nyuuzyou/wikis huggingface dataset:
Newton's Method

== Proof Technique ==
Newton's Method is a method of solving an equation expressed as a real function for which there may be no convenient closed form solution.
The derivative of the function has to be known in order to use Newton's Method.
Let the equation to be solved be of the form:
:$y = \map f x$
Let the value of $x$ be required for a given $y$.
Let $x_1$ be an initial guess.
Then an iterative improvement on $x_1$ is of the form:
:$x_2 = x_1 - \dfrac {\map f {x_1} - y} {\map {f'} {x_1} }$
where $\map {f'} {x_1}$ is the derivative of $f$ $x$ evaluated at $x_1$.
== Proof ==
The function $\map f x$ can be expanded about $x_1$ using Taylor's Theorem:
:$\map f x = \map f {x_1} + \map {f'} {x_1} \paren {x - x_1} + \dfrac 1 2 \map {f} {x_1} \paren {x - x_1}^2 + \dotsb$
This series can be rewritten with an error term as follows:
:$\map f x = \map f {x_1} + \map {f'} {x_1} \paren {x - x_1} + \epsilon$
This gives:
For $\epsilon$ small enough the final term can be neglected:
:$x_2 = x_1 - \dfrac {\map f {x_1} - y} {\map {f'} {x_1} }$
== Graphical Illustration ==
== Example ==
== Also see ==
* Sequence of Approximations Converges Quadratically
== Sources ==
* 
Category:Newton's Method
Category:Calculus
Category:Numerical Analysis
Newton's Method/Example/x^3 - 2 x - 5 = 0

== Example of Use of Newton's Method ==
The real root of the cubic:
:$x^3 - 2 x - 5 = 0$
can be found by using Newton's Method.
Its approximate value is:
:$2 \cdotp 09455 \, 1$
== Proof ==
We are to solve $\map f x = x^3 - 2 x - 5 = 0$ numerically.
By Newton's Method, we are to iterate the solution using:
:$x_{n + 1} = x_n - \dfrac {\map f {x_n}} {\map {f'} {x_n}}$
We have $\map {f'} x = 3 x^2 - 2$.
Hence:
We can start with any number that is close to the solution.
We have:
:$\map f 2 = -1 < 0$
:$\map f 3 = 16 > 0$
By Intermediate Value Theorem a root exists between $2$ and $3$.
For convenience we set $x_1 = 2$.
Then:
at $x_4$ we already have the root accurate to $9$ decimal places.
This process can be continued to obtain increasingly accurate results.
== Also see ==
* Definition:Wallis's Number
Category:Newton's Method
Newton's Method/Sequence of Approximations Converges Quadratically

== Theorem ==
Let $\map f x$ be a real function.
Let $\alpha$ be a root of $\map f x$.
Let $\epsilon > 0$ be a positive real number, and $I = \closedint {\alpha - \epsilon} {\alpha + \epsilon}$.
Let $f$ have a continuous second derivative on $I$.
Let $\ds M = \map \sup {\size {\frac {\map {f' '} s} {\map {f'} t} } }$ over all $s, t \in I$.
:For this to be well-defined, it is also necessary that $\map {f'} x$ is non-vanishing on $I$.
Suppose $\epsilon < \dfrac 2 M$.
Then the sequence generated by Newton's Method, starting with any initial guess $x_0 \in I$, converges to $\alpha$ with order $2$.
== Proof ==
Suppose that the sequence is produced up to $x_n$.
Suppose also that $x_n \in I$.
First:
Therefore $x_{n + 1} \in I$.
But as $x_0 \in I$, by Principle of Mathematical Induction, every $x_i \in I$.
Next, apply the Principle of Recursive Definition to define the sequence:
Trivially:
:$\size {\alpha - x_0} \le \epsilon_0$
Assuming $\size {\alpha - x_n} \le \epsilon_n$, it follows that:
By Principle of Mathematical Induction, every $\size {\alpha - x_n} \le \epsilon_n$.
But:
Therefore, the sequence $\sequence {x_n}$ converges to $\alpha$ with order $2$, by definition.
== Sources ==
* 
Category:Newton's Method
Newton's Second Law of Motion/Also known as

== Newton's Second Law of Motion: Also known as ==
Newton's Second Law of Motion is also often referred to as just Newton's Second Law.
Some refer to it as just Newton's Law, on the grounds that it is the most significant of all of Newton's Laws of Motion, but there is more than one of those.
== Sources ==
* : Chapter $8$: The System of the World: Newton
* 
Category:Newton's Laws of Motion
Newton's Second Law of Motion/Also presented as

== Newton's Second Law of Motion: Also presented as ==
Newton's Second Law of Motion is also seen presented in the form:
which is not its most general form, as it assumes constant mass.
Indeed, as himself put it:
:The acceleration produced by a particular force acting on a body is directly proportional to the magnitude of the force and inversely proportional to the mass of the body.
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $3$: The Laws of Motion: Mass
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(1)$
* 
* 
* 
* 
* 
* 
* : Chapter $8$: The System of the World: Newton
* 
* 
Category:Newton's Laws of Motion
Newton's Third Law of Motion/Also known as

== Newton's Third Law of Motion: Also known as ==
Newton's Third Law of Motion is also referred to as:
:the Law of Action and Reaction
or:
:the Principle of Action and Reaction.
Hence its usual wording:
:To every action there is an equal and opposite reaction.
Some sources suggest that it may also be called the Law of Interaction.
It is also often referred to as just Newton's Third Law.
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : $\text {I}$: Motion, Sound and Heat: Chapter $3$: The Laws of Motion: Action and Reaction
* 
* 
* 
Category:Newton's Laws of Motion
Newton's Three-Eighths Rule

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {a = x_0, x_1, x_2, \ldots, x_{3 n - 1}, x_{3 n} = b}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, 3 n}: x_r - x_{r - 1} = \dfrac {b - a} {3 n}$
Then the definite integral of $f$ $x$ from $a$ to $b$ can be approximated as:
where:
:$h = \dfrac {b - a} {3 n}$
== Proof ==
== Also known as ==
== Sources ==
* 
* 
Category:Newton's Three-Eighths Rule
Category:Definite Integrals
Newton-Girard Identities

== Theorem ==
Let $X$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Define:
The Newton-Girard Identities are:
=== Recurrence Formula ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
* 
* : $\S 1.2.9$: Generating Functions: Exercise $10$
* 
* 
* 
* 
Category:Newton-Girard Identities
Category:Elementary Symmetric Functions
Newton-Girard Identities/Examples/Order 1

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{a \mathop \le i \mathop \le b} x_i = S_1$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$
== Proof ==
From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac {\paren {\paren {-1}^{m + 1} S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.
Setting $m = 1$, and setting $x_i := x_{j_1}$:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Examples of Newton-Girard Identities
Newton-Girard Identities/Examples/Order 2

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{a \mathop \le i \mathop < j \mathop \le b} x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 - \paren {\sum_{i \mathop = a}^b {x_i}^2} }$
== Proof ==
From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac {\paren {\paren {-1}^{m + 1} S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.
Setting $m = 2$, and setting $x_i := x_{j_1}, x_j := x_{j_2}$:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Examples of Newton-Girard Identities
Newton-Girard Identities/Examples/Order 3

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{a \mathop \le i \mathop < j \mathop < k \mathop \le b} x_i x_j x_k = \dfrac { {S_1}^3} 6 - \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$
== Proof ==
From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\paren {\paren {-1}^{m + 1} S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.
Setting $m = 3$, and setting $x_i := x_{j_1}, x_j := x_{j_2}, x_k := x_{j_3}$:
We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that:
:$k_1 + 2 k_2 + 3 k_3 = 3$
Thus $\tuple {k_1, k_2, k_3}$ can be:
:$\tuple {3, 0, 0}$
:$\tuple {1, 1, 0}$
:$\tuple {0, 0, 1}$
Hence:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Examples of Newton-Girard Identities
Newton-Girard Identities/Examples/Order 4

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < j_3 \mathop < j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} - \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 - \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$.
== Proof ==
From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\paren {\paren {-1}^{m + 1} S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_j = \ds \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$.
Setting $m = 4$:
We need to find all ordered quadruples of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that:
:$k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$
Thus $\tuple {k_1, k_2, k_3, k_4}$ can be:
:$\tuple {4, 0, 0, 0}$
:$\tuple {2, 1, 0, 0}$
:$\tuple {0, 2, 0, 0}$
:$\tuple {1, 0, 1, 0}$
:$\tuple {0, 0, 0, 1}$
Hence:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Examples of Newton-Girard Identities
Newton-Girard Identities/Historical Note

== Historical Note on Newton-Girard Identities ==
The Newton-Girard Identities were discovered by in $1629$.
 rediscovered them for himself in $1666$, apparently not having known about 's work.
He published them in his of $1707$.
== Sources ==
* 
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Historical Notes
Newton-Girard Identities/Lemma 1

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time, excluding repetitions.
Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
Then:
== Proof ==
This is an instance of Generating Function for Elementary Symmetric Function.
Category:Newton-Girard Identities
Newton-Girard Identities/Lemma 2

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time, excluding repetitions.
For $r \in \Z_{> 0}$, let:
:$S_r = \ds \sum_{j \mathop = a}^b {x_j}^r$
Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
Then:
== Proof ==
Category:Newton-Girard Identities
Newton-Girard Identities/Proof 1

== Theorem ==
== Proof ==
=== Outline ===
The proof is divided into three cases: $k < n$, $k = n$ and $k > n$.
The tools are Viète's Formulas, Recursion Property of Elementary Symmetric Function, telescoping sums and homogeneous functions of degree $k$.
=== Details ===
Discovery of formulas (1)-(2) has humble beginnings:
Create $n$ equations all with zero by substitution $x = x_j$ in (3), $1 \le j \le n$.
:$\ds 0 = \sum_{i \mathop = 0}^n \paren {-1}^{n-i} {\map {e_{n - i} } X} x_j^i$
Then multiply the $j$th equation by $x_j^r$ and add the $n$ equations, for $1 \le j \le n$:
Case $r = 0$ in (4) gives (1) for $k = n$, by isolating term $n \map {e_n} X$.
Case $r > 0$ in (4) gives (2) by re-indexing: $m = k - n + i$ with $m = k - n$ to $k$.
Case $k < n$ in (1), not yet discussed, does not use multiply and add as in (4).
The key ingredient:
The of (1) is:
It remains to prove (7) matches (1):
Define:
:$\map {y_i} t = t\, x_i, 1 \le i \le n$
:$X_t = \set {\map {y_1} t, \ldots, \map {y_n} t}$
:$\map f t = \map {e_k} {tx_1, \ldots, tx_n}$.
Then there are two equations (9)-(10) for $\map {f'} t$:
Let $t = 1$ in (9) and (10) to prove (8).
Category:Newton-Girard Identities
Newton-Girard Identities/Proof 2

== Theorem ==
== Proof ==
=== Outline ===
Calculus is used to prove identities (1) and (2) in a single effort. 
The tools are Viète's Formulas, the calculus derivative of powers $x^n$ and logarithm $\ln \size x$, Maclaurin series expansion coefficients, mathematical induction, and Leibniz's Rule in One Variable.
=== Lemma 1 ===
Proof of Lemma 1
Begin with:
Change variables in $(11)$: 
:$x = -1/z$
Details: Generating Function for Elementary Symmetric Function.
=== Lemma 2 ===
Denote by $D^k \map f z$ the $k$th derivative of $\map f z$.
Let:
Then:
Proof of Lemma 2
Then identity $(12)$ holds by Maclaurin series expansion applied to polynomial $G$.
Identity $(13)$ will be proved after mathematical induction establishes $(14)$ infra. 
Let $\map {\mathbf P} m$ be the statement:
Basis for the induction: $m=0$
By calculus and the definition of $G$:
Then $\map {\mathbf P} 0$ is true.
Induction step $\map {\mathbf P} m$ implies $\map {\mathbf P} {m+1}$:
Simplify to prove $\map {\mathbf P} {m + 1}$ is true. 
The induction is complete.
To prove equation (13), first let $z = 0$ in equation (14). 
Divide by $m!$ to isolate $\map {p_{m + 1} } X$, which proves (13).
=== Lemma 3 ===
Proof of Lemma 3
Begin with $D \map G z = {\map F z} {\map G z}$ and differentiate $m$ times on variable $z$:
Proof of the Theorem
To prove (1), start with equation (15) in Lemma 3.
Change indices via equations $m + 1 = k$, $r + 1 = i$. 
The summation is from $i = 0 + 1$ to $i = m + 1$, which gives range $i = 1$ to $k$. 
Subscript $m - r$ equals $k - 1- i + 1$, which simplifies to $k - i$.
Then:
To prove (2), assume $k > n \ge 1$ and $X = \set {x_1, \ldots, x_n}$.
Equation (16) implies:
Then (2) holds.
Category:Newton-Girard Identities
Newton-Girard Identities/Recurrence Formula

== Theorem ==
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time, excluding repetitions.
For $r \in \Z_{> 0}$, let:
:$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$
A recurrence relation for $h_n$ can be given as:
for $n \ge 1$.
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Newton-Girard Identities
Newton-Mercator Series/Examples/2

== Theorem ==
The Newton-Mercator Series for $x = 1$ converges to the natural logarithm of $2$:
This real number is known as Mercator's constant.
== Proof ==
From the definition of the Newton-Mercator Series:
This is valid for $-1 < x \le 1$.
Setting $x = 1$:
The fact that it does indeed converge is shown in Alternating Harmonic Series is Conditionally Convergent.
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.14$
* : $0 \cdotp 693 \, 147 \, 180 \, 559 \, 945 \, 309 \, 417 \, 232 \, 121 \, 458 \, 176 \, 568 \, 075 \, 500 \, 134 \, 360 \ldots$
* : $\S 1.3.2$: Power series
* : $0 \cdotp 69314 \, 71805 \, 59945 \, 30941 \, 72321 \, 21458 \, 17656 \, 80755 \, 00134 \, 360 \ldots$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.14.$
Category:Examples of Power Series
Category:Logarithms
Category:Reciprocals
Category:Newton-Mercator Series
Newton Divided Difference Interpolation Formula/Examples/f of 0, 1

== Example of Newton Divided Difference Interpolation Formula ==
:$f \sqbrk {0, 1} = \dfrac {\map f {x_1} - \map f {x_0} } {x_1 - x_0}$
== Sources ==
* 
* 
Category:Examples of Newton Divided Difference Interpolation Formula
Newton Divided Difference Interpolation Formula/Table

== Newton Divided Difference Interpolation Formula: Table ==
When manual calculation was the only game in town, it was de rigueur to build the following triangular table of coefficients, to reduce repetitive calculation.
For example, for $n = 3$ the table would have the form:
:$\begin {array} {l|llll} x_0 & f \sqbrk {x_0} & & \\
x_1 & f \sqbrk {x_1} & f \sqbrk {x_0, x_1} & \\
x_2 & f \sqbrk {x_2} & f \sqbrk {x_1, x_2} & f \sqbrk {x_0, x_1, x_2} & & \\
x_3 & f \sqbrk {x_3} & f \sqbrk {x_2, x_3} & f \sqbrk {x_1, x_2, x_3} & f \sqbrk {x_0, x_1, x_2, x_3} & & \\
\end {array}$
to which the computer would calculate each entry from the ones immediately $\gets$ and $\nwarrow$.
This of course can be translated into whatever contemporary software structure is constructed in the software library of your computer language of choice.
== Sources ==
* 
* 
Category:Examples of Newton Divided Difference Interpolation Formula
Newtonian Potential/Examples/Electric Charge

== Example of Newtonian Potential ==
The electric potential field arising from a point charge is an instance of a Newtonian potential.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $9$. The Vector Field $\map \grad {k / r}$
Category:Newtonian Potentials
Category:Electrostatics
Newtonian Potential/Examples/Gravitational Field

== Example of Newtonian Potential ==
The gravitational field arising from a point mass is an instance of a Newtonian potential.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $9$. The Vector Field $\map \grad {k / r}$
Category:Newtonian Potentials
Category:Gravitation
Newtonian Potential/Examples/Point Sink of Fluid Flow

== Example of Newtonian Potential ==
A point sink in a body of incompressible fluid acts as a Newtonian potential.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $9$. The Vector Field $\map \grad {k / r}$
Category:Newtonian Potentials
Category:Fluid Mechanics
Newtonian Potential satisfies Laplace's Equation

== Theorem ==
Let $R$ be a region of space.
Let $S$ be a Newtonian potential over $R$ defined as:
:$\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$
where:
:$\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
:$r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
:$k$ is some predetermined constant.
Then $S$ satisfies Laplace's equation:
:$\nabla^2 S = 0$
where $\nabla^2$ denotes the Laplacian.
== Proof ==
From Gradient of Newtonian Potential:
:$\grad S = -\dfrac {k \mathbf r} {r^3}$
where $\grad$ denotes the gradient operator.
Then:
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $9$. The Vector Field $\map \grad {k / r}$: $(5.12)$
Category:Laplace's Equation
Category:Newtonian Potentials
Neyman-Pearson Lemma

== Theorem ==
The best critical region of size $\alpha$ for testing a simple null hypothesis $H_0$ against a simple alternative hypothesis $H_1$ based on a likelihood ratio is given by
== Proof ==
== Also see ==
* Definition:Hypothesis Testing
== Historical Note ==
== Sources ==
* 
* 
Category:Neyman-Pearson Lemma
Category:Hypothesis Testing
Nicely Normed Alternative Algebra is Normed Division Algebra

== Theorem ==
$A = \struct {A_F, \oplus}$ be a nicely normed $*$-algebra which is also an alternative algebra.
Then $A$ is a normed division algebra.
== Proof ==
Let $a, b \in A$.
Then all of $a, b, a^*, b^*$ can be generated by $\map \Im a$ and $\map \Im b$.
So as $A$ is an alternative algebra, it follows that $\oplus$ is associative for $a, b, a^*, b^*$.
So:
== Sources ==
* 2002: : The Octonions: 2.2 The Cayley-Dickson Construction
Category:Nicely Normed Star-Algebras
Category:Alternative Algebras
Category:Normed Division Algebras
Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is a nicely normed alternative algebra $A$ is a nicely normed associative algebra.
== Proof ==
Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
The context will make it clear which is meant.
Let $A$ be a nicely normed associative algebra.
Then:
Similarly:
Thus it can be seen that:
:$\paren {\tuple {a, b} \tuple {a, b} } \tuple {c, d} = \tuple {a, b} \paren {\tuple {a, b} \tuple {c, d} }$
Similarly it can be shown that:
:$\paren {\tuple {c, d} \tuple {a, b} } \tuple {a, b} = \tuple {c, d} \paren {\tuple {a, b} \tuple {a, b} }$
and so $A'$ is seen to be an alternative algebra.
It follows from reversing the chain of equalities that if $A'$ is a nicely normed and alternative algebra then $A$ has to be a nicely normed associative algebra.
Then from Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, we have that $A'$ is a nicely normed algebra $A$ is also a nicely normed algebra.
Hence the result.
== Sources ==
* : The Octonions (2002): 2.2 The Cayley-Dickson Construction: Proposition $4$
Category:Cayley-Dickson Construction
Category:Associative Algebras
Category:Nicely Normed Star-Algebras
Category:Alternative Algebras
Nicomachus's Theorem

== Theorem ==
In general:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
In particular, the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.
== Proof 1 ==
== Proof 2 ==
Category:Algebra
Category:Sums of Sequences
Category:Nicomachus's Theorem