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Perlin Noise: Part 2

There is very few documents on the Web and few books that explain the Perlin noise method in an intuitive way; but finding information on how to compute Perlin noise derivatives (especially in an analytical way) is even harder. Though, for those of you who don't know what these derivatives are and why they are useful, let's first go through a quick introduction on derivatives.

## A Quick Introduction to (Partial) Derivatives

Figure 1: computing the noise function variation along the horizontal and vertical axis using a discrete approach.

Derivatives of any function (whether it is a one-, two- or three-dimensional function) are very useful. But before we give an example, let's first review what they are. If you create an image of a 2D noise and apply some sort of regular grid on top of that image, then we may want to know by how much the noise function varies along the x and y direction at each point of the grid (figure 1). Maybe this idea sounds familiar already? Remember that in the previous chapter, we used the result of a 2D noise function to displace a mesh. But let's get back to what we are trying to achieve here: how do know the rate of change of our 2D noise function along the x- or y-axis? A very simple solution to this problem consists of taking the value of the noise at the point where you want to compute this variation (let's call this point $Gn_x$), the value of the noise at the point a step further to the right from $Gn_x$ (let's call this second point $Gn_{x+1}$), and then subtract the second value from the first. Example: if at the grid position $Gn_{11}$ the noise is equal to 0.1 and at that the grid position $Gn_{12}$ the noise value is equal to 0.7, then we can assume that the noise has varied from $Gn_{11}$ to $Gn_{12}$ (along the x-axis) by 0.6 (figure 1). In equation form, we would write:

$$\Delta_x Gn_{11} = Gn_{12} - Gn_{11}.$$
Of course to compute the rate of change along the y-axis, we would be doing the same thing but move a step downward. Our equation would look like $\Delta_y Gn_{11} = Gn_{21} - Gn_{11}$.

Technically it is best to normalize this difference so that we get consistent results regardless of the distance that separates two points on the grids (if the results are normalised, measurements made with different grid spacing can then be compared to each other). To normalize this result, we just need to divide the difference by the distance between $G_{11}$ and $G_{12}$. So if the distance between two points on the grid is 2 for example, then we would need to write:

$$\Delta_x Gn_{11} = {\dfrac{Gn_{12} - Gn_{11}}{2}}.$$

In mathematics this technique is called a forward difference. Forward because we take the next computed point and subtract the value at the current point from the value at the next point.

Backward differencing is also possible: you use the previous point instead of the next point. You can also use a central difference.

Mathematically we can formalize this concept with the following equation:

$$f'(x) = \lim_{h \to 0}{\dfrac{f(x+h) - f(x)}{h}}.$$
The notation $f'(x)$ means that this is the derivative of the function $f(x)$.

Figure 2: the smaller the spacing, the more accurate the computation of the rate of change.

This means that we can compute the derivative of the function $f(x)$ using the forward difference technique that we just introduced but that the value of this derivative will become more and more accurate as the distance between the two points becomes smaller (in theory $h$ tends toward 0). When the spacing is large you get some sort of very crude value for what the derivative is at a given point $x$ but as the spacing becomes small this approximation improves. In the case of our noise image, the spacing of the grid is pretty large, so in fact, you would get a much better approximation of the variation of the noise function at each point on the grid, if the grid spacing was smaller (figure 2).

Figure 3: computing the tangent at P using central difference works better if we take smaller values for $h$.

This concept is more easily understood with a 1D example. Figure 3 shows the profile of one-dimensional function. Let's assume that we now want to know by how much this function varies within the proximity of P. Using the principle of forward differencing, we can take a point further down along the x-axis such as for example $x_1$, compute the value of the function at that point and then subtract $f(x_1)$ to $f(x)$. Note that that when we trace a line from $f(x)$ to$f(x_1)$ that line is tangent to the function at $x$. That's because in fact, the derivative of a one-dimensional function gives the slope of the line tangent to the function where the function's derivative is being computed. So now that we know how to geometrically interpret the derivative of a function, you can easily see that if we take a point further away than $x_1$ such as for example $x_2$, then the line between $x$ and $x_2$ is not "as tangent" to $x$ than is the line $x$-$x_1$. Conclusion, if you use forward difference to compute the derivative of a function, then the smaller the distance between $x$ and $x+h$ the better.

What have we learned so far?

• We learned that the derivative $f'(x)$ of a one-dimensional function $f(x)$ can be interpreted as the slope of the line tangent to the function $f(x)$ at $x$
• Beside the slope, the derivative $f'(x)$ can also be considered as the instantaneous rate of change of the function $f(x)$ at $x$.
• We also learned that we could use a technique called forward difference (the general method is called finite difference) to compute a "approximation" of that slope, but also that this approximation gets better as $h$ in the forward difference equation gets smaller.

Figure 4: visual representation of the function derivative or slope.

There is something really important to understand in order to make sense of how we will be using derivatives (actually partial derivatives) later on this chapter. So far, we explained that the derivative of a function can be interpreted as the slope of the function $f(x)$ at any value of $x$. The way we trace the tangent at $x$ (where we computed the derivative of the function $f(x)$) is by simply drawing the line $y = mx$ at the point on the function where we computed the derivative of the function. The value $m$ here is of course the slope of the function derivative we computed. Why is this important? It's important because note that when $x=1$ then $y=m$. This means that using this observation, we can say that the 2D vector tangent to the curve at the point where we evaluated the derivative is equal to Vec2f(1, m). Do you agree? Of course you then need to normalize this vector, but nonetheless note how this vector is tangent to the point where we evaluated the function's derivative (and that's what we want you to remember). It is important you understand this idea (which is illustrated in figure 4).

Figure 5: computing the normal by taking the cross product of the tangent and bi-tangent.

You can see the 3D Perlin noise function as two 2D functions perpendicular to each other at the point where the derivative is computed. So you will have a 1D function to compute the derivative of the 2D function in the xy plane if you wish, and another 1D function to compute the derivative of 2D noise function in the yz plane as showed in figure 5. Now if we apply the technique we just learned to compute the tangent at each one of these functions, note that the two obtained tangents are perpendicular to each other. But more interestingly by now taking the cross product of these two vectors you get a vector which is in fact perpendicular to the plane tangent to the point where the 2D noise function derivative was originally computed (as shown again in figure 5). This vector is the normal of our function at P. Hopefully by now you start to get it. These derivatives are going to be useful to compute the normals of our mesh displaced by a 3D or 2D noise function.

This is an important result, because this is how we are going to compute the normal of a mesh displaced by a 3D noise function. We will evaluate the derivative of the function along the x-axis at a given point (say P), then compute the vector tangent to P along the x-axis (let's call this vector $T_x$). Then we will compute the derivative of the noise function along the z-axis at P, from there compute the tangent to P along the z-axis (let's call this vector $T_z$). Finally we will use the two vectors in a cross product to compute the normal at P: $$N_P = T_z \times T_x.$$

This is simple and elegant. Now one remark and one question.

Remark: note that we don't really compute the derivative of the noise function here. We sort of cheat by computing a derivative of the function along the x-axis and then another derivative along the z-axis. What's interesting when we do that is that only one of the quantities varies. For example when compute the derivative of the 3D noise function along the x-axis then of course the value we get doesn't change because of a variation of the noise along the z-axis (since we evaluate the noise function in the plane xy - aka there's no variation in z). In mathematics when you have a function with several variables but that you compute its derivative with respect to one of its variables only, with the others held constant, then we say that we compute the function's partial derivatives. Let's take an example, if you have the function:

$$f(x,y) = x^2 + xy + y^2$$

Then if you wish to compute the derivative of the function while holding $y$ constant then you get:

$${\dfrac{f(x,y)}{\partial x}} = 2x + y,$$

which you can read as the function $f(x,y)$ partial derivative with respect to $x$. In other words, we ignore all the terms in which $x$ doesn't show up such as $y^2$ in our example (including any constant term), then compute the derivative of $x$ for each term in which $x$ shows up. For example, if in one of the terms we have a $x^2y$ then we replace it in the partial derivative by $2xy$. If we have $xy$, then we replace the term in the partial derivative by $y$. Simple?

Question: how do we compute these partial derivatives then?

Figure 6: xx.

One method consists of using the forward difference technique (which is why we learned about it at the beginning of this chapter). If we take the example of our displaced meshed, then we can compute the derivate along the x-axis at the vertex $V_{x,z}$ by subtracting the derivate from the noise value at the vertex $V_{x+1,z}$ from the noise value at the vertex $V_{x,z}$ (figure 6). In other words we can write:

$$\partial Nx = N_{Vx+1,z} - N_{Vx,z}.$$

Where $\partial Nx$ is the partial derivative of the noise function along the x-axis. Note that at this point in time, this is a real value, not a vector. We can similarly compute the partial derivative of the noise function along the z-axis:

$$\partial Nz = N_{Vx,z+1} - N_{Vx,z} .$$

The question is now, how do we transform this real value (in other words a float) into a vector (the vector tangent to the noise function at the point where the derivative is computed along the x- and z-axis). Well this is simple. Remember that when we compute the partial derivative along the x-axis we work in the xy plane. Thus the z-coordinate of the vector tangent to the noise function along the x-axis is necessarily 0:

$$T_x = \{?, ?, 0\}.$$

To compute the other two coordinates, you need to look at figure 4 again where we explained that to compute the tangent to the function in a plane, you need to set the x coordinate to 1 and the y coordinate of the vector to function partial derivative value (if you want to compute the tangent of the vector in the yz plane then you need to set the z-coordinate of the tangent to 1, the y-coordinate of the vector to the function partial derivative with respect to z, and then the x-coordinate of the tangent vector to 0). Finally we have:

$$\begin{array}{l} T_x = \{1, Nx, 0\}\\ T_z = \{0, Nz, 1\}. \end{array}$$

Finally to compute the normal of the vertex, all you need to do now is to compute the cross product of these two vectors (the result of the cross product will be correct even of the two input vectors are not normalized: the resulting vector will be perpendicular to the two input vectors, though it might not be normalized itself):

$$Normal_{Vx,z} = T_z \times T_x.$$

This technique works great but:

• First what happens when we want to compute the derivatives for the vertices at the edges of the grid? Well you can't.
• We also explained (figure 3) that the smaller the space between two samples when we compute the derivative using a forward difference, the more accurate the result. In other words the larger the space between the vertices, the less accurate the computation of the partial derivatives. Later on in this chapter we will show the difference between the partial derivative computed with a forward difference and the analytical solution that we are now going to study.

## Analytical Partial Derivatives of the Perlin Noise Function

So there is a better way of computing these partial derivatives. This technique only relies on maths and provides an "accurate" solution (in the mathematical sense of the term). A quick reminder: the partial derivative of the following equation: $$f(x,y) = x^2 + xy + y^2$$

with respect to x is:

$${\dfrac{f(x,y)}{\partial x}} = 2x + y.$$

Now let's re-write the Noise function a little. Let's first replace all the dot products with letters (as shown below):

a = dot(c000, p000) b = dot(c100, p100) c = dot(c010, p010) d = dot(c110, p110) e = dot(c001, p001) f = dot(c101, p101) g = dot(c011, p011) h = dot(c111, p111)

For the sake of the exercise let's recall that the parameters $u$, $v$ and $w$ are computed as follows (we use the smoothstep function):

u = tx * tx * (3 - 2 * tx) v = ty * ty * (3 - 2 * ty) w = tz * tz * (3 - 2 * tz)

Let's now write the different interpolations of the Perlin noise function into one single line (see the first chapter):

lerp( lerp( lerp(a, b, u), lerp(c, d, u), v) lerp( lerp(e, f, u), lerp(g, h, u), v), w)

Let's now replace the call to the lerp(a, b, t) function with its actual code (a(1-t)+bt) and develop:

((a(1 - u) + bu)(1 - v) + (c(1 - u) + du)v)(1 - w) + ((e(1 - u) + fu)(1 - v) + (g(1 - u) + hu)v)w ((a - au + bu)(1 - v) + (c - cu + du)v)(1 - w) + ((e - eu + fu)(1 - v) + (g - gu + hu)v)w (a - au + bu - av + auv - buv + cv - cuv + duv)(1 - w) + (e - eu + fu - ev + euv - fuv + gv - guv + huv)w a - au + bu - av + auv - buv + cv - cuv + duv - aw + auw - buw + avw - auvw + buvw - cvw + cuvw - duvw + ew - euw + fuw - evw + euvw - fuvw + gvw - guvw + huvw

And then finally regroup the terms as follows:

a + u(b - a) + v(c - a) + w(e - a) + uv(a + d - b - c) + uw(a + f - b - e) + vw(a + g - c - e) + uvw(b + c + e + h - a - d - f - g)

As you can see (and as expected) this is a function of three variables: $u$, $v$ and $w$. If we apply the technique we learned to compute the partial derivative of a function with respect to one of its variable, we need to remove all the terms that do not contain the variable in question, and then replace the variable by its derivative in the remaining terms. For example, if we wish to compute the noise function partial derivative with respect to $u$ we get:

u'(b - a) + u'v(a + d - b - c) + u'w(a + f - b - e) + u'vw(b + c + e + h - a - d - f - g) u'((b - a) + v(a + d - b - c) + w(a + f - b - e) + vw(b + c + e + h - a - d - f - g))

Similarly, the partial derivatives with respect to $v$ and $w$ are:

// partial derivative with respect to v v'(c - a) + uv'(a + d - b - c) + v'w(a + g - c - e) + uv'w(b + c + e + h - a - d - f - g) v'((c - a) + u(a + d - b - c) + w(a + g - c - e) + uw(b + c + e + h - a - d - f - g)) // partial derivative with respect to w w'(e - a) + uw'(a + f - b - e) + vw'(a + g - c - e) + uvw'(b + c + e + h - a - d - f - g) w'((e - a) + u(a + f - b - e) + v(a + g - c - e) + uv(b + c + e + h - a - d - f - g))

The remaining question is what is the derivative of $u'$, $v'$ and $w'$? Well simple $u$, $v$ and $w$ are computed as follows:

$$\begin{array}{l} u = 3tx^2 - 2tx^3\\ v = 3ty^2 - 2ty^3\\ w = 3tz^2 - 2tz^3\\ \end{array}$$

Thus the derivatives of these functions are:

$$\begin{array}{l} u' = 6tx - 6tx^2\\ v' = 6ty - 6ty^2\\ w' = 6tz - 6tz^2\\ \end{array}$$

Et voila! All you need to do now is compute these derivatives, and then construct the vectors tangent to the point where the function is evaluated using the technique we provided above. Here is a modified version of our eval function that evaluates the 3D noise function and its partial derivatives at a given location:

## Analytical Solution vs Forward Difference

We can now compete two versions of the displaced mesh, one using the geometric solution to compute the vertex normals of the mesh, and one using the analytic solution. To code to compute either one of these solutions is as follows:

int main(int argc, char **argv) { PerlinNoise noise; PolyMesh *poly = createPolyMesh(3, 3, 30, 30); // displace and compute analytical normal using noise function partial derivatives Vec3f derivs; for (uint32_t i = 0; i < poly->numVertices; ++i) { Vec3f p((poly->vertices[i].x + 0.5), 0, (poly->vertices[i].z + 0.5)); // displace current vertex poly->vertices[i].y = noise.eval(p, derivs); #if ANALYTICAL_NORMALS Vec3f tangent(1, derivs.x, 0); // tangent Vec3f bitangent(0, derivs.z, 1); // bitangent // equivalent to bitangent.cross(tangent) poly->normals[i] = Vec3f(-derivs.x, 1, -derivs.z); poly->normals[i].normalize(); #endif } #if !ANALYTICAL_NORMALS // compute face normal if you want for (uint32_t k = 0, off = 0; k < poly->numFaces; ++k) { uint32_t nverts = poly->faceArray[k]; const Vec3f &va = poly->vertices[poly->verticesArray[off]]; const Vec3f &vb = poly->vertices[poly->verticesArray[off + 1]]; const Vec3f &vc = poly->vertices[poly->verticesArray[off + nverts - 1]]; Vec3f tangent = vb - va; Vec3f bitangent = vc - va; poly->normals[poly->verticesArray[off]] = bitangent.cross(tangent); poly->normals[poly->verticesArray[off]].normalize(); off += nverts; } #endif poly->exportToObj(); delete poly; return 0; }

Here is an image of these two meshes with their associate vertex normals (displayed on the right):

Note that the vertex normals are not defined at the edge of the mesh whose normals were computed using the forward difference method (geometric solution). Note also the directions of the vertex normals are significantly different between the two meshes (even though their shape is the same). This obviously causes the shading of the two meshes to be noticeably different as well (remember that normals are used in shading).

Note that you can replace the code:
poly->normals[i] = bitangent.cross(tangent);
with:
poly->normals[i] = Vec3f(-derivs.x, 1, -derivs.z);
Why? Because we know that: $$\begin{array}{l} Tx=\{1,Nx,0\}\\ Tz=\{0,Nz,1\}. \end{array}$$ If you replace these two vectors in the equation of the cross product: $$\begin{array}{l} x = va.y * vb.z - va.z * vb.y\\ y = va.z * vb.y - va.x * vb.z\\ z = va.x * vb.x - va.y * vb.x, \end{array}$$ you will see that you will end up with: $$N = Tz \times Tx = \{-Nx, 1, -Nz\}$$


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# Mechanics - Work and Kinetic Energy question

1. Jan 18, 2009

### thehammer

1. The problem statement, all variables and given/known data
A 3kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x axis along the track is applied to the body. The force F is applied to the body at t = 0 when the body is at the origin. A stroboscope records the position of the body at 0.5 s intervals. The readings are as follows.

Interval (I) 1: 0.04 m
I2: 0.2 m
I3: 0.44 m
I4: 0.8 m

How much work is done on the body by the applied force F between t = 0 and t = 2s?

2. Relevant equations
K = 1/2mv^2
W = $$\Delta$$K
s = ut + 1/2at^2
v^2 = u^2 + 2as

3. The attempt at a solution

I really have no idea why I'm not getting the right answer. I can't see an error in what I've done.

1) Find a
u = 0

a = 2s/t^2
a = 2 x 0.04 / 0.5^2 = 0.32 ms^-2

2) Find v^2

v^2 = ((0 ms^-1)^2) + 2(0.32 ms^-1 x 0.8m)

3) Find the change in kinetic energy and hence the work done

$$\Delta$$K = 1/2mv^2 = 1/2(3kg)(0.512m^2 s^-2) = 0.768 J

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The answer is meant to be 0.96J. I think I might have done this question months ago but I cannot see where I have gone wrong in the least. Please point out any glaring mistakes :P.

2. Jan 18, 2009

### Staff: Mentor

You didn't make any mistakes, but the data seems to be off a bit. (The acceleration is not constant.) Try finding the acceleration using the I4 data.

3. Jan 18, 2009

### thehammer

I did what you suggested and got the question right. Thank you Doc Al. I've a few of questions about this problem.

If the acceleration is not constant then it follows that the force must be variable. This contradicts the first part of the question in which it is stated that the force is constant. Such a situation would require the use of calculus.

In this instance, I did not have to use calculus and got the correct answer. I initially got the answer wrong because I use the first time interval to work out the acceleration, which I assumed was constant. Why does this method work if the acceleration is nonlinear? Is it the average acceleration that is found using this method?

4. Jan 18, 2009

### Staff: Mentor

Correct. I think the explanation is simply that whoever made up this problem messed up the data. For example, if the data were:
I1 = 0.05
I2 = 0.2
I3 = 0.45
I4 = 0.8
then the acceleration would be a constant 0.4 m/s^2.

What textbook is this from?

5. Jan 18, 2009

### thehammer

The question is from "Fundamentals of Physics, 7th Edition," Halliday, Resnick and Walker.

6. Jan 18, 2009

### Staff: Mentor

What chapter and problem number?

7. Jan 18, 2009

### thehammer

Chapter 7, question 9. I sincerely hope I didn't write the distance down incorrectly, haha. I'm fairly sure I didn't.

8. Jan 18, 2009

### Staff: Mentor

Looks like I don't have that particular edition, but I suspect a typo was involved.

9. Jan 18, 2009

### thehammer

I hope so. This problem was really getting at me earlier when it shouldn't have been because of its simplicity!

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Composition Series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:12:50Z http://mathoverflow.net/feeds/question/104911 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104911/composition-series Composition Series solovei 2012-08-17T13:00:50Z 2012-08-18T08:15:15Z
Which finite groups have uniqueness for the ordered sequence of composition factors (up to isomorphism)?
 http://mathoverflow.net/questions/104911/composition-series/104947#104947 Answer by Mark Sapir for Composition Series Mark Sapir 2012-08-17T21:14:20Z 2012-08-18T08:15:15Z
Here is a characterization. A group \$G\$ has two different composition series if and only if it has a factor \$H/K\$ which is a direct product of two non-isomorphic simple subgroups, where \$H\$ is a subnormal subgroup of \$G\$, \$K\$ is a normal subgroup of \$H\$. Indeed, if such \$H/K\$ exists, then clearly there are two different composition series. Conversely, suppose that there are two different composition series \$A_0=1 < A_1 < A_2 < ... < A_n = G\$ and \$B_0=1 < B_1 < B_2 < ... < B_n=G\$. Let \$j\$ be the last index with \$A_{j-1}\ne B_{j-1}\$ (non-isomorphic), \$n\ge j\ge 1\$. Let \$H=A_j=B_j\$. Note that \$A_{j-1}\$ and \$B_{j-1}\$ are normal in \$H\$. Hence \$A_{j-1}B_{j-1}\$ is normal in \$H\$. Since it is bigger than \$A_{j-1}\$, we have \$H=A_{j-1}B_{j-1}\$. Hence \$K=A_{j-1}\cap B_{j-1}\$ is normal in \$H\$ and \$A_{i-1}/K\$ is isomorphic to \$H/B_{j-1}\$ hence simple. Similarly \$B_{j-1}/K\$ is simple. Now \$H/K\$ has two different normal simple subgroups \$A_{j-1}/K\$ and \$B_{j-1}/K\$ with trivial intersection, so \$A_{j-1}/K\$ and \$B_{j-1}/K\$ commute and form a direct product. Therefore \$H/K\$ is isomorphic to the direct product of two non-isomorphic simple groups \$A_{j-1}/K\$ and \$B_{j-1}/K\$.


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ISSN 0439-755X
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Special Issue on “Psychological Characteristics and Behaviors of Chinese People in Response to Crisis and Challenges”
 Zhong-yong action self as a contributing factor to COVID-19 crisis management YANG Chung Fang 2023, 55 (3):  355-373.  doi: 10.3724/SP.J.1041.2023.00355 Abstract ( 199 )   This paper adopts an indigenous approach to explain why China can contain the COVID-19 Crisis swiftly and efficiently. For this purpose, it proposes a new conceptualization for studying the Chinese self—the Zhong- yong action self.The action self refers to the self, activated by the situation an individual is facing, based on which the actor thinks about and decides the proper action to take. During the COVID-19 Crisis, beside the individuated self (the small self), many other more inclusive selves (the large selves), such as the family self, the community self, and the country self, are being mobilized at the same time, all of which demand the actor to exercise self-control and to help others to achieve the common goal—defeating the virus. This concerted effort thereby creates strength and flexibility in managing the crisis.In every-day life situation, the many selves activated may demand conflicting actions from the actor. An adoption of the Zhong-yong deliberation process negotiate the most appropriate action, to help maintain inner peace and outer harmony with others and the flux environment. The author hopes that this new formulation will lead to new directions to the study of “the Chinese self.”
 Social norm modulates the enhancement effect of behavioral visibility on altruistic preference HUANG Xinru, LI Jian, NI Yinmei 2023, 55 (3):  481-495.  doi: 10.3724/SP.J.1041.2023.00481 Abstract ( 38 )   HTML ( 10 )   PDF (374KB) ( 19 )   In social economic decisions, people not only care about their own payoffs but also the payoffs of others, a tendency termed altruistic preference. Numerous studies have shown that the sheer sense of being observed is sufficient to augment subjects’ altruistic choices. However, whether subjects’ altruistic behavior can be modulated by revealing their decision processes to other stake-holders (i.e., the receiver) remains unclear. In this study, we provide experimental evidence about the effects of both decision visibility to others and social norms on participants’ altruistic preference in two studies respectively. First, we confirmed that receiver visibility affected deciders’ altruistic preference in Study 1. In Study 2, we further showed that social norms modulated the effect of behavior visibility on deciders’ altruistic preference, suggesting a potential avenue via which social norms influences the relationship between behavioral visibility and altruistic preference.Study 1 implemented a 2 (Visibility: Visible vs. Invisible) × 2 (Reaction type: Choice vs. Rating) × 2 (Inequity type: AI vs. DI) within-subject design. We recruited 38 participants and they were required to either choose from two reward allocation options (Choice task) with another partner, or rate how satisfied concerning a particular allocation (Rating task) in a dictator game (DG). Participants’ behavior was either observed by their “partners” (Visible condition) or remained private (Invisible condition) (Figure 1). We provided both model-free and model-based evidence for the effects of visibility on altruistic preference. Specifically, we calculated the mean absolute differences between self- and other-payoffs in the chosen trials in Choice condition, and the mean rating scores in Rating condition, as model-free indicators (indicators were calculated in advantageous domain and disadvantageous domain separately, Table 1). As for model-based measures, we utilized the inequity aversion model $(U={{M}_{s}}-\alpha \cdot \max \{{{M}_{s}}-{{M}_{o}},0\}-$ $\beta \cdot \max \{{{M}_{o}}-{{M}_{s}},0\})$ to quantify social preferences (Table 2). Repeated measures ANOVA on model-free and model-based measures revealed similar patterns: Compared to the Invisible condition, participants exhibited greater altruistic preference when their behavior were visible to the receivers (partners), indicated by increased advantageous inequity aversion (AIA) and decreased disadvantageous inequity aversion (DIA) in the Visible condition. This tendency was significant across both Choice and Rating tasks. In addition, participants cared more about allocation efficiency in the Choice task than in the Rating task, indicated by decreased AIA and increased DIA. Finally, Visibility alleviated the behavioral discrepancies between the Rating and Choice tasks, indicating that social preference and choice strategies tend to converge in the Visible condition (Table 3 & 4). Study 2 implemented a 2 (Visibility: Visible vs. Invisible) × 2 (Social norm: Altruistic vs. Non-altruistic) within-subject design. 53 participants took part in the choice task with altruistic or non-altruistic social norms. Different social norms were manipulated by the proportions of unfair options chosen by supposedly previous participants (Altruistic social norm vs. Non-altruistic norm) (Figure 2). Measures remained the same as the choice condition in Study 1, except that we calculated the mean other-payoffs in the chosen trials as an additional measure of altruistic behavior (Table 5 & 6). ANOVA results exhibited an interaction between Visibility and Social norm: In the Altruistic social norm condition, visibility significantly increased participants’ altruistic preference; However, such effect diminished in the Non-altruistic social norm condition (Table 7 & 8).Our study revealed that deciders’ behavioral visibility to receivers increased altruistic preference and promoted altruistic behavior. Furthermore, altruistic social norm played a modulatory role on the visibility effect, supporting the signaling hypothesis of altruistic preference.


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# What is the derivative of y=2(x^(3)-1)(3x^(2)+1)^4?

Aug 11, 2015

${y}^{'} = 6 x \left(3 {x}^{2} + 1\right) \cdot \left(11 {x}^{3} + x - 8\right)$

#### Explanation:

You can differentiate this function by using the product rule and the chain rule.

If you write $y$ as

$y = f \left(x\right) \cdot g \left(x\right)$

then you can say that the derivative of $y$ is

$\frac{d}{\mathrm{dx}} \left(y\right) = 2 \cdot \left[\left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)\right]$

In your case, $f \left(x\right) = \left({x}^{3} - 1\right)$ and $g \left(x\right) = {\left(3 {x}^{2} + 1\right)}^{4}$.

${y}^{'} = 2 \left[\frac{d}{\mathrm{dx}} \left({x}^{3} - 1\right) \cdot {\left(3 {x}^{2} + 1\right)}^{4} + \left({x}^{3} - 1\right) \cdot \frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 1\right)}^{4}\right]$

To determine $\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$, use the chain rule for

• $g \left(u\right) = {u}^{4}$, with $u = 3 {x}^{2} + 1$

$\frac{d}{\mathrm{dx}} \left(g \left(u\right)\right) = \frac{d}{\mathrm{du}} \left({u}^{4}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(g \left(u\right)\right) = 4 {u}^{3} \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 1\right)$

$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 1\right) = 4 {\left(3 {x}^{2} + 1\right)}^{3} \cdot 6 x$

Your target derivative will thus be

${y}^{'} = 2 \cdot \left[3 {x}^{2} \cdot {\left(3 {x}^{2} + 1\right)}^{4} + \left({x}^{3} - 1\right) \cdot 24 x \cdot {\left(3 {x}^{2} + 1\right)}^{3}\right]$

${y}^{'} = 2 \cdot 3 x \cdot \left(3 {x}^{2} + 1\right) \cdot \left[x \cdot \left(3 {x}^{2} + 1\right) + \left({x}^{3} - 1\right) \cdot 8\right]$

${y}^{'} = 6 x \left(3 {x}^{2} + 1\right) \cdot \left(3 {x}^{3} + x + 8 {x}^{3} - 8\right)$

${y}^{'} = \textcolor{g r e e n}{6 x \left(3 {x}^{2} + 1\right) \cdot \left(11 {x}^{3} + x - 8\right)}$


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30
Dec

## definite integration formulas

An arbitrary domain value x. is selected in each subinterval and its corresponding subinterval length is calculated and these ‘n’ products are added to calculate their sums. With this Definite Integration Formulas list, you can learn definition, properties of definite Integrals, summation of series by intergration, and some other important formulas to solve complicated problems. Required fields are marked * Your rating. The definite integral is expressed as. $\int_0^1\frac{dx}{x^x}=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^3}+\cdots$, $\int_{-a}^a (a+x)^{m-1}(a-x)^{n-1}dx=(2a)^{m+n-1}\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. Standard formula: (a) ∫xn dx =. Let $f(x)$ be a defined integral in an interval $a\leq x\leq b$. ∫ = (− +). For example, if (fx) is greater than 0 on [a,b] then the Riemann sum will be the positive real number and if (fx) is lesser than 0 on [a,b], then the Riemann sum will be the negative real number. The definite integral f(k) is a number that denotes area under the curve f(k) from k = a and k = b. $+f(a+(n-1)\Delta x)\Delta x$. The upper limit (1) and the lower limit (-1) are the opposites. Get strong fundamentals of Definite Integration by using the available Definite Integration Formula cheat sheet. Pro Lite, Vedantu The introduction of the concept of a  definite integral of a given function initiates with a function f(x) which is continuous on a closed interval (a,b). Pages: 33. Sometimes we need to compute integral with a definite range of values, called Definite integrals. In this section, students will learn the main indefinite and definite integration formulas as well as some main properties of integration. This is called the mean value theorem for definite integrals and is valid if $f(x)$ is continuous in $a \leq x \leq b$. P 4 : ∫0→a f (x)dx = ∫0→a f (a - x) dx. Since the sequence () is decreasing and bounded below by 0, it converges to a non-negative limit. $\int_a^b f(x)\ dx\approx \frac{h}{2}(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n)$ Get strong fundamentals of Definite Integration by using the available Definite Integration Formula cheat sheet. This should explain the similarity in the notations for the indefinite and definite integrals. $\int^b_a f(x)\ dx=$ An arbitrary domain value xi is selected in each subinterval and its corresponding subinterval length is calculated and these ‘n’ products are added to calculate their sums. Revise all Integration Formulas (Definite Integration & Indefinite Integration) in 1 Shot By Neha Ma’am | Vedantu. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Integration is an important part of mathematics that was introduced earlier to differentiation. Therefore we have to apply the formula of integration by parts. Definite Integration by Parts Formula Definite integration by parts formula is generally used to integrate the product of two functions. In the above Definite integration by parts formula. As per the formula, we have to consider, dv/dx as one function and u as another function. P 3 : ∫a→b f (x) dx= ∫a→b f (a + b - x) dx. Explain the Meaning of Definite Integral and Indefinite Integral. This is a generalization of the previous one and is valid if $f(x)$ and $g(x)$ are continuous in $a\leq x\leq b$ and $g(x)\geq 0$. +8x 3 and -2x are odd functions. P 1 : ∫a→b f (x) dx = -∫b→a f (x) dx .In particular, ∫a→a f (x) dx = 0. Properties of definite integration. Introduction to Integral formulas: The list of integral calculus formula is here with all the rules which are needed to solve integration. The sum is known as Riemann sum and may be either positive, negative or zero relies upon the behavior of the function on a closed interval. So you can use the above formulas. So the odd functions, +8x 3 and -2x, are removed. In Maths, integration is a process of summing up parts to determine the whole. Evaluate $\int_{1}^{2}$ xdx/(x² + 2)³, The limit of integration can be transformed from x values to their corresponding  values. A Riemann integral is a definite integral where x is restricted to lie on the real line. It is also called antiderivative. Math Formulas and cheat sheet generator for definite integrals of trigonometric functions. When x= 1,u = 3 and when x =2 , u = 6, find, $\int_{1}^{2}$ xdx/(x² + 2)³ = ½ $\int_{3}^{6}$ du/u³, It is important to note that the substitution method is used to calculate definite integrals and it is not necessary to return back to the original variable if the limit of integration is transformed to the new variable values.’, 2. We will assume knowledge of the following well-known, basic indefinite integral formulas : , where is a constant , where is a constant Most of the following problems are average. Indefinite integral generally provides a general solution to the differential equation. Def. The definite integral of the function $$f\left( x \right)$$ over the interval $$\left[ {a,b} \right]$$ is defined as the limit of the integral sum (Riemann sums) as the maximum length … All letters are considered positive unless otherwise indicated. In fact, for all ≥: >, because it is an integral of a non-negative continuous function which is not identically zero; − + = ∫  − ∫ +  = ∫ () (− ) >, again because the last integral is of a non-negative function. The concept of definite integrals is frequently used for the real-world problems because it helps to measure or calculate the finite area in a plenty of cases. Some of the solved examples of definite integrals are given below: 1. 5x 4, -6x 2, and +7 are even functions. Indefinite integrals are antiderivative functions. With this Definite Integration Formulas list, you can learn definition, properties of definite Integrals, summation of series by intergration, and some other important formulas to … for and the right-hand endpoint formula is equation (**) for . Basic integration formulas on different functions are mentioned here. $\int_{\phi_1(\alpha)}^{\phi_2(\alpha)}\frac{\partial F}{\partial\alpha}dx+F(\phi_2,\alpha)\frac{d\phi_1}{d\alpha}-F(\phi_1,\alpha)\frac{d\phi_2}{d\alpha}$, In the following the interval from $x = a$ to $x = b$ is subdivided into $n$ equal parts by the points $a=x_0, x_2, . Integration by Parts: Knowing which function to call u and which to call dv takes some practice. Given that, $\int_{0}^{3}$ x² dx = 8 , solve $\int_{0}^{3}$ 4x² dx, $\int_{0}^{3}$ 4x² dx = 4 $\int_{0}^{3}$x² dx, 3. Some of the reduction formulas in definite integration are: Reduction formula for sin - Sinn x dx = -1/n cos x sinn-1 x + n-1/n $\int$ sinn-2 x dx, Reduction formula for cos = Cosn x dx = -1/n sin x cosn-1 x + n-1/n $\int$ cosn-2 x dx, Reduction formula for natural logarithm - $\int$ (In x)n dx = x(In x)n – n $\int$ (In x)n-1 dx. If the upper and lower limits of a definite integral are the same, the integral is zero: $${\large\int\limits_a^a\normalsize} {f\left( x \right)dx} = 0$$ Reversing the limits of integration changes the sign of the definite integral: Simpsonâs formula (or parabolic formula) for$n$even$\int_a^b f(x)\ dx\approx \frac{h}{3}(y_0+4y_1+2y_2+4y_3+\cdots+2y_{n-2}+4y_{n-1}+y_n)$,$\int_0^\infty\frac{dx}{x^2+a^2}=\frac{\pi}{2a}$,$\int_0^\infty\frac{x^{p-1}dx}{1+x}=\frac{\pi}{\sin p\pi}$,$0< p<1$,$\int_0^\infty\frac{x^mdx}{x^n+a^n}=\frac{\pi a^{m+1-n}}{n\sin[(m+1)\frac{\pi}{n}]}$,$0< m+1< n$,$\int_0^\infty\frac{x^mdx}{1+2x\cos\beta+x^2}=\frac{\pi}{\sin m\pi}\frac{\sin m\beta}{\sin\beta}$,$\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\frac{\pi}{2}$,$\int_0^a\sqrt{a^2-x^2}\ dx=\frac{\pi a^2}{4}$,$\int_0^a x^m(a^n-x^n)^p\ dx=\frac{a^{m+1+np}\Gamma\left[\frac{m+1}{n}\right]\Gamma(p+1)}{n\Gamma\left[\frac{m+1}{n}+p+1\right]}$,$\int_0^a \frac{x^m dx}{(x^n+a^n)^r}=\frac{(-1)^{r-1}\pi a^{m+1-nr}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](r-1)!\Gamma\left[\frac{m+1}{n}-r+1\right]}0< m+1< nr$. Integration is used to determine the summation under a very large scale. And then finish with dx to mean the slices go in the x direction (and approach zero in width). To carry out the indicated integration, and if it is a definite integral, substitute the limits of integration. Integration by Parts with a definite integral Previously, we found$\displaystyle \int x \ln(x)\,dx=x\ln x - \tfrac 1 4 x^2+c$. . Be the first to review “MHCET- Definite Integration Formulas 2017 PDF Free Download” Cancel reply. For example, if (fx) is greater than 0 on [a,b] then the Riemann sum will be the positive real number and if (fx) is lesser than 0 on [a,b], then the Riemann sum will be the negative real number. It is also asked frequently in competitive exams too like JEE or AIEEE etc. a and m are constants. It is applied in economics, finance, engineering, and physics. A few are challenging. Integration is the process of finding a function with its derivative. If the interval is infinite or if$f(x)$has a singularity at some point in the interval, the definite integral is called an The interval which is given is divided into “n” subinterval is that, although not mandatory can be considered of equal lengths(Δx). . The interval which is given is divided into “n” subinterval is that, although not mandatory can be considered of equal lengths(Δx). Definite integration by parts formula is generally used to integrate the product of two functions. ., x_{n - 1}, x_n=b$ and we let $y_0=f(x_0), y_1=f(x_1), y_2=f(x_2),...,$ After the Integral Symbol we put the function we want to find the integral of (called the Integrand). Your email address will not be published. Reduction Formula in Definite Integration. There are two different types of integration namely: This article delivers information about the concepts of definite integrals, definite integrals equations, properties of definite integrals, definite integration by parts formula, reduction formulas in definite integration etc. $\int_0^1\frac{\ln x}{1+x}dx=-\frac{\pi^2}{12}$, $\int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$, $\int_0^1\frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}$, $\int_0^1\frac{\ln(1-x)}{x}dx=-\frac{\pi^2}{6}$, $\int_0^1\ln x\ln(1+x)\ dx=2-2\ln2-\frac{\pi^2}{12}$, $\int_0^1\ln x\ln(1-x)\ dx=2-\frac{\pi^2}{6}$, $\int_0^\infty\frac{x^{p-1}\ln x}{1+x}dx=-\pi^2\csc p\pi\cot p\pi$   $0< p<1$, $\int_0^1\frac{x^m-x^n}{\ln x}dx=\ln\frac{m+1}{n+1}$, $\int_0^\infty e^{-x^2}\ln x\ dx=-\frac{\sqrt{\pi}}{4}(\gamma+2\ln2)$, $\int_0^\infty\ln\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{\pi^2}{4}$, $\int_0^\frac{\pi}{2}\ln\sin x\ dx=\int_0^\frac{\pi}{2}\ln\cos x\ dx=-\frac{\pi}{2}\ln2$, $\int_0^\frac{\pi}{2}(\ln\sin x)^2\ dx=\int_0^\frac{\pi}{2}(\ln\cos x)^2\ dx=\frac{\pi}{2}(\ln2)^2+\frac{\pi^3}{24}$, $\int_0^\pi x\ln\sin x\ dx=-\frac{\pi^2}{2}\ln2$, $\int_0^\frac{\pi}{2} \sin x\ln\sin x\ dx=\ln2-1$, $\int_0^{2\pi} \ln(a+b\sin x)\ dx=\int_0^{2\pi} \ln(a+b\cos x)\ dx=2\pi\ln(a+\sqrt{a^2-b^2})$, $\int_0^\pi \ln(a+b\cos x)\ dx=x\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right)$, $\int_0^\pi \ln(a^2-2ab\cos x+b^2)\ dx=\left\{\begin{array}{lr}2\pi\ln a,\quad a\geq b>0\\ 2\pi\ln b,\quad b\geq a>0\end{array}\right.$, $\int_0^\frac{\pi}{4} \ln(1+\tan x)\ dx=\frac{\pi}{8}\ln2$, $\int_0^\frac{\pi}{2} \sec x\ln\left(\frac{1+b\cos x}{1+a\cos x}\right)\ dx=\frac{1}{2}\left\{(\cos^{-1}a)^2-(\cos^{-1}b)^2\right\}$, $\int_0^a\ln\left(2\sin\frac{x}{2}\right)\ dx=-\left(\frac{\sin a}{1^2}+\frac{\sin2a}{2^2}+\frac{\sin3a}{3^2}+\cdots\right)$, $\int_0^\infty\frac{\sin ax}{\sinh bx}dx=\frac{\pi}{2b}\tanh\frac{a\pi}{2b}$, $\int_0^\infty\frac{\cos ax}{\cosh bx}dx=\frac{\pi}{2b}sech\frac{a\pi}{2b}$, $\int_0^\infty\frac{x\ dx}{\sinh ax}=\frac{\pi^2}{4a^2}$, $\int_0^\infty\frac{x^n\ dx}{\sinh ax}=\frac{2^{n+1}-1}{2^na^{n+1}}\Gamma(n+1)\left\{\frac{1}{1^{n+1}}+\frac{1}{2^{n+1}}+\frac{1}{3^{n+1}}+\cdots\right\}$. The limit will certainly exist if $f(x)$ is piecewise continuous. The definite integral can be interpreted to represent the area under the graph. Published Year: 2017. So, to evaluate a definite integral the first thing that we’re going to do is evaluate the indefinite integral for the function. Definite Integral: How to Solve The integral of (5x 4 - 6x 2 + 7) is 5⋅[1/5]x 5 - 6⋅[1/3]x 3 + 7x. Integration by Parts: Knowing which function to call u and which to call dv takes some practice. 1. Published Month: January. The definite integrals is also used to bring forth operations on functions such as calculating arc, length, volume, surface areas and many more. Your review * Name * Email * Additional information. First write 2. This integration can either be indefinite or definite type. ==Definite integrals involving rational or irrational expressions== ∫ 0 ∞ x m d x x n + a n = π a m − n + 1 n sin  ( m + 1 n π ) for 0 < m + 1 < n {\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin \left({\dfrac {m+1}{n}}\pi \right)}}\quad {\mbox{for }}00\\ 0\quad p=0\\ -\frac{\pi}{2}\quad p<0\end{array}\right.$,$\int_0^\infty\frac{\sin px\cos qx}{x}dx=\left\{\begin{array}{lr} 0 \qquad p>q>0\\ \frac{\pi}{2}\quad 0< p< q\\ \frac{\pi}{4}\quad p=q>0\end{array}\right.$,$\int_0^\infty\frac{\sin px\sin qx}{x^2}dx=\left\{\begin{array}{lr}\frac{\pi p}{2}\quad0< p\leq q\\ \frac{\pi q}{2}\quad p\geq q>0 \end{array}\right.$,$\int_0^\infty\frac{\sin^2 px}{x^2}\ dx=\frac{\pi p}{2}$,$\int_0^\infty\frac{1-\cos px}{x^2}\ dx=\frac{\pi p}{2}$,$\int_0^\infty\frac{\cos px-\cos qx}{x}\ dx=\ln\frac{q}{p}$,$\int_0^\infty\frac{\cos px-\cos qx}{x^2}\ dx=\frac{\pi(q-p)}{2}$,$\int_0^\infty\frac{\cos mx}{x^2+a^2}\ dx=\frac{\pi}{2a}e^{-ma}$,$\int_0^\infty\frac{x\sin mx}{x^2+a^2}\ dx=\frac{\pi}{2}e^{-ma}$,$\int_0^\infty\frac{\sin mx}{x(x^2+a^2)}\ dx=\frac{\pi}{2a^2}(1-e^{-ma})$,$\int_0^{2\pi}\frac{dx}{a+b\sin x}=\frac{2\pi}{\sqrt{a^2-b^2}}$,$\int_0^{2\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$,$\int_0^\frac{\pi}{2}\frac{dx}{a+b\cos x}=\frac{\cos^{-1}\left(\frac{b}{a}\right)}{\sqrt{a^2-b^2}}$,$\int_0^{2\pi}\frac{dx}{(a+b\sin x)^2}=\int_0^{2\pi}\frac{dx}{(a+b\cos x)^2}=\frac{2\pi a}{(a^2-b^2)^\frac{3}{2}}$,$\int_0^{2\pi}\frac{dx}{1-2a\cos x+a^2}=\frac{2\pi}{1-a^2},\qquad 0< a<1$,$\int_0^{\pi}\frac{x\sin x\ dx}{1-2a\cos x+a^2}=\left\{\begin{array}{lr}\left(\frac{\pi}{a}\right)\ln(1+a)\quad |a|<1\\ \pi\ln\left(1+\frac{1}{a}\right)\quad |a|>1\end{array}\right.$,$\int_0^{\pi}\frac{\cos mx\ dx}{1-2a\cos x+a^2}=\frac{\pi a^m}{1-a^2},\quad a^2<1,\quad m=0,1,2,\cdots$,$\int_0^\infty\sin ax^2\ dx=\int_0^\infty\cos ax^2\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}$,$\int_0^\infty\sin ax^n\ dx=\frac{1}{na^{\frac{1}{n}}}\Gamma\left(\frac{1}{n}\right)\sin\frac{\pi}{2n}$,$n>1$,$\int_0^\infty\cos ax^n\ dx=\frac{1}{na^{\frac{1}{n}}}\Gamma\left(\frac{1}{n}\right)\cos\frac{\pi}{2n}$,$n>1$,$\int_0^\infty\frac{\sin x}{\sqrt{x}}dx=\int_0^\infty\frac{\cos x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$,$\int_0^\infty\frac{\sin x}{x^p}dx=\frac{\pi}{2\Gamma(p)\sin\left(\frac{p \pi}{2}\right)}$,$0< p<1$,$\int_0^\infty\frac{\cos x}{x^p}dx=\frac{\pi}{2\Gamma(p)\cos\left(\frac{p \pi}{2}\right)}$,$0< p<1$,$\int_0^\infty\sin ax^2\cos2bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos\frac{b^2}{a}-\sin\frac{b^2}{a}\right)$,$\int_0^\infty\cos ax^2\cos2bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos\frac{b^2}{a}+\sin\frac{b^2}{a}\right)$,$\int_0^\infty\frac{\sin^3 x}{x^3}dx=\frac{3\pi}{8}$,$\int_0^\infty\frac{\sin^4 x}{x^4}dx=\frac{\pi}{3}$,$\int_0^\infty\frac{\tan x}{x}dx=\frac{\pi}{2}$,$\int_0^\frac{\pi}{2}\frac{dx}{1+\tan^mx}=\frac{\pi}{4}$,$\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=2\left\{\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots\right\}$,$\int_0^1\frac{\tan^{-1}x}{x}dx=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots$,$\int_0^1\frac{\sin^{-1}x}{x}dx=\frac{\pi}{2}\ln2$,$\int_0^1\frac{1-\cos x}{x}dx-\int_1^\infty\frac{\cos x}{x}dx=\gamma$,$\int_0^\infty\left(\frac{1}{1+x^2}-\cos x\right)\frac{dx}{x}=\gamma$,$\int_0^\infty\frac{\tan^{-1}px-\tan^{-1}qx}{x}dx=\frac{\pi}{2}\ln\frac{p}{q}$,$\int_0^\infty e^{-ax}\cos bx\ dx=\frac{a}{a^2+b^2}$,$\int_0^\infty e^{-ax}\sin bx\ dx=\frac{b}{a^2+b^2}$,$\int_0^\infty \frac{e^{-ax}\sin bx}{x}\ dx=\tan^{-1}\frac{b}{a}$,$\int_0^\infty \frac{e^{-ax}-e^{-bx}}{x}\ dx=\ln\frac{b}{a}$,$\int_0^\infty e^{-ax^2}\ dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$,$\int_0^\infty e^{-ax^2}\cos bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}$,$\int_0^\infty e^{-(ax^2+bx+c)} dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{\frac{(b^2-4ac)}{4a}}\ \text{erfc}\frac{b}{2\sqrt{a}}$,$\text{erfc}(p)=\frac{2}{\pi}\int_p^\infty e^{-x^2}dx$,$\int_{-\infty}^\infty e^{-(ax^2+bx+c)} dx=\sqrt{\frac{\pi}{a}}e^{\frac{(b^2-4ac)}{4a}}$,$\int_0^\infty x^n e^{-ax}\ dx=\frac{\Gamma(n+1)}{a^{n+1}}$,$\int_0^\infty x^m e^{-ax^2}\ dx=\frac{\Gamma\left[\frac{m+1}{2}\right]}{2a^\frac{m+1}{2}}$,$\int_0^\infty e^{-(ax^2+\frac{b}{x^2})} dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$,$\int_0^\infty\frac{x\ dx}{e^x-1}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6}$,$\int_0^\infty\frac{x^{n-1}\ dx}{e^x-1}=\Gamma(n)\left(\frac{1}{1^n}+\frac{1}{2^n}+\frac{1}{3^n}+\cdots\right)$, For even$n$this can be summed in terms of Bernoulli numbers.$\int_0^\infty\frac{x\ dx}{e^x+1}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots=\frac{\pi^2}{12}$$\int_0^\infty\frac{x^{n-1}\ dx}{e^x+1}=\Gamma(n)\left(\frac{1}{1^n}-\frac{1}{2^n}+\frac{1}{3^n}-\cdots\right), For some positive integer values of n the series can be summed.\int_0^\infty\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth\frac{m}{2}-\frac{1}{2m}$$\int_0^\infty\left(\frac{1}{1+x}-e^{-x}\right)\frac{dx}{x}=\gamma$,$\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}dx=\frac{1}{2}\gamma$,$\int_0^\infty\left(\frac{1}{e^x-1}-\frac{e^{-x}}{x}\right)dx=\gamma$,$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x\sec px}dx=\frac{1}{2}\ln\left(\frac{b^2+p^2}{a^2+p^2}\right)$,$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x\csc px}dx=\tan^{-1}\frac{b}{p}-\tan^{-1}\frac{a}{p}$,$\int_0^\infty\frac{e^{-ax}(1-\cos x)}{x^2}dx=\cot^{-1}a-\frac{a}{2}\ln(a^2+1)$,$\int_0^1x^m(\ln x)^n\ dx=\frac{(-1)^n n! 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text:
Sunday, May 31, 2015

 Lógica Matemáticas Astronomía y Astrofísica Física Química Ciencias de la Vida Ciencias de la Tierra y Espacio Ciencias Agrarias Ciencias Médicas Ciencias Tecnológicas Antropología Demografía Ciencias Económicas Geografía Historia Ciencias Jurídicas y Derecho Lingüística Pedagogía Ciencia Política Psicología Artes y Letras Sociología Ética Filosofía

## Monodromy of codimension 1 subfamilies of universal curves

1) La descarga del recurso depende de la página de origen
2) Para poder descargar el recurso, es necesario ser usuario

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### Detalles del recurso

Pertenece a: Project Euclid (Hosted at Cornell University Library)

Descripción: Suppose that $g\ge3$ , that $n\ge0$ , and that $\ell\ge1$ . The main result is that if $E$ is a smooth variety that dominates a codimension $1$ subvariety $D$ of $\mathcal{M}_{g,n}[\ell]$ , the moduli space of $n$ -pointed, genus $g$ , smooth, projective curves with a level $\ell$ structure, then the closure of the image of the monodromy representation $\pi_{1}(E,e_{o})\to {\mathrm{Sp}}_{g}(\widehat{ \mathbb{Z}})$ has finite index in ${\mathrm{Sp}}_{g}(\widehat{ \mathbb{Z}})$ . A similar result is proved for codimension $1$ families of principally polarized abelian varieties.

Autor(es): Hain, Richard -

Id.: 55259222

Idioma: English  -

Versión: 1.0

Tipo:  application/pdf -

Palabras clave14D05 -

Tipo de recurso: Text  -

Tipo de Interactividad: Expositivo

Nivel de Interactividad: muy bajo

Audiencia: Estudiante  -  Profesor  -  Autor  -

Estructura: Atomic

Coste: no

: Copyright 2012 Duke University Press

Formatos:  application/pdf -

Requerimientos técnicos:  Browser: Any -

Relación: [References] 0012-7094
[References] 1547-7398

Fecha de contribución: 25-jul-2012

Contacto:

Localización:
* Duke Math. J. 161, no. 7 (2012), 1351-1378
* doi:10.1215/00127094-1593299

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3. On the fundamental groups of normal varieties
4. Combining Shape-from-Shading and Stereo using Gaussian-Markov Random Fields In this paper we present a method of combining stereo and shape-from-shading information, taking acc...
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text:
# FDL requirements for original author

hwe hwe at fsfe.org
Wed Feb 6 20:42:52 UTC 2008

First, thanks a lot for the constructive information.

Second, to set the record straight: It's not about texts from me
personally, I release the .tex.  I just cut "the professor whom I've
suggested the FDL for lecture notes" to "I", as it's irrelevant for
the answer.  (Now he' and I' are precise.)

> It depends on what is it you like about FDL.

Foremost freedom 0, freedom 1, freedom 2, and freedom 3. Then the
copyleft.  And the clear delineation of modifications.  So I don't
think it's fully fair to say he "does not want his work to be free."

Please keep in mind this is about text.  Contrary to a program, no
information is "hidden" in the source.  Anyone who knows LaTeX can
easily tell the few \section, \emph or \footnote tags -- the only
opaque words.

Even without the mentioned tools to edit pdfs, most of the text can
simply be copied and pasted; only with formulae problems are likely.
Thus it's not so much a matter of freedom, but of convenience.

> No-derives since, hiding the source seems to be directed
> towards that aim
It's not, also freedom 3 should be included.  I fully agree the .tex
/should/ be published, to not complicate derivatives.  (Frankly, I'm
not sure why showing the .tex is an issue.  I guess it's just not
written as nicely as everything else he shows.)

The problem is that the license choice will likely apply to all works
(other than papers to publish), potentially for several professors,
and mostly the source /would/ be available.  That's why I hesitate to
suggest a CC license instead of the FDL.
Any recommendations under this circumstances?

Finally, when GNU PDF [1] is done, will pdf qualify as transparent?

Thanks again for plenty of great input,
hwe

PS: What if a transparent original never existed?  What to suggest to
those who still use MS Office?  (Other than switching.)
(Also, a hand-written manuscript under FDL is not meaningless I guess.)

> disappointed by more non-free-software discussion on FSFE's list
@MJ Ray: I've read your opinion on the FDL, and actually in some
respects I agree.  Yet the FDL is part of the GNU project.  If GNU is
not to be discussed at FSFE, what is?

[1] http://gnupdf.org/


text:
# Wise decision using game theory?

There are 2 persons and two bags of oranges present in system.A bag is assigned to person.Each bag contains some oranges in range 1-10.After opening each person has been asked if they want to trade or not if they both say yes then trading will happen,if they contradict they will get whatever present in their respective bags(no exchange).What is the maximum number of oranges for which either player says yes in a Nash equilibrium?

-
I would never say yes (but I am not sure I'm in Nash equilibrium ;-)) Ok, I if have one orange in my bag I might say yes but I don't expect to get more than an orange back. –  Fabian Apr 14 '12 at 21:05
I'll say no with ten in my bag, and so will the other. So if I have nine ... –  Mark Bennet Apr 14 '12 at 21:24
Do the players like oranges? –  joriki Apr 14 '12 at 22:02
@Matt: It's obvious that you say "no" for $10$ only because you have nothing to gain from saying "yes". But that's not enough to say something about all Nash equilibria; you might still say "yes" for $10$ if you have nothing to lose from doing so. –  joriki Apr 14 '12 at 22:27
How are payoffs calculated? Expected payoffs are not really defined without a prior and you need them for the definition of equilibrium. –  Michael Greinecker Apr 15 '12 at 7:03

There is a two-dimensional continuum of Nash equilibria in which both players have an arbitrary probability to say "yes" for $1$ and zero probability for all other numbers.
Now assume that at least one player has a non-zero probability of saying "yes" for at least one number other than $1$. If the other player never says "yes", she could improve her strategy by always saying "yes" for $1$. Thus both players sometimes say "yes". Let $n_i$ be the highest number for which the probability of player $i$ saying "yes" is non-zero. If $n_1\gt n_2$, player $1$ could improve her strategy by never saying "yes" for $n_1$. Thus $n_1=n_2=n$. If either player also had a non-zero probability of saying "yes" for some number less than $n$, the other player could improve her strategy by never saying "yes" for $n$. Thus both players have a non-zero probability of saying "yes" only for $n$. If $n\ne1$, then either player could improve her strategy by always saying "yes" for some number less than $n$. Thus $n=1$, in contradiction to the assumption.
The Nash equilibria with zero probability for all numbers other than $1$ are therefore the only Nash equilibria.
The probability of saying yes by other person can be proportional to $(10-number\ of\ oranges\ in\ his\ bag)/10$ so in this way we can associate probability and with each player but i'm not able to get payoff matrix. –  user997704 Apr 15 '12 at 8:16
@user997704: I don't understand that. What do you mean by "can be"? The question was specifically about what it can be in a Nash equilibrium, and I showed that it can't be non-zero for any number other than $1$ in a Nash equilibrium. –  joriki Apr 15 '12 at 9:02


text:
# Definition:Field Norm/Definition 1

## Definition

Let $K$ be a field and $L / K$ a finite field extension of $K$.

Let $\alpha\in L$.

By Vector Space on Field Extension is Vector Space, $L$ is naturally a finite dimensional vector space over $K$.

Let $\theta_\alpha$ be the linear operator:

$\theta_\alpha: L \to L: \beta \mapsto \alpha \beta$

The field norm $\map {N_{L/K} }\alpha$ of $\alpha$ is the determinant of $\theta_\alpha$.


text:
# Exact and asymptotic properties of δ-records in the linear drift model

The study of records in the Linear Drift Model (LDM) has attracted much attention recently due to applications in several fields. In the present paper we study δ-records in the LDM, defined as observations which are greater than all previous observations, plus a fixed real quantity δ. We give analytical properties of the probability of δ-records and study the correlation between δ-record events. We also analyse the asymptotic behaviour of the number of δ-records among the first n observations and give conditions for convergence to the Gaussian distribution. As a consequence of our results, we solve a conjecture posed in J. Stat. Mech. 2010, P10013, regarding the total number of records in a LDM with negative drift. Examples of application to particular distributions, such as Gumbel or Pareto are also provided. We illustrate our results with a real data set of summer temperatures in Spain, where the LDM is consistent with the global-warming phenomenon.

## Authors

• 1 publication
• 1 publication
• 1 publication
• 1 publication
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• ### Records for Some Stationary Dependent Sequences

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07/01/2018 ∙ by Michael Falk, et al. ∙ 0

• ### Records for Some Stationary Dependence Sequences

For a zero-mean, unit-variance second-order stationary univariate Gaussi...
07/01/2018 ∙ by Michael Falk, et al. ∙ 0

• ### The Pareto Record Frontier

For iid d-dimensional observations X^(1), X^(2), ... with independent Ex...
01/17/2019 ∙ by James Allen Fill, et al. ∙ 0

• ### Variational Bayes survival analysis for unemployment modelling

Mathematical modelling of unemployment dynamics attempts to predict the ...
02/03/2021 ∙ by Pavle Boškoski, et al. ∙ 0

• ### Bayesian Estimations for Diagonalizable Bilinear SPDEs

The main goal of this paper is to study the parameter estimation problem...
05/29/2018 ∙ by Ziteng Cheng, et al. ∙ 0

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## 1 Introduction.

Extreme values and records have attracted large efforts and attention since the beginnings of statistics and probability, due to their intrinsic interest and their mathematical challenges. An important motivation for studying records comes from their connections with other interesting problems and, of course, from their countless practical applications in different fields such as climatology [1, 2, 3, 4], sports [5, 6, 7], finance [8, 9] or biology [10]. Moreover, records have been used in statistical inference because, in some contexts, data is inherently composed of record observations [11, 12, 13, 14]. The classical probabilistic setting of independent and identically distributed random observations (iid) observations has been profusely studied. Main results in this framework can be found in the monographs [15, 16, 17]. In the last years there has been an increasing interest in the study of records in correlated observations such as random walks or time series [18, 19, 20, 21, 22, 23].

An interesting departure from the iid model, which introduces time-dependence between observations, results from adding a deterministic linear trend to the iid observations, thus obtaining the so named Linear Drift Model (LDM). This model was first introduced in [24] and later developed in [25, 26, 27]. The model was also considered in [28], under a wide range of scenarios, and has proven particularly useful in the study of global warming phenomena [4, 29]. Furthermore, the importance of this model is not only related to applications but also to its mathematical structure. For instance, the study of records in the LDM model can be helpful in determining whether the underlying distribution is heavy-tailed or not [30, 31].

Also, different generalizations of the notion of record, such as near-records [32, 33, 34] or -exceedance records [35, 36] have been proposed recently. We will work with -records, first introduced in [37], which are observations greater than all previous entries, plus a fixed quantity . In the iid setting, the distribution [38, 39], process structure [40] and asymptotic properties [41] of -records have been studied. In the case , where -records are more numerous than records, their use in statistical inference has been recently proposed and positively assessed; see [41, 42, 43].

In this work, we study -records from observations obeying the LDM, while revisiting some open questions about records. We analyse the positivity and continuity of the asymptotic -record probability as a function of and of the trend parameter

. We also obtain a law of large numbers and a central limit theorem for the counting process of

-records, thus extending the corresponding results in [24]. Furthermore, we completely characterize the finiteness of the number of -records and, in particular, we solve a conjecture posed in [28], about the finiteness of the number of usual records in the LDM with negative trend.

We assess the effect of on the -record probabilities and correlations, for explicitly solvable models. Some of the results obtained in these examples are new and shed light on the behaviour of record events, when the underlying distribution is heavy-tailed. Finally we illustrate our results by analyzing a real dataset of temperatures, which fits the LDM with a trend parameter consistent with the global-warming phenomenon.

## 2 δ-records in the linear drift model

Our objects of interest in this paper are -records, formally defined as follows: given a sequence of observations and a parameter, is defined conventionally as -record and, for , is a -record if .

Note that -records are just (upper) records, if . If , a -record is necessarily a record and -records are a subsequence of records. On the other hand, if , a -record can be smaller than the current maximum, so records are a subsequence of -records.

Throughout this paper we assume that the

are random variables obeying the LDM, that is,

can be represented as

 Yn=Xn+cn,n≥1, (1)

where is the trend parameter and

is a sequence of iid random variables, with (absolutely continuous) cumulative distribution function (cdf)

and probability density function (pdf)

. Another important parameter of the model is the right-tail expectation of the , defined as

 μ+=∫∞0xf(x)dx.

For simplicity, we assume the existence of an interval of real numbers , with , such that , for all , and otherwise. Note that and .

Let denote the indicator of the event is a -record. That is, if and otherwise. So, the number of -records up to index is computed as .

Under the LDM, the probability of { is a -record} is easily computed by conditioning, as

 pj,δ:=E[1j,δ]=∫∞−∞j−1∏i=1F(x+ci−δ)f(x)dx,

where denotes the mathematical expectation. Moreover, the asymptotic -record probability is given by the formula

 pδ:=limn→∞pn,δ=∫∞−∞∞∏i=1F(x+ci−δ)f(x)dx, (2)

which is mathematically justified by the monotone convergence theorem for integrals.

In what follows we occasionally write , etc. to emphasize the dependence on the trend parameter .

## 3 Properties of the δ-record probabilities

We begin with a simple property about the asymptotic -record probability of an affine transformation of the LDM. Let , with , , and . If is the -record probability in this model, then it holds

 ~pδ(c)=pδb(cb).

We consider next some analytical properties of and , as functions of and . We note first that both are increasing in and decreasing in . Moreover, it is easy to see that is decreasing in and continuous in , converging to 1 as . The continuity of is less clear because of the infinite product within the integral in (2).

### 3.1 Positivity of pδ(c)

We show that the positivity of depends on and and on the right-tail behaviour of . We consider two cases depending on :

1. . In this case , for all .

To justify this claim, we show that for all .

If the conclusion is immediate because , as .

If , we note that implies and so, . Thus .

Finally, if , we note that implies , which in turn implies . This follows from the definition of and from Taylor’s expansion of .

Distributions with can be considered as “right-heavy-tailed” and we observe that, for such distributions, the linear trend has no impact on the asymptotic probability of a -record. This class of distributions includes the Pareto and Fréchet, with shape parameter .

2. . As in the previous case, we have three situations depending on the sign of .

For , , for all , since , for all .

If ,

 pδ(0)=∫∞−∞∞∏j=1F(x−δ)f(x)dx=∫∞x++δf(x)dx, (3)

which is positive if and only if and .

Finally, if , then if and only if . Indeed, note that, if , then , for all , and so, only the first observation (by convention) is a -record. Conversely, if , then the interval is nonempty and, for every , we have , for all . Now, since as , and , we have , which implies and, so .

Summarizing the above findings, we state

###### Theorem 1

if and only if and one of the following conditions holds

1. ,

2. .

### 3.2 Continuity of pδ(c)

As commented at the beginning of this section, the continuity of is not obvious. However, thanks to Theorem 1 we can restrict attention to distributions with finite right-tail expectation since, otherwise, vanishes and continuity is trivial. Thus, we assume throughout this section that .

A first interesting fact, which is rigorously proved in Proposition 6 of the Appendix, is that is continuous at every , for every , such that . Then, thanks to the bounded convergence theorem of integration, we conclude that is continuous, at every .

The continuity at is subtler to establish and depends of the sign of and the finiteness of , the right-end point of . Note that, for every and , we have

 ∞∏j=1F(x−δ)≤∞∏j=1F(x+cj−δ)≤N∏j=1F(x+cj−δ).

Then, taking the limit as in the above inequalities,

 ∞∏j=1F(x−δ)≤limc→0+∞∏j=1F(x+cj−δ)≤F(x−δ)N.

Therefore, is 0, if , and 1 otherwise. Then, by the dominated convergence theorem,

 limc→0+pδ(c)=∫∞−∞limc→0+∞∏j=1F(x+cj−δ)f(x)dx=∫∞x++δf(x)dx.

Thus, is right-continuous at by (3). Regarding left-continuity at 0, recall that for . So, is discontinuous at 0 if and only if and .

We now show the continuity of as a function of . The result is trivial if , since , for all . For , note that, by (3), , which is continuous since is a continuous function.

If and is a sequence converging to , we prove that

 limn→∞∞∏i=1F(x+ci−δn)=∞∏i=1F(x+ci−δ), (4)

for all , . Indeed, let , then yielding . Also for large enough and (4) follows. Let now and such that . Then, for large enough, we have and

 −∞∑i=1logF(x+ci−δn)≤−∞∑i=1logF(x+ci−(δ+ε))<∞,

since . So (4) holds, and continuity follows.

In the following theorem we summarize conditions for continuity of .

###### Theorem 2

The asymptotic -record probability , as a function of , is

• continuous at every and right-continuous at , for all ;

• discontinuous at if and only if , and

• continuous in , for all .

## 4 Exactly solvable models

In general it is not possible to compute exactly the probabilities or . We show below explicit results for the Gumbel distribution and for particular instances of the Dagum family of distributions.

### 4.1 The Gumbel distribution

Let , for , be the Gumbel distribution. Note that . Then, if ,

 n−1∏j=1F(x+cj−δ)=F(x)∑n−1j=1e−cj+δ=F(x)eδe−c−e−nc1−e−c

and, if , . So, from (2) we get

 pn,δ(c)=∫∞−∞F(x)eδe−c−e−nc1−e−cf(x)dx=1−e−c1−e−c+eδ(e−c−e−nc),

if , and

 pn,δ(0)=1(n−1)eδ+1.

Note that, taking limits as , in the above formulas, we obtain

 pδ(c)=1−e−ceδe−c+1−e−c=11+e−c1−e−ceδ,

if and , if , as expected from Theorem 1.

Also, for every , decreases with as a logistic function of . Figure 1 shows the behaviour of as a function of and .

### 4.2 The Dagum family of distributions

The random variables in the Dagum family of distributions have cdf given by , for , and , for , where are positive parameters. Two important distributions in the Dagum family are the Loglogistic, with parameters , , and the Pareto (up to a shift), with , . For simplicity, in this example we limit our attention to the case , which has .

By Theorem 1 we know that , for every , so we chose to analyse the speed of convergence of to 0, for some values of . To that end, observe that the formula for takes the manageable form

 pn,δ(c)=∫∞(δ−c)+n−1∏i=1(x+ci−δx+b+ci−δ)qf(x)dx, (5)

which becomes simpler if we further assume that (that is, the trend parameter of the LDM is equal to the scale parameter of the distribution). From (5) we get

 pn,δ(c)=∫∞(δ−c)+(x+c−δx+cn−δ)qf(x)dx. (6)

Note that the Pareto(1,1) distribution, taking , is included as a particular case. This distribution will be studied at the end of this example and later, in section 5, in the context of -record correlations.

We introduce the notation to make explicit the dependence of on . First, for records () we have,

 p(q)n,0(c)= cq∫∞0xq−1(x+cn)−q(x+c)−1dx = qn−q∫10tq−1(1−t(n−1)/n)−qdt (7) = q(n−1)q∫n1(y−1)q−1ydy, (8)

where the second equality follows from the change of variable and the third from .

Observe that (7) and (8) do not depend on and so, for the sake of simplicity, we write . Moreover, from formula (7) we see that

 p(q)n,0=n−q 2F1(q,q;q+1;(n−1)/n),

where is the Gauss hypergeometric function.

Also, from (8) and using the binomial expansion, for , we readily obtain

 p(q)n,0=q(n−1)q((−1)q−1logn+q−1∑k=1(q−1k)(−1)q−1−kk(nk−1)). (9)

The asymptotic behaviour of , for any , can be obtained from (8). For , (9) yields . For , the leading term in the integral in (8) is , so . For , the integral in (8) converges and, using formula 3.191.2 in [44], we get

 p(q)n,0∼n−qq∫∞1(y−1)q−1ydy=n−qqΓ(1−q)Γ(q).

Thus,

 p(q)n,0∼⎧⎪⎨⎪⎩n−qqΓ(1−q)Γ(q),%if01. (10)

It is interesting to observe that the limiting behaviour of , as a function of the power of the tail , seems to match the asymptotic behaviour of when is the Fréchet distribution (, ) and the tuning parameter is the trend , studied in [27].

We now consider and investigate whether , as . This result can be expected since, as , the variables take very large values, so may have little influence on the probability of -record, in the long term.

From (6) we may evaluate , for any , although the computation becomes lengthy as grows. We have carried out the computation with values of from 1 to 7, and obtained

 p(1)n,δ∼log(n)n,p(q)n,δ∼qq−11n,q=2,…,7.

So, from (10) we have , at least for .

For noninteger values of , the limit behaviour of (6) is harder to analyse. To get a tractable expression, we impose . Proceeding as above, we have, for ,

 p(q)n,δ=q(n−1)q(n−2)2q∫n−11(y−1)2q−1yq+1dy.

Therefore, we have

 p(q)n,δ∼⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩n−qΓ(2q)Γ(1−q)Γ(q),if 01.

So, under the above stated conditions, , for , but this is not the case if .

To conclude this example, we study the particular case of the Pareto distribution, that is, , and take . The probability of -record is explicitly computed as:

 pn,δ=∫∞δ∨1x−δx2(x+n−1−δ)dx=1(n−1−δ)2((n−1)log(n−min{1,δ}max{1,δ})−min{1,δ}(n−1−δ)), (11)

if and , if . Figure 2 shows the behaviour of as a function of and .

## 5 Correlations

The indicators of -records are in general not independent in the case of iid random variables, see [41]. In [31] the authors study the dependence of record events in the LDM, by means of the following dependence index ( in their case)

 ln(c,δ):=P[obs. n and n+1 are δ-records]P[obs. n is δ-record% ]P[obs. n+1 is δ-record]=E[1n,δ1n+1,δ]E[1n,δ]E[1n+1,δ].

If the events are independent, then . Otherwise, values greater or smaller than indicate positive or negative correlation, respectively. That is, neighbouring -records tend to attract or repel each other, if or .

In order to manipulate we consider the decomposition

 E[1n,δ1n+1,δ]=E[1n,δ1n+1,δ1{YnYn+1}], (12)

which, for , can be written as

 E[1n,δ1n+1,δ] =∞∫−∞(∞∫s−cn−1∏j=1F(s+cj−δ)f(t)dt+s−c∫s−c+δn∏j=2F(t+cj−δ)f(t)dt)f(s)ds =∞∫−∞((1−F(s−c))n−1∏j=1F(s+cj−δ)+s−c∫s−c+δn∏j=2F(t+cj−δ)f(t)dt)f(s)ds, (13)

and, for ,

 E[1n,δ1n+1,δ] =∞∫−∞∞∫s−c+δn−1∏j=1F(s+cj−δ)f(t)dtf(s)ds =∞∫−∞(1−F(s−c+δ))n−1∏j=1F(s+cj−δ)f(s)ds, (14)

since the second term in (12) vanishes.

As for , it is not possible to explicitly compute , in general. Nevertheless, it is still possible to describe the behaviour of the dependence index in some particular cases.

### 5.1 The Gumbel distribution

Let and the Gumbel distribution, as in section 4.1. When and , elementary but lengthy computations yield

 limn→∞E[1n,δ1n+1,δ]=(ec−1)2(ec−eδ+1)(ec+eδ−1)(e2c+eδ−1)

and

 l∞(c,δ):=limn→∞ln(c,δ)=(ec+eδ−1)(ec−eδ+1)(e2c+eδ−1).

By differentiating with respect to , we see that is decreasing in and bounded below by 1, since . With respect to we find that the derivative vanishes at

 δ=log(1−e2c+√e4c−e2c),

and then, for any ,

 maxδ<0l∞(c,δ)=2e2c(√e2c(e2c−1)−e2c+1)√e2c(e2c−1)=2(e2c−√2e3csinh(c)).

Note also that .

For ,

 limn→∞E[1n,δ1n+1,δ]=ec(ec−1)2(ec+eδ−1)(ec+δ−ec+e2c−eδ+e2δ)

and

 l∞(c,δ)=ec(ec+eδ−1)ec+δ−ec+e2c−eδ+e2δ.

We note that , , if , which results in the asymptotic independence of consecutive record indicators in the LDM. Also, there are no critical points for the index when . So, in this case is increasing in with , and decreasing in , with , as can be seen in Figure 3. Gathering these results, we conclude that if and only if . The asymptotic independence for records () was proved in [26]; we have shown here that -records attract each other for and repel each other for .

### 5.2 The Pareto distribution

Let be the Pareto distribution and . The probability of -record is given in section 4.2. Computations of are cumbersome and the explicit expression of can be found in Appendix 10.5.

We have and , for every . Also, , for all , that is, -record-attraction grows unboundedly, as increases. Moreover, it can be proved that as , where is a constant depending on .

The sublinear growth of as increases can be observed in the right panel of Figure 4, for different values of , as well as the decrease in . Also, for fixed (left panel of Figure 4), there is a negative value of where the correlation reaches a maximum, as in the Gumbel case. Note that, for negative and small positive values of , , while, for big values of , .

## 6 Asymptotic behaviour of Nn,δ

In sections 3 and 4 we have presented properties of the probability that observation is a -record. In this section we analyse the random variable , defined as the number of -records among the first observations, and study its behaviour as .

Depending on , and , it might be the case that only finitely many -records are observed. We give necessary and sufficient conditions for this to happen. On the other hand, if grows to infinity, we investigate if the ratio converges (in a certain stochastic sense) to and, in that case, how the oscillations of around are distributed.

Recall that, in the classical record model (), the number of records grows to infinity, and there are universal results ensuring that, for any continuous , converges to 1, almost surely (a.s.) and has, asymptotically, a standard Gaussian distribution. However, when , results in [41] and [45] for the model with , show that may grow to a finite limit and, when it diverges, the corresponding limit laws depend both on and . We begin by analyzing the situation where has a finite limit.

### 6.1 Finiteness of the total number of δ-records

Let be the total number of -records along the sequence . In this section we find necessary and sufficient conditions for the finiteness of and .

Clearly, these questions are related to the asymptotic behaviour of . If , then we can expect . On the other hand, if , it may happen that grows sublinearly to or . Since, by Theorem 1, the positivity of is linked to the finiteness of , we split the analysis in two cases:

1. . In this situation, for any .

To check this assertion, we first prove that . Observe that implies and

 ∞∑n=1P[Yn>a]=∞∑n=1P[Xn>a−cn]=∞∑n=1(1−F(a−cn))=∞, ∀a∈R. (15)

From (15) and the second Borel-Cantelli lemma, we conclude that infinitely often (i.o.), for any , and so, , with probability one. This fact clearly implies . Now, since, for , , we get . On the other hand, for , the event

 {Xn+(c−δ)n>max1≤j≤n−1{Xj+(c−δ)j}} % implies {Xn+cn>max1≤j≤n−1{Xj+cj}+δ},

that is, . Therefore, .

2. . We distinguish three scenarios depending on the sign of .

If , we first assume . In this case, we have and is an immediate consequence of the law of large numbers in Theorem 5 below. If , only the first observation will be a -record as shown in section 3.1, so .

If and , then , since . If and , the situation is more complicated. In fact, if and only if

 ∫∞01−F(x+δ)(1−F(x))2f(x)dx<∞,

which is also equivalent to . This is shown in Proposition 7 of the Appendix, by relating this question to the counting process of geometric records, as studied in [45].

If , we proceed as in (15) to obtain

where the last inequality follows from . Thus, the first Borel-Cantelli lemma ensures that , for all , so . Then, there exists a random variable such that and, consequently, . In this case, we can also prove that ; see Proposition 9.

Summarizing the above, we give a complete characterization of the (almost sure) finiteness of the number of -records in the next theorem.

###### Theorem 3

if and only if one of the following conditions holds

1. and ,

2. , and