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21373
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\section{Simple Graph of Maximum Size is Complete Graph}
Tags: Simple Graph of Maximum Size is Complete Graph, Complete Graphs, Simple Graphs
\begin{theorem}
Let $G$ be a simple graph of order $n$ such that $n \ge 1$.
Let $G$ have the largest size of all simple graphs of order $n$.
Then:
:$G$ is the complete graph $K_n$
:its size is $\dfrac {n \paren {n - 1} } 2$.
\end{theorem}
\begin{proof}
By definition, $K_n$ is the simple graph of order $n$ such that every vertex of $K_n$ is adjacent to all other vertices.
So, let $G$ have the largest size of all simple graphs of order $n$.
Then by definition of largest size, it is not possible for another edge to be added to $G$.
The only way that could be is if all the vertices of $G$ are already adjacent to all other vertices.
Hence $G = K_n$ by definition.
The size of $G$ then follows from Size of Complete Graph.
{{qed}}
\end{proof}
|
21374
|
\section{Simple Graph where All Vertices and All Edges are Adjacent}
Tags: Simple Graphs
\begin{theorem}
Let $G$ be a simple graph in which:
:every vertex is adjacent to every other vertex
and:
:every edge is adjacent to every other edge.
Then $G$ is of order no greater than $3$.
\end{theorem}
\begin{proof}
It is seen that examples exist of simple graphs which fulfil the criteria where the order of $G$ is no greater than $3$:
:400px
The cases where the order of $G$ is $1$ or $2$ are trivial.
When the order of $G$ is $3$, the criteria can be verified by inspection.
Let the order of $G = \struct {V, E}$ be $4$ or more.
Let $v_1, v_2, v_3, v_4 \in V$.
Suppose every vertex is adjacent to every other vertex.
As $v_1$ is adjacent to $v_2$, there exists the edge $v_1 v_2$.
As $v_3$ is adjacent to $v_4$, there exists the edge $v_3 v_4$.
But $v_1 v_2$ and $v_3 v_4$ both join two distinct pairs of vertices.
Thus $v_1 v_2$ and $v_3 v_4$ are not adjacent, by definition.
So when there are $4$ or more vertices in $G$, it cannot fulfil the criteria.
{{qed}}
\end{proof}
|
21375
|
\section{Simple Graph whose Vertices Incident to All Edges}
Tags: Graph Theory
\begin{theorem}
Let $G = \struct {V, E}$ be a simple graph whose vertices are incident to all its edges.
Then $G$ is either:
:the star graph $S_2$, which is also the complete graph $K_2$
:an edgeless graph of any order.
\end{theorem}
\begin{proof}
If $G$ has no edges, then all the vertices are incident to all the edges vacuously.
So any of the edgeless graphs $N_n$ for order $n \in \Z_{\ge 0}$ fulfils the criterion.
Suppose $G$ has more than $2$ vertices $v_1, v_2, v_3$ and at least one edge.
{{WLOG}}, let one edge be $v_1 v_2$.
But $v_3$ cannot be incident to edge $v_1 v_2$.
So $G$ can have no more than $2$ vertices.
Furthermore, there can be only one edge joining those two vertices.
The result follows from inspection of $K_2$ and $S_2$.
{{qed}}
\end{proof}
|
21376
|
\section{Simple Graph whose Vertices all Incident but Edges not Adjacent}
Tags: Simple Graphs
\begin{theorem}
Let $G = \struct {V, E}$ be a simple graph such that:
:every vertex is incident with at least one edge
:no two edges are adjacent to each other.
Then $G$ has an even number of vertices.
\end{theorem}
\begin{proof}
Suppose there exists a set of $3$ vertices that are connected.
Then at least one of these vertices has at least $2$ edges.
That would mean that at least $2$ edges were incident with the same vertex.
That is, that at least $2$ edges were adjacent to each other.
So, for a simple graph to fulfil the conditions, vertices can exist only in $2$-vertex components.
So such a simple graph must have an even number of vertices.
{{qed}}
\end{proof}
|
21377
|
\section{Simple Graph with Finite Vertex Set is Finite}
Tags: Simple Graphs, Graph Theory
\begin{theorem}
Let $G$ be a simple graph.
Suppose that the vertex set of $G$ is finite.
Then $G$ is a finite graph.
That is to say, its edge set is also finite.
\end{theorem}
\begin{proof}
Since $G$ is simple, it can have at most one edge between each two vertices.
Therefore, the mapping:
:$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$
which assigns to each edge its endvertices $a_e, b_e$, is an injection.
From Power Set of Finite Set is Finite and Domain of Injection Not Larger than Codomain, it follows that $\map E G$ is finite.
Hence $G$ is finite.
{{qed}}
Category:Graph Theory
Category:Simple Graphs
\end{proof}
|
21378
|
\section{Simple Group of Order Less than 60 is Prime}
Tags: Simple Groups, Groups of Order 60
\begin{theorem}
Let $G$ be a simple group.
Let $\order G < 60$, where $\order G$ denotes the order of $G$.
Then $G$ is a prime group.
\end{theorem}
\begin{proof}
First it is noted that Prime Group is Simple.
We also note from Alternating Group is Simple except on 4 Letters that the alternating group $A_5$, which is of order $60$, is simple.
Hence the motivation for the result.
It remains to be shown that all groups of composite order such that $\order G < 60$ are not simple.
Let $S$ be the set:
:$S = \set {n \in \Z: 0 < n < 60: \text { there exists a simple group of order $n$ such that $n$ is composite} }$
The aim is to show that $S$ is empty.
From Abelian Group is Simple iff Prime, all abelian groups of composite order are not simple.
Thus any simple group must be non-abelian.
Let $p$ and $q$ be prime.
From Prime Power Group has Non-Trivial Proper Normal Subgroup, no group of order $p^n$ is simple.
Thus:
:$\forall n \in \Z_{>0}: p^n \notin S$
Thus, with the primes and prime powers eliminated, we have:
:$S \subseteq \set {6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58}$
From Group of Order $p q$ has Normal Sylow $p$-Subgroup:
:$p q \notin S$
The set of non-square semiprimes less than $60$ is:
:$\set {6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58}$
Thus we have so far:
:$S \subseteq \set {12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56}$
From Group of Order $p^2 q$ is not Simple:
:$p^2 q \notin S$
This eliminates:
{{begin-eqn}}
{{eqn | l = 12
| r = 2^2 \times 3
}}
{{eqn | l = 18
| r = 3^2 \times 2
}}
{{eqn | l = 20
| r = 2^2 \times 5
}}
{{eqn | l = 28
| r = 2^2 \times 7
}}
{{eqn | l = 44
| r = 2^2 \times 11
}}
{{eqn | l = 45
| r = 3^2 \times 5
}}
{{eqn | l = 50
| r = 5^2 \times 2
}}
{{eqn | l = 52
| r = 2^2 \times 13
}}
{{end-eqn}}
Thus we are left with:
:$S \subseteq \set {24, 30, 36, 40, 42, 48, 54, 56}$
From Normal Subgroup of Group of Order 24, a group of order $24$ has a normal subgroup either of order $4$ or order $8$.
Hence $24 \notin S$.
From Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup, a group of order $56$ has at least one normal subgroup.
Hence $56 \notin S$.
From Group of Order 30 is not Simple:
:$30 \notin S$
Thus we are left with:
:$S \subseteq \set {36, 40, 42, 48, 54}$
$40$ is eliminated by Group of Order 40 has Normal Subgroup of Order 5.
$42$ is eliminated by Group of Order 42 has Normal Subgroup of Order 7.
$54$ is eliminated by Group of Order 54 has Normal Subgroup of Order 27.
Thus remains:
:$S \subseteq \set {36, 48}$
{{finish}}
\end{proof}
|
21379
|
\section{Simple Infinite Continued Fraction Converges}
Tags: Continued Fractions
\begin{theorem}
Let $C = (a_0, a_1, \ldots)$ be a simple infinite continued fraction in $\R$.
Then $C$ converges.
\end{theorem}
\begin{proof}
We need to show that for any SICF its sequence of convergents $\sequence {C_n}$ always tends to a limit.
Several techniques can be used here, but a quick and easy one is to show that $\sequence {C_n}$ is a Cauchy sequence.
Let $\epsilon > 0$.
For $m > n \ge \max \set {5, \dfrac 1 \epsilon}$:
{{begin-eqn}}
{{eqn | l = \size {C_m - C_n}
| o = \le
| r = \size {C_m - C_{m - 1} } + \cdots + \size {C_{n + 1} - C_n}
| c = Triangle Inequality
}}
{{eqn | r = \frac 1 {q_m q_{m - 1} } + \cdots + \frac 1 {q_{n + 1} q_n}
| c = Difference between Adjacent Convergents of Simple Continued Fraction
}}
{{eqn | o = <
| r = \frac 1 {m \paren {m - 1} } + \cdots + \frac 1 {\paren {n + 1} n}
| c = Lower Bounds for Denominators of Simple Continued Fraction
}}
{{eqn | r = \sum_{k \mathop = n}^{m - 1} \frac 1 {k \paren {k + 1} }
}}
{{eqn | r = \sum_{k \mathop = n}^{m - 1} \paren {\frac 1 k - \frac 1 {k + 1} }
}}
{{eqn | r = \frac 1 n - \frac 1 m
| c = Telescoping Series/Example 1
}}
{{eqn | o = <
| r = \frac 1 n
}}
{{eqn | o = \le
| r = \epsilon
}}
{{end-eqn}}
So $\sequence {C_n}$ is indeed a Cauchy sequence.
Hence the result.
{{qed}}
Category:Continued Fractions
\end{proof}
|
21380
|
\section{Simple Infinite Continued Fraction Converges to Irrational Number}
Tags: Continued Fractions
\begin{theorem}
The value of any simple infinite continued fraction in $\R$ is irrational.
\end{theorem}
\begin{proof}
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a simple infinite continued fraction.
Note that by Simple Infinite Continued Fraction Converges, a simple infinite continued fraction is indeed convergent, say to $x \in \R$.
Let $p_0, p_1, \ldots$ and $q_0, q_1, \ldots$ be its numerators and denominators.
Let $C_0, C_1, \ldots$ be its convergents, so that $C_n = p_n/q_n$ for $n \geq 0$.
For all $n \geq 0$, from Accuracy of Convergents of Convergent Simple Infinite Continued Fraction:
:$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n q_{n + 1} }$
Suppose $x$ is rational.
That is, let $x = \dfrac r s$ where $r, s \in \Z$ such that $s > 0$.
Then:
:$0 < \left|{\dfrac r s - \dfrac {p_n} {q_n}}\right| = \dfrac {\left|{r q_n - s p_n}\right|} {s q_n} < \dfrac 1 {q_n q_{n+1}}$
(Note that $\dfrac r s \ne \dfrac {p_n} {q_n}$ or otherwise the continued fraction would be finite.)
So:
:$0 < \left|{r q_n - s p_n}\right| < \dfrac s {q_{n+1}}$
But Denominators of Simple Continued Fraction are Strictly Increasing.
That means we can choose $n$ so that $q_{n+1} > s$.
But then $\left|{r q_n - s p_n}\right|$ would be an integer lying strictly between $0$ and $1$, which cannot happen.
So no such integers $r, s$ exist.
Thus $x$ must be irrational.
{{qed}}
Category:Continued Fractions
\end{proof}
|
21381
|
\section{Simple Infinite Continued Fraction is Uniquely Determined by Limit}
Tags: Continued Fractions
\begin{theorem}
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be simple infinite continued fractions in $\R$.
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same limit.
Then they are equal.
\end{theorem}
\begin{proof}
Note that by Simple Infinite Continued Fraction Converges, they do indeed have a limit.
The result will be achieved by the Second Principle of Mathematical Induction.
Suppose $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{b_0, b_1, b_2, \ldots}\right]$ have the same value.
First we note that if $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{b_0, b_1, b_2, \ldots}\right]$ then $a_0 = b_0$ since both are equal to the integer part of the common value.
{{explain|a result proving the above}}
This is our basis for the induction.
Now suppose that for some $k \ge 1$, we have:
:$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$.
Then all need to do is show that $a_{k+1} = b_{k+1}$.
Now:
:$\left[{a_0, a_1, a_2, \ldots}\right] = \left[{a_0, a_1, \ldots, a_k, \left[{a_{k+1}, a_{k+2}, \ldots}\right]}\right]$
and similarly
:$\left[{b_0, b_1, b_2, \ldots}\right] = \left[{b_0, b_1, \ldots, b_k, \left[{b_{k+1}, b_{k+2}, \ldots}\right]}\right]$.
{{explain|this needs to be proved}}
As these have the same value and have the same first $k$ partial quotients, it follows that:
:$\left[{a_{k+1}, a_{k+2}, \ldots,}\right] = \left[{b_{k+1}, b_{k+2}, \ldots}\right]$.
But now $a_{k+1} = b_{k+1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.
Hence the result.
{{qed}}
\end{proof}
|
21382
|
\section{Simple Order Product of Pair of Ordered Sets is Ordered Set}
Tags: Order Products, Simple Order Product
\begin{theorem}
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is also an ordered set.
\end{theorem}
\begin{proof}
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
By definition of simple product:
:$T := S_1 \times S_2$ where $\times$ denotes Cartesian product
:$\forall \tuple {a, b}, \tuple {c, d} \in T: \tuple {a, b} \preccurlyeq_s \tuple {c, d} \iff a \preccurlyeq_1 c \text { and } b \preccurlyeq_2 d$
We check in turn each of the criteria for an ordering:
\end{proof}
|
21383
|
\section{Simple Order Product of Totally Ordered Sets may not be Totally Ordered}
Tags: Order Products, Totally Ordered Sets, Simple Order Product, Total Orderings
\begin{theorem}
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be totally ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then it is not necessarily the case that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is also a totally ordered set.
\end{theorem}
\begin{proof}
From Simple Order Product of Pair of Ordered Sets is Ordered Set, we do have that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is an ordered set.
Let us take the simple product of the ordered set that is the natural numbers under the usual ordering with itself:
:$\struct {\N, \le} \otimes^s \struct {\N, \le}$
We have from the Well-Ordering Principle that $\struct {\N, \le}$ is a well-ordered set, and so a fortiori a totally ordered set.
Consider the ordered pairs of natural numbers:
:$s = \tuple {1, 4}$
:$t = \tuple {2, 3}$
Both $s$ and $t$ are elements of $\N \times \N$.
We have that:
:$1 \le 2$
but:
:$4 \not \le 3$
So it is not the case either that:
:$1 \le 2$ and $4 \le 3$
or:
:$2 \le 1$ and $3 \le 4$
Hence $s$ and $t$ are non-comparable elements.
The result follows by definition of totally ordered set.
{{qed}}
\end{proof}
|
21384
|
\section{Simple Variable End Point Problem}
Tags: Calculus of Variations
\begin{theorem}
Let $y$ and $F$ be mappings.
{{explain|Define their domain and codomain}}
Suppose the endpoints of $y$ lie on two given vertical lines $x = a$ and $x = b$.
Suppose $J$ is a functional of the form
:$(1): \quad J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
and has an extremum for a certain function $\hat y$.
Then $y$ satisfies the system of equations
:$\begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$
\end{theorem}
\begin{proof}
From Condition for Differentiable Functional to have Extremum we have
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$
The variation exists if $J$ is a differentiable functional.
We will start from the increment of a functional:
{{explain|make the above link point to a page dedicated to the appropriate definition}}
{{begin-eqn}}
{{eqn | l = \Delta J \sqbrk {y; h}
| r = J \sqbrk {y + h} - J \sqbrk y
| c = definition
}}
{{eqn | r = \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x
| c = $(1)$
}}
{{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x
}}
{{end-eqn}}
{{explain|"definition" in the above -- point to what it is a definition of}}
Using multivariate Taylor's theorem, one can expand $\map F {x, y + h, y' + h'}$ with respect to $h$ and $h'$:
:$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial \map F {x, y + h, y' + h'} } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$
Substitute this back into the integral.
Note that the first term in the expansion and the negative one in the integral will cancel out:
:$\ds \Delta J \paren {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
{{explain|What do the subscripted $y$ and $y'$ mean in the above?}}
Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Now we expand:
:$\ds \int_a^b\map \OO {h^2, h h', h'^2} \rd x$
Every term in this expansion will be of the form:
:$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial {y'}^n} h^m h'^n \rd x$
where $m, n \in \N$ and $m + n \ge 2$
{{Explain|How to convert powers of $h'$ into $h$? Integration by parts is the only obvious candidate, but how precisely? Also check, if this is necessary}}
By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional.
:$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:
{{begin-eqn}}
{{eqn | l = \delta J \sqbrk {y; h}
| r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigintlimits {F_{y'} \map h x} {x \mathop = a} {x \mathop = b}
}}
{{eqn | r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigvalueat {F_{y'} } {x \mathop = b} \map h b - \bigvalueat {F_{y'} } {x \mathop = a} \map h a
}}
{{end-eqn}}
Then, for arbitrary $\map h x$, $J$ has an extremum if:
:$ \begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0\\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$
{{qed}}
\end{proof}
|
21385
|
\section{Simple Variable End Point Problem/Endpoints on Curves}
Tags: Calculus of Variations, Definitions: Calculus of Variations
\begin{theorem}
Let $y$, $F$, $\phi$ and $\psi$ be smooth real functions.
Let $J = J \sqbrk y$ be a functional of the form:
:$\ds J \sqbrk y = \int_{x_0}^{x_1} \map F {x, y, y'} \rd x$
Let $P_0$, $P_1$ be the endpoints of the curve $y$.
Suppose $P_0$, $P_1$ lie on curves $y = \map {\phi} x$, $y = \map {\psi} x$.
Then the extremum of $J \sqbrk y$ is a curve which satisfies the following system of Euler and transversality equations:
{{begin-eqn}}
{{eqn | l = F_y - \dfrac {\d} {\d x} F_{y'}
| r = 0
}}
{{eqn | l = \bigvalueat {\paren {F + \paren {\phi' - y'} F_{y'} } } {x \mathop = x_0}
| r = 0
}}
{{eqn | l = \bigvalueat {\paren {F + \paren {\psi' - y'} F_{y'} } } {x \mathop = x_1} = 0
| r = 0
}}
{{end-eqn}}
{{explain|meaning of square brackets}}
\end{theorem}
\begin{proof}
By general variation of integral functional with $n = 1$:
:$\ds \delta J \sqbrk{y; h} = \int_{x_0}^{x_1} \intlimits {\paren {F_y - \dfrac \d {\d x} F_{y'} } \map h x + F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$
Since the curve $y = \map y x$ is an extremum of $\map J y$, the integral term vanishes:
:$\ds \delta J \sqbrk {y; h} = \bigintlimits {F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$
According to Taylor's Theorem:
:$\delta y_0 = \paren {\map {\phi'} {x_0} + \epsilon_0} \delta x_0$
:$\delta y_1 = \paren {\map {\phi'} {x_1} + \epsilon_1} \delta x_1$
{{explain|what do square brackets mean?}}
where $\epsilon_0 \to 0$ as $\delta x_0 \to 0$ and $\epsilon_1 \to 0$ as $\delta x_1 \to 0$.
{{Stub| add a figure; explain better}}
Substitution of $\delta y_0$ and $\delta y_1$ into $\delta J$ leads to:
:$\delta J \sqbrk {y; h} = \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_1} \delta x_1 - \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_0} \delta x_0$
$y = \map y x$ is an extremal of $J \sqbrk y$, thus $\delta J = 0$.
$\delta x_0$ and $\delta x_1$ are independent increments.
Hence both remaining terms in $\delta J$ vanish independently.
{{qed}}
\end{proof}
|
21386
|
\section{Simplest Variational Problem with Subsidiary Conditions}
Tags: Calculus of Variations
\begin{theorem}
Let $J \sqbrk y$ and $K \sqbrk y$ be (real) functionals, such that
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
:$\ds K \sqbrk y = \int_a^b \map G {x, y, y'} \rd x = l$
where $l$ is a constant.
Let $y = \map y x$ be an extremum of $F \sqbrk y$, and satisfy boundary conditions:
:$\map y a = A$
:$\map y b = B$
Then, if $y = \map y x$ is not an extremal of $K \sqbrk y$, there exists a constant $\lambda$ such that $y = \map y x$ is an extremal of the functional:
:$\ds \int_a^b \paren {F + \lambda G} \rd x$
or, in other words, $y = \map y x$ satisfies:
:$F_y - \dfrac {\d} {\d x} F_{y'} + \lambda \paren {G_y - \dfrac {\d} {\d x} G_{y'} } = 0$
\end{theorem}
\begin{proof}
Let $J \sqbrk y$ be a functional, for which $y = \map y x$ is an extremal with the boundary conditions $\map y a = A$ and $\map y b = B$.
Choose two points, $x_1$ and $x_2$ from the interval $\closedint a b$.
Let $\delta_1 \map y x$ and $\delta_2 \map y x$ be functions, different from zero only in the neighbourhood of $x_1$ and $x_2$ respectively.
Then we can exploit the definition of variational derivative in the following way:
:$\Delta J \sqbrk {y; \, \delta_1 \map y x + \delta_2 \map y x} = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2$
where:
:$\ds \Delta \sigma_1 = \int_a^b \delta_1 \map y x$
:$\ds \Delta \sigma_2 = \int_a^b \delta_2 \map y x$
and $\epsilon_1, \epsilon_2 \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.
{{Explain|support above statement with an actual calculation}}
We now require that the varied curve $y^* = \map y x + \delta_1 \map y x + \delta_2 \map y x$ satisfies the condition $K \sqbrk {y^*} = K \sqbrk y$.
This way we limit arbitrary varied curves to those who still satisfy the given functional constraint.
Similarly, write down $\Delta K \sqbrk y$ as:
:$\ds \Delta K \sqbrk y = K \sqbrk{y^*} - K \sqbrk y = \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} + \epsilon_1'} \Delta \sigma_1 + \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2'} \Delta\sigma_2$
where $\epsilon_1', \epsilon_2' \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.
Suppose $x_2$ is such that:
:$\valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} \ne 0$
Such a point exists, because by assumption $\map y x$ is not an extremal of $K \sqbrk y$.
Since $\Delta K = 0$, the previous equation can be rewritten as
:$\Delta \sigma_2 = -\paren {\dfrac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1$
where $\epsilon' \to 0$ as $\Delta \sigma_1 \to 0$.
{{Explain|how?}}
Set :
:$\lambda = -\dfrac {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} }$
Substitute this into the formula for $\Delta J$:
{{begin-eqn}}
{{eqn | l = \Delta J \sqbrk {y; \delta_1 \map y x + \delta_2 \map y x}
| r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2
| c =
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {-\valueat {\lambda \frac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2} \paren {-\paren {\frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1}
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} \epsilon' - \epsilon_2 \frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } - \epsilon_2 \epsilon'} \Delta \sigma_1
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma_1 + \epsilon \Delta \sigma_1
}}
{{end-eqn}}
where $\epsilon \to 0$ as $\Delta \sigma_1 \to 0$.
Then the variation of the functional $J \sqbrk y$ at the point $x_1$ is:
:$\delta J = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma$
A necessary condition for $\delta J$ to vanish for any $\Delta \sigma$ and arbitrary $x_1$ is:
:$\dfrac {\delta F} {\delta y} + \lambda \dfrac {\delta G} {\delta y} = F_y - \dfrac \d {\d x} F_{y'} + \lambda \paren {G_y - \dfrac \d {\d x} G_{y'} } = 0$
{{qed}}
\end{proof}
|
21387
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\section{Simplest Variational Problem with Subsidiary Conditions for Curve on Surface}
Tags: Calculus of Variations
\begin{theorem}
Let $J \sqbrk {y, z}$ be a (real) functional of the form:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, z, y', z'} \rd x$
Let there exist admissible curves $y, z$ lying on the surface:
:$\map g {x, y, z} = 0$
which satisfy boundary conditions:
:$\map y a = A_1, \map y b = B_1$
:$\map z a = A_2, \map z b = B_2$
Let $J \sqbrk {y, z}$ have an extremum for the curve $y = \map y x, z = \map z x$.
Let $g_y$ and $g_z$ not simultaneously vanish at any point of the surface $g = 0$.
Then there exists a function $\map \lambda x$ such that the curve $y = \map y x, z = \map z x$ is an extremal of the functional:
:$\ds \int_a^b \paren {F + \map \lambda x g} \rd x$
In other words, $y = \map y x$ satisfies the differential equations:
{{begin-eqn}}
{{eqn | l = F_y + \lambda g_y - \frac \d {\d x} F_{y'}
| r = 0
}}
{{eqn | l = F_z + \lambda g_z - \frac \d {\d x} F_{z'}
| r = 0
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let $J \sqbrk y$ be a functional, for which the curve $y = \map y x, z = \map z x$ is an extremal with the boundary conditions $\map y a = A, \map y b = B$ as well as $\map g {x, y, z} = 0$.
Choose an arbitrary point $x_1$ from the interval $\closedint a b$.
Let $\delta \map y x$ and $\delta \map z x$ be functions, different from zero only in the neighbourhood of $x_1$.
Then we can exploit the definition of variational derivative in a following way:
:$\Delta J \sqbrk {y; \delta_1 \map y x + \delta_2 \map y x} = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\dfrac {\delta F} {\delta z} } {x \mathop = x_1} + \epsilon_2} \Delta \sigma_2$
where:
:$\ds \Delta \sigma_1 = \int_a^b \delta \map y x$
:$\ds \Delta \sigma_2 = \int_a^b \delta \map z x$
and $\epsilon_1, \epsilon_2 \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.
{{Explain|support above statement with an actual calculation}}
We now require that the varied curve $y^* = \map y x + \map {\delta_y} x$, $z^* = \map y x + \map {\delta_z} x$ satisfies the condition $\map g {x, y^*, z^*} = 0$.
This condition limits arbitrary varied curves only to those which still satisfy the original constraint on the surface.
By using constraints on $g$, we can follow the following chain of equalities:
{{begin-eqn}}
{{eqn | l = 0
| r = \int_a^b \paren {\map g {x, y^*, z^*} - \map g {x, y, z} } \rd x
| c =
}}
{{eqn | r = \int_a^b \paren {\overline {g_y} \delta y + \overline {g_z} \delta z}
}}
{{eqn | r = \paren {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} \Delta \sigma_1 + \paren {\bigvalueat {g_z} {x \mathop = x_1} + \epsilon'_2} \Delta \sigma_2
}}
{{end-eqn}}
where:
:$\epsilon_1', \epsilon_2' \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$
and overbar indicates that corresponding derivatives are evaluated along certain intermediate curves.
{{Explain|how?}}
By hypothesis, either $\bigvalueat {g_y} {x \mathop = x_1}$ or $\bigvalueat {g_z} {x \mathop = x_1}$ is nonzero.
Suppose $\bigvalueat {g_z} {x \mathop = x_1} \ne 0$.
Then the previous result can be rewritten as:
{{begin-eqn}}
{{eqn | l = \Delta \sigma_2
| r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} + \epsilon'_2}
| c =
}}
{{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} \paren {1 + \frac {\epsilon'_2} {\bigvalueat {g_z} {x \mathop = x_1} } } }
}}
{{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } \sum_{n \mathop = 0}^\infty \paren {\frac {\epsilon_2'} {\bigvalueat {g_z} {x \mathop = x_1} } }^n
| c = Sum of Infinite Geometric Sequence, holds for $\size {\epsilon_2'} \to 0$
}}
{{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } - \Delta \sigma_1 \frac {\epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } - \Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } \sum_{n \mathop = 1}^\infty \paren { {\frac {\epsilon_2'} {\bigvalueat {g_z} {x \mathop = x_1} } } }^n
}}
{{eqn | ll= \leadsto
| l = \Delta \sigma_2
| r = -\paren {\frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } + \epsilon'} \Delta \sigma_1
| c =
}}
{{end-eqn}}
where $\epsilon'\to 0$ as $\Delta \sigma_1 \to 0$.
Substitute this back into the equation for $\Delta J \sqbrk {y, z}$
{{begin-eqn}}
{{eqn | l = \Delta J
| r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} + \epsilon_2} \Delta \sigma_2
| c =
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} + \epsilon_2} \paren {-\paren {\frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } + \epsilon'} \Delta \sigma_1}
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\frac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1 + \paren {\epsilon_1 - \epsilon_2 \frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } - \epsilon' \valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} - \epsilon_2 \epsilon'} \Delta \sigma_1
}}
{{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\frac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1 + \epsilon \Delta \sigma_1
}}
{{end-eqn}}
where $\epsilon \to 0$ as $\Delta \sigma_1 \to 0$.
Then the variation of the functional $J \sqbrk y$ at the point $x_1$ is:
:$\delta J = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\dfrac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1$
A necessary condition for $\delta J$ vanish for any $\Delta \sigma$ and arbitrary $x_1$ is:
:$\dfrac {\delta F} {\delta y} - \dfrac {g_y} {g_z} \dfrac {\delta F} {\delta z} = F_y - \dfrac \d {\d x} F_{y'} - \dfrac {g_y} {g_z} \paren {F_z - \dfrac \d {\d x} F_{z'} } = 0$
The latter equation can be rewritten as
:$\dfrac {F_y - \dfrac \d {\d x} F_{y'} } {g_y} = \dfrac {F_z - \dfrac \d {\d x} F_{z'} } {g_z}$
If we denote this ratio by $-\map \lambda x$, then this ratio can be rewritten as two equations presented in the theorem.
{{qed}}
\end{proof}
|
21388
|
\section{Simpson's Formulas/Cosine by Sine}
Tags: Sine Function, Simpson's Formula for Cosine by Sine, Simpson's Formulas, Cosine Function
\begin{theorem}
:$\cos \alpha \sin \beta = \dfrac {\map \sin {\alpha + \beta} - \map \sin {\alpha - \beta} } 2$
where $\cos$ denotes cosine and $\sin$ denotes sine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | o=
| r=\frac {\sin \left({\alpha + \beta}\right) - \sin \left({\alpha - \beta}\right)} 2
}}
{{eqn | r=\frac {\left({\sin \alpha \cos \beta + \cos \alpha \sin \beta}\right) - \left({\sin \alpha \cos \beta - \cos \alpha \sin \beta}\right)} 2
| c=Sine of Sum and Sine of Difference
}}
{{eqn | r=\frac {2 \cos \alpha \sin \beta} 2
}}
{{eqn | r=\cos \alpha \sin \beta
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
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\end{proof}
|
21389
|
\section{Simson Line Theorem}
Tags: Triangles
\begin{theorem}
Let $\triangle ABC$ be a triangle.
Let $P$ be a point on the circumcircle of $\triangle ABC$.
Then the feet of the perpendiculars drawn from $P$ to each of the sides of $\triangle ABC$ are collinear.
:300px
This line is called the '''Simson Line'''.
\end{theorem}
\begin{proof}
In the figure above, construct the lines $BP$ and $CP$.
:300px
By the converse of Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $EPDB$ is cyclic.
By the converse of Angles in Same Segment of Circle are Equal, $EPCF$ is cyclic.
{{WIP|Looking for the pages for the above converses and the (simple) unlinked theorem below. The theorem can be split trivially into Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles and Two Angles on Straight Line make Two Right Angles, but the same cannot be said for those converses. <br> Category:Cyclic Quadrilaterals is of no help}}
Therefore:
{{begin-eqn}}
{{eqn | l = \angle DEP
| r = \angle DBP
| c = Angles in Same Segment of Circle are Equal: $EPDB$ is cyclic
}}
{{eqn | r = \angle ACP
| c = The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle: $ABPC$ is cyclic
}}
{{eqn | r = 180^\circ - \angle PEF
| c = Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles: $EPCF$ is cyclic
}}
{{end-eqn}}
This gives:
:$\angle DEP + \angle PEF = 180^\circ$
hence $DEF$ is a straight line.
{{qed}}
{{Namedfor|Robert Simson|cat = Simson}}
Category:Triangles
\end{proof}
|
21390
|
\section{Simultaneous Equation With Two Unknowns}
Tags: Linear Algebra
\begin{theorem}
A pair of simultaneous linear equations of the form:
{{begin-eqn}}
{{eqn | l = a x + b y
| r = c
}}
{{eqn | l = d x + e y
| r = f
}}
{{end-eqn}}
where $a e \ne b d$, has as its only solution:
{{begin-eqn}}
{{eqn | l = x
| r = \frac {c e - b f} {a e - b d}
}}
{{eqn | l = y
| r = \frac {a f - c d} {a e - b d}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l=ax+by
| r=c
}}
{{eqn | ll=\implies
| l=x
| r=\frac{c-by}{a}
| c=Rearranging
}}
{{eqn | l=dx+ey
| r=f
}}
{{eqn | ll=\implies
| l=d(\frac{c-by}{a})+ey
| r=f
| c=Substituting $x=\frac{c-by}{a}$
}}
{{eqn | ll=\implies
| l=\frac{cd-bdy}{a}+ey
| r=f
| c=Multiplying out brackets
}}
{{eqn | ll=\implies
| l=cd-bdy+aey
| r=af
| c=Multiplying by $a$
}}
{{eqn | ll=\implies
| l=aey-bdy
| r=af-cd
| c=Subtracting $cd$
}}
{{eqn | ll=\implies
| l=y(ae-bd)
| r=af-cd
| c=Factorising
}}
{{eqn | ll=\implies
| l=y
| r=\frac{af-cd}{ae-bd}
| c=Dividing by $ae-bd$
}}
{{end-eqn}}
The solution for $x$ can be found similarly.
{{qed}}
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\end{proof}
|
21391
|
\section{Simultaneous Homogeneous Linear First Order ODEs/Examples/y' - 3y + 2z = 0, y' + 4y - z = 0}
Tags: Examples of Systems of Differential Equations, Examples of Linear First Order ODEs
\begin{theorem}
Consider the system of linear first order ordinary differential equations with constant coefficients:
{{begin-eqn}}
{{eqn | n = 1
| l = \dfrac {\d y} {\d x} - 3 y + 2 z
| r = 0
}}
{{eqn | n = 2
| l = \dfrac {\d x} {\d z} + 4 y - z
| r = 0
}}
{{end-eqn}}
The general solution to $(1)$ and $(2)$ consists of the linear combinations of the following:
{{begin-eqn}}
{{eqn | l = y
| r = C_1 e^{5 x} + C_2 e^{-x}
}}
{{eqn | l = z
| r = -C_1 e^{5 x} + 2 C_2 e^{-x}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Using the technique of Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients, we calculate the roots of the quadratic equation:
:$\paren {k + a} \paren {k + d} - b c = 0$
where:
{{begin-eqn}}
{{eqn | l = a
| r = -3
}}
{{eqn | l = b
| r = 2
}}
{{eqn | l = c
| r = 4
}}
{{eqn | l = d
| r = -1
}}
{{end-eqn}}
That is:
:$\paren {k - 3} \paren {k - 1} - 8 = 0$
or:
:$k^2 - 4 k - 5 = 0$
This has roots:
{{begin-eqn}}
{{eqn | l = k_1
| r = 5
}}
{{eqn | l = k_2
| r = -1
}}
{{end-eqn}}
We also obtain:
{{begin-eqn}}
{{eqn | l = \paren {k - 3} A + 2 B
| r = 0
}}
{{eqn | l = 4 A + \paren {k - 1} B
| r = -1
}}
{{end-eqn}}
When $k = 5$ we get that $A + B = 0$.
When $k = -1$ we get that $2 A - B = 0$.
This provides us with the solutions:
{{begin-eqn}}
{{eqn | l = y
| r = e^{5 x}
}}
{{eqn | l = z
| r = e^{-5 x}
}}
{{end-eqn}}
or:
{{begin-eqn}}
{{eqn | l = y
| r = e^{-x}
}}
{{eqn | l = z
| r = 2 e^{-x}
}}
{{end-eqn}}
From these, the general solution is constructed:
{{begin-eqn}}
{{eqn | l = y
| r = C_1 e^{5 x} + C_2 e^{-x}
}}
{{eqn | l = z
| r = -C_1 e^{5 x} + 2 C_2 e^{-x}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21392
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\section{Sine Exponential Formulation}
Tags: Sine Function, Trigonometric Functions, Trigonometry, Analysis, Sine Exponential Formulation
\begin{theorem}
For any complex number $z$:
:$\sin z = \dfrac {\map \exp {i z} - \map \exp {-i z} } {2 i}$
:$\exp z$ denotes the exponential function
:$\sin z$ denotes the complex sine function
:$i$ denotes the inaginary unit.
\end{theorem}
\begin{proof}
{{tidy}}
Recall the definition of the sine function:
: $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
Recall the definition of the exponential as a power series:
: $\displaystyle e^x = \sum_{n \mathop = 0}^\infty \frac {x^n}{n!} = 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots$
Then, starting from the RHS:
{{begin-eqn}}
{{eqn|ll = |l = |r = \frac 1 2 i \left( e^{-ix} - e^{ix} \right) |rr = |c = |cc = |o = }}
{{eqn|ll = |l = |r = \frac 1 2 i \left( \sum_{n \mathop = 0}^\infty \frac {(-ix)^n}{n!} - \sum_{n \mathop = 0}^\infty \frac {(ix)^n}{n!} \right) |rr = |c = |cc = }}
{{eqn|ll = |l = |r = \frac 1 2 i \sum_{n \mathop = 0}^\infty \left( \frac {(-ix)^n - (ix)^n}{n!} \right) |rr = |c = split into even and odd $n$|cc = }}
{{eqn|ll = |l = |r = \frac 1 2 i \sum_{n \mathop = 0}^\infty \left( \frac {(-ix)^{2n} - (ix)^{2n} }{(2n)!} + \frac {(-ix)^{2n+1} - (ix)^{2n+1} }{(2n+1)!} \right) |rr = |c = $(-ix)^{2n} = (ix)^{2n}$|cc = }}
{{eqn|ll = |l = |r = \frac 1 2 i \sum_{n \mathop = 0}^\infty \frac {(-ix)^{2n+1} - (ix)^{2n+1} }{(2n+1)!} |rr = |c = $(-1)^{2n+1} = -1$|cc = }}
{{eqn|ll = |l = |r = \frac 1 2 i \sum_{n \mathop = 0}^\infty \frac {-2 (ix)^{2n+1} }{(2n+1)!} |rr = |c = cancel 2 |cc = }}
{{eqn|ll = |l = |r = i \sum_{n \mathop = 0}^\infty \frac {- (ix)^{2n+1} }{(2n+1)!} |rr = |c = $i^{2n+1} = i (-1)^n $ |cc = }}
{{eqn|ll = |l = |r = i \sum_{n \mathop = 0}^\infty \frac {- i (-1)^n x^{2n+1} }{(2n+1)!} |rr = |c = $i^2 = -1$ |cc = }}
{{eqn|ll = |l = |r = \sum_{n \mathop = 0}^\infty (-1)^n \frac {x^{2n+1} }{(2n+1)!} |rr = |c = |cc = }}
{{end-eqn}}
{{qed}}
\end{proof}
|
21393
|
\section{Sine Function is Absolutely Convergent}
Tags: Sine Function, Analysis
\begin{theorem}
Let $x \in \R$ be a real number.
Let $\sin x$ be the sine of $x$.
Then:
:$\sin x$ is absolutely convergent for all $x \in \R$.
\end{theorem}
\begin{proof}
Recall the definition of the sine function:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
For:
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
to be absolutely convergent we want:
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } = \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$
to be convergent.
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$
is just the terms of:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!}$
for odd $n$.
Thus:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!} < \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!}$
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!} = \exp \size x$
from the Taylor Series Expansion for Exponential Function of $\size x$, which converges for all $x \in \R$.
The result follows from the Squeeze Theorem.
{{qed}}
\end{proof}
|
21394
|
\section{Sine Function is Odd}
Tags: Sine Function, Odd Functions, Examples of Odd Functions, Analysis
\begin{theorem}
For all $z \in \C$:
:$\map \sin {-z} = -\sin z$
That is, the sine function is odd.
\end{theorem}
\begin{proof}
Recall the definition of the sine function:
{{begin-eqn}}
{{eqn | l = \sin z
| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}
}}
{{eqn | r = z - \frac {z^3} {3!} + \frac {z^5} {5!} - \cdots
}}
{{end-eqn}}
From Sign of Odd Power, we have that:
:$\forall n \in \N: -\paren {z^{2 n + 1} } = \paren {-z}^{2 n + 1}$
The result follows directly.
{{qed}}
\end{proof}
|
21395
|
\section{Sine Inequality}
Tags: Sine Function, Named Theorems
\begin{theorem}
:$\size {\sin x} \le \size x$
for all $x \in \R$.
\end{theorem}
\begin{proof}
Let $\map f x = x - \sin x$.
By Derivative of Sine Function:
:$\map {f'} x = 1 - \cos x$
From Real Cosine Function is Bounded we know $\cos x \le 1$ for all $x$.
Hence $\map f x \ge 0$ for all $x$.
From Derivative of Monotone Function, $\map f x$ is increasing.
By Sine of Zero is Zero, $\map f x = 0$.
It follows that $\map f x \ge 0$ for all $x \ge 0$.
{{qed|lemma}}
Now let $\map g x = x^2 - \sin^2 x$.
{{begin-eqn}}
{{eqn | l = \map {g'} x
| r = 2 x - 2 \sin x \cos x
| c = Chain Rule for Derivatives and Derivative of Sine Function
}}
{{eqn | r = 2 x - \sin 2 x
| c = Double Angle Formula for Sine
}}
{{eqn | o = \ge
| r = 0
| c = for $x \ge 0$, as shown above
}}
{{end-eqn}}
From Derivative of Monotone Function, $\map g x$ is increasing for $x \ge 0$.
By Sine of Zero is Zero, $\map g x = 0$.
It follows that $\map g x \ge 0$ for all $x \ge 0$.
Observe that $\map g x$ is an even function.
This implies $\map g x \ge 0$ for all $x \in \R$.
Finally note that $\sin^2 x \le x^2 \iff \size {\sin x} \le \size x$.
{{qed}}
Category:Sine Function
Category:Named Theorems
\end{proof}
|
21396
|
\section{Sine Integral Function of Zero}
Tags: Sine Integral Function
\begin{theorem}
:$\map \Si 0 = 0$
where $\Si$ denotes the sine integral function.
\end{theorem}
\begin{proof}
By Sine Integral Function is Odd, $\Si$ is an odd function.
Therefore, by Odd Function of Zero is Zero:
:$\map \Si 0 = 0$
{{qed}}
\end{proof}
|
21397
|
\section{Sine Plus Cosine times Tangent Plus Cotangent}
Tags: Trigonometric Identities
\begin{theorem}
:$\paren {\sin x + \cos x} \paren {\tan x + \cot x} = \sec x + \csc x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {\sin x + \cos x} \paren {\tan x + \cot x}
| r = \paren {\sin x + \cos x} \paren {\sec x \csc x}
| c = Sum of Tangent and Cotangent
}}
{{eqn | r = \frac {\sin x + \cos x} {\sin x \cos x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac 1 {\cos x} + \frac 1 {\sin x}
}}
{{eqn | r = \sec x + \csc x
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
21398
|
\section{Sine and Cosine are Periodic on Reals}
Tags: Periodic Functions, Sine Function, Analysis, Cosine Function
\begin{theorem}
The sine and cosine functions are periodic on the set of real numbers $\R$:
:$(1): \quad \map \cos {x + 2 \pi} = \cos x$
:$(2): \quad \map \sin {x + 2 \pi} = \sin x$
:800px
\end{theorem}
\begin{proof}
From Cosine of Zero is One we have that $\cos 0 = 1$.
From Cosine Function is Even we have that $\cos x = \map \cos {-x}$.
As the Cosine Function is Continuous, it follows that:
:$\exists \xi > 0: \forall x \in \openint {-\xi} \xi: \cos x > 0$
{{AimForCont}} $\cos x$ were positive everywhere on $\R$.
From Derivative of Cosine Function:
:$\map {D_{xx} } {\cos x} = \map {D_x} {-\sin x} = -\cos x$
Thus $-\cos x$ would always be negative.
Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$.
But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$.
By Differentiable Bounded Concave Real Function is Constant, $\cos x$ would then be a constant function.
This contradicts the fact that $\cos x$ is not a constant function.
Thus by Proof by Contradiction $\cos x$ can not be positive everywhere on $\R$.
Therefore, there must exist a smallest positive $\eta \in \R_{>0}$ such that $\cos \eta = 0$.
By definition, $\cos \eta = \map \cos {-\eta} = 0$ and $\cos x > 0$ for $-\eta < x < \eta$.
Now we show that $\sin \eta = 1$.
From Sum of Squares of Sine and Cosine:
:$\cos^2 x + \sin^2 x = 1$
Hence as $\cos \eta = 0$ it follows that $\sin^2 \eta = 1$.
So either $\sin \eta = 1$ or $\sin \eta = -1$.
But $\map {D_x} {\sin x} = \cos x$.
On the interval $\closedint {-\eta} \eta$, it has been shown that $\cos x > 0$.
Thus by Derivative of Monotone Function, $\sin x$ is increasing on $\closedint {-\eta} \eta$.
Since $\sin 0 = 0$ it follows that $\sin \eta > 0$.
So it must be that $\sin \eta = 1$.
Now we apply Sine of Sum and Cosine of Sum:
:$\map \sin {x + \eta} = \sin x \cos \eta + \cos x \sin \eta = \cos x$
:$\map \cos {x + \eta} = \cos x \cos \eta - \sin x \sin \eta = -\sin x$
Hence it follows, after some algebra, that:
:$\map \sin {x + 4 \eta} = \sin x$
:$\map \cos {x + 4 \eta} = \cos x$
Thus $\sin$ and $\cos$ are periodic on $\R$ with period $4 \eta$.
{{qed}}
\end{proof}
|
21399
|
\section{Sine in terms of Cosine}
Tags: Trigonometric Functions, Sine Function, Cosine Function
\begin{theorem}
Let $x$ be an real number.
Then:
{{begin-eqn}}
{{eqn | l = \sin x
| r = +\sqrt {1 - \cos ^2 x}
| c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$
}}
{{eqn | l = \sin x
| r = -\sqrt {1 - \cos ^2 x}
| c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$
}}
{{end-eqn}}
where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \cos^2 x + \sin^2 x
| r = 1
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
| l = \sin^2 x
| r = 1 - \cos^2 x
}}
{{eqn | ll= \leadsto
| l = \sin x
| r = \pm \sqrt {1 - \cos^2 x}
}}
{{end-eqn}}
Then from Sign of Sine:
{{begin-eqn}}
{{eqn | l = \sin x
| o = >
| r = 0
| c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$
}}
{{eqn | l = \sin x
| o = <
| r = 0
| c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21400
|
\section{Sine in terms of Secant}
Tags: Trigonometric Functions, Sine Function, Secant Function, Cosine Function
\begin{theorem}
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = \sin x
| r = + \frac {\sqrt{\sec ^2 x - 1} } {\sec x}
| c = if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$
}}
{{eqn | l = \sin x
| r = - \frac {\sqrt{\sec ^2 x - 1} } {\sec x}
| c = if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$
}}
{{end-eqn}}
where $\sin$ denotes the sine function and $\sec$ denotes the secant function.
\end{theorem}
\begin{proof}
For the first part, if there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$:
{{begin-eqn}}
{{eqn | l = \tan x
| r = +\sqrt {\sec^2 x - 1}
| c = Tangent in terms of Secant
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\cos x}
| r = +\sqrt {\sec^2 x - 1}
| c = Tangent is Sine divided by Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x \sec x
| r = +\sqrt {\sec^2 x - 1}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x
| r = +\frac {\sqrt {\sec^2 x - 1} } {\sec x}
}}
{{end-eqn}}
For the second part, if there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$:
{{begin-eqn}}
{{eqn | l = \tan x
| r = -\sqrt {\sec^2 x - 1}
| c = Tangent in terms of Secant
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\cos x}
| r = -\sqrt {\sec^2 x - 1}
| c = Tangent is Sine divided by Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x \sec x
| r = -\sqrt {\sec^2 x - 1}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x
| r = -\frac {\sqrt {\sec^2 x - 1} } {\sec x}
}}
{{end-eqn}}
When $\cos x = 0$, $\sec x$ is undefined.
{{qed}}
\end{proof}
|
21401
|
\section{Sine in terms of Tangent}
Tags: Trigonometric Functions, Sine Function, Tangent Function
\begin{theorem}
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = \sin x
| r = +\frac {\tan x} {\sqrt {1 + \tan^2 x} }
| c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$
}}
{{eqn | l = \sin x
| r = -\frac {\tan x} {\sqrt {1 + \tan^2 x} }
| c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$
}}
{{end-eqn}}
where $\sin$ denotes the real sine function and $\tan$ denotes the real tangent function.
\end{theorem}
\begin{proof}
For the first part, if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$:
{{begin-eqn}}
{{eqn | l = \cos x
| r = +\frac 1 {\sqrt {1 + \tan^2 x} }
| c = Cosine in terms of Tangent
}}
{{eqn | ll= \leadsto
| l = \frac 1 {\paren {\frac 1 {\cos x} } }
| r = +\frac 1 {\sqrt {1 + \tan^2 x} }
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }
| r = +\frac 1 {\sqrt {1 + \tan^2 x} }
| c = multiplying denominator and numerator by $\sin x$
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\tan x}
| r = + \frac 1 {\sqrt {1 + \tan^2 x} }
| c = Tangent is Sine divided by Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x
| r = + \frac {\tan x} {\sqrt {1 + \tan^2 x} }
}}
{{end-eqn}}
For the second part, if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$:
{{begin-eqn}}
{{eqn | l = \cos x
| r = -\frac 1 {\sqrt {1 + \tan^2 x} }
| c = Cosine in terms of Tangent
}}
{{eqn | ll= \leadsto
| l = \frac 1 {\paren {\frac 1 {\cos x} } }
| r = -\frac 1 {\sqrt {1 + \tan^2 x} }
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }
| r = -\frac 1 {\sqrt {1 + \tan^2 x} }
| c = multiplying denominator and numerator by $\sin x$
}}
{{eqn | ll= \leadsto
| l = \frac {\sin x} {\tan x}
| r = -\frac 1 {\sqrt {1 + \tan^2 x} }
| c = Tangent is Sine divided by Cosine
}}
{{eqn | ll= \leadsto
| l = \sin x
| r = -\frac {\tan x} {\sqrt {1 + \tan^2 x} }
}}
{{end-eqn}}
When $\cos x = 0$, $\tan x$ is undefined.
{{qed}}
\end{proof}
|
21402
|
\section{Sine is of Exponential Order Zero}
Tags: Sine Function, Exponential Order, Sine is of Exponential Order Zero
\begin{theorem}
Let $\sin t$ be the sine of $t$, where $t \in \R$.
Then $\sin t$ is of exponential order $0$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \left \vert \sin t \right\vert
| o = \le
| r = 1
| c = Boundedness of Sine and Cosine
}}
{{eqn | ll = \implies
| l = \left \vert \sin t \right\vert
| o = <
| r = 2
}}
{{eqn | r = 2e^{0t}
| c = Exponential of Zero
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
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\end{proof}
|
21403
|
\section{Sine n x over Pi x Delta Sequence}
Tags: Examples of Delta Sequences, Dirac Delta Distribution
\begin{theorem}
thumb500pxThe graph of the $\ds \frac {\map \sin {nx}} {\pi x}$ delta sequence. As $n$ grows, the central peak becomes thinner and taller. The area between each curve and the horizontal axis is equal to $1$. Note that that the area below the line contribute negatively.
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\ds \map {\delta_n} x := \frac {\map \sin {n x} }{\pi x}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
\end{theorem}
\begin{proof}
From Integral to Infinity of Sine p x over x for $n \in \N_{> 0}$ we have that:
:$\ds \int_0^\infty \frac {\map \sin {n x} } x \rd x = \frac \pi 2$
Furthermore:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac {\map \sin {n x} } x \rd x
| r = \int_0^{-\infty} \frac {\map \sin {-n y} } {-y} \paren {- \rd y}
| c = $x = - y$, Integration by Substitution
}}
{{eqn | r = \int_{-\infty}^0 \frac {\map \sin {n y} } y \rd y
}}
{{eqn | r = \frac \pi 2
}}
{{end-eqn}}
Hence:
:$\ds \int_{-\infty}^0 \frac {\map \sin {n y} } y \rd y + \int_0^\infty \frac {\map \sin {n x} } x \rd x = \pi$
Let $a,b \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \map {\delta_n} x \rd x
| r = \int_a^b \frac {\map \sin {n x} } {\pi x} \rd x
}}
{{eqn | r = \int_{a n}^{b n} \frac {\map \sin y } {\pi y} \rd y
| c = $y = n x$, Integration by Substitution
}}
{{end-eqn}}
Suppose $0 < a < b$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac {\map \sin y } {\pi y} \rd y
| r = \lim_{n \mathop \to \infty} \int_0^{b n} \frac {\map \sin y } {\pi y} \rd y - \lim_{n \mathop \to \infty} \int_0^{a n} \frac {\map \sin y } {\pi y} \rd y
| c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_0^\infty \frac {\map \sin y } {\pi y} \rd y - \int_0^\infty \frac {\map \sin y } {\pi y} \rd y
}}
{{eqn | r = \frac \pi 2 - \frac \pi 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Analogously, suppose $a < b < 0$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac {\map \sin y } {\pi y} \rd y
| r = \lim_{n \mathop \to \infty} \int_{a_n}^0 \frac {\map \sin y } {\pi y} \rd y - \lim_{n \mathop \to \infty} \int_{b_n}^0 \frac {\map \sin y } {\pi y} \rd y
| c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_{-\infty}^0 \frac {\map \sin y } {\pi y} \rd y - \int_{-\infty}^0 \frac {\map \sin y } {\pi y} \rd y
}}
{{eqn | r = \frac \pi 2 - \frac \pi 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Let $\epsilon \in \R_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \frac {\map \sin {n x} }{\pi x} \rd x
| r = \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \map \phi x \frac {\map \sin {n x} }{\pi x} \rd x + \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \map \phi x \frac {\map \sin {n x} }{\pi x} \rd x + \lim_{n \mathop \to \infty} \int_\epsilon^\infty \map \phi x \frac {\map \sin {n x} }{\pi x} \rd x
}}
{{eqn | r = \map \phi {\xi_-} \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \frac {\map \sin {n x} }{\pi x} \rd x + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \frac {\map \sin {n x} }{\pi x} \rd x + \map \phi {\xi_+} \lim_{n \mathop \to \infty} \int_{\epsilon}^\infty \frac {\map \sin {n x} }{\pi x} \rd x
| c = Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-\infty} {-\epsilon}$, $\xi_+ \in \hointr \epsilon \infty$
}}
{{eqn | r = 0 + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-n \epsilon}^{n \epsilon} \frac {\map \sin {y} }{\pi y} \rd y + 0
}}
{{eqn | r = \map \phi {\xi_\epsilon}
}}
{{end-eqn}}
$\epsilon$ is an arbitrary positive real number.
Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.
Suppose $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
:$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
Then {{WRT}} $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.
But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.
Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.
{{Proofread|Check the rigour of the last few lines. Make the argument stricter.}}
{{qed}}
\end{proof}
|
21404
|
\section{Sine of 144 Degrees}
Tags: Sine Function
\begin{theorem}
:$\sin 144 \degrees = \cos \dfrac {4 \pi} 5 = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}$
where $\sin$ denotes sine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin 144 \degrees
| r = \sin \paren {180 \degrees - 36 \degrees}
| c =
}}
{{eqn | r = \sin 36 \degrees
| c = Sine of Supplementary Angle
}}
{{eqn | r = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}
| c = Sine of $36 \degrees$
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
\end{proof}
|
21405
|
\section{Sine of 18 Degrees}
Tags: Sine Function
\begin{theorem}
:$\sin 18 \degrees = \sin \dfrac \pi {10} = \dfrac {\sqrt 5 - 1} 4$
where $\sin$ denotes the sine function.
\end{theorem}
\begin{proof}
From Sine of $90 \degrees$:
:$\map \sin {5 \times 18 \degrees} = \sin 90 \degrees = 1$.
Consider the equation:
:$\sin 5x = 1$
where $x = 18 \degrees$ is one of the solutions.
From Quintuple Angle Formula of Sine:
:$16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$
Let $s = \sin \theta$:
:$16 s^5 - 20 s^3 + 5s - 1 = 0$
That is:
:$\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$
Therefore, either:
:$s = 1$
or by the Quadratic Formula:
:$s = \dfrac 1 4 \paren {\pm \sqrt 5 - 1}$
Since $0 < \sin 18 \degrees < 1$:
:$\sin 18 \degrees = \dfrac {\sqrt 5 - 1} 4$
{{qed}}
Category:Sine Function
\end{proof}
|
21406
|
\section{Sine of 1 Degree}
Tags: Sine Function
\begin{theorem}
{{begin-eqn}}
{{eqn | l = \sin 1 \degrees = \sin \dfrac {\pi} {180}
| r = \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
}}
{{eqn | o =
| ro= -
| r = \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
}}
{{end-eqn}}
where $\sin$ denotes the sine function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \sin {3 \times 1 \degrees}
| r = 3 \sin 1 \degrees - 4 \sin^3 1 \degrees
| c = Triple Angle Formula for Sine
}}
{{eqn | l = \sin 3 \degrees
| r = 3 \sin 1 \degrees - 4 \sin^3 1 \degrees
}}
{{eqn | l = 4 \sin^3 1 \degrees - 3 \sin 1 \degrees + \sin 3 \degrees
| r = 0
}}
{{end-eqn}}
This is in the form:
:$a x^3 + b x^2 + c x + d = 0$
where:
:$x = \sin 1 \degrees$
:$a = 4$
:$b = 0$
:$c = -3$
:$d = \sin 3 \degrees$
From Cardano's Formula:
:$x = S + T$
where:
:$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
:$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$
where:
{{begin-eqn}}
{{eqn | l = Q
| r = \dfrac {3 a c - b^2} {9 a^2}
| c =
}}
{{eqn | r = \dfrac {3 \times 4 \times \paren {-3} - 0^2} {9 \times 4^2}
| c =
}}
{{eqn | r = -\dfrac 1 4
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = R
| r = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}
| c =
}}
{{eqn | r = \dfrac {9 \times 4 \times 0 \times \paren {-3} - 27 \times 4^2 \times \sin 3 \degrees - 2 \times 0^3} {54 \times 4^3}
| c =
}}
{{eqn | r = - \dfrac {\sin 3 \degrees} 8
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \sin 1 \degrees
| r = S + T
| c = putting $x = S + T$
}}
{{eqn | r = \sqrt [3] {R + \sqrt {Q^3 + R^2} } + \sqrt[3] {R - \sqrt {Q^3 + R^2} }
| c = substituting for $S$ and $T$
}}
{{eqn | r = \sqrt [3] {R + \sqrt {\paren {-\dfrac 1 4}^3 + R^2} } + \sqrt [3] {R - \sqrt {\paren {-\dfrac 1 4}^3 + R^2} }
| c = substituting for $Q$
}}
{{eqn | r = \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\paren {-\dfrac 1 4}^3 + \dfrac {\sin^2 3 \degrees} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\paren {-\dfrac 1 4}^3 + \dfrac {\sin^2 3 \degrees} {64} } }
| c = substituting for $R$
}}
{{eqn | r = \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\dfrac {\sin^2 3 \degrees - 1} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\dfrac {\sin^2 3 \degrees - 1} {64} } }
| c =
}}
{{eqn | r = \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\dfrac {-\cos^2 3 \degrees} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\dfrac {-\cos^2 3 \degrees} {64} } }
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8 } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - i \dfrac {\cos 3 \degrees} 8 }
| c = $i$ is a square root of $-1$, that is, $i = \sqrt {-1}$.
}}
{{eqn | r = \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8 } + \sqrt [3] {-1 \paren {\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8 } }
| c =
}}
{{eqn | r = \dfrac 1 2 \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } - \dfrac 1 2 \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees }
| c =
}}
{{eqn | r = \dfrac 1 2 \sqrt [3] {-\cos 87 \degrees + i \sin 87 \degrees } - \dfrac 1 2 \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees }
| c = Sine of Complement equals Cosine
}}
{{eqn | r = \dfrac 1 2 \sqrt [3] {\cos 93 \degrees + i \sin 93 \degrees } - \dfrac 1 2 \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees }
| c =
}}
{{eqn | r = \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees } - \sqrt [3] {\cis 87 \degrees } }
| c = KEY EQUATION
}}
{{end-eqn}}
:$\sqrt [3] {\cis 93 \degrees }$ has $3$ unique cube roots.
$\tuple { \cis 31 \degrees, \cis 151 \degrees, \cis 271 \degrees }$
:$\sqrt [3] {\cis 87 \degrees }$ also has $3$ unique cube roots.
$\tuple { \cis 29 \degrees, \cis 149 \degrees, \cis 269 \degrees }$
We need to investigate $9$ differences and find solutions where the complex part vanishes.
{{begin-eqn}}
{{eqn | ll = 1)
| l = \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees } - \sqrt [3] {\cis 87 \degrees } }
| r = \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }
| c = No. $\paren {\sin 31 \degrees - \sin 29 \degrees } > 0$
}}
{{eqn | ll = 2)
| r = \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }
| c = Yes. $\paren {\sin 31 \degrees - \sin 149 \degrees } = 0$
}}
{{eqn | ll = 3)
| r = \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }
| c = No. $\paren {\sin 31 \degrees - \sin 269 \degrees } > 0$
}}
{{eqn | ll = 4)
| r = \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }
| c = Yes. $\paren {\sin 151 \degrees - \sin 29 \degrees } = 0$
}}
{{eqn | ll = 5)
| r = \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }
| c = No. $\paren {\sin 151 \degrees - \sin 149 \degrees } < 0$
}}
{{eqn | ll = 6)
| r = \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }
| c = No. $\paren {\sin 151 \degrees - \sin 269 \degrees } > 0$
}}
{{eqn | ll = 7)
| r = \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }
| c = No. $\paren {\sin 271 \degrees - \sin 29 \degrees } < 0$
}}
{{eqn | ll = 8)
| r = \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }
| c = No. $\paren {\sin 271 \degrees - \sin 149 \degrees } < 0$
}}
{{eqn | ll = 9)
| r = \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }
| c = Yes. $\paren {\sin 271 \degrees - \sin 269 \degrees } = 0$
}}
{{end-eqn}}
In solution $2)$ above, we rotate $\sqrt [3] {\cis 87 \degrees }$ by $120 \degrees$ (Multiply by $\paren { -\dfrac 1 2 + i \dfrac {\sqrt {3} } 2 } $ )
In solution $4)$ above, we rotate $\sqrt [3] {\cis 93 \degrees }$ by $120 \degrees$ (Multiply by $\paren { -\dfrac 1 2 + i \dfrac {\sqrt {3} } 2 } $ )
In solution $9)$ above, we would rotate BOTH $\sqrt [3] {\cis 93 \degrees }$ and $\sqrt [3] {\cis 87 \degrees }$ by $240 \degrees$ (Multiply by $\paren { -\dfrac 1 2 - i \dfrac {\sqrt {3} } 2 } $ )
For the remainder of this proof, we will arbitrarily choose solution $9)$
{{begin-eqn}}
{{eqn | l = \sin 1 \degrees
| r = \paren { -\dfrac 1 2 - i \dfrac {\sqrt {3} } 2 } \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees } - \sqrt [3] {\cis 87 \degrees } }
| c = Solution $9)$ above - rotating by $240 \degrees$
}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\cos 87 \degrees + i \sin 87 \degrees } - \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees } }
| c =
}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } - \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees } }
| c = Sine of Complement equals Cosine
}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\sin 3 \degrees + i \sqrt { 1 - \sin^2 3 \degrees } } - \sqrt [3] {\sin 3 \degrees + i \sqrt {1 - \sin^2 3 \degrees } } }
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} }^2 } } }
| c = Sine of 3 Degrees
}}
{{eqn | o =
| ro= -
| r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} }^2 } } }
}}
{{end-eqn}}
Squaring the Numerator of the Sine of 3 Degrees, we get...
{{begin-eqn}}
{{eqn | l = \paren {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} }^2
| r = 30 + 10\sqrt 3 - 6\sqrt 5 - 2\sqrt {15} - 6 \sqrt {50 + 10 \sqrt 5} + 2 \sqrt {150 + 30\sqrt 5}
| c =
}}
{{eqn | o =
| ro= +
| r = 10\sqrt 3 + 10 - 2\sqrt {15} - 2\sqrt 5 - 2 \sqrt {150 + 30\sqrt 5} + 2 \sqrt {50 + 10 \sqrt 5}
}}
{{eqn | o =
| ro= -
| r = 6\sqrt 5 - 2\sqrt {15} + 6 + 2\sqrt 3 + 6 \sqrt {10 + 2 \sqrt 5} - 2 \sqrt {30 + 6\sqrt 5}
}}
{{eqn | o =
| ro= -
| r = 2\sqrt {15} - 2\sqrt 5 + 2\sqrt 3 + 2 + 2 \sqrt {30 + 6 \sqrt 5} - 2 \sqrt {10 + 2\sqrt 5}
}}
{{eqn | o =
| ro= -
| r = 6 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {150 + 30\sqrt 5} + 6 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {30 + 6 \sqrt 5} + \paren {60 + 12 \sqrt 5 } - 4 \sqrt {90 + 30\sqrt 5}
}}
{{eqn | o =
| ro= +
| r = 2 \sqrt {150 + 30\sqrt 5} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {30 + 6\sqrt 5} - 2 \sqrt {10 + 2\sqrt 5} - 4 \sqrt {90 + 30\sqrt 5} + \paren {20 + 4 \sqrt 5 }
}}
{{eqn | r = 128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30\sqrt 5}
| c = Aggregating terms
}}
{{end-eqn}}
Bringing that result back into our main result, we get...
{{begin-eqn}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30\sqrt 5} } {256} } } } }
| c = Sine of 3 Degrees
}}
{{eqn | o =
| ro= -
| r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30\sqrt 5} } {256} } } } }
}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {128 - 24 \sqrt 3 + 8 \sqrt {15} + 8 \sqrt {50 + 10 \sqrt 5} - 8 \sqrt {10 + 2 \sqrt 5} + 8 \sqrt {90 + 30\sqrt 5} } {256} } } } }
| c = $\paren {1 - \dfrac {128} {256} } = \dfrac {128} {256}$
}}
{{eqn | o =
| ro= -
| r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {128 - 24 \sqrt 3 + 8 \sqrt {15} + 8 \sqrt {50 + 10 \sqrt 5} - 8 \sqrt {10 + 2 \sqrt 5} + 8 \sqrt {90 + 30\sqrt 5} } {256} } } } }
}}
{{end-eqn}}
Canceling $4$ from the numerator and denominator, we get...
{{begin-eqn}}
{{eqn | r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } {64} } } } }
| c =
}}
{{eqn | o =
| ro= -
| r = \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } {64} } } } }
}}
{{end-eqn}}
Moving an $8$ to the outside, we get...
{{begin-eqn}}
{{eqn | r = \paren { -\dfrac 1 8 - i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
| c =
}}
{{eqn | o =
| ro= -
| r = \paren { -\dfrac 1 8 - i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
}}
{{end-eqn}}
Flipping the top and bottom lines, we get our main result:
{{begin-eqn}}
{{eqn | r = \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
| c =
}}
{{eqn | o =
| ro= -
| r = \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} } + i \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } } }
}}
{{end-eqn}}
We can verify that this is correct by noting that...
{{begin-eqn}}
{{eqn | l = 8 \sin 3 \degrees
| r = \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {2} }
}}
{{eqn | l = 8 \cos 3 \degrees
| r = \sqrt {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} }
}}
{{eqn | l = \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 }
| r = \dfrac 1 4 \cis 60 \degrees
}}
{{end-eqn}}
Plugging these into the main result, we see...
{{begin-eqn}}
{{eqn | r = \dfrac 1 4 \cis 60 \degrees \paren {\paren { \sqrt [3] {8 \sin 3 \degrees + i 8 \cos 3 \degrees } } - \paren { \sqrt [3] {-8 \sin 3 \degrees + i 8 \cos 3 \degrees } } }
| c =
}}
{{eqn | r = \dfrac 1 2 \cis 60 \degrees \paren {\paren { \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees } } - \paren { \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } } }
| c = Moving the $8$ outside
}}
{{eqn | r = \dfrac 1 2 \cis 60 \degrees \paren {\paren { \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees } } - \paren { \sqrt [3] {\cos 93 \degrees + i \sin 93 \degrees } } }
| c = Sine of Complement equals Cosine
}}
{{eqn | r = \dfrac 1 2 \cis 60 \degrees \paren { \cis 29 \degrees - \cis 31 \degrees }
| c = Roots of Complex Number
}}
{{eqn | r = \dfrac 1 2 \paren { \cis 89 \degrees - \cis 91 \degrees }
| c = Complex Multiplication as Geometrical Transformation
}}
{{eqn | r = \dfrac 1 2 \paren { \paren {\cos 89 \degrees + i \sin 89 \degrees } - \paren {\cos 91 \degrees + i \sin 91 \degrees } }
| c =
}}
{{eqn | r = \dfrac 1 2 \paren { 2 \paren {\cos 89 \degrees } }
| c = Sine of Supplementary Angle and Cosine of Supplementary Angle
}}
{{eqn | r = \dfrac 1 2 \paren { 2 \paren {\sin 1 \degrees } }
| c = Sine of Complement equals Cosine
}}
{{eqn | r = \sin 1 \degrees
| c =
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
\end{proof}
|
21407
|
\section{Sine of 30 Degrees}
Tags: Sine Function
\begin{theorem}
:$\sin 30 \degrees = \sin \dfrac \pi 6 = \dfrac 1 2$
where $\sin$ denotes the sine function.
\end{theorem}
\begin{proof}
:200px
Let $\triangle ABC$ be an equilateral triangle of side $r$.
By definition, each angle of $\triangle ABC$ is equal.
From Sum of Angles of Triangle equals Two Right Angles it follows that each angle measures $60^\circ$.
Let $CD$ be a perpendicular dropped from $C$ to $AB$ at $D$.
Then $AD = \dfrac r 2$ while:
: $\angle ACD = \dfrac {60 \degrees} 2 = 30 \degrees$
So by definition of sine function:
: $\sin \paren {\angle ACD} = \dfrac {r / 2} r = \dfrac 1 2$
{{qed}}
\end{proof}
|
21408
|
\section{Sine of 36 Degrees}
Tags: Sine Function, Golden Mean
\begin{theorem}
:$\sin 36^\circ = \sin \dfrac \pi 5 = \dfrac {\sqrt {\sqrt 5 / \phi} } 2 = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}$
where $\phi$ denotes the golden mean.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin 36^\circ
| r = \sqrt {1 - \cos^2 36^\circ}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sqrt {1 - \frac {\phi^2} 4}
| c = Cosine of $36^\circ$
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \phi^2}
| c =
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 2 + \frac {\sqrt 5} 2}^2}
| c = {{Defof|Golden Mean|index = 2}}
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 4 + \frac 5 4 + 2 \cdot \frac 1 2 \cdot \frac {\sqrt 5} 2} }
| c = Square of Sum
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 3 2 + \frac {\sqrt 5} 2} }
| c =
}}
{{eqn | r = \frac 1 2 \sqrt {\frac 5 2 - \frac {\sqrt 5} 2}
| c =
}}
{{eqn | r = \sqrt {\frac 5 8 - \frac {\sqrt 5} 8}
| c =
}}
{{end-eqn}}
We also have:
{{begin-eqn}}
{{eqn | l = \frac {\sqrt {\sqrt 5 / \phi} } 2
| r = \frac 1 2 \sqrt {\sqrt 5 \paren {\phi - 1} }
| c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = \frac 1 2 \sqrt {\sqrt 5 \paren {\frac 1 2 + \frac {\sqrt 5} 2 - 1} }
| c = {{Defof|Golden Mean|index = 2}}
}}
{{eqn | r = \frac 1 2 \sqrt {\frac 5 2 - \frac {\sqrt 5} 2}
}}
{{eqn | r = \sqrt {\frac 5 8 - \frac {\sqrt 5} 8}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21409
|
\section{Sine of 3 Degrees}
Tags: Sine Function
\begin{theorem}
:$\sin 3^\circ = \sin \dfrac \pi {60} = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}$
where $\sin$ denotes the sine function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin 3^\circ
| r = \sin \left({75^\circ - 72^\circ}\right)
}}
{{eqn | r = \sin 75^\circ \cos 72^\circ - \cos 75^\circ \sin 72^\circ
| c = Sine of Difference
}}
{{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \cos 75^\circ \sin 72^\circ
| c = Sine of $75^\circ$, Cosine of $72^\circ$ ...
}}
{{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt 5} } 4
| c = ... Cosine of $75^\circ$, Sine of $72^\circ$
}}
{{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt 6 + \sqrt 2}\right) \times \left({\sqrt 5 - 1}\right) - \left({\sqrt 6 - \sqrt 2}\right) \times \left({\sqrt {10 + 2 \sqrt 5} }\right) }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2}\right) - \left({\sqrt {60 + 12 \sqrt 5} - \sqrt {20 + 4 \sqrt 5} }\right) }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - \sqrt {60 + 12 \sqrt5 } + \sqrt {20 + 4 \sqrt 5} }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2 \sqrt {5 + \sqrt 5} }\right)
}}
{{eqn | r = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2\sqrt {5 + \sqrt 5} } {16}
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
\end{proof}
|
21410
|
\section{Sine of Angle in Cartesian Plane}
Tags: Trigonometry, Sine Function, Analytic Geometry
\begin{theorem}
Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\sin \theta = \dfrac y r$
where $\sin$ denotes the sine of $\theta$.
\end{theorem}
\begin{proof}
:500px
Let a unit circle $C$ be drawn with its center at the origin $O$.
Let $Q$ be the point on $C$ which intersects $OP$.
$\angle OSP = \angle ORQ$, as both are right angles.
Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$.
By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.
By definition of similarity:
Then:
{{begin-eqn}}
{{eqn | l = \frac y r
| r = \frac {SP} {OP}
| c =
}}
{{eqn | r = \frac {RQ} {OQ}
| c = {{Defof|Similar Triangles}}
}}
{{eqn | r = RQ
| c = $OP$ is Radius of Unit Circle
}}
{{eqn | r = \sin \theta
| c = {{Defof|Sine|subdef = Definition from Circle}}
}}
{{end-eqn}}
When $\theta$ is obtuse, the same argument holds.
When $\theta = \dfrac \pi 2$ we have that $x = 0$.
Thus $y = r$ and $\sin \theta = 1 \dfrac y r$.
Thus the relation holds for $\theta = \dfrac \pi 2$.
When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.
In this case, both $\sin \theta$ and $y$ are negative.
Thus the relation continues to hold.
When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\sin \theta = 0 = \dfrac y r$.
Hence the result.
{{qed}}
\end{proof}
|
21411
|
\section{Sine of Angle of Triangle by Semiperimeter}
Tags: Triangles, Sine Function
\begin{theorem}
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
: $\sin A = \dfrac 2 {b c} \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $\sin$ denotes sine and $s$ is the semiperimeter: $s = \dfrac {a + b + c} 2$.
\end{theorem}
\begin{proof}
Let $Q$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Two Sides and Angle:
:$Q = \dfrac {b c \sin A} 2$
From Heron's Formula:
:$Q = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
Equating the two:
:$\dfrac {b c \sin A} 2 = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
from which follows the result.
{{qed}}
\end{proof}
|
21412
|
\section{Sine of Angle plus Full Angle/Corollary}
Tags: Sine Function
\begin{theorem}
Let $n \in \Z$ be an integer.
Then:
:$\map \sin {x + 2 n \pi} = \sin x$
\end{theorem}
\begin{proof}
From Sine of Angle plus Full Angle:
:$\map \sin {x + 2 \pi} = \sin x$
The result follows from the General Periodicity Property:
If:
:$\forall x \in X: \map f x = \map f {x + L}$
then:
:$\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$
{{qed}}
Category:Sine Function
\end{proof}
|
21413
|
\section{Sine of Complement equals Cosine}
Tags: Trigonometry, Sine Function, Sine of Complement equals Cosine, Cosine Function
\begin{theorem}
:$\sin \left({\dfrac \pi 2 - \theta}\right) = \cos \theta$
where $\sin$ and $\cos$ are sine and cosine respectively.
That is, the cosine of an angle is the sine of its complement.
\end{theorem}
\begin{proof}
* From Sine and Cosine are Periodic on Reals, we have $\sin \left({x + \dfrac \pi 2}\right) = \cos x$;
* Also from Sine and Cosine are Periodic on Reals, we have $\sin \left({x + \pi}\right) = \cos \left({x + \dfrac \pi 2}\right) = -\sin x$;
* From Basic Properties of Sine Function, we have $\sin \left({x + \dfrac \pi 2}\right) = - \sin \left({- x - \dfrac \pi 2}\right)$.
So:
{{begin-eqn}}
{{eqn | l=\sin \left({\frac \pi 2 - x}\right)
| r=- \sin \left({- \pi + \frac \pi 2 - x}\right)
| c=
}}
{{eqn | r=- \sin \left({- x - \frac \pi 2}\right)
| c=
}}
{{eqn | r=\sin \left({x + \frac \pi 2}\right)
| c=
}}
{{eqn | r=\cos x
| c=
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21414
|
\section{Sine of Full Angle}
Tags: Sine Function
\begin{theorem}
:$\sin 360^\circ = \sin 2 \pi = 0$
where $\sin$ denotes the sine function and $360^\circ = 2 \pi$ is the full angle.
\end{theorem}
\begin{proof}
A direct implementation of Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
In this case, $n = 2$ and so:
:$\sin 2 \pi = 0$
{{qed}}
\end{proof}
|
21415
|
\section{Sine of Half-Integer Multiple of Pi}
Tags: Sine Function, Analysis
\begin{theorem}
Let $x \in \R$ be a real number.
Let $\sin x$ be the sine of $x$.
Then:
:$\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$
or:
{{begin-eqn}}
{{eqn | q = \forall m \in \Z
| l = \map \sin {2 m + \dfrac 1 2} \pi
| r = 1
}}
{{eqn | q = \forall m \in \Z
| l = \map \sin {2 m - \dfrac 1 2} \pi
| r = -1
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From the discussion of Sine and Cosine are Periodic on Reals:
:$\map \sin {x + \dfrac \pi 2} = \cos x$
The result then follows directly from the Cosine of Multiple of Pi.
{{qed}}
Category:Sine Function
\end{proof}
|
21416
|
\section{Sine of Half Angle for Spherical Triangles}
Tags: Spherical Trigonometry, Half Angle Formulas for Spherical Triangles
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\sin \dfrac A 2 = \sqrt {\dfrac {\map \sin {s - b} \, \map \sin {s - c} } {\sin b \sin c} }$
where $s = \dfrac {a + b + c} 2$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \cos a
| r = \cos b \cos c + \sin b \sin c \cos A
| c = Spherical Law of Cosines
}}
{{eqn | r = \cos b \cos c + \sin b \sin c \paren {1 - 2 \sin^2 \dfrac A 2}
| c = Double Angle Formula for Cosine: Corollary 2
}}
{{eqn | r = \map \cos {b - c} - 2 \sin b \sin c \sin^2 \dfrac A 2
| c = Cosine of Difference
}}
{{eqn | ll= \leadsto
| l = \map \cos {b - c} - \cos a
| r = 2 \sin b \sin c \sin^2 \dfrac A 2
| c = rearranging
}}
{{eqn | ll= \leadsto
| l = 2 \sin \dfrac {a + \paren {b - c} } 2 \sin \dfrac {a - \paren {b - c} } 2
| r = 2 \sin b \sin c \sin^2 \dfrac A 2
| c = Prosthaphaeresis Formula for Cosine minus Cosine
}}
{{eqn | ll= \leadsto
| l = 2 \, \map \sin {\dfrac {a + b + c} 2 - c} \, \map \sin {\dfrac {a + b + c} 2 - b}
| r = 2 \sin b \sin c \sin^2 \dfrac A 2
| c =
}}
{{eqn | ll= \leadsto
| l = \map \sin {s - c} \, \map \sin {s - b}
| r = \sin b \sin c \sin^2 \dfrac A 2
| c = setting $s = \dfrac {a + b + c} 2$ and simplifying
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
21417
|
\section{Sine of Half Side for Spherical Triangles}
Tags: Half Side Formulas for Spherical Triangles
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\sin \dfrac a 2 = \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\sin B \sin C} }$
where $S = \dfrac {A + B + C} 2$.
\end{theorem}
\begin{proof}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.
From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:
:not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
:but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
Let $s' = \dfrac {a' + b' + c'} 2$.
We have:
{{begin-eqn}}
{{eqn | l = \cos \dfrac {A'} 2
| r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\sin b' \sin c'} }
| c = Cosine of Half Angle for Spherical Triangles
}}
{{eqn | ll= \leadsto
| l = \cos \dfrac {\pi - a} 2
| r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\map \sin {\pi - B} \, \map \sin {\pi - C} } }
| c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle
}}
{{eqn | ll= \leadsto
| l = \map \cos {\dfrac \pi 2 - \dfrac a 2}
| r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\sin B \sin C} }
| c = Sine of Supplementary Angle
}}
{{eqn | ll= \leadsto
| l = \sin \dfrac a 2
| r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\sin B \sin C} }
| c = Sine of Complement equals Cosine
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = s' - a'
| r = \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2 - \paren {\pi - A}
| c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle
}}
{{eqn | r = \dfrac \pi 2 - \dfrac {A + B + C} 2 + A
| c = simplifying
}}
{{eqn | r = \dfrac \pi 2 - \paren {S - A}
| c = where $S = \dfrac {A + B + C} 2$
}}
{{eqn | ll= \leadsto
| l = \map \sin {s' - a'}
| r = \map \sin {\dfrac \pi 2 - \paren {S - A} }
| c =
}}
{{eqn | r = \map \cos {S - A}
| c = Sine of Complement equals Cosine
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = s'
| r = \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2
| c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle
}}
{{eqn | r = \dfrac {3 \pi} 2 - \dfrac {A + B + C} 2
| c = simplifying
}}
{{eqn | r = \dfrac {3 \pi} 2 - S
| c = where $S = \dfrac {A + B + C} 2$
}}
{{eqn | ll= \leadsto
| l = \sin s'
| r = \map \sin {\dfrac {3 \pi} 2 - S}
| c =
}}
{{eqn | ll= \leadsto
| l = \sin s'
| r = -\map \sin {\dfrac \pi 2 - S}
| c = Sine of Angle plus Straight Angle
}}
{{eqn | r = -\cos S
| c = Sine of Complement equals Cosine
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
21418
|
\section{Sine of Integer Multiple of Argument/Formulation 1}
Tags: Sine Function, Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \sin \theta \paren {\paren {2 \cos \theta}^{n - 1} - \dbinom {n - 2} 1 \paren {2 \cos \theta}^{n - 3} + \dbinom {n - 3} 2 \paren {2 \cos \theta}^{n - 5} - \cdots}
| c =
}}
{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$
\end{proof}
|
21419
|
\section{Sine of Integer Multiple of Argument/Formulation 1/Lemma}
Tags: Sine Function, Sine of Integer Multiple of Argument, Cosine Function
\begin{theorem}
:For $n \in \Z$:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta} \map \sin {\theta}
| r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta} \map \sin {\theta}
| r = \map \cos {n \theta} \map \sin {\theta}
}}
{{eqn | r = \paren {\map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta} \map \cos {\theta} } + \map \cos {n \theta} \map \sin {\theta}
| c = add zero
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \paren {\map \sin {n \theta} \map \cos {\theta} - \map \cos {n \theta} \map \sin {\theta} }
| c = regroup
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta - \theta}
| c = Sine of Difference
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1} \theta}
| c = simplification
}}
{{end-eqn}}
{{qed}}
Category:Sine of Integer Multiple of Argument
\end{proof}
|
21420
|
\section{Sine of Integer Multiple of Argument/Formulation 2}
Tags: Sine Function, Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \cos^n \theta \paren {\dbinom n 1 \paren {\tan \theta} - \dbinom n 3 \paren {\tan \theta}^3 + \dbinom n 5 \paren {\tan \theta}^5 - \cdots}
| c =
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
By De Moivre's Formula:
:$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$
As $n \in \Z_{>0}$, we use the Binomial Theorem on the {{RHS}}, resulting in:
:$\ds \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$
When $k$ is odd, the expression being summed is imaginary.
Equating the imaginary parts of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ odd, gives:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\cos^{n - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}
| c =
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
| c = factor out $\cos^n \theta$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21421
|
\section{Sine of Integer Multiple of Argument/Formulation 3}
Tags: Sine Function, Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \sin \theta \cos^{n - 1} \theta \paren {1 + 1 + \frac {\cos 2 \theta} {\cos^2 \theta} + \frac {\cos 3 \theta} {\cos^3 \theta} + \cdots + \frac {\cos \paren {n - 1} \theta} {\cos^{n - 1} \theta} }
| c =
}}
{{eqn | r = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop \ge 0} \frac {\cos k \theta} {\cos^k \theta}$
\end{proof}
|
21422
|
\section{Sine of Integer Multiple of Argument/Formulation 4}
Tags: Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z$:
{{begin-eqn}}
{{eqn | l = \map \sin {n \theta}
| r = \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 2 } \theta}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
To proceed, we will require the following lemma:
\end{proof}
|
21423
|
\section{Sine of Integer Multiple of Argument/Formulation 5}
Tags: Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \paren {\sin \frac {n \pi } 2 } \paren {\sin \theta } + \paren {2 \cos \theta } \paren {\map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 3 } \theta} + \map \sin {\paren {n - 5 } \theta} - \cdots}
| c =
}}
{{eqn | r = \paren {\sin \frac {n \pi } 2 } \paren {\sin \theta } + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n } \paren {\sin \frac {k \pi } 2 } \map \sin {\paren {n - k } \theta} }
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \paren {\sin \frac {n \pi } 2 } \paren {\sin \theta } + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n } \paren {\sin \frac {k \pi } 2 } \map \sin {\paren {n - k } \theta} }$
\end{proof}
|
21424
|
\section{Sine of Integer Multiple of Argument/Formulation 7}
Tags: Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | l = \sin n \theta
| r = \paren {\sin^2 \frac {n \pi } 2 } \paren {\sin \theta } + \paren {2 \sin \theta } \paren {\map \cos {\paren {n - 1 } \theta} + \map \cos {\paren {n - 3 } \theta} + \map \cos {\paren {n - 5 } \theta} - \cdots}
| c =
}}
{{eqn | r = \paren {\sin^2 \frac {n \pi } 2 } \paren {\sin \theta } + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1 } \paren {\sin^2 \frac {k \pi } 2 } \map \cos {\paren {n - k } \theta} }
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \paren {\sin^2 \frac {n \pi } 2 } \paren {\sin \theta } + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1 } \paren {\sin^2 \frac {k \pi } 2 } \map \cos {\paren {n - k } \theta} }$
\end{proof}
|
21425
|
\section{Sine of Integer Multiple of Argument/Formulation 8}
Tags: Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
\end{theorem}
\begin{proof}
From Sine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \map \sin {n \theta}
| r = \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 2 } \theta}
| c =
}}
{{eqn | r = \map \sin {\paren {n - 1 } \theta} \paren {\paren {2 \cos \theta } - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} } }
| c = factoring out $\map \sin {\paren {n - 1 } \theta}$
}}
{{end-eqn}}
Therefore $a_0 = 2 \cos \theta$
Once again, from Sine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {n \theta} } {\map \sin {\paren {n - 1 } \theta} }
| r = 2 \cos \theta - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} }
| c = dividing both sides by $\map \sin {\paren {n - 1 } \theta}$
}}
{{eqn | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - 1 } \theta} } {\map \sin {\paren {n - 2 } \theta} } }
| c = Move the numerator to the denominator
}}
{{end-eqn}}
In the equations above, let $n = n - k$:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - k - 1 } \theta} }
| r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - k - 1 } \theta} } {\map \sin {\paren {n - k - 2 } \theta} } }
| c =
}}
{{eqn | l = \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - \paren {k + 1} } \theta} }
| r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - \paren {k + 1 } } \theta} } {\map \sin {\paren {n - \paren {k + 2 } } \theta} } }
| c =
}}
{{end-eqn}}
Therefore $a_1 = a_2 = \cdots = a_{n-3} = 2 \cos \theta$
Finally, let $k = n - 3$, then:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {3 \theta} } {\map \sin {2 \theta } }
| r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {2 \theta} } {\map \sin {\theta} } }
| c =
}}
{{eqn | r = 2 \cos \theta - \cfrac 1 {\cfrac {2 \map \sin {\theta} \map \cos {\theta} } {\map \sin {\theta} } }
| c = Double Angle Formula for Sine
}}
{{eqn | r = 2 \cos \theta - \dfrac 1 {2 \map \cos \theta }
| c =
}}
{{end-eqn}}
Therefore $a_{n - 2} = 2 \cos \theta$
Therefore:
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
{{qed}}
\end{proof}
|
21426
|
\section{Sine of Integer Multiple of Argument/Formulation 9}
Tags: Sine of Integer Multiple of Argument
\begin{theorem}
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \cos {\paren {n - 1} \theta} \paren {a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n - r} } } } } }$
where:
:$r = \begin {cases} 2 & : \text {$n$ is even} \\ 1 & : \text {$n$ is odd} \end {cases}$
:$a_k = \begin {cases} 2 \sin \theta & : \text {$k$ is even} \\ -2 \sin \theta & : \text {$k$ is odd and $k < n - 1$} \\ \sin \theta & : k = n - 1 \end {cases}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \sin {n \theta}
| r = \paren {2 \sin \theta } \map \cos {\paren {n - 1 } \theta} + \map \sin {\paren {n - 2 } \theta}
| c = Line 1: Sine of Integer Multiple of Argument/Formulation 6
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { \paren {2 \sin \theta } + \frac {\map \sin {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} } }
| c = Line 2: Factor out $\map \cos {\paren {n - 1 } \theta}$
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {\cfrac {\map \cos {\paren {n - 1 } \theta} } {\map \sin {\paren {n - 2 } \theta} } } }
| c = Line 3: Move the numerator to the denominator
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {\cfrac {\paren {-2 \sin \theta } \map \sin {\paren {n - 2 } \theta} + \map \cos {\paren {n - 3 } \theta} } {\map \sin {\paren {n - 2 } \theta} } } }
| c = Line 4: Cosine of Integer Multiple of Argument/Formulation 6
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac {\map \cos {\paren {n - 3 } \theta} } {\map \sin {\paren {n - 2 } \theta} } } }
| c = Line 5: Simplify expression
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {\cfrac {\map \sin {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 3 } \theta} } } } }
| c = Line 6: Move the numerator to the denominator
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {\cfrac {\paren {2 \sin \theta } \map \cos {\paren {n - 3 } \theta} + \map \sin {\paren {n - 4 } \theta} } {\map \cos {\paren {n - 3} \theta} } } } }
| c = Line 7: Sine of Integer Multiple of Argument/Formulation 6
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {2 \sin \theta + \cfrac {\map \sin {\paren {n - 4 } \theta} } {\map \cos {\paren {n - 3} \theta} } } } }
| c = Line 8: Simplify expression
}}
{{end-eqn}}
By comparing Line 2 to Line 8, we see that:
{{handwaving|All I see on line 2 is an expression concerning $\map \cos {\paren {n - 1} \theta}$, and all I see on line 8 is another more complicated expression concerning $\map \cos {\paren {n - 1} \theta}$. Nothing about $\dfrac {\map \sin {\paren {n - \paren {2k } } \theta} } {\map \cos {\paren {n - \paren {2k - 1 } } \theta} }$.}}
{{begin-eqn}}
{{eqn | l = \frac {\map \sin {\paren {n - \paren {2k } } \theta} } {\map \cos {\paren {n - \paren {2k - 1 } } \theta} }
| r = \paren {\cfrac 1 {-2 \sin \theta + \cfrac 1 {2 \sin \theta + \cfrac {\map \sin {\paren {n - 2\paren {k + 1 } } \theta} } {\map \cos {\paren {n - \paren {2\paren {k + 1 } - 1 } } \theta} } } } }
| c =
}}
{{end-eqn}}
Therefore, the terminal denominator will be:
{{begin-eqn}}
{{eqn | r = \paren {2 \sin \theta} + \frac {\map \sin {\paren {n - \paren {2k } } \theta} } {\map \cos {\paren {n - \paren {2k - 1 } } \theta} }
| c =
}}
{{end-eqn}}
Assume $n$ even
$n = 2k$
{{begin-eqn}}
{{eqn | r = \paren {2 \sin \theta} + \frac {\map \sin {\paren {2k - \paren {2k } } \theta} } {\map \cos {\paren {2k - \paren {2k - 1 } } \theta} }
| c =
}}
{{eqn | r = \paren {2 \sin \theta} + \frac {\map \sin {0} } {\map \cos {\theta} }
| c =
}}
{{eqn | r = \paren {2 \sin \theta}
| c =
}}
{{end-eqn}}
Assume $n$ odd
$n = 2k - 1$
{{begin-eqn}}
{{eqn | r = \paren {2 \sin \theta} + \frac {\map \sin {\paren {2k - 1 - \paren {2k } } \theta} } {\map \cos {\paren {2k - 1 - \paren {2k - 1 } } \theta} }
| c =
}}
{{eqn | r = \paren {2 \sin \theta} + \frac {\map \sin {-\theta} } {\map \cos {0} }
| c =
}}
{{eqn | r = \paren {2 \sin \theta} - \frac {\map \sin {\theta} } {\map \cos {0} }
| c =
}}
{{eqn | r = \paren {\sin \theta}
| c =
}}
{{end-eqn}}
Therefore:
:$\sin n \theta = \map \cos {\paren {n - 1} \theta} \paren {a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n - r} } } } } }$
where:
:$r = \begin {cases} 2 & : \text {$n$ is even} \\ 1 & : \text {$n$ is odd} \end {cases}$
:$a_k = \begin {cases} 2 \sin \theta & : \text {$k$ is even} \\ -2 \sin \theta & : \text {$k$ is odd and $k < n - 1$} \\ \sin \theta & : k = n - 1 \end {cases}$
{{qed}}
\end{proof}
|
21427
|
\section{Sine of Integer Multiple of Pi}
Tags: Sine Function, Analysis
\begin{theorem}
Let $x \in \R$ be a real number.
Let $\sin x$ be the sine of $x$.
Then:
:$\forall n \in \Z: \sin n \pi = 0$
\end{theorem}
\begin{proof}
This is established in Zeroes of Sine and Cosine.
{{qed}}
\end{proof}
|
21428
|
\section{Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean}
Tags: Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean, Fibonacci Numbers, Golden Mean
\begin{theorem}
Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the golden mean
:$F_n$ denotes the $n$th Fibonacci number.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin n z
| r = \sin \left({\dfrac {n \pi} 2 + i n \ln \phi}\right)
| c =
}}
{{eqn | r = \frac {e^{i \left({\left({n \pi / 2}\right) + i n \ln \phi}\right)} - e^{-i \left({\left({n \pi / 2}\right) + i n \ln \phi}\right)} } {2 i}
| c = Sine Exponential Formulation
}}
{{eqn | r = \frac {e^{i n \pi / 2} e^{-n \ln \phi} - e^{-i n \pi / 2} e^{n \ln \phi} } {2 i}
| c =
}}
{{eqn | r = \frac {e^{-n \ln \phi} \left({\cos \frac {n \pi} 2 + i \sin \frac {n \pi} 2}\right) - e^{n \ln \phi} \left({\cos \left({-\frac {n \pi} 2}\right) + i \sin \left({-\frac {n \pi} 2}\right)}\right)} {2 i}
| c = Euler's Formula and Corollary
}}
{{eqn | r = \frac {e^{-n \ln \phi} \left({i \sin \frac {n \pi} 2}\right) - e^{n \ln \phi} \left({i \sin \left({-\frac {n \pi} 2}\right)}\right)} {2 i}
| c = Cosine of Half-Integer Multiple of Pi
}}
{{eqn | r = \frac {i^n e^{-n \ln \phi} - \left({-i}\right)^n e^{n \ln \phi} } {2 i}
| c = Sine of Half-Integer Multiple of Pi and simplification
}}
{{eqn | r = \frac {i^{n - 1} \left({e^{-n \ln \phi} + e^{n \ln \phi} }\right)} 2
| c = simplification
}}
{{eqn | r = i^{n - 1} \frac {\phi^n + \frac 1 {\phi^n} } 2
| c = Exponential of Natural Logarithm
}}
{{eqn | r = i^{n - 1} \frac {\phi^n - \left({-\frac 1 {\phi^n} }\right)} 2
| c =
}}
{{eqn | r = i^{n - 1} \frac {\phi^n - \hat \phi^n} 2
| c = Reciprocal Form of One Minus Golden Mean
}}
{{end-eqn}}
Setting $n = 1$:
:$\sin z = i^0 \frac {\phi^1 - \hat \phi^1} 2 = \frac {\phi - \hat \phi} 2$
Thus:
{{begin-eqn}}
{{eqn | l = \dfrac {\sin n z} {\sin z}
| r = \dfrac {i^{n - 1} \frac {\phi^n - \hat \phi^n} 2} {\frac {\phi - \hat \phi} 2}
| c =
}}
{{eqn | r = i^{n - 1} \dfrac {\phi^n - \hat \phi^n } {\phi - \hat \phi}
| c =
}}
{{eqn | r = i^{n - 1} \frac {\phi^n - \hat \phi^n} {\sqrt 5} } {\frac {\phi - \hat \phi} {\sqrt 5} }
| c = Euler-Binet Formula
}}
{{eqn | r = i^{n - 1} \dfrac {F_n} {F_1}
| c = Euler-Binet Formula
}}
{{end-eqn}}
{{qed}}
360216
360212
2018-07-09T15:57:28Z
Prime.mover
59
360216
wikitext
text/x-wiki
\end{proof}
|
21429
|
\section{Sine of Right Angle}
Tags: Sine Function
\begin{theorem}
:$\sin 90 \degrees = \sin \dfrac \pi 2 = 1$
where $\sin$ denotes the sine function.
\end{theorem}
\begin{proof}
A direct implementation of Sine of Half-Integer Multiple of Pi:
:$\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$
In this case, $n = 0$ and so:
:$\sin \dfrac 1 2 \pi = \paren {-1}^0 = 1$
{{qed}}
\end{proof}
|
21430
|
\section{Sine of Straight Angle}
Tags: Sine Function
\begin{theorem}
:$\sin 180 \degrees = \sin \pi = 0$
where $\sin$ denotes the sine function and $180 \degrees = \pi$ is the straight angle.
\end{theorem}
\begin{proof}
A direct implementation of Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
In this case, $n = 1$ and so:
:$\sin \pi = 0$
{{qed}}
\end{proof}
|
21431
|
\section{Sine of X over X as Infinite Product}
Tags: Trigonometric Identities, Proofs by Induction, Examples of Infinite Products, Infinite Products
\begin{theorem}
Let $z$ be a non-zero Complex Number.
Then:
:$\ds \frac {\sin z} z = \cos \frac z 2 \cos \frac z 4 \cos \frac z 8 \cdots = \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$
where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
\end{theorem}
\begin{proof}
First we prove that:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
for $n \in \N$.
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
\end{proof}
|
21432
|
\section{Sine of X over X is not Continuous at 0}
Tags: Examples of Continuous Real Functions
\begin{theorem}
Let $f$ be the real function defined as:
:$\map f x := \dfrac {\sin x} x$
Then $f$ is not continuous at $x = 0$.
\end{theorem}
\begin{proof}
For $f$ to be continuous at $x = 0$ it is necessary that it be defined there.
But at the point $x = 0$, we have that $\map f x = \dfrac {\sin 0} 0$.
Division by $0$ is not defined.
Hence $f$ is not continuous at $x = 0$.
{{qed}}
\end{proof}
|
21433
|
\section{Sine of Zero is Zero}
Tags: Sine Function, Analysis
\begin{theorem}
:$\sin 0 = 0$
where $\sin$ denotes the sine.
\end{theorem}
\begin{proof}
Recall the definition of the sine function:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
Thus:
:$\ds \sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$
{{qed}}
\end{proof}
|
21434
|
\section{Sine of i}
Tags: Sine Function, Sine of i, Complex Numbers
\begin{theorem}
:$\sin i = \paren {\dfrac e 2 - \dfrac 1 {2 e} } i$
where $\sin$ denotes the complex sine function and $i$ is the imaginary unit.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | o=
| r=\cos i + i \sin i
}}
{{eqn | r=e^{i \times i}
| c=Euler's Formula
}}
{{eqn | r= e^{-1}
| c=Definition of imaginary number
}}
{{eqn | r=\frac 1 e
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | o=
| r=\cos \left({i}\right) - i \sin \left({i}\right)
}}
{{eqn | r=\cos \left({-i}\right) + i \sin \left({-i}\right)
| c=Cosine Function is Even and Sine Function is Odd
}}
{{eqn | r=e^{i \times \left({-i}\right)}
| c=Euler's Formula
}}
{{eqn | r= e^1
| c=Definition of imaginary number
}}
{{eqn | r=e
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | n=1
| l=\cos i + i \sin i
| r=\frac 1 e
}}
{{eqn | n=2
| l=\cos i - i \sin i
| r=e
}}
{{end-eqn}}
Then from $\left({1}\right) - \left({2}\right)$:
{{begin-eqn}}
{{eqn | l=2i \sin i
| r=\frac 1 e - e
}}
{{eqn | l=\sin i
| r=\frac {1} {2i} \left({ \frac 1 e - e }\right)
}}
{{eqn | l=\sin i
| r=\left({ \frac e 2 - \frac 1 {2e} }\right) i
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Complex Numbers
219735
219733
2015-06-04T07:25:14Z
Kc kennylau
2331
219735
wikitext
text/x-wiki
{{refactor}}
\end{proof}
|
21435
|
\section{Sine of x minus Cosine of x/Cosine Form}
Tags: Sine Function, Cosine Function
\begin{theorem}
:$\sin x - \cos x = \sqrt 2 \, \map \cos {x - \dfrac {3 \pi} 4}$
where $\sin$ denotes sine and $\cos$ denotes cosine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin x - \cos x
| r = \sqrt 2 \, \map \sin {x - \dfrac \pi 4}
| c = Sine of x minus Cosine of x: Sine Form
}}
{{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - \paren {x - \dfrac \pi 4} }
| c = Cosine of Complement equals Sine
}}
{{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - x + \dfrac \pi 4}
| c = simplifying
}}
{{eqn | r = \sqrt 2 \, \map \cos {\frac {3 \pi} 4 - x}
| c = simplifying
}}
{{eqn | r = \sqrt 2 \, \map \cos {x - \frac {3 \pi} 4}
| c = Cosine Function is Even
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
\end{proof}
|
21436
|
\section{Sine of x minus Cosine of x/Sine Form}
Tags: Sine Function, Cosine Function
\begin{theorem}
:$\sin x - \cos x = \sqrt 2 \sin \left({x - \dfrac \pi 4}\right)$
where $\sin$ denotes sine and $\cos$ denotes cosine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin x - \cos x
| r = \sin x - \sin \left({\frac \pi 2 - x}\right)
| c = Sine of Complement equals Cosine
}}
{{eqn | r = 2 \cos \left({\frac {x + \left({\frac \pi 2 - x}\right)} 2}\right) \sin \left({\frac {x - \left({\frac \pi 2 - x}\right)} 2}\right)
| c = Prosthaphaeresis Formula for Sine minus Sine
}}
{{eqn | r = 2 \cos \frac \pi 4 \sin \left({x - \frac \pi 4}\right)
| c = simplifying
}}
{{eqn | r = \sqrt 2 \sin \left({x - \frac \pi 4}\right)
| c = Cosine of $\dfrac \pi 4$
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
\end{proof}
|
21437
|
\section{Sine of x plus Cosine of x/Cosine Form}
Tags: Sine Function, Cosine Function
\begin{theorem}
:$\sin x + \cos x = \sqrt 2 \, \map \cos {x - \dfrac \pi 4}$
where $\sin$ denotes sine and $\cos$ denotes cosine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin x + \cos x
| r = \sin x + \map \sin {\frac \pi 2 - x}
| c = Sine of Complement equals Cosine
}}
{{eqn | r = 2 \, \map \sin {\frac {x + \paren {\frac \pi 2 - x} } 2} \map \cos {\frac {x - \paren {\frac \pi 2 - x} } 2}
| c = Prosthaphaeresis Formula for Sine plus Sine
}}
{{eqn | r = 2 \sin \frac \pi 4 \, \map \cos {x - \frac \pi 4}
| c = simplifying
}}
{{eqn | r = \sqrt 2 \, \map \cos {x - \frac \pi 4}
| c = Sine of $\dfrac \pi 4$
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
\end{proof}
|
21438
|
\section{Sine of x plus Cosine of x/Sine Form}
Tags: Sine Function, Cosine Function
\begin{theorem}
:$\sin x + \cos x = \sqrt 2 \sin \left({x + \dfrac \pi 4}\right)$
where $\sin$ denotes sine and $\cos$ denotes cosine.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin x + \cos x
| r = \sqrt 2 \cos \left({x - \dfrac \pi 4}\right)
| c = Sine of x plus Cosine of x: Cosine Form
}}
{{eqn | r = \sqrt 2 \sin \left({x - \frac \pi 4 + \frac \pi 4}\right)
| c = Sine of Angle plus Right Angle
}}
{{eqn | r = \sqrt 2 \sin \left({x + \frac \pi 4}\right)
| c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
\end{proof}
|
21439
|
\section{Singleton Class can be Formed from Set}
Tags: Singleton Classes
\begin{theorem}
Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then the singleton class $\set a$ can be formed, which is a subclass of $V$.
\end{theorem}
\begin{proof}
Using the axiom of specification, let $A$ be the class defined as:
:$A := \set {x: x \in V \land x = a}$
That is:
:$A = \set a$
By the axiom of extension, $\set a$ is the only such class which has $a$ as an element.
\end{proof}
|
21440
|
\section{Singleton Class of Empty Set is Supercomplete}
Tags: Singleton Classes, Supercomplete Classes, Empty Set
\begin{theorem}
Let $\O$ denote the empty set.
Then the singleton $\set \O$ is supercomplete.
\end{theorem}
\begin{proof}
Let $x \in \set \O$ be any element of $\set \O$.
Then it has to be the case that $x = \O$.
Then every element of $\O$ is an element of $\set \O$ vacuously.
That is, $\set \O$ is swelled.
There is one element of $\set \O$, and that is $\O$.
This is a subclass of $\set \O$.
That is, $\set \O$ is transitive.
The result follows by definition of supercomplete class.
{{qed}}
\end{proof}
|
21441
|
\section{Singleton Class of Set is Set}
Tags: Singletons, Singleton Classes, Axiom of Pairing
\begin{theorem}
Let $x$ be a set.
Then the singleton class $\set x$ is likewise a set.
\end{theorem}
\begin{proof}
Let $x$ and $y$ be sets.
Let $x = y$.
From Doubleton Class of Equal Sets is Singleton Class, the doubleton class $\set {x, y}$ is the singleton class $\set x$.
From the axiom of pairing, the doubleton class $\set {x, y}$ is a set when $x$ and $y$ are sets.
Hence $\set x$ is a set.
{{qed}}
\end{proof}
|
21442
|
\section{Singleton Classes are Equal iff Sets are Equal}
Tags: Singleton Classes
\begin{theorem}
Let $a$ and $b$ be sets.
Let $\set a$ and $\set b$ denote the singleton classes of $a$ and $b$.
Then:
:$\set a = \set b \iff a = b$
\end{theorem}
\begin{proof}
Let $a = b$.
Then $\set a$ and $\set b$ contain the same elements.
Hence by the axiom of extension it follows that $\set a = \set b$.
Let $\set a = \set b$.
We have that $a \in \set a$.
As $\set a = \set b$ we also have that $a \in \set b$.
But $b$ is the only element of $\set b$.
Hence it must be the case that $a = b$.
{{qed}}
\end{proof}
|
21443
|
\section{Singleton Partition yields Indiscrete Topology}
Tags: Partition Topology, Indiscrete Topology
\begin{theorem}
Let $S$ be a set which is not empty.
Let $\PP$ be the (trivial) singleton partition $\set S$ on $S$.
Then the partition topology on $\PP$ is the indiscrete topology.
\end{theorem}
\begin{proof}
By definition, the partition topology on $\PP$ is the set of all unions from $\PP$.
This is (trivially, and from Union of Empty Set) $\set {\O, S}$ which is the indiscrete topology on $S$ by definition.
{{qed}}
\end{proof}
|
21444
|
\section{Singleton Point is Isolated}
Tags: Singletons, Isolated Points
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then $x$ is an isolated point of the singleton set $\set x$, but not necessarily an isolated point of $T$.
\end{theorem}
\begin{proof}
Let $U \in \tau$ be an open set of $T$ such that $x \in T$.
The fact that such a $U$ exists follows from the fact that:
:from {{OpenSetAxiom|3}}, $S$ is open in $T$
:$x \in S$.
Hence:
:$\set x \subseteq S$
and so from Intersection with Subset is Subset:
$\set x \cap U = \set x$
So by definition, $x$ is an isolated point of $\set x$.
From Topological Space is Discrete iff All Points are Isolated, unless $T$ is the discrete space on $S$, not all elements of $T$ are isolated points of $T$.
{{qed}}
Category:Isolated Points
Category:Singletons
\end{proof}
|
21445
|
\section{Singleton Set in Discrete Space is Compact}
Tags: Compact Spaces, Singleton Set in Discrete Space is Compact, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Let $x \in S$.
Then $\set x$ is compact.
\end{theorem}
\begin{proof}
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\left\{{x}\right\}$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\left\{{x}\right\}$ equals its closure.
As $\left\{{\left\{{x}\right\}}\right\}$ is (trivially) an open cover of $\left\{{x}\right\}$, it follows by definition that $\left\{{x}\right\}$ is compact.
{{qed}}
Category:Compact Spaces
Category:Discrete Topology
316631
110924
2017-09-09T16:51:14Z
Barto
3079
316631
wikitext
text/x-wiki
\end{proof}
|
21446
|
\section{Singleton Set is Nowhere Dense in Rational Space}
Tags: Singletons, Rational Number Space
\begin{theorem}
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then every singleton subset of $\Q$ is nowhere dense in $\struct {\Q, \tau_d}$.
\end{theorem}
\begin{proof}
Let $x \in \Q$.
By definition of nowhere dense, we need to show that:
:$\paren {\set x^-}^\circ = \O$
where $S^-$ denotes the closure of a set $S$ and $S^\circ$ denotes its interior.
By Real Number is Closed in Real Number Line, $\set x$ is closed in $\struct {\R, \tau_d}$.
$\struct {\Q, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$.
Thus by Closed Set in Topological Subspace, $\set x$ is closed in $\struct {\Q, \tau_d}$.
From Closed Set Equals its Closure, it follows that:
:$\set x^- = \set x$
From Interior of Singleton in Real Number Line is Empty:
:$\set x^\circ = \O$
Hence:
:$\paren {\set x^-}^\circ = \set x^\circ = \O$
{{qed}}
Category:Rational Number Space
Category:Singletons
\end{proof}
|
21447
|
\section{Singleton Set is not Dense-in-itself}
Tags: Singletons, Denseness
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x \in S$.
Then the singleton set $\left\{{x}\right\}$ is not dense-in-itself.
\end{theorem}
\begin{proof}
From Singleton Point is Isolated, $x$ is isolated in $\left\{{x}\right\}$.
So by definition $\left\{{x}\right\}$ is not dense-in-itself.
{{qed}}
Category:Denseness
Category:Singletons
\end{proof}
|
21448
|
\section{Singleton fulfils Naturally Ordered Semigroup Axioms 1 to 3}
Tags: Naturally Ordered Semigroup
\begin{theorem}
Let $S$ be a singleton:
:$S = \set s$
for an arbitrary object $s$.
Let $+$ be the operation on $S$ defined as:
:$\forall s \in S: s + s = s$
Let $\le$ be the relation defined on $S$ as:
:$s \le s$
Then the algebraic structure:
:$\struct {S, +, \le}$
is an ordered semigroup which fulfils the axioms:
:{{NOSAxiom|1}}
:{{NOSAxiom|2}}
:{{NOSAxiom|3}}
but:
:does not fulfil {{NOSAxiom|4}}
:$\struct {S, +}$ is not isomorphic to $\struct {\N, +}$.
\end{theorem}
\begin{proof}
Recall the axioms:
{{:Axiom:Naturally Ordered Semigroup Axioms}}
\end{proof}
|
21449
|
\section{Singleton in Normed Vector Space is Closed}
Tags: Definitions: Normed Vector Spaces
\begin{theorem}
Let $X$ be a vector space over $\R$ or $\C$.
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $\set x \subseteq X$ be a singleton.
Then $\set x$ is closed.
\end{theorem}
\begin{proof}
Note that either $\set x = X$ or $\set x \subset X$.
Suppose, $X = \set x$.
Then:
:$X \setminus \set x = \empty$
We have that Empty Set is Open in Normed Vector Space.
$\set x$ is a complement of $\empty$ in $X$.
By definition, $x$ is closed.
Suppose, $X \ne \set x$, i.e. $\set x \subset X$.
Define $U := X \setminus \set x$.
Let $y \in U$.
Let $\epsilon := \norm {x - y}$.
Suppose, $\epsilon > 0$.
By Normed Vector Space is Open in Itself:
:$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq X$
Suppose, $z \in \map {B_\epsilon} y$.
Then:
:$\norm {y - z} < \epsilon$.
We have that:
{{begin-eqn}}
{{eqn | l = \norm {z - x}
| o = \ge
| r = \norm {x - y} - \norm {y - z}
| c = Triangle inequality
}}
{{eqn | o = \ge
| r = \epsilon - \norm {y - z}
}}
{{eqn | o = >
| r = \epsilon - \epsilon
}}
{{eqn | r = 0}}
{{end-eqn}}
Hence, $z \ne x$.
Then $z$ is in the complement of $\set x$ in $X$, i.e. $z \in U$.
In other words:
:$\forall y \in U : \exists \epsilon = \norm {x - y} > 0 : \map {B_\epsilon} x \subseteq U$
So $U$ is open, and $\set x = X \setminus U$ is closed.
{{qed}}
\end{proof}
|
21450
|
\section{Singleton is Chain}
Tags: Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x \in S$.
Then $\left\{ {x}\right\}$ is a chain of $\left({S, \preceq}\right)$.
\end{theorem}
\begin{proof}
It suffices to prove that
:$\left\{ {x}\right\}$ is connected
Let $y, z \in \left\{ {x}\right\}$
By definition of singleton:
:$y = x$ and $z = x$
By definition of reflexivity;
:$y \preceq z$
Thus
:$y \preceq z$ or $z \preceq y$
{{qed}}
\end{proof}
|
21451
|
\section{Singleton is Connected in Topological Space}
Tags: Connected Spaces, Singletons
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then the singleton $\set{x}$ is connected.
\end{theorem}
\begin{proof}
Let $A = \set{x}$.
From definition $3$ of a connected set, $A$ is connected in $T$ {{iff}} the subspace $\struct {A, \tau_A}$ is a connected space.
From Topology on Singleton is Indiscrete Topology, $\tau_A$ is the indiscrete topology.
From Indiscrete Space is Connected, $\struct {A, \tau_A}$ is a connected space.
{{qed}}
Category:Connected Spaces
Category:Singletons
\end{proof}
|
21452
|
\section{Singleton is Convex Set}
Tags: Singletons, Vector Spaces, Convex Sets (Vector Spaces)
\begin{theorem}
Let $V$ be a vector space over $\R$ or $\C$, and let $v \in V$.
Then the singleton $S = \set v$ is a convex set.
\end{theorem}
\begin{proof}
For any $x, y \in S$, we have $x = y = v$.
It follows that:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y = v \in S$
Hence $S$ is a convex set.
{{qed}}
\end{proof}
|
21453
|
\section{Singleton is Convex Set (Order Theory)}
Tags: Convex Sets (Order Theory), Singletons, Convex Sets, Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x \in S$.
Then the singleton $\left\{{x}\right\}$ is convex.
\end{theorem}
\begin{proof}
Let:
:$a, c \in \left\{{x}\right\}$
:$b \in S$
:$a \preceq b \preceq c$
Then $a = c = x$.
Thus $x \preceq b \preceq x$.
Since $\preceq$ is a ordering, it is antisymmetric.
Thus $b = x$, so $b \in \left\{{x}\right\}$.
Since this holds for the only such triple, $\left\{{x}\right\}$ is convex.
{{qed}}
Category:Convex Sets (Order Theory)
Category:Singletons
\end{proof}
|
21454
|
\section{Singleton is Dependent implies Rank is Zero}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Let $\set x$ be dependent.
Then:
:$\map \rho {\set x} = 0$
where $\rho$ denotes the rank function of $M$.
\end{theorem}
\begin{proof}
By definition of a dependent subset:
:$\set x \notin \mathscr I$
Then:
{{begin-eqn}}
{{eqn | l = \map \rho {\set x}
| r = \max \set{\size A : A \in \powerset {\set x} \land A \in \mathscr I}
| c = {{Defof|Rank Function (Matroid)|Rank Function}}
}}
{{eqn | r = \max \set {\size A : A \in \set {\O, \set x} \land A \in \mathscr I}
| c = Power Set of Singleton
}}
{{eqn | r = \max \set {\size \O}
| c = As $\set x \notin \mathscr I$ and Matroid axiom $(\text I 1)$ : $\O \in \mathscr I$
}}
{{eqn | r = \max \set 0
| c = Cardinality of Empty Set
}}
{{eqn | r = 0
| c = {{Defof|Max Operation}}
}}
{{end-eqn}}
{{qed}}
Category:Matroid Theory
\end{proof}
|
21455
|
\section{Singleton is Dependent implies Rank is Zero/Corollary}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop {{iff}} $\map \rho {\set x} = 0$
where $\rho$ denotes the rank function of $M$.
\end{theorem}
\begin{proof}
By definition of a loop:
:$x$ is a loop {{iff}} $\set x \notin \mathscr I$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
It follows that:
:$\set x \notin \mathscr I$ {{iff}} $\map \rho {\set x} = 0$
{{qed}}
\end{proof}
|
21456
|
\section{Singleton is Directed and Filtered Subset}
Tags: Preorder Theory
\begin{theorem}
Let $\struct {S, \precsim}$ be a preordered set.
Let $x$ be an element of $S$.
Then $\set x$ is directed and filtered subset of $S$.
\end{theorem}
\begin{proof}
Let $y, z \in \set x$.
By definition of singleton:
:$ y = x \land z = x$
By definition of reflexivity:
:$y \precsim x \land z \precsim x$
Thus:
:$\exists h \in \set x: y \precsim h \land z \precsim h$
Thus by definition:
:$\set x$ is a directed subset of $S$.
$\set x$ is a filtered subset of $S$ by mutatis mutandis.
{{qed}}
\end{proof}
|
21457
|
\section{Singleton is Finite}
Tags: Singletons
\begin{theorem}
Let $x$ be arbitrary.
Then $\set x$ is a finite set.
\end{theorem}
\begin{proof}
Define a mapping $f: \set x \to \N_{< 1}$:
:$\map f x = 0$
By definition of singleton:
:$\forall y, z \in \set x: \map f y = \map f z \implies y = z$
By definition:
:$f$ is an injection.
By definition of initial segment of natural numbers:
:$\N_{< 1} = \set 0$
By definition of $f$:
:$\forall n \in N_{< 1}: \exists z \in \set x: \map f z = n$
By definition:
:$f$ is a surjection.
By definition:
:$f$ is a bijection.
By definition of set equivalence:
:$\set x \sim \N_{< 1}$
Thus by definition:
:$\set x$ is a finite set.
{{qed}}
\end{proof}
|
21458
|
\section{Singleton is Independent implies Rank is One}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Let $\set x$ be independent.
Then:
:$\map \rho {\set x} = 1$
where $\rho$ denotes the rank function of $M$.
\end{theorem}
\begin{proof}
From Rank of Independent Subset Equals Cardinality:
:$\map \rho {\set x} = \size {\set x}$
From Cardinality of Singleton:
:$\size {\set x} = 1$
The result follows.
{{qed}}
Category:Matroid Theory
\end{proof}
|
21459
|
\section{Singleton is Independent implies Rank is One/Corollary}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$\set x$ is an independent subset {{iff}} $\map \rho {\set x} = 1$
where $\rho$ denotes the rank function of $M$.
\end{theorem}
\begin{proof}
By definition of an independent subset:
:$x$ is an independent subset {{iff}} $\set x \notin \mathscr I$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
It follows that:
:$\set x \in \mathscr I$ {{iff}} $\map \rho {\set x} = 1$
{{qed}}
Category:Matroid Theory
\end{proof}
|
21460
|
\section{Singleton is Linearly Independent}
Tags: Linear Algebra
\begin{theorem}
Let $K$ be a division ring.
Let $\struct {G, +_G}$ be a group whose identity is $e$.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space whose zero is $0_K$.
Let $x \in G: x \ne e$.
Then $\set x$ is a linearly independent subset of $G$.
\end{theorem}
\begin{proof}
The only sequence of distinct terms in $\set x$ is the one that goes: $x$.
Suppose $\exists \lambda \in K: \lambda \circ x = e$.
From Zero Vector Space Product iff Factor is Zero it follows that $\lambda = 0$.
Hence the result from definition of linearly independent set.
{{qed}}
\end{proof}
|
21461
|
\section{Singleton is Terminal Object of Category of Sets}
Tags: Singletons, Category Theory
\begin{theorem}
Let $\mathbf {Set}$ be the category of sets.
Let $S = \set x$ be any singleton set.
Then $S$ is a terminal object of $\mathbf {Set}$.
\end{theorem}
\begin{proof}
Let $T$ be a set, and let $f: T \to S$ be a mapping.
Then since for all $t \in T$, we have $\map f t \in S$, it follows that:
:$\forall t \in T: \map f t = x$
By Equality of Mappings, there is precisely one such mapping $f: T \to S$.
Hence the result, by definition of terminal object.
{{qed}}
\end{proof}
|
21462
|
\section{Singleton of Power Set less Empty Set is Minimal Subset}
Tags: Power Set
\begin{theorem}
Let $S$ be a set which is non-empty.
Let $\CC = \powerset S \setminus \O$, that is, the power set of $S$ without the empty set.
Let $x \in S$.
Then $\set x$ is a minimal element of the ordered structure $\struct {\CC, \subseteq}$.
\end{theorem}
\begin{proof}
Let $y \in \CC$ such that $y \subseteq \set x$.
We have that $\O \notin \CC$.
Therefore:
:$\exists z \in S: z \in y$
But as $y \subseteq \set x$ it follows that:
:$z \in \set x$
and so by definition of singleton:
:$z = x$
and so:
:$y = \set x$
and so:
:$y = x$
Thus, by definition, $\set x$ is a minimal element of $\struct {\CC, \subseteq}$.
{{qed}}
\end{proof}
|
21463
|
\section{Singleton of Set is Filter in Lattice of Power Set}
Tags: Power Set
\begin{theorem}
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \subseteq}$ be an inclusion lattice of power set of $X$.
Then $\set X$ is a filter on $L$.
\end{theorem}
\begin{proof}
By Singleton is Directed and Filtered Subset:
:$\set X$ is filtered.
We will prove that
:$\set X$ is an upper set.
Let $x \in \set X$, $y \in \powerset X$ such that:
:$x \subseteq y$
By definition of singleton:
:$x = X$
By definition of power set:
:$y \subseteq X$
By definition of set equality:
:$y = X$
Thus:
:$y \in \set X$
{{qed|lemma}}
Thus by definition of filter in ordered set:
:$\set X$ is a filter.
{{qed}}
\end{proof}
|
21464
|
\section{Singular Random Variable is not Absolutely Continuous}
Tags: Singular Random Variables
\begin{theorem}
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a singular random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ is not absolutely continuous.
\end{theorem}
\begin{proof}
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\lambda$ be the Lebesgue measure for $\struct {\R, \map \BB \R}$.
From the definition of an absolutely continuous random variable, we have that $X$ is absolutely continuous {{iff}}:
:$P_X$ is absolutely continuous with respect to $\lambda$.
That is:
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.
Since $X$ is singular:
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map \Pr {X \in B} = 1$.
That is, from the definition of probability distribution:
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map {P_X} B = 1$.
So:
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
does not hold and so $X$ is not absolutely continuous.
{{qed}}
Category:Singular Random Variables
\end{proof}
|
21465
|
\section{Six Sevenths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 6 7$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
21466
|
\section{Size of Complete Graph}
Tags: Proofs by Induction, Complete Graphs
\begin{theorem}
Let $K_n$ denote the complete graph of order $n$ where $n \ge 0$.
The size of $K_n$ is given by:
:$\size {K_n} = \dfrac {n \paren {n - 1} } 2$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\size {K_n} = \dfrac {n \paren {n - 1} } 2$
First we explore the degenerate case $\map P 0$:
{{begin-eqn}}
{{eqn | l = \size {K_0}
| r = 0
| c = as $K_0$ is the null graph
}}
{{eqn | r = \dfrac {0 \paren {0 - 1} } 2
| c =
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
21467
|
\section{Size of Conjugacy Class is Index of Normalizer}
Tags: Conjugacy Classes, Conjugacy, Normalizers
\begin{theorem}
Let $G$ be a group.
Let $x \in G$.
Let $\conjclass x$ be the conjugacy class of $x$ in $G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ be the index of $\map {N_G} x$ in $G$.
The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$.
\end{theorem}
\begin{proof}
The number of elements in $\conjclass x$ is the number of conjugates of the set $\set x$.
From Number of Distinct Conjugate Subsets is Index of Normalizer, the number of distinct subsets of a $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$.
The result follows.
{{qed}}
\end{proof}
|
21468
|
\section{Size of Cycle Graph equals Order}
Tags: Graph Theory
\begin{theorem}
The size of a cycle graph equals its order.
\end{theorem}
\begin{proof}
Follows immediately from:
:The fact that a cycle graph is $2$-regular
:The result Number of Edges of Regular Graph which says that an $r$-regular graph with $n$ vertices has $\dfrac {n r} 2$ edges.
{{qed}}
Category:Graph Theory
\end{proof}
|
21469
|
\section{Size of Regular Graph in terms of Degree and Order}
Tags: Regular Graphs
\begin{theorem}
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.
\end{theorem}
\begin{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 the corollary to the Handshake Lemma, this is not possible
Thus for $G$ to exist, it is necessary for $p r$ to be even.
The result follows.
{{Qed}}
\end{proof}
|
21470
|
\section{Size of Star Graph}
Tags: Star Graphs
\begin{theorem}
Let $S_n$ denote the star graph of order $n$ where $n > 0$.
The size of $S_n$ is given by:
:$\size {S_n} = n - 1$
\end{theorem}
\begin{proof}
By definition of star graph, one vertex is adjacent to all the $n - 1$ other vertices.
So the size of $S_n$ is at least $n - 1$.
All the other vertices are of degree $1$, and the edges they are incident with are the ones joining them to the distinguished vertex.
Those edges have already been counted.
So the size of $S_n$ is exactly $n - 1$.
{{qed}}
Category:Star Graphs
Category:Star Graphs
\end{proof}
|
21471
|
\section{Size of Tree is One Less than Order}
Tags: Size of Tree is One Less than Order, Tree Theory, Graph Theory, Trees
\begin{theorem}
Let $T$ be a connected simple graph of order $n$.
Then $T$ is a tree {{Iff}} the size of $T$ is $n-1$.
\end{theorem}
\begin{proof}
By definition:
:the order of a tree is how many nodes it has
and:
:its size is how many edges it has.
\end{proof}
|
21472
|
\section{Size of Tree is One Less than Order/Necessary Condition}
Tags: Size of Tree is One Less than Order, Tree Theory, Trees
\begin{theorem}
Let $T$ be a tree of order $n$.
Then the size of $T$ is $n-1$.
\end{theorem}
\begin{proof}
By definition, the order of a tree is how many nodes it has, and its size is how many edges it has.
Suppose $T$ is a tree with $n$ nodes. We need to show that $T$ has $n - 1$ edges.
{{questionable|This is actually a Proof by Complete Induction}}
Proof by induction:
Let $T_n$ be a tree with $n$ nodes.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition that a tree with $n$ nodes has $n-1$ edges.
\end{proof}
|
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