X to the x is not of Exponential Order/Lemma 0 52223 357277 357276 2018-05-27T05:21:45Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $f: \R \to \R$ be defined on $\left [{0 \,.\,.\, \to} \right)$ with $f \left({x}\right) = x^x$. Suppose there exist strictly positive real constants $M, K, a \in \R_{> 0}$ such that: :$\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{a t}$ Then there exists a constant $C$ such that: :$\forall t > C: \left\vert {f \left({t}\right)} \right \vert > K e^{a t}$ </onlyinclude> == Proof == By the definition of power: :$f \left({t}\right) = \exp \left({t \ln t}\right)$  By Exponential of Real Number is Strictly Positive, we can reduce the lemma into the existence of $C$ such that: :$\forall t > C: f \left({t}\right) > K e^{a t}$ We will divide into two cases. === Case 1: $K > 1$ === Assume that $t > K e^a$. {{begin-eqn}} {{eqn | l = a       | o = >       | r = 0       | c =  }} {{eqn | l = e^a       | o = >       | r = e^0       | c = Exponential is Strictly Increasing }} {{eqn | l = e^a       | o = >       | r = 1       | c = Exponential of Zero }} {{eqn | l = K e^a       | o = >       | r = 1       | c = As $K > 1$ }} {{eqn | l = t       | o = >       | r = 1       | c = As $t > K e^a$ }} {{eqn | n = 1       | l = t \ln K       | o = >       | r = \ln K       | c =  }} {{eqn | l = t       | o = >       | r = K e^a       | c = Assumption }} {{eqn | l = \ln t       | o = >       | r = \ln \left({K e^a}\right)       | c = Logarithm is Strictly Increasing }} {{eqn | l = \ln t       | o = >       | r = a + \ln K       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + t \ln K       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + \ln K       | c = from $(1)$ }} {{eqn | l = \exp \left({t \ln t}\right)       | o = >       | r = \exp \left({a t + \ln K}\right)       | c = Exponential is Strictly Increasing }} {{eqn | l = f \left({t}\right)       | o = >       | r = K e^{a t}       | c = Exponential is Strictly Increasing }} {{end-eqn}} Here, $C = K e^a$. === Case 2: $K \le 1$ === Assume that $t > e^a$. {{begin-eqn}} {{eqn | l = \ln K       | o = \le       | r = \ln 1       | c = Logarithm is Strictly Increasing }} {{eqn | n = 1       | l = \ln K       | o = \le       | r = 0       | c = Logarithm of 1 is 0 }} {{eqn | l = t       | o = >       | r = e^a       | c = by assumption }} {{eqn | l = \ln t       | o = >       | r = a       | c = Logarithm is Strictly Increasing }} {{eqn | l = t \ln t       | o = >       | r = a t       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + \ln k       | c = from $(1)$ }} {{eqn | l = \exp \left({t \ln t}\right)       | o = >       | r = \exp \left({a t + \ln K}\right)       | c = Exponential is Strictly Increasing }} {{eqn | l = f \left({t}\right)       | o = >       | r = K e^{a t}       | c = Exponential is Strictly Increasing }} {{end-eqn}} Here, $C = e^a$. {{qed}} Category:Exponential Order tnwa4ejjlezlzh9pfl6um4dtvf4c64u


Schanuel's Conjecture

Schanuel's Conjecture 0 52233 440312 353764 2019-12-20T13:58:41Z Prime.mover 59 wikitext text/x-wiki == Conjecture == Let $z_1, \cdots, z_n$ be complex numbers that are linearly independent over the rational numbers $\Q$. Then: :the extension field $\map \Q {z_1, \cdots, z_n, e^{z_1}, \cdots, e^{z_n} }$ has transcendence degree at least $n$ over $\Q$ where $e^z$ is the complex exponential of $z$. {{Namedfor|Stephen Hoel Schanuel|cat = Schanuel}} Category:Transcendental Number Theory Category:Unproven Hypotheses Category:Schanuel's Conjecture aw8yqn1akkivluwqxl046e7ci2s6qy4


Characterization of Cosine Integral Function 0 52237 417225 405915 2019-08-08T14:18:35Z Caliburn 3218 wikitext text/x-wiki == Definition == Let $\Ci: \R_{>0}: \R$ denote the cosine integral function: :$\map \Ci x = \displaystyle \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {\cos t} t \rd t$ Then: :$\map \Ci x = -\gamma - \ln x + \displaystyle \int_{t \mathop \to 0}^{t \mathop = x} \frac{1 - \cos t} t \rd t$ where $\gamma$ is the Euler-Macheroni constant. == Proof == {{proof wanted|Laplace transform to the one, then invert to the other?}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Cosine Integral Function|next = Power Series Expansion for Cosine Integral Function plus Logarithm}}: $\S 35$: Miscellaneous Special Functions: Cosine Integral: $35.14$ * {{MathWorld|Cosine Integral|CosineIntegral}} Category:Cosine Integral Function py644aus0b491ucur1l08l2rlwh2zqg


Sum of Terms of Magic Square/Sequence

Sum of Terms of Magic Square/Sequence 0 52261 433514 382814 2019-11-01T14:01:42Z Prime.mover 59 wikitext text/x-wiki == Sequence of Sums of Terms of Magic Squares == <onlyinclude> The sequence of the sum totals of all the entries in magic squares of order $n$ begins: :$1, \paren {10,} \, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10 \, 440, 14 \, 365, 19 \, 306, 25 \, 425, 32 \, 896, \ldots$ However, note that while $10 = \dfrac {2^2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist. </onlyinclude> {{OEIS|A037270}} Category:Magic Squares oqru1w4drzmzya0javj3gt9qjay04bc


Magic Constant of Magic Square/Sequence

Magic Constant of Magic Square/Sequence 0 52274 433416 382848 2019-11-01T12:58:36Z Prime.mover 59 wikitext text/x-wiki == Sequence of Magic Constants of Magic Squares == <onlyinclude> The sequence of the magic constants of magic squares of order $n$ begins: :$1, (5,) \, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, \ldots$ However, note that while $5 = \dfrac {2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist. </onlyinclude> {{OEIS|A006003}} Category:Magic Squares b0xwscvu10axivs1fnjg6ulrk1283ei


Schanuel's Conjecture Implies Transcendence of Pi to the power of Euler's Number/Lemma 0 52293 442788 363086 2020-01-07T13:01:25Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let Schanuel's Conjecture be true. Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$. Then, $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$. == Proof == Assume the truth of Schanuel's Conjecture. Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Equivalently, they are linearly independent over the integers $\Z$. Let $a, b \in \Z$ such that: :$a z_1 + b z_3 = 0$ Substituting: :$a \ln \ln \pi + b \ln \pi = 0$ Applying the exponential function to both sides: :$\paren {\ln \pi}^a \pi^b = 1$ By Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals, the above equation only has solution when $a = b = 0$. Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Since $z_5$ is wholly imaginary, $z_1$, $z_3$, and $z_5$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, e^{z_1}, e^{z_3}, e^{z_5} }$ has transcendence degree at least $3$ over the rational numbers $\Q$. That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, \ln \pi, \pi, e^{i \pi} }$ has transcendence degree at least $3$ over $\Q$. However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$. Therefore, $\ln \ln \pi$, $\ln \pi$, and $i \pi$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $z_5$, and $1$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, 1, e^{z_1}, e^{z_3}, e^{z_5}, e}$ has transcendence degree at least $4$ over the rational numbers $\Q$. That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, 1, \ln \pi, \pi, e^{i \pi}, e}$ has transcendence degree at least $4$ over $\Q$. However, $1$ is algebraic. Moreover, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$. Therefore, $\ln \ln \pi$, $\ln \pi$, $i \pi$, and $e$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, $z_5$, and $e$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $e z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$. That is, $z_1$, $z_3$, $z_4$, and $z_5$ must be algebraically independent over $\Q$. Therefore, $z_1$, $1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$. Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$. That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$. {{qed|lemma}} Category:Transcendental Numbers Category:Logarithms Category:Pi Category:Euler's Number Category:Schanuel's Conjecture bx4wk4y7r4gw7tfasu9nwmzru2ii363


Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number/Lemma 0 52294 353794 280546 2018-05-04T18:15:23Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks|or rather, correct the existing links}} == Lemma == Let: :$z_1 = \ln \ln 2$ :$z_2 = 1 + \ln \ln 2$ :$z_3 = \ln 2$ :$z_4 = e \ln 2$ Let Schanuel's Conjecture be true. Then $z_1$, $z_2$, $z_3$ and $z_4$ are linearly independent over the rational numbers $\Q$. == Proof == Assume the truth of Schanuel's Conjecture. $2$ is algebraic. Hence, by the Corollary of the weaker Hermite-Lindemann-Weierstrass theorem, $\ln 2$ is transcendental. Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Equivalently, by Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers, they are linearly independent over the integers $\Z$. Let $a, b \in \Z$ such that: :$a z_1 + b z_3 = 0$ Substituting: :$a \ln \ln 2 + b \ln 2 = 0$ Applying the exponential function to both sides: :$\left({\ln 2}\right)^a 2^b = 1$ Since $\ln 2$ is transcendental, the above equation only has solution when $a = b = 0$. Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, e^{z_1}, e^{z_3}}\right)$ has transcendence degree at least $2$ over the rational numbers $\Q$. That is, the extension field $\Q \left({\ln \ln 2, \ln 2, \ln 2, 2}\right)$ has transcendence degree at least $2$ over $\Q$. However, $2$ is algebraic. Therefore, $\ln \ln 2$ and $\ln 2$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, and $1$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, 1, e^{z_1}, e^{z_3}, e}\right)$ has transcendence degree at least $3$ over the rational numbers $\Q$. That is, the extension field $\Q \left({\ln \ln 2, \ln 2, 1, \ln 2, 2, e}\right)$ has transcendence degree at least $3$ over $\Q$. However, $1$ and $2$ are algebraic. Therefore, $\ln \ln 2$, $\ln 2$, $e$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, and $e$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $e z_3$ must be algebraically independent over rational numbers $\Q$. That is, $z_1$, $z_3$, $z_4$ must be algebraically independent over $\Q$. Therefore, $z_1$, $1$, $z_3$, $z_4$ must be linearly independent over $\Q$. Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$ must be linearly independent over $\Q$. That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$ are linearly independent over the rational numbers $\Q$. {{qed|lemma}} Category:Transcendental Numbers Category:2 Category:Euler's Number Category:Schanuel's Conjecture bafqcvc586o7i7lkv4u9rm9xwz2e8sp


Divisor Counting Function/Examples/12

Divisor Counting Function/Examples/12 0 52622 451779 292229 2020-03-02T09:43:28Z Prime.mover 59 Prime.mover moved page Tau Function/Examples/12 to Divisor Counting Function/Examples/12 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({12}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$12 = 2^2 \times 3$ Thus: :$\tau \left({12}\right) = \tau \left({2^2 \times 3^1}\right) = \left({2 + 1}\right) \left({1 + 1}\right) = 6$ The divisors of $12$ can be enumerated as: :$1, 2, 3, 4, 6, 12$ {{qed}} Category:Tau Function Category:12 hhyg9izsflqs51fkdhzqv7zjj2p2zhl


Differential of Differentiable Functional is Unique/Lemma 0 52631 282479 282477 2017-01-17T23:18:53Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Lemma == <onlyinclude> Let $\phi \left[{y; h}\right]$ be a linear functional w.r.t. $h$. Let: :$\displaystyle \lim_{\left \vert{h}\right\vert \to 0} \frac {\phi \left[{y; h}\right]} {\left\vert{h}\right\vert} = 0$ Then : :$\phi \left[{y; h}\right] = 0$ </onlyinclude> == Proof == This will be a proof by contradiction. {{AimForCont}} there exists a linear functional satisfying $\phi \left[{y; h_0}\right] \ne 0$ for some $h_0 \ne 0$. Also suppose: :$\displaystyle \lim_{\left \vert{h_0}\right\vert \to 0} \frac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert} = 0$ Now, define: :$h_n = \dfrac {h_0} n$ and: :$m = \dfrac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}$ Notice that $\left\vert{h_n}\right\vert \to 0$ as $n \to \infty$.  Hence, from the assumption of the limit it should hold that: :$\displaystyle \lim_{n \to \infty} \frac{\phi \left[{y; h_n}\right]}{\left\vert{h_n}\right\vert} = \lim_{\left\vert{h_n}\right\vert \to 0} \frac{ \left[{y; h_n}\right]} {\left\vert{h_n}\right\vert} = 0$ However, using the linearity of $\phi \left[{y; h_0}\right]$ w.r.t. $h_0$: {{begin-eqn}} {{eqn | l = \lim_{n \to \infty} \frac {\phi \left[{y; h_n}\right]} {\left\vert{h_n}\right\vert}       | r = \lim_{n \to \infty} \frac {\phi \left[{y; \frac {h_0} n}\right]} { \left\vert{\frac {h_0} n}\right\vert}       | c = Definition of $h_n$ }} {{eqn | r = \lim_{n \to \infty} \frac {n \, \phi \left[{y; h_0}\right]} {n \, \left\vert{h_0}\right\vert}       | c = extract of $n$ through linearity }} {{eqn | r = \lim_{n \to \infty} \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}       | c = cancel $n$ }} {{eqn | r = \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}       | c = the limit does not depend on $n$ }} {{eqn | r = m       | c = definition of $m$ }} {{end-eqn}} However, by hypothesis: : $m \ne 0$ Hence, the contradiction is achieved and the initial statement of the lemma holds. {{qed}} == Sources == * {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin}}: $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum Category:Calculus of Variations 818us82q2tkfdakeemo573ft46fcnqt


Continued Fraction Expansion of Irrational Square Root/Examples/13 0 52897 470666 470627 2020-05-26T07:57:26Z Prime.mover 59 wikitext text/x-wiki == Example of Continued Fraction Expansion of Irrational Square Root == <onlyinclude> The continued fraction expansion of the square root of $13$ is given by: :$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$ </onlyinclude> {{OEIS|A010122}} === Convergents === {{:Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents}} == Proof == Let $\sqrt {13} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$ From Partial Quotients of Continued Fraction Expansion of Irrational Square Root, the partial quotients of this continued fraction expansion can be calculated as: :$a_r = \floor {\dfrac {\floor {\sqrt {13} } + P_r} {Q_r} }$ where: :$P_r = \begin {cases} 0 & : r = 0 \\ a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$ :$Q_r = \begin {cases} 1 & : r = 0 \\ \dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$ {{PartialQuotientCalculator-Start | n = 13}} {{PartialQuotientCalculator | n = 13 | r = 1 | ar-1 = 3 | Qr-1 = 1 | Pr-1 = 0}} {{PartialQuotientCalculator | n = 13 | r = 2 | ar-1 = 1 | Qr-1 = 4 | Pr-1 = 3}} {{PartialQuotientCalculator | n = 13 | r = 3 | ar-1 = 1 | Qr-1 = 3 | Pr-1 = 1}} {{PartialQuotientCalculator | n = 13 | r = 4 | ar-1 = 1 | Qr-1 = 3 | Pr-1 = 2}} {{PartialQuotientCalculator | n = 13 | r = 5 | ar-1 = 1 | Qr-1 = 4 | Pr-1 = 1}} |} and the cycle is complete: :$\sequence {1, 1, 1, 1, 6}$ {{qed}} Category:Examples of Continued Fractions Category:13 7armlnz95rqtjn6vf9wz1v5rpx5spzv


Continued Fraction Expansion of Irrational Square Root/Examples/29 0 52898 470654 470633 2020-05-26T06:45:22Z Prime.mover 59 wikitext text/x-wiki == Examples of Continued Fraction Expansion of Irrational Square Root == <onlyinclude> The continued fraction expansion of the square root of $29$ is given by: :$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$ </onlyinclude> {{OEIS|A010128}} === Convergents === {{:Continued Fraction Expansion of Irrational Square Root/Examples/29/Convergents}} == Proof == Let $\sqrt {29} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$ From Partial Quotients of Continued Fraction Expansion of Irrational Square Root, the partial quotients of this continued fraction expansion can be calculated as: :$a_r = \floor {\dfrac {\floor {\sqrt {29} } + P_r} {Q_r} }$ where: :$P_r = \begin {cases} 0 & : r = 0 \\ a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$ :$Q_r = \begin {cases} 1 & : r = 0 \\ \dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$ {{PartialQuotientCalculator-Start | n = 29}} {{PartialQuotientCalculator | n = 29 | r = 1 | ar-1 = 5 | Qr-1 = 1 | Pr-1 = 0}} {{PartialQuotientCalculator | n = 29 | r = 2 | ar-1 = 2 | Qr-1 = 4 | Pr-1 = 5}} {{PartialQuotientCalculator | n = 29 | r = 3 | ar-1 = 1 | Qr-1 = 5 | Pr-1 = 3}} {{PartialQuotientCalculator | n = 29 | r = 4 | ar-1 = 1 | Qr-1 = 5 | Pr-1 = 2}} {{PartialQuotientCalculator | n = 29 | r = 5 | ar-1 = 2 | Qr-1 = 4 | Pr-1 = 3}} |} and the cycle is complete: :$\sequence {2, 1, 1, 2, 10}$ {{qed}} Category:Examples of Continued Fractions Category:29 876zglxnx6xicqdyytjb2cwqplr12yr


Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents 0 52899 470657 470629 2020-05-26T06:50:08Z Prime.mover 59 wikitext text/x-wiki == Convergents to Continued Fraction Expansion of $\sqrt {13}$ == <onlyinclude> The sequence of convergents to the continued fraction expansion of the square root of $13$ begins: :$\dfrac 3 1, \dfrac 4 1, \dfrac 7 2, \dfrac {11} 3, \dfrac {18} 5, \dfrac {119} {33}, \dfrac {137} {38}, \dfrac {256} {71}, \dfrac {393} {109}, \dfrac {649} {180}, \ldots$ </onlyinclude> {{OEIS-Numerators|A041018}} {{OEIS-Denominators|A041019}} == Proof == Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion. Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators. Then the $n$th convergent is $\dfrac {p_n} {q_n}$. By definition: :$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$ :$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$ From Continued Fraction Expansion of $\sqrt {13}$: :$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$ Thus the convergents are assembled: {{ConvergentCalculator-Start | a0 = 3 | a1 = 1}} {{ConvergentCalculator | k = 2 | ak = 1 | pk-1 = 4 | pk-2 = 3 | qk-1 = 1 | qk-2 = 1}} {{ConvergentCalculator | k = 3 | ak = 1 | pk-1 = 7 | pk-2 = 4 | qk-1 = 2 | qk-2 = 1}} {{ConvergentCalculator | k = 4 | ak = 1 | pk-1 = 11 | pk-2 = 7 | qk-1 = 3 | qk-2 = 2 }} {{ConvergentCalculator | k = 5 | ak = 6 | pk-1 = 18 | pk-2 = 11 | qk-1 = 5 | qk-2 = 3}} {{ConvergentCalculator | k = 6 | ak = 1 | pk-1 = 119 | pk-2 = 18 | qk-1 = 33 | qk-2 = 5}} {{ConvergentCalculator | k = 7 | ak = 1 | pk-1 = 137 | pk-2 = 119 | qk-1 = 38 | qk-2 = 33}} {{ConvergentCalculator | k = 8 | ak = 1 | pk-1 = 256 | pk-2 = 137 | qk-1 = 71 | qk-2 = 38}} {{ConvergentCalculator | k = 9 | ak = 1 | pk-1 = 393 | pk-2 = 256 | qk-1 = 109 | qk-2 = 71}} |} {{qed}} Category:Continued Fractions Category:13 6k36tbcv7u9lpzztikxt314un6ggl5x


Continued Fraction Expansion of Irrational Square Root/Examples/29/Convergents 0 52904 470653 470635 2020-05-26T06:44:59Z Prime.mover 59 wikitext text/x-wiki == Convergents to Continued Fraction Expansion of $\sqrt {29}$ == <onlyinclude> The sequence of convergents to the continued fraction expansion of the square root of $29$ begins: :$\dfrac 5 1, \dfrac {11} 2, \dfrac {16} 3, \dfrac {27} 5, \dfrac {70} {13}, \dfrac {727} {135}, \dfrac {1524} {283}, \dfrac {2251} {418}, \dfrac {3775} {701}, \dfrac {9801} {1820}, \ldots$ </onlyinclude> {{OEIS-Numerators|A041046}} {{OEIS-Denominators|A041047}} == Proof == Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion. Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators. Then the $n$th convergent is $p_n / q_n$. By definition: :$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1\end {cases}$ :$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1\end {cases}$ From Continued Fraction Expansion of $\sqrt {29}$: :$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$ Thus the convergents are assembled: {{ConvergentCalculator-Start | a0 = 5 | a1 = 2}} {{ConvergentCalculator | k = 2 | ak =  1 | pk-1 =   11 | pk-2 =    5 | qk-1 =   2 | qk-2 =   1}} {{ConvergentCalculator | k = 3 | ak =  1 | pk-1 =   16 | pk-2 =   11 | qk-1 =   3 | qk-2 =   2}} {{ConvergentCalculator | k = 4 | ak =  2 | pk-1 =   27 | pk-2 =   16 | qk-1 =   5 | qk-2 =   3}} {{ConvergentCalculator | k = 5 | ak = 10 | pk-1 =   70 | pk-2 =   27 | qk-1 =  13 | qk-2 =   5}} {{ConvergentCalculator | k = 6 | ak =  2 | pk-1 =  727 | pk-2 =   70 | qk-1 = 135 | qk-2 =  13}} {{ConvergentCalculator | k = 7 | ak =  1 | pk-1 = 1524 | pk-2 =  727 | qk-1 = 283 | qk-2 = 135}} {{ConvergentCalculator | k = 8 | ak =  1 | pk-1 = 2251 | pk-2 = 1524 | qk-1 = 418 | qk-2 = 283}} {{ConvergentCalculator | k = 9 | ak =  2 | pk-1 = 3775 | pk-2 = 2251 | qk-1 = 701 | qk-2 = 418}} |} {{qed}} Category:Continued Fractions Category:29 3yesbydecvf35st0npptdo09kpr9pum


Lifting The Exponent Lemma/Lemma

Lifting The Exponent Lemma/Lemma 0 52990 457792 312888 2020-03-27T08:04:40Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $x, y \in \Z$ be distinct integers. Let $p$ be an odd prime. Let: :$p \divides x - y$ and: :$p \nmid x y$. Then  :$\map {\nu_p} {x^p - y^p} = \map {\nu_p} {x - y} + 1$ </onlyinclude> where $\nu_p$ denotes $p$-adic valuation. == Proof == Let $\map {\nu_p} {x - y} = k$. Then $x=p^k m + y$ where $p \nmid m$. We have: {{begin-eqn}} {{eqn | l = x^p - y^p       | r = (p^k m + y)^p - y^p       | c =  }} {{eqn | r = \sum_{i \mathop = 0}^p \paren {\binom p i \paren {p^k m}^{p - i} y^i} - y^p       | c = Binomial Theorem }} {{eqn | r = \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + \binom p {p - 1} \paren {p^k m} y^{p - 1}       | c = picking out the last two terms from the summation }} {{eqn | r = \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + p^{k + 1} m y^{p - 1}       | c =  }} {{end-eqn}} Note that all terms in the above expression have a factor of $p$ to the order at least $k+1$. So, $p^{k + 1} \mid x^p - y^p$. Also note that all terms in the summation have a factor of $p$ to the order at least $k + 2$. But in the term $p^{k + 1} m y^{p - 1}$, since $p \nmid m$ and $p \nmid y$, we have: :$p^{k + 2} \nmid p^{k + 1} m y^{p - 1}$ So: :$p^{k + 2} \nmid x^p - y^p$ So by definition of $p$-adic valuation: :$\map {\nu_p} {x^p - y^p} = k + 1$ {{qed}} == Sources == * {{citation|date = July 1904|title = On the Integral Divisors of $a^n - b^n$|journal = Annals of Mathematics|abbr = Ann. Math.|volume = 5|startpage = 173|endpage = 180|jstor = 2007263|author = George David Birkhoff|author2 = Harry Schultz Vandiver|jstorcat = yes}}: Theorem $\text{III}$. Category:P-adic Valuations d4e0h0xni7916lg5nyycr7wdwh3khto


Solutions of Pythagorean Equation/Sequence 0 53030 283072 283069 2017-01-24T23:19:35Z Prime.mover 59 wikitext text/x-wiki == Sequence == <onlyinclude> The sequence of solutions of the Pythagorean equation can be tabulated as follows: :$\begin{array} {r r | r r | r r r | c} m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 5 & \text{Primitive} \\ \hline 3 & 1 & 9 & 1 & 6 & 8 & 10 \\ 3 & 2 & 9 & 4 & 12 & 5 & 13 & \text{Primitive} \\ \hline 4 & 1 & 16 & 1 & 8 & 15 & 17 & \text{Primitive} \\ 4 & 2 & 16 & 4 & 16 & 12 & 20 \\ 4 & 3 & 16 & 9 & 24 & 7 & 25 & \text{Primitive} \\ \hline 5 & 1 & 25 & 1 & 10 & 24 & 26 \\ 5 & 2 & 25 & 4 & 20 & 21 & 29 & \text{Primitive} \\ 5 & 3 & 25 & 9 & 30 & 16 & 34 \\ 5 & 4 & 25 & 16 & 40 & 9 & 41 & \text{Primitive} \\ \hline 6 & 1 & 36 & 1 & 12 & 35 & 37 & \text{Primitive} \\ 6 & 2 & 36 & 4 & 24 & 32 & 40 \\ 6 & 3 & 36 & 9 & 36 & 27 & 45 \\ 6 & 4 & 36 & 16 & 48 & 20 & 52 \\ 6 & 5 & 36 & 25 & 60 & 11 & 61 & \text{Primitive} \\ \hline 7 & 1 & 49 & 1 & 14 & 48 & 50 \\ 7 & 2 & 49 & 4 & 28 & 45 & 53 & \text{Primitive} \\ 7 & 3 & 49 & 9 & 42 & 40 & 58 \\ 7 & 4 & 49 & 16 & 56 & 33 & 65 & \text{Primitive} \\ 7 & 5 & 49 & 25 & 70 & 24 & 74 \\ 7 & 6 & 49 & 36 & 84 & 13 & 85 & \text{Primitive} \\ \hline \end{array}$ </onlyinclude> Category:Pythagorean Triples Category:Solutions of Pythagorean Equation dqkjdu0c3nvufai4wklffejw24x8wb2


Pythagorean Triangle/Examples/5-12-13

Pythagorean Triangle/Examples/5-12-13 0 53032 478775 478750 2020-07-18T12:13:55Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $5$, $12$ and $13$ is a primitive Pythagorean triangle. </onlyinclude> :400px == Proof == {{begin-eqn}} {{eqn | l = 5^2 + 12^2       | r = 25 + 144       | c =  }} {{eqn | r = 169       | c =  }} {{eqn | r = 13^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $5$, $12$ and $13$ form a Pythagorean triple. Note that $5$ and $12$ are coprime. Hence, by definition, $5$, $12$ and $13$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Smallest Pythagorean Triangle is 3-4-5|next = Pythagorean Triangle/Examples/6-8-10}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Smallest Pythagorean Triangle is 3-4-5|next = Pythagorean Triangle/Examples/6-8-10}}: $13$ Category:Examples of Pythagorean Triangles 3tbh5ki2lweolv39z8bwoob7o159qdm


Dirichlet Series Convergence Lemma/Lemma 0 53046 357559 350951 2018-05-28T17:18:39Z AliceInNumberland 3357 Fixed the proof, it follows the broad strokes of the cited source, with a few modifications to make it more elementary, as that is the objective of proving it in the ordinary sense wikitext text/x-wiki == Lemma to Dirichlet Series Convergence Lemma == <onlyinclude> Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty \frac{a_n}{n^s}$ be a Dirichlet series. Suppose that for some $s_0 = \sigma_0 + i t_0 \in \C$, $f \left({s_0}\right)$ has bounded partial sums: :$(1): \quad \displaystyle \left\vert{\sum_{n \mathop = 1}^N a_n n^{-s_0} }\right\vert \le M$ for some $M \in \R$ and all $N \ge 1$. Then for every $s = \sigma + i t \in \C$ with $\sigma > \sigma_0$: :$\displaystyle \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert \le 2 M m^{\sigma_0 - \sigma} \left( 1+ \frac {\left\vert s - s_0\right\vert} {\sigma-\sigma_0} \right)$ </onlyinclude> == Proof == We have the Summation by Parts formula: :$\displaystyle \sum_{n \mathop = m}^N f_n g_n = f_N G_N - f_m G_{m-1} - \sum_{n \mathop = m}^{N - 1} G_n \left({f_{n+1} - f_n}\right)$ We let $g_n = a_n n^{-s_0}$ and $f_n = n^{s_0 - s}$.  For $N \ge 1$, the quantities $G_N$ are the partial sums $(1)$ Thus $G_N \le M$ for all $N \ge 1$. We have: {{begin-eqn}} {{eqn | l = \left\vert{\sum_{n \mathop = m}^N \frac {a_n} {n^s} }\right\vert       | r = \left\vert{\sum_{n \mathop = m}^N f_n g_n}\right\vert }} {{eqn | r = \left\vert{f_N G_N}\right\vert + \left\vert{f_m G_{m-1} }\right\vert + \sum_{n \mathop = m}^{N - 1} \left\vert{G_n \left({f_{n+1} - f_n}\right)}\right\vert       | o = \le       | c = using partial summation and the Triangle Inequality }} {{eqn | r = M \left\vert{N^{s_0 - s} }\right\vert + M \left\vert{m^{s_0 - s} }\right\vert + M \sum_{n \mathop = m}^{N - 1} \left\vert{\left({\left({n + 1}\right)^{s_0 - s} - n^{s_0 - s} }\right) }\right\vert       | o = \le       | c = using the given bound on the partial sums }} {{eqn |r = M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{ \left(s - s_0 \right) \int_n^{n+1} t^{s_0 - s -1} }\right\vert }} {{eqn |o= \le       |r = M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{ s - s_0 }\right\vert \int_n^{n+1} \left\vert{ t^{s_0 - s -1} }\right\vert       |c = Modulus of Complex Integral }} {{eqn |r =  M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{s - s_0 }\right\vert \int_n^{n+1}  t^{\sigma_0 - \sigma -1} }} {{eqn |r =  M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \left\vert{ s - s_0 }\right\vert \int_m^{N}  t^{\sigma_0 - \sigma -1} }} {{eqn |r =  M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \left( m^{\sigma_0 - \sigma}- N^{\sigma_0 - \sigma} \right) }} {{eqn |o= \le       |r =  M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \left( m^{\sigma_0 - \sigma} + N^{\sigma_0 - \sigma} \right) }} {{end-eqn}} Finally, because $N \ge m$ and $\sigma_0 - \sigma < 0$, we have: :$ N^{\sigma_0 - \sigma}  +  m^{\sigma_0 - \sigma} \le 2 m^{\sigma_0 - \sigma}$ Therefore: {{begin-eqn}} {{eqn |l= \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert       |o= \le       |r= M N^{\sigma_0 - \sigma}  + M m^{\sigma_0 - \sigma} + M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \left( m^{\sigma_0 - \sigma} + N^{\sigma_0 - \sigma} \right) }} {{eqn |o=\leq       |r=   2 M m^{\sigma_0 - \sigma} + 2 M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} m^{\sigma_0 - \sigma} }} {{eqn |r = 2 M m^{\sigma_0 - \sigma} \left(1 + \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \right) }} {{end-eqn}} Hence the result. {{qed|lemma}} == Sources == {{SourceReview|wrt Dirichlet Series Convergence Lemma}} * {{BookReference|Introduction to Analytic Number Theory|1976|Tom M. Apostol}}: $\S 11.6$: Lemma $2$, Theorem $11.8$ Category:Dirichlet Series jqn0qz0s10dk2k2et5rlpomi63xtihe


Pythagorean Triangle/Examples/6-8-10

Pythagorean Triangle/Examples/6-8-10 0 53055 478773 478752 2020-07-18T12:12:27Z Prime.mover 59 wikitext text/x-wiki == Example of Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $6$, $8$ and $10$ is a Pythagorean triangle. This is not a primitive Pythagorean triangle. </onlyinclude> :300px == Proof == {{begin-eqn}} {{eqn | l = 6^2 + 8^2       | r = 2^2 \times 3^2 + 2^2 \times 4^2       | c =  }} {{eqn | r = 4 \times \paren {9 + 16}       | c =  }} {{eqn | r = 4 \times 25       | c =  }} {{eqn | r = 2^2 \times 5^2       | c =  }} {{eqn | r = 10^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $6$, $8$ and $10$ form a Pythagorean triple. Note that $6$ and $8$ are not coprime as $\gcd \set {6, 8} = 2$. Hence, by definition, $6$, $8$ and $10$ do not form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Pythagorean Triangle/Examples/5-12-13|next = Pythagorean Triangle/Examples/7-24-25}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Pythagorean Triangle/Examples/5-12-13|next = Pythagorean Triangle/Examples/7-24-25}}: $13$ Category:Examples of Pythagorean Triangles fxd28zo105jua0jquhuvakzl9v7vg0d


Convergence of Complex Sequence in Polar Form/Corollary 0 53060 407029 362409 2019-06-08T13:31:11Z Prime.mover 59 wikitext text/x-wiki == Corollary to Convergence of Complex Sequence in Polar Form == <onlyinclude> Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers. Let $z \ne 0$ be a complex number with modulus $r$ and argument $\theta$. Let $I$ be a real interval of length at most $2 \pi$ that contains $\theta$. Suppose $\theta$ is not an endpoint of $I$. Suppose each $z_n$ admits an argument $\theta_n \in I$. Let $r_n$ be the modulus of $z_n$. Then $z_n$ converges to $z$ {{iff}} $r_n$ converges to $r$ and $\theta_n$ converges to $\theta$. </onlyinclude> == Proof == Suppose $r_n \to r$ and $\theta_n \to \theta$. Then by Convergence of Complex Sequence in Polar Form, $z_n \to z$. Conversely, suppose $z_n \to z$. By Convergence of Complex Sequence in Polar Form, we have: :$(1): \quad r_n \to r$ :$(2): \quad$ There exists a sequence $\left\langle{k_n}\right\rangle$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$. It remains to be proved that $\theta_n \to \theta$. Let $N \in \N$ such that $\left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi / 2$ for all $n \ge N$. By the Triangle Inequality for Real Numbers: :$\left\vert{2 k_n \pi - 2 k_N \pi}\right\vert \le \left\vert{\theta_n + 2 k_n \pi - \theta}\right\vert + \left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi$ for all $n \ge N$. Thus $\left\vert{k_n - k_N}\right\vert \le 1 / 2$, so $k_n = k_N$ for all $n \ge N$. By the Triangle Inequality for Real Numbers: :$\left\vert{2 \pi k_N}\right\vert \le \left\vert{\theta_n - \theta}\right\vert + \left\vert{\theta_n + 2 \pi k_N - \theta}\right\vert$ for all $n \in \N$. Because $\theta_n \in I$ and $\theta$ is not an endpoint of $I$: :$\left\vert{\theta_n - \theta}\right\vert < 2 \pi$ for all $n \in \N$. Because $\theta_n + 2 \pi k_N - \theta \to 0$: :$\left\vert{2 \pi k_N}\right\vert < 2 \pi$ Thus $k_n = 0$ for all $n \ge N$. Thus $\theta_n \to \theta$. {{qed}} Category:Complex Analysis i7cch7fbmgc5k27c7k67jp1ygqqypw0


Pythagorean Triangle/Examples/693-1924-2045 0 53074 478760 478754 2020-07-18T12:06:32Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $693$, $1924$ and $2045$ is a primitive Pythagorean triangle. </onlyinclude> :700px == Proof == {{begin-eqn}} {{eqn | l = 693^2 + 1924^2       | r = 480 \, 249 + 3 \, 701 \, 776       | c =  }} {{eqn | r = 4 \, 182 \, 025       | c =  }} {{eqn | r = 2045^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $693$, $1924$ and $2045$  form a Pythagorean triple. Note that $693$ and $1924$ are coprime. Hence, by definition, $693$, $1924$ and $2045$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Pythagorean Triangle whose Area is Half Perimeter|next = Pythagorean Triangles whose Areas are Repdigit Numbers}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Pythagorean Triangle whose Area is Half Perimeter|next = Pythagorean Triangles whose Areas are Repdigit Numbers}}: $13$ Category:Examples of Pythagorean Triangles h1hz8c88k0y8dp6iz2daoaftzcivnpz


693

693 0 53097 478771 306362 2020-07-18T12:10:45Z Prime.mover 59 wikitext text/x-wiki {{NumberPageLink|prev = 692|next = 694}} == Number == $693$ ('''six hundred and ninety three''') is: :$3^2 \times 7 \times 11$ :One of the legs of the Pythagorean triangle $693-1924-2045$, whose area is $666 \, 666$. == Also see == * Pythagorean Triangles whose Areas are Repdigit Numbers Category:Specific Numbers Category:693 lrnrx7zujqpwoskv5ajj4nun7p1y6me


1924

1924 0 53099 478770 303430 2020-07-18T12:10:34Z Prime.mover 59 wikitext text/x-wiki {{NumberPageLink|prev = 1923|next = 1925}} == Number == $1924$ ('''one thousand nine hundred and twenty four''') is: :$2^2 \times 13 \times 37$ :One of the legs of the Pythagorean triangle $693-1924-2045$, whose area is $666 \, 666$. == Also see == * Pythagorean Triangles whose Areas are Repdigit Numbers Category:Specific Numbers Category:1924 ny180aji87241vumpmtwksdgvejdo3i


111

111 0 53106 478342 461296 2020-07-16T05:33:30Z Prime.mover 59 wikitext text/x-wiki {{NumberPageLink|prev = 110|next = 112}} == Number == $111$ ('''one hundred and eleven''') is: :$3 \times 37$ :The magic constant of the smallest prime magic square. :The $3$rd repuint after $1$, $11$ :The $4$th of the $17$ positive integers for which the value of the Euler $\phi$ function is $72$: ::$73$, $91$, $95$, $111$, $117$, $135$, $146$, $148$, $152$, $182$, $190$, $216$, $222$, $228$, $234$, $252$, $270$ :The $6$th positive integer after $1$, $2$, $7$, $11$, $101$ whose cube is palindromic: ::$111^3 = 1 \, 367 \, 631$ :The magic constant of a magic square of order $6$, after $1$, $(5)$, $15$, $34$, $65$: ::$111 = \displaystyle \dfrac 1 6 \sum_{k \mathop = 1}^{6^2} k = \dfrac {6 \paren {6^2 + 1} } 2$ :The $7$th palindromic lucky number: ::$1$, $3$, $7$, $9$, $33$, $99$, $111$, $\ldots$ :The $8$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$ whose square is also palindromic integer ::$111^2 = 12 \, 321$ :The $15$th Zuckerman number after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $11$, $12$, $15$, $24$, $36$: ::$111 = 111 \times 1 = 111 \times \paren {1 \times 1 \times 1}$ :The $24$th lucky number: ::$1$, $3$, $7$, $9$, $13$, $15$, $21$, $25$, $31$, $33$, $37$, $43$, $49$, $51$, $63$, $67$, $73$, $75$, $79$, $87$, $93$, $99$, $105$, $111$, $\ldots$ :The $35$th semiprime: ::$111 = 3 \times 37$ == Also see == * Magic Constant of Smallest Prime Magic Square * {{NumberPageLink|prev = 11|next = 1111|type = Repunit|cat = Repunits}} * {{NumberPageLink|prev = 36|next = 112|type = Zuckerman Number|cat = Zuckerman Numbers}} * {{NumberPageLink|prev = 65|next = 175|result = Magic Constant of Magic Square}} * {{NumberPageLink|prev = 95|next = 117|result = Numbers with Euler Phi Value of 72}} * {{NumberPageLink|prev = 99|next = 141|result = Sequence of Palindromic Lucky Numbers}} * {{NumberPageLink|prev = 101|next = 121|result = Square of Small-Digit Palindromic Number is Palindromic}} * {{NumberPageLink|prev = 101|next = 1001|result = Sequence of Integers whose Cube is Palindromic}} * {{NumberPageLink|prev = 105|next = 115|type = Lucky Number|cat = Lucky Numbers}} * {{NumberPageLink|prev = 106|next = 115|type = Semiprime Number|cat = Semiprimes}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Magic Constant of Smallest Prime Magic Square|next = 112/Historical Note}}: $111$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Magic Constant of Smallest Prime Magic Square|next = Sequence of Palindromic Lucky Numbers}}: $111$ Category:Specific Numbers Category:111 nj1uqqd71djtmooa94n27t3kops38hp


1001

1001 0 53108 478349 461059 2020-07-16T05:49:34Z Prime.mover 59 wikitext text/x-wiki {{NumberPageLink|prev = 1000|next = 1002}} == Number == $1001$ ('''one thousand and one''') is: :$7 \times 11 \times 13$ :The $4$th pentagonal number after $1$, $5$, $22$ which is also palindromic: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ :The $7$th positive integer after $1$, $2$, $7$, $11$, $101$, $111$ whose cube is palindromic: ::$1001^3 = 1 \, 003 \, 003 \, 001$ :The $11$th pentatope number after $1$, $5$, $15$, $35$, $70$, $126$, $210$, $330$, $495$, $715$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{11} \dfrac {k \paren {k + 1} \paren {k + 2} } 6 = \dfrac {11 \paren {11 + 1} \paren {11 + 2} \paren {11 + 3} } {24}$ :The $12$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$, $111$, $121$, $202$, $212$ whose square is also palindromic integer ::$1001^2 = 1 \, 002 \, 001$ :The $26$th pentagonal number after $1$, $5$, $12$, $22$, $35$, $\ldots$, $477$, $532$, $590$, $651$, $715$, $782$, $852$, $925$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ :The $51$st generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$, $\ldots$, $610$, $651$, $672$, $715$, $737$, $782$, $805$, $852$, $876$, $925$, $950$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ == Also see == * {{NumberPageLink|prev = 22|next = 2882|result = Sequence of Palindromic Pentagonal Numbers}} * {{NumberPageLink|prev = 101|next = 10,001|result = Prime Factors of One More than Power of 10}} * {{NumberPageLink|prev = 111|next = 2201|result = Sequence of Integers whose Cube is Palindromic}} * {{NumberPageLink|prev = 212|next = 1111|result = Square of Small-Digit Palindromic Number is Palindromic}} * {{NumberPageLink|prev = 715|next = 1365|type = Pentatope Number|cat = Pentatope Numbers}} * {{NumberPageLink|prev = 925|next = 1080|type = Pentagonal Number|cat = Pentagonal Numbers}} * {{NumberPageLink|prev = 950|next = 1027|type = Generalized Pentagonal Number|cat = Generalized Pentagonal Numbers}} == Historical Note == {{:1001/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 1000|next = Divisibility Test for 7, 11 and 13}}: $1001$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 1000|next = Divisibility Test for 7, 11 and 13}}: $1001$ Category:Specific Numbers Category:1001 25posi7kny0a9yy14labtbcg1ve4mdt


Brahmagupta-Fibonacci Identity/General

Brahmagupta-Fibonacci Identity/General 0 53203 283820 283813 2017-01-29T09:29:18Z Prime.mover 59 wikitext text/x-wiki == General version of Brahmagupta-Fibonacci Identity == <onlyinclude> Let $a, b, c, d, n$ be numbers. :$\left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right) = \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2$ </onlyinclude> === Corollary === {{:Brahmagupta-Fibonacci Identity/General/Corollary}} === Extension === {{:Brahmagupta-Fibonacci Identity/General/Extension}} == Proof == {{begin-eqn}} {{eqn | o =        | r = \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2       | c =  }} {{eqn | r = \left({\left({a c}\right)^2 + 2 \left({a c}\right) \left({n b d}\right) + \left({n b d}\right)^2}\right) + n \left({\left({a d}\right)^2 - 2 \left({a b}\right) \left({c d}\right) + \left({b c}\right)^2}\right)       | c = Square of Sum, Square of Difference }} {{eqn | r = a^2 c^2 + 2 n a b c d + n^2 b^2 d^2 + n a^2 d^2 - 2 n a b c d + n b^2 c^2       | c = multiplying out }} {{eqn | r = a^2 c^2 + n a^2 d^2 + n b^2 c^2 + n^2 b^2 d^2       | c = simplifying }} {{eqn | r = \left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right)       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Brahmagupta|name2 = Leonardo Fibonacci}} Category:Brahmagupta-Fibonacci Identity 1usmy94pvju94p1q87n512l2liguijm


Brahmagupta-Fibonacci Identity/Extension/General 0 53211 437759 283952 2019-12-04T16:28:06Z Prime.mover 59 wikitext text/x-wiki == Extension to Brahmagupta-Fibonacci Identity == <onlyinclude> Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, m$ be integers. Then: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. That is: the set of all integers of the form $a^2 + m b^2$ is closed under multiplication. </onlyinclude> == Proof == The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. $\map P 1$ is the trivial case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^1 \paren { {a_j}^2 + m {b_j}^2}       | r = {a_1}^2 + m {b_1}^2       | c =  }} {{eqn | r = c^2 + m d^2       | c = setting $c = a_1$ and $d = b_1$ }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^2 \paren { {a_j}^2 + m {b_j}^2}       | r = \paren { {a_1}^2 + m {b_1}^2} \paren { {a_2}^2 + m {b_2}^2}       | c =  }} {{eqn | r = \paren {a_1 a_2 + m b_1 b_2}^2 + m \paren {a_1 b_2 - b_1 a_2}^2       | c = General Brahmagupta-Fibonacci Identity }} {{eqn | r = c^2 + m d^2       | c = setting $c = a_1 a_2 + m b_1 b_2$ and $d = a_1 b_2 - b_1 a_2$ }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. from which it is to be shown that: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + m {b_j}^2}       | r = \prod_{j \mathop = 1}^k \paren { {a_j}^2 + m {b_j}^2} \paren { {a_{k + 1} }^2 + m {b_{k + 1} }^2}       | c =  }} {{eqn | r = \paren { {c'}^2 + m {d'}^2} \paren { {a_{k + 1} }^2 + m {b_{k + 1} }^2}       | c = Induction Hypothesis: for some $c', d' \in \Z$ }} {{eqn | r = c^2 + m d^2       | c = Basis for the Induction: for some $c, d \in \Z$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \Z_{> 0}$: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. {{qed}} Category:Brahmagupta-Fibonacci Identity 6fkhzc2kgy79r4dx5r34g6nzf2kxisq


Brahmagupta-Fibonacci Identity/Extension/Proof 2 0 53226 283951 283948 2017-01-29T15:26:25Z Prime.mover 59 wikitext text/x-wiki == Extension to Brahmagupta-Fibonacci Identity == {{:Brahmagupta-Fibonacci Identity/Extension}} == Proof == <onlyinclude> The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^n \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. $P \left({1}\right)$ is the trivial case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^1 \left({ {a_j}^2 + {b_j}^2}\right)       | r = {a_1}^2 + {b_1}^2       | c =  }} {{eqn | r = c^2 + d^2       | c = setting $c = a_1$ and $d = b_1$ }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. === Basis for the Induction === $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^2 \left({ {a_j}^2 + {b_j}^2}\right)       | r = \left({ {a_1}^2 + {b_1}^2}\right) \left({ {a_2}^2 + {b_2}^2}\right)       | c =  }} {{eqn | r = \left({a_1 a_2 + b_1 b_2}\right)^2 + \left({a_1 b_2 - b_1 a_2}\right)^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = c^2 + d^2       | c = setting $c = a_1 a_2 + b_1 b_2$ and $d = a_1 b_2 - b_1 a_2$ }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. from which it is to be shown that: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^{k + 1} \left({ {a_j}^2 + {b_j}^2}\right)       | r = \prod_{j \mathop = 1}^k \left({ {a_j}^2 + {b_j}^2}\right) \left({ {a_{k + 1} }^2 + {b_{k + 1} }^2}\right)       | c =  }} {{eqn | r = \left({ {c'}^2 + {d'}^2}\right) \left({ {a_{k + 1} }^2 + {b_{k + 1} }^2}\right)       | c = Induction Hypothesis: for some $c', d' \in \Z$ }} {{eqn | r = c^2 + m d^2       | c = Basis for the Induction: for some $c, d \in \Z$ }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \Z_{> 0}$: :$\displaystyle \prod_{j \mathop = 1}^n \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. {{qed}} </onlyinclude> Category:Brahmagupta-Fibonacci Identity ezs3g9lp2x0ml1h4z22zb4yg5j92d8a


Magic Square/Examples/Order 4

Magic Square/Examples/Order 4 0 53567 310268 300669 2017-08-10T05:40:04Z Prime.mover 59 wikitext text/x-wiki == Examples of Order $4$ Magic Squares == <onlyinclude> There are many order $4$ magic squares. ==== Dürer's Order $4$ Magic Square ==== This is one of the more famous ones, due to {{AuthorRef|Albrecht Dürer}}: {{:Magic Square/Examples/Order 4/Dürer}} ==== Moessner's Order $4$ Magic Square ==== This one, created by {{AuthorRef|Alfred Moessner}}, has extra interesting properties: {{:Magic Square/Examples/Order 4/Alfred Moessner}}</onlyinclude> == Also see == * Magic Constant of Order 4 Magic Square * Number of Magic Squares of Order 4 Category:Magic Squares q21ghuew2eagljs5i7h9xx2q5uuvfhz


Magic Square/Examples/Order 4/Alfred Moessner 0 53573 381665 381661 2018-12-12T23:57:37Z Prime.mover 59 wikitext text/x-wiki == Example of Order $4$ Magic Square == This example of an order $4$ magic square is due to {{AuthorRef|Alfred Moessner}}: <onlyinclude> :$\begin{array}{|c|c|c|c|} \hline 12 & 13 & 1 & 8 \\ \hline 6 & 3 & 15 & 10 \\ \hline 7 & 2 & 14 & 11 \\ \hline 9 & 16 & 4 & 5 \\ \hline \end{array}$ </onlyinclude> == Also see == * Properties of Moessner's Order $4$ Magic Square == Historical Note == {{:Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}} == Sources == * {{citation|author = Alfred Moessner|title = A Curious Magic Square|journal = Scripta Mathematica|volume = 13|date = 1947?|startpage = ???|endpage = ???}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Magic Square/Examples/Order 4/Dürer/Historical Note|next = Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}}: $16$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Magic Square/Examples/Order 4/Dürer/Historical Note|next = Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}}: $16$ Category:Magic Squares 30gjmsv91qxdg3xjo5bcb8iiqyk24oo


Euler Lucky Number/Examples/41

Euler Lucky Number/Examples/41 0 53646 382719 291883 2018-12-17T06:43:57Z Prime.mover 59 wikitext text/x-wiki == Example of Euler Lucky Number == <onlyinclude> The expression: :$n^2 + n + 41$ yields primes for $n = 0$ to $n = 39$. It also generates the same set of primes for $n = -1 \to n = -40$. These are not the only primes generated by this formula. No other quadratic function of the form $x^2 + a x + b$, where $a, b \in \Z_{>0}$ and $a, b < 10000$ generates a longer sequence of primes. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 0^2 + 0 + 41       | r = 0 + 0 + 41       | rr= = 41       | c = which is prime }} {{eqn | l = 1^2 + 1 + 41       | r = 1 + 1 + 41       | rr= = 43       | c = which is prime }} {{eqn | l = 2^2 + 2 + 41       | r = 4 + 2 + 41       | rr= = 47       | c = which is prime }} {{eqn | l = 3^2 + 3 + 41       | r = 9 + 3 + 41       | rr= = 53       | c = which is prime }} {{eqn | l = 4^2 + 4 + 41       | r = 16 + 4 + 41       | rr= = 61       | c = which is prime }} {{eqn | l = 5^2 + 5 + 41       | r = 25 + 5 + 41       | rr= = 71       | c = which is prime }} {{eqn | l = 6^2 + 6 + 41       | r = 36 + 6 + 41       | rr= = 83       | c = which is prime }} {{eqn | l = 7^2 + 7 + 41       | r = 49 + 7 + 41       | rr= = 97       | c = which is prime }} {{eqn | l = 8^2 + 8 + 41       | r = 64 + 8 + 41       | rr= = 113       | c = which is prime }} {{eqn | l = 9^2 + 9 + 41       | r = 81 + 9 + 41       | rr= = 131       | c = which is prime }} {{eqn | l = 10^2 + 10 + 41       | r = 100 + 10 + 41       | rr= = 151       | c = which is prime }} {{eqn | l = 11^2 + 11 + 17       | r = 121 + 11 + 17       | rr= = 173       | c = which is prime }} {{eqn | l = 12^2 + 12 + 41       | r = 144 + 12 + 41       | rr= = 197       | c = which is prime }} {{eqn | l = 13^2 + 13 + 17       | r = 169 + 13 + 17       | rr= = 223       | c = which is prime }} {{eqn | l = 14^2 + 14 + 41       | r = 196 + 14 + 41       | rr= = 251       | c = which is prime }} {{eqn | l = 15^2 + 15 + 41       | r = 225 + 15 + 41       | rr= = 281       | c = which is prime }} {{eqn | l = 16^2 + 16 + 41       | r = 256 + 16 + 41       | rr= = 313       | c = which is prime }} {{eqn | l = 17^2 + 17 + 41       | r = 289 + 17 + 41       | rr= = 347       | c = which is prime }} {{eqn | l = 18^2 + 18 + 41       | r = 324 + 18 + 41       | rr= = 383       | c = which is prime }} {{eqn | l = 19^2 + 19 + 41       | r = 361 + 19 + 41       | rr= = 421       | c = which is prime }} {{eqn | l = 20^2 + 20 + 41       | r = 400 + 20 + 41       | rr= = 461       | c = which is prime }} {{eqn | l = 21^2 + 21 + 41       | r = 441 + 21 + 41       | rr= = 503       | c = which is prime }} {{eqn | l = 22^2 + 22 + 41       | r = 484 + 22 + 41       | rr= = 547       | c = which is prime }} {{eqn | l = 23^2 + 23 + 41       | r = 529 + 23 + 41       | rr= = 593       | c = which is prime }} {{eqn | l = 24^2 + 24 + 41       | r = 576 + 24 + 41       | rr= = 641       | c = which is prime }} {{eqn | l = 25^2 + 25 + 41       | r = 625 + 25 + 41       | rr= = 691       | c = which is prime }} {{eqn | l = 26^2 + 26 + 41       | r = 676 + 26 + 41       | rr= = 743       | c = which is prime }} {{eqn | l = 27^2 + 27 + 41       | r = 729 + 27 + 41       | rr= = 797       | c = which is prime }} {{eqn | l = 28^2 + 28 + 41       | r = 784 + 28 + 41       | rr= = 853       | c = which is prime }} {{eqn | l = 29^2 + 29 + 41       | r = 841 + 29 + 41       | rr= = 911       | c = which is prime }} {{eqn | l = 30^2 + 30 + 41       | r = 900 + 30 + 41       | rr= = 971       | c = which is prime }} {{eqn | l = 31^2 + 31 + 41       | r = 961 + 31 + 41       | rr= = 1033       | c = which is prime }} {{eqn | l = 32^2 + 32 + 41       | r = 1024 + 32 + 41       | rr= = 1097       | c = which is prime }} {{eqn | l = 33^2 + 33 + 41       | r = 1089 + 33 + 41       | rr= = 1163       | c = which is prime }} {{eqn | l = 34^2 + 34 + 41       | r = 1156 + 34 + 41       | rr= = 1231       | c = which is prime }} {{eqn | l = 35^2 + 35 + 41       | r = 1225 + 35 + 41       | rr= = 1301       | c = which is prime }} {{eqn | l = 36^2 + 36 + 41       | r = 1296 + 36 + 41       | rr= = 1373       | c = which is prime }} {{eqn | l = 37^2 + 37 + 41       | r = 1369 + 37 + 41       | rr= = 1447       | c = which is prime }} {{eqn | l = 38^2 + 38 + 41       | r = 1444 + 38 + 41       | rr= = 1523       | c = which is prime }} {{eqn | l = 39^2 + 39 + 41       | r = 1521 + 39 + 41       | rr= = 1601       | c = which is prime }} {{eqn | l = 40^2 + 40 + 41       | r = 1600 + 40+ 41       | rr= = 1681       | c = which is not prime: $1681 = 41^2$ }} {{end-eqn}} {{OEIS|A005846}} Then we have: {{begin-eqn}} {{eqn | l = \paren {-\paren  {n + 1} }^2 + \paren {-\paren {n + 1} }       | r = n^2 + 2 n + 1 - \paren {n + 1}       | c =  }} {{eqn | r = n^2 + n       | c =  }} {{end-eqn}} and so replacing $0$ to $39$ with $-1$ to $-40$ yields exactly the same sequence of primes. We note in addition the example: :$581^2 + 581 + 41 = 338 \, 183$ which is prime. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 41|next = Prime-Generating Quadratic of form x squared - 79 x + 1601}}: $41$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 41|next = Euler Lucky Number/Examples/41/Mistake}}: $41$ Category:Euler Lucky Numbers Category:41 lk4bfvl2uaiw5v6frkad6sln0swla02


Feit-Thompson Conjecture

Feit-Thompson Conjecture 0 53680 346749 285703 2018-03-11T10:32:43Z Prime.mover 59 wikitext text/x-wiki == Conjecture == There exist no distinct prime numbers $p$ and $q$ such that: :$\dfrac {p^q - 1} {p - 1}$ divides $\dfrac {q^p - 1} {q - 1}$ === Stronger Feit-Thompson Conjecture === {{:Feit-Thompson Conjecture/Stronger}} {{Namedfor|Walter Feit|name2 = John Griggs Thompson|cat = Feit|cat2 = Thompson}} Category:Prime Numbers Category:Divisors Category:Unproven Hypotheses Category:Feit-Thompson Conjecture mdebsx8acjyb9iigtnnvjvbmi14qoh6


Factorial/Examples/11

Factorial/Examples/11 0 53752 307317 285906 2017-07-30T05:19:59Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$11! = 39 \, 916 \, 800$ </onlyinclude> == Proof == From Examples of Factorials: :$10! = 3 \, 628 \, 800$ Then: <pre>  3 628 800 x       11 ----------  3 628 800 36 288 000 ---------- 39 916 800 </pre>{{qed}} Category:Factorials/Examples Category:11 6ra9v68ayfjn5rouvqw5o43wmkd2r37


Factorial/Examples/12

Factorial/Examples/12 0 53753 307318 285905 2017-07-30T05:20:19Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$12! = 479 \, 001 \, 600$ </onlyinclude> == Proof == From $11$ factorial: :$11! = 39 \, 916 \, 800$ Then: <pre>  39 916 800 x        12 -----------  79 833 600 399 168 000 ----------- 479 001 600 </pre>{{qed}} Category:Factorials/Examples Category:12 c3mh9jjeduydt0onxjery7csw6wgrbe


Factorial/Examples/13

Factorial/Examples/13 0 53754 307319 285911 2017-07-30T05:20:39Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$13! = 6 \, 227 \, 020 \, 800$ </onlyinclude> === Prime Factors of $13!$ === {{:Prime Factors of 13 Factorial}} == Proof == From $12$ factorial: :$12! = 479 \, 001 \, 600$ Then: <pre>   479 001 600 x          13 ------------- 1 437 004 800 4 790 016 000 ------------- 6 227 020 800 </pre>{{qed}} Category:Factorials/Examples Category:13 7d7m48618hj5aze0mz9oc74dnppukox


Factorial/Examples/14

Factorial/Examples/14 0 53755 307320 285903 2017-07-30T05:20:57Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$14! = 87 \, 178 \, 291 \, 200$ </onlyinclude> == Proof == From $13$ factorial: :$13! = 6 \, 227 \, 020 \, 800$ Then: <pre>  6 227 020 800 x           14 --------------- 24 908 083 200 62 270 208 000 -------------- 87 178 291 200 </pre>{{qed}} Category:Factorials/Examples Category:14 0n3ojsabmqnri974kjbmklppvl0jc9q


Factorial/Examples/15

Factorial/Examples/15 0 53757 307321 285907 2017-07-30T05:21:10Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$15! = 1 \, 307 \, 674 \, 368 \, 000$ </onlyinclude> == Proof == From $14$ Factorial: :$14! = 87 \, 178 \, 291 \, 200$ Then: <pre>    87 178 291 200 x              15 ------------------   435 891 456 000   871 782 912 000 ----------------- 1 307 674 368 000 </pre>{{qed}} Category:Factorials/Examples Category:15 lqtdq70q03acamfxx2281pfzhlhabvk


Factorial/Examples/16

Factorial/Examples/16 0 53758 307322 285908 2017-07-30T05:21:22Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$16! = 20 \, 922 \, 789 \, 888 \, 000$ </onlyinclude> == Proof == From $15$ Factorial: :$15! = 1 \, 307 \, 674 \, 368 \, 000$ Then: <pre>  1 307 674 368 000 x               16 ------------------  7 846 046 208 000 13 076 743 680 000 ------------------ 20 922 789 888 000 </pre>{{qed}} Category:Factorials/Examples Category:16 d1yze2qeyc9jpylsn7didmwujg4epuj


Factorial/Examples/17

Factorial/Examples/17 0 53759 307323 285909 2017-07-30T05:21:32Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$17! = 355 \, 687 \, 428 \, 096 \, 000$ </onlyinclude> == Proof == From $16$ Factorial: :$16! = 20 \, 922 \, 789 \, 888 \, 000$ Then: <pre>  20 922 789 888 000 x                17 ------------------- 146 459 529 216 000 209 227 898 880 000 ------------------- 355 687 428 096 000 </pre>{{qed}} Category:Factorials/Examples Category:17 aa7clqqtqsra93zliurbbr6ut34uy58


Factorial/Examples/18

Factorial/Examples/18 0 53760 307324 285910 2017-07-30T05:21:42Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$ </onlyinclude> == Proof == From $17$ Factorial: :$17! = 355 \, 687 \, 428 \, 096 \, 000$ Then: <pre>   355 687 428 096 000 x                  18 --------------------- 2 845 499 424 768 000 3 556 874 280 960 000 --------------------- 6 402 373 705 728 000 </pre>{{qed}} Category:Factorials/Examples Category:18 iy07acl27x5ararkfo5bnxgb1fezbbe


Factorial/Examples/19

Factorial/Examples/19 0 53761 307325 285915 2017-07-30T05:21:53Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$ </onlyinclude> == Proof == From $18$ Factorial: :$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$ Then: <pre>   6 402 373 705 728 000 x                    19 -----------------------  57 621 363 351 552 000  64 023 737 057 280 000 ----------------------- 121 645 100 408 832 000 </pre>{{qed}} Category:Factorials/Examples Category:19 8bw062ee8izvjerlllqkr0j857z6fw4


Factorial/Examples/20

Factorial/Examples/20 0 53762 382273 307326 2018-12-14T22:42:48Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$ </onlyinclude> === Prime Factors of $20!$ === {{:Prime Factors of 20 Factorial}} == Proof == From $19$ Factorial: :$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$ Then: <pre>   121 645 100 408 832 000 x                      20 ------------------------ 2 432 902 008 176 640 000 </pre>{{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Cross-Sections of Leech Lattice|next = Factorial/Examples/450/Historical Note}}: $24$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Factorial|next = Factorial/Examples/1,000,000}}: $24$ Category:Factorials/Examples Category:20 sflj3v0q5ogye6so4mmykwgq7oynkkw


Factorial/Examples/21

Factorial/Examples/21 0 53836 307327 286137 2017-07-30T05:22:17Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$ </onlyinclude> == Proof == From $20$ Factorial: :$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$ Then: <pre>   2 432 902 008 176 640 000 x                        21 ---------------------------   2 432 902 008 176 640 000  48 658 040 163 532 800 000 ---------------------------  51 090 942 171 709 440 000    </pre>{{qed}} Category:Factorials/Examples Category:21 0a57vhpp7r4z699dh61a3eb8dtpaaw8


Factorial/Examples/22

Factorial/Examples/22 0 53837 307328 286139 2017-07-30T05:22:27Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$ </onlyinclude> == Proof == From $21$ Factorial: :$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$ Then: <pre>    51 090 942 171 709 440 000 x                          22 -----------------------------   102 181 884 343 418 880 000 1 021 818 843 434 188 800 000 ----------------------------- 1 124 000 727 777 607 680 000    </pre>{{qed}} Category:Factorials/Examples Category:22 s6zkqhpng4z50u0nvcp4kzeyag2zrkm


Factorial/Examples/23

Factorial/Examples/23 0 53838 307329 286140 2017-07-30T05:22:38Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$ </onlyinclude> == Proof == From $22$ Factorial: :$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$ Then: <pre>  1 124 000 727 777 607 680 000 x                           23 ------------------------------  3 372 002 183 332 823 040 000 22 480 014 555 552 153 600 000 ------------------------------ 25 852 016 738 884 976 640 000    </pre>{{qed}} Category:Factorials/Examples Category:23 fpy1sfyi1q72rxgk957fgb4wwlyhqyc


Factorial/Examples/24

Factorial/Examples/24 0 53839 307330 286141 2017-07-30T05:22:50Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$ </onlyinclude> == Proof == From $23$ Factorial: :$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$ Then: <pre>  25 852 016 738 884 976 640 000 x                            24 ------------------------------- 103 408 066 955 539 906 560 000 517 040 334 777 699 532 800 000 ------------------------------- 620 448 401 733 239 439 360 000    </pre>{{qed}} Category:Factorials/Examples Category:24 4tg05bm160modl1t8ynopwd8so39p4d


Factorial/Examples/25

Factorial/Examples/25 0 53840 307331 286143 2017-07-30T05:23:00Z Prime.mover 59 wikitext text/x-wiki == Example of Factorial == <onlyinclude> :$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$ </onlyinclude> == Proof == From $24$ Factorial: :$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$ Then: <pre>    620 448 401 733 239 439 360 000 x                               25 ----------------------------------  3 102 242 008 666 197 196 800 000 12 408 968 034 664 788 787 200 000 ---------------------------------- 15 511 210 043 330 985 984 000 000    </pre>{{qed}} Category:Factorials/Examples Category:25 3rsp94dts04c0kf6b2axggtlhyfstfn


Square of Small-Digit Palindromic Number is Palindromic/Examples/11 0 53853 383304 382082 2018-12-20T06:39:14Z Prime.mover 59 wikitext text/x-wiki == Example of Square of Small-Digit Palindromic Number is Palindromic == <onlyinclude> ::$11^2 = 121$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 11^2       | r = \left({10 + 1}\right)^2       | c =  }} {{eqn | r = 10^2 + 2 \times 10 \times 1 + 1^2       | c =  }} {{eqn | r = 100 + 20 + 1       | c =  }} {{eqn | r = 121       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/22|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/111}}: $22$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 121|next = 121 is Square Number in All Bases greater than 2}}: $121$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/22|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/111}}: $22$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 121|next = 121 is Square Number in All Bases greater than 2}}: $121$ Category:Square of Small-Digit Palindromic Number is Palindromic Category:11 Category:121 k0ar8p9ci8onod77ll6uy8e8c8s2snc


Square of Small-Digit Palindromic Number is Palindromic/Examples/121 0 53859 382085 298747 2018-12-14T06:17:48Z Prime.mover 59 wikitext text/x-wiki == Example of Square of Small-Digit Palindromic Number is Palindromic == <onlyinclude> ::$121^2 = 14641$ </onlyinclude> == Proof == <pre>   121 x 121 -----   121  2420 12100 ------- 14641 </pre>{{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/1111|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/212}}: $22$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/1111|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/212}}: $22$ Category:Square of Small-Digit Palindromic Number is Palindromic Category:121 tdhaur6apnd1st1t64fjvusyxar35xc


Sigma Function of Square-Free Integer/Examples/70 0 53870 492965 317776 2020-10-06T18:06:59Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\map \sigma {70} = 144$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof 1 == {{:Sigma Function of Square-Free Integer/Examples/70/Proof 1}} == Proof 2 == {{:Sigma Function of Square-Free Integer/Examples/70/Proof 2}} Category:Sigma Function of Square-Free Integer Category:70 Category:Sigma Function of Square-Free Integer/Examples/70 ld8kh8u994m7leko1i8hf74cwmhm918


Birthday Paradox

Birthday Paradox 0 53960 478611 478607 2020-07-17T10:49:09Z Prime.mover 59 wikitext text/x-wiki == Paradox == Let there be $23$ or more people in a room. The probability that at least $2$ of them have the same birthday is greater than $50 \%$. == Proof == Let there be $n$ people in the room. Let $\map p n$ be the probability that no two people in the room have the same birthday. For simplicity, let us ignore leap years and assume there are $365$ days in the year. Let the birthday of person $1$ be established. The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$. Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$. Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$. And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$. Hence the total probability that none of the $n$ people share a birthday is given by: :$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$ {{begin-eqn}} {{eqn | l = \map p n       | r = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}       | c =  }} {{eqn | r = \dfrac {365!} {365^n} \binom {365} n       | c =  }} {{end-eqn}} Setting $n = 23$ and evaluating the above gives: :$\map p {23} \approx 0.493$ Hence the probability that at least $2$ people share a birthday is $1 = 0.492 = 0.507 = 50.7 \%$ {{qed}} == Conclusion == This is a veridical paradox. Counter-intuitively, the probability of a shared birthday amongst such a small group of people is surprisingly high. == General Birthday Paradox == {{:Birthday Paradox/General}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 23 is Largest Integer not Sum of Distinct Perfect Powers/Mistake|next = Harmonic Series is Divergent}}: $23$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 23 is Largest Integer not Sum of Distinct Perfect Powers/Mistake|next = Birthday Paradox/General/3}}: $23$ Category:Probability Theory Category:Combinatorics Category:Veridical Paradoxes Category:Birthday Paradox n82iwsg3p01ongoa3zlquh7ap1exnev


Legendre's Condition/Lemma 1

Legendre's Condition/Lemma 1 0 54016 496498 457855 2020-10-25T13:41:04Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Lemma == Let $y = \map y x$ be a real function, such that: :$\map y a = A$ :$\map y b = B$ Let $J \sqbrk y$ be a functional, such that: :$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$ where: :$F \in C^2 \closedint a b$  {{WRT}} all its variables. Then: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$ where: {{begin-eqn}} {{eqn | l = \map P {x, \map y x}       | r = \frac 1 2 F_{y'y'} }} {{eqn | l = \map Q {x, \map y x}       | r = \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy} } }} {{end-eqn}} == Proof == The minimising function $y$ has fixed end-points. Therefore, consider an increment of a functional with $h$ such that: :$h \in C^1 \closedint a b: \paren {\map h a = 0} \land \paren {\map h b = 0}$ Then: {{begin-eqn}} {{eqn | l = \Delta J \sqbrk {y; h}       | r = J \sqbrk {y + h} - J \sqbrk y       | c = {{Defof|Increment of Functional}} }} {{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x       | c = form of $J$ }} {{eqn | r = \int_a^b \paren {F + \paren {F_y h + F_{y'} h'} + \frac 1 2 \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} - F} \rd x       | c = Taylor's Theorem }} {{eqn | r = \int_a^b \paren {F_y h + F_{y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} \rd x       | c = cancel $F$ }} {{end-eqn}} where omitted variables are $\paren {x, y, y'}$, and the overbar indicates derivatives being taken along some intermediate curves: {{begin-eqn}} {{eqn | l = \overline {\map {F_{yy} } {x,y,y'} }       | r = \map {F_{yy} } {x, y + \theta h, y' + \theta h'} }} {{eqn | l = \overline {\map {F_{yy'} } {x,y,y'} }       | r = \map {F_{yy'} } {x, y + \theta h, y' + \theta h'} }} {{eqn | l = \overline {\map {F_{y'y'} } {x,y,y'} }       | r = \map {F_{y'y'} } {x, y + \theta h, y' + \theta h'} }} {{end-eqn}} with $0 < \theta < 1$. If $\overline F_{yy}$, $\overline F_{yy'}$, $\overline F_{y'y'} $ are to be replaced by $F_{yy}$, $F_{yy}$, $F_{y'y'}$ evaluated at the point $\tuple {x, \map y x, \map {y'} x}$, then: :$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map {F_y} {x, y, y'} h + \map {F_{y'} } {x, y, y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\map {F_{yy} } {x, y, y'} h^2 + 2 \map {F_{yy'} } {x, y, y'} h h'+ \map {F_{y'y'} } {x, y, y'} h'^2} \rd x + \epsilon$ where: :$\ds \epsilon = \int_a^b \paren {\epsilon_1 h^2 + \epsilon_2 h h' + \epsilon_3 h'^2}$ By continuity of $F_{yy}$, $F_{yy}$, $F_{y'y'}$: :$\size h_1 \to 0 \implies \epsilon_1, \epsilon_2, \epsilon_3 \to 0$ {{explain|What does $\size h_1$ mean?}} Thus, $\epsilon$ is an infinitesimal of the order higher than 2 {{WRT}} $\size h$. {{Stub|Expand on steps including $ \epsilon $}} The first and second term on the {{RHS}} of $\Delta J \sqbrk {y; h}$ are $\delta J \sqbrk {y; h}$ and $\delta^2 J \sqbrk {y; h}$ respectively. Integrate the second term of $\delta^2 J \sqbrk {y; h}$ by parts: {{begin-eqn}} {{eqn | l = \int_a^b 2 F_{yy'} h h' \rd x       | r = \int_a^b 2 F_{yy'} h \rd h }} {{eqn | r = \int_a^b F_{yy'} \rd h^2 }} {{eqn | r = \bigintlimits {F_{yy'} h^2} {x \mathop = a} {x \mathop = b} - \int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x }} {{eqn | r = -\int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x }} {{end-eqn}} {{explain|Why does $\bigintlimits {F_{yy'} h^2} {x \mathop {{=}} a} {x \mathop {{=}} b}$ vanish?}} Therefore: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\frac 1 2 F_{y'y'} h'^2 + \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy'} } h^2} \rd x$ {{qed|lemma}} {{explain|Review use of square brackets. If they are being used purely for parenthesis, better to replace with round ones, as square ones have conventional meanings, so use of them for parenthesis may confuse.}} == Mistake == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin}}: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition p. 102 states that :$P = \dfrac 1 2 F_{y'y'} \quad Q = \dfrac 1 2 \paren {F_{yy'} - \dfrac \d {\d x} F_{yy'} }$ This is a mistake, since the second variation should contain both pure and mixed partial derivatives of the order 2. However, $F_{yy} $ is missing and could not have been lost during derivation of the proof. It should be: :$Q = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$ == Sources == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin|prev = Legendre's Condition|next = Legendre's Condition/Lemma 2}}: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition  Category:Calculus of Variations 4f4ujy87m15vkm3s39frexhfs87749v


Legendre's Condition/Lemma 2

Legendre's Condition/Lemma 2 0 54021 458074 458070 2020-03-28T22:13:00Z Julius 3095 wikitext text/x-wiki {{MissingLinks}} == Lemma == Let $ h$ be a real function such that: :$h \in C^1 \openint a b$ :$\map h a = 0$ :$\map h b = 0$ Let: :$\displaystyle \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$ where $P \in C^0 \closedint a b$.  Then a necessary condition for: :$\delta^2 J \sqbrk {y; h} \ge 0$  is: :$\forall x\in\closedint a b: \map P {x, \map y x} \ge 0$ == Proof == Assume that above is not true. Then: :$\paren {\exists x_0 \in \closedint a b} \land \paren {\exists \beta \in \R_{<0} }: \map P {x_0} = -2 \beta$ $P$ is continuous. Thus: :$\exists \alpha \in \R_{>0}: \paren {a \ge x_0 - \alpha} \land \paren {x_0 + \alpha \ge b}$ and: :$\forall x \in \openint {x_0 - \alpha} {x_0 + \alpha}: \map P x < -\beta$ In other words: $\map P x \begin {cases} = 0 & : x \in \closedint a {x_0 - \alpha} \lor \closedint {x_0 + \alpha} b \\ < 0 & : x \in \closedint {x_0 - \alpha} {x_0 + \alpha} \end {cases}$ Let :$h = \begin {cases} \sin^2 \paren {\dfrac {\map \pi {x - x_0} } \alpha} & : x_0 - \alpha \ge x \ge x_0 + \alpha \\ 0 & : \text {otherwise} \end{cases}$ It belongs to $C^1 \openint a b$ because: {{explain|The derivation and significance of all the below: what they mean, what depends on what, etc.}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 - \alpha + 0^+} h       | r = \lim_{x \to x_0 - \alpha + 0^+} \map {\sin^2} {\frac {\map \pi {x - x_0} } \alpha} }} {{eqn | r = \map {\sin^2} {\map \pi {\frac {0^+} \alpha - 1} } }} {{eqn | r = 0 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 - \alpha + 0^+}h'       | r = \lim_{x \to x_0 - \alpha + 0^+} \map \sin {\frac {2 \map \pi {x - x_0} } \alpha} \frac \pi \alpha }} {{eqn | r = \map \sin {2 \map \pi {\frac {0^+} \alpha - 1} } \frac \pi \alpha }} {{eqn | r = 0 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 - \alpha + 0^+}h''       | r = \lim_{x \to x_0 - \alpha + 0^+} \map \cos {\frac {2 \map \pi {x - x_0} } \alpha} \frac {2 \pi^2} {\alpha^2} }} {{eqn | r = \map \cos {2 \map \pi {\frac {0^+} \alpha - 1} } \frac {2 \pi^2} {\alpha^2} }} {{eqn | r = \frac {2 \pi^2} {\alpha^2} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 + \alpha + 0^-} h       | r = \lim_{x \to x_0 + \alpha + 0^-} \map {\sin^2} {\frac {\map \pi {x - x_0} } \alpha} }} {{eqn | r = \map {\sin^2} {\map \pi {\frac {0^-} \alpha + 1} } }} {{eqn | r = 0 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 + \alpha + 0^-}h'       | r = \lim_{x \to x_0 - \alpha + 0^-} \map \sin {\frac {2 \map \pi {x - x_0} } \alpha} \frac \pi \alpha }} {{eqn | r = \map \sin {2 \map \pi {\frac {0^-} \alpha + 1} } \frac \pi \alpha }} {{eqn | r = 0 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \to x_0 + \alpha + 0^-}h''       | r = \lim_{x \to x_0 - \alpha + 0^-} \map \cos {\frac {2 \map \pi {x - x_0} } \alpha} \frac {2 \pi^2} {\alpha^2} }} {{eqn | r = \map \cos {2 \map \pi {\frac {0^-} \alpha + 1} } \frac {2 \pi^2} {\alpha^2} }} {{eqn | r = \frac {2\pi^2} {\alpha^2} }} {{end-eqn}} In other words, only $h$ and $h'$ are continuous in $\closedint a b$ Then: {{begin-eqn}} {{eqn | l = \int_a^b \paren {P h'^2 + Q h^2} \rd x       | r = \int_{x_0 - \alpha}^{x_0 + \alpha} P \frac {\pi^2} {\alpha^2} \map {\sin^2} {\frac {2 \map \pi {x - x_0} } \alpha} \rd x + \int_{x_0 - \alpha}^{x_0 + \alpha} Q \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha} \rd x }} {{eqn | o = <       | r = -\beta \frac {\pi^2} {\alpha^2} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^2} {\frac {2 \map \pi {x - x_0} } \alpha} \rd x + \max_{a \le x \le b} \size {\map Q x} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha}\rd x       | c = as $\displaystyle \map P x < -\beta, \quad \map Q x < \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$ }} {{eqn | r = -\beta \frac {\pi^2} \alpha + \max_{a \mathop \le x \mathop \le b} \size {\map Q x} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha} \rd x }} {{eqn | r = -\beta \frac {\pi^2} \alpha + \frac 3 4 \alpha M  }} {{end-eqn}} where: :$\displaystyle M = \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$ For sufficiently small $\alpha$ the {{RHS}} is negative. Hence, $\delta^2 J$ is negative for the corresponding $h$. To conclude, it has been shown that :$P \ge 0 \quad \neg \forall x \in \closedint a b \implies \delta^2 J<0$ Then, by contrapositive statement this is equivalent to: :$\forall x \in \closedint a b: \delta^2 J \ge 0 \implies P \ge 0$ {{qed}} == Sources == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin|prev = Legendre's Condition/Lemma 1|next = Definition:Conjugate Point}}: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition  Category:Calculus of Variations r3a5e2emmox1zng2irquahrotbcr76s


Sigma Function of Integer/Examples/20

Sigma Function of Integer/Examples/20 0 54025 317610 317522 2017-09-16T11:58:00Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({20}\right) = 42$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$20 = 2^2 \times 5$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({20}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1}       | c =  }} {{eqn | r = \frac 7 1 \times \frac {6 \times 4} 4       | c =  }} {{eqn | r = 7 \times 6       | c =  }} {{eqn | r = 42       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:20 svrer95xyqtzsimn8y4q3fhlh5itsor


Sigma Function of Non-Square Semiprime/Examples/35 0 54077 484228 352264 2020-08-30T21:18:49Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Non-Square Semiprime == <onlyinclude> :$\map \sigma {35} = 48$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$35 = 5 \times 7$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \map \sigma {35}       | r = \paren {5 + 1} \paren {7 + 1}       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 6 \times 8       | c =  }} {{eqn | r = 48       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Non-Square Semiprime Category:35 f563vp1dfsoqnxbfeuiwl1ldgq3e6hn


Binomial Theorem/Examples/Cube of Sum

Binomial Theorem/Examples/Cube of Sum 0 54109 412151 393428 2019-07-08T08:23:28Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Binomial Theorem == <onlyinclude> :$\paren {x + y}^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$ </onlyinclude> === Corollary === {{:Cube of Sum/Corollary}} == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 3$. {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Difference|next = Cube of Difference}}: $\S 2$: Special Products and Factors: $2.3$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Sum/Algebraic Proof 2|next = Fourth Power of Sum}}: $\S 20$: Binomial Series: $20.6$ * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Square of Sum|next = Binomial Theorem/Examples/4th Power of Sum}}: Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem Category:Examples of Use of Binomial Theorem Category:Third Powers jux8rm3q0livgh17bgq7gn9mq7opwy1


Taylor Expansion for Polynomials/Order 1 0 54559 288840 2017-03-15T22:09:30Z Barto 3079 Created page with "== Theorem == <onlyinclude> Let $R$ be a commutative ring with unity.  Let $f(X)\in R[X]$ be a Definition:Polynomial (Abstract Al..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $R$ be a commutative ring with unity. Let $f(X)\in R[X]$ be a polynomial. Let $a\in R$. Then there exists a polynomial $g(X)\in R[X]$ such that: :$f(X+a) = f(X) + a f'(X) +a^2 g(X)$ where $f'$ denotes the formal derivative of $f$. </onlyinclude> == Proof == By linearity, it suffices to prove this for $f(X)=X^n$. This is now a direct consequence of the Binomial Theorem. {{qed}} Category:Polynomial Theory 2y2stiza6omv9tu8mo9wsygmj8frhvf


Smallest Set of Weights for One-Pan Balance 0 54967 481405 390665 2020-08-11T21:23:51Z Prime.mover 59 wikitext text/x-wiki == Classic Problem == Consider a set of balance scales for determining the weight of a physical object. Let this set of scales be such that weights may be placed in one of the pans. What is the smallest set of weights needed to weigh any given integer weight up to a given amount? == Solution == A set of $m$ weights in the sequence $\sequence {2^n}$: :$1, 2, 4, 8, 16, \ldots$ allows one to weigh any given integer weight up to $2^m - 1$. == Proof == This is equivalent to the statement that every positive integer can be expressed uniquely in binary notation. This in turn is an application of the Basis Representation Theorem. {{qed}} == Examples == {{:Smallest Set of Weights for One-Pan Balance/Examples}} == Also see == * Smallest Set of Weights for Two-Pan Balance == Historical Note == {{:Smallest Set of Weights for One-Pan Balance/Historical Note}} == Sources == * {{BookReference|Mathematical Recreations and Essays|1974|W.W. Rouse Ball|author2 = H.S.M. Coxeter|ed = 12th|edpage = Twelfth Edition}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 18,446,744,073,709,551,615|next = Smallest Set of Weights for One-Pan Balance/Historical Note}}: $31$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance}}: Bachet: $108$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 18,446,744,073,709,551,615|next = Smallest Set of Weights for One-Pan Balance/Historical Note}}: $31$ Category:Classic Problems Category:Binary Notation Category:Smallest Set of Weights for One-Pan Balance pep184wcvd7st290r60wp69n9qxgoza


Smallest Set of Weights for Two-Pan Balance 0 54969 481412 390754 2020-08-11T22:01:32Z Prime.mover 59 wikitext text/x-wiki == Classic Problem == Consider a set of balance scales for determining the weight of a physical object. Let this set of scales be such that weights may be placed in either of the two pans. What is the smallest set of weights needed to weigh any given integer weight up to a given amount? == Solution == A set of weights up to $3^m$ in the sequence $\sequence {3^n}$: :$1, 3, 9, 27, \ldots$ allows one to weigh any given integer weight up to $\dfrac {3^{m + 1} - 1} 2$. == Proof == Place the item to be weighed in the left hand pan of the balance. Let it weigh $n$. Let $n$ be expressed in balanced ternary representation. From Representation of Integers in Balanced Ternary, $n$ can be uniquely so represented. With $m$ digits, we can count up to $\dfrac {3^{m + 1} - 1} 2$. We use the balanced ternary representation to model how to place the weights. Let the digits of $n$ in such a representation be numbered $0, 1, \ldots, m$ from the least significant digit to the most significant digit. The $k$th digit represents the weight which weighs $3^k$. When the $k$th digit is $1$, place weight $3^k$ in the right hand pan. When the $k$th digit is $\underline 1$, place weight $3^k$ in the left hand pan. When the $k$th digit is $$, place weight $3^k$ in neither pan. This will make the scales balance. {{qed}} == Examples == {{:Smallest Set of Weights for Two-Pan Balance/Examples}} == Also see == * Smallest Set of Weights for One-Pan Balance == Historical Note == {{:Smallest Set of Weights for Two-Pan Balance/Historical Note}} == Sources == * {{BookReference|Problèmes Plaisans et Delectables qui se font par les Nombres|1612|Claude-Gaspar Bachet}} * {{BookReference|Mathematical Recreations and Essays|1974|W.W. Rouse Ball|author2 = H.S.M. Coxeter|ed = 12th|edpage = Twelfth Edition}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance/Historical Note}}: $31$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Smallest Set of Weights for One-Pan Balance|next = Book:Claudio Gaspare Bacheto/Diophanti Alexandrini Arithmeticorum}}: Bachet: $108$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance/Historical Note}}: $31$ Category:Classic Problems Category:Ternary Notation Category:Smallest Set of Weights for Two-Pan Balance ck5ksbidrvet8wqobildwk75wly3x9a


Sigma Function of 217

Sigma Function of 217 0 55223 352260 317842 2018-05-01T14:11:33Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Non-Square Semiprime == <onlyinclude> :$\sigma \left({217}\right) = 256$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$217 = 7 \times 31$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \sigma \left({217}\right)       | r = \left({7 + 1}\right) \left({31 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 8 \times 32       | c =  }} {{eqn | r = 2^3 \times 2^5       | c =  }} {{eqn | r = 2^8       | c =  }} {{eqn | r = \left({2^4}\right)^2       | c =  }} {{eqn | r = 16^2       | c =  }} {{eqn | r = 256       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Non-Square Semiprime Category:217 t1dzyhch4vk8p05ofatyl44vay5sw3p


Sigma Function of 214

Sigma Function of 214 0 55226 392679 352261 2019-02-18T14:54:06Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Non-Square Semiprime == <onlyinclude> :$\map \sigma {214} = 324$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$214 = 2 \times 107$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \map \sigma {214}       | r = \paren {2 + 1} \paren {107 + 1}       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 3 \times 108       | c =  }} {{eqn | r = 3 \times \paren {2^2 \times 3^3}       | c =  }} {{eqn | r = 2^3 \times 3^4       | c =  }} {{eqn | r = \paren {2 \times 3^2}^2       | c =  }} {{eqn | r = 18^2       | c =  }} {{eqn | r = 324       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Non-Square Semiprime Category:214 llhx6cmvjx6219ej3u92ldhuco7nalq


Sigma Function of 210

Sigma Function of 210 0 55249 444501 444500 2020-01-22T15:27:26Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\map \sigma {210} = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$210 = 2 \times 3 \times 5 \times 7$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {210}       | r = \paren {2 + 1} \paren {3 + 1} \paren {5 + 1} \paren {7 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 4 \times 6 \times 8       | c =  }} {{eqn | r = 3 \times 2^2 \times \paren {2 \times 3} \times 2^3       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \paren {2^3 \times 3}^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:210 dcu9qi73ill3ucy6qxc63z0usb33us4


Sigma Function of 265

Sigma Function of 265 0 55252 317910 317908 2017-09-16T13:26:59Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Non-Square Semiprime == <onlyinclude> :$\sigma \left({265}\right) = 324$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$265 = 5 \times 53$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({265}\right)       | r = \left({5 + 1}\right) \left({53 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 6 \times 54       | c =  }} {{eqn | r = \left({2 \times 3}\right) \times \left({2 \times 3^3}\right)       | c =  }} {{eqn | r = 2^2 \times 3^4       | c =  }} {{eqn | r = \left({2 \times 3^2}\right)^2       | c =  }} {{eqn | r = 18^2       | c =  }} {{eqn | r = 324       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Non-Square Semiprime Category:265 27lfvdolzh8xkyhjnkvfr2k2qer0quv


Sigma Function of 282

Sigma Function of 282 0 55254 439920 317859 2019-12-17T13:28:01Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\map \sigma {282} = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$282 = 2 \times 3 \times 47$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {282}       | r = \paren {2 + 1} \paren {3 + 1} \paren {47 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 4 \times 48       | c =  }} {{eqn | r = 3 \times 2^2 \times \paren {2^4 \times 3}       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \paren {2^3 \times 3}^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:282 mxl5x6554hbloriiuetd6t7yb5ft1f2


Sigma Function of 310

Sigma Function of 310 0 55257 317871 317869 2017-09-16T13:20:45Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\sigma \left({310}\right) = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$310 = 2 \times 5 \times 31$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({310}\right)       | r = \left({2 + 1}\right) \left({5 + 1}\right) \left({31 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 6 \times 32       | c =  }} {{eqn | r = 3 \times \left({2 \times 3}\right) \times 2^5       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:310 fgx8h5izqq9d9m76ftnwn651hr8qtts


Sigma Function of 322

Sigma Function of 322 0 55260 317874 317872 2017-09-16T13:21:13Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\sigma \left({322}\right) = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$322 = 2 \times 7 \times 23$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({322}\right)       | r = \left({2 + 1}\right) \left({7 + 1}\right) \left({23 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 8 \times 24       | c =  }} {{eqn | r = 3 \times 2^3 \times \left({2^3 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:322 7tw0yluzqzp8h0teneku39m8ua1bta0


Sigma Function of 343

Sigma Function of 343 0 55263 383904 383903 2018-12-22T16:29:10Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Power of Prime == <onlyinclude> :$\map \sigma {343} = 400$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Power of Prime: :$\map \sigma {p^k} = \dfrac {p^{k + 1} - 1} {p_i - 1}$ We have that: :$343 = 7^3$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {343}       | r = \frac {7^4 - 1} {7 - 1}       | c =  }} {{eqn | r = \frac {2400} 6       | c =  }} {{eqn | r = 400       | c =  }} {{eqn | r = 20^2       | c =  }} {{end-eqn}} Thus we have that: :$7^0 + 7^2 + 7^2 + 7^3 = 20^2$ {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 400|next = Numbers whose Sigma is Square/Examples/400}}: $400$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 400|next = Numbers whose Sigma is Square/Examples/400}}: $400$ Category:Sigma Function of Power of Prime Category:343 Category:400 ipq88t9fu4afs6dd70zbupj0gzvi375


Sigma Function of 345

Sigma Function of 345 0 55267 317877 317875 2017-09-16T13:21:35Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\sigma \left({345}\right) = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$345 = 3 \times 5 \times 23$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({345}\right)       | r = \left({3 + 1}\right) \left({5 + 1}\right) \left({23 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 4 \times 6 \times 24       | c =  }} {{eqn | r = 2^2 \times \left({2 \times 3}\right) \times \left({2^3 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:345 k9hdfy09zns4sbi7wj6r2bxagpmucpf


Sigma Function of 382

Sigma Function of 382 0 55282 352262 318092 2018-05-01T14:12:21Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Non-Square Semiprime == <onlyinclude> :$\sigma \left({382}\right) = 576$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$382 = 2 \times 191$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \sigma \left({382}\right)       | r = \left({2 + 1}\right) \left({191 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 3 \times 192       | c =  }} {{eqn | r = 3 \times \left({2^6 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Non-Square Semiprime Category:382 8acbm1vdulf64bmj1lrz6oshkujbmsa