hcju/proofwikips huggingface dataset:
Up-Complete Product/Lemma 1
Up-Complete Product/Lemma 1 0 47356 256329 2016-05-19T16:27:54Z GrzegorzBancerek 2860 Created page with "== Theorem == {{:Up-Complete Product}}  Let $X$ be a directed subset of $S$.  Let $Y$ be a Definition:Directed Subset|directed su..." wikitext text/x-wiki == Theorem == {{:Up-Complete Product}}  Let $X$ be a directed subset of $S$. Let $Y$ be a directed subset of $T$. Then $X \times Y$ is also a directed subset of $S \times T$.  == Proof == Let $\left({s_1, t_1}\right)$, $\left({s_2, t_2}\right) \in X \times Y$. By definition of Cartesian product: :$s_1, s_2 \in X$ and $t_1, t_2 \in Y$ By definition of directed subset: :$\exists h_1 \in X: s_1 \preceq_1 h_1 \land s_2 \preceq_1 h_1$ and :$\exists h_2 \in X: t_1 \preceq_2 h_2 \land t_2 \preceq_2 h_2$ By definition of Cartesian product of ordered sets: :$\exists \left({h_1, h_2}\right) \in X \times Y: \left({s_1, t_1}\right) \preceq \left({h_1, h_2}\right) \land \left({s_2, t_2}\right) \preceq \left({h_1, h_2}\right)$ Thus by definition: :$X \times Y$ is a directed subset of $S \times T$.  == Sources == * {{BookReference|A Compendium of Continuous Lattices|1980|G. Gierz|author2 = K.H. Hofmann|author3 = K. Keimel|author4 = J.D. Lawson|author5 = M.W. Mislove|author6 = D.S. Scott}} * {{Mizar|link = yellow_3|sublink = FC9|display = YELLOW_3:funcreg 9}}
Dirichlet's Theorem on Arithmetic Sequences/Lemma 1
== Lemma for Dirichlet's Theorem on Arithmetic Sequences == Let $a, q$ be coprime integers. Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$. {{MissingLinks|trivial character}}  Let $\chi$ be a Dirichlet character modulo $q$. Let: :$\displaystyle \map f s = \sum_p \map \chi p p^{-s}$ If $\chi$ is non-trivial then $\map f s$ is bounded as $s \to 1$. If $\chi$ is the trivial character then: :$\map f s \sim \map \ln {\dfrac 1 {s - 1} }$ as $s \to 1$.  == Proof == By Logarithm of Dirichlet L-Functions: :$(1):\quad \displaystyle \sum_p \map \chi p p^{-s} = \ln \map L {s, \chi} - \sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} }$ If $\chi$ is non-trivial, then by L-Function does not Vanish at One, $\ln \map L {s, \chi}$ is bounded as $s \to 1$. If $\chi$ is trivial, then by Analytic Continuation of Dirichlet L-Function, $\map L {s, \chi}$ has a simple pole at $s = 1$. Therefore, in this case: :$\map L {s, \chi} \sim \dfrac \lambda {s - 1}$ where $\lambda$ is the residue of $\map L {s, \chi}$ at $1$, and: :$\ln \map L {s, \chi} \sim \map \ln {\dfrac \lambda {s - 1} } \sim \map \ln {\dfrac 1 {s - 1} }$ Thus if we can show that the second term of $(1)$ is bounded, the result holds. On $\map \Re s > 1$: {{begin-eqn}} {{eqn | l = \size {\sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} } }       | o = \le       | r = \sum_p \sum_{n \mathop \ge 2} \frac 1 {\size {p^s}^n} }} {{eqn | r = \sum_p \frac 1 {\size {p^{2 s} } \, \size {p^s - 1}^n}       | c = Sum of Infinite Geometric Sequence }} {{eqn | o = \le       | r = \sum_p \frac 1 {p^2}       | c = because $\map \Re s > 1$ }} {{eqn | o = \le       | r = \sum_n \frac 1 {n^2} }} {{end-eqn}} This last is $\map \zeta 2$ where $\zeta$ is the Riemann zeta function, so is finite.
Dirichlet's Theorem on Arithmetic Sequences/Lemma 2
== Lemma for Dirichlet's Theorem on Arithmetic Sequences == Let $a, q$ be coprime integers. Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$. Define: :$\eta_{a, q} : n \mapsto \begin{cases} 1 & : n \equiv a \pmod q\\ 0 & : \text{otherwise} \end{cases}$  Let $G = \paren {\Z / q \Z}^\times$. Let $G^*$ be the dual group of characters on $G$. {{explain|Explain the notation $\paren {\Z / q \Z}^\times$}} Then for all $n \in \N$: :$\displaystyle \map {\eta_{a, q} } n = \sum_{\chi \mathop \in G^*} \frac {\map {\overline \chi} a} {\map \phi q} \, \map \chi n$  == Proof == There is only one $x \in G$ such that $\map \eta x \ne 0$, and this equals $\map \eta a = 1$. {{explain|Back up the above statement by a link or an explanation. Are we sure that $\map \eta x$ means $\map {\eta_{a, q} } x$ here?}} So: :$\displaystyle \sum_{x \mathop \in G} \map {\eta_{a, q} } x \, \map {\overline \chi} x = \map {\overline \chi} a$ Therefore, by Discrete Fourier Transform on Abelian Group we have for all $x \in G$: :$\displaystyle \map \eta x = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \map {\overline \chi} a \, \map \chi x$ as required.
Squaring the Circle
== Classic Problem == '''Squaring the Circle''' is creating a square with the same area as that of a given circle. == Solution == The problem of '''squaring the circle''' reduces to: :starting with a line segment of given length ($1$, say) :constructing a line segment of length $\sqrt \pi$ of the first. == Fallacious Proofs == {{:Squaring the Circle/Fallacious Proofs}} == Also see == * Squaring the Circle by Archimedean Spiral * Squaring the Circle by Compass and Straightedge Construction is Impossible * Trisecting the Angle * Doubling the Cube == Historical Note == {{:Squaring the Circle/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Doubling the Cube|next = Trisecting the Angle/Historical Note}}: Chapter $\text{II}$: Modern Minds in Ancient Bodies * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Compass and Straightedge Construction|next = Squaring the Circle by Compass and Straightedge Construction is Impossible/Historical Note}}: Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Positive Square Root|next = Squaring the Circle/Historical Note|entry = squaring the circle}}
Liouville's Constant is Transcendental/Corollary
== Corollary to Liouville's Constant is Transcendental ==  All real numbers of the form: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }       | r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots       | c =  }} {{end-eqn}} where :$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$ are transcendental.  == Proof == Let $n \in \N$. For $n = 1$, let $p = a_1$ and $q = 10$. Then: :$\size {L - \dfrac p q} = \displaystyle \sum_{k \mathop \ge 2} \dfrac {a_k} {10^{k!} } < \dfrac 1 {10} = \dfrac 1 q$ For $n > 1$, let $q = 10^{n!}$ and write: :$\displaystyle L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }$ for some suitable $p \in \Z$. Then: {{begin-eqn}} {{eqn | l = \size {L - \frac p q}       | r = \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }       | c =   }} {{eqn | r = \sum_{k \mathop = n + 1}^\infty \frac 9 {10^{k!} }       | o = \le       | c =  }} {{eqn | r = \frac {18} {10^{\paren {n + 1}!} }       | o = \le       | c =  }} {{eqn | r = \frac {18} {q^{n + 1} }       | c =  }} {{eqn | r = \frac 1 {q^n}       | o = <       | c = as $q \ge 100$ for all $n > 1$ }} {{end-eqn}} Thus, by definition, $L$ is a Liouville number. Therefore, by Liouville's Theorem, $L$ is transcendental.  == Sources == * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Liouville's Constant is Transcendental|next = Euler's Number is Transcendental/Historical Note}}: Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Product of Sums/Corollary
== Corollary to Product of Sums ==  Let $\displaystyle \sum_{i \mathop \in X} a_{ij}$ be absolutely convergent sequences for all $j \in Y$. Then: :$\displaystyle \prod_{j \mathop \in Y} \left({\sum_{i \mathop \in X} a_{ij}}\right) = \sum_{f: Y \mathop \to X} \left( \prod_{j \mathop \in Y} a_{f \left({j}\right) j} \right)$ where $f$ runs over all mappings from $Y$ to $X$.  == Proof == We will prove the case $X = Y = \N$ to avoid the notational inconvenience of enumerating the elements of $Y$ as $j_1, j_2, j_3 \dots$. The general case where $X, Y$ are arbitrary sets has the same proof, but with more indices and notational distractions. Consider that by the main theorem: :$\displaystyle \prod_{j \mathop = 1, 2} \left({\sum_{i \mathop \in \N} a_{ij} }\right) = \sum_{x, y \mathop \in \N} a_{x_1}a_{y_2}$ and continuing in this vein: :$\displaystyle \prod_{j \mathop = 1, 2, 3} \left({\sum_{i \mathop \in \N} a_{ij} }\right) = \left({\sum_{x, y \mathop \in \N} a_{x_1} a_{y_2} }\right) \left({\sum_{z \mathop \in \N} a_{z_3} }\right) = \sum_{x, y, z \mathop \in \N} a_{x_1} a_{y_2} a_{z_3}$ For an inductive proof of this concept for finite $n$, we assume that for some $n \in \N$: :$\displaystyle \prod_{j \mathop = 1}^n \left({ \sum_{i \mathop \in \N} a_{ij} }\right)  = \sum_{u, v, \dots, x, y \mathop \in \N} a_{u_1} a_{v_2}\dots a_{x_{(n-1)}} a_{y_n}$ Then: :$\displaystyle \prod_{j \mathop = 1}^{n+1} \left({ \sum_{i \mathop \in \N} a_{ij} }\right) = \left({ \sum_{u, v, \dots, x, y  \mathop \in \N} a_{u_1} a_{v_2}\dots a_{x_{(n-1)}} a_{y_n} }\right) \left({\sum_{z \mathop \in \N} a_{z_n} }\right)$ which by Product of Sums is simply: :$\displaystyle \sum_{u, v, \ldots, x, y, z \mathop \in \N} a_{u_1} a_{v_2} \ldots a_{x_{(n-1}} a_{y_n} a_{z_{(n+1)}}$ completing the induction for finite $n$.
Real Function/Examples/Square Root
== Example of Real Function ==  The '''(real) square root function''' is the real function $f: \R \to \R$ defined on the positive real numbers as: :$\forall x \in \R_{\ge 0}: \map f x = \sqrt x$  :400px == Also see == * Domain of Real Square Root Function * Image of Real Square Root Function == Sources == * {{BookReference|Limits and Continuity|1964|William K. Smith|prev = Image of Real Square Function|next = Domain of Real Square Root Function}}: $\S 2.2$: Functions
Real Number to Negative Power/Positive Integer
Real Number to Negative Power/Positive Integer 0 48238 260871 2016-06-29T17:14:12Z Prime.mover 59 Created page with "== Theorem ==  Let $r \in \R_{> 0}$ be a positive real number.  Let $n \in \Z_{\ge 0}$ be a Definition:Positive Integer|posi..." wikitext text/x-wiki == Theorem == Let $r \in \R_{> 0}$ be a positive real number.  Let $n \in \Z_{\ge 0}$ be a positive integer. Let $r^n$ be defined as $r$ to the power of $n$. Then: :$r^{-n} = \dfrac 1 {r^n}$  == Proof == Proof by induction on $m$: For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$r^{-n} = \dfrac 1 {r^n}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = r^{-0}       | r = r^0       | c =  }} {{eqn | r = 1       | c = Definition of Integer Power }} {{eqn | r = \dfrac 1 1       | c =  }} {{eqn | r = \dfrac 1 {r^0}       | c = Definition of Integer Power }} {{end-eqn}} === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = r^{-1}       | r = \dfrac {r^{-1 + 1} } r       | c = Definition of Integer Power }} {{eqn | r = \dfrac {r^0} r       | c =  }} {{eqn | r = \dfrac 1 r       | c = Definition of Integer Power }} {{eqn | r = \dfrac 1 {r^1}       | c = Definition of Integer Power }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$r^{- k} = \dfrac 1 {r^k}$ Then we need to show: :$r^{- \left({k + 1}\right)} = \dfrac 1 {r^{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = r^{- \left({k + 1}\right)}       | r = \dfrac {r^{-\left({k + 1}\right) + 1} } r       | c = Definition of Integer Power }} {{eqn | r = \dfrac {r^{-k} } r       | c = simplification }} {{eqn | r = \dfrac 1 {r^k \times r}       | c = Induction Hypothesis }} {{eqn | r = \dfrac 1 {r^{k + 1} }       | c = Definition of Integer Power }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: r^{-n} = \dfrac 1 {r^n}$
Sum of Summations over Overlapping Domains/Example
== Example of Sum of Summations over Overlapping Domains ==  :$\displaystyle \sum_{1 \mathop \le j \mathop \le m} a_j + \sum_{m \mathop \le j \mathop \le n} a_j = \left({\sum_{1 \mathop \le j \mathop \le n} a_j}\right) + a_m$  == Proof == Let $R \left({j}\right)$ be the propositional function $1 \mathop \le j \mathop \le m$. Let $S \left({j}\right)$ be the propositional function $m \mathop \le j \mathop \le n$. Then we have: {{begin-eqn}} {{eqn | l = R \left({j}\right) \lor S \left({j}\right)       | r = \left({1 \mathop \le j \mathop \le m}\right) \lor \left({m \mathop \le j \mathop \le n}\right)       | c =  }} {{eqn | r = \left({1 \mathop \le j \mathop \le n}\right)       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = R \left({j}\right) \land S \left({j}\right)       | r = \left({1 \mathop \le j \mathop \le m}\right) \land \left({m \mathop \le j \mathop \le n}\right)       | c =  }} {{eqn | r = \left({j = m}\right)       | c =  }} {{end-eqn}} The result follows from Sum of Summations over Overlapping Domains.  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Sum of Summations over Overlapping Domains|next = Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 1}}: $\S 1.2.3$: Sums and Products: $(12)$
Factorial/Examples
== Examples of Factorials ==  The factorials of the first few positive integers are as follows: $\begin{array}{r|r} n & n! \\ \hline 0 & 1 \\ 1 & 1 \\ 2 & 2 \\ 3 & 6 \\ 4 & 24 \\ 5 & 120 \\ 6 & 720 \\ 7 & 5 \, 040 \\ 8 & 40 \, 320 \\ 9 & 362 \, 880 \\ 10 & 3 \, 628 \, 800 \\ \end{array}$  {{OEIS|A000142}} === Factorial of $0$ === {{:Factorial/Examples/0}} === Factorial of $1$ === {{:Factorial/Examples/1}} === Factorial of $10$ === {{:Factorial/Examples/10}} === Factorial of $11$ === {{:Factorial/Examples/11}} === Factorial of $12$ === {{:Factorial/Examples/12}} === Factorial of $13$ === {{:Factorial/Examples/13}} === Factorial of $14$ === {{:Factorial/Examples/14}} === Factorial of $15$ === {{:Factorial/Examples/15}} === Factorial of $16$ === {{:Factorial/Examples/16}} === Factorial of $17$ === {{:Factorial/Examples/17}} === Factorial of $18$ === {{:Factorial/Examples/18}} === Factorial of $19$ === {{:Factorial/Examples/19}} === Factorial of $20$ === {{:Factorial/Examples/20}} === Factorial of $21$ === {{:Factorial/Examples/21}} === Factorial of $22$ === {{:Factorial/Examples/22}} === Factorial of $23$ === {{:Factorial/Examples/23}} === Factorial of $24$ === {{:Factorial/Examples/24}} === Factorial of $25$ === {{:Factorial/Examples/25}} === Prime Factors of $39!$ === {{:Prime Factors of 39 Factorial}} === Factorial of $52$ === {{:Factorial/Examples/52}} === Factorial of $450$ === {{:Factorial/Examples/450}} === Factorial of $1\,000$ === {{:Factorial/Examples/1000}} === Factorial of $1\,000\,000$ === {{:Factorial/Examples/1,000,000}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Prime Number/Sequence|next = Reciprocals of Prime Numbers}}: Tables: $5$ The Factorials of the Numbers $1$ to $20$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Factorial/Definition 1|next = Factorial/Examples/1000}}: $\S 1.2.5$: Permutations and Factorials * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Prime Number/Sequence|next = Reciprocals of Prime Numbers}}: Tables: $5$ The Factorials of the Numbers $1$ to $20$
Factorial/Examples/0
== Example of Factorial ==  The factorial of $0$ is $1$: :$0! = 1$  == Proof == From the definition of factorial: :$n! = \displaystyle \prod_{k \mathop = 1}^n k$ where $\prod$ denotes product notation. When $n = 0$ we have: :$0! = \displaystyle \prod_{k \mathop = 1}^0 k$ Hence the result, by definition of vacuous product.  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Cardinality of Power Set of Finite Set/Proof 1|next = Definition:Factorial/Definition 2}}: $\S 19$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Factorial/Definition 2|next = Binomial Theorem/Integral Index}}: $\S 3$: The Binomial Formula and Binomial Coefficients: $3.2$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Definition:Factorial/Definition 2|next = Quintuple Angle Formulas/Cosine/Proof 2}}: $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21$ * {{BookReference|Special Functions of Mathematics for Engineers|1992|Larry C. Andrews|ed = 2nd|edpage = Second Edition|prev = Definition:Factorial/Definition 1|next = Binomial Theorem/Integral Index}}: $\S 1.2.4$: Factorials and binomial coefficients * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Factorial/Definition 2|next = Definition:Factorial/Definition 1}}: $\S 1.2.5$: Permutations and Factorials: $(5)$
Common Logarithm/Examples/e
== Example of Common Logarithm ==  The common logarithm of Euler's number $e$ is: :$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$  {{OEIS|A002285}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Common Logarithm of 3|next = Common Logarithm of Pi}}: $\S 1$: Special Constants: $1.15$ * {{BookReference|Les Nombres Remarquables|1983|François Le Lionnais|author2 = Jean Brette|prev = Definition:Bloch's Constant/Historical Note|next = Change of Base of Logarithm/Base e to Base 10/Form 1}}: $0,43429 44819 \ldots$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Envelope Problem|next = One Half as Pandigital Fraction}}: $0 \cdotp 434 \, 294 \, 481 \, 903 \, 251 \, 827 \, 651 \, 128 \, 918 \, 916 \, 605 \, 082 \, 294 \, 397 \, 005 \, 803 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Envelope Problem|next = One Half as Pandigital Fraction}}: $0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$
Common Logarithm/Examples/pi
== Example of Common Logarithm ==  The common logarithm of $\pi$ is: :$\log_{10} \pi = 0.49714 \, 98726 \, 94133 \, 85435 \, 12683 \ldots$  {{OEIS|A053511}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Common Logarithm of e|next = Natural Logarithm of 10}}: $\S 1$: Special Constants: $1.16$
Natural Logarithm/Examples/10
== Example of Natural Logarithm ==  The natural logarithm of $10$ is approximately: :$\ln 10 \approx 2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 7991 \ldots$  {{OEIS|A002392}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Common Logarithm of Pi|next = Natural Logarithm of 2}}: $\S 1$: Special Constants: $1.17$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square Root of 5|next = Definition:Stirling's Constant}}: $2 \cdotp 302 \, 585 \, 092 \, 994 \, 045 \, 684 \, 017 \, 991 \, 454 \, 684 \, 364 \, 207 \, 601 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Wallis's Number/Historical Note|next = Definition:Stirling's Constant}}: $2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 79914 \, 54684 \, 36420 \, 7601 \ldots$
Common Logarithm/Examples/2
== Example of Common Logarithm ==  The common logarithm of $2$ is: :$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$  {{OEIS|A007524}} == Also see == * Logarithm Base 10 of 2 is Irrational == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Euler's Number to Power of Itself|next = Common Logarithm of 3}}: $\S 1$: Special Constants: $1.13$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39|next = Number of Digits in Power of 2}}: $0 \cdotp 301 \, 029 \, 995 \, 663 \, 981 \ldots$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Base of Logarithm|next = Sum of Logarithms/General Logarithm}}: $\S 1.2.2$: Numbers, Powers, and Logarithms: $(10)$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39|next = Number of Digits in Power of 2}}: $0 \cdotp 30102 \, 99956 \, 63981 \ldots$
De Polignac's Formula/Examples/3 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $3$ appears in $1000!$ to the power of $498$. That is: :$3^{498} \divides 1000!$ but: :$3^{499} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $3$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 3} + \floor {\frac {1000} 9} + \floor {\frac {1000} {27} } + \floor {\frac {1000} {81} } + \floor {\frac {1000} {243} } + \floor {\frac {1000} {729} }       | c =  }} {{eqn | r = 333 + 111 + 37 + 12 + 4 + 1       | c =  }} {{eqn | r = 498       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = De Polignac's Formula|next = De Polignac's Formula/Technique}}: $\S 1.2.5$: Permutations and Factorials
Negated Upper Index of Binomial Coefficient/Corollary 2
Negated Upper Index of Binomial Coefficient/Corollary 2 0 48908 417835 391833 2019-08-11T14:05:13Z Danhook 3709 Remove incorrectly duplicated comment wikitext text/x-wiki == Corollary to Negated Upper Index of Binomial Coefficient == Let $n, m \in \Z$. Then:  :$\dbinom n m = \paren {-1}^{n - m} \dbinom {-\paren {m + 1} } {n - m}$  where $\dbinom n m$ is a binomial coefficient. == Proof == {{begin-eqn}} {{eqn | l = \dbinom r k       | r = \paren {-1}^k \dbinom {k - r - 1} k       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | ll= \leadsto       | l = \dbinom n {n - m}       | r = \paren {-1}^{n - m} \dbinom {\paren {n - m} - n - 1} {n - m}       | c = setting $r = n$ and $k = n - m$ }} {{eqn | ll= \leadsto       | l = \dbinom n m       | r = \paren {-1}^{n - m} \dbinom {-m - 1} {n - m}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \paren {-1}^{n - m} \dbinom {- \paren {m + 1} } {n - m} }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Alternating Sum and Difference of r Choose k up to n/Proof 1|next = Product of r Choose m with m Choose k/Proof 1}}: $\S 1.2.6$: Binomial Coefficients: $(19)$
Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary
== Corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$ == Let $r \in \Z_{\ge 0}, n \in \Z$. Then:  :$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$ where $\delta_{n r}$ is the Kronecker delta.  == Proof == From Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$: :$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$ which holds for $s \in \R, r \in \Z_{\ge 0}, n \in \Z$. Setting $s = 0$: :$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \binom 0 {n - r}$ We have by definition of binomial coefficient that: :$\dbinom 0 0 = 1$ and: :$\forall n \in \Z_{\ne 0}: \dbinom 0 n = 0$ So, using Iverson's convention: :$\dbinom 0 {n - r} = \sqrbk {n = r}$ The result follows by definition of the Kronecker delta.  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Proof|next = Sum over k of r Choose k by -1^r-k by Polynomial/Proof 1}}: $\S 1.2.6$: Binomial Coefficients: $(33)$