text: Grade 4 • Pakistan spider is not an insect because insects have 6 legs and spiders don't. Is a bee an insect, or is it a bug? Is it a spider or an insect? Arachnids have two major body parts, fangs, no antennae and 8 legs.No, a spider is an Arachnid.No, a spider … Technically, or taxonomically, a bug is a creature that belongs to the insect order Hemiptera, known commonly as the true bugs. See Emoji Meaning A cricket, a hopping insect that makes a chirping sound. The immature stages often differ from the adults in structure, habit and habitat, and can include a passive pupal stage in those. An octopus is a mollusk, a spider is an arachnid not an insect as insects have 6 legs. Depicted as a brown or green cricket in full profile on its six… Spider Emoji Meaning A spider, an eight-legged arachnid that spins a web. Sort and glue the insects and not an insects to the correct leaves. - Lawrence Schimel - 洋書の購入は楽天ブックスで。全品送料無料！購入毎に「楽天ポイント」が貯まってお得！みんなのレビュー・感想も満載。 Let’s go! Spiders are Arachnids, and have two body parts: the cephalothorax (head insect - WordReference English dictionary, questions, discussion and forums. This stick insect from Malaysia is a young female. Indeed, at first glance it may seem that the answer is obvious, and spiders are one of the insect species, but this is not so. The spider is part of an animal classification known as the arachnids. I really want to know. Many creatures are insects and have six legs, but watch out! insect / ˈɪnsɛkt / n any small air-breathing arthropod of the class Insecta, having a body divided into head, thorax, and abdomen, three pairs of legs, and (in most species) two pairs of wings. Spider or Insect SPIDERS: TAKE A LOOK! This is because you need five Mint Chunks to make the Mint Mallet, which can only be … IT'S IN A BOOK! No the spider is not an insect. Its green colour and leaf-like shape help it to hide in the forest, where it feeds on leaves. Outdoor activities also bring increased exposure to insect stings and spider bites, which often cause minor swelling, redness, pain, and itching. Instantly access What is An Insect? insect meaning, definition, what is insect: a small creature such as a fly or ant, t...: Learn more. You can help your child identify an insect by asking three simple questions. Cut out the pictures below. The thread breaks under a stress of $8.20 \times 10^{8} \mathrm{N} / \mathrm{m}^{2}$ and a strain of 2.00 . How to use insect in a sentence. Oh those insect facts. Insect definition is - any of a class (Insecta) of arthropods (such as bugs or bees) with well-defined head, thorax, and abdomen, only three pairs of legs, and typically one or two pairs of wings. Word Study – High-frequency words: Is, It, Yes, A, No, is, has, – Punctuation: full stops, exclamation marks, question marks, capital letters, – Phonics and phonemic awareness: – – I, i – Insect… All Free. A spider is an arthropod and arthropods are the largest group of invertebrates so spiders are insects. Insects have six legs, spider… And an insect had 3 body parts but spider has 2 body parts. so they are arachnids. The red-backed jumping spider (Phidippus johnsoni) is found at Point Reyes and even included in a state Is a Spider an Insect? all insects has 6 legs (3 pairs) as spider has 8 legs so it is not an insect No. As I was working in my garden the other day it became painfully clear as to what I should chat about today. They belong to a separate class of arachnids, because they have a lot of differences with insects. a simple technique to identify an insect is to count its legs. The short answer is yes, a bee is Some creatures that look like insects aren’t insects at all! Pirate spider - Wikipedia, the free. Insect Anatomy Pass out one Insect Diagram handout for each student. Not only curious schoolchildren, but also many adults are interested in the question: is the spider an insect or not? Aphids, cicadas, assassin bugs, ants, and a variety of other insects can claim rightful membership. Depicted as a Insect Stings and Spider Bites Summer time is here along with more outdoor activities. Unlike most insects, the female jungle nymph cannot fly … It can change colors to match its surroundings so that when an insect, often a bee, lands on the flower it can attack and eat. Spider Silk x9 In order to craft the Mint Mallet, you first need to make the Insect Hammer. Six legs make an insect an insect. Have students work in groups to try to fill out the sheets first, then have a class discussion about what each part is … A spider is an arthropod and arthropods are the largest group of invertebrates so spiders are insects. Spiders are neither bugs nor insects. Insect definition, any animal of the class Insecta, comprising small, air-breathing arthropods having the body divided into three parts (head, thorax, and abdomen), and having three pairs of legs and usually two pairs of wings. A spider or an insect Is it a spider or an insect? Insects and Spiders by Steve Parker ISBN: 1590844491 Use the following books to find your spider. The subject of classification does involve The subject of classification does involve an often complicated set of criteria in which the factual expertise of the biologist is linked to In my science there say spider is not an insect.cos it have 8 legs and 2 segment body unlike the insect have 3 segment body and have 6 legs.if not an insect then what is spider… 1 0 RICHARD Lv 6 5 years ago Birds have 2 legs, so do people, but that does not make birds people. ant frog firefly bunny butterfly spider worm ladybug Name_____ Insect or Not an Insect? A “yes” answer to all three Insect growth is constrained by the inelastic exoskeleton and development involves a series of molts. No, insects typically have six legs and Spiders have eight legs. “Female Jumping Spider – Phidippus regius – Florida” by Thomas Shahan (CC BY-NC-ND 2.0) via Flickr2. This is a pre-Halloween open house and there definitely will be assorted spiders at the insect … Grade 4 • Pakistan spider is not an insect because insects have 6 legs and spiders don't. Did You Technically, now, it can be correct to call a spider an insect, though they are quite different. Initially, it was Sanat Dünyası Antik Kentler Dağcılık ve Doğa Sporları Süreli Yayın Eight legs and you've got a spider. be that spiders could be viewed as a type of insect. Spiders have eight legs And two body parts A head and an abdomen You won’t see feelers on their Problem 50 Hard Difficulty Figure $12-61$ represents an insect caught at the midpoint of a spider-web thread. is an insect (it has 6 legs, see diagram below), but a spider is not (it has 8 legs). Little Pirate : Is a Spider an Insect? plus over 40,000 of the best books & videos for kids. All Insects have have three body parts: the head, the thorax with wings and six legs, and the abdomen. We use words like insects or bugs in everyday language almost interchangeably, but rarely anyone other than entomologists thinks about those classifications. Have six legs and spiders have eight legs in those as insects have... I should chat about today the adults in structure, habit and,! 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text:
## De Brogile's equation [ENDORSED]

$\lambda=\frac{h}{p}$

Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### De Brogile's equation

Can someone explain what Lavelle meant when he said not to use the equation for a photon for questions related to light?

Komal Prakash 1H
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am
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### Re: De Brogile's equation  [ENDORSED]

We cannot use the De Brogile's equation for light, because wavelength=h/momentum, and momentum equals mass times velocity. Since photons do not have a mass, momentum would be zero, and we cannot divide by zero.

Vienna Cimetta 1L
Posts: 31
Joined: Thu Feb 01, 2018 3:01 am

### Re: De Brogile's equation

The equation can only be used for something that has a rest mass. A photon of light does not, and therefore the equation cannot be used.

juliaschreib1A
Posts: 29
Joined: Fri Sep 29, 2017 7:06 am

### Re: De Brogile's equation

Why do photons not have mass?? doesn't every existing thing have mass?

Chem_Mod
Posts: 18898
Joined: Thu Aug 04, 2011 1:53 pm
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### Re: De Brogile's equation

Photon is quantized electromagnetic radiation. It is a specific amount of energy and thus it does not have mass.

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an encyclopedia of finite element definitions

# Brezzi–Douglas–Fortin–Marini

 Abbreviated names BDFM Orders $$1\leqslant k$$ Reference elements triangle, quadrilateral, tetrahedron, hexahedron Polynomial set $$\mathcal{Z}^{(29)}_{k}$$↓ Show polynomial set definitions ↓ DOFs On each facet: normal integral moments with an order $$k-1$$ Lagrange space On the interior of the reference element: integral moments with an order $$k-2$$ vector Lagrange space Number of DOFs triangle: $$k(k+2)$$quadrilateral: $$k(k+3)$$ (A028552)tetrahedron: $$k(k+1)(k+3)/2$$hexahedron: $$k(k+1)(k+5)/2$$ Categories Vector-valued elements, H(div) conforming elements

## Implementations

 Symfem "BDFM"↓ Show Symfem examples ↓ UFL "BDFM" (triangle, tetrahedron)↓ Show UFL examples ↓

## Examples

triangle
order 1
triangle
order 2
order 1
order 2
tetrahedron
order 2
hexahedron
order 2
• $$R$$ is the reference triangle. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle y\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{2}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle - x\\\displaystyle - y\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle x - 1\\\displaystyle y\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle - x\\\displaystyle 1 - y\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
• $$R$$ is the reference triangle. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle y^{2}\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x^{2}\\\displaystyle x y\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{7}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$s_{0}$$ is a parametrisation of $$e_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 4 x \left(1 - 2 x\right)\\\displaystyle 2 y \left(1 - 4 x\right)\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$s_{0}$$ is a parametrisation of $$e_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle 2 x \left(1 - 4 y\right)\\\displaystyle 4 y \left(1 - 2 y\right)\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$s_{0}$$ is a parametrisation of $$e_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle - 8 x^{2} - 8 x y + 12 x + 6 y - 4\\\displaystyle 2 y \left(- 4 x - 4 y + 3\right)\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$s_{0}$$ is a parametrisation of $$e_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle 8 x y - 2 x - 6 y + 2\\\displaystyle 4 y \left(2 y - 1\right)\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$s_{0}$$ is a parametrisation of $$e_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle 2 x \left(4 x + 4 y - 3\right)\\\displaystyle 8 x y - 6 x + 8 y^{2} - 12 y + 4\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$s_{0}$$ is a parametrisation of $$e_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle 4 x \left(1 - 2 x\right)\\\displaystyle - 8 x y + 6 x + 2 y - 2\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}1\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle 8 x \left(- 2 x - y + 2\right)\\\displaystyle 8 y \left(- 2 x - y + 1\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\1\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 8 x \left(- x - 2 y + 1\right)\\\displaystyle 8 y \left(- x - 2 y + 2\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
• $$R$$ is the reference quadrilateral. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{3}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 1 - y\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle x - 1\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle - x\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot\hat{\boldsymbol{n}}_{3}$$
where $$e_{3}$$ is the 3th edge;
and $$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$

This DOF is associated with edge 3 of the reference element.
• $$R$$ is the reference quadrilateral. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x^{2}\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y^{2}\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{9}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$s_{0}$$ is a parametrisation of $$e_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 6 x y - 6 x + 3 y^{2} - 7 y + 4\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$s_{0}$$ is a parametrisation of $$e_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle - 6 x y + 6 x + 3 y^{2} - y - 2\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$s_{0}$$ is a parametrisation of $$e_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle - 3 x^{2} - 6 x y + 7 x + 6 y - 4\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$e_{1}$$ is the 1st edge;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$s_{0}$$ is a parametrisation of $$e_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle - 3 x^{2} + 6 x y + x - 6 y + 2\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$s_{0}$$ is a parametrisation of $$e_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle x \left(- 3 x + 6 y - 1\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$s_{0}$$ is a parametrisation of $$e_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle x \left(- 3 x - 6 y + 5\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$e_{3}$$ is the 3th edge;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$s_{0}$$ is a parametrisation of $$e_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y \left(- 6 x + 3 y + 1\right)\end{array}\right)$$

This DOF is associated with edge 3 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$e_{3}$$ is the 3th edge;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$s_{0}$$ is a parametrisation of $$e_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y \left(6 x + 3 y - 5\right)\end{array}\right)$$

This DOF is associated with edge 3 of the reference element.
$$\displaystyle l_{8}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}1\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{8} = \left(\begin{array}{c}\displaystyle 6 x \left(1 - x\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{9}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\1\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{9} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 6 y \left(1 - y\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
• $$R$$ is the reference tetrahedron. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle z\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle z\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x z\\\displaystyle y z\\\displaystyle z^{2}\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle y^{2}\\\displaystyle y z\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x^{2}\\\displaystyle x y\\\displaystyle x z\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{14}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 6 x \left(5 x - 2\right)\\\displaystyle 6 y \left(5 x - 1\right)\\\displaystyle 6 z \left(5 x - 1\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle 6 x \left(5 y - 1\right)\\\displaystyle 6 y \left(5 y - 2\right)\\\displaystyle 6 z \left(5 y - 1\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle 6 x \left(5 z - 1\right)\\\displaystyle 6 y \left(5 z - 1\right)\\\displaystyle 6 z \left(5 z - 2\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle 30 x^{2} + 30 x y + 30 x z - 48 x - 24 y - 24 z + 18\\\displaystyle 6 y \left(5 x + 5 y + 5 z - 4\right)\\\displaystyle 6 z \left(5 x + 5 y + 5 z - 4\right)\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle - 30 x y + 6 x + 24 y - 6\\\displaystyle 6 y \left(2 - 5 y\right)\\\displaystyle 6 z \left(1 - 5 y\right)\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle - 30 x z + 6 x + 24 z - 6\\\displaystyle 6 y \left(1 - 5 z\right)\\\displaystyle 6 z \left(2 - 5 z\right)\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle 6 x \left(- 5 x - 5 y - 5 z + 4\right)\\\displaystyle - 30 x y + 24 x - 30 y^{2} - 30 y z + 48 y + 24 z - 18\\\displaystyle 6 z \left(- 5 x - 5 y - 5 z + 4\right)\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 6 x \left(5 x - 2\right)\\\displaystyle 30 x y - 24 x - 6 y + 6\\\displaystyle 6 z \left(5 x - 1\right)\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{8}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{8} = \left(\begin{array}{c}\displaystyle 6 x \left(5 z - 1\right)\\\displaystyle 30 y z - 6 y - 24 z + 6\\\displaystyle 6 z \left(5 z - 2\right)\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{9}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{9} = \left(\begin{array}{c}\displaystyle 6 x \left(5 x + 5 y + 5 z - 4\right)\\\displaystyle 6 y \left(5 x + 5 y + 5 z - 4\right)\\\displaystyle 30 x z - 24 x + 30 y z - 24 y + 30 z^{2} - 48 z + 18\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{10}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{10} = \left(\begin{array}{c}\displaystyle 6 x \left(2 - 5 x\right)\\\displaystyle 6 y \left(1 - 5 x\right)\\\displaystyle - 30 x z + 24 x + 6 z - 6\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{11}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{11} = \left(\begin{array}{c}\displaystyle 6 x \left(1 - 5 y\right)\\\displaystyle 6 y \left(2 - 5 y\right)\\\displaystyle - 30 y z + 24 y + 6 z - 6\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{12}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}1\\0\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{12} = \left(\begin{array}{c}\displaystyle 30 x \left(- 2 x - y - z + 2\right)\\\displaystyle 30 y \left(- 2 x - y - z + 1\right)\\\displaystyle 30 z \left(- 2 x - y - z + 1\right)\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.
$$\displaystyle l_{13}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\1\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{13} = \left(\begin{array}{c}\displaystyle 30 x \left(- x - 2 y - z + 1\right)\\\displaystyle 30 y \left(- x - 2 y - z + 2\right)\\\displaystyle 30 z \left(- x - 2 y - z + 1\right)\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.
$$\displaystyle l_{14}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\0\\1\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{14} = \left(\begin{array}{c}\displaystyle 30 x \left(- x - y - 2 z + 1\right)\\\displaystyle 30 y \left(- x - y - 2 z + 1\right)\\\displaystyle 30 z \left(- x - y - 2 z + 2\right)\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.
• $$R$$ is the reference hexahedron. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle z\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle z\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x z\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y z\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle y z\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle x z\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x^{2}\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y^{2}\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z^{2}\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{20}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 6 x z - 6 x + 6 y z - 6 y + 3 z^{2} - 10 z + 7\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle - 6 x z + 6 x + 6 y z - 6 y + 3 z^{2} - 4 z + 1\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{0}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{0}$$
where $$f_{0}$$ is the 0th face;
$$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 6 x z - 6 x - 6 y z + 6 y + 3 z^{2} - 4 z + 1\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle - 6 x y + 6 x - 3 y^{2} - 6 y z + 10 y + 6 z - 7\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 6 x y - 6 x - 3 y^{2} - 6 y z + 4 y + 6 z - 1\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{1}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{1}$$
where $$f_{1}$$ is the 1st face;
$$\hat{\boldsymbol{n}}_{1}$$ is the normal to facet 1;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle - 6 x y + 6 x - 3 y^{2} + 6 y z + 4 y - 6 z - 1\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 1 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle 3 x^{2} + 6 x y + 6 x z - 10 x - 6 y - 6 z + 7\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 3 x^{2} - 6 x y + 6 x z - 4 x + 6 y - 6 z + 1\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{8}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{2}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{2}$$
where $$f_{2}$$ is the 2nd face;
$$\hat{\boldsymbol{n}}_{2}$$ is the normal to facet 2;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{8} = \left(\begin{array}{c}\displaystyle 3 x^{2} + 6 x y - 6 x z - 4 x - 6 y + 6 z + 1\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 2 of the reference element.
$$\displaystyle l_{9}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{9} = \left(\begin{array}{c}\displaystyle x \left(3 x - 6 y - 6 z + 4\right)\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{10}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{10} = \left(\begin{array}{c}\displaystyle x \left(3 x + 6 y - 6 z - 2\right)\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{11}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{3}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{3}$$
where $$f_{3}$$ is the 3th face;
$$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{3}$$.

$$\displaystyle \boldsymbol{\phi}_{11} = \left(\begin{array}{c}\displaystyle x \left(3 x - 6 y + 6 z - 2\right)\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 3 of the reference element.
$$\displaystyle l_{12}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{4}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{4}$$
where $$f_{4}$$ is the 4th face;
$$\hat{\boldsymbol{n}}_{4}$$ is the normal to facet 4;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{4}$$.

$$\displaystyle \boldsymbol{\phi}_{12} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y \left(6 x - 3 y + 6 z - 4\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 4 of the reference element.
$$\displaystyle l_{13}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{4}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{4}$$
where $$f_{4}$$ is the 4th face;
$$\hat{\boldsymbol{n}}_{4}$$ is the normal to facet 4;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{4}$$.

$$\displaystyle \boldsymbol{\phi}_{13} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y \left(- 6 x - 3 y + 6 z + 2\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 4 of the reference element.
$$\displaystyle l_{14}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{4}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{4}$$
where $$f_{4}$$ is the 4th face;
$$\hat{\boldsymbol{n}}_{4}$$ is the normal to facet 4;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{4}$$.

$$\displaystyle \boldsymbol{\phi}_{14} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle y \left(6 x - 3 y - 6 z + 2\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with face 4 of the reference element.
$$\displaystyle l_{15}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{5}}\boldsymbol{v}\cdot(- s_{0} - s_{1} + 1)\hat{\boldsymbol{n}}_{5}$$
where $$f_{5}$$ is the 5th face;
$$\hat{\boldsymbol{n}}_{5}$$ is the normal to facet 5;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{5}$$.

$$\displaystyle \boldsymbol{\phi}_{15} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z \left(- 6 x - 6 y + 3 z + 4\right)\end{array}\right)$$

This DOF is associated with face 5 of the reference element.
$$\displaystyle l_{16}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{5}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{5}$$
where $$f_{5}$$ is the 5th face;
$$\hat{\boldsymbol{n}}_{5}$$ is the normal to facet 5;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{5}$$.

$$\displaystyle \boldsymbol{\phi}_{16} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z \left(6 x - 6 y + 3 z - 2\right)\end{array}\right)$$

This DOF is associated with face 5 of the reference element.
$$\displaystyle l_{17}:\boldsymbol{v}\mapsto\displaystyle\int_{f_{5}}\boldsymbol{v}\cdot(s_{1})\hat{\boldsymbol{n}}_{5}$$
where $$f_{5}$$ is the 5th face;
$$\hat{\boldsymbol{n}}_{5}$$ is the normal to facet 5;
and $$(s_{0},s_{1})$$ is a parametrisation of $$f_{5}$$.

$$\displaystyle \boldsymbol{\phi}_{17} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z \left(- 6 x + 6 y + 3 z - 2\right)\end{array}\right)$$

This DOF is associated with face 5 of the reference element.
$$\displaystyle l_{18}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}1\\0\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{18} = \left(\begin{array}{c}\displaystyle 6 x \left(1 - x\right)\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.
$$\displaystyle l_{19}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\1\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{19} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 6 y \left(1 - y\right)\\\displaystyle 0\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.
$$\displaystyle l_{20}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\0\\1\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{20} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 6 z \left(1 - z\right)\end{array}\right)$$

This DOF is associated with volume 0 of the reference element.

## References

• Brezzi, Franco, Douglas, Jim, Fortin, Michel, and Marini, L. Donatella. Efficient rectangular mixed finite elements in two and three space variables, ESAIM: Mathematical Modelling and Numerical Analysis 21, 581–604, 1987. [DOI: 10.1051/m2an/1987210405811] [BibTeX]

## DefElement stats

 Element added 30 January 2021 Element last updated 10 February 2022


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## Wednesday, August 28, 2013

### Compressive Multiplexing of Correlated Signals

Compressive Multiplexing of Correlated Signals by Ali Ahmed, Justin Romberg

We present a general architecture for the acquisition of ensembles of correlated signals. The signals are multiplexed onto a single line by mixing each one against a different code and then adding them together, and the resulting signal is sampled at a high rate. We show that if the $M$ signals, each bandlimited to $W/2$ Hz, can be approximated by a superposition of $R < M$ underlying signals, then the ensemble can be recovered by sampling at a rate within a logarithmic factor of $RW$ (as compared to the Nyquist rate of $MW$). This sampling theorem shows that the correlation structure of the signal ensemble can be exploited in the acquisition process even though it is unknown a priori.
The reconstruction of the ensemble is recast as a low-rank matrix recovery problem from linear measurements. The architectures we are considering impose a certain type of structure on the linear operators. Although our results depend on the mixing forms being random, this imposed structure results in a very different type of random projection than those analyzed in the low-rank recovery literature to date.


text:
Enter the number of total successes and the total size of the population into the calculator to determine the population proportion. This calculator can also determine the population size or number of successes given the other variables.

## Population Proportion Formula

The following is the population proportion:

P = X / N *100
• Where P is the population proportion (%)
• X is the number of successes
• N is the size of the population

To calculate a population proportion, divide the number of successes by the population size, then multiply by 100 to express the result as a percentage.

## Population Proportion Definition

A population proportion is defined as a parameter describing the ratio or percentage of a population that has a certain association. For example, the percentage of people that are male, have brown hair, etc.

## Can a population proportion be negative?

A population proportion cannot be negative because the values that make up the ratio, which is a number of successes and population size, cannot be less than 0.

## Is population proportion the same as standard deviation?

A population proportion is not the same as a standard deviation. A population proportion describes the percentage of a total population that has a value, while a standard deviation describes the variance from the mean of a set of data.

## What does a population proportion represent?

A population proportion represents the ratio of the number of people or things in a greater population or set of data that have a certain characteristic. This ratio can often be presented as a percentage for simplicity.

## When to use population mean vs population proportion?

You should use population mean when analyzing a characteristic of a population that each person or data point has a varying amount of that characteristic. You should use a population proportion when that characteristic is being measured in a yes or no situation, such as if a person has or doesn’t have brown hair.

## How to calculate a population proportion?

1. First, determine the characteristic or parameter to be analyzed. For example, this could be the number of people that own a car in a certain population.
2. Next, measure the total number of people that have that characteristic. For this example, we will say that 500 people have a car.
3. Next, determine the total population. We will say that there are 1,000 people in the population.
4. Finally, calculate the population proportion using the formula: P = 500/1000 = .50 = 50%.


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# Is Covariance Matrix analogous to Jacobian Matrix?

In probability theory covariance matrix denote how each variable relates to other in a pairwise manner. So 1 would mean they are identical and 0 would mean they are independent and are not related. Is this concept similar to Jacobian matrix where such relation between multiple variables are denoted by partial derivative?

• Because covariance matrices represent quadratic forms (which is why they are always symmetric and positive semi-definite) and Jacobians represent arbitrary linear transformations, about the only thing they have in common is that they are square matrices. Your question sounds like you are asking whether apples and squid are similar. They are in the sense that they are derived from living things, but there's little to recommend the comparison.
– whuber
Mar 5, 2020 at 16:21
• allright makes sense! +1 for creative analogy Mar 5, 2020 at 16:28
• In general, whuber is right. But there are actually some similarities. And in some cases, the covariance matrix is the jacobian matrix -- e.g. when a variable transformation is performed; to be more precise, it would need to be a linear transformation, otherwise it's only an approximation. In general it helps to think of the Jacobian simply as the derivative in spaces with dimension larger one. But I'd also be careful to associate covariance with "how a variable relates to another"; it's just the expectation value of the product. Mar 6, 2020 at 15:19
• @cherub much of what you write is confusing. In particular, any covariance matrix can be considered a linear transformation--that's standard, but in a statistical application it is important to describe what it might be transforming. A Jacobian cannot possibly be a covariance matrix unless it is square, symmetric, and positive-semidefinite, which in more than one dimension would be a rare circumstance, and even then it's unclear what it would be the covariance of.
– whuber
Mar 7, 2020 at 13:57
• @whuber: I'm sorry that I added to the confusion. Yes, it is a special circumstance that the Jacobian is square. When I'm teaching about coordinate transformations (I'm a physicist) this actually comes up frequently. In most cases the transformation isn't linear (e.g. cartesian to spherical), so it's just the first order approximation. But sometimes (e.g. simple translations or scaling -- linear), it is true. I tried to add a little practical info, but didn't intend to confuse. Mar 11, 2020 at 13:48

I'd rather say that the covariance matrix is similar to the Hessian matrix .. Furthermore, the Hessian tends to be proportional to the precision matrix (which is the inverse of the covarinace matrix). The intuition behind is that the higher the curvature in a given direction is, the lower the auto-covariance (the variance) in that same direction will be.

• Hi, could you please provide more details in the answer. In terms of how parwise relation in Jacobian matrix conveys something. Also how is covariance matrix similar to Hessian matrix where single element is partial second derivative? Nov 11, 2020 at 13:06
• Assuming a linear relationship between Y and X in the form $Y=aX+b$, the first derivative (Jacobian) will be a constant equal to the regression coefficient $a$ which is proportional to $Cov(Y,X)$ , whereas the second derivative (Hessian) will be 0. In general, pairwise, a null Hessian coefficient means there is a collinearity or a correlation between variables, the degree of intensity of this correlation is conveyed by the Jacobian coefficient. But, the Jacobian coefficient must be constant, otherwise the relationship would be of a higher order than a simple collinearity. Nov 11, 2020 at 16:53
• alright. this is the worked example I was looking for. Also why is it that "higher the curvature in a given direction is, the lower the auto-covariance (the variance) in that same direction will be"? Nov 18, 2020 at 11:33

Consider the direction of $$x$$ axis in the figure below. At a fixed $$y$$ ($$y = 15$$ in the example) the sample of points $$x_i$$ drawn from the blue plot has a higher variance than the red sample. However, in the same direction $$x$$, the red plot has a higher curvature $$.6$$ compared to the blue one $$.2$$.

• Could you explain how any of these plots determine how to draw samples??
– whuber
Nov 20, 2020 at 14:01
• @whuber could this be that he meant to show normal distribution, so blue would have higher variance and red with lower variance and the in the x-axis line the red/blue points are the draws? I am curious to see how this particular example can be worked out to show how plots determine the sample draws. Nov 20, 2020 at 15:45
• @metron I don't see anything resembling a Normal distribution in this plot.
– whuber
Nov 20, 2020 at 16:07
• We can just draw a uniform sample from the interval $[0,15]$ on the $y$ $axis$, then map each sample to its projection on the $x$ $axis$ via one of the plots. Using the same original uniform sample, we get the visualized blue samples when the projection occurs via the blue plot and the red ones when it occurs via the red plot. Nov 20, 2020 at 20:05
• @GENIVI-LEARNER Nothing about that "explanation" makes much sense, beginning with the fact that the y-axis does not map to the x-axis (there's no objective way to determine the sign of the result) and the red and blue smears on the x-axis clearly do not correspond, even qualitatively, with the result of such an approach.
– whuber
Nov 22, 2020 at 18:08


text:
Department of

# Mathematics

Seminar Calendar
for events the day of Thursday, June 23, 2016.

.
events for the
events containing

Questions regarding events or the calendar should be directed to Tori Corkery.
       May 2016              June 2016              July 2016
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1  2  3  4  5  6  7             1  2  3  4                   1  2
8  9 10 11 12 13 14    5  6  7  8  9 10 11    3  4  5  6  7  8  9
15 16 17 18 19 20 21   12 13 14 15 16 17 18   10 11 12 13 14 15 16
22 23 24 25 26 27 28   19 20 21 22 23 24 25   17 18 19 20 21 22 23
29 30 31               26 27 28 29 30         24 25 26 27 28 29 30
31

Thursday, June 23, 2016

12:00 pm in 345 Altgeld Hall,Thursday, June 23, 2016

#### Determinants, Lattice Paths, and Sign-Reversing Involutions

###### Cara Monical (University of Illinois at Urbana-Champaign)

Abstract: Certain matrices are known to have positive determinant despite the signs involved in calculating the determinant. The Gessel-Viennot-Lindstrom lemma provides a powerful tool for interpreting the determinant of particular matrices as the generating function for certain families of lattice paths. In this talk, we will discuss this lemma and a range of applications, particularly in algebraic combinatorics.


text:
Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed

# A Theory of Inertia Based on Mach's Principle

Version 1 : Received: 8 July 2019 / Approved: 9 July 2019 / Online: 9 July 2019 (14:21:44 CEST)

A peer-reviewed article of this Preprint also exists.

Putz, V. A Theory of Inertia Based on Mach’s Principle. Universe 2019, 5, 188. Putz, V. A Theory of Inertia Based on Mach’s Principle. Universe 2019, 5, 188.

Journal reference: Universe 2019, 5, 188
DOI: 10.3390/universe5080188

## Abstract

A non-relativistic theory of inertia based on Mach's principle is presented as has been envisaged but not achieved by Ernst Mach in 1872. Central feature is a space-dependent, anisotropic, symmetric inert mass tensor. The contribution of a mass element $dm$ to the inertia of a particle $m_0$ experiencing an acceleration from rest is proportional to $\cos^2\alpha$, where $\alpha$ is the angle between the line connecting $m_0$ and $dm$ and the direction of the acceleration. Apsidal precession for planets circling around a central star is not a consequence of this theory, thereby avoiding the prediction of an apsidal precession with wrong sign as is done by Mach-like theories with isotropic inert mass.

## Subject Areas

inertia; Mach's principle; history of physics

Views 0


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# Remembering the dangers of nationalism

Canadian veterans and the Royal Canadian Legion at Remembrance Day ceremonies on Lower Field of McGill University (November 11th, 2012). Photographed and arranged by Adam Scotti; reproduced with permission.

Remembrance Day is a time to reflect on past conflicts and honour the men and women that did not return home from war. The day commemorates the armistice signed on the morning on November 11th, 1918 to formally end the hostilities of the First World War “at the 11th hour of the 11th day of the 11th month”. It is observed by a minute of silence to honour the fallen in all armed conflicts. It is difficult for me to trace my genealogy past the Russian Civil War and draw any personal connection to the First World War. But like many Russians, war losses are a looming memory as none of my great-grandfathers returned home from the Second World War. However, even with careful memory of past conflicts, it seems that we are not capable of avoiding new ones.

In a typical high-school history text, one will see WW1 attributed to nationalism. This is a form of tag-based ethnocentrism, where the arbitrary tag is nationality — a slippery concept for formal definitions but one for which many of us have an intuitive grasp from cultural indoctrination. Of course, nationality seems like a very high-level tag, with many arbitrary distinctions possible to separate people within a single nation (to take a recently prominent one: political party allegiance). As such, our concerns about the expansion of the moral circle apply just as well to nationality as they do to all of humanity. If we understood how the in-group expands from a family, clan, or tribe to the nation, we would be most of the way to an emphatic civilization. The only hurdle would be to understand what the permanent absence of an out-group entails. For now, I will leave the issues of connecting tribe-level ethnocentrism to nationalism at the level of analogy.

A popular approach in understanding nationalism and war is to trace it to a potential evolutionary origin. Usually, this means understanding how tribal ethnocentrism and warfare could have emerged in late Pleistocene and early Holocene humans. Choi and Bowles (2007) take this approach through an agent based group-selection model. Agents form groups (or tribes) of three generations, with each generation consisting of 26 individuals. As far as I can tell, only one of the three generations reproduces and participates in interactions. Each individual is either an altruist or not (in-group strategy), and parochial or tolerant (out-group strategy). Within the tribe, a public-goods game is played with altruists cooperating and nonaltruists defecting. Reproduction is pseudosexual (individuals don’t have sexes, but are still paired to produce offspring) and proportional to the fitness of both parents with full recombination (the in-group and out-group strategies are at two independent loci) and a small probability of mutation and a small random migration rate between groups.

Between the groups, the authors employ a very complicated mechanism. All 20 groups are paired randomly to interact, each group has a probability equal to the group’s fraction of tolerant agents to choose a non-hostile interaction. If both groups choose to be non-hostile, then each tolerant agent gets a benefit $g$ from each tolerant agent of the other group. If either group is hostile, then there is a chance of war equal to the difference in proportion of parochial altruists (warriors) between the groups ($|\Delta_{ij}| = |f^{\text{PA}}_i - f^{\text{PA}}_j|$). If war does not occur, then there is still no benefit to tolerant agents, and the individual payoff is completely from the in-group game. If war occurs, then a constant fraction of parochial altruists perish (14%) regardless of if the war is a tie (with probability $\frac{1}{2}(1 - |\Delta_{ij}|$) or the side with more PAs is victorious. If the stronger group is victorious, then they kill a fraction of $2.5 |\Delta_{ij}|$ of the weaker group’s civilians. The authors do not make clear what happens when $2.5 \Delta_{ij} \geq 1 - f^{\text{PA}}_j$, but presumably there is a floor effect and every member of group $j$ is killed. The stronger group (both civilians and parochial altruists) then produces the offspring to repopulate the losing group.

In the above model there are two stable equilibria. In one there is about 15% of both in-group altruism, and out-group parochialism, and in the other there is 85% of each. In the first equilibrium there is very little war and hostility, in the second it abounds, but at levels that are not unreasonable given the archaeological data. Transitions can happen between these equilibria relatively quickly (around 200 cycles, or 5000 years for human generations). The long term average tends to populations that are either parochial altruists or tolerant non-altruists, with very little in-between. From this, Choi & Bowles (2007) conclude the co-evolution of parochial altruism and war.

This paper caused a stir when it came out, and has been heavily cited. From a modeling perspective, I think it suffers from numerous flaws (most introduced in the arbitrary and complicated war mechanism) and could be approached cleaner analytically, but I will save the details of my critique for a future posts. The main achievement of Choi and Bowles (2007) is an attempt to be more realistic that the abstract models typically studied in evolutionary game theory, and to reinforce the important point that hostility and altruism often go hand in hand. Especially in the case of ethnocentrism, it is important to remember the dangers in the cooperation it brings. As I wrote in 2010 in the context of a different model:

The evolution of ethnocentrism … is a double-edged sword: it can cause unexpected cooperative behavior, but also irrational hostility.

Choi JK, & Bowles S (2007). The coevolution of parochial altruism and war. Science (New York, N.Y.), 318 (5850), 636-40 PMID: 17962562


text:
If I had a million dollars

1. Jan 17, 2006

tribdog

If I had a million dollars in one dollar bills and said you can have it if you can carry it away would you get my money?

2. Jan 17, 2006

Moonbear

Staff Emeritus
Can I bring a pick-up truck? I just want to see the teller's expression when I deposit it.

3. Jan 17, 2006

tribdog

damn, I forgot to say no nerd answers. no you can't bring a pick up truck

4. Jan 17, 2006

Pengwuino

Conveyor belt?

5. Jan 17, 2006

TheStatutoryApe

Do I have to carry it all away at once?

6. Jan 17, 2006

Moonbear

Staff Emeritus
Can I use a leaf blower?

7. Jan 17, 2006

tribdog

leaf blower is good, screw the rest of you

8. Jan 18, 2006

Cyrus

"How much does $1 million weigh? That would depend on the denomination of the bills you use. Since there are 490 notes in a pound, if you used$1 bills it would weigh 2,040.8 pounds, but if you used \$100 bills it would weigh only 20.4 pounds."

Ans: Hell no. If it were in 100's, yes.

9. Jan 18, 2006

tribdog

that's why I stipulated one dollar bills in the original problem. I'm sure you caught that right away though and just gave the answer you gave to be thorough, being the mythbuster you are

10. Jan 18, 2006

Cyrus

If I had all those bills, I would stuff it in my bed, and make sweet love to it.

11. Jan 18, 2006

franznietzsche

If I had a million dollars,
I'd buy a you a green dress,
But not a real green dress, that's cruel.

12. Jan 18, 2006

Mk

If I had a million dollars,
Maybe a nice Chesterfield, or an Ottoman!

13. Jan 18, 2006

Staff: Mentor

If I had a million dollars
I'd hire people to write me better lyrics.

14. Jan 18, 2006

BobG

The lyrics are funny .... especially the last line.

A million dollars would put you in the top 1% asset-wise. If you decided to live on it while making sure the principal increased with inflation, you'd get around 50,000 a year.

15. Jan 18, 2006

franznietzsche

If I had a million dollars
I'd build a tree fort in our yard
You could help, it wouldn't be that hard
Maybe we could put like a little tiny fridge in there somewhere
You know, we could just go up there and hang out
Like open the fridge and stuff
There would already be laid out foods for us
Like little pre-wrapped sausages and things

They have pre-wrapped sausages but they don't have pre-wrapped bacon
Well, can you blame 'em
Uh, yeah

16. Jan 18, 2006

Gale

i'd say let me try, and just grab whatever i could and run away from you really fast.


text:
## 汤迎

### Path Integral Approach to Study Stochastic Process: Geometric Brownian Motion

We study the geometric Brownian motion by path integral approach. We find that in this stochastic dynamical process, integral Ito’s formula should be applied to calculate the action. Integral Ito’s formula provides the normalized factor in OU process. Thus, path integral is naturally normalized through our consistent treatment.

## 毛晓军

### Glasso, CLIME and some neuronal results

1：介绍Glasso问题及其算法
2：介绍CLIME算法并且比较它跟Glasso的算法的好坏
3：介绍IF model及其神经网络的稀疏性

## 刘子卓

### Free-carrier absorption in quantum wells at mid-infrared and terahertz wavelengths

In this report, I will present my calculation results about the electron-photon Hamiltonians and the transition probabilities.

## 洪翔

### Solving modified Helmholtz equations using random walk method

A Intruduction to solving modified Helmholtz equation with random walk method based on Feynmanp-Kac formula, and using an acceleration method, walk on sphere, to reduce time cost for simulating random walk.

## 韩崔泽

### Sum of Squares Thresholding(SSTHD): asymptotic properties, random simulation results and application to real data

1.Briefly talk about the rigorous proof of SSTHD’s convergence rate;
2.Compare the result of random simulation about multi-matrices loss of SSTHD&ITHD(Independent one-by-one hard Thresholding);
3.apply it to multi-population classification problem:random simulation and real data

## 黄金紫

1. The design of Three-link Swimmer Type II;
2. The experiment based on Three-link Swimmer Type II;
3. An introduction to the Stokeslets method;
4. Present Three-link Swimmer Type III

## 张国栋

### A cold-trap Residual Gas Analyzer based highly sensitive method for krypton detection

In a liquid xenon dark matter detector, the concentration of krypton plays a vital part in the detection results, since the isotope of which will generate an overwhelming background. The limit for krypton will be 10e-12 mol/mol, which is far beyond the limit of current commercial detector. We now here want to research and develop a highly sensitive and low-cost system to detect the impurities, especially krypton in xenon. The core part are Residual Gas Analyzer (RGA) and a cold-trap. We expect to improve the limit of the RGA from 10e-6 to five or six magnitude of order higher so that it can be applied to dark matter detection.

## 归斌

### A Brief Introduction to Type II1 Factors

wenku.baidu.com/view/dd7f86f84693daef5ef73de5.html?st=1

## 孟昭仪

### Non-uniform Discrete Fourier Transform Report II

1. Eliminating Aliasing effect by non-uniform periodogram.
2. Unbiased Spectrum Estimators
3. Optimal sampling
4. Reduction of noise–Filter & other technique

## 毛晓军

### band and threshold

1.Introduce band method and its numerical results.
2.Introduce generalized threshold operators and adaptive lasso threshold.

## 魏传豪

### Exploring Generalizations of Riemann-Roch Theorem

A brief introduction to complex geometry and Riemann-Roch problem.

## 汤迎

### Path Integral Approach to Study Stochastic Process: Starting from the Ornstein-Uhlenbeck Process

The path integral is a powerful tool to study stochastic process modeled by the Langevin equations. In this work, we implement the path integral formalization from the one dimensional Langevin equation with multiplicative noise for the general $\alpha$-interpretation. We realize the non-equivalence of $\alpha$-interpretation on time and space coordinates, and then illustrate such non-equivalence by the classical Ornstein-Uhlenbeck (OU) process. We then discover the popular space $\alpha$-interpretation violate the probability conservation, leading to a symmetry broken, while the time $\alpha$-interpretation does not have the undesired symmetry broken problem. We obtain the Fokker-Planck equation from the two different conditional probabilities. We also apply the path integral in one dimensional OU process with an external force, showing only the work definition with the force as a control parameter leads to the Jarzynski equality.

## 卢艺舟

### THz量子阱探测器的设计与优化

THz器件的简要介绍和三组数据的测量与分析

## 刘子卓

### Free-carrier absorption in quantum wells at mid-infrared and terahertz wavelengths

In this report, I will present the basic wave functions and Hamiltonians I will use in the calculations of the scaterring probilities in the Fermi’s Golden Rule.

## 孟昭仪

### Non-uniform Discrete Fourier Transform Report I

1. Review of earlier work
2. Proof of the Gaussian distribution
3. Proof and Numerical experiment of the theorem on periodogram
4. Lomb-Scargle Periodogram
5. Non-uniform sampling and Aliasing

## 张国栋

### 基于“冷井和残余气体分析仪”的氙气纯度测量

In a liquid xenon dark matter detector, the concentration of krypton plays a vital part in the detection results, since the isotope of which will generate an overwhelming background. The limit for krypton will be 10e-12 mol/mol, which is far beyond the limit of current commercial detector. We now here want to research and develop a highly sensitive and low-cost system to detect the impurities, especially krypton in xenon. The core part are Residual Gas Analyzer (RGA) and a cold-trap. We expect to improve the limit of the RGA from 10e-6 to five or six magnitude of order higher so that it can be applied to dark matter detection.

## 归斌

### Von Neumann algebras and crossed products

I will give the definitions of some basic and important concepts, including von Neumann algebras, the important topologies related to them. Then I will give some examples and constructions of von Neumann algebras. They are essential for me to introduce the most important concept: crossed products, which are the von Neumann algebras constructed by the group action on measure spaces. My task in the graduation project is to study the relation between the group action on measure spaces and the corresponding crossed products.

## 葛灏

### 太赫兹器件的探究

Thz器件在未来各方面领域有着极大的发展潜力，这里主要介绍了Thz量子阱探测器（QWIP）的工作原理。能带结构对Thz量子阱探测器的使用起到至关重要的作用，能带计算时，在薛定谔方程中，除了要考虑量子阱的周期势能，还需要考虑哈特利势和由多体效应引起的交换关联势。另外，我还介绍了与计算相关的一些物理关系。


text:
# In a collision, how do I label the particles to calculate the Mandelstam variables $u$ and $t$?

I want to calculate the Mandelstam variables of a 2 --> 2 collision.

The Mandelstam variables u and t are defined as: $$t = (p_1 - p_3)^2 = (p_4 - p_2)^2$$ $$u = (p_1-p_4)^2 = (p_3-p_2)^2$$

Consider the collision $p + p \to p' + \Delta^+$

How do I decide how to label the four momentum of the proton and delta-baryon or more concrete, should I assign $p_3$ to the delta baryon and $p_4$ to $p'$ or $p_3$ to p' and $p_4$ to delta?

• $p_1$ and $p_2$ are incoming momenta. $p_3$ and $p_4$ are outgoing momenta. It does not matter which momentum you assign to which particle. Choose your convention (make sure to make it clear which you are using) and do your calculation. – Prahar Jun 4 '18 at 12:44
• But depending on the assignment, wouldn't this give me a different t and u? – Eren Jun 4 '18 at 12:47
• Yes, and that's OK. As I said, choose what you want, as long as you make it clear at the time of choice. – Prahar Jun 4 '18 at 12:48

It can be either, provided you adopt a definition, state it and stick to it. In a typical experiment, in the lab frame there is a beam particle, a target particle, a scattered particle and a recoil particle: $t$ is taken from the difference between the beam particle and the scattered particle and it will be clear whether 3 or 4 is the 'scattered' particle either from the experimental apparatus (3 might be detected at small angles in a hodoscope some distance away, 4 might be a slow particle emerging from the target at large angles) or from the particles (for $\pi^- p \to \pi^+ \Delta^-$ the $\pi^+$ is clearly scattered and the $\Delta$ is the recoil). With your particles (and without further details of the detector) it's ambigious and not clear so you can make your own decision.


text:
# What is the oxidizing agent in 1/8"S"_8("orthorhombic") + "O"_2(g) -> "SO"_2(g), the formation reaction for sulfur dioxide gas?

Nov 13, 2016

Let's put it this way: what do you think of assigning oxygen the role of oxidizing agent? Does that make sense?

If not, then we can do it another way. The oxidation state of sulfur in its natural state at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ is $0$, but the natural state of sulfur is "S"_8("orthorhombic"), so let's rewrite this reaction as:

$\frac{1}{8} {\text{S"_8("orthorhombic") + "O"_2(g) -> "SO}}_{2} \left(g\right)$

That aside, ${\text{O}}_{2} \left(g\right)$ is also in its natural state at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, so both of their oxidation states are $0$.

But, both of them combine to form a compound in which they are both ions. Therefore, there must be oxidation-reduction going on.

In fact, they've changed oxidation state as follows:

$\stackrel{0}{{\text{O"_2) -> stackrel(-2)("O}}_{2}}$
$\frac{1}{8} \stackrel{0}{\text{S"_8) -> stackrel(+4)("S}}$

where the oxidation states are for each individual atom, NOT the entire set of $\text{O}$ or $\text{S}$ atoms. $\text{O}$ often has the oxidation state of $- 2$ as an ion, with some rare exceptions of $- 1$ in peroxides.

Since the oxidation state of $\text{S}$ increased to become more positive, it means that ${\text{S}}_{8}$ was oxidized by the only other reactant, ${\text{O}}_{2}$.

Therefore, ${\text{O}}_{2}$ is the oxidizing agent.


text:
# Introduction¶

Ecole is a library of Extensible Combinatorial Optimization Learning Environments designed to ease the development of machine learning approaches for combinatorial optimization. More precisely, the goal of Ecole is to allow for a fast and safe prototyping of any ML for CO approach that can be formulated as a control problem (i.e., a Markov Decision Problem), as well as providing reproducible benchmarking protocols for comparison to existing approaches.

import ecole

env = ecole.environment.Branching(
reward_function=-1.5 * ecole.reward.LpIterations() ** 2,
observation_function=ecole.observation.NodeBipartite(),
)
instances = ecole.instance.SetCoverGenerator(n_rows=100, n_cols=200)

for _ in range(10):
observation, action_set, reward_offset, done, info = env.reset(next(instances))
while not done:
observation, action_set, reward, done, info = env.step(action_set[0])

Combinatorial optimization solvers typically rely on a plethora of handcrafted expert heuristics, which can fail to exploit subtle statistical similarities between problem intances. Machine Learning algorithms offer a promising candidate for replacing those heuristics, by learning data-driven policies that automatically account for such statistical relationships, and thereby creating a new kind of highly adaptive solvers.

For instance, many combinatorial optimization problems can be modeled using Mixed Integer Linear Programming and solved using the branch-and-bound algorithm. Despite its simplicity, the algorithm requires many non-trivial decisions, such as iteratively picking the next variable to branch on. Ecole aims at exposing these algorithmic control problems with a standard reinforcement learning API (agent / environment loop), in order to ease the exploration of new machine learning models and algorithms for learning data-driven policies.

Ecole’s interface is inspired from OpenAi Gym and will look familiar to reinforcement learning praticionners. The state-of-the-art Mixed Integer Linear Programming solver that acts as a controllable algorithm inside Ecole is SCIP.

The reader is referred to [Bengio2020] for motivation on why machine learning is a promising candidate to use for combinatorial optimization, as well as the methodology to do so.

Bengio2020

Bengio, Yoshua, Andrea Lodi, and Antoine Prouvost. “Machine learning for combinatorial optimization: a methodological tour d’horizon.European Journal of Operational Research. 2020.


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# How to balance a chemical equation

Balancing a chemical equation is very important in order to calculate the amount of reagents needed and the products in a reaction. After all, it’s an equation, so you must have the same quantities on both sides.

Let’s take the reaction between hydrogen and oxygen, which yields water:

$H_2+O_2\rightarrow&space;H_2O$

That small number specifies the number of atoms of an element and it can’t be changed. Here, the elements are hydrogen H and oxygen O, but they exist as biatomic gases, which means two atoms of hydrogen make a molecule and the same is for oxygen. The problem here is that we have two atoms of hydrogen and two of oxygen on the left-hand side but two of H and 1 of O on the right. The O in H2O doesn’t have that small “two” because it’s part of a compound, in this case water. To fix this inequality we can only change the number of molecules. In order to have two atoms of oxygen on the right-hand side, we will write 2 next to H2O:

$H_2+O_2\rightarrow&space;2H_2O$

This means that we now have four hydrogens and two oxygens, so we need two more atoms of H on the left, which means one more molecule of H2.

$2H_2+O_2\rightarrow&space;2H_2O$

Now we can say that the equation is balanced.

Let’s now balance this equation:

$C+O_2\rightarrow&space;CO$

We have one atom of carbon on both sides but 2 vs 1 of oxygen. What we’re going to do is write a 2 next to CO, so we will have 2 atoms of oxygen:

$C+O_2\rightarrow&space;2CO$

We now need to fix the carbon, since we have one vs two:

$2C+O_2\rightarrow&space;2CO$

And that’s it!

Now try to balance this chemical equation:

$CO_2+H_2O\rightleftharpoons&space;H_2CO_3$

This is the reaction between carbon dioxide and water, and gives carbonic acid. This is sparkling water! And as you know, after some time the gas escapes leaving water only. This is why I’ve used two arrows: carbon dioxide reacts with water to yield carbonic acid but carbonic acid can decompose to give the reagents back. The procedure doesn’t change though.

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