nyuuzyou/wikis huggingface dataset:
Logarithm of One plus x in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\map \ln {1 + x} = x \map F {1, 1; 2; -x}$
where:
:$x$ is a real number with $\size x < 1$
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Logarithm of One plus x over One minus x in terms of Gaussian Hypergeometric Function
== Also presented as ==
Some sources give this as:
:$\map F {1, 1; 2; -x} = \dfrac {\map \ln {1 + x} } x$
== Sources ==
* : $\S 31$: Hypergeometric Functions: Special Cases: $31.4$
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Category:Logarithms
Logarithm of One plus x over One minus x in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\map \ln {\dfrac {1 + x} {1 - x} } = 2 x \map F {\dfrac 1 2, 1; \dfrac 3 2; x^2}$
where:
:$x$ is a real number with $\size x < 1$
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Logarithm of One plus x in terms of Gaussian Hypergeometric Function
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Category:Logarithms
Logarithm of Power/General Logarithm

== Theorem ==
Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\log_a x$ be the logarithm to the base $a$ of $x$.
Then:
:$\map {\log_a} {x^r} = r \log_a x$
== Proof ==
Let $y = r \log_a x$.
Then:
The result follows by taking logs base $a$ of both sides.
== Sources ==
* : $\S 7$: Laws of Logarithms: $7.12$
* : Chapter $1$: Computations Using Logarithms
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(12)$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $14$
* 
Category:Logarithm of Power
Logarithm of Power/Natural Logarithm

== Theorem ==
Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^r} = r \ln x$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* 
Category:Logarithm of Power
Logarithm of Power/Natural Logarithm/Integer Power

== Theorem ==
Let $x \in \R$ be a strictly positive real number.
Let $n \in \R$ be any integer.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^n} = n \ln x$
== Proof ==
From Logarithm of Power/Natural Logarithm/Natural Power, the theorem is already proven for positive integers.
Let $j \in \Z_{<0}$.
Let $-j = k \in Z_{>0}$.
Then:
Category:Logarithm of Power
Logarithm of Power/Natural Logarithm/Natural Power

== Theorem ==
Let $x \in \R$ be a strictly positive real number.
Let $n \in \Z_{\ge 0}$ be any natural number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^n} = n \ln x$
== Proof ==
Proof by Mathematical Induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\forall x \in \R_{>0}: \map \ln 1 = 0$
from Logarithm of 1 is 0.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall x \in \R_{>0}: \map \ln {x^k} = k \ln x$
Then we need to show:
:$\forall x \in \R_{>0}: \map \ln {x^{k + 1} } = \paren {k + 1} \ln x$
=== Induction Step ===
This is our induction step:
Fix $x \in \R_{>0}$.
Then:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$
Category:Logarithm of Power
Logarithm of Power/Natural Logarithm/Proof 1

== Theorem ==
== Proof ==
Consider the function $\map f x = \map \ln {x^r} - r \ln x$.
Then from:
:The definition of the natural logarithm
:The Fundamental Theorem of Calculus
:The Power Rule for Derivatives
:The Chain Rule for Derivatives:
:$\forall x > 0: \map {f'} x = \dfrac 1 {x^r} r x^{r-1} - \dfrac r x = 0$
Thus from Zero Derivative implies Constant Function, $f$ is constant:
:$\forall x > 0: \map \ln {x^r} - r \ln x = c$
To determine the value of $c$, put $x = 1$.
From Logarithm of 1 is 0:
:$\ln 1 = 0$
Thus:
:$c = \ln 1 - r \ln 1 = 0$
== Sources ==
* : $\S 14.2 \ \text{(ii)}$
* 
Category:Logarithm of Power
Logarithm of Power/Natural Logarithm/Rational Power

== Theorem ==
Let $x \in \R$ be a strictly positive real number.
Let $r \in \R$ be any rational number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^r} = r \ln x$
== Proof ==
Let $r = \dfrac s t$, where $s \in \Z$ and $t \in \Z_{>0}$.
First:
Thus:
Category:Logarithm of Power
Logarithm of Reciprocal

== Theorem ==
Let $x, b \in \R$ be strictly positive real numbers such that $b > 1$.
Then:
:$\map {\log_b} {\dfrac 1 x} = -\log_b x$
where $\log_b$ denotes the logarithm to base $b$.
== Proof ==
Hence the result.
Category:Logarithms
Category:Reciprocals
Logarithm on Positive Real Numbers is Group Isomorphism

== Theorem ==
Let $\struct {\R_{>0}, \times}$ be the multiplicative group of positive real numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $b$ be any real number such that $b > 1$.
Let $\log_b: \struct {\R_{>0}, \times} \to \struct {\R, +}$ be the mapping:
:$x \mapsto \map {\log_b} x$
where $\log_b$ is the logarithm to base $b$.
Then $\log_b$ is a group isomorphism.
== Proof ==
From Sum of Logarithms we have:
:$\forall x, y \in \R_{>0}: \map {\log_b} {x y} = \map {\log_b} x + \map {\log_b} y$
That is $\log_b$ is a group homomorphism.
From Change of Base of Logarithm, $\log_b$ is a constant multiplied by the natural logarithm function.
Then we have that Logarithm is Strictly Increasing.
From Strictly Monotone Real Function is Bijective, it follows that $\log_b$ is a bijection.
So $\log_b$ is a bijective group homomorphism, and so a group isomorphism.
== Sources ==
* : $\S 47$. Homomorphisms and their elementary properties: Illustration
Category:Examples of Group Isomorphisms
Category:Logarithms
Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function

== Theorem == 
Let $z \in \C$ with a strictly positive real part and $\size {\arg z} < \dfrac \pi 2$.
Then: 
:$\ds \map \Ln {\map \Gamma z} = \paren {z - \dfrac 1 2} \map \Ln z - z + \dfrac {\ln 2 \pi} 2 + \sum_{n \mathop = 1}^d \frac {B_{2 n} } {2 n \paren {2 n - 1} z^{2 n - 1} } + \OO \paren {\dfrac 1 {z^{2 d + 1} } }$
where: 
:$\Gamma$ is the gamma function
:$\Ln$ is the principal branch of the complex logarithm
:$B_{2 n}$ is the $2 n$th Bernoulli number
:$\OO$ is big-$\OO$ notation.
== Proof ==
From Binet's Formula for Logarithm of Gamma Function Formulation 2, we have:
:$\ds \Ln \map \Gamma z = \paren {z - \dfrac 1 2} \Ln z - z + \dfrac 1 2 \ln 2 \pi + 2 \int_0^\infty \dfrac {\map \arctan {t / z} } {e^{2 \pi t} - 1} \rd t$
:Let $A = \paren {z - \dfrac 1 2} \Ln z - z + \dfrac 1 2 \ln 2 \pi$
Then:
Turning our attention now to the remainder term:
:$\ds \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2} } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t$
Let $K_z = \sup {\set {\size {\dfrac {z^2} {u^2 + z^2} } : u \in \R} }$.
Then: 
Therefore:
We now observe that if $K_z$ is bounded above, then:
:$\ds \lim_{\size z \mathop \to +\infty} \dfrac {K_z B_{2 \paren {d + 1} } } {2 \paren {d + 1} \paren {2 d + 1} \size{z}^{2 d + 1} } = 0$
Next, we observe that:
=== Lemma 1 ===
We note here that if $\size {\arg z} \le \dfrac \pi 4 $, then $\ds \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2} } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t$ is less than the $\paren {n + 1}$th term.
That is:
:$\ds \dfrac {B_{2 \paren {d + 1} } } {2 \paren {d + 1} \paren {2 d + 1} \size{z}^{2 d + 1} }$
Finally, we observe that:
=== Lemma 2 ===
From the definition of big-$\OO$, we have:
:$\map f z = \map \OO {\map g z}$ as $\cmod z \to \infty$
implies:
:$\cmod {\map f z} \le c \cdot \cmod {\map g z}$
for all $z$ in a neighborhood of infinity in $\C$.
We have here:
:$\ds \size {\dfrac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2} } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t } \le \frac {K_z B_{2 \paren {d + 1} } } {2 \paren {d + 1} \paren {2 d + 1} } \size {\dfrac 1 {z^{2 d + 1} } }$
Therefore:
:$\ds \dfrac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2} } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t = \map \OO {\dfrac 1 {z^{2 d + 1} } }$ as $\cmod z \to \infty$
and we finally arrive at:
== Also see == 
* Stirling's Formula for Gamma Function
== Sources == 
* : $12$: The Transcendental Functions
* : Chapter $7$. Sums of Powers, Bernoulli Numbers and the Gamma Function
* : Chapter $1$. The Gamma and Beta Functions
Category:Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function
Category:Stirling's Formula
Category:Gamma Function
Category:Asymptotic Expansions
Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function/Lemma 1

== Theorem ==
Let $z \in \C$ with a strictly positive real part.
Let $K_z = \sup {\set {\size {\dfrac {z^2} {u^2 + z^2} } : u \in \R} }$.
Then:
:If $\size {\arg z} \le \dfrac \pi 4 $, then $K_z = 1$
== Proof ==
Category:Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function
Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function/Lemma 2

== Theorem ==
Let $z \in \C$ with a strictly positive real part.
Let $K_z = \sup {\set {\size {\dfrac {z^2} {u^2 + z^2} } : u \in \R} }$.
Then:
:If $\dfrac \pi 4 < \size {\arg z} < \dfrac \pi 2 $, then $K_z = \map \csc {2 \size {\arg z} }$
== Proof ==
Category:Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function
Logarithmic Derivative of Infinite Product of Analytic Functions

== Theorem ==
Let $D \subseteq \C$ be open.
Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$.
Let none of the $f_n$ be identically zero on any open subset of $D$.
Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.
Then:
:$\ds \dfrac {f'} f = \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$
and the series converges locally uniformly in $D \setminus \set {z \in D : \map f z = 0}$.
== Outline of Proof ==
Using Corollary to Logarithm of Infinite Product of Complex Functions we can write $\log f$ locally as $\ds \sum_{n \mathop = 1}^\infty \log f_n$ up to a constant and some terms.
By Derivative of Uniform Limit of Analytic Functions we can differentiate term-wise.
== Proof ==
Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic.
Let $z_0 \in D$ with $\map f {z_0} \ne 0$.
By the Corollary to Logarithm of Infinite Product of Complex Functions, there exist $n_0 \in \N$, $k \in \Z$ and a open neighborhood $U$ of $z_0$ such that:
:$\map {f_n} z \ne 0$ for $n \ge n_0$ and $z \in U$
:The series $\ds \sum_{n \mathop = n_0}^\infty \log f_n$ converges uniformly on $U$ to $\log g + 2 k \pi i$, where $g = \ds \prod_{n \mathop = n_0}^\infty f_n$.
We have, for $z \in U$:
and
and the series converges locally uniformly on $U$.
Thus $\dfrac {f'} f = \ds \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$ converges locally uniformly on $U \setminus \set {z \in U : \map f z = 0}$.
== Also see ==
* Derivative of Infinite Product of Analytic Functions
Category:Analytic Functions
Category:Infinite Products
Logarithmic Derivative of Product of Analytic Functions

== Theorem ==
Let $D \subset \C$ be open.
Let $f, g: D \to \C$ be analytic.
Let $f g$ be their pointwise product.
Let $z \in D$ with $\map f z \ne 0 \ne \map g z$.
Then:
:$\dfrac {\map {\paren {f g}' } z} {\map {\paren {f g} } z} = \dfrac{\map {f'} z} {\map f z} + \dfrac {\map {g'} z} {\map g z}$
== Proof ==
Follows directly from Product Rule for Complex Derivatives.
== Also see ==
* Product of Analytic Functions is Analytic
* Product Rule for Complex Derivatives
* Logarithmic Derivative of Infinite Product of Analytic Functions
Category:Complex Analysis
Logarithmic Derivative of Riemann Zeta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function:
:$\ds \forall s \in \C: \map \Re s > 1: \map \zeta s = \sum_{n \mathop \ge 1} n^{-s}$
Then for all $s \in \C$ with $\map \Re s > 1$:
:$\ds -\frac {\map {\zeta'} s} {\map \zeta s} = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}$
where $\Lambda$ is von Mangoldt's function.
== Proof ==
By Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where $p$ ranges over the primes.
From Laws of Logarithms:
:$\ds \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$
Taking the derivative:
Also, by the definition of $\Lambda$:
:$\ds \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$
and the proof is complete.
Category:Riemann Zeta Function
Logarithmic Differentiation/Examples/Continued Product

== Example of Logarithmic Differentiation ==
Let $y = x \paren {1 + 2 x} \paren {1 + 3 x}$.
Then:
:$\dfrac {\d y} {\d x} = \paren {1 + 2 x} \paren {1 + 3 x} + 2 x \paren {1 + 3 x} + 3 x \paren {1 + 2 x}$
== Proof ==
== Sources ==
* 
Category:Examples of Logarithmic Differentiation
Logarithmic Integral and Eulerian Logarithmic Integral Differ by Constant

== Theorem ==
Let $x \in \R$ be a real number such that $x > 2$.
Let $\map \li x$ denote the logarithmic integral of $x$.
Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$.
Then $\map \li x - \map \Li x$ is a constant.
== Proof ==
== Sources ==
* 
Category:Logarithmic Integral
Category:Eulerian Logarithmic Integral
Logarithmic Integral as Non-Convergent Series

== Theorem ==
The logarithmic integral can be defined in terms of a non-convergent series.
That is:
:$\ds \map \li z = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z} = \frac z {\ln z} \paren {\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }$
== Proof ==
From the definition of the logarithmic integral:
:$\ds \map \li z = \int_0^z \frac {\d t} {\ln t}$
Using Integration by Parts:
This sequence can be continued indefinitely.
We will consider the nature of the terms outside and inside the integral, after a number $n$ of iterations of integration by parts.
Let $u_n$ be the term outside the integral.
Let $v_n$ be the term inside the integral.
After $n$ iterations of Integration by Parts as above, we have:
:$\ds \map \li z = u_n + \int_0^z v_n \rd t$
:$u_0 = 0$
:$v_0 = \dfrac 1 {\ln t}$
It follows that:
which gives us the recurrence relations:
By recurrence on $n$, with the following recurrence hypothesis:
:$\text{R.H.}: \quad v_n = \dfrac {n!} {\ln^{n + 1} t}$
When $n = 0$, we have:
:$v_0 = \dfrac 1 {\ln t} = \dfrac {0!} {\ln^{0 + 1} t}$
which verifies the hypothesis.
By supposing true at $n$, we have at $n + 1$:
So $\text{R.H.}$ is verified at $n + 1$ if it is verified at $n$.
So it is proved for every $n \in \N$ (since it is true at $n = 0$):
:$(3): \quad v_n = \dfrac {n!} {\ln^{n + 1} t}$
By taking $(1)$, and substituting from $(3)$, a new expression for $u_{n + 1}$ in function of $u_n$ (recursive expression):
That is, we can write by expanding:
:$(4): \quad u_{n + 1} = \ds \sum_{i \mathop = 0}^n \frac {z \, i!} {\ln^{i + 1} z}$
Category:Logarithmic Integral
Logarithmic Integral is Asymptotic to Prime-Counting Function

== Theorem ==
Let $x \in \R$ be a real number such that $x > 2$.
Let $\map \li x$ denote the logarithmic integral of $x$.
Let $\map \pi x$ denote the prime-counting function of $x$.
Then $\map \li x$ is asymptotically equal to $\map \pi x$.
== Proof ==
== Sources ==
* 
Category:Logarithmic Integral
Category:Prime-Counting Function
Logarithmic Scale/Examples/Base 10

== Example of Logarithmic Scale ==
A commonly used logarithmic scale has points marked at equal intervals corresponding to values of $\ldots, 1, 10, 100, 1000, \ldots$, whose common logarithms are $\ldots, 0, 1, 2, 3, \ldots$
== Sources ==
* 
Category:Examples of Logarithmic Scales
Logarithmic Spiral is Equiangular

== Theorem ==
The logarithmic spiral is equiangular, in the following sense:
Let $P = \polar {r, \theta}$ be a point on a logarithmic spiral $S$ expressed in polar coordinates as:
:$r = a e^{b \theta}$
Then the angle $\psi$ that the tangent makes to the radius vector of $S$ is constant.
== Proof ==
Consider the logarithmic spiral $S$ expressed as:
:$r = a e^{b \theta}$
:
Let $\psi$ be the angle between the tangent to $S$ and the radius vector.
The derivative of $r$ $\theta$ is:
:$\dfrac {\d r} {\d \theta} = a b e^{b \theta} = b r$
and thus:
Thus for a given logarithmic spiral, $\psi$ is constant and equal to $\operatorname{arccot} b$.
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Logarithmic Spiral
Logical Matrix/Examples/Loop-Digraph

== Example of Logical Matrix ==
The adjacency matrix for a loop-digraph is a logical matrix:
: $\qquad \begin{pmatrix}
1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 \\
\end{pmatrix}$
Category:Examples of Logical Matrices
Logical Matrix/Examples/Relation

== Example of Logical Matrix ==
A logical matrix can be used as an explicit definition of a binary relation.
The binary relation is a divisor of on the set of natural numbers $\set {1, 2, 3, 4}$ consists of the set of ordered pairs:
:$\set {\tuple {1, 1}, \tuple {1, 2}, \tuple {1, 3}, \tuple {1, 4}, \tuple {2, 2}, \tuple {2, 4}, \tuple {3, 3}, \tuple {4, 4} }$
This relation can be represented by the following logical matrix:
:$\begin {pmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end {pmatrix}$
Category:Examples of Logical Matrices
Logistic Regression/Examples/Insecticide

== Example of Logistic Regression ==
Consider a logistic regression experiment to investigate the efficacy of a certain insecticide.
The treatment level $x_i$ can be amended according to the response of interest $Y_i$ where:
:$Y_i = 1$ denotes death of the insect
:$Y_i = 0$ denotes survival of the insect.
The odds of death for the $i$th individual are then:
:$\theta_i = \dfrac {p_i} {1 - p_i}$
== Sources ==
* 
* 
Category:Examples of Logistic Regression
Loglinear Model/Examples/2x2 Table

== Example of Loglinear Model ==
Let $\CC$ be a $2 \times 2$ contingency table:
:$\begin{array}{r|cc|c}
& \text {A} & \text {B} & \text {Totals} \\
\hline
\text {P} & m_{11} & m_{12} & m_{11} + m_{12} \\
\text {Q} & m_{21} & m_{22} & m_{21} + m_{22} \\
\hline
\text {Totals} & m_{11} + m_{21} & m_{12} + m_{22} 
\end{array}$
Then under independence of rows and columns:
:$\map \ln {\dfrac {m_{11} m_{22} } {m_{12} m_{21} } } = 0$
which is in accordance with the condition for independence predicted by the odds ratio:
:$\theta = \dfrac {m_{11} m_{22} } {m_{12} m_{21} } = 1$
== Sources ==
* 
* 
Category:Examples of Loglinear Models
Long Period Prime/Examples

== Examples of Long Period Primes ==
=== $7$ is Long Period Prime ===
=== $17$ is Long Period Prime ===
=== $19$ is Long Period Prime ===
=== $23$ is Long Period Prime ===
=== $29$ is Long Period Prime ===
=== $47$ is Long Period Prime ===
=== $59$ is Long Period Prime ===
=== $61$ is Long Period Prime ===
Category:Examples of Long Period Primes
Long Period Prime/Examples/131

== Theorem ==
The prime number $131$ is a long period prime:
:$\dfrac 1 {131} = 0 \cdotp \dot 00763 \, 35877 \, 86259 \, 54198 \, 47328 \, 24427 \, 48091 \, 60305 \, 34351 \, 14503 \, 81679 \, 38931 \, 29770 \, 99236 \, 64122 \, 13740 \, 45801 \, 52671 \, 75572 \, 51908 \, 39694 \, 65648 \, 85496 \, 18320 \, 61068 \, 7022 \dot 9$
It also contains an equal number ($13$) of each of the digits from $0$ to $9$.
== Proof ==
From Reciprocal of $131$:
Counting the digits, it is seen that this has a period of recurrence of $130$.
Inspecting the expansion and counting the digits, we find that each one appears exactly $13$ times.
Hence the result.
== Sources ==
* : $61$
Category:131
Category:Examples of Long Period Primes
Long Period Prime/Examples/17

== Theorem ==
The prime number $17$ is a long period prime:
:$\dfrac 1 {17} = 0 \cdotp \dot 05882 \, 35294 \, 11764 \, \dot 7$
== Proof ==
From Reciprocal of $17$:
Counting the digits, it is seen that this has a period of recurrence of $16$.
Hence the result.
== Sources ==
* : $17$
* : $0,588,235,294,117,647$
* : $17$
* : $0,588,235,294,117,647$
Category:17
Category:Examples of Long Period Primes
Long Period Prime/Examples/19

== Theorem ==
The prime number $19$ is a long period prime:
:$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$
== Proof ==
From Reciprocal of $19$:
Counting the digits, it is seen that this has a period of recurrence of $18$.
Hence the result.
== Sources ==
* : $19$
* : $052,631,578,947,368,421$
* : $19$
* : $052,631,578,947,368,421$
Category:19
Category:Examples of Long Period Primes
Long Period Prime/Examples/23

== Theorem ==
The prime number $23$ is a long period prime:
:$\dfrac 1 {23} = 0 \cdotp \dot 04347 \, 82608 \, 69565 \, 21739 \, 1 \dot 3$
== Proof ==
From Reciprocal of $23$:
Counting the digits, it is seen that this has a period of recurrence of $22$.
Hence the result.
== Sources ==
* : $23$
* : $0,434,782,608,695,652,173,913$
* : $23$
* : $0,434,782,608,695,652,173,913$
Category:23
Category:Examples of Long Period Primes
Long Period Prime/Examples/29

== Theorem ==
The prime number $29$ is a long period prime:
:$\dfrac 1 {29} = 0 \cdotp \dot 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 93 \dot 1$
== Proof ==
From Reciprocal of $29$:
Counting the digits, it is seen that this has a period of recurrence of $28$.
Hence the result.
Category:29
Category:Examples of Long Period Primes
Long Period Prime/Examples/47

== Theorem ==
The prime number $47$ is a long period prime:
:$\dfrac 1 {47} = 0 \cdotp \dot 02127 \, 65957 \, 44680 \, 85106 \, 38297 \, 87234 \, 04255 \, 31914 \, 89361 \, \dot 7$
== Proof ==
From Reciprocal of $47$:
Counting the digits, it is seen that this has a period of recurrence of $46$.
Hence the result.
Category:47
Category:Examples of Long Period Primes
Long Period Prime/Examples/59

== Theorem ==
The prime number $59$ is a long period prime:
:$\dfrac 1 {59} = 0 \cdotp \dot 01694 \, 91525 \, 42372 \, 88135 \, 59322 \, 03389 \, 83050 \, 84745 \, 76271 \, 18644 \, 06779 \, 66 \dot 1$
== Proof ==
From Reciprocal of $59$:
Counting the digits, it is seen that this has a period of recurrence of $58$.
Hence the result.
Category:59
Category:Examples of Long Period Primes
Long Period Prime/Examples/61

== Theorem ==
The prime number $61$ is a long period prime:
:$\dfrac 1 {61} = 0 \cdotp \dot 01639 \, 34426 \, 22950 \, 81967 \, 21311 \, 47540 \, 98360 \, 65573 \, 77049 \, 18032 \, 78688 \, 5245 \dot 9$
It also contains an equal number ($6$) of each of the digits from $0$ to $9$.
== Proof ==
From Reciprocal of $61$:
Counting the digits, it is seen that this has a period of recurrence of $60$.
Inspecting the expansion and counting the digits, we find that each one appears exactly $6$ times.
Hence the result.
== Sources ==
* : $61$
Category:61
Category:Examples of Long Period Primes
Long Period Prime/Examples/7

== Theorem ==
$7$ is the smallest long period prime:
:$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$
:
== Proof ==
From Reciprocal of $7$:
The reciprocals of $1$, $2$, $4$ and $5$ do not recur:
while those of $3$ and $6$ do recur, but with the non-maximum period of $1$:
== Sources ==
* : $142,857$
* : $142,857$
Category:7
Category:Examples of Long Period Primes
Long Period Prime/Examples/97

== Theorem ==
The prime number $97$ is a long period prime:
:$\dfrac 1 {97} = 0 \cdotp \dot 01030 \, 92783 \, 50515 \, 46391 \, 75257 \, 73195 \, 87628 \, 86597 \, 93814 \, 43298 \, 96907 \, 21649 \, 48453 \, 60824 \, 74226 \, 80412 \, 37113 \, 40206 \, 18556 \, \dot 7$
== Proof ==
From Reciprocal of $97$:
Counting the digits, it is seen that this has a period of recurrence of $96$.
Hence the result.
== Sources ==
* : $97$
* : $97$
Category:97
Category:Examples of Long Period Primes