   0   0
f(0) =  0  -1   0
0   0  -1

and

       -1   0   0
f(1) =  0   1   0
0   0   1

We define

       -1       0           0
f(t) =  0  -cos(pi*t)  -sin(pi*t)
0   sin(pi*t)  -cos(pi*t)

det(f(t)) = -cos2 (pi*t) - sin2 (pi*t) = -1 =/= 0, so f is, in fact a map from [0,1] to GL(3,R), and each component of f is continuous, so f is continuous. Finally,

       -1      0        0     -1   0   0
f(0) =  0  -cos(0)  -sin(0) =  0  -1   0
0   sin(0)  -cos(0)    0   0  -1

and

       -1      0         0      -1   0   0
f(1) =  0  -cos(pi)  -sin(pi) =  0   1   0
0   sin(pi)  -cos(pi)    0   0   1

So f is a continuous transformation (in topology, actually called a path) from the first matrix to the second.

Intuitively, what's happening is that you start with one matrix and multiply it by the other matrix. Letting the angle of the rotation matrix continuously change from 0 to pi, you end up with a matrix in which two of the eigenvalues have changed sign. If you tracked what happened to the eigenvalues in the meantime, you'd find that they avoided 0 by rotating through the complex plane which is perfectly fine since they're always complex conjugates, so the determinant is always real.

[–] 0 points1 point  (4 children)

sorry, this has been archived and can no longer be voted on

I understand this, but quoting you: "You only need to flip one vector, hence a mirror-image." and I simply don't see how you only need to flip one vector, instead of an odd number of eigenvectors to have have a negative determinant. (unless I misunderstood your whole statement from the first comment)

[–]Mathematical Physics 1 point2 points  (3 children)

sorry, this has been archived and can no longer be voted on

By flipping a vector, I mean skipping from eigenvalue 1 to eigenvalue -1, or similar. There's no continuous path from

 1   0   0
0   1   0
0   0   1

to

-1   0   0
0   1   0
0   0   1

But if you allow only one such jump, the rest of the transformation from the identity matrix to an arbitrary invertible matrix can be continuous. The continuous part of the transformation corresponds to a matrix with positive determinant, which means that it can be constructed entirely out of rotations and stretches (and skews, which I sort of skipped over).

It's also worth pointing out that flipping an eigenvector doesn't do anything, since if v is an eigenvector then so is -v. What you're actually talking about is flipping an eigenvalue.

What I'm talking about is transforming the matrix by multiplying it by a single flip, followed by some sequence of rotations, stretches and skews. The negative determinant indicates that to perform the corresponding transformation using these smaller steps, you need to include one flip. An odd number of flips is fine, but only one is ever necessary.

[–] 0 points1 point  (1 child)

sorry, this has been archived and can no longer be voted on

Wouldn't f(t) = [cos(pi*t) 0 0 0 1 0 0 0 1]

Be a continuous map from A to B?

[–]Mathematical Physics 0 points1 point  (0 children)

sorry, this has been archived and can no longer be voted on

That isn't a map from [0,1] to GL(3,R), because det(f(1/2))=0. Remember, GL(3,R) is the group of invertible 3x3 matrices.

The reason GL(n,R) is disconnected is that det(M)=0, M in GL(n,R) defines an n2 - 1 dimensional subspace of the overall n2 dimensional space. You can think of it like drawing a curve in the plane (with no endpoints). To get from one side of the curve to the other, you have to intersect the curve at some point.

[–] 0 points1 point  (0 children)

sorry, this has been archived and can no longer be voted on

Yeah, I meant an eigenvalue, sorry. I think I just misunderstood your first statement in the cotext of the OP you were replying to. Thanks for taking your time to explain it, I appreciate.


text:
## Gravitational dipole radiations from binary systems

We investigate the possibility of generating sizeable dipole radiations in relativistic theories of gravity. Optimal parameters to observe their effects through the orbital period decay of binary star systems are discussed. Constraints on gravitational couplings beyond general relativity are derived.

Published in:
Phys. Rev. D, 66, 024040
Year:
2002
Keywords:
Laboratories:


text:
## Sunday, August 31, 2014

### Gillies’ Philosophical Theories of Probability, Chapter 3

Chapter 3 of Donald Gillies’ Philosophical Theories of Probability (2000) deals with Keynes’ logical theory of probability, which was taken up by the Vienna circle and logical positivists like Carnap (Gillies 2000: 25).

Keynes’ logical theory was partly inspired by the lectures of W. E. Johnson at Cambridge university, which were also attended by Harold Jeffreys who later formulated his own logical theory of probability (Gillies 2000: 25).

Keynes had finished the proofs of his Treatise on Probability in 1913, but it was not published until 1921.

Keynes’ views on probability were influenced by the intellectual climate at Cambridge university, particularly the ethical work of G. E. Moore and Bertrand Russell’s logicist work on mathematics (Gillies 2000: 27). Gillies (2000: 27) sees Keynes’ attempts to provide a “logical” foundation for probability, and particularly inductive reasoning, as inspired by Russell and Whitehead’s attempts to found mathematics on logic.

For Keynes, if the evidence h justifies a conclusion a to some degree α, then there is a probability relation of degree α between a and h, so that inductive probability is a degree of partial entailment or degree of rational belief (Gillies 2000: 31)

Keynes thought that we have knowledge of this probability relation by logical intuition or direct acquaintance, but this is a problematic part of Keynes’ theory (Gillies 2000: 31–32).

Gillies (2000: 32–33) argues that underlying Keynes’ idea that probability is objective and known by logical intuition is a problematic concept derived from Platonic ontology: that the probability relation is objective and belongs to some Platonic realm.

Keynes also thought that mathematical probabilities are expressible as numbers on a range in the interval [0, 1]; he thought that not every probability had a numerical value, and also that some probabilities can only be arranged in an ordinal ranking, while others cannot even be compared at all (Gillies 2000: 33–34).

Numerical “point” probabilities are possible when the relevant outcomes involved are finite, exclusive and equiprobable (Gillies 2000: 35).

The “principle of indifference,” which was introduced by Bernoulli as the “principle of non-sufficient reason,” is an a priori principle invoked by Keynes to determine when cases are equiprobable (Gillies 2000: 35–36). Gillies (2000: 37–46) sees insuperable difficulties with the a priori principle of indifference, such as the book paradox and wine–water paradox, which Keynes did not really solve.

While the “principle of indifference” might in some cases be a useful heuristic, it cannot be a sound logical principle, and so, since Keynes’ logical theory of probability requires it to be a successful theory, Gillies (2000: 48–49) concludes that Keynes’ overall logical theory is flawed, though important aspects of it may be salvaged, provided that the theory’s Platonist view of how the probability relation is understood and the a priori principle of indifference are abandoned (Gillies 2000: 35, 49; Runde 1994).

“Bibliography on Keynes’s Theory of Probability (Updated),” July 6, 2014.

“Interpretations of Probability,” Stanford Encyclopedia of Philosophy, 2002 (rev. 2011)
http://plato.stanford.edu/entries/probability-interpret/

BIBLIOGRAPHY
Gillies, D. A. 2000. Philosophical Theories of Probability. Routledge, London.

Runde, J. 1994. “Keynes After Ramsey: In Defence of A Treatise on Probability,” Studies in History and Philosophy of Science 25.1: 97–121.

## Saturday, August 30, 2014

### Gillies’ Philosophical Theories of Probability, Chapter 2

Chapter 2 of Donald Gillies’ Philosophical Theories of Probability (2000) deals with the Classical interpretation of probability theory.

As Gillies notes, the “Classical” interpretation was the earliest theory of probability and its most important statement was by Pierre-Simon Laplace (1749–1827) in his Essai Philosophique sur les Probabilités [A Philosophical Essay on Probabilities] (1814). However, it is largely of historical interest now, and has no supporters today (Gillies 2000: 3).

Laplace’s Essai Philosophique sur les Probabilités (1814) made the assumption of universal determinism on the basis of Newtonian mechanics (Gillies 2000: 16). Laplace argued that an agent with perfect knowledge of Newtonian mechanics and all matter could predict the future state of the universe. It is only human ignorance that prevents perfect forecasting, and leads us to calculate probabilities (Gillies 2000: 17). Thus probability, according to Laplace, is a measure of human ignorance (Gillies 2000: 21).

Laplace’s formula for calculating probabilities is the familiar one where the probability P(E) of any event E in a finite sample space S, where all outcomes are equally likely, is the number of outcomes for E divided by the total number of outcomes in S.

But there is an obvious limitation with this, as pointed out by the later advocates of the frequency theory of probability like Richard von Mises: what if our outcomes are not equiprobable? (Gillies 2000: 18). Thus the Classical interpretation of probability has a serious shortcoming.

As probability theory came to be increasingly applied to phenomena in the natural and social sciences in the 19th century, its limiting assumption of equiprobable outcomes was exposed as a problem, and the relative frequency approach was developed as a new and alternative theory.

BIBLIOGRAPHY
Gillies, D. A. 2000. Philosophical Theories of Probability. Routledge, London.


text:
## ccmom82 Group Title Jim Moore opened a new savings account. He deposited $12,000 at 12 percent compounded semiannually. At the start of the fourth year, Jim deposits an additional$50,000 that is also compounded semiannually at 12 percent. At the end of 6 years, the balance in Jim Moore's account is: (Use the tables in the handbook) one year ago one year ago

• This Question is Open
1. mike91335

$12000\times ( 1+.06)^{12} +50000\times ( 1+.06)^{8}$\$103,838.76

2. Sophie123

It is 12000 x (1.06)^12 + 50000 x (1.06)^6 = 95,072.31 start of 4th year to end of 6th year = 6 semi-annual periods where interest is compounded for the second deposit


text: It’s raining cats and dogs outside, which means that your car is going to have a hard time gripping the slick asphalt. It’s best to stay home and enjoy some hot cocoa while you watch the storm roll in (or at least wait until it passes). But what if you can’t? What if you’re on a deadline or there is an emergency? You may need to take your chances with Mother Nature’s fury. When driving in extreme weather conditions, kinetic energy becomes more important than ever! The Kinetical Energy Equation tells us how much work we will need to do when our tires are slipping all over the place. So, let’s break down this equation so that you know what it looks like and how you can use it. -The Kinetical Energy Equation is written as: $$W = \frac{½mv^²}{μ}$$ where $$\mu$$ represents a coefficient of friction and $$v$$ refers to the velocity or speed at which your car is travelling with respect to some surface. Note that in this equation, work equals kinetic energy so if we want our tires on the road then we need to have more kinetic energy! -Useful numbers might be: Weight (kg) 1500; Velocity(km/h) 0; Coefficient Of Friction (.50); Work done (Joules), 14076400 J – The amount of work done by an object equals its


text:
# Close range ultrasonic fluid proximity level sensor

I'm looking for a way to sense a fluid's level from above, in a small package, with a relatively low dead zone. Most commonly available ultrasonic sensors (which I would see as the best option) have dead zone's of at least 5 inches/13 cm like this one.

What I'm wondering is if that dead zone of around 13 cm comes from the IC's used in the package, or the physical limit of the transducer.

1. In the first case, are there packages out there that do overcome this, or would it be an option to purchase an individual transducer and hook it up to my own circuitry, optimized for a very low dead zone, and if so, any examples out there on how to go about building such a sensor out of a transducer?

OR

1. In the second case, are there any alternatives to a ultrasonic sensor that would help me achieve this goal? The liquid in the container or the container itself may in this application not be affected.

First, some physics.

• In dry air at 20 °C (68 °F), the speed of sound is 343.2 metres per second.
• At 42 kHz (the operating frequency of the device you linked) the wavelength, $\lambda$, can be calculated:

$$\lambda = \frac {v}{f} = \frac {360}{42k} = 8.6~mm$$

To send out a "ping" to the target we will have to send at least several cycles. We don't know the response of the transducer but let's assume ten cycles. When the last wavefront is leaving the transducer the first is 86 mm towards the target.

It should be getting clear now that since the transducer can't be in receive mode and transmitting simultaneously that there will be some minimum distance that the device can handle. The manufacturer has stated this to be 130 mm.

In the first case, are there packages out there that do overcome this, or would it be an option to purchase an individual transducer and hook it up to my own circuitry, optimized for a very low dead zone, ...

You could look for a high-frequency transducer. If you could find one that operated at 10 times that frequency it might work with 1/10 minimum range.

... and if so, any examples out there on how to go about building such a sensor out of a transducer?

Search, I'm afraid.

In the second case, are there any alternatives to a ultrasonic sensor that would help me achieve this goal?

Figure 1. Common methods of fluid level measurement include various float, pressure, and capacitive schemes as well as ultrasonic. Image: Clemson University.

Capacitive would be worth a look.

What I'm wondering is if that dead zone of around 13 cm comes from the IC's used in the package, or the physical limit of the transducer.

The brochure says 6" may be the limit and this generally will be down to switching the transducer from transmitting the outgoing acoustic pulse to being able to receive a reflection coming from the target. It will take time for the ultrasound transducer to stop "ringing" and this will be the main problem because while it is ringing it is producing a "fake" receive signal that masks the wanted return signal.

Part of the rest of your question is really about shopping and shopping questions are off-limit on this site.

are there any alternatives to a ultrasonic sensor that would help me achieve this goal?

Maybe someone else can answer this because I'm no expert in this area.


text:
# The current I in the given network is

Free Practice With Testbook Mock Tests

## Options:

1. 0 A

2. 2.38 ∠ - 96.37° A

3. 2.38 ∠ 143.63° A

4. 2.38 ∠ -23.63°A

### Correct Answer: Option 3 (Solution Below)

This question was previously asked in

GATE EC 2020 Official Paper

## Solution:

Since the voltage sources are in parallel with Z, I1 and I2 will be calculated as:

$${I_1} = \frac{{120\angle - 90^\circ }}{Z}$$

Z = 80 – j35

In polar form, this can be written as:

Z = 87.32 -23.63

$${I_1} = \frac{{120\angle 90^\circ }}{{87.32\angle - 23.63}}$$

I1 = 1.374 -90 + 23.63°

I1 = 1.374 -66.37°

Similarly,

$${I_2} = \frac{{120\angle - 30^\circ }}{{87.32\angle - 23.63}}$$

I2 = 1.374 -6.37°

Applying KCL at node A, we get:

I + I1 + I2 = 0

I = -(I1 + I2)

I = -(1.374 < -66.37 + 1.374 -6.37°)

Converting back to coordinate axis:

I = -(0.55 – 1.258 j + 1.365 – 0.252 j)

I = (2.915 – 1.41 j)

I = -2.915 + 1.47 j

$$I = \sqrt {{{\left( { - 1.916} \right)}^2} + {{\left( {1.41} \right)}^2}} \angle {\tan ^{ - 1}}\left( {\frac{{1.41}}{{ - 1.915}}} \right)$$

I = 2.38 -36.36

I = 2.38 180 36.36

I = 2.38 143.63


text:
# Counting and summing over solutions of a Diophantine equation

Say I have a Diophantine equation of the form $a_1 x_1 + a_2 x_2 + ... + a_m x_m = n$ such that the $a_is$ are all co-prime to each other. And I also have a function say $f$ which depends only on the $x_i's$ (and will be evaluated on solutions of the equations)

• Is there a general method or simple examples of summing over the values of $f$ evaluated on the non-negative integral solutions of the equation?

• Is there a way to count the number of non-negative integral solutions of such Diophantine equations? (...I am aware that it is trivially doable in some special cases like when all the $a_is$ are equal to $1$ or when $a_i = i$ and $m=n$...)

-
Do you want all the a_i positive? Also the f, I assume it is a function of the 'solutions' so x_i not a_i, or am I confusing something. –  quid Apr 11 '11 at 19:47
Yes. I want all the a_i's to be positive. The typos with f has been corrected. It depends on the solutions and not $a_i$s. Thanks for pointing out! –  Anirbit Apr 11 '11 at 20:00
What kind of functions $f$ are you interested in? –  Gjergji Zaimi Apr 12 '11 at 3:09
In the specific case I am concerned with $f$ will be the number of solution of the Polya problem for the cyclic group and with $m$ colours, $x_1$ balls of the first colour, $x_2$ balls of the second colour etc. –  Anirbit Apr 12 '11 at 4:45

The number of solutions of $a_1x_1+\dots+a_mx_m=n$ in non-negative integers $x_1,\dots,x_m$, call it $d(n;a_1,\dots,a_m)$, is called the $\it denumerant$. This goes back to Sylvester, On the partition of numbers, Quart J Pure Appl Math 1 (1857) 141-152. Much is known. For example, Schur proves that if $\gcd(a_1,\dots,a_m)=1$ and $P_m=\prod a_i$ then $d(n;a_1,\dots,a_m)$ is asymptotic to $P_m^{-1}n^{m-1}/(m-1)!$ as $n\to\infty$. (The reference is Zur additiven zahlentheorie, Sitzungsberichte Preussiche Akad Wiss Phys Math Kl (1926) 488-495.)

This and more is in Chapter 4 of J L Ramirez Alfonsin, The Diophantine Frobenius Problem, published by Oxford.

-
@Gerry! By tinkering around I had guessed this result you say is by Schur! But haven't been able to give a precise proof. Does Ramirez's book explain this? –  Anirbit Apr 12 '11 at 6:31
@Anirbit : Schur's theorem is also proved in the book Generatingfunctionology by H.S. Wilf (section 3.15) math.upenn.edu/~wilf/DownldGF.html –  François Brunault Apr 12 '11 at 6:53
@Anirbit, yes, Ramirez Alfonsin gives a proof - and many more results, besides. If you're interested in this problem I strongly suggest you track down a copy of the book. –  Gerry Myerson Apr 12 '11 at 7:02
@Gerry Thanks for the references. And any hope of an exact solution? And any ideas about doing the sum over solutions part of the question? –  Anirbit Apr 12 '11 at 7:09
@Anirbit, any hope of you trying to read the book? –  Gerry Myerson Apr 12 '11 at 13:03

This is no way a complete answer, but it shows that one cannot ask for too much, or might have to impose some additional conditions, perhaps relative size assumptions.

Let us just say we want to know if there is at least one or no solution. Given $a_i$ it follows easily that for all sufficiently large $n$ there is at least some solution, and we can thus answer this question in case $n$ is 'large'. However, what does 'large' mean exactly? The problem of determining the precise threshold is known as the Frobenius problem (or also Coin problem).

This is a well-investigated and difficult problem (except for only two $a_i$). In general, no 'formula' is known; yet good algorithms to compute the exact threshold are known if the number of $a_i$ s is fixed; if not, this is not so.

So, even to decide whether such an equation has a solution or not can be a very challenging question if $n$ is not 'large'.

-
Even if I assume that my $n$ is large enough for solutions to exist then is there a way to count the number of non-negative integral solutions? Asymptotically? –  Anirbit Apr 11 '11 at 20:26
@Anirbit, perhaps yes, as one can count the number of suitable related congruences, but I am not sure this helps. I will think a bit about it and report if I find something. (But I cannot do so within the next ca 16 hours.) –  quid Apr 11 '11 at 22:34
Since meanwhile another answer appeared, I consider this as obsolete. –  quid Apr 12 '11 at 13:33


text:
# Simple Line Integral — Just making sure I'm right

Compute the line integral $\int_C xe^{z^2} ds$ where C is the piecewise linear path from $(0,0,1)$ to $(0,2,0)$ to $(1,1,1)$.

I started out by parametrizing the curve and got:

$C_1: x = 0,\space y = 2t,\space z = 1-t$

$C_2: x = t, \space y = 2 -t,\space z = t$

$0 \leq t \leq 1$

I then set up two integrals:

$\int_{C_1} xe^{z^2} ds \space + \int_{C_2} xe^{z^2} ds$

Found that the first integral is $0$ because $x=0$ in the parametrization.

I then evaluate the second integral and got $\frac12(e-1)$. Did I do this right?

$C_1: \ \ \ x=0+0t, \ \ \ \ y = 0+2t, \ \ \ \ z=1+0t$
$C_1:x=0, y=2t, z=1$
I think the second integral should be $\frac{1}{2}(1-e)$ instead?, wait no you're right but your parameterization for $z$ on $C_2$ should be $z=t$ not $z=1-t$.
As to the parametrization, I got the same answer as yours, and I can confidently say that your parametrization is correct. Obviously the first line integral is 0. For the second line integral, the magnitude of the tangent vector is always $\sqrt 3$, and then plug in expressions for x, y, z in terms of t: $$\sqrt 3\int^1_{t=0}t*e^{t^2}dt$$ Mathematica yields the result $\frac {\sqrt 3}2(e-1)$, so the value for your second line integral that you gave is wrong.


text:
Here is a very old high school exam question I am trying to solve (purely for interest only):

If $a,b,c$ are real numbers such that $-1 \le ax^2+bx+c \le 1$ for $-1 \le x \le 1$ prove that $-4 \le 2ax+b \le4$ for $-1 \le x \le 1$ (Hint: Consider the functions at the end-points and at the mid-point of the interval).

I can see (graphically) that if $a>0$ then, as $2ax+b$ is the gradient function, the max and min gradients will occur when the parabola passes through the end-points $(-1,1)$ and $(1,1)$ and the mid-point $(0,-1)$. This gives 3 equations with 3 unknowns and is solved to give $a=2, b=0$ and $c=-1$. The required result follows easily from this. Due to symmetry, a<0 gives the same result.

We first note that by setting $x=0$ in our initial inequality we get: $$-1\le c\le1 \\ |c|\le1.$$
If we set $x=\pm1$ in our given inequality we will get: $$\text{(i)}\;\;-1\le a+b+c\le 1 \;\;\;\text{and}\;\;\;\text{(ii)}\;\;-1\le a-b+c\le 1.$$ Adding these two inequalities together will give us: $$-2\le 2a+2c\le2 \\ -1\le a+c\le1 \\ -2\le-c-1\le a\le-c+1\le2 \\ |a|\le2.$$ Thus, considering only $-1\le x\le1$, we see that if $a\gt 0$ then: $$2ax+b \;\le\; 2a+b \;=\; a+b+c + (a-c)\;\le\;1 + (a-c) \;\le\;1+1+2\;=\;4,$$ and $$-4 \;=\; -2-1 -1\;\le\; (-a+c) - 1\;\le\; (-a+c) - a+b-c\;\le\; -2a+b \;\le\;2ax+b.$$
Similarily, if $a\lt0$ then: $$2ax+b \;\ge\; 2a+b \;=\; a+b+c + (a-c)\;\ge\;-1 + (a-c) \;\ge\;-1-1-2\;=\;-4,$$ and similarily: $$4 \;=\; 2+1 +1\;\ge\; (-a+c) + 1\;\ge\; (-a+c) - a+b-c\;\ge\; -2a+b \;\ge\;2ax+b.$$


text:
# zbMATH — the first resource for mathematics

On the relationship between fractal dimension and fractal index for stationary stochastic processes. (English) Zbl 0798.60035
The real number $$d$$ is said to be the fractal dimension of a sample path on an interval of some stationary square-integrable stochastic process $$X_ t$$ if $$d$$ is the Hausdorff dimension of the path as a subset in $$\mathbb{R}^ 2$$. On the other hand the fractional index $$\alpha$$ of this process is defined by $\alpha = \sup \bigl\{ \beta \mid 1-\text{cov} (X_ 0, X_ t) = O (t^ \beta) \quad \text{as} \quad t \downarrow 0 \bigr\}.$ The authors give sufficient conditions that $$d=2 - \alpha/2$$ is valid provided that $$X_ t = g(Z_ t)$$ for a smooth function $$g$$ and $$Z_ t$$ is a stationary Gaussian process. It may be happen that $$X_ t$$ is non-Gaussian.

##### MSC:
 60G10 Stationary stochastic processes 60G15 Gaussian processes 62G05 Nonparametric estimation
Full Text:


text:
# Reaction force exerted by pulley's support

Consider an in-extensible string passing over an ideal pulley connected to a block of mass $$m$$ as shown:

The ends are pulled with a force $$F$$

Am I correct in saying that the block will experience a reaction force $$\sqrt{2}F$$? The forces acting are perpendicular to each other and the support of the pulley is inclined at an angle of $$45^\circ$$ with horizontal.

In other words can the forces acting on the string be vectorially added to find the resultant reaction force exerted by pulley?

• I encountered this specific scenario while solving an entirely different question. That is the reason I havent given any context. Im not sure about why this is tagged as a homework question. This is a purely conceptual doubt – newbie105 Jan 2 at 16:27


text:
appsignal

# Revamped dashboards!

Tom de Bruijn on

We're launching a big update for the custom metrics dashboards. We combined dashboards and custom metrics in the navigation, improved the dashboard editor and updated the dashboard definition YAML.

## Combining dashboards and custom metrics

The "Dashboard" and "Metrics" sections had grown so similar, it made sense to combine them in the navigation. All dashboards will now appear together in the sidebar as per our recent navigation change.

## Enhanced editor

Before today, all your dashboards were editable in one big YAML definition. This YAML became too large to easily manage in our online editor as the number of dashboards increased for an app. It was difficult to keep focus on which dashboard the YAML belonged to and syntax errors were difficult to spot.

The dashboard editor is now scoped per dashboard. This keeps focus on the dashboard that's being created or updated. We've also outfitted the editor with new features such as syntax highlighting and better validation so it's easier to track down where an error is coming from.

The new one is so much better, we're scared to show you the 'before' image 😉

## Updated dashboard definition YAML

In order to support more features now, and in the future, we have made some changes to the YAML format of the dashboard definitions. Don't worry, we've updated all your existing dashboards to the new format automatically. There's no need to update these yourself.

The new format is explained in detail on our dashboards definition YAML documentation page.

These are some notable changes in the format:

• The data on which the lines in a graph are based are now grouped under the metrics key on every graph. This makes it easier to configure fields and tags per metric.
• We've combined the old fields and filter keys into the name key on the metrics items. This still supports the wildcards from the filter key like before: name: 'my_metric_*_key'.
• The kind key has been renamed to fields. (The kind key was an old name we used for fields.) This now accepts an array of fields which makes it easy for measurements to show the mean, 90th percentile and 95th percentile in the same graph.
# Old YAML definition
title: 'Example dashboard'
graphs:
-
title: 'Query duration'
line_label: '%namespace% - mean'
format: duration
kind: measurement
fields:
- query_duration
tags:
namespace: "*"

# New YAML definition
title: 'Example dashboard'
graphs:
-
title: 'Query duration'
line_label: '%namespace% - %field%'
format: duration
metrics:
-
name: query_duration
fields:
- MEAN
- P90
- P95
tags:
namespace: "*"

We think these changes make it easier to edit the custom metrics dashboards and customize them even more! Let us know if and how we can do even better!

If you have feedback about this or any other feature, want to suggest changes or share something you like, please let us know. Your input is extremely valuable to us!


text:
## algorithm analysis – \$Phi_1=1\$ or \$Phi_1=2\$ for the dynamic \$text{Table-Insert}\$ , where \$Phi_i\$ is the potential function after \$i\$ th operation, as per CLRS

The following is the section from Introduction to Algorithms by Cormen. et. al. in the Dynamic Tables section.

In the following pseudocode, we assume that $$T$$ is an object representing the table. The field $$table(T)$$ contains a pointer to the block of storage representing the table. The field $$num(T)$$ contains the number of items in the table, and the field $$size(T)$$ is the total number of slots in the table. Initially, the table is empty: $$num(T) = size(T) = 0$$.

$$text{Table-Insert(T,x)}$$

$$1quad text{if size(T)=0}$$

$$2quadquad text{then allocate table(T) with 1 slot}$$

$$3quadquad size(T) leftarrow 1$$

$$4quadtext{if } num(T) =size(T)$$

$$5quadquadtext{then allocate new-table with 2 • size(T) slots}$$

$$6quadquadquadtext{insert all items in table(T) into new-table}$$

$$7quadquadquadtext{free table(T)}$$

$$8quadquadquad table(T) leftarrow new-table$$

$$9quadquadquad size(T) leftarrow 2 • size(T)$$

$$10quad text{insert x into table(T)}$$

$$11quad num(T) leftarrow num(T) + 1$$

For the amortized analysis for the a sequence of $$n$$ $$text{Table-Insert}$$ the potential function which they choose is as follows,

$$Phi(T) = 2.num(T)-size(T)$$

To analyze the amortized cost of the $$i$$th $$text{Table-Insert}$$ operation, we let $$num_i$$ denote the number of items stored in the table after the $$i$$ th operation, $$size_i$$ denote the total size of the table after the $$i$$ th operation, and $$Phi_i$$ denote the potential after the $$i$$th operation.

Initially, we have $$num_0 = 0, size_0 = 0$$, and $$Phi_0 = 0$$.

If the $$i$$ th Table-Insert operation does not trigger an expansion, then we have $$size_i = size_{i-i}$$ and $$num_i=num_{i-1}+1$$, the amortized cost of the operation is $$widehat{c_i}$$ is the amortized cost and $$c_i$$ is the total cost.

$$widehat{c_i}=c_i+Phi_i- Phi_{i-1} = 3 text{ (details not shown)}$$

If the $$i$$ th operation does trigger an expansion, then we have $$size_i = 2 . size_{i-1}$$ and $$size_{i-1} = num_{i-1} = num_i —1$$,so again,

$$widehat{c_i}=c_i+Phi_i- Phi_{i-1} = 3 text{ (details not shown)}$$

Now the problem is that they do not make calculations for $$widehat{c_1}$$ the situation for the first insertion of an element in the table (line 1,2,3,10,11 of code only gets executed).

In that situation, the cost $$c_1=1$$, $$Phi_0=0$$ and $$num_1=size_1=1 implies Phi_1 = 2.1-1 =1$$

We see that $$Phi_1=1 tag 1$$

So, $$widehat{c_1}=c_1+Phi_1-Phi_0=2$$

But the text says that the amortized cost is $$3$$, (I feel they should have said the amortized cost is at most $$3$$, from what I can understand)

Moreover in the plot below,

The text represents graphically the $$Phi_1=2$$ which sort of contracts $$(1)$$, but as per the graph if we assume $$Phi_1=2$$ then $$widehat{c_i}=3, forall i$$

I do not quite get where I am making the fault.

## encryption – Wrap key operation in Azure Key Vault – symmetric keys

Could anyone explain why the bolded part of the wrap key description?

Wraps a symmetric key using a specified key. The WRAP operation
supports encryption of a symmetric key using a key encryption key that
has previously been stored in an Azure Key Vault. The WRAP operation
is only strictly necessary for symmetric keys stored in Azure Key
Vault since protection with an asymmetric key can be performed using
the public portion of the key.
This operation is supported for
asymmetric keys as a convenience for callers that have a key-reference
but do not have access to the public key material. This operation
requires the keys/wrapKey permission.

AFAIK, all the keys in Azure Key Vault are stored at rest in HSM modules. Why is key wrapping necessary for symmetric keys? What does ‘protection’ mean in this case? Using a public key to encrypt data?

If HSM are securing all the keys in Key Vault (using its built-in symmetric key), then why would encrypting a symmetric key be necessary as quoted?

## symbolic – How to define a function based on an operation with parameters

I would like to define a function as the output of another operation depending on a few parameters. For example

``````s[t1_,t2_]:= Integrate[t^n,{t,t1,t2}]
s[0,1]
``````

I only get back an echo s[0,1] rather than a function dependent on $$a, n$$. What is the correct format for the definition?

I tried the following.

``````s[n_,t1_,t2_]:= Integrate[t^n,{t,t1,t2}]
s[2,0,1]
``````

Edit:
I got a conditional expression after a long wait. How do I specify $$n$$ to be a real number for this integral?

## linear algebra – Matrix operation to get a ‘reflected’ upper-triangular matrix from of an \$n!times! n\$ matrix

If I have an $$n!times! n$$ matrix $$mathbf{A}$$ with non-zero entries, is there a matrix operation (or a set of matrix operations) that results in the matrix $$mathbf{A’}$$ such that $$A’_{ij} = A_{ij}$$ if $$(i!+!j) leq (n!+!1)$$ and $$A’_{ij}=0$$ otherwise? For example, if $$mathbf{A}$$ is a $$3!times! 3$$ matrix given by
$$mathbf{A} = left(begin{matrix} A_{11} & A_{12} & A_{13} \A_{21} & A_{22} & A_{23} \A_{31} & A_{32} & A_{33} end{matrix}right),$$
are there matrix operations to get
$$mathbf{A’} = left(begin{matrix} A_{11} & A_{12} & A_{13} \A_{21} & A_{22} & 0 \A_{31} & 0 & 0 end{matrix}right)?$$
Apologies for my use of the phrase “reflected upper-triangular”, as I did not know how else to refer to this kind of a matrix.

## nt.number theory – Maximum number by AND operation

Help me to solve this

You are given an array of N numbers say (A1,A2,⋯An). Let us define a function
F(x)=∑i=1 to n (Ai&x)
where & is bitwise AND operator. We needs to find the number of different values of x for which the value of F(x) is maximised. But there is a constraint for x that it must have exactly L bits-set in its binary representation. Further, print -1 if infinite number of X possible that are fulfilling the constraint

For example. if array is (3,5,7,1,4) and L=2 then answer is 2 because 5 and 6 are numbers with 2 set bits in their binary representation and gives maximum value after perform (A1&x)+(A2&x)+(A3&x)+⋯+(An&x). So the Values for x are 5 and 6 and answer is 2.

Another example is array is (3,5,7,1,4) and L=1 then answer is 1 because only x=4 satisfied the constraints.

## javascript – Minimize sum of array by applying specific operation on x carefully chosen elements

First let me point out a few details.

Use `const` intead of `let` unless you are going to modify the value after initialization. You use it for the `ofNumber` variable, but there are more that deserve it.

But actually there’s often no need to define a variable at all if it is used only once. Similarily, storing a return value to a variable and immediately returning that variable is redundant, just return the value returned by the function directly.

``````function operation(max) {
return Math.floor(max / 10);
}
``````

But you could also stay consistent with the `ofNumber` callback like this:

``````const operation = (max) => Math.floor(max / 10);
``````

Another thing is that we usualy use `for` loop in theese cases.

``````for (let ops = x; ops > 0; --ops) {...}
``````

Now, let’s analyse the big-O time complexity of your algorithm.
Adding a comment above each statement in your code. `n=nums.length`.

``````function minSum(nums, x) {
// O(1)
if (nums.length === 0) {
return false;
}
function operation(max) {
// O(1)
let redcuedMax = Math.floor(max / 10);
return redcuedMax
}
let ops = x;
// O(x * inner)
while (ops > 0) {
// O(n)
let max = Math.max(...nums);
// O(1)
const ofNumber = (element) => element >= max ;
// O(n)
let maxIndex = nums.findIndex(ofNumber)
// O(1)
let operated = operation(max);
// O(1)
nums(maxIndex) = operated;
// O(1)
ops--
}
// O(n)
return nums.reduce((prev,next) => prev + next, 0)
}
``````

That makes for `O(x * n)` in time. `O(1)` in space of course, because you are never making any copies of the array.

How can we optimize this?

Well first thing i see is that there are 2 `O(n)` operations in the loop body.
Maybe we can find the maximum’s element index in `O(n)`.And if we do, we can access the maximum element in `O(1)`. This optimization will be less efective if the input is sorted or almost sorted in descendant order, because the second `O(n)` operation is basically `O(1)` for such sorted input.

Another thing is that after the loop, there is another `O(n)` operation. Maybe we can keep track of the sum (updating it in `O(1)` time) ever since the first time we needed to scan the entire array. Although, this optimization is `x` times less significant, for small `x` it may help.

Of course, the most significant improvement can only arise from changing the entire algorithm’s big-O complexity from `O(x * n)` to something with slower rate of change. Even if it costs us increased memory complexity to say `O(n)`.

To do that we have to leave the code for now and let’s think about the problem itself.

You wrote:

As a result, you have to perform the operation on the elements with the highest values first.

Good. But is there more? How many of the highest elements will you actually need?

At most `x`, right? Either the highest element divided by 10 remains the highest element, in which case you continue with that one, or the next highest element will become the current highest. So maybe we dont want to track just 1 highest element, but `x` of them. This may raise our memory complexity to `O(min(x,n))`, but that would be still a good tradeoff.

Well, and I think I will break off at this point. I don’t wanna write it for you. I hope I gave you enough hints to come up with a faster solution on your own. Just one more thing to say, don’t be afraid to use your own specialized loops in such optimizations even if it means your code will grow. It’s always trade-offs. Time, space, readability/code size, … you improve one, you loose on the other… well sometimes not if you got it very wrong on the first shot 😀 (not saying this is the case:)).

## nist – How to deside what model should be picket for security operation center, design and implimentation?

To pick the right model for design and implement a Security Operation Center, it should pick a most suitable model that is for the business, that could be capable to be tailored.

What are the differences between, Best practices, standards, or frameworks in SOC design?

## summation – Donald Knuth and change of variable operation on sum

I am reading the very first Donald Knuth book on Algorithms, chapter 1.2.3. “Sum and Products” (p.30).

Donald Knuth introduces operations on sum. But I can’t fully understand the 1 example that utilizes two operations:

• 2. change of variable (particularly confused by)
• 4. manipulating the domain

An example of change of variable that fully understand the explanation:

$$sumlimits_{R(i)}a_i = sumlimits_{R(j)}a_j = sumlimits_{R(p(j))}a_{p(j)}$$

However, I can’t understand the very first example given in the book: how did we come up with $$2j$$ and how this stores all the variables by changing the variable?
The steps are as follows:

$$sumlimits_{0le jle n}a_j=sumlimits_{0le jle n,j,even}a_j + sumlimits_{0<=j<=n ,j,odd}a_j$$ and then $$sumlimits_{0<=2j<=n j,even}a_{2j} + sumlimits_{0<=2j+1<=n,j,odd}a_{2j+1}$$
$$=sumlimits_{0<=j<=n/2,j,even}a_{2j} + sumlimits_{0<=j<=n/2 2j+1,j,odd}a_{2j+1}$$

## reference request – Algebraic geometry additionally equipped with field automorphism operation. Ideal Membership?

I am looking for some facts on theory, which is essentially algebraic geometry but with field automorphisms added as ‘basic’ operations. (Precisely, I mean universal algebraic geometry for (universal) algebra being a field $$mathbb{F}$$ and operations being polynomials and field automorphisms) I am also interested in computational aspects.

I am mainly interested in case of field of multivariate rational functions, i.e. $$mathbb{F} = K(x,y)$$.
An algebraic set in such setting could be $${(f,g) in (K(x,y))^2 mid x^2-y^2-f cdot g(x=frac{x+y}{2}, y = frac{x-y}{2}) = 0}.$$ and ideals are additionally closed under field automorphisms.

I would be grateful for reference to any source that considers theory like this, also the computational aspects.

## What is a system operation and how do they relate to use cases?

This is in a Systems Analysis and Design context. I did search google and everything on the first 2 pages returned results for “operating systems”.


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Classes of meromorphically multivalent functions associated with the generalized hypergeometric function. (English) Zbl 1049.30008
Making use of a linear operator, which is defined here by means of a Hadamard product (or convolution) involving the generalized hypergeometric function, the authors introduce and investigate two novel classes of meromorphically multivalent functions. In recent years, many important properties and charatcteristics of various interesting subclasses of the class ${{\Sigma }}_{p}$ of meromorphically $p$-valent functions were investigated extensively by (among others) Aouf et al. Joshi and Srivastava, Kulkarni et al., Liu and Srivastava, Owa et al., Srivastava et al., Uralegaddi and Somanatha, and Yang. The main object of this paper is to present several inclusion and other properties of functions in the classes ${{\Omega }}_{p,q,s}\left({\alpha }_{1};A,B\right)$ and ${{\Omega }}_{p,q,s}^{+}\left({\alpha }_{1};A,B\right)$ which the authors introduced here. They also apply the familiar concept of neighborhoods of analytic functions to meromorphically $p$-valent functions in the class ${{\Sigma }}_{p}$. In the first results the authors apply the well-known Jack’s Lemma.
##### MSC:
 30C45 Special classes of univalent and multivalent functions 33C20 Generalized hypergeometric series, ${}_{p}{F}_{q}$