nyuuzyou/wikis huggingface dataset:
Fourier's Theorem/Lemma 1/Mistake 1

== Source Work ==
* :
** Chapter Two: $\S 2$. Some Important Limits
== Mistake ==
:$\ds \int^b \map \psi u \sin N u \rd u = \sum_{r \mathop = 0}^{m - 1} \int_{a_r}^{a_{r + 1} } \map \psi u \sin N u \rd u$
The lower limit is missing from the integral.
It should be:
:$\ds \int_a^b \map \psi u \sin N u \rd u = \sum_{r \mathop = 0}^{m - 1} \int_{a_r}^{a_{r + 1} } \map \psi u \sin N u \rd u$
== Sources ==
* : Chapter Two: $\S 2$. Some Important Limits
Category:Mistakes/Fourier Series/Sneddon
Fourier's Theorem/Lemma 1/Mistake 2

== Source Work ==
* :Chapter Two: $\S 2$. Some Important Limits
== Mistake ==
:... If $M$ is the greatest of the finite numbers $1 M_0 1, 1 M_1 1, \ldots, 1 M_{m - 1} 1$ we have
::$\ds 1 \int_a^b \map \psi u \sin N u \rd u 1 < \dfrac {M m} N$
This appears to be an error of transcription from the manuscript to the typeset copy, where absolute value indicators have been interpreted as instances of the number $1$.
This should read:
:... If $M$ is the greatest of the finite numbers $\size {M_0}, \size {M_1}, \ldots, \size {M_{m - 1} }$ we have
::$\ds \size {\int_a^b \map \psi u \sin N u \rd u} < \dfrac {M m} N$
== Sources ==
* : Chapter Two: $\S 2$. Some Important Limits
Category:Mistakes/Fourier Series/Sneddon
Fourier's Theorem/Lemma 2

== Lemma for Fourier's Theorem ==
Let $\psi$ be a real function defined on a half-open interval $\hointl 0 a$.
Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\hointl 0 a$.
Let $\map \psi u$ have a right-hand derivative at $u = 0$.
Then:
:$\ds \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$
where $\map \psi {0^+}$ denotes the limit of $\psi$ at $0$ from the right.
== Proof ==
We have:
:$\map \psi u = \map \psi {0^+} + \paren {\map \psi u - \map \psi {0^+} }$
from which:
:$(1): \quad \ds \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$
where:
:$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$
Let $\xi = N u$.
Then:
We have that $\map \psi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$.
Hence it follows that $\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ is also piecewise continuous with one-sided limits on $\hointl 0 a$.
We also have that $\map \psi u$ has a right-hand derivative at $u = 0$.
It follows that $\map \phi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$.
Thus from Lemma 1 for Fourier's Theorem:
:$\ds \lim_{N \mathop \to \infty} \int_0^a \map \phi u \sin N u \rd u = 0$
and letting $N \to \infty$ in $(1)$ above:
:$\ds \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$
== Sources ==
* : Chapter Two: $\S 2$. Some Important Limits: Lemma $(2)$
Category:Fourier's Theorem
Fourier's Theorem/Lemma 2/Mistake

== Source Work ==
* : Chapter Two: $\S 2$. Some Important Limits
== Mistake ==
:we find that:
::$\ds \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u + \int_0^a \map \phi u \sin N u \rd u$
:where:
::$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$.
The variable of integration is missing from the middle integral.
It should be:
:$\ds \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$
== Sources ==
* : Chapter Two: $\S 2$. Some Important Limits
Category:Mistakes/Fourier Series/Sneddon
Fourier's Theorem/Lemma 3

== Lemma for Fourier's Theorem ==
Let $\psi$ be a real function defined on an open interval $\openint a b$.
Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\openint a b$.
Let $\map \psi u$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \openint a b$.
Then:
:$\ds \lim_{N \mathop \to \infty} \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$
where:
:$\map \psi {x^+}$ denotes the limit of $\psi$ at $x$ from the right
:$\map \psi {x^-}$ denotes the limit of $\psi$ at $x$ from the left.
== Proof ==
From Sum of Integrals on Adjacent Intervals for Integrable Functions, we have:
:$\ds \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u + \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u$
Let $u = x - \xi$.
Then by Integration by Substitution:
:$\ds \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi$
where:
:$\map \phi \xi = \map \psi {u - \xi}$
By Fourier's Theorem: Lemma 2:
Similarly, substituting $u = x + \eta$:
:$\ds \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta$
where:
:$\map \chi \xi = \map \psi {x + \eta}$
By Fourier's Theorem: Lemma 2:
The result follows by adding the two limits.
== Sources ==
* : Chapter Two: $\S 2$. Some Important Limits: Lemma $(3)$
Category:Fourier's Theorem
Fourier Cosine 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:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then 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 ==
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
:$a_n = \dfrac 1 \lambda \ds \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x$
From Cosine Function is Even:
:$\cos a = \map \cos {-a}$
for all $a$.
By Even Function Times Even Function is Even, $\map f x \cos \dfrac {n \pi x} \lambda$ is even.
Thus:
== Also see ==
* Fourier Sine Coefficients for Even Function over Symmetric Range
* Fourier Cosine Coefficients for Odd Function over Symmetric Range
* Fourier Sine Coefficients for Odd Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions: $(5)$
Category:Even Functions
Category:Fourier Series
Fourier Cosine 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:
:$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
Then for all $n \in \Z_{\ge 0}$:
:$a_n = 0$
== Proof ==
As suggested, let the Fourier series of $\map f x$ be expressed as:
:$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
By definition of Fourier series:
From Cosine Function is Even:
:$\cos a = \map \cos {-a}$
for all $a$.
By Odd Function Times Even Function is Odd, $\map f x \cos \dfrac {n \pi x} \lambda$ is odd.
The result follows from the corollary to Definite Integral of Odd Function:
:$\ds \dfrac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x = 0$
== Also see ==
* Fourier Sine Coefficients for Odd Function over Symmetric Range
* Fourier Sine Coefficients for Even Function over Symmetric Range
* Fourier Cosine Coefficients for Even Function over Symmetric Range
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions
Category:Odd Functions
Category:Fourier Series
Fourier Series/1 over -1 to 0, Cosine of Pi x over 0 to 1

== Theorem ==
Let $\map f x$ be the real function defined on $\openint {-1} 1$ as:
:
:$\map f x = \begin{cases}
1 & : -1 < x < 0 \\
\map \cos {\pi x} & : 0 < x < 1
\end{cases}$
Then its Fourier series can be expressed as:
:$\map f x \sim \displaystyle \dfrac 1 2 + \frac {\cos \pi x} 2 + \sum_{r \mathop = 1}^\infty \paren {\dfrac {4 r \sin 2 r \pi x} {\pi \paren {2 r + 1} \paren {2 r - 1} } - \dfrac {2 \map \sin {2 r + 1} \pi x } {\pi \paren {2 r + 1} } }$
== Proof ==
By definition of Fourier series:
:$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n \pi x + b_n \sin n \pi x}$
where:
for all $n \in \Z_{>0}$.
Thus:
For $n > 0$:
Splitting this up into bits:
For $0 < x < 1$, there are two cases that need to be addressed: when $n = 1$ and when $n > 1$.
First let $n > 1$.
We have:
Now let $n = 1$.
We have:
There is only one non-vanishing term:
:$a_1 = \dfrac 1 2$
Now for the $\sin n \pi x$ terms:
For $-1 < x < 0$:
For $0 < x < 1$, there are two cases that need to be addressed: when $n = 1$ and when $n > 1$.
First let $n > 1$.
Now let $n = 1$.
In summary:
:$\begin {cases} a_0 = 1 \\ a_1 = \dfrac 1 2 \\ b_{2 r} = \dfrac {4 r} {\pi \paren {2 r + 1} \paren {2 r - 1} } \\ b_{2 r + 1} = \dfrac {-2} {\pi \paren {2 r + 1} } \end {cases}$
Finally:
== Sources ==
* : Chapter One: $\S 3$. Other Types of Whole-Range Series: Example $3$
Category:Examples of Fourier Series
Fourier Series/1 over -1 to 0, Cosine of Pi x over 0 to 1/Mistake

== Source Work ==
* : Chapter One: $\S 3$. Other Types of Whole-Range Series: Example $3$
== Mistake ==
:By formulae $(3)$ the Fourier coefficients are:
::$a_0 = \ds \int_{-1}^0 \rd x + \int_0^1 \rd x = 1$
::$a_n = \ds \int_{-1}^0 \map \cos {n \pi x} \rd x + \int_0^1 \map \cos {n \pi x} \map \cos {\pi x} \rd x = 0$
::$b_n = \ds \int_{-1}^0 \map \sin {n \pi x} \rd x + \int_0^1 \map \sin {n \pi x} \map \cos {\pi x} \rd x$
:$= \dfrac {\map \cos {-n \pi} - 1} {n \pi} + \dfrac 1 {2 \pi} \set {\dfrac {1 - \map \cos {n + 1} \pi} {n + 1} + \dfrac {1 - \map \cos {n - 1} \pi} {n - 1} }$
== Correction ==
The author appears to have omitted to note that the term $a_1$ is non-vanishing:
This is the graph of $\map f x$ and its expansion to $b_7$ according to the correct analysis:
:
In comparison, this is the corresponding graph as a result of 's analysis, that is, without the $a_1$ term.
The missing $a_1 \cos \pi x$ component has been presented in red, to illustrate the contribution it makes:
:
== Sources ==
* : Chapter One: $\S 3$. Other Types of Whole-Range Series: Example $3$
Category:Mistakes/Fourier Series/Sneddon
Fourier Series/4 minus x squared over Range of 2

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 2$ as:
:
:$\map f x = 4 - x^2$
Then its Fourier series can be expressed as:
:$\map f x \sim \ds \frac 8 3 - \frac 4 {\pi^2} \sum_{n \mathop = 1}^\infty \frac {\cos n \pi x} {n^2} + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\sin n \pi x} n$
== Proof ==
By definition of Fourier series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n \pi x + b_n \sin n \pi x}$
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 $\sin n \pi x$ terms:
Splitting this up into bits:
Reassembling $b_n$ from the remaining non-vanishing terms:
Finally:
== Sources ==
* : Chapter One: $\S 3$. Other Types of Whole-Range Series: Example $2$
Category:Examples of Fourier Series
Fourier Series/Absolute Value Function over Symmetric Range

== Theorem ==
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint {-\lambda} \lambda \to \R$ be the absolute value function on the open real interval $\openint {-\lambda} \lambda$:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = \size x$
The Fourier series of $f$ over $\openint {-\lambda} \lambda$ can be given as:
== Proof ==
From Absolute Value Function is Even Function, $\map f x$ is an even function.
By Fourier Series for Even Function over Symmetric Range, we have:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$
where:
Hence from Half-Range Fourier Cosine Series for Identity Function:
:$\map f x \sim \dfrac \lambda 2 - \ds \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$
Hence the result.
Category:Absolute Value Function
Category:Fourier Series for Absolute Value Function
Fourier Series/Absolute Value of x over Minus Pi to Pi

== Theorem ==
For $x \in \openint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$
== Proof 1 ==
== Proof 2 ==
Category:Fourier Series for Absolute Value Function
Category:Fourier Series/Absolute Value of x over Minus Pi to Pi
Fourier Series/Absolute Value of x over Minus Pi to Pi/Proof 1

== Theorem ==
== Proof ==
By definition, the absolute value function is an even function:
:$\size {-x} = x = \size x$
Thus by Fourier Series for Even Function over Symmetric Range, $\size x$ can be expressed as:
:$\ds \size x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{\ge 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \size x \cos n x \rd x$
On the real interval $\openint 0 \pi$:
:$\size x = x$
and so for all $n \in \Z_{\ge 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi x \cos n x \rd x$
Thus Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$ can be applied directly.
So for $x \in \openint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$
Category:Fourier Series/Absolute Value of x over Minus Pi to Pi
Fourier Series/Absolute Value of x over Minus Pi to Pi/Proof 2

== Theorem ==
== Proof ==
By Fourier Series for Absolute Value Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = \size x$
has a Fourier series:
:$\map f x \sim \dfrac \lambda 2 - \ds \dfrac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$
Substituting for $\lambda = \pi$ gives:
:$\size x = \dfrac \pi 2 - \ds \dfrac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$
as required. 
Category:Fourier Series/Absolute Value of x over Minus Pi to Pi
Fourier Series/Cosine of x over Minus Pi to Zero, Minus Cosine of x over Zero to Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
:
:$\map f x = \begin {cases}
\cos x & : -\pi < x < 0 \\
-\cos x & : 0 < x < \pi \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
It is apparent by inspection that $\map f x$ is an odd function over $\openint {-\pi} \pi$.
It follows from Fourier Series for Odd Function over Symmetric Range:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$
for all $n \in \Z_{>0}$.
Thus by definition of $f$:
:$\ds b_n = -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$
When $n \ne 1$, we have:
When $n = 1$, we have:
Hence:
When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so:
When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so:
== Sources ==
* : Exercises on Chapter $\text I$: $3$.
* : $\S 23$: Special Fourier Series and their Graphs: $23.13$
Category:Examples of Fourier Series
Fourier Series/Exponential of x over Minus Pi to Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\R$ as:
:
:$\map f x = \begin{cases}
e^x & : -\pi < x \le \pi \\
\map f {x + 2 \pi} & : \text{everywhere} \end{cases}$
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 by definition of $f$:
For $n > 0$:
Now for the $\sin n x$ terms:
Finally:
== Sources ==
* : Exercises on Chapter $\text I$: $4$.
Category:Examples of Fourier Series
Fourier Series/Fourth Power of x over Minus Pi to Pi

== Theorem ==
For $x \in \openint {-\pi} \pi$:
:$\ds x^4 = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi \cos n x$
 
== Proof ==
Since $x^4 = \paren {-x}^4$, $x^4$ is an even function.
By Fourier Series for Even Function over Symmetric Range, the Fourier series of $\map f x$ can be expressed as:
:$x^4 \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
This gives:
:$x^4 \sim \dfrac {\pi^4} 5 + \ds \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi \cos n x$
Category:Examples of Fourier Series
Fourier Series/Identity Function over Minus Pi to Pi

== Theorem ==
For $x \in \openint {-\pi} \pi$:
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
== Proof 1 ==
== Proof 2 ==
Category:Fourier Series for Identity Function
Fourier Series/Identity Function over Minus Pi to Pi/Proof 1

== Theorem ==
== Proof ==
From Odd Power is Odd Function, $x$ is an odd function.
By Fourier Series for Odd Function over Symmetric Range, we have:
:$\ds x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where:
Substituting for $b_n$ in $(1)$: 
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
as required. 
Category:Fourier Series for Identity Function
Fourier Series/Identity Function over Minus Pi to Pi/Proof 2

== Theorem ==
== Proof ==
By Fourier Series for Identity Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = x$
has a Fourier series:
:$\map f x \sim \dfrac {2 \lambda} \pi \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
Substituting for $\lambda = \pi$ gives:
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
as required. 
Category:Fourier Series for Identity Function
Fourier Series/Identity Function over Symmetric Range

== Theorem ==
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint {-\lambda} \lambda \to \R$ be the identity function on the open real interval $\openint {-\lambda} \lambda$:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = x$
The Fourier series of $f$ over $\openint {-\lambda} \lambda$ can be given as:
:$\map f x \sim \dfrac {2 \lambda} \pi \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
== Proof ==
From Identity Function is Odd Function, $\map f x$ is a odd function.
By Fourier Series for Odd Function over Symmetric Range, we have:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where:
Substituting for $b_n$ in $(1)$: 
:$\ds x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
as required. 
Category:Identity Mappings
Category:Fourier Series for Identity Function
Fourier Series/Minus Pi over 0 to Pi, x minus Pi over Pi to 2 Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:
:$\map f x = \begin{cases}
-\pi & : 0 < x \le \pi \\
x - \pi & : \pi < x < 2 \pi \end{cases}$
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 $n > 0$:
Now for the $\sin n x$ terms:
Finally:
== Sources ==
* : Exercises on Chapter $\text I$: $1$.
Category:Examples of Fourier Series
Fourier Series/Pi Squared minus x Squared over Minus Pi to Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
:
:$\map f x = \pi^2 - x^2$
$f$ can be expressed as a half-range Fourier cosine series thus:
== Proof ==
We have that:
:$\pi^2 - \paren {-x}^2 = \pi^2 - x^2$
and so $\map f x$ is even on $\openint {-\pi} \pi$.
It follows from Fourier Series for Even Function over Symmetric Range:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x$
Thus by definition of $f$:
Then for $n > 0$:
Splitting this up into two:
Finally:
== Sources ==
* : Exercises on Chapter $\text I$: $6 \, \text{(i)}$.
Category:Examples of Half-Range Fourier Series
Fourier Series/Pi minus x over 0 to 2 Pi

== Theorem ==
For $x \in \openint 0 {2 \pi}$:
:$\ds \pi - x = 2 \sum_{n \mathop = 1}^\infty \frac {\sin n x} n$
== Proof ==
By definition of Fourier series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
where:
for all $n \in \Z_{>0}$.
Thus:
For $n > 0$:
Now for the $\sin n x$ terms:
Finally:
== Sources ==
* : $\S 23$: Special Fourier Series and their Graphs: $23.10$
Category:Examples of Fourier Series
Fourier Series/Sawtooth Wave

== Theorem ==
Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
x & : x \in \openint {-l} l \\
\map S {x + 2 l} & : x < -l \\
\map S {x - 2 l} & : x > +l \end {cases}$
where $l$ is a given real constant.
Then its Fourier series can be expressed as:
== Proof ==
Let $\map f x: \openint {-l} l \to \R$ denote the identity function on the open interval $\openint {-l} l$:
:$\map f x = x$
From Fourier Series for Identity Function over Symmetric Range, $\map f x$ can immediately be expressed as:
:$\ds \map f x \sim \frac {2 l} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \dfrac {n \pi x} l$
== Special Cases ==
Category:Sawtooth Waves
Category:Fourier Series for Sawtooth Wave
Fourier Series/Sawtooth Wave/Special Cases/Half Interval Pi

== Special Case of Fourier Series for Sawtooth Wave ==
Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
x & : x \in \openint {-\pi} \pi \\
\map S {x + 2 \pi} & : x < -\pi \\
\map S {x - 2 \pi} & : x > +\pi \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
From Fourier Series for Sawtooth Wave, the sawtooth wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
x & : x \in \openint {-l} l \\
\map S {x + 2 l} & : x < -1 \\
\map S {x - 2 l} & : x > +1 \end {cases}$
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: $23.9$
Category:Fourier Series for Sawtooth Wave
Fourier Series/Sawtooth Wave/Special Cases/Unit Half Interval

== Special Case of Fourier Series for Sawtooth Wave ==
Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
x & : x \in \openint {-1} 1 \\
\map S {x + 2} & : x < -1 \\
\map S {x - 2} & : x > +1 \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
From Fourier Series for Sawtooth Wave, the sawtooth wave defined on the real numbers $\R$ as:
:$\forall x \in \R: \map S x = \begin {cases}
x & : x \in \openint {-l} l \\
\map S {x + 2 l} & : x < -l \\
\map S {x - 2 l} & : x > +l \end {cases}$
has a Fourier series which can be expressed as:
The result follows by setting $l = 1$.
Category:Fourier Series for Sawtooth Wave
Fourier Series/Sixth Power of x over Minus Pi to Pi

== Theorem ==
For $x \in \openint {-\pi} \pi$:
:$\ds x^6 = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 + 1440} {n^6} \cos n \pi \cos n x$
 
== Proof ==
Since $x^6 = \paren {-x}^6$, $x^6$ is an even function.
By Fourier Series for Even Function over Symmetric Range, the Fourier series of $\map f x$ can be expressed as:
:$x^6 \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
This gives:
:$x^6 \sim \dfrac {\pi^6} 7 + \ds \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 + 1440} {n^6} \cos n \pi \cos n x$
== Also see ==
* Fourier Series of Fourth Power of x
Category:Examples of Fourier Series