avewright/proofwiki-math huggingface dataset:
Theorem: Function of Two Discrete Random Variables

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $g: \R^2 \to \R$ be a real-valued function.
Then $Z = \map g {X, Y}$, defined as:
:$\forall \omega \in \Omega: \map Z \omega = \map g {\map X \omega, \map Y \omega}$
is also a discrete random variable.

Proof:
As $\Img X$ and $\Img Y$ are countable, then so is $\Img {\map g {X, Y} }$.
Now consider $\map {g^{-1} } Z$.
We have that:
* $\forall x \in \R: \map {X^{-1} } x \in \Sigma$.
* $\forall y \in \R: \map {Y^{-1} } x \in \Sigma$.
We also have that:
: $\ds \forall z \in \R: \map {g^{-1} } z = \bigcup_{\tuple {x, y}: \map g {x, y} = z} \set {\tuple {x, y} }$
But $\Sigma$ is a sigma-algebra and therefore closed for unions.
Thus $\forall z \in \R: \map {g^{-1} } z \in \Sigma^2$.
Now consider $\tuple {v, w} \in \map {g^{-1} } Z$.
As $\Sigma$ is a sigma-algebra it follows directly that:
*$\forall x \in v: \map {X^{-1} } x \in \Sigma$
*$\forall y \in w: \map {Y^{-1} } y \in \Sigma$
Hence the result.
{{qed}}
Theorem: Expectation of Function of Joint Probability Mass Distribution

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ be the expectation of $X$.
Let $g: \R^2 \to \R$ be a real-valued function
Let $p_{X, Y}$ be the joint probability mass function of $X$ and $Y$.
:$\ds \expect {\map g {X, Y} } = \sum_{x \mathop \in \Omega_X} \sum_{y \mathop \in \Omega_Y} \map g {x, y} \map {p_{X, Y} } {x, y}$
whenever the sum is absolutely convergent.

Proof:
Let $\Omega_X = \Img X = I_X$ and $\Omega_Y = \Img Y = I_Y$.
Let $Z = \map g {X, Y}$.
Thus $\Omega_Z = \Img Z = g \sqbrk {I_X, I_Y}$.
So:
{{begin-eqn}}
{{eqn | l = \expect Z
      | r = \sum_{z \mathop \in g \sqbrk {I_X, I_Y} } z \map \Pr {Z = z}
      | c = 
}}
{{eqn | r = \sum_{z \mathop \in g \sqbrk {I_X, I_Y} } z \sum_{ {x \mathop \in I_X, y \mathop \in I_Y} \atop {\map g {x, y} \mathop = z} } \map \Pr {X = x, Y = y}
      | c = Probability Mass Function of Function of Discrete Random Variable
}}
{{eqn | r = \sum_{x \mathop \in I_X} \sum_{y \mathop \in I_Y} \map g {x, y} \map \Pr {X = x, Y = y}
      | c = 
}}
{{eqn | r = \sum_{x \mathop \in I_X} \sum_{y \mathop \in I_Y} \map g {x, y} \map {p_{X, Y} } {x, y}
      | c = 
}}
{{end-eqn}}
From the definition of expectation, this last sum applies only when the last sum is absolutely convergent.
{{qed}}
Theorem: Expectation is Linear

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be integrable random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \, \expect X + \beta \, \expect Y$

Proof:
=== Discrete Random Variable ===
{{:Expectation is Linear/Discrete}}
Theorem: Condition for Independence of Discrete Random Variables

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ and $Y$ are independent {{iff}} there exist functions $f, g: \R \to \R$ such that the joint mass function of $X$ and $Y$ satisfies:
:$\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$

Proof:
We have by definition of joint mass function that:
:$x \notin \Omega_X \implies \map {p_{X, Y} } {x, y} = 0$
:$y \notin \Omega_Y \implies \map {p_{X, Y} } {x, y} = 0$
Hence we only need to worry about values of $x$ and $y$ in their appropriate $\Omega$ spaces.
Theorem: Condition for Independence from Product of Expectations

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then $X$ and $Y$ are independent {{iff}}:
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.

Proof:
=== Sufficient Condition ===
{{AimForCont}} $X$ and $Y$ are ''not'' independent.
That is:
:$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$
for some $a, b \in \R$.
Now, suppose that:
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.
Let us define $g$ and $h$ as examples of such functions as follows:
:<nowiki>$(1): \quad \map g x := \begin{cases}
1 & : x = a \\
0 & : x \ne a
\end{cases}$</nowiki>
:<nowiki>$(2): \quad \map h y := \begin{cases}
1 & : y = b \\
0 & : y \ne b
\end{cases}$</nowiki>
where $a \in \R$ and $b \in \R$ are arbitrary real numbers.
Then:
{{begin-eqn}}
{{eqn | l = \expect {\map g x \map h y}
      | r = \map \Pr {X = a, Y = b}
      | c =
}}
{{eqn | l = \expect {\map g X} \expect {\map h Y}
      | r = \map \Pr {X = a} \map \Pr {Y = b}
      | c =
}}
{{end-eqn}}
So {{hypothesis}}:
:$\map \Pr {X = a, Y = b} = \map \Pr {X = a} \map \Pr {Y = b}$
But also {{hypothesis}}:
:$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$
This contradicts our supposition that:
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.
So, if the above supposition holds, then $X$ and $Y$ have to be independent.
{{qed|lemma}}
Theorem: PGF of Sum of Independent Discrete Random Variables

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be independent discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $Z$ be a discrete random variable such that $Z = X + Y$.
Then:
:$\map {\Pi_Z} s = \map {\Pi_X} s \, \map {\Pi_Y} s$
where $\map {\Pi_Z} s$ is the probability generating function of $Z$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\Pi_Z} s
      | r = \expect {s^Z}
      | c = {{Defof|Probability Generating Function}}
}}
{{eqn | r = \expect {s^{X + Y} }
      | c = Definition of $Z$ (see above)
}}
{{eqn | r = \expect {s^X s^Y}
      | c = Exponential of Sum
}}
{{eqn | r = \expect {s^X} \expect {s^Y}
      | c = Condition for Independence from Product of Expectations
}}
{{eqn | r = \map {\Pi_X} s \, \map {\Pi_Y} s
      | c = {{Defof|Probability Generating Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Brahmagupta Theorem

If a cyclic quadrilateral has diagonals which are perpendicular, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.
Specifically:
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular, crossing at $M$.
Let $EF$ be a line passing through $M$ and crossing opposite sides $BC$ and $AD$ of $ABCD$.
Then $EF$ is perpendicular to $BC$ {{iff}} $F$ is the midpoint of $AD$.

Proof:
:420px
Theorem: Crossbar Theorem

Let $\triangle ABC$ be a triangle.
Let $D$ be a point in the interior of $\triangle ABC$.
Then there exists a point $E$ such that $E$ lies on both $AD$ and $BC$.

Proof:
:380px
{{AimForCont}} $BC$ does not meet ray $\overrightarrow {AD}$.
Either $BC$ meets line $\overleftrightarrow {AD}$ or it does not.
If it meets $\overleftrightarrow {AD}$, by the Line Separation Property it must meet the ray opposite to $\overrightarrow {AD}$ at a point $E \ne A$.
According to Proposition 3.8 (b), $E$ is not in the interior of $\angle CAB$.
Point $B$ does not lie on $\overleftrightarrow {AD}$; this is because $D \in \operatorname {int} \angle CAB$, so $D$ and $C$ lie on the same side of $\overleftrightarrow {AB}$, so $D \notin \set {\overleftrightarrow {AB} }$, so $B \notin \set {\overleftrightarrow {AB} }$.
Thus, since $E \in \set {\overleftrightarrow {AD} }$, we have:
:$E \ne B$
By the same reasoning with $C$ and $B$ interchanged, we have:
:$E \ne C$
Since $E \in BC$ and $E$ is not an endpoint, we have:
:$B ∗ E ∗ C$
Thus by Proposition 3.7, $E$ is in the interior of $\angle CAB$, which is a contradiction.
Thus $\overleftrightarrow {AD}$ does not meet $BC$ at all; that is, $B$ and $C$ are on the same side of $\overleftrightarrow {AD}$.
By B-2, we have a point $E$ such that:
:$C ∗ A ∗ E$
By Lemma 3.2.2, $C$ and $E$ are on opposite sides of $\overleftrightarrow {AD}$.
Thus, by B-4(iii), E and B are on opposite sides of $\overleftrightarrow {AD}$.
But by Proposition 3.8(c), $B$ is on the interior of $\angle DAE$, so $E$ and $B$ are on the same side of $\overleftrightarrow {AD}$
This is a contradiction.
Thus $\overrightarrow {AD}$ meets $BC$.
{{MissingLinks}}
Category:Triangles
Category:Named Theorems
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Theorem: Clavius's Law

If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
{{:Clavius's Law/Formulation 1}}
=== Formulation 2 ===
{{:Clavius's Law/Formulation 2}}

Proof:
{{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{EndTableau|qed}}
{{LEM||3}}
Theorem: Clavius's Law

If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
{{:Clavius's Law/Formulation 1}}
=== Formulation 2 ===
{{:Clavius's Law/Formulation 2}}

Proof:
{{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{Assumption|2|p \implies \bot}}
{{SequentIntro|3|2|\neg p|2|Negation as Implication of Bottom}}
{{ModusPonens|4|1,2|p|1|3}}
{{Implication|5|1|(p \implies \bot) \implies p|2|4}}
{{SequentIntro|6|1|p|5|Peirce's Law}}
{{EndTableau|qed}}
{{LEM|Peirce's Law|3}}
Theorem: Clavius's Law

If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
{{:Clavius's Law/Formulation 1}}
=== Formulation 2 ===
{{:Clavius's Law/Formulation 2}}

Proof:
{{BeginTableau|\vdash \paren {\neg p \implies p} \implies  p}}
{{Premise|1|\neg p \implies p}}
{{SequentIntro|2|1|p|1|Clavius's Law: Formulation 1}}
{{Implication|3||\paren {\neg p \implies p} \implies  p|1|2}}
{{EndTableau|qed}}
{{LEM|Clavius's Law/Formulation 1|3}}
Theorem: Clavius's Law

If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
{{:Clavius's Law/Formulation 1}}
=== Formulation 2 ===
{{:Clavius's Law/Formulation 2}}

Proof:
{{BeginTableau|\vdash \paren {\neg p \implies p} \implies  p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{Implication|7||\paren {\neg p \implies p} \implies  p|1|6}}
{{EndTableau|qed}}
{{LEM||3}}
Theorem: Clavius's Law

If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
{{:Clavius's Law/Formulation 1}}
=== Formulation 2 ===
{{:Clavius's Law/Formulation 2}}

Proof:
We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations.
:<nowiki>$\begin{array}{|cccc|c|c|} \hline
(\neg & p & \implies & p) & \implies & p \\
\hline
\T & \F & \F & \F & \T & \F \\
\F & \T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
{{qed}}
Theorem: Viète's Formulas

Let $P_n$ be a polynomial of degree $n$ with real or complex coefficients:
{{begin-eqn}}
{{eqn | l = \map {P_n} x
      | r = \sum_{i \mathop = 0}^n a_i x^i
}}
{{eqn | r = a_n x^n + a_{n - 1} x^{n - 1} + \dotsb + a_1 x + a_0
}}
{{end-eqn}}
where $a_n \ne 0$.
Let $z_1, \ldots, z_n$ be the roots of $P_n$ (be they real or complex), not assumed distinct.
Then:
{{begin-eqn}}
{{eqn | q = \forall k \in \set {0, 1, \ldots, n}
      | l = \paren {-1}^k \dfrac {a_{n - k} } {a_n}
      | r = \map {e_k} {\set {z_1, \ldots, z_n} }
}}
{{eqn | r = \sum_{\substack {1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n \\ 1 \mathop \le k \mathop \le n} } z_{i_1} \dotsm z_{i_k} 
      | c = 
}}
{{end-eqn}}
where $\map {e_k} {\set {z_1, \ldots, z_n} }$ denotes the elementary symmetric function of degree $k$ on $\set {z_1, \ldots, z_n}$.

Proof:
First we note that from the Polynomial Factor Theorem:
:$\ds \map {P_n} x = a_n \prod_{k \mathop = 1}^n \paren {x - z_k}$
To shorten the exposition, let us define:
:$E_{\tuple {r, s} } := \map {e_r} {\set {z_1, \ldots, z_s} }$
for $r, s \in \Z_{\ge 1}$.
It is sufficient to consider the case $a_n = 1$, in which case:
:$\ds \map {P_n} x = \prod_{k \mathop = 1}^n \paren {x - z_k}$
The proof then proceeds by induction as follows.
Let $\map {\Bbb P} n$ be the statement:
{{begin-eqn}}
{{eqn | q = \forall \set {z_1, z_2, \ldots, z_n} 
      | l = \prod_{j \mathop = 1}^n \paren {x - z_j} 
      | r = \sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j}
}}
{{eqn | r = E_{\tuple {0, n} } x^n - E_{\tuple {1, n} } x^{n - 1} + E_{\tuple {2, n} } x^{n - 2} - \cdots + \paren {-1}^n E_{\tuple {n, n} }
}}
{{end-eqn}}
Theorem: Inscribed Angle Theorem

An inscribed angle is equal to half the angle that is subtended by that arc.
:300px
Thus, in the figure above:
:$\angle ABC = \frac 1 2 \angle ADC$
{{EuclidSaid}}
:''In a circle the angle at the center is double of the angle at the circumference, when the angles have the same circumference as base.''
{{EuclidPropRef|III|20}}

Proof:
:380px
Let $ABC$ be a circle, let $\angle BEC$ be an angle at its center, and let $\angle BAC$ be an angle at the circumference.
Let these angles have the same arc $BC$ at their base.
Let $AE$ be joined and drawn through to $F$.
Since $EA = EB$, then from Isosceles Triangle has Two Equal Angles we have that $\angle EBA = \angle EAB$.
So $\angle EBA + \angle EAB = 2 \angle EAB$.
But from Sum of Angles of Triangle equals Two Right Angles we have that $\angle BEF = \angle EBA + \angle EAB$.
That is, $\angle BEF = 2 \angle EAB$.
For the same reason, $\angle FEC = 2 \angle EAC$.
So adding them together, we see that $\angle BEC = 2 \angle BAC$.
Now consider the point $D$, from which we have another angle $\angle BDC$.
Let $DE$ be joined and produced to $G$.
Similarly, we prove that $\angle GEC = 2 \angle EDC$.
Then $\angle GEB = 2 \angle EDB$.
Therefore $\angle BEC$ which remains is equal to $2 \angle BDC$.
Hence the result.
{{Qed}}
Theorem: Inscribed Angle Theorem

An inscribed angle is equal to half the angle that is subtended by that arc.
:300px
Thus, in the figure above:
:$\angle ABC = \frac 1 2 \angle ADC$
{{EuclidSaid}}
:''In a circle the angle at the center is double of the angle at the circumference, when the angles have the same circumference as base.''
{{EuclidPropRef|III|20}}

Proof:
Consider the simplest case that occurs when $AC$ is a diameter of the circle:
:300px
Because all lines radiating from $D$ to the circumference are radii and thus equal:
:$AD = BD = CD$
Hence the triangles $\triangle ADB$ and $\triangle BDC$ are isosceles.
Therefore from Isosceles Triangle has Two Equal Angles:
:$\angle DBC = \angle DCB$.
From Sum of Angles of Triangle equals Two Right Angles:
:$\angle BDC$ is a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$.
From Thales' Theorem, $\angle ABC$ is right.
By similar reasoning $\angle DAB$ is the complement of $\angle DCB$.
If $\angle BDC$ is the supplement of twice the complement of $\angle DAB$, then $\angle BDC = 2 \angle DAB$.
That proves the theorem for this case.
{{qed|lemma}}
The general case is illustrated below.
:300px
A diameter is drawn from $A$ through the center $D$ to $E$. 
By the previous logic:
:$\angle BDE = 2 \angle BAE$
:$\angle CDE = 2 \angle CAE$
Subtracting the latter from the former equation obtains the general result. 
{{qed}}
Theorem: Linearly Independent Solutions to 1st Order Systems

The 1st-order homogeneous linear system of differential equations $x' = A \left({t}\right) x$, expressed with the vectors $x', x: \R \to \R^n$ and the matrix function $A: \R \to M_{n \times n} \left({\R}\right)$, has $n$ linearly independent solutions, and if $ \phi_1 , \phi_2, \dots, \phi_n$ are $n$ linearly independent solutions, then $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n$, where $c_i$ are constants, is a general solution.

Proof:
Let $v_1, v_2, \dots, v_n$ be linearly independent vectors in $\R^n$, and let $\phi_i$ be solutions to the IVPs $x' = A \left({t}\right) x, \, x \left({t_0}\right) = v_i$ for $i = 1, 2, \dots, n$.
Suppose the solutions are not independent, i.e. $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n = 0$ for some constants $c_i$ not all zero.
Then:
: $c_1 \phi_1 \left({t_0}\right) + c_2 \phi_2 \left({t_0}\right) + \cdots c_n \phi_n \left({t_0}\right) = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0$
meaning the vectors $v_i$ are linearly dependent, a contradiction, so the solutions $\phi_i$ must be linearly independent.
By linearity of the system, every vector function of the form $ x = c_1 \phi_1 + \cdots + c_n \phi_n$ is a solution.
Let $z$ be an arbitrary solution of the system.
Since $\phi_i \left({t_0}\right)$ are linearly independent and count $n$ in number, they form a basis for $\R^n$, hence $z \left({t_0}\right)$ must be a linear combination of those solutions, and then by uniqueness of solutions $z$ is a linear combination of the vector functions $\phi_i$.
This proves this is a general solution.
{{qed}}
Category:Ordinary Differential Equations
e8ruc4vwedz8lj7on6a708g9e0wpcv7
Theorem: Liouville's Theorem (Differential Equations)

Let $\map \Phi t$ be a solution to the matrix differential equation:
:$X' = \map A t X$
with $\map A t$ continuous on the interval $I$ such that $t_0 \in I$.
Then:
:$\ds \det \map \Phi t = \map \exp {\int_{t_0}^t \map \tr {\map A s} \rd s} \det \map \Phi {t_0}$

Proof:
{{ProofWanted}}
{{Namedfor|Joseph Liouville|cat = Liouville}}
Category:Differential Equations
Category:Liouville's Theorem
k2u4k0nv2vl7jnxo5h4qryjh3xu8658
Theorem: Maximum Modulus Principle

Let $D$ be an open region of the complex plane $\C$.
Let $f: D \to \C$ be a non-constant holomorphic function.
Then $\cmod f$ does not have any maximum points in the interior of $D$.
That is, for each $z \in D$ and $\delta > 0$, there exists some $\omega \in \map {B_\delta} z \cap D$, such that:
:$\cmod {\map f \omega} > \cmod {\map f z}$

Proof:
Pick some $r > 0$ such that $\map {B_r} z \subset D$.
By the Mean Value Theorem for Holomorphic Functions:
:$\ds \map f z = \dfrac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {\map f z}
      | o = \le
      | r = \frac 1 {2 \pi} \int_0^{2 \pi} \cmod {\map f {z + r e^{i \theta} } } \rd \theta
}}
{{eqn | o = \le
      | r = \max_\theta \cmod {\map f {z + r e^{i \theta} } }
      | c = Darboux's Theorem
}}
{{end-eqn}}
So it must be that there exists $\omega \in \map {C_r} z$ such that:
:$\cmod {\map f z} \le \cmod {\map f \omega}$
where $\map {C_r} z$ is the circle of radius $r$ centered at $z$.
Note that equality is only obtained when $\cmod f$ is constant on $\map {C_r} z$.
However, since this holds for all sufficiently small $r > 0$, $\cmod f$ would be constant in $\map {B_r} z$.
Then $f$ must be constant in $D$, contradicting our assumption.
It follows that there exists $\omega \in \map {C_r} z$ such that: 
:$\cmod {\map f z} < \cmod {\map f \omega}$
{{qed}}
Theorem: Euler Polyhedron Formula

For any convex polyhedron with $V$ vertices, $E$ edges, and $F$ faces: 
:$V - E + F = 2$

Proof:
{{ProofWanted|There should be a proof that the net of the polyhedron is a planar graph. The result then follows from Euler's Theorem for Planar Graphs.}}
Theorem: Line Joining Centers of Two Circles Touching Internally

Let two circles touch internally.
Then the straight line joining their centers passes through the point where they touch.
{{:Euclid:Proposition/III/11}}

Proof:
Let the circles $ABC$ and $ADE$ touch internally at $A$.
Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$.
We are to show that the straight line joining $F$ to $G$ passes through $A$.
:400px
Suppose, as in the diagram above, that it does not.
Let $FG$ fall on $H$ instead.
It will also pass through $D$ on its way, which lies on circle $ADE$.
Join $AF$ and $AG$.
From Sum of Two Sides of Triangle Greater than Third Side $AG + GF$ is greater than $AF$.
Therefore $AG + GF$ is greater than $FH$ as $F$ is the center and both $AF$ and $FH$ are radii.
So, subtract $FG$ from both $AG + GF$ and $FH$.
It follows that $AG$ is greater than $GH$, and hence also greater than $GD$.
But as $G$ is the center of $ADE$, it follows that $AG$ is supposed to be equal to $GD$.
So the $AG + GF$ can not be greater than $AF$ and so must be coincident with it.
Hence the result.
{{Qed}}
{{Euclid Note|11|III}}
Theorem: PGF of Sum of Random Number of Independent Discrete Random Variables

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let:
:$N, X_1, X_2, \ldots$
be independent discrete random variables such that the $X$'s have the same probability distribution.
Let:
:$\map {\Pi_N} s$ be the PGF of $N$
:$\map {\Pi_X} s$ be the PGF of each of the $X$'s.
Let:
:$Z = X_1 + X_2 + \ldots + X_N$
Then:
:$\map {\Pi_Z} s = \map {\Pi_N} {\map {\Pi_X} s}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\Pi_Z} s
      | r = \expect {s^{X_1 + X_2 + \cdots + X_N} }
      | c = {{Defof|Probability Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} \expect {s^{X_1 + X_2 + \cdots + X_N} \mid N = n} \map \Pr {N = n}
      | c = Total Expectation Theorem
}}
{{eqn | r = \sum_{n \mathop \ge 0} \expect {s^{X_1 + X_2 + \cdots + X_n} } \map \Pr {N = n}
      | c = 
}}
{{eqn | r = \sum_{n \mathop \ge 0} \map {\Pi_X} s^n \map \Pr {N = n}
      | c = PGF of Sum of Independent Discrete Random Variables
}}
{{eqn | r = \map {\Pi_N} {\map {\Pi_X} s}
      | c = Definition of $\map {\Pi_N} s$
}}
{{end-eqn}}
{{qed}}
Theorem: Floquet's Theorem

Let $\mathbf A \left({t}\right)$ be a continuous matrix function with period $T$.
Let $\Phi \left({t}\right)$ be a fundamental matrix of the Floquet system $\mathbf x' = \mathbf A \left({t}\right) \mathbf x$.
Then $\Phi \left({t + T}\right)$ is also a fundamental matrix.
Moreover, there exists:
: A nonsingular, continuously differentiable matrix function $\mathbf P \left({t}\right)$ with period $T$
: A constant (possibly complex) matrix $\mathbf B$ such that:
::$\Phi \left({t}\right) = \mathbf P \left({t}\right) e^{\mathbf Bt}$

Proof:
We assume the two hypotheses of the theorem.
We have that:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d t} } {\map \Phi {t + T} }
      | r = \map {\Phi'} {t + T}
      | c = 
}}
{{eqn | r = \map {\mathbf A} {t + T} \map \Phi {t + T}
      | c = 
}}
{{eqn | r = \map {\mathbf A} t \map \Phi {t + T}
      | c = 
}}
{{end-eqn}}
So the first implication of the theorem holds, that is: that $\map \Phi {t + T}$ is a fundamental matrix.
Because $\map \Phi t$ and $\map \Phi {t + T}$ are both fundamental matrices, there must exist some matrix $\mathbf C$ such that:
:$\map \Phi {t + T} = \map \Phi t \mathbf C$
Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that:
:$\mathbf C = e^{\mathbf BT}$
Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that:
{{begin-eqn}}
{{eqn | l = \map {\mathbf P} {t + T}
      | r = \map \Phi {t + T} e^{-\mathbf B t - \mathbf B T}
      | c = 
}}
{{eqn | r = \map \Phi t C e^{-\mathbf B T} e^{-\mathbf B t}
      | c = 
}}
{{eqn | r = \map \Phi t e^{-\mathbf B t}
      | c = 
}}
{{eqn | r = \map {\mathbf P} t
      | c = 
}}
{{end-eqn}}
and hence $\map {\mathbf P} t$ is periodic with period $T$.
As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.
{{qed}}