OpenLLM-France/wikimedia huggingface dataset:
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg17
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 17: Thur, 10Oct09

### Page 17-1

Linearity $\Rightarrow \ \$ superposition $y=y_{H}+y_{P}\$
Homogeneous Solution $y_{H}\$
- Euler Equations
- Trial solution (undefined coefficient)
- Reduction of order method 2: Undetermined factor
Homogeneous L2_ODE_VC: cf. Eq.(1) P.3-1
| $\displaystyle {\begin{aligned}y''+a_{1}(x)y'+a_{0}y=0\end{aligned}}$ | (1) |
Where $a_{0}(x)\$ can be substituted for $a_{1}(x)\$
or $a_{0}y\$ in Eq(1)
Given one homogeneous solution $u_{1}(x)\$ known
Find second homogeneous solution $u_{2}(x)\$ such that
| $\displaystyle {\begin{aligned}y_{H}(x)=k_{1}u_{1}(x)+k_{2}u_{2}(x)\end{aligned}}$ | (3) |
Where$k_{1},k_{2}\$ are constants
Assume full homogeneous solution
| $\displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)\end{aligned}}$ | (2) |
Where$U(x)\$ is an unknown to be determined
Where$u_{1}(x)\$ is known

### Page 17-2

"Full" = includes $u_{2}(x)\$
Add the following: $a_{0}(x)\left[y=Uu_{1}\right]\$
and $a_{1}(x)\left[y'=Uu_{1}'+U'u_{1}\right]$
and $\left[y''=Uu_{1}''+2U'u_{1}'+U''u_{1}\right]$
To get $a_{0}y+a_{1}y'+y''=U\left[a_{0}u_{1}+a_{1}u_{1}'+u_{1}''\right]+U'\left[a_{1}u_{1}+2u_{1}'\right]+U''u_{1}=0\$ by Eq(1) p17-1
Reduce to $u_{1}''+a_{1}u_{1}'+a_{0}u_{1}=0\$
Since $u_{1}\$ is a homogeneous solution $\Rightarrow \ 0=U'(a_{1}u_{1}+2u_{1}')+U''u_{1}\$, NOTE missing dependent variable U in front of $U'\$ term
Let $Z:=U'\Rightarrow \ \$ homogeneous L1_ODE_VC for Z

### Page 17-3

| $\displaystyle {\begin{aligned}\Rightarrow \ u_{1}(x)Z'+(a_{1}(x)u_{1}(x)+2u_{1}'(x)Z=0\end{aligned}}$ | (1) |
Solve for Z,
- integration factorial method (HW)
- Direct integration (because Eq(1) is homogeneous)
$\Rightarrow \ {\frac {Z'}{Z}}+(a_{1}+{\frac {2u_{1}'}{u_{1}}})=0\$
Where $a_{1},{\frac {2u_{1}'}{u_{1}}}\$ are known
Integrate $\log \left|Z\right\vert +2\log \left|u_{1}\right\vert +\int _^{x}a_{1}(s)\,ds=k$, where k is a constant
| $\displaystyle {\begin{aligned}Z(x)={\frac {c}{(u_{1})^{2}}}e^{-\int _^{x}a_{1}(s)\,ds}=U'(x)\end{aligned}}$ | (2) |
where $c=e^{k}\$
$\Rightarrow \ U(x)=\int _^{x}{\frac {c}{(u_{1}(t))^{2}}}e^{-\int _^{t}a_{1}(s)\,ds}\,dt+{\tilde {c}}\$
where ${\bar {a}}\ _{1}(t)=\int _^{t}a_{1}(s)\,ds\$

### Page 17-4

Homogeneous solution
| $\displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)=cu_{1}\int _^{x}{\frac {1}{(u_{1}(t))^{2}}}e^{-{\bar {a}}\ _{1}(t)}\,dt+{\tilde {c}}u_{1}={\tilde {c}}u_{1}+cu_{2}\end{aligned}}$ | (1) |
where ${\tilde {c}}=k_{1}\$ and $c=k_{2}\$
| $\displaystyle {\begin{aligned}\Rightarrow u_{2}=u_{1}\int _^{x}{\frac {1}{(u_{1}(t))^{2}}}e^{-{\bar {a}}\ _{1}(t)}\,dt\end{aligned}}$ | (2) |
HW: obtain Eq.(2) P.17-3 $Z(x)\$ using the integrating factor method
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg3

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 3: Thur, 27 Aug 09

### Page 3-1

To find class homepage, go to Wikiversity: Main Page (http://en.wikiversity.org/wiki/Wikiversity:Main_Page) and search for--> user:egm6321.f09
My Wiki address is: http://clesm.mae.ufl.edu/wiki.vq/index.php/Main_Page
From Eq.(1)p.2-3: If $P(x)\neq \ 0\ \forall x$, divide throughout by $P(x)\$ to get:
| $\displaystyle {\begin{aligned}1y''+{\frac {Q}{P}}y'+{\frac {R}{P}}y={\frac {F}{P}}\end{aligned}}$ | (1) |
Where:
$\forall \$ is defined as "for all"
$a_{2}(x)=1\$,
$a_{1}(x)={\frac {Q}{P}}$,
$a_{0}(x)={\frac {R}{P}}$, and
$f(x)={\frac {F}{P}}$
$\forall x_{0}\$ such that $P(x_{0})\neq \ 0\$ then $x_{0}\$ is a regular point
Any $x_{0}\$ such that $P(x_{0})=0\$ is a regular point

### Page 3-2

2nd order--> need 2 conditions to solve for 2 constraints
Boundary Value Problem (BVP)
Prescribe:
| $\displaystyle {\begin{aligned}y(a)=\alpha \ \end{aligned}}$ | (1) |
| $\displaystyle {\begin{aligned}y(b)=\beta \ \end{aligned}}$ | (1) |
where $\alpha \ \$ and $\beta \ \$ are known values
Initial Value Problem (IVP)
Prescribe:
| $\displaystyle {\begin{aligned}y(a)=\alpha \ \end{aligned}}$ | (2) |
| $\displaystyle {\begin{aligned}y'(a)=\beta \ \end{aligned}}$ | (2) |
where $\alpha \ \$ and $\beta \ \$ are known values
Solve IVP by ODE from p3-1 Eq(1) or initial condition p3-2 Eq(2)
Two points:
1) Existence and uniqueness of solution

### Page 3-3

2) Superposition based on linearity of differential operation L(.)
| $\displaystyle {\begin{aligned}L_{2}(.)={\frac {d^{2}(.)}{dx^{2}}}+a_{1}{\frac {d(.)}{dx}}+a_{0}(.)\end{aligned}}$ | (1) |
| $\displaystyle {\begin{aligned}L_{2}(y)=y''+a_{1}y'+a_{0}y\end{aligned}}$ | (2) |
Where the 2 in $L_{2}(y)\$ is defined as 2nd order
| Linearity of $\displaystyle {\begin{aligned}L(.)\end{aligned}}$ | (3) |
Example: Matrix Algebra
$\mathbf {A} \epsilon \ \mathbb {R} \ ^{nxm}$ matrix with n rows and m columns of real numbers
$\forall \mathbf {u} ,\mathbf {v} \epsilon \ \mathbb {R} \ ^{mx1}\$ is a column matrix
$\forall \alpha \,\beta \,\epsilon \ \mathbb {R} \ \$

### Page 3-4

Clearly: $\mathbf {A} (\alpha \ \mathbf {u} +\beta \ \mathbf {v} )=\alpha \ \mathbf {A} \mathbf {u} +\beta \ \mathbf {A} \mathbf {v}$
Example:
${\frac {d}{dx}}(.)\$ is a linear operation
$(\alpha \ u+\beta \ v)'=\alpha \ u'+\beta \ v'\$
linearity allows the use of superposition
$y=y_{H}+y_{P}\$
$L(y)=L(y_{H})+L(y_{P})\$, where the subscripts H and P stand for homogeneous and particular in respective order.
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg13
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 13: Tues, 22Sept09

### Page 13-1

Euler integrating factor method Eq.(1) P.12-3
| $\displaystyle {\begin{aligned}(x^{m}y^{n})\left[{\sqrt {x}}y''+2xy'+3y\right]=0\end{aligned}}$ | (1) |
Where: $(x^{w}y^{n})=h(x,y)\$
HW: Find (m,n) such that Eq(1) is exact.
Result: A first integration is:
| $\displaystyle {\begin{aligned}\phi \ (x,y,p)=xp+(2x^{\frac {3}{2}}-1)y+k_{1}=k_{2}\end{aligned}}$ | (2) |
Where $p=y'\$ and $k_{1}\$ and $k_{2}\$ are constants
HW: Solve Eq(2) for $y(x)\$ HINT: L1_ODE_VC (integrating factor)
A class of exact L2_ODE_VC (how to invent more exact L2_ODE_VC)
HW: Find mathematical structure of $\phi \ \$ that yields the above class

### Page 13-2

$F={\frac {d\phi \ }{dx}}=\phi \ _{x}(x,y,p)+\phi \ _{y}p+\phi \ _{p}p'=P(x)y''+Q(x)y'+R(x)y\$
Where $p=y'\$ and $p'=y''\$
and $P(x)=\phi \ _{p}\$ and $Q(x)=\phi \ _{y}\$ and $R(x)y=\phi \ _{x}\$
Answer:
| $\displaystyle {\begin{aligned}\phi \ (x,y,p)=P(x)p+T(x)y+k\end{aligned}}$ | (1) |
Eq.(2) P.13-1 : $P(x)=x\$ and $T(x)=2x^{\frac {3}{2}}-1\$ and $k=k_{1}\$
Exact Nn_ODE's, where N means nonlinear and n means nth order
| $\displaystyle {\begin{aligned}F(x,y^{(0)}...y^{(n)}=0\end{aligned}}$ | (2) |
Where $y^{(0)}=y\$ and $y^{(1)}=y'\$ and $y^{(n)}={\frac {d^{n}y}{dx^{n}}}\$

### Page 13-3

Condition 1 for exactness:
| $\displaystyle {\begin{aligned}F={\frac {d\phi \ }{dx}}(x,y^{(0)},...,y^{n-1})=\phi \ _{x}+\phi \ _{y(0)}y^{(1)}+...+\phi \ _{y(n-1)}y^{(n)}\end{aligned}}$ | (1) |
Where $\phi \ _{x}+\phi \ _{y(0)}y^{(1)}+...+\phi \ _{y(n-1)}y^{(n)}\$ is related term by term to $x,y^{(0)},...,y^{n-1}\$
Condition 2 of exactness: $f_{i}:={\frac {\partial F}{\partial y^{(i)}}}\$, where $i=1,...,n\$
| $\displaystyle {\begin{aligned}f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}-...+(-1)^{n}{\frac {d^{n}f_{n}}{dx^{n}}}=0\end{aligned}}$ | (2) |
HW: Case $n=1\$ (N1_ODE)
$F(x,y,y')=0={\frac {d}{dx}}\phi \ (x,y)\$
$f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \phi \ _{xy}=\phi \ _{yx}\$
HINT: $f_{1}=\phi \ _{y}\$

### Page 13-4

Case $n=2\$ (N2_ODE)
$f(x,y,y',y'')=0={\frac {d\phi \ }{dx}}(x,y,y')\$
$f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}=0\$
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg19
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 19: Tues, 5Oct09

### Page 19-1

HW: Legendre differential Eq.(1) P.14-2 with $n=0\$, such that homogeneous solution $u_{1}(x)=1\$.
Use reduction of order method 2 (undetermined factor) to find $u_{2}(x)\$, second homgenous solution
HW: K. p28, pb. 1.1.b.
Variation of parameters (continued) P.18-4
Use expression for $y'\$ Eq.(2) P.18-4 and $y''\$ Eq.(3) P.18-4 in non-homogeneous L2_ODE_VC Eq.(1) P.3-1
| $\displaystyle {\begin{aligned}c_{1}{\cancel {(u_{1}''+a_{1}u_{1}'+a_{0}u_{1})}}+c_{2}{\cancel {(u_{2}''+a_{1}u_{2}'+a_{0}u_{2})}}+c_{1}'u_{1}'+c_{2}'u_{2}'=f\end{aligned}}$ | (1) |
Where $u_{1}''+a_{1}u_{1}'+a_{0}u_{1}\Rightarrow 0\$ , because $u_{1}\$ is a homogeneous solution
Where $u_{2}''+a_{1}u_{2}'+a_{0}u_{2}\Rightarrow 0\$ , because $u_{2}\$ is a homogeneous solution

### Page 19-2

2 equations Eq.(1) P.18-4 and Eq.(1) P.19-1 for two unknowns ${\begin{Bmatrix}c_{1}'\\c_{2}'\end{Bmatrix}}\$
In matrix form: ${\begin{bmatrix}u_{1}&u_{2}\\u_{1}'&u_{2}'\end{bmatrix}}{\begin{Bmatrix}c_{1}'\\c_{2}'\end{Bmatrix}}={\begin{Bmatrix}0\\f\end{Bmatrix}}\$
Where ${\begin{bmatrix}u_{1}&u_{2}\\u_{1}'&u_{2}'\end{bmatrix}}\$ is the Wronskian matrix designated as ${\underline {W}}\$
The Wronskian, W, is the determinant of ${\underline {W}}\$
$W=det{\underline {W}}\$
If $W\neq \ 0\$, then ${\underline {W}}^{-1}\$ exists and ${\begin{Bmatrix}c_{1}'\\c_{2}'\end{Bmatrix}}={\underline {W}}^{-1}{\begin{Bmatrix}0\\f\end{Bmatrix}}\$
Theorem: $\forall \!\,$ $u_{1},u_{2}\$ (function of x) are linearly independant if ${\underline {W}}\neq \ 0\$ , where $0=\$ zero function.

### Page 19-3

| $\displaystyle {\begin{aligned}{\underline {W}}^{-1}={\frac {1}{W}}{\begin{bmatrix}u_{2}'&-u_{2}\\-u_{1}'&u_{1}\end{bmatrix}}\end{aligned}}$ | (1) |
| $\displaystyle {\begin{aligned}{\begin{Bmatrix}c_{1}'\\c_{2}'\end{Bmatrix}}={\underline {W}}^{-1}{\begin{Bmatrix}0\\f\end{Bmatrix}}={\frac {1}{W}}{\begin{Bmatrix}-u_{2}f\\u_{1}f\end{Bmatrix}}\end{aligned}}$ | (2) |
Where ${\begin{Bmatrix}-u_{2}f\\u_{1}f\end{Bmatrix}}\$ are known
| $\displaystyle {\begin{aligned}c_{1}(x)=\int _^{x}{\frac {u_{2}(s)f(s)}{W_{s}}}\,ds+A\end{aligned}}$ | (3) |
Where $\int _^{x}{\frac {u_{2}(s)f(s)}{W_{s}}}\,ds=d_{1}(x)\$
| $\displaystyle {\begin{aligned}c_{2}(x)=\int _^{x}{\frac {u_{1}(s)f(s)}{W_{s}}}\,ds+B\end{aligned}}$ | (4) |
Where $\int _^{x}{\frac {u_{1}(s)f(s)}{W_{s}}}\,ds=d_{2}(x)\$
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg23

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 23: Thurs, 15Octt09

### Page 23-1

HW: Non-homogeneous L2_ODE_VC (cf. K. p28, problem 1.1.ab)
a) $(x-1)y''-xy'+y=x\$
b) $xy''+2y'+xy=x\$
using:
1) Variation of parameters (after knowing $u_{1}(x),u_{2}(x))\$
2) Alternative method: P.21-2 and P.21-3
Engineering application: Motivate L2_ODE_VC
Ware equation: Drums=vibrating membrane
Rectangular membrane
Circular membrane (Bessel functions)
Bessel differential equation
Chladni: Experimental method for vibrating modes of plates

### Page 23-2

Wave eq: (PDE) solution by separation of variables
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg21
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 21: Thurs, 8Oct09

### Page 21-1

P.20-4 (continued)
$\Rightarrow ax^{-1}\$ is a homgenous solution $\forall \!\,a\$
$u_{1}(x)=c_{1}x^{-1}\$
2) $b=2\Rightarrow ax^{2}\$ is another homogeneous solution since $b^{2}-b-2=0\$
$u_{2}(x)=c_{2}x^{2}\$
(Verify $u_{1}\$ and $u_{2}\$ are linearly independant components of $W\$
3) $b=6,a={\frac {1}{4}}\Rightarrow \$ left hand side of Eq(1) p20-4 $=7x^{4}\$ , where $=7x^{4}\$ is the 1st term on the right hand side
for $b=6,a={\frac {1}{4}}\Rightarrow {\frac {1}{4}}x^{6}\$
4) $b=5,a={\frac {1}{6}}\Rightarrow \$ left hand side of Eq(1) p20-4 $=3x^{3}\$ , where $=3x^{3}\$ is the 2nd term on the right hand side
for $b=5,a={\frac {1}{6}}\Rightarrow {\frac {1}{6}}x^{5}\$

### Page 21-2

Llinearity of ordinary differential equation $\Rightarrow \$ superposition
$y_{P}(x)={\frac {1}{4}}x^{6}+{\frac {1}{6}}x^{5}\$
$y(x)=c_{1}x^{-1}+c_{2}x^{2}+y_{P}(x)\$ , where $c_{1}x^{-1}+c_{2}x^{2}=y_{H}(x)\$
Alternative method to obtain full solution for non-homogeneous L2_ODE_VC knowing only one homogeneous solution (e.g. obtained by trial solution) (bypassing reduction of order method2-undertermined factor for $u_{2}\$ and variation of parameter method)
Eq.(1) P.3-1 = $y''+a_{1}(x)y'+a_{0}(x)y=f(x)\$
Assume having found $u_{1}(x)\$, a homogeneous solution: $u_{1}''+a_{1}(x)u_{1}'+a_{0}(x)u_{1}=0\$
Consider: $y(x)=U(x)u_{1}(x)\$ , where $U(x)\$ is an undetermined factor

### Page 21-3

Follow the same argument as on P.17-2 to obtain:
| $\displaystyle {\begin{aligned}f(x)=U'(a_{1}u_{1}+2u_{1}')+U''u_{1}\end{aligned}}$ | (1) |
NOTE: this equation is missing the dependant variable $U\$ in front of $U'\$ term due to reduction of order method $\phi \ \$
| $\displaystyle {\begin{aligned}Z(x):=U'(x)\end{aligned}}$ | (2) |
| $\displaystyle {\begin{aligned}\Rightarrow u_{1}(x)Z'+\left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]Z=f(x)\end{aligned}}$ | (3) |
where $u_{1}(x)\$ and $\left[a_{1}(x)u_{1}(x)+2u_{1}'(x)\right]\$ are known
Non-homogeneous L1_ODE_VC solution for $Z(x)\$ : Eq.(4) P.8-2
| $\displaystyle {\begin{aligned}U(x)=\int _^{x}Z(s)\,ds\end{aligned}}$ | (4) |
| $\displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)\end{aligned}}$ | (5) |
ref: K p.28, problem 1.1ab
a) $u_{1}(x)=e^{x}\$ , $(x-1)y''-xy'+y=0\$
Trial solution $y(x)=e^{rx}\$ , where $r=\$ constant

### Page 21-4

Find $r_{1},r_{2}\$
How many valid homogeneous solutions to $u_{1}=e^{r_{1}x}\$, find $u_{2}\$ using undetermined factor method
# Software Design/Code smells/Function is used in many places

Checklist questions:
This code smell is a lead to applying practice Don't repeat logic in several places.

## Why

A non-library function used in many different places may be a sign that some abstractions are missing, or the repetition is not removed vigorously enough, for example, in unit test code.
If the calls to the function are refactored in two parallel commits, it leads to large, not automatically mergeable conflict.

## Why not

Widely used library functions, especially operations on collections.
# Software Design/Call function's return variable "result"

Checklist questions:
* Function's result variable is called result?
* Function's result variable name starts with result-?

## Why

When there is a variable called result (or res, r, etc.) in a function, it's harder to make a mistake by accidentally returning a wrong variable because it's expected that the return statement will look like return result.
When this practice is a convention in a codebase which is followed consistently, always naming the result variable result becomes a form of structure in the code. Readers can grasp functions' logic quicker by recognizing appearances of result in the code. This convention is used by Martin Fowler.

## Why not

As described in practice Avoid a variable called "result", naming the return variable result is a missed opportunity to attach semantic information to the variable. In relatively longer functions, readers may need to deduce or to return to the function's declaration and the documentation to recall what is returned from the function. Thus, a return variable called result makes the code of the function less obvious and makes readers navigate more.

## Combining the approaches

A compromise solution is to call functions' result variables resultSemanticName, for example, resultDiscountDecimal in getDiscountPlatinumLoyalityForAccessories function in the opposite practice.
# Software Design/Catch the most specific type of exception
## Examples

Rather than catch (e: Exception) { ... } or Throwable.

## Why

Code robustness: less likely that bugs in the code wrapped with exception handling would be processed as if they were expected exceptional conditions (such as a network I/O error).

## Why not

Duplication of logic between handlers of different specific types of exceptions which leads to change amplification, higher probability of inserting bugs in the exception handling logic itself (rather than the logic being wrapped with exception handling), etc. Eliminating this duplication (e. g. via extracting a function) has its own drawbacks.
# Software Design/Code size
## Relevant practices

* Don't hide an object access chain

## Sources

* Boswell, Dustin; Foucher, Trevor (2011). The Art of Readable Code. ISBN 978-0596802295. Chapter 13. "Writing Less Code"
# Software Design/Change amplification

Change amplification is defined by John Ousterhout in A Philosophy of Software Design as a symptom of complexity which is that a seemingly simple change requires code modifications in many different places.
As a code quality, change amplification (or repetitiveness) may be defined as the expected (or the average) number of places and volume of code in the codebase that needs to be modified in a logically atomic change to the software.

## Relations to other qualities

Note that the definition of complexity as a high-level quality in A Philosophy of Software Design differs from the definition of complexity as a primary quality in the Software Design project. Change amplification has little to do with complexity as a primary quality. Rather, it is related to the following qualities:
* Easiness of change: modifying more code takes more effort.
* How much navigation is needed while working with code: to modify code in different places, the developer should first navigate to those places.
* Code robustness: often, there are no compile-time checks that guarantee that all places in the codebase that should be modified as part of the change are actually modified. In this situation, it's easy to forget to modify some of the places and thus introduce defects.

## Relevant practices

* Don't repeat yourself
Code smells:
* Object access chain: both exposing and hiding object access chains may contribute to change amplification in different circumstances.
# Software Design/Code clarity
## Relations to other qualities

* Code robustness (induces)
* Code size (induced by)
# Software Design/Choose programming language or framework similar to the language of the problem domain
## Examples

* A functional language for a compiler with transformations between immutable representations: source code, IR, machine code.
* Object-oriented language for a game with characters.
* A declarative language or framework for static descriptions, e. g. of a user interface, or configuration.
* Actor platform or framework for situations when actors and state machines match the problem domain language.

## Why

When the program's language is similar to the problem domain's language, the code is more natural to read, like a DSL.

## Related

* Use a domain-specific language
* Transform data with pipeline operator

## Sources

* David Thomas; Andrew Hunt (2019). The Pragmatic Programmer (20th Anniversary ed.). Addison-Wesley Professional. ISBN 978-0135957059.
# System

Systems are usually defined either in the context of thermodynamics or cybernetics. However, a more general system theory is being developed that includes categorical dynamics, quantum computers, fuzzy systems and economical or social systems.
* An information system is usually a database.
* A computer system is just a programmed automaton
* A philosophical system is a general framework for philosophical discussion, such as ontological, epistemological, metaphysical, etc.
* An economic system is usually defined by a corresponding ideology and is in theory controlled by different types of societies, such as : medieval, primitive, utopian, modern, and so on.
* A quantum system is non-deterministic, whereas a classical system is deterministic--with a strict causality
* A dynamic system is usually represented by a 'system' of differential equations (ODE's) that admit integrable solutions
* A chaotic system is neither quantum nor classical, although like the classical system it is still deterministic
* An anticipatory system is characteristic of organisms and also of some societies
* A complex system--unlike a simple one-- cannot either be decomposed into its components without ceasing to function or cannot be completely simulated by computer models.
An interesting exception is the Environment as it has no strict boundaries, only a horizon, and therefore it cannot be represented as a system.
The term system may be used in many other contexts, e.g. Solar System and System of Government.
# Syntactics/Quiz

Syntactics is a stand-alone lecture, an outgrowth of the dominant group project. It has been included in the course elements of terminology. It is part of a series of lectures on terminology.
Try this quiz to see if your native knowledge is already top-of-the-line or better. You are free to take this quiz at any time and as many times as you wish. It is part of Wikiversity's dedication to learning by doing.
To improve your scores, read and study the lecture, the links contained within, listed under See also, External links, and in the {{terminology resources}} template. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
Suggestion: Have the lecture available in a separate window.
To master the information and use only your memory while taking the quiz, try rewriting the information from more familiar points of view, or be creative with association.
Enjoy learning by doing!

## Hypotheses

* The syntax of dominant group suggests a group that injures others deliberately.
# Synthetic division

Synthetic division is a shortcut method of performing polynomial long division.
These notes constitute an overview of the topic.

## Introduction

Synthetic division is commonly used when dividing a polynomial (such as $x^{3}+8x^{2}-3x+6$) by a binomial (such as $x-2$). The first step is to set the binomial equal to zero and solve for $x$ ($x-2=0$, $x=2$). The next step is to set up the synthetic division using this solution and the coefficients of the polynomial. In this case, the setup would look like this:
It is important to make sure all the numbers are correctly positive or negative depending on the polynomial. Also, if the exponents of the polynomial do not decrease by one from one term to the next, a coefficient of zero must be included. For example, the setup for $x^{3}+8x^{2}-3x+6$ would use the coefficients 1, 8, -3, and 6.
The next step is to bring down the first coefficient, like this:
After that, the number at the bottom is multiplied by the solution from the binomial (in this case, $2*1=2$):
The two numbers above each other are then added:
Again, the new number at the bottom is multiplied by the solution of the binomial:
The two numbers are added:
Again, multiplication:
And addition:
These numbers are the coefficients of the new polynomial. The highest exponent of this new polynomial is one less than that of the original polynomial, so since the original polynomial started with $x^{3}$, the new one will begin with $x^{2}$. The final term is obtained in a slightly different way. The remaining coefficient (in this case, 40) is divided by the original binomial to obtain the last term. Therefore, the final answer is
${\frac {x^{3}+8x^{2}-3x+6}{x-2}}=x^{2}+10x+17+{\frac {40}{x-2}}$
This may look like a long process, but in practice it is much faster and simpler than long polynomial division.

## Uses

One major use of synthetic division is finding the roots of a polynomial. Synthetic division is an easy way to do this. In the polynomial from the previous example, $x=2$ was not a root of the polynomial, because there was a remainder left over after the division. However, if we, for example, take the polynomial $x^{3}+3x^{2}-3x+4$ and divide it by the binomial $x+4$ (when $x=-4$)
Since the remainder of this equation is 0, we have found $x=-4$ to be a root of the polynomial. We can then take the remaining quadratic $x^{2}-x+1$ and solve for the remaining two roots using the quadratic formula.