avewright/proofwiki-math huggingface dataset:
Theorem: Exterior Angle of Regular Polygon

Let $P$ be a regular polygon with $n$ sides.
Then each of the exterior angles of $P$ is equal to $\dfrac {360 \degrees} n$.

Proof:
From Sum of External Angles of Polygon equals Four Right Angles, the sum of all $n$ exterior angles of $P$ equals $360 \degrees$.
We have {{afortiori}} that the interior angles of $P$ are equal.
Hence the exterior angles of $P$ are also equal.
Hence each exterior angle of $P$ equals $\dfrac {360 \degrees} n$.
{{qed}}
Theorem: Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic

Let $Q$ be a quadrilateral whose sides are $a$, $b$, $c$ and $d$.
Let $\AA$ be the area of $Q$.
Then:
:$Q$ is a cyclic quadrilateral
{{iff}}:
:$\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.

Proof:
From Bretschneider's Formula:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$
where $s$ is the semiperimeter of $Q$.
Hence $\AA$ is greatest exactly when:
:$\map {\cos^2} {\dfrac {\alpha + \gamma} 2} = 0$
That is, when:
:$\map \cos {\dfrac {\alpha + \gamma} 2} = 0$
When this happens:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
which is Brahmagupta's Formula.
That is, $\AA$ is the area of a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$.
This happens exactly when:
:$\dfrac {\alpha + \gamma} 2 = \dfrac \pi 2$
That is, when:
:$\alpha + \gamma = \pi = 180 \degrees$
Hence the result from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles.
{{qed}}
Category:Cyclic Quadrilaterals
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Theorem: Regular Polygon is Cyclic

Let $P$ be a regular polygon.
Then $P$ is a cyclic polygon.

Proof:
Let $P$ be a regular polygon.
{{AimForCont}} $P$ is not cyclic.
Let $AB$, $BC$ and $CD$ be sides of $P$ such that $A$, $B$ and $C$ are on the circumference of a circle $K$ such that $D$ is not on that circumference.
This has to be possible, or all vertices of $P$ would lie on $K$.
That would make $P$ cyclic which contradicts our supposition about $P$.
Let the center of $K$ be $O$.
As $D$ is not on the circumference of that circle, then $OC \ne OD$.
Hence $\angle ABC \ne \angle BCD$ and so $P$ is not equiangular.
Hence {{afortiori}} $P$ is not a regular polygon.
This contradicts our assertion about $P$.
Hence by Proof by Contradiction it follows that $P$ is cyclic.
{{qed}}
Theorem: Long Radius of Regular Polygon equals Radius of Circumcircle

Let $P$ be a regular polygon.
Let $C$ be the circumcircle of $P$.
The long radius of $P$ is equal to the radius of $C$.

Proof:
By definition of circumcircle, $C$ is the circle such that all vertices of $P$ are on $C$.
By definition, the center of $P$ is defined as the center of $C$.
Hence:
:$OA$ is the radius of $C$
and also:
:$OA$ is the long radius of $P$.
Hence the result.
{{qed}}
Theorem: Apothem of Regular Polygon equals Radius of Incircle

Let $P$ be a regular polygon.
Let $C$ be the incircle of $P$.
The apothem of $P$ is equal to the radius of $C$.

Proof:
By definition of incircle, $C$ is the circle such that all sides of $P$ are tangent to $C$.
From Regular Polygon can be Circumscribed around Circle, it is established that the center of $P$ and the center of $C$ are the same point.
Hence:
:the radius of $C$
is the same thing as:
:the perpendicular distance from $C$ of $P$.
Hence the result.
{{qed}}
Theorem: Regular Polygon can be Circumscribed around Circle

Let $P$ be a regular polygon.
Then it is possible to circumscribe $P$ around a circle $C$, whose center is the same as the center of $P$.

Proof:
Let $P$ be a regular polygon.
{{AimForCont}} it is not possible to circumscribe $P$ around a circle $C$, whose center is the same as the center of $P$.
Let $AB$, $BC$ and $CD$ be sides of $P$ such that $AB$ and $BC$ are on the tangent to a circle $K$ such that $CD$ is not tangent to that circumference.
There can only be one such circle tangent to both $AB$ and $BC$.
This has to be possible, or all sides of $P$ would be tangent to $K$.
That would make $P$ cyclic which contradicts our supposition about $P$.
Let the center of $K$ be $O$.
As $CD$ is not tangent to $K$, then $OC \ne OD$.
Hence $\angle ABC \ne \angle BCD$ and so $P$ is not equiangular.
Hence {{afortiori}} $P$ is not a regular polygon.
This contradicts our assertion about $P$.
Hence by Proof by Contradiction it follows that $P$ is cyclic.
{{qed}}
Category:Regular Polygons
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Theorem: Dougall's Hypergeometric Theorem

:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} }$

Proof:
By definition of the generalized hypergeometric function
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {\dfrac n 2 + 1}^{\overline k} n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} } {\paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} } \dfrac {1^k} {k!}$
{{ProofWanted}}
Theorem: Kummer's Hypergeometric Theorem

:$\map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }$

Proof:
From Kummer's Quadratic Transformation, we have:
:$\ds \map F {a, b; 1 + a - b; z} = \paren {1 - z}^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; \dfrac {-4 z} {\paren {1 - z }^2} }$
Let $z \to -1$ and we have:
:$\ds \map F {a, b; 1 + a - b; -1} = 2^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; 1 }$
From Gauss's Hypergeometric Theorem, we have:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
Therefore, the {{RHS}} becomes:
{{begin-eqn}}
{{eqn | l = 2^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; 1 }
      | r = 2^{-a} \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\paren {1 + a - b} - \dfrac a 2 - \paren {\dfrac {1 + a} 2 - b} } } {\map \Gamma {\paren {1 + a - b} - \dfrac a 2} \map \Gamma {\paren {1 + a - b} - \paren {\dfrac {1 + a} 2 - b} } }
      | c = Gauss's Hypergeometric Theorem
}}
{{eqn | r = \dfrac {\map \Gamma {1 + a - b} } {\map \Gamma {1 + \dfrac a 2 - b} } \paren {2^{-a} \dfrac {\map \Gamma {\dfrac 1 2 } } {\map \Gamma {\dfrac {1 + a} 2 } } }
      | c = simplifying
}}
{{eqn | r = \dfrac {\map \Gamma {1 + a - b} } {\map \Gamma {1 + \dfrac a 2 - b} } \paren {\dfrac {\map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {a + 1 } } }
      | c = Legendre's Duplication Formula
}}
{{eqn | r = \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {1 + \dfrac a 2 - b} \map \Gamma {a + 1 } }
      | c = simplifying
}}
{{end-eqn}}
Substituting $a = n$ and $b = -x$, we obtain:
{{begin-eqn}}
{{eqn | l = \map F {a, b; 1 + a - b; -1}
      | r = \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {1 + \dfrac a 2 - b} \map \Gamma {a + 1 } }
      | c = before substitution
}}
{{eqn | ll= \leadsto
      | l = \map F {n, -x; 1 + n + x; -1}
      | r = \dfrac {\map \Gamma {1 + n + x} \map \Gamma {\dfrac n 2 + 1 } } {\map \Gamma {1 + \dfrac n 2 + x} \map \Gamma {n + 1 } }
      | c = after substitution
}}
{{end-eqn}}
{{qed}}
Theorem: Dougall-Ramanujan Identity

{{begin-eqn}}
{{eqn | l = \map { {}_7 \operatorname F_6} { { {n, 1 + \dfrac n 2, -x, -y, -z, -u, x + y + z + u + 2n + 1} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1, u + n + 1, -x - y - z - u - n} } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1}  } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} }
}}
{{eqn | o = 
      | ro= \times
      | r = \dfrac {\map \Gamma {u + n + 1} \map \Gamma {x + z + u + n + 1} \map \Gamma {y + z + u + n + 1} \map \Gamma {x + y + u + n + 1}   } {\map \Gamma {x + u + n + 1} \map \Gamma {z + u + n + 1} \map \Gamma {y + u + n + 1} \map \Gamma {x + y + z + u + n + 1} }
}}
{{end-eqn}}

Proof:
By definition of generalized hypergeometric function of $1$:
:$\ds \map { {}_7 \operatorname F_6} { { {n, 1 + \dfrac n 2, -x, -y, -z, -u, x + y + z + u + 2n + 1} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1, u + n + 1, -x - y - z - u - n} } \, \middle \vert \, 1} = \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {1 + \dfrac n 2}^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} \paren {-u}^{\overline k} \paren {x + y + z + u + 2n + 1}^{\overline k} } { \paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} \paren {u + n + 1}^{\overline k} \paren {-x - y - z - u - n}^{\overline k} } \dfrac {1^k} {k!}$
{{ProofWanted}}
Theorem: Existence of Set of Ordinals leads to Contradiction

The existence of the set of all ordinals leads to a contradiction.

Proof:
Suppose that the collection of all ordinals is a set.
Let this set be denoted as $\On$.
From {{Corollary|Ordinals are Well-Ordered}}, it is seen that $\Epsilon {\restriction_{\On} }$ is a strict well-ordering on $\On$.
By Element of Ordinal is Ordinal, it is seen that $\On$ is transitive.
Hence $\On$ is itself an ordinal.
Since $\On$ is an ordinal, it follows {{hypothesis}} that $\On \in \On$.
Since $\On$ is an ordinal, it follows from Ordinal is not Element of Itself that $\On \notin \On$.
This is a contradiction.
{{qed}}
Theorem: Essential Singularity may or may not be Isolated

Let $f$ be a complex function with an essential singularity at $z_0 \in \C$.
Then $z_0$ may or may not be an '''isolated singularity'''.

Proof:
{{ProofWanted|An example of each. Motivation: to remove confusion and establish this basic fact. The author of this page is infodumping an encyclopedia entry and cannot remember the slightest inkling of his Complex Analysis studies except for the fact that nothing is intuitive and everything needs to be investigated with great care.}}
Category:Essential Singularities
Category:Isolated Singularities
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Theorem: Derivative of Concatenation of Complex Paths

Let $\gamma_0 : \closedint {a_0}{b_0} \to \C$ be an injective complex-differentiable function.
Let $\gamma_1 : \closedint {a_1}{b_1} \to \C$ be a complex-differentiable function such that $\map {\gamma_0}{b_0} = \map {\gamma_1}{a_1}$.
Let $t_0 \in \openint {a_0}{b_0}$.
Define the concatenation $\gamma_0 * \gamma_1 : \closedint 0 1 \to \C$ as:
:$\map {\gamma_0 * \gamma_1}{ s } = \begin{cases} \map {\gamma_0}{2 s \paren{b_0 - a_0} + a_0} & \forall s \in \closedint 0 {\dfrac 1 2} \\ \map {\gamma_1}{ \paren{2 s - 1} \paren{b_1 - a_1} + a_1} & \forall s \in \closedint {\dfrac 1 2} 1 \end{cases}$
Then there exists $s_0 \in \openint 0 {\dfrac 1 2}$ and $C \in \R_{>0}$ such that:
{{begin-eqn}}
{{eqn | l = \map {\gamma_0}{t_0} 
      | r = \map {\gamma_0 * \gamma_1}{s_0}
}}
{{eqn | l = C \map {\gamma_0 '}{t_0}
      | r = \map { \paren{ \gamma_0 * \gamma_1 }' }{s_0}
}}
{{end-eqn}}

Proof:
Set $s_0 := \dfrac {t_0 - a_0}{2 \paren{ b_0 - a_0 } }$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\gamma_0 * \gamma_1}{s_0}
      | r = \map{\gamma_0}{ \dfrac{2 \paren{ t_0 - a_0 } \paren{ b_0 - a_0 } }{2 \paren{ b_0 - a_0 } } + a_0 }
      | c = as $t_0 - a_0 < b_0 - a_0$, so $\dfrac {t_0 - a_0}{2 \paren{ b_0 - a_0 } } < \dfrac 1 2$.
}}
{{eqn | r = \map {\gamma_0}{t_0} 
}}
{{end-eqn}}
For all $s \in \openint 0 {\dfrac 1 2}$, it follows that:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\d \paren{\gamma_0 * \gamma_1} } {\d s} } s 
      | r = \map {\dfrac {\d {\gamma_0} } {\d s} }{ 2 s \paren{b_0 - a_0} + a_0 } \dfrac{\d}{\d s} \paren{ 2 s \paren{b_0 - a_0} + a_0}
      | c = by Derivative of Complex Composite Function
}}
{{eqn | r = 2 \paren{ b_0 - a_0} \map {\dfrac {\d {\gamma_0} } {\d s} }{ 2 s \paren{b_0 - a_0} + a_0 }
      | c = by Derivative of Complex Polynomial
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map { \paren{ \gamma_0 * \gamma_1 }' }{s_0}
      | r = 2 \paren{ b_0 - a_0} \map {\gamma_0 '}{ \dfrac{2 \paren{ t_0 - a_0 } \paren{ b_0 - a_0 } }{2 \paren{ b_0 - a_0 } } + a_0 }
}}
{{eqn | r = 2 \paren{ b_0 - a_0} \map {\gamma_0 '}{t_0}
}}
{{end-eqn}}
As $\closedint {a_0}{b_0}$ is a closed real interval, it follows that $a_0 < b_0$, so $2 \paren {b_0 - a_0} > 0$.
{{qed}}
Category:Complex Differential Calculus
dvcxwdc24q7a0dm7f5if9yzjqg1tptw
Theorem: Equivalent Norms on Lipschitz Space/Shift of Finite Type

Let $\struct {X, \sigma}$ be a shift of finite type.
Let $F_\theta$ be the Lipschitz space on $X$.
Let $x_0 \in F_\theta$.
Then the following norms on $F_\theta$ are equivalent:
:$(1): \quad$ Lipschitz norm $\norm\cdot_\theta$
:$(2): \quad \forall f \in F_\theta : \norm f '_\theta := \cmod {\map f {x_0} } + \size f_\theta$

Proof:
{{ProofWanted}}
Category:Functional Analysis
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Theorem: Transfer Operator with respect to One-Sided Shift Space of Finite Type is Linear Bounded Operator

Let $\struct {X ^+, \sigma}$ be a one-sided shift of finite type.
Let $\struct {\map C {X ^+, \C}, \norm \cdot_\infty}$ be the continuous mapping space.
Let $f \in \map C {X ^+, \C}$.
Let $\LL_f : \map C {X ^+, \C} \to \map C {X ^+, \C}$ the transfer operator.
Then $\LL_f$ is a bounded linear operator.

Proof:
The linearity is clear by definition of $\LL_f$ since Summation is Linear.
Let $\mathbf A$ be the $k\times k$ logical matrix of $X ^+$.
Let $g \in \map C {X ^+, \C}$.
For each $x \in X^+$:
{{begin-eqn}}
{{eqn | l = \cmod {\map {\LL_f g} x}
      | r = \cmod {\sum_{\map \sigma y = x} e^{\map f y} \map g y}
}}
{{eqn | o = \le
      | r = \sum_{\map \sigma y = x} e^{\norm f_\infty} \norm g_\infty
}}
{{eqn | r = \card {\set {y_0 : \map {\mathbf A} {y_0, x_0} = 1} } e^{\norm f_\infty} \norm g_\infty
}}
{{eqn | o = \le
      | r = k e^{\norm f_\infty} \norm g_\infty
}}
{{end-eqn}}
Let $c := k e^{\norm f_\infty}$.
Then:
:$\forall g \in \map C {X ^+, \C} : \norm {\LL_f g}_\infty \le c \norm g_\infty$
{{finish|Details}}
Category:Ergodic Theory
Category:Functional Analysis
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Theorem: Ruelle-Perron-Frobenius Theorem

Let $\struct {X_\mathbf A ^+, \sigma}$ be a one-sided shift of finite type.
Let $\map C {X_\mathbf A ^+} := \map C {X_\mathbf A ^+, \C}$ be the continuous mapping space.
Let $F_\theta ^+$ be the space of Lipschitz functions on $X_\mathbf A ^+$.
Let $f \in F_\theta ^+$ be a real-valued function.
Let $\LL_f : \map C {X_\mathbf A ^+} \to \map C {X_\mathbf A ^+}$ be the transfer operator.
Let $\beta $ be the spectral radius of $\LL_f$.
Suppose $\mathbf A$ is irreducible.
Then:
:$(1): \quad \beta$ is a simple eigenvalue of $\LL_f$ with a strictly positive eigenfunction $h \in F_\theta ^+$, i.e.:
::::$\set {g \in \map C {X_\mathbf A ^+} : \LL_f g = \beta g} = \C \cdot h$
Furthermore, if $\mathbf A$ is aperiodic, then:
:$(2): \quad$ There is an $r \in \openint 0 \beta$ such that:
::::$\map \sigma {\LL_f} \setminus \set \beta \subseteq \map B {0, r}$
:::where:
::::the {{LHS}} denotes a closed disk
::::$\map \sigma {\LL_f}$ denotes the spectrum of $\LL_f : F_\theta ^+ \to F_\theta ^+$
:$(3): \quad$ There is a Borel probability measure $\mu$ such that:
::::$\ds \forall v \in \map C {X_\mathbf A ^+} : \int \LL_f v \rd \mu = \beta \int v \rd \mu$
:$(4): \quad \ds \forall v \in \map C {X_\mathbf A ^+} : \lim _{n \mathop \to \infty} \norm {\beta^{-n} \LL_f ^n v - h \int v \rd \mu}_\infty = 0$
:$(5): \quad \ds \int h \rd \mu = 1$

Proof:
{{ProofWanted}}
{{Namedfor|David Pierre Ruelle|name2 = Oskar Perron|name3 = Ferdinand Georg Frobenius|cat = Ruelle|cat2 = Perron|cat3 = Frobenius}}
Category:Ergodic Theory
Category:Functional Analysis
3mx6hi111yd4ulawcq0olmbotwsh3ta
Theorem: Primitive of x by Cosine of x

:$\ds \int x \cos x \rd x = \cos x + x \sin x + C$

Proof:
From Primitive of $x \cos a x$:
:$\ds \int x \cos a x \rd x = \frac {\cos a x} {a^2} + \frac {x \sin a x} a + C$
The result follows on setting $a = 1$.
{{qed}}
Theorem: Primitive of x by Cosine of x

:$\ds \int x \cos x \rd x = \cos x + x \sin x + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Derivative of Identity Function
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \cos x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sin x
      | c = Primitive of $\cos x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \cos x \rd x
      | r = x \sin x - \int \paren {\sin x} \times 1 \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \sin x - \int \sin x \rd x + C
      | c = simplifying
}}
{{eqn | r = x \sin x - \paren {-\cos x} + C
      | c = Primitive of $\sin x$
}}
{{eqn | r = \cos x + x \sin x + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Cosine of x

:$\ds \int x \cos x \rd x = \cos x + x \sin x + C$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\cos x + x \sin x}
      | r = \map {\dfrac \d {\d x} } {\cos x} + \map {\dfrac \d {\d x} } {x \sin x}
      | c = Sum Rule for Derivatives
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\cos x} + x \map {\dfrac \d {\d x} } {\sin x} + \map {\dfrac \d {\d x} } x \sin x
      | c = Product Rule for Derivatives
}}
{{eqn | r = -\sin x + x \cos x + \sin x
      | c = Derivative of Cosine Function, Derivative of Sine Function, Derivative of Identity Function
}}
{{eqn | r = x \cos x
      | c = simplifying
}}
{{end-eqn}}
Hence the result by definition of primitive.
{{qed}}
Theorem: Primitive of Power of a x + b/Proof 3

{{:Primitive of Power of a x + b}}
where $n \ne 1$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} }
      | r = \dfrac {\paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a} \map {\dfrac \d {\d x} } {a x + b}
      | c = Power Rule for Derivatives, Chain Rule for Derivatives
}}
{{eqn | r = \dfrac {a \paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a}
      | c = Power Rule for Derivatives
}}
{{eqn | r = \paren {a x + b}^n
      | c = simplifying
}}
{{end-eqn}}
The result follows by definition of primitive.
{{qed}}
Theorem: Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale/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 discrete-time $\sequence {\FF_n}_{n \mathop \ge 0}$-adapted stochastic process.
Then $\sequence {X_n}_{n \mathop \ge 0}$ is a $\sequence {\FF_n}_{n \mathop \ge 0}$-martingale {{iff}} it is a $\sequence {\FF_n}_{n \mathop \ge 0}$-supermartingale and a $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.

Proof:
For each $n \in \Z_{\ge 0}$, we have: 
:$\expect {X_{n + 1} \mid \FF_n} = X_n$ almost surely
{{iff}}:
:$\expect {X_{n + 1} \mid \FF_n} \le X_n$ almost surely
and:
:$\expect {X_{n + 1} \mid \FF_n} \ge X_n$ almost surely.
That is: 
:$\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \mathop \ge 0}$-martingale {{iff}} it is a $\sequence {\FF_n}_{n \mathop \ge 0}$-supermartingale and a $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.
{{qed}}
Theorem: Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale/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 continuous-time $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
Then $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale {{iff}} it is a $\sequence {\FF_t}_{t \ge 0}$-supermartingale and a $\sequence {\FF_t}_{t \ge 0}$-submartingale.

Proof:
For each $t, s \in \hointr 0 \infty$ with $0 \le s < t$, we have: 
:$\expect {X_t \mid \FF_s} = X_s$ almost surely
{{iff}}:
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely
and:
:$\expect {X_t \mid \FF_s} \ge X_s$ almost surely.
That is: 
:$\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale {{iff}} it is a $\sequence {\FF_t}_{t \ge 0}$-supermartingale and a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
{{qed}}
Category:Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale
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Theorem: Adapted Stochastic Process is Supermartingale iff Negative is Submartingale/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 $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
Then $\sequence {X_t}_{t \ge 0}$ is a supermartingale {{iff}} $\sequence {-X_t}_{t \ge 0}$ is a submartingale.

Proof:
Since $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process:
:$X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$.
From Pointwise Scalar Multiple of Measurable Function is Measurable:
:$-X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$.
So $\sequence {-X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
We then just need to check conditional expectations.
Let $s, t \in \hointr 0 \infty$ with $0 \le s < t$. 
From Conditional Expectation is Linear, we have: 
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely {{iff}} $\expect {-X_t \mid \FF_s} \ge -X_s$ almost surely.
That is: 
:$\sequence {X_t}_{t \ge 0}$ is a supermartingale {{iff}} $\sequence {-X_t}_{t \ge 0}$ is a submartingale.
{{qed}}