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I am aware of the facts that every totally bounded metric space is separable and a metric space is compact iff it is totally bounded and complete but I wanted to know, is every totally bounded metric space is locally compact or not. If not, then give an example of a metric space that is totally bounded but not locally compact.
Follow this question on the given link
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Metric space A is said to be a totally bounded if every Cauchy sequence in A has convergent sub sequence
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I am trying to solve the differential equation. I was able to solve it when the function P is constant and independent of r and z. But I am not able to solve it further when P is a function of r and z or function of r only (IMAGE 1).
Any general solution for IMAGE 2?
Kindly help me with this. Thanks
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check out this paper using a Laplace transformation for solving nonlinear nonhomogenous partial equations
the solutions are not trivial which is different if your coefficients and the pressure P are constant
Hopefully it helps
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Is the reciprocal of the inverse tangent $\frac{1}{\arctan x}$ a (logarithmically) completely monotonic function on the right-half line?
If $\frac{1}{\arctan x}$ is a (logarithmically) completely monotonic function on $(0,\infty)$, can one give an explicit expression of the measure $\mu(t)$ in the integral representation in the Bernstein--Widder theorem for $f(x)=\frac{1}{\arctan x}$?
These questions have been stated in details at the website https://math.stackexchange.com/questions/4247090
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It seems that a correct proof for this question has been announced at arxiv.org/abs/2112.09960v1.
Qi’s conjecture on logarithmically complete monotonicity of the reciprocal of the inverse tangent function
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Hello
Can someone help me to solve this?
Because I really don't know about these problems and still can't solve it until now
But I am still curious about the solutions
Hopefully you can make all the solutions
Sincerely
Wesley
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Hi In what area was the issue raised? Euclidean space, Hilbert space, Banach space?
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Many proposals for solving RH have been suggested, but has it been splved? What do you
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Interesting question for possible future discussions.
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  1. I would like to post this question to clarify my doubts as there were two different answers seems to be correct. Two different experts (1. faculty members in applied mathematics department and (ii) and faculty in pure mathematics department ) have different opinion . Question: Find the limits of integration in the double integral over R , where R is the region in the first quadrant (i) bounded by x=1, y=1 and y^2=4x Problem 2 (ii) bounded by x=1, y=0 and y^2=4x.
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If we consider your problems as MCQ problems (I) and (ii) for undergraduate students, the correct answers are :
(a) for problem(i) and (d) for problem(ii).
PS. Observe that we have several correct choices in each case, but they are not included in the offered choices.
Regards
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More precisely, if the Orlik-Solomon algebras A(A_1) and A(A_2) are isomorphic in such a way that the standard generators in degree 1, associated to the hyperplanes, correspond to each other, does this imply that the corresponding Milnor fibers $F(A_1)$ and $F(A_2)$ have the same Betti numbers ?
When A_1 and A_2 are in C^3 and the corresponding line arrangements in P^2 have only double and triple points, the answer seems to be positive by the results of Papadima and Suciu.
See also Example 6.3 in A. Suciu's survey in Rev. Roumaine Math. Pures Appl. 62 (2017), 191-215.
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Your question is very interesting,
Regards and the best wishes,
Mirjana
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We study about some laws for group theory and ring theory in algebra but where it is used.
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an application of ring theoty is geometry, for example check the geometrical properties of complex numbers ring
or neutrosophic numbers
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Can we apply the theoretical computer science for proofs of theorems in Math?
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By dynamical systems, I mean systems that can be modeled by ODEs.
For linear ODEs, we can investigate the stability by eigenvalues, and for nonlinear systems as well as linear systems we can use the Lyapunov stability theory.
I want to know is there any other method to investigate the stability of dynamical systems?
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An alternative method of demonstrating stability is given by Vasile Mihai POPOV, a great scientist of Romanian origin, who settled in the USA.
The theory of hyperstability (it has been renamed the theory of stability for positive systems) belongs exclusively to him ... (1965).
See Yakubovic-Kalman-Popov theorem, Popov-Belevitch-Hautus criterion, etc.
If the Liapunov (1892) method involves "guessing the optimal construction" of the Liapunov function to obtain a domain close to the maximum stability domain, Popov's stability criterion provides the maximum stability domain for nonlinearity parameters in the system (see Hurwitz , Aizerman hypothesis, etc.).
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A careful reading of THE ABSOLUTE DIFFERENTIAL CALCULUS, by Tullio Levi-Civita published by Blackie & Son Limited 50 Old bailey London 1927 together Plato's cosmology strongly suggest that gravity is actually a real world mathematics or in another words is gravitation a pure experimental mathematics?
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Dear Javad Fardaei .
Sorry for the delay.Good question. I think this is a matter for the future.
Greetings,
Sergey Klykov
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Please share your Opinion.
Mathematics is the queen of Sciences. It deals with the scientific approach of getting useful solutions in multifarious fields. It is the back bone of modern science. Ever since its inception it is going into manifold directions. Now in these days of advanced development, it is interlinked with every important branch of technical and modern science. Pure mathematics and Applied mathematics are two eyes of Mathematics. Both are having and playing an equal and significant role in the field of research.
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Mathematics is a backbone of all branches of knowledge.
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Compute nontrivial zeros of Riemann zeta function is an algebraically complex task. However, if someone able to prove such an iterative formula can be used to get all approximate nontrivial using an iterative formula, then its value is limitless.How ever to prove such an iterative formula is kind of a huge challenge. If somebody can proved such a formula what kind of impact will produce to Riemann hypothesis? . Also accuracy of approximately calculated non trivial accept as close calculation to non trivial zeros ?
Here I have been calculated and attached first 50 of approximate nontrivial using an iterative such formula that I have been proved. Also it is also can be produce millions of none trivial zeros. But I am very much voirie about its appearance of its accuracy !!. Are these calculations Is ok?
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In a paper that can be found on arXiv or at , LeClair gives a reasonably accurate algorithm to estimate the non-trivial zeros up to 10^200=Googol^2. My paper that can be found at Cogent Mathematics, on arxiv or RG
gives an estimate that bounds the n'th zero and checks LeClairs result for the number Googol. Although both these are not iterative, and work only for non-trivial zeros that sit on the critical line, they are predictive and easily calculated. Once a zero is estimated, or bounded, it's accurate value can then be found from formula given.
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What are your opinions and suggestions regarding my research (Development of Mathematical Model for accounting system in the Stocks)
Is our use of pure mathematics in this research a shift in the field of modern accounting
And continuous development of analytical accounting curricula
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best wishes
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Is there a difference between pure and applied mathematics?
In Wikipédia, we can find the following definition :
Number theory (or arithmetic or higher arithmetic in older usage) is a branch of pure mathematics devoted primarily to the study of the integers. But Number Theory is mostly applided, for example, in modern Data Encryption Techniques (cryptography).
So, Is there really a difference between pure and applied mathematics?
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The division of “Mathematics” into two “kinds” is fictitious and misleading. A particular mathematical technique may turn out to be useful in practical situations and, conversely, the application of mathematics to the solution of a practical problem may lead to new mathematical discoveries. In either case, the adjectives “pure” and “applied” are not descriptors of the technique itself, but to its context.
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Dear All,
I am hoping that someone of you have the First Edition of this book (pdf)
Introduction to Real Analysis by Bartle and Sherbert
The other editions are already available online. I need the First Edition only.
It would be a great help to me!
Thank you so much in advance.
Sarah
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Search by the elements of real analysis by Bartle and Sherbert.
(For its first edition )
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I'm teacher and suffering a lot to complete my MS. I need to write an MS level research thesis. I can work in Decision Making (Preference relations related research work), Artificial Intelligence, Semigroups or Γ-semigroups, Computing, Soft Computing, Soft Sets, MATLAB related project etc. Kindly help me. I would be much grateful to you for this. Thanks.
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The answers to the question for this thread are excellent. There is a bit more to add.
Before starting either a M.Sc. or Ph.D. thesis, it is very important to read published theses by others. Here are examples:
Source of M.Sc. and Ph.D. theses:
Another source of theses:
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The function assumes a direct and reverse law. What do we know about the inverse function? Never mind. This is just the shadow of the direct function. Why don't we use the inverse function, as well as direct? ------------- I propose the concept of an unrelated function as extended concept of reverse function. ------------ There is a sum of intervals, on each of which the function is reversible (strictly monotonic) -nondegenerate function. ---------- For any sum of intervals, there is an interval where the function is an irreversible-degenerate function.
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@ Vasiliy Knyshev ,
What is your question ?
Kindly state and explain clearly the Newton's Second Law of the third order.
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Mathematics has been always one of the most active field for researchers but the most attentions has gone to one or few subjects in one time for several years or decades. I'd like to know what are the most active research areas in mathematics today?
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Mathematics is a science that creates, models, describes, explains, applies, and of course its areas of research are always new. Ask about a branch in recent development and of interest to the scientific community is to prepare to find countless answers. Of course, the investigator's self-interest will guide him to appropriate topics.
I have read some answers to this question, published in this medium and I am surprised. They do not do science a favor with them.
What is the reason for writing " Physics will beat mathematics — look at my reform! " . What reform? Somebody knows about that reform?
Physics is a great science, and from its observations mathematics has developed very serious theories (Fourier - Heat, Gauss - sound). They were times of illustration. And conversely, Physics has found that without the development of mathematics there are many observations that could not be modeled.
But, returning to the initial question, I think you should look for an answer, and in that I agree with @ Mirjana Vukovic , if your interest is in pure or applied mathematics.
Greetings from Venezuela
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Is the canonical unit 2 standard probability simplex, the convex hull of the equilateral triangle in the three dimensional Cartesian plane whose vertices are (1,0,0) , (0,1,0) and (0,0,1) in euclidean coordinates, closed under all and only all convex combinations of  probability vector 
that is the set of all non negative triples/vectors of three real numbers that are non negative and sum to 1, ?
Do any  unit probability vectors, set of three non negative three numbers at each pt, if conceveid as a probability vector space,  go missing; for example <p1=0.3, p2=0.2, p3=0.5>may not be an element of the domain if the probability  simplex in barry-centric /probabilty coordinate s a function of p1, p2, p3 .
where y denotes p2, and z denotes p3,  is not  constructed appropriately?
and the pi entries of each vector,  p1, p2 p3 in <p1, p2,p3> p1+p2+p3=1 pi>=0
in the  x,y,z plane where x =m=1/3 for example, denotes the set of probability vectors whose first entry is 1/3 ie < p1=1/3, p2, p3> p2+p3=2/3 p1, p2, p3>=0; p1=1/3 +p2+p3=1?
p1=1/3, the coordinates value of all vectors whose first entry is x=p1=m =1/3 ie
Does using absolute barry-centric coordinates rule out this possibility? That vector going missing?
where <p1=0.3, p2=0.2, p3=0.5> is the vector located at p1, p2 ,p3 in absolute barycentric coordinates.
Given that its convex hull, - it is the smallest such set such that inscribed in the equilateral such that any subset of its not closed under all convex combinations of the vertices (I presume that this means all and only triples of non negative pi that sum to 1 are included, and because any subset may not include the vertices etc). so that the there are no vectors with negative entries
every go missing in the domain  in the , when its traditionally described in three coordinates, as the convex hull of three standard unit vectors  (1,0 0) (0,0,1 and (0,1,0), the equilateral triangle in Cartesian coordinates x,y,z in three dimensional euclidean spaces whose vertices are    Or can this only be guaranteed by representing in this fashion.
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Of course it is, its what it is by definition. I probably should have thought more about this, back when I originally posted. If it isnt, then nothing is. Its the "closed" convex hull of its vertices, or all points in [0,1]^3, that can be expressed by convex combinations of its vertices:, (1,0,0), (0,0,1), (0,1,0).
So that any vector (x,y,z), x+y+z=1 , or rather point, (x,y,z), in [0,1]^3, where x+y+z=1,and x >=0,y,>=0z>=0 can be simply expressed by x* (1,0,0)+(0,1,0)+z*(0,0,1)=(x,y,z), as closed under all convex combinations of the vertices convexity (of this form) simply means/ requires that its contains all points in [0,1]^3, that can be expressed by-negative c1, c2,c3 ; c1*(1,0,0)+c2*(0,1,0) +c3*(0,0,1) where c1+c2+c3 =1, it just is the set of 3 all and only non-negative coordinates, that sum to one . And clearly we can set c1=x, c2=y, and c3=y and these are non-negative and sum to one. So its all, and "only" probably triples (x,y,z),x+y+z=1 .
I should have thought this through.
The main questions are (1) and (2):
(1): does the canonical 2 probability simplex, contain the canonical 2 simplex of the sums at each point x+y, x+z, z+y, clearly at each point (x,y,z) in the canonical 2 probability simplex, x+y,in [0,1] x+z\in [0,1], z+yin [0,1] and their sum (x+y+x+z+z+y=2(x+y+z)=2*1=1, but does it contain at "every" point, every such convex combination of (x+y=l, x+z=g, z+y=h), 1>=h>=0,1>=g>=0, 1>=l>=0 & l+g, g+h, h+l in where h+g+l=2 at some point (x,y,z)?
Id say yes, because if there were a (l,g,h), l+g+h=2, 1>=h>=0,1>=g>=0, 1>=l>=0, but with no accompanying (x,y,z) x+y+z=1, (x,y,z)>=0, l= x+y, g=x+z, h=z+y
, g=x+h-y, y=x+h-g, so l=x+x+h-g, l=2x+h-: x=1/2*(l-h+g), which is uniquely determined where l-h+g<=l+g+h=2, as h, g, l>=0, and clearly x+y+z=1/2(2x+2y+2x)1/2(l+g+h)=1, the only way that point could not be in the simplex is if, for example x<0
which in this case of l+g<h, as h<=1, g<=1, l<l, as per the conditions above, that we have a contradiction as (l+g)+h=2 where, l+g<h gives (l+g)+(l+g)<(l+g)+h=2, to 2(l+g)<2, L+g<1 but h<=1, so 1+h<=1+1=2, so, 1+h<=2,
L+g<1 gives (l+g)+h< 1+h
and (l+g)+h<1+h<=2, L+g+h<2, ie l+g+h<1
1/2*(1-g+h)<=1/2*2=1
And moreover, (2):
(3)Does it contain all such h+g+l=1, where the l, g, h, are the sums of the first 2, the first and third, second and third coordinates, respectively of 3 "distinct" cartesian points in the simplex, (x1,y1,z1), (x2,y2,z2), and (x3, z3, y3), l=x1+y1, g=x2+z2, h= y3+z3, h+g+l=1, where none of the x1,x2,x3, x2,y2,z2, x3,z3, y3 are 0?.
Clearly for any m, in [0,1] there is a point in the simplex <x,y,z>, x=m.
So as x is an element of a point, in the simplex, then at those points, where the x coordinate is m, m=x>=0, and clearly x=m<=1.
As, 1>=m>=0 entails that 0<=1-m<=1, then (1-m) is in [0,1],. Then as, for any real in [0,1], there are pts in the simplex, where the x coordinate attains that real, then there is a pt, whose x coordinate is x1, where, (the x coordinate), assumes the value, x1=1-x=1-m.
<x1,y1,z1>, x1+y1+z1=1 so, y1+z1=1-x1=1-(1-m)=m, so x+z assumes the value m,
As the y and z coordinates can assume any value in [0,1] as well, then we have all non negative coordinate of three points, in [0,1]^3:, p1 =(x1,y1,z1), p2=(x2,y2,z2), and p3=(x3, z3, y3), where x1+y2+y3<=3 where x1,y2,y3, are all in [0,1], and thus, the subset, comprising all, sets of three points, p1, p2, p3 in [0,1]^3 where x1+y2+z3=2 , where x1,y2, z3 are non-negative and in [0,1,] are in the simplex.
As, at any pt, in the simplex, and at all pts , x+y+z=1, so: (x1+z1+y1)+)x2+y2+z2)+(x3+y3+z3)=3, and as x1+y2+z3=2 ,
so( x1+y1)+(y2+z2)+(z3+x3)= 3-(x1+y2+z3)= 1,
ie for any real in [0,1] x=m at(x,y,z) if and only there is a point (x1,y1,z1), x1+z1= m
(ie not on the edges of the simplex, where, aeprt from the vertices, only two of the coordinate of a point are non-zero and not 1 ?
At the edges we set y1=0, x1=x1+y1=0+x1= l,
x2=0, so as x2+y2+z2=1, x2+z2 in [0,1], so we set z2=g, z2+x2=z2+0=z2=h so,
z2+x2=g, and at the third point. Let z3=0, h=y3 so y3+z3=0+y3=y3=h
I suppose so, there was a counter-example there wou as if it did not there would have to be no point in the simplex (x,y,z), where 0<=y+z=l=1-g-h<=1 for some, (h,g, l), h+g+l=1 h>=0, g>=0, l>=0, where of course, this gives that 0=1-1<=1-(g+h)=l>=1-0=0 , It also has to express all such combinations because if did not for some l, g, h.
So as long as the the set of all probability triples, 2 canonical probability simplex , contain on its edges, (the 2 positive entries,), the set of call of all probability doubles, that is, the "canonical 1 probability simplex/ the set of 2 non-negative points in R^2 whose sum is 1 (all convex combinations of 2 non-negatives that sum to 1, then the it will contain some such three such edge points, meeting the above sum constraintm(x1,y1,z1), (x2,y2,z2), and (x3, z3, y3), l=x1+y1, g=x2+z2, h= y3+z3, h+g+l=1.
Where along the edges, the double in the canonical 1 probability simplex, which any triple on the edges of canonical 2 probability simplex, identifies, are those doubles, formed by the, 2 elements, of the 3 elements at that point (x,y,z), which happen to be positive (non-zero). SO, setting the first positive coordinate, in said triple (which will be x, or y) to be the x coordinate of the double, and the second positive entry (which will be z or y) to y, ie (x=1/6, y=0, z=5/6) goes to (x=1/6, y=5/6) which will be unique except at the vertices (where one can use both (1,0) or (0,1), and x+y=1 as the other point z will be zero, as the point was taken from the edge of the set of canonical 2 simplex (x,y, z), where precisely one of (x,y,z), is 0, (except at the vertices, themselves, where 2 of them are of course)
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In QFT, computations are done with plane-wave free solutions of the Dirac equations: if one were to consider full solutions of the Dirac equation in interaction with its own electrodynamic field, even without field quantization, what would one obtain? Does anybody know of full solutions, even at a classical level?
NOTE: the question is of purely mathematical interest, so I am not interested in reading that we commonly do not do that in standard computations, I would like to know what would happen if considering the problem with mathematical rigour.
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Igor, is it possible that you missed the Volkov solutions? These are used routinely to study matter-field interaction in the strong e.m. field limit.
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People usually say that the number greater than any assignable quantity is infinity and probably same in the case of -ve ∞.
We are dealing with infinity ∞ in our mathematical or statistical calculations, sometimes we assume, sometimes we come up with it. But whats the physical significance of infinity.
Or
Anyone with some philosophical comments?
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Mathematicians have a precise definition of infinity that can be used to prove theorems about it. A set S has infinite size if it is possible to create a 1-to-1 correspondence of the elements of S with the elements of a proper subset of S. For example, the positive integers S = {1,2,3,...} is an infinite set because there is a 1-to-1 correspondence between S and the even integers in S: 1 <-> 2, 2 <-> 4, 3 <-> 6, 4 <-> 8, ... Such a 1-to-1 correspondence is impossible for finite sets.
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Today, every educational field or domain contains several branches. You first choose one branch of your field to prepare your Master and Ph.D degrees. What is preferable system for you:
1- study the same branch at both Master and Ph.D degrees
or
2-study different branch at Ph.D degrees from Master degree
And, Why?
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I believe that the continuation of research begun in the Masters preferred.
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All of us have a different view of point when we prepare study on some topics. Some of us try to study only the depth results in the work, whereas other think it must to investigate all results that associated to the topic regardless its difficulty or ease. On the other hand, some scholars focus on both.
What is your opinion?
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This is an important question. In addition to the excellent answer given by @ Md. Sarfaraz Alam , there is a bit more to add.
Quality is the most important feature of good papers. A paper with very high quality will stand the test of time and will be remembered long after it is written. By quality, I mean that precise definitions and detailed examples are given. Appropriate definitions lead to theorems. The need for the highest quality is needed in proofs of theorems. The quality of proofs of theorems is measured in terms of concise but accurate deductions from the definitions,
relations and other theorems that precede each proof.
Quantity is definitely not an appropriate goal in writing good papers. It is better to let the size of a paper be a function of the need for narrative and examples that illustrate abstract ideas in a paper.
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it is pure mathematics.
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<geometers don't like groups>:
In his influential 'Erlanger Programm' Felix Klein characterized geometries as the theorie of associated group invariants.
<algebrists don't like fields>:
Galois solved the most urgent algebra problem of his time (general solvability of polynomial equations of order n) by studying properties of fields.
So my rudimentary ideas concerning the history of mathemathics suggest just the contrary of what Claude cames up with.
Difffuse questions get diffuse answers!
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State dependent additivity and state independent additivity? ;
akin to more to cauchy additivity versus local  kolmorgov additivity/normalization of subjective credence/utility,  in a simplex representation of subjective probability or utility ranked by objective probability  distinction? Ie in the 2 or more unit simplex (at least three atomic outcomes on each unit probability vector, finitely additive space) where every events is ranked globally within vectors and between distinct vectors by < > and especially '='
i [resume that one is mere representability and the other unique-ness
the distinction between the trival
(1)x+y+z x,y,z mutually exclusive and exhaustive F(x)+F(y)+F(z)=1
(2)or F(x u y) = F(x)+F(y) xu y ie F(A V B)=F(A)+F(B)=F(A)+ F(B) A, B disjoint
disjoint on samevector)
(3)F(A)+F(AC)=1 disjoint and mutually excluisve on the same unit vector
and more like this or the properties below something more these
to these (3a, 3b, 3C) which are uniqueness properties
forall x,y events in the simplex
(3.A) F(x+y)=F(x)+F(y) cauchy addivity(same vector or probability state, or not)
This needs no explaining
aritrarily in the simplex of interest (ie whether they are on the same vector or not)
or(B)  x+y =z+m=F(z)+F(m) (any arbitary two or more events with teh same objective sum must have the same credence sum, same vector or not) disjoint or not (almost jensens equality)
or (C)F(1-x-y)+F(x)+F(y)=1 *(any arbitrary three events in the simplex, same vector or not, must to one in credence if they sum to one in objective chance)
(D) F(1-x)+F(x)=1 any arbitary two events whose sum is one,in chance must sum to 1 in credence same probability space,/state/vector or not
global symmetry (distinct from complement additivity) it applies to non disjoint events on disitnct vectors to the equalities in the rank. 'rank equalities, plus complement addivitity' gives rise to this in a two outcome system, a
. It seems to be entailed by global modal cross world rank, so long as there at least three outcome, without use of mixtures, unions or tradeoffs. Iff ones domain is the entire simplex
that is adding up function values of sums of evenst on distinct vectors to the value of some other event on some non commutting (arguably) probability vector
 F(x+y)=F(x)+F(y)
In the context of certain probabilistic and/or utility unique-ness theorems,where one takes one objective probability function  and tries to show that any other probability function, given ones' constraints, must be the same function.
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and F(x+y)=F(x)+F(y)
In the context of certain probabilistic and/or utility unique-ness theorems,where one takes one objective probability function  and tries to show that any other probability function, given ones' constraints, must be the same function.
what is meant by state dependent additivity does that mean that instead of F(x U y)=F(x)+F(y) x,y disjoint (lie on the same vector) that (same finite probability triple) ; or instead of F(x)+F(y)+F(z)=1 iff x,y,z are elements of the very same vector (same triple)
in the simplex one literally instead has that F(x+y)=F(x)+F(y) over the entire domain of the function. ie adding up (arguably non commuting) elements of distinct vectors,   or F(x)+F(y)+F(1-x-y)=1 arbitarily over the simplex; or domain of interest, where the only restriction is that one can only add up elements as many times as they are present in the domain. Whilst with cauchy additivity one if one domain is merely a single vector. <1,3, 1,6, 1,2, unit event=1 > one so long 1/6 is in the domain (supposing that entire domain, probability vector space, just is that vector dom(F)={1,3 1,6, 1,2 1),  where if F(1)=1 one can arbitrarily add up 1=F(1)=F(1/6+1/6+1/6 ) six times= 6F(1/6), so F(1/6)=1/6.
I presume however, that if ones domain is the entire simplex there would not be any relevant difference, between outright Cauchy additivity and state independent additivity; and thus to presume would be outright presumptuous.  Or this a name for cross world global rank which entails its long as there at least three atomic events on each simplex (so long as the simplex is well constructed, and the rank is global and modal), even if finite local standard additivity is presumed). As one can transfer values of equiprobable events onto other vectors where they are disjoint?
if by state independent addivity  this means one can arbitrary add up the function values of F(1/6) for objective probability six times, to F(1)=1 the function value lets say at chance=1 to attain that F(1/6)=6 so long as those events are ranked equal and are present at least six times somewhere or other, even if in distinct state, or vectors (in the same system).
or does this apply to local additivity, where one has a global modally transitive rank over the simplex where n>=2 (ie the number of elements in each triple is at least three) because a cross world rank, with equalities, will entail this in any case, if justified. So if one can derive that cross world additivity must hold given finite additivity and global modal rank including equalities cross world/vector, on pain of either either local addivitity failing (probabilism)  or ones justified global and local  total rank  is violated, justified for whatever reason is violated)  including equalities must hold (for whatever reason) does this count as presumptious.
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Given a1, a2, ..., an positive real numbers, and defined
p= a\prod { i = 1 to n, i <> k} [(ai)- (ak)2]
how to prove that \sum \frac{1}{pi} is positive?
I had an idea of proof, but not sure it would work....
I have the idea written in the attached .png file.
EDIT: See the .png file here.
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Thanks Viera answer I've got that the proof for even  n  is quite sufficient. Thank you Viera for keeping calm while giving interesting answers to interesting questions:)
Best regards, Joachim
PS. Meanwhile I have realized, that replacing   ak2  by  ck  and and assuming increasing ordering (without losses, as Viera has noticed) , we are getting for the sum S of the question the following expression with the use of divided difference of order  n-1:
               (n-1)! (-1)n-1 S = (n-1)! [ c1, c2, . . . , cn :  f(c) ], 
which equals the value   f(n-1)(b)  of the derivative  of  f  of  order  n-1  at some point  b  from [c1 ,  cn ],  where  in this case   f(c) = 1/\sqrt{c}.
  For the calculus of divided differences and their reprepresentation see e.g.
Having this and the sign changes of the derivative of the inverse square root one gets positive value of   S, for any choice of   positive   ck -s .
Further conclusion is the  this holds for every negative power of  c  put in place of  f(c),  and also for every Laplace transform  of a positive measure on  R+  since then the derivatives of order  n  have  sign equal   (-1)n  (cf. the William Feller Bible on probability about the completely monotone functions)  JoD
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In order to get a homogeneous population by inspecting two conditions and filtering  the entire population (all possible members) according these two conditions, then used the all remaining filtered members in the research, Is it still population? or it is a sample ( what is called?).
working on mathematical equation by adding other part to it then find the solution and applying it on the real world. can we generalize its result to other real world?
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Rula -
I am not sure I understand your process, but if your 'sample' is really just a census of a special part of your population, then you can get descriptive statistics on it, but you cannot do inference to the entire population from it. 
You might find the following instructive and entertaining.  I think it is quite good.
Ken Brewer's Waksberg Award article: 
 Brewer, K.R.W. (2014), “Three controversies in the history of survey sampling,” Survey Methodology,
(December 2013/January 2014), Vol 39, No 2, pp. 249-262. Statistics Canada, Catalogue No. 12-001-X.
Cheers - Jim
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It will be of immense help for me if you can suggest me some papers and books related to the same.
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Dear Alka Munjal,
Perhaps you can see 
Summability Theory And Its Applications
 Author(s): Feyzi Basar
From the cover
''The theory of summability has many uses throughout analysis and applied mathematics. Engineers and physicists working with Fourier series or analytic continuation will also find the concepts of summability theory valuable to their research. The concepts of summability have been extended to the sequences of fuzzy numbers and also to the theorems of ergodic theory. This ebook explains various aspects of summability and demonstrates applications in a coherent manner. The content can readily serve as a useful series of lecture notes on the subject. This ebook comprises of 8 chapters starting from classical sequence spaces and covering matrix transformations and fuzzy numbers. An accompanying bibliography with extensive references makes this a valuable source of information for readers interested in summability theory as well as other branches of science.''
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(1)  How  can we   find    the partial sum    of   n1000  instantly ?  
(2)  is  there  is  a  simple method to find  partial sum  of the  sequence  f(n)  ?
(3)  Any general method   to compute partial   sum of sequence   ?
(4)  What is the  value  of   Method , if we have   good  approximation  for all differentable  sequence   ?
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@ Juan Weisz  
That  is more than that   .  It  is   a little secret   have little secret     of  analytic continuity   and discreetness  of  integers   
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In basic numerical analysis, it is shown that Aitken's method improves on the basic iteration method in speed of convergence in the asymptotic sense (see detail below if desired). Now it seems that this should be meaningless in practice, giving no guarantees of 'faster' for any finite number of iterations. I found that this is not only my feeling, but that this concern is echoed in the related Wikipedia article:
Although strictly speaking, a limit does not give information about any finite first part of the sequence, this concept is of practical importance in dealing with a sequence of successive approximations for an iterative method, as then typically fewer iterations are needed to yield a useful approximation if the rate of convergence is higher. This may even make the difference between needing ten or a million iterations insignificant.
My questions then are
1. Barring empirical evidence, is there ANY formal way of turning the asymptotic result into a result in terms of finite iterations? Even at least probabilistically? Even when conditions are added? This would be an example of what I have in mind: Given a function with condition such and such (smooth, etc.), the convergence is indeed faster in nn iterations with probability p(n)p(n).
2. If there are such results, can you point me to some of them? 
3. If there are no such results, should there be no interest in trying to find them? If not why not?
4. Doesn't this state of affairs 'bother' numerical analysts? If not shouldn't it?
5.Do people reading this have their own 'intuitions' about when the speed of convergence holds in practice? What are these intuitions? Why not try to formalise them?
Detail
When the series {x_n} generated using x_i=f(x_i−1) converges under the usual conditions, Aitken's method, generating the series {x′_n} using 
x′n=xn−(xn+1−xn)^2 / (xn+2−2xn+1+xn)
converges faster in the sense that, with s being the solution f(s)=sf(s)=s, we have
(x′n−s) / (xn−s) → 0, n→∞.
 
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Hi, I think in practice there is a real interest in using this kind of acceleration scheme. I have implemented in the past Richardson scheme (an alternative of Aitken) and the convergence speed was improved in practice. I think the asymptotic methematical result cannot be converted in a probabilistic condition. In order to really measure the benefit of using an acceleration scheme you can perform Monte Carlo runs, changing the initial conditions in ordrer to quantify the acceleration of convergence.
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I was wondering if there are any set of n; n>=3continuous or somewhat smooth functions (certain polynomials), all of which have the same domain [0,1]
f^n(min,max):[0,1] \to[min,max]; where max>min, min, max\in [0,1] that have these properties
1.\forall v \in [0,1]\sum_n=1\toN(f^n(v)=1; for all values in the domain [0,1], the sum of the 'function values' =1at all points in this domain, and for all possible min and max values of the ranges of the n functions
2. The n functions have   range continuously between [min,max], max>=min, both in [0,1]; and will all reach their maximum at some point, but only once (if possible).
3. These functions must reach their global maximum value only  at the same point on the domain st,t the other distinct functions attain their global minimum value. These global minima for each fi,  may occur at multiple points in the domain [0,1] .
4. There is, for each point in the domain, under which any distinct function fi\neqfj,( fj j\neqi) reaches a global maxima, at  least one point (the same point) in the domain  s.t that  any fi attains its global  minima. Correspondingly, each function fi has (or at least) n-1 points (in the domain) under which it attains its global minima, if each of the n functions has a singular point in the domain under which it attains its global maxima (the n-1 points corresponding to the distinct points in the domain under which the n-1 distinct functions, fj\neqfi attain their global maxima .
That is, for each, of the (n-1) global maximum values of the range of the other functions, f^i;kneqi, where there are n functions  ; the  value of the domain in [0,1] when one function reaches its maximum, is the same value of the domain wherein other functions at which they all reach their absolute minimum . W
5. The max values of the range of said functions,  f^n= 1-\sum_{i;i\neqn}(min(ranf^i)). Ie the maximum value of the functions, f^i for all i\in{1,n} for n functions, is set by  1- sum (of the min function value/range of the other 'distinct' n-1 functions)
5. It must be such that whenever the minimum values are set so as to sum to one. ie min of any function= max of that same function for all such functions; that all  the functions values become a flat line (ie, ie min=max, for such functions)
6.This should be possible for all possible combinations of min values of all 'n' f's (functiions), So each combination of n, non negative values in [0,1] inclusive that sum to one. So one should be able to set the min value of any given value of a function to any value in [0,1], so long as the conjoint sum, sums to one. Likewise presumable if any such function is set such that min=max, all of them are, and these will mins will add to one.
It must be be such that these functions should have to change, to make this work.So that for all min, max range values for the n functions and all elements in their domain (which is common to all and is [0,1])  these function values \forall(n, min function values)\forall(v in [0,1])\sumf^i(v) sum to one
6. Moreover and most importantly they must also allow for the non trivial case where the function minimums do not sum to one (ie are not all flat lines); but such that the sum of the values of the function of all such functions sum to one for all values in their common domain ,v=[0,1](which just is their domain, as their domains;f^i:[0,1]->[min,max] are the same).
Ie so that the functions will range between a maximum and minimum values (which are not equal) in a continuous and smooth fashion (no gaps, steps or spikes).
In this case it must be such that it possible for the the sum of the minimum values, to sum to some positive value smaller then one, although not greater than one, or smaller then zero; for all or 'some' possible combinations (obviously values may not be possible. although, not in virtue of the sum of the functions at some point not being one, but because the max has been set to be smaller then the min for some function; that is the sum of the mins is a particular combination greater then one; (ie 7 function min=0.16....., so that max of fi=1-1=0<0.166=min fi; that is if they sum to more then one;
7. I was wondering if there for any n>3,4,5,6,7,,,,four sets of such sets of 3,4,5,6,7 functions that will do this.
And then given this; it must also be such that the function can be modified so that a function only ever has a minimum of 0 if the maximum is 0; without having to set th sum of the mins of the function to 1.
And likewise such that the function only ever reaches a max of one if its min is one, unless under certain circumstancess; unless that is, it is possible to do this,for the same reason; without doing this having to set the sum of the mins to one. If indeed one does want some of the functions to range between. If possible, Although, What is most important, is that if any function has a min and max set to one, all of the rest of the functions values sit at zero for all v in [0,1].
It also must be such that if the maximum range value for any given function f1 is larger than that of another function f2 in the set, then f1's minimum value will be larger then that of f2's minimum range value; and conversely.
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I agree peter, I apologize. Nonetheless, However,  i must insist,logic entails that one can leave A>T out of the conjunction; you already contradicted yourself (you already said 'can' ,NOT 'must').
It does not entail that one must leave it. Consequentially it does not entail it. What is necessarily possible, is not 'necessarily, necessarily necessary, and that which is entailed or necessary, are only statements which are Necessarily Necessarily
Secondly, I never said that I did not originally say it max>=min, when I later said A>B that was a correction.
Secondly I repeat, with reference even back to the original statement, max>=Min; that I never said it was NOT Tautologous, despite the fact that I DID SAY THAT IT SOUNDED TAUTOLOGOUS.
I did not say this for example,
" it only " merely" "sounded tautologous', so in what sense did I contradict what you said? ."
  I agree, that it is tautologous, and I meant that back then, then just as do now. With relation to what I said originally. The original claim 'It sounds tautologous' does not contradict it being nor being interpreted as saying 'it is tautological' in the strict sense.  technically speaking does not contradict the possible interpretation that it is, and that perhaps I  meantthat IT IS TAUTOLOGOUS.
In the  the literal sense, at least.
It only contradicts that claim if you apply an interpretation involving common sense and the maxim of pleasant  communication. Nonetheless I apologize i wont be like this in the future. I am just pointing out again and again, that back then, nor now, did I ever say, that I was not tautologous. Common sense, appears to give that implication though so I wont do that in the future
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Are there sets of three, or rather for any N>3  sets of  N sur-jective uniformly continuous functions, for all N>3, where n denotes the number of function in the sets, such that each function in the set has the same domain [0,1] and the same range [0,1],such that these  functions, in a given set, sum to one on all points in the domain [0,1]. ie \forallv\in[0,1] [\sum_{i=1}^{i=1n)fi(v)} =1 in [0,1] .Moro-ever, Are there arbitrarily many such functions for all n>=3.
By non trivial, I mean, -non linear, (and presumably not quadratic) functions which just so happen to be that that their sum give value 1, in [0,1] or perhaps for any domain, and not by way of the algebraic sum cancelling out to a constant 1 as in the case of x, 1-x (ie error correcting) .Presumably due to the nature of the derivatives
That is, so that the functions, would sum to one, regardless of the domain [0,1] or least for if the  domain [0,1] is held fixed, and the functions are weighted;  without it being a mere artefact that the algebraic sum to cancels out to give a constant,1,for any x; that is error-correcting functions. That is three (or n>3) functions all of which have a maximum range value of one, miniimum range value of zero, and which sum to one for all points in the domain [0,1]; and which are surjective and uniformly continuous, ie for every value in the range ri\in [0,1], these functions have some (at least one)  value, ci in the domain [0,1] such that f(ci)=ri such that takes on that .
Where these maximum values and min values for all fi coincide (the same element of the domain for which fi corresponds to 1, is such that other n-1/2 functions correspond to zero etc); where obviously these three/n max points such that fi(c)=1,(one, and only one for each function, fi,i1<=i<=n for n>=3, such functions) are distinct (correspond to distinct elements of the domain), although they correspond to the same element of the domain for for which each of the other 2/n-1 are at their minimum value (ie 0)
So for n=3 there is one one max point for each function, and at least , and presumably only, two/n-1 min points), s.t, when one functions fi(c) hits 1, fi(c)=1,the other 2/n-1 functions fj(c),jneqi take the value 0, fj(c)=0 for all j.And are there at least two such such sets for all sets of N>=3 such functions. So there are  three distinct  domain points in [0,1] corresponding to a n/3-tuple of function values,f1,f2,f3(c)= <1,0,0>,,f1,f2,f3(c1)=<0,1,0>f1,f2,f3(c2)=,<0,0,1>; corresponding the three distinct elements/points in the domain [0,1] , c\neqc1\neqc2, c,c1,c2,\in  [0,1] . Where the first function is at its maximum value here 1, and the other two are their minimum, zero at c, the second is at its maximum at c1 (1), and first and third are a their mimimum (zero here) at c1 etc
And are continuous  (no gaps, and uniformly continuous, ie no spikes).
As one could not make use of these such functions, if one wanted them to be weighted otherwise; if said functions have sums which either (A) as just cancel out to be constant, or two (are such that it their sums are do not cancel out to a constant, but just so happen to line up because the domain is [0,1]
Likwise, the functions a similar form; so that one does not want, one having two maximums whilst the other two for example have one maxima, and two minima. Perhaps Berstein polynomials could be so weighted, but I do not know; the linear forms cancel out but their weighted bezauir forms seem to be a little unstable from what I have read.
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I passed through
  1. Pythagorean theorem in 2D,3D,...nD [or norm of unit vectors in Rn]
  2. unit circle, unit sphere, unit sphere in Rn
  3. transformation to generalized spherical coordinates in Rn
A set of uniformly continuous surjective functions f1,...,fn from [0,1] to [0,1] is enclosed.
In particular, for n=3:  
f1(x) = cos2(2pi.x).cos2(4pi.x),
f2(x) = cos2(2pi.x).sin2(4pi.x),
f3(x) = sin2(2pi.x).
The construction shows that many other convenient sets of functions exist.
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Dear professors:
Good afternoon. I am researching about teaching triangle inequality. Are there papers about theorems´production (or formulation) by secondary level's students? Which theoretical framework) in Mathematics Education) may be suitable to study conditions to construct triangle with three segments?
Best regards of Peru!
Luis
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This is an excellent question with many possible answers.    In addition to the helpful answers already given, there is a bit more to add.
A good paper on the straightforward formulation of a theorem and its proof (useful for those starting out in getting comfortable with theorems and their proofs) is given in
The beauty of this paper on a simple proof of a well-known algebraic curves theorem is its piecewise approach with the use of a number of lemmas.   For related papers, see also:
For an introduction to theorem proving paradigms, see
A new proof of the Pythagorean theorem is given in
An elementary proof of the converse of the mean value theorem is given in
A very good discussion of the discovery (and invention) of theorems is given in
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It is possible to write a set of quaternionic partial differential equations that are similar to Maxwell equations. For example:
The quaternionic nabla ∇ acts like a multiplying operator. The (partial) differential ∇ ψ represents the full first order change of field ψ. 
      ϕ = ∇ ψ = ϕᵣ + 𝟇 = (∇ᵣ + 𝞩 ) (ψᵣ + 𝟁) = ∇ᵣ ψᵣ − ⟨𝞩,𝟁⟩ + ∇ᵣ 𝟁 + 𝞩 ψᵣ ±𝞩 × 𝟁
The terms at the right side show the components that constitute the full first order change.
They represent subfields of field ϕ and often they get special names and symbols.
𝞩 ψᵣ is the gradient of ψᵣ
⟨𝞩,𝟁⟩ is the divergence of 𝟁.
𝞩 × 𝟁 is the curl of 𝟁
The equation is a quaternionic first order partial differential equation.
      ϕᵣ = ∇ᵣ ψᵣ − ⟨𝞩,𝟁⟩ (This is not part of Maxwell equations!)
      𝟇 = ∇ᵣ 𝟁 + 𝞩 ψᵣ ±𝞩 × 𝟁
      𝜠 = −∇ᵣ 𝟁 − 𝞩 ψᵣ
      𝜝 = 𝞩 × 𝟁
From the above formulas follows that the Maxwell equations do not form a complete set.
Physicists use gauge equations to make Maxwell equations more complete.
      χ = ∇* ∇ ψ = (∇ᵣ − 𝞩 )(∇ᵣ + 𝞩 ) (ψᵣ + 𝟁) = (∇ᵣ ∇ᵣ + ⟨𝞩,𝞩⟩) ψ
and
      ζ = (∇ᵣ ∇ᵣ − ⟨𝞩,𝞩⟩) ψ
are quaternionic second order partial differential equations. 
      χ = ∇* ϕ
and
      ϕ = ∇ ψ
split the first second order partial differential equation into two first order partial differential equations.
The other second order partial differential equation cannot be split into two quaternionic first order partial differential equations. This equation offers waves as parts of its set of solution. For that reason it is also called a wave equation.
In odd numbers of participating dimensions both second order partial differential equations offer shape keeping fronts as part of its set of solutions.
After integration over a sufficient period the spherical shape keeping front results in the Green’s function of the field under spherical conditions.
      𝔔 = (∇ᵣ ∇ᵣ − ⟨𝞩,𝞩⟩) is equivalent to d'Alembert's operator.
      ⊡ = ∇* ∇ = ∇ ∇* = (∇ᵣ ∇ᵣ + ⟨𝞩,𝞩⟩ describes the variance of the subject
Maxwell equations must be extended by gauge equations in order to derive the second order partial wave equation.
Maxwell equations use coordinate time, where quaternionic differential equations use proper time. In terms of quaternions the norm of the quaternion plays the role of coordinate time. These time values are not used in their absolute versions. Thus, only time intervals are used.
The quaternionic nabla obeys some other pure mathematical relations:
      ⟨𝞩 × 𝞩, 𝟁⟩=0
      𝞩 × (𝞩 × 𝟁) = 𝞩⟨𝞩,𝟁 ⟩ − ⟨𝞩,𝞩⟩ 𝟁
       (𝞩𝞩) ψ = (𝞩 × 𝞩) ψ − ⟨𝞩,𝞩⟩ ψ = (𝞩 × 𝞩) 𝟁 − ⟨𝞩,𝞩⟩ ψ = 𝞩⟨𝞩,𝟁 ⟩ − 2 ⟨𝞩,𝞩⟩ ψ + ⟨𝞩,𝞩⟩ ψᵣ
The term (𝞩 × 𝞩) ψ indicates the curvature of field ψ.
The term ⟨𝞩,𝞩⟩ ψ indicates the stress of the field ψ.
(𝞩 × 𝞩) ψ + ⟨𝞩,𝞩⟩ ψ = 𝞩⟨𝞩,𝟁 ⟩ − ⟨𝞩,𝞩⟩ ψᵣ
Einstein equations for general relativity use the curvature tensor and the stress tensor. Above is shown that some terms of the partial differential equations relate to terms in Einstein's equations.
The advantage of writing equations with nabla based operators instead of with the help of tensors is that these PDE's are more compact and therefore easier comprehensible. The disadvantage is that the quaternionic PDE's enforce you to work in an Euclidean space-progression structure instead of in a spacetime structure that has a Minkowski signature.
Personally I consider the Euclidean structure as an advantage, but the Minkowski signature is more in concordance with mainstream physics.
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Stefano,
I undertake the conversion enterprise because for me the PDE's are better comprehensible than tensor equations. For similar reasons I use quaternions instead of Clifford algebras.
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I have triangle mesh and calculate normal of triangles then calculate vertex normal and do some calculations on it and want to calculate vertex coordinates from this vertex normal after do calculations.
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Look at this doc it may be helpful for your topic. Good luck.
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In this figure called concentric circles, the circumference on A less than circumference on B. and so on. i.e, The distance between neighbor circles are the same, s.
circA < circB < circC < circD < circE
When XA finishes moving round the circumference A, it moves (transits) to the next circle, B, to help XB complete moving round the circumference. When XA and XB complete the movement round the circumference B, they now make a transition to circumference C and help XC complete the movement round the circumference C. After the completion, they also transit to join XC on circumference C and so on.
The question is this. How can this be presented mathematically (arithmetically)?
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   Your statement of the problema is not clear. One can interpret it in two ways:
   If you have only one person (or object) moving and n circles, the answer is:
D(n) = n(n+1)pi + (n-1)s.
   If you have n persons moving (one on each circle), the answer is:
D(n) = (ns/2){[(n+1)(2n+1)pi/3] + n-1}. 
I got this by doing D(n) = 2pi.s (sum of i squared from 1 to n) + s .(sum of i from 1 to n-1).
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I am interested in prime number generation. Apart from the 2p-1 formula generated so many years ago by the French mathematician are there known formula for determining the next prime?
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Suppose $u(n)$ is the Lie algebra of the unitary group $U(n)$, why the dual vector space of $u(n)$ can be identified with $\sqrt{-1}u(n)$?
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Hi Pan,
$u(n)$ is a real Lie algebra, and in particular a real vector space. Using a non-degenerate symmetric bilinear form on $u(n)$, you can identify $u(n)$ with its dual vector space. The $\sqrt{-1}$ is not that important in a sense, and probably comes from using a non-degenerate pairing between skew-hermitian and hermitian matrices (instead of a non-degenerate symmetric bilinear form). $u(n)$ is the space of skew-hermitian $n$ by $n$ matrices, and $\sqrt{-1} u(n)$ is the space of hermitian $n$ by $n$ matrices by the way.
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Any mathematical expert can see my attachment I have highlighted few mathematical symbols , what that symbol signifies  how to understand that can anyone tell
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Naveen, you probably did not have good teachers who explain basic ideas of hydrodynamics "on fingers". Imagine water over rigid bottom. If it is incompressible, Laplace equation is valid in every internal point. \psi is velocity potential, and \eta is deviation of free surface from equilibrium. Due to non-compressivility, the total volume of water is preserved, and thus an integral of deviation function over unperturbed surface is zero. The condition on bottom shows that the velocity is locally parallel to the bottom (given by differentiable function); its normal component is zero.
To understand notations. Recall that a symbol resembling Euro symbol means: element belongs to a set. R is a set of real numbers, while other sets in your case are some subsets of R1(real line) or R2 (plane).
You can look at equations (1) in my attached article to see what is what and what one can do with that: https://www.researchgate.net/publication/275581998_Evolution_of_long_nonlinear_waves_on_shelves
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As we know, an elliptic curve defined over Fq with a rational 2-torsion subgroup can be expressed in the special form (up to twists). Accordingly a natural question arises about the number of distinct (up to isomorphism) elliptic curves over Fq in the family.
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Dear Davood,
Here are links and attached files in subject.
-On the number of distinct elliptic curves in some families - ResearchGate
-On the number of distinct elliptic curves in some families | SpringerLink
-On the isomorphism classes of Legendre elliptic curves ... - Springer
Best regards
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Let HT denotes the statement of Hindman’s theorem. Within RCA0 one can prove that:
1. HT implies ACA0
2. HT can be proved in ACA0+.
An open question is the strength of Hindman’s theorem.
Is HT equivalent to ACA0+ , or to ACA0, or does it lie strictly between them?
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This is a very good question.
A good place to start in answering this question is
Unfortnately, this paper is not available on RG but it is available a University of Connecticut web page at
See also
where the issue of the strength of Hindman's theorem is raised.
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Let f(x)+g(x) = h(x). Here, h(x) is minimum at the points(a1,a2,...,ak). For which condition , we can say that f(x) is also minimum at the points(a1,a2,...,ak)?
Thanks in advance for your idea and please give any reference
.
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My answer from above gives a sufficient condition (on g) for the desired conclusion on f, and it is independent on any differentiability hypothesis. It follows by simple handling inequalities. Hence this implication holds and one can say something on this subject.
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Let $p(.)$ be an equivalent norm to the usual norm on $\ell_1$ such that
$$\limsup\limits_{n\to\infty} p(x_n+x)=\limsup\limits_{n\to\infty}p(x_n)+p(x)$$ for every $w^*-$null sequence $(x_n)$ and for all $x\in\ell_1,$ moreover, let $$\rho_{k}(x)=p(x)+\lambda\gamma_{k}\sum\limits_{n=k}^{\infty}|x_n|,$$ where, $(\gamma_{k})$ be any non-decreasing sequence in $(0,1)$ and $\lambda >0$. I'd like to prove for every $w^*-$null sequence $(x_n)$ and for all $x\in\ell_1,$
$\limsup\limits_{n\to\infty}\rho_k(x_n+x)=\limsup\limits_{n\to\infty}\rho_k(x_n)+\rho(x)$ .
**My attempt is the following**
\begin{align}
\limsup\limits_{n\to\infty}\rho_k(x_n+x)
=\limsup\limits_{n\to\infty} p(x_n+x) +\limsup\limits_{n\to\infty}\lambda\gamma_{k}\sum\limits_{n=k}^{\infty}|x_n+x| \\
=\limsup\limits_{n\to\infty}p(x_n)+p(x) +\limsup\limits_{n\to\infty}\lambda\gamma_{k}\sum\limits_{n=k}^{\infty}|x_n+x|\\
\end{align}
Now I could not proceed to prove, any ideas or hints would be greatly appreciated.
Thanks in advance
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First of all, I suppose that you consider $\ell_1$ as dual to $c_0$, so by $w^*-$null sequence in $\ell_1$ you mean such a sequence $x_n$ that is norm-bounded and goes to zero coordinate-wise. In your question you are using letter $x_n$ in two different senses: as elements of $\ell_1$ and as coordinates of an element $x$. This leads to a confusion. If you denote coordinates of $x$ as $x_n$ and consider vectors $y_n = (y_{n,1}, y_{n,2}, \ldots)$, then the formula you are going to demonstrate is $\limsup\limits_{n\to\infty}\rho_k(y_n+x)=\limsup\limits_{n\to\infty}\rho_k(y_n)+\rho(x)$
 The hint to this exercise is: if a sequence of vectors $y_n \in \ell_1$ is norm-bounded, goes to zero coordinate-wise and if there exists the limit $\lim_{n \to \infty} \|y_n\|$, then there exists $\lim_{n \to \infty} \|x +y_n\|$, and
 $$\lim_{n \to \infty} \|x +y_n\| = \lim_{n \to \infty} \sum_{j=1}^\infty |x_j + y_{n,j}| = \|x\| + \lim_{n \to \infty} \|y_n\|.$$
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Yes. Pure mathematics is useful for theoretical physics. Mathematics is nothing but  logical expression of physics and physical things. It is much more clear than physical logic if the methodology is right. A physical concept has to be converted to mathematical equation and the mathematics will drag it to a final equation which can explain a new concept which is not visible to the conceptual logic. Conceptual logic must provide a physical phenomena which can be observed and verified physically. Otherwise that mathematics will create complicated and chaotic outputs in theoretical physics..
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Every natural number n can be written as:
n= a_0 + a_1 *(10)^1 + a_2 * (10)^2 +... a_i between 0 and 9. how can we generate a new method to find the divisors of n apart from the well known method of prime factorization. If so, we can provide a new method to calculate the sum of divisors function.
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Dear Hanifa;
Could you please write down an example
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In some cases, learners find it easy to deal with decimal fractions than proper and improper fractions. Looking at the complex formation of fractions when adding or subtracting seems harder and almost impossible.
e.g.
0.5 + 3.3 = 3.8
1/2 + 33/10 =  38/10
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Since children had an intensive experience with numbers, hence from my experience working with children in the primary level, adding decimal numbers is not a difficult concept for them to comprehend.  The challenge is just to help children extend their understand regarding place value of, one tenth, one hundreth etc... However, fraction is a relatively new concept for children to grasp.  Hence, I feel that operations with fractions should be taught later, once their are comfortable with the concept of fractions itself. 
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A mathematical colleague and I are working on an article which uses the pure mathematical analysis for equilibrium in equity crowdfunding. We inspired from the model of consumer-product (brand) preference to build up a mathematical model for investor-project preference on a crowdfunding platform. A common point in consumer-product and investor-project relations is that an agent has to choose among different options with optimal efficiency.
We presented a first draft recently at a conference.Non mathematician researchers had difficulty to follow and understand our paper.How can we make such a mathematical reasoning more understandable for non-mathematicians? Do you any article as model? What is your adive? 
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By giving evidence or reference,
Who first discovered the base of the natural logarithm: e?
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Does anyone know something about first sciences, especially mathematics.               1. My question is: when and which science have emerged as first ?                     2. The place and role of mathematics among the oldest sciences.
I will start with a few remarks about mathematics:
1. Already its paradigmatic place in the domain of human knowledge, independent of all other valid reasons, mathematics deserves a special place.
2. The oldest known thinkers of antique civilization have been characteristic way of mathematical form of knowledgeand since then it deserves as a model of scientific value and measures of  exactness of the overall knowledge.
3. Already in the middle ages, mathematics in his former division accounted for two of the seven skills which was dedicated to the study of the traditional University (geometry and arithmetic) in quadriviumu. And the third one-seventh, logic, the trivium of today would be the relevant part, in the form of mathematical logic, also regarded as one of the domains of mathematics.
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A well-known result of B.M. Levitan and T.V. Avadhani asserts that the Riesz-summability of order k of the eigenfunction expansion f (P) of f (P) from L2 (D) at the point P =Po from D depend only on the behaviour of f (P) in the neighbourhood of Po if k > (n-1)/2, i. e. is a local property of f(P) at the considered point Po if k > (n-1)/2. Is it possible to prove (applying Parseval’s formula) the analogue of Avadhani's theorem for Avakumovic’s G - method of summability. A crucial step in the proof of this theorem is to find a function g that lead us to the core of Avakumovic's summability which is more complex than the core of Riesz’s summability.
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Dear Cenap Özel,
If you have a problem with russian presentaton of my work  "About one application Parseval's formula to the Avakumović's G - method of summability of eigenfunction expansion" I want inform you that it soon publish in Sarajevo Journal of Mathematics in number dedicated to the memory of my professor Academician Mahmud Bajraktarevic,
Sincerely, Mirjana Vukovic
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Since it is difficult to write mathematical formulae please consider the attached file.
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${F_{{5^k}n}}(q) \equiv 0\bmod {[{5^k}]_q}$ is equivalent with ${F_{{5^k}}}(q) \equiv 0\bmod {[{5^k}]_q}$ 
because ${q^{{5^k} + n}} \equiv {q^{{5^k}}}\bmod {[{5^k}]_q}$ and therefore  ${F_{{5^k}(n + 1)}}(q) \equiv {F_{{5^k}n}}(q){F_{{5^k}n + 1}}(q) \equiv 0\bmod {[{5^k}]_q}.$
Therefore it suffices to show that
${F_{{5^k}}}(q) \equiv 0\bmod {[{5^k}]_q}.$
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We define a factoriangular number (Ftn) as the sum of a factorial and its corresponding triangular number, that is, Ftn = n! + n(n+1) / 2. If both n and m are natural numbers greater than or equal to 4, is there an Ftn that is a divisor of Ftm? Please also see the article provided in the link below, specifically Conjecture 2 on pp. 8-9.
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Let m=Ftn if Ftn is odd, and m=2Ftn if Ftn is even. Then Ftn divides Ftm.
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My function is nonlinear with respect to a scalar \alpha .
However, the calculation of objective function is very time consuming, making optimization also very time consuming. Also, I have to do it for 1/2 millon voxels (3d equivalent of pixels). I plan to do it using “lsqnonlin” of matlab. 
Rather than optimizing over all possible real values, I plan to search over preselected 60 values. My variable \alpha  (or flip angle error) could be anything between 0-35%; but, I want to pass only linearly spaced points as candidates (i.e. 0:005:0.35). In other words, I want lsqnonlin to choose possible solution only from (0:005:0.35). Since I can pre-calculate objective values for these, it would be very fast. In other words, I need to restrict search space.
Here, I am talking about single voxel; though I performs lsqnonlin over multivoxel and corresponding \alpha is mapped accordingly to a column vector. 
I can not do grid search over preselected value as I plan to perform spatial smoothing in 3D. Some guidance would be highly appreciated.
Regards, Dushyant
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My two cents to the question:
1) What you want to do is the discrete optimization (see the link). The discrete coded (binary) genetic algorithm, e.g., could do the trick.
2) You could perform the factorial design of experiments (factorial d.o.e.) with a number of levels for each parameter you want, but the fraction should be very small to calculate it (see below, why).
1/2 millon voxels as optimization variables is somewhat difficult to optimize, even assuming varying them on 60 levels. It is probably possible to run on supercomputer, but first consider if the problem is formulated correctly and the results would be significant, if it is worth trying.
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For example,
{2},
{3,5,7},
{11,13},
{17,19}, etc.
`A second interested question would be that if any such patterns terminate at any level then "Does the cardinality of such sets follow any pattern?" 
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Dear U. Dreher,
about the conjecture on the 'twin numbers' let me inform you that I proved this conjecture as a particular case of the 'de Polignac’s conjecture'.
You can find solutions of all the Landau problems in the following my papers:
These solutions could answer also to the Shahid's question.
My best regards
Agostino,
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It is believed that there are the bijection relationships between Infinite Natural Number Set and Infinite Rational Number Set, but following simple story tells us that Infinite Rational Number Set has far more elements than that of Infinite Natural Number Set:
The elements of a tiny portion of rational numbers from Infinite Rational Number Set (the sub set : 0, 1, 1/2, 1/3, 1/4, 1/5, 1/6, …, 1/n …) map and use up (bijective) all the numbers in Infinite Natural Number Set (0,1, 2, 3, 4, 5, 6, …, n …); so,infinite rational numbers (at least 2,3,4,5,6,…n,…) from Infinite Rational Number Set are left in the “one—to—one element mapping between Infinite Rational Number Set and Infinite Natural Number Set (not the integer set )------- Infinite Rational Number Set has infinite more elements than Infinite Natural Number Set.
This is the truth of a one-to-one corresponding operation and its result between two infinite sets: Infinite Rational Number Set and Infinite Natural Number Set. This is the business just between the elements’ quantity of two infinite sets and it can be nothing to do with the term of “proper subset, CARDINAL NUMBER, DENUMERABLE or INDENUMERABLE”.
Can we have many different bijection operations (proofs) with different one-to-one corresponding results between two infinite sets? If we can, what operation and conclusion should people choose in front of two opposite results, why?”
Such a question needs to be thought deeply: there are indeed all kinds of different infinite sets in mathematics, but what on earth make infinite sets different?
There is only one answer: unique elements contained in different infinite sets -------the characteristics of their special properties, special conditions of existence, special forms, special relationships as well as very special quantitative meaning! However, studies have shown that, due to the lack of the whole “carriers’ theory” in the foundation of present classical infinite theory, it is impossible for mathematicians to study and cognize those unique characteristics of elements operationally and theoretically in present classical set theory. So, it is impossible to carry out effectively the quantitative cognitions to the elements in various different infinite set scientifically -------a newly constructed Quantum Mathematics.
The article《On the Quantitative Cognitions to “Infinite Things” (IX) ------- "The Infinite Carrier Gene”, "The Infinite Carrier Measure" And "Quantum Mathematics”》 has been up loaded onto RG introducing the working ideas. https://www.researchgate.net/publication/344722827_On_the_Quantitative_Cognitions_to_Infinite_Things_IX_---------_The_Infinite_Carrier_Gene_The_Infinite_Carrier_Measure_And_Quantum_Mathematics
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Dear Geng
Lets use your own reasoning in a different way:
Just take a tiny portion of the Natural Number Set (2,4,6, ... ,2n, ...) and they map very well on the set of all Natural Numbers  (1,2,3,...,n,...). So a lot of natural numbers are left behind in this one -to-one mapping (2n onto n) from the Natural Numbers onto the Natural Numbers, Hence your conclusion should be: the Natural Number Set has far more elements than the Natural Number set. Think about this.
Best regards, Joseph.
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Dear RG friends:
In two weeks time, I am all set to conduct a technical session on Analysis.
I plan to deliver a long lecture on "Fixed Point Theorems". Of course, Banach fixed point theorem is useful to establish the local existence and uniqueness of solutions of ODEs, and contraction mapping ideas are also useful to develop some simple numerical methods for solving nonlinear equations. Are there any other interesting science / engineering applications?
Kindly let me know! Thank you for the kind help.
With best wishes,
Sundar
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As applications :   logic programming and fuzzy logic programming
Nash equilibrium and game theory.
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Let q be an odd positive integer, and let Nq denote the number of integers a such that 0 < a < q/4 and gcd(a, q) = 1. How do I see that Nq is odd if and only if q is of the form pk with k a positive integer and p a prime congruent to 5 or 7 modulo 8?
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I see that you, too, follow the Putnam Exam.
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As the title suggests, how do i see that for any n, the covering map S^{2n} → RP^{2n} induces 0 in integral homology and cohomology, except in dimension 0?
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Maybe you would like to see geometrically how the n-th homology of  S^n can be killed when projected to RP^n? Take a triangulation of the n-sphere which is invariant under the antipodal map  -I , e.g. the generalized octahedron (which exists in every dimension). The projection is a triangulation of the projective space RP^n, and each of its simplices is hit twice by the projection map S^n \to RP^n. However, when n is even, the antipodal map -I on R^{n+1} reverses orientation (determinant -1) but preserves the exterior normal vector field of S^n, thus it reverses the orientation on S^n. Hence each simplex contributes with both signs and you get extinction, while in odd dimensions you obtain a factor 2 (the mapping degree of the projection map).
Best regards
Jost
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Say a definition to be self-referential provided that contains either an occurrence of the defined object or a set containing it. For instance,
Example 1) n := (n∈ℕ)⋀(n = n⁴)⋀(n > 0)
This is a definition for the positive integer 1, and it is self-referential because contains occurrences of the defined object denoted by n.
Example 2) Def := "The member of ℕ which is the smaller odd prime."
Def is a self-referential definition, because contains an occurrence of the set ℕ containing the defined object.
Now, let us consider the following definition.
Def := "The set K of all non-self-referential definitions."
If Def is not a self-referential definition, then belongs to K, hence it is self-referential. By contrast, if Def is self-referential does not belong to K, therefore it is non-self-referential. Can you solve this paradox?
Take into account that non-self-referential definitions are widely used in math.
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Juan-Esteban,
That's a nice way to avoid the so-called Russell paradox at any finite stage of the process. It feels as if you avoid the paradox by rejecting infinity. If you read my post carefully, you will see that the problem is solved in a more fundamental way with or without  considerations of infinite sets. Formal logic tells you that there cannot be x such that
(all y) ( not P(y,y) <--> P(y,x) ),
whatever you mean by P(y,x) and whatever your universe of discourse may be. If you take the "barber definition" (anyone  shaving all those people not shaving themselves), even in an imaginary infinite society of humans, it turns out that there is no such barber. In fact, it is not a definition because it deals with nothing. Let me illustrate this with a more obvious contradiction:
Remarkable_Weather := weather with rain and yet without rain.
Such a "definition" is verbosity with (literally) no subject, hence with no meaning.
You said: A paradox either leads to a contradiction or it is not a paradox. With your quoted statement at hand, my "claim" that there is Remarkable_Weather deserves being called a paradox (it leads to a contradiction, even stronger: it is a contradiction). It's too cheap, don't you think?
I prefer Webster's definition of "paradox":
A tenet or proposition contrary to received opinion; an
assertion or sentiment seemingly contradictory, or opposed to
common sense; that which in appearance or terms is absurd,
but yet may be true in fact.
The only difference between having a Remarkable_Weather and the Russell paradox is, that the latter involves a slightly more hidden logical contradiction.
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Two numbers a and b are elements of the set of real numbers exclusive of the set of rational numbers; they are irrational.  Are there cases where  a times b, or where a divided by b, yield a member of the set of integers? How rare or commonplace is the condition of irrational number products yielding rational values?
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If a is irrational and n is an integer, then b:=n/a is irrational (!) and ab =n.
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I was working on 2 papers on statistics when I recalled a study I’d read some time ago: “On ‘Rethinking Rigor in Calculus...,’ or Why We Don't Do Calculus on the Rational Numbers’”. The answer is obviously trivial, and the paper was really in response to another suggesting that we eliminate certain theorems and their proofs from elementary collegiate calculus courses. But I started to wonder (initially just as a thought exercise) whether one could “do calculus” on the rationals and if so could the benefits outweigh the restrictions? Measure theory already allows us to construct countably infinite sample spaces. However, many researchers who regularly use statistics haven’t even taken undergraduate probability courses, let alone courses on or that include rigorous probability. Also, even students like engineers who take several calculus courses frequently don’t really understand the real number line because they’ve never taken a course in real analysis.
The rationals are the only set we learn about early on that have so many of the properties the reals do, and in particular that of infinite density. So, for example, textbook examples of why integration isn’t appropriate for pdfs of countably infinite sets typically use examples like the binomial or Bernoulli distributions, but such examples are clearly discrete. Other objections to defining the rationals to be continuous include:
1) The irrational numbers were discovered over 2,000 years ago and the attempts to make calculus rigorous since have (almost) always taken as desirable the inclusion of numbers like pi or sqrt(2). Yet we know from measure theory that the line between distinct and continuous can be fuzzy and that we can construct abstract probability spaces that handle both countable and uncountable sets.
2) We already have a perfectly good way to deal with countably infinite sets using measure theory (not to mention both discrete calculus and discretized calculus). But the majority of those who regularly use statistics and therefore probability aren’t familiar with measure theory.
The third and most important reason is actually the question I’m asking: nobody has bothered to rigorously define the rationals to be continuous to allow a more limited application of differential and integral calculi because there are so many applications which require the reals and (as noted) we already have superior ways for dealing with any arbitrary set.
Yet most of the reasons we can’t e.g., integrate over the rationals in the interval [0,1] have to do with the intuitive notion that it contains “gaps” where we know irrational numbers exist even though the rationals are infinitely dense. It is, in fact, possible to construct functions that are continuous on the rationals and discontinuous on the reals. Moreover, we frequently use statistical methods that assume continuity even though the outcomes can’t ever be irrational-valued. Further, the Riemann integral is defined in elementary calculus and often elsewhere as an integer-valued and thus a countable set of summed "terms" (i.e., a function that is Riemann integrable over the interval [a,b]  is integrated by a summation from i=1 to infinity of f(x*I)Δx, but whatever values the function may take, by definition the terms/partitions are ordered by integer multiples of i). As for the gaps, work since Cantor in particular (e.g., the Cantor set) have demonstrated how the rationals “fill” the entire unit interval such that one can e.g., recursively remove infinitely many thirds from it equal to 1 yet be left with infinitely many remaining numbers. In addition to objections mostly from philosophers that even the reals are continuous, we know the real number line has "gaps" in some sense anyway; how many "gaps" depends on whether or not one thinks that in addition to sqrt(-1) the number line should include hyperreals or other extensions of R1. Finally, in practice (or at least application) we never deal with real numbers anyway (we can only approximate their values).
Another potential use is educational: students who take calculus (including multivariable calculus and differential equations) never gain an appreciable understanding of the reals because they never take courses in which these are constructed. Initial use of derivatives and integrals defined on the rationals and then the reals would at least make clear that there are extremely nuanced, conceptually difficult properties of the reals even if these were never elucidated.
However, I’ve been sick recently and my head has been in a perpetual fog from cold medicines, so the time I have available to answer my own question is temporarily too short. I start thinking about e.g., the relevance of the differences between uncountable and countable sets, compact spaces and topological considerations, or that were we to assume there are no “gaps” where real numbers would be we'd encounter issues with e.g., least upper bounds, but I can't think clearly and I get nowhere: the medication induced fog won't clear. So I am trying to take the lazy, cowardly way out and ask somebody else to do my thinking for me rather than wait until I am not taking cough suppressants and similar meds. 
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David: We actually do integrate over the rational numbers. Probably the most essential integration formula is that of the integral of x^n over the interval [0,1]. The value of this can be established entirely over the rationals. You can have a look at my Famous Math Problems10 video at my channel (user njwildberger). 
There are some of us that don't believe in the infinite-precision dream which supports the `real numbers'. If you are interested in why, my recent seminar: `A Socratic look at logical weaknesses in modern pure mathematics' gives some reasons. Also at my YouTube channel.
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I have come out with my own equation(I have no idea whether it is new) for pie : π = √2/2 x n x √((1-cos(dΘ)), where n is the number of triangles in the circle, and  dΘ is the angle of the triangle that is -->0. Ok, now, say, I put n = 1440, so dΘ will be 360/1440 = 0.25, and put it into the equation, i will get  π= 3.141590118…; if I put n=2880, so dΘ will be 0.125, putting it into the equation, I will get π=3.141591603.., if I put n=5760, so dΘ will be 0.0625, putting it into the equation, i will get π= 3.141592923..., we know π= 3.141592654..,but i can never really get the n and dΘ to give that answer. Anyone can come out with some good idea??
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Dear Mason,
I dont know. I took this from you.
Best Regards,
Henri.
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If a polynomial P(z) of degree n omits w in |z|<1, show that P(z)+(1-e^{ih})zP'(z)/n also omits w in |z|<1 for every real h. I know at least two proofs of this result, one follows by using Laguerre's theorem concerning the polar derivative of a polynomial. I want to find the direct proof of this result with out using any known Theorem.
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It seems that you should give more details. Is this known
result? Is the hypothesis that h is real? For example, the set
$1-e^{ih}: h \in R$ is the circle $K(1;1)$.
One can try to use induction  or Bernstein inequality with Argument Principle.
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If a1,a2,...,aare given positive integers in strictly increasing order, what would be the best possible lower bound of
|1+za1+za2+...+zan| for |z|>1?
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It seems that 0 is exactly this lower bound in the general case.
Indeed, the product of all roots of the polynomial 1+...+z^{a(m)} is 1. Therefore, it is impossible that all these roots lie within the unit circle.
Hence some roots are at least on |z|=1 or even outside the unit circle.
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A (single variable) function is differentiable iff some other function is continuous. Can we get similar characterizations for (different kinds of) multifunctions? How to express: a multifunction is X-differentiable iff some other (multi)function is continuous?
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Hello dear friends
You know we can put such type questions in pure classification field and there is no reason for  thinking about any applications (may be it could be find many applications in future). But, it is not bad we could get some sights about the question and we could get some imagines. Anyway, I interest more answers from members of this group. Finally, may be it will be interest for somebody that some researchers are investigating some systems of fractional differential inclusions.
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PARITY is about whether a unary predicate of a structure has even numbers of elements in it. 
If a kind of logic can define PARITY, then there is a formula of this logic so that:
PARITY return True on a structure iff this structure is a model of this formula.
We have known that logics with counting can easily define PARITY.
But what about others without counting?
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Right Peter. I just wanted to avoid such things as "there is a set" or "there is a function", so I proposed a first order formula containing given predicates and a function symbol. Better I make it formally here (for Arthur much  more than for Peter)
All finite models of the following axiom have even cardinality, and every set of even cardinality can be expanded to a model of the following axiom.
Language L = { A(.), f(.) } (one unary predicate and one unary function)
Axiom:
for all x, y [ (A(x) ---> not A(f(x))) and (not A(x) ---> A(f(x))) and (x =/= y ---> f(x) =/= f(y) ]
Let M be a finite model. The function f : M ---> M is injective. Let A \subset M be the set of realizations of A(x) and let M\ A be the set of realizations of not A(x).  So f(A) \subset M\A and f(M\A) \subset A. By injectivity of f follows | A | < = | M\A | from the first and | M \ A | < = | A | from the second. So | M \ A | = | A | , so |M| is even.
The theory is not complete! f can be an involution, or not. For example, M = {1, 2, 3, 4}, A = {1, 3} and f: 1 --> 2 --> 3 --> 4 --> 1 is not an involution; 1 --> 2 --> 1, 3 -->4 -->3 is an involution. The proposition "for all x f^2(x) = x"  is true in the second model but false in the first. 
However every finite set with even cardinality can be expanded to such a model in many ways. 
The problem might be that the negation of this axiom does not imply that the models are of odd cardinality. OK, I guess that at this point one would really need second order predicate logic! [If there do not exist subset A and function f such that the axiom is fulfilled, then the cardinality must be odd indeed]
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I am just wondering. For example, for series 1+2+3+4+5+6+.... the Tn is k, while the Sn is (n)(n+1)/2. I do know there are rules to reach the summation for each case (for example, for series k2, the summation is n(n+1)(2n+1)/2, etc), so is there a more general way to convert the summation Sn to Tn just like the the case of derivation and integration in Calculus?
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T(1) = S(1) and T(n) = S(n) - S(n-1) for n>1?  Perhaps I just don't understand what you mean by T(n) and S(n).  I'll keep quiet.
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Equation: $(eiaXf)=f(ax)$, where $X$ may be unbounded operator and $a in R>0$. I have found something but I am not convinced. The important point: Operator $X$ is not in terms of $a$ and the Hilbert space L^2(R>0, dx/x).
Thank you in advance. 
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