Product of Sums/Corollary

Product of Sums/Corollary 0 47758 353398 329927 2018-05-03T08:01:18Z Prime.mover 59 wikitext text/x-wiki == Corollary to Product of Sums == <onlyinclude> 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 $X$. </onlyinclude> == 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 $n$. {{finish}} Category:Number Theory Category:Analysis 3j7c7lbps4uu24cf3jzt9ytaxdr66d3


Real Function/Examples/Square Root

Real Function/Examples/Square Root 0 48059 465506 465501 2020-05-02T15:00:47Z Prime.mover 59 wikitext text/x-wiki == Example of Real Function == <onlyinclude> The '''(real) square root function''' is the real function $f: \R \to \R$ defined on the positive real numbers as: :$\forall x \in \R_{\ge 0}: \map f x = \sqrt x$ </onlyinclude> :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 Category:Examples of Real Functions j51cw9n2k2c634qth6wol2nx83s8aqd


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. <onlyinclude> 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. <onlyinclude> Let $n \in \Z_{\ge 0}$ be a positive integer. Let $r^n$ be defined as $r$ to the power of $n$. Then: :$r^{-n} = \dfrac 1 {r^n}$ </onlyinclude> == 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: :$\forall n \in \Z_{\ge 0}: r^{-n} = \dfrac 1 {r^n}$ {{qed}} Category:Powers hwjw4kv1xbs5pk3wtufm0089r6bd0yu


Sum of Summations over Overlapping Domains/Example 0 48365 361114 361002 2018-07-28T11:26:35Z Prime.mover 59 wikitext text/x-wiki == Example of Sum of Summations over Overlapping Domains == <onlyinclude> :$\displaystyle \sum_{1 \mathop \le j \mathop \le m} a_j + \sum_{m \mathop \le j \mathop \le n} a_j = \left({\sum_{1 \mathop \le j \mathop \le n} a_j}\right) + a_m$ </onlyinclude> == 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. {{qed}} == 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: $(12)$ Category:Sum of Summations over Overlapping Domains 90i5vzypn2dvf587wlc21bcxt9gl4th


Factorial/Examples

Factorial/Examples 0 48835 433303 385347 2019-11-01T09:02:03Z Prime.mover 59 wikitext text/x-wiki == Examples of Factorials == <onlyinclude> The factorials of the first few positive integers are as follows: $\begin{array}{r|r} n & n! \\ \hline 0 & 1 \\ 1 & 1 \\ 2 & 2 \\ 3 & 6 \\ 4 & 24 \\ 5 & 120 \\ 6 & 720 \\ 7 & 5 \, 040 \\ 8 & 40 \, 320 \\ 9 & 362 \, 880 \\ 10 & 3 \, 628 \, 800 \\ \end{array}$ </onlyinclude> {{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 $20$ Category:Factorials/Examples d7h41nrft4lnbuea0crnzgroxevvwqd


Factorial/Examples/0

Factorial/Examples/0 0 48836 492902 394672 2020-10-06T09:58:31Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> The factorial of $0$ is $1$: :$0! = 1$ </onlyinclude> == 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. {{qed}} == 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: $(5)$ Category:Factorials/Examples lngw6wmaj8gpip8lsyzya4wm9s5cf22


Common Logarithm/Examples/e

Common Logarithm/Examples/e 0 48843 433234 433024 2019-11-01T08:09:29Z Prime.mover 59 wikitext text/x-wiki == Example of Common Logarithm == <onlyinclude> The common logarithm of Euler's number $e$ is: :$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$ </onlyinclude> {{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}}: $0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$ Category:Examples of Common Logarithms Category:Euler's Number 5o0xza90mr8tpyg38zeer7glp0ow4ps


Common Logarithm/Examples/pi

Common Logarithm/Examples/pi 0 48845 433237 433235 2019-11-01T08:10:12Z Prime.mover 59 wikitext text/x-wiki == Example of Common Logarithm == <onlyinclude> The common logarithm of $\pi$ is: :$\log_{10} \pi = 0.49714 \, 98726 \, 94133 \, 85435 \, 12683 \ldots$ </onlyinclude> {{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: $1.16$ Category:Examples of Common Logarithms f2x26ka1sftbcfjmv4v8j8a2qps5198


Natural Logarithm/Examples/10

Natural Logarithm/Examples/10 0 48849 433418 432995 2019-11-01T12:59:21Z Prime.mover 59 wikitext text/x-wiki == Example of Natural Logarithm == <onlyinclude> The natural logarithm of $10$ is approximately: :$\ln 10 \approx 2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 7991 \ldots$ </onlyinclude> {{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}}: $2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 79914 \, 54684 \, 36420 \, 7601 \ldots$ Category:Examples of Natural Logarithms sqzz4hegej02mjzduxzvjdoysvr5ugq


Common Logarithm/Examples/2

Common Logarithm/Examples/2 0 48852 433232 432975 2019-11-01T08:08:56Z Prime.mover 59 wikitext text/x-wiki == Example of Common Logarithm == <onlyinclude> The common logarithm of $2$ is: :$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$ </onlyinclude> {{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}}: $0 \cdotp 30102 \, 99956 \, 63981 \ldots$ Category:Examples of Common Logarithms rg51xodj34e98e8q0kyr4c0nhhsomwt


De Polignac's Formula/Examples/3 in 1000 0 48858 366427 298881 2018-09-19T12:42:27Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $3$ appears in $1000!$ to the power of $498$. That is: :$3^{498} \divides 1000!$ but: :$3^{499} \nmid 1000!$ </onlyinclude> == 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}} {{qed}} == 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 Category:De Polignac's Formula cxmcv5bldadrqcq4nkiaz91z076af8t


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: <onlyinclude> :$\dbinom n m = \paren {-1}^{n - m} \dbinom {-\paren {m + 1} } {n - m}$ </onlyinclude> 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}} {{qed}} == 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: $(19)$ Category:Binomial Coefficients Category:Negated Upper Index of Binomial Coefficient 1rdl6gyubi8v4fhw6qqxn3qmnv6bi4f


Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary 0 48983 496845 357669 2020-10-28T22:07:25Z Prime.mover 59 wikitext text/x-wiki == 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: <onlyinclude> :$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$ where $\delta_{n r}$ is the Kronecker delta. </onlyinclude> == 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. {{qed}} == 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: $(33)$ Category:Sum over k of r Choose k by s+k Choose n by -1^r-k endpvg3aubb87ul64mh39oqtkmwx386


Condition for Infimum of Subset to equal Infimum of Set 0 49120 265908 2016-08-02T09:05:03Z Ivar Sand 2302 Created page with "== Lemma ==  Let $S$ be a real set.  Let $T$ be a subset of $S$.  Let $S$ and $T$ admit Definition:Infimum of Subset of Real..." wikitext text/x-wiki == Lemma == Let $S$ be a real set. Let $T$ be a subset of $S$. Let $S$ and $T$ admit infima. Then: : $\inf S = \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$ == Proof == === Necessary Condition === Let $\inf S = \inf T$. The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$. We have: {{begin-eqn}} {{eqn | l = \inf S       | r = \inf T }} {{eqn | ll = \implies       | l = \inf S       | o = \ge       | r = \inf T }} {{eqn | ll = \iff       | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon       | o = >       | r = t       | c = Infima of two Real Sets }} {{end-eqn}} === Sufficient Condition === Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$. The aim is to establish that $\inf S = \inf T$. We have: {{begin-eqn}} {{eqn | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon       | o = >       | r = t }} {{eqn | ll = \iff       | l = \inf S       | o = \ge       | r = \inf T       | c = Infima of two Real Sets }} {{eqn | ll = \iff       | l = \inf S       | o = \ge       | r = \inf T \ge \inf S       | c = as $\inf T \ge \inf S$ is true by Infimum of Subset }} {{eqn | ll = \iff       | l = \inf S       | r = \inf T       | c =  }} {{end-eqn}} {{qed}} Category:Real Analysis dv10hz54fb12g0za3wp8mwk2s3tdzx3


Limit of Monotone Real Function/Increasing/Corollary 0 49157 454676 354386 2020-03-16T06:18:14Z Prime.mover 59 wikitext text/x-wiki == Corollary to Limit of Increasing Function == <onlyinclude> Let $f$ be a real function which is increasing on the open interval $\openint a b$. If $\xi \in \openint a b$, then: :$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist and: :$\map f x \le \map f {\xi^-} \le \map f \xi \le \map f {\xi^+} \le \map f y$ provided that $a < x < \xi < y < b$. </onlyinclude> == Proof == $f$ is bounded above on $\openint a b$ by $\map f \xi$. By Limit of Increasing Function, the supremum is $\map f {\xi^-}$. So it follows that: :$\forall x \in \openint a \xi: \map f x \le \map f {\xi^-} \le \map f \xi$ A similar argument for $\openint \xi b$ holds for the other inequalities. {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Limit of Decreasing Function|next = Derivative of Monotone Function}}: $\S 12.5$ Category:Limits of Functions hhyl57m26twizx6dylbouij3p8zx4l7


Limit of Monotone Real Function/Decreasing/Corollary 0 49159 447658 354387 2020-02-09T08:47:12Z Prime.mover 59 wikitext text/x-wiki == Corollary to Limit of Decreasing Function == <onlyinclude> Let $f$ be a real function which is decreasing on the open interval $\openint a b$. If $\xi \in \openint a b$, then: :$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist and: :$\map f x \ge \map f {\xi^-} \ge \map f \xi \ge \map f {\xi^+} \ge \map f y$ provided that $a < x < \xi < y < b$. </onlyinclude> == Proof == $f$ is bounded below on $\openint a \xi$ by $\map f \xi$. By Limit of Decreasing Function, the infimum is $\map f {\xi^-}$. So it follows that: :$\forall x \in \openint a \xi: \map f x \ge \map f {\xi^-} \ge \map f \xi$ A similar argument for $\openint \xi b$ holds for the other inequalities. {{qed}} Category:Limits of Functions h4jsquhqhqxxu35tt2yubcvdpl9pfi2


Prime Factors of 52 Factorial

Prime Factors of 52 Factorial 0 49189 475913 447243 2020-06-25T07:01:05Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> The prime decomposition of $52!$ is given as: :$52! = 2^{49} \times 3^{23} \times 5^{12} \times 7^8 \times 11^4 \times 13^4 \times 17^3 \times 19^2 \times 23^2 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47$ </onlyinclude> == Proof == For each prime factor $p$ of $52!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 52!$ :$p^{a_p + 1} \nmid 52!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {52} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 2} + \floor {\frac {52} 4} + \floor {\frac {52} 8 } + \floor {\frac {52} {16} } + \floor {\frac {52} {32} }       | c =  }} {{eqn | r = 26 + 13 + 6 + 3 + 1       | c =  }} {{eqn | r = 49       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {52} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 3} + \floor {\frac {52} 9} + \floor {\frac {52} {27} }       | c =  }} {{eqn | r = 17 + 5 + 1       | c =  }} {{eqn | r = 23       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {52} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 5} + \floor {\frac {52} {25} }       | c =  }} {{eqn | r = 10 + 2       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {52} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 7} + \floor {\frac {52} {49} }       | c =  }} {{eqn | r = 7 + 1       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{11}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {11} }       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{13}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {13} }       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{17}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {17^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {17} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{19}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {19^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {19} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{23}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {23^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {23} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Similarly: :$a_{29} = 1$ :$a_{31} = 1$ :$a_{37} = 1$ :$a_{41} = 1$ :$a_{43} = 1$ :$a_{47} = 1$ Hence the result. {{qed}} Category:Factorials/Examples Category:52 dydr7sxaqe3lc78afr39ts81wcply67


Prime Factors of 13 Factorial

Prime Factors of 13 Factorial 0 49190 475909 307335 2020-06-25T06:59:05Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> The prime decomposition of $13!$ is given as: :$13! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$ </onlyinclude> == Proof == For each prime factor $p$ of $13!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 13!$ :$p^{a_p + 1} \nmid 13!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {13} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 2} + \floor {\frac {13} 4} + \floor {\frac {13} 8}       | c =  }} {{eqn | r = 6 + 3 + 1       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {13} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 3} + \floor {\frac {13} 9}       | c =  }} {{eqn | r = 4 + 1       | c =  }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {13} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 5}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {13} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 7}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Similarly: :$a_{11} = 1$ :$a_{13} = 1$ Hence the result. {{qed}} Category:Factorials/Examples Category:13 nawfhubalbhphu2ruckw39pdtg157lr


Prime Factors of 39 Factorial

Prime Factors of 39 Factorial 0 49191 475912 453485 2020-06-25T07:00:40Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> The prime decomposition of $39!$ is given as: :$39! = 2^{35} \times 3^{18} \times 5^8 \times 7^5 \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29 \times 31 \times 37$ </onlyinclude> == Proof == For each prime factor $p$ of $39!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 39!$ :$p^{a_p + 1} \nmid 39!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {39} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 2} + \floor {\frac {39} 4} + \floor {\frac {39} 8 } + \floor {\frac {39} {16} } + \floor {\frac {39} {32} }       | c =  }} {{eqn | r = 19 + 9 + 4 + 2 + 1       | c =  }} {{eqn | r = 35       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {39} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 3} + \floor {\frac {39} 9} + \floor {\frac {39} {27} }       | c =  }} {{eqn | r = 13 + 4 + 1       | c =  }} {{eqn | r = 18       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {39} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 5} + \floor {\frac {39} {25} }       | c =  }} {{eqn | r = 7 + 1       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {39} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 7}       | c =  }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{11}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {11} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{13}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {13} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{17}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {17^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {17} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{19}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {19^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {19} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Similarly: :$a_{23} = 1$ :$a_{29} = 1$ :$a_{31} = 1$ :$a_{37} = 1$ Hence the result. {{qed}} Category:Factorials/Examples Category:39 kj1lusrp5gnw9m7w8puaq3gvq5xprdp


Lucas' Theorem/Corollary

Lucas' Theorem/Corollary 0 49210 484319 465328 2020-08-31T10:46:07Z Prime.mover 59 wikitext text/x-wiki == Corollary to Lucas' Theorem == Let $p$ be a prime number. Let $n, k \in \Z$. <onlyinclude> Let the representations of $n$ and $k$ to the base $p$ be given by: :$n = a_r p^r + \cdots + a_1 p + a_0$ :$k = b_r p^r + \cdots + b_1 p + b_0$ Then: :$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$ </onlyinclude> where $\dbinom n k$ denotes a binomial coefficient. == Proof == Consider the representations of $n$ and $k$ to the base $p$: :$n = a_r p^r + \cdots + a_1 p + a_0$ :$k = b_r p^r + \cdots + b_1 p + b_0$ Let: :$n_1 = \floor {n / p}$ :$k_1 = \floor {k / p}$ We have that: :$n \bmod p = a_0$ :$k \bmod p = b_0$ :$n_1 = a_r p^{r - 1} + a_{r - 1} p^{r - 2} \cdots + a_1$ :$k_1 = b_r p^{r - 1} + b_{r - 1} p^{r - 2} \cdots + b_1$ It follows from Lucas' Theorem that: :$\dbinom n k \equiv \dbinom {n_1} {k_1} \dbinom {a_0} {b_0} \pmod p$ Now we do the same again to the representation to the base $p$ of $n_1$ and $n_2$. Thus: :$\dbinom n k \equiv \dbinom {\floor {n_1 / p}} {\floor {k_1 / p} } \dbinom {a_1} {b_1} \dbinom {a_0} {b_0} \pmod p$ and so on until: :$\floor {n_r / p}$ and: :$\floor {k_r / p}$ Hence the result. {{qed}} {{Namedfor|François Édouard Anatole Lucas|cat = Lucas}} == Sources == * {{citation|date = 1947|title = Binomial Coefficients Modulo a Prime|journal = American Mathematical Monthly|abbr = Amer. Math. Monthly|volume = 54|startpage = 589|endpage = 592|author = N.J. Fine|jstor = 2304500}} * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Lucas' Theorem|next = Kummer's Theorem}}: $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(f)}$ Category:Binomial Coefficients Category:Number Theory Category:Prime Numbers a9gkjsyp1fk0vetvl25z9ir6ks8mecg


De Polignac's Formula/Examples/2 in 1000 0 49319 447241 298883 2020-02-08T00:17:04Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $2$ appears in $1000!$ to the power of $994$. That is: :$2^{994} \divides 1000!$ but: :$2^{995} \nmid 1000!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $2$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 2} + \floor {\frac {1000} 4} + \floor {\frac {1000} 8} + \floor {\frac {1000} {16} } + \floor {\frac {1000} {32} }       | c =  }} {{eqn | o =        | ro= +       | r = \floor {\frac {1000} {64} } + \floor {\frac {1000} {128} } + \floor {\frac {1000} {256} } + \floor {\frac {1000} {512} }       | c =  }} {{eqn | r = 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1       | c =  }} {{eqn | r = 994       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula hyh7qxmaoutkcvc1jokk2e0hz3mw8uf


De Polignac's Formula/Examples/5 in 1000 0 49320 371895 298893 2018-10-19T08:48:45Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $5$ appears in $1000!$ to the power of $249$. That is: :$5^{249} \divides 1000!$ but: :$5^{250} \nmid 1000!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $5$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 5} + \floor {\frac {1000} {25} } + \floor {\frac {1000} {125} } + \floor {\frac {1000} {625} }       | c =  }} {{eqn | r = 200 + 40 + 8 + 1       | c =  }} {{eqn | r = 249       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula 3qhncqeqbifm0j1axs8y8kc7x5lpvoi


De Polignac's Formula/Examples/7 in 1000 0 49321 457791 298892 2020-03-27T07:59:04Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $7$ appears in $1000!$ to the power of $164$. That is: :$7^{164} \divides 1000!$ but: :$7^{165} \nmid 1000!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $7$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 7} + \floor {\frac {1000} {49} } + \floor {\frac {1000} {343} }       | c =  }} {{eqn | r = 142 + 20 + 2       | c =  }} {{eqn | r = 164       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula 3uy57p7sp7vx5rkrfwudk3v311e6odb


De Polignac's Formula/Examples/11 in 1000 0 49322 376170 298887 2018-11-12T14:41:11Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $11$ appears in $1000!$ to the power of $98$. That is: :$11^{98} \divides 1000!$ but: :$11^{99} \nmid 1000!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $11$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {11} } + \floor {\frac {1000} {121} }       | c =  }} {{eqn | r = 90 + 8       | c =  }} {{eqn | r = 98       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula pejdiyzux29elx765xubzzatd3st0ru


De Polignac's Formula/Examples/13 in 1000 0 49323 457790 298886 2020-03-27T07:58:15Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $13$ appears in $1000!$ to the power of $81$. That is: :$13^{81} \divides 1000!$ but: :$13^{82} \nmid 1000!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $13$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {13} } + \floor {\frac {1000} {169} }       | c =  }} {{eqn | r = 76 + 5       | c =  }} {{eqn | r = 81       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula asv117cxyp8yjv4onegahw8c5c8g1fu


Prime Factors of 1000 Factorial

Prime Factors of 1000 Factorial 0 49324 475910 437932 2020-06-25T06:59:38Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> The prime decomposition of $1000!$ is given as: {{begin-eqn}} {{eqn | l = 1000!       | r = 2^{994} \times 3^{498} \times 5^{249} \times 7^{164} \times 11^{98} \times 13^{81} \times 17^{61} \times 19^{54} \times 23^{44} \times 29^{35} \times 31^{33} \times 37^{27} }} {{eqn | o =        | ro= \times       | r = 41^{24} \times 43^{23} \times 47^{21} \times 53^{18} \times 59^{16} \times 61^{16} \times 67^{14} \times 71^{14} \times 73^{13} \times 79^{13} \times 83^{12} \times 89^{11}       | c =  }} {{eqn | o =        | ro= \times       | r = 97^{10} \times 101^9 \times 103^9 \times 107^9 \times 109^9 \times 113^8 \times 127^7 \times 131^7 \times 137^7 \times 139^7 \times 149^6 \times 151^6 }} {{eqn | o =        | ro= \times       | r = 157^6 \times 163^6 \times 167^5 \times 173^5 \times 179^5 \times 181^5 \times 191^5 \times 193^5 \times 197^5 \times 199^5 \times 211^4 \times 223^4 }} {{eqn | o =        | ro= \times       | r = 227^4 \times 229^4 \times 233^4 \times 239^4 \times 241^4 \times 251^3 \times 257^3 \times 263^3 \times 269^3 \times 271^3 \times 277^3 \times 281^3 }} {{eqn | o =        | ro= \times       | r = 283^3 \times 293^3 \times 307^3 \times 311^3 \times 313^3 \times 317^3 \times 331^3 \times 337^2 \times 347^2 \times 349^2 \times 353^2 \times 359^2 }} {{eqn | o =        | ro= \times       | r = 367^2 \times 373^2 \times 379^2 \times 383^2 \times 389^2 \times 397^2 \times 401^2 \times 409^2 \times 419^2 \times 421^2 \times 431^2 \times 433^2 }} {{eqn | o =        | ro= \times       | r = 439^2 \times 443^2 \times 449^2 \times 457^2 \times 461^2 \times 463^2 \times 467^2 \times 479^2 \times 487^2 \times 491^2 \times 499^2 \times 503 }} {{eqn | o =        | ro= \times       | r = 509 \times 521 \times 523 \times 541 \times 547 \times 557 \times 563 \times 569 \times 571 \times 577 \times 587 \times 593 }} {{eqn | o =        | ro= \times       | r = 599 \times 601 \times 607 \times 613 \times 617 \times 619 \times 631 \times 641 \times 643 \times 647 \times 653 \times 659 }} {{eqn | o =        | ro= \times       | r = 661 \times 673 \times 677 \times 683 \times 691 \times 701 \times 709 \times 719 \times 727 \times 733 \times 739 \times 743 }} {{eqn | o =        | ro= \times       | r = 751 \times 757 \times 761 \times 769 \times 773 \times 787 \times 797 \times 809 \times 811 \times 821 \times 823 \times 827 }} {{eqn | o =        | ro= \times       | r = 829 \times 839 \times 853 \times 857 \times 859 \times 863 \times 877 \times 881 \times 883 \times 887 \times 907 \times 911 }} {{eqn | o =        | ro= \times       | r = 919 \times 929 \times 937 \times 941 \times 947 \times 953 \times 967 \times 971 \times 977 \times 983 \times 991 \times 997       | c =  }} {{end-eqn}} </onlyinclude> == Proof == For each prime factor $p$ of $1000!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 1000!$ :$p^{a_p + 1} \nmid 1000!$ Taking the prime factors in turn: === Multiplicity of $2$ === {{:De Polignac's Formula/Examples/2 in 1000}} === Multiplicity of $3$ === {{:De Polignac's Formula/Examples/3 in 1000}} === Multiplicity of $5$ === {{:De Polignac's Formula/Examples/5 in 1000}} === Multiplicity of $7$ === {{:De Polignac's Formula/Examples/7 in 1000}} === Multiplicity of $11$ === {{:De Polignac's Formula/Examples/11 in 1000}} === Multiplicity of $13$ === {{:De Polignac's Formula/Examples/13 in 1000}} And so on: for the remaining factors: {{begin-eqn}} {{eqn | l = a_{17}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {17^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {17} } + \floor {\frac {1000} {289} }       | c =  }} {{eqn | r = 58 + 3       | c =  }} {{eqn | r = 61       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{19}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {19^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {19} } + \floor {\frac {1000} {361} }       | c =  }} {{eqn | r = 52 + 2       | c =  }} {{eqn | r = 54       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{23}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {23^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {23} } + \floor {\frac {1000} {529} }       | c =  }} {{eqn | r = 43 + 1       | c =  }} {{eqn | r = 44       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{29}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {29^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {29} } + \floor {\frac {1000} {841} }       | c =  }} {{eqn | r = 34 + 1       | c =  }} {{eqn | r = 35       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{31}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {31^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {31} } + \floor {\frac {1000} {961} }       | c =  }} {{eqn | r = 32 + 1       | c =  }} {{eqn | r = 33       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{37}       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {37^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {37} }       | c =  }} {{eqn | r = 27       | c =  }} {{end-eqn}} The pattern is clear: the remaining indices are calculated in the same way. {{qed}} Category:Factorials/Examples Category:1000 stulnrpncxqchhg3iv0vt628gszvv2g


Integral Form of Gamma Function equivalent to Euler Form/Lemma 0 49351 300866 266937 2017-06-12T11:29:10Z Prime.mover 59 wikitext text/x-wiki == Lemma for Integral Form of Gamma Function equivalent to Euler Form == <onlyinclude> Let $0 \le t \le m$. Then: :$0 \le e^{-t} - \left({1 - \dfrac t m}\right)^m \le t^2 \dfrac {e^{-t} } m$ </onlyinclude> == Proof == From Exponential of x not less than 1+x: :$1 + x \le e^x$ Let $x = \pm \dfrac t m$. Then: {{begin-eqn}} {{eqn | l = 1 - \dfrac t m       | o = \le       | r = e^{-t/m} }} {{eqn | ll= \leadsto       | l = \left({1 - \dfrac t m}\right)^m       | o = \le       | r = e^{-t}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = 1 + \dfrac t m       | o = \le       | r = e^{t/m} }} {{eqn | ll= \leadsto       | l = \left({1 + \dfrac t m}\right)^m       | o = \le       | r = e^t       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = e^{-t}       | o = \ge       | r = \left({1 - \dfrac t m}\right)^m }} {{eqn | r = e^t \left({1 - \dfrac t m}\right)^m e^{-t}       | c =  }} {{eqn | o = \ge       | r = e^{-t} \left({1 - \dfrac t m}\right)^m \left({1 + \dfrac t m}\right)^m       | c =  }} {{eqn | r = e^{-t} \left({1 - \dfrac {t^2} {m^2} }\right)^m       | c = Difference of Two Squares }} {{eqn | o = \ge       | r = e^{-t} \left({1 - \dfrac {t^2} {m^2} }\right)       | c = Corollary to Bernoulli's Inequality }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = e^{-t} - \left({1 - \dfrac t m}\right)^m }} {{eqn | o = \le       | r = t^2 \dfrac {e^{-t} } m       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Partial Gamma Function expressed as Integral/Lemma|next = Integral Form of Gamma Function equivalent to Euler Form/Proof 2}}: $\S 1.2.5$: Permutations and Factorials: Exercise $20$ Category:Integral Form of Gamma Function equivalent to Euler Form b37ssvfo7qtkuriuvsk52ms7298rtzq


Faà di Bruno's Formula/Example/0

Faà di Bruno's Formula/Example/0 0 49370 267024 267023 2016-08-17T19:14:08Z Prime.mover 59 wikitext text/x-wiki == Example of use of Faà di Bruno's Formula: $n = 0$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 0$ we have: <onlyinclude> :$D_x^0 w = w$ </onlyinclude> == Proof == {{:Faà di Bruno's Formula/Example/0/Proof}} Category:Faà di Bruno's Formula rrbp367adksn7fc0otxjxc9ykz4b9yq


Faà di Bruno's Formula/Example/1

Faà di Bruno's Formula/Example/1 0 49372 267044 267035 2016-08-17T21:00:53Z Prime.mover 59 wikitext text/x-wiki == Example of use of Faà di Bruno's Formula: $n = 1$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 1$ we have: <onlyinclude> :$D_x^1 w = D_u^1 w D_x^1 u$ </onlyinclude> == Proof == {{:Faà di Bruno's Formula/Example/1/Proof}} Category:Faà di Bruno's Formula orpdb3msjh87gcr1juse7hv8af8k8q7


Faà di Bruno's Formula/Example/2

Faà di Bruno's Formula/Example/2 0 49374 267042 2016-08-17T20:45:57Z Prime.mover 59 Created page with "== Example of use of Faà di Bruno's Formula: $n = 2$ ==  Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}}  When $n = 2$ we have: <onlyinclude> :$D_x^2 w..." wikitext text/x-wiki == Example of use of Faà di Bruno's Formula: $n = 2$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 2$ we have: <onlyinclude> :$D_x^2 w = D_u^2 w \left({D_x^1 u}\right)^2 + D_u^1 w D_x^2 u$ </onlyinclude> == Proof == {{:Faà di Bruno's Formula/Example/2/Proof}} Category:Faà di Bruno's Formula frd5i1ungjfiggnq778rl14v58spnjp


Ceiling of x+m over n/Corollary

Ceiling of x+m over n/Corollary 0 49513 316257 267646 2017-09-08T16:29:41Z Caliburn 3218 cat wikitext text/x-wiki == Corollary to Ceiling of $\frac {x + m} n$ == <onlyinclude> Let $n \in \Z$ such that $n > 0$. Let $x \in \R$. Then: :$\left \lceil{\dfrac x n}\right \rceil = \left \lceil{\dfrac {\left \lceil{x}\right \rceil} n}\right \rceil$ where $\left\lceil{x}\right\rceil$ denotes the ceiling of $x$. </onlyinclude> == Proof == This is a special case of Ceiling of $\dfrac {x + m} n$: :$\left \lceil{\dfrac {x + m} n}\right \rceil = \left \lceil{\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rceil$ where $m = 0$. == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Ceiling of x+m over n|next = Ceiling of x+m over n/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ Category:Ceiling Function Category:Ceiling of x+m over n el4z1w1znfboid67kkdqimm9yqbclef


Floor of x+m over n/Corollary

Floor of x+m over n/Corollary 0 49514 496877 346261 2020-10-29T22:07:19Z Prime.mover 59 wikitext text/x-wiki == Corollary to Floor of $\frac {x + m} n$ == <onlyinclude> Let $n \in \Z$ such that $n > 0$. Let $x \in \R$. Then: :$\floor {\dfrac x n} = \floor {\dfrac {\floor x} n}$ where $\floor x$ denotes the floor of $x$. </onlyinclude> == Proof == This is a special case of Floor of $\dfrac {x + m} n$: :$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$ where $m = 0$. == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Floor of x+m over n|next = Floor of x+m over n/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ Category:Floor Function Category:Floor of x+m over n if4kgg6dh24exg2tdlh26agh3y7y4ib


Euclidean Algorithm/Demonstration

Euclidean Algorithm/Demonstration 0 49575 493848 390997 2020-10-11T06:58:28Z Prime.mover 59 wikitext text/x-wiki == Example of use of Euclidean Algorithm == <onlyinclude> Using the '''Euclidean Algorithm''', we can investigate in detail what happens when we apply the Division Theorem repeatedly to $a$ and $b$. {{begin-eqn}} {{eqn | l = a       | r = q_1 b + r_1       | c =  }} {{eqn | l = b       | r = q_2 r_1 + r_2       | c =  }} {{eqn | l = r_1       | r = q_3 r_2 + r_3       | c =  }} {{eqn | l = \cdots       | o =        | c =  }} {{eqn | l = r_{n - 2}       | r = q_n r_{n - 1} + r_n       | c =  }} {{eqn | l = r_{n - 1}       | r = q_{n + 1} r_n + 0       | c =  }} {{end-eqn}} From the Division Theorem, we know that the remainder is always strictly less than the divisor. That is, in $a = q b + r$: :$0 \le r < \size b$ So we know that: :$b > r_1 > r_2 > \ldots > r_{n - 2} > r_{n - 1} > r_n > 0$ So the algorithm ''has'' to terminate. {{qed}} </onlyinclude> == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Euclidean Algorithm|next = Existence of Greatest Common Divisor/Proof 4}}: $\text {2-2}$ Divisibility: Theorem $\text {2-2}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Euclidean Algorithm|next = Definition:Integer Combination}}: $\S 12$: Highest common factors and Euclid's algorithm * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Euclidean Algorithm|next = Definition:Linearly Independent Set}}: $\text{A}.1$: Number theory Category:Euclidean Algorithm gqtjjtsmaz86hb0cj0udaq6807ouum7


Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection 0 49666 268458 2016-08-30T05:32:36Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $\left({S, \circ}\right)$ be a finite semigroup.  Let $a \in S$ be Definition:Right Cancellabl..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $\left({S, \circ}\right)$ be a finite semigroup. Let $a \in S$ be right cancellable. Then the right regular representation $\rho_a$ of $\left({S, \circ}\right)$ with respect to $a$ is a bijection. </onlyinclude> == Proof == By Right Cancellable iff Right Regular Representation Injective, $\rho_a$ is an injection. By hypothesis, $S$ is finite. From Injection from Finite Set to Itself is Surjection, $\rho_a$ is a surjection. Thus $\rho_a$ is injective and surjective, and therefore a bijection. {{qed}} Category:Semigroups Category:Regular Representations Category:Cancellability 3dwl17i5fe551prpra9brqpnd0whzgb


Euler Phi Function/Examples/9

Euler Phi Function/Examples/9 0 49689 487588 357187 2020-09-13T21:27:28Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> :$\map \phi 9 = 6$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ function. == Proof == From Euler Phi Function of Prime Power: :$\map \phi {3^k} = 2 \times 3^{k - 1}$ Thus: :$\map \phi 9 = \map \phi {3^2} = 2 \times 3 = 6$ They can be enumerated as: :$1, 2, 4, 5, 7, 8$ {{qed}} == Sources == * {{BookReference|Introduction to the Theory of Finite Groups|1964|Walter Ledermann|ed = 5th|edpage = Fifth Edition|prev = Definition:Euler Phi Function|next = Euler Phi Function of Prime}}: Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$ Category:Examples of Euler Phi Function Category:9 p5i3t2xd8nx03vkk15bcgmdgnhfbckz


Group of Rotation Matrices Order 4/Cayley Table 0 49776 370971 370969 2018-10-14T05:06:03Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Group of Rotation Matrices Order $4$ == Consider the group of rotation matrices order $4$ :$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} }$ $R_4$ can be described completely by showing its Cayley table. Let: :$r_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ :$r_2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ :$r_3 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ Then we have: <onlyinclude> :$\begin{array}{r|rrrr} \times & r_0 & r_1 & r_2 & r_3 \\ \hline r_0 & r_0 & r_1 & r_2 & r_3 \\ r_1 & r_1 & r_2 & r_3 & r_0 \\ r_2 & r_2 & r_3 & r_0 & r_1 \\ r_3 & r_3 & r_0 & r_1 & r_2 \\ \end{array}$ </onlyinclude> Category:Examples of Cayley Tables h310fikahfzikm7xwk68bhgocwhub0j


Multiplicative Group of Reduced Residues Modulo 5/Cayley Table 0 49801 385903 371167 2019-01-02T22:00:46Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Multiplicative Group of Reduced Residues Modulo 5 == The multiplicative group of reduced residues modulo $5$: :$\Z'_5 = \set {\eqclass 1 5, \eqclass 2 5, \eqclass 3 5, \eqclass 4 5}$ can be described completely by showing its Cayley table: <onlyinclude> :$\begin{array}{r|rrrr} \times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\ \hline \eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\ \eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 1 5 & \eqclass 3 5 \\ \eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 4 5 & \eqclass 2 5 \\ \eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 2 5 & \eqclass 1 5 \\ \end{array}$ By arranging the rows and columns into a different order, its cyclic nature becomes clear: :$\begin{array}{r|rrrr} \times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\ \hline \eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\ \eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 \\ \eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 \\ \eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Modulo Addition/Cayley Table/Modulo 4|next = Modulo Multiplication is Well-Defined}}: Chapter $2$: Maps and relations on sets: Exercise $5$ Category:Examples of Cayley Tables Category:Multiplicative Group of Reduced Residues Modulo 5 9uajnfql30wlvvgzmphmjtvembcikit


Group Generated by Reciprocal of z and 1 minus z/Cayley Table 0 49814 487572 371069 2020-09-13T21:16:39Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Group Generated by Reciprocal of $1 / z$ and $1 - z$ == We have: {{begin-eqn}} {{eqn | l = \map {f_1} z       | r = z }} {{eqn | l = \map {f_2} z       | r = \dfrac 1 {1 - z} }} {{eqn | l = \map {f_3} z       | r = \dfrac {z - 1} z }} {{eqn | l = \map {f_4} z       | r = \dfrac 1 z }} {{eqn | l = \map {f_5} z       | r = 1 - z }} {{eqn | l = \map {f_6} z       | r = \dfrac z {z - 1} }} {{end-eqn}} Hence from Group Generated by Reciprocal of z and 1 minus z: <onlyinclude> :$\begin{array}{r|rrrrrr} \circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ f_2 & f_2 & f_3 & f_1 & f_6 & f_4 & f_5 \\ f_3 & f_3 & f_1 & f_2 & f_5 & f_6 & f_4 \\ f_4 & f_4 & f_5 & f_6 & f_1 & f_2 & f_3 \\ f_5 & f_5 & f_6 & f_4 & f_3 & f_1 & f_2 \\ f_6 & f_6 & f_4 & f_5 & f_2 & f_3 & f_1 \\ \end{array}$ Expressing the elements in full: :$\begin{array}{c|cccccc} \circ & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \hline z & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \dfrac 1 {1 - z} & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z & \dfrac z {z - 1} & \dfrac 1 z & 1 - z \\ \dfrac {z - 1} z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} & 1 - z & \dfrac z {z - 1} & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z \\ 1 - z & 1 - z & \dfrac z {z - 1} & \dfrac 1 z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} \\ \dfrac z {z - 1} & \dfrac z {z - 1} & \dfrac 1 z & 1 - z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|Introduction to the Theory of Finite Groups|1964|Walter Ledermann|ed = 5th|edpage = Fifth Edition|prev = Group Generated by Reciprocal of z and 1 minus z|next = Definition:Congruence (Number Theory)/Integers/Integer Multiple}}: Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(ii)}$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Group Generated by Reciprocal of z and 1 minus z|next = Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters}}: Introduction Category:Examples of Cayley Tables t51dc10j2jycz7bfmz2foh7jztds4aw


Group Generated by Reciprocal of z and Minus z/Cayley Table 0 49823 372457 371010 2018-10-22T06:21:26Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Group Generated by Reciprocal of $1 / z$ and $-z$ == We have: :$\map {f_1} z = z$ :$\map {f_2} z = -z$ :$\map {f_3} z = \dfrac 1 z$ :$\map {f_4} z = -\dfrac 1 z$ Hence from Group Generated by Reciprocal of z and Minus z: <onlyinclude> :$\begin{array}{r|rrrr} \circ & f_1 & f_2 & f_3 & f_4 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 \\ f_2 & f_2 & f_1 & f_4 & f_3 \\ f_3 & f_3 & f_4 & f_1 & f_2 \\ f_4 & f_4 & f_3 & f_2 & f_1 \\ \end{array}$ Expressing the elements in full: :$\begin{array}{c|cccc} \circ & z & -z & \dfrac 1 z & -\dfrac 1 z \\ \hline z & z & -z & \dfrac 1 z & -\dfrac 1 z \\ -z & -z & z & -\dfrac 1 z & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & -\dfrac 1 z & z & -z \\ -\dfrac 1 z & -\dfrac 1 z & \dfrac 1 z & -z & z \\ \end{array}$ </onlyinclude> Category:Examples of Cayley Tables Category:Group Generated by Reciprocal of z and Minus z to60cmgij1pjl5e8utjqi95ilhndakp


Group of Reflection Matrices Order 4/Cayley Table 0 49826 372448 370987 2018-10-22T06:17:56Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Group of Reflection Matrices Order $4$ == Consider the group of reflection matrices order $4$ :$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} }$ $R_4$ can be described completely by showing its Cayley table. Let: :$r_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_1 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ :$r_2 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_3 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ Then we have: <onlyinclude> :$\begin{array}{r|rrrr} \times & r_0 & r_1 & r_2 & r_3 \\ \hline r_0 & r_0 & r_1 & r_2 & r_3 \\ r_1 & r_1 & r_0 & r_3 & r_2 \\ r_2 & r_2 & r_3 & r_0 & r_1 \\ r_3 & r_3 & r_2 & r_1 & r_0 \\ \end{array}$ </onlyinclude> Category:Examples of Cayley Tables dgdam5yfreyiuqlpsfcllilq96gisd4


Multiplicative Group of Reduced Residues Modulo 8/Cayley Table 0 49828 371171 371165 2018-10-14T09:38:19Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Multiplicative Group of Reduced Residues Modulo 8 == The multiplicative group of reduced residues modulo $8$: :$\Z'_8 = \set {\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8}$ can be described completely by showing its Cayley table: <onlyinclude> :$\begin{array}{r|rrrr} \times_8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\ \hline \eqclass 1 8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\ \eqclass 3 8 & \eqclass 3 8 & \eqclass 1 8 & \eqclass 7 8 & \eqclass 5 8 \\ \eqclass 5 8 & \eqclass 5 8 & \eqclass 7 8 & \eqclass 1 8 & \eqclass 3 8 \\ \eqclass 7 8 & \eqclass 7 8 & \eqclass 5 8 & \eqclass 3 8 & \eqclass 1 8 \\ \end{array}$ </onlyinclude> Category:Examples of Cayley Tables Category:Multiplicative Group of Reduced Residues Modulo 8 tkrj64cqoh4o7k1sjya20y02wwf7qde


Klein Four-Group/Cayley Table

Klein Four-Group/Cayley Table 0 49829 379885 374806 2018-12-03T05:45:56Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Klein $4$-Group == The Klein $4$-group can be described completely by showing its Cayley table: <onlyinclude> :$\begin{array}{c|cccc}   & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$ </onlyinclude> {{NamedforDef|Felix Christian Klein}} == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Klein Four-Group|next = Symmetric Group on 3 Letters/Group Presentation}}: $\S 25$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Exponential on Real Numbers is Group Isomorphism/Proof 1|next = Definition:Klein Four-Group}}: $\S 1.5$: Example $15$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Definition:Klein Four-Group|next = Klein Four-Group is Group}}: Chapter $2$: The Definition of Group Structure: $\S 26 \iota$ * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Definition:Klein Four-Group|next = Reduced Residue System under Multiplication forms Abelian Group/Corollary}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $6$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Klein Four-Group|next = Definition:Symmetry Group of Rectangle}}: $\S 44.5$ Some consequences of Lagrange's Theorem Category:Klein Four-Group Category:Examples of Cayley Tables rmuh1ij005q9d890mjwk5plu3oio1hv


Complex Roots of Unity/Examples/Cube Roots 0 49894 451602 451599 2020-03-01T14:12:50Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Roots of Unity == <onlyinclude> The '''complex cube roots of unity''' are the elements of the set: :$U_3 = \set {z \in \C: z^3 = 1}$ They are: {{begin-eqn}} {{eqn | m = e^{0 i \pi / 3}       | o =       | r = 1       | mo= =       | c =  }} {{eqn | l = \omega        | mo= =       | m = e^{2 i \pi / 3}       | r = -\frac 1 2 + \frac {i \sqrt 3} 2 }} {{eqn | l = \omega^2        | mo= =       | m = e^{4 i \pi / 3}       | r = -\frac 1 2 - \frac {i \sqrt 3} 2 }} {{end-eqn}} The notation $\omega$ for, specifically, the complex '''cube''' roots of unity is conventional. </onlyinclude> === Conjugate Form === {{:Complex Roots of Unity/Examples/Cube Roots/Conjugate Form}} == Proof == {{:Complex Roots of Unity/Examples/Cube Roots/Proof}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Sum of Powers of Primitive Complex Roots of Unity|next = Difference of Two Cubes/Corollary}}: $\S 3$. Roots of Unity * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Roots of Complex Number/Examples/Cube Roots|next = Roots of Complex Number/Corollary/Examples/Cube Roots}}: Introduction * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Cube Root|next = Definition:Cubic (Geometry)|entry = cube root of unity}} Category:Examples of Complex Roots of Unity Category:Cube Roots of Unity plam55cl80oep7dnkq3wkwwvzv93wau


Primitive Complex Roots of Unity/Examples/Cube Roots 0 49902 431571 363692 2019-10-16T01:33:03Z Caliburn 3218 wikitext text/x-wiki == Examples of Primitive Complex Roots of Unity == <onlyinclude> The '''primitive complex cube roots of unity''' are: {{begin-eqn}} {{eqn | l = \omega        | mo= =       | m = e^{2 \pi i / 3}       | r = -\frac 1 2 + \frac {i \sqrt 3} 2 }} {{eqn | l = \omega^2        | mo= =       | m = e^{4 \pi i / 3}       | r = -\frac 1 2 - \frac {i \sqrt 3} 2 }} {{end-eqn}} </onlyinclude> == Proof == There are $3$ (complex) cube roots of unity: :$1, \omega, \omega^2$ We have: {{begin-eqn}} {{eqn | l = \omega        | r = e^{2 \pi i / 3} }} {{eqn | l = \omega^2        | r = e^{4 \pi i / 3} }} {{eqn | l = \omega^3        | r = 1       | c = Cube Roots of Unity }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \omega^2       | r = e^{4 \pi i / 3} }} {{eqn | l = \paren {\omega^2}^2       | r = \omega^3 \times \omega }} {{eqn | r = 1 \times \omega       | c = Cube Roots of Unity }} {{eqn | r = \omega       | c =  }} {{eqn | l = \paren {\omega^2}^3        | r = \paren {\omega^3}^2       | c = Cube Roots of Unity }} {{eqn | r = 1 \times 1       | c = Cube Roots of Unity }} {{eqn | r = 1       | c =  }} {{end-eqn}} Trivially, $1$ is not a primitive complex cube root of unity because you cannot make either $\omega$ or $\omega^2$ by multiplying $1$ by itself as many times as you like. Hence the result. {{qed}} Category:Cube Roots of Unity thejark721e79ld0xl97nug8xzhwxdl


Symmetric Group on 3 Letters

Symmetric Group on 3 Letters 0 49917 388523 388518 2019-01-19T12:30:30Z Prime.mover 59 wikitext text/x-wiki == Group Example == <onlyinclude> Let $S_3$ denote the set of permutations on $3$ letters. The '''symmetric group on $3$ letters''' is the algebraic structure: :$\struct {S_3, \circ}$ where $\circ$ denotes composition of mappings. </onlyinclude> It is usually denoted, when the context is clear, without the operator: $S_3$. === Cycle Notation === It can be expressed in the form of permutations given in cycle notation as follows: {{:Symmetric Group on 3 Letters/Cycle Notation}} === Cayley Table === {{:Symmetric Group on 3 Letters/Cayley Table}} === Group Presentation === Its group presentation is: {{:Symmetric Group on 3 Letters/Group Presentation}} == Order of Elements == {{:Symmetric Group on 3 Letters/Order of Elements}} == Subgroups == {{:Symmetric Group on 3 Letters/Subgroups}} == Normal Subgroups == {{:Symmetric Group on 3 Letters/Normal Subgroups}} == Generators == {{:Symmetric Group on 3 Letters/Generators}} == Centralizers == {{:Symmetric Group on 3 Letters/Centralizers}} == Normalizers of Subgroups == {{:Symmetric Group on 3 Letters/Normalizers}} == Center == {{:Symmetric Group on 3 Letters/Center}} == Conjugacy Classes == {{:Symmetric Group on 3 Letters/Conjugacy Classes}} == Also see == * Symmetric Group is Group, which demonstrates that this is a (finite) group. Category:Groups of Order 6 Category:Symmetric Group on 3 Letters Category:Symmetric Groups h2sn5oa3n3faf7jeos3g7n2x850c8ve


Solution to Card Game with Bluffing

Solution to Card Game with Bluffing 0 50132 270689 270687 2016-09-13T06:35:01Z Prime.mover 59 wikitext text/x-wiki == Solution to Card Game with Bluffing == {{:Definition:Card Game with Bluffing}} == Proof == <onlyinclude> From the payoff table: {{:Definition:Card Game with Bluffing/Payoff Table}} The solution is: : $A$ takes strategy $A_1$ for $2/3$ of the time, and $A_2$ for $1/3$ of the time. : $B$ takes strategy $B_1$ for $2/3$ of the time, and $B_2$ for $1/3$ of the time. {{explain|It is not made clear in the source work why this is. You just get: "This is indeed a solution, because no player can do better if the other sticks to his strategy, but could gain or lose more than $1/3$ (which is what $A$ gains now) if he departed from the optimal solution and his opponent took advantage of it." This is handwaving. A source work with a higher level of precision is needed here.}} </onlyinclude> == Sources == * {{BookReference|The Theory of Games and Linear Programming|1956|Steven Vajda|prev = Definition:Card Game with Bluffing/Payoff Table|next = Two-Person Zero-Sum Game with Multiple Solutions}}: Chapter $\text{I}$: An Outline of the Theory of Games: $3$ Category:Examples of Two-Person Zero-Sum Games 4ksn9gq6ze2pq47senvjeub2xvbd4sp


Induction of Finite Set

Induction of Finite Set 0 50174 414435 270828 2019-07-24T17:03:06Z Prime.mover 59 wikitext text/x-wiki == Theorem Scheme == Let $A$ be finite set. Let $\map P -$ be a predicate. Let $\map P \O$. Let :$\forall B \subseteq A, x \in A: \paren {\map P B \implies \map P {B \cup \set x} }$ Then: :$\map P A$ == Proof == We will prove the result by induction on cardinality of argument. === Base Case === :$\forall X \subseteq A: \paren {\size X = 0 \implies \map P X}$ Let $X \subseteq A$ such that: :$\size X = 0$ By Cardinality of Empty Set: :$X = \O$ Thus by assumption: :$\map P X$ === Induction Hypothesis === :$\forall X \subseteq A: \paren {\size X = n \implies \map P X}$ === Induction Step === :$\forall X \subseteq A: \paren {\size X = n + 1 \implies \map P X}$ Let $X \subseteq A$ such that: :$\size X = n + 1$ By definition of cardinality: :$X = \set {x_1, \dots, x_n, x_{n + 1} }$ By Union of Unordered Tuples: :$X = \set {x_1, \dots, x_n} \cup \set {x_{n + 1} }$ By definition of cardinality: :$\size {\set {x_1, \dots, x_n} } = n$ By Set is Subset of Union: :$\set {x_1, \dots, x_n} \subseteq X \subseteq A$ Then by Induction Hypothesis: :$\map P {\set {x_1, \dots, x_n} }$ By definition of subset: :$x_{n + 1} \in A$ Thus by assumption: :$\map P X$ {{qed|lemma}} By the Principle of Mathematical Induction: :$\forall X \subseteq A: \paren {\size X = \size A \implies \map P X}$ Hence: :$\map P A$ {{qed}} == Sources == * {{Mizar|link = finset_1|sublink = S2|display = FINSET_1:sch 2}} Category:Set Theory kvjbq876k6dqr3q8pdqldwexmetyrrf


Infimum of Real Subset

Infimum of Real Subset 0 50378 407208 271659 2019-06-10T07:08:22Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $S$ be a set of extended real numbers. Let $S$ be bounded below (in $\R$). Let $T = S \cap \R$. Then: :$S$ admits an infimum (in $\R$) if and only if $T$ admits an infimum (in $\R$) and, if $\inf S$ and $\inf T$ exist as real numbers: :$\inf S = \inf T$ == Proof == We observe that $T$ constitutes the real numbers of $S$. Since there is a real number that is a lower bound for $S$, $-\infty$ is not an element of $S$. Accordingly, $\infty$ is the only possible element of $S \setminus T$. Therefore: :$S$ is a subset of $T \cup \set \infty$ First, we show that $S$ and $T$ have the same set of lower bounds. Let $b$ be a lower bound (in $\R$) for $S$. Then $b$ is a lower bound for $T$ as $T$ is a subset of $S$. Therefore: :the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$ Assume that $c$ is a lower bound (in $\R$) for $T$. Then $c$ is a lower bound for $T \cup \left\{{\infty}\right\}$ as well since $c < \infty$. Accordingly, $c$ is a lower bound for $S$ since $S$ is a subset of $T \cup \set \infty$. Therefore: :the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$ We have: :the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$ :the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$ Therefore: :the set of lower bounds for $T$ equals the set of lower bounds for $S$ by definition Next, we show that $S$ and $T$ have the same infima. We have that $S$ and $T$ have the same set of lower bounds. Therefore, $S$ and $T$ have the same greatest lower bound in $\overline \R$. Accordingly, as a corollary, if one of the sets $S$ and $T$ admits an infimum (in $\R$), so does the other. Furthermore, these infima are equal. {{qed}} Category:Extended Real Numbers kx4t3fbsz4b7rckt7fjm04bbiubtkgz


Infimum of Set of Oscillations on Set

Infimum of Set of Oscillations on Set 0 50466 407204 272152 2019-06-10T06:32:13Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $f: D \to \R$ be a real function where $D \subseteq \R$. Let $x$ be a point in $D$. Let $S_x$ be a set of real sets that contain (as an element) $x$. Let: :$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in S_x}$ where $\map {\omega_f} I$ denotes the oscillation of $f$ on the set $I$: :$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ Then: :$\map {\omega_f} x \in \R$ {{iff}} $\set {\map {\omega_f} I: I \in S_x}$ contains a real number. == Proof == Let: :$S = \set {\map {\omega_f} I: I \in S_x}$ We observe that: :$\inf S = \map {\omega_f} x$ === Necessary Condition === Let $\inf S \in \R$. We need to prove that $S$ contains a real number. Note that $S$ is non-empty as the empty set does not admit an infimum (in $\R$). Therefore, $S$ has at least one element. Accordingly, there is an $I \in S_x$ such that $\map {\omega_f} I \in S$. Let $I \in S_x$. Therefore, $x \in I$. From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I$ is an extended real number. Therefore $S$ is a set of extended real numbers as $S = \set {\map {\omega_f} I: I \in S_x}$. Accordingly, $S$ contains a real number by Infimum of Subset of Extended Real Numbers is Arbitrarily Close as $\inf S \in \R$. {{qed|lemma}} === Sufficient Condition === Let $S$ contain a real number. We need to prove that $\inf S \in \R$. We have: :$S \cap \R$ is non-empty as $S$ contains a real number. Let $I \in S_x$. Therefore, $x \in I$. From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I \in \overline \R_{\ge 0}$. Therefore: :$S$ is a subset of $\overline \R_{\ge 0}$ as $S = \set {\map {\omega_f} I: I \in S_x}$ Accordingly: :$S$ is bounded below. From this follows that: :$S \cap \R$ is bounded below as $SR$ is a subset of $S$ We have: :$S \cap \R$ is bounded below :$S \cap \R$ is not empty Therefore: :$\inf S \cap \R \in \R$ Continuum Property We have: :$S$ is a set of extended real numbers as $S$ is a subset of $\overline \R_{\ge 0}$ :$S$ is bounded below Therefore: :$\inf S \in \R$ by Infimum of Real Subset as $\inf S \cap \R \in \R$ {{qed}} Category:Real Analysis Category:Oscillation 6csqfmfq0gbqeu6e9pqlkytunxctepo


Infimum of Set of Oscillations on Set is Arbitrarily Close 0 50507 272883 272273 2016-10-10T07:37:43Z Ivar Sand 2302 Generalized wikitext text/x-wiki == Lemma == Let $f: D \to \R$ be a real function where $D \subseteq \R$. Let $x$ be a point in $D$. Let $S_x$ be a set of real sets that contain (as an element) $x$. Let: :$\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$: :$\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ Let $\epsilon \in \R_{>0}$. Let $\omega_f \left({x}\right) \in \R$. Then an $I \in S_x$ exists such that: :$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ == Proof == Let $\epsilon \in \R_{>0}$. Let $\omega_f \left({x}\right) \in \R$. We need to prove that an $I \in S_x$ exists such that: :$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ We have that $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in S_x$ by Oscillation on Set is an Extended Real Number. Therefore: :$\left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ is a subset of $\overline{\R}$ We have also: :$\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} \in \R$ as $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} = \omega_f \left({x}\right)$ Therefore, an $I \in S_x$ exists such that: {{begin-eqn}} {{eqn | l = \omega_f \left({I}\right) - \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}       | o = <        | r = \epsilon       | c = by Infimum of Subset of Extended Real Numbers is Arbitrarily Close }} {{eqn | ll = \iff       | l = \omega_f \left({I}\right) - \omega_f \left({x}\right)       | o = <        | r = \epsilon       | c = as $\omega_f \left({x}\right) = \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}$ }} {{end-eqn}} {{qed|lemma}} Category:Real Analysis ez6vb772zpbteefj77vvv2y8o5rk4rh


Reciprocal/Examples/Euler's Number

Reciprocal/Examples/Euler's Number 0 50525 433471 379617 2019-11-01T13:37:32Z Prime.mover 59 wikitext text/x-wiki == Example of Reciprocal == The reciprocal of Euler's Number $e$ is approximately: <onlyinclude> :$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$ </onlyinclude> {{OEIS|A068985}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Reciprocal/Examples/Pi|next = Envelope Problem}}: $0 \cdotp 367 \, 879 \, 441 \, 171 \, 442 \, 321 \, 595 \, 523 \, 770 \, 161 \, 460 \, 867 \, 445 \, 811 \, 131 \, 031$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = One Third as Quotient of Sequences of Odd Numbers|next = Euler's Number as Sum of Egyptian Fractions}}: $0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031$ Category:Examples of Reciprocals Category:Euler's Number fkt7k64j5ggfv11l4uh6q12mhgl4a9m


Continued Fraction Expansion of Irrational Square Root/Examples/2 0 50722 470631 433258 2020-05-26T06:33:17Z Prime.mover 59 Prime.mover moved page Continued Fraction Expansion of Irrational Square Root/Example/2 to Continued Fraction Expansion of Irrational Square Root/Examples/2 wikitext text/x-wiki == Examples of Continued Fraction Expansion of Irrational Square Root == <onlyinclude> The continued fraction expansion of the square root of $2$ is given by: :$\sqrt 2 = \sqbrk {1, \sequence 2}$ </onlyinclude> {{OEIS|A040000}} == Proof == {{begin-eqn}} {{eqn | l = \sqrt 2       | r = 1 + \paren {\sqrt 2 − 1}       | c =  }} {{eqn | r = 1 + \frac {\paren {\sqrt 2 − 1} \paren {\sqrt 2 + 1} } {\sqrt 2 + 1}       | c = multiplying top and bottom by $\sqrt 2 + 1$ }} {{eqn | r = 1 + \frac {\paren {\sqrt 2}^2 − 1^2} {\sqrt 2 + 1}       | c = Difference of Two Squares }} {{eqn | r = 1 + \frac 1 {1 + \sqrt 2}       | c = as $\paren {\sqrt 2}^2 − 1^2 = 2 - 1 = 1$ }} {{end-eqn}} Thus it is possible to replace $\sqrt 2$ recursively: {{begin-eqn}} {{eqn | l = \sqrt 2       | r = 1 + \frac 1 {1 + \sqrt 2}       | c =  }} {{eqn | r = 1 + \frac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {1 + \sqrt 2} }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } } }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {2 + \cfrac 1 {1 + \sqrt 2} } }       | c =  }} {{end-eqn}} The pattern repeats indefinitely, producing the continued fraction expansion: :$\sqrt 2 = \sqbrk {1, 2, 2, 2, \ldots} = \sqbrk {1, \sequence 2}$ {{handwaving}} {{qed}} Category:Continued Fractions 4kyf0xltted86u8fkvn1ttmfsyps42z


Riemann Zeta Function at Even Integers/Examples/2 0 50830 433362 393171 2019-11-01T09:37:43Z Prime.mover 59 wikitext text/x-wiki == Example of Riemann Zeta Function at Even Integers == <onlyinclude> The Riemann zeta function of $2$ is given by: {{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots       | c =  }} {{eqn | r = \dfrac {\pi^2} 6       | c =  }} {{eqn | o = \approx       | r = 1 \cdotp 64493 \, 4066 \ldots       | c =  }} {{end-eqn}} </onlyinclude> {{OEIS|A013661}} == Proof == {{:Basel Problem/Proof 6}} The decimal expansion can be found by an application of arithmetic. == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Wythoff's Game|next = Square Root/Examples/3}}: $1 \cdotp 644 \, 934 \, 066 \ldots$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Riemann Zeta Function at Even Integers|next = Riemann Zeta Function of 4}}: $\S 1.2.7$: Harmonic Numbers: $(7)$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Wythoff's Game|next = Zeta of 2 as Product of Fractions with Prime Numerators}}: $1 \cdotp 64493 \, 4066 \ldots$ Category:Riemann Zeta Function at Even Integers 4ddaa3tqvt1zokkd9f2ykl46o2pmv2k


Axiom:Real Number as Complex Number

Axiom:Real Number as Complex Number 100 51043 274515 274511 2016-10-30T14:33:07Z Prime.mover 59 wikitext text/x-wiki {{refactor|obviously needs refactoring}} {{MissingLinks|Needs a category}} == Axiom == Let $x \in \R$ be a real number. Then: : $x = \left({x, 0}\right)$ where $\left({a, b}\right)$ where $a, b \in R$ denotes a complex number. Also: : $x = x + 0 i$ where $a + b i$ where $a, b \in R$ denotes a complex number. 0ceu0zey8cofgzlvd94lyogii14qmr8


Magic Square/Examples/Order 3

Magic Square/Examples/Order 3 0 51180 446278 424228 2020-02-04T15:47:05Z Prime.mover 59 wikitext text/x-wiki == Example of Order $3$ Magic Square == Order $3$ magic square: <onlyinclude> :$\begin{array}{|c|c|c|} \hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end{array}$ </onlyinclude> == Also known as == {{:Magic Square/Examples/Order 3/Also known as}} == Also see == * Smallest Magic Square is of Order 3 * Magic Constant of Order 3 Magic Square == Historical Note == {{:Magic Square/Examples/Order 3/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Hilbert-Waring Theorem/Particular Cases/4/Historical Note|next = Magic Constant of Order 3 Magic Square}}: $9$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Hilbert-Waring Theorem/Particular Cases/4/Historical Note|next = Magic Constant of Order 3 Magic Square}}: $9$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 3/Also known as|entry = magic square}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 3/Also known as|entry = magic square}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 4/Dürer|entry = magic square}} Category:Magic Squares izm9s929wxgdxosf4i8k13lsxtsnq01


Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary 0 51241 468131 459088 2020-05-16T09:47:50Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions == Let $f$ be a real function which is Darboux integrable on any closed interval $\mathbb I$. <onlyinclude> Let $a_0, a_1, \ldots, a_n$ be real numbers, where $n \in \N$ and $n \ge 2$. Then: :$\displaystyle \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$ </onlyinclude> == Proof == Proof by induction: === Basis for the Induction === According to Sum of Integrals on Adjacent Intervals for Integrable Functions, $n = 2$ holds. This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\displaystyle \int_{a_0}^{a_k} \map f t \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$ Now we need to show true for $n = k + 1$: :$\displaystyle \int_{a_0}^{a_{k + 1} } \map f t \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \int_{a_0}^{a_{k + 1} } \map f t \rd t       | r = \int_{a_0}^{a_k} \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t       | c =  }} {{end-eqn}} The result follows by induction. {{qed}} == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Piecewise Continuous Function with One-Sided Limits is Darboux Integrable|next = Definition:Real Right-Hand Derivative}}: Chapter Two: $\S 1$. Piecewise-Continuous Functions: $(5)$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Improper Integral on Open Below Interval|next = Definite Integral of Constant Multiple of Real Function}}: $\S 15$: General Formulas involving Definite Integrals: $15.7$ Category:Proofs by Induction Category:Definite Integrals k1fa60kmb57gjadcvaqfg3nbpzzr9dn


Integration by Substitution/Corollary

Integration by Substitution/Corollary 0 51242 483482 468279 2020-08-28T12:48:52Z Prime.mover 59 wikitext text/x-wiki == Corollary to Integration by Substitution == <onlyinclude> Let $f : \R \to \R$ be a real function. Let $f$ be integrable. Let $a$, $b$, and $c$ be real numbers. Then: :$\displaystyle \int_{a - c}^{b - c} \map f t \rd t = \int_a^b \map f {t - c} \rd t$ </onlyinclude> == Proof == Let $\map \phi u = u - c$. By Sum Rule for Derivatives, Derivative of Identity Function, and Derivative of Constant, we have: :$\map {\phi'} u = 1$ By Integration by Substitution: {{begin-eqn}} {{eqn | l = \int_{\map \phi a}^{\map \phi b} \map f t \rd t       | r = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u       | c =  }} {{eqn | ll = \leadsto       | l = \int_{a - c}^{b - c} \map f t \rd t       | r = \int_a^b \map f {u - c} \paren 1 \rd u       | c =  }} {{eqn | r = \int_a^b \map f {u - c} \rd u       | c =  }} {{eqn | r = \int_a^b \map f {t - c} \rd t       | c =  }} {{end-eqn}} {{qed}} Category:Integration by Substitution 1ipmzu7snocnfv8io33p1jmkhas10ci


Finite Ordinal Times Ordinal/Lemma

Finite Ordinal Times Ordinal/Lemma 0 51259 275440 2016-11-06T09:54:20Z Kc kennylau 2331 Created page with "== Lemma ==  Let $m$ be a finite ordinal.  Let $m \ne 0$, where $0$ is the zero ordinal.   Then:  : $m \times \omeg..." wikitext text/x-wiki == Lemma == Let $m$ be a finite ordinal. Let $m \ne 0$, where $0$ is the zero ordinal. Then: : $m \times \omega = \omega$ where $\omega$ denotes the minimal infinite successor set. == Proof == {{begin-eqn}} {{eqn | ll = \forall n \in \omega       | l = m \times n       | o = \in       | r = \omega       | c = Natural Number Multiplication is Closed }} {{eqn | l = \bigcup_{n \mathop \in \omega} \left({ m \times n }\right)       | o = \le       | r = \omega       | c = Supremum Inequality for Ordinals }} {{eqn | ll = \implies       | l = m \times \omega       | o = \le       | r = \omega       | c = Definition of Ordinal Multiplication }} {{end-eqn}} Also, $\omega \le \left({ m \times \omega }\right)$ by Subset is Right Compatible with Ordinal Multiplication. The lemma follows from the definition of equality. {{qed|lemma}} Category:Ordinal Arithmetic Category:Transfinite Arithmetic Category:Finite Ordinals Category:Minimal Infinite Successor Set 18ma4g26xnn9ic9e0uzk74435l9nfut


Binary Logarithm/Examples/10

Binary Logarithm/Examples/10 0 51349 433259 433010 2019-11-01T08:32:16Z Prime.mover 59 wikitext text/x-wiki == Example of Binary Logarithm == <onlyinclude> The binary logarithm of $10$ is: :$\log_2 10 \approx 3 \cdotp 32192 \, 80948 \, 87362 \, 34787 \, 0319 \ldots$ </onlyinclude> {{OEIS|A020862}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square Root/Examples/10|next = Number of Binary Digits in Power of 10}}: $3 \cdotp 321 \, 928 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square Root/Examples/10|next = Number of Binary Digits in Power of 10}}: $3 \cdotp 32192 \, 8 \ldots$ Category:Examples of Binary Logarithms 9g9r9j5lxwxupvxo71z87icnax3idki


Bound for Analytic Function and Derivatives 0 51357 435939 435768 2019-11-21T10:51:19Z Ivar Sand 2302 Refactor template removed wikitext text/x-wiki == Lemma == Let $f$ be a complex function. Let $z_0$ be a point in $\C$. Let $r$ be a real number in $\R_{>0}$. Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$. Let $f$ be analytic on $\Gamma$ and its interior. Let $t \in \C$ be such that $\cmod {t - z_0} < r$. Then a real number $M$ exists such that, for every $n \in \N$: :$\displaystyle \cmod {\map {f^{\paren n} } t} \le \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^\paren {n + 1} }$ == Proof == === Lemma (Analytic Function Bounded on Circle) === {{:Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle}}{{qed|lemma}} We have: :$f$ is analytic on $\Gamma$ and its interior :$t$ is in the interior of $\Gamma$ Therefore: :$\displaystyle \map {f^{\paren n} } t = \frac {n!} {2 \pi i} \int_\Gamma \frac {\map f z} {\paren {z - t}^{\paren {n + 1} } } \rd z$ by Cauchy's Integral Formula for Derivatives where $\Gamma$ is traversed counterclockwise. We have that $f$ is bounded on $\Gamma$ by Lemma (Analytic Function Bounded on Circle). Therefore, there is a positive real number $M$ that satisfies: :$M \ge \cmod {\map f z}$ for every $z$ on $\Gamma$ We have $\cmod {t - z_0} < r$. Therefore: :$0 < r - \cmod {t - z_0}$ We observe that $r - \cmod {t - z_0}$ is the minimum distance between $t$ and $\Gamma$. Therefore: :$\paren {r - \cmod {t - z_0} } \le \cmod {z - t}$ for every $z$ on $\Gamma$ We get: {{begin-eqn}} {{eqn | l = \cmod {\map {f^{\paren n} } t}       | r = \cmod {\frac {n!} {2 \pi i} \int_\Gamma \frac {\map f z} {\paren {z - t}^{\paren {n + 1} } } \rd z} }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac {\cmod {\map f z} } {\cmod {z - t}^{\paren {n + 1} } } \cmod {\d z} }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac M {\cmod {z - t}^{\paren {n + 1} } } \cmod {\d z}       | c = as $M \ge \cmod {\map f z}$ for every $z$ on $\Gamma$ }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } \cmod {\d z}       | c = as $0 < \paren {r - \cmod {t - z_0} } \le \cmod {z - t}$ for every $z$ on $\Gamma$ }} {{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } \int_\Gamma \cmod {\d z} }} {{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } 2 \pi r }} {{eqn | r = \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } }} {{end-eqn}} {{qed}} {{explain|The notation $\cmod {\d z}$ has not been raised before on {{ProofWiki}} -- we need to do something about that}} Category:Complex Analysis kfj9fmwm0855gzst2c2sez0pfyzc49o


Pythagorean Triangle/Examples/3-4-5

Pythagorean Triangle/Examples/3-4-5 0 51433 478808 478785 2020-07-18T13:12:02Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $3$, $4$ and $5$ is a primitive Pythagorean triangle. </onlyinclude> :300px == Proof == {{begin-eqn}} {{eqn | l = 3^2 + 4^2       | r = 9 + 16       | c =  }} {{eqn | r = 25       | c =  }} {{eqn | r = 5^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $3$, $4$ and $5$ form a Pythagorean triple. Note that $3$ and $4$ are coprime. Hence, by definition, $3$, $4$ and $5$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} == Also see == * Smallest Pythagorean Triangle is 3-4-5 * Pythagorean Triangle whose Area is Half Perimeter: its area and semiperimeter are both $6$ * Pythagorean Triangle with Sides in Arithmetic Sequence == Historical Note == {{:Pythagorean Triangle/Examples/3-4-5/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Pythagorean Triangle|next = Pythagoras's Theorem}}: $5$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 6/Historical Note|next = Definition:Perfect Number}}: $6$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Definition:Pythagorean Triple|next = Babylonian Mathematics/Examples/Sliding Ladder}}: Pythagorean Triples * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Pythagorean Triangle|next = Pythagoras's Theorem}}: $5$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 6/Historical Note|next = Definition:Perfect Number}}: $6$ Category:Examples of Pythagorean Triangles lwzjl2zodgtbfkortdr19jog0vbvdmh


Volume of Unit Hypersphere/Sequence

Volume of Unit Hypersphere/Sequence 0 51628 381091 276914 2018-12-10T23:29:46Z Prime.mover 59 wikitext text/x-wiki == Sequence of Volumes of Unit Hyperspheres == <onlyinclude> The sequence of volumes of the unit sphere in $n$-dimensional space begins as follows: {{begin-eqn}} {{eqn | l = n = 1       | o = :       | r = \map V 1 = 2 }} {{eqn | l = n = 2       | o = :       | r = \map V 2 = 3.1 }} {{eqn | l = n = 3       | o = :       | r = \map V 3 = 4.2 }} {{eqn | l = n = 4       | o = :       | r = \map V 4 = 4.9 }} {{eqn | l = n = 5       | o = :       | r = \map V 5 = 5.264 }} {{eqn | l = n = 6       | o = :       | r = \map V 6 = 5.2 }} {{eqn | l = n = 7       | o = :       | r = \map V 7 = 4.7 }} {{end-eqn}} </onlyinclude> == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Maximum Volume of Unit Radius Sphere in Fractional Dimensions|next = 6}}: $5 \cdotp 256 \, 946 \, 404 \, 860 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Maximum Volume of Unit Radius Sphere in Fractional Dimensions|next = 6}}: $5 \cdotp 25694 \, 64048 \, 60 \ldots$ Category:Spheres 2132urzjbd5jd3ru1q589926yp8zeyw


Complex Power by Complex Exponential is Analytic 0 51901 405960 353765 2019-05-30T07:10:27Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\psi, \eta \in \C$ be constant. Let $\map f z = z^\psi \, \map \exp {-\eta z}$, where: :$z^\psi$ denotes $z$ to the power of $\psi$, defined on its principal branch :$\map \exp {-\eta z}$ denotes the complex exponential function. Then $f$ is analytic on any simply connected domain that does not contain the origin nor any points on the negative real axis. == Proof == Let $z$ be written in exponential form: :$z = r \map \exp {i \theta}$  where: :$r > 0$ :$\theta \in \hointl {-\pi} \pi$ Let $\psi = a + i b, \eta = c + i d$. By the definition of $f$: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta} }       | r = z^\psi \, \map \exp {-\eta z} }} {{eqn | r = \paren {r e^{i \theta} }^\psi \, \map \exp {-\eta r \, \map \exp {i \theta} } }} {{eqn | r = \map \exp {\psi \, \map \Log {r \, \map \exp {i \theta} } } \, \map \exp {-\eta r \paren {\cos \theta + i \sin \theta} } }} {{eqn | r = \map \exp {\psi \ln r + i \psi \theta - \eta r \cos \theta - i \eta r \sin \theta} }} {{eqn | r = \map \exp {\paren {a + i b} \ln r + i \paren {a + i b} \theta - \paren {c + i d} r \cos \theta - i \paren {c + i d} r \sin \theta} }} {{eqn | r = \map \exp {a \ln r + i b \ln r + i a \theta - b \theta - c r \cos \theta - i d r \cos \theta - i c r \sin \theta + d r \sin \theta} }} {{eqn | r = \map \exp {a \ln r - b \theta - c r \cos \theta + d r \sin \theta} \, \map \exp {i \paren {b \ln r + a \theta - d r \cos \theta - c r \sin \theta} } }} {{end-eqn}} Define: {{begin-eqn}} {{eqn | l = \map g {r, \theta}       | r = a \ln r - b \theta - c r \cos \theta + d r \sin \theta }} {{eqn | l = \map h {r, \theta}       | r = b \ln r + a \theta - d r \cos \theta - c r \sin \theta }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta} }       | r = \map \exp g \, \map \exp {i h} }} {{eqn | r = \map \exp g \, \map \cos h + i \map \exp g \, \map \sin h }} {{end-eqn}} Define: {{begin-eqn}} {{eqn | l = \map u {r, \theta}       | r = \map \exp g \, \map \cos h }} {{eqn | l = \map v {r, \theta}       | r = \map \exp g \, \map \sin h }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta } }        | r = u + iv }} {{end-eqn}} We check the Polar Form of Cauchy-Riemann Equations. As a preliminary:  {{begin-eqn}} {{eqn | l = \frac {\partial g} {\partial r}       | r = \frac a r - c \cos \theta + d \sin \theta }} {{eqn | l = \frac {\partial h} {\partial r}       | r = \frac b r - d \cos \theta - \sin \theta }} {{eqn | l = \frac {\partial g} {\partial \theta}       | r = -b + c r \sin \theta + d r \cos \theta }} {{eqn | l = \frac {\partial h} {\partial \theta}       | r = a + d r \sin \theta - c r \cos \theta }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \frac {\partial u} {\partial r}       | r = \map \exp g \frac {\partial g} {\partial r} \map \cos h - \map \exp g \, \map \sin f \frac {\partial h} {\partial r} }} {{eqn | r = \frac 1 r \map \exp g \paren {\paren {a - r c \cos \theta + r d \sin \theta} \, \map \cos h + \map \sin h \paren {- b + r d \cos \theta + r c \sin \theta} } }} {{eqn | r = \frac 1 r \map \exp g \paren {\frac {\partial h} {\partial \theta} \map \cos h + \, \map \sin h \frac {\partial g} {\partial \theta} } }} {{eqn | r = \frac 1 r \frac {\partial v} {\partial \theta} }} {{eqn | l = \frac {\partial u} {\partial \theta}       | r = \map \exp g \frac {\partial g} {\partial \theta} \, \map \cos h - \map \exp g \, \map \sin h \frac {\partial h} {\partial \theta} }} {{eqn | r = -r \, \map \exp g \paren {\paren {\frac b r - c \sin \theta - d \cos \theta} \, \map \cos h + \map \sin h \paren {\frac a r + d \sin \theta - c  \cos \theta} }       | c = Sine Function is Odd }} {{eqn | r = -r \, \map \exp g \paren {\frac {\partial h} {\partial r} \, \map \cos h + \map \sin h \frac {\partial g} {\partial r} } }} {{eqn | r = -r \frac {\partial v} {\partial r} }} {{end-eqn}} {{qed}} Category:Complex Analysis oaqckgvsft6v52oxrbqf9p5bjzs3sz8