nyuuzyou/wikis huggingface dataset:
Minkowski Functional of Convex Absorbing Set is Sublinear

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let $A$ be a convex absorbing set. 
Let $\mu_A$ be the Minkowski functional of $A$. 
Then for each $x, y \in X$ we have: 
:$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$ 
== Proof ==
Let $\epsilon > 0$. 
By the definition of infimum, we can pick $t > 0$ such that:
:$\map {\mu_A} x \le t \le \map {\mu_A} x + \epsilon$
and $t^{-1} x \in A$.
We can also pick $s > 0$ such that: 
:$\map {\mu_A} y \le s \le \map {\mu_A} y + \epsilon$
and $s^{-1} y \in A$. 
We have: 
:$\ds \frac t {s + t} + \frac s {s + t} = 1$
So, since $A$ is convex, we have: 
:$\ds \frac t {s + t} \paren {\frac x t} + \frac s {s + t} \paren {\frac y s} = \frac {x + y} {s + t} \in A$ 
So, we have: 
:$\map {\mu_A} {x + y} \le s + t \le \map {\mu_A} x + \map {\mu_A} y + 2 \epsilon$
Since this inequality holds for all $\epsilon > 0$, we have: 
:$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$ 
== Sources ==
* : $1.34$: Theorem
Category:Minkowski Functionals
Minkowski Functional of Convex Absorbing Set is Sublinear Functional

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $A$ be a convex absorbing set. 
Let $\mu_A$ be the Minkowski functional of $A$. 
Then $\mu_A$ is a sublinear functional.
== Proof ==
This follows from Minkowski Functional of Convex Absorbing Set is Positive Homogeneous and Minkowski Functional of Convex Absorbing Set is Sublinear.
Category:Minkowski Functionals
Category:Sublinear Functionals
Minkowski Functional of Open Ball with respect to Seminorm is Seminorm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let: 
:$B = \set {x \in X : \map p x < 1}$ 
be the open ball centred at ${\mathbf 0}_X$ with radius $1$ in the pseudometric induced by $p$. 
Let $\mu_B$ the Minkowski functional of $B$. 
Then $p = \mu_B$.
== Proof ==
From Open Ball with respect to Seminorm is Convex, Balanced and Absorbing, $B$ is convex and absorbing.
Hence the definition is valid.
Let $x \in X$.
From :
:$\forall s > \map p x: \map p {\dfrac x s} = \dfrac 1 s \map p x < 1$ 
Then we have: 
:$s \in \set {t > 0 : t^{-1} x \in B}$
so that: 
:$\map {\mu_B} x \le s$ 
Taking the infimum over $s > \map p x$, we have: 
:$\map {\mu_B} x \le \map p x$ 
Let $\map p x = 0$.
Then:
:$\map p x \le \map {\mu_B} x$ 
Let $\map p x \ne 0$. 
Let $0 < t \le \map p x$.
Then: 
:$\map p {\dfrac x t} = \dfrac 1 t \map p x \ge 1$
So $t^{-1} x \notin B$.
So:
:$\forall x \in X: \map {\mu_B} x \ge \map p x$
We therefore obtain: 
:$\forall x \in X: \map {\mu_B} x = \map p x$
== Sources ==
* : $1.34$: Theorem
Category:Seminorms
Category:Minkowski Functionals
Minkowski Functional of Open Convex Set containing Zero Vector in Topological Vector Space recovers Set

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $C \subseteq X$ be an open convex set with ${\mathbf 0}_X \in C$. 
From Convex Subset of Topological Vector Space containing Zero Vector in Interior is Absorbing Set, $C$ is absorbing.
Let $\mu_C$ be the Minkowski functional of $C$. 
Then we have: 
:$C = \set {x \in X : \map {\mu_C} x < 1}$
== Proof ==
From Convex Absorbing Set contained between Sets in terms of Minkowski Functional, we have: 
:$\set {x \in X : \map {\mu_C} x < 1} \subseteq C$
Conversely, suppose that $x \in C$. 
From Multiple of Vector in Topological Vector Space Converges, we have: 
:$\paren {1 + \dfrac 1 n} x \to x$
From the definition of a convergent sequence, we have: 
:$\paren {1 + \dfrac 1 N} x \in C$
for some $N \in \N$. 
Then: 
:$x \in \dfrac 1 {1 + \dfrac 1 N} C$
So, by the definition of the Minkowski functional, we have: 
:$\map {\mu_A} x \le \dfrac 1 {1 + \dfrac 1 N} < 1$
Category:Minkowski Functionals
Minkowski Functional of Open Convex Set in Normed Vector Space is Bounded

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then there exists a real number $c > 0$ such that: 
:$0 \le \map {p_C} x \le c \norm x$
for each $x \in X$.
== Proof ==
From the definition of the Minkowski functional, we have: 
:$\map {p_C} x = \inf \set {t > 0 : t^{-1} x \in C}$
We have:
:$t \ge 0$ for every $t \in \set {t > 0 : t^{-1} x \in C}$
so:
:$\inf \set {t > 0 : t^{-1} x \in C} \ge 0$
from the definition of infimum. 
If $x = 0$, we have: 
:$t^{-1} x \in C$
for each $t > 0$, so:
:$\set {t > 0 : t^{-1} x \in C} = \openint 0 \infty$
so:
:$\inf \set {t > 0 : t^{-1} x \in C} = 0$ 
So the desired inequality holds for $x = 0$ for all real numbers $c > 0$. 
Now take $x \ne 0$.
Since $C$ is open, there exists $\delta > 0$ such that for all $x \in X$ with: 
:$\norm x < \delta$
we have $x \in C$.
In Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined, it is shown that: 
:$\dfrac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in C}$
So from the definition of infimum, we have: 
:$\ds \inf \set {t > 0 : t^{-1} x \in C} \le \frac {2 \norm x} \delta$
That is: 
:$\ds \map {p_C} x \le \frac 2 \delta \norm x$
Note that $\delta > 0$ depended only on $C$, so we have: 
:$\ds 0 \le \map {p_C} x \le \frac 2 \delta \norm x$ 
for all $x \in X$, hence the result.
== Sources ==
* $21.1$: The Minkowski Functional
Category:Minkowski Functionals in Normed Vector Spaces
Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then $p_C$ is a sublinear functional.
== Proof ==
We will show that: 
:$(1): \quad \map {p_C} {\lambda x} = \lambda \map {p_C} x$ for each $x \in X$ and $\lambda \in \R_{\ge 0}$
:$(2): \quad \map {p_C} {x + y} \le \map {p_C} x + \map {p_C} y$ for each $x, y \in X$.
=== Proof of $(1)$ ===
If $\lambda = 0$, then $(1)$ follows immediately since: 
:$\map {p_C} 0 = 0$
as shown in Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined.
Now take $\lambda \ne 0$. 
We then have, for each $x \in X$: 
:$t^{-1} x \in C$ $\paren {\lambda t}^{-1} \paren {\lambda x} \in C$
So:
:$t \in \set {t > 0 : t^{-1} x \in C}$ $\lambda t \in \set {t > 0 : t^{-1} \paren {\lambda x} \in C}$
giving: 
:$\set {t > 0 : t^{-1} \paren {\lambda x} \in C} = \lambda \set {t > 0 : t^{-1} x \in C}$
So, from Multiple of Infimum, we have: 
:$\inf \set {t > 0 : t^{-1} \paren {\lambda x} \in C} = \lambda \inf \set {t > 0 : t^{-1} x \in C}$
Then from the definition of the Minkowski functional, we have: 
:$\map {p_C} {\lambda x} = \lambda \map {p_C} x$
=== Proof of $(2)$ ===
==== Lemma ====
Now let $x, y \in X$. 
We show that: 
:$\map {p_C} {x + y} \le \map {p_C} x + \map {p_C} y$
Let $\epsilon > 0$. 
Pick $\alpha > 0$ such that: 
:$\ds \map {p_C} x < \alpha < \map {p_C} x + \frac \epsilon 2$
and pick $\beta$ such that: 
:$\ds \map {p_C} y < \beta < \map {p_C} y + \frac \epsilon 2$
Then we have: 
:$\alpha^{-1} x \in C$
and:
:$\beta^{-1} y \in C$ 
Note that we have: 
:$\ds \frac \alpha {\alpha + \beta} + \frac \beta {\alpha + \beta} = 1$ 
So, from convexity, we have: 
:$\ds \frac \alpha {\alpha + \beta} \paren {\alpha^{-1} x} + \frac \beta {\alpha + \beta} \paren {\beta^{-1} y} \in C$
That is: 
:$\ds \frac {x + y} {\alpha + \beta} \in C$ 
So:
:$\alpha + \beta \in \set {t > 0 : t^{-1} \paren {x + y} \in C}$
From the definition of infimum, we have: 
:$\map {p_C} {x + y} \le \alpha + \beta < \map {p_C} x + \map {p_C} y + \epsilon$
Since $\epsilon > 0$ was arbitrary, we have: 
:$\map {p_C} {x + y} \le \map {p_C} x + \map {p_C} y$
Since $(1)$ and $(2)$ hold, we have that: 
:$p_C$ is a sublinear functional.
== Sources ==
* $21.1$: The Minkowski Functional
Category:Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional
Category:Minkowski Functionals in Normed Vector Spaces
Category:Sublinear Functionals
Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional/Lemma

== Lemma ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Then iIf:
:$\lambda \in \set {t > 0 : t^{-1} x \in C}$ 
we have: 
:$\hointr \lambda \infty \subseteq \set {t > 0 : t^{-1} x \in C}$
Further: 
:$\openint {\map {p_C} x} \infty \subseteq \set {t > 0 : t^{-1} x \in C}$
where $p_C$ is the Minkowski functional of $C$. 
== Proof ==
Let: 
:$t \in \set {t > 0 : t^{-1} x \in C}$ 
Then:
:$t^{-1} x \in C$ 
Then from convexity, for all $\alpha \ge 1$ we have: 
:$\alpha^{-1} t^{-1} x + \paren {1 - \alpha^{-1} } \times 0 \in C$
that is: 
:$\alpha t \in \set {t > 0 : t^{-1} x \in C}$
so:
:$\hointr t \infty \subseteq \set {t > 0 : t^{-1} x \in C}$
Now let: 
:$\lambda \in \openint {\map {p_C} x} \infty$
Then:
:$\map {p_C} x < \lambda$
From the definition of infimum, there exists: 
:$\alpha \in \set {t > 0 : t^{-1} x \in C}$
such that: 
:$\map {p_C} x < \alpha < \lambda$
Then, from the above computation we have: 
:$\openint \alpha \infty \subseteq \set {t > 0 : t^{-1} x \in C}$
And, since $\alpha < \lambda$, we have: 
:$\lambda \in \openint \alpha \infty$
so:
:$\lambda \in \set {t > 0 : t^{-1} x \in C}$
So, we obtain: 
:$\openint {\map {p_C} x} \infty \subseteq \set {t > 0 : t^{-1} x \in C}$
Category:Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional
Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Then, for each $x \in X$:
:$\set {t > 0 : t^{-1} x \in C} \ne \O$
and so the Minkowski functional of $C$ is well-defined.
== Proof ==
If $x = 0$, then: 
:$t^{-1} x \in C$
for all $t > 0$, so:
:$\set {t > 0 : t^{-1} x \in C} = \openint 0 \infty$
Now take $x \ne 0$. 
Since $C$ is open, there exists $\delta > 0$ such that for all $x \in X$ with: 
:$\norm x < \delta$
we have $x \in C$.
Note that we have, from positive homogeneity: 
:$\ds \norm {\frac \delta {2 \norm x} x} = \frac \delta 2$ 
so:
:$\dfrac \delta {2 \norm x} x \in C$
We therefore have:
:$\dfrac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in C}$
so in particular: 
:$\set {t > 0 : t^{-1} x \in C} \ne \O$
We also have that:
:$t \ge 0$ for all $t \in \set {t > 0 : t^{-1} x \in C}$
So, from the Continuum Property:
:$\set {t > 0 : t^{-1} x \in C}$ has an infimum.
So:
:$\ds \inf \set {t > 0 : t^{-1} x \in C}$ is well-defined.
Since $x \in X$ was arbitrary, we have: 
:the Minkowski functional of $C$ is well-defined
Category:Minkowski Functionals in Normed Vector Spaces
Minkowski Functional of Open Convex Set in Normed Vector Space recovers Set

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then: 
:$C = \set {x \in X : \map {p_C} x < 1}$
== Proof ==
We first show that: 
:$\set {x \in X : \map {p_C} x < 1} \subseteq C$
Suppose that: 
:$x \in \set {x \in X : \map {p_C} x < 1}$
Then: 
:$\inf \set {t > 0 : t^{-1} x \in C} < 1$
from the definition of a Minkowski functional.
So, there exists $0 < \alpha < 1$ such that: 
:$\alpha^{-1} x \in C$ 
Since $C$ is convex and $0 \in C$, we have: 
:$\alpha \paren {\alpha^{-1} x} + \paren {1 - \alpha} \times 0 \in C$
so:
:$x \in C$
So we have: 
:$\set {x \in X : \map {p_C} x < 1} \subseteq C$
Now we show that: 
:$C \subseteq \set {x \in X : \map {p_C} x < 1}$
Let:
:$x \in C$ 
Since $C$ is open, there exists $\epsilon > 0$ such that for all $y \in X$ with: 
:$\norm {y - x} < \epsilon$
we have $y \in C$. 
So, we have: 
:$\ds \paren {1 + \frac \epsilon 2} x \in C$
So:
:$\ds \frac 2 {2 + \epsilon} \in \set {t > 0 : t^{-1} x \in C}$
So from the definition of infimum, we have: 
:$\ds \map {p_C} x \le \frac 2 {2 + \epsilon}$
So:
:$\map {p_C} x < 1$ 
We therefore have: 
:$C \subseteq \set {x \in X : \map {p_C} x < 1}$
So, from the definition of set equality, we have: 
:$C = \set {x \in X : \map {p_C} x < 1}$
== Sources ==
* $21.1$: The Minkowski Functional
Category:Minkowski Functionals in Normed Vector Spaces
Minkowski Functional of Symmetric Convex Absorbing Set in Real Vector Space is Seminorm

== Theorem ==
Let $X$ be a vector space over $\R$.
Let $A \subseteq X$ be a set that is symmetric, convex and absorbing.
Let $\mu_A$ be the Minkowski functional of $A$. 
Then $\mu_A$ is a seminorm. 
== Proof ==
From Minkowski Functional of Convex Absorbing Set is Positive Homogeneous, $\mu_A$ is a sublinear functional.
Hence we have: 
:$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$ for all $x, y \in X$
and hence .
We also have: 
:$\map {\mu_A} {r x} = r \map {\mu_A} x$ for all $r \ge 0$ and $x \in X$.
To establish , it remains to show that: 
:$\map {\mu_A} {r x} = \size r \map {\mu_A} x$ for all $r < 0$ and $x \in X$. 
For this we first show that: 
:$\map {\mu_A} {-x} = \map {\mu_A} x$ for all $x \in X$.
Since $A$ is symmetric, we have $A = -A$. 
Then we have: 
Now for $r < 0$ and $x \in X$ we have:
showing .
Category:Minkowski Functionals
Category:Seminorms
Minor Axis of Ellipsoid is Axis of Symmetry

== Theorem ==
Let $\EE$ be an ellipsoid.
The minor axis of $\EE$ is an axis of symmetry of $\EE$.
== Proof ==
Let us arrange $\EE$ such that the minor axis is aligned with the $x$-axis, say.
By Standard Equation of Ellipsoid, $\EE$ can be expressed by the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} + \dfrac {z^2} {c^2} = 1$
where $2 a$ is the length of the minor axis.
Let $R$ denote the space rotation of $180 \degrees$ about the $x$-axis, that is, the minor axis of $\EE$.
Then:
Hence $R$ is demonstrated to be a symmetry, and the result follows.
== Sources ==
* 
* 
Category:Minor Axis of Ellipsoid
Category:Symmetries (Geometry)
Minor Trigonometric Identities

== Theorem ==
=== Sum of Tangent and Cotangent ===
=== Tangent times Tangent plus Cotangent ===
=== Secant Minus Cosine ===
=== Square of Tangent Minus Square of Sine ===
=== Difference of Fourth Powers of Cosine and Sine ===
=== Cosecant Minus Sine ===
=== Cotangent Minus Tangent ===
=== Sum of Cosecant and Cotangent ===
=== Sum of Squares of Secant and Cosecant ===
=== Difference of Fourth Powers of Secant and Tangent ===
=== Reciprocal of One Plus Sine ===
=== Reciprocal of One Minus Sine ===
=== Sum of Reciprocals of One Plus and Minus Sine ===
=== Difference of Reciprocals of One Plus and Minus Sine ===
=== Reciprocal of One Plus Cosine ===
=== Reciprocal of One Minus Cosine ===
=== Sum of Secant and Tangent ===
=== Cosine over Sum of Secant and Tangent ===
=== Secant Plus One over Secant Squared ===
=== Sine Plus Cosine times Tangent Plus Cotangent ===
=== Tangent over Secant Plus One ===
=== Squares of Linear Combination of Sine and Cosine ===
=== Reciprocal of One Minus Secant ===
=== Reciprocal of One Plus Cosecant ===
Category:Trigonometric Identities
Minor is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line.
Let a straight line $BC$ such that:
:$BC$ is incommensurable in square with $AB$
:$AB^2 + BC^2$ is rational
:the rectangle contained by $AB$ and $BC$ is medial
be cut off from $AB$.
We have that:
:$AB^2 + BC^2$ is rational
while:
:$2 \cdot AB \cdot BC$ is medial.
Therefore $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
From:
:
and:
:
it follows that:
:$AB^2 + BC^2$ is incommensurable with $AC^2$.
But $AB^2 + BC^2$ is rational.
Therefore $AC^2$ is irrational.
Therefore $AC$ is irrational.
Such a straight line is known as a minor.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Mittag-Leffler's Expansion Theorem

== Theorem == 
 
Let $f$ be a meromorphic function that: 
:has only simple poles 
:is continuous, or has a removable singularity, at $0$. 
Let $X$ be the set of poles of $f$. 
For $N \in \N$, let $C_N$ be a disk, centred at the origin, of radius $R_N$ where: 
:$R_N \to \infty$ as $N \to \infty$ 
:$\partial C_N$ contains no poles of $f$ for any $N$. 
:there exists a real number $M > 0$ independent of $N$ such that for all $z \in \partial C_N$, $\cmod {\map f z} < M$ , for all $N \in \N$.
Then:
:$\ds \map f z = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {z - n} + \frac 1 n}$ 
where:
:$\Res f n$ is the residue of $f$ at $n$
:$z$ is not a pole of $f$
:$\ds \map f 0 = \lim_{z \mathop \to 0} \map f z$ if $f$ has a removable singularity at $0$.
 
== Proof == 
Let $\zeta \in \C \setminus X$. 
Then: 
:$\ds \frac {\map f z} {z - \zeta}$ 
has simple poles for $z \in X \cup \set \zeta$. 
Let $X_N$ be the set of poles contained within $C_N$. 
Then: 
Setting $\zeta = 0$, we obtain: 
:$\ds \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$ 
So: 
:$\ds \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \mathop \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
It remains to show that the integral on the vanishes as $N \to \infty$.
We have:
Letting $N \to \infty$ gives: 
:$\ds 0 = \map f \zeta + \sum_{n \mathop \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
Giving:
:$\ds \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$ 
== Sources == 
* : $7.9$: Mittag-Leffler's Expansion Theorem
* : Corollary $10.24$: Theorem of Mittag-Leffler
Category:Mittag-Leffler Expansions
Mittag-Leffler Expansion for Cosecant Function

== Theorem == 
:$\ds \pi \cosec \pi z = \frac 1 z + 2\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 - n^2}$
where:
:$z \in \C$ is not an integer
:$\cosec$ is the cosecant function.
=== Real Domain ===
Let $\alpha \in \R$ be a real number which is specifically not an integer.
Then:
== Proof == 
== Sources ==
* : $\S 37$: Partial Fraction Expansions: $37.2$
Category:Mittag-Leffler Expansion for Cosecant Function
Category:Cosecant Function
Category:Mittag-Leffler Expansions
Mittag-Leffler Expansion for Cosecant Function/Real Domain

== Theorem ==
Let $\alpha \in \R$ be a real number which is specifically not an integer.
Then:
:$\pi \cosec \pi \alpha = \dfrac 1 \alpha + \ds 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}$
== Proof ==
From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$:
:$\ds \cos \alpha x \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\alpha^2 - n^2} }$
Setting $x = 0$:
== Sources ==
* : Exercises on Chapter $\text {II}$: $1$.
Category:Mittag-Leffler Expansion for Cosecant Function
Mittag-Leffler Expansion for Cotangent Function

== Theorem == 
:$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$
where:
:$z \in \C$ is not an integer
:$\cot$ is the cotangent function.
=== Real Domain ===
Let $\alpha \in \R$ be a real number which is specifically not an integer.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Partial Fractions Expansion of Cotangent
== Sources ==
* : $\S 37$: Partial Fraction Expansions: $37.1$
Category:Cotangent Function
Category:Mittag-Leffler Expansions
Category:Mittag-Leffler Expansion for Cotangent Function
Mittag-Leffler Expansion for Cotangent Function/Proof 1

== Theorem ==
== Outline of proof ==
Informally, we can say:
:$\pi \cot \pi z = \map {\dfrac \d {\d z} } {\ln \map \sin {\pi z} }$.
We then use the Euler Formula for Sine Function to write $\map \sin {\pi z}$ as an infinite product and differentiate its logarithm.
Formally, we work with logarithmic derivatives and use Logarithmic Derivative of Infinite Product of Analytic Functions.
== Proof ==
Let $\LL$ denote the logarithmic derivative.
On the open set $\C \setminus \Z$ we have:
Category:Mittag-Leffler Expansion for Cotangent Function
Mittag-Leffler Expansion for Cotangent Function/Proof 2

== Theorem ==
== Proof ==
Let $\map \zeta s$ be the Riemann zeta function.
Let $\ds \map g z = \sum_{n \mathop = 1}^\infty z^n \map \zeta {2 n}$ be the generating function of $\map \zeta {2 n}$
By Power Series Expansion for Cotangent Function, for $\size z < 1$:
By Riemann Zeta Function at Even Integers:
:$\map \zeta {2 n} = \dfrac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} } {\paren {2 n}!}$
Thus:
By Analytic Continuation of Generating Function of Dirichlet Series and Uniqueness of Analytic Continuation: 
:$\ds \dfrac {\pi z \map \cot {\pi z} - 1} {-2} = \sum_{n \mathop = 1}^\infty \dfrac {z^2} {n^2 - z^2}$
for all of $\C$, as this is the overlap of their domains.
Thus:
Category:Mittag-Leffler Expansion for Cotangent Function
Mittag-Leffler Expansion for Cotangent Function/Real Domain

== Theorem ==
Let $\alpha \in \R$ be a real number which is specifically not an integer.
Then:
:$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \pi \cot \pi \alpha$
== Proof ==
From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$:
:$\ds \cos \alpha x \sim \frac {2 \alpha \sin \pi \alpha} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\alpha^2 - n^2} }$
Setting $x = \pi$:
== Sources ==
* : Exercises on Chapter $\text {II}$: $2 \ \text {(ii)}$
Category:Mittag-Leffler Expansion for Cotangent Function
Mittag-Leffler Expansion for Hyperbolic Cosecant Function

== Theorem == 
 
:$\ds \pi \map \csch {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 + n^2}$
where:
:$z \in \C$ is not an integer multiple of $i$
:$\csch$ is the hyperbolic cosecant function. 
 
== Proof ==
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.8$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Hyperbolic Cosecant Function
Mittag-Leffler Expansion for Hyperbolic Cotangent Function

== Theorem == 
:$\ds \pi \map \coth {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 + n^2}$
where: 
:$z \in \C$ is not an integer multiple of $i$
:$\coth$ is the hyperbolic cotangent function. 
== Proof == 
== Sources ==
* : $\S 37$: Partial Fraction Expansions: $37.7$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Hyperbolic Cotangent Function
Mittag-Leffler Expansion for Hyperbolic Secant Function

== Theorem == 
:$\ds \pi \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1}^2 + 4 z^2}$
where: 
:$z \in \C$ is not a half-integer multiple of $i$ 
:$\sech$ is the hyperbolic secant function. 
 
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.9$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Hyperbolic Secant Function
Mittag-Leffler Expansion for Hyperbolic Tangent Function

== Theorem == 
:$\ds \pi \map \tanh {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {4 z^2 + \paren {2 n + 1}^2}$
where:
:$z \in \C$ is not a half-integer multiple of $i$
:$\tanh$ is the hyperbolic tangent function.
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.10$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Hyperbolic Tangent Function
Mittag-Leffler Expansion for Secant Function

== Theorem == 
 
:$\ds \pi \map \sec {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1}^2 - 4 z^2}$
where: 
:$z \in \C$ is not a half-integer
:$\sec$ is the secant function.
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.3$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Secant Function
Mittag-Leffler Expansion for Square of Cosecant Function

== Theorem ==
:$\ds \pi^2 \map {\csc^2} {\pi z} = \frac 1 {z^2} + 2 \sum_{n \mathop = 1}^\infty \frac {z^2 + n^2} {\paren {z^2 - n^2}^2}$
where:
:$z$ is a complex number that is not a integer
:$\csc$ denotes the cosecant function.
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.6$
Category:Mittag-Leffler Expansions
Category:Cosecant Function