avewright/proofwiki-math huggingface dataset:
Theorem: Expected Value of Martingale is Constant in Time/Discrete Time

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a discrete-time filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a martingale.
Then: 
:$\expect {X_n} = \expect {X_0}$
for each $n \in \Z_{\ge 0}$.

Proof:
From Definition 2 of a discrete time martingale, we have:
:$\expect {X_n \mid \FF_0} = X_0$ almost surely.
So:
:$\expect {\expect {X_n \mid \FF_0} } = \expect {X_0}$
{{explain|Why does "almost surely equal" imply directly that expectations are equal? It is intuitively plausible to a person coming new to this page, but it needs to be made rigorous.}}
From Expectation of Conditional Expectation, we have:
:$\expect {\expect {X_n \mid \FF_0} } = \expect {X_n}$
So:
:$\expect {X_n} = \expect {X_0}$
for each $n \in \Z_{\ge 0}$.
{{qed}}
Category:Expected Value of Martingale is Constant in Time
8sy0h3wgm8rlbbcka42b28v8kghm50f
Theorem: Inradius of Triangle in Terms of Exradii

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then we have the following relation:
:$\dfrac 1 r = \dfrac 1 {\rho_a} + \dfrac 1 {\rho_b} + \dfrac 1 {\rho_c}$
where:
:$r$ is the inradius
:$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ {{WRT}} $a$, $b$ and $c$ respectively.

Proof:
The area $\AA$ of $\triangle ABC$ is given by:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \rho_a \paren {s - a}
      | c = Area of Triangle in Terms of Exradius
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\rho_a}
      | r = \frac {s - a} \AA
      | c = rearranging
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \quad \quad \frac 1 {\rho_b}
      | r = \frac {s - b} \AA
}}
{{eqn | l = \frac 1 {\rho_c}
      | r = \frac {s - c} \AA
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | o =
      | r = \frac 1 {\rho_a} + \frac 1 {\rho_b} + \frac 1 {\rho_c}
}}
{{eqn | r = \frac {s - a} \AA + \frac {s - b} \AA + \frac {s - c} \AA
}}
{{eqn | r = \frac {3 s - a - b - c} \AA
}}
{{eqn | r = \frac {3 s - 2 s} \AA
      | c = {{Defof|Semiperimeter}}
}}
{{eqn | r = \frac s {r s}
      | c = Area of Triangle in Terms of Inradius
}}
{{eqn | r = \frac 1 r
}}
{{end-eqn}}
{{qed}}
Category:Incircles of Triangles
Category:Excircles of Triangles
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Theorem: Expected Value of Supermartingale is Decreasing in Time/Continuous Time

Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-supermartingale.
Let $t, s \in \hointr 0 \infty$ with $0 \le s < t$.
Then, we have: 
:$\expect {X_t} \le \expect {X_s}$

Proof:
From the definition of a supermartingale, we have: 
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely.
From Expectation is Monotone, we have: 
:$\expect {\expect {X_t \mid \FF_s} } \le \expect {X_s}$
From Expectation of Conditional Expectation, we have: 
:$\expect {X_t} \le \expect {X_s}$ 
{{qed}}
Theorem: Expected Value of Submartingale is Increasing in Time/Continuous Time

Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
Let $t, s \in \hointr 0 \infty$ with $0 \le s < t$.
Then, we have: 
:$\expect {X_s} \le \expect {X_t}$

Proof:
From the definition of a submartingale, we have: 
:$\expect {X_t \mid \FF_s} \ge X_s$ almost surely.
From Expectation is Monotone, we have: 
:$\expect {\expect {X_t \mid \FF_s} } \ge \expect {X_s}$
From Expectation of Conditional Expectation, we have: 
:$\expect {X_s} \le \expect {X_t}$
{{qed}}
Theorem: Conditional Expectations of Integrable Random Variable with respect to Filtration forms Martingale/Continuous Time

Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $Z$ be an integrable random variable.
For each $t \in \hointr 0 \infty$, let $\expect {Z \mid \FF_t}$ be a version of the conditional expectation of $Z$ given $\FF_t$. 
For each $t \in \hointr 0 \infty$ set:
:$X_t = \expect {Z \mid \FF_t}$
Then $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale.

Proof:
From the definition of the conditional expectation of $Z$ given $\FF_t$, we have that: 
:$X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$. 
So $\sequence {X_t}_{t \ge 0}$ is $\sequence {\FF_t}_{t \ge 0}$-adapted.
Now let $s, t \in \hointr 0 \infty$ have $s \le t$. 
Then we have: 
{{begin-eqn}}
{{eqn	| l = \expect {X_t \mid \FF_s}
		| r = \expect {\expect {Z \mid \FF_t} \mid \FF_s}
}}
{{eqn	| r = \expect {Z \mid \FF_s}
		| c = applying Tower Property of Conditional Expectation, since $\FF_s \subseteq \FF_t$
}}
{{eqn	| r = X_s
}}
{{end-eqn}}
So $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale.
{{qed}}
Theorem: Linear Span is Linear Subspace

Let $V$ be a vector space over a division ring $K$.
Let $S \subseteq V$ be a subset of $V$.
Then the linear span $\map \span S$ is a subspace of $V$.

Proof:
First, suppose that $S = \O$.
By definition of linear combination of empty set, it follows that $\map \span \O = \set \bszero$, where $\bszero$ denotes the zero vector of $V$.
From Zero Subspace is Subspace, it follows that the trivial vector space $\set \bszero$ is a subspace of $V$.
Suppose instead that $S$ is non-empty.
All $v \in \map \span S$ are of the form:
:$v = \ds \sum_{k \mathop = 1}^n \lambda_k v_k$
where $n \in \N$, $\lambda_1, \ldots, \lambda_n \in K$, and $v_1, \ldots, v_n \in S$.
We use the Two-Step Vector Subspace Test to prove that $\map \span S$ is a subspace of $V$.
For any $\lambda \in K$, it follows from {{Vector-space-axiom|1}} that:
:$\ds \lambda v = \lambda \paren {\sum_{k \mathop = 1}^n \lambda_k v_k} = \sum_{k \mathop = 1}^n \paren {\lambda \lambda_k} v_k$
which shows that $\lambda v \in \map \span S$, which is the first condition of the Two-Step Vector Subspace Test.
For any $w \in \map \span S$, let $w$ be of the form:
:$w = \ds \sum_{l \mathop = 1}^m \mu_l w_k$
where $m \in \N$, $\mu_1, \ldots, \mu_m \in K$, and $w_1, \ldots, w_m \in S$.
Define the three sets:
{{begin-eqn}}
{{eqn | o = :=
      | l = N
      | r = \set {1 \le k \le n: v_k \notin \set {w_1, \ldots, w_m} }
}}
{{eqn | o = :=
      | l = M
      | r = \set {1 \le l \le m: w_l \notin \set {v_1, \ldots, v_n} }
}}
{{eqn | o = :=
      | l = \OO
      | r = \set {\tuple {k,l}: 1 \le k \le n, 1 \le l \le m, v_k = w_l}
}}
{{end-eqn}}
It follows that:
{{begin-eqn}}
{{eqn | l = v + w
      | r = \sum_{k \mathop = 1}^n \lambda_k v_k + \sum_{l \mathop = 1}^m \mu_l w_k
}}
{{eqn | r = \sum_{k \mathop \in N} \lambda_k v_k + \sum_{l \mathop \in M} \mu_l w_l + \sum_{\tuple {k,l} \mathop \in \OO} \paren {\lambda_k + \mu_l} v_k
      | c = {{Vector-space-axiom|2}}
}}
{{end-eqn}}
which shows that $v + w \in \map \span S$, which is the second condition of the Two-Step Vector Subspace Test.
{{qed}}
Theorem: Linear Span is Linear Subspace

Let $V$ be a vector space over a division ring $K$.
Let $S \subseteq V$ be a subset of $V$.
Then the linear span $\map \span S$ is a subspace of $V$.

Proof:
This is a special case of Generated Submodule is Linear Combinations.
As such, the statement follows immediately from that theorem.
{{qed}}
Theorem: Characterization of Stopping Times with respect to Right-Limit Filtration

Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $T : \Omega \to \closedint 0 \infty$ be a random variable.
Let $\sequence {\GG_t}_{t \ge 0}$ be the right-limit filtration associated with $\sequence {\FF_t}_{t \ge 0}$.
{{TFAE}}
:$(1) \quad$ $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$
:$(2) \quad$ for each $t \in \hointr 0 \infty$, we have $\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$
:$(3) \quad$ for each $t \in \hointr 0 \infty$, the pointwise minimum $T \wedge t$ is $\FF_t$-measurable.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose $T$ is a stopping time with respect to $\sequence {\GG_t}_{t \ge 0}$ and let $t \in \hointr 0 \infty$.
Then, for each $n \in \N$ we have: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \GG_{t - \frac 1 n}$
Since by the definition of the right-limit filtration, we have: 
:$\ds \GG_{t - \frac 1 n} = \bigcap_{s > t - \frac 1 n} \FF_s$
we have: 
:$\GG_{t - \frac 1 n} \subseteq \FF_s$ for all $\ds s > t - \frac 1 n$
and in particular:
:$\GG_{t - \frac 1 n} \subseteq \FF_t$ 
and hence:
:$\ds \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$
Note that: 
:$\ds \set {\omega \in \Omega : \map T \omega < t} = \bigcup_{n \mathop = 1}^\infty \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n}$ 
Since $\FF_t$ is closed under countable union, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \set {\omega \in \Omega : \map T \omega \le t - \frac 1 n} \in \FF_t$
and hence:
:$\set {\omega \in \Omega : \map T \omega < t} \in \FF_t$
{{qed|lemma}}
Theorem: Chu-Vandermonde Identity/Falling Factorial Variant

:$\ds \sum_{k \mathop = 0}^n \dbinom n k r^{\underline k} s^{\underline {n-k} } = \paren {r + s}^{\underline n}$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \dbinom n k r^{\underline k} s^{\underline {n - k} }
      | r = \sum_{k \mathop = 0}^n \paren {\dfrac {n!} {k! \paren{n - k}!} } \paren{ \dfrac {r!} {\paren {r - k}!} } \paren{ \dfrac {s!} {\paren {s - \paren {n - k} }!} } 
      | c = {{Defof|Binomial Coefficient}} and {{Defof|Falling Factorial}}
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n {\dbinom r k} \dbinom s {n - k} 
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = n! \binom {r + s} n
      | c = Chu-Vandermonde Identity
}}
{{eqn | r = n! \dfrac {\paren {r + s}!} {n! \paren {r + s - n}!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \paren{r + s}^{\underline n}
      | c = {{Defof|Falling Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Chu-Vandermonde Identity/Rising Factorial Variant

:$\ds \sum_{k \mathop = 0}^n \dbinom n k r^{\overline k} s^{\overline {n-k} } = \paren {r + s}^{\overline n}$

Proof:
From Rising Factorial as Factorial by Binomial Coefficient, we have:
{{begin-eqn}}
{{eqn | l = r^{\overline k}
      | r = k! \dbinom {r + k - 1} k
      | c = 
}}
{{eqn | l = s^{\overline {n - k} }
      | r = \paren{n - k}! \dbinom {s + n - k - 1} {n - k}
      | c = 
}}
{{eqn | l = \paren{r + s}^{\overline n}
      | r = n! \dbinom {r + s + n - 1} n
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \dbinom n k r^{\overline k} s^{\overline {n - k} }
      | r = \sum_{k \mathop = 0}^n \paren {\dfrac {n!} {k! \paren{n - k}!} } \paren{ {k! \dbinom {r + k - 1} k} } \paren{ {\paren{n - k}! \dbinom {s + n - k - 1} {n - k} } }
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n {\dbinom {r + k - 1} k} \dbinom {s + n - k - 1} {n - k} 
      | c = $k!$ and $\paren{n - k}!$ cancel
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n {\paren{-1}^k \dbinom {-r} k} \paren{-1}^{n - k} \dbinom {-s} {n - k} 
      | c = Negated Upper Index of Binomial Coefficient/Corollary 2
}}
{{eqn | r = {\paren{-1}^n n! \sum_{k \mathop = 0}^n \dbinom {-r} k} \dbinom {-s} {n - k} 
      | c = Product of Powers $\paren{-1}^k \times \paren{-1}^{n - k} = \paren{-1}^n$
}}
{{eqn | r = \paren{-1}^n n! \binom {-\paren{r + s} } n
      | c = Chu-Vandermonde Identity
}}
{{eqn | r = n! \dbinom {r + s + n - 1} n
      | c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \paren{r + s}^{\overline n}
      | c = Rising Factorial as Factorial by Binomial Coefficient
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Connected Manifold

Let $M$ be a topological manifold.
{{TFAE|def = Connected Manifold}}

Proof:
=== Definition 1 implies Definition 2 ===
Let $M$ be connected.
From Topological Manifold is Locally Path-Connected:
:$M$ is locally path-connected.
From Connected and Locally Path-Connected Implies Path-Connected:
:$M$ is path-connected.
{{qed|lemma}}
Theorem: Topological Manifold is Locally Path-Connected

Let $M$ be a topological manifold.
Then $M$ is a locally path-connected space.

Proof:
By definition of manifold:
:$M$ is a locally Euclidean space
The result follows from Locally Euclidean Space is Locally Path-Connected
{{qed}}
Theorem: Locally Euclidean Space is Locally Path-Connected

Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is locally path-connected.

Proof:
Let $m \in M$ be arbitrary.
From Locally Euclidean Space has Countable Local Basis Homeomorphic to Open Balls:
:there exists a local basis $\family{U_n}_{n \in \N}$ of $m$ where each $U_n$ is the homeomorphic image of an open ball of $\R^d$.
For all $n \in \N$, let:
:$U_n = \phi_n \sqbrk {B_n}$
where $B_n$ is an open ball of $\R^d$ and $\phi_n: B_n \to U_n$ is a homeomorphism.
From Open Ball in Normed Vector Space is Path-Connected:
:$B_n$ is path-connected.
From Continuous Image of Path-Connected Set is Path-Connected: Metric Space:
:$U_n$ is path-connected.
Hence $m$ has a local basis of path-connected sets. 
Since $m$ was arbitrary, it follows that $M$ is locally path-connected by definition.
{{qed}}
Category:Locally Euclidean Spaces
Category:Path-Connected Spaces
9lp1b93tdggkbhswy9dejscudohk0hs
Theorem: Locally Euclidean Space is First-Countable

Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is first-countable.

Proof:
Follows immediately from Locally Euclidean Space has Countable Local Basis Homeomorphic to Open Balls.
{{qed}}
Category:Locally Euclidean Spaces
Category:First-Countable Spaces
39t7sdcwcgxewea5urhcbnimoh6crb4
Theorem: Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints/Complex Plane

Let $\C$ denote the complex plane.

Proof:
By definition of reparameterization, there exists a bijective differentiable strictly increasing real function $\phi: \closedint c d \closedint a b$ such that $\sigma = \gamma \circ \phi$.
As $\map {\phi^{-1} }{a} \in \closedint c d$:
: $c \le \map {\phi^{-1} }{a}$
As $\phi$ is strictly increasing:
: $\map \phi c \le \map \phi { \map {\phi^{-1} } a } = a$
As $\map \phi c \in \closedint a b$:
:$\map \phi c = a$
Hence:
:$\map \sigma c = \map {\gamma \circ \phi} c = \map \gamma a$
As $\map {\phi^{-1} } b \in \closedint c d$:
: $d \ge \map {\phi^{-1} } b$
As $\phi$ is strictly increasing:
: $\map \phi d \ge \map \phi { \map {\phi^{-1} } b } = b$
As $\map \phi d \in \closedint a b$:
: $\map \phi d = b$
Hence:
:$\map \sigma d = \map {\gamma \circ \phi} d = \map \gamma b)$
{{qed}}
Theorem: Arcsine in terms of Arctangent

:$\arcsin x = \map \arctan {\dfrac x {\sqrt {1 - x^2} } }$

Proof:
Let: 
:$\theta = \arcsin x$
Then by the definition of arcsine:
:$x = \sin \theta$
and: 
:$-\dfrac \pi 2 < \theta < \dfrac \pi 2$
Then: 
{{begin-eqn}}
{{eqn | l = \map \arctan {\dfrac x {\sqrt {1 - x^2} } }
      | r = \map \arctan {\dfrac {\sin \theta} {\sqrt {1 - \sin^2 \theta} } }
}}
{{eqn | r = \map \arctan {\dfrac {\sin \theta} {\sqrt {\cos^2 \theta} } }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \map \arctan {\tan \theta}
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \theta
      | c = {{Defof|Real Arctangent}}
}}
{{eqn | r = \arcsin x
}}
{{end-eqn}}
{{qed}}
Theorem: Arctangent in terms of Arcsine

:$\arctan x = \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }$

Proof:
Let: 
:$\theta = \arctan x$
Then by the definition of arctangent:
:$x = \tan \theta$
Then: 
{{begin-eqn}}
{{eqn | l = \map \arcsin { \dfrac x {\sqrt {1 + x^2} }  }
      | r = \map \arcsin { \dfrac {\tan \theta} {\sqrt {1 + \tan^2 \theta} }  }
}}
{{eqn | r = \map \arcsin { \dfrac {\tan \theta} {\sqrt {\sec^2 \theta} }  }
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \map \arcsin { \dfrac {\sin \theta} { \cos \theta \dfrac 1 {\cos \theta } }  }
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \map \arcsin {\sin \theta }
      | c = 
}}
{{eqn | r = \theta
      | c = {{Defof|Real Arcsine}}
}}
{{eqn | r = \arctan x
}}
{{end-eqn}}
{{qed}}
Theorem: Arctangent in terms of Arcsine

:$\arctan x = \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }$

Proof:
From Pfaff's Transformation:
:$\ds \map F {a, b; c; x} = \paren {1 - x}^{-a} \map F {a, c - b; c; \dfrac x {x - 1} }$
where $\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$.
We have:
{{begin-eqn}}
{{eqn | l = \map \arctan x
      | r = x \map F {\dfrac 1 2, 1; \dfrac 3 2; -x^2}
      | c = Arctangent Function in terms of Gaussian Hypergeometric Function
}}
{{eqn | r = x \paren {1 - \paren {-x^2} }^{-\frac 1 2} \map F {\dfrac 1 2, \dfrac 3 2 - 1; \dfrac 3 2; \dfrac {\paren {-x^2} } {\paren {-x^2} - 1} }
      | c = Pfaff's Transformation
}}
{{eqn | r = \dfrac x {\sqrt {1 + x^2} } \map F {\dfrac 1 2, \dfrac 1 2; \dfrac 3 2; \dfrac {x^2} {1 + x^2} }
      | c = simplifying
}}
{{eqn | r = \map \arcsin {\dfrac x {\sqrt {1 + x^2} }  }
      | c = Arcsine Function in terms of Gaussian Hypergeometric Function
}}
{{end-eqn}}
Therefore:
:$\map \arctan x = \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }$
{{qed}}
Theorem: Derivative of Gaussian Hypergeometric Function

:$\map {\dfrac \d {\d x} } {\map F {a, b; c; x} } = \dfrac {a b} c \map F {a + 1, b + 1; c + 1; x} $

Proof:
{{begin-eqn}}
{{eqn | l = \map F {a, b; c; x}
      | r = 1 + \dfrac {a b} c x + \dfrac {a \paren {a + 1} b \paren {b + 1} } {c \paren {c + 1} } \dfrac {x^2} {2!} + \cdots + \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^k} {k!}
      | c = {{Defof|Gaussian Hypergeometric Function}}
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\map F {a, b; c; x} }
      | r = \dfrac {a b} c + \dfrac {a \paren {a + 1} b \paren {b + 1} } {c \paren {c + 1} } x + \cdots + \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^{k - 1} } {\paren {k - 1}!}
      | c = Derivative of Power of Function
}}
{{eqn | r = \dfrac {a b} c \paren {1 + \dfrac {\paren {a + 1} \paren {b + 1} } {\paren {c + 1} } x + \cdots + \dfrac {\paren {a + 1}^{\overline k} \paren {b + 1}^{\overline k} } {\paren {c + 1}^{\overline k} } \dfrac {x^k } {k!} }
      | c = factor out $\dfrac {a b} c$
}}
{{eqn | r = \dfrac {a b} c \map F {a + 1, b + 1; c + 1; x}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Convex Real Function Composed with Increasing Convex Real Function is Convex

Let $I$ be real interval.
Let $f : I \to \R$ be a convex function.
Let $J$ be a real interval containing the image of $f$.
Let $g : J \to \R$ be a increasing convex function.
Then $g \circ f : I \to \R$ is a convex function.

Proof:
Let $x, y \in I$ and $t \in \closedint 0 1$.
Since $f$ is convex, we have: 
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$ 
Since $g$ is increasing, we then have: 
:$\map {\paren {g \circ f} } {t x + \paren {1 - t} y} \le \map g {t \map f x + \paren {1 - t} \map f y}$
Since $g$ is convex, we then have: 
:$\map g {t \map f x + \paren {1 - t} \map f y} \le t \map g {\map f x} + \paren {1 - t} \map g {\map f y}$ 
so that: 
:$\map {\paren {g \circ f} } {t x + \paren {1 - t} y} \le t \map {\paren {g \circ f} } x + \paren {1 - t} \map {\paren {g \circ f} } y$
That is: 
:$g \circ f : I \to \R$ is a convex function.
{{qed}}
Category:Convex Real Functions
cvlyen3u5run1kgf3xoj64iyne1il46
Theorem: Power Function is Convex Real Function

Let $p \ge 1$ be a real number.
Define $f : \hointr 0 \infty \to \hointr 0 \infty$ by: 
:$\map f x = x^p$
for each $x \in \hointr 0 \infty$. 
Then $f$ is a convex function.

Proof:
Applying Derivative of Power twice, we have that: 
:$f$ is twice differentiable on $\openint 0 \infty$
with:
:$\map {f' '} x = p \paren {p - 1} x^{p - 2}$
for each $x \in \openint 0 \infty$.
Since $p \ge 1$, we have: 
:$p \paren {p - 1} \ge 0$
and so:
:$\map {f' '} x \ge 0$
for each $x \in \openint 0 \infty$.
From Real Function with Positive Derivative is Increasing:
:$f'$ is increasing on $\openint 0 \infty$
and so from Real Function is Convex iff Derivative is Increasing:
:$f$ is convex on $\openint 0 \infty$.
It remains to assure ourselves that:
:$\map f {t x + \paren {1 - t} 0} \le t \map f x + \paren {1 - t} \map f 0 = t \map f x$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$. 
We have:
:$\map f {t x} = \paren {t x}^p = t^p x^p$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$.
Since $p \ge 1$, we have:
:$t^p \le t$ for $t \in \closedint 0 1$.
Hence we have:
:$t^p x^p \le t x^p$
Hence:
:$\map f {t x} \le t \map f x$
Hence:
:$f$ is convex.
{{qed}}
Category:Convex Real Functions
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Theorem: Power of Absolute Value is Convex Real Function

Let $p \ge 1$ be a real number.
Define $f : \R \to \R$ by: 
:$\map f x = {\size x}^p$
for each $x \in \R$. 
Then $f$ is a convex function.

Proof:
From Absolute Value Function is Convex:
:$x \mapsto \size x$ is a convex function.
Note now that: 
:$x \mapsto x^p$ is increasing on $\hointr 0 \infty$ 
From Power Function is Convex Real Function, we also have: 
:$x \mapsto x^p$ is convex.
Since $f$ is the composition of the maps $x \mapsto \size x$ and $x \mapsto x^p$, we have:
:$f$ is convex
by Convex Real Function Composed with Increasing Convex Real Function is Convex.
{{qed}}
Category:Convex Real Functions
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Theorem: Components of Vector between two Points

Let $A, B$ be points in the Euclidean space $\R^n$.
Let their Cartesian coordinates be given by:
{{begin-eqn}}
{{eqn | l = A
      | r = \tuple {a_1 , a_2, \ldots, a_n }
}}
{{eqn | l = B
      | r = \tuple {b_1 , b_2, \ldots, b_n }
}}
{{end-eqn}}
Let $\vec {AB}$ be the vector quantity that represents the directed line segment with startpoint $A$ and endpoint $B$.
Then the vector components of $\vec {AB}$ are:
:$\vec {AB} = \tuple {b_1 - a_1, b_2 - a_2, \ldots, b_n - a_n}$

Proof:
Let $O$ denote the origin of $\R^n$.
Let $\vec {OA}$ and $\vec {OB}$ be the positions vectors of $A$ and $B$.
By definition of positions vectors, their components are:
{{begin-eqn}}
{{eqn | l = \vec {OA}
      | r = \tuple {a_1 , a_2, \ldots, a_n }
}}
{{eqn | l = \vec {OB}
      | r = \tuple {b_1 , b_2, \ldots, b_n }
}}
{{end-eqn}}
for some $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R$.
Let the components of $\vec {AB}$ be denoted as:
:$\vec {AB} := \tuple {v_1 , v_2, \ldots, v_n}$
By the definition of Triangle Law of Vector Addition, it follows that $\vec {OB}$ can be represented as:
:$\vec {OB} = \vec{OA} + \vec{AB}$
From Equivalence of Definitions of Vector Sum, it follows that:
:$\tuple {b_1 , b_2, \ldots, b_n} = \tuple {a_1 , a_2, \ldots, a_n} + \tuple {v_1 , v_2, \ldots, v_n}$
By the definition of vector addition of components, it follows that:
:$\tuple {b_1 - a_1, b_2 - a_2, \ldots, b_n - a_n} = \tuple {v_1 , v_2, \ldots, v_n}$
Hence the result.
{{qed}}
Category:Vectors
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Theorem: Regular Polygon is composed of Isosceles Triangles

Let $P$ be a regular $n$-gon.
Let $O$ be the center of $P$.
Then there exists a triangulation of $P$ into $n$ congruent isosceles triangles.
The three vertices of each triangle are $O$ and two adjacent vertices of $P$.

Proof:
From Regular Polygon is Cyclic, it follows that $O$ is the center of the circumcircle of $P$.
Let $V_1 , V_2$ be adjacent vertices of $P$.
By definition of circumcircle, $V_1 O$ and $V_2 O$ are radii of the circumcircle.
As $V_1 O$ and $V_2 O$ have the same length, it follows that $\triangle V_1 V_2 O$ is an isosceles triangles.
As there are $n$ pairs of adjacent vertices in $P$, corresponding to the $n$ sides of $P$, it follows that there are $n$ triangles of the type $\triangle V_1 V_2 O$.
All of these triangles consist of one side of $P$, as well as two sides of the type $V_1 O, V_2 O$ which they share with one of the adjacent triangles.
Hence, the triangles form a triangulation of $P$.
By definition of regular polygon, all sides of $P$ have the same length.
From Triangle Side-Side-Side Congruence, it follows that all the triangles are congruent.
{{qed}}
Category:Regular Polygons
Category:Isosceles Triangles
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