hcju/proofwikips huggingface dataset:
Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries
== Example of Ideal of Ring ==  Let $R$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ with $x, y, z \in \R$. Let $S$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & 0 \end{pmatrix}$ with $x, y \in \R$. Then $R$ is a ring and $S$ is an ideal of $R$.  === Corollary === {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Corollary}} == Proof 1 == {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 1}} == Proof 2 == {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 2}}
Cyclotomic Ring/Examples/5th
== Examples of Cyclotomic Rings ==  The '''$5$th cyclotomic ring''' is the algebraic structure: :$\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ where $\Z \sqbrk {i \sqrt 5}$ is the set $\set {a + i b \sqrt 5: a, b \in \Z}$. $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring.  == Proof == We have that $\Z \sqbrk {i \sqrt 5}$ is a subset of the Field of Complex Numbers $\struct {\C, +, \times}$. So to prove that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring it is sufficient to demonstrate that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ fulfils the conditions of the Subring Test. First we note that setting $a = 1, b = 0$ we have that $1 + 0 i \in \Z \sqbrk {i \sqrt 5}$ and so $\Z \sqbrk {i \sqrt 5} \ne \O$. Let $z_1 = a_1 + i b_1 \sqrt 5$ and $z_2 = a_2 + i b_2 \sqrt 5$ be arbitrary elements of $\Z \sqbrk {i \sqrt 5}$ Then: {{begin-eqn}} {{eqn | l = z_1 - z_2       | r = \paren {a_1 + i b_1 \sqrt 5} - \paren {a_2 + i b_2 \sqrt 5}       | c =  }} {{eqn | r = \paren {a_1 - a_2} + i \paren {b_1 - b_2} \sqrt 5       | c = {{Defof|Complex Addition}} }} {{eqn | o = \in       | r = \Z \sqbrk {i \sqrt 5}       | c = as $a_1 - a_2$ and $b_1 - b_2$ are both integers }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = z_1 z_2       | r = \paren {a_1 + i b_1 \sqrt 5} \paren {a_2 + i b_2 \sqrt 5}       | c =  }} {{eqn | r = \paren {a_1 a_2 - 5 b_1 b_2} + i \sqrt 5 \paren {a_1 b_2 + a_2 b_1}       | c = {{Defof|Complex Multiplication}} }} {{eqn | o = \in       | r = \Z \sqbrk {i \sqrt 5}       | c = as $a_1 a_2 - 5 b_1 b_2$ and $a_1 b_2 + a_2 b_1$ are both integers }} {{end-eqn}} The Subring Test is satisfied, and so $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Field Norm of Complex Number is Multiplicative Function|next = Field Norm on 5th Cyclotomic Ring}}: Chapter $9$: Rings: Exercise $19$
Matrix Entrywise Addition/Examples/Real 2 x 2
Matrix Entrywise Addition/Examples/Real 2 x 2 0 76177 399417 2019-04-07T13:43:52Z Prime.mover 59 Created page with "== Example of Matrix Entrywise Addition ==  Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf..." wikitext text/x-wiki == Example of Matrix Entrywise Addition ==  Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} w & x \\ y & z \end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix sum of $\mathbf A$ and $\mathbf B$ is given by: :$\mathbf A + \mathbf B = \begin {pmatrix} p + w & q + x \\ r + y & s + z \end {pmatrix}$  == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Square Matrix/Examples/Real 2 x 2|next = Matrix Product (Conventional)/Examples/2 x 2 Real Matrices}}: Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$
Matrix Product (Conventional)/Examples/2 x 2 Real Matrices
== Example of (Conventional) Matrix Product ==  Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} w & x \\ y & z \end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix product of $\mathbf A$ with $\mathbf B$ is given by: :$\mathbf A \mathbf B = \begin {pmatrix} p w + q y & p x + q z \\ r w + s y & r x + s z \end {pmatrix}$  == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Matrix Entrywise Addition/Examples/Real 2 x 2|next = Ring of Square Matrices over Real Numbers/Examples/2 x 2}}: Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$
Irreducible Polynomial/Examples/8 x^3 - 6 x - 1 in Rationals
== Examples of Irreducible Polynomials ==  Consider the polynomial: :$\map P x = 8 x^3 - 6 x - 1$ over the ring of polynomials $\Q \sqbrk X$ over the rational numbers. Then $\map P x$ is irreducible.  == Proof ==  $\map P x$ has proper factors. Then one of these has to be of degree $1$. Thus from Factors of Polynomial with Integer Coefficients have Integer Coefficients we have: :$8 x^3 - 6 x - 1 = \paren {a x + b} \paren {c^2 + d x + e}$ for some $a, b, c, d, e \in \Z$. Hence: {{begin-eqn}} {{eqn | l = a c       | r = 8       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \divides       | r = 8       | c =  }} {{eqn | l = b e       | r = -1       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \divides       | r = 1       | c =  }} {{end-eqn}} The only possible degree $1$ factors with integer coefficients are: :$x \pm 1, 2 x \pm 1, 4 x \pm 1, 8 x \pm 1$ By trying each of these possibilities, it is determined that no integer value of $d$ gives the correct values. Hence the result.  == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Polynomial which is Irreducible over Integers is Irreducible over Rationals|next = Square Root of Prime is Irrational/Proof 2}}: Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Example $59$
Vector Space over Division Subring/Examples/Real Numbers in Complex Numbers
== Example of Vector Space over Division Subring ==  Consider the field of complex numbers $\struct {\C, +, \times}$, which is a ring with unity whose unity is $1$. Consider the field of real numbers $\struct {\R, +, \times}$, which is a division subring of $\struct {\C, +, \circ}$ such that $1 \in \R$. Then $\struct {\C, +, \times_\R}_\R$ is an $\R$-vector space, where $\times_\R$ is the restriction of $\times$ to $\R \times \C$. $\struct {\C, +, \times_\R}_\R$ is of dimension $2$. The set $\paren {1 + 0 i, 0 + i}$ forms a basis of $\struct {\C, +, \times_\R}_\R$, as do any two complex numbers which are not real multiples of each other.  == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Vector Space over Division Subring is Vector Space|next = Definition:Generator of Vector Space}}: Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $64$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Dimension of Vector Space of Polynomial Functions|next = Linearly Independent Set is Contained in some Basis/Proof 2}}: Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $68$
Smallest Field containing Subfield and Complex Number/Examples/Numbers of Type Rational a plus b root 2
== Example of Smallest Field containing Subfield and Complex Number ==  Let $\Q \sqbrk {\sqrt 2}$ denote the set: :$\Q \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Q}$ that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers. Then $\Q \sqbrk {\sqrt 2}$ is the smallest field containing $\Q$ and $\sqrt 2$.  {{refactor|The following material needs to go on a page explaining splitting fields.}} Formally, $\Q \sqbrk {\sqrt 2}$ is the field extension of $\Q$ for the minimal polynomial of $\sqrt 2$, the second-degree polynomial $x^2 - 2$. Therefore, $\Q \sqbrk {\sqrt 2}$ is the vector space of dimension $2$ isomorphic to $\Q \sqbrk x / \left\langle x^2 - 2 \right\rangle$. == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Definition:Smallest Field containing Subfield and Complex Number/General Definition|next = Smallest Field containing Subfield and Complex Number/Examples/Complex Numbers}}: Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $72$ * {{BookReference|Contemporary Abstract Algebra|2017|Joseph A. Gallian|ed = 9th|edpage = Ninth Edition|prev = Definition:Vector Space|next = Definition:Algebraic Field Extension}}: Chapter $20$: Extension Fields: $\S 1$. Splitting Fields
Algebraic Number/Examples/Root of (2 plus Root 3)
Algebraic Number/Examples/Root of (2 plus Root 3) 0 76862 403850 2019-05-02T17:55:58Z Prime.mover 59 Created page with "== Example of Algebraic Number ==  :$\sqrt {2 + \sqrt 3}$ is an algebraic number. ..." wikitext text/x-wiki == Example of Algebraic Number ==  :$\sqrt {2 + \sqrt 3}$ is an algebraic number.  == Proof == Let $x = \sqrt {2 + \sqrt 3}$. We have that: {{begin-eqn}} {{eqn | l = x^2       | r = 2 + \sqrt 3       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 - 2}^2       | r = 3       | c =  }} {{eqn | ll= \leadsto       | l = x^4 - 4 x^2 + 4       | r = 3       | c =  }} {{eqn | ll= \leadsto       | l = x^4 - 4 x^2 + 1       | r = 0       | c =  }} {{end-eqn}} Thus $\sqrt {2 + \sqrt 3}$ is a root of $x^4 - 4 x^2 + 1 = 0$. Hence the result by definition of algebraic number.   == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Algebraic Number/Examples/Imaginary Unit|next = Pi is Transcendental}}: Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$
Group of Order 30 is not Simple
Group of Order 30 is not Simple 0 76938 404179 2019-05-04T14:22:49Z Prime.mover 59 Created page with "== Theorem ==  Let $G$ be a group of order $30$.  Then $G$ is not simple. </onlyinc..." wikitext text/x-wiki == Theorem ==  Let $G$ be a group of order $30$. Then $G$ is not simple.  == Proof == From Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$. Hence the result, by definition of simple group.  == Sources == * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Prime Group is Simple|next = Group of Order p^2 q is not Simple}}: Chapter $2$: The Sylow Theorems: $\S 59 \epsilon$
Bessel Function of the First Kind/Instances/Order 0
== Specific Instance of Bessel Functions of the First Kind == Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.  {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac x 2}^{2 k}       | c =  }} {{eqn | r = 1 - \dfrac {x^2} {2^2} + \dfrac {x^4} {2^2 \times 4^2} - \dfrac {x^6} {2^2 \times 4^2 \times 6^2} + \dotsb       | c =  }} {{end-eqn}}  == Proof == From Series Expansion of Bessel Function of the First Kind: {{begin-eqn}} {{eqn | l = \map {J_n} x       | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}       | c =  }} {{end-eqn}} where $n$ is not a (strictly) negative integer. $0$ fits that category, and so: {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {0 + k + 1} } \paren {\dfrac x 2}^{0 + 2 k}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1} } \paren {\dfrac x 2}^{2 k}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac x 2}^{2 k}       | c = Gamma Function Extends Factorial }} {{end-eqn}} Or working directly upon the terms themselves: {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \dfrac {x^0} {2^0 \, \map \Gamma {0 + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 \times 0 + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 \times 0 + 2} \paren {2 \times 0 + 4} } - \cdots}       | c =  }} {{eqn | r = \dfrac 1 {1 \times \map \Gamma 1} \paren {1 - \dfrac {x^2} {2 \times 2} + \dfrac {x^4} {2 \times 4 \times 2 \times 4} - \cdots}       | c =  }} {{eqn | r = 1 - \dfrac {x^2} {2^2} + \dfrac {x^4} {2^2 \times 4^2} - \cdots       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Bessel Function of the First Kind for Negative Non-Integer Order is Unbounded|next = Bessel Function of the First Kind/Instances/Order 1}}: $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.5$
Laplace Transform of Dirac Delta Function/Lemma
Laplace Transform of Dirac Delta Function/Lemma 0 77407 406282 2019-06-03T06:43:29Z Prime.mover 59 Created page with "== Lemma for Laplace Transform of Dirac Delta Function ==  Let $F_\epsilon: \R \to \R$ be the real function defined as:  :$\map {..." wikitext text/x-wiki == Lemma for Laplace Transform of Dirac Delta Function ==  Let $F_\epsilon: \R \to \R$ be the real function defined as: :$\map {F_\epsilon} t = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le t \le \epsilon \\ 0 & : t > \epsilon \end{cases}$ Then: :$\laptrans {\map {F_\epsilon} t} = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}$  == Proof == {{begin-eqn}} {{eqn | l = \laptrans {\map {F_\epsilon} t}       | r = \int_0^\infty e^{-s t} \map {F_\epsilon} t \rd t       | c =  }} {{eqn | r = \int_0^\epsilon e^{-s t} \map {F_\epsilon} t \rd t + \int_\epsilon^\infty e^{-s t} \map {F_\epsilon} t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^\epsilon e^{-s t} \dfrac 1 \epsilon \rd t + \int_\epsilon^\infty  e^{-s t} \times 0 \rd t       | c = Definition of $F_\epsilon$ }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon e^{-s t} \rd t       | c = simplification }} {{eqn | r = \dfrac 1 \epsilon \intlimits {\dfrac {e^{-s t} } {-s} } 0 \epsilon       | c = Primitive of $e^{a x}$ }} {{eqn | r = \dfrac 1 \epsilon \paren {\dfrac {e^{-s \epsilon} - e^{-s \times 0} } {-s} }       | c =  }} {{eqn | r = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}       | c = simplification }} {{end-eqn}}  == Sources == * {{BookReference|Theory and Problems of Laplace Transforms|1965|Murray R. Spiegel|prev = Laplace Transform of Heaviside Step Function/Proof 2|next = Laplace Transform of Dirac Delta Function/Proof 2}}: Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $41$
Infimum of Subset of Real Numbers/Examples/Example 3
== Example of Infimum of Subset of Real Numbers ==  The subset $V$ of the real numbers $\R$ defined as: :$V := \set {x \in \R: x > 0}$ admits an infimum: :$\inf V = 0$  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Supremum of Subset of Real Numbers/Examples/Example 3|next = Definition:Supremum of Subset of Real Numbers}}: $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(iii)}$
Real Null Sequence/Examples/n^alpha x^n/Lemma
== Lemma for Real Null Sequence: $n^\alpha x^n$ == Let $\alpha \in \Q$ be a (strictly) positive rational number. Let $x \in \R$ be a real number such that $\size x < 1$.  There exists $N \in \N$ such that: :$\paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$  == Proof == : :$\forall n \in \N: \paren {1 + \dfrac 1 n}^{\alpha + 1} \, \size x > 1$ Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N:       | l = \dfrac 1 n       | o = >       | r = \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} } - 1       | c =  }} {{eqn | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} }       | o = >       | r = 1       | c = as $\dfrac 1 n > 0$ for all $n \in \N$ }} {{end-eqn}} But this contradicts Sequence of Powers of Reciprocals is Null Sequence. Hence by Proof by Contradiction: :$\exists N \in \N: \paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Power over Factorial|next = Real Null Sequence/Examples/n^alpha x^n}}: $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (5)$
Hero's Method/Lemma 1
== Lemma for Hero's Method == Let $a \in \R$ be a real number such that $a > 0$. Let $x_1 \in \R$ be a real number such that $x_1 > 0$. Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by: :$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$ Then:  :$\forall n \in \N_{>0}: x_n > 0$  == Proof == The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$x_n > 0$ === Basis for the Induction === $\map P 1$ is the case: :$x_1 > 0$ which is assumed. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$x_k > 0$ from which it is to be shown that: :$x_{k + 1} > 0$ === Induction Step === This is the induction step: We have that: :$x_{k + 1} = \dfrac {x_k + \dfrac a {x_k} } 2$ But as $x_k > 0$ and $a > 0$, it follows that: :$\dfrac a {x_k} > 0$ Then as $x_k > 0$ and $\dfrac a {x_k} > 0$, it follows that: :$\dfrac 1 2 \paren {x_k + \dfrac a {x_k} } > 0$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: x_n > 0$  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Hero's Method|next = Hero's Method/Lemma 2}}: $\S 5$: Subsequences: $\S 5.5$: Example
Hero's Method/Lemma 2
== Lemma for Hero's Method == Let $a \in \R$ be a real number such that $a > 0$. Let $x_1 \in \R$ be a real number such that $x_1 > 0$. Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by: :$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$ Then:  :$\forall n \ge 2: x_n \ge \sqrt a$  == Proof == === Lemma 1 === {{:Hero's Method/Lemma 1}}{{qed|lemma}} We have: {{begin-eqn}} {{eqn | l = x_{n + 1}       | r = \frac {x_n + \dfrac a {x_n} } 2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 2 x_n x_{n + 1}       | r = x_n^2 + a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x_n^2 - 2 x_n x_{n + 1} + a       | r = 0       | c =  }} {{end-eqn}} This is a quadratic equation in $x_n$. We know that this equation must have a real solution with respect to $x_n$, because $x_n$ has been explicitly constructed by the iterative process. Thus its discriminant is $b^2 - 4 a c \ge 0$, where: :$a = 1$ :$b = -2 x_{n + 1}$ :$c = a$ Thus $x_{n + 1}^2 \ge a$. From Lemma 1: :$x_{n + 1} > 0$  It follows that: :$\forall n \ge 1: x_{n + 1} \ge \sqrt a$ for $n \ge 1$ Thus: :$\forall n \ge 2: x_n \ge \sqrt a$ for $n \ge 2$  == Sources == {{SourceReview|username = Prime.mover}} * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Hero's Method/Lemma 1|next = Hero's Method/Proof 1}}: $\S 5$: Subsequences: $\S 5.5$: Example
Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 1
== Example of Convergent Real Sequence == Let $\sequence {x_n}$ be the real sequence defined as: :$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$ where: :$0 < k < \dfrac 1 4$ :$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation $x^2 - x + k = 0$. Then:  :$\forall n \in \N_{>0}: a < x_n < b$  == Proof == The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$a < x_n < b$ === Basis for the Induction === $\map P 1$ is the case: :$a < x_1 < b$ By assertion: :$a < h < b$ and: :$x_1 = h$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$a < x_k < b$ from which it is to be shown that: :$a < x_{k + 1} < b$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = x_{k + 1} - a       | r = {x_k}^2 + k - a       | c = Definition of $x_{k + 1}$ }} {{eqn | o = >       | r = a^2 - a + k       | c = as $x_k > a$ }} {{eqn | r = 0       | c = Definition of $a$ }} {{eqn | ll= \leadsto       | l = x_{k + 1}       | o = >       | r = a       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = x_{k + 1} - b       | r = {x_k}^2 + k - b       | c = Definition of $x_{k + 1}$ }} {{eqn | o = <       | r = b^2 - b + k       | c = as $x_k < b$ }} {{eqn | r = 0       | c = Definition of $b$ }} {{eqn | ll= \leadsto       | l = x_{k + 1}       | o = <       | r = b       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: a < x_n < b$  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k|next = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$
Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2
== Example of Convergent Real Sequence ==  Let $a$ and $b$ be the roots of the quadratic equation: :$(1): \quad x^2 - x + k = 0$ Let: :$0 < k < \dfrac 1 4$ Then $a$ and $b$ are both strictly positive real numbers.  == Proof == First we investigate the consequences of the condition $k < \dfrac 1 4$. By Solution to Quadratic Equation with Real Coefficients: :In order for the quadratic equation $a x^2 + b x + c$ to have real roots, its discriminant $b^2 - 4 a c$ needs to be strictly positive. The discriminant $D$ of $(1)$ is: {{begin-eqn}} {{eqn | l = D       | r = \paren {-1}^2 - 4 \times 1 \times k       | c =  }} {{eqn | r = 1 - 4 k       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = k       | o = <       | r = \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = 1 - 4 k       | o = >       | r = 0       | c =  }} {{end-eqn}} and so when $k < \dfrac 1 4$, $(1)$ has real roots. {{qed|lemma}} Next we investigate the consequences of the condition $0 < k$. By Solution to Quadratic Equation: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}       | c = where $b = -1$, $a = 1$, $c = k$ in $(1)$ }} {{eqn | r = \dfrac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times k} } {2 \times 1}       | c =  }} {{eqn | r = \dfrac {1 \pm \sqrt {1 - 4 k} } 2       | c =  }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = 1 - 4 k       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = +\sqrt {1 - 4 k}       | c =  }} {{eqn | l = -1       | o = <       | r = -\sqrt {1 - 4 k}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {1 \pm \sqrt {1 - 4 k} } 2       | o = >       | r = 0       | c =  }} {{end-eqn}} That is, when $0 < k$ both roots of $(1)$ are strictly positive. {{qed|lemma}} Hence when $0 < k < \dfrac 1 4$, both roots of $(1)$ are strictly positive real numbers.  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 1|next = Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$