OpenLLM-France/wikimedia huggingface dataset:
# Abstract Algebra/Polynomial Rings

The degree of a polynomial $a_{0}+a_{1}X+...+a_{n}X^{n}$ is defined to be $n$. If $R$ is a field, and $f$ and $g$ are polynomials of $R[X]$, then we can divide $f$ by $g$ to get $f=gq+r$. However, we can also do this for any arbitrary ring if the leading coefficient of $g$ is 1.
Take real numbers R for the ring and adjoin two indeterminants X and Y. The free algebra R<X,Y> over R is the collection of sums and products involving X, Y, and real numbers. The polynomial ring R[X,Y] is the algebra reduced by XY=YX, commutativity of the two indeterminants. In terms of quotient rings, with ideal generated by XY−YX, R[X,Y] = R<X,Y>/(XY−YX). The polynomial ring is central to commutative algebra, such as in the discussion of bibinarions and tessarines to follow.
In non-commutative algebra, the anti-commutative property XY=−YX is illustrated in the quaternions. So-called "imaginary units" correspond to irreducible binomials XX+1 and YY+1. The elements of the quotient algebra R<X,Y>/(XY+YX, XX+1, YY+1) multiply as quaternions.
Exercises: What are the common names of these quotients ?
* Polynomial quotient R[X]/(XX−1)
* Free algebra quotient R<X,Y>/(XY+YX, XX+1, YY−1).

## Application: Bibinarions & Tessarines

Commutative four-dimensional hypercomplex number systems were put forward in the nineteenth century. James Cockle presented the tessarines and Corrado Segre the bicomplex numbers, called Bibinarions in the study of Associative Composition Algebra. The isomorphism of these systems can be demonstrated through quotients of polynomial rings:
Consider the polynomial ring R[X,Y], where XY = YX. The ideal $A=(X^{2}+1,\ Y^{2}-1)$ then provides a quotient ring representing tessarines. In this quotient ring approach, elements of the tessarines correspond to cosets with respect to the ideal A. Similarly, the ideal $B=(X^{2}+1,\ Y^{2}+1)$ produces a quotient representing bicomplex numbers.
A generalization of this approach uses the free algebra R⟨X,Y⟩ in two non-commuting indeterminates X and Y. Consider these three second degree polynomials $X^{2}+1,\ Y^{2}-1,\ XY-YX$. Let A be the ideal generated by them. Then the quotient ring R⟨X,Y⟩/A is isomorphic to the ring of tessarines.
To see that $(XY)^{2}+1\in A$ note that
    $XY^{2}X=X(Y^{2}-1)X+(X^{2}+1)-1,$ so that
    $XY^{2}X+1=X(Y^{2}-1)X+(X^{2}+1)\in A.$ But then
    $XY(XY-YX)+XY^{2}X+1\in A.$ as required.
Now consider the alternative ideal B generated by $X^{2}+1,\ Y^{2}+1,\ XY-YX$. In this case one can prove $(XY)^{2}-1\in B$. The ring isomorphism R⟨X,Y⟩/A ≅ R⟨X,Y⟩/B involves a change of basis exchanging $Y\leftrightarrow XY$.
Alternatively, suppose the field C of ordinary complex numbers is presumed given, and C[X] is the ring of polynomials in X with complex coefficients. Then the quotient C[X]/(X$^2$ + 1) is another presentation of bicomplex numbers.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/The Wöhler Curve

The Wöhler curve, also referred to as the S-N curve, describes the function $\sigma _{a}(N_{f})$ or $\sigma _{m}(N_{f})$. It is based on empirical results and often represents the median of the data scatter.
A significant interval of the curve can be approximated by the Basquin relation
$\sigma _{a}^{m}N=C$
Where $C$ is a constant specific to the test case.
The Basquin relation is often presented in the form
$\Delta \sigma =\Delta \sigma _{C}({\frac {N_{C}}{N}})^{\frac {1}{m}}$
where
* $N_{C}=2\cdot 10^{6}$ cycles
* $\Delta \sigma _{C}=$ is the so called detail category number.

## Detail Category Number

The fatigue detail number defines the Basquin relation and specifies a Wöhler curve. The property is often denoted FAT, C, or in mathematical expressions: $\Delta \sigma _{C}$.
If there are more than one FAT tied to a geometry-type, they refer to different specifics such as weld quality, loading direction etc. Moreover, the FAT normally corresponds to a fatigue life of $N=2\cdot 10^{6}$ cycles.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Influence of Mean Stress/Haigh Diagram

Haigh diagrams depict fatigue failure curves in $\sigma _{a},\sigma _{m}$ -space. The failure curves correspond to specific life lengths, $n$.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Load Spectra/Multiple Loads

* Introduction
* Synchronized Loads
* Non-synchronized Loads
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Load Spectra

* Introduction
* Miner-Palmgren rule
* Cycle Counting
* Multiple Loads

# Introduction

The goal of this section is to lay out some common methods for assessing the total fatigue damage effect of varying loads.
It is not unusual to have several varying loads affecting an engineering product. Unfortunately, textbooks on fatigue analysis seldom offer adequate descriptions of how such systems should be analyzed. Section Multiple Loads presents a discussion on this matter.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading

* Introduction
* Influence of Mean Stress
* Load Spectra

# Introduction

Fatigue life is influenced by virtually every thinkable aspect of the applied loading history, such as:
* Stress amplitude
* Mean stress
* Loading frequency
* Waveform
* Loading direction
* Combined effect of complex stress filed
* Chronology
The effect of some of these influences can be quantified with fairly good accuracy whilst the effect of other influences must be regarded in a more qualitative sense.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Influence of Mean Stress

* Introduction
* Algebraic Models
* Haigh Diagram

# Introduction

The mean stress during a load cycle can be compressive, zero or tensile. A compressive mean stress reduces the rate of micro crack growth, and consequently increases the fatigue life. A tensile mean stress on the other hand has the reversed effect on fatigue life. The impact of mean stress can be very significant.
There are several potential causes of non-zero mean stress, including: biased external loading, residual stresses and structural nonlinearities.
One frequently used term is the stress ratio $R={\frac {\sigma _{min}}{\sigma _{max}}}$. This measure offers a convenient way to classify different stress constellations, as follows:
* $R=1\Rightarrow$ constant stress
* $R=\pm \infty \Rightarrow$ no tension, only cyclic compression
* $R=0\Rightarrow$ no compression, only cyclic tension
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/The Similarity Principle

Wöhler curves relate directly to specific test specimens and environments. Test specimens for assessment of the fatigue properties of materials are often small unnotched peaces of material with smooth surfaces. Thus, the corresponding Wöhler curve represents something that is very dissimilar, and therefore not immediately pertinent to many structural elements in real engineering applications.
One important concept in expanding the applicability domain of fatigue test data is the Similarity Principle which implies that high cycle fatigue at a local surface is governed by the local stress field. Consequently, according to the similarity principle, the Wöhler curve of an unnotched material can be directly applied to a notched specimen of the same material, if the analysis is based on the local stress field of the considered regions. For instance.
$\sigma _{fNotched}={\frac {\sigma _{f}}{K_{t}}}$
Where:
$\sigma _{fNotched}=$ the nominal fatigue limit of a notched specimen
$\sigma _{f}=$ the fatigue limit of the unnotched reference specimen
$K_{t}=$ the stress concentration factor of the notch
It should be noted that experiments indicate that the Similarity Principle is conservative.
Clearly, detailed finite element analysis may yield the local stress level directly, and thus eliminate the need to assess any stress concentration factors.
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Influence of Mean Stress/Algebraic Models

The effect of mean stress, $\sigma _{m}$,can be modeled by the following relations:
* Soderberg: $\sigma _{a}=\sigma _{a,\sigma _{m}=0}(1-{\frac {\sigma _{m}}{\sigma _{y}}})$; which is conservative for most engineering alloys
* Modified Goodman: $\sigma _{a}=\sigma _{a,\sigma _{m}=0}(1-{\frac {\sigma _{m}}{\sigma _{u}}})$; which is nonconservative when $\sigma _{m}<0$. It is however, a good approximation for brittle materials and conservative for ductile alloys, when $\sigma _{m}\leq 0$
* Gerber: $\sigma _{a}=\sigma _{a,\sigma _{m}=0}(1-{\frac {\sigma _{m}}{\sigma _{u}}}^{2})$; which is only valid when $\sigma _{m}\geq 0$, as a result of the quadratic term. It is a good approximation for ductile alloys
# Advanced Structural Analysis/Part I - Theory/Failure Modes/Fatigue/Crack Initiation/Loading/Load Spectra/Miner-Palmgren rule

Load histories in real applications are often complex and irregular. The total fatigue effect of such loads can be estimated by the Miner-Palmgren rule, see the equation below. Fatigue failure is assumed to occur when the Miner-Palmgren equation is true.
$D=\sum {\frac {n_{i}}{N_{i}}}$
Where: $D=$ a constant, often ranging between 0.33 and 3
$n_{i}=$ the number of occurrences of a specific load cycle progression, $i$ $N_{i}=$ the fatigue life of a specific load cycle progression, $i$

## Shortcomings of Miner's rule

The Miner-Palmgren rule is rather rough and neglects several important qualities of a general load history. Among these properties are the order and shape of the individual cycles.
# Circuit Theory/2Source Excitement

Examples worked out:
* LRC circuit #1
* parallel RLC network with extra resistor in L leg
Working on this example:
This circuit was earlier analyzed using Kirchhoff. The goal is to repeat the analysis using Node and Mesh ... then find Thevenin's voltage, Norton's current and the equivalent impedance of the circuit from R$_3$'s point of view.
The Kirchhoff goal is to find everything:
    Given that the voltage source is defined by $V_{s}(t)=10cos(1000t+30^{\circ })$ and the current source is defined by $I_{s}(t)=0.5sin(1000t+45^{\circ })$, find all other voltages, currents and power of the sources.
The Node and Mesh analysis goal does not find everything. Node voltages and mesh currents are the first things found. Then everything else can be found.
Thevenin and Norton look at the rest of the circuit from just one component (or group of components) point of view. So even fewer voltages and currents are found.
# Circuit Theory/1Source Excitement/Example 7/code

In Matlab:
In MuPad and MatLab (have to cut and paste)
# Circuit Theory/2Source Excitement/Example45
## Particular/Steady State solution

Inductor short, cap open, V$_s$ = 5 μ(t),find i$_r$

## Homogeneous/Transient Solution

Loop equation:
    $V_{s}(t)=V_{L}(t)+V_{CR}(t)$
    $V_{s}(t)=L{di_{t} \over dt}+V_{RC}$
    $V_{CR}=i_{R}*R$
    $i_{C}=C{dV_{CR} \over dt}$
    $i_{t}=i_{R}+i_{C}={\frac {V_{CR}}{R}}+C{dV_{CR} \over dt}$
    $V_{s}(t)=L({d({\frac {V_{CR}}{R}}) \over dt}+C{d^{2}(V_{CR}) \over dt^{2}})+V_{CR}$
Differential equation that needs to be solved:
    $0=L{d({\frac {V_{CR}}{R}}) \over dt}+LC{d^{2}(V_{CR}) \over dt^{2}}+V_{CR}$
Guess:
    $V_{CR}=Ae^{st}$
Substitute to check if possible:
    $0=L{d({\frac {Ae^{-st}}{R}}) \over dt}+LC{d^{2}(Ae^{-st}) \over dt^{2}}+Ae^{-st}$
    $0=L(s{\frac {Ae^{-st}}{R}})+LC(s^{2}Ae^{-st})+Ae^{-st}$
    $0=L(s{\frac {A}{R}})+LC(s^{2}A)+A$
    $0={\frac {L}{R}}s+LCs^{2}+1$
So the answer is going to be second order, thus guess was wrong, but can guess more accurately now by computing roots of the above second order equation:
    $s^{2}+2s+1$
    $s_{1,2}=-1,-1$
Both roots are negative and equal, so the new guess is:
    $V_{CR}=Ae^{-t}+Bte^{-t}$
Checking again by plugging into s$^2$ + 2s + 1 = 0:
    $(Ae^{-t}+Bte^{-t}-Be^{-t}-Be^{-t})+2(-Ae^{-t}-Bte^{-t}+Be^{-t})+Ae^{-t}+Bte^{-t}=0$
Yes it equals zero now! So can go on. Have to add a constant to the differential equation solution so V$_{cr}$ is:
    $V_{CR}=Ae^{-t}+Bte^{-t}+C_{1}$

## Without Initial Conditions .. Finding the Constants

Have initial conditions: V$_{CR}$(0+) = 0 since initially cap is a short and impedance times the derivative of the inductor current i$_t$(0+) = 5. Turning this into an equation:
    $V_{CR}(0_{+})=0=A+C_{1}$
The final voltage across the parallel RC combination is going to be 5 volts (after a very long time) because the capacitor opens and the inductor shorts.
    $V_{CR}(\infty )=C_{1}=5\Rightarrow A=-5$
This is the matlab code that computes the limit:
Only B is unknown now:
    $V_{CR}=-5e^{-t}+Bte^{-t}+5$
The initial voltage across the inductor is going to be 5 volts. But this does not lead to the value of B. Another initial condition is that the initial current through the capacitor (even though it is initially a short) is zero because the inductor is initially an open. This leads to B:
    $i_{c}=C{dV_{CR} \over dt}=5e^{-t}+Be^{-t}-Bte^{-t}$
    $i_{c}(0_{+})=5+B=0\Rightarrow B=-5$
Now V$_{CR}$ is:
    $V_{CR}=5(1-e^{-t}-te^{-t})$
Which means that i$_r$ is:
    $i_{R}={\frac {V_{CR}}{R}}=10(1-e^{-t}-te^{-t})$

## Without C_1 constant

Trying to do this problem without the C_1 constant ends in something like this:
    $V_{L}(0_{+})=5=B(2-2)$
Which has no solution. Or it can lead to 5=5 where the constant disappears from the equation without finding a number for it. Or it can lead to A or B equaling infinity. Any of these non-answers means a mistake was made somewhere.
# Circuit Theory/2Source Excitement/Example46

Find the time domain expression for i$_o$ given that I$_s$ = 1μ(t) amp.

## Steady State Particular Solution

After a long time, the cap opens and inductor shorts putting the two resistors in parallel and splitting the current making i$_o$ 1/2 amp.

## Transient Particular Solution

Start writing a node equation:
    $I_{s}=i_{c}+i_{r}+i_{o}$
Substitute voltage terminal relationships:
    $I_{s}=C{dV \over dt}+{\frac {V}{R}}+i_{o}$
Find V in terms of i$_o$ through the L R branch:
    $V=L{di_{o} \over dt}+Ri_{o}$
Substitute to get I$_s$ in terms of i$_o$:
    $I_{s}=C{d(L{di_{o} \over dt}+Ri_{o}) \over dt}+{\frac {L{di_{o} \over dt}+Ri_{o}}{R}}+i_{o}$
    $I_{s}=CL{d^{2}i_{o} \over dt^{2}}+CR{di_{o} \over dt}+{\frac {L}{R}}{di_{o} \over dt}+i_{o}+i_{o}$
Substituting numbers from the problem:
    $I_{s}={d^{2}i_{o} \over dt^{2}}+2{di_{o} \over dt}+2i_{o}$

## Time constant

Guess: $i_{o}(t)=Ae^{st}$ Substituting:
    $I_{s}={d^{2}Ae^{-st} \over dt^{2}}+2{dAe^{-st} \over dt}+2Ae^{-st}$
    $I_{s}=s^{2}Ae^{-st}+2sAe^{-st}+2Ae^{-st}$
Does this equal zero?
    $s^{2}Ae^{st}-2sAe^{st}+2Ae^{st}=0$
    $s^{2}+2s+2=0$
No. Rats. Need to evaluate the above quadratic in order to guess another solution.
    $s_{1,2}=-1-j,-1+j=\sigma \pm j\omega$
    $\sigma =-1$
    $\omega =1$
So the next guess is:
    $i_{o}=e^{-t}(A_{1}\cos t+A_{2}\sin t)+C_{1}$

## Finding the Constants

After a very long time, the capacitor is going to open and the inductor is going to short. This leaves two equal resistors in parallel that are going to split the current in half.
    $i_{o}(\infty )={\frac {1}{2}}=C_{1}$
So now the expression for i$_o$ is:
    $i_{o}=e^{-t}(A_{1}\cos t+A_{2}\sin t)+{\frac {1}{2}}$
Initially the current through the conductor is 0, so i$_o$(0+) = 0:
    $i_{o}(o_{+})={\frac {1}{2}}+A_{1}=0$
Which means that:
    $A_{1}=-{\frac {1}{2}}$
The other initial condition affecting i$_o$ is the voltage across the inductor .. which is zero. We can find an expression for V$_L$:
    $V_{L}=L{di_{0}(t) \over dt}=L(-e^{-t}(A_{1}\cos t+A_{2}\sin t)+e^{-t}(-A_{1}\sin t+A_{2}\cos t))$
Setting all this equal to 0 at t=0 yields:
    $0=-A_{1}+A_{2}$
So:
    $A_{2}=A_{1}=-{\frac {1}{2}}$
Thus i$_o$ is:
    $i_{o}={\frac {1}{2}}+e^{-t}(-{\frac {1}{2}}\cos t-{\frac {1}{2}}\sin t)={\frac {1}{2}}(1-e^{-t}(\cos t+\sin t))$
This solution is used to find i$_o$ for a complicated source using the convolution integral.
# Circuit Theory/1Source Excitement/Example 7

Given:
    $V_{s}(t)=120{\sqrt {2}}cos(377t+120^{\circ })$

## Prior Work calculating Steady State/Particular Solution

Have already found the steady state/Particular solution:
    $i(t)=I_{m}cos(\omega t+\alpha )$
Where:
    $I_{m}={\frac {V_{m}}{\sqrt {R^{2}+(L\omega )^{2}}}}$
    $\alpha =\phi -\angle arctan({\frac {L\omega }{R}})$
Or numerically:
    $i(t)_{s_{P}}=15.9cos(377t+1.73)$

## Calculating the transient/Homogeneous Solution

Need to find the transient/Homogeneous Solution to:
    $R*i(t)+L*{d \over dt}i(t)=0$
There is no V$_S$ ... this makes the homogeneous solution easy!
Guess:
    $i(t)_{s_{H}}=A*e^{\frac {-t}{\tau }}$
Finding the time constant:
    $R*A*e^{\frac {-t}{\tau }}+{\frac {LA}{-\tau }}e^{\frac {-t}{\tau }}=0$
    $R-{\frac {L}{\tau }}=0$
    $\tau =L/R=.001$
Now find see if it works:
    $RAe^{\frac {-t}{L/R}}+LA(-{\frac {R}{L}})*e^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }}{=}}0$
    divide through by A, cancel L's
    $Re^{\frac {-t}{L/R}}-Re^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }}{=}}0$
It works, therefore it must be the solution:
    $i(t)=i(t)_{s_{P}}+i(t)_{s_{H}}=599\cos(377t+3.30)+Ae^{\frac {-t}{0.0001}}$
Now must find the initial conditions.

## Determining the Constants

There are two constants. $A$ and $C$ come from any homogeneous solution to a non-homogeneous differential equation equation. These were not ignored in the steady state phasor solutions earlier, the fact that they were not being computed was pointed out.
    $i(t)_{s}=i(t)_{s_{P}}+i(t)_{s_{H}}$
    $i(t)_{s}=15.9cos(377t+1.73)+A*e^{\frac {-t}{.001}}+C$
There are two initial conditions that have to be true:
* initial source voltage has a value at t=0: $120*{\sqrt {2}}\cos({\frac {2\pi }{3}})$
* initial current through the inductor has at t=0 has to be 0, thus the current throughout the entire series circuit is 0 at t=0

## Finding two initial conditions

Two equations are necessary to find A and C.
Initially the current through the inductor and the entire circuit is going to be zero:
    $i(0_{-})=0$, thus $i(0_{+})=0$.
This means that setting t=0, have one equation:
    $i(t)_{s}=15.9cos(377t+1.73)+A*e^{\frac {-t}{.001}}+C=0$
Evaluating this at t=0:
    $A+C-2.58=0$
The second equation comes from the loop:
    $v_{r}(t)+v_{L}(t)-V_{s}=0$
    $R*i(t)_{s}+L*{di(t)_{s} \over dt}-V_{s}=0$
Substituting for i(t)$_S$ and V(t)$_S$, taking the differential and then evaluating at t=0, get:
    $1.86*10^{-9}-10.0*C=0$
So solving get:
    $C=0$
    $A=2.5781$

## Summary

    $i(t)=15.9cos(377t+1.73)+2.58*e^{\frac {-t}{.001}}$
This agrees with the Laplace solution and simulation.
# Circuit Theory/2Source Excitement/Node and Mesh

The first step is to convert everything to phasors and impedances in symbolic form if possible:
    $\mathbb {V} _{1}=5*({\sqrt {3}}+j)$
    $\mathbb {I} _{1}={\frac {1-j}{2*{\sqrt {2}}}}$
    $Z_{C1}={\frac {1}{j\omega C_{1}}}=-10j$
    $Z_{C2}={\frac {1}{j\omega C_{2}}}=-5j$
    $Z_{L}=j\omega L_{1}=L_{2}=1j$

## Node Analysis

    ${\frac {\mathbb {V} _{1}-\mathbb {V} _{a}}{R_{1}+j\omega L_{1}}}+\mathbb {I} _{1}-{\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}=0$
    ${\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}-{\frac {\mathbb {V} _{b}}{R_{3}}}-{\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}=0$
    ${\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}-\mathbb {I} _{1}-{\frac {\mathbb {V} _{c}}{j\omega L_{2}}}=0$
Results using matlab:
    $\mathbb {V} _{a}=-6.5-7i\Rightarrow V_{a}(t)=9.55cos(1000t-2.32)$
    $\mathbb {V} _{b}=-3.08-3.31i\Rightarrow V_{b}(t)=4.52cos(1000t-2.32)$
    $\mathbb {V} _{c}=0.327+0.385i\Rightarrow V_{c}(t)=0.506cos(1000t+0.866)$

## Mesh Analysis

    $\mathbb {I} _{1}R_{2}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{2}}{j\omega C_{1}}}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{3}}{j\omega C_{2}}}-\mathbb {V} _{Is}=0$
    $\mathbb {V} _{1}+\mathbb {I} _{2}R_{1}+{\frac {\mathbb {I} _{2}+\mathbb {I} _{1}}{j\omega C_{1}}}+(\mathbb {I} _{2}-\mathbb {I} _{3})R_{3}+\mathbb {I} _{2}j\omega L_{1}=0$
    ${\frac {\mathbb {I} _{3}+\mathbb {I} _{1}}{j\omega C_{2}}}+\mathbb {I} _{3}j\omega L_{2}+(\mathbb {I} _{3}-\mathbb {I} _{2})R_{3}=0$
Results using Matlab:
    $\mathbb {I} _{2}=-0.239+0.234i\Rightarrow I_{2}(t)=0.334cos(1000t+2.37)$
    $\mathbb {I} _{3}=-0.238+0.235i\Rightarrow I_{3}(t)=0.334cos(1000t+2.36)$
    $\mathbb {V} _{Is}=175-179i\Rightarrow V_{Is}(t)=250cos(1000t-0.795)$
# Circuit Theory/1st Order Filters

* Low-pass filter
* High-pass filter
* Band-pass filter
* Band-reject filter
# An Introduction to Molecular Biology/Glossary/Aliphatic compounds

Aliphatic compounds are acyclic or cyclic, non-aromatic carbon compounds. Thus, aliphatic compounds are opposite to aromatic compounds.
# An Introduction to Molecular Biology/Glossary/Alpha helix

A common motif in the secondary structure of proteins, the alpha helix (α-helix) is a right-handed coiled or spiral conformation, in which every backbone N-H group donates a hydrogen bond to the backbone C=O group of the amino acid four residues earlier.
# An Introduction to Molecular Biology/Glossary/Asparagine

Asparagine (abbreviated as Asn or N) is one of the 20 most common natural amino acids on Earth. It has carboxamide as the side chain's functional group. It is not an essential amino acid.
# An Introduction to Molecular Biology/Glossary/Aminoacids

Amino acids are molecules containing an amine group, a carboxylic acid group and a side chain that varies between different amino acids. The key elements of an amino acid are carbon, hydrogen, oxygen, and nitrogen.
# Arabic/1stPersonSing

أنا
(ana) Although the spelling of أنا ('ana) indicates a long ا (aa) sound. It is always pronounced ('ana), The use of (anaa) has a special usage (particularly found in the reciting of sacred religious texts (i.e. the Quran). So just say ('a-na). That's the Arabic word for "I". "I" and أنا are the singular, first person, nominative pronoun.
أنا ولد
'ana walad(un) 'a-na-wa-lad I am a boy.(I'm a boy)
أنا ولد والكتاب كبير
'ana waladun wa_al_kitaabu kabiir(un) 'a-na-wa-la-dun-wal-ki-taa-bu-ka-biir I am a boy and the book is big.
أنا بنت والكتاب كبير جدا
'ana bintun wa_al_kitaabu kabiirun jiddan 'a-na-bin-tun-wal-ki-taa-bu-ka-bii-run-jid-dan I'm a girl and the book is very big.
أنا من كندا
'ana min kanada 'a-na-min-ka-na-da I'm from Canada.
أنا من لبنان
'ana min lubnaan 'a-na-min-lub-naan I'm from Lebanon.
أنا ولد من العاصمة
'ana waladun min al-`aaSimah(-ti) 'a-na-wa-la-dun-mi-nal-`aa-Si-mah I am a boy from the capital.
أنا ذهبت إلى الـمدرسة
'ana dhahabtu ila al_madrasah(-ti) 'a-na-dha-hab-tu-i-lal-mad-ra-sah
I went to school.
# Appendix

* Basic Phrases
* Meeting people: a dialog
* Intermediate Phrases
* Being polite
* Directions
* Useful Words
* Professions