nyuuzyou/wikis huggingface dataset:
Approximate Motion of Simple Pendulum

== Theorem ==
Consider a simple pendulum consisting of a bob whose mass is $m$, at the end of a rod of negligible mass of length $a$.
Let the bob be pulled to one side so that the rod is at a small angle $\alpha$ (less than about $10 \degrees$ or $15 \degrees$) from the vertical and then released.
Let $T$ be the period of the pendulum, that is, the time through which the bob takes to travel from one end of its path to the other, and back again.
Then:
:$T \approx 2 \pi \sqrt {\dfrac a g}$
where $g$ is the Acceleration Due to Gravity.
== Proof ==
At a time $t$, let:
:the rod be at an angle $\theta$ to the the vertical
:the bob be travelling at a speed $v$
:the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $s$.
:
From Motion of Simple Pendulum, the equation of motion of the bob is given by:
:$\dfrac {a^2} 2 \paren {\dfrac {\d \theta} {\d t} }^2 = g a \paren {a \cos \theta - a \cos \alpha}$
Differentiating $t$ and simplifying yields:
:$a \dfrac {\d^2 \theta} {\d t^2} = -g \sin \theta$
For small angles, $\sin \theta \approx \theta$ and so:
:$\dfrac {\d^2 \theta} {\d t^2} + \dfrac g a \theta = 0$
This is an instance of a $2$nd order linear ODE.
From Linear Second Order ODE: $y' ' + k^2 y = 0$, this has the solution:
:$\theta = c_1 \sin \sqrt {\dfrac g a} t + c_1 \cos \sqrt {\dfrac g a} t$
We have at $t = 0$ that $\theta = \alpha$ and $\dfrac {\d \theta} {\d t} = 0$.
This yields:
:$\theta = \alpha \cos \sqrt {\dfrac g a} t$
and so the period of this simple pendulum is:
:$T = 2 \pi \sqrt {\dfrac a g}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
* 
* 
* 
* : Chapter $9$: Patterns in Nature: Differential equations
Category:Simple Pendulums
Category:Mechanics
Approximate Relations between Pi and Euler's Number

== Approximate Relations between $\pi$ (pi) and Euler's number $e$ ==
=== Fanelli's Approximation ===
=== Gelfond's Constant minus $\pi$ ===
=== $e^{\pi \sqrt {163} }$ ===
Category:Approximate Relations between Pi and Euler's Number
Category:Euler's Number
Category:Pi
Category:Examples of Approximations
Approximate Relations between Pi and Euler's Number/Fanelli's Approximation

== Approximate Relation between $\pi$ (pi) and Euler's number $e$ ==
This approximation to $\pi$ is accurate to $5$ decimal places:
:$\sqrt [9] {10 e^8} = 3 \cdotp 14159 \, 828 \ldots$
== Historical Note ==
== Sources ==
* : $2 \cdotp 71828 \, 18284 \, 59045 \, 23536 \, 02874 \, 71352 \, 66249 \, 77572 \, 47093 \, 69995 \ \ldots$
Category:Euler's Number
Category:Pi
Category:Approximate Relations between Pi and Euler's Number
Approximate Relations between Pi and Euler's Number/Fanelli's Approximation/Historical Note

== Historical Note on Approximate Relations between Pi and Euler's Number/Fanelli's Approximation ==
Fanelli's approximation is the name coined by to the approximation $\sqrt [9] {10 e^8} = 3 \cdotp 14159 \, 828 \ldots$.
 abandoned further work on establishing similar correspondences, and in more recent times has contributed towards the literature on the Riemann Hypothesis.
Category:Historical Notes
Approximate Relations between Pi and Euler's Number/e to the pi root 163

== Approximate Relation between $\pi$ (pi) and Euler's number $e$ ==
This expression relating $e$ and $\pi$ comes within $10^{-12}$ of an integer:
:$e^{\pi \sqrt {163} } \approx 262 \, 537 \, 412 \, 640 \, 768 \, 743 \cdotp 99999 \, 99999 \, 99250 \ldots$
== Historical Note ==
== Sources ==
* : $163$
* : $163$
Category:Euler's Number
Category:Pi
Category:163
Category:Approximate Relations between Pi and Euler's Number
Approximate Relations between Pi and Euler's Number/e to the pi root 163/Historical Note

== Historical Note on Approximate Relations between $\pi$ and $e$: $e^{\pi \sqrt {163} }$ ==
The fact that $e^{\pi \sqrt {163} }$ comes so close to an integer was noticed by , who considered the property "amusing".
== Sources ==
* : $163$
* : $163$
Category:Historical Notes
Approximate Size of Sum of Harmonic Series

== Theorem ==
Let $H_n$ denote the sum of the harmonic series:
:$H_n = \ds \sum_{k \mathop = 1}^n \frac 1 k$
Then $H_n$ can be approximated as follows:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
where:
:$\gamma$ denotes the Euler-Mascheroni constant: $\gamma \approx 0 \cdotp 57721 \, 56649 \, \ldots$
:$0 < \epsilon < \dfrac 1 {252 n^6}$
== Proof ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: $(3)$
Category:Harmonic Numbers
Category:Harmonic Series
Approximate Value of Nth Prime Number

== Theorem ==
The $n$th prime number is approximately $n \ln n$.
== Proof ==
This will be demonstrated by showing that:
:$\ds \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n} = 1$
where $p_n$ denotes the $n$th prime number.
By definition of prime-counting function:
:$\map \pi {p_n} = n$
The Prime Number Theorem gives:
:$\ds \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x} = 1$
Thus:
:$\ds \lim_{x \mathop \to \infty} \dfrac n {p_n / \ln p_n} = 1$
== Sources ==
* : Chapter $\text {B}.16$: The Sequence of Primes
Category:Prime Number Theorem
Category:Prime Numbers
Category:Analytic Number Theory
Approximation/Examples/22 over 7

== Theorem ==
$\dfrac {22} 7$ is a convenient approximation to $\pi$:
:$\dfrac {22} 7 = 3 \cdotp \dot 14285 \dot 7$
== Proof ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* 
Category:Approximations to Pi
Category:Examples of Approximations
Approximation/Examples/Sine of x

== Theorem ==
For small values of $x$ measured in radians:
:$\sin x \approx x$
== Proof ==
From Limit of Sine of X over X at Zero:
:$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
Hence the result.
== Sources ==
* 
Category:Sine Function
Category:Examples of Approximations
Approximation by Taylor Polynomials

== Definition ==
Let $a < b$ be real numbers.
Let $f : \closedint a b \to \R$ be a real function.
Let $n \in \Z_{>0}$.
Let $\xi \in \closedint a b$.
Let $f$ be $n$ times differentiable at $\xi$, where all derivatives are one-sided if $\xi = a$ or $\xi = b$.
Let $T_n$ be the Taylor polynomial of degree $n$ for $f$ about $\xi$, that is:
:$\ds \map {T_n} x = \sum_{i \mathop = 0}^n \frac {\map {f^{\paren i}} \xi} {i!} \paren {x - \xi}^i$
Then:
:$\ds \lim_{x \mathop \to \xi} \frac {\map f x - \map {T_n} x} {\paren {x - \xi}^n} = 0$
where the limit is one-sided if $\xi = a$ or $\xi = b$.
== Proof ==
== Sources ==
* : $4$.Differentiation
Category:Calculus
Category:Taylor's Theorem
Approximation to 2n Choose n

== Theorem ==
:$\ds \lim_{n \mathop \to \infty} \dbinom {2 n} n = \dfrac {4^n} {\sqrt {n \pi} }$
== Proof ==
From Approximation to $\dbinom {x + y} y$:
:$\ds \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y$
Thus:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $46$
Category:Binomial Coefficients
Approximation to Binary Logarithm from Natural and Common Logarithm

== Theorem ==
The binary logarithm $\lg x$ can be approximated, to within $1 \%$, by the expression:
:$\lg x \approx \ln x + \log_{10} x$
That is, by the sum of the natural logarithm and common logarithm.
== Proof ==
We have that:
:$\ln 2 = 0 \cdotp 69347 \ 1805 \ldots$
:$\log_{10} e = 0 \cdotp 43429 \ 44819 \ldots$
Thus:
Hence:
:$\lg x + \log_{10} x \approx 0 \cdotp 994 \ln x$
That is, the approximation is $99.4 \%$ accurate.
== Historical Note ==
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $22$
Category:Logarithms
Approximation to Binary Logarithm from Natural and Common Logarithm/Historical Note

== Historical Note on Approximation to Binary Logarithm from Natural and Common Logarithm ==
The useful Approximation to Binary Logarithm from Natural and Common Logarithm was observed by .
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $22$
Category:Historical Notes
Approximation to Golden Rectangle using Fibonacci Squares

== Theorem ==
An approximation to a golden rectangle can be obtained by placing adjacent to one another squares with side lengths corresponding to consecutive Fibonacci numbers in the following manner:
:
It can also be noted, as from Sequence of Golden Rectangles, that an equiangular spiral can be approximated by constructing quarter circles as indicated.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Golden Mean
Category:Approximation to Golden Rectangle using Fibonacci Squares
Approximation to Golden Rectangle using Fibonacci Squares/Proof 1

== Theorem ==
== Proof ==
Let the last two squares to be added have side lengths of $F_{n - 1}$ and $F_n$.
Then from the method of construction, the sides of the rectangle generated will be $F_n$ and $F_{n + 1}$.
From Continued Fraction Expansion of Golden Mean it follows that the limit of the ratio of the side lengths of the rectangle, as $n$ tends to infinity, is the golden section $\phi$.
Hence the result.
Category:Approximation to Golden Rectangle using Fibonacci Squares
Approximation to Golden Rectangle using Fibonacci Squares/Proof 2

== Theorem ==
== Proof ==
From Sum of Sequence of Squares of Fibonacci Numbers:
:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
Hence the result.
Category:Approximation to Golden Rectangle using Fibonacci Squares
Approximation to Power of 7 by Power of 10

== Theorem ==
:$7^{510} \approx 1 \cdotp 00000 \, 09377 \, 76536 \ldots \times 10^{431}$
This is the closest known approximation of a power of $7$ by a power of $10$.
== Proof ==
An intuition is given as follows:
Suppose for some $m, n \in \N$:
:$7^m = c \cdot 10^n$, where $c$ is very close to $1$.
Taking common logarithm:
:$m \log 7 = \log c + n$
Which leads to:
:$\log 7 = \dfrac n m + \dfrac {\log c} m$
where $\dfrac {\log c} m$ is very close to $0$.
To make a good approximation is to minimize $\size {\log c}$.
One may calculate the continued fraction of $\log 7$.
We use continued fractions because Convergents are Best Approximations.
One obtains $\sqbrk {0; 1, 5, 2, 5, 6, 1, 4813, 1, 1, 2, 2, \dots}$.
Let $\dfrac {p_n} {q_n}$ be the convergent $\sqbrk {0; a_1, \dots, a_n}$.
Then:
Thus, at least intuitively, the larger the value of the next $a_i$, the smaller $\size {\log c}$ is.
$4813$ is the largest number in the expansion among at least the first $100$ numbers.
Note that $\sqbrk {0; 1, 5, 2, 5, 6, 1} = \dfrac {431} {510}$
This gives the estimate $7^{510} \approx 10^{431}$.
== Sources ==
* : $510$
Category:Powers of 7
Category:Powers of 10
Approximation to Reciprocal times Derivative of Gamma Function

== Theorem ==
Let $\Gamma$ denote the gamma function.
For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:
:$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\OO_\epsilon} {z^{-1} }$
where:
:$\map \OO {z^{-1} }$ denotes big-$\OO$ notation
:the implied constant depends on $\epsilon$.
== Proof ==
From Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function:
:$\ln \map \Gamma z = \paren {z - \dfrac 1 2} \ln z - z + \dfrac {\ln 2 \pi} 2 + \map \OO {z^{-1} }$
Taking the derivative $z$:
:$(1): \quad \dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z - \dfrac 1 {2 z} + \dfrac \d {\d z} \map \OO {z^{-1} }$
Since there exists $\map c \epsilon > 0$ such that:
:$\forall \size z > 1: -\dfrac c {\size {z^{-1} } } < \size {\map \OO {z^{-1} } } < \dfrac c {\size {z^{-1} } }$
it follows directly that the third term in $(1)$ is $\map \OO {z^{-1} }$.
Category:Gamma Function
Category:Reciprocals
Approximation to Stirling's Formula for Gamma Function

== Theorem ==
Let:
:$D_\epsilon = \set {z \in \C : \cmod {\Arg z} < \pi - \epsilon,\ \cmod z > 1}$
where:
:$\cmod {\Arg z}$ denotes the absolute value of the principal argument of $z$
:$\cmod z$ denotes the modulus of $z$
:$\epsilon \in \R_{>0}$.
Then for all $z \in D_\epsilon$, the gamma function of $z$ satisfies:
:$\map \Gamma z = \sqrt {\dfrac {2 \pi} z} \paren {\dfrac z e}^z \paren {1 + \map \OO {z^{-1} } }$
where $\map \OO {z^{-1} }$ denotes big-O of $z^{-1}$.
== Proof ==
From Gamma Function is Unique Extension of Factorial:
for $0 < y \le 1$ and $n \in \N$.
Let $x$ be given.
Let $n + 1$ be the largest natural number such that $n + 1 \le x$.
Let $x = y + n + 1$, and thus $0 < y \le 1$.
Then:
Similarly for the .
The result follows from Gamma Function Extends Factorial.
== Also see ==
* Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function
== Sources ==
* : $\S 17.7 \ (2)$
Category:Gamma Function
Category:Stirling's Formula
Approximation to x+y Choose y

== Theorem ==
:$\ds \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y$
== Proof ==
It can be assumed that both $x$ and $y$ are integers.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $46$
Category:Binomial Coefficients
Category:Stirling's Formula
Approximations to Equilateral Triangles by Heronian Triangles

== Theorem ==
The sequence of best approximations to an equilateral triangle by a Heronian triangle begins:
:The $\tuple {3, 4, 5}$ triangle, with area $6$
:The $\tuple {13, 14, 15}$ triangle, with area $84$, where $14 = 4^2 - 2$
:The $\tuple {193, 194, 195}$ triangle, where $194 = 14^2 - 2$
:The $\tuple {37 \, 633, 37 \, 634, 37 \, 635}$ triangle, where $37 \, 634 = 194^2 - 2$
and so on.
== Proof ==
Suppose a triangle with side lengths $\tuple {a - 1, a, a + 1}$ is Heronian.
By Heron's Formula, the area of this triangle is:
By , $\dfrac a 4 \sqrt {3 a^2 - 12} \in \N$.
Hence it is necessary for $a$ to be even.
Writing $a = 2 n$:
:$\dfrac a 4 \sqrt {3 a^2 - 12} = \dfrac {2 n} 4 \sqrt {3 \paren {2 n}^2 - 12} = n \sqrt {3 n^2 - 3}$
So we need to find $n$ such that $\sqrt {3 n^2 - 3}$ is an integer.
This corresponds to a Pell-like equation $x^2 - 3 n^2 = -3$.
This equation has a trivial solution $\tuple {x, n} = \tuple {0, 1}$.
The smallest $n$ with integer $x$ solution to $x^2 - 3 n^2 = 1$ are:
:$\tuple {n, x} = \tuple {0, 1}$ (trivial), $\tuple {1, 2}$
Then a family of solutions can be found by:
:$x^2 - 3 y^2 = 1 = \paren {2^2 - 3 \times 1^2}^m$ for all $m \in \N$
Factorising gives:
:$\paren {x - \sqrt 3 y} \paren {x + \sqrt 3 y} = \paren {2 - \sqrt 3}^m \paren {2 + \sqrt 3}^m$
The system of equations:
:$x - \sqrt 3 y = \paren {2 - \sqrt 3}^m$
:$x + \sqrt 3 y = \paren {2 + \sqrt 3}^m$
Has solutions:
:$x = \dfrac {\paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m} 2$
:$y = \dfrac {\paren {2 + \sqrt 3}^m - \paren {2 - \sqrt 3}^m} {2 \sqrt 3}$
Suppose that $p^2 - 3 q^2 = -3, r^2 - 3 s^2 = 1$.
Then:
showing that $\tuple {p r \pm 3 q s, p s \pm q r}$ is a solution for $x^2 - 3 y^2 = -3$.
In particular:
:$\tuple {\dfrac {\sqrt 3 \paren {\paren {2 + \sqrt 3}^m - \paren {2 - \sqrt 3}^m}} 2, \dfrac {\paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m} 2}$ is a solution.
From our discussion above:
:$a = \paren {2 - \sqrt 3}^m + \paren {2 + \sqrt 3}^m$ is a family of solutions for the almost-equilateral Heronian triangles.
The sequence above is a subfamily of the solution above, as seen by:
Some other solutions include:
:$m = 3: \tuple {51, 52, 53}$, with area $\sqrt {78 \times 27 \times 26 \times 25} = 1170$
:$m = 5: \tuple {723, 724, 725}$, with area $\sqrt {1086 \times 361 \times 362 \times 363} = 226 \, 974$
== Also see ==
These triangles are known as Fleenor-Heronian triangles.
== Sources ==
* 
* : $14$
* : $14$
Category:Fleenor-Heronian Triangles
Category:Equilateral Triangles
Apéry's Constant in terms of Central Binomial Coefficients

== Theorem ==
Apéry's constant can be expressed in terms of binomial coefficients as:
:$\map \zeta 3 = \ds \dfrac 5 2 \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n - 1} } {n^3 \dbinom {2 n} n}$
== Proof ==
== Sources ==
* : $1,20205 69 \ldots$
* 
Category:Apéry's Constant
Apéry's Theorem

== Theorem ==
Apéry's constant:
:$\map \zeta 3 = \ds \sum_{n \mathop = 1}^\infty \frac 1 {n^3}$
is irrational.
== Proof ==
We have:
:$\dfrac 6 {\map \zeta 3} = 5 - \cfrac {1^6} {117 - \cfrac {2^6} {535 - \cfrac {\dotsb} {\dotsb - \cfrac {n^6} {34 n^3 + 51 n^2 + 27 n + 5} - \dotsb } } }$
== Also known as ==
Some sources present the name of this without the accent: Apery's theorem.
== Sources ==
* 
* : $1,20205 69 \ldots$
* : $1 \cdotp 202 \, 056 \ldots$
* 
* : $1 \cdotp 20205 \, 6 \ldots$
* 
Category:Apéry's Constant
Arbitrary Cyclic Group of Order 4

== Theorem ==
Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
:$\begin{array}{r|rrrr}
\circ & 2 & 3 & 4 & 1 \\
\hline
2 & 2 & 3 & 4 & 1 
\\
3 & 3 & 4 & 1 & 2
\\
4 & 4 & 1 & 2 & 3
\\
1 & 1 & 2 & 3 & 4
\\
\end{array}$
Then $\struct {S, \circ}$ is a group.
Specifically, $\struct {S, \circ}$ is the cyclic group of order $4$.
== Proof ==
Let $S' = \set {0, 1, 2, 3}$.
Let $\phi: S \to S'$ be the bijection:
By inspection, the Cayley table presented above is in the same form as the Cayley table for the cyclic group of order $4$:
:$\begin{array}{r|rrrr}
+_4 & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 1 & 2 & 3 
\\
1 & 1 & 2 & 3 & 0
\\
2 & 2 & 3 & 0 & 1
\\
3 & 3 & 0 & 1 & 2
\\
\end{array}$
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{B v}$
Category:Cyclic Group of Order 4
Arbitrary Power of Complex Number

== Theorem ==
Let $z = a + i b$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
== Proof ==
=== Lemma ===
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds z^n = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j a^{n - j} b^j}$
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds z^k = \paren {\sum_{\substack {0 \mathop \le j \mathop \le k \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom k j a^{k - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le k \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom k j a^{k - j} b^j}$
from which it is to be shown that:
:$\ds z^{k + 1} = \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom {k + 1} j a^{k + 1 - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom {k + 1} j a^{k + 1 - j} b^j}$
=== Induction Step ===
This is the induction step:
From :
where $z^k = u_k + i v_k$.
Taking the real part:
Taking the imaginary part:
Thus we have shown:
$\ds z^{k + 1} = \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom {k + 1} j a^{k + 1 - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom {k + 1} j a^{k + 1 - j} b^j}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds z^n = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j a^{n - j} b^j}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.22$
Category:Arbitrary Power of Complex Number
Category:Complex Numbers
Category:Powers
Arbitrary Power of Complex Number/Lemma

== Lemma for Arbitrary Power of Complex Number ==
Let $z = a + i b$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n = u_n + i v_n$.
Then $z^{n + 1} = u_{n + 1} + i v_{n + 1}$ where:
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.23$
Category:Arbitrary Power of Complex Number
Arborescence/Examples/English Sentence

== Example of Arborescence ==
The parsing structure of a sentence in English can be presented as an arborescence in the following format:
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
& & & \meta {sentence} \ar[dl] \ar[dr] \\
& & \meta {subject} \ar[d] & & \meta {predicate} \ar[dr] \\
& & \meta {noun \ phrase} \ar[dl] \ar[d] & & & \meta {verb \ phrase} \ar[dl] \ar[d] \\
& \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] & & \meta {verb} \ar[dl] & \meta {adverbial \ phrase} \ar[dl] \ar[d] \\
\mathtt {the} & \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] & \mathtt {jumps} & \meta {preposition} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] \\
\mathtt {quick} & \meta {adjective} \ar[d] & \meta {noun \ phrase} \ar[d] & \mathtt {over} & \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d]\\
& \mathtt {brown} & \meta {noun} \ar[d] & \mathtt {the} & \meta {adjective} \ar[d] & \meta {noun \ phrase} \ar[d]\\
& & \mathtt {fox} & & \mathtt {lazy} & \meta {noun} \ar[d] \\
& & & & & \mathtt {dog}
} \end {xy}$
=== Types of Node ===
== Sources ==
* : Chapter $1$: Preliminaries: $1.2$ Graphs and Trees: Trees
Category:Examples of Arborescences
Arborescence/Examples/English Sentence/Types of Node

== Example of Arborescence: Types of Node ==
Consider the parsing structure of the following sentence in English, presented as an arborescence:
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
& & & \meta {sentence} \ar[dl] \ar[dr] \\
& & \meta {subject} \ar[d] & & \meta {predicate} \ar[dr] \\
& & \meta {noun \ phrase} \ar[dl] \ar[d] & & & \meta {verb \ phrase} \ar[dl] \ar[d] \\
& \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] & & \meta {verb} \ar[dl] & \meta {adverbial \ phrase} \ar[dl] \ar[d] \\
\mathtt {the} & \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] & \mathtt {jumps} & \meta {preposition} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d] \\
\mathtt {quick} & \meta {adjective} \ar[d] & \meta {noun \ phrase} \ar[d] & \mathtt {over} & \meta {adjective} \ar[dl] & \meta {noun \ phrase} \ar[dl] \ar[d]\\
& \mathtt {brown} & \meta {noun} \ar[d] & \mathtt {the} & \meta {adjective} \ar[d] & \meta {noun \ phrase} \ar[d]\\
& & \mathtt {fox} & & \mathtt {lazy} & \meta {noun} \ar[d] \\
& & & & & \mathtt {dog}
} \end {xy}$
:The node with the label $\meta {verb}$ is a child of the node with the label $\meta {verb \ phrase}$
:The node with the label $\meta {verb \ phrase}$ is the parent of the node with the label $\meta {verb}$
:The node with the label $\mathtt {dog}$ is a decendant of:
::itself
::the node with the label $\meta {verb \ phrase}$
::the node with the label $\meta {sentence}$
::$6$ other nodes, including the nodes with the labels $\meta {adverbial \ phrase}$ and $\meta {predicate}$.
:The nodes with the labels consisting of English words are the leaf nodes
:The nodes with the labels consisting of English parts of speech enclosed in angle brackets: $\meta{\, \cdot \,}$ are the interior nodes.
== Sources ==
* : Chapter $1$: Preliminaries: $1.2$ Graphs and Trees: Trees
Category:Examples of Arborescences
Arc-Connected Space is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is arc-connected.
Then $T$ is path-connected.
== Proof ==
Let $T = \struct {S, \tau}$ be arc-connected.
Then $\forall x, y \in S$, there exists a continuous injection $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = y$.
As $f$ is a continuous injection, it is also simply a continuous mapping.
The result follows from the definition of path-connectedness.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Path-Connected Spaces
Category:Arc-Connected Spaces