nyuuzyou/wikis huggingface dataset:
Value of Vacuum Permeability

== Theorem ==
The value of the vacuum permeability is calculated as:
:$\mu_0 = 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \, \mathrm H \, \mathrm m^{-1}$ (henries per metre)
with a relative uncertainty of $1 \cdotp 5 \times 10^{-10}$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Value of Vacuum Permittivity
=== Interconnection between Vacuum Permittivity and Vacuum Permeability ===
Category:Value of Vacuum Permeability
Category:Vacuum Permeability
Value of Vacuum Permeability/Proof 1

== Theorem ==
== Proof ==
The vacuum permeability is the physical constant denoted $\mu_0$ defined as:
:$\mu_0:= \dfrac {2 \alpha h} {e^2 c}$
where:
:$e$ is the elementary charge
:$\alpha$ is the fine-structure constant
:$h$ is Planck's constant
:$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$
$e$ is defined precisely as:
:$e = 1 \cdotp 60217 \, 6634 \times 10^{-19} \, \mathrm C$ (coulombs)
$\alpha$ has the value determined experimentally as:
:$\alpha \approx 0 \cdotp 00729 \, 73525 \, 693 (11)$
$h$ is defined precisely as:
:$h = 6 \cdotp 62607 \, 015 \times 10^{-34} \, \mathrm J \, \mathrm s$ (joule seconds)
$c$ is defined precisely as:
:$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$ (metres per second)
Hence $\mu_0$ can be calculated as:
Category:Value of Vacuum Permeability
Value of Vacuum Permeability/Proof 2

== Theorem ==
== Proof ==
The vacuum permeability is the physical constant denoted $\mu_0$ defined as:
:$\mu_0 := \dfrac 1 {\varepsilon_0c^2}$
where:
:$\varepsilon_0$ is the vacuum permittivity defined in $\mathrm F \, \mathrm m^{-1}$ (farads per metre)
:$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$
$\varepsilon_0$ has the value determined as:
:$\varepsilon_0 \approx 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$
$c$ is defined precisely as:
:$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$
Hence $\mu_0$ can be calculated as:
Category:Value of Vacuum Permeability
Value of Vacuum Permittivity

== Theorem ==
The value of the vacuum permittivity is calculated as:
:$\varepsilon_0 = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$ (farads per metre)
with a relative uncertainty of $1 \cdotp 5 \times 10^{-10}$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Value of Vacuum Permeability
=== Interconnection between Vacuum Permittivity and Vacuum Permeability ===
Category:Value of Vacuum Permittivity
Category:Vacuum Permittivity
Value of Vacuum Permittivity/Proof 1

== Theorem ==
== Proof ==
The vacuum permittivity is the physical constant denoted $\varepsilon_0$ defined as:
:$\varepsilon_0 := \dfrac {e^2} {2 \alpha h c}$
where:
:$e$ is the elementary charge
:$\alpha$ is the fine-structure constant
:$h$ is Planck's constant
:$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$
$e$ is defined precisely as:
:$e = 1 \cdotp 60217 \, 6634 \times 10^{-19} \, \mathrm C$ (coulombs)
$\alpha$ has the value determined experimentally as:
:$\alpha \approx 0 \cdotp 00729 \, 73525 \, 693 (11)$
$h$ is defined precisely as:
:$h = 6 \cdotp 62607 \, 015 \times 10^{-34} \, \mathrm J \, \mathrm s$ (joule seconds)
$c$ is defined precisely as:
:$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$ (metres per second)
Hence $\varepsilon_0$ can be calculated as:
Category:Value of Vacuum Permittivity
Value of Vacuum Permittivity/Proof 2

== Theorem ==
== Proof ==
The vacuum permittivity is the physical constant denoted $\varepsilon_0$ defined as:
:$\varepsilon_0 := \dfrac 1 {\mu_0 c^2}$
where:
:$\mu_0$ is the vacuum permeability defined in $\mathrm H \, \mathrm m^{-1}$ (henries per metre)
:$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$
$\mu_0$ has the value determined experimentally as:
:$\mu_0 \approx 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \, \mathrm H \, \mathrm m^{-1}$
$c$ is defined precisely as:
:$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$
Hence $\varepsilon_0$ can be calculated as:
Category:Value of Vacuum Permittivity
Value of Vandermonde Determinant/Formulation 1

== Theorem ==
Let $V_n$ be the Vandermonde determinant of order $n$ defined as the following formulation:
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* 
* 
* : Preface to first edition: Prerequisites: $\text {(i)}$
* 
Category:Value of Vandermonde Determinant
Value of Vandermonde Determinant/Formulation 1/Proof 3

== Theorem ==
== Proof ==
Let:
:$V_n = \begin {vmatrix}
 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
 1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1}
\end {vmatrix}$
Start by replacing number $x_n$ in $V_n$ with the unknown $x$.
Thus $V_n$ is made into a function of $x$:
:$\map P x = \begin{vmatrix}
 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
 1 & x & x^2 & \cdots & x^{n - 2} & x^{n - 1}
\end{vmatrix}$
Let $x$ equal a value from the set $\set {x_1, \ldots, x_{n-1} }$.
Then determinant $\map P x$ has equal rows, giving:
Perform row expansion by the last row.
Then $\map P x$ is seen to be a polynomial of degree $n-1$:
:$\map P x = \begin {vmatrix}
 x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
 x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}
\end{vmatrix} + 
\begin{vmatrix}
 1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
 1 & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}
\end {vmatrix} x + \cdots + 
\begin {vmatrix}
 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2}
\end{vmatrix} x^{n - 1}$
By the Polynomial Factor Theorem:
:$\map P x = \map C {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$
where $C$ is the leading coefficient (with $x^{n - 1}$ power).
Thus:
:$\map P x = V_{n - 1} \paren {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$
which by evaluating at $x = x_n$ gives: 
:$V_n = V_{n - 1} \paren {x_n - x_1} \paren {x_n - x_2} \dotsm \paren {x_n - x_{n - 1} }$
Repeating the process:
which establishes the solution.
Category:Value of Vandermonde Determinant
Value of Vandermonde Determinant/Formulation 1/Proof 4

== Theorem ==
== Proof ==
Let:
:$V_n = \begin {vmatrix}
1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1}
\end {vmatrix}$
Let $\map f x$ be an arbitrary monic polynomial of degree $n - 1$:
:$\ds \map f x = x^{n - 1} + \sum_{i \mathop = 0}^{n - 2} a_i x^i$
Apply elementary column operations to $V_n$ repeatedly to show:
:$V_n = W$
where:
:$W = \begin {vmatrix}
1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & \map f {x_1} \\
1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & \map f {x_2} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & \map f {x_n}
\end {vmatrix}$
Select a specific degree $n - 1$ monic polynomial:
:$\ds \map f x = \prod_{k \mathop = 1}^{n - 1} \paren {x - x_k}$
The selected polynomial is zero at all values $x_1, \ldots, x_{n - 1}$.
Then the last column of $W$ is all zeros except the entry $\map f {x_n}$.
Expand $\map \det W$ by cofactors along the last column to prove:
For $n \ge 2$, let $\map P n$ be the statement:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
Mathematical induction will be applied.
=== Basis for the Induction ===
By definition, determinant $V_1 = 1$.
To prove $\map P 2$ is true, use equation $(1)$ with $n = 2$:
:$V_2 = \paren {x_2 - x_1} V_1$
This is the basis for the induction.
=== Induction Step ===
This is the induction step:
Let $\map P n$ is be assumed true.
We are to prove that $\map P {n + 1}$ is true.
As follows:
Thus $\map P {n + 1}$ has been shown to be true.
The induction is complete.
Category:Value of Vandermonde Determinant
Value of Vandermonde Determinant/Formulation 2

== Theorem ==
Let $V_n$ be the Vandermonde determinant of order $n$ defined as the following formulation:
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le j \mathop \le n} x_j \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
== Proof ==
The proof follows directly from that for Value of Vandermonde Determinant/Formulation 1 and the result Determinant with Row Multiplied by Constant.
== Sources ==
Category:Value of Vandermonde Determinant
Value of b for b by Logarithm Base b of x to be Minimum

== Theorem ==
Let $x \in \R_{> 0}$ be a (strictly) positive real number.
Consider the real function $f: \R_{> 0} \to \R$ defined as:
:$\map f b := b \log_b x$
$f$ attains a minimum when
:$b = e$
where $e$ is Euler's number.
== Proof ==
From Derivative at Maximum or Minimum, when $f$ is at a minimum, its derivative $\dfrac \d {\d b} f$ will be zero.
Let $y = \map f b$.
We have:
Thus:
To determine that $f$ is a minimum at this point, we differentiate again $b$:
Setting $b = e$ gives:
:$\valueat {\dfrac {\d^2 y} {\d b^2} } {b \mathop = e} = \dfrac {\ln x} e \dfrac {\paren {1 - 2 \paren 0} } 1$
which works out to be (strictly) positive.
From Real Function with Strictly Positive Second Derivative is Strictly Convex, $f$ is strictly convex at this point.
Thus $f$ is a minimum.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $29$
Category:Logarithms
Values of n for which Integer Solutions to n = x^3 + y^3 + 2 z^3 are Unknown

== Open Question ==
For the following values of $n$, it is not known whether the indeterminate equation:
:$n = x^3 + y^3 + 2 z^3$
has any integer solutions:
:$76, 148, 183, 230, 356, 418, 428, 445, 482, 491, 580, 671, 788, 931, 967$
== Historical Note ==
== Sources ==
* 
* : $76$
Category:Open Questions
Values of n for which Integer Solutions to n = x^3 + y^3 + 2 z^3 are Unknown/Historical Note

== Historical Note on Values of $n$ for which Integer Solutions to $n = x^3 + y^3 + 2 z^3$ are Unknown ==
 reports that the given list of positive integers are the only ones known under $1000$.
The author of this webpage has been unable to determine whether this statement is still accurate.
Category:Historical Notes
Vandermonde Determinant/Examples/Order 4

== Example of Vandermonde Determinant ==
Consider the Vandermonde determinant of order $4$:
:$V_n = \begin {vmatrix}
 1 & 1 & 1 & 1 \\
 x_1 & x_2 & x_3 & x_4 \\
 {x_1}^2 & {x_2}^2 & {x_3}^2 & {x_4}^2 \\
 {x_1}^3 & {x_2}^3 & {x_3}^3 & {x_4}^3
\end {vmatrix}$
Its value is:
:$\paren {x_2 - x_1} \paren {x_3 - x_1} \paren {x_3 - x_2} \paren {x_4 - x_1} \paren {x_4 - x_2} \paren {x_4 - x_3}$
== Sources ==
* 
Category:Examples of Vandermonde Determinants
Vandermonde Matrix Identity for Cauchy Matrix

== Theorem ==
Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct in matrix
Then:
Definitions of Vandermonde matrices $V_x$, $V_y$ and diagonal matrices $P$, $Q$:
:$V_x = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & \vdots & \ddots & \vdots \\
{x_1}^{n - 1} & {x_2}^{n - 1} & \cdots & {x_n}^{n - 1} \\
\end {pmatrix}, \quad 
V_y = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
y_1 & y_2 & \cdots & y_n \\
\vdots & \vdots & \ddots & \vdots \\
{y_1}^{n - 1} & {y_2}^{n - 1} & \cdots & {y_n}^{n - 1} \\
\end {pmatrix}$ Vandermonde matrices
:$P = \begin {pmatrix}
\map {p_1} {x_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map {p_n} {x_n} \\
\end {pmatrix}, \quad
Q = \begin {pmatrix}
\map p {y_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map p {y_n} \\
\end {pmatrix}$ Diagonal matrices
Definitions of polynomials $p, p_1, \ldots, p_n$:
:$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}$
:$\ds \map {p_k} x = \dfrac {\map p x} {x - x_k} = \prod_{i \mathop = 1, i \mathop \ne k}^n \paren {x - x_i}$, $1 \mathop \le k \mathop \le n$
== Proof ==
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.
For $1 \le i \le n$ express polynomial $p_i$ as:
:$\ds \map {p_i} x = \sum_{k \mathop = 1}^n a_{i k} x^{k - 1}$
Then:
Use second equation $\map {p_i} {y_j} = \dfrac {\map p {y_j} } {y_j - x_i}$:
== Examples ==
== Also see ==
* Definition:Vandermonde Determinant
* Inverse of Vandermonde Matrix
* Hilbert Matrix is Cauchy Matrix
* Inverse of Cauchy Matrix
* Inverse of Hilbert Matrix
* Value of Cauchy Determinant
* Sum of Elements in Inverse of Cauchy Matrix
* Sum of Elements in Inverse of Hilbert Matrix
== Historical Note ==
== Sources ==
* : Chapter $\text{VI}$. $47$: Alternant Matrices and Determinants
* 
* : Matrix Algebra Ch2 and Determinants Ch3
Category:Vandermonde Matrix Identity
Category:Cauchy Matrices
Vandermonde Matrix Identity for Cauchy Matrix/Examples/3x3

== Example of Vandermonde Matrix Identity for Cauchy Matrix ==
Illustrate $3 \times 3$ case for Vandermonde Matrix Identity for Cauchy Matrix and Value of Cauchy Determinant.
Let $C$ denote the Cauchy matrix of order $3$:
:$C = \begin {pmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \dfrac 1 {x_1 - y_3} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \dfrac 1 {x_2 - y_3} \\
\dfrac 1 {x_3 - y_1} & \dfrac 1 {x_3 - y_2} & \dfrac 1 {x_3 - y_3} \\
\end{pmatrix}$
where the values in $\set {x_1, x_2, x_3, y_1, y_2, y_3}$ are assumed to be distinct.
Then:
== Proof ==
Define Vandermonde matrices
Define polynomials:
 
Define invertible diagonal matrices:
Then:
Determinant of Diagonal Matrix gives:
Value of Vandermonde Determinant gives:
Determinant of Matrix Product and Definition:Inverse Matrix give:
Then:
Insert the four determinant equations and simplify to obtain the equation for $3 \times 3$ $\map \det C$.
Category:Examples of Vandermonde Matrix Identity
Vandermonde Matrix Identity for Cauchy Matrix/Examples/nxn

== Example of Vandermonde Matrix Identity for Cauchy Matrix ==
The methods of the $3 \times 3$ example apply unchanged for the general $n \times n$ Cauchy matrix:
Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct. Then:
:$\map \det {\begin{smallmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \cdots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
\end{smallmatrix} } = 
\paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_i - y_j} }
{\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i - y_j} }$ Value of Cauchy Determinant
Assume values $\set {x_1, \ldots, x_n, -y_1, \ldots, -y_n}$ are distinct, then replace in the preceding equation $y_i$ by $-y_i$, $1 \le i \le n$:
:$\map \det {\begin{smallmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \cdots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\
\end{smallmatrix} } =
\paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_j - y_i} }
{\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i + y_j} }$ Value of Cauchy Determinant
Category:Examples of Vandermonde Matrix Identity
Vandermonde Matrix Identity for Hilbert Matrix

== Theorem ==
Define polynomial root sets $\set {1, 2, \ldots, n}$ and $\set {0, -1, \ldots, -n + 1}$ for Definition:Cauchy Matrix.
Let $H$ be the Hilbert matrix of order $n$:
:$H = \begin {pmatrix}
 1 & \dfrac 1 2 & \cdots & \dfrac 1 n \\
 \dfrac 1 2 & \dfrac 1 3 & \cdots & \dfrac 1 {n + 1} \\
 \vdots & \vdots & \cdots & \vdots \\
 \dfrac 1 n & \dfrac 1 {n + 1} & \cdots & \dfrac 1 {2 n - 1} \end {pmatrix}$
Then from Vandermonde Matrix Identity for Cauchy Matrix and Hilbert Matrix is Cauchy Matrix:
:$H = -P V_x^{-1} V_y Q^{-1}$
where $V_x$, $V_y$ are Vandermonde matrices:
:$V_x = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
1 & 2 & \cdots & n \\
\vdots & \vdots & \ddots & \vdots \\
1 & 2^{n - 1} & \cdots & n^{n -1 } \\
\end {pmatrix}, \quad 
V_y = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
0 & -1 & \cdots & -n + 1 \\
\vdots & \vdots & \ddots & \vdots \\
0 & \paren {-1}^{n - 1} & \cdots & \paren {-n + 1}^{n - 1} \\
\end {pmatrix}$
and $P$, $Q$ are diagonal matrices:
:$P = \begin {pmatrix}
\map {p_1} 1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map {p_n} n \\
\end {pmatrix}, \quad
Q = \begin {pmatrix}
\map p 0 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map p {-n + 1} \\
\end {pmatrix}$
Definitions of polynomials $p$, $p_1$, $\ldots$, $p_n$:
:$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - i}$
:$\ds \map {p_k} x = \dfrac {\map p x} {x - k} = \prod_{i \mathop = 1, i \mathop \ne k}^n \, \paren {x - i}$, $1 \mathop \le k \mathop \le n$
== Proof ==
Apply Vandermonde Matrix Identity for Cauchy Matrix and Hilbert Matrix is Cauchy Matrix.
Matrices $V_x$ and $V_y$ are invertible by Inverse of Vandermonde Matrix.
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.
== Examples ==
== Also see ==
* Hilbert Matrix is Cauchy Matrix
* Definition:Vandermonde Determinant
* Inverse of Vandermonde Matrix
* Inverse of Hilbert Matrix
* Sum of Elements in Inverse of Hilbert Matrix
== Sources ==
* : Chapter $\text{VI}$. $47$: Alternant Matrices and Determinants
* 
Category:Vandermonde Matrix Identity
Category:Hilbert Matrices
Vandermonde Matrix Identity for Hilbert Matrix/Examples/3x3

== Example of Vandermonde Matrix Identity for Hilbert Matrix ==
Define polynomial root sets $\set {1, 2, 3}$ and $\set {0, -1, -2}$ for Definition:Cauchy Matrix because Hilbert Matrix is Cauchy Matrix.
Illustrate $3\times 3$ case for Vandermonde Matrix Identity for Hilbert Matrix and value of Hilbert matrix determinant:
 
Then:
Details: 
Define Vandermonde matrices
:$\ds V_x = \begin{pmatrix}
1 & 1 & 1 \\
1 & 2 & 3 \\
1^2 & 2^2 & 3^2 \\
\end{pmatrix}, \quad 
V_y = \begin {pmatrix}
1 & 1 & 1 \\
0 & -1 & -2 \\
0 & \paren {-1}^2 & \paren {-2}^2 \\
\end {pmatrix}$
Define polynomials:
:$\ds \map p x = \paren {x - 1} \paren {x - 2} \paren {x - 3}$
:$\ds \map {p_1} x = \paren {x - 2} \paren {x - 3},
\quad \map {p_2} x = \paren {x - 1} \paren {x - 3},
\quad \map {p_3} x = \paren {x - 1} \paren {x - 2}$
 
Define invertible diagonal matrices:
:$\ds P = \begin {pmatrix}
\map {p_1} 1 & 0 & 0 \\
0 & \map {p_2} 2 & 0 \\
0 & 0 & \map {p_3} 3 \\
\end {pmatrix}, \quad
Q = \begin {pmatrix}
\map p 0 & 0 & 0 \\
0 & \map p {-1} & 0 \\
0 & 0 & \map p {-2} \\
\end {pmatrix}$
Then:
:$\ds P = \begin {pmatrix}
\paren {1 - 2} \paren {1 - 3} & 0 & 0 \\
0 & \paren {2 - 1} \paren {2 - 3} & 0 \\
0 & 0 & \paren {3 - 1} \paren {3 - 2} \\
\end {pmatrix}, \quad
Q = \begin {pmatrix}
 \paren {0 - 1} \paren {0 - 2} \paren {0 - 3} & 0 & 0 \\
0 & \paren {-1 - 1} \paren {-1 - 2} \paren {-1 - 3} & 0 \\
0 & 0 & \paren {-2 - 1} \paren {-2 - 2} \paren {-2 - 3} \\
\end {pmatrix}$
Determinant of Diagonal Matrix implies
:$\ds \map \det P = \paren {1 - 2} \paren {1 - 3} \paren {2 - 1}\paren {2 - 3} \paren {3 - 1} \paren {3 - 2} = -4$
:$\ds \map \det Q = \paren {0 - 1} \paren {0 - 2} \paren {0 - 3} \paren {-1 - 1} \paren {-1 - 2} \paren {-1 - 3} \paren {-2 - 1} \paren {-2 - 2} \paren {-2 - 3} = -8640$
Value of Vandermonde Determinant gives:
:$\ds \map \det {V_x} = \paren {3 - 2} \paren {3 - 1} \paren {2 - 1} = 2$
:$\ds \map \det {V_y} = \paren {-2 - \paren {-1} } \paren {-2 - 0} \paren {-1 - 0} = -2$
Determinant of Matrix Product and Definition:Inverse Matrix imply
:$\map \det {V_x^{-1} } = \dfrac 1 {\map \det {V_x} }$
:$\map \det {Q^{-1} } = \dfrac 1 {\map \det Q}$
Then:
Category:Examples of Vandermonde Matrix Identity
Vanishing Distributional Derivative of Distribution implies Distribution is Constant

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Let $\mathbf 0$ be the zero distribution.
Suppose the distributional derivative of $T$ vanishes:
:$\ds \dfrac \d {\d x} T = \mathbf 0$
Then $T$ is a constant distribution.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
Then:
Hence:
:$\set {\phi' : \phi \in \map \DD \R} \subseteq \ker T$
where $\ker$ denotes the kernel.
Let $\mathbf 1$ be a constant mapping such that $\mathbf 1 : \R \to 1$.
Then the associated distribution reads:
:$\ds \map {T_{\mathbf 1}} \phi = \int_{-\infty}^\infty \map \phi x \rd x$
Furthermore:
We have that Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space.
Let $V = \map \DD \R$, $L = T$ and $\ell = T_{\mathbf 1}$.
By Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional‎:
:$\exists c \in \C : T = c T_{\mathbf 1}$ 
By definition of multiplication of a distribution by a smooth function:
:$c T_{\mathbf 1} = T_c$
== Sources ==
* : Chapter $\S 6.2$: Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Vanishing Ideal of Larger Subset of Affine Space is Smaller

== Theorem ==
Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $S \subseteq T \subseteq \mathbb A^n_k$.
Then:
:$\map I S \supseteq \map I T$
where $\map I S$ and $\map I T$ denote the vanishing ideals of $S$ and $T$, respectively.
== Proof ==
Category:Algebraic Geometry
Vanishing Ideal of Zero Locus of Ideal is Radical

== Theorem ==
Let $k$ be an algebraically closed field.
Let $n \ge 0$ be a natural number.
Let $k \sqbrk {X_1, \ldots, X_n}$ be the polynomial ring in $n$ variables over $k$.
Let $\mathfrak a \subseteq k \sqbrk {X_1, \ldots, X_n}$ be an ideal.
Then:
:$\map I {\map V {\mathfrak a} } = \map \Rad {\mathfrak a}$
where:
:$\map V \cdot$ denotes the zero locus
:$\map I \cdot$ denotes the vanishing ideal
:$\map \Rad \cdot$ denotes the radical
== Proof ==
This is exactly Hilbert's Nullstellensatz.
Category:Algebraic Geometry
Vanishing of Quasi-Coherent Sheaf Cohomology of Affine Scheme

== Theorem ==
Let $X = \Spec A$ be the spectrum of a commutative ring $A$.
Let $\FF$ be a quasi-coherent sheaf on $X$.
Then for all $i \in \Z$ with $i > 0$ the $i$-th sheaf cohomology $\map {H^i} {X, \FF} = 0$.
== Proof ==
== Sources ==
* $(1.3.1)$
* Remark $\text{III}.3.5.1$
Category:Algebraic Geometry
Category:Schemes
Category:Sheaf Cohomologies