hcju/proofwikips huggingface dataset:
Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals/Corollary
== Corollary to Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals ==  {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \left({j + 2}\right)}       | r = \frac 1 {1 \times 3} + \frac 1 {2 \times 4} + \frac 1 {3 \times 5} + \frac 1 {4 \times 6} + \cdots       | c =  }} {{eqn | r = \frac 3 4       | c =  }} {{end-eqn}}  == Proof == {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \left({j + 2}\right)}       | r = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \left({j + 2}\right)}       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \left({\frac 3 4 - \frac {2 n + 3} {2 \left({n + 1}\right) \left({n + 2}\right)} }\right)       | c = Sum of Sequence of Products of Consecutive Odd Reciprocals }} {{eqn | r = \frac 3 4 - \lim_{n \mathop \to \infty} \frac {\frac 2 n + \frac 3 {n^2} } {2 \left({1 + \frac 1 n}\right) \left({1 + \frac 2 n}\right)}       | c = dividing top and bottom by $n^2$ }} {{eqn | r = \frac 3 4       | c = Basic Null Sequences }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sum of Sequence of Products of Consecutive Odd Reciprocals/Corollary|next = Sum of Sequence of Products of Squares of Consecutive Odd Reciprocals}}: $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.31$
Binomial Theorem/Examples/4th Power of Sum
== Example of Use of Binomial Theorem ==  :$\paren {x + y}^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$  == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 4$.  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Difference|next = Fourth Power of Difference}}: $\S 2$: Special Products and Factors: $2.5$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Sum|next = Power Series Expansion of Reciprocal of 1 + x/Proof 2}}: $\S 20$: Binomial Series: $20.7$ * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Binomial Theorem/Examples/Cube of Sum|next = Definition:Pascal's Triangle/Historical Note}}: Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
Power Series Expansion of Reciprocal of Square of 1 + x
Power Series Expansion of Reciprocal of Square of 1 + x 0 68177 349714 2018-04-03T15:07:34Z Prime.mover 59 Created page with "== Theorem ==  Let $x \in \R$ such that $-1 < x < 1$.  Then:  {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty..." wikitext text/x-wiki == Theorem ==  Let $x \in \R$ such that $-1 < x < 1$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) x^k       | c =  }} {{eqn | r = 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - \cdots       | c =  }} {{end-eqn}}  == Proof 1 == {{:Power Series Expansion of Reciprocal of Square of 1 + x/Proof 1}} == Proof 2 == {{:Power Series Expansion of Reciprocal of Square of 1 + x/Proof 2}}
Harmonic Numbers/Examples/H1
== Example of Harmonic Number ==  :$H_1 = 1$  where $H_1$ denotes the first harmonic number. == Proof == {{begin-eqn}} {{eqn | l = H_1       | r = \sum_{k \mathop = 1}^1 \frac 1 k       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \frac 1 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Harmonic Numbers/Examples/H0|next = Harmonic Numbers/Examples/H2}}: $\S 1.2.7$: Harmonic Numbers: Exercise $1$
Harmonic Numbers/Examples/H2
== Example of Harmonic Number ==  :$H_2 = \dfrac 3 2$  where $H_2$ denotes the second harmonic number. == Proof == {{begin-eqn}} {{eqn | l = H_2       | r = \sum_{k \mathop = 1}^2 \frac 1 k       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \frac 1 1 + \frac 1 2       | c =  }} {{eqn | r = \dfrac 3 2       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Harmonic Number H1|next = Upper Bound for Harmonic Number}}: $\S 1.2.7$: Harmonic Numbers: Exercise $1$
Sum of Arithmetic Sequence/Examples/Sum of j from m to n
== Example of Sum of Arithmetic Sequence ==  {{begin-eqn}} {{eqn | l = \sum_{j \mathop = m}^n j       | r = m \paren {n - m + 1} + \frac 1 2 \paren {n - m} \paren {n - m + 1}       | c =  }} {{eqn | r = \frac {n \paren {n + 1} } 2 - \frac {\paren {m - 1} m} 2       | c =  }} {{end-eqn}}  == Proof 1 == {{:Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 1}} == Proof 2 == {{:Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 2}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Sum of Geometric Sequence/Examples/One Seventh from 1 to n|next = General Distributivity Theorem/Examples/Sum of j from m to n by Sum of k from r to s}}: $\S 1.2.3$: Sums and Products: Exercise $13$
Factorial/Examples/1
Factorial/Examples/1 0 69333 357503 2018-05-28T07:04:05Z Prime.mover 59 Created page with "== Example of Factorial ==  The factorial of $1$ is $1$: :$1! = 1$   == Proof ==  From the definit..." wikitext text/x-wiki == Example of Factorial ==  The factorial of $1$ is $1$: :$1! = 1$  == Proof == From the definition of factorial: :$n! = \displaystyle \prod_{k \mathop = 1}^n k$ where $\prod$ denotes product notation. When $n = 1$ we have: :$1! = \displaystyle \prod_{k \mathop = 1}^1 k$ {{begin-eqn}} {{eqn | l = 1!       | r = \prod_{k \mathop = 1}^1 k       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}}
Factors of Binomial Coefficient/Corollary 1
Factors of Binomial Coefficient/Corollary 1 0 69484 358308 2018-06-04T21:51:29Z Prime.mover 59 Created page with "== Theorem ==  For all $r \in \R, k \in \Z$: :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ from which: :$\dbinom r k = \dfrac r {r - k} \dbinom {r -..." wikitext text/x-wiki == Theorem ==  For all $r \in \R, k \in \Z$: :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ from which: :$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$)  == Proof == {{begin-eqn}} {{eqn | l = r \binom {r - 1} k       | r = r \frac {\left({r - 1}\right) \left({\left({r - 1}\right) - 1}\right) \cdots \left({\left({r - 1}\right) - k + 2}\right) \left({\left({r - 1}\right) - k + 1}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \frac {r \left({r - 1}\right) \left({r - 2}\right) \cdots \left({r - k + 1}\right) \left({r - k}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \left({r - k}\right) \frac {r \left({r - 1}\right) \left({r - 2}\right) \cdots \left({r - k + 1}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \left({r - k}\right) \binom r k       | c =  }} {{end-eqn}} {{qed|lemma}} Then: :$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ follows from the  :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ by dividing both sides by $r - k$.  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Factors of Binomial Coefficient|next = Pascal's Rule/Real Numbers}}: $\S 1.2.6$: Binomial Coefficients: $\text{C}$
Abel's Lemma/Formulation 1
== Lemma == Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$. Then:  :$\displaystyle \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$  Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$. === Corollary === {{:Abel's Lemma/Formulation 1/Corollary}} == Proof == {{begin-eqn}} {{eqn | l = \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k}        | r = \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_k b_k        | c = }} {{eqn | r = \sum_{k \mathop = m}^n a_k b_{k + 1} - \paren {a_m b_m + \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1} - a_{n + 1} b_{n + 1} }        | c = }} {{eqn | r = a_{n + 1} b_{n + 1} - a_m b_m + \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1}        | c = }} {{eqn | r = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}        | c = }} {{end-eqn}}  == Also reported as == Some sources give this as: :$\displaystyle \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n a_{k + 1} \paren {b_{k + 1} - b_k}$ which is obtained from the main result by interchanging $a$ and $b$. Others take the upper index to $n - 1$: :$\displaystyle \sum_{k \mathop = m}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_m b_m - \sum_{k \mathop = m}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$ == Also known as == {{:Abel's Lemma/Also known as}} {{Namedfor|Niels Henrik Abel}}
Abel's Lemma/Formulation 2
== Lemma == Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.  Let $\displaystyle A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$. Then: :$\displaystyle \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$  Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$. === Corollary === {{:Abel's Lemma/Formulation 2/Corollary}} == Proof 1 == {{:Abel's Lemma/Formulation 2/Proof 1}} == Proof 2 == {{:Abel's Lemma/Formulation 2/Proof 2}} == Also known as == {{:Abel's Lemma/Also known as}} {{Namedfor|Niels Henrik Abel}} == Sources == * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Symmetric Group is not Abelian|next = Definition:Abel Summation Method|entry = Abel's partial summation formula}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Abel's Limit Theorem|next = Definition:Abel Summation Method|entry = Abel's partial summation formula}}
Sum of Angles between Straight Lines at Point form Four Right Angles
== Corollary to Two Angles on Straight Line make Two Right Angles ==  If any number of straight lines are drawn from a given point, the sum of the consecutive angles so formed is $4$ right angles.  == Proof == Let $OA_1, OA_2, \ldots, OA_n$ be straight lines drawn from a point $O$ to points $A_1, A_2, \ldots, A_n$. Let $OA_1$ be produced past $O$ to $B$. Then $OB$ either coincides with $OA_j$ for some $j$ between $1$ and $n$, or $OB$ divides angle $A_j O A_k$ for some $j, k$ between $1$ and $n$. First suppose $OB$ coincides with $OA_j$. {{finish|etc.}} == Sources == * {{BookReference|Problems & Solutions in Euclidean Geometry|1968|M.N. Aref|author2 = William Wernick|prev = Two Angles on Straight Line make Two Right Angles|next = Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular}}: Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.1$: Corollary $1$
Characterization of Strictly Increasing Mapping on Woset
 == Lemma == Let $J$ and $E$ be well-ordered sets. Let $h: J \to E$ be a mapping. Let $S_\alpha$ denote an initial segment determined by $\alpha$.  :$(1):\quad$ $h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$ :$(2):\quad$ $\forall \alpha \in J: h\left({\alpha}\right) = \min \left({E\setminus h\left[{S_\alpha}\right]}\right)$, and $h[S_\alpha] = S_{h(\alpha)}$ where: :$h\left[{S_\alpha}\right]$ denotes the image of $S_\alpha$ under $h$ :$\min$ denotes the smallest element of the set. == Proof == === $(1)$ implies $(2)$ === Suppose $h$ satisfies: :$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$ Then for any $x,y \in J$: {{begin-eqn}} {{eqn | l = x       | o = \prec       | r = y }} {{eqn | ll = \implies       | l = h(x)       | o = \prec       | r = h(y)       | c = {{Defof|Strictly Increasing Mapping|strictly increasing}} }} {{eqn | l = h[S_y]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y) } \right\} }} {{eqn | r = S_{h(y)} }} {{eqn | l = \min\left({E \setminus h\left[{S_y}\right] }\right)       | r = \min\left({E \setminus S_{h(y)} }\right) }} {{eqn | r = h(y)       | c = {{Defof|Smallest Element|smallest}} and of initial segment }} {{end-eqn}} {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose $h$ satisfies: :$h(\alpha)  = \min\left({E \setminus h\left[{S_\alpha}\right] }\right)$ By the Principle of Recursive Definition for Well-Ordered Sets, $h$ is thus uniquely determined. Then: {{begin-eqn}} {{eqn | l = h(y)       | r = \min\left({E \setminus h\left[{S_y}\right] }\right) }} {{eqn | l = h[S_y]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) =              \min\left({E \setminus h\left[{S_y }\right] }\right)} \right\} }} {{eqn | r =  \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y)} \right\} }} {{eqn | r = S_{h(y)}       | c = {{Defof|Initial Segment|initial segment}} }} {{end-eqn}} Thus for every $x \in S_y$, we have that $h(x) \in S_{h(y)}$. Therefore $h$ is an strictly increasing mapping. Furthermore, the image set of $h$ is the union of initial segments in $E$. By Union of Initial Segments is Initial Segment or All of Woset, $h[J]$ is an initial segment of $E$ or all of $E$.   == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercise $1.2$
Equality to Initial Segment Imposes Well-Ordering
== Lemma == Let $X$ be a set. Let $\AA$ be the set of all ordered pairs $\struct {A, <}$ such that $A$ is a subset of $X$ and $<$ is a strict well-ordering of $A$. Define $\prec$ as: :$\struct {A, <} \prec \struct {A', <'}$   :$\struct {A, <}$ equals an initial segment of $\struct {A', <'}$. Let $\BB$ be a set of ordered pairs in $\AA$ such that $\BB$ is ordered by $\prec$. Let $B'$ be the union of the sets $B$ for all $\struct {B, <} \in \BB$. Let $<'$ be the union of the relations $<$ for all $\struct {B, <}$. Then $\struct {B', <'}$ is strictly well-ordered set. == Proof == If the set $X$ considered is empty or a singleton, the lemma holds vacuously or trivially. Thus assume $X$ contains at least two elements. We first prove that $\prec$ is a strict partial ordering on $\AA$. From the definition of initial segment, no $\struct {A, <}$ can equal an initial segment of itself. Thus $\prec$ is antireflexive. Suppose $\struct {A, <_A}$ equals an initial segment of $\struct {B, <_B}$ and $\struct {B, <_B}$ equals an initial segment of $\struct {C, <_C}$. Then $\struct {A, <_A}$ equals an initial segment of $\struct {C, <'}$ from Equality is Transitive. Thus $\prec$ is a strict partial ordering on $\AA$. We then prove that any $\struct {B', <'}$ is a strictly well-ordered set. Let $x_1,x_2 \in B'$. That is, let $x_i \in \struct {A_i, <_i}$ for $i = 1, 2$. Suppose $x_2 \prec x_1$. That is, $\struct {A_2, <_2}$ equals an initial segment in $\struct {A_1, <_1}$ By the definition of initial segment, both $x_1$ and $x_2$ are in $\struct {A_1, <_1}$. Thus $<'$ is connected, as all $<_i$ are strict well-orderings by hypothesis. For any $x_i \in \struct {A_i, <_i}$, $x_i \nprec x_i$ as all $<_i$ are strict well-orderings by hypothesis. Thus $<'$ is antireflexive. To show that $<'$ is transitive, consider $x_i \in \struct {B', <'}$ for $i = 1, 2, 3$. Suppose $x_1 <' x_2 <' x_3$. Then $x_1 <_1 x_2$ and $x_2 <_2 x_3$, from the definition of $<'$ as a union of relations $<_i$. Then $\struct {A_j, <_j}$ is an initial segment of $\struct {A_i, <_i}$ for $j  = 1, 2; j < i$ Thus $x_1 <_i x_2 <_i x_3$. Then $x_1 <_i x_3$, as all $<_i$ as all $<_i$ are strict well-orderings by hypothesis. Conclude that $<'$ is itself a strict ordering. It remains to be shown that $<'$ is a well-ordering. Let $A$ be an arbitrary non-empty subset of $B'$. Let $x \in A$ and $x \in \struct {B, <}$ for $\struct {B, <} \in \BB$. Then for all $y \in A$, $y <' x$  $y < x$ and $y \in B$. As $<$ is a well-ordering, $\struct {B \cap A, <}$ has a smallest element $b$. This $b$ is then a smallest element of $<'$ in $A$. Conclude that $<'$ is a strict well-ordering on $B'$.  == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercises $1.5$
Euler-Binet Formula/Corollary 1
== Corollary to Euler-Binet Formula ==  :$F_n = \dfrac {\phi^n} {\sqrt 5}$ rounded to the nearest integer  where: :$F_n$ denotes the $n$th Fibonacci number :$\phi$ denotes the golden mean. == Proof == By definition of $n$th Fibonacci number, $F_n$ is an integer. From Euler-Binet Formula: :$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} = \dfrac {\phi^n } {\sqrt 5} - \dfrac {\hat \phi^n} {\sqrt 5}$ But $\size {\dfrac {\hat \phi^n} {\sqrt 5} } < \dfrac 1 2$ for all $n \ge 0$. Thus $\dfrac {\phi^n } {\sqrt 5}$ differs from $F_n$ by a number less than $\dfrac 1 2$. Thus the nearest integer to $\dfrac {\phi^n } {\sqrt 5}$ is $F_n$.  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Euler-Binet Formula/Historical Note|next = Euler-Binet Formula/Corollary 1/Mistake}}: $5$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Golden Mean/One Minus Golden Mean/Decimal Expansion|next = Summation over k to n of Product of kth with n-kth Fibonacci Numbers}}: $\S 1.2.8$: Fibonacci Numbers: $(15)$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Euler-Binet Formula/Historical Note|next = Cassini's Identity}}: $5$
Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1
Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1 0 69897 360163 2018-07-08T11:52:00Z Prime.mover 59 Created page with "== Theorem ==  :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{..." wikitext text/x-wiki == Theorem ==  :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} F_k = F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} {k - 1}_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} = 0$  where: :$\dbinom m k_\mathcal F$ denotes a Fibonomial coefficient :$F_{n + k}$ denotes the $n + k$th Fibonacci number :$\left\lceil{\, \cdot \,}\right\rceil$ denotes the ceiling function == Proof ==  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers|next = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $30$: Solution: $\text{(a)}$
Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2
Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2 0 69904 360188 2018-07-08T13:24:12Z Prime.mover 59 Created page with "== Theorem ==  :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{..." wikitext text/x-wiki == Theorem ==  :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} \left({-1}\right)^k F_{m - k} = \left({-1}\right)^m F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - 1 - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} = 0$  where: :$\dbinom m k_\mathcal F$ denotes a Fibonomial coefficient :$F_{n + k}$ denotes the $n + k$th Fibonacci number :$\left\lceil{\, \cdot \,}\right\rceil$ denotes the ceiling function == Proof ==  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1|next = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Examples/m = 3}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $30$: Solution: $\text{(b)}$
Addition of 1 in Golden Mean Number System
== Algorithm == Let $x \in \R$ be a real number. The following algorithm performs the operation of addition of $1$ to $x$ in the golden mean number system. Let $S$ be the representation of $x$ in the golden mean number system in its simplest form. :'''Step $1$''': Is the digit immediately to the left of the radix point a zero? :: If '''Yes''', replace that $0$ with $1$. Go to '''Step $4$'''. :: If '''No''', set $m = 2$ and go to '''Step $2$'''. :'''Step $2$''': Does the $m$th place after the radix point contain $0$? :: If '''Yes''', expand the $100$ in the $3$ places ending in the $m$th place with $011$. Subtract $2$ from $m$. Go to '''Step $3$'''. :: If '''No''', add $2$ to $m$. Repeat '''Step $2$'''. :'''Step $3$''': Is $m = 0$? :: If '''Yes''', set the digit immediately to the left of the radix point from $0$ to $1$. Go to '''Step $4$'''. :: If '''No''', go to '''Step $2$'''. :'''Step $4$''': Convert $S$ to its simplest form. '''Stop'''. == Proof == The above constitutes an algorithm, for the following reasons: === Finiteness === The only case in which it is possible for the process not to terminate is if the $m$th place never contains $0$. This can only happen if $S$ ends in an infinite string $01010101 \ldots$ But if this is the case, $S$ is not in its simplest form. === Definiteness === Each step can be seen to be precisely defined. === Inputs === The only input to the algorithm is the representation $S$ of $x$. === Outputs === The only output from the algorithm is the representation $S$ of $x + 1$. All operations that change $S$ are of the following nature: :$(1): \quad$ Simplification of $S$, which does not change $x$, which happens if at all in '''Step $4$. :$(2): \quad$ Expansion of $S$, which does not change $x$, which happens if at all in ''Step $2$. :$(3): \quad$ Setting the digit corresponding to $\phi^0$ to $1$ from $0$, which happens either in '''Step $1$''' or in '''Step $3$'''. :::In either step, it happens only once, after which the algorithm terminates. === Effective === Each step is basic enough to be done exactly and predictably.  == Sources == * {{citation|author = George Bergman|title = Number System with an Irrational Base|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 31|issue = 2|date = 1957|startpage = 98|endpage = 110|jstor = 3029218}}
Fibonacci String/Examples/S3
Fibonacci String/Examples/S3 0 69980 360465 2018-07-14T08:42:36Z Prime.mover 59 Created page with "== Example of Fibonacci Strings ==  The Fibonacci string $S_3$ is $\text{ba}$.   ==..." wikitext text/x-wiki == Example of Fibonacci Strings ==  The Fibonacci string $S_3$ is $\text{ba}$.  == Proof == By definition of Fibonacci string: {{begin-eqn}} {{eqn | l = S_1       | r = \text{a} }} {{eqn | l = S_2       | r = \text{b} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = S_3       | r = S_2 S_1       | c =  }} {{eqn | r = \text{ba}       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Fibonacci String|next = Fibonacci String/Examples/S4}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $36$