nyuuzyou/wikis huggingface dataset:
Differential Equation/Examples/Second Order Second Degree Non-Linear Ordinary

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $2$nd order and $2$nd degree:
:$\paren {1 + \paren {\dfrac {\d y} {\d x} }^2}^{3/2} = 3 \dfrac {\d^2 y} {\d x^2}$
== Proof ==
Squaring both sides gives:
:$\paren {1 + \paren {\dfrac {\d y} {\d x} }^2}^3 = 9 \paren {\dfrac {\d^2 y} {\d x^2} }^2$
It is seen that the term in $\dfrac {\d^2 y} {\d x^2}$ is squared.
Hence the result by definition of degree.
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (3)$
Category:Examples of Differential Equations
Differential Equation/Examples/Second Order Second Degree Non-Linear Partial

== Example of Differential Equation ==
Partial non-linear differential equation of the $2$nd order and $2$nd degree in $2$ independent variables:
:$\dfrac {\partial^2 z} {\partial x^2} \cdot \dfrac {\partial^2 z} {\partial y^2} - \paren {\dfrac {\partial^2 x} {\partial x \partial y} }^2 = 0$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/Third Order Linear Ordinary

== Example of Differential Equation ==
Ordinary linear differential equation of the $3$rd order:
:$2 \dfrac {\d^3 y} {\d x^3} + 3 \dfrac {\d^2 y} {\d x^2} + \dfrac {\d y} {\d x} - 10 y = e^{-3 x} \sin 5 x$
== Sources ==
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (2)$
Category:Examples of Differential Equations
Differential Equation/Examples/Third Order Second Degree Non-Linear Ordinary

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $3$rd order and $2$nd degree:
:$\paren {\dfrac {\d^3 y} {\d x^3} }^2 + \paren {\dfrac {\d^2 y} {\d x^2} }^4 + \dfrac {\d y} {\d x} = x$
== Proof ==
The term in $\dfrac {\d^3 y} {\d x^3}$ is raised to the $3$nd degree and so the equation is non-linear.
The result follows by definition of degree.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.17)$
Category:Examples of Differential Equations
Differential Equation defining Confocal Conics

== Theorem ==
Consider the equation:
:$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$
where $a^2 > b^2$ and $-\lambda < a^2$.
defining the set of confocal conics whose foci are at $\tuple {\pm \sqrt {a^2 - b^2}, 0}$.
The differential equation defining these confocal conics is:
:$x y \paren {\paren {y'}^2 - 1} + \paren {x^2 - y^2 - a^2 + b^2} y' = 0$
== Proof ==
Then we have:
and similarly:
Eliminating $\lambda$:
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.201$ The Differential Equation of a Family of Confocal Conics
Category:Confocal Conics
Differential Equation governing First-Order Reaction

== Theorem ==
Let a substance decompose spontaneously in a first-order reaction.
The differential equation which governs this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
where:
:$x$ determines the quantity of substance at time $t$.
:$k \in \R_{>0}$.
== Proof ==
From the definition of a first-order reaction, the rate of change of the quantity of the substance is proportional to the quantity of the substance present at any time.
As the rate of change is a decrease, this rate will be negative.
Thus the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for some $k \in \R_{>0}$.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate
Category:First-Order Reactions
Differential Equation governing Simple Harmonic Motion

== Theorem ==
Let $B$ be a body undergoing simple harmonic motion in a straight line.
Then its motion can be described using the equation:
:$\dfrac {\d^2 x} {\d x^2} = -\omega^2 x$
== Proof ==
Consider the equation governing simple harmonic motion:
:$(1): \quad x = A \map \cos {\omega t + \phi}$
Differentiating $2$ times $t$:
We have arrived at a ordinary differential equation of order $2$:
:$\dfrac {\d^2 x} {\d t^2} = -\omega^2 x$
== Sources ==
* 
* 
Category:Simple Harmonic Motion
Differential Equation of Family of Linear Combination of Functions is Linear

== Theorem ==
Consider the one-parameter family of curves:
:$(1): \quad y = C \map f x + \map g x$
The differential equation that describes $(1)$ is linear and of first order.
== Proof ==
Differentiating $(1)$ $x$ gives:
:$(2): \quad \dfrac {\d y} {\d x} = C \map {f'} x + \map {g'} x$
Rearranging $(1)$, we have:
:$C = \dfrac {y - \map g x} {\map f x}$
Substituting for $C$ in $(2)$:
which leaves:
:$\dfrac {\d y} {\d x} - \dfrac {\map {f'} x} {\map f x} y = \map g x \paren {1 - \dfrac {\map {f'} x} {\map f x} }$
which is linear and of first order.
== Sources ==
* : $\S 2.10$: Problem $6$
Category:Linear First Order ODEs
Differential Equation of Perpendicular Pursuit Curve

== Therorem ==
Let a $R$ be situated at the origin of a cartesian plane.
Let a $D$ be situated at the point $\tuple {c, 0}$.
Let $R$ start running up the $y$-axis with speed $a$.
At the same instant, let $D$ start pursuing $R$ by running directly towards it at speed $b$.
Then the differential equation describing the path taken by $D$ is:
:$\dfrac {\d y} {\d x} = \dfrac 1 2 \paren {\paren {\dfrac x c}^k - \paren {\dfrac c x}^k}$
where $k = \dfrac a b$.
== Proof ==
:
Let $C$ be the curve traced out by $D$.
Let the time $t$ be measured from the instant $R$ and $D$ start running.
At $t$, $R$ will be at the point $\tuple {0, a t}$.
Let $D$ be at $\tuple {x, y}$ at $t$.
By the nature of the motion of $D$, the line $DR$ is tangent to $C$.
Hence:
:$\dfrac {\d y} {\d x} = \dfrac {y - a t} x$
or:
:$(1): \quad x y' - y = - a t$
Differentiating $(1)$ $x$:
:$(2): \quad x y' ' = -a \dfrac {\d t} {\d x}$
Let $s$ be the length of $C$ between $\tuple {c, 0}$ and $\tuple {x, y}$ at time $t$.
Then:
:$\dfrac {\d s} {\d t} = b$
so:
:$(3): \quad \dfrac {\d t} {\d x} = \dfrac {\d t} {\d s} \dfrac {\d s} {\d x} = -\dfrac 1 b \sqrt {1 + \paren {y'}^2}$
The minus sign is there because $s$ increases as $x$ decreases.
When $(2)$ and $(3)$ are combined, the differential equation describing $C$ is seen to be:
:$(4): \quad x y' ' = k \sqrt {1 + \paren {y'}^2}$
where $k = \dfrac a b$.
Let $p = y'$.
Then $(4)$ becomes:
When $x = c$ we have $p = 0$, so:
Hence:
Solving for $p$ reveals:
:$p = \dfrac {\d y} {\d x} = \dfrac 1 2 \paren {\paren {\dfrac x c}^k - \paren {\dfrac c x}^k}$
== Sources ==
* : $\S 2.12$: The Hanging Chain. Pursuit Curves: Example $(3)$
Category:Pursuit Curves
Differential Equation satisfied by Weierstrass's Elliptic Function

== Theorem == 
The differential equation: 
:$\paren {\dfrac {\d f} {\d z} }^2 = 4 f^3 - g_2 f - g_3$
where: 
:$\ds g_2 = 60 \sum_{\tuple {n, m} \mathop \in \Z^2 \setminus \tuple {0, 0} } \frac 1 {\paren {2 m \omega_1 + 2 n \omega_2}^4}$
and: 
:$\ds g_3 = 140 \sum_{\tuple {n, m} \mathop \in \Z^2 \setminus \tuple {0, 0} } \frac 1 {\paren {2 m \omega_1 + 2 n \omega_2}^6}$
has the general solution: 
:$\map f z = \map \wp {z + C; \omega_1, \omega_2}$
where:
:$\wp$ is Weierstrass's elliptic function
:$C$ is an arbitrary constant
:$\omega_1$, $\omega_2$ are constants independent of $z$.
 
== Proof == 
 
== Sources == 
* : $20.22$: The differential equation satisfied by $\map \wp z$
Category:Differential Equations
Category:Weierstrass's Elliptic Function
Differential Equations for Shortest Path on 3d Sphere/Cartesian Coordinates

== Theorem ==
Let $M$ be a $3$-dimensional Euclidean space.
Let $S$ be a sphere embedded in $M$.
Let $\gamma$ be a curve on $S$.
Let the chosen coordinate system be Cartesian.
Let $\gamma$ begin at $\paren {x_0, y_0, z_0}$ and terminate at $\paren {x_1, y_1, z_1}$.
Let $\map y x$, $\map z x$ be real functions.
Let $\gamma$ connecting both endpoints be of minimum length.
Then $\gamma$ satisfies the following equations of motion:
:$2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} } = 0$
:$2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt {1 + y'^2 + z'^2} } = 0$
== Proof ==
In $3$-dimensional Euclidean space the length of the curve is:
:$\ds \int_{x_0}^{x_1} \sqrt {1 + y'^2 + z'^2} \rd x$
The sphere satisfies the following equation:
Consider its partial derivatives $y$ and $z$.
:$\dfrac {\partial g} {\partial y} = 2y$
:$\dfrac {\partial g} {\partial z} = 2z$
$g_y$ and $g_z$ vanish for $y = 0$ and $z = 0$ respectively.
Substitution of this into the sphere equation tells us that $x^2 = a^2$.
Therefore, the following analysis should exclude points with $x = \pm a$.
By Simplest Variational Problem with Subsidiary Conditions for Curve on Surface, the length functional is replaced by the following auxiliary functional:
:$\ds \int_{x_0}^{x_1} \sqbrk {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } \rd x$
It follows that:
:$\map {\dfrac {\partial} {\partial y'} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} }$
:$\map {\dfrac {\partial} {\partial y} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = 2 y \lambda$
Analogous relations hold for $\map z x$. 
Then by Euler's Equations the following equations of motion hold:
:$2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$
:$2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$
== Sources ==
* : $\S 2.12$: Variational Problems with Subsidiary Conditions
Category:Calculus of Variations
Differential Form of Arc Length on Curved Surface

== Theorem ==
The first fundamental form for the element of arc length on a curved surface is given by:
:$\d s^2 = E \rd u^2 + 2 F \d u \rd v + G \rd v^2$
where $E$, $F$ and $G$ are the coefficients of the first fundamental form.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.25$: Gauss ()
* : Chapter $\text {A}.32$: Riemann ()
Category:Differential Geometry
Differential of Differentiable Functional is Unique

== Theorem ==
The differential of a differentiable functional is unique.
== Proof ==
=== Lemma ===
Let $J \sqbrk y$ be a differentiable functional.
Suppose the differential of $J \sqbrk y$ is not uniquely defined. 
Then at least $2$ different forms of this exist:
:$\Delta J \sqbrk {y; h} = \phi_1 \sqbrk {y; h} + \epsilon_1 \size h$
:$\Delta J \sqbrk {y; h} = \phi_2 \sqbrk {y; h} + \epsilon_2 \size h$
where:
:$\phi_1 \sqbrk {y; h}$ and $\phi_2 \sqbrk {y; h}$ are linear functionals
:$\epsilon_1, \epsilon_2 \to 0$ as $\size h \to 0$.
Because $\size h \to 0$, the whole expression $\to 0$ faster than $\size h$ due to constraints on $\epsilon$.
Hence $\epsilon_1 \size h$, $\epsilon_2 \size h$ and $\paren {\epsilon_1 - \epsilon_2} \size h$ are all infinitesimals of order higher than $1$ relative to $\size h$.
Consequently, subtraction of one from the other leads to:
:$(1): \quad \phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = \paren {\epsilon_2 - \epsilon_1} \size h$
Therefore, $\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h}$ is an infinitesimal of order higher than $1$ relative to $\size h$.
Since each of the members are linear functionals, the whole term keeps the same property. 
Rearranging the terms of $(1)$:
:$\dfrac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = \epsilon_2 - \epsilon_1$
Taking the limit and recalling the constraint on both $\epsilon_1$ and $\epsilon_2$:
:$\ds \lim_{\size h \mathop \to 0} \frac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = 0$
The given limit with the linearity of the term in the numerator allows us to use the lemma. 
Hence:
:$\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = 0$
That is, there is only one form the differential of a differentiable functional can take.
== Sources ==
* : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Category:Calculus of Variations
Differential of Differentiable Functional is Unique/Lemma

== Lemma ==
Let $\phi \sqbrk {y; h}$ be a linear functional $h$.
Let:
:$\ds \lim_{\size h \mathop \to 0} \frac {\phi \sqbrk {y; h} } {\size h} = 0$
Then:
:$\phi \sqbrk {y; h} = 0$
== Proof ==
This will be a Proof by Contradiction.
 there exists a linear functional satisfying $\phi \sqbrk {y; h_0} \ne 0$ for some $h_0 \ne 0$.
Also suppose:
:$\ds \lim_{\size {h_0} \mathop \to 0} \frac{\phi \sqbrk {y; h_0} } {\size {h_0} } = 0$
Now, define:
:$h_n = \dfrac {h_0} n$
and:
:$m = \dfrac {\phi \sqbrk {y; h_0} } {\size {h_0} }$
Notice that $\size {h_n} \to 0$ as $n \to \infty$. 
Hence, from the assumption of the limit it should hold that:
:$\ds \lim_{n \mathop \to \infty} \frac {\phi \sqbrk {y; h_n} } {\size {h_n} } = \lim_{\size {h_n} \mathop \to 0} \frac {\sqbrk {y; h_n} } {\size {h_n} } = 0$
However, using the linearity of $\phi \sqbrk {y; h_0}$ $h_0$:
However, :
:$m \ne 0$
Hence, the contradiction is achieved and the initial statement of the lemma holds.
== Sources ==
* : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Category:Calculus of Variations
Differential of Real-Valued Function/Examples/Function of 2 Variables

== Example of Differential of Real-Valued Function ==
Let $f: \R^2 \to \R$ be a real-valued function of $2$ variables.
Let $z = \map f {x, y}$ for all $\tuple {x, y} \in \R^2$.
Then the differential of $z$ is given by:
:$\d z := \dfrac {\partial f} {\partial x} \d x + \dfrac {\partial f} {\partial y} \d y$
== Sources ==
* 
* 
Category:Examples of Differentials
Differentiation of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x wrt x as Nonsingular Matrix

== Theorem ==
Let $a, b, x \in \R$ be real numbers.
Suppose $a \ne 0 \ne b$.
Let $f_1$ and $f_2$ be the real functions defined as:
Let $\map \CC \R$ be the space of continuous real-valued functions.
Let $\struct {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.
Let $S = \span \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.
Let $D : S \to S$ be the derivative $x$.
Then, basis $\tuple {f_1, f_2}$, $D$ is expressible as:
:$\mathbf D = \begin {bmatrix} a & b \\ -b & a \end {bmatrix}$
and is nonsingular.
== Proof ==
Thus:
Furthermore:
:$\map \det {\mathbf D} = a^2 + b^2$.
By Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $\mathbf D$ is nonsingular.
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Derivatives involving Exponential Function
Category:Derivatives involving Sine Function
Category:Derivatives involving Cosine Function
Category:Inverse Matrices
Differentiation of Polynomials induces Well-Founded Relation

== Theorem ==
Let $P$ be the set of all polynomials over $\R$ in one variable with real coefficients.
Let $\DD$ be a relation on $P$ defined as:
:$\forall p_0, p_1 \in P: \tuple {p_0, p_1} \in \DD$ $p_0$ is the derivative of $p_1$.
Then $\DD$ is a well-founded relation on $P$.
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $7$
Category:Examples of Well-Founded Relations
Differentiation of Real Power Series

== Theorem ==
Let $\xi \in \R$ be a real number.
Let $\sequence {a_n}$ be a sequence in $\R$.
Let $\ds \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ be the power series in $x$ about the point $\xi$.
Then within the interval of convergence:
:$\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \paren {x - \xi}^{m - n}$
where $m^{\underline n}$ denotes the falling factorial.
=== Corollary ===
== Proof ==
First we can make the substitution $z = x - \xi$ and convert the expression into:
:$\ds \dfrac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m z^m$
We then use $n$th Derivative of $m$th Power:
:$\dfrac {\d^n} {\d z^n} z^m = \begin{cases} m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$
$x$ is within the interval of convergence.
It follows from Abel's Theorem that:
:$\ds \frac {\d^n} {\d z^n} \sum_{m \mathop \ge 0} a_m z^m = \sum_{m \mathop \ge n} a_m m^{\underline n} z^{m - n}$
Then from Derivative of Identity Function and others, we have:
:$\map {\dfrac \d {\d x} } {x - \xi} = 1$
The result follows from the Chain Rule for Derivatives.
Category:Differentiation of Real Power Series
Category:Differential Calculus
Category:Real Power Series
Differentiation of Real Power Series/Corollary

== Corollary to Differentiation of Real Power Series ==
Let $\xi \in \R$ be a real number.
Let $\sequence {a_n}$ be a sequence in $\R$.
Let $\ds \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ be the power series in $x$ about the point $\xi$.
The value of $\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ at $x = \xi$ is:
:$\ds \valueat {\frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m} {x \mathop = \xi} = a_n n!$
== Proof ==
From Differentiation of Real Power Series:
:$\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \paren {x - \xi}^{m - n}$
where $m^{\underline n}$ denotes the falling factorial.
When $x = \xi$, all the terms in $\paren {x - \xi}^{m - n}$ vanish except when $m = n$.
When $m = n$, from Nth Derivative of Nth Power we have:
:$\ds \frac {\d^n} {\d x^n} a_m \paren {x - \xi}^m = a_m n!$
But then $m = n$ and the result follows.
Category:Differentiation of Real Power Series
Category:Differential Calculus
Category:Real Power Series
Differentiation of Vector-Valued Function Componentwise

== Theorem ==
Let:
:$\mathbf r: t \mapsto \tuple {\map {r_1} t, \map {r_2} t, \ldots, \map {r_n} t}$ 
be a differentiable vector-valued function.
The derivative of a vector-valued function can be calculated by differentiating each of its component functions:
:$\dfrac {\d \map {\mathbf r} t} {\d t} = \tuple {\dfrac \d {\d t} \map {r_1} t, \dfrac \d {\d t} \map {r_2} t, \ldots, \dfrac \d {\d t} \map {r_n} t}$
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $1$. Scalar Differentiation: $(3.1)$
* : Chapter $\text I$: Vector Algebra: $\S 7$
Category:Differential Calculus
Category:Vector Calculus
Differentiation on Polynomials is Linear Operator

== Theorem ==
Let $\map P \R$ be the vector space of all polynomial functions on the real number line $\R$.
Then the differentiation operator $D$ on $\map P \R$ is a linear operator.
== Proof ==
Let $\map f x, \map g x$ be real functions which are differentiable on $\R$.
Then from Linear Combination of Derivatives:
:$\forall x \in I: \map D {\lambda \map f x + \mu \map g x} = \lambda D \map f x + \mu D \map g x$
It follows from Real Polynomial Function is Differentiable that $\lambda D \map f x + \mu D \map g x$ is differentiable on $\R$.
The result follows from Condition for Linear Transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.6$
Category:Linear Operators
Category:Differential Calculus
Category:Real Polynomial Functions
Diffuse Measure of Countable Set

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Suppose that for all $x \in X$, the singleton $\set x$ is in $\Sigma$.
Suppose further that $\mu$ is a diffuse measure.
Let $E \in \Sigma$ be a countable measurable set.
Then $\map \mu E = 0$.
== Proof ==
It holds trivially that:
:$\ds E = \bigcup_{e \mathop \in E} \set e$
and in particular, this union is countable.
Also, $\map \mu {\set e} = 0$ for all $e \in E$ as $\mu$ is diffuse.
Hence Null Sets Closed under Countable Union applies to yield:
:$\map \mu E = 0$
== Sources ==
* : $\S 6$: Problem $5 \ \text{(iii)}$
Category:Diffuse Measures
Digamma Additive Formula

== Theorem == 
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Let $z \in \C \cap z \notin \set {-\dfrac m n: m \in \N}$
Then: 
:$\ds \map \psi {n z} = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \map \psi {z + \frac k n} + \ln n$
 
where:
:$\psi$ is the digamma function
:$\ln$ is the complex natural logarithm.
=== Corollary ===
== Proof ==
== Also see ==
* Gauss Multiplication Formula
* Product Formula for Sine
* Cotangent Additive Formula
* General Harmonic Number Additive Formula
== Sources ==
* : Chapter $8$. Analogues of the Gamma Function
* 
Category:Digamma Additive Formula
Category:Digamma Function
Digamma Additive Formula/Corollary

== Corollary to Digamma Additive Formula == 
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Then: 
:$\ds \sum_{k \mathop = 1}^{n - 1} \map \psi {\frac k n} = -\paren {n - 1} \gamma - n \ln n$
where:
:$\psi$ is the digamma function
:$\ln$ is the complex natural logarithm. 
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
== Sources ==
* : Chapter $8$. Analogues of the Gamma Function
Category:Digamma Additive Formula
Category:Euler-Mascheroni Constant
Digamma Function/Examples

== Examples of Digamma Function ==
=== Example: $\map \psi {\dfrac 1 6}$ ===
=== Example: $\map \psi {\dfrac 1 4}$ ===
=== Example: $\map \psi {\dfrac 1 3}$ ===
=== Example: $\map \psi {\dfrac 1 2}$ ===
=== Example: $\map \psi {\dfrac 2 3}$ ===
=== Example: $\map \psi {\dfrac 3 4}$ ===
=== Example: $\map \psi {\dfrac 5 6}$ ===
=== Example: $\map \psi 1$ ===
=== Example: $\map \psi {\dfrac 3 2}$ ===
=== Example: $\map \psi {\dfrac 4 3}$ ===
=== Example: $\map \psi {\dfrac 5 4}$ ===
=== Example: $\map \psi {\dfrac 7 6}$ ===
=== Example: $\map \psi 2$ ===
Category:Examples of Digamma Function
Digamma Function/Examples/Digamma Function of Seven Sixths

== Example of Use of Recurrence Relation for Digamma Function ==
:$\map \psi {\dfrac 7 6} = -\gamma - 2 \ln 2 - \dfrac 3 2 \ln 3 - \dfrac {\pi \sqrt 3} 2 + 6$
where $\psi$ denotes the digamma function.
== Proof ==
Category:Examples of Digamma Function
Digamma Function in terms of General Harmonic Number

== Corollary to Reciprocal times Derivative of Gamma Function ==
Let $z \in \C \setminus \Z_{\le 0}$.
Then:
:$\ds \map \psi {z + 1} = -\gamma + \harm 1 z$
where:
:$\psi$ is the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant
:$\harm 1 z$ denotes the general harmonic number of order $1$ evaluated at $z$.
== Proof ==
Category:Reciprocal times Derivative of Gamma Function
Category:Digamma Function
Category:General Harmonic Numbers
Category:Euler-Mascheroni Constant
Digamma Function of Five Sixths

== Theorem ==
:$\map \psi {\dfrac 5 6} = -\gamma - 2 \ln 2 - \dfrac 3 2 \ln 3 + \dfrac {\pi \sqrt 3} 2$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
Digamma Function of One Fourth

== Theorem ==
:$\map \psi {\dfrac 1 4} = -\gamma - 3 \ln 2 - \dfrac \pi 2$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
Category:Digamma Function of One Fourth
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
Digamma Function of One Half

== Theorem ==
:$\map \psi {\dfrac 1 2} = -\gamma - 2 \ln 2$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
Category:Digamma Function of One Half
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
Digamma Function of One Sixth

== Theorem ==
:$\map \psi {\dfrac 1 6} = -\gamma - 2 \ln 2 - \dfrac 3 2 \ln 3 - \dfrac {\pi \sqrt 3} 2$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
Category:Digamma Function of One Sixth
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
Digamma Function of One Third

== Theorem ==
:$\map \psi {\dfrac 1 3} = -\gamma - \dfrac 3 2 \ln 3 - \dfrac \pi {2 \sqrt 3}$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
Category:Digamma Function of One Third
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
Digamma Function of Three Fourths

== Theorem ==
:$\map \psi {\dfrac 3 4} = -\gamma - 3 \ln 2 + \dfrac \pi 2$
where:
:$\psi$ denotes the digamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant