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# Modular Elliptic Curves and Fermat's Last Theorem

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... In his work on modularity of elliptic curves and Fermat's last theorem, Wiles [17] discovered a numerical criterion for certain noetherian local rings A to be complete intersections. Diamond [8] generalized Wiles' result by establishing a criterion for modules M over A to be free and for A to be a complete intersection; see the discussion below for the precise statements of their results. ...
... In Section 5, we give a more streamlined proof of a formula for δ A (A), formulated (albeit in the setting of certain derived rings) by Venkatesh [16], and proved in [10], in terms of certain André-Quillen cohomology modules. We end by explaining how this formula gives another proof of the isomorphism criterion for maps between complete intersection rings due to Wiles [17] and Lenstra [13]. ...
... The question of comparing congruence modules for A and M has been studied extensively, in the context of the theory of congruences between modular forms. This theory plays a key role in the breakthrough work of Wiles [17]. As recalled above, one studies a Hecke algebra T acting on H 1 (X 0 (N), O) m that is isomorphic to M⊕M where M is T-module that is finite f lat over O, and hence of positive depth, and of generic rank one as a T-module. ...
... The patching method introduced in the work of Wiles [49] and Taylor-Wiles [46] on modularity lifting theorems, and instrumental in Wiles' proof of Fermat's Theorem, has undergone intense development. In the original approach of Wiles, patching was combined with level raising results, and a numerical criterion [49,Appendix,Proposition], extended by Lenstra [34], for surjective maps between rings R → T, with T finite flat over O, to be isomorphisms of complete intersections. ...
... The patching method introduced in the work of Wiles [49] and Taylor-Wiles [46] on modularity lifting theorems, and instrumental in Wiles' proof of Fermat's Theorem, has undergone intense development. In the original approach of Wiles, patching was combined with level raising results, and a numerical criterion [49,Appendix,Proposition], extended by Lenstra [34], for surjective maps between rings R → T, with T finite flat over O, to be isomorphisms of complete intersections. In applications in [49], the ring R is a deformation ring for Galois representations and T a Hecke algebra. ...
... In the original approach of Wiles, patching was combined with level raising results, and a numerical criterion [49,Appendix,Proposition], extended by Lenstra [34], for surjective maps between rings R → T, with T finite flat over O, to be isomorphisms of complete intersections. In applications in [49], the ring R is a deformation ring for Galois representations and T a Hecke algebra. Diamond [20,Theorem 2.4] developed the results of Wiles and Lenstra by proving a numerical criterion for freeness of R-modules M finite flat over O, in terms of an augmentation λ : R → O supported on M , that is to say, M p = 0 for p the kernel of λ. ...
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We define a congruence module $\Psi_A(M)$ associated to a surjective $\mathcal{O}$-algebra morphism $\lambda\colon A \to \mathcal{O}$, with $\mathcal{O}$ a discrete valuation ring, $A$ a complete noetherian local $\mathcal{O}$-algebra regular at $\mathfrak{p}$, the kernel of $\lambda$, and $M$ a finitely generated $A$-module. We establish a numerical criterion for $M$ to have a free direct summand over $A$ of positive rank. It is in terms of the lengths of $\Psi_A(M)$ and the torsion part of $\mathfrak{p}/\mathfrak{p}^2$. It generalizes results of Wiles, Lenstra, and Diamond, that deal with the case when the codimension of $\mathfrak{p}$ is zero. Number theoretic applications include integral (non-minimal) $R=\mathbb T$ theorems in situations of positive defect conditional on certain standard conjectures. Here $R$ is a deformation ring parametrizing certain Galois representations and $\mathbb T$ is a Hecke algebra. An example of a positive defect situation is that of proving modularity lifting for 2-dimensional $\ell$-adic Galois representations over an imaginary quadratic field. The proofs combine our commutative algebra results with a generalization due to Calegari and Geraghty of the patching method of Wiles and Taylor--Wiles and level raising arguments that go back to Ribet. The results provide new evidence in favor of the intriguing, and as yet fledgling, torsion analog of the classical Langlands correspondence.
... Wiles [5], and Taylor and Wiles [4], used Galois representations, Frey's elliptic curves and the modular forms associated with them, and Wiles' demonstration of the validity of the Shimura-Taniyama Conjecture, to prove the theorem [ The proof by contradiction I offer below employs algebra that would have been available to Fermat in the 17th century. Using an alternative to the standard binomial expansion, (a + b) n = a n + b n i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I ...
... (5) In order for equation (5) to sum to zero, however, the negative term must be equal in value to the sum of all of the other terms. Thus the last line of equation (5), with its terms regrouped, must sum to less than zero, ...
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Fermat's Last Theorem: A Proof by Contradiction Benson Schae�er* Portland, OR, USA Abstract In this paper I o�er an algebraic proof by contradiction of Fermat's Last Theorem. Using an alternative to the standard binomial expansion, (a+b)n = an + b Pn i=1 an􀀀i(a + b)i􀀀1, a and b nonzero integers, n a positive integer, I show that a simple rewrite of the equation stating the theorem, Ap + (A + b)p = (2A + b 􀀀 c)p; A; b and c positive integers, entails the contradiction of two positive integers that sum to less than zero, (2f + g)(f + g)(f + g + b) Xp􀀀2 i=1 (2f + g)p􀀀2􀀀i(3f + 2g + b)i􀀀1 + (f + b)(f + g)(3f + 2g + b)p􀀀2 + fb(3f + 2g + b)p􀀀2 < 0; f and g positive integers. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat's Last Theorem. AMS 2020 subject classi�cation: Diophantine equations, Fermat's equation
... Number theory has been developed leaps and bounds by the immense contributions by a lot of mathematicians. Finally, after 350 years, the theorem was completely proved by Prof. Andrew Wiles, using highly complicated mathematical tools and advanced number theory [2], [3]. ...
... )( )( ) 3 ...
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Fermat's Last Theorem states that it is impossible to find positive integer A, B and C satisfying the equation A n + B n = C n where n is any integer > 2. Taking the proofs of Fermat for the index n = 4, and Euler for n = 3, it is sufficient to prove the theorem for n = p, any prime > 3 [1]. We hypothesize that all r, s and t are non-zero integers in the equation r p + s p = t p and establish contradiction, and prove that st = 0. Just for supporting the proof in the above equation, we have another equation x 3 + y 3 = z 3 Without loss of generality, we assert that both x and y as non-zero integers; z 3 a non-zero integer; z and z 2 irrational. By trial and error method, we create the transformed equations to the above two equations, solving by which we prove the theorem.
... The most famous example of a Diophantine equation appears in Fermat's Last Theorem. This is the statement, asserted by Fermat in 1637 without proof, that the Diophantine equation a n + b n = c n has no solutions in whole numbers when n is at least 3, other than the trivial solutions which arise when abc = 0. Andrew Wiles famously proved the Fermat's Last Theorem in 1995 in his paper "Modular elliptic curves and Fermat's Last Theorem" [34]. The proof is by contradiction employing techniques from algebraic geometry and number theory to prove a special case of the modularity theorem for elliptic curves, which together with Ribet's level lowering theorem gives the long-waited result. ...
... In each of the cases, we can associate a so-called Frey elliptic curve E a,b,c /Q and let ρ E,p be its mod p Galois representation, where E = E a,b,c . Then ρ E,p is irreducible by Mazur [18] and modular by Wiles and Taylor [34] and [29]. Applying Ribet's level lowering theorem [21] one gets that that ρ E,p arises from a weight 2 newform of level 32 for (i) and level 27 for (ii). ...
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Let $K$ be a totally real number field and consider a Fermat-type equation $Aa^p+Bb^q=Cc^r$ over $K$. We call the triple of exponents $(p,q,r)$ the signature of the equation. We prove various results concerning the solutions to the Fermat equation with signature $(p,p,2)$ and $(p,p,3)$ using a method involving modularity, level lowering and image of inertia comparison. These generalize and extend the recent work of I\c{s}ik, Kara and Ozman. For example, consider $K$ a totally real field of degree $n$ with $2 \nmid h_K^+$ and $2$ inert. Moreover, suppose there is a prime $q\geq 5$ which totally ramifies in $K$ and satisfies $\gcd(n,q-1)=1$, then we know that the equation $a^p+b^p=c^2$ has no primitive, non-trivial solutions $(a,b,c) \in \mathcal{O}_K^3$ with $2 | b$ for $p$ sufficiently large.
... When the residual representationρ f is absolutely irreducible and p-distinguished, it follows from the results of [Wil95] and [Hid88] that C(f ) is generated by a nonzero element η f ∈ I. ...
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Let $E_{/\mathbf{Q}}$ be a CM elliptic curve and $p>3$ a prime of good ordinary reduction for $E$. Assume $L(E,1)=0$ with sign $+1$ (so ${\rm ord}_{s=1}L(E,s)\geq 2$) and $E$ has a rational point of infinite order. We prove that if ${\rm Sel}(\mathbf{Q},V_pE)$ is $2$-dimensional, then a certain generalised Kato class $\kappa_p\in{\rm Sel}(\mathbf{Q},V_pE)$ is nonzero; and conversely, the nonvanishing of $\kappa_p$ implies that ${\rm Sel}(\mathbf{Q},V_pE)$ is $2$-dimensional. For non-CM elliptic curves, a similar result was proved in a joint work with M.-L. Hsieh. The proof in this paper is completely different, and should extend to other settings. Moreover, combined with work of Rubin, we can exhibit explicit bases for $2$-dimensional ${\rm Sel}(\mathbf{Q},V_pE)$.
... The modular degree m E of E is the minimum degree of all modular parametrizations φ : X 0 (N ) → E over Q. The modularity Theorem [24,20,2] implies that it is well-defined. In 2002 Watkins [23] conjectured that for every elliptic curve E over Q we have r ≤ ν 2 (m E ), where ν 2 denotes the 2-adic valuation and r := rank Z (E(Q)). ...
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In 2002 Watkins conjectured that given an elliptic curve defined over $\mathbb{Q}$, its Mordell-Weil rank is at most the $2$-adic valuation of its modular degree. We consider the analogous problem over function fields of positive characteristic, and we prove it in several cases. More precisely, every modular semi-stable elliptic curve over $\mathbb{F}_q(T)$ after extending constant scalars, and every quadratic twist of a modular elliptic curve over $\mathbb{F}_q(T)$ by a polynomial with sufficiently many prime factors satisfy the analogue of Watkins' conjecture. Furthermore, for a well-known family of elliptic curves with unbounded rank due to Ulmer, we prove the analogue of Watkins' conjecture.
... Now let E 0 /Q be an elliptic curve of conductor N, E 0 : y 2 + e 1 xy + e 3 y = x 3 + e 2 x 2 + e 4 x + e 6 , e i ∈ Z, which is modular by [8], [26], [30]. Assume further that the modular parametrization µ : X 0 (N) → E 0 is optimal, i.e., E 0 is a strong Weil curve. ...
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We analyze log-algebraic power series identities for formal groups of elliptic curves over $\mathbb{Q}$ which arise from modular parametrizations. We further investigate applications to special values of elliptic curve $L$-functions.
... This gives an algorithm to list all the solutions of an arbitrary genus 1 equation. In particular, by the genus-degree formula (35), this result covers all 2-variable cubic equations. While Baker's bounds are enormous and the corresponding algorithm is impractical, a practical method for finding all integer solutions to genus 1 equations was later developed by Stroeker and Tzanakis [31]. ...
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This paper reports on the current status of the project in which we order all polynomial Diophantine equations by an appropriate version of "size", and then solve the equations in that order. We list the "smallest" equations that are currently open, both unrestricted and in various families, like the smallest open symmetric, 2-variable or 3-monomial equations. All the equations we discuss are amazingly simple to write down but some of them seem to be very difficult to solve.
... The most important progress in the field of the Diophantine equations has been with Wiles' proof of Fermat's Last Theorem [18,20]. His proof is based on deep results about Galois representations associated to elliptic curves and modular forms. ...
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Let k be a positive integer. In this paper, using the modular approach, we prove that if k≡0(mod4), 30<k<724 and 2k−1 is an odd prime power, then under the GRH, the equation x2+(2k−1)y=kz has only one positive integer solution (x,y,z)=(k−1,1,2). The above results solve some difficult cases of Terai’s conecture concerning this equation.
... Hence NAFL requires that FLT, if true, must necessarily have an elementary proof in NPA. Wiles's proof of FLT [9,10] assumes the consistency of classical theories stronger than PA and hence cannot be formalized in NPA. ...
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Non-Aristotelian finitary logic (NAFL) is a finitistic paraconsistent logic that redefines finitism. It is argued that the existence of nonstandard models of arithmetic is an artifact of infinitary classical semantics, which must be rejected by the finitist, for whom the meaning of finite'' is not negotiable. The main postulate of NAFL semantics defines formal truth as time-dependent axiomatic declarations of the human mind, an immediate consequence of which is the following metatheorem. If the axioms of an NAFL theory T are pairwise consistent, then T is consistent. This metatheorem, which is the more restrictive counterpart of the compactness theorem of classical first-order logic, leads to the diametrically opposite conclusion that T supports only constructive existence, and consequently, nonstandard models of T do not exist, which in turn implies that infinite sets cannot exist in consistent NAFL theories. It is shown that arithmetization of syntax, Godel's incompleteness theorems and Turing's argument for the undecidability of the halting problem, which lead classically to nonstandard models, cannot be formalized in NAFL theories. The NAFL theories of arithmetic and real numbers are defined. Several paradoxical phenomena in quantum mechanics, such as, quantum superposition, entanglement, the quantum Zeno effect and wave-particle duality, are shown to be justifiable in NAFL, which provides a logical basis for the incompatibility of quantum mechanics and infinitary (by the NAFL yardstick) relativity theory. Finally, Zeno's dichotomy paradox and its many variants, which pose a problem for classical infinitary reasoning, are shown to be resolvable in NAFL.
... This fact follows from results of Deligne [13] and Deligne-Serre [14]. In particular, landmark work on modularity in [7,48,51] shows that if E/Q is an elliptic curve, then L(s, E) is an analytic L-function. This property was leveraged by Fiorilli [16] to obtain so-called highly biased prime number races in the context of elliptic curves under certain hypotheses. ...
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The Riemann Hypothesis, one of the Millennium Prize Problems, was formulated by Bernhard Riemann in 1859 as part of his attempts to understand how prime numbers are distributed along the number line. In this expository article, we delve into the Deep Riemann Hypothesis for the general linear group $\mathrm{GL}_{n}$, which was posed by the third author and evinces the convergence of partial Euler products of $L$-functions on the critical line. As applications of the Deep Riemann Hypothesis, one can improve on the error term in the Prime Number Theorem obtained from the Riemann Hypothesis and verify a similar phenomenon to Chebyshev's bias for Satake parameters. The aim of introducing the notion of the Deep Riemann Hypothesis is to rid ourselves of a quagmire of the Riemann Hypothesis.
... Which is in contradiction with Fermat's Last Theorem [7]. Therefore, for all p ≥ 3 there is no operation ν on N * , such that, H θp (ν) = +. ...
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The aim of this article is to revolutionize completely the way of understanding the different links between the structures of two given non-empty sets $G$ and $G^{'}$. Indeed, for any operation $\tau$ defined on a non empty set $G$, and for all $\theta$ an injective mapping from $G$ into $G^{'}$, there exists a \textit{unique operation} $\mathcal{H}_{\theta}(\tau)$ on $\theta(G)$ that makes $\theta$ a \textit{homomorphism} from $(G,\tau)$ to $(\theta(G),\mathcal{H}_{\theta}(\tau))$. Hence, if $\theta$ is bijective, therefore, $G$ is isomorphic to $G^{'}$. We establish that the symmetric group of $G$ \textit{acts} on the set of all operations on $G$, then, we can define the \textit{geometry of operations} for any nonempty set $G$. If, there is an operation $\tau$ on $G$, such that, for all $\theta$ in $\mathcal{S}_{G}, \,\mathcal{H}_{\theta}(\tau)=\tau$, then, $\mathcal{S}_{G}$ becomes the group of \textit{automorphisms} of $G$.
... Thus, we may assume that x ≥ 3. By Wiles' solution of Fermat's last theorem [23] it follows that x = y. ...
... Let E be an elliptic curve defined over Q. The modularity theorem [26,24,4] ensures the existence of a non-constant morphism φ : X 0 (N ) → E defined over Q. Denote by φ E the morphism, up to sign, which has minimal degree and which sends the cusp i∞ to the neutral point of E. The modular degree m E of E is the degree of φ E . There are many relevant conjectures in Number Theory about this invariant. ...
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Watkins' conjecture asserts that the rank of an elliptic curve is upper bounded by the $2$-adic valuation of its modular degree. We show that this conjecture is satisfied when $E$ is any quadratic twist of an elliptic curve with rational $2$-torsion and prime power conductor. Furthermore, we give a lower bound of the congruence number for elliptic curves of the form $y^2=x^3-dx$, with $d$ a biquadratefree integer.
... The celebrated Fermat's Last Theorem was proven in [22] [19] by Wiles and Taylor-Wiles. Ever since then, it has been natural to attempt to use the same methods to tackle more general forms of the Fermat equation, particularly the still unresolved Beal conjecture [16]. ...
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We prove Fermat's Last Theorem over ${\mathbb Q}(\sqrt{5})$ and ${\mathbb Q}(\sqrt{17})$ for prime exponents $p \ge 5$ in certain congruence classes modulo $48$ by using a combination of the modular method and Brauer-Manin obstructions explicitly given by quadratic reciprocity constraints. The reciprocity constraint used to treat the case of ${\mathbb Q}(\sqrt{5})$ is a generalization to a real quadratic base field of the one used by Chen-Siksek. For the case of ${\mathbb Q}(\sqrt{17})$, this is insufficient, and we generalize a reciprocity constraint of Bennett-Chen-Dahmen-Yazdani using Hilbert symbols from the rational field to certain real quadratic fields.
... This formula follows Poitou-Tate duality and the global Euler-Poincaré characteristic formula of Tate. This particular statement is a special case of [23, 8.7.9], which is itself a generalization of a formula of Wiles [41]. We note that, in the enumeration of places, each pair of complex embeddings of F corresponds to one place in V 0 . ...
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We study the distribution of fixed point Selmer groups in the twist family of a given Galois module over a number field. Building off the work on higher Selmer groups in the author's preceding paper, we find conditions under which we can compute the distribution of the $\ell^\infty$-Selmer groups for a given degree $\ell$ twist family. Along the way, we show that the average rank in the quadratic twist family of any given abelian variety over a number field is bounded.
... As everyone knows, the study of the existence of solutions for Fermat type equations has always been an important and interesting problem. The famous Fermat's Last Theorem has attracted the attention of many mathematical scholars [1,2]. About 60 years ago or even earlier, Montel [3] and Gross [4] had considered the equation f m + g m = 1 and obtained that the entire solutions of f 2 + g 2 = 1 are f = cos ζðzÞ, g = sin ζðzÞ for the case m = 2, where ζðzÞ is an entire function, and this equation does not admit any nonconstant entire solution for any positive integer m > 2. ...
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Our main aim is to describe the entire solutions of several systems of α 1 f 1 z 2 + α 2 f 2 z + c 2 = 1 , β 1 f 2 z 2 + β 2 f 1 z + c 2 = 1 , α 1 ∂ f 1 / ∂ z 1 n 1 + α 2 f 2 z + c m 1 = 1 , β 1 ∂ f 2 / ∂ z 1 n 2 + β 2 f 1 z + c m 2 = 1 , and α 1 ∂ f 1 / ∂ z 1 2 + α 2 f 2 z + c 2 = 1 , β 1 ∂ f 2 / ∂ z 1 2 + β 2 f 1 z + c 2 = 1 , α 1 ∂ f 1 / ∂ z 1 2 + α 2 f 2 z + c + α 3 f 1 z 2 = 1 , β 1 ∂ f 2 / ∂ z 1 2 + β 2 f 1 z + c + β 3 f 2 z 2 = 1 , where α j , β j j = 1 , 2 , 3 are nonzero constants in ℂ and m j , n j j = 1 , 2 are positive integers. We obtain several theorems on the existence and the forms of solutions for these systems, which are some improvements and supplements of the previous theorems given by Xu and Cao, Gao, and Liu and Yang. Moreover, we give some examples to explain the existence of solutions for such systems.
... Notably, important results in number theory have recently become so loaded with complicated techniques that mathematicians have begun to question whether the proofs extrapolated Peano's axioms. This is the case of Fermat's last theorem and the weak Goldbach conjecture, proved respectively by Andrew Wiles [21,22] and by Harald Helfgott [23]. This type of question is akin to the program of reverse mathematics and has drawn the attention of mathematicians like Harvey Friedman. ...
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We consider the foundational relation between arithmetic and set theory. Our goal is to criticize the construction of standard arithmetic models as providing grounds for arithmetic truth. Our method is to emphasize the incomplete picture of both theories and to treat models as their syntactical counterparts. Insisting on the incomplete picture will allow us to argue in favor of the revisability of the standard-model interpretation. We start briefly characterizing the expansion of arithmetic ‘truth’ provided by the interpretation in a set theory. Interpreted versions of an arithmetic theory into set theories generally have more theorems than the original. This theorem expansion is not complete however. Using this, the set theoretic multiversalist concludes that there are multiple legitimate standard models of arithmetic. We suggest a different multiversalist conclusion: while there is a single arithmetic structure, its interpretation in each universe may vary or even not be possible. We continue by defining the coordination problem. We consider two independent communities of mathematicians responsible for deciding over new axioms for ZF and PA. How likely are they to be coordinated regarding PA’s interpretation in ZF? We prove that it is possible to have extensions of PA not interpretable in a given set theory ST. We further show that the number of extensions of arithmetic is uncountable, while interpretable extensions in ST are countable. We finally argue that this fact suggests that coordination can only work if it is assumed from the start.
... This result does imply that the elliptic curve given above is modular. Therefore, proving Fermat's Last Theorem [20,21]. Many mathematicians are still heavenly involved on studying Fermat's Last Theorem [22,23,24]. ...
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We construct sequences of triples of circulant matrices with positive integers as entries which are solutions of the equation. We introduce Mouanda's choice function for matrices which allows us to construct galaxies of sequences of triples of circulant matrices with positive integers as entries. We give many examples of galaxies of circulant matrices. The characterization of the matrix solutions of the equation allows us to show that the equation 2) has no circulant matrix with positive integers as entries solutions. This allows us to prove that, in general, the equation 3) has no circulant matrix with positive integers as entries solutions. We prove Fermat's Last Theorem for eigenvalues of circulant matrices. Also, we prove Fermat's Last Theorem for complex polynomials over D associated to circulant matrices.
... We also note that the theta blocks associated to elliptic curves given in Theorem 4.16 have first Taylor coefficients whose square when evaluated at certain CM points are essentially the central L-value of the L-function of the elliptic curve as shown in [24]. When we say a weight 2 newform corresponds to an elliptic curve it is meant via the modularity theorem of Taylor and Wiles [29,31]. ...
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We define Jacobi forms with complex multiplication. Analogous to modular forms with complex multiplication, they are constructed from Hecke characters of the associated imaginary quadratic field. From this construction we obtain a Jacobi form which specializes to $\eta(\tau)^{26}$ which we present to highlight an open question of Dyson and Serre. We give other examples and applications of Jacobi forms with complex multiplication including constructing theta blocks associated to elliptic curves with complex multiplication and new families of congruences and cranks for certain partition functions.
... The celebrated Fermat's last theorem [17] elaborates that it do not exist nonzero rational numbers x, y and an integer n ≥ 3 such that x n + y n = 1. The equation x 2 + y 2 = 1 does admit nontrivial rational solutions. ...
Article
In this paper, we investigate some systems of the Fermat type differential-difference equations with polynomial coefficients and obtain the condition for the existence of finite order transcendental entire solutions and the expression for the entire solutions. We also give some corresponding examples.
... The well-known "Fermat's Last Theorem", which was proved by Wiles [1] and Taylor and Wiles [2], states that there do not exist non-zero rational numbers x and y and an integer m > 2, such that x m + y m = 1. Analogous to this result of number theory, there have been similar function theory investigations. ...
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We describe the entire solutions for Fermat type functional equations with functional coefficients in Cn , i.e., h f p +k g q = 1, where p, q 2 are two integers. We then apply the result to obtain that entire function solutions f , g of f 2 + g 2 = 1 in n are constant if D f −1 (0) ⊆ D g −1 (0) with ignoring multiplicities, where D := n j=1 z j ∂ ∂ z j is the Euler operator. Meromorphic function solutions of f 3 + g 3 = 1 in n and applications to nonlinear (ordinary and partial) differential equations are also discussed.
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We consider the Kisin variety associated to an n-dimensional absolutely irreducible mod p Galois representation ρ ¯ of a p-adic field K together with a cocharacter μ. Kisin conjectured that the Kisin variety is connected in this case. We show that Kisin’s conjecture holds if K is totally ramified with n = 3 or μ is of a very particular form. As an application, we get a connectedness result for the deformation ring associated to ρ ¯ of given Hodge–Tate weights. We also give counterexamples to show Kisin’s conjecture does not hold in general.
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Mathematical proofs are both paradigms of certainty and some of the most explicitly-justified arguments that we have in the cultural record. Their very explicitness, however, leads to a paradox, because the probability of error grows exponentially as the argument expands. When a mathematician encounters a proof, how does she come to believe it? Here we show that, under a cognitively-plausible belief formation mechanism combining deductive and abductive reasoning, belief in mathematical arguments can undergo what we call an epistemic phase transition: a dramatic and rapidly-propagating jump from uncertainty to near-complete confidence at reasonable levels of claim-to-claim error rates. To show this, we analyze an unusual dataset of forty-eight machine-aided proofs from the formalized reasoning system Coq, including major theorems ranging from ancient to 21st Century mathematics, along with five hand-constructed cases including Euclid, Apollonius, Hernstein's Topics in Algebra, and Andrew Wiles's proof of Fermat's Last Theorem. Our results bear both on recent work in the history and philosophy of mathematics on how we understand proofs, and on a question, basic to cognitive science, of how we justify complex beliefs.
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We prove under mild hypotheses the three-variable Iwasawa Main Conjecture for p-ordinary modular forms base changed to an imaginary quadratic field K in which p splits in the indefinite setting (in the definite setting this is a result due to Skinner–Urban). Being in a setting encompassing Heegner points and their variation in p-adic families, our main result has new applications to Greenberg's nonvanishing conjecture for central derivatives of p-adic L-functions of Hida families with root number −1.
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Two well-studied Diophantine equations are those of Pythagorean triples and elliptic curves, for the first we have a parametrization through rational points on the unit circle, and for the second we have a structure theorem for the group of rational solutions. Recently, Yekutieli discussed a connection between these two problems, and described the group structure of Pythagorean triples and the number of triples for a given hypotenuse. In arXiv:2112.03663 we generalized these methods and results to Pell's equation. We find a similar group structure and count on the number of solutions for a given $z$ to $x^2 + Dy^2 = z^2$ when $D$ is 1 or 2 modulo 4 and the class group of $\mathbb{Q}[\sqrt{-D}]$ is a free $\mathbb{Z}_2$ module, which always happens if the class number is at most 2. In this paper, we discuss the main results of arXiv:2112.03663 using some concrete examples in the case of $D=105$.
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We study ring structure of the big ordinary Hecke algebra T with the modular deformation ρT:Gal(Q¯/Q)→GL2(T) of an induced Artin representation IndFQφ from a real quadratic field F with a fundamental unit ε, varying a prime p≥3 split in F. Under mild assumptions (H0)–(H3) given in the Introduction (on the prime p), we prove that T is an integral domain free of even rank e>0 over Λ for the weight Iwasawa algebra Λ étale outside Spec(Λ/p(⟨ε⟩-1)) for ⟨ε⟩:=(1+T)logp(ε)/logp(1+p)∈Zp[[T]]⊂Λ. If p∤e, T is shown to be a normal noetherian domain of dimension 2 with ramification locus exactly given by (⟨ε⟩-1). Moreover, only under p-distinguishedness (H0), we prove that any modular specialization of weight ≥2 of ρT is indecomposable over the inertia group at p (solving a conjecture of Greenberg without exception).
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We present a new Java package, named binMeta, for the development and the study of meta-heuristic searches for global optimization. The solution space for our optimization problems is based on a discrete representation, but it does not restrict to combinatorial problems, for every representation on computer machines finally reduces to a sequence of bits. We focus on general purpose meta-heuristics, which are not tailored to any specific subclass of problems. Although we are aware that this is not the first attempt to develop one unique tool implementing more than one meta-heuristic search, we are motivated by the following three main research lines on meta-heuristics. First, we plan to collect several implementations of meta-heuristic searches, developed by several programmers under the common interface of the package, where a particular attention is given to the common components of the various meta-heuristics. Second, the discrete representation for the solutions that we employ allows the user to perform a preliminary study on the degrees of freedom that is likely to give a positive impact on the performance of the meta-heuristic searches. Third, the choice of Java as a programming language is motivated by its flexibility and the use of a high-level objective-oriented paradigm. Finally, an important point in the development of binMeta is that a meta-heuristic search implemented in the package can also be seen as an optimization problem, where its parameters play the role of decision variables.
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Functions and Tamagawa Numbers of Motives, The Grothendieck Festschrift
• S Bloch
• K Kato
S. BLOCH and K. KATO, L-Functions and Tamagawa Numbers of Motives, The Grothendieck Festschrift, Vol. 1 (P. Cartier et al. eds.), Birkhduser, 1990. [BLR]
Linear Groups with an Exposition of the Galois Field Theory, Teub-ner, Leipzig, 1901. This content downloaded from 130.127.238.233 on Wed
• L E Dickson
L. E. DICKSON, Linear Groups with an Exposition of the Galois Field Theory, Teub-ner, Leipzig, 1901. This content downloaded from 130.127.238.233 on Wed, 25 Sep 2013 07:52:30 AM All use subject to JSTOR Terms and Conditions MODULAR ELLIPTIC CURVES AND FERMAT'S LAST THEOREM 549 [Dr]
Theory of p-adic Hecke algebras and Galois representations
• Hi
Hi2], Theory of p-adic Hecke algebras and Galois representations, Sugaku Ex-positions 2-3 (1989), 75-102.


text:
# workforce5.py

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### workforce5.py

#!/usr/bin/env python3.7

# Copyright 2022, Gurobi Optimization, LLC

# Assign workers to shifts; each worker may or may not be available on a
# particular day. We use multi-objective optimization to solve the model.
# The highest-priority objective minimizes the sum of the slacks
# (i.e., the total number of uncovered shifts). The secondary objective
# minimizes the difference between the maximum and minimum number of
# shifts worked among all workers.  The second optimization is allowed
# to degrade the first objective by up to the smaller value of 10% and 2 */

import gurobipy as gp
from gurobipy import GRB
import sys

# Sample data
# Sets of days and workers
Shifts = [
"Mon1", "Tue2", "Wed3", "Thu4", "Fri5", "Sat6", "Sun7", "Mon8", "Tue9",
"Wed10", "Thu11", "Fri12", "Sat13", "Sun14"
]

Workers = ["Amy", "Bob", "Cathy", "Dan", "Ed", "Fred", "Gu", "Tobi"]

# Number of workers required for each shift
S = [3, 2, 4, 4, 5, 6, 5, 2, 2, 3, 4, 6, 7, 5]
shiftRequirements = {s: S[i] for i, s in enumerate(Shifts)}

# Worker availability: 0 if the worker is unavailable for a shift
A = [
[0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1],
[1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0],
[0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1],
[1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1],
[0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1],
[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
]

availability = {(w, s): A[j][i]
for i, s in enumerate(Shifts)
for j, w in enumerate(Workers)}

try:
# Create model with a context manager. Upon exit from this block,
# model.dispose is called automatically, and memory consumed by the model
# is released.
#
# The model is created in the default environment, which will be created
# automatically upon model construction.  For safe release of resources
# tied to the default environment, disposeDefaultEnv is called below.
with gp.Model("workforce5") as model:

# Initialize assignment decision variables:
# x[w][s] == 1 if worker w is assigned to shift s.
# This is no longer a pure assignment model, so we must
# use binary variables.
x = model.addVars(availability.keys(), ub=availability,
vtype=GRB.BINARY, name='x')

# Slack variables for each shift constraint so that the shifts can
# be satisfied
slacks = model.addVars(Shifts, name='Slack')

# Variable to represent the total slack

# Variables to count the total shifts worked by each worker
totShifts = model.addVars(Workers, name='TotShifts')

# Constraint: assign exactly shiftRequirements[s] workers
# to each shift s, plus the slack
(x.sum('*', s) + slacks[s] == shiftRequirements[s] for s in Shifts),
name='shiftRequirement')

# Constraint: set totSlack equal to the total slack
model.addConstr(totSlack == slacks.sum(), name='totSlack')

# Constraint: compute the total number of shifts for each worker
model.addConstrs((totShifts[w] == x.sum(w, '*') for w in Workers),
name='totShifts')

# Constraint: set minShift/maxShift variable to less/greater than the
# number of shifts among all workers

# Set global sense for ALL objectives
model.ModelSense = GRB.MINIMIZE

# Set up primary objective
model.setObjectiveN(totSlack, index=0, priority=2, abstol=2.0,
reltol=0.1, name='TotalSlack')

# Set up secondary objective
model.setObjectiveN(maxShift - minShift, index=1, priority=1,
name='Fairness')

# Save problem
model.write('workforce5.lp')

# Optimize
model.optimize()

status = model.Status

if status in (GRB.INF_OR_UNBD, GRB.INFEASIBLE, GRB.UNBOUNDED):
print('Model cannot be solved because it is infeasible or unbounded')
sys.exit(0)

if status != GRB.OPTIMAL:
print('Optimization was stopped with status ' + str(status))
sys.exit(0)

# Print total slack and the number of shifts worked for each worker
print('')
print('Total slack required: ' + str(totSlack.X))
for w in Workers:
print(w + ' worked ' + str(totShifts[w].X) + ' shifts')
print('')

except gp.GurobiError as e:
print('Error code ' + str(e.errno) + ": " + str(e))
except AttributeError as e:
print('Encountered an attribute error: ' + str(e))
finally:
# Safely release memory and/or server side resources consumed by
# the default environment.
gp.disposeDefaultEnv()

format: plain


text:
# High school Math: confusion about the basic probability

I am confused about the following two scenarios:

Out of a bag of 3 apples and 3 oranges, you pick 2 items.

1) What is the probability that you will have 2 apples?

2) What is the probability that you will have 1 apple and 1 orange?

My attempt:

1) \begin{aligned} P(\mbox{2 apples}) &= P(\mbox{1st apple}) \times P(\mbox{2nd apple}) \\ &= \frac{3}{6} \times \frac{2}{5}. \end{aligned}

2) \begin{aligned} P(\mbox{1 apple and 1 orange}) &= P(\mbox{1st apple}) \times P(\mbox{2nd orange}) \\ & + P(\mbox{1st orange}) \times P(\mbox{2nd apple}) \\ &= \frac{3}{6} \times \frac{2}{5} \times 2. \end{aligned}

My confusion is with case number 1: why you don't need to multiply the result by 2? Since your first pick could be apple #1, #2, #3.

• Shouldn't your second example be (3/6)*(3/5)*2? There are still 3 of the second fruit after successfully drawing the first. – TripeHound Nov 16 '15 at 17:18

Consider all of the $6\times 5$ ways to pick two pieces of fruit.   That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} \\ A_2 & \color{green}{A_2,A_1} & \times & \color{green}{A_2,A_3} & \color{blue}{A_2,O_1} & \color{blue}{A_2,O_2} & \color{blue}{A_2,O_3} \\ A_3 & \color{green}{A_3,A_1} & \color{green}{A_3,A_2} & \times & \color{blue}{A_3,O_1} & \color{blue}{A_3,O_2} & \color{blue}{A_3,O_3} \\ \hdashline O_1 & \color{indigo}{O_1,A_1} & \color{indigo}{O_1,A_2} & \color{indigo}{O_1,A_3} & \times & \color{red}{O_1,O_2} & \color{red}{O_1,O_3} \\ O_2 & \color{indigo}{O_2,A_1} & \color{indigo}{O_2,A_2} & \color{indigo}{O_2,A_3} & \color{red}{O_2,O_1} & \times & \color{red}{O_2,O_3} \\ O_3 & \color{indigo}{O_3,A_1} & \color{indigo}{O_3,A_2} & \color{indigo}{O_3,A_3} & \color{red}{O_3,O_1} & \color{red}{O_3,O_2} & \times \\ \hline \end{array}}$$

The ways to pick two apples are in the green quarter (upper left).   There are $3\times 2$ of them; that is $6$ of $30$

The ways to pick an apple and an orange are in the blue quarter, but also in the indigo quarter (upper-right and lower-left).   There are $3\times 3+3\times 3$ of them; that's $18$ of $30$

The final quarter are ways to pick two oranges.   Again, just $3\times 2$ of these; that's $6$ of $30$.

• +1 for your coloured picture, which must have necessitated some time! – Greek - Area 51 Proposal Nov 17 '15 at 0:17

The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking.

Case 1) $\frac{3}{6}*\frac{2}{5}$

Case 2) $\frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5}$

Or, like I said, you could use a $*2$ in the second case, but only as an arithmetic simplification. Not because of any rules of probability.

When you state that the chance of getting an apple first is $3/6$ you are assuming that all the apples are interchangeable. If you want to label the apples you can. You can then say the chance of getting apple 1 followed by apple 2 is $(1/6)(1/5)=1/30$. Now you can find that there are six different permutations of two apples, so the total chance of getting two apples is $6(1/30)=1/5$, the same result as you give.

I think you are way over complicating the second one. The first pick is irrelevant, because no matter what, you will get one thing you need and leave five pieces of fruit, three of which are the second thing you need. Which leaves you with 3/5's.

• Care to explain downvote? – Kevin Nov 17 '15 at 6:18
• I'm not the downvoter. I will say this, though... although your approach looks fine to me, please do not ever accuse someone of "overcomplicating" a probability problem. I have seen far too many cases where attempting to look at things in a simple way leads to an erroneous result. Probability problems must be handled with care, and doing everything the long, complex way is one way to ensure that you're doing it right. – ajb Nov 17 '15 at 6:34
• @ajb so you can never over-complicate your approach to a probability problem? – Kevin Nov 17 '15 at 13:32

This answer is a sort of side-shot way of counting, based on Kemp's answer.

We know that we have $3$ oranges and $3$ apples. We assume that they are randomly chosen without replacement, with equal probability. Much like the binomial theorem, consider the following multiplication:

$$1 = 1\cdot 1 = \Big(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\Big)\Big(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\Big)$$

What does this represent? It is all the possible combinations we could have involving different ways of picking apples and oranges. This is a direct explanation regarding the probabilities. That is, the each of the $1/6$ in the first brackets tells us about the probability of choosing an apple or an orange, whereas the second group tell us the probability of picking an apple or an orange after picking the first.

Noting that there are $6$ terms on the left and $5$ on the right, we will have $30$ total terms before cancellation (as mentioned above). Likewise $3$ of the $6$ fractions, $1/6$ correspond to apples and $2$ of the fractions $1/5$ would correspond to an apple after first picking an orange. Thus, there are $3 \cdot 2$ of the fractions $\frac{1}{30} = \frac{1}{5 \cdot 6}$ which correspond to apples begin picked twice.

Although this amounts to the exact same things as Kemp mentioned, it is interesting to see that the sample space is preserved--this is indeed, always, the case. In addition, it then amounts to a simple counting argument; remembering that multiplication is commutative, we could have pick an apple out of a bag with three oranges and two apples then a bag with three of each. It doesn't make a difference, therefore, we don't need to multiply by a factor of two.


text:
# Let’s build a fast, slick and customizable rich text editor with Slate.js and React — Part II

This is a fun series about building a rich text editor with React and Slate. Here’s part one where we left off. You’ll want to read it first if you haven’t already!

The editor has to look sharp. Who wants to stare at and use a scruffy editor? So why don’t we add some quick CSS to make it look more spectacular.

In a traditional create-react-app, we have the index.css file inside the src folder. index.css is used for applying “global” styles for our app. By global styles I’m keeping in mind DOM elements which apply to all pages/documents. For example body and html.

Applying system fonts, background colors, paddings, margins, colors and the width for our editor.

If you’re curious about what kind of “shenanigans” are going on with the font-family — listen up. We are defaulting to the font(s) the user has installed on their computer. This has many performance boosts since there is no need to download the fonts.

Check out this clever explanation by css-tricks for more in detail.

System Font Stack | CSS-Tricks

We should end up with something smarter. 😅

What are some of the key features a rich text editor has? Hmm — let’s see.

• Images
• Code highlighting
• Embedded Videos
• Text formatting (alignment, bullet list, ordered list, etc)
• Save the text to a database for safe keeping

Let’s start by adding the images.

Rich text editors have a toolbar which eases the efforts of editing. Take a look at Medium’s toolbar.

Medium’s toolbar for formatting

When we click on a selected text, a popup toolbar pops up from the ashes which allows us to format our text.

It makes sense to start implementing the toolbar first and then hook it up with the appropriate functionality.

Let’s grab a small library which gives us access to a huge array of icons.

The icons we want to use are feather icons.

I propose creating a component which holds all the formatting logic and styles in one place. Call it FormatToolbar.js and place it inside components.

The FormatToolbar is a stateless component which renders its content thanks to using props.children.

A stateless component is a component which doesn’t make use of the this.state mechanism, but rather lets its parent figure out the state.

Think of it this way: we want to separate concerns as much as possible. A stateless component handles rendering the view (divs, spans, images) and only the view.

A stateful container takes care of its internal state and passes it as props to the stateless components.

props.children allows you nest components inside a component.

If you’re unsure about props.children — this article is good at explaining the concept behind children props in React. ✌

Next up, we will import FormatToolbar.js component to our TextEditor.js.

Importing our icons to the TextEditor component

And finally, consume our FormatToolbar like this:

Passing our toolbar component to the editor render function

Oh and I also added some styles to make it look neat. Remember, the .format-toolbar is our container component which holds all the icons. We set it to flex so they’re all nicely aligned in a row and start from the left. We also removed the default border from our button since we’re going for the clean/minimalist approach.

Applying styles to our tooltip

Terminology: The toolbar is the full row and the tooltip is a single item/collection.

What gets rendered on the screen

Woah! Starting to look like an actual rich text editor. Well done! Allow yourself a fresh coffee and let’s get right back to tackling our app.

There’s no shortage of remarkable ideas, what’s missing is the will to execute them. — Seth Godin

### One finds limits by pushing them

Clicking on the icons has no affect since we didn’t add a single event listener to our icons.

Time to change that. One of the goals of this tutorial is to teach you how to approach problems and how to solve them. Solving complex problems is very challenging. An ubiquitous approach to problem solving is breaking the problem apart into small solvable pieces.

Think about it — what are we missing? We have the formatting functionality in place, the icons, and the styling.

You guessed it, we need to bundle icons and text together. By clicking on the icon, we want to format our text.

Let’s start by adding a event listener on the button. Which event listener should we use? There are a bazillion options out there like onClick, onMouseDown, touchend and so on.

So, what’s the best choice? Good question and I’m glad you asked. We’re going to use a new event listener which was introduced in React version 16.4. The event listener has a particular name — pointer event.

I wrote this brief introduction to pointer events. In a nutshell, instead of writing an event listener for the mouse, touch surface, and the pen, we use a different approach. That approach uses the 3-in-1 solution, compatible with all devices.

The pointer is like any event, just attach it to your button like so:

Adding pointer events to our buttons

The onPointerDown is fired just like onMouseDown.

Notice we added a callback with the name onMarkClick. The first argument is the event object and the second argument type is the formatting we want to apply.

onMarkClick method

Don’t be startled, the comments make it seem bloated. The only functionality we applied was the following:

1. Cancel the browser’s default behavior

2. Get the this.state.value — remember this is the initial value we passed to Slate. The slate model has built in functions, just like mongoose.

3. Apply the formatting on the selected text with the desired format.

4. Finally call an update on the component to display the newly made changes.

Clicking on the icon changes our text

Marvelous!

Let’s finish up by adding some final touches. We’ll be adding code formatting, lists, and underlining functionality. Try following our examples and finish it up.

onKeyDown the left and renderMark the right

importing icons

And finally, creating our elements inside the render

Woohoo! We made it. Time for the long-awaited dopamine hit.

The editor already looks like a tool we use everyday. Notice the elements being placed and removed in the vDOM — React takes care of this for you.

Congratulations for making it to the end. If you would like me to add more features, let me know on Twitter or in the discussion section.

wesharehoodies/slate-react-rich-text-editor


text:
Category:Deleted Integer Topology

This category contains results about the deleted integer topology.

Let $\mathcal P$ be the set:

$\mathcal P = \left\{{\left({n - 1 \,.\,.\, n}\right): n \in \Z_{> 0}}\right\}$

that is, the set of all open real intervals of the form:

$\left({0 \,.\,.\, 1}\right), \left({1 \,.\,.\, 2}\right), \left({2 \,.\,.\, 3}\right), \ldots$

Let $S$ be the set defined as:

$S = \displaystyle \bigcup \mathcal P = \R_{\ge 0} \setminus \Z$

that is, the positive real numbers minus the integers.

Let $T = \left({S, \tau}\right)$ be the partition topology whose basis is $\mathcal P$.

Then $T$ is called the deleted integer topology.

Pages in category "Deleted Integer Topology"

The following 6 pages are in this category, out of 6 total.


text:
# Do Hilbert Space Isomorphism Map Dense Sets to Dense Sets?

## Main Question or Discussion Point

Suppose that H, K are Hilbert spaces, and A : H -> K is a bounded linear operator and an isomorphism.

If X is a dense set in H, then is A(X) a dense set in K?

Any references to texts would also be helpful.


text:
# Thread: Calculate how much invested

1. ## Calculate how much invested

What of solve it and why that answer?
Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was $1,500. At the end of the year, she had earned$106.40 in interest. How much had Mrs. Ming invested in the account paying 6%?
$117$680
$760$820

2. ## Re: Calculate how much invested

Originally Posted by AKM17
Mrs. Ming invested an amount of money in two accounts for one year.
She invested some at 8% interest and the rest at 6% interest.
Her total amount invested was 1,500.
At the end of the year, she had earned 106.40 in interest.
How much had Mrs. Ming invested in the account paying 6%?
117
680
760
820
Hint: x = amount at 6% : .06x + .08(1500 - x) = 106.40


text:
# Items tagged with integrationintegration Tagged Items Feed

### a few integrals in Maple...

June 13 2015
1 6

Trying a few integrals in Maple -- doesn't seem to handle them very well. Any workarounds please?

f1(p,b):= 1/(p^2 + b^2)^2 ;

f2(p,b):= exp(-p/b)^2;

T1pbm := Int(q*f1(q, b)*ln((p+q)^2+m^2), q = 0 .. infinity);

T2pbm := Int(q*f1(q, b)*ln((p-q)^2+m^2), q = 0 .. infinity);

with(IntegrationTools):

PT1 := evalf(Parts(T1pbm, q*f1(q, b), ln((p+q)^2+m^2)));

PT2 := evalf(Parts(T2pbm, q*f1(q, b), ln((p-q)^2+m^2)));

Ev := evalf(-pi*c*Int*(p*f1(p, b)*(PT1-PT2), p = 0 .. infinity))

### Integarion of function...

June 04 2015
1 4

hello all.

trying to integrate sach equation:

Fun := Diff(f(x), x);

int(Fun, x = 2 .. 7);

waiting that maple give my something like this f(7)-f(2). But it write only integration form.

What i need to have f(7)-f(2)?

Thanks.

### How to speed up numerical integration...

June 02 2015
0 2

Hello all.

Need help.

My project are too slow calculate integrals. What can i do to speed up numerical integration?

Thanks all.
MP_1.mw

June 01 2015
0 5

Int(ln(1+sqrt((x+1)/x)),x);

value(%);

Does not work.

Int(ln(2+sqrt((x+1)/x)),x);
value(%);

Works.

But with some effect,

Could this possibly be improved in the next release?

### Integrating an equation with unknown constants....

May 05 2015
1 2

This might seem easy but i am a total beginner in Maple and i really need to know how to do this. If i have an expression like this:

X^2 + 3*y + 3*z + X^7

where (y) and (z) are constants of unknown values. I need to integrate the expression over an interval for example (from 1 to 10) so that the answer would still be in terms of (y) and (z).

### A maple/maths question to be answered by YOU ;-)...

May 01 2015
2 10

$\int_{a}^{b} f(x) \, dx \approx \tfrac{3h}{8}\left[f(x_0) + 3f(x_1) + 3f(x_2) + 2f(x_3) + 3f(x_4) + 3f(x_5) + 2f(x_6) + ... + f(x_n)\right] .$

A Fantastic Good morning Ladies and Gentelmen,

You have the mission to find the problem in that code/ give a better code?
I want to write a procedure for the simpson 3/8 rule using the above formula I took from wikipedia :
(I don't want to use any super maple command that estimate the simpson 3/8, but just using this formula and starting from scratch, I am a beginner so feel free to laugh at my code hehehe)
here is the code I managed to write after 365 hours :D

restart:
simp38:=proc(f,x0,xn,n)
local h,summ,i,x,Integral:
h=evalf((xn-x0)/n);               #h is the distance between 2 points
summ:=f(x0)+f(xn);              #initialise the summ

for i from 1 to n-1 by 3 do
x[i]:=x0+i*h:
summ:=evalf(summ+3*f(x[i])):
od:
for i from 2 to n-1 by 3 do
x[i]:=x0+i*h:
summ:=evalf(summ+3*f(x[i])):
od:
for i from 3 to n-1 by 3 do
x[i]:=x0+i*h:
summ:=evalf(summ+2*f(x[i])):
od:

Integral:=3*h/8*summ;
end:

simp38(x->(x^5-5.15*x^2+8.55*x-4.05045),1,5,6);   #testing it      --->2481.087090
evalf(int(x^5-5.15*x^2+8.55*x-4.05045,x=1..5));      #comparing it --->2477.531533

### An unknown error int command to use...

February 19 2015
2 4

Hi number of officials,

As I enter the command defined 'int' I get an error message.

What can I do. Can you help me.

Commands Entered:

int(x^2+3*x, x);

ansvers:

Error, (in int) wrong number (or type) of arguments: invalid option value passed to indefinite integration: {}

### posterior plot animation...

February 15 2015
0 4

Greetings, seeking an expert to animate a plot.

see worksheet.posterior_graphs_(encapsulted)_1D.mw

before they play each other, each have a law (a normal distribution) plot-output 6.

after DD defeats CC, and a numerical integration is performed the new laws are given by plot-output 18.

as you can see, the laws of DD and CC are closer together.

if the calc was repeated (DD defeats CC again), the laws would be closer again.

so what i require is an animation of the new laws from game 1 to (say) game 6 (DD defeats CC every time). seeing the red and blue distributions merging would be ideal.

as an aside I heard maples FFT could simplify the complicated integration. any suggestions?

cheers

### four dimensional integration...

February 14 2015
1 1

Hi,

I am trying to discretize a kernel of the form $K(x,y,t,s)$. I want to evaluate a four dimensional integral of the form

\int\int\int\int K(x,y,t,s)*h_m(x)*h_n(y)*h_p(t)*h_q(s) dsdtdxdy, where limits of integration are from 0 to 1.

$h_m()$ are function of one variable.

please suggest how to evaluate this.

thanks

### How do i solve integration of odes using composite...

February 13 2015
1 5

hai everyone. i am currently trying to solve an integration of the following ∫g(η)dη . integrate from 0 to 10.

from the following odes.

f ''' +1-(f ')2 +ff ''=0,

g''-gf'+fg'=0,

with boundary conditions f(0)=0, f'(0)=λ, f'(∞)=1, g(0)=1,g(∞)=0

First, i solve the odes using the shooting method. then i used the trapezoidal rule to solve for the integration of g(eta) using the following codes

> with(student);
> trapezoid(g(eta), eta = 0 .. 10, 10);
> evalf(%);

it seems that it can not read the data from the shooting method. can anyone suggest why it is happening?

thank you verymuch for your concern :)

### maple 18 does not integrate...

February 01 2015
1 3

Hello everybody!

My name is Mathew and I am a new member of this forum, and this is my first post, so please be lenient towards me:) I wanted to ask for help since my adventure with Maple 18 was supposed to start yesterday, unfortunately I encountered a problem I couldn't solve myself so far- my Maple 18 does not integrate. At all. It has no problem with differentiating, adding, subtracting and such, unfortunately any type of integral is unsolvable by Maple. I tried typing the easiest functions, such as e2xdx under the integral, unfortunately it is met with following message:

Error, (in int) wrong number (or type) of arguments: invalid option value passed to indefinite integration: {}.

I understood from several tutorials on the internet that this program should not have any problems with dealing with such form of formulas. What could be wrong? How to make  it work as it is supposed to?

Please help me. As I mentioned in the beginning, I am a new member, so in case I did something wrong or placed this post in wrong category- please forgive me;)

### Speeding Up Solving Equation Numerically...

January 26 2015
0 4

I am currently running into an issue where the numerical solution to an equation (involving an integral, yes, but the value I am solving for is simly a constant in the integral) is taking significantly longer than I would hope it would to solve. I am solving a similar equation (with a simpler expression) and it is significantly easier to solve, and I am hoping for that kind of speed.

On the last two lines in the attachment, I have two expressions. The penultimate expression is the baseline speed that I would like to match. The last expression is the fsolve I would like to speed up.

Is there any way to numerically speed up the process? I found that when I did tracelast after halting the process, there were HUGE numbers being added and subtracted, multiplied and divided. Not only did this significantly slow down the proecss but it also adds much numerical instability, which I would also like to avoid.

All help would be appreciated.

pole-dragging-mapleprimes.mw

### Numerica multiple integration...

January 24 2015
1 11

Hi, all

I use INT to calculate multiple integration as below. It runs more than 20 hours without results. I wander is there any problem in my codes.

restart;
A := sin(k*Pi*(x-h*cos(theta))/a)*sin(l*Pi*(y-h*sin(theta))/b)*sin(k[0]*h)*sin(k*Pi*x/a)*sin(l*Pi*y/b);

W := evalf[5](int(int(int(int(A, h = 0 .. (x-a)/cos(theta)), theta = Pi+arctan((b-y)/(a-x)) .. 3*Pi*(1/2)), x = 0 .. a), y = 0 .. b, numeric))

Thanks

### Error, (in IntegrationTools:-Combine) invalid subs...

January 04 2015
1 4

according to help on Combine

The Combine command combines integrals using linearity. The parameter v is any expression involving definite or indefinite integrals.

So, why I get an error from the following?

restart;
expr:=Int(sin(x),x) + Int(cos(x),x);

IntegrationTools:-Combine(expr);

Error, (in IntegrationTools:-Combine) invalid subscript selector

interface(version);
Standard Worksheet Interface, Maple 18.02, Windows 7, October  20 2014

### Integral of natural logarithm finction...

December 16 2014
0 4

Hello everyone,

I'm using Maple18, I tried to integrate the function including natural logarithm:

Is there any simple way to directly convert the answer to the kind of form we want? I cannot not finish the conversion:

So the toolbox "IntegrationTools" was used, but finally, we couldn't compute the integral:

However, we can get the correct answer by manually inputting the formula.

Using IntegrationTools is pretty nasty and not very convenient, such as the problem I mentioned.

Does anyone have another solution?

 1 2 3 4 5 6 7 Last Page 1 of 11


text:
# Program for calculating and ranking scores

I've recently completed a vanilla JS challenge . It is a method for calculating and ranking scores. If possible, I would love to make it better. Any suggestions are welcome.

Challenge Directions: The input array contains a series of objects that hold user scores. Scores are ranked both individually (within the object) as well as against the other score objects to find overall ranking.

MY CODE:

//STEP 1 - CALCULATE TOTAL SCORE & SCORE RANKS FOR EACH OBJECT
function calculateResults(input) {
return input.map((scoreSeries) => {
let final_rank = 0;
calculatedScoreSeries = {...scoreSeries}

Object.keys(scoreSeries).forEach((key) => {
if (key !== 'category') {
final_rank = final_rank + scoreSeries[key]  //Adding up the score.
calculatedScoreSeries[key] = calculateSourceRank(scoreSeries, key) //calculating the rank
}})
calculatedScoreSeries.final_rank = final_rank //adding total_score prop to the new object
return calculatedScoreSeries
})
}
//to calculate the ranking of each source
function calculateSourceRank(scoreSeries, key) {
let rank = 1;
for (let i = 1; i < Object.keys(scoreSeries).length; i++) {
if (scoreSeries[Object.keys(scoreSeries)[i]] > scoreSeries[key]) {
rank++
}
}
return rank
}

//STEP 2 - COMPARE OBJECTS FOR FINAL RANK
function calculateFinalRank(calculatedResults) {
return calculatedResults.map((calculatedScoreSeries) => {
let rankedScoreSeries = {...calculatedScoreSeries}
let rank = 1;
for (let i = 0; i < calculatedResults.length; i++) {
if (calculatedResults[i].final_rank > calculatedScoreSeries.final_rank) {
rank++
}
rankedScoreSeries.final_rank = rank
}
return rankedScoreSeries
})
}

let output = calculateFinalRank(calculateResults(input))
console.log(output)

TEST INPUTS W EXPECTED OUTPUTS:

//each object's score sources are ranked (if two are equal they share the higher ranking).
//then all objects are compared for the overall rnking (if two are equal they share the higher ranking)
input = [ // test input 1
{
"category":"test3",
"source1":50,
"source2":100,
"source3":30,
"source4":10,
"source5":10,
},

{
"category":"test4",
"source1":100,
"source2":30,
"source3":10,
"source4":10,
"source5":50,
},

{
"category":"test5",
"source1":5,
"source2":10,
"source3":10,
"source4":40,
"source5":5,
},
]

output = [
{
"category":"test3",
"source1":2
"source2":1
"source3":3
"source4":4
"source5":4
"final_rank":1
},

{
"category":"test4",
"source1":1
"source2":3
"source3":4
"source4":4
"source5":2
"final_rank":1
},

{
"category":"test5",
"source1":4,
"source2":2,
"source3":2,
"source4":1,
"source5":4,
"final_rank":3
},
]

input = [ // test input 2
{
"category":"cat1",
"src1":20,
"src2":30,
"src3":40,
"src4":50,
"src5":50
},

{
"category":"cat1",
"src1":10,
"src2":0,
"src3":20,
"src4":20,
"src5":100
},
]

output = [
{
"category":"cat1",
"src1":5,
"src2":4,
"src3":3,
"src4":1,
"src5":1,
"final_rank":1
},

{
"category":"cat1",
"src1":4,
"src2":5,
"src3":2,
"src4":2,
"src5":1,
"final_rank":2
},
]

1 - Category will always have at least one 'category'. More than 1 catgeory will be saved as an array under the category key. Category will always be the first key.

2 - each object in the input array will have an equal number of sources (although this number can be any amount)

3 - if a total score is equal to anothers, then they will share the higher ranking (same as individual score ranking)

• If this is a programming challenge from an online site it would be nice if the text of the challenge as well as a link to the challenge was posted as part of the question. – pacmaninbw Jan 27 '20 at 15:11
• @pacmaninbw this is all the info, not an online one I can link to, unfortunately. The idea was to look at the input data and determine what is happening yourself, and then build business logic based on your findings (which I've outlined for my particular case). – gracie catherine Jan 27 '20 at 16:51

## Good things

I like the functional approach taken with this code, and that some features like arrow functions are used.

## Suggestions

### const vs let

It would be wise to default to using const for any variable that doesn't need to be re-assigned. If you later determine a value should be re-assigned then switch it to using let. This helps prevent accidental re-assignment in the future.

### Use consistent indentation

Some lines appear to be indented with two spaces, while others are indented with four. It is wise to use uniform indentation throughout the code.

### Use consistent line terminators

Many lines are terminated with a semi-colon but some are not. Unless you completely understand rules of Automatic semicolon insertion or are using a compiler/module bundler it is best to include line terminators.

### multiple calls to Object.keys() in loop

Let us take a look at the following block:

for (let i = 1; i < Object.keys(scoreSeries).length; i++) {
if (scoreSeries[Object.keys(scoreSeries)[i]] > scoreSeries[key]) {
rank++
}
}

For each iteration of the loop, Object.keys() is called twice. That function could be called once if the result is stored in a variable.

const scoreKeys = Object.keys(scoreSeries)
for (let i = 1; i < scoreKeys.length; i++) {
if (scoreSeries[scoreKeys[i]] > scoreSeries[key]) {
rank++
}
}

The syntax could be simplified using a for...of loop:

const scoreKeys = Object.keys(scoreSeries)
for (const scoreKey of scoreKeys) {
if (scoreSeries[scoreKey] > scoreSeries[key]) {
rank++
}
}

The whole function could be simplified with a more function approach using .filter():

//to calculate the ranking of each source
function calculateSourceRank(scoreSeries, key) {
const scoreKeys = Object.keys(scoreSeries)
return 1 + scoreKeys.filter(scoreKey => scoreSeries[scoreKey] > scoreSeries[key]).length
}

A similar approach could be applied to calculateFinalRank()


text:
org.jamesframework.core.problems.constraints

## Interface PenalizingConstraint<SolutionType extends Solution,DataType>

• Type Parameters:
SolutionType - solution type to which the constraint is applied, required to extend Solution
DataType - underlying data type used for validation
All Superinterfaces:
Constraint<SolutionType,DataType>

public interface PenalizingConstraint<SolutionType extends Solution,DataType>
extends Constraint<SolutionType,DataType>

Interface of a penalizing constraint that assigns a penalty to a solution's evaluation if it does not satisfy the constraint. This penalty is taken into account when evaluating the corresponding solution: in case evaluations are maximized the penalty is subtracted from the solution's evaluation, in case of minimization the penalty is added to the evaluation. When a solution satisfies the constraint, no penalty (zero) should be assigned.

It is required to provide a full validation by implementing validate(Solution, Object). If desired, an efficient delta validation can also be provided by overriding the default behaviour of validate(Move, Solution, Validation, Object) which (1) applies the move, (2) performs a full validation and (3) undoes the move.

Author:
Herman De Beukelaer
• ### Method Summary

All Methods
Modifier and Type Method and Description
default <ActualSolutionType extends SolutionType>PenalizingValidation validate(Move<? super ActualSolutionType> move, ActualSolutionType curSolution, Validation curValidation, DataType data)
Validates a move that will be applied to the current solution of a local search (delta validation).
PenalizingValidation validate(SolutionType solution, DataType data)
Validates a solution given the underlying data.
• ### Method Detail

• #### validate

PenalizingValidation validate(SolutionType solution,
DataType data)
Validates a solution given the underlying data. Returns an object of type PenalizingValidation. The assigned penalty can be retrieved by calling PenalizingValidation.getPenalty() on this object.
Specified by:
validate in interface Constraint<SolutionType extends Solution,DataType>
Parameters:
solution - solution to validate
data - underlying data used for validation
Returns:
penalizing validation that indicates the assigned penalty
• #### validate

default <ActualSolutionType extends SolutionType> PenalizingValidation validate(Move<? super ActualSolutionType> move,
ActualSolutionType curSolution,
Validation curValidation,
DataType data)

Validates a move that will be applied to the current solution of a local search (delta validation). The result corresponds to the validation of the modified solution that would be obtained by applying the given move to the current solution. A default implementation is provided that (1) applies the move, (2) performs a full validation by calling validate(Solution, Object) and (3) undoes the applied move.

It is often possible to provide a custom, much more efficient delta validation based on the current validation and the changes that will be made when applying the move to the current solution. This can be done by overriding this method. It is usually required to cast the received move to a specific type so that this constraint can only be used in combination with neighbourhoods that generate moves of this type (or a subtype). If an incompatible move type is received, an IncompatibleDeltaValidationException may be thrown.

Given that both this method and the full validation (validate(Solution, Object)) return validations of the same type, it is guaranteed that curValidation will also be of this specific type and it is safe to perform a cast, if required.

Specified by:
validate in interface Constraint<SolutionType extends Solution,DataType>
Type Parameters:
ActualSolutionType - the actual solution type of the problem that is being solved; a subtype of the solution types of both the constraint and the applied move
Parameters:
move - move to validate
curSolution - current solution
curValidation - validation of current solution
data - underlying data used for validation
Returns:
penalizing validation of modified solution obtained when applying the move to the current solution
Throws:
IncompatibleDeltaValidationException - if the provided delta validation is not compatible with the received move type


text:
Linearization of a scheduling objective function

I am trying to maximize the workload per employee.

An example:

• $$e$$ the index of an employee
• $$j$$ the index of a project
• decision variable: $$x_{e,j} \in \mathbb{Z}$$ and $$0 \leq x_{e,j} \leq 100$$ deciding with how much percent an employee may work on a project. (Yes it is possible that an employee works on a project with 40%. The remaining 60% should get covered from someone else and all projects should get covered but this is not a part of this problem.)
• decision variable: $$y_e \in \{0,1\}$$ deciding if employee e is working

I had in mind something like the following as a part of the objective function $$\max\sum\limits_{e,j}\frac{x_{e,j}}{y_e}$$

The problem is that this is not linear.

Example: $$\frac{160}{3} < \frac{160}{2}$$ thus 2 employees whould be preferred instead of 3.

Note:

The following constraint has to hold,

$$\sum\limits_e x_{e,j} = 1 \ \ \forall \ \ j$$

implying that all projects have to get covered.

• Is that the same as $\min\sum{y_e}$? – Stradivari Oct 14 '19 at 14:18
• @Stradivari I think you are right. I'll just test your suggestion. – Georgios Oct 14 '19 at 14:38
• Check this objective function. You wrote $\frac{x_{e,j}}{y_e}$ but If $y_e=0$ you will have a division by zero. Thus, there is no possible linearization. I guess your objective function is: (maximize workload per employee) $$\max \frac{\sum_{e}\sum_{j} x_{e,j} }{\sum_{e}y_e}$$ And you should add a constraint to guarantee more than zero employees. $$\sum_{e}y_e \geq 1$$ This problem needs more constraints yet. Probably, you need to describe constraints for work capacity per employee. – Alexandre Frias Oct 14 '19 at 18:12
• @AlexandreFrias This case will never occur since I already mentioned in the description that all projects $j$ should get fully covered. Thus, a constraint is implicitly implemented and not mentioned here avoiding the case of $\sum\limits_e y_{e} = 0$. – Georgios Oct 15 '19 at 10:25
• @Georgios, $y_e = 0$ for certain index $e$. Thus, your objective function has this problem yet. The term of the summation $\frac{x_{e,j}}{y_e}$ can be undefined. – Alexandre Frias Oct 15 '19 at 16:11

You can linearize the objective function, but at the cost of more binary variables and big-M type constraints.

Let's assume that we know a priori that the number of employees used will be between 0 (or maybe 1) and $$N$$. Introduce binary variables $$z_0, \dots, z_N$$ with the constraint $$\sum_{k=0}^N z_k = 1.$$The $$z$$ variables will determine how many employees are being used, via the constraint $$\sum_e y_e = \sum_k kz_k.$$

Next, introduce a continuous variable $$W$$ representing the objective value (to be maximized) and continuous variables $$S_0, \dots, S_k$$, along with the linear constraints $$S_k =\frac{1}{k}\sum_{e,j} x_{e,j}\quad \forall k$$(so that $$S_k$$ is the objective value you want if $$k$$ employees are used) along with the constraints $$S_k - \underline{M}_k (1-z_k) \le W \le S_k + \overline{M}_k (1-z_k)\quad \forall k,$$where $$\underline{M}_k$$ and $$\overline{M}_k$$ are sufficiently large constants (the sizes of which, as the notation suggests, may vary as $$k$$ varies). This ensures that $$W$$, the thing you are maximizing, equals $$S_k$$ if and only if $$z_k=1$$ (i.e., you are using $$k$$ workers).

Another approach to model the objective function could be the following where $$M$$ is a big number that forces the model to choose as few as possible numbers of operators.

$$\max \sum_{e,j} x_{e,j}-M\times \sum_e y_e$$

• When the question defines the profits, $d_e$ and $c_{e,j}$, associated to the variables $y_e$ and $x_{e,j}$ I agree with you. But, I do not understand the meaning of this objective function. In this case, what are you maximizing? – Alexandre Frias Oct 14 '19 at 16:29
• I feel that in his case maybe $\sum_{e,j} x_{e,j}$ is constant. – Stradivari Oct 14 '19 at 19:59
• @Stradivari , maybe $\sum_{e} x_{e,j} =1 \forall j$ but, only Georgios knows that. – Alexandre Frias Oct 15 '19 at 4:40
• I want to say " All project $j$ must be completed." – Alexandre Frias Oct 15 '19 at 4:44
• @AlexandreFrias yes this is exactly what I meant. I will add that to the description. – Georgios Oct 15 '19 at 10:30

Suppose that, the following non-linear objective arises (MAX/MIN): $$$$\max\frac{\sum\limits_{j} a_{j} x_{j}}{\sum\limits_{j} b_{j} x_{j}}$$$$

1) Replace the expression $$\dfrac{1}{\sum\limits_{j} b_{j} x_{j}}$$ by a variable $$t$$.

2) Represent the products $$x_{j} t$$ by variables $$w_{j}$$. The objective now becomes:

$$$$\max \sum_{j} a_{j} w_{j}.$$$$ Introduce a constraint: $$$$\sum_{j} b_{j} w_{j}=1$$$$ in order to satisfy condition 1.

Convert the original constraints of the form: \begin{align} \sum_{j} d_{j} x_{j} &\leqq e \\ \sum_{j} d_{j} w_{j}-e t &\leqq 0 \end{align} It must be pointed out that this transformation is only valid if the denominator $$\sum\limits_{j} b_{j} x_{j}$$ is always of the same sign and non-zero. If necessary (and it is valid), an extra constraint must be introduced to ensure this. If $$\sum\limits_{j} b_{j} x_{j}$$ always be negative the directions of the inequalities in the constraints above must, of course, be reversed.