# Applied Mathematics/Printable version

Applied Mathematics

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# The Basics of Theory of The Fourier Transform

## The Basics

The two most important things in Theory of The Fourier Transform are "differential calculus" and "integral calculus". The readers are required to learn "differential calculus" and "integral calculus" before studying the Theory of The Fourier Transform. Hence, we will learn them on this page.

## Differential calculus

Differentiation is the process of finding a derivative of the function ${\displaystyle f(x)}$ in the independent input x. The differentiation of ${\displaystyle f(x)}$ is denoted as ${\displaystyle f'(x)}$ or ${\displaystyle {\frac {d}{dx}}f(x)}$. Both of the two notations are same meaning.

Differentiation is manipulated as follows:
${\displaystyle {\frac {d}{dx}}(x^{3}+1)}$
${\displaystyle =3x^{3-1}}$
${\displaystyle =3x^{2}}$

As you see, in differentiation, the number of the degree of the variable is multiplied to the variable, while the degree is subtracted one from itself at the same time. The term which doesn't have the variable ${\displaystyle x}$ is just removed in differentiation.

### Examples

${\displaystyle {\frac {d}{dx}}(x^{5}+x^{2}+28)}$
${\displaystyle =5x^{5-1}+2x^{2-1}}$ 28 doesn't have the variable x, so 28 is removed
${\displaystyle =5x^{4}+2x^{1}}$
${\displaystyle =5x^{4}+2x}$

${\displaystyle {\frac {d}{dx}}(x^{7}+x^{4}+x+7)}$
${\displaystyle =7x^{7-1}+4x^{4-1}+1x^{1-1}}$ 7 doesn't have the variable x, so 7 is removed
${\displaystyle =7x^{6}+4x^{3}++1x^{0}}$
${\displaystyle =7x^{6}+4x^{3}+1}$

### Practice problems

(1)${\displaystyle {\frac {d}{dx}}(x^{4}+15)=}$

(2)${\displaystyle {\frac {d}{dx}}(x^{5}+x^{3}+x)=}$

## Integral calculus

If you differentiate ${\displaystyle f(x)=x^{3}}$ or ${\displaystyle f(x)=x^{3}-2}$, each of them become ${\displaystyle f'(x)=3x^{2}}$. Then let's think of the opposite case. A function is provided, and when the function is differentiated, the function became ${\displaystyle f'(x)=3x^{2}}$. What's the original function? To find the original function, the integral calculus is used. Integration of ${\displaystyle f(x)}$ is denoted as ${\displaystyle \int f(x)dx}$.

Integration is manipulated as follows:
${\displaystyle \int 3x^{2}dx}$
${\displaystyle ={\frac {3x^{2+1}}{2+1}}+C}$
${\displaystyle =x^{3}+C}$
${\displaystyle C}$ denotes Constant in the equation.

More generally speaking, the integration of f(x) is defined as:

${\displaystyle \int x^{n}dx={\frac {x^{n+1}}{n+1}}+C}$

### Definite integral

Definite integral is defined as follows:
${\displaystyle \int _{a}^{b}f(x)dx}$
${\displaystyle =[F(x)]_{a}^{b}}$
${\displaystyle =F(b)-F(a)}$
where ${\displaystyle F(x)=\int f(x)dx}$

### Examples

(1)${\displaystyle \int 4xdx}$
${\displaystyle ={\frac {4x^{1+1}}{1+1}}+C}$
${\displaystyle =2x^{2}+C}$

(2)${\displaystyle \int _{1}^{2}(2x+1)dx}$
${\displaystyle =[x^{2}+x]_{1}^{2}}$
${\displaystyle =(4+2)-(1+1)}$
${\displaystyle =4}$

### Practice problems

(1)${\displaystyle \int 6xdx=}$
(2)${\displaystyle \int _{0}^{2}(3x^{2}+3)dx=}$

## Euler's number "e"

Euler's number ${\displaystyle e}$ (also known as Napier's constant) has special features in differentiation and integration:

${\displaystyle \int e^{x}dx=e^{x}+C}$
${\displaystyle {\frac {d}{dx}}e^{x}=e^{x}}$

By the way, in Mathematics, ${\displaystyle exp(x)}$ denotes ${\displaystyle e^{x}}$.

# Fourier Series

For the function ${\displaystyle f(x)}$, Taylor expansion is possible.

${\displaystyle f(x)=f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}+\cdots }$

This is the Taylor expansion of ${\displaystyle f(x)}$. On the other hand, more generally speaking, ${\displaystyle f(x)}$ can be expanded by also Orthogonal f

## Fourier series

For the function ${\displaystyle f(x)}$ which has ${\displaystyle 2\pi }$ for its period, the series below is defined:

${\displaystyle {\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin nx)\cdots (1)}$

This series is referred to as Fourier series of ${\displaystyle f(x)}$. ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$ are called Fourier coefficients.

${\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cos(nx)dx}$
${\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)dx}$

where ${\displaystyle n}$ is natural number. Especially when the Fourier series is equal to the ${\displaystyle f(x)}$, (1) is called Fourier series expansion of ${\displaystyle f(x)}$. Thus Fourier series expansion is defined as follows:

${\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin nx)}$

# General Fourier Transform

## Fourier Transform

The Fourier transform relates the function's time domain, shown in red, to the function's frequency domain, shown in blue. The component frequencies, spread across the frequency spectrum, are represented as peaks in the frequency domain.

Fourier Transform is to transform the function which has certain kinds of variables, such as time or spatial coordinate, ${\displaystyle f(t)}$ for example, to the function which has variable of frequency.

## Definition

${\displaystyle {\hat {f}}(\xi )=\int _{-\infty }^{\infty }f(t)\ e^{-i2\pi t\xi }\,dt}$...(1)

This integral above is referred to as Fourier integral, while ${\displaystyle {\hat {f}}(\xi )}$ is called Fourier transform of ${\displaystyle f(t)}$. ${\displaystyle t}$ denotes "time". ${\displaystyle \xi }$ denotes "frequency".

On the other hand, Inverse Fourier transform is defined as follows:

${\displaystyle f(t)=\int _{-\infty }^{\infty }{\hat {f}}(\xi )\ e^{i2\pi t\xi }\,d\xi }$ ...(2)

In the textbooks of universities, the Fourier transform is usually introduced with the variable Angular frequency ${\displaystyle \omega }$. In other word, ${\displaystyle \xi \rightarrow \omega =2\pi \xi }$ is substituted to (1) and (2) in the books. In that case, the Fourier transform is written in two different ways.

1.

${\displaystyle {\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{-i\omega t}dt}$
${\displaystyle f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\hat {f}}(t)e^{i\omega t}d\omega }$

2.

${\displaystyle {\hat {f}}(\omega )={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-i\omega t}dt}$
${\displaystyle f(t)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }{\hat {f}}(t)e^{i\omega t}d\omega }$

# Fourier Sine Series

## Fourier sine series

The series below is called Fourier sine series.

${\displaystyle f(x)=\sum _{n=1}^{\infty }b_{n}\sin {\frac {n\pi x}{L}}}$

where

${\displaystyle b_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\sin {\frac {n\pi x}{L}}dx}$

# Fourier Cosine Series

## Fourier cosine series

The series below is called Fourier cosine series.

${\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }a_{n}\cos {\frac {n\pi x}{L}}}$

where

${\displaystyle a_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\cos {\frac {n\pi x}{L}}dx}$

# Fourier Integral Transforms

Let ${\displaystyle f(x)=(-\infty ,\infty )}$ and

suppose

${\displaystyle \int _{-\infty }^{\infty }|f(t)|dt\leq M}$.

Then we have the functions below.

${\displaystyle f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\hat {f}}(\omega )e^{i\omega t}d\omega }$

This function ${\displaystyle f(t)}$ is referred to as Fourier integral.

${\displaystyle {\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{-i\omega t}dt}$

This function ${\displaystyle {\hat {f}}(\omega )}$ is referred to as fourier transform as we previously learned.

# Parseval's Theorem

## Parseval's theorem

${\displaystyle \int _{-\infty }^{\infty }|x(t)|^{2}\,dt=\int _{-\infty }^{\infty }|X(f)|^{2}\,df}$

where ${\displaystyle X(f)={\mathcal {F}}\{x(t)\}}$ represents the continuous Fourier transform of x(t) and f represents the frequency component of x. The function above is called Parseval's theorem.

## Derivation

Let ${\displaystyle {\bar {X}}(f)}$ be the complex conjugation of ${\displaystyle X(f)}$.

${\displaystyle X(-f)=\int _{-\infty }^{\infty }x(-t)e^{-ift}}$
${\displaystyle =\int _{-\infty }^{\infty }x(t)e^{ift}}$
${\displaystyle ={\bar {X}}(f)}$
${\displaystyle \int _{-\infty }^{\infty }|X(f)|^{2}\,df}$

Here, we know that ${\displaystyle X(f)}$ is eqaul to the expansion coefficient of ${\displaystyle x(t)}$ in fourier transforming of ${\displaystyle x(t)}$.
Hence, the integral of ${\displaystyle |X(f)|^{2}}$ is

${\displaystyle \int _{-\infty }^{\infty }{\bar {X}}(f)X(f)\,df}$
${\displaystyle =\int _{-\infty }^{\infty }\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t)e^{ift}dt\right)\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t')e^{-ift'}dt'\right)df}$
${\displaystyle =\int _{-\infty }^{\infty }x(t)x(t')\left({\frac {1}{2\pi }}e^{-if(t-t')}df\right)dtdt'}$
${\displaystyle =\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x(t)x(t')\delta (t-t')dtdt'}$
${\displaystyle =\int _{-\infty }^{\infty }|x(t)|^{2}dt}$

Hence

${\displaystyle \int _{-\infty }^{\infty }|x(t)|^{2}\,dt=\int _{-\infty }^{\infty }|X(f)|^{2}\,df}$

# Bessel Functions

## Bessel Functions

The equation below is called Bessel's differential equation.

${\displaystyle x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+(x^{2}-n^{2})y=0}$

The two distinctive solutions of Bessel's differential equation are either one of the two pairs: (1)Linear combination of Bessel function(also known as Bessel function of the first kind) and Neumann function(also known as Bessel function of the second kind) (2)Linear combination of Hankel function of the first kind and Hankel function of the second kind. Bessel function (of the first kind) is denoted as ${\displaystyle \displaystyle J_{n}(x)}$. Bessel function is defined as follow:

${\displaystyle J_{n}(x)={\frac {x^{n}}{2^{n}\Gamma (1-n)}}(1-{\frac {x^{2}}{2(2n+2)}}+{\frac {x^{4}}{2\cdot 4(2n+2)(2n+4)}}-\cdots )}$
${\displaystyle =\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\,\Gamma (m+n+1)}}{\left({\frac {x}{2}}\right)}^{2m+n}}$

where

${\displaystyle H_{n}^{(1)}(x)=J_{n}(x)+iN_{n}(x)}$
${\displaystyle H_{n}^{(2)}(x)=J_{n}(x)-iN_{n}(x)}$
${\displaystyle N_{n}(x)={\frac {J_{n}(x)\cos(n\pi )-J_{-n}(x)}{\sin(n\pi )}}}$

Γ(z) is the gamma function. ${\displaystyle i}$ is the imaginary unit. ${\displaystyle N_{n}(x)}$ is the Neumann function(or Bessel function of the second kind). ${\displaystyle H_{n}(x)}$ is the Hankel functions. If n is an integer, the Bessel function of the first kind is an entire function.

# Laplace Transforms

The Laplace transform is an integral transform which is widely used in physics and engineering. Laplace transform is denoted as ${\displaystyle \displaystyle {\mathcal {L}}\left\{f(t)\right\}}$.

The Laplace transform is named after mathematician and astronomer Pierre-Simon Laplace.

## Definition

For a function f(t), using Napier's constant"e" and complex number "s", the Laplace transform F(s) is defined as follow:

${\displaystyle F(s)={\mathcal {L}}\left\{f(t)\right\}(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

The parameter s is a complex number:

${\displaystyle s=\sigma +i\omega ,\,}$ with real numbers σ and ω.

This ${\displaystyle F(s)}$ is the Laplace transform of f(t).

## Examples of Laplace transform

Examples of Laplace transform
function result of Laplace transform
${\displaystyle C}$ (constant) ${\displaystyle {\frac {C}{s}}}$
${\displaystyle t}$ ${\displaystyle {\frac {1}{s^{2}}}}$
${\displaystyle t^{n}}$ (n is natural number) ${\displaystyle {\frac {n!}{s^{n+1}}}}$
${\displaystyle {\frac {t^{n-1}}{(n-1)!}}}$ ${\displaystyle {\frac {1}{s^{n}}}}$
${\displaystyle e^{at}}$ ${\displaystyle {\frac {1}{s-a}}}$
${\displaystyle e^{-at}}$ ${\displaystyle {\frac {1}{s+a}}}$
${\displaystyle {\rm {cos}}\ \omega t}$ ${\displaystyle {\frac {s}{s^{2}+{\omega }^{2}}}}$
${\displaystyle {\rm {sin}}\ \omega t}$ ${\displaystyle {\frac {\omega }{s^{2}+{\omega }^{2}}}}$
${\displaystyle {\frac {t^{n-1}}{\Gamma (n)}}}$ ${\displaystyle {\frac {1}{s^{n}}}}$ (n>0)
${\displaystyle \delta (t-a)}$ (Delta function) ${\displaystyle e^{-as}}$
${\displaystyle H(t-a)}$ (Heaviside function) ${\displaystyle {\frac {e^{-as}}{s}}}$

## Examples of calculation

(1)Suppose ${\displaystyle f(t)=C}$ (C = constant)
${\displaystyle \int _{0}^{\infty }e^{-st}C\,dt}$
${\displaystyle ={\frac {C}{s}}}$
${\displaystyle =F(s)}$

(2)Suppose ${\displaystyle f(t)=e^{-at}}$
${\displaystyle \int _{0}^{\infty }e^{-st}\cdot e^{-at}\,dt}$
${\displaystyle =\int _{0}^{\infty }e^{-(s+a)t}\,dt}$
${\displaystyle =\left[{\frac {-e^{-(s+a)t}}{s+a}}\right]_{0}^{\infty }}$
${\displaystyle ={\frac {1}{s+a}}}$

# Complex Integration

## Complex integration

On the piecewise smooth curve ${\displaystyle C:z=z(t)}$ ${\displaystyle (a\leqq t\leqq b)}$, suppose the function f(z) is continuous. Then we obtain the equation below.

${\displaystyle \int _{C}f(z)dz=\int _{a}^{b}f\{z(t)\}\ {\frac {dz(t)}{dt}}dt}$

where ${\displaystyle f(z)}$ is the complex function, and ${\displaystyle z}$ is the complex variable.

## Proof

Let

${\displaystyle f(z)=u(x,y)+iv(x,y)}$
${\displaystyle dz=dx+idy}$

Then

${\displaystyle \int _{C}f(z)dz}$
${\displaystyle =\int _{C}(u+iv)(dx+idy)}$
${\displaystyle =\left(\int _{C}udx-\int _{C}vdy\right)+i\left(\int _{C}vdx-\int _{C}udy\right)}$

The right side of the equation is the real integral, therefore, according to calculus, the relationship below can be applied.

${\displaystyle \int _{x_{1}}^{x_{2}}f(x)dx=\int _{t_{1}}^{t_{2}}f(x){\frac {dx}{dt}}dt}$

Hence

${\displaystyle \int _{C}f(z)dz}$
${\displaystyle =\left(\int _{a}^{b}u{\frac {dx}{dt}}dt-\int _{a}^{b}v{\frac {dy}{dt}}dt\right)+i\left(\int _{a}^{b}v{\frac {dx}{dt}}dt-\int _{a}^{b}u{\frac {dy}{dt}}dt\right)}$
${\displaystyle =\left(\int _{a}^{b}ux'(t)dt-\int _{a}^{b}vy'(t)dt\right)+i\left(\int _{a}^{b}vx'(t)dt-\int _{a}^{b}uy'(t)dt\right)}$
${\displaystyle =(u+iv)\left(x'(t)+iy'(t)\right)dt}$
${\displaystyle =\int _{a}^{b}f\{z(t)\}\ z'(t)dt}$

This completes the proof.

# The Basics

## The Basics of linear algebra

${\displaystyle \mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}&a_{12}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\\end{bmatrix}}.}$

A matrix is composed of a rectangular array of numbers arranged in rows and columns. The horizontal lines are called rows and the vertical lines are called columns. The individual items in a matrix are called elements. The element in the i-th row and the j-th column of a matrix is referred to as the i,j, (i,j), or (i,j)th element of the matrix. To specify the size of a matrix, a matrix with m rows and n columns is called an m-by-n matrix, and m and n are called its dimensions.

### Basic operation[1]

Operation Definition Example
Addition The sum A+B of two m-by-n matrices A and B is calculated entrywise:
(A + B)i,j = Ai,j + Bi,j, where 1 ≤ im and 1 ≤ jn.

${\displaystyle {\begin{bmatrix}1&3&1\\1&0&0\end{bmatrix}}+{\begin{bmatrix}0&0&5\\7&5&0\end{bmatrix}}={\begin{bmatrix}1+0&3+0&1+5\\1+7&0+5&0+0\end{bmatrix}}={\begin{bmatrix}1&3&6\\8&5&0\end{bmatrix}}}$

Scalar multiplication The scalar multiplication cA of a matrix A and a number c (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of A by c:
(cA)i,j = c · Ai,j.
${\displaystyle 2\cdot {\begin{bmatrix}1&8&-3\\4&-2&5\end{bmatrix}}={\begin{bmatrix}2\cdot 1&2\cdot 8&2\cdot -3\\2\cdot 4&2\cdot -2&2\cdot 5\end{bmatrix}}={\begin{bmatrix}2&16&-6\\8&-4&10\end{bmatrix}}}$
Transpose The transpose of an m-by-n matrix A is the n-by-m matrix AT (also denoted Atr or tA) formed by turning rows into columns and vice versa:
(AT)i,j = Aj,i.
${\displaystyle {\begin{bmatrix}1&2&3\\0&-6&7\end{bmatrix}}^{\mathrm {T} }={\begin{bmatrix}1&0\\2&-6\\3&7\end{bmatrix}}}$

### Practice problems

(1) ${\displaystyle {\begin{bmatrix}5&7&3\\1&2&9\end{bmatrix}}+{\begin{bmatrix}4&0&5\\8&3&0\end{bmatrix}}=}$
(2) ${\displaystyle 4{\begin{bmatrix}-1&0&-5\\7&9&-6\end{bmatrix}}=}$
(3) ${\displaystyle {\begin{bmatrix}-2&5&7\\0&0&9\end{bmatrix}}^{\mathrm {T} }=}$

## Matrix multiplication

Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B[2]

${\displaystyle [\mathbf {AB} ]_{i,j}=A_{i,1}B_{1,j}+A_{i,2}B_{2,j}+\cdots +A_{i,n}B_{n,j}=\sum _{r=1}^{n}A_{i,r}B_{r,j}}$[3]

Schematic depiction of the matrix product AB of two matrices A and B.

### Example

${\displaystyle {\begin{bmatrix}-2&0\\3&2\end{bmatrix}}{\begin{bmatrix}1&2\\3&-1\end{bmatrix}}}$

${\displaystyle ={\begin{bmatrix}-2+0&-4+0\\3+6&6+(-2)\end{bmatrix}}}$

${\displaystyle ={\begin{bmatrix}-2&-4\\9&4\end{bmatrix}}}$

### Practice Problems

(1) ${\displaystyle {\begin{bmatrix}1&0\\2&2\end{bmatrix}}{\begin{bmatrix}4\\2\end{bmatrix}}=}$

(2) ${\displaystyle {\begin{bmatrix}1&2\\2&3\end{bmatrix}}{\begin{bmatrix}2&3\\1&4\end{bmatrix}}=}$

## Dot product

A row vector is a 1 × m matrix, while a column vector is a m × 1 matrix.

Suppose A is row vector and B is column vector, then the dot product is defined as follows;

${\displaystyle A\cdot B=|A||B|cos\theta }$

or

${\displaystyle \mathbf {A} \cdot \mathbf {B} ={\begin{pmatrix}a_{1}&a_{2}&\cdots &a_{n}\end{pmatrix}}{\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}=\sum _{i=1}^{n}a_{i}b_{i}}$

Suppose ${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{1}&a_{2}&a_{3}\end{pmatrix}}}$ and ${\displaystyle \mathbf {B} ={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}}$ The dot product is

${\displaystyle \mathbf {A} \cdot \mathbf {B} ={\begin{pmatrix}a_{1}&a_{2}&a_{3}\end{pmatrix}}{\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}$

### Example

Suppose ${\displaystyle \mathbf {A} ={\begin{pmatrix}2\\1\\3\end{pmatrix}}}$ and ${\displaystyle \mathbf {B} ={\begin{pmatrix}7\\5\\4\end{pmatrix}}}$

${\displaystyle \mathbf {A} \cdot \mathbf {B} ={\begin{pmatrix}2&1&3\end{pmatrix}}{\begin{pmatrix}7\\5\\4\end{pmatrix}}}$

${\displaystyle =2\cdot 7+1\cdot 5+3\cdot 4}$
${\displaystyle =14+5+12}$

${\displaystyle =31}$

### Practice problems

(1) ${\displaystyle \mathbf {A} ={\begin{pmatrix}3\\2\\5\end{pmatrix}}}$ and ${\displaystyle \mathbf {B} ={\begin{pmatrix}1\\4\\3\end{pmatrix}}}$

${\displaystyle \mathbf {A} \cdot \mathbf {B} =}$

(2) ${\displaystyle \mathbf {A} ={\begin{pmatrix}1\\0\\3\end{pmatrix}}}$ and ${\displaystyle \mathbf {B} ={\begin{pmatrix}6\\9\\2\end{pmatrix}}}$

${\displaystyle \mathbf {A} \cdot \mathbf {B} =}$

## Cross product

Cross product is defined as follows:

${\displaystyle A\times B=|A||B|sin\theta }$

Or, using detriment,

${\displaystyle \mathbf {A\times B} ={\begin{vmatrix}e_{x}&e_{y}&e_{z}\\a_{x}&a_{y}&a_{z}\\b_{x}&b_{y}&b_{z}\\\end{vmatrix}}=(a_{y}b_{z}-a_{z}b_{y},a_{z}b_{x}-a_{x}b_{z},a_{x}b_{y}-a_{y}b_{x})}$

where ${\displaystyle e}$ is unit vector.

## References

1. Sourced from Matrix (mathematics), Wikipedia, 28th March 2013.
2. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013.
3. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013.

# Lagrange Equations

## Lagrange Equation

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {x}}}}\right)-{\frac {\partial L}{\partial x}}=0}$

where ${\displaystyle {\dot {x}}={\frac {dx}{dt}}}$
The equation above is called Lagrange Equation.

Let the kinetic energy of the point mass be ${\displaystyle T}$ and the potential energy be ${\displaystyle U}$.
${\displaystyle T-U}$ is called Lagrangian. Then the kinetic energy is expressed by

${\displaystyle T={\frac {1}{2}}m{\dot {x}}^{2}+{\frac {1}{2}}m{\dot {y}}^{2}}$
${\displaystyle ={\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2})}$

Thus

${\displaystyle T=T({\dot {x}},{\dot {y}})}$
${\displaystyle U=U(x,y)}$

Hence the Lagrangian ${\displaystyle L}$ is

${\displaystyle L=T-U}$
${\displaystyle =T({\dot {x}},{\dot {y}})-U(x,y)}$
${\displaystyle ={\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2})-U(x,y)}$

Therefore ${\displaystyle T}$ relies on only ${\displaystyle {\dot {x}}}$ and ${\displaystyle {\dot {y}}}$. ${\displaystyle U}$ relies on only ${\displaystyle x}$ and ${\displaystyle y}$. Thus

${\displaystyle {\frac {\partial L}{\partial {\dot {x}}}}={\frac {\partial T}{\partial {\dot {x}}}}=m{\dot {x}}}$
${\displaystyle {\frac {\partial L}{\partial {\dot {y}}}}={\frac {\partial T}{\partial {\dot {y}}}}=m{\dot {y}}}$

In the same way, we have

${\displaystyle {\frac {\partial L}{\partial x}}=-{\frac {\partial U}{\partial x}}}$
${\displaystyle {\frac {\partial L}{\partial y}}=-{\frac {\partial U}{\partial y}}}$

# The State Equation

## Ideal gas law

As an experimental rule, about gas, the equation below is found.

${\displaystyle PV=nRT}$...(1)

where

${\displaystyle \ p}$ = Pressure (absolute)
${\displaystyle \ V}$ = Volume
${\displaystyle \ n}$ = Number of moles of a substance
${\displaystyle \ T}$ = Absolute temperature
${\displaystyle \ R}$ = Gas constant
${\displaystyle R\simeq 8.31J(Kmol)^{-1}}$

The gas which strictly follows the law (1) is called ideal gas. (1) is called ideal gas law. Ideal gas law is the state equation of gas. In high school, the ideal gas constant below might be used:

${\displaystyle R=0.082LatmK^{-1}mol^{-1}}$

But, in university, the gas constant below is often used:

${\displaystyle R=8.31JK^{-1}mol^{-1}}$