Science topics: AnalysisReal and Complex Analysis

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# Real and Complex Analysis - Science topic

For discussion about the analytic properties of real and complex sequences and functions.

Questions related to Real and Complex Analysis

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

The question details are contained in the attached pdf file.

I'm looking for optimizing multivalued vector valued function.

Is there exist at least 9 distinct

**a bi-j-ective, diffeomorphic& homoeomorphic.****analytic**functions F(x,y), dom i of two variables in the x,y, cartesian planeF(x,y); dom F:[0,1]\times[0,srt(3)/2],\toΔ^2, unit 2 probability simplex

CO-DOM(F)=IM(F)=delta_2: {p(i)=<p_1,p_2,p_3>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0\in [0,1]\forall (x1,x2,x3)\in [0,1]

F(0,0)=(1,0,0)

F(1,0)=(0,1,0)

F(1/2, sqrt(3)/2)=(0,0,1)

F(x=1/2,y=sqrt(3)/6))=(1/3,1/3,1/3)_(1/2, sqrt(3)/6) =

The inverse function being

i=(x,y)=F-1(<p1i,p2i,p3i>_i=(x,y))= <x=[2p2+p3+1]/2,y= p3 *[sqrt(3)/2]]

F(x=1/2,y=sqrt(3)/6))=(1/3,1/3,1/3)_(1/2, sqrt(3)/6) ;

Incidentally it also has to accomodate the 8 element boolean algebra of events on each pt as well <p1,p2,p3>, pi1+p2+p3=1 pi>=0 omega=1, emptyset =0,

PR(A or B)=PR(A)+PR(B)= p1+p2 >=0

and <p1+p2, p2+p3, p3+p4)

ie F(x,y)= <p1,p2,p3>,\to <1,0, p1, p2, p3, p1+p2, p2+p3, p3+p1}

and as every element of the simplex must be present at very least six times,

It actually must consist of at least six identical simplexes, that where the euclidean

F(x,y, {1.2,3.4.5.6})=Δ^2\cup_i=1-6, these can be the same simplex but with the order, of each pi in each <p1, p2, p3> interchanged

<Omega={A,B,C}, F= {{Omega},{ emptyset}, {A}, {B}, {C}, {A V B}, {AVC}, {B VD)

PR(A)=p1

PR(B)=p2 PR(C)=p3

PR(A or B)=PR(A)+PR(B)= p1+p2 >=0

PR(A or C)=PR(A)+PR(C)=p2+p3

PR(B or C)=PR(B)+PR(C)=p1+p3

where in addition there is a further triangle probability function that is ranked by this chances. The triangle frame function G(x,y,1,6)={{1,0, g1,g2,g3, g1+g2, g2+g3, g1+g3}, 1<=gi, g1+g2, g2+g3, g3+g1>=0, g1+g2 +g3=1;

such that for F(x,y,{i,6] on the same coordinates, \forall Ei\in(A, B, C,A or B, A or C, B or C)

{{1,0, g1,g2,g3, g1+g2, g2+g3, g1+g3}= [G(x,y,{1,,,6,1), G(x,y,{1,,,6},2), G(x,y,{1,6},3), G(x,y,{1,,,6},4), G(x,y,{1,,,6},5),.....

**G(x,y,{1,,,6},8),]****; G(x,y,{1,,6},1)=1, G(x,y,{1,,,6},2)=0**

**1>G(x,y,{1,,6},3)=g1,>0 in the interior**

**1>G(x,y,{1,,6},4)=g2 >0,**

**1>G(x,y,{1,,6},5)=g3>0**

**1>G(x,y,{1,,6},5)=g1+g2>0, >{g1,g2}**

**1>G(x,y,{1,,6},7)=g2>0**

**1>G(x,y,{1,,6},8)=g1+g3>0, g1+g3>g3, g1+g3>g1**

**1>G(x,y,{1,,6},7)=g2+g3>0, g2+g3> (g2, g3)**

**g1+g2+g3=1**

**1>G(x,y,{1,,6},5)=g1+g2>0**

\forall i in [i in 6} G(0,0,i,)=(0,0,1),

G(0,0,i\in [1,6},3)=G(0,0,i\in [1,6},, 1),=G(0,0,i\in [1,6},5)=G(0,0,i\in [1,6},8}=1

G(0,0,i\in [1,6},2)=G(0,0,i, 4),=G(0,0,i\in [1,6},{7}=,G(0,0,i\in [1,6},{8})=0

\forall i in [i in 6} G(1,0)=(0,1,0)

G(1/2, sqrt(3)/2)=(0,0,1)=

G(x=1/2,y=sqrt(3)/6, {1,,,6}))=(1/3,1/3,1/3)=(1,0,1/3.1/3.1/3. 2/3,2/3.2/3}

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, G(x,y,i,j)=gj>1/3, iff F(x,y,i,j)=pj>1/3,

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, G(x,y,i,j)=gj>1/3, iff F(x,y,i,j)=pj>1/3,

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, G(x,y,i,j)=gj=1/2, iff F(x,y,i,j)=pj=1/2,

\forall {x,(y)}\forall,l {i....6},\forall j \in {1,,,8}, F(x,y,i,j)=pt<|=|>F(x1,y,1i1,t)=pt[\forany {x1,(y1)}\foranyl,l1 {i....6},forany t \in {1,,,8}, iff G(x,y,i,j)=gj|<|=|> G(x1,y1,i1,t)=gt, ,

\forall {x,(y)}\forall,l {i....6},\forall t \in {1,,,8}, \forall {x1,(y1)}\forall,i1 {i....6},\forall t_1 \in {1,,,8}, such that

F(x,y,i,,t1)+F(x1,y1,i1,t2)=p_t(y,x,i)+p_t1(x1,x1,i1)<|=|> F(x2,y2,i2, t3)+F(x3,y3,i3,t4) =p_t3(x2,x2,i2)+ p_t4(x3,x3,i3)\forany{x2,(y2),i3},(x3,y3,i3}dom(F) such that, forany (t2,t3) \in {1,,,8}, where t2 @,sigma F(x2,y2,i3_, t3 in sigma @,F( x3,y3, t3) iff

G(x,y,i,,t1)+G(x1,y1,i1,t2)=g_t1(y,x,i)+g_t2(x1,x1,i1)<|=|> G(x2,y2,i2,j,t3)+ G(x3,y3,i, t4)=g_t3(y2,x2,i2)+g_t4(x3,x3,i3)

where

, iff F(x,y,i,j)=pj=1/2,

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, 0<G(x,y,i,j)=gj<1/3, iff 0<F(x,y,i,j)=pj<1/3,

\forall {x,(y),l {i....6},\forall j \in {1,,,8},2/3 <G(x,y,i,j)=gj>1/3, iff 2/3>F(x,y,i,j)=pj>1/3,

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, G(x,y,i,j)=gj=2/3 iff F(x,y,i,j)=pj=2/3}

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, 1>G(x,y,i,j)=gj>2/3 iff 1>F(x,y,i,j)>2/3

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, 1>G(x,y,i,j)=gj>0iff 1>F(x,y,i,j)>0

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, 1>G(x,y,i,j)=gj= 0 iff F(x,y,i,j)=0

\forall {x,(y),l {i....6},\forall j \in {1,,,8}, 1>G(x,y,i,j)=gj= 1 iff F(x,y,i,j)=1

forall

_(1/2, sqrt(3)/6) =

{{1,0, p1,p2,p3, p1+p2, p2+p3, p1+p3}

\forall Et\in(A, B, C,A or B, A or C, B or C), \forall Ej\in(A, B, C,A or B, A or C, B or C)

ie for t\in {1,......8), j in {1,,,,,8}

G(,x,y,{i,,,,6},t,) @ x,y,i>G(x,y,{1,,,6},j) or G(,x,y,{i,,,,6},t,)=G(x,y,{1,,,6,j,) or G(,x,y,{i,,,,6},t,) @ x,y,i<G(x,y,{1,,,6},j)

G(,x,y,{i,,,,6},t,)_t @ x,y,i>G(x,y,{1,,,6},j) iff F((x,y, {i,6},t)>F(x,y,{i,,6},j)= PR(F(x,y,{i,,6})>PR(x,y,{i,,,6}

G(Ei)=G(Ej) iff P(Ei)>PR(Ej) ie g1=g2 iff p1=p2,

g1+g2= g3 iff p3=p1+p2,

G(Ei<G(Ej) iff P(Ei)>PR(Ej)

G(0,0)=(1,0,0)

G(1,0)=(0,1,0)

G(1/2, sqrt(3)/2)=(0,0,1)=

F(x=1/2,y=sqrt(3)/6))=(1/3,1/3,1/3)_(1/2, sqrt(3)/6) =

where on each vector, it is subject to the same constraints F-1{x,y,i} G(A)+G(B)+G(C)=1, G(A v B)=G(A)+G(B), G(sigma)=1, G(emptyset)=0 etc whenver G-1(x,y, i \in {1,,6})=F-1(x,y,{i,6}

for all of the 8 elements in the sigma algebra of each of the uncountably many vectors in the interior of each of the six simplexes of uncountably many vectors

and all elements F_i in the algebra of said vector in each in simplex, except omega, 0, G(

Such in addition every element of Δ^1, the unit one probability simplex, set of all two non non negative numbers which sum to one, are present and within the image of the function; described by triples like (0, p, 1-p) on the edges of the triangle in cartesian coordinates

**to, the unit 2 probability simplex**

**consisting of every triple of three real non-negative numbers, which sum to 1. Is the equilateral triangle, ternary plot representation using cartesian coordinates over a euclidean triangle bi-jective and convex hull. Do terms p[probability triples go missing.**

**I have been told that in the iso-celes representations (ie the marshak and machina triangle) that certain triple or convex combinations of three non -negative values that sum to one are not present.**

**Simply said, does there exist a bijective, homeomorphic (and analytic) function F(x,y)of two variables x,y, from the x-y plane to to the probability 2- simplex; delta2 where delta2, the set ofi each and every triple of three non negative numbers which sum to one <p1, p2, p3> 1>p1, p2 p3>=0; p1+p2+p3=1**

**F(x,y)=<p1, p2, p3> where F maps each (x,y) in dom(F) subset R^2 to one and only to element of the probability simplex delta2 subset (R>=o)^3; and where the inverse function, F-1 maps each and every element of delta 2**

**<p1, p2, p3>;p1+p2+p3 P1. p2. p3. >=0 , that is in the ENTIRE probability simplex, delta 2 uniquely to every element of the dom(F), the prescribed Cartesian plane.**

Apparently one generally has to use a euclidean triangle, with side lengths of one in Cartesian coordinates, often an altitude of one however is used as well according to the book attached attached, la

**st attachment p 169.****(which suggests that certain elements of the simplex will go missing there will be no pt in Dom (F), such F(x,y)=<p1,p2, p3> for some <p1,p2,p3> in S the probability simplex**

**is in-vertible and has a unique inverse, such that there exists no <p1, p2, p3> in the simplex such that there is no element (xi,yi)of dom(F) such that F(xi,yi)= <p1, p2, p3> in**

**F(x,y), that is continuous and analytic**

**map to every vector in the simplex, ie there exists no set of three non negative three numbers p1, p2 p3 where p1+p2 +p3=1 such that**

**ie for each of the nine F,**

**Where CODOM(F)=IM(F)=delta_2: {p(i)=<p_1i,p_2i,p_3i>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0}**

**where p(i( described a triple and i whose Cartesian index is i= (x,y), ie F(x,y)=p(i)<p_1i,p_2i,p_3i>_i).**

**and**

,

**CO-DOM(F)=IM(F)=delta_2: {p(i)=<p_1,p_2,p_3>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0\in [0,1]\forall (x1,x2,x3)\in [0,1**]

F(0,0)=(1,0,0)

F(1,0)=(0,1,0)

F(1/2, sqrt(3)/2)=(0,0,1)

(with a continuous inverse)he car-tesian plane, incribed within an equilateral triangle to the delta 2,

**coDOM(F)=IM(F)=delta_2: {p(i)=<p_1,p_2,p_3>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0\in [0,1]\forall (x1,x2,x3)\in [0,1]**

**where, no triple goes missing, and where delta 1, the unit 1 probability simplex subspace, (the set of all 2=real non negative numbers probability doubles which sum to one, described as triples with a single zero entry),**

**delta _1 subset IM(F)=codom(F)=Dela_2**

**and each probability value in [0,1] , that is each and every real number in [0,1] occurs infinitely times many for each of the p1-i, p2_2, p3_3 , on some such vector,**

**1. and one each degenerate double**

**2, And which contains, as a proper subset, the unit 1 probability simplex, delta 1 (set of all probability doubles)contained within the IM(F)=dom(F)=delta2; in the form of a set of degenerate triples, delta_1*, subset delta 2=IM(F)=codom(F) ( the subset of vectors in the unit 2 (triple) probability simplex with one and only one, 0, entry),**

**ie <0.6, 0.4,0>, <0, 0.6, 0.4>**

**3. where for each pi /evctor (degenerate triple) in the degenerate subset delta 1 * of delta 2; the map delta1*=delta 1 is the identity (that is no double goes missing). The unit 2 simplex (set of all real non negative triples= im(F) must contain along the edges of the equilateral, every element of delta 1, every set of two number which sum 1).**

**4. And where among-st these degenerate vectors in delta 1* (the doubles inscribed as triples with a single 0), (not the vertices), must contain, for each, and every of the two convex, combinations, or positive real numbers in the unit 1 probability simplex (those which sum to one) at each of them, at least three times, such that every real number value p in (0,1), such that :**

**p1 +1-p =1, occurs at least six times, among-st six distinct degenerate, double vectors <p1, p2,. p3}{i\in {1...6}**

**that for all p in (0,1)and there exists six distinct degenerate triple vectors mapped to six distinct points in the plane**

**5. in addition in must contain the unit 2 simplex as the sum of the entries in each triple.**

**ie among-st the triples <p1, p2, p3> in IM (F) it must be that \for reals, r, in (0,1) and for each possible value of p1+p2 assumes that value , infinitinely many times,**

**p2+p3 on a distinct vector assumes that value r in (0,1) infinitely many times.**

**p1+p3 must assume that value r in (0,1) infinitely many times, , Corresponding there must not exist some real value in (0,1) such that one of p1, p2 p3 assumes that value e, and moreover, not infintiely times, and but no sum value on some vector (ie element of F(p1+p2, p2+p3, p3+p1) that also assumes that value and infinitely times. The entire unit interval of values must for each such rin [0,1]and for each of three distinct sums in Fsimplex must be contained and assumed individuallyt infinitely many**

**6, Finally for every element of a 2or more distinct vectors such that the two elements (p1, p2 , p3) sumto one**

**such that for any given p1\in vector 1 , p2\in vector 2 ,p1+p2=1**

**for any given p1 in vector 1 p3 in vector 2, p1+p3=1**

**for any given p1 in vector 1 p1in vector 2, p1+p1=1**

such that for any given p2\in vector 1 , p2\in vector 2 ,p2+p2=1

for any given p2 in vector 1 p3 in vector 2, p2+p3=1 ,

for any given p3 in vector 1 p3 in vector 2, p3+p3=1

6, Finally for every 3 elements (,p1 p2 , p3, p1, +p2, p3+p1, p2+ p3) in 2 or 3or more distinct vectors v1, v2, v3, such that the three elements in the three distinct vectors,in sum to any of these numbers \forall n \in {\forall n\in {1....48}n/28,1, 4/3, 1.25, 1.5, 4/3. 1.75, 1.85, 2}, or such that or any 2 elements (p1, p2 , p3, p1 +p2 p1+p3, p2+p3) in common vector v1 and another for all possible other elements in (p1, p2 , p3, p1 +p2 p1+p3, p2+p3) in distinct vector v3 sum to those values these must be present

Moreover it must also at least extend to any given 4/5/6/7/8 /9/10distinct elements of (p1, p2 , p3, p1 +p2 p1+p3, p2+p3) that sum to for all possible combinations of being in up to 10 distinct vectors, 9 distinct vectors two elements in common,,,,,,,,,,,,,,,,, such that for any such one of all such combinations there must be uncountable many versions of each element of p1, p2 , p3, p1 +p2 p1+p3, p2+p3) in each combination individually for each of the above sum values & for each of the above cardinalities of entries,.

More over the entire simplex must be present for each of foralll of the ten possible combinations or sum term number values,10

and for all r each different combinations of elements that sum to that those values in (p1, p2 p3, p1+p2.....)

and for each of the distinct number of distinct vectors that could be present in that sum upl to ten

and, that could sum to each of all those approximately 70 distinct values. that could to those values, and for each of the different number of terms in each sum,. the entire simplex must be present and every such value in [0,1[ must be assumed individually for every term in every

(1)for each of the term length sum (any given 2 that sum to one, any given three which sum to one, any given four that sum one m any given five which sum to one, any three which sum to 2, any given four which sum to 2 any five which sum to 2, any given four which sum to three, any five which sum to 3

(2)for all of the 70 or so , values mentioned d

(3)for all number of distinct vectors of which those terms are in that sum are associated

(4) for all 6 distinct terms types of in the sum p2 p3, p1+p2.....)

forall n \in {\forall n\in {1....48}n/28,1/2, 2/3, 1,1.125, 4/3, 1.25, 1.5, 5/3. 1.75, 1.877,11/6, 2,2.25, 2.33, 2.5, 2.66, 2.75, 3, 3.25, 3.33 ,3.5,3.666.3.75, 4, 3.333, 4.5, 4.666, 5, 5.5, 6, 6.5, 7 7.333, 7.5, 7,666, 8, }

IE there vcant be any

**GAPS** elements elements such that in four or five distinct vectors with no elments in common

, three or four distinct vectors, two elements in a common vector,the other three/2 being in distinct vectors when there are fouir elements

three distinct vectors, with two elements in two common vectors, or three elements in one vector common vector, and two and the other two in either one or two common vectors and the other elements or in 2 elements in one common vector and one in a common vector, 2 distinct vectors with 2 elements common to each of either one or /two of the vectors, 2 vectors with 3 elements in one vectors and 2 in the other

n a distinct vectors,

such p1 vector 1 +p2 in vector 2 +p3 in vector 3=1

such p1 vector 1 +p3 in vector 2 +p2 in vector 3=1

such p1 vector 1 +p2 in vector 2 +p2 in vector 3=1

such p1 vector 1 +p3 in vector 2 +p2 in vector 3=1

such p1 vector 1 +p1 in vector 1 +p1 in vector 2

such p1 vector 1 +p1 in vector 1 +p3 in vector 2

such p1 vector 1 +p1 in vector 1 +p2 in vector 2

such p1 vector 1 +p1 in vector 1 +p2 in vector 3

such p1 vector 1 +p1 in vector 2 +p2 in vector 3

such p1 vector 1 +p1 in vector 3 +p2 in vector 3

such p1 vector 1 +p3 in vector 1 +p2 in vector 3

such p1 vector 1 +p2 in vector 1 +p2 in vector 3

such p1 vector 1 +p3 in vector 1 +p2 in vector 3=1

such p1 vector 1 +p2 in vector 2 +p2 in vector 3=1

such p1 vector 1 +p2 in vector 3 +p3 in vector 3=1

such p1 vector 1 +p1 in vector 2 +p1 in vector 3=1

such p1 vector 1 +p1 in vector 2 +p1 in vector 3=1

such p2 vector 1 +p2 in vector 2 +p2 in vector 3=1

such p3 vector 1 +p3 in vector 2 +p3 in vector 3=1

such p3 vector 1 +p3 in vector 1 +p3 in vector 3=1

such p3 vector 1 +p3 in vector 2 +p3 in vector 2=1

such p3 vector 1 +p3 in vector 3 +p3 in vector 3=1

such p3 vector 1 +p3 in vector 2 +p3 in vector 2=1

such p3 vector 1 +p3 in vector 3 +p3 in vector 3=1

such p2 vector 1 +p1 in vector 2 +p3 in vector 3=1

such p2 vector 1 +p1 in vector 2 +p1 in vector 3=1

such p2 vector 1 +p3 in vector 2 +p2 in vector 3=1

such p2 vector 1 +p2 in vector 2 +p2 in vector 2=1

such p1 vector 1 +p2 in vector 1 +p2 in vector 2=1

such p1 vector 1 +p2 in vector 1 +p3 in vector 3=1

such p1 vector 1 +p2 in vector 1 +p3 in vector 3=1

such p1 vector 1 +p2 in vector 1 +p3 in vector 3=1

uch that for any given p1\in vector 1 , p2\in vector 2 ,p1+p2=1

for any given p1 in vector 1 p3 in vector 2, p1+p3=1

for any given p1 in vector 1 p1in vector 2, p1+p1=1

such that for any given p2\in vector 1 , p2\in vector 2 ,p2+p2=1

for any given p2 in vector 1 p3 in vector 2, p2+p3=1 ,

for any given p3 in vector 1 p3 in vector 2, p3+p3=1

**for any given p1 in vector 1 p1+p2 in vector 2 p1+(p1+p2)=1**

**for any given p1 in vector 1 p1+p3 in vector 2 p1+(p1+p3)=1**

**for any given p1 in vector 1 p2+p3 in vector 2 p1+(p2+p3)=1**

for any given p2 in vector 1 p1+p2 in vector 2; p2+(p1+p2)=1

for any given p2 in vector 1 p1+p3 in vector 2; p2+(p1+p3)=1

for any given p2 in vector 1 p2+p3 in vector 2 p2+(p2+p3)=1

for any given p3 in vector 1 p2+p3 in vector 2 p3+(p2+p3)=1

for any given p3 in vector 1 p2+p3 in vector 2 p3+(p2+p3)=1

for any given p3 in vector 1 p2+p3 in vector 2 p3+(p2+p3)=1

**for any given p1+p2 in vector 1 p1+p2 in vector 2 p1+p2)+(p1+p2)=1**

**for any given p1+p2 in vector 1 p1+p3 in vector 2 p1+p2)+(p1+p3)=1**

**for any given p1+p2 in vector 1 p2 +p3in vector 2( p1+p2)+(p2+p3)=1**

**for any given p2+p3 in vector 1 p3 +p1in vector 2 (p2+p3)+(p3+p2)=1**

**for any given p2+p3 in vector 1 p2 +p3in vector 2**

**for any given p1+p3 in vector 1 p3+p1in vector 2**

**for any given p2+p1 in vector 1 p3 +p2in vector 2**

**for any given p2+p1 in vector 1 p3 in vector 2**

such that for any given p2\in vector 1 , p1\in vector 2 ,p1+p2=1

for any given p1 in vector 1 p2 in vector 2, p1+p3=1

for any given p1 in vector 1 p3in vector 2, p1+p1=1

plus each of the

such that for any given p3\in vector 1 , p2\in vector 2 ,p1+p2=1

for any given p3 in vector 1 p1 in vector 2, p1+p3=1

for any given p3 in vector 1 p3n vector 2, p1+p1=1

**for all of the 36 or distinct sombination such that p1 in one vector and one of (p1, p2, p3,p1+p2 p2+p3, p3+p1) on a disitnct vector sum to one , for all reals in [0,1]each of these combination must obtain infinitely many times, each oif these combinations must be surjective/ bijective with regard to the unit 1 probability simplex, and for each such combination listed, each of the two terms must assume individually ,for each of the uncountably many real values i m [0,1],, uncountably many times.**

**for each of the way, and each of the 36 values in each of that one the 36 and obtain infinitely many time, assume each value within in [0,1] infinitely many time,, and**

**In additions**

**p2\in vector 2 , p3\in vector 2**

**, p3 p1**

**, p3+p1=1, p2+p1=1, p3+p2=1,**

**p2+p3=1**

**where these are entries on distinct vectors each of these entries must contain infinitely many distinct probability**

**to either 1, 2, 3, or the**

**Moreover, it must that both horizontally and vertically, the entire set of element in the sum to 2 non negative simplex**

**Where CODOM(F)=IM(F)=delta 1, it must be such that the for bijective function G which for each element of Im (F) , G: maps <p1,p2, p3> \in IM (F)=delta 2 ,to G(<p1+p2, p2+p3, p3+p1)>**

**, ie F(x,y)=<p1,p2, p3) then G[(x,y))=G(F-1(p1,p2, p3)])=<p1+p2, p2+p3, p1+p3>**

**such that G is also a bi-jective and analytic diffeomorphism onto**

**delta_2: {p(i)=<p_1i,p_2i,p_3i>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=2, 1>=p_1i,x_p2i,x_3_i>=0}, the set of all three real number that sum to 2.**

**ie dom (G)=dom (F) and and thus for any<p1, p2 p3,> domain we compute F-1(,p1, p2, p3) to get the cartesian coordinates of that vector and feed them into G, where G computes the probabilities of the disjunctive events**

**(unit 2 probability simplex)**

**that delta_2: {p(i)=<p_1i,p_2i,p_3i>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=2, 1>=p_1i,x_p2i,x_3_i>=0}**

**in these sums in this sense.**

**Moreover, it also be the case, that there must exist**

**, p2+p3, p1+p3**

**see (4) below**

**i=(x,y), and i(2)=(x2,y2); . i(3)=(x3, y3), i(4)=(x4, y4), i(5)=(x5, y5), i(6)=(x6,y6);**

**(x6,y6)\neq (x5, y5)\neq (x4, y4)\neq(x3,y3)\neq(x2,y2)\neq (x,y)**

**where**

**F(x2, y2)= p(i(2)=<p_1(i2)=p, p_2(i2)=0, p_3_(i2)=1-p>{i2}; p1+p3=1; 0>(p3, p1)<1, p2=0, p1=p**

**F(x3, y3)= p(i(3)=<p_1(i3)=p, p_2(i3)=1-p, p_3_(i3)=0>^{i3}; 0>(p1, p2) <1, p1+p2=1, p3=0, p1=p**

**F(x4, y4)= p(i(4)=<p_1(i4)=0, p_2(i4)=p, p_3_(i4)=1-p>{i4); 0>(p3, p2) <1, p3+p2=1, p1=0, p2=p**

**F(x5, y5)= p(i(5)=<p_1(i5)=1-p, p_2(i5)=p, p_3_(i5)=0>^{i5) 0>(p1, p2) <1, p1+p2=1, p3=0, p2=p**

**F(x6, y6)= p(i(6)=<p_1(i6)=1-p, p_2(i6)=0, p_3_(i6)=p>{i6) 0>(p1, p3) <1, p1+p3=1, p2=0, p3=p**

**F(x, y)= p(i)=<p_1(i)=0, p_2(i)=1-p, p_3_(i)=p>{i);**

**0>(p3, p2) <1; p3+p2=1, p1=0, p3=p**

**where for all i\in {i,i(1)...i(5)} and p_t1(m)+p_t2(m) +pt3(m)1 etc**

,p1(i)+p2(i) +p3(i)=1

p1(i2)+p2(i2) +p3(i2)=1,

such that p_j_i(2)\in p(i(2); p_j_i(2)\=p

\oplus p2 oplus p3 =0, and

**where ONE and only of p1, p2, p3 =0, where that precise values occurs at least twice in the first entry of two distinct vectors,**

**<0.6, 0.4, 0>**

**<0.6, 0, 0.4>, at least twice in the second entry, p2, here p2=0.6**

**<0, 0.6, 0.4)**

**<0.4, 0.6, 0)**

**and in the third entry p3, p3=0.6; at least twice**

**<0.4,6, 0.6)**

**<0, 0.4, 6)**

**. In other words \forall (p1)\in (0,1) and for all p2=in \in (0,1), amd for all p3 \in (0,1), where p2+p3=1, there exists two distinct vectors (if only in name) such that <p1=0, p2, p3=1-p2), and < p1=0, p2=1-p3, p3>**

**there exists**

**\forall p\in [0,1] ; 0<p1, p2<1; <p, p2=1-p>**

**possible degenerate triple combinations, for each degenerate convex combination in for all real positive p ; 0<p<1 p, 1-p**

**<0.6, 0.4, 0> <0.4, 0.6, 0> <0.6, 0.4, 0>, <0, 0.6, 0.4)**

**[delta_1* ]subset delta2=IM(F)=codom(F)= {p(i)=<p_1,p_2,p_3>_i;**|\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0,

**& \exists in p(i), one and only one(p2_i, p1_i, p3_i)=0; where the other two entries \neq 0, s.y1< p1 p2>0}**

**for all convex combinations in delta 1 , all possible probability doubles, tuples element of [0,1]^2, of non negative real numbers which sum to 1**

**where map G(delta1*)=delta1 is the identity**

or some subset of I^2\subset R^2 to the unit probability simplex, (the triangle simplex of all triples of non-negative numbers <=1, which sum to one.?

Are such functions convex, that those which use absolute bary-centric coordinates over the probability simplex, when defined over an equilateral triangle with unit length in the Cartesian plane.

seehttps://en.wikipedia.org/wiki/Affine_space#Affine_coordinates

I presume that such function are hardly homogeneous in that infinitely many possitive triples will not be present?

F:I^2, to {<x_1,x_2,x_3>_m; x1+x2+x3=1, (x_1,x_2,x_3)\in [0,1]\forall (x1,x2,x3)\in [0,1]}

from the set of all or triples {<x_1,x_2,x_3>_m; x1+x2+x3=1, (x_1,x_2,x_3)\in [0,1]\forall (x1,x2,x3)\in [0,1]} to a unique index m,\in

**I^2, a real interval in the cartesian plane?****dom(F):[0,1]\times[0,srt(3)/2]**

**IM(F)=F: {p(i)=<p_1,p_2,p_3>_i; |\forall p(i)\in [0,1]^3, s.t: p1_i+p2_i+p3_i=1, 1>=p_1i,x_p2i,x_3_i>=0\in [0,1]\forall (x1,x2,x3)\in [0,1]}**

**where**

**F(0,0)=(1,0,0)**

**F(1,0)=(0,1,0)**

**F(1/2, sqrt(3)/2)=(0,0,1)**

**F(x=1/2,y=sqrt(3)/6))=(1/3,1/3,1/3)_(1/2, sqrt(3)/6) =**

**ie x=2p2+p3]/2,= [2 times 1/3 +1/3]/=1/2**

**y=srt(3)/2-sqrt(3)/2p1-sqrt(3)/2 times p2= sqrt(3)/2*(p3)= sqrt(3)/2*(1/3)=sqrt(3)/6**

**i=(x,y)=F-1(<p1i,p2i,p3i>_i=(x,y))= <x=[2p2+p3+1]/2,y= p3 *[sqrt(3)/2]]**

**, ill have to check the properties there a lot of other roles that it has to fulfill then just this.**

**Where in addition, every no value of x1+x2, x2+x3, x3+x1 can be missing these must assume each value in [0,1], prefererably infinitely many if positive and <1, and cannot assume a value, that is not assumed by one of the x1,x2, x3, somewhere in the structure,**

**Preferably this must property contain the unit 1 simplex, as a function x1, x2+x3, where every convex combination which sum to one, of two values must be assumed, by x1, 1-x, on distinct vectors <x1,x2,x3>m**

**x2, 1-x2=x1+x3, <x1,x2,x3>m_1 ,m_1\neq m**

**x3,1-x3 =x2+x3=<x1,x2,x3_m2, m2\neq m1\neq m,**

**There also cannot be any mismatched between element of the domain on distinct vectors, ie diagonal or vertical sums, where any two of them sum to one, any three of that sum to 1, or 2,**

<

**0.6**, 0.25, 0,15> <

**0.4**, 0.32, 0.28**>**<

**0.26**, 0.4,**0.34**><0.3

**0.4**,**0.3**><

**0.26**, 0.38,**0.36**> <0.35, 0.35, 0.3>, <0.26,0.4, 28>I presume if its convex it would contain the doubly stochastic matrices or the permutation matrics

,

<

**0.7,**0.25, 0,15> <

**0.8**, 0.32, 0.28> <

**0.5**, 0.32, 0.28> must be a vectors <0.3=1-0.7x1, 0.5=1-x2=0,5, x1=1-0.8=0.2) and conversely for <x1,x2,x3> there must be triad of three distinct vectors, such that one elements =1-x1, , another =1-x2, and another =1-x3

<x1,x2,x3, >

**<y1,y2,y3>, where one of y1,y2,y3, = 1-x1, one of z1,z2,z3, =1-x2,**

**<z1,z2,z3>**

**there must be distinct vectors such that <x1,x2,x3>, x1=25+0.32,**

**as well for x2, x3, x1+x2, x3+x2, x1+x3,**

**and which sum to 0.15+0.28, and common vectors, where all of the six events, or rather 12 events , whose collective sum <= 1, lie on a common vector as atomic events <x1,x2,x3> or disjunctive events <x1+x2,x2+x3, x3+x1>**

**ie a vectors <0.4, 0.25, 0.35>, and one where <x1=0.4, x2, x3> where x1+x2=0.6**

**< 0.4, 15, 0.45>, <0.6, 0.28,0.12>**

**<x1,x2,x3> where x1+x2=0.4**

**, any 'two sums';, three, sum of two elements, to one, or any three sum sum to one**

**, or three elements of distinct vectors which sum to 2,**

**and the set of three non positive numbers which sum to 2, as s**

**Where m denote a Cartesian pair of points in the x,y, plane which uniquely denotes a specific vector, built over an equilateral triangle in Cartesian coordinates,**

**and with unit length in side,in Cartesian coordinates**

**-where this is distance in Cartesian coordinates (x,y) in euclidean norm of each Cartesian coordinate probability vectors vertex = F-1(1,0,0),F-1(0,1,0),F-1(0,0,1)=1, where the respective euclidean norm in probability coordinates is clearly sqrt (2) sqrt sqrt(1-0)^2+ (1-0)^2+(0-0)^2)in**

**sqrt (2) in probability coordinates, in 2- norm,1, in 1 norm,**

**distance from the triangle center/circum-centre and vertexes,**(the Cartesian coordinates of the mid spaced probability vector (1/3, 1,3, 1.3)whose distance from each vertex as a euclidean norm in probability =2/3=equal to the 1-norm distance between value in the triple and its relevant vertex)=2/3, where the overall 1-norm difference in probability between the cent-roid and each vertex =4/3,

**= 1/sqrt (3), in Cartesian coordinates in euclidean norm,**

**all three altitudes=medians (distance from each vertex in Cartesian coordinates, to center of each opposing side of the triangle)-in 2-norm again=**

**=**,sqrt(3)/2

**the probability vertices whose Im(F)=(1,0,0),(0,1,0),(0,0,1) whose untoward cartesian coordinates are given above.**

**and the three, apothem**(2-norm distance from the cir-cum center, the Cartesian coordinates (1/2,sqrt(3)/6) of the centroids/probability vector, (1/3,1.3, 1.3),) and the Cartesian coordinates of the mid point of each side of the equilateral triangle,

**= 1/(2 * sqrt(3))**

**where the Centro id (1/3,1/3,1/3 )is the vector in n simplex, entries are just the average , n-pt average of the unit (the only vector with three values precisely the same, and the sums precisely the same 2,3,2.3,2,3)**

**mid probability vector all entries 1/n here 1/3, whose Cartesian coordinate are the circumcentre of the triangle, the point were all three medians cross, (ie the Cartesian point equidistant from each vertex.**

**that being the circumcentre , (1/2, sqrt(3)/6) =in 2-norm of the (the pt whose probability coordinates are just the average for an n simplex of a unit vector (1/n, 1/n,1.n)**

**denoting the distance between the cen are**

**F-1(1/n, 1/n, 1/n) here F-1(1/3, 1.3, 1.3)**

**probability/bary-centric coordinates, with side lengths, between the vertics of sqrt (2)**

**length sqrt(2) in probability coodinates (euclidean norm ) ie sqrt ([1-0]^2+[0-1)^2+ [0,-0,]^2)=sqrt(2), and have side lengths =1 in cartesian coordinates, with distance from the centre 1/sqrt (3), andall median/altitudes /angular bisectors/perpendicular bisectors=sqrt(3)/2, area=sqrt(3)/4 and apothem=1/2sqrt(3)**

**F(x,y)=<x-1,x_2,x_3>m=(x,y)**

I was wondering if someone knows whether Mathematica allows one to plot another 'probability function over the unit 2 simplex (it can be expressed as a function of two arguments subject to certain contrainsts function .

Where I am taking the domain to be of the function to be over vectors in the 2- standard simplex itself as it were,subject to certain contrain'ts.

Does it actually have a closed form expression as a function x,y coordinates. as a function of two arguments. I presume mathematica can allows you plot it .and optimize certain functions over tenary plot, ternary graph, triangle plot, simplex plot, G? Is that correct

Is the following function F:[0,1] to [0,1]

F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two

(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1

(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1

(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1

Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term). F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y)

.

I

Is the series in the picture convergent? If it is convergent, what is the sum of the series?

I has made proposition and prove that (f o f ')(x)= (f 'o f )(x) if f is additive real function and (f o f) (x)=f (x) but I still difficult get example non-trivial. Could you help me? example trivial is an identity function.

I apologize for the inconvenience this

I was looking for examples of first order sentences written in the language of fields, true in Q (field of rational numbers) and C (field of complex numbers) but false in R (field of real numbers). I found the following recipe to construct such sentences. Let a be a statement true in C but false in R and let b be a statement true in Q but false in R. Then the statement z = a \/ b is of course true in Q and C, but false in R.

Using this method, I found the following z:=

(Ex x^2 = 2) ---> (Au Ev v^2 = u)

which formulated in english sounds as "If 2 has a square-root in the field, then all elements of the field have square roots in the field." Of course, in Q the premise is false, so the implication is true. In C both premise and conclusion are true, so the implication is true. In R, the premise is true and the conclusion false, so the implication is false. Bingo.

However, this example is just constructed and does not really contain too much mathematical enlightment. Do you know more interesting and more substantial (natural) examples? (from both logic and algebraic point of view)

If we accept the classical differentiability of spacetime, for all cosmological time, a critical value of that expansion may prevent the sea of virtual particles from ever recombining. This would convert "virtual" mass to "measurable" mass, keeping the critical density constant and thus solve the horizon problem of the Standard Model.

This possibility may also admit an explanation of why the CMBR is so uniform. Galaxies would not "wink out" one by one, and a "time averaged" Cosmological Constant and Gravitational Constant may both be necessary consequences.

Is there a formula to enumerate the lattice paths from (1,1) to (m,n), confined to the region: y=x/2 and y=2x contruct to a triangle, which step set is {(1,0),(0,1)}?

A NSWE-path is a path consisting of North, South, East and West steps of length 1 in the plane. Define a weight w for the paths by w(N)=w(E)=1 and w(S)=w(W)=t. Define the height of a path as the y-coordinate of the endpoint. For example the path NEENWSSSEENN has length 12, height 1 and its weight is t^4.

Let B(n,k) be the weight of all non-negative NSEW-paths of length n (i.e. those which never cross the x-axis) with endpoint on height k.

With generating functions it can be shown that for each n the identity

(*) B(n,0)+(1+t)B(n,1)+…+(1+t+…+t^n)B(n,n)=(2+2t)^n

holds. The right-hand side is the weight of all paths of length n.

Is there a combinatorial proof of this identity?

For example B(2,0)=1+3t+t^2 because the non-negative paths of length 2 with height 0 are EE with weight 1, EW+WE+NS with weight 3t, and WW with weight t^2.

B(2,1)=2+2t because the non-negative paths are NE+EN with weight 2 and NW+WN with weight 2t. And B(2,2)=1 because w(NN)=1.

In this case we get the identity

B(2,0)+(1+t)B(2,1)+(1+t+t^2)B(2,2)=(2+2t)^2.

Let f

_{1}:ℕ⭢ℕ be the identity, that is, ∀n∈ℕ: f_{1}(n)=n. For each n in ℕ, let f_{n:}ℕ⭢ℕ be the map defined recursively as follows.f

_{n}(m) = f_{n-1}(m+1) if f_{n-1}(m) = 2f

_{n}(m+1) = f_{n-1}(m) = 2 if f_{n-1}(m) = 2f

_{n}(m) = f_{n-1}(m) otherwise.Since every map f

_{n}is defined recursively with a transposition in the imageof f

_{n-1}, the map f_{n}is again a bijection that satisfies the following properties.1) ∀n ≤ m: f

_{m}(n)≠ 2.2) ∀n ≤ m the map f

_{n}:ℕ⭢ℕ is a bijection.If the recursion process never ends, both properties are compatible.

However, assuming the actual infinity existence, every infinite process can be completed. Under this assumption the recursive proces give rise to the following properties for m =∞.

1) ∀n∈ℕ: f

_{∞}(n)≠ 2.2) The map f

_{∞}:ℕ⭢ℕ is a bijection.which is a contradiction, unless we assume that every inifinity is potential, and

n keeps always finite.

What are completely monotonic functions on an interval $I$? See the picture 1.png

What is the Bernstein-Widder theorem for completely monotonic functions on the infinite interval $I=(0,\infty)$? See the picture 1.png

My question is: is there an anology on the finite interval $I=(a,b)$ of the Bernstein-Widder theorem for completely monotonic functions on the infinite interval $I=(0,\infty)$? In other words, if $f(x)$ is a completely monotonic function on the finite interval $I=(a,b)$, is there an integral representation like (1.2) in the picture 1.png for the completely monotonic function $f(x)$ on the finite interval $(a,b)$?

The answer to this question is very important for me. Anyway, thank everybody who would provide answers and who would pay attention on this question.

See the attached file for mathematical description of a PDE system steaming from Flash Photolysis. I'm interested in analytical (symbolic) solution but if that is not possible then numerical method would suffice. Thanks.

Greeting and salutation!

I have a initial value differential equation with a unknown parameter, how can i solve it With Matlab ODE solver or other software????

With Best Regards

Hamed

We have at least two different definitions of “infinite” (“actual infinite” and “potential infinite”) since antiquity, and these two different “infinites” with different natures unavoidably become the foundation of present classical infinite related science theory system dominating all the infinite related contents as well as all of our infinite related cognizing activities since then.

When one faces an infinite related content in mathematical analysis, is it in “actual infinite mathematical analysis” or “potential infinite mathematical analysis”?

The infinite small (few) related errors and paradox families (such as the newly discovered Harmonic Series Paradox) and the infinite big (many) related errors and paradox families (such as the newly discovered Cantor’s ideas and operating process of mistaken diagonal proof of “the elements in real number set are more infinite than that in natural number set”) are typical examples of “master pieces of confusing potential infinite and actual infinite”.

I have a 4x4 density matrix (Trace 1) whose elements are nonzero. Its form is

**a b c d**

**b* e f g**

**c* f* h j**

**d* g* j* k**

where

**a+e+h+k=1.****Is there a simple way to find eigenvalues and eigenvectors of this matrix? I calculated in Mathematicai Maple, MATLAB (There are too much terms). More simple way?**

Dear all

What and where is the formula for higher derivatives of a product of many functions? In other words, how to compute higher derivatives of a product of $n$ functions? Concretely speaking, what is the answer (a general formula) to

$$

\frac{d^m}{d x^m}\left[\prod_{k=1}^n f_k(x)\right]=?

$$

where $m,n\in\mathbb{N}$. Could you please show me a reference containing the answer? Thank a lot!

Best regards

Feng Qi (F. Qi)

In two-dimensional case, we can follow Kohn's definition of type by using holomorphic tangent vector field and the Levi function to define infinite type(cite H. Kang, Holomorphic automorphisms of certain class of domains of infinite type, Tohoku Math. J. (2) 46 (1994), 435–442. MR 95f:32041). But when we consider in higher dimensions, can we still use this method to define infinite type?

How to compute the limit of a complex function below? Thanks.

For $b>a>0$, $x\in(-\infty,-a)$, $r>0$, $s\in\mathbb{R}$, and $i=\sqrt{-1}\,$, let

\begin{equation*}

f_{a,b;s}(x+ir)=

\begin{cases}

\ln\dfrac{(x+ir+b)^s-(x+ir+a)^s}s, & s\ne0;\\

\ln\ln\dfrac{x+ir+b}{x+ir+a}, & s=0.

\end{cases}

\end{equation*}

Compute the limit

\begin{equation*}

\lim_{r\to0^+}f_{a,b;s}(x+ir).

\end{equation*}

How to define and compute the power-exponential function of a complex variable? For example, how to define and compute the complex function $(\ln z)^{\sin z}$? where $z=x+iy$ is a complex variable. Thank you for your help.

Hello dear reader,

Remark : they are several "logics" to make sense additively with simple recognition shapes for instance like dashes. If you want to add shapes in a compositionnal way you get the following equation : One thing that is an horizontal dash ( ___ ) " + " (drawing) one other thing that is a vertical dash ( / ) can build by composition a cross like the additive symbole (+) that is either one thing as this symbole or 5 things ( 4 segments + a central point.) So you could have 1"+"1=1 or 1"+"1=5. Immediately you see these "counting logics" make fun of the most elementary arithmetical habits !?

Waiting for your feedback,

JYTA

As I understand that for differentiable and monotone functions we can partition the period and find the total variation, but what about the case when it's not differentiable ?

For example in this article http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=1083433 the authors have essentially mentioned the total variation of signum type function is 4 . But how is it done ? In general it looks to be 2.

I have a doubt regarding completeness of C[0, infinity) with respect to sup norm. Is the space Banach Space? Is it a Banach lattice?

For example, Circle is not a function. Can any one explain with this inverse and implicit function theorem?

Hello;

I have ended up with a quadratic function of X in the form of:

f(X)= X

^{T}AX-X^{T}BX+C^{T}X with A and B as positive definite matrices. Note that the second term has a negative sign. Clearly the function is not necessarily convex, but based on some experimentation and prior calculations, it should convex. I wonder to know if there is any analytical or famous experimental method to prove the convexity of this quadratic function.Thank you

The real number definition requires de limit concept corresponding to the standard topology. If we forget the topological structure of

**R**, then its definition vanishes. I think that, handling non-defined objects it is not an accurate method.Nevertheless, it is not difficult to find mathematical proofs using the topological structure of

**R**while it is assuming that the result is concerned to the underlying set of**R**exclusively, that is to say, the structure-free real number set.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.

I need these measurement for the diversity analysis of population of vectors

Consider the cubic equation:

$ t^3 = 3pt + 2 q \qquad (1)$.

We introduce two variables $u$ and $v$ linked by the condition

$ u+v=t\,$

and substitute this in the depressed cubic (1), giving

$u^3+v^3+(3uv-3p)(u+v)-2q=0 \qquad (2)\,.$

At this point Cardano imposed a second condition for the variables $u$ and $v$:

$3uv-3p=0\,$, ie $uv=p$.

As the first parenthesis vanishes in (2), we get $u^3+v^3=2q$ and $u^3v^3=p^3$. Thus $u^3$ and $v^3$ are the two roots of the equation

$ z^2 -2 qz + p^3 = 0\,.$

$ (z-q)^2 =q^2 - p^3=D$. Hence $z= q + \sqrt{q^2 - p^3} $. Denote by $a_1$ and $a_2$ two roots of $D$ and set $

**z**_k=q +a_k$.Solutions of equations $w^3=1$ are $1,e^{2\pi i/3},e^{-2\pi i/3}$.

Denote by $\underline{A}=\{u_1,u_2,u_3 \}$ and $\underline{B}=\{v_1,v_2,v_3\}$ solutions of equations $u^3=

**z_1**$ and $v^3=**z_2**.$Check that $u_k v_j \in \{p , p e^{2\pi i/3},p e^{-2\pi i/3} \}$. Set $p_1=p , p_2= p e^{2\pi i/3}$, and $p_3=p e^{-2\pi i/3}$.

In discussion with my students (Svetlik,Knezevic,Stankovic,Avalic), we prove the following:

Proposition 1.

The set $\underline{X}= \{u_k + v_j:1\leq k,j \leq 3

\}$ has $9$ elements if $q + \sqrt{q^2 - p^3} \neq 0 $ and six elements if $D=0$ and $q\neq 0$.

If $u_0\in A$, $v_0\in B$, then $\underline{X}= \{\omega^k u_0 + \omega^l v_0:k,l=0,1,2\}$ and

the points of $\underline{X}$ forms three (respectively two) equilateral triangles if $D\neq 0$(respectively $D=0$ and $q\neq 0$).

Presently, we did not find origin of this result in the literature.

What are possible generalizations of this results?

I study the existence of germs of J-holomorphic curves inside a compact real analytic hypersurface embedding in a J-almost complex manifold (with J analytic). In general for fourth dimensional manifold, we have only two possibilities : the hypersurface is foliated or does not contains any germ of J-holomorphic curves. In dimension bigger than four, is it true that for a Stein J-almost complex manifold, a compact real analytic hypersurface does not contain any germ of J-holomorphic curve ?

Suppose we have an analytic function f(z) of a complex variable z and that its Taylor expansion possesses only even order terms. Is the function g(z) = f(sqrt(z)) analytic?

How to construct the system of equations by susbtituting the assumed series solution in the nonlinear partial differential equation? Kindly find the attachment.

I am wondering whether the relative interior of a linear subspace which is not closed is empty or not ? I work in a general Banach space.

If K and L are convex bodies whose radial functions have the same distribution function, i.e. equi-measurable, what can be said? In particular, does this mean that the two radial functions(of the sphere) are equal up to composition with some measure-preserving transformation? If so, does the convexity of the two bodies imply any regularity of this measure-preserving transformation of the sphere? If possible, how much regularity must the bodies possess to force this transformation to be a rotation or orthonormal linear transformation of the sphere?

If we have a matrix and remove for example its last row and column, is

there a condition under which the new matrix has different minimum

eigenvalue than the original matrix?

By Cauchy theorem they might be equal.

Looman–Menchoff theorem states that a continuous complex-valued function defined in an open set of the complex plane is analytic if and only if it satisfies the Cauchy–Riemann equations.

My Question is: Can we drop the assumption of continuity in Looman-Menchoff theorem, if not please provide an example.

As we may know, in the standard approach to the $q$-calculus there are two types of $q$-exponential functions $e_{q}=\sum_{n=0}^{\infty}\frac{z^{n}}{\left[ n\right]_{q}!}$ and $E_{q}(z)=\sum_{n=0}^{\infty}\frac{q^{\frac{1}{2}n\left(n-1\right) }z^{n}}{\left[ n\right]_{q}!}$. Based on these $q$-exponential functions, also, some new functions are defined whose most of their properties are similar to the exponential function in calculus. Despite having interesting properties similar to the exponential function in calculus, for none of them the above property holds. So, can we define a new $q$-exponential function with similar characteristics to the exponential function in calculus which satisfies the aforesaid property?Any help is appreciated.

I created a model for a coil located over a conductor, coil is a multiturn coil with driving current of 1 amper. I modeled the half of geometry and the quarter of geometry. By decreasing the geometry from half to quarter the coil inductance value is getting half but this does not happen for the coil resistance. In fact I get a wrong value for the resistance of the coil. Could anybody take a look at these models and help me to find the mistake?

We focus on the “deep structural relationship” between “nonstandard one” and “standard one”. Let’s exam following facts:

1, as “monad of infinitesimals” has much to do with analysis; nonstandard analysis is much more a way of thinking about analysis, as a different analysis------simpler than standard one.

2, CONSERVATIVE is the nature and a must for Nonstandard Analysis or Nonstandard Mathematics, it is called a conservative extension of the standard one.

3, because of the “deep structural CONSERVATIVE”, the “provable” equivalence are guaranteed.

If there are “no defects” in the “standard one”, the “CONSERVATIVE guaranteed nonstandard” work would be really meaningful.

Now the problem is “nonstandard one” inherits all the fundamental defects disclosed by “infinite related paradoxes” from “standard one” since Zeno’s time 2500years ago------guaranteed by the “deep structural CONSERVATIVE” .

Theoretically and operationally, “nonstandard one” is exactly the same as those of “standard one” with suspended infinite related defects in nature.Simpler or not weights nothing here.

after importing geometry file (I tried with parasolid, iges, iam, stp), the structure is imported in 5 bodies separately with contacts. but it can be see that exists some separations that DM shows. then when I try to mesh, there is no conformal association between nodes and meshing.

Does anyone know an example of a Cauchy continuous function on a bounded subset of R?

By space of measurable functions, I mean L_0(m) where m is a non-atomic sigma-finite measure space.

I have one equation as shown in the following. I also discussed the integration region for 6.21. What I am interested to know about is the integration region of 6.22.

X is an Inner product space and U is a subset(space) of X. U^\perp is the space of all vertices orthogonal to all the elements of U. Give an example of a space U^\perp\perp is not a subset of U. Can it be from l^2?

If we have an n by n matrix called A. How do we know if there is an inverse matrix A^-1 such that the product A * A^-1 is the n by n identity matrix?

Assume the underlying measure is a probability measure. I think I've heard this is true but I could be wrong.

This looks "reasonable" from a geometrical point of view. If one of the conditions does not hold, there are obvious counterexamples.

In the paper by Hwa Kil Kim published in June 14, 2012. What is the meaning of (R^D) and the meaning of (J:(R^D)--->(R^D) is a matrix satisfying (

(J_v )_|_v) for all v in (R^D) ). The name of the paper is:Moreau-Yosida approximation and convergence of Hamiltonian systems on Wasserstein space, and it is on RG.

It's related to wavelet analysis and I want to know what problem arises in those usual basis.

We have the following two equations, where A_1, A_2, B_1 and B_2 are the coefficients that we are interested in to find the values. We need to calculate them so that, for example, p_1(R) and A_1 p_1(q) + B_1 p_2(q) represent the same function. We could do this by requiring that, at some point R_0, both sides of this equation have the same value and the same derivative.

Can we define a set of complete Jacobi SN orthonormal functions?

By using comparison tests, how we can explain the convergence and divergence of these two integrals?

The problem is that the scientific community places more value on absolute answers, but wicked problems are unanswerable because solutions involve tradeoffs and compromised objectives. There is no point of finality in the solution of wicked problems (Kunz & Rittel, 1972).

Refs:

Kunz, Werner, & Rittel, Horst. (1972). Information science: on the structure of its problems. Information Storage Research, 8, 95-98. doi: 10.1016/0020-0271(72)90011-3

We have a waveguide as shown in the attached figure.

At point A it has the singularity that gives the divergence of the integral related to this wave-guide. Now in order to make this integral convergence we have the following three ways.

1-Circumvent the singularity by a contour inside the wave-guide.

2-Circumvent the singularity by a contour outside the wave-guide.

3-Pass the singularity but use the Cauchy principal value.

These three methods can be used to get the convergent result of the integral related to this wave-guide.

Now I am confused about these three methods, I know Cauchy Principal value, but I have less information about the other method. One more thing that I wanted to know is that method one is physical realistic, that is why it is used by them, but I want some argument as to why the other two are not used.

In the attached file, there are two inequalities. When will these two different inequalities be true for real value of w?

I need information about Pridmore-Brown equation. How one can solve such equations?