nyuuzyou/wikis huggingface dataset:
Derivatives of Moment Generating Function of Logistic Distribution

== Theorem ==
Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
for some $\mu \in \R, s \in \R_{> 0}$.
Let $\size t < \dfrac 1 s$.
Let $M_X$ denote the moment generating function of $X$.
The $n$th derivative of $M_X$ is given by:
:$\ds {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ 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:
:$\ds {M_X}^{\paren k} = \map \exp {\mu t} \sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
from which it is to be shown that:
:$\ds {M_X}^{\paren {k + 1} } = \map \exp {\mu t} \sum_{r \mathop = 0}^{k + 1} \paren {-1}^r \dbinom {k + 1} r \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
Category:Moment Generating Function of Logistic Distribution
Derivatives of PGF of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin{cases} p & : k = 1 \\ 0 & : k > 1 \end{cases}$
== Proof 1 ==
== Proof 2 ==
Category:Bernoulli Distribution
Category:Derivatives of PGFs
Category:Derivatives of PGF of Bernoulli Distribution
Derivatives of PGF of Bernoulli Distribution/Proof 1

== Theorem ==
== Proof ==
The Probability Generating Function of Bernoulli Distribution is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
We have that for a given Bernoulli distribution, $p$ and $q$ are constant.
So, from Derivative of Constant, Sum Rule for Derivatives, Derivative of Identity Function and Derivative of Constant Multiple:
:$\dfrac {\d} {\d s} \map {\Pi_X} s = p$
Again, $p$ is constant, so from Derivative of Constant:
:$\dfrac {\d} {\d s} p = 0$
Higher derivatives are also of course zero, also from Derivative of Constant.
Category:Derivatives of PGF of Bernoulli Distribution
Derivatives of PGF of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin {cases}
n^{\underline k} p^k \paren {q + p s}^{n-k} & : k \le n \\
0 & : k > n
\end {cases}$
where:
:$n^{\underline k}$ is the falling factorial
:$q = 1 - p$
== Proof ==
The Probability Generating Function of Binomial Distribution is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Derivatives of Function of $a x + b$:
:$\map {\dfrac {\d^k} {\d s^k} } {\map f {q + p s} } = p^k \dfrac {\d^k} {\d z^k} \paren {\map f z}$
where $z = q + p s$.
Here we have that $\map f z = z^n$.
From Nth Derivative of Mth Power:
:$\dfrac {\d^k} {\d z^k} z^n = \begin {cases}
n^{\underline k} z^{n - k} & : k \le n \\
0 & : k > n
\end {cases}$
So putting it together:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin {cases}
n^{\underline k} p^k \paren {q + p s}^{n - k} & : k \le n \\
0 & : k > n
\end {cases}$
== Sources ==
* 
* 
Category:Binomial Distribution
Category:Derivatives of PGFs
Derivatives of PGF of Discrete Uniform Distribution

== Theorem ==
Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the derivatives of the PGF of $X$ $s$ are:
$\quad\dfrac {\d^m} {\d s^m} \map {\Pi_X} s = \begin{cases}
\ds \dfrac 1 n \sum_{k \mathop = m}^n k^{\underline m} s^{k - m} & : m \le n \\
0 & : k > n
\end{cases}$
where $k^{\underline m}$ is the falling factorial.
== Proof ==
The Probability Generating Function of Discrete Uniform Distribution is:
:$\map {\Pi_X} s = \dfrac {s \paren {1 - s^n} } {n \paren {1 - s} } = \dfrac 1 n \ds \sum_{k \mathop = 1}^n s^k$
From Nth Derivative of Mth Power:
$\quad\dfrac {\d^k} {\d s^k} s^n = \begin {cases}
n^{\underline k} s^{n - k} & : k \le n \\
0 & : k > n
\end {cases}$
The result follows.
Category:Discrete Uniform Distribution
Category:Derivatives of PGFs
Derivatives of PGF of Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = \dfrac {q p^n n!} {\paren {1 - p s}^{n + 1} }$
where $q = 1 - p$.
== Proof ==
The Probability Generating Function of Geometric Distribution is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Derivatives of Function of $a x + b$, we have that:
:$\map {\dfrac {\d^n} {\d s^n} } {\map f {1 - p s} } = \paren {-p}^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = 1 - ps$.
Here we have that $\map f z = \dfrac q z$.
From Nth Derivative of Reciprocal of Mth Power:
:$\dfrac {\d^n} {\d z^n} \dfrac q z = q \dfrac {\d^n} {\d z^n} \dfrac 1 z = q \dfrac {\paren {-1}^n n!} {z^{n + 1} }$
where $n!$ denotes $n$ factorial.
So putting it together:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = q \paren {-p}^n \dfrac {\paren {-1}^n n!} {\paren {1 - p s}^{n + 1} }$
whence (after algebra):
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = \dfrac {q p^n n!} {\paren {1 - p s}^{n + 1} }$
Category:Geometric Distribution
Category:Derivatives of PGFs
Derivatives of PGF of Negative Binomial Distribution/Type 1

== Theorem ==
Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = ...$
== Proof ==
The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given type $1$ negative binomial distribution , $n, p$ and $q$ are constant.
Category:Derivatives of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Derivatives of PGFs
Derivatives of PGF of Negative Binomial Distribution/Type 2

== Theorem ==
Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n + k}$
where:
:$n^{\overline k}$ is the rising factorial: $n^{\overline k} = n \paren {n + 1} \paren {n + 2} \cdots \paren {n + k - 1}$
:$q = 1 - p$
== Proof ==
Proof by induction:
The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\ds \map {\Pi_X} s = \paren {\frac q {1 - p s} }^n$
For all $k \in \N_{> 0}$, let $\map P k$ be the proposition:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n + k}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\dfrac \d {\d s} \map {\Pi_X} s = \dfrac {n^{\overline 1} p} q \paren {\dfrac q {1 - p s} }^{n + 1}$
which is proved in First Derivative of PGF of Negative Binomial Distribution: Type 2.
Note that from Number to Power of One Rising is Itself:
:$n^\overline 1 = n$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j + 1}$ is true.
So this is our induction hypothesis:
:$\dfrac {\d^j} {\d s^j} \map {\Pi_X} s = \dfrac {n^{\overline j} p^j} {q^j} \paren {\dfrac q {1 - p s} }^{n + j}$
Then we need to show:
:$\dfrac {\d^{j + 1} } {\d s^{j + 1} } \map {\Pi_X} s = \dfrac {n^{\overline {j + 1}} p^{j + 1} } {q^{j + 1} } \paren {\dfrac q {1 - p s} }^{n + j + 1}$
=== Induction Step ===
This is our induction step:
So $\map P j \implies \map P {j + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall k \in \N_{> 0}: \dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n + k}$
Category:Derivatives of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Derivatives of PGFs
Derivatives of PGF of Poisson Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\dfrac {d^k} {\d s^k} \, \map {\Pi_X} s = \lambda^k e^{- \lambda \paren {1 - s} }$
== Proof ==
The Probability Generating Function of Poisson Distribution is:
We have that for a given Poisson distribution, $\lambda$ is constant.
From Higher Derivatives of Exponential Function, we have that:
:$\dfrac {\d^k} {\d s^k} \paren {e^{\lambda s} } = \lambda^k e^{\lambda s}$
Thus we have:
Hence the result.
Category:Poisson Distribution
Category:Derivatives of PGFs
Derivatives of PGF of Shifted Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the derivatives of the PGF of $X$ $s$ are:
:$\map {\dfrac {\d^n} {\d s^n} } {\map {\Pi_X} s} = \dfrac {p q^{n - 1} \paren {n - 1}!} {\paren {1 - q s}^{n + 1} }$
where $q = 1 - p$.
== Proof ==
The Probability Generating Function of Shifted Geometric Distribution is:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
First we need to obtain the first derivative:
From Derivatives of Function of $a x + b$:
:$\map {\dfrac {\d^n} {\d s^n} } {\map f {1 - q s} } = \paren {-q}^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = 1 - q s$.
Here we have that:
:$\map f z = p \dfrac 1 {z^2}$
From Nth Derivative of Reciprocal of Mth Power:
:$\dfrac {\d^{n - 1} } {\d z^{n - 1} } \dfrac 1 {z^2} = \dfrac {\paren {-1}^{n - 1} 2^{\overline {n - 1} } } {z^{\paren {n - 1} + 2} }$
where $\overline {n - 1}$ denotes the rising factorial.
Note that we consider the $n-1$th derivative because we have already taken the first one.
Also note that $2^{\overline {n - 1} } = 1^{\overline {n - 1} } = \paren {n - 1}!$
So putting it together:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = p \paren {-q}^{n - 1} \dfrac {\paren {-1}^{n - 1} \paren {n - 1}!} {\paren {1 - q s}^{n + 1} }$
whence (after algebra):
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = \dfrac {p q^{n - 1} \paren {n - 1}!} {\paren {1 - q s}^{n + 1} }$
Category:Geometric Distribution
Category:Derivatives of PGFs
Derivatives of Probability Generating Function at One

== Theorem ==
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the $n$th derivative of $\map {\Pi_X} s$ at $s = 1$ is given by:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$
for $n = 1, 2, \ldots$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\dfrac \d {\d s} \map {\Pi_X} 1 = \expect X$
This is demonstrated in Expectation of Discrete Random Variable from PGF.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1})$ is true.
So this is our induction hypothesis:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k + 1} }$
Then we need to show:
:$\dfrac {\d^{k + 1} } {\d s^{k + 1} } \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: \dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$
== Sources ==
* : $\S 4.3$: Moments: Theorem $4 \ \text{B}$
* 
* 
Category:Probability Generating Functions
Derivatives of Real Area Hyperbolic Functions

== Theorem ==
This page gathers together derivatives of real area hyperbolic functions.
=== Derivative of Real Area Hyperbolic Sine ===
=== Derivative of Real Area Hyperbolic Cosine ===
=== Derivative of Real Area Hyperbolic Tangent ===
=== Derivative of Real Area Hyperbolic Cotangent ===
=== Derivative of Real Area Hyperbolic Secant ===
=== Derivative of Real Area Hyperbolic Cosecant ===
Category:Derivatives
Category:Derivatives of Inverse Hyperbolic Functions
Derivatives of Trigonometric Functions

== Theorem ==
This page gathers together derivatives of trigonometric functions.
=== Derivative of Sine Function ===
=== Derivative of Cosine Function ===
=== Derivative of Tangent Function ===
=== Derivative of Cotangent Function ===
=== Derivative of Secant Function ===
=== Derivative of Cosecant Function ===
Category:Derivatives
Category:Derivatives of Trigonometric Functions
Derivatives of Unit Vectors in Polar Coordinates

== Theorem ==
Consider a particle $p$ moving in the plane.
Let the position of $p$ be given in polar coordinates as $\polar {r, \theta}$.
Let:
:$\mathbf u_r$ be the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ be the unit vector in the direction of the angular coordinate of $p$
Then the derivative of $\mathbf u_r$ and $\mathbf u_\theta$ $\theta$ can be expressed as:
== Proof ==
By definition of sine and cosine:
where $\mathbf i$ and $\mathbf j$ are the unit vectors in the $x$-axis and $y$-axis respectively.
:
Differentiating $(1)$ and $(2)$ $\theta$ gives:
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation:
Category:Polar Coordinates
Derived Set Preserves Set Inclusion

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $B \subseteq A \subseteq S$.
Then:
:$B' \subseteq A'$
where $A'$ and $B'$ are the derived sets in $T$ of $A$ and $B$ respectively.
== Proof ==
Let: 
:$x \in B'$
By the definition of derived set:
:$x$ is a limit point of $B$.
Note that $B \subseteq A$.
From Limit Point of Subset is Limit Point of Set:
:$x$ is a limit point of $A$.
So, $x \in A'$.
That is:
:$B' \subseteq A'$
Category:Limit Points
Derived Set in T1 Space is Closed

== Theorem ==
Let $\struct {X, \tau}$ be a $T_1$ space.
Let $S \subseteq X$.
Let $S'$ be the derived set of $S$.
Then $S'$ is closed.
== Proof ==
Let $x \in S$.
Let $U$ be an open neighborhood of $x$.
Then by the definition of derived set, $U$ contains an element $y$ of $S'$ such that $x \ne y$.
Then $U \setminus \set x$ is an open neighborhood of $y$ by the definition of a $T_1$ space.
The definition of derived set is applied once more to see that $U \setminus \set x$ contains an element $z$ of $S$.
Thus $z \in U \cap S$ and $z \ne x$.
As this holds for all open neighborhoods $U$ of $x$, it follows that $x \in S'$.
This holds for all $x \in S$.
It follows by definition of subset that $S \subseteq S'$.
Thus $S'$ is closed.
Category:Topology
Derived Subgroup is Normal

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $\sqbrk {G, G}$ denote the derived subgroup of $G$.
Then $\sqbrk {G, G}$ is a normal subgroup of $G$.
== Proof ==
Recall the definition of $\sqbrk {G, G}$:
:$\sqbrk {G, G}$ is the subgroup of $G$ generated by all its commutators.
Recall also the definition of the commutator of $g, h \in G$:
:$\sqbrk {g, h} = g^{-1} h^{-1} g h$
From Derived Subgroup is Subgroup we note that $\sqbrk {G, G}$ is indeed a subgroup of $G$.
Let $g, h \in G$ be arbitrary.
Let $x \in \sqbrk {G, G}$ be arbitrary.
By definition of generated subgroup:
:$\sqbrk {g, h} x \in \sqbrk {G, G}$
and:
:$\sqbrk {h, g} x \in \sqbrk {G, G}$
Hence also by definition of generated subgroup:
:$\sqbrk {g, h} x \sqbrk {h, g} \in \sqbrk {G, G}$
and:
:$\sqbrk {h, g} x \sqbrk {g, h} \in \sqbrk {G, G}$
But from Inverse of Group Commutator:
:$\sqbrk {g, h}^{-1} = \sqbrk {h, g}$
So:
:$\sqbrk {g, h} x \sqbrk {g, h}^{-1} \in \sqbrk {G, G}$
and:
:$\sqbrk {g, h}^{-1} x \sqbrk {g, h} \in \sqbrk {G, G}$
As $x \in \sqbrk {G, G}$ is arbitrary, by definition of coset:
:$\forall \sqbrk {g, h} \in G: \sqbrk {g, h} \sqbrk {G, G} \sqbrk {g, h}^{-1} \subseteq \sqbrk {G, G}$
and:
:$\forall \sqbrk {g, h} \in G: \sqbrk {g, h}^{-1} \sqbrk {G, G} \sqbrk {g, h} \subseteq \sqbrk {G, G}$
Hence the result by definition of normal subgroup.
== Sources ==
* 
Category:Derived Subgroups
Category:Normal Subgroups
Derived Subgroup is Subgroup

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $\sqbrk {G, G}$ denote the derived subgroup of $G$.
Then $\sqbrk {G, G}$ is indeed a subgroup of $G$.
== Proof ==
Recall the definition of $\sqbrk {G, G}$:
:$\sqbrk {G, G}$ is the subgroup of $G$ generated by all its commutators.
Recall also the definition of the commutator of $g, h \in G$:
:$\sqbrk {g, h} = g^{-1} h^{-1} g h$
We note that from Commutator of Group Element with Identity is Identity:
:$\sqbrk {e, e} = e$
and so $e \in \sqbrk {G, G}$
Thus:
:$\sqbrk {G, G} \ne \O$
Let $g, h \in G$ be arbitrary.
We have by definition of inverse element that:
:$\sqbrk {g, h} = \sqbrk {g, h} e$
and so $\sqbrk {g, h} \in \sqbrk {G, G}$.
Let $g_1, h_1, g_2, h_2 \in G$ be arbitrary.
By definition of $\sqbrk {G, G}$:
:$\sqbrk {g_1, h_1} \in \sqbrk {G, G}$
and also:
:$\sqbrk {g_2, h_2} \in \sqbrk {G, G}$
Hence by definition of generated subgroup:
:$\sqbrk {g_1, h_1} \sqbrk {g_2, h_2} \in \sqbrk {G, G}$
Hence by the Two-Step Subgroup Test, $\sqbrk {G, G}$ is a subgroup of $G$.
Category:Derived Subgroups
Derived Subgroup is Trivial iff Group is Abelian

== Theorem ==
Let $G$ be a group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Then:
:$G'$ is the trivial group $\set e$
:
:$G$ is abelian.
== Proof ==
=== Sufficient Condition ===
This is proved in Derived Subgroup of Abelian Group is Trivial.
=== Necessary Condition ===
This is proved in Group whose Derived Subgroup is Trivial is Abelian.
Category:Derived Subgroups
Category:Abelian Groups
Category:Trivial Group
Derived Subgroup of Abelian Group is Trivial

== Theorem ==
Let $G$ be a abelian group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Then $G'$ is the trivial group $\set e$.
== Proof ==
For $g, h \in G$, let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$:
:$\sqbrk {g, h} := g^{-1} \circ h^{-1} \circ g \circ h$
Let $C$ be the set of the commutators of $G$:
:$C = \set {\sqbrk {g, h}: g, h \in G}$
From Commutators are Identity iff Group is Abelian, $C = \set e$.
From Trivial Subgroup is Subgroup, $C = \set e$ is a subgroup of $G$.
From Subgroup Generated by Subgroup:
:$\gen C = C$
where $\gen C$ denotes the subgroup generated by $C$.
From the definition of derived subgroup:
:$G' := \gen C$
Hence the result:
:$G' = \set e$
== Also see ==
* Group whose Derived Subgroup is Trivial is Abelian
Category:Derived Subgroups
Category:Abelian Groups
Category:Trivial Group
Desargues' Theorem

== Theorem ==
Let $\triangle ABC$ and $\triangle A'B'C'$ be triangles lying in the same or different planes.
Let the lines $AA'$, $BB'$ and $CC'$ intersect in the point $O$.
Then:
:$BC$ meets $B'C'$ in $L$
:$CA$ meets $C'A'$ in $M$
:$AB$ meets $A'B'$ in $N$
where $L, M, N$ are collinear.
=== Converse ===
== Proof ==
:
Let $\triangle ABC$ and $\triangle A'B'C'$ be in different planes $\pi$ and $\pi'$ respectively.
Since $BB'$ and $CC'$ intersect in $O$, it follows that $B$, $B'$, $C$ and $C'$ lie in a plane.
Thus $BC$ must meet $B'C'$ in a point $L$.
By the same argument, $CA$ meets $C'A'$ in a point $M$ and $AB$ meets $A'B'$ in a point $N$.
These points $L, M, N$ are in each of the planes $\pi$ and $\pi'$.
By Two Planes have Line in Common they are therefore collinear on the line where $\pi$ and $\pi'$ meet.
Now let $\triangle ABC$ and $\triangle A'B'C'$ be in the same plane $\pi$.
Let $OPP'$ be any line through $O$ which does not lie in $\pi$.
Then since $PP'$ meets $AA'$ in $O$, the four points $P, P', A, A$ are coplanar.
Thus $PA$ meets $P'A'$ at a point $A' '$.
Similarly $PB$ meets $P'B'$ at a point $B' '$, and $PC$ meets $P'C'$ at a point $C' '$.
The lines $BC, B'C'$ and $B' 'C' '$ are the three lines of intersection of the three planes $PBC$, $P'B'C'$ and $\pi$ taken in pairs.
So $BC$, $B'C'$ and $B' 'C' '$ meet in a point $L$.
Similarly $CA$, $C'A'$ and $C' 'A' '$ meet in a point $M$ and $AB$, $A'B'$ and $A' 'B' '$ meet in a point $N$.
The two triangles $\triangle ABC$ and $\triangle A' 'B' 'C' '$ are in different planes, and $AA' '$, $BB' '$ and $CC' '$ meet in $P$.
Thus $L$, $M$ and $N$ are collinear by the first part of this proof.
== Also known as ==
== Also see ==
* Pascal's Theorem
* Pappus's Hexagon Theorem
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.3$: Desargues' Theorem
* : $10$
* 
* : $10$
* 
* 
* 
Category:Desargues' Theorem
Category:Projective Geometry
Desargues' Theorem/Converse

== Theorem ==
Let $\triangle ABC$ and $\triangle A'B'C'$ be triangles lying in the same or different planes.
Let:
:$BC$ meet $B'C'$ in $L$
:$CA$ meet $C'A'$ in $M$
:$AB$ meet $A'B'$ in $N$
where $L, M, N$ are collinear.
Then the lines $AA'$, $BB'$ and $CC'$ intersect in the point $O$.
== Proof ==
:
Let $L$, $M$ and $N$ be collinear .
Then $\triangle BB'N$ and $\triangle CC'M$ are perspective with center $L$ ($L = BC \cap B'C' \cap MN$)
From Desargues' Theorem:
:$O = BB' \cap CC'$
:$A = BN \cap CM$
:$A' = C'M \cap B'N$
are collinear.
Thus:
:$AA' \cap BB' \cap CC' = O$
Hence $\triangle ABC$ and $\triangle A'B'C'$ are perspective with center $O$.
== Also see ==
* Desargues' Theorem
== Sources ==
* 
* 
* 
Category:Desargues' Theorem
Descartes's Rule of Signs/Examples/Arbitrary Example 1

== Example of Use of Descartes's Rule of Signs ==
Consider the polynomial equation $\map f x$ over real numbers:
:$x^5 + x^4 - 2 x^3 + x^2 - 1 = 0$
This has three variations in sign:
:from $x^4$ to $-2 x^3$, where it goes from positive to negative
:from $-2 x^3$ to $x^2$, where it goes from negative to positive
:from $x^2$ to $-1$, where it goes from positive to negative.
Hence $\map f x$ has no more than $3$ positive real roots.
Replacing $x$ with $-x$ in $\map f x$ gives us the polynomial equation $\map {f'} x$:
:$-x^5 + x^4 + 2 x^3 + x^2 - 1 = 0$
This has two variations in sign:
:from $-x^5$ to $x^4$, where it goes from negative to positive
:from $x^2$ to $-1$, where it goes from positive to negative.
Hence $\map f x$ has no more than $2$ negative real roots.
== Sources ==
* 
* 
Category:Descartes's Rule of Signs
Descartes's Rule of Signs/Original Statement

== Theorem ==
Descartes's Rule of Signs was originally stated in the following form:
Let $\map f x$ be a polynomial equation over the real numbers:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 = 0$
where $a_j \in \R$.
The number of positive real roots of $\map f x$ cannot be greater than the number of variations in sign of $\map f x$ (although it may be fewer).
=== Negative Roots ===
== Also known as ==
== Sources ==
* 
* 
Category:Descartes's Rule of Signs