OpenLLM-France/wikimedia huggingface dataset:
# Elasticity/Stress functions

Stress functions are scalar or vector potential functions that satisfy the equilibrium and compatibility conditions. The stresses can be derived from these functions by taking partial derivatives of the potentials with respect to the spatial coordinates.

## Potential functions

You have probably encountered potential functions in introductory physics. Forces can often be calculated assuming the existence of a scalar potential field. One takes the derivative of the potential function with respect to spatial coordinates to get a vector field.
An example is the gravitational potential field.
Consider the gravitational force ($\mathbf {f}$) between two masses ($M_{1},~M_{2}$) separated by a distance $r$.
* Gravitational force between two masses
    $\mathbf {f} =-G~{\cfrac {M_{1}~M_{2}}{r}}~{\widehat {\mathbf {r} }}$
The gravitational potential energy ($\textstyle U$) is a scalar given by
    $U=\mathbf {f} \bullet {\widehat {\mathbf {r} }}=-G{\cfrac {M_{1}M_{2}}{r}}$
The gravitational field ($\textstyle \mathbf {g}$) (due to the effect of many masses), at a point (per unit mass $\textstyle M_{0}$) is
    $\mathbf {g} ={\cfrac {\mathbf {f} }{M_{0}}}$
The gravitational potential ($\textstyle V$) is a scalar function of position
    $V={\cfrac {U}{M_{0}}}$
The work done by the gravitational force equals the decrease in gravitational potential energy.
The scalar gravitational potential can be used to find the vector gravitational field.
    $g_{i}=-{\cfrac {\partial V}{\partial x_{i}}}$

## Stress functions

Stress functions are very similar to other scalar potential functions such as the gravitational potential. However, instead of delivering a vector field upon differentiation, stress functions deliver a second-order tensor field.
Here are some facts about most stress functions:
* Scalar potential functions that can be used to determine the stress field.
* No obvious physical meaning other than their use in defining stress components in terms of derivatives.
* The only restriction on the choice of a stress function is that the stress components must be defined in terms of the second derivatives of a scalar.
* This is because the second derivatives of a scalar form the components of a second-order tensor.
* Why is such a restriction necessary? To preserve the rules for coordinate transformations.

## Examples of stress functions

Airy stress function

## Related Content

Introduction to Elasticity
# Elasticity/Torsion of rectangular sections
## Sample homework problem
### Torsion of a bar with a rectangular cross section

Given:
A cylinder with a rectangular cross sections $a\times b$ under torison.
* A bar with a rectangular cross section
Show:
Why a function of the form
    $\varphi =m\left({\frac {x_{1}^{2}}{a^{2}}}-1\right)\left({\frac {x_{2}^{2}}{b^{2}}}-1\right)$
cannot be used as a Prandtl stress function for this cross section.

### Solution

For compatibility, we need
    $\nabla ^{2}{\varphi }=C$
However, for the given $\varphi$,
    $\nabla ^{2}{\varphi }=-{\frac {2m}{a^{2}b^{2}}}\left[a^{2}+b^{2}-(x_{1}^{2}+x_{2}^{2})\right]$
which is not constant. Hence, the given function cannot be used as a Prandtl stress function.
# Elasticity/Stress example 2
## Example 2

Given: A homogeneous stress field with components in the basis $(\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})\,$ given by
    $\left[{\boldsymbol {\sigma }}\right]={\begin{bmatrix}3&1&1\\1&0&2\\1&2&0\end{bmatrix}}{\text{(MPa)}}$
Find:
* The traction ($\mathbf {t} \,$) acting on a surface with unit normal ${\widehat {\mathbf {n} }}=({\widehat {\mathbf {e} }}_{2}+{\widehat {\mathbf {e} }}_{3})/{\sqrt {2}}$.
* The normal traction ($\mathbf {t} _{n}\,$) acting on a surface with unit normal ${\widehat {\mathbf {n} }}=({\widehat {\mathbf {e} }}_{2}+{\widehat {\mathbf {e} }}_{3})/{\sqrt {2}}$.
* The projected shear traction ($\mathbf {t} _{s}$) acting on a surface with unit normal ${\widehat {\mathbf {n} }}=({\widehat {\mathbf {e} }}_{2}+{\widehat {\mathbf {e} }}_{3})/{\sqrt {2}}$.
* The principal stresses.
* The principal directions of stress.

### Solution

Here's how you can solve this problem using Maple.
with(linalg):
sigma := linalg[matrix](3,3,[3,1,1,1,0,2,1,2,0]);
    $\sigma :={\begin{bmatrix}3&1&1\\1&0&2\\1&2&0\end{bmatrix}}$
e2 := linalg[matrix](3,1,[0,1,0]);
    ${\mathit {e2}}:={\begin{bmatrix}0\\1\\0\end{bmatrix}}$
e3 := linalg[matrix](3,1,[0,0,1]);
    ${\mathit {e3}}:={\begin{bmatrix}0\\0\\1\end{bmatrix}}$
n := evalm((e2+e3)/sqrt(2));
    $n:={\begin{bmatrix}0\\{\frac {\sqrt {2}}{2}}\\[2ex]{\frac {\sqrt {2}}{2}}\end{bmatrix}}$
sigmaT := transpose(sigma);
    ${\mathit {sigmaT}}:={\begin{bmatrix}3&1&1\\1&0&2\\1&2&0\end{bmatrix}}$
t := evalm(sigmaT&*n);
    $\mathbf {t} :={\begin{bmatrix}{\sqrt {2}}\\{\sqrt {2}}\\{\sqrt {2}}\end{bmatrix}}~~~~{\text{Solution for Part 1}}$
tT := transpose(t);
    ${\mathit {tT}}:={\begin{bmatrix}{\sqrt {2}}&{\sqrt {2}}&{\sqrt {2}}\end{bmatrix}}$
N := evalm(tT&*n);
    $N:={\begin{bmatrix}2\end{bmatrix}}~~~~{\text{Solution for Part 2}}$
tdott := evalm(tT&*t);
    ${\mathit {tdott}}:={\begin{bmatrix}6\end{bmatrix}}$
S := sqrt(tdott[1,1] - N[1,1]^2);
    $S:={\sqrt {2}}~~~~{\text{Solution for Part 3}}$
sigPrin := eigenvals(sigma);
    ${\mathit {sigPrin}}:=1,\,-2,\,4~~~~{\text{Solution for Part 4}}$
dirPrin := eigenvects(sigma);
    ${\mathit {dirPrin}}:=[1,\,1,\,\{[-1,\,1,\,1]\}],\,[-2,\,1,\,\{[0,\,-1,\,1]\}],\,[4,\,1,\,\{[2,\,1,\,1]\}]$
dirPrin[1];
    $[1,\,1,\,\{[-1,\,1,\,1]\}]~~~~{\text{Solution for Part 5}}$
dirPrin[2];
    $[-2,\,1,\,\{[0,\,-1,\,1]\}]~~~~{\text{Solution for Part 5}}$
dirPrin[3];
    $[4,\,1,\,\{[2,\,1,\,1]\}]~~~~{\text{Solution for Part 5}}$
# Elasticity/Stress example 4
## Example 4

Given:
The octahedral plane is the plane that is equally inclined to the directions of the three principal stresses. For any given stress of state there are eight such planes.
Show:
* The normal traction on an octahedral plane is given by
    $\mathbf {t} _{n}^{\text{oct}}={\cfrac {1}{3}}\left(\sigma _{1}+\sigma _{2}+\sigma _{3}\right)={\cfrac {1}{3}}~I_{\boldsymbol {\sigma }}$.
* The projected shear traction on an octahedral plane is given by
    $\mathbf {t} _{s}^{\text{oct}}={\cfrac {1}{3}}{\sqrt {\left(\sigma _{1}-\sigma _{2}\right)^{2}+\left(\sigma _{2}-\sigma _{3}\right)^{2}+\left(\sigma _{3}-\sigma _{2}\right)^{2}}}={\cfrac {1}{3}}{\sqrt {2~I_{\boldsymbol {\sigma }}^{2}-6~II_{\boldsymbol {\sigma }}}}.$
Here $(\sigma _{1},\sigma _{2},\sigma _{3})\,$ are the principal stresses and $(I_{\boldsymbol {\sigma }},II_{\boldsymbol {\sigma }})\,$ are the first two invariants of the stress tensor (${\boldsymbol {\sigma }}\,$).

### Solution

Let us take the basis as the directions of the principal stresses ${\widehat {\mathbf {n} }}{1}$, ${\widehat {\mathbf {n} }}{2}$, ${\widehat {\mathbf {n} }}{3}$. Then the stress tensor is given by
    $\left[{\boldsymbol {\sigma }}\right]={\begin{bmatrix}\sigma _{1}&0&0\\0&\sigma _{2}&0\\0&0&\sigma _{3}\end{bmatrix}}$
If ${\widehat {\mathbf {n} }}_{o}$ is the direction of the normal to an octahedral plane, then the components of this normal with respect to the principal basis are $n_{o1}\,$, $n_{o2}\,$, and $n_{o3}\,$. The normal is oriented in such a manner that it makes equal angles with the principal directions. Therefore, $n_{o1}=n_{o2}=n_{o3}=n_{o}\,$. Since $n_{o1}^{2}+n_{o2}^{2}+n_{o3}^{2}=1\,$, we have $n_{o}=1/{\sqrt {}3)$.
The traction vector on an octahedral plane is given by
    $\mathbf {t} _{o}={\widehat {\mathbf {n} }}_{o}\bullet \left[{\boldsymbol {\sigma }}\right]={n_{o}\sigma _{1},n_{o}\sigma _{2},n_{o}\sigma _{3}}$
The normal traction is,
    $N=\mathbf {t} _{o}\bullet {\widehat {\mathbf {n} }}_{o}=n_{o}^{2}\sigma _{1}+n_{o}^{2}\sigma _{2}+n_{o}^{2}\sigma _{3}$
Now, $I_{\sigma }=(\sigma _{1}+\sigma _{2}+\sigma _{3})\,$. Therefore,
    $N=(1/3)(\sigma _{1}+\sigma _{2}+\sigma _{3})=(1/3)I_{\sigma }\,$
The projected shear traction is given by
    $S={\sqrt {\mathbf {t} _{o}\bullet \mathbf {t} _{o}-N^{2}}}$
Therefore,
    $S={\sqrt {n_{o}^{2}\sigma _{1}^{2}+n_{o}^{2}\sigma _{2}^{2}+n_{o}^{2}\sigma _{3}^{2}-(1/9)(\sigma _{1}+\sigma _{2}+\sigma _{3})^{2}}}$
Also,
    $II_{\sigma }=\sigma _{1}\sigma _{2}+\sigma _{2}\sigma _{3}+\sigma _{3}\sigma _{1}\,$
If you do the algebra for S, you will get the required relations.
# Elasticity/Static admissibility
## Statically admissible stress field

This type of field is used in the definition of the principle of minimum complementary energy.
A statically admissible stress field is one that satisfies equilibrium (with the prescribed body force) and traction boundary conditions but not compatibility or the displacement boundary conditions.
# Elasticity/Torsion of circular cylinders
## Torsion of Circular Cylinders

* Torsion of a cylinder with a circular cross section

### About the problem:

* Circular Cylinder.
* Centroidal axis thru the center of each cross section (c.s.)
* Length $L$, Outer radius $c$.
* Applied torque $T$.
* Angle of twist $\phi$.

#### Assumptions:

* Each c.s. remains plane and undistorted.
* Each c.s. rotates through the same angle.
* No warping or change in shape.
* Amount of displacement of each c.s. is proportional to distance from end.

#### Find:

* Shear strains in the cylinder ($\gamma$).
* Shear stress in the cylinder ($\tau$).
* Relation between torque ($T$) and angle of twist ($\phi$).
* Relation between torque ($T$) and shear stress ($\tau$).

### Solution:

If $\gamma$ is small, then
    ${\text{(1)}}\qquad L\gamma =r\phi ~~\Rightarrow ~~{\gamma ={\frac {r\phi }{L}}}$
Therefore,
    ${\text{(2)}}\qquad \gamma _{\text{max}}={\frac {c\phi }{L}}~~\Rightarrow ~~\gamma ={\frac {r}{c}}\gamma _{\text{max}}$
If the material is linearly elastic,
    ${\text{(3)}}\qquad \tau =G\gamma ~~\Rightarrow ~~{\tau ={\frac {r\phi G}{L}}}$
Therefore,
    ${\text{(4)}}\qquad \tau _{\text{max}}={\frac {c\phi G}{L}}~~\Rightarrow ~~\tau ={\frac {r}{c}}\tau _{\text{max}}$
The torque on each c.s. is given by
    ${\text{(5)}}\qquad T=\int _{A}\tau rdA={\frac {\phi G}{L}}\int _{A}r^{2}dA={\frac {G\phi J}{L}}$
where $J$ is the polar moment of inertia of the c.s.
    ${\text{(6)}}\qquad J={\begin{cases}{\frac {1}{2}}\pi c^{4}&{\text{solid circular c.s.}}\\{\frac {1}{2}}\pi (c_{2}^{~4}-c_{1}^{~4})&{\text{ annular circular c.s.}}\end{cases}}$
Therefore,
    ${\text{(7)}}\qquad {\tau ={\frac {Tr}{J}}}~~\Rightarrow \tau _{\text{max}}={\frac {Tc}{J}}$
and
    ${\text{(8)}}\qquad {\phi ={\frac {TL}{JG}}}$
# English-Chinese/Hanzi
## English

Hanzi is a written form in Chinese language. Another Chinese written form is Pinyin.

## Hanzi

汉字 是 中文 的 一种 书写 形式。另一种 中文 书写 形式 是 拼音。

## Pinyin

Hànzì shì Zhōngwén de yīzhǒng shūxiě xíngshì. Lìngyīzhǒng Zhōngwén shūxiě xíngshì shì Pīnyīn.
# English-Chinese/Hair loss in some people causes psychological distress

Hair loss in some people causes psychological distress.
有些人脱发导致心理困扰。
# English-Chinese/Halloween

Halloween is the festival on October 31, the eve of All Saints' Day. It is mainly celebrated in the US, Canada, the UK, Australia, New Zealand, and etc.
Halloween 是 10月31日 万圣节 前夕 的 节日。 它 主要 在 美国，Canada，英国，Australia，New Zealand 等地 庆祝。
# English-Chinese/Hallucination

A hallucination is defined as sensory perception in the absence of external stimuli.
幻觉 被 定义 为 在 没有 外部 刺激 的 情况 下 的 感官 感觉。
# English-Chinese/Hanzi word segmentation
## English

Although Hanzi represent single syllables, Chinese is a polysyllabic language. Spacing in Hanzi word segmentation is based on words but not single Hanzi.

## Hanzi

虽然 汉字 代表 单音节，但 汉语 是 一种 多音节 语言。汉字 词切分 的 间隔 是 基于 词语 而 不是 单个 汉字。

## Pinyin

Suīrán Hànzì dàibiǎo dānyīnjié, dàn Hànyǔ shì yīzhǒng duōyīnjié yǔyán. Hànzì cíqiēfēn de jiàngé shì jīyú cíyǔ ér bùshì dāngè Hànzì.
# English-Chinese/Hawaii

Hawaii is a U.S. state and the only U.S. state which is in Oceania. It is the last state which joined the United States. Hawaii became a state on August 21, 1959. It is the only state made only of islands. Hawaii is also the name of the largest island. The capital and largest city of Hawaii is Honolulu on the Oahu Island. Hawaii's population is 1.2 million.
夏威夷是美国的州，也是唯一一个在大洋洲的美国州。这是加入美国的最后一个州。夏威夷于1959年8月21日成为一个州。它是唯一一个仅由岛屿组成的州。夏威夷也是最大岛屿的名称。夏威夷首府和最大城市是瓦胡岛上的檀香山。夏威夷人口为120万。
# English-Chinese/Hammer

A hammer is a tool for putting nails into objects such as wood. It has a long handle, and a tip with a flat end for hitting nails. The most common type of hammer (sometimes called a claw hammer) has a curved end (on the other side of the hitting end) for pulling nails out.
锤子 是 将 钉子 钉入 木材等 物体的 工具。它有 一个 长手柄，一个 头儿，用于 击打 钉子。最常见的 锤子类型（有时 称为 羊角锤）具有 弯曲的 末端（在 击打端的 另一侧），用于 将 钉子 拔出。
# English-Chinese/Hair loss is the loss of hair from a part of the head or body

Hair loss is the loss of hair from a part of the head or body.
脱发是头部或身体的一部分失去毛发。
# English-Chinese/Hair loss typically at least the head is involved

Hair loss typically at least the head is involved.
脱发通常至少涉及头部。
# English-Chinese/Hallucinations may command a person to do something which may be dangerous when combined with delusions

Hallucinations may command a person to do something which may be dangerous when combined with delusions.
幻觉可能指示一个人做出与妄想相结合可能有危险的事情。
# Linux/Installation/CentOS

CentOS is an Enterprise-class Linux Distribution derived from sources freely provided to the public by a prominent North American Enterprise Linux vendor. CentOS conforms fully with the upstream vendors redistribution policy and aims to be 100% binary compatible. (CentOS mainly changes packages to remove upstream vendor branding and artwork.) CentOS is free.
CentOS is developed by a small but growing team of core developers. In turn the core developers are supported by an active user community including system administrators, network administrators, enterprise users, managers, core Linux contributors and Linux enthusiasts from around the world.
CentOS has numerous advantages over some of the other clone projects including: an active and growing user community, quickly rebuilt, tested, and QA'ed errata packages, an extensive mirror network, developers who are contactable and responsive, multiple free support avenues including IRC Chat, Mailing Lists, Forums, a dynamic FAQ.
http://www.centos.org/
# Linux/Distributions

A Linux distribution is a collection of software, built upon the Linux kernel, that creates a Unix-like operating system. This resource is here to provide a means for participants who wish to study the similarities and differences between the hundereds of distros available.
NOTE: This page is a worksheet for Topic:Linux content development on Wikiversity. It may evolve into a table to be placed in the main Linux learning resource.

## Short list

| Distribution | w: | notes |
| Debian | w | Wikiversity resource started |
| Fedora | w | Wikiversity resource started |
| Gentoo | w | Wikiversity resource planned |
| Knoppix | w | Wikiversity resource planned |
| Mandriva | w | |
| Red Hat | w | |
| Slackware | w | Wikiversity resource planned |
| SUSE | w | |
| Ubuntu | w | Wikiversity resource started |

## Other distributions

Tiny Core Linux
# Linux/Filesystems permissions

File system permissions controls the ability of the users to view, change, navigate, and execute the contents of the file system.

## Objetives

* Understand and format file systems
* Understand file systems permissions
* Understand setuid and setgid bits
* Learn chmod commands: chmod ug+s

## Setuid

setuid and setgid (short for "set user ID" and "set group ID") are Unix access rights flags that allow users to run an executable with the permissions of the executable's owner or group respectively and to change behaviour in directories. The setuid and setgid flags only have effect on binary executable files. Set these bits on scripts, like bash, perl or python do not have any effect.

## chmod examples

chmod ug+s file.txt

## Hidden and system attributes

Linux does not regard file systems' hidden and system attributes like Windows does. Instead, file and directory entry names starting with a dot (.) are considered hidden, and can optionally be set to be revealed in file managers and command line tools.
# Linux/Searching compressed archives

Find out which file within an archive contains a text string:
The redirection of stderr to stdout using 2>/dev/stdout is necessary since 7z prints file names to stderr (prepended by the word "Extracting", hence "(Extracting|search term)"), but it needs to be in stdout to be picked up by grep through the pipe. Optionally, append |less (read: Wikipedia: less (Unix)).
7z is used since it apparently supports the widest range of archive formats, more than zgrep, bzgrep, and xzgrep.
If anyone has a better solution, feel free to add it.
If the archive is a TAR wrapped within a format such as GZip, BZip2, or XZip, the following command can be used:
# Linux/Compiling the Linux kernel

Compiling the Linux Kernel from the source code will allow you to get a great understanding of the internal workings of the kernel. Obtaining the source code, modifying, and then compiling and installing it is explained in this article. All examples are done on a Fedora 11 virtual machine.

## Kernel Versions

The developers of the linux kernel use a numbering system to track the newest versions of the source code. The current major revision is 2.6. Numbers following this version, (example, 2.6.32.1) indicate minor revisions and bug fixes. More details can be found on the Wikipedia article for the linux kernel. Linux Kernel Version numbering

## Obtaining the latest Linux Kernel version

The latest kernel source code is always available from www.kernel.org. To start with we need to obtain the latest stable release of the Linux kernel. This should be downloaded and extracted into the right directory on our linux machine. Typically, the kernel will be kept in the /usr/src/kernels directory, so we will use this in this guide.
Assuming the kernel package has been downloaded to your Download folder, the following command will copy this to our working directory.
To then extract the files we use the 'tar' program.
This will leave the original compressed file in your /usr/src/kernels directory. We can delete this using the rm command, leaving just our uncompressed directory.
You can now browse the entire Linux source code at your leisure.

## Compiling

The Linux Kernel is distributed with it's 'Makefile'. This file is a set of rules which simplifies the procedure of compiling the kernel. To use the Makefile to build the kernel, there are two commands.
You can expect this to be a time consuming process. How long it will take exactly depends on your machines hardware. To install the kernel you just compiled, you can use the command,
This will install the Kernel to your system and add the record to your grub loader. Once you reboot your system, make sure to select your new kernel version from the grub menu.

## Modifying

The Makefile can also be used to modify aspects of the kernel before you compile it. To access the tool used to modify our kernel, we use
This will bring up the interface used for our modifications. This requires the 'ncurses-devel' library to be installed. Use your package manager to install this if required.
The kernel can be compiled with a huge amount of different options set. Explaining each is beyond the scope of this document, but if it is not self-explanatory from the title, then an Internet search should yield enough results to help explain what that option does.
The Linux kernel is primarily written in the C language. To experiment with making modifications to the source code, you can open up the relevant .c file and make your changes. Once you wish to compile the kernel after your changes, you can run this make command, which will only compile what you have changed, saving a lot of time rather than compiling the whole kernel again.
# Linux/Installation/Ubuntu
## Step by Step Guide to Installing Ubuntu

* Get a LiveCD.
** Visit the download page or check your local ISP mirror.
** If you are using a desktop pc select the default distribution (Ubuntu 10.10 Maverick Meerkat) and select the closest download location to you. The closer the location, the faster you will be able to download the cd image. If you are unable to download full cd images you can order ubuntu online for free here
** Burn the iso image to a cdrom using your favorite cd burning software.
* Reboot your computer and insert the ubuntu cd. Make sure your BIOS is set to read from optical drives.
* Choose the default menu option (boot into live desktop).
* Wait for Ubuntu to load fully
* Explore Ubuntu!
* When you are ready, double click the install icon on the desktop
* Follow the prompts. Information about partitioning can be found here
* When you have entered all the information you need, grab a coffee and wait for Ubuntu to install fully.
* Reboot your computer.
* By default the GRUB bootloader selects Ubuntu as the operating system to boot into after approximately 10 seconds.
* Log in and enjoy!
# Legendre differential equation

The Legendre differential equation is the second order ordinary differential equation (ODE) which can be written as:
    $(1-x^{2})d^{2}y/dx^{2}-2xdy/dx+l(l+1)y=0\,$
which when rearranged to:
    ${d \over dx}[(1-x^{2}){dy \over dx}]+l(l+1)y=0\,$ is called Legendre differential equation of order $l$, where the quantity $l$ is a constant.
    $Ly=0\,$
where $L\,$ is the Legendre operator:
    $L={d \over dx}[(1-x^{2}){d \over dx}]+l(l+1)\,$
In principle, $l$ can be any number, but it is usually an integer.
We use the Frobenius method to solve the equation in the region $|x|\leq 1$. We start by setting the parameter p in Frobenius method zero.
    $y=\sum _{n=0}^{\infty }a_{n}x^{n}$,
    $y'=\sum _{n=0}^{\infty }na_{n}x^{n-1}$,
    $y''=\sum _{n=0}^{\infty }n(n-1)a_{n}x^{n-2}$.
Substituting these terms into the original equation, one obtains
    $0=Ly\,$ $={\big1-x^{2})y''-2xy'+l(l+1)y$ $=(1-x^{2})\sum _{n=0}^{\infty }n(n-1)a_{n}x^{n-2}-2x\sum _{n=0}^{\infty }na_{n}x^{n-1}+l(l+1)\sum _{n=0}^{\infty }a_{n}x^{n}$ $=\sum _{n=0}^{\infty }\left[-n(n-1)-2n+l(l+1)\right]a_{n}x^{n}+\sum _{n=0}^{\infty }n(n-1)a_{n}x^{n-2}$ $=\sum _{n=0}^{\infty }\left[l^{2}-n^{2}+l-n\right]a_{n}x^{n}+\sum _{n=-2}^{\infty }(n+2)(n+1)a_{n+2}x^{n}$ $=\sum _{n=0}^{\infty }\left[(l+n+1)(l-n)a_{n}+(n+2)(n+1)a_{n+2}\right]x^{n}$.
Thus
    $a_{2}=-{l(l+1) \over 2}a_{0}$,
and in general,
    $a_{n+2}=-{(l+n+1)(l-n) \over (n+2)(n+1)}a_{n}$.
This series converges when
    $\lim _{n\to \infty }\left|{a_{n+2}x^{n+2} \over a_{n}x^{n}}\right|<1$.
Therefore the series solution has to be cut by choosing:
    $n=l{\mbox{ or }}n=-(l+1)\,$.
The series cut in specific integers $l$ and $l+1$ produce polynomials called Legendre polynomials.
# Legacy cycle

The Legacy Cycle is a process students go through in order to solve a problem. We plan on introducing our students to the process as early on during their education as possible so that they may eventually use this process habitually when solving problems. In this process, the student is first given the problem and asked to generate ideas for this problem and converse with others to improve their ideas by focusing on the “best” answers. Next, a solution is chosen, researched, and modified. This solution is then tested and perfected such that the final answer can be presented to the class. The student will then be asked to evaluate their own work and provide suggestions for others who may be presented with the same problem in the future.

## Curriculum Development Process

Legacy Cycles Challenge Based Instruction modules will be designed such that the instructor will
* Analyze course and module objectives
* Develop the taxonomy of knowledge for the course
* Analyze student needs and obstacles to learning
* Design instruction for a legacy cycle for each challenge, methods for students to generate ideas, materials for multiple perspectives, and research and revise activities
* Develop formative assessment
* Develop summative assessment instruments utilizing all 3 levels of questioning