avewright/proofwiki-math huggingface dataset:
Theorem: Order and Size of Graph do not determine Degrees of Vertices

Let $G = \struct {V, E}$ be a simple graph.
Let both the order $\card V$ and size $\card E$ of $G$ be given.
Then it is not always possible to determine the degrees of each of the vertices of $G$.

Proof:
The following $2$ simple graphs both have order $4$ and size $4$:
:400px
But:
:the graph on the left has vertices with degrees $1, 2, 2, 3$
:the graph on the right has vertices with degrees $2, 2, 2, 2$.
{{qed}}
Theorem: Size of Regular Graph in terms of Degree and Order

Let $G = \struct {V, E}$ be an $r$-regular graph of order $p$.
Let $q$ denote the size of $G$.
Then:
:$q = \dfrac {p r} 2$
when such an $r$-regular graph exists.
If an $r$-regular graph of order $p$ does exist, then $p r$ is an even integer.

Proof:
Let $G$ be of order $p$, size $q$ and $r$-regular.
Then by definition $G$ has:
:$p$ vertices each of which is of degree $r$.
:$q$ edges.
Thus:
{{begin-eqn}}
{{eqn | l = q
      | r = \dfrac 1 2 \sum_{v \mathop \in V} \map \deg v
      | c = Handshake Lemma
}}
{{eqn | r = \dfrac 1 2 \sum_{v \mathop \in V} r
      | c = {{Defof|Regular Graph}}
}}
{{eqn | r = \dfrac 1 2 p r
      | c = {{Defof|Order of Graph}}
}}
{{end-eqn}}
If $p r$ is odd, that means $p$ is odd and $r$ is odd.
That means there is an odd number of odd vertices.
From {{Corollary|Handshake Lemma}}, this is not possible.
Thus for $G$ to exist, it is necessary for $p r$ to be even.
The result follows.
{{Qed}}
Theorem: Same Degrees of Vertices does not imply Graph Isomorphism

Let $G = \struct {\map V G, \map E G}$ and $H = \struct {\map V H, \map E H}$ be graphs such that:
:$\card {\map V G} = \card {\map V H}$
where $\card {\map V G}$ denotes the order of $G$.
Let $\phi: G \to H$ be a mapping which preserves the degrees of the vertices:
:$\forall v \in \map V G: \map {\deg_H} {\map \phi v} = \map {\deg_G} v$
Then it is not necessarily the case that $\phi$ is an isomorphism.

Proof:
Proof by Counterexample:
:400px
Consider a bijection $\phi: \map V {G_1} \to \map V {G_2}$, where $G_1$ is the graph on the left and $G_2$ is the graph on the right.
The vertices $v_1$, $v_2$ and $v_5$ of $G_2$ are each adjacent to both of the others.
Because $\phi$ is a bijection, it must map $3$ vertices of $G_1$ to $v_1$, $v_2$ and $v_5$.
For $\phi$ to be an isomorphism, two of the vertices of $G_1$ are adjacent {{iff}} the two image vertices in $G_2$ udner $\phi$ are also adjacent.
So the $3$ vertices of $G_1$ which map to $v_1$, $v_2$ and $v_5$ of $G_2$ must also each be adjacent to both of the others.
But $G_1$ does not contain $3$ such vertices.
It follows that there is no such isomorphism from $\map V {G_1}$ to $\map V {G_2}$.
That is, $G_1$ and $G_2$ are not isomorphic.
{{qed}}
Theorem: Order 1 Simple Graph is Unique up to Isomorphism

Let $G_1 = \struct {\map V {G_1}, \map E {G_1} }$ and $G_2 = \struct {\map V {G_2}, \map E {G_2} }$ be simple graphs of order $1$.
Then $G_1$ and $G_2$ are isomorphic.

Proof:
There is only one bijection from $\map V {G_1}$ to $\map V {G_2}$.
There are no vertices adjacent to the sole vertex in $\map V {G_1}$
There are no vertices adjacent to the sole vertex in $\map V {G_2}$.
Hence the result, vacuously.
{{qed}}
Theorem: Edgeless Graphs of Order n are Isomorphic

Let $n \in \Z_{>0}$ be a positive integer.
Let $G_1 = \struct {\map V {G_1}, \map E {G_1} }$ and $G_2 = \struct {\map V {G_2}, \map E {G_2} }$ be edgeless graphs of order $n$.
Then $G_1$ and $G_2$ are isomorphic.

Proof:
Let $\phi$ be a bijection from $\map V {G_1}$ to $\map V {G_2}$.
This is possible because $\card {\map V {G_1} } = \card {\map V {G_2} } = n$.
Let $v_a \in \map V {G_1}$.
By definition, $v_a$ is adjacent to no other vertices of $G_1$.
Similarly by definition, $\map \phi {v_a}$ is adjacent to no other vertices of $G_2$.
Thus, vacuously, $v_a$ and $v_b$ are adjacent in $G_1$ {{iff}} $\map \phi {v_a}$ and $\map \phi {v_b}$ are adjacent in $G_2$.
Hence the result.
{{qed}}
Category:Graph Isomorphisms
l4uf2dh89r03u17clwi3axsy0o01c1i
Theorem: Isomorphism Classes for Order 2 Simple Graphs

There are $2$ equivalence classes for simple graphs of order $2$ under graph isomorphism:
:the edgeless graph of order $2$
and
:the complete graph of order $2$.

Proof:
{{ProofWanted|Too tedious to contemplate.}}
Theorem: Isomorphism Classes for Order 4 Size 3 Simple Graphs

There are $3$ equivalence classes for simple graphs of order $4$ and size $3$ under isomorphism:
:400px

Proof:
{{tidy}}
{{MissingLinks}}
The fact that the $3$ graphs given are not isomorphic follows from Vertex Condition for Isomorphic Graphs.
The vertices have degrees as follows:
:Graph $1$: $2, 2, 1, 1$
:Graph $2$: $3, 1, 1, 1$
:Graph $3$: $2, 2, 2, 0$
The fact that there are no more isomorphism classes of such graphs can be proved constructively.
Let the $4$ vertices be named $A, B, C$ and $D$.
Lemma: There must be intersections among the edges.
Proof: If there were no intersection at all, it requires at least $3 \times 2 = 6$ vertices.
{{WLOG}}, let $2$ of the edges be $AB$ and $AC$.
To place the last edge, there are $\dbinom 4 2 = 6$ potential choices:
:$AB$: this makes the graph not simple
:$AC$: this makes the graph not simple
:$AD$: this is isomorphic to graph $2$
:$BC$: this is isomorphic to graph $3$
:$BD$: this is isomorphic to graph $1$
:$CD$: this is isomorphic to graph $1$.
Hence, by Proof by Cases, these $3$ are the only isomorphism classes.
{{qed}}
Theorem: Differential Equation governing First-Order Reaction

Let a substance decompose spontaneously in a '''first-order reaction'''.
The differential equation which governs this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
where:
:$x$ determines the quantity of substance at time $t$.
:$k \in \R_{>0}$.

Proof:
From the definition of a first-order reaction, the rate of change of the quantity of the substance is proportional to the quantity of the substance present at any time.
As the rate of change is a decrease, this rate will be negative.
Thus the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for some $k \in \R_{>0}$.
{{qed}}
Theorem: Formula for Radiocarbon Dating

Let $Q$ be a quantity of a sample of dead organic material (usually wood) whose time of death is to be determined.
Let $t$ years be the age of $Q$ which is to be determined.
Let $r$ denote the ratio of the quantity of carbon-14 remaining in $Q$ after time $t$ to the quantity of carbon-14 in $Q$ at the time of its death.
Then the number of years that have elapsed since the death of $Q$ is given by:
:$t = -8060 \ln r$

Proof:
Let $x_0$ denote the ratio of carbon-14 to carbon-12 in $Q$ at the time of its death.
Let $x$ denote the ratio of carbon-14 to carbon-12 in $Q$ after time $t$.
Thus:
:$r = \dfrac x {x_0}$
It is assumed that the rate of decay of carbon-14 is a first-order reaction.
Hence we use:
{{begin-eqn}}
{{eqn | l = x
      | r = x_0 e^{-k t}
      | c = First-Order Reaction
}}
{{eqn | ll= \leadsto
      | l = \dfrac x {x_0}
      | r = e^{-k t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -k t
      | r = \map \ln {\dfrac x {x_0} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -k t
      | r = \map \ln r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t
      | r = -\dfrac {\ln r} k
      | c = 
}}
{{end-eqn}}
Chemical analysis tells us that after $10$ years, $99.876 \%$ of the carbon-14 that was present in the organic matter still remains.
Thus at time $t = 10$, we have:
{{begin-eqn}}
{{eqn | l = 0.99876 x_0
      | r = x_0 e^{-10 k}
      | c = substituting for $t$ and $x$ in $(1)$
}}
{{eqn | ll= \leadsto
      | l = \ln 0.99876
      | r = -10 k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k
      | o = \approx
      | r = 0.000124
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 k
      | o = \approx
      | r = 8060
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Period of Reciprocal of Prime

Consider the decimal expansion of the reciprocal $\dfrac 1 p$ of a prime $p$.
If $p \nmid a$, the decimal expansion of $\dfrac 1 p$ is periodic in base $a$ and its period of recurrence is the order of $a$ modulo $p$.
If $p \divides a$, the decimal expansion of $\dfrac 1 p$ in base $a$ terminates.

Proof:
=== Case $1$: $p \divides a$ ===
Let $q = \dfrac a p$.
Then $\dfrac 1 p = \dfrac q a$.
So the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.q$ and terminates.
{{qed|lemma}}
Theorem: Maximum Period of Reciprocal of Prime

Let $p$ be a prime number such that $p$ is not a divisor of $10$.
The maximum period of recurrence of the reciprocal of $p$ when expressed in decimal notation is $p - 1$.

Proof:
First we dispose of the case where $p$ ''is'' a divisor of $10$:
When $p \divides 10$, $\dfrac 1 p$ expressed in decimal notation is a terminating decimal:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2
      | r = 0 \cdotp 5
}}
{{eqn | l = \dfrac 1 5
      | r = 0 \cdotp 2
}}
{{end-eqn}}
{{qed|lemma}}
So, let $p$ be such that $p \nmid 10$.
From Period of Reciprocal of Prime, the period of recurrence is the order of $10$ modulo $p$.
That is, it is the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
From Fermat's Little Theorem:
:$10^{p - 1} \equiv 1 \pmod p$
Hence the maximum period of recurrence occurs when the order of $10$ modulo $p$ is $p - 1$.
{{qed}}
Theorem: Maximum Abscissa for Loop of Folium of Descartes

Consider the folium of Descartes defined in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
:500px
The point on the loop at which the $x$ value is at a maximum occurs when $t = \sqrt [3] {\dfrac 1 2}$, corresponding to the point $P$ defined as:
:$P = \tuple {2^{2/3} a, 2^{1/3} a}$

Proof:
We calculate the derivative of $x$ {{WRT|Differentiation}} $t$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d x} {\d t}
      | r = \map {\dfrac \d {\d t} } {\dfrac {3 a t} {1 + t^3} }
      | c = 
}}
{{eqn | r = \dfrac {\paren {1 + t^3} \times 3 a - 3 a t \paren {3 t^2} } {\paren {1 + t^3}^2}
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \dfrac {3 a - 6 a t^3} {\paren {1 + t^3}^2}
      | c = simplifying
}}
{{end-eqn}}
Thus $x$ is stationary when:
{{begin-eqn}}
{{eqn | l = 3 a - 6 a t^3
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t
      | r = \paren {\dfrac 1 2}^{1/3}
      | c = 
}}
{{end-eqn}}
From Behaviour of Parametric Equations for Folium of Descartes according to Parameter, it is clear from the geometry that $x$ is a local maximum for this value of $t$.
Then we have:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + \paren {\paren {1/2}^{1/3} }^3}
      | c = 
}}
{{eqn | r = \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + 1/2}
      | c = 
}}
{{eqn | r = 2 a \times \paren {\dfrac 1 2}^{1/3}
      | c = 
}}
{{eqn | r = \paren {\dfrac {2^3} 2}^{1/3} a
      | c = 
}}
{{eqn | r = 2^{2/3} a
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y
      | r = \dfrac {3 a \times \paren {\paren {1/2}^{1/3} }^2} {1 + \paren {\paren {1/2}^{1/3} }^3}
      | c = 
}}
{{eqn | r = \dfrac {3 a \times \paren {1/2}^{2/3} } {1 + 1/2}
      | c = 
}}
{{eqn | r = 2 a \times \paren {\dfrac 1 2}^{2/3}
      | c = 
}}
{{eqn | r = \paren {\dfrac {2^3} {2^2} }^{1/3} a
      | c = 
}}
{{eqn | r = 2^{1/3} a
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Folium of Descartes
36vvg81m23h2dtjw42t9p2jjtb49jzn
Theorem: Third Derivative of Natural Logarithm Function

Let $\ln x$ be the natural logarithm function.
Then:
:$\map {D^3_x} {\ln x} = \dfrac 2 {x^3}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {D^3_x} {\ln x}
      | r = \map {D_x} {\map {D^2_x} {\ln x} }
      | c = 
}}
{{eqn | r = \map {D_x} {-\dfrac 1 {x^2} }
      | c = Second Derivative of Natural Logarithm Function
}}
{{eqn | r = -\paren {-2 \paren {\dfrac 1 {x^3} } }
      | c = Power Rule for Derivatives: Integer Index
}}
{{eqn | r = \dfrac 2 {x^3}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Denesting of Square Root/Lemma

Let $a, b, c, d \in \Q_{\ge 0}$.
Suppose $\sqrt b \notin \Q$.
Then:
:$\sqrt {a + \sqrt b} = \sqrt {c + \sqrt d} \implies a = c, b = d$

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt {a + \sqrt b}
      | r = \sqrt {c + \sqrt d}
}}
{{eqn | ll= \leadsto
      | l = a + \sqrt b
      | r = c + \sqrt d
}}
{{eqn | ll= \leadsto
      | l = a - c
      | r = \sqrt d - \sqrt b
}}
{{eqn | ll= \leadsto
      | l = \sqrt d - \sqrt b
      | o = \in
      | r = \Q
      | c = Rational Subtraction is Closed
}}
{{end-eqn}}
{{AimForCont}} $b \ne d$.
Then:
{{begin-eqn}}
{{eqn | l = \sqrt d - \sqrt b
      | o = \ne
      | r = 0
}}
{{eqn | l = d - b
      | r = \paren {\sqrt d - \sqrt b} \paren {\sqrt d + \sqrt b}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = \sqrt d + \sqrt b
      | r = \dfrac {d - b} {\sqrt d - \sqrt b} \in \Q
      | c = Rational Division is Closed
}}
{{eqn | ll= \leadsto
      | l = 2 \sqrt b
      | r = \paren {\sqrt d + \sqrt b} - \paren {\sqrt d - \sqrt b} \in \Q
      | c = Rational Subtraction is Closed
}}
{{eqn | ll= \leadsto
      | l = \sqrt b
      | o = \in
      | r = \Q
      | c = Rational Division is Closed
}}
{{end-eqn}}
But this contradicts our assertion that $\sqrt b \notin \Q$.
Hence our supposition that $b \ne d$ must be false.
Therefore we must have $b = d$. 
Consequently:
:$a - c = \sqrt d - \sqrt b = 0$
This implies $a = c$.
{{qed}}
Category:Condition for Denesting of Square Root
bn0zgpws24168r8vhlgmpc23beune07
Theorem: First Order ODE/y' + y = 0

The first order ODE:
:$\dfrac {\d y} {\d x} + y = 0$
has the general solution:
:$y = C e^{-x}$
where $C$ is an arbitrary constant.

Proof:
This first order ODE is in the form:
:$\dfrac {\d y} {\d x} + k y = 0$
where $k = 1$.
From First Order ODE: $\d y = k y \rd x$, this has the solution:
:$y = C e^{-x}$
Theorem: Linear Second Order ODE/y'' = 1 over 1 - x^2

The second order ODE:
:$(1): \quad y'' = \dfrac 1 {1 - x^2}$
has the general solution:
:$y = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D$

Proof:
{{begin-eqn}}
{{eqn | l = y''
      | r = \dfrac 1 {1 - x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \dfrac {\d^2 y} {\d x^2} \rd x
      | r = \int \frac 1 {1 - x^2} \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d x}
      | r = \tanh^{-1} x + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: Inverse Hyperbolic Tangent Form
}}
{{eqn | ll= \leadsto
      | l = \int \dfrac {\d y} {\d x} \rd x
      | r = \int \paren {\tanh^{-1} x + C} \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D
      | c = Primitive of $\tanh^{-1} x$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/y' = e^x

The first order ODE:
:$y' = e^x$
has the general solution:
:$y = e^x + C$
where $C$ is an arbitrary constant.

Proof:
{{begin-eqn}}
{{eqn | l = y'
      | r = e^x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \d y
      | r = \int e^x \rd x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = y
      | r = e^x + C
      | c = Primitive of Constant, Primitive of Exponential Function
}}
{{end-eqn}}
{{qed}}
Theorem: Semilattice Homomorphism is Order-Preserving

Let $\struct {S, \circ}$ and $\struct {T, *}$ be semilattices.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a semilattice homomorphism.
Let $\preceq_1$ be the ordering on $S$ defined by:
:$a \preceq_1 b \iff \paren {a \circ b} = b$
Let $\preceq_2$ be the ordering on $T$ defined by:
:$x \preceq_2 y \iff \paren {x * y} = y$
Then:
:$\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is order-preserving.

Proof:
{{begin-eqn}}
{{eqn | l = a \preceq_1 b
      | o = \leadstoandfrom
      | r = a \circ b = b
      | c = Definition of the ordering $\preceq_1$
}}
{{eqn | o = \leadsto
      | r = \map \phi { a \circ b} = \map \phi b
}}
{{eqn | o = \leadstoandfrom
      | r = \map \phi a * \map \phi b = \map \phi b
      | c = {{Defof|Semilattice Homomorphism}}
}}
{{eqn | o = \leadstoandfrom
      | r = \map \phi a \preceq_2 \map \phi b
      | c = Definition of the ordering $\preceq_2$
}}
{{end-eqn}}
{{qed}}
Theorem: Order-Preserving Mapping Not Always Semilattice Homomorphism

Let $\struct {S, \circ}$ and $\struct {T, *}$ be semilattices.
Let $\preceq_1$ be the ordering on $S$ defined by:
:$a \preceq_1 b \iff \paren {a \circ b} = b$
Let $\preceq_2$ be the ordering on $T$ defined by:
:$x \preceq_2 y \iff \paren {x * y} = y$
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order-preserving mapping.
Then:
:$\phi: \struct {S, \circ} \to \struct {T, *}$ may not be a semilattice homomorphism

Proof:
Let $S = \set {a, b, c}$.
Let $\circ : S \times S \to S$ be defined by:
:$a \circ b = b \circ a = c$
:$a \circ c = c \circ a = c$
:$b \circ c = c \circ b = c$
:$a \circ a = a$
:$b \circ b = b$
:$c \circ c = c$
Then a manual check shows that $\circ$ is:
:closed
:associative
:commutative
:idempotent
So $\struct {S, \circ}$ is a semilattice with ordering:
:$\preceq_1 = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$
Let $T = \set{p, q, r, s}$.
Let $* : S \times S \to S$ be defined by:
:$p * q = q * p = r$
:$p * r = r * p = r$
:$q * r = r * q = r$
:$p * s = s * p = s$
:$q * s = s * q = s$
:$r * s = s * r = s$
:$p * p = p$
:$q * q = q$
:$r * r = r$
:$s * s = s$
Then a manual check shows that $*$ is:
:closed
:associative
:commutative
:idempotent
So $\struct {T, *}$ is a semilattice with ordering
:$\preceq_2 = \set {\tuple {p, p}, \tuple {q, q}, \tuple {r, r}, \tuple {s, s}, \tuple {p, r}, \tuple {q, r}, \tuple {p, s}, \tuple {q, s}, \tuple {r, s} }$
Let $\phi : S \to T$ be defined by:
:$\map \phi a = p$
:$\map \phi b = q$
:$\map \phi c = s$
Then a manual check shows that $\phi$ is order-preserving.
Now:
{{begin-eqn}}
{{eqn | l = \map \phi a * \map \phi b
      | r = p * q
}}
{{eqn | r = r
}}
{{eqn | o = \ne
      | r = s
}}
{{eqn | r = \map \phi c
}}
{{eqn | r = \map \phi {a \circ b}
}}
{{end-eqn}}
Hence $\phi$ is not a semilattice homomorphism by definition.
{{qed}}
Theorem: Schwarz's Lemma/Lemma

Let $D$ be the unit disk centred at $0$.
Let $g : D \to \C$ be a complex function such that: 
:$\map g z = \begin {cases} \dfrac {\map f z} z & z \ne 0 \\ \\ \map {f'} 0 & z = 0\end {cases}$
Then $g$ is holomorphic on $D$.

Proof:
By Differentiable Function is Continuous, $f$ is continuous.
So by Quotient Rule for Continuous Complex Functions:
:$g$ is continuous on $D \setminus \set 0$.
We aim to show that $f$ is continuous on $D$. 
Note that since $f$ is holomorphic on $D$ and $0 \in D$ we have, by the definition of the complex derivative: 
:$\ds \lim_{z \mathop \to 0} \frac {\map f z - \map f 0} z = \map {f'} 0 \in \C$
Since $\map f 0 = 0$, we furthermore have: 
:$\ds \map {f'} 0 = \lim_{z \mathop \to 0} \frac {\map f z} z$
That is: 
:$\ds \map g 0 = \lim_{z \mathop \to 0} \map g z$
so $g$ is continuous at $0$.
Since $f$ is holomorphic on $D$, by the Quotient Rule for Continuous Complex Functions:
:$g$ is differentiable on $D \setminus \set 0$.
It remains to show that $g$ is differentiable at $0$. 
Take $z \ne 0$ and consider: 
:$\dfrac {\map g z - \map g 0} z$
We have: 
{{begin-eqn}}
{{eqn	| l = \frac {\map g z - \map g 0} z
	| r = \frac {\frac {\map f z} z - \map {f'} 0} z
	| c = as $\map g z = \dfrac {\map f z} z$ for $z \ne 0$ and $\map g 0 = \map {f'} 0$
}}
{{eqn	| r = \frac {\map f z - z \map {f'} 0} {z^2}
}}
{{end-eqn}}
Since $f$ is holomorphic on $D$, by Holomorphic Function is Analytic, there exists a positive real number $R$ such that the series: 
:$\ds \sum_{n \mathop = 0}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n$
converges to $\map f z$ on $\cmod z < R$.
Note that since $\map f 0 = 0$, the first term of this series is zero.
With that, we have: 
{{begin-eqn}}
{{eqn	| l = \frac {\map f z - z \map {f'} 0} {z^2}
	| r = \frac {\sum_{n \mathop = 1}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n - z \map {f'} 0} {z^2}
}}
{{eqn	| r = \frac {z \map {f'} 0 + \frac {z^2} 2 \map {f' '} 0 + \sum_{n \mathop = 3}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n - z \map {f'} 0} {z^2}
}}
{{eqn	| r = \frac 1 2 \map {f' '} 0 + \sum_{n \mathop = 3}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^{n - 2}
}}
{{end-eqn}}
Taking $z \to 0$ we have: 
:$\ds \lim_{z \mathop \to 0} \frac {\map g z - \map g 0} z = \frac 1 2 \map {f' '} 0$
so $g$ is indeed differentiable at $0$.
Hence by definition $g$ is holomorphic on $D$.
{{qed}}
Category:Complex Analysis
Category:Schwarz's Lemma
ru81dh4sn4zm1qnr3y2876d67uxjvr5
Theorem: Factorial as Product of Consecutive Factorials/Lemma 2

Let $n \in \N$.
Then $\paren {2 n - 2}! \, \paren {2 n - 1}! > \paren {3 n - 1}!$ for all $n \ge 7$.

Proof:
We prove the result by induction on $n$.
Theorem: Factorial as Product of Consecutive Factorials/Lemma 1

Let $n \in \N$.
Then $\paren {2 n - 1}! \, \paren {2 n}! > \paren {3 n - 1}!$ for all $n > 1$.

Proof:
Let $n, k \in \N_{> 0}$.
Suppose $n > 1$ and $n > k$.
We show that $\paren {k + 1} \paren {2 n - k} > 2 n + k$.
For $k = 1$:
:$2 \paren {2 n - 1} = 4 n - 2 \ge 2 n + 2 > 2 n + 1$
For $k > 1$:
{{begin-eqn}}
{{eqn | l = \paren {k + 1} \paren {2 n - k}
      | r = 2 n k + 2 n - k^2 - k
}}
{{eqn | o = >
      | r = 2 k^2 + 2 n - k^2 - k
      | c = because $n > k$
}} 
{{eqn | r = k^2 + 2 n - k
}}
{{eqn | o = >
      | r = 2 k + 2 n - k
      | c = because $k \ge 2$
}} 
{{eqn | r = 2 n + k
}} 
{{end-eqn}}
Therefore we have:
{{begin-eqn}}
{{eqn | l = \paren {2 n - 1}! \, \paren {2 n}!
      | r = \paren {2 n}! \prod_{k \mathop = 1}^{2 n - 1} k
      | c = {{Defof|Factorial}}
}}
{{eqn | r = \paren {2 n}! \paren {\prod_{k \mathop = 1}^n k} \paren {\prod_{k \mathop = n + 1}^{2 n - 1} k}
}}
{{eqn | r = \paren {2 n}! \paren {\prod_{k \mathop = 0}^{n - 1} \paren {k + 1} } \paren {\prod_{k \mathop = 1}^{n - 1} \paren {2 n - k} }
}}
{{eqn | r = \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {k + 1} \paren {2 n - k}
      | c = $0 + 1 = 1$
}}
{{eqn | o = >
      | r = \paren {2 n}! \prod_{k \mathop = 1}^{n - 1} \paren {2 n + k}
}}
{{eqn | r = \paren {3 n - 1}!
}} 
{{end-eqn}}
{{qed}}
Category:Factorials
Category:Factorial as Product of Consecutive Factorials
8ee0wpjl6r7bkpljmuyx6kg9n3tf3a0
Theorem: Entire Function with Bounded Real Part is Constant

Let $f : \C \to \C$ be an entire function. 
Let the real part of $f$ be bounded.
That is, there exists a positive real number $M$ such that: 
:$\cmod {\map \Re {\map f z} } < M$
for all $z \in \C$, where $\map \Re {\map f z}$ denotes the real part of $\map f z$.
Then $f$ is constant.

Proof:
Let $g : \C \to \C$ be a complex function with:
:$\ds \map g z = e^{\map f z}$
By Derivative of Complex Composite Function, $g$ is entire with derivative:
:$\ds \map {g'} z = \map {f'} z e^{\map f z}$
We have:
{{begin-eqn}}
{{eqn	| l = \cmod {\map g z}
	| r = e^{\map \Re {\map f z} }
	| c = Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part
}}
{{eqn	| o = \le
	| r = e^{\cmod {\map \Re {\map f z} } }
	| c = Exponential is Strictly Increasing
}}
{{eqn	| o = <
	| r = e^M
	| c = Exponential is Strictly Increasing
}}
{{end-eqn}}
So $g$ is a bounded entire function.
By Liouville's Theorem, $g$ is therefore a constant function.
We therefore have, by Derivative of Constant: Complex:
:$\map {g'} z = 0$
for all $z \in \C$.
That is:
:$\map {f'} z e^{\map f z} = 0$
Since the exponential function is non-zero, we must have: 
:$\map {f'} z = 0$
for all $z \in \C$.
From Zero Derivative implies Constant Complex Function, we then have that $f$ is constant on $\C$. 
{{qed}}
Category:Complex Analysis
1g5fd8xm6jcre09hht6q1kgwu4in3fd