nyuuzyou/wikis huggingface dataset:
Fourier Series/Square Wave

== Theorem ==
Let $\map S x$ be the square wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
1 & : x \in \openint 0 l \\
-1 & : x \in \openint {-l} 0 \\
\map S {x + 2 l} & : x < -l \\
\map S {x - 2 l} & : x > +l \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
Let $\map f x$ be the function defined as:
:$\forall x \in \openint {-l} l: \begin{cases} -1 & : -l < x < 0 \\ 1 & : 0 < x < l \end {cases}$
By inspection we see that $\map f x$ is an odd function.
Hence from Fourier Series for Odd Function over Symmetric Range we can express $f$ by a half-range Fourier sine series:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n \pi x} l$
where for all $n \in \Z_{> 0}$:
:$\ds b_n = \frac 2 l \int_0^l \map f x \sin \dfrac {n \pi x} l \rd x$
over the interval $\openint 0 l$.
In that interval $\openint 0 l$, we have:
:$\map S x = 1$
Hence:
The result follows.
== Special Cases ==
Category:Square Waves
Category:Fourier Series for Square Wave
Fourier Series/Square Wave/Special Cases/Half Interval Pi

== Special Case of Fourier Series for Square Wave ==
Let $\map S x$ be the square wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
1 & : x \in \openint 0 \pi \\
-1 & : x \in \openint {-\pi} 0 \\
\map S {x + 2 \pi} & : x < -\pi \\
\map S {x - 2 \pi} & : x > +\pi \end {cases}$
Then its half-range Fourier sine series can be expressed as:
== Proof ==
From Fourier Series for Square Wave, the real function $\map f x$ defined on the open interval $\openint {-l} l$ as:
:$\map f x = \begin {cases} -1 & : x \in \openint {-l} 0 \\ 1 & : x \in \openint 0 l \end {cases}$
has a half-range Fourier sine series which can be expressed as:
The result follows by setting $l = \pi$.
== Sources ==
* : $\S 23$: Fourier Series: Special Fourier Series and their Graphs $23.7$
Category:Fourier Series for Square Wave
Fourier Series/Square Wave/Special Cases/Unit Half Interval

== Special Case of Fourier Series for Square Wave ==
Let $\map S x$ be the square wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
1 & : x \in \openint 0 1 \\
-1 & : x \in \openint {-1} 0 \\
\map S {x + 2} & : x < -1 \\
\map S {x - 2} & : x > +1 \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
From Fourier Series for Square Wave, the real function $\map f x$ defined on the open interval $\openint {-l} l$ as:
:$\map f x = \begin {cases} -1 & : x \in \openint {-l} 0 \\ 1 & : x \in \openint 0 l \end {cases}$
has a Fourier series which can be expressed as:
The result follows by setting $l = 1$.
Category:Fourier Series for Square Wave
Fourier Series/Square of x minus pi, Square of pi

== Theorem ==
:
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:$\map f x = \begin{cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end{cases}$
Then its Fourier series can be expressed as:
== Proof ==
By definition of Fourier series:
:$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
where:
for all $n \in \Z_{>0}$.
Thus:
For $n > 0$:
Splitting this up into bits:
Reassembling $a_n$ from the remaining non-vanishing terms:
Now for the $\sin n x$ terms:
Splitting this up into bits:
Reassembling $b_n$ from the remaining non-vanishing terms:
Finally:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: Example $1$
Category:Examples of Fourier Series
Fourier Series/Square of x minus pi, Square of pi/Mistake

== Source Work ==
* : Chapter One: $\S 2$. Fourier Series: Example $1$
== Mistake ==
:$\map f x \sim \ds \frac 2 3 \pi^2 + 2 \sum_{n \mathop = 1}^\infty \sqbrk {\frac {\cos n x} {n^2} + \set {\frac {\paren {-1}^n \pi} n - \frac {2 \paren {1 - \paren {-1}^n} } {\pi n^3} } \sin n x}$
== Correction ==
The author appears to have gained a spurious multiple of $2$ in the $\sin n x$ term, apparently by mistakenly including the $\sin n x$ terms within the coefficient $2$ which applies to only the $\cos n x$ terms.
The correct expression, according to Fourier Series: $\paren {x - \pi}^2$, $\pi^2$, should be:
:$\map f x \sim \ds \frac 2 3 \pi^2 + \sum_{n \mathop = 1}^\infty \sqbrk {\frac {2 \cos n x} {n^2} + \set {\frac {\paren {-1}^n \pi} n - \frac {2 \paren {1 - \paren {-1}^n} } {\pi n^3} } \sin n x}$
This is the graph of $\map f x$ and its expansion to $a_5$ and $b_5$ according to the correct analysis:
:
In comparison, this is the corresponding graph as a result of 's analysis:
:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: Example $1$
Category:Mistakes/Fourier Series/Sneddon
Fourier Series/Triangle Wave

== Theorem ==
<onlyinclude>
Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map T x = \begin {cases}
\size x & : x \in \closedint {-l} l \\
\map T {x + 2 l} & : x < -l \\
\map T {x - 2 l} & : x > +l \end {cases}$
where:
:$l$ is a given real constant
:$\size x$ denotes the absolute value of $x$.
Then its Fourier series can be expressed as:
== Proof ==
Let $\map f x: \openint {-l} l \to \R$ denote the absolute value function on the open interval $\openint {-l} l$:
:$\map f x = \size x = \begin{cases}
x & : x > 0 \\
0 & : x = 0 \\
-x & : x < 0
\end{cases}$
From Fourier Series for Absolute Value Function over Symmetric Range, $\map f x$ can immediately be expressed as:
:$\ds \map f x \sim \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$
== Special Cases ==
Category:Triangle Waves
Category:Fourier Series for Triangle Wave
Fourier Series/Triangle Wave/Special Cases/Half Interval Pi

== Special Case of Fourier Series for Triangle Wave ==
Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map T x = \begin {cases}
\size x & : x \in \closedint {-\pi} \pi \\
\map T {x + 2 \pi} & : x < -\pi \\
\map T {x - 2 \pi} & : x > +\pi \end {cases}$
where $\size x$ denotes the absolute value of $x$.
Then its Fourier series can be expressed as:
== Proof ==
From Fourier Series for Triangle Wave, the real function $\map T x$ defined on the open interval $\openint {-l} l$ as:
:$\map T x = \size x$
has a Fourier series which can be expressed as:
The result follows by setting $l = \pi$.
== Sources ==
* : $\S 23$: Special Fourier Series and their Graphs: $23.8$
Category:Fourier Series for Triangle Wave
Fourier Series/Triangle Wave/Special Cases/Unit Half Interval

== Special Case of Fourier Series for Triangle Wave ==
Let $\map T x$ be the triangle wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map T x = \begin {cases}
\size x & : x \in \closedint {-1} 1 \\
\map T {x + 2} & : x < -1 \\
\map T {x - 2} & : x > +1 \end {cases}$
where $\size x$ denotes the absolute value of $x$.
Then its Fourier series can be expressed as:
== Proof ==
From Fourier Series for Triangle Wave, the real function $\map T x$ defined on the open interval $\openint {-l} l$ as:
:$\map T x = \size x$
has a Fourier series which can be expressed as:
The result follows by setting $l = 1$.
Category:Fourier Series for Triangle Wave
Fourier Series/x over 0 to 2, x-2 over 2 to 4

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 4$ as:
$\quad \map f x = \begin{cases}
x & : 0 < x \le 2 \\
x - 2 & : 2 < x < 4 \end{cases}$
Then its Fourier series can be expressed as:
:$\ds \map f x \sim 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \paren {1 + \frac {4 \paren {-1}^r} {\paren {2 r - 1} \pi} } \cos \frac {\paren {2 r - 1} \pi x} 4$
== Proof ==
Let $\map f x$ be the function defined as:
$\quad \forall x \in \openint 0 4: \begin{cases}
x & : 0 < x \le 2 \\
x - 2 & : 2 < x < 4 \end{cases}$
Let $f$ be expressed by a half-range Fourier cosine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} 4$
where for all $n \in \Z_{> 0}$:
:$\ds a_n = \frac 2 l \int_0^l \map f x \cos \frac {n \pi x} l \rd x $
In this context, $l = 4$ and so this can be expressed as:
First the case when $n = 0$:
When $n \ne 0$:
Splitting it up into two:
On the other hand:
The above expression is $0$ when $n$ is even, by Sine of Multiple of Pi.
Thus we may substitute $n = 2 r - 1$.
Then:
Combining the results:
== Sources ==
* : Chapter One: $\S 6$. Half-Range Cosine Series: Example $5$
Category:Examples of Fourier Series
Fourier Series/x over 0 to 2, x-2 over 2 to 4/Mistake

== Source Work ==
* : Chapter One: $\S 6$. Half-Range Cosine Series: Example $5$
== Mistake ==
:Find the half-range cosine series for
::$\map f x = \begin{cases}
1 , 0 < x < 2 \\
x - 2 , 2 < x < 4 \end{cases}$
:for the half-range $0 < x < 4$.
... and the required series is
::$\map S x = \ds 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \set {1 + \frac {4 \paren {-1}^r} {\paren {2 r - 1} \pi} } x \cos \frac {\paren {2 r - 1} \pi x} 4$.
== Correction ==
The subsequent analysis is performed for the function:
:$\map f x = \begin{cases}
x & : 0 < x < 2 \\
x - 2 & : 2 < x < 4 \end{cases}$
and the required series is actually:
:$\map S x = \ds 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \set {1 + \frac {4 \paren {-1}^r} {\paren {2 r - 1} \pi} } \cos \frac {\paren {2 r - 1} \pi x} 4$.
== Sources ==
* : Chapter One: $\S 6$. Half-Range Cosine Series: Example $5$
Category:Mistakes/Fourier Series/Sneddon
Fourier Series/x squared over Minus Pi to Pi

== Theorem ==
For $x \in \openint {-\pi} \pi$:
:$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$
== Proof ==
From Even Power is Even Function, $x^2$ is an even function.
By Fourier Series for Even Function over Symmetric Range, we have:
:$\ds x^2 \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where:
Then: 
Substituting for $a_n$ in $(1)$: 
:$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$
as required.
== Sources ==
* : $\S 23$: Special Fourier Series and their Graphs: $23.14$
Category:Examples of Fourier Series
Fourier Series for Even Function over Symmetric Range

== Theorem ==
Let $\map f x$ be an even real function defined on the interval $\openint {-\lambda} \lambda$.
Then the Fourier series of $\map f x$ can be expressed as:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$
where for all $n \in \Z_{\ge 0}$:
:$a_n = \dfrac 2 \lambda \ds \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$
== Proof ==
By definition of the Fourier series for $f$:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
From Fourier Cosine Coefficients for Even Function over Symmetric Range:
:$a_n = \ds \dfrac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
From Fourier Sine Coefficients for Even Function over Symmetric Range
:$b_n = 0$
for all $n \in \Z_{\ge 0}$:
== Also see ==
* Fourier Series for Odd Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions: $(6)$
Category:Even Functions
Category:Fourier Series
Fourier Series for Logarithm of Sine of x over 0 to Pi

== Theorem ==
:$\ds \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$
where $0 < x < \pi$. 
== Proof 1 ==
== Proof 2 ==
Category:Examples of Fourier Series
Category:Fourier Series for Logarithm of Sine of x over 0 to Pi
Fourier Series for Logarithm of Sine of x over 0 to Pi/Proof 1

== Theorem ==
== Proof ==
We find the Half-Range Fourier Cosine Series over $\openint 0 {\dfrac \pi 2}$ for $\map \ln {\sin x}$.
By definition:
:$\ds \map \ln {\sin x} \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x$
where for all $n \in \Z_{\ge 0}$:
:$\ds a_n = \frac 4 \pi \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x$
By Definite Integral from 0 to Half Pi of Logarithm of Sine x:
:$a_0 = \dfrac 4 \pi \paren {-\dfrac \pi 2 \ln 2} = -2 \ln 2$
By Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx:
:$a_n = \dfrac 4 \pi \paren {-\dfrac \pi {4 n} } = -\dfrac 1 n$
Therefore:
Category:Fourier Series for Logarithm of Sine of x over 0 to Pi
Fourier Series for Odd Function over Symmetric Range

== Theorem ==
Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.
Then the Fourier series of $\map f x$ can be expressed as:
:$\map f x \sim \ds \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
== Proof ==
By definition of the Fourier series for $f$:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
From Fourier Cosine Coefficients for Odd Function over Symmetric Range:
:$a_n = 0$
for all $n \in \Z_{\ge 0}$.
From Fourier Sine Coefficients for Odd Function over Symmetric Range
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{> 0}$:
== Also see ==
* Fourier Series for Even Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions: $(8)$
Category:Odd Functions
Category:Fourier Series
Fourier Series over General Range from Specific

== Theorem ==
Let $a, b \in \R$ be real numbers.
Let $f: \R \to \R$ be a function such that $\ds \int_a^b \map f x \rd x$ converges absolutely.
Then $f$ can be expressed by a Fourier series of the form:
:$\ds \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$
where:
== Proof ==
Consider the Fourier series:
:$(1): \quad \ds \map S x = \frac {A_0} 2 + \sum_{m \mathop = 1}^\infty \paren {A_m \cos \frac {2 m \pi \paren {x - a} } {b - a} + B_m \sin \frac {2 m \pi \paren {x - a} } {b - a} }$
Let $\xi = \dfrac {2 \pi \paren {x - a} } {b - a}$.
Then:
:$\dfrac {\d \xi} {\d x} = \dfrac {2 \pi} {b - a}$
Then:
and:
Substituting $\xi$ for $x$ in $(1)$:
:$(2): \quad \map S \xi = \dfrac {A_0} 2 + \ds \sum_{m \mathop = 1}^\infty \paren {A_m \cos m \xi + B_m \sin m \xi}$
which is a Fourier series for a real function $\map g \xi$ such that $\ds \int_0^{2 \pi} \map g \xi \rd x$ converges absolutely.
The Fourier coefficients of $\map g \xi$ are given by:
Substituting $x$ for $\xi$:
and:
Thus while $A_n$ and $B_n$ are exactly the Fourier coefficients of $\map g \xi$, they are also exactly the Fourier coefficients of $\map S x$.
Thus $\map S x$ is the Fourier series for $\map f x$.
== Sources ==
* : Chapter One: $\S 7$. Fourier Series over a General Range $\tuple {a, b}$
Category:Fourier Series
Fourier Sine Coefficients for Even Function over Symmetric Range

== Theorem ==
Let $\map f x$ be an even real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{> 0}$:
:$b_n = 0$
== Proof ==
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
From Sine Function is Odd:
:$\sin a = -\map \sin {-a}$
for all $a$.
By Odd Function Times Even Function is Odd, $\map f x \sin \dfrac {n \pi x} l$ is odd.
The result follows from the corollary to Definite Integral of Odd Function:
:$\ds \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x = 0$
== Also see ==
* Fourier Cosine Coefficients for Even Function over Symmetric Range
* Fourier Sine Coefficients for Odd Function over Symmetric Range
* Fourier Cosine Coefficients for Odd Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions
Category:Even Functions
Category:Fourier Series
Fourier Sine Coefficients for Odd Function over Symmetric Range

== Theorem ==
Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.
Let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
== Proof ==
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
:$b_n = \ds \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x$
From Sine Function is Odd:
:$\sin a = -\map \sin {-a}$
for all $a$.
By Odd Function Times Odd Function is Even, $\map f x \sin \dfrac {n \pi x} \lambda$ is even.
Thus:
== Also see ==
* Fourier Cosine Coefficients for Odd Function over Symmetric Range
* Fourier Cosine Coefficients for Even Function over Symmetric Range
* Fourier Sine Coefficients for Even Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions: $(7)$
Category:Odd Functions
Category:Fourier Series
Fourier Transform of 1

== Theorem ==
Let:
:$\map f x = 1$
Then:
:$\map {\hat f} s = \map \delta s$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.
== Proof ==
By the definition of a Fourier transform:
== Sources ==
* 
Category:Examples of Fourier Transforms
Fourier Transform of 1-Lebesgue Space Function is Bounded

== Theorem ==
Let $n \in \N_{>0}$.
Let $\map {L^1} {\R^n}$ be the complex-valued Lebesgue $1$-space with respect to the Lebesgue measure.
Let $f \in \map {L^1} {\R^n}$.
Let $\hat f$ be the Fourier transform of $f$.
Then:
:$\ds \sup_{\mathbf s \mathop \in \R^n} \cmod {\map {\hat f} {\mathbf s} } \le \norm f_{\map {L^1} {\R^n} }$
== Proof ==
For each $\mathbf s \in \R^n$:
Category:Fourier Transforms
Fourier Transform of Derivative of Tempered Distribution

== Theorem ==
Let $T \in \map {\SS'} \R$ be a tempered distribution.
Let $\xi \in \R$ be a real number.
Let the hat denote the Fourier transform.
Then in the distributional sense it holds that:
:$\hat {\paren{T'} } = 2 \pi i \xi \hat T$
== Proof ==
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Then:
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Category:Fourier Transforms
Fourier Transform of Dirac Delta Distribution

== Theorem ==
Let $\delta \in \map {\SS'} \R$ be the Dirac delta distribution.
Let $\mathbf 1 : \map \SS \R \to \R$ be the constant tempered distribution such that for all $\phi \in \map \SS \R$ we have:
:$\ds \map {\mathbf 1} \phi = \int_{-\infty}^\infty 1 \cdot \map \phi x \rd x$
Then in the distributional sense it holds that:
:$\hat \delta = \mathbf 1$
where the hat denotes the Fourier transform of a tempered distribution.
== Theorem ==
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Then:
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Category:Fourier Transforms
Fourier Transform of Exponential Function

== Theorem ==
Let $\map f x$ be defined as the real exponential function where the absolute value of the input is used in the exponent and the exponent is scaled by a factor of $-2 \pi a$:
:$\map f x = e^{-2 \pi a \size x}: \R \to \R$
Then:
:$\map {\hat f} s = \dfrac 1 \pi \dfrac a {a^2 + s^2}$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.
== Proof ==
By the definition of a Fourier transform:
Therefore:
:$\map {\hat f} s = \dfrac 1 \pi \dfrac a {a^2 + s^2 }$
== Sources == 
* 
Category:Examples of Fourier Transforms
Fourier Transform of Normal Function

== Theorem ==
Let $\map f x$ be defined as $\sqrt \pi$ times the normal probability density function where $\mu = 0$ and $\sigma = \dfrac {\sqrt 2} 2$:
Then:
:$\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s }^2}$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.
== Proof ==
By the definition of a Fourier transform:
Taking the derivative with respect to $s$, we have:
Integrating by parts, we have:
Let $u = e^{-2 \pi i x s}$ and $\rd v = x e^{-x^2} \rd x$
Then:
:$\rd u = -2 \pi i s e^{-2 \pi i x s} \rd x$ and $v = -\dfrac 1 2 e^{-x^2}$
Hence:
We now have the following:
We solve for $A$ by setting $s = 0$:
Therefore:
:$\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s}^2}$
== Sources == 
* 
Category:Examples of Fourier Transforms
Fourier Transform of Tempered Distribution on 1-Lebesgue Space equals Tempered Distribution of Fourier Transform of defining Function

== Theorem ==
Let $\map {L^1} \R$ be the Lebesgue $1$-space.
Let $f \in \map {L^1} \R$.
Let $T_f \in \map {\SS'} \R$ be a tempered distribution associated with $f$.
Then:
:$\hat T_f = T_{\hat f}$
where the hat denotes the Fourier transform of the tempered distribution $T_f$ and of the real function $f$ respectively.
== Proof ==
Let $\phi \in \map \SS \R$ be a Schwartz test function.
We have that the Fourier Transform of 1-Lebesgue Space Function is Bounded.
Then, from Lebesgue Infinity-Space is Subset of Tempered Distribution Space it follows that $\ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi$ is a tempered distribution.
That is, by definition:
:$\ds \ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi = T_{\hat f}$
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Category:Fourier Transforms
Fourth Apotome/Example

== Example of Fourth Apotome==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a fourth apotome :
:$(1): \quad a \in \Q$
:$(2): \quad \dfrac {\sqrt {a^2 - b^2}} a \notin \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = 3$ and $b = \sqrt 2$.
Then:
Therefore $3 - \sqrt 2$ is a fourth apotome.
Category:Examples of Apotome