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22073
|
\section{Substitution in Big-O Estimate/Real Analysis}
Tags: Asymptotic Notation
\begin{theorem}
Let $f$ and $g$ be real-valued or complex-valued functions defined on a neighborhood of $+ \infty$ in $\R$.
Let $f = \map \OO g$, where $\OO$ denotes big-O notation.
Let $h$ be a real-valued defined on a neighborhood of $+ \infty$ in $\R$.
Let $\ds \lim_{x \mathop \to +\infty} \map h x = +\infty$.
Then:
:$f \circ h = \map \OO {g \circ h}$ as $x \to +\infty$.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Asymptotic Notation
\end{proof}
|
22074
|
\section{Substitution in Big-O Estimate/Sequences}
Tags: Asymptotic Notation
\begin{theorem}
Let $(a_n)$ and $(b_n)$ be sequences of real or complex numbers.
Let $a_n = O(b_n)$ where $O$ denotes big-O notation.
Let $(n_k)$ be a diverging sequence of natural numbers.
Then $a_{n_k} = O(b_{n_k})$.
\end{theorem}
\begin{proof}
Because $a_n = O(b_n)$, there exists $M\geq0$ and $n_0 \in\N$ such that $|a_n| \leq M \cdot |b_n|$ for $n\geq n_0$.
Because $n_k$ diverges, there exists $k_0\in\N$ such that $n_k\geq n_0$ for $k\geq k_0$.
Then $|a_{n_k}| \leq M\cdot |b_{n_k}|$ for $k\geq k_0$.
Thus $a_{n_k} = O(b_{n_k})$.
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22075
|
\section{Substitution of Constant yields Primitive Recursive Function}
Tags: Primitive Recursive Functions
\begin{theorem}
Let $f: \N^{k + 1} \to \N$ be a primitive recursive function.
Then $g: \N^k \to \N$ given by:
:$\map g {n_1, n_2, \ldots, n_k} = \map f {n_1, n_2, \ldots, n_{i - 1}, a, n_i \ldots, n_k}$
is primitive recursive.
\end{theorem}
\begin{proof}
Let $n = \tuple {n_1, n_2, \ldots, n_{i - 1}, n_i \ldots, n_k}$.
We see that:
:$\map g {n_1, n_2, \ldots, n_k} = \map f {\map {\pr^k_1} n, \map {\pr^k_2} n, \ldots, \map {\pr^k_{i - 1} } n, \map {f_a} n, \map {\pr^k_i} n, \ldots, \map {\pr^k_k} n}$
We have that:
* $\pr^k_j$ is a basic primitive recursive function for all $j$ such that $1 \ne j \le k$
* $f_a$ is a primitive recursive function.
So $g$ is obtained by substitution from primitive recursive functions and so is primitive recursive.
{{qed}}
Category:Primitive Recursive Functions
\end{proof}
|
22076
|
\section{Substitution of Elements}
Tags: Set Theory
\begin{theorem}
Let $a$, $b$, and $x$ be sets.
:$a = b \implies \left({a \in x \iff b \in x}\right)$
\end{theorem}
\begin{proof}
By the Axiom of Extension:
: $a = b \implies \left({a \in x \implies b \in x}\right)$
Equality is Symmetric, so also by the Axiom of Extension:
: $a = b \implies \left({b \in x \implies a \in x}\right)$
{{qed}}
\end{proof}
|
22077
|
\section{Substitutivity of Class Equality}
Tags: Zermelo-Fraenkel Class Theory, Class Theory
\begin{theorem}
Let $A$ and $B$ be classes.
Let $\map P A$ be a well-formed formula of the language of set theory.
Let $\map P B$ be the same proposition $\map P A$ with all instances of $A$ replaced with instances of $B$.
Let $=$ denote class equality.
:$A = B \implies \paren {\map P A \iff \map P B}$
\end{theorem}
\begin{proof}
{{NotZFC}}
By induction on the well-formed parts of $\map P A$.
The proof shall use $\implies$ and $\neg$ as the primitive connectives.
\end{proof}
|
22078
|
\section{Substitutivity of Equality}
Tags: Set Theory, Equality
\begin{theorem}
Let $x$ and $y$ be sets.
Let $\map P x$ be a well-formed formula of the language of set theory.
Let $\map P y$ be the same proposition $\map P x$ with some (not necessarily all) free instances of $x$ replaced with free instances of $y$.
Let $=$ denote set equality.
:$x = y \implies \paren {\map P x \iff \map P y}$
\end{theorem}
\begin{proof}
By induction on the well-formed parts of $\map P x$.
The proof shall use $\implies$ and $\neg$ as the primitive connectives.
\end{proof}
|
22079
|
\section{Substructure of Entropic Structure is Entropic}
Tags: Entropic Structures
\begin{theorem}
Let $\struct {S, \odot}$ be an entropic structure:
:$\forall a, b, c, d \in S: \paren {a \odot b} \odot \paren {c \odot d} = \paren {a \odot c} \odot \paren {b \odot d}$
Let $\struct {T, \odot_T}$ be a substructure of $\struct {S, \odot}$.
Then $\struct {T, \odot_T}$ is also an entropic structure.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | q = \forall a, b, c, d \in T
| o =
| r = \paren {a \odot_T b} \odot_T \paren {c \odot_T d}
| c =
}}
{{eqn | r = \paren {a \odot b} \odot \paren {c \odot d}
| c = {{Defof|Operation Induced by Restriction}}
}}
{{eqn | r = \paren {a \odot c} \odot \paren {b \odot d}
| c = as $S$ is an entropic structure
}}
{{eqn | r = \paren {a \odot_T c} \odot_T \paren {b \odot_T d}
| c = {{Defof|Operation Induced by Restriction}}
}}
{{end-eqn}}
{{qed}}
Category:Entropic Structures
\end{proof}
|
22080
|
\section{Subtract Half is Replicative Function}
Tags: Replicative Functions
\begin{theorem}
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x - \dfrac 1 2$
Then $f$ is a replicative function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {x - \frac 1 2 + \frac k n}
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \frac {n \paren {n - 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = n x - \frac n 2 + \frac n 2 - \frac 1 2
| c =
}}
{{eqn | r = n x - \frac 1 2
| c =
}}
{{eqn | r = \map f {n x}
| c =
}}
{{end-eqn}}
Hence the result by definition of replicative function.
{{qed}}
\end{proof}
|
22081
|
\section{Subtraction has no Identity Element}
Tags: Numbers, Examples of Identity Elements, Identity Elements, Subtraction
\begin{theorem}
The operation of subtraction on numbers of any kind has no identity.
\end{theorem}
\begin{proof}
{{AimForCont}} there exists an identity $e$ in one of the standard number systems $\F$.
{{begin-eqn}}
{{eqn | q = \forall x \in \F:
| l = x
| r = x - e
| c =
}}
{{eqn | r = e - x
| c =
}}
{{eqn | ll= \leadsto
| l = x + \paren {-e}
| r = e + \paren {-x}
| c =
}}
{{eqn | ll= \leadsto
| l = x + x
| r = e + e
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = e
| c =
}}
{{end-eqn}}
That is:
:$\forall x \in \F: x = e$
But from Identity is Unique, if $e$ is an identity then there can be only one such.
From Proof by Contradiction it follows that $\F$ has no such $e$.
{{qed}}
\end{proof}
|
22082
|
\section{Subtraction of Divisors obeys Distributive Law}
Tags: Algebra, Subtraction of Divisors obeys Distributive Law
\begin{theorem}
{{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$
\end{theorem}
\begin{proof}
Let $AB$ be that part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted.
We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted.
:500px
Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$.
Then from {{EuclidPropLink|book=VII|prop=5|title=Divisors obey Distributive Law}}, whatever part $AE$ is of $CF$, the same part also is $AB$ of $GF$.
But whatever part $AE$ is of $CF$, the same part also is $AB$ of $CD$, by hypothesis.
Therefore, whatever part $AB$ is of $GF$, the same pat is it of $CD$ also.
Therefore $GF = CD$.
Let $CF$ be subtracted from each.
Therefore $GC = FD$.
We have that whatever part $AE$ is of $CF$, the same part also is $EB$ of $CG$
Therefore whatever part $AE$ is of $CF$, the same part also is $EB$ of $FD$.
But whatever part $AE$ is of $CF$, the same part also is $AB$ of $CD$.
Therefore the remainder $EB$ is the same part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
{{qed}}
{{Euclid Note|7|VII}}
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\end{proof}
|
22083
|
\section{Subtraction of Multiples of Divisors obeys Distributive Law}
Tags: Divisors, Algebra, Subtraction of Multiples of Divisors obeys Distributive Law, Subtraction
\begin{theorem}
{{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$
\end{theorem}
\begin{proof}
Let the (natural) number $AB$ be the same parts of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted.
We need to show that $EB$ is also the same parts of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:400px
Let $GH = AB$.
Then whatever parts $GH$ is of $CD$, the same parts also is $AE$ of $CF$.
Let $GH$ be divided into the parts of $CD$, namely $GK + KH$, and $AE$ into the parts of $CF$, namely $AL + LE$.
Thus the multitude of $GK, KH$ will be equal to the multitude of $AL, LE$.
We have that whatever part $GK$ is of $CD$, the same part also is $AL$ of $CF$.
We also have that $CD > CF$.
Therefore $GK > AL$.
Now let $GM = AL$.
Then whatever part $GK$ is of $CD$, the same part also is $GM$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $MK$ is of the same part of the remainder $FD$ that the whole $GK$ is of the whole $CD$.
Again, we have that whatever part $KH$ is of $CD$, the same part also is $EL$ of $CF$.
We also have that $CD > CF$.
Therefore $HK > EL$.
Let $KN = EL$.
Then whatever part $KH$ is of $CD$, the same part also is $KN$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $NH$ is of the same part of the remainder $FD$ that the whole $KH$ is of the whole $CD$.
But $MK$ was proved to be the same part of $FD$ that $GK$ is of $CD$.
Therefore $MK + NH$ is the same parts of $DF$ that $HG$ is of $CD$.
But $MK + NH = EB$ and $HG = BA$.
Therefore $EB$ is the same parts of $FD$ that $AB$ is of $CD$.
{{Qed}}
{{Euclid Note|8|VII}}
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\end{proof}
|
22084
|
\section{Subtraction of Subring is Subtraction of Ring}
Tags: Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be an ring.
For each $x, y \in R$ let $x - y$ denote the subtraction of $x$ and $y$ in $R$.
Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.
For each $x, y \in S$ let $x \sim y$ denote the subtraction of $x$ and $y$ in $S$.
Then:
:$\forall x, y \in S: x \sim y = x - y$
\end{theorem}
\begin{proof}
Let $x, y \in S$.
Let $-x$ denote the ring negative of $x$ in $R$.
Let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.
Then:
{{begin-eqn}}
{{eqn | l = x \sim y
| r = x \mathbin {+ {\restriction_S} } \paren {\mathbin \sim y}
| c = {{Defof|Ring Subtraction}}
}}
{{eqn | r = x + \paren {\mathbin \sim y}
| c = {{Defof|Subring|Addition on Subring}}
}}
{{eqn | r = x + \paren {-y}
| c = Negative of Subring is Negative of Ring
}}
{{eqn | r = x - y
| c = {{Defof|Ring Subtraction}}
}}
{{end-eqn}}
{{qed}}
Category:Ring Theory
\end{proof}
|
22085
|
\section{Subtraction on Integers is Extension of Natural Numbers}
Tags: Integers, Subtraction
\begin{theorem}
Integer subtraction is an extension of the definition of subtraction on the natural numbers.
\end{theorem}
\begin{proof}
Let $m, n \in \N: m \le n$.
From natural number subtraction, $\exists p \in \N: m + p = n$ such that $n - m = p$.
As $m, n, p \in \N$, it follows that $m, n, p \in \Z$ as well.
However, as $\Z$ is the inverse completion of $\N$, it follows that $-m \in \Z$ as well, so it makes sense to express the following:
{{begin-eqn}}
{{eqn | l = \paren {n + \paren {-m} } + m
| r = n + \paren {\paren {-m} + m}
| c =
}}
{{eqn | r = n
| c =
}}
{{eqn | r = p + m
| c =
}}
{{eqn | r = \paren {n - m} + m
| c =
}}
{{end-eqn}}
Thus, as all elements of $\Z$ are cancellable, it follows that $n + \paren {-m} = n - m$.
So:
: $\forall m, n \in \Z, m \le n: n + \paren {-m} = n - m = n -_\N m$
and the result follows.
{{qed}}
Category:Integers
Category:Subtraction
\end{proof}
|
22086
|
\section{Subtraction on Numbers is Anticommutative/Integral Domains}
Tags: Numbers, Subtraction on Numbers is Anticommutative, Subtraction
\begin{theorem}
The operation of subtraction on the numbers is anticommutative.
That is:
:$a - b = b - a \iff a = b$
\end{theorem}
\begin{proof}
Let $a, b$ be elements of one of the standard number sets: $\Z, \Q, \R, \C$.
Each of those systems is an integral domain, and so is closed under the operation of subtraction.
\end{proof}
|
22087
|
\section{Subtraction on Numbers is Anticommutative/Natural Numbers}
Tags: Subtraction on Numbers is Anticommutative, Natural Numbers, Subtraction
\begin{theorem}
The operation of subtraction on the natural numbers $\N$ is anticommutative, and defined only when $a = b$:
That is:
:$a - b = b - a \iff a = b$
\end{theorem}
\begin{proof}
$a - b$ is defined on $\N$ only if $a \ge b$.
If $a > b$, then although $a - b$ is defined, $b - a$ is not.
So for $a - b = b - a$ it is necessary for both to be defined.
This happens only when $a = b$.
Hence the result.
Category:Subtraction on Numbers is Anticommutative
\end{proof}
|
22088
|
\section{Subtraction on Numbers is Not Associative}
Tags: Numbers, Examples of Associative Operations, Associativity, Subtraction
\begin{theorem}
The operation of subtraction on the numbers is not associative.
That is, in general:
:$a - \paren {b - c} \ne \paren {a - b} - c$
\end{theorem}
\begin{proof}
By definition of subtraction:
{{begin-eqn}}
{{eqn | l = a - \paren {b - c}
| r = a + \paren {-\paren {b + \paren {-c} } }
| c =
}}
{{eqn | r = a + \paren {-b} + c
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a - b} - c
| r = \paren {a + \paren {-b} } + \paren {-c}
| c =
}}
{{eqn | r = a + \paren {-b} + \paren {-c}
| c =
}}
{{end-eqn}}
So we see that:
:$a - \paren {b - c} = \paren {a - b} - c \iff c = 0$
and so in general:
:$a - \paren {b - c} \ne \paren {a - b} - c$
{{qed}}
\end{proof}
|
22089
|
\section{Succeed is Dual to Precede}
Tags: Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$a$ succeeds $b$
:$a$ precedes $b$
\end{theorem}
\begin{proof}
By definition, $a$ succeeds $b$ {{iff}}:
:$b \preceq a$
The dual of this statement is:
:$a \preceq b$
by Dual Pairs (Order Theory).
By definition, this means $a$ precedes $b$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}}
\end{proof}
|
22090
|
\section{Successive Solutions of Phi of n equals Phi of n + 2}
Tags: Euler Phi Function
\begin{theorem}
Let $\phi$ denote the Euler $\phi$ function.
$7$ and $8$ are two successive integers which are solutions to the equation:
:$\map \phi n = \map \phi {n + 2}$
\end{theorem}
\begin{proof}
From Euler Phi Function of Prime:
:$\map \phi 7 = 7 - 1 = 6$
From Euler Phi Function of Prime Power:
:$\map \phi 9 = \map \phi {3^2} = 2 \times 3^{2 - 1} = 6 = \map \phi 7$
From the corollary to Euler Phi Function of Prime Power:
:$\map \phi 8 = \map \phi {2^3} = 2^{3 - 1} = 4$
From Euler Phi Function of Integer:
:$\map \phi {10} = \map \phi {2 \times 5} = 10 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 5} = 10 \times \dfrac 1 2 \times \dfrac 4 5 = 4 = \map \phi 8$
{{qed}}
\end{proof}
|
22091
|
\section{Successor Mapping is Inflationary}
Tags: Inflationary Mappings, Successor Mapping
\begin{theorem}
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
Let $s: \omega \to \omega$ denote the successor mapping on $\omega$.
Then $s$ is an inflationary mapping.
\end{theorem}
\begin{proof}
By definition of the von Neumann construction:
:$n^+ = n \cup \set n$
from which it follows that:
:$n \subseteq n^+$
{{qed}}
\end{proof}
|
22092
|
\section{Successor Mapping of Peano Structure has no Fixed Point}
Tags: Peano's Axioms
\begin{theorem}
Let $\PP = \struct {P, s, 0}$ be a Peano structure.
Then:
:$\forall n \in P: \map s n \ne n$
That is, the successor mapping has no fixed points.
\end{theorem}
\begin{proof}
Let $T$ be the set:
:$T = \set {n \in P: \map s n \ne n}$
We will use {{PeanoAxiom|5}} to prove that $T = P$.
\end{proof}
|
22093
|
\section{Successor Mapping on Natural Numbers has no Fixed Element}
Tags: Natural Numbers
\begin{theorem}
Let $\N$ denote the set of natural numbers.
Then:
:$\forall n \in \N: n + 1 \ne n$
\end{theorem}
\begin{proof}
Consider the set of natural numbers as defined by the von Neumann construction.
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\N$ is a minimally inductive class under the successor mapping.
Let $s: \N \to \N$ denote the successor mapping:
:$\forall x \in \N: \map s x := x + 1$
{{AimForCont}} $\exists n \in \N: n = n + 1$
From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $n$ is the greatest element of $\N$.
From Minimally Inductive Class with Fixed Element is Finite it follows that $\N$ is a finite set.
This contradicts the fact that the natural numbers are by definition countably infinite.
{{qed}}
\end{proof}
|
22094
|
\section{Successor Mapping on Natural Numbers is not Surjection}
Tags: Surjections
\begin{theorem}
Let $f: \N \to \N$ be the successor mapping on the natural numbers $\N$:
:$\forall n \in \N: \map f n = n + 1$
Then $f$ is not a surjection.
\end{theorem}
\begin{proof}
There exists no $n \in \N$ such that $n + 1 = 0$.
Thus $\map f 0$ has no preimage.
The result follows by definition of surjection.
{{qed}}
\end{proof}
|
22095
|
\section{Successor Set of Ordinal is Ordinal}
Tags: Ordinals
\begin{theorem}
Let $S$ be an ordinal.
Then its successor set $S^+ = S \cup \set S$ is also an ordinal.
\end{theorem}
\begin{proof}
Since $S$ is transitive, it follows by Successor Set of Transitive Set is Transitive that $S^+$ is transitive.
We now have to show that $S^+$ is strictly well-ordered by the epsilon restriction $\Epsilon \! \restriction_{S^+}$.
So suppose that a subset $A \subseteq S^+$ is non-empty.
Then:
{{begin-eqn}}
{{eqn | l = A
| r = A \cap \paren {S \cup \set S}
| c = Intersection with Subset is Subset and {{Defof|Successor Set}}
}}
{{eqn | r = \paren {A \cap S} \cup \paren {A \cap \set S}
| c = Intersection Distributes over Union
}}
{{end-eqn}}
Let us refer to the above equation by the symbol $\paren 1$.
We need to show that $A$ has a smallest element.
We first consider the case where $A \cap S$ is empty.
By equation $\paren 1$, it follows that $A \cap \set S$ is non-empty (because $A$ is non-empty).
Therefore, $S \in A$. That is, $\set S \subseteq A$.
By Union with Empty Set and Intersection with Subset is Subset, equation $\paren 1$ implies that $A \subseteq \set S$.
Therefore, $A = \set S$ by the definition of set equality.
So $S$ is the smallest element of $A$.
We now consider the case where $A \cap S$ is non-empty.
By Intersection is Subset, $A \cap S \subseteq S$; by the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap S$.
Let $y \in A$.
If $y \in S$, then $y \in A \cap S$; therefore, by the definition of the smallest element, either $x \in y$ or $x = y$.
Otherwise, $y = S$, and so $x \in S = y$.
That is, $x$ is the smallest element of $A$.
{{qed}}
\end{proof}
|
22096
|
\section{Successor Set of Transitive Set is Transitive}
Tags: Set Theory
\begin{theorem}
Let $S$ be a transitive set.
Then its successor set $S\,^+ = S \cup \left\{{S}\right\}$ is also transitive.
\end{theorem}
\begin{proof}
Suppose that $x \in S\,^+$.
Then either $x \in S$ or $x = S$.
If $x \in S$, it follows by the transitivity of $S$ that $x \subseteq S$.
If $x = S$, then $x = S \subseteq S$ because a set is a subset of itself.
Since $S \subseteq S\,^+$, it follows by the transitivity of the subset relation that $x \subseteq S\,^+$.
{{qed}}
Category:Set Theory
\end{proof}
|
22097
|
\section{Successor Sets of Linearly Ordered Set Induced by Convex Component Partition}
Tags: Linearly Ordered Spaces
\begin{theorem}
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
:$M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Then each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$, the set of strict upper bounds for $A_\alpha$.
Similarly for $B_\beta$.
That immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
\end{theorem}
\begin{proof}
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$.
This belongs to the complement in $S$ of the closed set $\paren {B^*}^-$.
Hence there exists a neighborhood $\openint x y$ of $p$ which is disjoint from $\paren {B^*}^-$.
Then:
:$\openint x y \cap S_\alpha \ne \O$
and so:
:$\openint p y \ne \O$
But $\openint p y$ is disjoint from both $A^*$ and $B^*$.
Thus there must exist some $C_\gamma$ which contains $\openint p y$.
{{qed}}
\end{proof}
|
22098
|
\section{Successor in Limit Ordinal}
Tags: Ordinals
\begin{theorem}
Let $x$ be a limit ordinal.
Let $y \in x$.
Then $y^+ \in x$ where $y^+$ denotes the successor set of $y$:
:$\forall y \in x: y^+ \in x$
\end{theorem}
\begin{proof}
Because $x$ is a limit ordinal:
:$x \ne y^+$
Moreover, by Successor of Element of Ordinal is Subset:
:$y \in x \implies y^+ \subseteq x$
Therefore by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$y^+ \subset x$ and $y^+ \in x$
{{qed}}
Category:Ordinals
\end{proof}
|
22099
|
\section{Successor is Less than Successor}
Tags: Ordinals
\begin{theorem}
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Then, $x \in y \iff x^+ \in y^+$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x \in y
| o = \implies
| r = x^+ \in y^+
| c = Subset is Compatible with Ordinal Successor
}}
{{eqn | l = x \in y
| o = \impliedby
| r = x^+ \in y^+
| c = Sufficient Condition
}}
{{eqn | ll= \leadsto
| l = x \in y
| o = \iff
| r = x^+ \in y^+
| c =
}}
{{end-eqn}}
{{qed}}
Category:Ordinals
\end{proof}
|
22100
|
\section{Successor is Less than Successor/Sufficient Condition/Proof 1}
Tags: Ordinals
\begin{theorem}
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$
\end{theorem}
\begin{proof}
Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$.
{{qed}}
Category:Ordinals
\end{proof}
|
22101
|
\section{Successor is Less than Successor/Sufficient Condition/Proof 2}
Tags: Ordinals
\begin{theorem}
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$
\end{theorem}
\begin{proof}
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$.
Category:Ordinals
\end{proof}
|
22102
|
\section{Successor of Element of Ordinal is Subset}
Tags: Ordinals
\begin{theorem}
Let $x$ and $y$ be ordinals.
Then:
:$x \in y \iff x^+ \subseteq y$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y^+
| c = Successor is Less than Successor
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y
| c = {{Defof|Successor Set}}
}}
{{eqn | lo= \lor
| l = x^+
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \subsetneq
| r = y
| c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal
}}
{{eqn | lo = \lor
| l = x^+
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \subseteq
| r = y
| c =
}}
{{end-eqn}}
{{qed}}
Category:Ordinals
\end{proof}
|
22103
|
\section{Successor to Natural Number}
Tags: Natural Numbers: 1-Based
\begin{theorem}
Let $\N_{> 0}$ be the 1-based natural numbers:
:$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$
Let $<$ be the ordering on $\N_{> 0}$:
:$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$
Let $a \in \N_{>0}$.
Then there exists no natural number $n$ such that $a < n < a + 1$.
\end{theorem}
\begin{proof}
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.
Then by the definition of ordering on natural numbers:
{{begin-eqn}}
{{eqn | l = a + x
| r = n
| c = Definition of Ordering on Natural Numbers: $a < n$
}}
{{eqn | l = n + y
| r = a + 1
| c = Definition of Ordering on Natural Numbers: $a < n$
}}
{{eqn | ll= \implies
| l = \left({a + x}\right) + y
| r = a + 1
| c = substitution for $n$
}}
{{eqn | ll= \implies
| l = a + \left({x + y}\right)
| r = a + 1
| c = Natural Number Addition is Associative
}}
{{eqn | ll= \implies
| l = x + y
| r = 1
| c = Addition on $1$-Based Natural Numbers is Cancellable
}}
{{end-eqn}}
By Axiom $D$, either:
:$y = 1$
or:
:$y = t + 1$ for some $t \in \N_{>0}$
Then either:
:$x + 1 = 1$ when $y = 1$
or:
:$x + \left({t + 1}\right) = \left({x + t}\right) + 1 = 1$ when $y = t + 1$
Both of these conclusions violate Natural Number is Not Equal to Successor.
Hence the result.
{{qed}}
\end{proof}
|
22104
|
\section{Sufficient Condition for 5 to divide n^2+1}
Tags: Modulo Arithmetic
\begin{theorem}
Let:
{{begin-eqn}}
{{eqn | l = 5
| o = \nmid
| r = n - 1
}}
{{eqn | l = 5
| o = \nmid
| r = n
}}
{{eqn | l = 5
| o = \nmid
| r = n + 1
}}
{{end-eqn}}
where $\nmid$ denotes non-divisibility.
Then:
:$5 \divides n^2 + 1$
where $\divides$ denotes divisibility.
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = 5
| o = \nmid
| r = n - 1
}}
{{eqn | ll= \leadsto
| l = n - 1
| o = \not \equiv
| r = 0
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n
| o = \not \equiv
| r = 1
| rr= \pmod 5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 5
| o = \nmid
| r = n
}}
{{eqn | ll= \leadsto
| l = n
| o = \not \equiv
| r = 0
| rr= \pmod 5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 5
| o = \nmid
| r = n + 1
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = \not \equiv
| r = 0
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n
| o = \not \equiv
| r = 4
| rr= \pmod 5
}}
{{end-eqn}}
So either:
:$n \equiv 2 \pmod 5$
or:
:$n \equiv 3 \pmod 5$
and so:
{{begin-eqn}}
{{eqn | l = n
| o = \equiv
| r = 2
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n^2
| o = \equiv
| r = 4
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n^2 + 1
| o = \equiv
| r = 0
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = 5
| o = \divides
| r = n^2 + 1
}}
{{end-eqn}}
or:
{{begin-eqn}}
{{eqn | l = n
| o = \equiv
| r = 3
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n^2
| o = \equiv
| r = 3^2
| rr= \pmod 5
}}
{{eqn | o = \equiv
| r = 4
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = n^2 + 1
| o = \equiv
| r = 0
| rr= \pmod 5
}}
{{eqn | ll= \leadsto
| l = 5
| o = \divides
| r = n^2 + 1
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22105
|
\section{Sufficient Condition for Quotient Group by Intersection to be Abelian}
Tags: Quotient Groups
\begin{theorem}
Let $G$ be a group.
Let $N$ and $K$ be normal subgroups of $G$.
Let the quotient groups $G / N$ and $G / K$ be abelian.
Then the quotient group $G / \paren {N \cap K}$ is also abelian.
\end{theorem}
\begin{proof}
From Intersection of Normal Subgroups is Normal, we have that $N \cap K$ is normal in $G$.
We are given that $G / N$ and $G / K$ are abelian.
Hence:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = \sqbrk {x, y}
| o = \in
| r = N
| c = Quotient Group is Abelian iff All Commutators in Divisor
}}
{{eqn | lo= \land
| l = \sqbrk {x, y}
| o = \in
| r = K
| c =
}}
{{eqn | ll= \leadsto
| q = \forall x, y \in G
| l = \sqbrk {x, y}
| o = \in
| r = N \cap K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| l = G / \paren {N \cap K}
| o =
| r = \text {is abelian}
| c = Quotient Group is Abelian iff All Commutators in Divisor
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22106
|
\section{Sufficient Condition for Square of Product to be Triangular}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $2 n^2 \pm 1 = m^2$ be a square number.
Then $\paren {m n}^2$ is a triangular number.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {m n}^2
| r = \paren {2 n^2 \pm 1} \times n^2
| c =
}}
{{eqn | r = \dfrac {\paren {2 n^2 \pm 1} \paren {2 n^2} } 2
| c =
}}
{{end-eqn}}
That is, either:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2 - 1} \paren {2 n^2} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2 - 1}$
or:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2} \paren {2 n^2 + 1} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2}$
Hence the result.
{{Qed}}
\end{proof}
|
22107
|
\section{Sufficient Condition for Twice Differentiable Functional to have Minimum}
Tags: Calculus of Variations, Definitions: Calculus of Variations
\begin{theorem}
Let $J$ be a twice differentiable functional.
Let $J$ have an extremum for $y=\hat y$.
Let the second variation $\delta^2 J \sqbrk {\hat y; h}$ be strongly positive {{WRT}} $h$.
Then $J$ acquires the minimum for $y = \hat y$ .
\end{theorem}
\begin{proof}
By assumption, $J$ has an extremum for $y = \hat y$:
:$\delta J \sqbrk {\hat y; h} = 0$
The increment is expressible then as:
:$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
where $\epsilon \to 0$ as $\size h \to 0$.
By assumption, the second variation is strongly positive:
:$\delta^2 J \sqbrk {\hat y; h} \ge k \size h^2, \quad k \in \R_{>0}$
Hence:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2$
What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that {{RHS}} is always positive.
Since $\epsilon \to 0$ as $\size h \to 0$, there exist $c \in \R_{>0}$, such that:
:$\size h < c \implies \size \epsilon < \dfrac 1 2 k$
Choose $h$ such that this inequality holds.
Then
{{begin-eqn}}
{{eqn | l = \frac 1 2 k
| o = >
| r = \epsilon > -\frac 1 2 k
| c = $\big \vert + k$, by Membership is Left Compatible with Ordinal Addition
}}
{{eqn | l = \frac 3 2 k
| o = >
| r = k + \epsilon > \frac 1 2 k
| c = $\big \vert \cdot \size h^2$, by Membership is Left Compatible with Ordinal Multiplication
}}
{{eqn | l = \frac 3 2 k \size h^2
| o = >
| r = \paren {k + \epsilon} \size h^2 > \frac 1 2 k \size h^2
}}
{{end-eqn}}
{{Explain|What does this mean? : $\big \vert + k$ and $\big \vert \cdot \size h^2$}}
Therefore:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2 > \dfrac 1 2 k \size h^2 $
For $k \in \R_{>0}$ and $\size h \ne 0$ {{RHS}} is always positive.
Thus, there exists a neighbourhood around $y = \hat y$ where the increment is always positive:
:$\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$
and $J$ has a minimum for $y = \hat y$.
{{qed}}
\end{proof}
|
22108
|
\section{Sufficient Conditions for Uncountability}
Tags: Uncountable Sets, Infinite Sets, Set Theory
\begin{theorem}
Let $X$ be a set.
The following are equivalent:
:$(1): \quad X$ contains an uncountable subset
:$(2): \quad X$ is uncountable
:$(3): \quad $ Every sequence of distinct points $\sequence {x_n}_{n \mathop \in \N}$ in $X$ omits at least one $x \in X$
:$(4): \quad $ There is no surjection $\N \twoheadrightarrow X$
:$(5): \quad X$ is infinite and there is no bijection $X \leftrightarrow \N$
Assuming the Continuum Hypothesis holds, we also have the equivalent uncountability condition:
:$(6): \quad $There exist extended real numbers $a < b$ and a surjection $X \to \closedint a b$
\end{theorem}
\begin{proof}
Recall that $X$ is uncountable if there is no injection $X \hookrightarrow \N$.
\end{proof}
|
22109
|
\section{Sufficient Conditions for Weak Extremum}
Tags: Calculus of Variations
\begin{theorem}
Let $J$ be a functional such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
:$\map y a = A$
:$\map y b = B$
Let $y = \map y x$ be an extremum.
Let the strengthened Legendre's Condition hold.
Let the strengthened Jacobi's Necessary Condition hold.
{{explain|specific links to those strengthened versions}}
Then the functional $J$ has a weak minimum for $y = \map y x$.
\end{theorem}
\begin{proof}
By the continuity of function $\map P x$ and the solution of Jacobi's equation:
:$\exists \epsilon > 0: \paren {\forall x \in \closedint a {b + \epsilon}:\map P x > 0} \land \paren {\tilde a \notin \closedint a {b + \epsilon} }$
Consider the quadratic functional:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x - \alpha^2 \int_a^b h'^2 \rd x$
together with Euler's equation:
:$-\dfrac \rd {\rd x} \paren{\paren {P - \alpha^2} h'} + Q h = 0$
The Euler's equation is continuous {{WRT}} $\alpha$.
Thus the solution of the Euler's equation is continuous {{WRT}} $\alpha $.
{{ProofWanted|solution to continuous differential equation is continuous}}
Since:
:$\forall x \in \closedint a {b + \epsilon}: \map P x > 0$
$\map P x$ has a positive lower bound in $\closedint a {b + \epsilon}$.
Consider the solution with $\map h a = 0$, $\map {h'} 0 = 1$.
Then
:$\exists \alpha \in \R: \forall x \in \closedint a b: \map P x - \alpha^2 > 0$
Also:
:$\forall x \in \hointl a b: \map h x \ne 0$
{{Stub|seems to be Jacobi's condition where $P \to P - \alpha^2$}}
By Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:
:$\ds \int_a^b \paren {\paren {P - \alpha^2} h'^2 + Q h^2} \rd x > 0$
In other words, if $c = \alpha^2$, then:
:$(1): \quad \exists c > 0: \ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > c \int_a^b h'^2 \rd x$
Let $y = \map y x$ be an extremal.
Let $y = \map y x + \map h x$ be a curve, sufficiently close to $y = \map y x$.
By expansion of $\Delta J \sqbrk {y; h}$ from lemma $1$ of Legendre's Condition:
:$\ds J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$
where:
:$\ds \forall x \in \closedint a b: \lim_{\size h_1 \mathop \to 0} \set {\xi,\eta} = \set {0, 0}$
{{explain|$\size h_1$}}
and the limit is uniform.
{{Stub|why?}}
By Schwarz inequality:
{{begin-eqn}}
{{eqn | l = \map {h^2} x
| r = \paren {\int_a^x \map {h'} x \rd x}^2
}}
{{eqn | r = \int_a^x 1^2 d y \int_a^x \map {h'^2} x \rd x
}}
{{eqn | o = \le
| r = \paren {x - a} \int_a^x \map {h'^2} x \rd x
}}
{{eqn | o = \le
| r = \paren {x - a} \int_a^b \map {h'^2} x \rd x
| c = $h'^2 \ge 0$
}}
{{end-eqn}}
Notice that the integral on the right does not depend on $x$.
Integrate the inequality {{WRT|Integration}} $x$:
{{begin-eqn}}
{{eqn | l = \int_a^b \map{h^2} x \rd x
| o = \le
| r = \int_a^b \paren {x - a} \rd x \int_a^b \map {h'^2} x \rd x
}}
{{eqn | r = \frac {\paren {b - a}^2} 2 \int_a^b \map {h'^2} x \rd x
}}
{{end-eqn}}
Let $\epsilon \in \R_{>0}$ be a constant such that:
:$\size \xi \le \epsilon$, $\size \eta \le \epsilon$
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_a^b \paren {\xi h^2 + \eta h'^2} \rd x}
| o = \le
| r = \int_a^b \size \xi h^2 \rd x + \int_a^b \size \eta h'^2 \rd x
| c = Absolute Value of Definite Integral, Absolute Value of Product
}}
{{eqn | o = \le
| r = \epsilon \int_a^b h'^2 \rd x + \epsilon \frac {\paren {a - b}^2} 2 \int_a^b h'^2 \rd x
}}
{{eqn | n = 2
| r = \epsilon \paren {1 + \frac {\paren {b - a}^2} 2} \int_a^b h'^2 \rd x
}}
{{end-eqn}}
Thus, by $(1)$:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$
while by $(2)$:
:$\ds \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$
can be made arbitrarily small.
Thus, for all sufficiently small $\size h_1$, which implies sufficiently small $\size \xi$ and $\size \eta$, and, consequently, sufficiently small $\epsilon$:
:$J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x > 0$
Therefore, in some small neighbourhood $y = \map y x$ there exists a weak minimum of the functional.
{{qed}}
\end{proof}
|
22110
|
\section{Sufficient Conditions for Weak Stationarity of Order 2}
Tags: Stationary Stochastic Processes
\begin{theorem}
Let $S$ be a stochastic process giving rise to a time series $T$.
Let the mean of $S$ be fixed.
Let the autocovariance matrix of $S$ be of the form:
:$\boldsymbol \Gamma_n = \begin {pmatrix}
\gamma_0 & \gamma_1 & \gamma_2 & \cdots & \gamma_{n - 1} \\
\gamma_1 & \gamma_0 & \gamma_1 & \cdots & \gamma_{n - 2} \\
\gamma_2 & \gamma_1 & \gamma_0 & \cdots & \gamma_{n - 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\gamma_{n - 1} & \gamma_{n - 2} & \gamma_{n - 3} & \cdots & \gamma_0
\end {pmatrix} = \sigma_z^2 \mathbf P_n = \begin {pmatrix}
1 & \rho_1 & \rho_2 & \cdots & \rho_{n - 1} \\
\rho_1 & 1 & \rho_1 & \cdots & \rho_{n - 2} \\
\rho_2 & \rho_1 & 1 & \cdots & \rho_{n - 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\rho_{n - 1} & \rho_{n - 2} & \rho_{n - 3} & \cdots & 1
\end {pmatrix}$
Then $S$ is weakly stationary of order $2$.
\end{theorem}
\begin{proof}
Follows from the definition of weak stationarity.
\end{proof}
|
22111
|
\section{Sum Less Maximum is Minimum}
Tags: Algebra
\begin{theorem}
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \max \set {a, b} = \min \set {a, b}$.
\end{theorem}
\begin{proof}
From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \max \set {a, b} = \min \set {a, b}$ follows by subtracting $\max \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets.
However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
This is because the fact that $a + b \ge \max \set {a, b}$ follows immediately again from $a + b = \max \set {a, b} + \min \set {a, b}$.
{{qed}}
Category:Algebra
\end{proof}
|
22112
|
\section{Sum Less Minimum is Maximum}
Tags: Algebra
\begin{theorem}
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \min \set {a, b} = \max \set {a, b}$
\end{theorem}
\begin{proof}
From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \min \set {a, b} = \max \set {a, b}$ follows by subtracting $\min \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets.
However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
This is because the fact that $a + b \ge \min \set {a, b}$ follows immediately again from $a + b = \max \set {a, b} + \min \set {a, b}$.
{{qed}}
Category:Algebra
\end{proof}
|
22113
|
\section{Sum Over Divisors Equals Sum Over Quotients}
Tags: Number Theory, Divisors
\begin{theorem}
Let $n$ be a positive integer.
Let $f: \Z_{>0} \to \Z_{>0}$ be a mapping on the positive integers.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum of $\map f d$ over the divisors of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \map f d = \sum_{d \mathop \divides n} \map f {\frac n d}$.
\end{theorem}
\begin{proof}
If $d$ is a divisor of $n$ then $d \times \dfrac n d = n$ and so $\dfrac n d$ is also a divisor of $n$.
Therefore if $d_1, d_2, \ldots, d_r$ are all the divisors of $n$, then so are $\dfrac n {d_1}, \dfrac n {d_2}, \ldots, \dfrac n {d_r}$, except in a different order.
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{d \mathop \divides n} \map f {\frac n d}
| r = \map f {\frac n {d_1} } + \map f {\frac n {d_2} } + \cdots + \map f {\frac n {d_r} }
| c =
}}
{{eqn | r = \map f {d_1} + \map f {d_2} + \cdots + \map f {d_r}
| c =
}}
{{eqn | r = \sum_{d \mathop \divides n} \map f d
| c =
}}
{{end-eqn}}
{{qed}}
Category:Number Theory
Category:Divisors
\end{proof}
|
22114
|
\section{Sum Over Divisors of Multiplicative Function}
Tags: Multiplicative Functions, Number Theory
\begin{theorem}
Let $f: \Z_{>0} \to \Z_{>0}$ be a multiplicative function.
Let $n \in \Z_{>0}$.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum over the divisors of $n$.
Then $\ds \map F n = \sum_{d \mathop \divides n} \map f d$ is also a multiplicative function.
\end{theorem}
\begin{proof}
Let $\ds \map F n = \sum_{d \mathop \divides n} \map f d$.
Let $m, n \in \Z_{>0}: m \perp n$.
Then by definition:
:$\ds \map F {m n} = \sum_{d \mathop \divides m n} \map f d$
The divisors of $m n$ are of the form $d = r s$ where $r$ and $s$ are divisors of $m$ and $n$ respectively, from Divisors of Product of Coprime Integers.
It is noted that $r \perp s$, otherwise any common divisor of $r$ and $s$ would be a common divisor of $m$ and $n$.
Therefore:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f {r s}$
So, as $f$ is multiplicative:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f r \map f s$
But at the same time:
:$\ds \map F m \map F n = \paren {\sum_{r \mathop \divides m} \map f r} \paren {\sum_{s \mathop \divides n} \map f s}$
Multiplying out the product on the {{RHS}}, $\map F {m n}$ and $\map F m \map F n$ are seen to be the same.
{{qed}}
\end{proof}
|
22115
|
\section{Sum Over Divisors of von Mangoldt is Logarithm}
Tags: Von Mangoldt Function, Analytic Number Theory
\begin{theorem}
Let $\Lambda$ be von Mangoldt's function.
Then for $n \ge 1$:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$
\end{theorem}
\begin{proof}
Let $n \ge 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct primes and $e_1, \ldots, e_k > 0$.
Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.
By the definition of $\Lambda$, for such $d$ we have $\map \Lambda d \ne 0$ {{iff}} there is exactly one $i \in \set {1, \ldots, k}$ such that $f_i > 0$.
If this is the case, let $d = p_i^{f_i}$.
Then:
:$\map \Lambda d = \ln p_i$
Therefore:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \sum_{i \mathop = 1}^k e_i \ln p_i$
Also, we have:
{{begin-eqn}}
{{eqn | l = \ln n
| r = \ln (p_1^{e_1} \cdots p_k^{e_k})
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^k e_i \ln p_i
| c = Sum of Logarithms
}}
{{end-eqn}}
Thus we indeed have:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$
{{qed}}
Category:Analytic Number Theory
Category:Von Mangoldt Function
\end{proof}
|
22116
|
\section{Sum Rule for Complex Derivatives}
Tags: Complex Differential Calculus
\begin{theorem}
Let $\map f z, \map j z, \map k z$ be single-valued continuous complex functions in a domain $D \subseteq \C$, where $D$ is open.
Let $f$, $j$, and $k$ be complex-differentiable at all points in $D$.
Let $\map f z = \map j z + \map k z$.
Then:
:$\forall z \in D: \map {f'} z = \map {j'} z + \map {k'} z$
\end{theorem}
\begin{proof}
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {f'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {z_0 + h} + \map k {z_0 + h} } - \paren {\map j {z_0} +\map k {z_0} } } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {z_0 + h} + \map k {z_0 + h} - \map j {z_0} - \map k {z_0} } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {z_0 + h} - \map j {z_0} } + \paren {\map k {z_0 + h} - \map k {z_0} } } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \paren {\frac {\map j {z_0 + h} - \map j {z_0} } h + \frac {\map k {z_0 + h} - \map k {z_0} } h}
| c = Complex Multiplication Distributes over Addition
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {z_0 + h} - \map j {z_0} } h + \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c = Sum Rule for Limits of Complex Functions
}}
{{eqn | r = \map {j'} {z_0} + \map {k'} {z_0}
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | ll= \leadsto
| q = \forall z \in D
| l = \map {f'} z
| r = \map {j'} z + \map {k'} z
| c = {{Defof|Derivative of Complex Function}}
}}
{{end-eqn}}
{{qed}}
Category:Complex Differential Calculus
\end{proof}
|
22117
|
\section{Sum Rule for Counting}
Tags: combinatorics, Counting Arguments, counting arguments, Combinatorics
\begin{theorem}
Let there be:
:$r_1$ different objects in the set $S_1$
:$r_2$ different objects in the set $S_2$
:$\ldots$
:$r_m$ different objects in the set $S_m$.
Let $\ds \bigcap_{i \mathop = 1}^m S_i = \O$.
Then the number of ways to select an object from one of the $m$ sets is $\ds \sum_{i \mathop = 1}^m r_i$.
\end{theorem}
\begin{proof}
A direct application of Cardinality of Set Union.
{{qed}}
\end{proof}
|
22118
|
\section{Sum Rule for Derivatives}
Tags: Differential Calculus, calculu, Calculus, Sum Rule for Derivatives
\begin{theorem}
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x + \map k x$.
Then $f$ is differentiable at $\xi$ and:
:$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
:$\forall x \in I: \map {f'} x = \map {j'} x + \map {k'} x$
\end{theorem}
\begin{proof}
{{begin-equation}}
{{equation | l=<math>f^{\prime} \left({\xi}\right)</math>
| r=<math>\lim_{h \to 0} \frac {f \left({\xi+h}\right) - f \left({\xi}\right)} {h}</math>
| c=by the definition of the derivative
}}
{{equation | r=<math>\lim_{h \to 0} \frac {\left({j \left({\xi+h}\right) + k \left({\xi+h}\right)}\right) - \left({j \left({\xi}\right) + k \left({\xi}\right)}\right)} {h}</math>
| c=
}}
{{equation | r=<math>\lim_{h \to 0} \frac {j \left({\xi+h}\right) + k \left({\xi+h}\right) - j \left({\xi}\right) - k \left({\xi}\right)} {h}</math>
| c=
}}
{{equation | r=<math>\lim_{h \to 0} \frac {\left({j \left({\xi+h}\right) - j \left({\xi}\right)}\right) + \left({k \left({\xi+h}\right) - k \left({\xi}\right)}\right)} {h}</math>
| c=
}}
{{equation | r=<math>\lim_{h \to 0} \left({\frac {j \left({\xi+h}\right) - j \left({\xi}\right)}{h} + \frac {k \left({\xi+h}\right) - k \left({\xi}\right)} {h}}\right)</math>
| c=
}}
{{equation | r=<math>\lim_{h \to 0} \frac {j \left({\xi+h}\right) - j \left({\xi}\right)}{h} + \lim_{h \to 0} \frac {k \left({\xi+h}\right) - k \left({\xi}\right)} {h}</math>
| c=
}}
{{equation | r=<math>j^{\prime} \left({x}\right) + k^{\prime} \left({x}\right)</math>
| c=by the definition of the derivative
}}
{{end-equation}}
{{qed}}
Alternatively, it can be observed that this is an example of a Linear Combination of Derivatives with <math>\lambda = \mu = 1</math>.
\end{proof}
|
22119
|
\section{Sum Rule for Derivatives/General Result}
Tags: Differential Calculus, Sum Rule for Derivatives
\begin{theorem}
Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions all differentiable.
Then for all $n \in \N_{>0}$:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$
$\map P 1$ is true, as this just says:
:$\map {D_x} {\map {f_1} x} = \map {D_x} {\map {f_1} x}$
which is trivially true.
\end{proof}
|
22120
|
\section{Sum and Product of Discrete Random Variables}
Tags: Probability Theory
\begin{theorem}
Let $X$ and $Y$ be discrete random variables on the probability space $\left({\Omega, \Sigma, \Pr}\right)$.
\end{theorem}
\begin{proof}
To show that $U$ and $V$ are discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$, we need to show that:
:$(1)$ The image of $U$ and $V$ are countable subsets of $\R$;
:$(2)$ $\forall x \in \R: \left\{{\omega \in \Omega: U \left({\omega}\right) = x}\right\} \in \Sigma$ and $\left\{{\omega \in \Omega: V \left({\omega}\right) = x}\right\} \in \Sigma$.
\end{proof}
|
22121
|
\section{Sum from -m to m of 1 minus Cosine of n + alpha of theta over n + alpha}
Tags: Cosine Function
\begin{theorem}
For $0 < \theta < 2 \pi$:
:$\ds \sum_{n \mathop = -m}^m \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \sin {\alpha u} \dfrac {\sin \paren {m + \frac 1 2} u \rd u} {\sin \frac 1 2 u}$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}
| r = \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}
| c =
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}
| c =
}}
{{eqn | n = 1
| r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }
| c = Sum of Geometric Sequence
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }
| c = extracting factors
}}
{{eqn | r = e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }
| c = Exponential of Sum and some algebra
}}
{{eqn | r = e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }
| c = Sine Exponential Formulation
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}
| r = \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }
| c = Euler's Formula and simplifying
}}
{{eqn | n = 2
| ll= \leadsto
| l = \sum_{n \mathop = -m}^m \sin \paren {n + \alpha} \theta
| r = \map \sin {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta} } {\map \sin {\theta / 2} }
| c = equating imaginary parts
}}
{{end-eqn}}
Note that the {{RHS}} at $(1)$ is not defined when $e^{i u} = 1$.
This happens when $u = 2 k \pi$ for $k \in \Z$.
For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^\theta \sin \paren {\alpha + n} u \rd u
| r = \intlimits {\dfrac {-\cos \paren {n + \alpha} u} {n + \alpha} } {u \mathop = 0} {u \mathop = \theta}
| c = Primitive of $\sin a x$
}}
{{eqn | r = \paren {\dfrac {-\cos \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {-\cos \paren {n + \alpha} 0} {n + \alpha} }
| c =
}}
{{eqn | r = \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha}
| c = Cosine of Zero is One
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = -m}^m \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha}
| r = \sum_{n \mathop = -m}^m \int_0^\theta \sin \paren {\alpha + n} u \rd u
| c =
}}
{{eqn | r = \int_0^\theta \sum_{n \mathop = -m}^m \sin \paren {\alpha + n} u \rd u
| c = Linear Combination of Definite Integrals
}}
{{eqn | r = \int_0^\theta \map \sin {\alpha u} \frac {\map \sin {\paren {m + \frac 1 2} u} } {\map \sin {u / 2} } \rd u
| c = from $(2)$, changing the variable name
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22122
|
\section{Sum from -m to m of Sine of n + alpha of theta over n + alpha}
Tags: Sine Function
\begin{theorem}
For $0 < \theta < 2 \pi$:
:$\ds \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \cos {\alpha \theta} \dfrac {\sin \paren {m + \frac 1 2} \theta \rd \theta} {\sin \frac 1 2 \theta}$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}
| r = \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}
| c =
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}
| c =
}}
{{eqn | n = 1
| r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }
| c = Sum of Geometric Sequence
}}
{{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }
| c = extracting factors
}}
{{eqn | r = e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }
| c = Exponential of Sum and some algebra
}}
{{eqn | r = e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }
| c = Sine Exponential Formulation
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}
| r = \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }
| c = Euler's Formula and simplifying
}}
{{eqn | n = 2
| ll= \leadsto
| l = \sum_{n \mathop = -m}^m \cos \paren {n + \alpha} \theta
| r = \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }
| c = equating real parts
}}
{{end-eqn}}
Note that the {{RHS}} at $(1)$ is not defined when $e^{i \theta} = 1$.
This happens when $\theta = 2 k \pi$ for $k \in \Z$.
For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta
| r = \intlimits {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } {\theta \mathop = 0} {\theta \mathop = \theta}
| c = Primitive of $\cos a x$
}}
{{eqn | r = \paren {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {\sin \paren {n + \alpha} 0} {n + \alpha} }
| c =
}}
{{eqn | r = \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}
| c = Sine of Zero is Zero
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}
| r = \sum_{n \mathop = -m}^m \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta
| c =
}}
{{eqn | r = \int_0^\theta \sum_{n \mathop = -m}^m \cos \paren {\alpha + n} \theta \rd \theta
| c = Linear Combination of Definite Integrals
}}
{{eqn | r = \int_0^\theta \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} } \rd \theta
| c = from $(2)$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22123
|
\section{Sum of 2 Lucky Numbers in 4 Ways}
Tags: 34, Lucky Numbers
\begin{theorem}
The number $34$ is the smallest positive integer to be the sum of $2$ lucky numbers in $4$ different ways.
\end{theorem}
\begin{proof}
The sequence of lucky numbers begins:
:$1, 3, 7, 9, 13, 15, 21, 25, 31, 33, \ldots$
Thus we have:
{{begin-eqn}}
{{eqn | l = 34
| r = 1 + 33
| c =
}}
{{eqn | r = 3 + 31
| c =
}}
{{eqn | r = 9 + 25
| c =
}}
{{eqn | r = 13 + 21
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22124
|
\section{Sum of 2 Squares in 2 Distinct Ways}
Tags: Sum of 2 Squares in 2 Distinct Ways, Sums of Squares, Brahmagupta-Fibonacci Identity, Square Numbers
\begin{theorem}
Let $m, n \in \Z_{>0}$ be distinct positive integers that can be expressed as the sum of two distinct square numbers.
Then $m n$ can be expressed as the sum of two square numbers in at least two distinct ways.
\end{theorem}
\begin{proof}
Let:
:$m = a^2 + b^2$
:$n = c^2 + d^2$
Then:
{{begin-eqn}}
{{eqn | l = m n
| r = \paren {a^2 + b^2} \paren {c^2 + d^2}
| c =
}}
{{eqn | r = \paren {a c + b d}^2 + \paren {a d - b c}^2
| c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = \paren {a c - b d}^2 + \paren {a d + b c}^2
| c = Brahmagupta-Fibonacci Identity: Corollary
}}
{{end-eqn}}
It remains to be shown that if $a \ne b$ and $c \ne d$, then the four numbers:
:$a c + b d, a d - b c, a c - b d, a d + b c$
are distinct.
Because $a, b, c, d > 0$, we have:
:$a c + b d \ne a c - b d$
:$a d + b c \ne a d - b c$
We also have:
{{begin-eqn}}
{{eqn | l = a c \pm b d
| r = a d \pm b c
}}
{{eqn | ll= \leadstoandfrom
| l = a c \mp b c - a d \pm b d
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = c \paren {a \mp b} - d \paren {a \mp b}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {a \mp b} \paren {c \mp d}
| r = 0
}}
{{end-eqn}}
Thus $a \ne b$ and $c \ne d$ implies $a c \pm b d \ne a d \pm b c$.
The case for $a c \pm b d \ne a d \mp b c$ is similar.
{{qed}}
\end{proof}
|
22125
|
\section{Sum of 2 Squares in 2 Distinct Ways/Examples/145}
Tags: 145, Sum of 2 Squares in 2 Distinct Ways
\begin{theorem}
$145$ can be expressed as the sum of two square numbers in two distinct ways:
{{begin-eqn}}
{{eqn | l = 145
| r = 12^2 + 1^2
}}
{{eqn | r = 9^2 + 8^2
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We have that:
:$145 = 5 \times 29$
Both $5$ and $29$ can be expressed as the sum of two distinct square numbers:
{{begin-eqn}}
{{eqn | l = 5
| r = 1^2 + 2^2
}}
{{eqn | l = 29
| r = 2^2 + 5^2
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}
| c =
}}
{{eqn | r = \paren {1 \times 2 + 2 \times 5}^2 + \paren {1 \times 5 - 2 \times 2}^2
| c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = \paren {2 + 10}^2 + \paren {5 - 4}^2
| c =
}}
{{eqn | r = 12^2 + 1^2
| c =
}}
{{eqn | r = 144 + 1
| c =
}}
{{eqn | r = 145
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}
| c =
}}
{{eqn | r = \paren {1 \times 2 - 2 \times 5}^2 + \paren {1 \times 5 + 2 \times 2}^2
| c = Brahmagupta-Fibonacci Identity/Corollary
}}
{{eqn | r = \paren {2 - 10}^2 + \paren {5 + 4}^2
| c =
}}
{{eqn | r = \paren {8 - 2}^2 + \paren {5 + 4}^2
| c =
}}
{{eqn | r = 8^2 + 9^2
| c =
}}
{{eqn | r = 64 + 81
| c =
}}
{{eqn | r = 145
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22126
|
\section{Sum of 2 Squares in 2 Distinct Ways/Examples/50}
Tags: Sum of 2 Squares in 2 Distinct Ways, 50
\begin{theorem}
$50$ is the smallest positive integer which can be expressed as the sum of two square numbers in two distinct ways:
{{begin-eqn}}
{{eqn | l = 50
| r = 5^2 + 5^2
}}
{{eqn | r = 7^2 + 1^2
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The smallest two positive integers which can be expressed as the sum of two distinct square numbers are:
{{begin-eqn}}
{{eqn | l = 5
| r = 1^2 + 2^2
}}
{{eqn | l = 10
| r = 1^2 + 3^2
}}
{{end-eqn}}
We have that:
:$50 = 5 \times 10$
Thus:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {1^2 + 3^2}
| c =
}}
{{eqn | r = \paren {1 \times 1 + 2 \times 3}^2 + \paren {1 \times 3 - 2 \times 1}^2
| c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = \paren {1 + 6}^2 + \paren {3 - 2}^2
| c =
}}
{{eqn | r = 7^2 + 1^2
| c =
}}
{{eqn | r = 49 + 1
| c =
}}
{{eqn | r = 50
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {1^2 + 3^2}
| c =
}}
{{eqn | r = \paren {1 \times 1 - 2 \times 3}^2 + \paren {1 \times 3 + 2 \times 1}^2
| c = Brahmagupta-Fibonacci Identity: Corollary
}}
{{eqn | r = \paren {1 - 6}^2 + \paren {3 + 2}^2
| c =
}}
{{eqn | r = \paren {6 - 1}^2 + \paren {3 + 2}^2
| c =
}}
{{eqn | r = 5^2 + 5^2
| c =
}}
{{eqn | r = 25 + 25
| c =
}}
{{eqn | r = 50
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22127
|
\section{Sum of 2 Squares in 2 Distinct Ways/Examples/65}
Tags: Sum of 2 Squares in 2 Distinct Ways, 65
\begin{theorem}
$65$ can be expressed as the sum of two square numbers in two distinct ways:
{{begin-eqn}}
{{eqn | l = 65
| r = 8^2 + 1^2
}}
{{eqn | r = 7^2 + 4^2
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We have that:
:$65 = 5 \times 13$
Both $5$ and $13$ can be expressed as the sum of two distinct square numbers:
{{begin-eqn}}
{{eqn | l = 5
| r = 1^2 + 2^2
}}
{{eqn | l = 13
| r = 2^2 + 3^2
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 3^2}
| c =
}}
{{eqn | r = \paren {1 \times 2 + 2 \times 3}^2 + \paren {1 \times 3 - 2 \times 2}^2
| c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = \paren {2 + 6}^2 + \paren {3 - 4}^2
| c =
}}
{{eqn | r = \paren {2 + 6}^2 + \paren {4 - 3}^2
| c =
}}
{{eqn | r = 8^2 + 1^2
| c =
}}
{{eqn | r = 64 + 1
| c =
}}
{{eqn | r = 65
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 3^2}
| c =
}}
{{eqn | r = \paren {1 \times 2 - 2 \times 3}^2 + \paren {1 \times 3 + 2 \times 2}^2
| c = Brahmagupta-Fibonacci Identity: Corollary
}}
{{eqn | r = \paren {2 - 6}^2 + \paren {3 + 4}^2
| c =
}}
{{eqn | r = \paren {6 - 2}^2 + \paren {3 + 4}^2
| c =
}}
{{eqn | r = 4^2 + 7^2
| c =
}}
{{eqn | r = 16 + 49
| c =
}}
{{eqn | r = 65
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22128
|
\section{Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes}
Tags: Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes, 65, Sums of Squares, Sums of Cubes
\begin{theorem}
The smallest positive integer which is both the sum of $2$ square numbers in two distinct ways and also the sum of $2$ cube numbers is $65$:
{{begin-eqn}}
{{eqn | l = 65
| m = 16 + 49
| mo= =
| r = 4^2 + 7^2
| c =
}}
{{eqn | m = 1 + 64
| mo= =
| r = 1^2 + 8^2
| c =
}}
{{eqn | o =
| mo= =
| r = 1^3 + 4^3
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Sum of 2 Squares in 2 Distinct Ways, the smallest $2$ positive integer which are the sum of $2$ square numbers in two distinct ways are $50$ and $65$.
But $50$ cannot be expressed as the sum of $2$ cube numbers:
{{begin-eqn}}
{{eqn | l = 50 - 1^3
| r = 49
| c = which is not cubic
}}
{{eqn | l = 50 - 2^3
| r = 42
| c = which is not cubic
}}
{{eqn | l = 50 - 3^3
| r = 23
| c = which is not cubic
}}
{{eqn | l = 50 - 4^3
| r = -14
| c = and we have fallen off the end
}}
{{end-eqn}}
Hence $65$ is that smallest number.
{{qed}}
\end{proof}
|
22129
|
\section{Sum of 3 Squares in 2 Distinct Ways}
Tags: 27, Square Numbers
\begin{theorem}
$27$ is the smallest positive integer which can be expressed as the sum of $3$ square numbers in $2$ distinct ways:
{{begin-eqn}}
{{eqn | l = 27
| r = 3^2 + 3^2 + 3^2
}}
{{eqn | r = 5^2 + 1^2 + 1^2
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Can be performed by brute-force investigation.
\end{proof}
|
22130
|
\section{Sum of 3 Unit Fractions that equals 1}
Tags: Unit Fractions, Recreational Mathematics
\begin{theorem}
There are $3$ ways to represent $1$ as the sum of exactly $3$ unit fractions.
\end{theorem}
\begin{proof}
Let:
:$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$
where:
:$0 < a \le b \le c$
and:
{{AimForCont}} $a = 1$.
Then:
:$1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$
and so:
:$\dfrac 1 b + \dfrac 1 c = 0$
which contradicts the stipulation that $b, c > 0$.
So there is no solution possible when $a = 1$.
Therefore $a \ge 2$.
\end{proof}
|
22131
|
\section{Sum of 4 Unit Fractions that equals 1}
Tags: Unit Fractions, 1, Recreational Mathematics, Fractions
\begin{theorem}
There are $14$ ways to represent $1$ as the sum of exactly $4$ unit fractions.
\end{theorem}
\begin{proof}
Let:
:$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d$
where:
:$a \le b \le c \le d$
and:
:$a \ge 2$
\end{proof}
|
22132
|
\section{Sum of 714 and 715}
Tags: Prime Numbers, 714, 715
\begin{theorem}
The sum of $714$ and $715$ is a $4$-digit integer which has $6$ anagrams which are prime.
\end{theorem}
\begin{proof}
We have that:
:$714 + 715 = 1429$
Hence we investigate its anagrams.
We bother only to check those which do not end in either $2$ or $4$, as those are even.
{{begin-eqn}}
{{eqn | l = 1429
| o =
| c = is prime
}}
{{eqn | l = 1249
| o =
| c = is prime
}}
{{eqn | l = 4129
| o =
| c = is prime
}}
{{eqn | l = 4219
| o =
| c = is prime
}}
{{eqn | l = 2149
| r = 7 \times 307
| c = and so is not prime
}}
{{eqn | l = 2419
| r = 41 \times 59
| c = and so is not prime
}}
{{eqn | l = 9241
| o =
| c = is prime
}}
{{eqn | l = 9421
| o =
| c = is prime
}}
{{eqn | l = 2941
| r = 17 \times 173
| c = and so is not prime
}}
{{eqn | l = 2491
| r = 47 \times 53
| c = and so is not prime
}}
{{eqn | l = 4291
| r = 7 \times 613
| c = and so is not prime
}}
{{eqn | l = 4921
| r = 7 \times 19 \times 37
| c = and so is not prime
}}
{{end-eqn}}
Of the above, $6$ are seen to be prime.
{{qed}}
\end{proof}
|
22133
|
\section{Sum of Absolute Values on Ordered Integral Domain}
Tags: Integral Domains, Absolute Value Function
\begin{theorem}
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then:
:$\size {a + b} \le \size a + \size b$
\end{theorem}
\begin{proof}
Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
Let $N$ be the (strict) negativity property on $D$.
Let $a \in D$.
If $\map P a$ or $a = 0$ then $a \le \size a$.
If $\map N a$ then by Properties of Strict Negativity: $(1)$ and definition of absolute value:
:$a < 0 < \size a$
and hence by transitivity $<$ we have:
:$a < \size a$
By similar reasoning:
:$-a < \size a$
Thus for all $a, b \in D$ we have:
:$a \le \size a, b \le \size b$
As $<$ is compatible with $+$, we have:
:$a + b \le \size a + \size b$
and:
:$-\paren {a + b} = \paren {-a} + \paren {-b} \le \size a + \size b$
But either:
:$\size {a + b} = a + b$
or:
:$\size {a + b} = -\paren {a + b}$
Hence the result:
:$\size {a + b} \le \size a + \size b$
{{qed}}
\end{proof}
|
22134
|
\section{Sum of Absolutely Continuous Functions is Absolutely Continuous}
Tags: Absolutely Continuous Functions
\begin{theorem}
Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous functions.
Then $f + g$ is absolutely continuous.
\end{theorem}
\begin{proof}
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
Similarly, since $g$ is absolutely continuous, there exists real $\delta_2 > 0$ such that whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
Let:
:$\delta = \map \min {\delta_1, \delta_2}$
Then, for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
and:
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
We then have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {b_i} - \map {\paren {f + g} } {a_i} }
| r = \sum_{i \mathop = 1}^n \size {\paren {\map f {b_i} - \map f {a_i} } + \paren {\map g {b_i} - \map g {a_i} } }
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }
| c = Triangle Inequality for Real Numbers
}}
{{eqn | o = <
| r = \frac \epsilon 2 + \frac \epsilon 2
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
:$f + g$ is absolutely continuous.
{{qed}}
Category:Absolutely Continuous Functions
\end{proof}
|
22135
|
\section{Sum of Absolutely Convergent Series}
Tags: Absolute Convergence, Convergence, Series
\begin{theorem}
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$
\end{theorem}
\begin{proof}
Let $\epsilon \in \R_{>0}$.
From Tail of Convergent Series tends to Zero, it follows that there exists $M \in \N$ such that:
:$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$
and:
:$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$
For all $m \ge M$, it follows that:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n - \sum_{n \mathop = 1}^m \paren {a_n + b_n} }
| r = \cmod {\sum_{n \mathop = m + 1}^\infty a_n + \sum_{n \mathop = m + 1}^\infty b_n}
}}
{{eqn | o = \le
| r = \sum_{n \mathop = m + 1}^\infty \cmod {a_n} + \sum_{n \mathop = m + 1}^\infty \cmod {b_n}
| c = by Triangle Inequality
}}
{{eqn | o = \le
| r = \sum_{n \mathop = M + 1}^\infty \cmod {a_n} + \sum_{n \mathop = M + 1}^\infty \cmod {b_n}
}}
{{eqn | o = <
| r = \epsilon
}}
{{end-eqn}}
By definition of convergent series, it follows that:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n
| r = \lim_{m \mathop \to \infty} \sum_{n \mathop = 1}^m \paren {a_n + b_n}
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}
}}
{{end-eqn}}
To show that $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, note that:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty \cmod {a_n} + \sum_{n \mathop = 1}^\infty \cmod {b_n}
| r = \sum_{n \mathop = 1}^\infty \paren {\cmod {a_n} + \cmod {b_n} }
| c = as shown above
}}
{{eqn | o = \ge
| r = \sum_{n \mathop = 1}^\infty \cmod {a_n + b_n}
| c = by Triangle Inequality
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22136
|
\section{Sum of All Ring Products is Additive Subgroup}
Tags: Rings, Subset Products, Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S + T$ be defined as subset product.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then both $S + T$ and $S T$ are additive subgroups of $\struct {R, +, \circ}$.
\end{theorem}
\begin{proof}
As $\struct {R, +}$ is abelian (from the definition of a ring), we have:
:$S + T = T + S$
from Subset Product of Commutative is Commutative.
So from Subset Product of Subgroups it follows that $S + T$ is an additive subgroup of $\struct {R, +, \circ}$.
Let $x, y \in S T$.
We have that $\struct {S T, +}$ is closed.
So $x + y \in S T$.
So, if $\ds y = \sum s_i \circ t_i \in S T$, it follows that:
:$\ds -y = \sum \paren {-s_i} \circ t_i \in S T$
By the Two-Step Subgroup Test, we have that $S T$ is an additive subgroup of $\struct {R, +, \circ}$.
{{qed}}
\end{proof}
|
22137
|
\section{Sum of All Ring Products is Associative}
Tags: Rings, Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}, \struct {T, +}, \struct {U, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then:
:$\paren {S T} U = S \paren {T U}$
\end{theorem}
\begin{proof}
We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.
From Sum of All Ring Products is Closed under Addition, this set is closed under ring addition.
Therefore, so are $\paren {S T} U$ and $S \paren {T U}$.
Let $z \in \paren {S T} U$.
Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.
So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.
Therefore $z$ is a finite sum of elements in the form $\paren {s \circ t} \circ u$ where $s \in S, t \in T, u \in U$.
As $\struct {R, +, \circ}$ is a ring, $\circ$ is associative.
So $z$ is a finite sum of elements in the form $s \circ \paren {t \circ u}$ where $s \in S, t \in T, u \in U$.
So these elements all belong to $S \paren {T U}$.
Since $S \paren {T U}$ is closed under addition, $z \in S \paren {T U}$.
So:
:$\paren {S T} U \subseteq S \paren {T U}$
By a similar argument in the other direction:
:$S \paren {T U} \subseteq \paren {S T} U $
and so by definition of set equality:
:$\paren {S T} U = S \paren {T U}$
{{qed}}
\end{proof}
|
22138
|
\section{Sum of All Ring Products is Closed under Addition}
Tags: Rings, Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $\struct {S T, +}$ is a closed subset of $\struct {R, +}$.
\end{theorem}
\begin{proof}
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$
for some $s_i, t_i, j, k$, etc.
By renaming the indices, we can express $x_2$ as:
:$\ds x_2 = \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i$
and hence:
:$\ds x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$
So $x_1 + x_2 \in S T$ and $\struct {S T, +}$ is shown to be closed.
{{qed}}
\end{proof}
|
22139
|
\section{Sum of Angles of Triangle equals Two Right Angles}
Tags: Triangles, Sum of Angles of Triangle equals Two Right Angles
\begin{theorem}
In a triangle, the sum of the three interior angles equals two right angles.
{{:Euclid:Proposition/I/32}}
\end{theorem}
\begin{proof}
:300px
Let $\triangle ABC$ be a triangle.
Let $BC$ be extended to a point $D$.
From External Angle of Triangle equals Sum of other Internal Angles:
: $\angle ACD = \angle ABC + \angle BAC$
Bby by Euclid's Second Common Notion:
: $\angle ACB + \angle ACD = \angle ABC + \angle BAC + \angle ACB$
But from Two Angles on Straight Line make Two Right Angles, $ACB + ACD$ equals two right angles.
So by Euclid's First Common Notion, $\angle ABC + \angle BAC + \angle ACB$ equals two right angles.
{{qed}}
{{Euclid Note|32|I|Euclid's proposition $32$ consists of two parts, the first of which is External Angle of Triangle equals Sum of other Internal Angles, and the second part of which is this.|part = second}}
\end{proof}
|
22140
|
\section{Sum of Antecedent and Consequent of Proportion}
Tags: Ratios
\begin{theorem}
{{:Euclid:Proposition/V/25}}
That is, if $a : b = c : d$ and $a$ is the greatest and $d$ is the least, then:
:$a + d > b + c$
\end{theorem}
\begin{proof}
Let the four magnitudes $AB, CD, E, F$ be proportional, so that $AB : CD = E : F$.
Let $AB$ be the greatest and $F$ the least.
We need to show that $AB + F > CD + E$.
:250px
Let $AG = E, CH = F$.
We have that $AB : CD = E : F$, $AG = E, F = CH$.
So $AB : CD = AG : CH$.
So from Proportional Magnitudes have Proportional Remainders $GB : HD = AB : CD$.
But $AB > CD$ and so $GB > HD$.
Since $AG = E$ and $CH = F$, it follows that $AG + F = CH + E$.
We have that $GB > HD$.
So add $AG + F$ to $GB$ and $CH + E$ to $HD$.
It follows that $AB + F > CD + E$.
{{qed}}
{{Euclid Note|25|V}}
\end{proof}
|
22141
|
\section{Sum of Arccotangents}
Tags: Arccotangent Function, Inverse Cotangent
\begin{theorem}
:$\arccot a + \arccot b = \arccot \dfrac {a b - 1} {a + b}$
where $\arccot$ denotes the arccotangent.
\end{theorem}
\begin{proof}
Let $x = \arccot a$ and $y = \arccot b$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \cot x
| r = a
| c =
}}
{{eqn | n = 2
| l = \cot y
| r = b
| c =
}}
{{eqn | l = \map \cot {\arccot a + \arccot b}
| r = \map \cot {x + y}
| c =
}}
{{eqn | r = \frac {\cot x \cot y - 1} {\cot x + \cot y}
| c = Cotangent of Sum
}}
{{eqn | r = \frac {a + b} {1 - a b}
| c = by $(1)$ and $(2)$
}}
{{eqn | ll= \leadsto
| l = \arccot a + \arccot b
| r = \arccot \frac {a b - 1} {a + b}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22142
|
\section{Sum of Arcsecant and Arccosecant}
Tags: Inverse Cosecant, Inverse Trigonometric Functions, Arccosecant Function, Analysis, Inverse Secant, Arcsecant Function
\begin{theorem}
Let $x \in \R$ be a real number such that $\size x \ge 1$.
Then:
: $\arcsec x + \arccsc x = \dfrac \pi 2$
where $\arcsec$ and $\arccsc$ denote arcsecant and arccosecant respectively.
\end{theorem}
\begin{proof}
Let $y \in \R$ such that:
: $\exists x \in \R: \size x \ge 1$ and $x = \map \csc {y + \dfrac \pi 2}$
Then:
{{begin-eqn}}
{{eqn | l = x
| r = \map \sec {y + \frac \pi 2}
| c =
}}
{{eqn | r = -\csc y
| c = Secant of Angle plus Right Angle
}}
{{eqn | r = \map \csc {-y}
| c = Cosecant Function is Odd
}}
{{end-eqn}}
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write $-y = \arccsc x$.
But then $\map \csc {y + \dfrac \pi 2} = x$.
Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.
Hence $y + \dfrac \pi 2 = \arcsec x$.
That is, $\dfrac \pi 2 = \arcsec x + \arccsc x$.
{{qed}}
\end{proof}
|
22143
|
\section{Sum of Arcsine and Arccosine}
Tags: Inverse Trigonometric Functions, Analysis, Inverse Cosine, Inverse Hyperbolic Functions, Inverse Sine, Arcsine Function, Arccosine Function
\begin{theorem}
Let $x \in \R$ be a real number such that $-1 \le x \le 1$.
Then:
: $\arcsin x + \arccos x = \dfrac \pi 2$
where $\arcsin$ and $\arccos$ denote arcsine and arccosine respectively.
\end{theorem}
\begin{proof}
Let $y \in \R$ such that:
: $\exists x \in \left[{-1 \,.\,.\, 1}\right]: x = \cos \left({y + \dfrac \pi 2}\right)$
Then:
{{begin-eqn}}
{{eqn | l = x
| r = \cos \left({y + \frac \pi 2}\right)
| c =
}}
{{eqn | r = -\sin y
| c = Cosine of Angle plus Right Angle
}}
{{eqn | r = \sin \left({-y}\right)
| c = Sine Function is Odd
}}
{{end-eqn}}
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write $-y = \arcsin x$.
But then $\cos \left({y + \dfrac \pi 2}\right) = x$.
Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.
Hence $y + \dfrac \pi 2 = \arccos x$.
That is, $\dfrac \pi 2 = \arccos x + \arcsin x$.
{{qed}}
\end{proof}
|
22144
|
\section{Sum of Arctangent and Arccotangent}
Tags: Inverse Trigonometric Functions, Arccotangent Function, Inverse Tangent, Inverse Cotangent, Arctangent Function
\begin{theorem}
Let $x \in \R$ be a real number.
Then:
: $\arctan x + \operatorname{arccot} x = \dfrac \pi 2$
where $\arctan$ and $\operatorname{arccot}$ denote arctangent and arccotangent respectively.
\end{theorem}
\begin{proof}
Let $y \in \R$ such that:
: $\exists x \in \R: x = \cot \left({y + \dfrac \pi 2}\right)$
Then:
{{begin-eqn}}
{{eqn | l = x
| r = \cot \left({y + \frac \pi 2}\right)
| c =
}}
{{eqn | r = -\tan y
| c = Cotangent of Angle plus Right Angle
}}
{{eqn | r = \tan \left({-y}\right)
| c = Tangent Function is Odd
}}
{{end-eqn}}
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write $-y = \arctan x$.
But then $\cot \left({y + \dfrac \pi 2}\right) = x$.
Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.
Hence $y + \dfrac \pi 2 = \operatorname{arccot} x$.
That is, $\dfrac \pi 2 = \operatorname{arccot} x + \arctan x$.
{{qed}}
\end{proof}
|
22145
|
\section{Sum of Arctangents}
Tags: Arctangent Function, Inverse Tangent
\begin{theorem}
:$\arctan a + \arctan b = \arctan \dfrac {a + b} {1 - a b}$
where $\arctan$ denotes the arctangent.
\end{theorem}
\begin{proof}
Let $x = \arctan a$ and $y = \arctan b$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \tan x
| r = a
| c =
}}
{{eqn | n = 2
| l = \tan y
| r = b
| c =
}}
{{eqn | l = \map \tan {\arctan a + \arctan b}
| r = \map \tan {x + y}
| c =
}}
{{eqn | r = \frac {\tan x + \tan y} {1 - \tan x \tan y}
| c = Tangent of Sum
}}
{{eqn | r = \frac {a + b} {1 - a b}
| c = by $(1)$ and $(2)$
}}
{{eqn | ll= \leadsto
| l = \arctan a + \arctan b
| r = \arctan \frac {a + b} {1 - a b}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22146
|
\section{Sum of Arithmetic-Geometric Sequence}
Tags: Arithmetic-Geometric Sequences, Arithmetic-Geometric Progressions, Sum of Arithmetic-Geometric Sequence, Sums of Sequences, Sum of Arithmetic-Geometric Progression, Algebra
\begin{theorem}
Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \left({a + k d}\right) r^k = \frac {a \left({1 - r^n}\right)} {1 - r} + \frac {r d \left({1 - n r^{n - 1} + \left({n - 1}\right) r^n}\right)} {\left({1 - r}\right)^2}$
\end{proof}
|
22147
|
\section{Sum of Arithmetic Sequence}
Tags: Arithmetic Sequences, Sum of Arithmetic Sequence, Sum of Arithmetic Progression, Arithmetic Progressions, Sums of Sequences, Algebra
\begin{theorem}
Let $\sequence {a_k}$ be an arithmetic sequence defined as:
:$a_k = a + k d$ for $n = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \frac {n - 1} 2 d}
| c =
}}
{{eqn | r = \frac {n \paren {a + l} } 2
| c = where $l$ is the last term of $\sequence {a_k}$
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We have that:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = 2 \paren {a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d} }
}}
{{eqn | r = \paren {a + \paren {a + d} + \dotsb + \paren {a + \paren {n - 1} d} }
}}
{{eqn | ro= +
| r = \paren {\paren {a + \paren {n - 1} d} + \paren {a + \paren {n - 2} d} + \dotsb + \paren {a + d} + a}
}}
{{eqn | r = \paren {2 a + \paren {n - 1} d}_1 + \paren {2 a + \paren {n - 1} d}_2 + \dotsb + \paren {2 a + \paren {n - 1} d}_n
}}
{{eqn | r = n \paren {2 a + \paren {n - 1} d}
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {2 a + \paren {n - 1} d}
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = \frac {n \paren {2 a + \paren {n - 1} d} } 2
}}
{{eqn | r = \frac {n \paren {a + l} } 2
| c = {{Defof|Last Term of Arithmetic Sequence|Last Term}} $l$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22148
|
\section{Sum of Bernoulli Numbers by Binomial Coefficients Vanishes}
Tags: Bernoulli Numbers, Binomial Coefficients, Sum of Bernoulli Numbers by Binomial Coefficients Vanishes
\begin{theorem}
:$\forall n \in \Z_{>1}: \ds \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$
where $B_k$ denotes the $k$th Bernoulli number.
\end{theorem}
\begin{proof}
Take the definition of Bernoulli numbers:
:$\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
From the definition of the exponential function:
{{begin-eqn}}
{{eqn | l = e^x
| r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
| c =
}}
{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!}
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {e^x - 1} x
| r = \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!}
| c =
}}
{{eqn | r = 1 + \frac x {2!} + \frac {x^2} {3!} + \cdots
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = 1
| r = \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}
| c =
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }
| c =
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^n} {\paren {n + 1}!} }
| c = as both series start at zero
}}
{{end-eqn}}
By Product of Absolutely Convergent Series, we will let:
{{begin-eqn}}
{{eqn | l = a_n
| r = \frac {B_n x^n} {n!}
| c =
}}
{{eqn | l = b_n
| r = \frac {x^n} {\paren {n + 1}!}
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
| r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}
| rr= =1
| c =
}}
{{eqn | l = c_n
| r = \sum_{k \mathop = 0}^n a_k b_{n - k}
| c =
}}
{{eqn | l = c_0
| r = \frac {B_0 x^0} {0!} \frac {x^0} {\paren {0 + 1}!}
| rr= = 1
| c = as $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = 1}^\infty c_n
| r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n} - a_0 b_0
| rr= = 0
| c = subtracting $1$ from both sides
}}
{{eqn | r = c_1 x + c_2 x^2 + c_3 x^3 + \cdots
| rr= = 0
}}
{{eqn | ll= \leadsto
| q = \forall n \in \Z_{>0}
| l = c_n
| r = 0
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = c_1
| r = \frac {B_0 x^0} {0!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{0} } {\paren {0 + 1 }!}
| rr= = 0
| rrr= = a_0 b_1 + a_1 b_0
}}
{{eqn | l = c_2
| r = \frac {B_0 x^0} {0!} \frac {x^{2} } {\paren {2 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_2 x^2} {2!} \frac {x^{0} } {\paren {0 + 1 }!}
| rr= = 0
| rrr= = a_0 b_2 + a_1 b_1 + a_2 b_0
}}
{{eqn | l = \cdots
| r = \cdots
| rr= = 0
}}
{{eqn | l = c_n
| r = \frac {B_0 x^0} {0!} \frac {x^{n} } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {x^{0} } {\paren {0 + 1 }!}
| rr= = 0
| rrr= = a_0 b_n + a_1 b_{n - 1 } + a_2 b_{n - 2 } + \cdots + a_n b_0
}}
{{end-eqn}}
Multiplying $c_n$ through by $\paren {n + 1 }!$ gives:
{{begin-eqn}}
{{eqn | l = \paren {n + 1 }! c_n
| r = \frac {B_0 x^0} {0!} \frac {\paren {n + 1 }! x^n } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {\paren {n + 1 }! x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {\paren {n + 1 }! x^{0} } {\paren {0 + 1 }!}
| rr= = 0
| c =
}}
{{eqn | r = x^n \paren {\frac {\paren {n + 1 }! } {0! \paren {n + 1 }!} B_0 + \frac {\paren {n + 1 }! } {1! \paren {n - 1 + 1 }!} B_1 + \cdots + \frac {\paren {n + 1 }! } {n! \paren {0 + 1 }!} B_n }
| rr= = 0
| c = factoring out $x^n$
}}
{{end-eqn}}
But those coefficients are the binomial coefficients:
{{begin-eqn}}
{{eqn | l = \paren {n + 1 }! c_n
| r = \dbinom {n + 1 } 0 B_0 + \dbinom {n + 1 } 1 B_1 + \dbinom {n + 1 } 2 B_2 + \cdots + \dbinom {n + 1 } n B_n
| rr= = 0
| c =
}}
{{eqn | l = n! c_{n-1 }
| r = \dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1}
| rr= = 0
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22149
|
\section{Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient}
Tags: Bernoulli Numbers, Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient, Definitions: Bernoulli Numbers
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}
| r = \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots
| c =
}}
{{eqn | r = 2 n
| c =
}}
{{end-eqn}}
where $B_n$ denotes the $n$th Bernoulli number.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
:$\displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
\end{proof}
|
22150
|
\section{Sum of Big-O Estimates/Real Analysis}
Tags: Asymptotic Notation
\begin{theorem}
Let $c$ be a real number.
Let $f, g : \hointr c \infty \to \R$ be real functions.
Let $R_1 : \hointr c \infty \to \R$ be a real function such that $f = \map \OO {R_1}$.
Let $R_2 : \hointr c \infty \to \R$ be a real function such that $g = \map \OO {R_2}$.
Then:
:$f + g = \map \OO {\size {R_1} + \size {R_2} }$
\end{theorem}
\begin{proof}
Since:
:$f = \map \OO {R_1}$
there exists $x_1 \in \hointr c \infty$ and a real number $C_1$ such that:
:$\size {\map f x} \le C_1 \size {\map {R_1} x}$
for $x \ge x_1$.
Similarly, since:
:$g = \map \OO {R_2}$
there exists $x_2 \in \hointr c \infty$ and a real number $C_2$ such that:
:$\size {\map g x} \le C_2 \size {\map {R_2} x}$
Set:
:$x_0 = \max \set {x_1, x_2}$
and:
:$C = \max \set {C_1, C_2}$
Then, for $x \ge x_0$ we have:
{{begin-eqn}}
{{eqn | l = \size {\map f x + \map g x}
| o = \le
| r = \size {\map f x} + \size {\map g x}
| c = Triangle Inequality
}}
{{eqn | o = \le
| r = C_1 \size {\map {R_1} x} + C_2 \size {\map {R_2} x}
| c = since $x \ge x_1$ and $x \ge x_2$
}}
{{eqn | o = \le
| r = C \size {\map {R_1} x} + C \size {\map {R_2} x}
}}
{{eqn | r = C \size {\size {\map {R_1} x} + \size {\map {R_2} x} }
}}
{{end-eqn}}
That is, by the definition of big-O notation, we have:
:$f + g = \map \OO {\size {R_1} + \size {R_2} }$
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22151
|
\section{Sum of Big-O Estimates/Sequences}
Tags: Asymptotic Notation
\begin{theorem}
Let $\sequence {a_n},\sequence {b_n},\sequence {c_n},\sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map \OO {d_n}$
where $\OO$ denotes big-O notation.
Then:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$
\end{theorem}
\begin{proof}
Since:
:$a_n = \map \OO {b_n}$
there exists a positive real number $C_1$ and natural number $N_1$ such that:
:$\size {a_n} \le C_1 \size {b_n}$
for all $n \ge N_1$.
Similarly, since:
:$c_n = \map \OO {d_n}$
there exists a positive real number $C_2$ and natural number $N_2$ such that:
:$\size {c_n} \le C_2 \size {d_n}$
for all $n \ge N_2$.
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n \ge N$ we have:
{{begin-eqn}}
{{eqn | l = \size {a_n + c_n}
| o = \le
| r = \size {a_n} + \size {c_n}
| c = Triangle Inequality
}}
{{eqn | r = C_1 \size {b_n} + \size {c_n}
| c = since $n \ge N_1$
}}
{{eqn | r = C_1 \size {b_n} + C_2 \size {d_n}
| c = since $n \ge N_2$
}}
{{end-eqn}}
Let:
:$C = \max \set {C_1, C_2}$
Then:
{{begin-eqn}}
{{eqn | l = C_1 \size {b_n} + C_2 \size {d_n}
| r = C \size {b_n} + C \size {d_n}
}}
{{eqn | r = C \paren {\size {b_n} + \size {d_n} }
}}
{{eqn | r = C \size {\size {b_n} + \size {d_n} }
}}
{{end-eqn}}
So:
:$\size {a_n + c_n} \le C \size {\size {b_n} + \size {d_n} }$
for $n \ge N$.
So:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22152
|
\section{Sum of Binomial Coefficients over Lower Index/Corollary}
Tags: Binomial Coefficients, Sum of Binomial Coefficients over Lower Index
\begin{theorem}
:$\ds \forall n \in \Z_{\ge 0}: \sum_{i \mathop \in \Z} \binom n i = 2^n$
where $\dbinom n i$ is a binomial coefficient.
\end{theorem}
\begin{proof}
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\dbinom n i = 0$.
The result follows directly from Sum of Binomial Coefficients over Lower Index.
{{qed}}
Category:Binomial Coefficients
Category:Sum of Binomial Coefficients over Lower Index
\end{proof}
|
22153
|
\section{Sum of Bounded Linear Transformations is Bounded Linear Transformation}
Tags: Linear Transformations on Hilbert Spaces
\begin{theorem}
Let $\mathbb F \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$.
Let $A, B : \HH \to \KK$ be bounded linear transformations.
Let $\norm \cdot$ be the norm on the space of bounded linear transformations.
Then:
:$A + B$ is a bounded linear transformation
with:
:$\norm {A + B} \le \norm A + \norm B$
\end{theorem}
\begin{proof}
From Addition of Linear Transformations, we have that:
:$A + B$ is a linear transformation.
It remains to show that $A + B$ is bounded.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
Since $A$ is a bounded linear transformation, from Fundamental Property of Norm on Bounded Linear Transformation, we have:
:$\norm {A x}_\KK \le \norm A \norm x_\HH$
for all $x \in \HH$.
Similarly, since $B$ is a bounded linear transformation we have:
:$\norm {B x}_\KK \le \norm B \norm x_\HH$
for all $x \in \HH$.
Let $x \in \HH$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \norm {\paren {A + B} x}_\KK
| r = \norm {A x + B x}_\KK
}}
{{eqn | o = \le
| r = \norm {A x}_\KK + \norm {B x}_\KK
| c = {{Defof|Norm on Vector Space}}
}}
{{eqn | o = \le
| r = \norm A \norm x_\HH + \norm B \norm x_\HH
}}
{{eqn | r = \paren {\norm A + \norm B} \norm x_\HH
}}
{{end-eqn}}
So, taking $c = \norm A +\norm B$, we have:
:$\norm {\paren {A + B} x}_\KK \le c \norm x_\HH$
for all $x \in \HH$.
So:
:$A + B$ is a bounded linear transformation.
Note that:
:$\norm A + \norm B \in \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
while, by the definition of the norm, we have:
:$\norm {A + B} = \inf \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
So, by the definition of infimum:
:$\norm {A + B} \le \norm A + \norm B$
{{qed}}
Category:Linear Transformations on Hilbert Spaces
\end{proof}
|
22154
|
\section{Sum of Cardinals is Associative}
Tags: Cardinals
\begin{theorem}
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
\end{theorem}
\begin{proof}
Let $\mathbf a = \card A, \mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
Let $A, B, C$ be pairwise disjoint, that is:
:$A \cap B = \O$
:$B \cap C = \O$
:$A \cap C = \O$
Then we can define:
:$A \sqcup B := A \cup B$
:$B \sqcup C := B \cup C$
:$A \sqcup C := A \cup C$
where $A \sqcup B$ denotes the disjoint union of $A$ and $B$.
Then we have:
:$\mathbf a + \mathbf b = \card {A \sqcup B} = \card {A \cup B}$
:$\mathbf b + \mathbf c = \card {B \sqcup C} = \card {B \cup C}$
Then:
{{begin-eqn}}
{{eqn | l=\paren {A \cup B} \cap C
| r=\paren {A \cap C} \cup \paren {B \cap C}
| c=Intersection Distributes over Union
}}
{{eqn | r=\O \cup \O
| c=as $A \cap C = \O$ and $B \cap C = \O$
}}
{{eqn | r=\O
| c=Union with Empty Set
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l=\card {\paren {A \cup B} \cup C}
| r=\card {A \cup B} + \card C
| c=as $\paren {A \cup B} \cap C = \O$ from above
}}
{{eqn | r=\paren {\mathbf a + \mathbf b} + \mathbf c
| c={{Defof|Sum of Cardinals}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l=A \cap \paren {B \cup C}
| r=\paren {A \cap B} \cup \paren {A \cap C}
| c=Intersection Distributes over Union
}}
{{eqn | r=\O \cup \O
| c=as $A \cap B = \O$ and $A \cap C = \O$
}}
{{eqn | r=\O
| c=Union with Empty Set
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l=\card {A \cup \paren {B \cup C} }
| r=\card A + \card {B \cup C}
| c=as $A \cap \paren {B \cup C} = \O$ from above
}}
{{eqn | r=\mathbf a + \paren {\mathbf b + \mathbf c}
| c={{Defof|Sum of Cardinals}}
}}
{{end-eqn}}
Finally note that from Union is Associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
{{qed}}
\end{proof}
|
22155
|
\section{Sum of Cardinals is Commutative}
Tags: Cardinals
\begin{theorem}
Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
\end{theorem}
\begin{proof}
Let $\mathbf a = \map \Card A$ and $\mathbf b = \map \Card B$ for some sets $A$ and $B$ such that $A \cap B = \O$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
| r = \map \Card {A \cup B}
| c = {{Defof|Sum of Cardinals}}
}}
{{eqn | r = \map \Card {B \cup A}
| c = Union is Commutative
}}
{{eqn | r = \mathbf b + \mathbf a
| c = {{Defof|Sum of Cardinals}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22156
|
\section{Sum of Ceilings not less than Ceiling of Sum}
Tags: Ceiling Function, Floor and Ceiling
\begin{theorem}
Let $\ceiling x$ be the ceiling function.
Then:
:$\ceiling x + \ceiling y \ge \ceiling {x + y}$
The equality holds:
:$\ceiling x + \ceiling y = \ceiling {x + y}$
{{iff}} either:
:$x \in \Z$ or $y \in \Z$
or:
:$x \bmod 1 + y \bmod 1 > 1$
where $x \bmod 1$ denotes the modulo operation.
\end{theorem}
\begin{proof}
From the definition of the modulo operation, we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which we obtain:
:$x = \ceiling x - \sqbrk {x \notin \Z} + \paren {x \bmod 1}$
where $\sqbrk {x \notin \Z}$ uses Iverson's convention.
{{begin-eqn}}
{{eqn | l = \ceiling {x + y}
| r = \ceiling {\floor x + \paren {x \bmod 1} + \floor y + \paren {y \bmod 1} }
| c =
}}
{{eqn | r = \ceiling {\ceiling x - \sqbrk {x \notin \Z} + \paren {x \bmod 1} + \ceiling y - \sqbrk {y \notin \Z} + \paren {y \bmod 1} }
| c =
}}
{{eqn | r = \ceiling x + \ceiling y + \ceiling {\paren {x \bmod 1} + \paren {y \bmod 1} } - \sqbrk {x \notin \Z} - \sqbrk {y \notin \Z}
| c = Ceiling of Number plus Integer
}}
{{end-eqn}}
We have that:
:$x \notin \Z \implies x \bmod 1 > 0$
As $0 \le x \bmod 1 < 1$ it follows that:
:$\sqbrk {x \notin \Z} \ge x \bmod 1$
Hence the inequality.
The equality holds {{iff}}:
:$\ceiling {\paren {x \bmod 1} + \paren {y \bmod 1} } = \sqbrk {x \notin \Z} + \sqbrk {y \notin \Z}$
that is, {{iff}} one of the following holds:
:$x \in \Z$, in which case $x \bmod 1 = 0$
:$y \in \Z$, in which case $y \bmod 1 = 0$
:both $x, y \in \Z$, in which case $\paren {x \bmod 1} + \paren {y \bmod 1} = 0$
:both $x, y \notin \Z$ and $\paren {x \bmod 1} + \paren {y \bmod 1} > 1$.
{{qed}}
\end{proof}
|
22157
|
\section{Sum of Chi-Squared Random Variables}
Tags: Chi-Squared Distribution
\begin{theorem}
Let $n_1, n_2, \ldots, n_k$ be strictly positive integers which sum to $N$.
Let $X_i \sim {\chi^2}_{n_i}$ for $1 \le i \le k$, where ${\chi^2}_{n_i}$ is the chi-squared distribution with $n_i$ degrees of freedom.
Then:
:$\ds X = \sum_{i \mathop = 1}^k X_i \sim {\chi^2}_N$
\end{theorem}
\begin{proof}
Let $Y \sim {\chi^2}_N$.
By Moment Generating Function of Chi-Squared Distribution, the moment generating function of $X_i$ is given by:
:$\map {M_{X_i} } t = \paren {1 - 2 t}^{-n_i / 2}$
Similarly, the moment generating function of $Y$ is given by:
:$\map {M_Y} t = \paren {1 - 2 t}^{-N / 2}$
By Moment Generating Function of Linear Combination of Independent Random Variables, the moment generating function of $X$ is given by:
:$\ds \map {M_X} t = \prod_{i \mathop = 1}^k \map {M_{X_i} } t$
We aim to show that:
:$\map {M_X} t = \map {M_Y} t$
By Moment Generating Function is Unique, this ensures $X = Y$.
We have:
{{begin-eqn}}
{{eqn | l = \map {M_X} t
| r = \prod_{i \mathop = 1}^k \paren {1 - 2 t}^{-n_i / 2}
}}
{{eqn | r = \paren {1 - 2 t}^{-\paren {n_1 + n_2 + \ldots + n_k} / 2}
}}
{{eqn | r = \paren {1 - 2 t}^{-N / 2}
}}
{{eqn | r = \map {M_Y} t
}}
{{end-eqn}}
{{qed}}
Category:Chi-Squared Distribution
\end{proof}
|
22158
|
\section{Sum of Complex Conjugates}
Tags: Complex Analysis, Complex Conjugates, Complex Addition
\begin{theorem}
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\overline z$ denote the complex conjugate of the complex number $z$.
Then:
:$\overline {z_1 + z_2} = \overline {z_1} + \overline {z_2}$
\end{theorem}
\begin{proof}
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \overline {z_1 + z_2}
| r = \overline {\paren {x_1 + x_2} + i \paren {y_1 + y_2} }
| c =
}}
{{eqn | r = \paren {x_1 + x_2} - i \paren {y_1 + y_2}
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = \paren {x_1 - i y_1} + \paren {x_2 - i y_2}
| c = {{Defof|Complex Addition}}
}}
{{eqn | r = \overline {z_1} + \overline {z_2}
| c = {{Defof|Complex Conjugate}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22159
|
\section{Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 1}
Tags: Exponential Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
:$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
\end{theorem}
\begin{proof}
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} }
| r = e^{i \theta} \sum_{k \mathop = 0}^n e^{i k \alpha}
| c = factorising $e^{i \theta}$
}}
{{eqn | r = e^{i \theta} \paren {\frac {e^{i \paren {n + 1} \alpha} - 1} {e^{i \alpha} - 1} }
| c = Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$
}}
{{eqn | r = \frac {e^{i \theta} e^{i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = extracting factors
}}
{{eqn | r = e^{i \paren {\theta + n \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Exponential of Sum and some algebra
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Euler's Formula
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }
| c = Sine Exponential Formulation
}}
{{end-eqn}}
{{qed}}
Category:Exponential Function
\end{proof}
|
22160
|
\section{Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 2}
Tags: Exponential Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
:$\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
\end{theorem}
\begin{proof}
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} }
| r = e^{i \theta} e^{i \alpha} \sum_{k \mathop = 0}^{n - 1} e^{i k \alpha}
| c = factorising $e^{i \theta} e^{i \alpha}$
}}
{{eqn | r = e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha} - 1} {e^{i \alpha} - 1} }
| c = Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$
}}
{{eqn | r = e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha / 2} \paren {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } } {e^{i \alpha / 2} \paren {e^{i \alpha / 2} - e^{-i \alpha / 2} } } }
| c = extracting factors
}}
{{eqn | r = e^{i \paren {\theta + \paren {n + 1} \alpha / 2} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Exponential of Sum and some algebra
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Euler's Formula
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }
| c = Sine Exponential Formulation
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22161
|
\section{Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 3}
Tags: Exponential Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
:$\ds \sum_{k \mathop = p}^q e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \frac {\map \sin {\paren {q - p + 1} \alpha / 2} } {\map \sin {\alpha / 2} }$
\end{theorem}
\begin{proof}
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = p}^q e^{i \paren {\theta + k \alpha} }
| r = e^{i \theta} e^{i p \alpha} \sum_{k \mathop = 0}^{q - p} e^{i k \alpha}
| c = factorising $e^{i \theta} e^{i p \alpha}$
}}
{{eqn | r = e^{i \theta} e^{i p \alpha} \paren {\frac {e^{i \paren {q - p + 1} \alpha} - 1} {e^{i \alpha} - 1} }
| c = Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$
}}
{{eqn | r = e^{i \theta} e^{i p \alpha} \frac {e^{i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = extracting factors
}}
{{eqn | r = e^{i \paren {\theta + \paren {p + q} \alpha / 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Exponential of Sum and some algebra
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }
| c = Euler's Formula
}}
{{eqn | r = \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \frac {\map \sin {\paren {q - p + 1} \alpha / 2} } {\map \sin {\alpha / 2} }
| c = Sine Exponential Formulation
}}
{{end-eqn}}
{{qed}}
Category:Exponential Function
\end{proof}
|
22162
|
\section{Sum of Complex Indices of Real Number}
Tags: Powers
\begin{theorem}
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $\psi, \tau \in \C$ be complex numbers.
Let $r^\lambda$ be defined as the the principal branch of a positive real number raised to a complex number.
Then:
:$r^{\psi \mathop + \tau} = r^\psi \times r^\tau$
\end{theorem}
\begin{proof}
Then:
{{begin-eqn}}
{{eqn | l = r^{\psi \mathop + \tau}
| r = \map \exp {\paren {\psi + \tau} \ln r}
| c = {{Defof|Power (Algebra)/Complex Number/Principal Branch/Positive Real Base|Principal Branch of Positive Real Number raised to Complex Number}}
}}
{{eqn | r = \map \exp {\psi \ln r + \tau \ln r}
}}
{{eqn | r = \map \exp {\psi \ln r} \, \map \exp {\tau \ln r}
| c = Exponential of Sum
}}
{{eqn | r = r^\psi \times r^\tau
| c = {{Defof|Power (Algebra)/Complex Number/Principal Branch/Positive Real Base|Principal Branch of Positive Real Number raised to Complex Number}}
}}
{{end-eqn}}
{{qed}}
Category:Powers
\end{proof}
|
22163
|
\section{Sum of Complex Integrals on Adjacent Intervals}
Tags: Complex Analysis
\begin{theorem}
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Let $c \in \closedint a b$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$
\end{theorem}
\begin{proof}
From Continuous Complex Function is Complex Riemann Integrable, it follows that all three complex Riemann integrals are well defined.
From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions.
{{explain|Revisit the above link -- see if there is a more appropriate one to use so as not to invoke the concept of metric spaces}}
From Composite of Continuous Mappings is Continuous, it follows that $\Re \circ f: \R \to \R$ and $\Im \circ f: \R \to \R$ are continuous real functions.
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \map f t \rd t
| r = \int_a^b \map \Re {\map f t} \rd t + i \int_a^b \map \Im {\map f t} \rd t
| c = {{Defof|Complex Riemann Integral}}
}}
{{eqn | r = \int_a^c \map \Re {\map f t} \rd t + \int_c^b \map \Re {\map f t} \rd t + i \paren {\int_a^c \map \Im {\map f t} \rd t + \int_c^b \map \Im {\map f t} \rd t}
| c = Sum of Integrals on Adjacent Intervals for Continuous Functions
}}
{{eqn | r = \int_a^c \map \Re {\map f t} \rd t + i \int_a^c \map \Im {\map f t} \rd t + \int_c^b \map \Re {\map f t} \rd t + i \int_c^b \map \Im {\map f t} \rd t
}}
{{eqn | r = \int_a^c \map f t \rd t + \int_c^b \map f t \rd t
}}
{{end-eqn}}
{{qed}}
Category:Complex Analysis
\end{proof}
|
22164
|
\section{Sum of Complex Number with Conjugate}
Tags: Complex Analysis, Complex Conjugates
\begin{theorem}
Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Let $\map \Re z$ be the real part of $z$.
Then:
:$z + \overline z = 2 \, \map \Re z$
\end{theorem}
\begin{proof}
Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = z + \overline z
| r = \paren {x + i y} + \paren {x - i y}
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = 2 x
}}
{{eqn | r = 2 \, \map \Re z
| c = {{Defof|Real Part}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22165
|
\section{Sum of Complex Numbers in Exponential Form}
Tags: Complex Numbers, Complex Addition
\begin{theorem}
Let $z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$ be complex numbers expressed in exponential form.
Let $z_3 = r_3 e^{i \theta_3} = z_1 + z_2$.
Then:
:$r_3 = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$
:$\theta_3 = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2} {r_1 \cos \theta_1 + r_2 \cos \theta_2} }$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = r_1 e^{i \theta_1} + r_2 e^{i \theta_2}
| r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2}
| c = {{Defof|Polar Form of Complex Number}}
}}
{{eqn | r = \paren {r_1 \cos \theta_1 + r_2 \cos \theta_2} + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2}
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = {r_3}^2
| r = r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2}
| c = Complex Modulus of Sum of Complex Numbers
}}
{{eqn | ll= \leadsto
| l = r_3
| r = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2} }
| c =
}}
{{end-eqn}}
and similarly:
:$\theta_3 = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2} {r_1 \cos \theta_1 + r_2 \cos \theta_2} }$
{{qed}}
\end{proof}
|
22166
|
\section{Sum of Complex Numbers in Exponential Form/General Result}
Tags: Complex Addition
\begin{theorem}
Let $n \in \Z_{>0}$ be a positive integer.
For all $k \in \set {1, 2, \dotsc, n}$, let:
:$z_k = r_k e^{i \theta_k}$
be non-zero complex numbers in exponential form.
Let:
:$r e^{i \theta} = \ds \sum_{k \mathop = 1}^n z_k = z_1 + z_2 + \dotsb + z_k$
Then:
{{begin-eqn}}
{{eqn | l = r
| r = \sqrt {\sum_{k \mathop = 1}^n r_k + \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \map \cos {\theta_j - \theta_k} }
}}
{{eqn | l = \theta
| r = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let:
{{begin-eqn}}
{{eqn | l = r e^{i \theta}
| r = \sum_{k \mathop = 1}^n z_k
| c =
}}
{{eqn | r = z_1 + z_2 + \dotsb + z_k
| c =
}}
{{eqn | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2} + \dotsb + r_n \paren {\cos \theta_n + i \sin \theta_n}
| c = {{Defof|Complex Number}}
}}
{{eqn | r = r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}
| c = rerranging
}}
{{end-eqn}}
By the definition of the complex modulus, with $z = x + i y$, $r$ is defined as:
:$r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}$
Hence
{{begin-eqn}}
{{eqn | l = r
| r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}
| c =
}}
{{eqn | l = r
| r = \sqrt {\paren {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n }^2 + \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}^2 }
| c =
}}
{{end-eqn}}
In the above we have two types of pairs of terms:
{{begin-eqn}}
{{eqn | n = 1
| q = 1 \le k \le n
| l = {r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2
| r = {r_k}^2 \paren {\cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2}
| c =
}}
{{eqn | r = {r_k}^2
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | n = 2
| q = 1 \le j < k \le n
| l = 2 r_j r_k \cos \theta_j \cos \theta_k + 2 {r_j} {r_k} \sin \theta_j \sin \theta_k
| r = 2 r_j r_k \paren {\cos \theta_j \cos \theta_k + \sin \theta_j \sin \theta_k}
| c =
}}
{{eqn | r = 2 r_j r_k \map \cos {\theta_j - \theta_k}
| c = Cosine of Difference
}}
{{end-eqn}}
Hence:
:$\ds r = \sqrt {\sum_{k \mathop = 1}^n r_k + \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \map \cos {\theta_j - \theta_k} }$
Note that $r > 0$ since $r_k > 0$ for all $k$.
Hence we may safely assume that $r > 0$ when determining the argument below.
By definition of the argument of a complex number, with $z = x + i y$, $\theta$ is defined as any solution to the pair of equations:
:$(1): \quad \dfrac x {\cmod z} = \map \cos \theta$
:$(2): \quad \dfrac y {\cmod z} = \map \sin \theta$
where $\cmod z$ is the modulus of $z$.
As $r > 0$ we have that $\cmod z \ne 0$ by definition of modulus.
Hence we can divide $(2)$ by $(1)$, to get:
{{begin-eqn}}
{{eqn | l = \map \tan \theta
| r = \frac y x
| c =
}}
{{eqn | r = \frac {\map \Im z} {\map \Re z}
| c =
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \theta
| r = \map \arctan {\frac {\map \Im {r e^{i \theta} } } {\map \Re {r e^{i \theta} } } }
| c =
}}
{{eqn | r = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22167
|
\section{Sum of Components of Equal Ratios}
Tags: Ratios
\begin{theorem}
{{:Euclid:Proposition/V/12}}
That is:
:$a_1 : b_1 = a_2 : b_2 = a_3 : b_3 = \cdots \implies \left({a_1 + a_2 + a_3 + \cdots}\right) : \left({b_1 + b_2 + b_3 + \cdots}\right)$
\end{theorem}
\begin{proof}
Let any number of magnitudes $A, B, C, D, E, F$ be proportional, so that:
:$A : B = C : D = E : F$
etc.
:450px
Of $A, C, E$ let equimultiples $G, H, K$ be taken, and of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.
We have that $A : B = C : D = E : F$.
Therefore:
:$G > L \implies H > M, K > N$
:$G = L \implies H = M, K = N$
:$G < L \implies H < M, K < N$
So, in addition:
:$G > L \implies G + H + K > L + M + N$
:$G = L \implies G + H + K = L + M + N$
:$G < L \implies G + H + K < L + M + N$
It follows from Multiplication of Numbers is Left Distributive over Addition that $G$ and $G + H + K$ are equimultiples of $A$ and $A + C + E$.
For the same reason, $L$ and $L + M + N$ are equimultiples of $B$ and $B + D + F$.
The result follows from {{EuclidDefLink|V|5|Equality of Ratios}}.
{{qed}}
{{Euclid Note|12|V}}
\end{proof}
|
22168
|
\section{Sum of Consecutive Triangular Numbers is Square}
Tags: Polygonal Numbers, Triangle Numbers, Triangular Numbers, Sum of Consecutive Triangular Numbers is Square, Square Numbers
\begin{theorem}
The sum of two consecutive triangular numbers is a square number.
\end{theorem}
\begin{proof}
Let $T_{n - 1}$ and $T_n$ be two consecutive triangular numbers.
From Closed Form for Triangular Numbers, we have:
:$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$
:$T_n = \dfrac {n \paren {n + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = T_{n - 1} + T_n
| r = \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2
| c =
}}
{{eqn | r = \frac {\paren {n - 1 + n + 1} n} 2
| c =
}}
{{eqn | r = \frac {2 n^2} 2
| c =
}}
{{eqn | r = n^2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22169
|
\section{Sum of Cosecant and Cotangent}
Tags: Trigonometric Identities
\begin{theorem}
:$\csc x + \cot x = \cot {\dfrac x 2}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \csc x + \cot x
| r = \frac 1 {\sin x} + \frac {\cos x} {\sin x}
| c = {{Defof|Cosecant}} and {{Defof|Cotangent}}
}}
{{eqn | r = \frac {1 + \cos x} {\sin x}
| c =
}}
{{eqn | r = \frac {2 \cos^2 {\frac x 2} } {2 \sin {\frac x 2} \cos {\frac x 2} }
| c = Double Angle Formula for Sine and Double Angle Formula for Cosine
}}
{{eqn | r = \frac {\cos {\frac x 2} } {\sin {\frac x 2} }
| c =
}}
{{eqn | r = \cot {\frac x 2}
| c = {{Defof|Cotangent}}
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
22170
|
\section{Sum of Cosets of Ideals is Sum in Quotient Ring}
Tags: Ring Operations on Coset Space of Ideal, Ideal Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\powerset R$ be the power set of $R$.
Let $J$ be an ideal of $R$.
Let $X$ and $Y$ be cosets of $J$.
Let $X +_\PP Y$ be the sum of $X$ and $Y$, where $+_\PP$ is the operation induced on $\powerset R$ by $+$.
The sum $X +_\PP Y$ in $\powerset R$ is also their sum in the quotient ring $R / J$.
\end{theorem}
\begin{proof}
As $\struct {R, +, \circ}$ is a ring, it follows that $\struct {R, +}$ is an abelian group.
Thus by Subgroup of Abelian Group is Normal, all subgroups of $\struct {R, +, \circ}$ are normal.
So from the definition of quotient group, it follows directly that $X +_\PP Y$ in $\powerset R$ is also the sum in the quotient ring $R / J$.
{{qed}}
\end{proof}
|
22171
|
\section{Sum of Cosines of Arithmetic Sequence of Angles}
Tags: Cosine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
\end{theorem}
\begin{proof}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles:
:$\displaystyle \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
From Euler's Formula, this can be expressed as:
:$\displaystyle \paren {\sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$
Equating real parts:
:$\displaystyle \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}$
{{qed}}
\end{proof}
|
22172
|
\section{Sum of Cosines of Arithmetic Sequence of Angles/Formulation 1}
Tags: Cosine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha}
| r = \cos \theta + \map \cos {\theta + \alpha} + \map \cos {\theta + 2 \alpha} + \map \cos {\theta + 3 \alpha} + \dotsb
}}
{{eqn | r = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1:
:$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$
Equating real parts:
:$\ds \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}$
{{qed}}
\end{proof}
|
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