nyuuzyou/wikis huggingface dataset:
Liouville's Theorem (Complex Analysis)/Proof 1

== Theorem ==
== Proof ==
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$. 
For any $R \in \R: R > 0$, consider the function:
:$\map {f_R} z := \map f {R z}$
Using the Cauchy Integral Formula, we see that:
:$\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map {f_R} w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \cmod {\d w} = M$
where $\map {C_1} z$ denotes the circle of radius $1$ around $z$.
Hence:
:$\ds \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$
Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$.
Thus $f$ is constant.
Category:Liouville's Theorem (Complex Analysis)
Liouville's Theorem (Complex Analysis)/Proof 2

== Theorem ==
== Proof ==
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$. 
Let $r > 0$.
consider:
:$D_r = \set {z \in \C: \cmod z \le r}$
Then, consider the parameterization $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ given by:
:$\map {\gamma_r} t := r e^{2 \pi i t}$
For all $z \in \C$, we have:
where:
:$\map {g_z} w := \map f {w + z}$
Hence, for all $z \in \C$:
Thus it follows that $\map {f'} z = 0$ for all $z \in \C$.
By Zero Derivative implies Constant Complex Function, $f$ is constant.
Category:Liouville's Theorem (Complex Analysis)
Liouville's Theorem (Differential Equations)

== Theorem ==
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:
:$\det \map \Phi t = e^{\int_{t_0}^t \tr \map A s \rd s} \det \map \Phi {t_0}$
== Proof ==
Category:Differential Equations
Category:Liouville's Theorem
Liouville's Theorem (Number Theory)

== Theorem ==
Let $x$ be an irrational number that is algebraic of degree $n$.
Then there exists a constant $c > 0$ (which can depend on $x$) such that:
:$\size {x - \dfrac p q} \ge \dfrac c {q^n}$
for every pair $p, q \in \Z$ with $q \ne 0$.
=== Corollary ===
== Proof ==
Let $r_1, r_2, \ldots, r_k$ be the rational roots of a polynomial $P$ of degree $n$ that has $x$ as a root.
Since $x$ is irrational, it does not equal any $r_i$. 
Let $c_1 > 0$ be the minimum of $\size {x - r_i}$.
If there are no $r_i$, let $c_1 = 1$.
Now let $\alpha = \dfrac p q$ where $\alpha \notin \set {r_1, \ldots, r_k}$. 
Then:
Suppose:
:$\ds \map P x = \sum_{k \mathop = 0}^n a_k x^k$
Then:
Case 1: Let $\size {x - \alpha} \le 1$.
Then:
Therefore:
To summarize:
:$\size {\map P x - \map P \alpha} \le \size {x - \alpha} c_x$
where:
:$\ds c_x = \sum_{k \mathop = 1}^n \size {a_k} \paren {\paren {\size x + 1}^k - \size x^k}$
So for such $\alpha$:
:$\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$
Case 2: Let $\size {x - \alpha} > 1$.
Then:
:$\size {x - \alpha} > 1 \ge \dfrac 1 {q^n}$
== Examples ==
== Also see ==
* Definition:Liouville Number
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Liouville's Theorem (Number Theory)
Category:Number Theory
Category:Transcendental Number Theory
Category:Liouville's Theorem
Liouville's Theorem (Number Theory)/Corollary

== Corollary to Liouville's Theorem ==
Liouville numbers are transcendental.
== Proof ==
Let $x$ be a Liouville number.
 $x$ is an algebraic number.
By Liouville Numbers are Irrational, then there exists $c > 0$ and $n \in \N_{>0}$ such that:
:$\size {x - \dfrac p q} \ge \dfrac c {q^n}$
for every pair $p, q \in \Z$ with $q \ne 0$.
Let $r \in \N_{>0}$ such that $2^r \ge \dfrac 1 c$.
Since $x$ is a Liouville number, there exists $p, q \in \Z$ with $q > 1$ such that:
which is a contradiction.
Thus $x$ is transcendental.
 
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* 
* 
Category:Liouville's Theorem (Number Theory)
Category:Transcendental Number Theory
Category:Number Theory
Liouville's Theorem (Number Theory)/Examples/Root 2

== Example of Use of Liouville's Theorem (Number Theory) ==
== Example == 
Applying Liouville's Theorem (Number Theory) to $\sqrt 2$:
:$\size {\sqrt 2 - \dfrac p q} \ge \dfrac c {q^2} \implies 0 \lt c \le 6 - 4 \sqrt 2$ 
for every pair $p, q \in \Z$ with $q \ne 0$.
== Proof ==
We begin by deriving a formula that will calculate the constant c:
Next, we know that Convergents are Best Approximations to an irrational number
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
Then $p_i$ $q_i$ and $c_i$ are as follows:
:$\begin{array}{r|cccccccccc}
i & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
p_i & 1 & 3 & 7 & 17 & 41 & 99 & 239 \\
q_i & 1 & 2 & 5 & 12 & 29 & 70 & 169 \\
\paren {\dfrac {p_i} {q_i} } & 1.0000 & 1.5000 & 1.4000 & 1.4167 & 1.4138 & 1.4143 & 1.4142 \\
c_i & \sqrt 2 - 1 & 6 - 4 \sqrt 2 & 25 \sqrt 2 - 35 & 204 - 144 \sqrt 2 & 841 \sqrt 2 - 1189 & 6930 - 4900 \sqrt 2 & 28561 \sqrt 2 - 40391 \\
c_i & 0.4142 & 0.3431 & 0.3553 & 0.3532 & 0.3536 & 0.3535 & 0.3536 \\
\hline
\end{array}$
So for $\sqrt 2$, $c = 6 - 4 \sqrt 2$ (that is, the minimum constant $c_i$ of all of the convergents $\dfrac {p_i } {q_i }$ of $\sqrt 2$) will ensure that:
:$\size {\sqrt 2 - \dfrac p q} \ge \dfrac c {q^2}$
for every pair $p, q \in \Z$ with $q \ne 0$.
To further illustrate the concept, we review the graph of the function $\map f x = x^2 - 2$ in the diagram below.
For all convergents $\dfrac {p_i} {q_i}$ of $\sqrt 2$: 
:
Adding just one additional line to the table above:
:$\begin{array}{r|cccccccccc}
i & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
p_i & 1 & 3 & 7 & 17 & 41 & 99 & 239 \\
q_i & 1 & 2 & 5 & 12 & 29 & 70 & 169 \\
\paren {\dfrac {p_i } {q_i } } & 1.0000 & 1.5000 & 1.4000 & 1.4167 & 1.4138 & 1.4143 & 1.4142 \\
c_i & \sqrt 2 - 1 & 6 - 4 \sqrt 2 & 25 \sqrt 2 - 35 & 204 - 144 \sqrt 2 & 841 \sqrt 2 - 1189 & 6930 - 4900 \sqrt 2 & 28561 \sqrt 2 - 40391 \\
c_i & 0.4142 & 0.3431 & 0.3553 & 0.3532 & 0.3536 & 0.3535 & 0.3536 \\
\map {f'} x = \dfrac 1 {c_i} & \sqrt 2 + 1 & \sqrt 2 + \dfrac 3 2 & \sqrt 2 + \dfrac 7 5 & \sqrt 2 + \dfrac {17} {12} & \sqrt 2 + \dfrac {41} {29} & \sqrt 2 + \dfrac {99} {70} & \sqrt 2 + \dfrac {239} {169} \\
\hline
\end{array}$
Category:Liouville's Theorem (Number Theory)
Liouville Numbers are Irrational

== Theorem ==
Liouville numbers are irrational.
== Proof ==
Let $x$ be a Liouville number. 
 $x$ were rational, that is:
: $x = \dfrac a b$
with $a, b \in \Z$ and $b > 0$.
By definition of a Liouville number, for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that:
:$0 < \size {x - \dfrac p q} < \dfrac 1 {q^n}$
Let $n$ be a positive integer such that $2^{n - 1} > b$.
Let $p$ and $q$ be any integers with $q > 1$.
We have:
:$\size {x - \dfrac p q} = \dfrac {\size {a q - b p} } {b q}$
If $\size {a q - b p} = 0$, this would violate the first inequality.
If $\size {a q - b p} \ne 0$, then:
which would violate the second inequality.
Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities.
This is a contradiction.
Thus $x$ is irrational.
Category:Number Theory
Lipschitz Condition implies Uniform Continuity

== Theorem ==
Let $\struct {M_1, d_1}$ and $\struct {M_2, d_2}$ be metric spaces.
Let $g: M_1 \to M_2$ satisfy the Lipschitz condition.
Then $g$ is uniformly continuous on $M_1$.
== Proof ==
Let $\epsilon > 0$, $x,y \in M_1$. 
Let $K$ be a Lipschitz constant for $g$.
First, suppose that $K \le 0$.
Then:
:$\map {d_1} {x, y} \le 0 \map {d_2} {\map g x, \map g y}$ by the Lipschitz condition on $g$.
So $\map {d_1} {x, y} \le 0 \implies \map {d_1} {x, y} = 0 \implies x = y$ for all $x$ and $y$ in $M_1$.
Thus $g$ is a constant function, which is uniformly continuous.
Next, suppose that $K > 0$.
Take $\delta = \epsilon / K$.
Then if $\map {d_1} {x, y} < \delta$, we have:
:$K \map {d_1} {x, y} < \epsilon$
By the Lipschitz condition on $g$, we know that:
:$\map {d_2} {\map g x, \map g y} \le K \map {d_1} {x, y}$
These last two statements together imply $\map {d_2} {\map g x, \map g y} < \epsilon$.
Thus, $g$ is uniformly continuous on $M_1$.
Category:Metric Spaces
Category:Uniformly Continuous Mappings
Lipschitz Continuous Real Function is Absolutely Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f$ be Lipschitz continuous on $I$.
Then $f$ is absolutely continuous on $I$.
== Proof ==
By the definition of Lipschitz continuity, there exists $K \in \R$ such that: 
:$\size {\map f x - \map f y} \le K \size {x - y}$
for all $x, y \in I$.
If $K = 0$, then: 
:$\size {\map f x - \map f y} = 0$
for all $x, y \in I$. 
In this case, $f$ is constant.
Hence, by Constant Real Function is Absolutely Continuous:
:$f$ is absolutely continuous in the case $K = 0$.
Now, take $K > 0$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals.
We have: 
Let $\epsilon$ be a positive real number. 
Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon K$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is absolutely continuous in the case $K > 0$.
Hence the result.
== Sources ==
* 
Category:Lipschitz Continuity
Category:Absolutely Continuous Real Functions
Lipschitz Equivalence is Equivalence Relation

== Theorem ==
Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is Lipschitz equivalent to $d_2$
Then $\sim$ is an equivalence relation.
== Proof ==
Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $d_1$ be a metric on $A$.
Then trivially:
:$\forall x, y \in A: 1 \times \map {d_1} {x, y} \le \map {d_1} {x, y} \le 1 \times \map {d_1} {x, y}$
That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive.
=== Symmetry ===
Let $d_1 \sim d_2$.
That is, let $d_1, d_2$ be Lipschitz equivalent metrics on $A$.
Then by definition:
: $\forall x, y \in A: h \map {d_1} {x, y} \le \map {d_2} {x, y} \le k \map {d_1} {x, y}$
for some $h, k \in \R_{>0}$.
Then:
and:
That is:
:$\forall x, y \in A: \dfrac 1 h \map {d_2} {x, y} \le \map {d_1} {x, y} \le \dfrac 1 k \map {d_2} {x, y}$
for some $\dfrac 1 h, \dfrac 1 k \in \R_{>0}$.
That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be symmetric.
Note that in the definition of Lipschitz equivalent metrics, in the expression $d_1 \sim d_2$ it is not explicitly specified which of $d_1$ and $d_2$ goes in the middle of the defining statement.
This result demonstrates that it does not actually matter.
=== Transitivity ===
Let $d_1 \sim d_2$ and $d_2 \sim d_3$.
Then by definition:
:$\forall x, y \in A: h_1 \map {d_1} {x, y} \le \map {d_2} {x, y} \le k_1 \map {d_1} {x, y}$
:$\forall x, y \in A: h_2 \map {d_2} {x, y} \le \map {d_3} {x, y} \le k_2 \map {d_2} {x, y}$
for some $h_1, k_1, h_2, k_2 \in \R_{>0}$.
Then:
and:
So:
:$\forall x, y \in A: h_1 h_2 \map {d_1} {x, y} \le \map {d_3} {x, y} \le k_1 k_2 \map {d_1} {x, y}$
That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Proposal $2.4.3$
Category:Lipschitz Equivalence
Category:Examples of Equivalence Relations
Lipschitz Equivalence yields Lipschitz Equivalent Metric

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $f: M_1 \to M_2$ be a Lipschitz equivalence.
Let $d_3: A^2 \to \R$ be a mapping defined as:
:$\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$
Then $d_3$ yields a metric on $M_1$ which is Lipschitz equivalent to $d_1$.
== Proof ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics
Category:Lipschitz Equivalence
Lipschitz Equivalent Metric Spaces are Homeomorphic

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be Lipschitz equivalent.
Then $M_1$ and $M_2$ are homeomorphic.
== Proof ==
Let $M_1$ and $M_2$ be Lipschitz equivalent.
Then, by definition, $\exists h, k \in \R_{>0}$ such that:
:$\forall x, y \in A_1: h \map {d_1} {x, y} \le \map {d_2} {\map f x, \map f y} \le k \map {d_1} {x, y}$
From the definition of open $\epsilon$-ball:
and:
Thus:
:$\map {B_{h \epsilon} } {\map f x; d_2} \subseteq \map {B_\epsilon} {x; d_1}$
:$\map {B_{\epsilon / k} } {x; d_1} \subseteq \map {B_\epsilon} {\map f x; d_2}$
Now, suppose $U$ is $d_2$-open.
Let $x \in U$.
Then:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$.
Hence:
:$\map {B_{\epsilon / k} } {x; d_1} \subseteq U$
Thus $U$ is $d_1$-open.
Similarly, suppose $U$ is $d_1$-open.
Let $x \in U$.
Then:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$
Hence:
:$\map {B_{h \epsilon} } {\map f x; d_2} \subseteq U$
Thus $U$ is $d_2$-open.
The result follows by definition of homeomorphic metric spaces.
== Also see ==
* Lipschitz Equivalent Metrics are Topologically Equivalent
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Definition $2.4.8$
Category:Lipschitz Equivalence
Category:Homeomorphisms (Metric Spaces)
Lipschitz Equivalent Metrics are Topologically Equivalent

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be Lipschitz equivalent.
Then $d_1$ and $d_2$ are topologically equivalent.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Lipschitz Equivalent Metric Spaces are Homeomorphic
Category:Lipschitz Equivalence
Category:Topologically Equivalent Metrics
Category:Lipschitz Equivalent Metrics are Topologically Equivalent
Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 1

== Theorem ==
== Proof ==
Consider the identity mapping:
: $f: A \to A: \forall x \in A: f \left({x}\right) = x$
Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a Lipschitz equivalence.
The result then follows from Lipschitz Equivalent Metric Spaces are Homeomorphic.
Category:Lipschitz Equivalent Metrics are Topologically Equivalent
Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 2

== Theorem ==
== Proof ==
By definition of Lipschitz equivalence:
:$\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$
for some $h, k \in \R_{>0}$.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_{h \epsilon} } {x; d_1}$ denote the open $h \epsilon$-ball with respect to $d_1$ of $x \in A$.
Then:
Similarly:
Now suppose $U \subseteq A$ is $d_1$-open.
Let $x \in U$.
Then:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$
Thus:
:$\map {B_{\epsilon / k} } {x; d_2} \subseteq \map {B_\epsilon} {x; d_1} \subseteq U$
and so $U$ is $d_2$-open.
Mutatis mutandis, if $U \subseteq A$ is $d_2$-open, it follows that $U$ is $d_1$-open.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Proposition $2.4.4$
Category:Lipschitz Equivalent Metrics are Topologically Equivalent
Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 3

== Theorem ==
== Proof ==
By definition of Lipschitz equivalence:
$\exists K_1, K_2 \in \R_{>0}$ such that:
:$(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
:$(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$
By Identity Mapping between Metrics separated by Scale Factor is Continuous:
:the identity mapping from $M_1$ to $M_2$ is continuous
:the identity mapping from $M_2$ to $M_1$ is continuous.
Hence the result by definition of topological equivalence.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Corollary $7.9$
Category:Lipschitz Equivalent Metrics are Topologically Equivalent
List of Elements in Finite Cyclic Group

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $a \in G$ have finite order $\order a = k$.
Let $\gen a$ be the cyclic subgroup generated by $a$.
Then:
:$\set {a^0, a^1, a^2, \ldots, a^{k - 1} }$ is a complete repetition-free list of the elements of $\gen a$
That is, let $\closedint 0 {k - 1}$ be the integer interval from $0$ to $k - 1$.
Then the mapping:
:$\closedint 0 {k - 1} \to \gen a$:
:$n \mapsto g^n$
is a bijection.
== Proof ==
By Element to Power of Remainder, every power of $a$ is equal to one appearing in the list $a^0, a^1, a^2, \ldots, a^{k - 1}$.
This list has to be repetition free, otherwise it would contain $a^m = a^n$ with $0 \le m < n < k$ which contradicts the hypothesis.
== Sources ==
* : $\S 5.4$. Cyclic groups: $(2)$
* : $\S 1.4$: Theorem $3 \ (3)$
* : Chapter $2$: Subgroups and Cosets: $\S 41$
* : $0$: Some Conventions and some Basic Facts
* : $\S 38.4$ Period of an element: $\text{(iii)}$
* : $\S 3.4$: Cyclic groups
* 
* 
Category:Finite Cyclic Groups
List of Elements in Infinite Cyclic Group

== Theorem ==
Let $\gen a$ be the infinite cyclic subgroup generated by $a$.
Then:
:$\set {\ldots, a^{-2}, a^{-1}, a^0, a^1, a^2, \ldots}$
is a complete repetition-free list of the elements of $\gen a$.
That is:
:$G = \set {a^n: n \in \Z}$
== Proof ==
From Infinite Cyclic Group is Isomorphic to Integers:
:$\gen a \cong \struct {\Z, +}$
That is, elements of $\gen a$ correspond one-to-one with $\struct {\Z, +}$ by the bijection $\phi: \Z \to \gen a$ defined as:
:$\forall x \in \Z: \map \phi x = a^x$
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 43$
* 
* 
Category:Infinite Cyclic Group
List of Fundamental Groups for 2-Manifolds

== Theorem ==
For the following two-manifolds, the fundamental group for any point in $X$, written $\map {\pi_1} X$ is isomorphic to the listed group:
:$\map {\pi_1} {\Bbb S^1 \times \closedint 0 1} = \Z$
:$\map {\pi_1} {\Bbb S^2} = \set e$, the trivial group.
:$\map {\pi_1} {\Bbb T^2} = \map {\pi_1} {\Bbb S^1 \times \Bbb S^1} = \Z \times \Z$
:$\map {\pi_1} {\Bbb {RP}^2} = \Z_2$
== Proof == 
Category:Algebraic Topology
Little-O Estimate for Real Function/Examples

== Examples of Little-$\oo$ Estimates for Real Functions ==
=== Example: Sine Function at $+\infty$ ===
=== Example: $x = \map \oo {x^2}$ at $+\infty$ ===
=== Example: $x^2 + x = \map \oo {x^3}$ at $+\infty$ ===
=== Example: $x^2 + x = \map \oo 1$ at $x = 0$ ===
Category:Examples of Little-O Notation
Little-O Estimate for Real Function/Examples/Sine Function at Infinity

== Example of Little-$\oo$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sin x$
Then:
:$\map f x = \map \oo x$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = x$
Then we have that:
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.6$ Some notations
Category:Examples of Little-O Notation
Little-O Estimate for Real Function/Examples/x^2 + x is Little-O of 1 at 0

== Example of Little-$\oo$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2 + x$
Then:
:$\map f x = \map \oo 1$
as $x \to 0$.
== Proof ==
Let us consider the constant real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = 1$
Then we have that:
Hence the result.
== Sources ==
* 
Category:Examples of Little-O Notation
Little-O Estimate for Real Function/Examples/x^2 + x is Little-O of x^3 at Infinity

== Example of Little-$\oo$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2 + x$
Then:
:$\map f x = \map \oo {x^3}$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = x^3$
Then we have that:
Hence the result.
== Sources ==
* 
Category:Examples of Little-O Notation
Little-O Estimate for Real Function/Examples/x is Little-O of x^2 at Infinity

== Example of Little-$\oo$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x$
Then:
:$\map f x = \map \oo {x^2}$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = x^2$
Then we have that:
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.6$ Some notations
Category:Examples of Little-O Notation
Little-O Implies Big-O/General Result

== Theorem ==
Let $X$ be a topological space.
Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\norm {\,\cdot\,}$
Let $f, g: X \to V$ be mappings.
Let $x_0 \in X$.
Let $f = \map \oo g$ as $x \to x_0$, where $\oo$ denotes little-$\oo$ notation.
Then $f = \map \OO g$ as $x \to x_0$, where $\OO$ denotes big-$\OO$ notation.
== Proof ==
From the definition of little-$\oo$ notation:
:there exists a neighborhood $U$ of $x_0$ such that $\norm {\map f x} \le \norm {\map g x}$ for all $x \in U$.
By definition of big-$\OO$ notation, $f = \map \OO g$ as $x \to x_0$.
Category:Big-O Notation
Category:Little-O Notation
Little-O Implies Big-O/Sequences

== Theorem ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences of real or complex numbers.
Let $a_n = \map \oo {b_n}$ where $\oo$ denotes little-$\oo$ notation.
Then $a_n = \map \OO {b_n}$ where $\OO$ denotes big-$\OO$ notation.
== Proof ==
Because $a_n = \map \oo {b_n}$, there exists $n_0 \in \N$ such that $\size {a_n} \le 1 \cdot \size {b_n}$ for $n \ge n_0$.
Thus $a_n = \map \OO {b_n}$.
Category:Big-O Notation
Category:Little-O Notation
Little-O Times Big-O is Little-O/Sequences

== Theorem ==
Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map {\mathcal o} {d_n}$
where:
:$\OO$ denotes big-$\OO$ notation
:$\mathcal o$ denotes little-$\mathcal o$ notation.
Then:
:$a_n c_n = \map {\mathcal o} {b_n d_n}$
== Proof ==
Let $\epsilon \in \R_{> 0}$.
Since $a_n = \map \OO {b_n}$:
:$\exists c \in \R: c \ge 0: \exists n_0 \in \N: \paren {n \ge n_0 \implies \size {a_n} \le c \cdot \size {b_n} }$
Since $c_n = \map {\mathcal o} {d_n}$:
:$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {c_n} \le \dfrac \epsilon {c + 1} \cdot \size {d_n} }$
Thus for $n \ge \max \set {n_0, n_1}$:
Thus $a_n c_n = \map {\mathcal o} {b_n d_n}$.
Category:Big-O Notation
Category:Little-O Notation
Little Bézout Theorem

== Theorem ==
Let $\map {P_n} x$ be a polynomial of degree $n$ in $x$.
Let $a$ be a constant.
Then the remainder of $\map {P_n} x$ when divided by $x - a$ is equal to $\map {P_n} a$.
== Proof ==
By the process of Polynomial Long Division, we can express $\map {P_n} x$ as:
:$(1): \quad \map {P_n} x = \paren {x - a} \map {Q_{n - 1} } x + R$
where:
:$\map {Q_{n - 1} } x$ is a polynomial in $x$ of degree $n - 1$
:$R$ is a polynomial in $x$ of degree no greater than $0$; that is, a constant.
It follows that, by setting $x = a$ in $(1)$, we get $\map {P_n} a = R$.
Hence the result.
== Also known as ==
== Sources ==
* : Chapter $1$: Polynomials; The Remainder and Factor Theorems; Undetermined Coefficients; Partial Fractions: $1.2$. The remainder and factor theorems
* : Chapter $8$: Field Extensions: $\S 37$. Roots of Polynomials: Theorem $69$
* 
* 
* 
Category:Little Bézout Theorem
Category:Polynomial Theory
Category:Algebra
Lobachevsky Integral Formula

== Theorem == 
Let $f$ be a continuous real function, periodic with period $\pi$.
Then:
:$\ds \int_0^\infty \frac {\sin x} x \map f x \rd x = \int_0^{\frac \pi 2} \map f x \rd x$
== Proof == 
Category:Integral Calculus
Category:Definite Integrals involving Sine Function