text:
# From screen space to world space for a 2D game

I'm making a 2D game and I want the user to be able to position an object in the world using mouse input. After a while I found an answer to 'OpenGL Math - Projecting Screen space to World space coords' on Stack Overflow but even after working on it for multiple hours I couldn't get the right depth (z) value; after clicking the x and y coordinates of the square are correct for this z value, but as you can see the z value is way to big in this case and hence the size is too; It should be as big as all the other squares: (I clicked where the circle is).

My current handler is:

void mouse_button_callback(GLFWwindow *window, int button, int action, int mods)
{
if (button == GLFW_MOUSE_BUTTON_LEFT && action == GLFW_PRESS)
{
double x, y;
int width, height;
glfwGetCursorPos(window, &x, &y);
glfwGetWindowSize(window, &width, &height);

vec4 pos1(
static_cast<float>(x / (width / 2.f)) - 1,
1 - static_cast<float>(y / (height / 2.f)), 0 /*☃*/, 1);

pos1 = MVPinv * pos1;
pos1.w = 1 / pos1.w;
pos1.x *= pos1.w;
pos1.y *= pos1.w;
pos1.z *= pos1.w;

std::cout << pos1.x << " / " << pos1.y << " / " << pos1.z << " // " << pos1.w << std::endl;

objects[0]->position = vec3(pos1);
}
}

Here MVPinv is the inverse of the model-view-projection matrix (don't mind the model matrix, it is the identity matrix and if I understand this correctly I can probably get rid of it in this place). I'm using modern OpenGL/GLEW, GLFW and GLM. Tweaking the z value indicated by ☃ can give me the correct result, but only for a certain view-projection matrix.

All objects I draw currently have a z value of 0. My question is thus: what is the best way to get the right xy coordinates if I want z to be 0?

Note: I'm very new to all these concepts of view-projection matrices, the w value, OpenGL and game development in general.

Edit: This is how the matrix is generated (more or less copied from a tutorial):

const mat4 Projection = perspective(radians(45.f), 4.0f / 3.0f, 0.1f, 100.0f);
const mat4 View = lookAt(
vec3(0, 0, 10),
vec3(0, 0, 0),
vec3(0, 1, 0)
);
const mat4 Model = mat4(1.0f);

MVP = Projection * View * Model;
MVPinv = inverse(MVP);

My updated code after reading the answer:

vec2 vecTranslate;

float factScale;
vec2 vecScale;

mat4 matTranslation, matScaling, matTransform;

void updateTranslate()
{
matTranslation = translate(mat4(1), vec3(vecTranslate, 0));
}

void updateScale(const int viewWidth, const int viewHeight)
{
vecScale = vec2(factScale, factScale / viewHeight * viewWidth);
matScaling = scale(mat4(1), vec3(vecScale, 1));
matTransform = matTranslation * matScaling;
}

void updateViewport(const int width, const int height)
{
glViewport(0, 0, width, height);
updateScale(width, height);
}

void mouse_button_callback(GLFWwindow *window, const int button, const int action, const int mods)
{
if (button == GLFW_MOUSE_BUTTON_LEFT && action == GLFW_PRESS)
{
double x, y;
int width, height;
glfwGetCursorPos(window, &x, &y);
glfwGetWindowSize(window, &width, &height);

const vec2 view(x / (width / 2.f) - 1,
1 - y / (height / 2.f));

objects[0]->position = vec3((view - vecTranslate) / vecScale, 0);

/* Or:
const vec4 view(x / (width / 2.f) - 1,
1 - y / (height / 2.f), 0, 1);
const auto pos = inverse(matTransform) * view;
objects[0]->position = {pos.x / pos.w, pos.y / pos.w, 0};
*/
}
}

void window_size_callback(GLFWwindow *window, const int width, const int height)
{
updateViewport(width, height);
}

• You're in 2D, so the full generality of a projection matrix might be more than you need. If your camera doesn't rotate, then the mapping between world & screen space should just be a scale (multiply) and a shift (add). We'll need to see how you're mapping your world coordinates to the screen to be able to tell you how to reverse the operation. – DMGregory Aug 3 '18 at 14:46
• @DMGregory I added the way I generated the matrices. Does the method that you proposed also work with matrices? Could you provide an example? – SWdV Aug 3 '18 at 14:57

## 1 Answer

You don't even need to use Look At and projection matrices if you are in 2D. They are good for 3D, but needless complexity for 2D. Your camera view can be simply the multiplication of a rotation matrix, scaling matrix and translation matrix. If your camera doesn't rotate, you can remove the rotation matrix out of it. Then your view matrix can be:

mat4 translation_matrix;
mat4 scaling_matrix;
mat4 view_matrix;

translation_matrix = mat4_translation(view_x, view_y, 0.0);
scaling_matrix  mat4_scaling(1.0 / view_width, 1.0 / view_height, 1.0);

view_matrix = translation_matrix * scaling_matrix;

Knowing where the player is clicking, in world coordinates, is a matter of adding view_x/y with cursor_x/y. No need to work with matrices here.

All objects I draw currently have a z value of 0. My question is thus: what is the best way to get the right xy coordinates if I want z to be 0?

The whole Z problem seems to me caused by the use of the projection/look at matrices.

• I struggled a bit to get everything right but in the end it finally seems to work, thanks! I'll add my updated code to the question for reference. I still have a question: why does inverse(view_matrix) not seem to work here? – SWdV Aug 4 '18 at 13:27
• I'm not sure why it's not working in your case, because calculating the matrix inverse is not something so complex. My only suspect is your "look at" or perspective matrix could not be adequate, as there are multiple ways of creating those, each with slight adjustments. – Ferreira da Selva Aug 4 '18 at 16:44
• For example, the code mat4 translate = glm::translate(mat4(1), vec3(1, 1, 0)), scale = glm::scale(mat4(1), vec3(2, 2, 0)), inv = glm::inverse(translate * scale); produces a matrix inv with -nan(ind) in every cell. Any idea what I'm doing wrong? – SWdV Aug 4 '18 at 20:14
• Try passing vec3(2, 2, 1) to the scale matrix, instead of vec3(2, 2, 0). NaN is result of a division by zero. As far as I know, it's caused by 0.0 / 0.0. – Ferreira da Selva Aug 4 '18 at 20:58
• Makes sense... And now using the inverse to reverse the operation works too as with these matrices z stays 0. Thanks! – SWdV Aug 4 '18 at 21:35


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format: plain


text:
# Continuously differentiable

1. Jan 3, 2009

### johnson12

I'm having trouble with this inequality:

let f be (real valued) continuously differentiable on [0,1] with f(0)=0, prove that

sup$$_{x\in[0,1]}$$ $$\left|f(x)\right|$$ $$\leq$$ $$\int^{1}_{0}\left|f\acute{}(x)\right| dx$$

Thanks for any help.

2. Jan 3, 2009

### jgens

Clarifying questions: Do you intend to find the supremum of the function or the interval? Furthermore, is the second f(x) term an f(x) or f'(x)? (It looks like there is a little tick next to it but I cannot tell for sure)

3. Jan 3, 2009

### johnson12

Sorry for the discrepancy, the problem is to show that the supremum of |f(x)| over [0,1] is less than or equal to the integral from 0 to 1 of |f ' (x)|, where f ' (x) is the derivative of f.

4. Jan 4, 2009

### dirk_mec1

Look at this:

$$|f(t)| = \left|\int_0^t f'(x) \mbox{ d}x \right| \leq \int_0^t |f'(x)|\ \mbox{ d}x \leq \int_0^1 |f'(x)|\ \mbox{ d}x$$

Can you finish this?


text:
Multilinear form

In abstract algebra and multilinear algebra, a multilinear form on ${\displaystyle V}$ is a map of the type

${\displaystyle f:V^{k}\to K,}$

where ${\displaystyle V}$ is a vector space over the field ${\displaystyle K}$ (or more generally, a module over a commutative ring), that is separately K-linear in each of its ${\displaystyle k}$ arguments.[1] (The rest of this article, however, will only consider multilinear forms on finite-dimensional vector spaces.)

A multilinear k-form on ${\displaystyle V}$ over ${\displaystyle \mathbf {R} }$ is called a (covariant) k-tensor, and the vector space of such forms is usually denoted ${\displaystyle {\mathcal {T}}^{k}(V)}$ or ${\displaystyle {\mathcal {L}}^{k}(V)}$.[2]

Tensor product

Given k-tensor ${\displaystyle f\in {\mathcal {T}}^{k}(V)}$  and -tensor ${\displaystyle g\in {\mathcal {T}}^{\ell }(V)}$ , a product ${\displaystyle f\otimes g\in {\mathcal {T}}^{k+\ell }(V)}$ , known as the tensor product, can be defined by the property

${\displaystyle (f\otimes g)(v_{1},\ldots ,v_{k},v_{k+1},\ldots ,v_{k+\ell })=f(v_{1},\ldots ,v_{k})g(v_{k+1},\ldots ,v_{k+\ell }),}$

for all ${\displaystyle v_{1},\ldots ,v_{k+\ell }\in V}$ . The tensor product of multilinear forms is not commutative; however it is bilinear and associative:

${\displaystyle f\otimes (ag_{1}+bg_{2})=a(f\otimes g_{1})+b(f\otimes g_{2})}$ , ${\displaystyle (af_{1}+bf_{2})\otimes g=a(f_{1}\otimes g)+b(f_{2}\otimes g),}$

and

${\displaystyle (f\otimes g)\otimes h=f\otimes (g\otimes h).}$

If ${\displaystyle (v_{1},\ldots ,v_{n})}$  forms a basis for n-dimensional vector space ${\displaystyle V}$  and ${\displaystyle (\phi ^{1},\ldots ,\phi ^{n})}$  is the corresponding dual basis for the dual space ${\displaystyle V^{*}={\mathcal {T}}^{1}(V)}$ , then the products ${\displaystyle \phi ^{i_{1}}\otimes \cdots \otimes \phi ^{i_{k}}}$ , with ${\displaystyle 1\leq i_{1},\ldots ,i_{k}\leq n}$  form a basis for ${\displaystyle {\mathcal {T}}^{k}(V)}$ . Consequently, ${\displaystyle {\mathcal {T}}^{k}(V)}$  has dimensionality ${\displaystyle n^{k}}$ .

Examples

Bilinear forms

Main article: Bilinear forms

If ${\displaystyle k=2}$ , ${\displaystyle f:V\times V\to K}$  is referred to as a bilinear form. A familiar and important example of a (symmetric) bilinear form is the standard inner product (dot product) of vectors.

Alternating multilinear forms

Main article: Alternating multilinear maps

An important class of multilinear forms are the alternating multilinear forms, which have the additional property that[3]

${\displaystyle f(x_{\sigma (1)},\ldots ,x_{\sigma (k)})=\operatorname {sgn}(\sigma )f(x_{1},\ldots ,x_{k}),}$

where ${\displaystyle \sigma :\mathbf {N} _{k}\to \mathbf {N} _{k}}$  is a permutation and ${\displaystyle \operatorname {sgn}(\sigma )}$  denotes its sign (+1 if even, –1 if odd). As a consequence, alternating multilinear forms are antisymmetric with respect to swapping of any two arguments (i.e., ${\displaystyle \sigma (p)=q,\sigma (q)=p}$  and ${\displaystyle \sigma (i)=i,1\leq i\leq k,i\neq p,q}$ ):

${\displaystyle f(x_{1},\ldots ,x_{p},\ldots ,x_{q},\ldots ,x_{k})=-f(x_{1},\ldots ,x_{q},\ldots ,x_{p},\ldots ,x_{k}).}$

With the additional hypothesis that the characteristic of the field ${\displaystyle K}$  is not 2, setting ${\displaystyle x_{p}=x_{q}=x}$  implies as a corollary that ${\displaystyle f(x_{1},\ldots ,x,\ldots ,x,\ldots ,x_{k})=0}$ ; that is, the form has a value of 0 whenever two of its arguments are equal. Note, however, that some authors[4] use this last condition as the defining property of alternating forms. This definition implies the property given at the beginning of the section, but as noted above, the converse implication holds only when ${\displaystyle \operatorname {char} (K)\neq 2}$ .

An alternating multilinear k-form on ${\displaystyle V}$  over ${\displaystyle \mathbf {R} }$  is called a multicovector of degree k or k-covector, and the vector space of such alternating forms, a subspace of ${\displaystyle {\mathcal {T}}^{k}(V)}$ , is generally denoted ${\displaystyle {\mathcal {A}}^{k}(V)}$ , or, using the notation for the isomorphic kth exterior power of ${\displaystyle V^{*}}$ (the dual space of ${\displaystyle V}$ ), ${\textstyle \bigwedge ^{k}V^{*}}$ .[5] Note that linear functionals (multilinear 1-forms over ${\displaystyle \mathbf {R} }$ ) are trivially alternating, so that ${\displaystyle {\mathcal {A}}^{1}(V)={\mathcal {T}}^{1}(V)=V^{*}}$ , while, by convention, 0-forms are defined to be scalars: ${\displaystyle {\mathcal {A}}^{0}(V)={\mathcal {T}}^{0}(V)=\mathbf {R} }$ .

The determinant on ${\displaystyle n\times n}$  matrices, viewed as an ${\displaystyle n}$  argument function of the column vectors, is an important example of an alternating multilinear form.

Wedge product

The tensor product of alternating multilinear forms is, in general, no longer alternating. However, by summing over all permutations of the tensor product, taking into account the parity of each term, the wedge product (${\displaystyle \wedge }$ ) of multicovectors can be defined, so that if ${\displaystyle f\in {\mathcal {A}}^{k}(V)}$  and ${\displaystyle g\in {\mathcal {A}}^{\ell }(V)}$ , then ${\displaystyle f\wedge g\in {\mathcal {A}}^{k+\ell }(V)}$ :

${\displaystyle (f\wedge g)(v_{1},\ldots ,v_{k+\ell })={\frac {1}{k!\ell !}}\sum _{\sigma \in S_{k+\ell }}(\operatorname {sgn}(\sigma ))f(v_{\sigma (1)},\ldots ,v_{\sigma (k)})g(v_{\sigma (k+1)},\ldots ,v_{\sigma (k+\ell )}),}$

where the sum is taken over the set of all permutations over ${\displaystyle k+\ell }$  elements, ${\displaystyle S_{k+\ell }}$ . The wedge product is bilinear, associative, and anticommutative: if ${\displaystyle f\in {\mathcal {A}}^{k}(V)}$  and ${\displaystyle g\in {\mathcal {A}}^{\ell }(V)}$  then ${\displaystyle f\wedge g=(-1)^{k\ell }g\wedge f}$ .

Given a basis ${\displaystyle (v_{1},\ldots ,v_{n})}$  for ${\displaystyle V}$  and dual basis ${\displaystyle (\phi ^{1},\ldots ,\phi ^{n})}$  for ${\displaystyle V^{*}={\mathcal {A}}^{1}(V)}$ , the wedge products ${\displaystyle \phi ^{i_{1}}\wedge \cdots \wedge \phi ^{i_{k}}}$ , with ${\displaystyle 1\leq i_{1}<\cdots   form a basis for ${\displaystyle {\mathcal {A}}^{k}(V)}$ . Hence, the dimensionality of ${\displaystyle {\mathcal {A}}^{k}(V)}$  for n-dimensional ${\displaystyle V}$  is ${\textstyle {\tbinom {n}{k}}={\frac {n!}{(n-k)!\,k!}}}$ .

Differential forms

Main article: Differential forms

Differential forms are mathematical objects constructed via tangent spaces and multilinear forms that behave, in many ways, like differentials in the classical sense. Though conceptually and computationally useful, differentials are founded on ill-defined notions of infinitesimal quantities developed early in the history of calculus. Differential forms provide a mathematically rigorous and precise framework to modernize this long-standing idea. Differential forms are especially useful in multivariable calculus (analysis) and differential geometry because they possess transformation properties that allow them be integrated on curves, surfaces, and their higher-dimensional analogues (differentiable manifolds). One far-reaching application is the modern statement of Stokes' theorem, a sweeping generalization of the fundamental theorem of calculus to higher dimensions.

The synopsis below is primarily based on Spivak (1965)[6] and Tu (2011).[3]

Definition and construction of differential 1-forms

To define differential forms on open subsets ${\displaystyle U\subset \mathbf {R} ^{n}}$ , we first need the notion of the tangent space of ${\displaystyle \mathbf {R} ^{n}}$ at ${\displaystyle p}$ , usually denoted ${\displaystyle T_{p}\mathbf {R} ^{n}}$  or ${\displaystyle \mathbf {R} _{p}^{n}}$ . The vector space ${\displaystyle \mathbf {R} _{p}^{n}}$  can be defined most conveniently as the set of elements ${\displaystyle v_{p}}$  (${\displaystyle v\in \mathbf {R} ^{n}}$ , with ${\displaystyle p\in \mathbf {R} ^{n}}$  fixed) with vector addition and scalar multiplication defined by ${\displaystyle v_{p}+w_{p}:=(v+w)_{p}}$  and ${\displaystyle a\cdot (v_{p}):=(a\cdot v)_{p}}$ , respectively. Moreover, if ${\displaystyle (e_{1},\ldots ,e_{n})}$  is the standard basis for ${\displaystyle \mathbf {R} ^{n}}$ , then ${\displaystyle ((e_{1})_{p},\ldots ,(e_{n})_{p})}$  is the analogous standard basis for ${\displaystyle \mathbf {R} _{p}^{n}}$ . In other words, each tangent space ${\displaystyle \mathbf {R} _{p}^{n}}$  can simply be regarded as a copy of ${\displaystyle \mathbf {R} ^{n}}$  (a set of tangent vectors) based at the point ${\displaystyle p}$ . The collection (disjoint union) of tangent spaces of ${\displaystyle \mathbf {R} ^{n}}$  at all ${\displaystyle p\in \mathbf {R} ^{n}}$  is known as the tangent bundle of ${\displaystyle \mathbf {R} ^{n}}$  and is usually denoted ${\textstyle T\mathbf {R} ^{n}:=\bigcup _{p\in \mathbf {R} ^{n}}\mathbf {R} _{p}^{n}}$ . While the definition given here provides a simple description of the tangent space of ${\displaystyle \mathbf {R} ^{n}}$ , there are other, more sophisticated constructions that are better suited for defining the tangent spaces of smooth manifolds in general (see the article on tangent spaces for details).

A differential k-form on ${\displaystyle U\subset \mathbf {R} ^{n}}$  is defined as a function ${\displaystyle \omega }$  that assigns to every ${\displaystyle p\in U}$  a k-covector on the tangent space of ${\displaystyle \mathbf {R} ^{n}}$ at ${\displaystyle p}$ , usually denoted ${\displaystyle \omega _{p}:=\omega (p)\in {\mathcal {A}}^{k}(\mathbf {R} _{p}^{n})}$ . In brief, a differential k-form is a k-covector field. The space of k-forms on ${\displaystyle U}$  is usually denoted ${\displaystyle \Omega ^{k}(U)}$ ; thus if ${\displaystyle \omega }$  is a differential k-form, we write ${\displaystyle \omega \in \Omega ^{k}(U)}$ . By convention, a continuous function on ${\displaystyle U}$  is a differential 0-form: ${\displaystyle f\in C^{0}(U)=\Omega ^{0}(U)}$ .

We first construct differential 1-forms from 0-forms and deduce some of their basic properties. To simplify the discussion below, we will only consider smooth differential forms constructed from smooth (${\displaystyle C^{\infty }}$ ) functions. Let ${\displaystyle f:\mathbf {R} ^{n}\to \mathbf {R} }$  be a smooth function. We define the 1-form ${\displaystyle df}$  on ${\displaystyle U}$  for ${\displaystyle p\in U}$  and ${\displaystyle v_{p}\in \mathbf {R} _{p}^{n}}$  by ${\displaystyle (df)_{p}(v_{p}):=Df|_{p}(v)}$ , where ${\displaystyle Df|_{p}:\mathbf {R} ^{n}\to \mathbf {R} }$  is the total derivative of ${\displaystyle f}$  at ${\displaystyle p}$ . (Recall that the total derivative is a linear transformation.) Of particular interest are the projection maps (also known as coordinate functions) ${\displaystyle \pi ^{i}:\mathbf {R} ^{n}\to \mathbf {R} }$ , defined by ${\displaystyle x\mapsto x^{i}}$ , where ${\displaystyle x^{i}}$  is the ith standard coordinate of ${\displaystyle x\in \mathbf {R} ^{n}}$ . The 1-forms ${\displaystyle d\pi ^{i}}$  are known as the basic 1-forms; they are conventionally denoted ${\displaystyle dx^{i}}$ . If the standard coordinates of ${\displaystyle v_{p}\in \mathbf {R} _{p}^{n}}$  are ${\displaystyle (v^{1},\ldots ,v^{n})}$ , then application of the definition of ${\displaystyle df}$  yields ${\displaystyle dx_{p}^{i}(v_{p})=v^{i}}$ , so that ${\displaystyle dx_{p}^{i}((e_{j})_{p})=\delta _{j}^{i}}$ , where ${\displaystyle \delta _{j}^{i}}$  is the Kronecker delta.[7] Thus, as the dual of the standard basis for ${\displaystyle \mathbf {R} _{p}^{n}}$ , ${\displaystyle (dx_{p}^{1},\ldots ,dx_{p}^{n})}$  forms a basis for ${\displaystyle {\mathcal {A}}^{1}(\mathbf {R} _{p}^{n})=(\mathbf {R} _{p}^{n})^{*}}$ . As a consequence, if ${\displaystyle \omega }$  is a 1-form on ${\displaystyle U}$ , then ${\displaystyle \omega }$  can be written as ${\textstyle \sum a_{i}\,dx^{i}}$  for smooth functions ${\displaystyle a_{i}:U\to \mathbf {R} }$ . Furthermore, we can derive an expression for ${\displaystyle df}$  that coincides with the classical expression for a total differential:

${\displaystyle df=\sum _{i=1}^{n}D_{i}f\;dx^{i}={\partial f \over \partial x^{1}}\,dx^{1}+\cdots +{\partial f \over \partial x^{n}}\,dx^{n}.}$

[Comments on notation: In this article, we follow the convention from tensor calculus and differential geometry in which multivectors and multicovectors are written with lower and upper indices, respectively. Since differential forms are multicovector fields, upper indices are employed to index them.[3] The opposite rule applies to the components of multivectors and multicovectors, which instead are written with upper and lower indices, respectively. For instance, we represent the standard coordinates of vector ${\displaystyle v\in \mathbf {R} ^{n}}$  as ${\displaystyle (v^{1},\ldots ,v^{n})}$ , so that ${\textstyle v=\sum _{i=1}^{n}v^{i}e_{i}}$  in terms of the standard basis ${\displaystyle (e_{1},\ldots ,e_{n})}$ . In addition, superscripts appearing in the denominator of an expression (as in ${\textstyle {\frac {\partial f}{\partial x^{i}}}}$ ) are treated as lower indices in this convention. When indices are applied and interpreted in this manner, the number of upper indices minus the number of lower indices in each term of an expression is conserved, both within the sum and across an equal sign, a feature that serves as a useful mnemonic device and helps pinpoint errors made during manual computation.]

Basic operations on differential k-forms

The wedge product (${\displaystyle \wedge }$ ) and exterior differentiation (${\displaystyle d}$ ) are two fundamental operations on differential forms. The wedge product of a k-form and an -form is a ${\displaystyle (k+\ell )}$ -form, while the exterior derivative of a k-form is a ${\displaystyle (k+1)}$ -form. Thus, both operations generate differential forms of higher degree from those of lower degree.

The wedge product ${\displaystyle \wedge :\Omega ^{k}(U)\times \Omega ^{\ell }(U)\to \Omega ^{k+\ell }(U)}$  of differential forms is a special case of the wedge product of multicovectors in general (see above). As is true in general for the wedge product, the wedge product of differential forms is bilinear, associative, and anticommutative.

More concretely, if ${\displaystyle \omega =a_{i_{1}\ldots i_{k}}\,dx^{i_{1}}\wedge \cdots \wedge dx^{i_{k}}}$  and ${\displaystyle \eta =a_{j_{1}\ldots i_{\ell }}dx^{j_{1}}\wedge \cdots \wedge dx^{j_{\ell }}}$ , then

${\displaystyle \omega \wedge \eta =a_{i_{1}\ldots i_{k}}a_{j_{1}\ldots j_{\ell }}\,dx^{i_{1}}\wedge \cdots \wedge dx^{i_{k}}\wedge dx^{j_{1}}\wedge \cdots \wedge dx^{j_{\ell }}.}$

Furthermore, for any set of indices ${\displaystyle \{\alpha _{1}\ldots ,\alpha _{m}\}}$ ,

${\displaystyle dx^{\alpha _{1}}\wedge \cdots \wedge dx^{\alpha _{p}}\wedge \cdots \wedge dx^{\alpha _{q}}\wedge \cdots \wedge dx^{\alpha _{m}}=-dx^{\alpha _{1}}\wedge \cdots \wedge dx^{\alpha _{q}}\wedge \cdots \wedge dx^{\alpha _{p}}\wedge \cdots \wedge dx^{\alpha _{m}}.}$

If ${\displaystyle I=\{i_{1},\ldots ,i_{k}\}}$ , ${\displaystyle J=\{j_{1},\ldots ,j_{\ell }\}}$ , and ${\displaystyle I\cap J=\emptyset }$ , then the indices of ${\displaystyle \omega \wedge \eta }$  can be arranged in ascending order by a (finite) sequence of such swaps. Since ${\displaystyle dx^{\alpha }\wedge dx^{\alpha }=0}$ , ${\displaystyle I\cap J\neq \emptyset }$  implies that ${\displaystyle \omega \wedge \eta =0}$ . Finally, as a consequence of bilinearity, if ${\displaystyle \omega }$  and ${\displaystyle \eta }$  are the sums of several terms, their wedge product obeys distributivity with respect to each of these terms.

The collection of the wedge products of basic 1-forms ${\displaystyle \{dx^{i_{1}}\wedge \cdots \wedge dx^{i_{k}}\mid 1\leq i_{1}<\cdots   constitutes a basis for the space of differential k-forms. Thus, any ${\displaystyle \omega \in \Omega ^{k}(U)}$  can be written in the form

${\displaystyle \omega =\sum _{i_{1}<\cdots

where ${\displaystyle a_{i_{1}\ldots i_{k}}:U\to \mathbf {R} }$  are smooth functions. With each set of indices ${\displaystyle \{i_{1},\ldots ,i_{k}\}}$  placed in ascending order, (*) is said to be the standard presentation of ${\displaystyle \omega }$ .

In the previous section, the 1-form ${\displaystyle df}$  was defined by taking the exterior derivative of the 0-form (continuous function) ${\displaystyle f}$ . We now extend this by defining the exterior derivative operator ${\displaystyle d:\Omega ^{k}(U)\to \Omega ^{k+1}(U)}$  for ${\displaystyle k\geq 1}$ . If the standard presentation of k-form ${\displaystyle \omega }$  is given by (*), the ${\displaystyle (k+1)}$ -form ${\displaystyle d\omega }$  is defined by

${\displaystyle d\omega :=\sum _{i_{1}<\ldots

A property of ${\displaystyle d}$  that holds for all smooth forms is that the second exterior derivative of any ${\displaystyle \omega }$  vanishes identically: ${\displaystyle d^{2}\omega =d(d\omega )\equiv 0}$ . This can be established directly from the definition of ${\displaystyle d}$  and the equality of mixed second-order partial derivatives of ${\displaystyle C^{2}}$  functions (see the article on closed and exact forms for details).

Integration of differential forms and Stokes' theorem for chains

To integrate a differential form over a parameterized domain, we first need to introduce the notion of the pullback of a differential form. Roughly speaking, when a differential form is integrated, applying the pullback transforms it in a way that correctly accounts for a change-of-coordinates.

Given a differentiable function ${\displaystyle f:\mathbf {R} ^{n}\to \mathbf {R} ^{m}}$  and k-form ${\displaystyle \eta \in \Omega ^{k}(\mathbf {R} ^{m})}$ , we call ${\displaystyle f^{*}\eta \in \Omega ^{k}(\mathbf {R} ^{n})}$  the pullback of ${\displaystyle \eta }$  by ${\displaystyle f}$  and define it as the k-form such that

${\displaystyle (f^{*}\eta )_{p}(v_{1p},\ldots ,v_{kp}):=\eta _{f(p)}(f_{*}(v_{1p}),\ldots ,f_{*}(v_{kp})),}$

for ${\displaystyle v_{1p},\ldots ,v_{kp}\in \mathbf {R} _{p}^{n}}$ , where ${\displaystyle f_{*}:\mathbf {R} _{p}^{n}\to \mathbf {R} _{f(p)}^{m}}$  is the map ${\displaystyle v_{p}\mapsto (Df|_{p}(v))_{f(p)}}$ .

If ${\displaystyle \omega =f\,dx^{1}\wedge \cdots \wedge dx^{n}}$  is an n-form on ${\displaystyle \mathbf {R} ^{n}}$  (i.e., ${\displaystyle \omega \in \Omega ^{n}(\mathbf {R} ^{n})}$ ), we define its integral over the unit n-cell as the iterated Riemann integral of ${\displaystyle f}$ :

${\displaystyle \int _{[0,1]^{n}}\omega =\int _{[0,1]^{n}}f\,dx^{1}\wedge \cdots \wedge dx^{n}:=\int _{0}^{1}\cdots \int _{0}^{1}f\,dx^{1}\cdots dx^{n}.}$

Next, we consider a domain of integration parameterized by a differentiable function ${\displaystyle c:[0,1]^{n}\to A\subset \mathbf {R} ^{m}}$ , known as an n-cube. To define the integral of ${\displaystyle \omega \in \Omega ^{n}(A)}$  over ${\displaystyle c}$ , we "pull back" from ${\displaystyle A}$  to the unit n-cell:

${\displaystyle \int _{c}\omega :=\int _{[0,1]^{n}}c^{*}\omega .}$

To integrate over more general domains, we define an n-chain ${\textstyle C=\sum _{i}n_{i}c_{i}}$  as the formal sum of n-cubes and set

${\displaystyle \int _{C}\omega :=\sum _{i}n_{i}\int _{c_{i}}\omega .}$

An appropriate definition of the ${\displaystyle (n-1)}$ -chain ${\displaystyle \partial C}$ , known as the boundary of ${\displaystyle C}$ ,[8] allows us to state the celebrated Stokes' theorem (Stokes–Cartan theorem) for chains in a subset of ${\displaystyle \mathbf {R} ^{m}}$ :

If ${\displaystyle \omega }$  is a smooth ${\displaystyle (n-1)}$ -form on an open set ${\displaystyle A\subset \mathbf {R} ^{m}}$  and ${\displaystyle C}$  is a smooth ${\displaystyle n}$ -chain in ${\displaystyle A}$ , then${\displaystyle \int _{C}d\omega =\int _{\partial C}\omega }$ .

Using more sophisticated machinery (e.g., germs and derivations), the tangent space ${\displaystyle T_{p}M}$  of any smooth manifold ${\displaystyle M}$  (not necessarily embedded in ${\displaystyle \mathbf {R} ^{m}}$ ) can be defined. Analogously, a differential form ${\displaystyle \omega \in \Omega ^{k}(M)}$  on a general smooth manifold is a map ${\displaystyle \omega :p\in M\mapsto \omega _{p}\in {\mathcal {A}}^{k}(T_{p}M)}$ . Stokes' theorem can be further generalized to arbitrary smooth manifolds-with-boundary and even certain "rough" domains (see the article on Stokes' theorem for details).

2. ^ Many authors use the opposite convention, writing ${\displaystyle {\mathcal {T}}^{k}(V)}$  to denote the contravariant k-tensors on ${\displaystyle V}$  and ${\displaystyle {\mathcal {T}}_{k}(V)}$  to denote the covariant k-tensors on ${\displaystyle V}$ .
5. ^ Spivak uses ${\displaystyle \Omega ^{k}(V)}$  for the space of k-covectors on ${\displaystyle V}$ . However, this notation is more commonly reserved for the space of differential k-forms on ${\displaystyle V}$ . In this article, we use ${\displaystyle \Omega ^{k}(V)}$  to mean the latter.
7. ^ The Kronecker delta is usually denoted by ${\displaystyle \delta _{ij}=\delta (i,j)}$  and defined as ${\textstyle \delta :X\times X\to \{0,1\},\ (i,j)\mapsto {\begin{cases}1,&i=j\\0,&i\neq j\end{cases}}}$ . Here, the notation ${\displaystyle \delta _{j}^{i}}$  is used to conform to the tensor calculus convention on the use of upper and lower indices.
8. ^ The formal definition of the boundary of a chain is somewhat involved and is omitted here (see Spivak (1965), pp. 98–99 for a discussion). Intuitively, if ${\displaystyle C}$  maps to a square, then ${\displaystyle \partial C}$  is a linear combination of functions that maps to its edges in a counterclockwise manner. The boundary of a chain is distinct from the notion of a boundary in point-set topology.


text:
Further Applications of Johnson's SU-normal Distribution to Various Regression Models

Title & Authors
Further Applications of Johnson's SU-normal Distribution to Various Regression Models
Choi, Pilsun; Min, In-Sik;

Abstract
This study discusses Johnson's $\small{S_U}$-normal distribution capturing a wide range of non-normality in various regression models. We provide the likelihood inference using Johnson's $\small{S_U}$-normal distribution, and propose a likelihood ratio (LR) test for normality. We also apply the $\small{S_U}$-normal distribution to the binary and censored regression models. Monte Carlo simulations are used to show that the LR test using the $\small{S_U}$-normal distribution can be served as a model specification test for normal error distribution, and that the $\small{S_U}$-normal maximum likelihood (ML) estimators tend to yield more reliable marginal effect estimates in the binary and censored model when the error distributions are non-normal.ࠀ䀀@ࠀࠀЀက ࠀЀĀကĀ ကȀЀЀȀࠀ@ @
Keywords
$\small{S_U}$-normal distribution;skewness and kurtosis;normality test;
Language
English
Cited by
1.
Analysis of time headway distribution on suburban arterial, KSCE Journal of Civil Engineering, 2012, 16, 4, 644
References
1.
Andrews, D. W. K. (1997). A conditional Kolmogorov test. Econometrica, 65, 1097-1128

2.
Bollerslev, T. (1987). A conditionally heteroskedastic time series model for speculative prices and rates of return. The Review of Economics and Statistics, 69, 542-547

3.
Choi, P. (2002). Essays on modeling asymmetric and leptokurtic distributions of asset returns, Ph.D. Thesis, Texas A&M University

4.
Choi, P. and Min, I. (2007). Estimating endogenous switching regression model with a °exible parametric distribution function: application to Korean housing demand. Forthcoming in Applied Economics

5.
Choi, P. and Nam, K. (2007). Asymmetric and leptokurtic distribution for heteroscedas- tic asset returns: the \$S_U-normal\$ distribution. Forthcoming in Journal of Empirical Finance

6.
Fan, Y., Li, Q. and Min, I. (2006). A nonparametric bootstrap test of conditional distributions. Econometric Theory, 22, 587-613

7.
Hansen, B. E. (1994). Autoregressive conditional density estimation. International Economic Review, 35, 705-730

8.
Johnson, N. L. (1949). Systems of frequency curves generated by methods of translation. Biometrika, 36, 149-176

9.
Klaauw, B. and Konig, R. H. (2003). Testing the normality assumption in the sample selection model with an application to travel demand. Journal of Business and Economic Statistics, 21, 31-42

10.
van der Klaauw, B. and Konig, R. H. (2003). Testing the normality assumption in the sample selection model with an application to travel demand. Journal of Business and Economic Statistics, 21, 31-42

11.
McDonald, J. B. (1996). An application and comparison of some flexible parametric and semi-parametric qualitative response models. Economics Letters, 53, 145-152

12.
McDonald, J. B. and Yexiao, J. X. (1996). A comparison of semi-parametric and partially adaptive estimators of the censored regression model with possibly skewed and leptokurtic error distributions. Economics Letters, 53, 153-159

13.
Rockinger, M. and Jondeau, E. (2001). Entropy Densities with an application to autore- gressive conditional skewness and kurtosis. Journal of Econometrics, 106, 119-142

14.
Theodossiou, P. (1998). Financial data and the skewed generalized T distribution. Management Science, 44, 1650-1661

15.
Theodossiou, P. (2000). Distribution of financial asset prices, the skewed generalized error distribution, and the pricing of options. Working Paper, Rutgers University

16.
Zheng, J. X. (2000). A consistent test of conditional parametric distributions. Econometric Theory, 16, 667-691


text:
# The difference between the high and low temperatures one day was 26f. the low was 64f. write and solve a subtraction equation to find the high temperature. answer

###### Question:

The difference between the high and low temperatures one day was 26f. the low was 64f. write and solve a subtraction equation to find the high temperature. answer

### Help me on number 23 please

Help me on number 23 please...

### What is World Music? Please do not search it on Google I will give Brainliest

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### Help pls uyguguygyugyg

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### The evolving trend of vcr, dvd, and now dvr ownership is an example of which of the seven truths?

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### On Friday, a local restaurant served 200 customers. Of these, 168 people ordered soup. What percent of the customers ordered soup?

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### Which of these statements can the reader infer from this passage? A. John hated Richard because their father had given him all the land. B. After 1215, King John could steal the land from English noblemen. C. When someone was jailed, jailers had to show the king the body. D. After 1215, King John could not make the nobles do as he wanted. n 1215, King John of England was forced to give up some of his royal power with the adoption of the Magna Carta, which means “Great Charter” in Latin.

Which of these statements can the reader infer from this passage? A. John hated Richard because their father had given him all the land. B. After 1215, King John could steal the land from English noblemen. C. When someone was jailed, jailers had to show the king the body. D. After 1215, King...

### Which of the following molecules has a shape is trigonal pyramid?ccl4ch2oNh2oh2o

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### Lyon, a cash basis taxpayer, died on January 15 of the current year. During the current year, the estate executor made the required periodic distribution of $9,000 from estate income to Lyon's sole heir. The following information pertains to the estate's income and disbursements for the year. For the current calendar year, what was the estate's distributable net income (DNI)? Estate Income:$20,000 Taxable interest $10,000 Net long-term capital gains allocable to corpus Estate Disbursements:$5,

Lyon, a cash basis taxpayer, died on January 15 of the current year. During the current year, the estate executor made the required periodic distribution of \$9,000 from estate income to Lyon's sole heir. The following information pertains to the estate's income and disbursements for the year. For th...

### 2. How did they become familiar to you?

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### What single percentage change is equivalent to a 7% decrease followed by a 12% increase?

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### M is directly proportional to r cubed when r = 4, m = 160. Find the value of r when m = 540

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### Looking for a friend

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### A remote control only works when it is pointed at the television. What is the best explanation for this? A) The remote control has a weak signal and will only work over short distances. B) The remote control transmits information at the speed of light to the television. C) The remote control transmits information that is encoded and the television decodes it. D) The remote control transmits information through waves, such as infrared waves, to the television.

A remote control only works when it is pointed at the television. What is the best explanation for this? A) The remote control has a weak signal and will only work over short distances. B) The remote control transmits information at the speed of light to the television. C) The remote control trans...

### During meiosis some of the genes inherited from your mother get swapped onto the strand of DNA you inherited from your father and vice versa. the event is called crossing over. what is the ultimate result of crossing over

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### Question 21 of 25 Look at the protein below. Which could be its function? A. Storing energy O B. Transporting blood Ο Ο Ο Ο O C. Moving material out of cells acer O D. Forming connective tissue

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### The danger of fish toxins isa.easy to detect b.never serious c.equal for the fish and the person eating the fishd.gaining more widespread attention, as fish becomes a larger part of the American diet

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### What is the answer 3x+2= -55

What is the answer 3x+2= -55...


text:
# Solve for J -10-2J=10+8J

-10-2J=10+8J
Move all terms containing J to the left side of the equation.
Subtract 8J from both sides of the equation.
-10-2J-8J=10
Subtract 8J from -2J.
-10-10J=10
-10-10J=10
Move all terms not containing J to the right side of the equation.
Add 10 to both sides of the equation.
-10J=10+10
-10J=20
-10J=20
Divide each term by -10 and simplify.
Divide each term in -10J=20 by -10.
-10J-10=20-10
Cancel the common factor of -10.
Cancel the common factor.
-10J-10=20-10
Divide J by 1.
J=20-10
J=20-10
Divide 20 by -10.
J=-2
J=-2
Solve for J -10-2J=10+8J

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# Reference request: parameter inference on simulated distribution

Say I have some data $(x_{1},\ldots,x_{n})$ which I believe to be drawn from some distribution $\nu_{\theta}(x)$. I'm moderately familiar with estimation techniques for $\theta$ when I have some functional form for $\nu_{\theta}(x)$.

However I'm wondering if someone can give me an introductory reference (or at least the best words to google) for when all I can do is simulate from $\nu_{\theta}(x)$ and so want to do inference with respect to the estimated distribution $\hat \nu_{\theta}(x)$.

My naive approach would be to construct an empirical cdf from my data $\hat{F}(x)$ and then use the KS metric to determine the closest distribution. More specifically, I let $$\hat\theta=\text{argmin}_{\theta}\sup_{x}|\hat\nu_{\theta}(x)-\hat{F}(x)|$$ but then what are the statistical properties of this estimator?

When $\nu_\theta$ is the true cdf, this method is called minimum distance estimation and there are consistency results in the literature. Since you can approximate $\nu_\theta$ with an arbitrary precision by simulating enough points, the above results extend to this case.


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# Infinite-fermion condensed $\langle \psi^n \rangle$ superconductor as $U(1)_f$ QED

I find it puzzling to consider the relation between (1) and (2).

(1) The superconductor obtained from an infinite number of fermion condensation $\langle \psi^n \rangle$ with a $n \to \infty$ with a fermion $Z_n$ parity symmetry. The $n$ fermion condense superconductor has an order parameter $$<ψ^n> = <ψ ψ ψ .. ψ> \neq 0.$$ This means the vacuum (or ground state) of this universe (as a superconductor) can have $ψ^n$ $n$-fermions poping up from vacuum freely.

Basically it is induced by Higgs-Yukawa coupling term to as $ψ^n \Phi + U(\Phi)$ where $U(\Phi)$ gives a Higgs potential. There is a global symmetry $$ψ \to e^{i \frac{2 \pi j}{n}} ψ$$ where $j \in \mathbb{Z}_n$. So there are many $n$ fermions create and annihilate from vacuum due to this term $$<ψ^n> \Phi + <(ψ^\dagger)^n> \Phi^\dagger.$$

(2) QED (Quantum electrodynamics) with a fermion number conservation global symmetry $U(1)_f$. QED has this global symmetry $$ψ \to e^{i \theta} ψ$$ where the gauge field $A$ can also transform. The $U(1)_f$ global symmetry is part of the $U(1)$ gauge symmetry of QED.

Now, the fact that when $n=2$ we have a usual $Z_2$ Bardeen Cooper Schrieffer superconductor, that we make

$U(1)_f$ breaks down to $Z_{2f}$,

which sends the $U(1)$ symmetry ($ψ$ to $e^{i \theta} ψ$) to $Z_{2 f}$ symmetry ($ψ \to - ψ$) is OK.

However, when we have $n \to$ infinity, it looks to me that we can almost restore

$Z_{n f}$ to $U(1)_f$ for large $n$.

Question: Isn’t that $<ψ^n> = <ψ ψ ψ .. ψ>$ condensation relation to a $U(1)$ QED theory puzzling?

I mean of course, we know $U(1)_f$ is not exactly the same as $Z_{nf}$. But they are similar in some way when $n \to \infty$.

How would you understand their relations and comment these? [I suspect that there could be some topological differences between the twos.]

Note Add: Here the $<ψ^n> = <ψ \cdot dψ \cdot d^2ψ ..\cdot d^{\#} ψ>$ may involve higher derivative terms, in order to have the Grassman valued fields $ψ$ nonzero in $<ψ^n>$. For $n=2$, we only need the usual Cooper pairing coupling.

Note Add 2: Here the fermionic condensate $<ψ^n>$ may be also adjusted to bosonic condensate $<\varphi^n>$. For $n=2$, we only need the usual two boson $\varphi$ pair condensed. We don't need higher derivative terms for boson because $\varphi$ is not Grassman valued.

• I'm not sure I understand your gauge transform. To me, $j$ can be in N, not necessarily in Zn. If I'm correct your questions and concerns fade out, as far as I can see. – FraSchelle Sep 13 '17 at 23:42


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# assistance needed in RFID

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Not open for further replies.

#### hahawahaha

##### New Member
hi ... i would like to build a tag and a reader for RFID stock check/management.
it could be a simple one that needs around like 1-5cm of distance to detect it.
i don't know where or how to start with it. i would love to get advices on circuitry, chips to use and so on as im totally lost.

thank you !!

#### tresca

##### Member
building one from scratch is probably going to be alot harder than you might expect.
Your best bet is to get a reader from somewhere else, and then use its compatible cards/tags/fobs watever...

depending on the type of hardware you buy..you can connect it to your pc directly, or if its portable then to a microcontroller etc...

but if you are attempting to build this from scratch, and you already dont know where to start or what to do, then i would advice against it because it will be beyond you. I go to school for this kind of stuff and its above me too.

#### hahawahaha

##### New Member
hrmmm ... if thats the case what sort of hardware should i be buying?
im doing this for a project in school and i would probably think that they need us to build it ... if building it from scratch isnt sensible for a student, what do you reckon?

please let me know what sort of hardware i should be looking for ... the reader and the tag ... also what sort of software i should be using ... if possible where could i find it ...

now i have with me a few circuits where i dun even know wat they are for ... i've checked this site

Final Design Project - RFID Proximity Security System
but im not sure it is useful or not ...

#### dougy83

piece of cake! Well the card reader is anyway. You can also buy ready-made tags for less than $1 each. The reader is just a 125kHz (usually) oscillator feeding a series LC circuit. The LC junction point is fed into a peak detector (diode and RC) and this is AC-coupled into a comparator or bit-slicer. Output is then fed into a controller or PC. Or you can just buy one. Try google. #### Mr RB ##### Well-Known Member Here's a complete reader module for$24;

RFid Reader Extra Development Board | mikroElektronika

They have MikroC source code available and they have cards for $1.50 too. #### hahawahaha ##### New Member hi dougy ... thanks for sharing ... so from what you're saying ... i just need to find out how each of those circuits that u said works? for example how an oscillator feeding a series LC circuit looks like and the right values ... then look for peak detector which contains diode and RC and so on .... look for these circuits and then build them up from there?? do you have any references that i could refer them to ??? i just need to compare whether those circuits that i found and combined together would look the same as how the actual one looks like thanks alot for your information ... sounds very useful to me piece of cake! Well the card reader is anyway. You can also buy ready-made tags for less than$1 each.

The reader is just a 125kHz (usually) oscillator feeding a series LC circuit. The LC junction point is fed into a peak detector (diode and RC) and this is AC-coupled into a comparator or bit-slicer. Output is then fed into a controller or PC.

#### hahawahaha

##### New Member
hey mr rb,

thanks for the info, but i would need to build the reader from scratch ... but i might need to buy the tag from somewhere ... any advice on building the reader ??

thank you

Here's a complete reader module for $24; RFid Reader Extra Development Board | mikroElektronika They have MikroC source code available and they have cards for$1.50 too.

#### dougy83

##### Well-Known Member
... i just need to find out how each of those circuits that u said works? .... look for these circuits and then build them up from there??
Sure. None of the building blocks are difficult to understand (the microcontroller or PC used to decode the signal might need a bit of work).

Just have a look around for the circuit blocks you think you might need.
An example oscillator is at Circuit - CMOS INVERTER 125KHz LC OSCILLATOR - Circuits designed by David A. Johnson, P.E.
There's some RFID design info at https://www.electro-tech-online.com/custompdfs/2009/11/51115F.pdf (don't need to read the whole thing tho) including some reference designs. I would suggest using ASK tags and reader instead of FSK or PSK as they may be easier to understand and decode.

#### hahawahaha

##### New Member
hey dougy ...

ermm ... when you say decode, what do you mean ?? you mean decode what my tag is right ?? if that is so, what sort of program or software do i use ? i will have to create a program that does all these interfacing ?? or what ??

im really really bad in programming, pardon me for asking foolish questions if it seems so to you ...

apart from that, what sort of microcontroller should i be using?

thanks

Sure. None of the building blocks are difficult to understand (the microcontroller or PC used to decode the signal might need a bit of work).

Just have a look around for the circuit blocks you think you might need.
An example oscillator is at Circuit - CMOS INVERTER 125KHz LC OSCILLATOR - Circuits designed by David A. Johnson, P.E.
There's some RFID design info at https://www.electro-tech-online.com/custompdfs/2009/11/51115F-1.pdf (don't need to read the whole thing tho) including some reference designs. I would suggest using ASK tags and reader instead of FSK or PSK as they may be easier to understand and decode.

#### hahawahaha

##### New Member
hey dougy ..
so how do i do the programming?

#### dougy83

##### Well-Known Member
hahawahaha said:
what i would like my RFID project to do is, a reader ... from the DOOR of the room (store room) then detects how many tags are there left in the store for certain items. its fine if i could detect only ONE item instead of various for start up. then i guess i would have to send those code to the computer through whatever methods to keep track of how many tags are gone (lesser stocks) and indicates that its time to stock up. so probably yes, a sending those data to a spreadsheet.

Programming for the microcontroller: choose a microcontroller. I might suggest a PIC; for this project it doesn't really matter - maybe a PIC12F509 or a PIC16F628 or one of the 18F series. Choose a language: Basic & C are very popular. Get a compiler; you can get demos for free that will be abl to do what you want. Get/make programming hardware. Ask someone for a tutorial on how to do this.

Or if you buy a prebuilt reader with serial output the only programming you'll need to do is for the PC - you can use one of the programming languages from Visual Studio [express] for this (or many others).

#### hahawahaha

##### New Member
hey ... sorry for leaving this thread unattended for the past few weeks. was really busy with the project. now that i've decided to BUY a reader instead of making one, could you recommend any simulation softwares that has got to do with my stock checking rfid that has a rfid device attached to it.

as in the simulation software is only for the use of a certain device that is in the market. so i can try out the simulation software, and then go straight ahead to buy the device if i know how to use the simulator.

thanks

#### genius.split

##### New Member

Is there a way to read RFID labels in stack and know their order as well(position of each tag in stock). If someone offers me solution there is reward for that person!

Thank you.

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show · group.subgroup.diagram.lmfdb all knowls · up · search:

If $G$ is a group, its subgroup diagram is the Hasse diagram for the set of subgroups of $G$ under inclusion. Generally, subgroups of the same order are drawn at the same height in the diagram.

Subgroup diagrams in the LMFDB are different from typical subgroup diagrams in that it is the Hasse diagram on the set of conjugacy classes of subgroups ordered by inclusion. The number of subgroups in each conjugacy class is given as a left subscript if it is bigger then one. Hence, normal subgroups are those which have no left subscripts.

The subgroups can be dragged to make it so that one can see parts of the diagram more clearly.

Clicking on a subgroup will highlight it, and display information about the subgroup below the diagram. Hovering the mouse over a subgroup anywhere on the page will cause all instances of that subgroup to have a highlit background throughout the page.

Authors:
Knowl status:
• Review status: beta
• Last edited by John Jones on 2019-07-01 20:02:31
Referred to by:

Not referenced anywhere at the moment.

History: Differences


text:
+0

# Let a,b,c,d be real numbers such that limx→0 sin dx+ cx^5 + bx^3 + ax/x^5 = 0. If b= 4/3 then what is the value of 15ac+ 2?

0
185
1

Let a,b,c,d be real numbers such that limx→0 sin dx+ cx^5 + bx^3 + ax/x^5 = 0. If b= 4/3 then what is the value of 15ac+ 2?

Apr 19, 2022

#1
+9461
+1

We expand $$\sin(d\cdot x)$$ using its Taylor series at x = 0.

$$\begin{array}{rcl} \sin(d\cdot x) &=& \displaystyle\sum_{k = 0}^\infty \dfrac{(d\cdot x)^{2k+1}(-1)^k}{(2k + 1)!}\\ &=&\displaystyle\sum_{k = 0}^\infty \dfrac{d^{2k+1}(-1)^k}{(2k + 1)!} x^{2k+1}\\ &=&d\cdot x - \dfrac{d^3}{6}x^3+\dfrac{d^5}{120}x^5-\mathcal O (x^7) \end{array}$$

For $$\displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5}$$ to exist, the polynomial part must eliminate all terms with degrees lower than 5 of the Taylor expansion above.

Therefore, we have $$\begin{cases}a = -d\\b = \dfrac{d^3}6\\\end{cases}$$. But we also know that $$b = \dfrac43$$.

Comparing gives $$\dfrac{d^3}6 = \dfrac43$$. Solving gives $$d = 2$$.

Then since $$a = -d$$, we have $$a = -2$$.

Therefore, we have:

$$\begin{array}{rcl} \displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5} &=& \displaystyle\lim_{x\to 0}\dfrac1{x^5}\left(d\cdot x - \dfrac{d^3}6x^3+\dfrac{d^5}{120}x^5-\mathcal O(x^7)+cx^5 + \dfrac{d^3}6x^3 - d\cdot x\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{d^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{2^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{4}{15} + c - \mathcal O(x^2)\right)\\ &=& \dfrac{4}{15} + c \end{array}$$

But we also know that this limit is 0. Then:

$$\dfrac{4}{15} + c = 0\\ c = -\dfrac{4}{15}$$

Now we know the values of a and c, so

$$\begin{array}{rcl} 15ac + 2 &=& 15 \cdot (-2) \cdot \dfrac{-4}{15} + 2\\ &=& 10 \end{array}$$

Remark: If you are not familiar with the Big $$\mathcal O$$ notation, just treat it as $$\cdots$$.

Apr 19, 2022


text:
## Chemistry: Molecular Approach (4th Edition)

1 km = 0.62137 mi 1 L = 0.26417 gal Using dimensional analysis gives: $(17 \times 0.62137) / (1 \times 0.26417) = 40 miles / gallon$
1 km = 0.62137 mi 1 L = 0.26417 gal Using dimensional analysis gives: $(17 \times 0.62137) / (1 \times 0.26417) = 40 miles / gallon$


text:
Groupe d’études et de recherche en analyse des décisions

# The Geometry of Finite Dimensional Pseudomodules

## Edouard Wagneur

A semimodule M over an idempotent semiring P is also idempotent. When P is linearly ordered and conditionnally complete, we call it a pseudoring, and we say that M is a pseudomodule (p.m.) over P. The classification problem of the isomorphy classes of p.m.'s is a combinatorial problem which, in part, is related to the classificationof isomorphy classes of semilattices. We define the structural semilattice of a p.m., which is then used to introduce the concept of torsion. Then we show that every finitely generated p.m. may be canonically decomposed into the "sum" of a torsion free sub-p.m., and another one which contains all the elements responsible for the torsion of M. This decomposition is similar to the classical decomposition of a module over an integral domain into a free part, and a torsion part. It allows for a great simplification of the classification problem, since each part can be studied separately. In a sub-pseudomodule of the free p.m. over n generators, the torsion free part, also called semiboolean, is completely charecterized by a weighted oriented graph whose set of vertices is the structural semilattice of M. Partial results on the classification of the isomorphy class of a torsion sub-pseudomodule of Pn with n generators will also be presented.

, 19 pages


text:
1. Feb 25, 2017

Arman777

I was watching The Feymann online lectures and he talked about the arrow of time.And entrophy etc.
I have some questions.
1-Can ve say non-conservative force are time irreversible, but conservative force are time reversible ?
2-So From one , If a non-conservatice force acts on a system that system will lose energy and that energy has something with entrophy.(I think that way ).And It must be defined by some equation
$ΔS=∫\frac {Q} {T}$
here Q is energy lost or another way, $Δ(ME)=ΔW_{non-conservative force}=Q$
T is the tempeture of the system (I thought this relationship myself so it could be wrong or right )
Here Q be aslo something else like If two atoms colllide some energy will lost.That energy also can be Q.
3-In closed systems ΔS=0 would be zero ?
4-If closed system ΔS=0 then how could we say entrophy always increases (I know its not "always" its "almost" always, but in here is İn a closed system.Entrophy will be same for a long time.So entrophy didnt decreased so.. ?
5- Seems that theres a strict relationship between time and entrophy.Can we say with time some information lost ? (Or hidden information )

2. Feb 25, 2017

stevendaryl

Staff Emeritus
No, the real definition of $\Delta S$ is: $\frac{\Delta Q_{rev}}{T}$. Entropy is a function of state. So what that means is that the entropy change in going from one state to another is independent of how you made that transition. So to calculate the entropy of a new state, you choose a reversible path to go from the initial state to the final state.

So there is a standard example of how this makes a difference.

Suppose you have 1 liter of a gas, and you release it suddenly into an evacuated 2 liter bottle. It's insulated so that no heat goes in or out. What is the change in entropy?

Well, figure out the new state. Since no energy went in or out, we know that the new state has the same energy as the initial state. For an ideal gas, that means that the temperature didn't change, either. But the volume doubled. Since the temperature didn't change, that means that PV is a constant, so that the pressure is 1/2 what it was.

So we went from a state $P, V$ to a state $\frac{P}{2}, 2V$. To compute the entropy change, we have to figure out a reversible way to go from the initial state to the final state. A reversible way to do it would be to hold the temperature constant and let it slowly expand.

What would be the entropy change in that case? Well, we use:

$\delta U = \delta Q - P \delta V$

Since there is no energy change (constant temperature), then that means $\delta U = 0$ so $\delta Q = P \delta V$

Since $PV = NRT$, that means $P = \frac{NRT}{V}$ so $P \delta V = NRT \frac{delta V}{V} = NRT \delta (log(V))$

So $\delta Q = NRT \delta (log(V))$ and $\delta S = \delta Q/T = NR \delta (log(V))$

Since the volume doubled, then $\delta log(V) = log(V_{final}/V_{initial}) = log(2)$

So $\delta S = NR log(2) > 0$

So this is kind of a weird calculation. You're calculating the entropy change for a sequence of events that didn't actually happen. In this alternate reality, $\delta Q > 0$ (because a gas must take in heat in order to expand at constant temperature). In the real world, $\delta Q = 0$. But the entropy change is the same in the alternate reality as in the real world---the only thing that matters is what state you end up in.

3. Feb 26, 2017

Arman777

I didnt understand this part.We are going initial to final so entrophy will be positive.But in real world its zero ?

4. Feb 26, 2017

stevendaryl

Staff Emeritus
No, the change in entropy in the real world for an irreversible expansion of a gas to double its volume is: $NR log(2)$. It only depends on the initial volume and pressure and the final volume and pressure.

If the change had been done reversibly, the entropy change would have been the same, but it would have been compensated by a decrease in entropy of the surrounding system. If the volume of a gas increases reversibly, then the volume of the surrounding gas correspondingly increases. But in an irreversible change, the increase in entropy of one system is not accompanied by the decrease of entropy of another. So the total entropy of the universe increases.

The one-way nature of entropy increase is also illustrated by this example: A gas can irreversibly expand when released into a larger volume, but there isn't a corresponding case where gas spontaneously decreases its volume.

5. Feb 26, 2017

stevendaryl

Staff Emeritus
Just to clarify once more: In the real-world example of a gas doubling its volume without the input of any heat, $\Delta Q$ is zero, but $\Delta S$ is not. Q and S are only proportional for reversible changes.

6. Feb 26, 2017

Arman777

I guess I understand.Is my 1 st and 2 nd thinking way are correct ?

7. Feb 26, 2017

stevendaryl

Staff Emeritus
Yes, nonconservative forces lead to entropy increases. Any irreversible change does, and energy loss due to nonconservative forces are an example of an irreversible change. And the other way around, as well: Motion under conservative forces does not lead to entropy loss.

However, there is a little bit of complication involved in the second statement. Often nonconservative forces arise through an oversimplification of a much more complicated situation. For example, we treat a block sliding on a flat surface using friction, and call it a nonconservative force. However, at the microscopic level, friction is a matter of the particles that make up the block imparting energy to the particles in the surface through electrical interactions. If you could give a detailed analysis of this interaction, you would find that it's perfectly reversible, but extremely complicated. When we simplify the calculation by assuming that the interaction between block and surface is characterized by turning the kinetic energy of the block into heat, then we introduce irreversibility as part of the modeling process.

8. Feb 26, 2017

Arman777

I understand and it make sense,electromagnetic force is conservative force and friction is part of it.


text:
# Math Help - Tricky Double Integral with polar coordinates

1. ## Tricky Double Integral with polar coordinates

Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.

2. Originally Posted by craig
Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.
In general to convert, you do the following:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$

Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:

$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$

Hence:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$

Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.

Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$

Hence, in this particular situation:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

3. Hi

The area of integration is the part of the disk which is above the red line

And $dx dy$ becomes $r dr d\theta$

4. Originally Posted by craig
Evaluate the integral:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$

where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Tried drawing the region $D$ and it looks like I have to use polar coordinates.

Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.

I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.

Thanks in advance, any pointers would be greatly appreciated.
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html

If you draw the xy domain you can see we want the area to the left of the line but inside the circle,

If we convert this to polar co-ordinates this means

$\frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis. And $\frac{ 5 \pi }{4}$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$0 \le r \le \sqrt{3}$

We can now compute this integral,

$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$

$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$

$= \iint_Q rcos^2 \theta dr d \theta$

$= \int_0^{ \sqrt{3} } rdr \int_{ \frac{ \pi }{4} }^{ \frac{ 5 \pi }{4} } cos^2 \theta d \theta$

Which can be easily computed using double angle formulas

5. Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

Originally Posted by Mush
In general to convert, you do the following:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$

Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:

$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$

Hence:

$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

Originally Posted by Mush
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$

Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.

Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess

Originally Posted by Mush
Hence, in this particular situation:

$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

Anyway on with the integral:

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =

$\frac{3}{4}\bigg[\theta + \frac{1}{2}\sin{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ =

$\frac{3}{4}\bigg(\frac{5\pi}{4} + \frac{1}{2}\sin{\frac{5\pi}{2}} -\frac{pi}{4} - \frac{1}{2}\sin{\frac{\pi}{2}} \bigg)$

$\frac{3}{4}(\pi + 0)$ = $\frac{3\pi}{4}$

Thanks a lot for the help!

6. Originally Posted by running-gag
Hi

The area of integration is the part of the disk which is above the red line

And $dx dy$ becomes $r dr d\theta$
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!

Originally Posted by AllanCuz
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html

If you draw the xy domain you can see we want the area under the line but inside the circle,

If we convert this to polar co-ordinates this means

$0 \le \theta \le \frac{ \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$0 \le r \le \sqrt{3}$

We can now compute this integral,

$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$

$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$

$= \iint_Q rcos^2 \theta dr d \theta$

$= \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$

Which can be easily computed using double angle formulas
Thanks a lot for the reply, really helped.

7. Originally Posted by craig
Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess

** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

Anyway on with the integral:

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =

$\frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ = $\frac{3}{4}(\pi)$ = $\frac{3\pi}{4}$

Thanks a lot for the help!
I don't think you're performed that integration properly, have you?

The integral of $1 + \cos(2\theta)$ is $\theta + \frac{1}{2}\sin(2\theta)$.

8. Originally Posted by craig
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!

Thanks a lot for the reply, really helped.
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P

9. Originally Posted by Mush
I don't think you're performed that integration properly, have you?
Yes & no. I did the integration properly, just forgot to update the latex code...oops.

Edited my above working now, sorry about that.

10. Originally Posted by AllanCuz
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
Haha thought that was what you meant, thanks again for the reply.

11. Originally Posted by craig
Yes & no. I did the integration properly, just forgot to update the latex code...oops.

Edited my above working now, sorry about that.
Indeed. Good show.


text:
## Fast Approximation on the 2-Sphere by Optimally Localized Approximate Identities

• We introduce a method to construct approximate identities on the 2-sphere which have an optimal localization. This approach can be used to accelerate the calculations of approximations on the 2-sphere essentially with a comparably small increase of the error. The localization measure in the optimization problem includes a weight function which can be chosen under some constraints. For each choice of weight function existence and uniqueness of the optimal kernel are proved as well as the generation of an approximate identity in the bandlimited case. Moreover, the optimally localizing approximate identity for a certain weight function is calculated and numerically tested.

### Additional Services

Author: Volker Michel urn:nbn:de:hbz:386-kluedo-14710 Schriften zur Funktionalanalysis und Geomathematik (29) Preprint English 2006 2006 Technische Universität Kaiserslautern 2006/12/13 Approximative Identität; Konvergenz; schnelle Approximationapproximate identity; convergence; fast approximation Approximation ; Faltung; L2-Approximation ; Lokalisation; Multivariate Approximation ; Numerische Mathematik; Polynomapproximation ; Sphäre Fachbereich Mathematik 5 Naturwissenschaften und Mathematik / 51 Mathematik / 510 Mathematik 41-XX APPROXIMATIONS AND EXPANSIONS (For all approximation theory in the complex domain, see 30E05 and 30E10; for all trigonometric approximation and interpolation, see 42A10 and 42A15; for numerical approximation, see 65Dxx) / 41Axx Approximations and expansions / 41A35 Approximation by operators (in particular, by integral operators) 41-XX APPROXIMATIONS AND EXPANSIONS (For all approximation theory in the complex domain, see 30E05 and 30E10; for all trigonometric approximation and interpolation, see 42A10 and 42A15; for numerical approximation, see 65Dxx) / 41Axx Approximations and expansions / 41A55 Approximate quadratures 42-XX FOURIER ANALYSIS / 42Cxx Nontrigonometric harmonic analysis / 42C25 Uniqueness and localization for orthogonal series 65-XX NUMERICAL ANALYSIS / 65Dxx Numerical approximation and computational geometry (primarily algorithms) (For theory, see 41-XX and 68Uxx) / 65D15 Algorithms for functional approximation 86-XX GEOPHYSICS [See also 76U05, 76V05] / 86-08 Computational methods Standard gemäß KLUEDO-Leitlinien vor dem 27.05.2011

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