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# Real and Complex Analysis - Science topic

For discussion about the analytic properties of real and complex sequences and functions.
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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
Hi In what area was the issue raised? Euclidean space, Hilbert space, Banach space?
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The question details are contained in the attached pdf file.
W.H. Young, “A note on monotone functions,” The Quarterly Journal of Pure and Applied Mathematics (Oxford Ser.) 41 (1910), 79–87.
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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 plane
F(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, last 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
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)
Dear Professor Mateljevic,
When you say that the unit disk is a real analytic bijection to R2, and has a bijective analytic function rto R^2, does it have one ; is it b-ijective to the simplex; are all pts in (p1, p2, p3)  in IM(F) of the unit disk function.  Do any go missing. I presumed that there were certain issues, with the surface area of the sphere
F: dom (unit disk) to . I was wondering if a quadratic function of the northern surface of the unit sphere, would be bijective as defined by F:[0, 2pi times [0, pi]in R^2\to < [p1=sinpi* cosphi]^2, p2=[sin pi * sin phi]^2, p3=cos^2phi> in R^3where  p1+p2+p3=1, and should be non-negative due to the squares.
Whilst the triangular functions appear to be the other way around and maximize surface area
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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
If you plot is assigned to G, then type FullForm[G]. Pay attention to the head.If you see Graphics[stuff], then you can use the Show command to combine plots, Its you see Graph[stuff], it might still be possible, but is harder. If you code is not too long, I recommend attaching it to your question.
• asked a question related to Real and Complex Analysis
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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
t appears that the third equation is redundant and I presuming any further equation where they sum to one (any given four, any given five, which sum to one is derivable from the first two) at least given that the first two entail some form of cauchy additivity. Is that correct
e
the above equation for any (4)given four points in the domain which sum to one
Does this generalize to Cauchy equation over the unit triangle. Ithough that one had to show that it held for any arbitary number of components not just 2 and three (although it appeas that the equations just generalize
F:[0,1] to [0,1]
F(1)=1 F(0)=1
F strictly monotonic increasing
(1)ie F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1
(2)F homomorphic(2) x+y+z if and only F(x)+F(y)+F(m)= 1
(3)x+y+z=2 if and only if F(x)+F(y)+F(z)=2, and th
(4) x+y+m+z=1 if and only F(x)+F(y)+F(z)+F(M)=1
Are the latter two redundant, ie do the first two generalize to any arbitrary factor; as the first two appears to entail cauchy equations of the unit trial
F(x+y+M)+F(Z)=1 F(Z)+F(x+y)+F((M)=1
so F(X+y+M)=F(x+y)+F(M)
but also x+y+z+m=1 entails
((x+y)+(1-x+y)) entails
F(x)+F(Y)+F(1-x+y) by (1) and (2)
F(x+y)+F(1-x-y)=1, so  by cancelling out common terms F(X)+F(Y)=F(x+Y)
and so
F(x+y+M)=F(x+y)+F(M)=F(X)+F(y)+F(M)
z+ (x+y+M)=1 entials F(z)+F(x+y+M)=1 so F(x+y+m)=1-F(z)
so 1-F(Z)=F(X+Y)+M)=F(X)+F(Y)+F(M), and so F(X)+F(Y)+F(Z)+F(M)=1 x+y+z+m=1, appears in addition to a form of cauchy equation s
These appear to entail those first two fixed pints
These appear to entail F(x+y)=F(x)+F(y), over x,y,(x +y)\in in [0,1],;
I thought one required that any for any n=2.........infinity which sum to one. But I presume this is entailed by these first both collectively working, as any given four example which sum to one can always be express as (x+y) +z+m = 1F(x+y)+F(m)+F(z)=1;
F(x+y+M)+F(Z)=1 so F(x+y+M)=F(x+y)+F(M) so -F(x+y)=-F(X+y+
F(M)+F(Z)=1-F(x+y)
F(x+y+m,)+F(Z)=1,
F(x+y+M)=F(x+y)+F(M)
yet 1-F(1-(X+y+M)=F(Z) as  so 1-F(1-x+
x+y+M+z=1 z=1-(x+y+M), then F((x+y+M))+F(z)=1, F(z)=1-F(x+y+z) so as z=1-x-m-y F(z)=F(1-(x+y-+M)=1-F(x+y+M)=F(z)
so 1-F(x+y+M)=1-F(x+y)-F(M), F(1-x+y+=F(M_M= ;1-F(x+y)=F(M)+F(Z)
(2) appears to just say this ie F(1-x-y)=1-F(x)-F(y)
which given 2 entails F(0)=0, F(1)=1; ie (1) specifies F(0.5)=0.5, and F(0)+F(1)=1 and F(0.5)+F(0.5)=1
(2) entails that 0 +0.5+0.5=1 so that F(0.5)+F(0.5)+F(0)=1,
so that 2 times 0.5+F(0)=1; 1+F(0)=1,which gives, F(0)=0
which by subsitution into into one gives F(1)=
and that (2)  F(1/3)=1/3 however when one has both
Moreover if(A) (x+y) +z =1 then (B)x+y+z =1 so clearly by (1) on A, (2) on (B)
F(x+y)+F(z)=1 and F(x)+F(y)+F(z)=1,
F(x+y)+F(z)=F(x)+F(y)+F(z) but F(z) is common
so \forall (x)(y)\in [0,1] so long as x+y is in [0,1]F(x+y)=F(x)+F(y)
where F(1)=1 and F non negative and strictly monotone increasing. Is this a pair wise result. It appears to generalize to F(x)=x for all rationals as far i can see
ie F(10)=2F(20), as 10+10+80=1 so F(10)+F(10)+F(80)=1, so 2F(10)+F(80)=1 and by the first equation 20 +80=1 so F(20)+F(80)=1 so that
F(20)=2F(10)
and F(30)=F(10)+F(20) use 10+20+70=1=F(10)+F(20)+F(70)=1 and F(30)+F(70)=1 and one can 10 F(10)=1=0.1 and one can seeiming building this standard sequence, It appears to entail F(x)=x for all rations even x>0..5 and F(x+y) generalizes to arbitarily many components if x+y in [0,1[
Or is just a cauchy function on the restricted interval with with strictly monononitc and positive, on [0,1] to [0,1] then F(x)=x
forall(x,y)
They appear t
and otherwise x+y+z>1 then F(x)+F(y)+F(z)>1 and x+y+z<1 if and only if F(x)+F(z)+F(m)>1
x+y =1 iff F(x)+F(y)=1
where otherwise x+y>1 iff F(x+y)>1, and x+y<1 if and only F(x)+F(y)<1
x+y+z=1 if and only if F(x)+F(y)+F(z)=1
x+y+z>1 if and only F(x)+F(y)+F(z)>1; x+y+z+m<1 iff and only F(x)+F(y)+F(z)+F(M)<1
x+y+z+m iff F(x)+F(y)+F(z)+F(m)=1
otherwise x+y+m +z>1 if and only if F(x)+F(y)+F(M)>1
x+y+z=2 if and only F(x)+F(y)+F(z)=2
where x+y+z>2 if and only if F(x)+F(y)+F(z)>2
x+y+z<2 F(x)+F(z)+F(m)<2
where other x+y+z>z
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Is the series in the picture convergent? If it is convergent, what is the sum of the series?
Dear Joachim,
you've right. Just ignore the first part of my answer. May be the double series calculation will lead to some result.
Rgrds,
Tibor
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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
By "additive function" do you mean function on (-\infty, + \infty) that satisfies equation f(x+y) = f(x) + f(y)? Then all such continuous functions are of the form f(x) = cx.
Since in your question you are speaking about the derivative, then you must deal only with continuous functions. So, f(x) = cx and you cannot get other examples for the identity (f o f ')(x)= (f 'o f )(x) with additive functions.
There are only two functions of the form f(x) = cx that satisfy the second identity
(f o f) (x)=f (x), namely f(x) = x and f(x) = -x.
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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)
Something algebraic, implicitly talking about ordering:
"for every nonzero number x, x or -x is a square but not both."
This holds in R (it is essentially an axiom of real closed fields) but not in Q or C (x=2 is a counterexample for both). Now you can take the logical negation.
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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.
I am not sure I have seen prof. Nash's work. There were several such attempts in the past from Neo-classical (hydrodynamic) ones and those based on Nelson's original stochastic mechanics. Check for instance Ord's work http://www.math.ryerson.ca/~gord/research.html
Italian "emergent QM" school  http://www.emqm15.org (related with original T' Hooft's attempts for a semi-discrete interpretation http://arxiv.org/abs/1405.1548)
The strangest of all -and probably deeper- is I think the Geneva-Brussels school originating in Diederik Aert's work https://en.wikipedia.org/wiki/Hidden-measurements_interpretation https://en.wikipedia.org/wiki/Diederik_Aerts
True problem behind all this: where do "hidden" variables reside? To make matters worse, recent findings in optics by Christensen (Illinois) showed that not even the good old Bell inequalities are sufficient to fully characterize non-locality.
Possible hint: the so called "entanglement' and non-locality appear now to be two totally disproportionate if disparate qualities. For the abstract logician this could beg the question of how the heck does nature manages to fill a continuum with a set of countable combinatoric structures without an additional stabilizing axiom/"external" constraint? It seems more and more like asking from a field theory to behave as an "automaton" running a "program". It is simply impossible without a very delicate imposition of a rather peculiar initial/boundary condition at the least. But that would again bring about the awkward question on "externality". Who/ what/ how would ever be able for such a delicate fine tuning exactly at the BB event? Unless there is no such "exact' event that is!
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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)}?
Denote p(m,n) the number of paths from (1,1) to (m,n).
If the region was x>0, y>0, p(m,n) would copy the coefficients in the Pascal's triangle.
I don't know if an explicit formula already exists for the cone x>0, y>0 and 2x≥y≥x/2 (maybe you have one?).
Anyway, a recursive formula is possible:
1) p(0,0)=0, p(1,2)=p(2,1)=1
2) p(m,n)=p(m-1,n)+p(m,n-1) if m≤2n-2 and n≤2m-2
3) p(2n,n)=p(2n-1,n)=p(2n-2,n) and p(m,2m)=p(m,2m-1)=p(m,2m-2)
(see the attached figure)
Thanks to the symmetry of the region, 1),2),3) can be simplified.
A better simplification is obtained if we put p(m,n)=0 for (m,n) outside the cone:
3) may be omitted, 2) holds for any (m,n) in the cone.
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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.
In the meantime I have found a combinatorial proof in the literature: Naiomi T. Cameron and Asamoah Nkwanta, On some (pseudo) involutions in the Riordan group,  J. Integer Sequences   8 (2005), Article 05.3.7, proof of identity 1. https://www.researchgate.net/profile/Asamoah_Nkwanta/publications?sorting=newest
Instead of NSWE-paths they use bicolored Motzkin paths, but their proof can easily be translated to the situation of NSWE-paths. So my question has been answered.
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Let f1:ℕ⭢ℕ be the identity, that is, ∀n∈ℕ: f1(n)=n. For each n in ℕ, let fn:ℕ⭢ℕ be the map defined recursively as follows.
fn(m) = fn-1(m+1) if fn-1(m) = 2
fn(m+1) = fn-1(m) = 2 if fn-1(m) = 2
fn(m) = fn-1(m) otherwise.
Since every map fn is defined recursively with a transposition in the image
of fn-1, the map fn is again a bijection that satisfies the following properties.
1) ∀n ≤ m: fm(n)≠ 2.
2) ∀n ≤ m the map fn: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.
Taneli Huuskonen
I think that your answer is true in the scope of potential infinity, but my question requires the actual infinity axiom. Recall that, according to  actual infinity concept,  every process can be ended. Thus, actual infinity is not synonimous of endless. To avoid any wrong interpretation of my words, I write both axioms below.
Axiom 1. Actual infinity does not exist. What we call infinite is only the endless possibility of creating new objects no matter how many exist already.
Henri Pointcaré (1854-1912).
Axiom 2. Every mathematical construction can form an actual and completed totality.
G. Cantor (1845-1918)
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Any idea or comment is appreciated!
Dear all,
That is true dear Octav.
The sum can be computed for any number s > 0 where the question is put when s is 1/2, from a transform definition. The story of it is very important as the discrete Laplace like transform is used to generate beautiful sequences of difference equations including the Fibonacci sequence.
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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.
The formula (1.2) for completely monotonic function on an interval is not valid. For example, f(t) = e-t - e-1 is completely monotonic on [0, 1] with f(1) = 0, and the latter property is impossible for a non-zero function satisfying the integral representation (1.2). In order to have (1.2) you need a function that has completely monotonic extension on the semi-axes (0, \infty). There is really a lot of information about absolutely monotonic / completely monotonic functions in the original Bernstein's paper
S. N. Bernstein (1928). "Sur les fonctions absolument monotones". Acta Mathematica 52: 1–66. doi:10.1007/BF02592679
Maybe some of this information can be of use for you. S. N. Bernstein is one of the most famous mathematicians from my University, and it is always a pleasure for me to mention his works: they contain much more material than the textbooks citing his results.
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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.
Dear Mikael,
Your system admits kink-solutions (travelling waves which connect lower energy values to higher). See the attachment, which was produced in Mathematica.
Cheers,
Sotiris
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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
Dear Prof Abram
attached figure is attached for clarification; don't hesitate if more explain is necessary
Hamed
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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”.
Dear Geng Ouyang,
one possible answer is the following:
What is meant by 1 + 1/2 + 1/3 + 1/4 + ...
It is the limes of the sequence
1
1+ 1/2
1 + 1/2 + 1/3
...
Each of the elements of this sequence is a finite sum. And therefore in every pair of brackets you can place there can only be finitely many summands. Perhaps this is not an answer to the question you intended, but I was not yet able to read the papers I found on your profile.
Best regards
Johann Hartl
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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?
real eigenvalues exist
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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)
Dear Feng Qi, you can find it on:
Best regards, Viera
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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?
In higher codimension, we cannot define the regular infinite one type with  tangent vector field and the levy form directly like in C^2. But it is in some sense possible : see my paper with J.F Barraud  (at least for define finite type):
Barraud, Jean-François; Mazzilli, Emmanuel
Regular type of real hyper-surfaces in (almost) complex manifolds. (English) Zbl 1082.32017
Math. Z. 248, No. 4, 757-772 (2004). This is done for regular one type. Now for the D'angelo type : singular type, it is possible see the paper :
Barraud, Jean-François; Mazzilli, Emmanuel
Lie brackets and singular type of real hypersurfaces. (English) Zbl 1154.32009
Math. Z. 261, No. 1, 143-147 (2009). I do not know if it is exactly a think like you want but perhaps it is helpfull for you. regards E/.Mazzilli.
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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*}
Before computing the limit one has to fix the precise meaning of the expression. The functions $g(z) = z^s$ and $h(z) = \ln(z)$ are multi-valued, so the question will be correctly posed only after selecting appropriate branches of all expressions that  include these functions. Outside of this I see no difficulties.
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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.
Hi colleague Feng Qi
I just had the intention to help you, but the Professor Yao Liang Chung was faster than me.
Best regards and wishes,
Mirjana Vukovic
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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 !?
JYTA
Hello Jean-Yves Tallet
I think that it is a problem of unit here. You can devide any dash into any number of smaller dashes and points relating any two concecutive of them. On the other hand if we consider dashes and point as unit, the lenght of a dash makes the difference.
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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.
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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?
Actually the sup norm is not defined on C[0,\infty) since this space contains unbounded functions. If you consider the space C^b[0,\infty) of bounded continuous functions (or the space suggested yb Luiz) , then the sup norm is well defined, and this space is complete (basically since the uniform limit of continuous functions is continuous).
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For example, Circle is not a function. Can any one explain with this inverse and implicit function theorem?
1. The  unit  circle   x^2 + y^2=1   is not a function,  but locally  is  a graph of a function .Set  f(x,y)=x^2 + y^2-1.  Then the partial derivatives  are   f'_x=2x    and   f'-y=2y.Hence  if    (x_0,y_0) is a point on the unit circle    and for example  y_0     is not 0,  by the implicit function theorem  locally  there is  a function y=g(x)   which is solution of the equation    f(x,y)=0.Denote   by   g_0(x) the   square root of 1-x^2, for x between -1 and 1. Actually if y_0  >0 , y=g0(x)  is solution    and    if y_0 <0 , y=-g0(x) is solution.  In a similar way we can discuss the case if   x0  is not 0.
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Hello;
I have ended up with a quadratic function of X in the form of:
f(X)= XTAX-XTBX+CTX 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
i agree with Octav;
f is convex iff A-B is positive semidefinite,
f is strictly convex iff A-B is positive definite,
f is  concave iff A-B is negative semidefinite,
f is strictly concave iff A-B is negative definite.
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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.
Juan Esteban,
My pdf-file describes a construction of the full system of real numbers. Of course you have recognized some traditional features in it. At the same time, it may urge you into an additional sharpening of your criteria. Constructing |R on the fly as in
|R := { x : x in X and P(x) }    (with some available set X and some property P)
is not to be expected. Instead, it may involve intermediate constructions and logical formulae. Where lies the watershed between acceptable formulae/constructions and the ones you find in my file?
Nevertheless, your question hits a point. Both Dedekind cuts and Cauchy sequences connect with topology (order topology and metric topology) and both do assist  in constructing the reals. The real field is also characterized as the only complete and totally ordered field. A structure with such references is expected not to be constructed in a simple way.
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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.
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.
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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I need these measurement for the diversity analysis of population of vectors
The (vector) norm not the suitable measure in this problem.
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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?
Dear Miodrag:
Please allow me to add just a very simple remark (which does not contribute to your question). There is a more intuitive way to Cardano's formula: Why should someone put $u+v=t$ and assume $uv=p$? Here is the idea: We want to solve the cubic equation
$$t^3 = 3pt + 2q \eqno(1)$$
which is difficult. But other cubic equations are apparently easy:
$$(t-u)^3 = v^3 \eqno(2)$$
is solved by $(*)$ $t - u = v$ or $t = u+v$. Now the idea is to transform the easy equation (2) into the difficult'' form (1). In fact,
\begin{eqnarray}
(t-u)^3 &=& t^3 - 3ut^2 + 3u^2t - u^3 \cr
&=& t^3 - 3u(t-u)t -u^3 \cr
&=& t^3 - 3uvt - u^3 \nonumber
\end{eqnarray}
using $(*)$. Thus (2) is of the form (1) with
$$p = uv,\ \ \ 2q = v^3 + u^3.$$
Solving these equations for $u^3$, $v^3$ gives Cardano's formula.
Best regards
Jost
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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 ?
U r great dear profesor El Naschie.. thanks for your kind explanations
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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?
g(w)=tan(w)/w
I guess you first note that g has a removable singularity at 0, so becomes analytic on a neighbourhood of zero. Also noting that g is an even function, the coefficients of odd powers of w are all zero.
Since g(-w)=g(w) away from poles of g, g(\sqrt z) is well-defined except at squares of poles of g, and the discussions above show that g(\sqrt z) is analytic away from these points.
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How to construct the system of equations by susbtituting the assumed series solution in the nonlinear partial differential equation? Kindly find the attachment.
Dear Saravanan
First at all, your problem seems to be so easy, if you want to solve it Numerically. i.e. if all constants are given and the function fi(zeta) is explicit. then you can catch the values of the unknown values for a0, a1, d and b.
Furthermore, If you want an explicit form for a1, you can let it to be a polynomial(as example) of suitable degree with unknown constants to be determined.
Is my understanding for your problem is suitable?
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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.
Hellow everybody,
What is meant by relative interior ? If you mean the interior with respect to its own topology, it is of course non empty since any topological space is open in itself. If you mean the interior of a subset  A of a subspace F of a Banach space E with respect to the relative topology on F, then Thomas Korimort has given the answer.
Remember that : in any topological vector space, the only subspace which has a non empty interior is E itself, because the 0-neighbourhoods are absorbant..
Also a subset of a topological space can fail to be neither open nor closed.
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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?
Pure math is always ahead of the game because we pave the roads and identify the dead ends for applied sciences. Sometimes we pave the roads so far ahead of technology that our  results get lost or forgotten. I'm sure there are answers to many applied questions hidden away in decades old pure math publications but the applied scientists don't have the vocabulary to even do a useful search for this information. There needs to be more communication from both sides.
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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.
Actually the matrix is indefinite and not triangular
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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.
The Cauchy-Riemann equations themselves do not imply analyticity! Indeed,the function given by
f(z)=exp(-z-4) when z not equal to 0 and f(z)=0 when z=0
is readily seen to satisfy Cauchy-Riemann equations everywhere but falls to be analytic at origin.
A stronger result than Looman–Menchoff theorem which asserts the Analyticity of complex valued function states that
If
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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.
Unfortunately, if you are interested in continuous solutions of your functional equation F(x+y)=F(x)F(y), then the exponential functions exp(a x) are the only answer, see the links below for details (note that if F satisfies the above equation, f=ln(F) satisfies the Cauchy functional equation f(x+y)=f(x)+f(y)):
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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?
Hi Ehsan,
Thanks for the link to the comsol blog on the use of symmetry which I read with interest. This actually confirms what I thought might be happening. From what is described in the blog, the multi-turn coil feature in comsol appears to use what is usually called the 'current sheet' approximation. This replaces the current flowing through the multi-turn coil with a uniform current sheet that flows around the coil perimeter. This is an approximation which, once in place, permits the use of symmetry as a means to reduce the model size.
However, this approximation is only good for tightly wound coils and only for estimating the coil's inductance. The reason is that the inductance is primarily determined by the magnetic flux external to the coil (giving rise to the external inductance) which depends on the total current flowing in the wire. Therefore, using a uniform current sheet doesn't alter the result for the inductance too much from that obtained using the actual coil geometry.
However, the internal inductance (that due to the magnetic flux inside the wire) and - most importantly for what you are looking at - the coil's resistance will not be accurately determined by the current sheet approximation.
To get an accurate simulation of the resistance, the spatial distribution of current in the actual wire geometry must be detemined. In an isolated single turn of circular wire, this distribution of current will be cylindrically symmetric and follow the same kind of distibution as the electric field, that is, it will be concentrated near the surface of the wire and decay exponentially into the wire - the familiar 'skin effect'.
When additional turns of wire are added and brought close together (as in a tightly wound coil) the current distribution in the wire will be altered due to the 'proximity effect' which results from the Lorentz Force on the electrons in the wire due to the magnetic field in neighbouring parts of the coil. The net result is that the current gets concentrated in the parts of the wire cross section that are nearest to the other turns. The current is then no longer cylindrically symmetric within each turn of wire. This proximity effect therefore increases the resistance over and above that due to the skin effect since the current now flows in a reduced cross sectional area.
The proximity effect isn't modelled by the current sheet approximation. Therefore, one cannot utilise symmetry in the comsol multi-turn coil model for calculating resistance (or internal inductance).  A real multi-turn coil doesn't have symmetry with regard to the actual current distribution in the wire.
Regards,
Ray
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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.
Hi Akira, I think a firm theoretical bridge between discrete systems and continuous approximations to them may not be possible any time soon.  Again, if a physicist can find an excuse to treat a large discrete system (e.g., a container of gas) as a continuous entity, she will.  "Anything that gets answers."   As for supporting/refuting physical theories, only degrees of confidence are required to underpin belief.  I don't believe empirical data near as much as I believe a mathematical argument;  and the more complicated the argument, the more difficult it is for me to believe.
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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.
Hello,
Have you tried performing Boolean operations? In DesignModeler it is: create > booleans >  unite.
Hope it helps.
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Does anyone know an example of a Cauchy continuous function on a bounded subset of R?
Right! By the same arguement scetched above you hae the following theorem: let f: X -> Y be Cauchy continuous and asume that Y is a complete metric space as well as X is a precompact metric space (as for example a bounded subset of R).. Then if f is Cauchy continuous it is uniformly continuous. For the completion of X is compact and since f ix Cauchy continuous it possesses a unique extension to a continous function from the completion of X to Y. But the completion of X is compact, so the extensioin and a fortiory f itself is uniformly continuous.
Manfred
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By space of measurable functions, I mean L_0(m) where m is a non-atomic sigma-finite measure space.
In my opinion, you can start with the book by Martin Vaeth "Ideal Spaces" LNM 1664. But, of course, you should think about classes of measurable functions...
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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.
Integration region is the one kinds of metric distance.so it may be the error(comparison) between two things(images, cars,samples,...)
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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 U is a subspace, then U^{\perp\perp} is its closure (by the Hahn-Banach theorem), so you may take any subspace that is not closed as an example. For instance, if X=\ell^2, then you may take U to be the space of all finitely supported sequences.
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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?
Yes Jose Vegas is right. Infact, what you should have if det(A) is non-zero A.Inv(A) = Inv(A).A = I the identity matrix. However, for large value of n it is difficult to find det(A). If you apply, Gauss elimination method, then during elimintion process t some point your diagonal element becomes zero can not be made non-zero by elementary row exchange then the matrix is singular and the inverse does not exist.
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Union of sets
There is a condition called 'local finiteness': suppose that every point has a neighborhood that intersects only finitely many of those closed sets. In that case, the union of the closed sets will be closed.
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Assume the underlying measure is a probability measure. I think I've heard this is true but I could be wrong.
Dear Geoff Diestel
I am not used to play with weird measured spaces, but on the part of the set where the measure has no atoms, one may use a theorem of Liapounov (not the famous one who worked on ODE) for which Zvi Artstein has given a quite simple proof in the mid 70s (Look for the Extreme Points), and construct a sequence of characteristic functions converging weakly * in L^{\infty} to 1/2 (hence a sequence like the Rademacher one mentioned by Anton Schep).
Luc TARTAR, mathematician
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.
Perhaps the simplest illustration of the differences between the integrals of Riemann and Lebesgue is the following. Imagine that you have a lot of coins of different denominations and you need to count how much money you have. Riemann integral answers this question as follows. He consistently adds dignity of another coin to the amount already recorded. Lebesgue integral first splits the set of all coins on the sets of coins of the same denomination. Then calculates the cost of each of the resulting subsets. That is quite simply. And then finds the sum of the resulting values.
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This looks "reasonable" from a geometrical point of view. If one of the conditions does not hold, there are obvious counterexamples.
If $\Delta=\sum_i D_{ii}u$ I think you mean non-negative Laplacian, since any concave function has a non-positive Laplacian. However even with non-negative the answer is not. Take u radial in the unit ball in 2d such that $ru_r =\int_0^r f$ with positive f. Then $\Delta u=u_{rr}+u_r/r=f/r \ge 0$. However $u_{rr}=f/r-(1/r^2) \int_0^r f$ can be negative somewhere (take f with compact support in [a,b], then u_{rr} <0 for r >b) and u is not convex.
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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 looks like D is a natural number and the condition on J means that v and Jv are orthogonal for each vector v, i.e. their scalar product is 0.
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It's related to wavelet analysis and I want to know what problem arises in those usual basis.
Hamel bases, well, they exist. And are not really that useful. Except to demonstrate that continuity is, even in very weak forms, a useful condition.
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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.
Perhaps, you should define r_c^{+} and r_c^{-}. :)
If you have a differential equation with a set of fundamental solutions, then, of course, the Wronskian is a useful tool.
The constants arising in expressing solutions as linear combinations of fundamental solutions can quite often be obtained from normalization and/or boundary conditions.
Thus, it could well be that your sought relations for A_i,B_i can be obtained from some pointwise information.
However, you should be careful here: It seems that you are using your ODE in terms of two different coordinates q and R, i.e. the ODE has different coefficient functions upon transforming between q and R, which means that you cannot use the above argument so easily, having essentially two different ODEs and two different fundamental solution sets, one as a function of q and one as a function of R. I only can guess that your four functions are meant to be fundamental solutions.
Therefore, let me add some further remarks (hopefully, you will not misunderstand me :): no harm is intended, it is only a recommendation)
* While asking for help with/discussion of mathematical problems, it does not hurt to provide at least pointers to such properties of the functions under consideration ("is solution of the ODE given by ...") Please note also: The "pridmore-brown equation" is not uniquely defined, since there are also two-dimensional versions of it, and there is always the possibility of using a variety of coordinates and/or scaling factors etc.. Also, this ODE is not known to numerical analysts, say.
* It would also be very helpful, if you would use a different notation for the function p_1(q) and p_1(R), e.g. P_1(R) instead of p_1(R). The functions P_1: x -> P_1(x) and p_1: x -> p_1(x) differ by a factor x^3 as is plain from the series expansion. Mutatis mutandis this applies also to the functions P_2 and p_2. Often, a mathematically sound notation helps a lot to clarify the problems.
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Can we define a set of complete Jacobi SN orthonormal functions?
Shouldn't k be equal or less than 1?
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By using comparison tests, how we can explain the convergence and divergence of these two integrals?
First one obviously diverges as far as the function under the integral is equivalent to \frac{constant}/{1 - w} as w \to 1-0.
The second one converges iff 0 \neq [\alpha_1; \alpha_2]
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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
Soft systems / problem structuring methods / soft OR deal with messy problems. For a little primer, have a look at
Fran Ackermann, Problem structuring methods ‘in the Dock’: Arguing the case for Soft OR, European Journal of Operational Research, Volume 219, Issue 3, 16 June 2012, Pages 652-658
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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.
This integral is not just numerically, but mathematically divergent because of the singularity as $w \to 1$! Are you sure the integral is correct or do you want to integrate to $1 - \alpha^2$ for example? The $4l$ is a constant and integrates trivially to $4l(1 - \alpha^2)$ so it looks suspect, or I just don't understand the notation and $l$ depends on $w$ in some implicit way.
The good news is that if you integrate $w$ from $\alpha^2$ to $1- \epsilon$, and let $\epsilon \to 0$, then it diverges like $\int_\alpha^2^ 1 dw \sqrt{(1 - \alpha^2)(1 - 1/\alpha^2)}/(1-w) \approx \sqrt{(1 - \alpha^2)(1 - 1/\alpha^2)} \log(1/\epsilon))$.
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In the attached file, there are two inequalities. When will these two different inequalities be true for real value of w?
If I am not too wrong, in problem 1, there are three distinct cases to consider:
1) w < 0
2) w > 1/a²
3) 0 < w < a²
The sign of the denominator is < 0 for case (1) and (2) and > 0 for case (3). From there, we can simplify further the problem to N > 1 or N < 1, where N is the numerator.
Moreover, since (w-a²)(w-1/a²) = w² - (a² + 1/a²)w + 1 then equations such as N > 1 or N < 1 become straightforward to solve.
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