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beidouxing

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the original...

beidouxing 10
January 31 2015
0 0

@Axel Vogt 

eqn1 := (1/8)*w*(1/((x+y+z)^2+1)^(3/2)+1/((x+y)^2+1)^(3/2)+1/((x+z)^2+1)^(3/2)+1/((y+z)^2+1)^(3/2)+1/(x^2+1)^(3/2)+1/(y^2+1)^(3/2)+1/(z^2+1)^(3/2)+1)+x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1):

eqn2 := (1/8)*w*(1/((x+y+z)^2+1)^(3/2)+1/((x+y)^2+1)^(3/2)+1/((x+z)^2+1)^(3/2)+1/((y+z)^2+1)^(3/2)+1/(x^2+1)^(3/2)+1/(y^2+1)^(3/2)+1/(z^2+1)^(3/2)+1)+y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1):

eqn3 := (1/8)*w*(1/((x+y+z)^2+1)^(3/2)+1/((x+y)^2+1)^(3/2)+1/((x+z)^2+1)^(3/2)+1/((y+z)^2+1)^(3/2)+1/(x^2+1)^(3/2)+1/(y^2+1)^(3/2)+1/(z^2+1)^(3/2)+1)+z/((x+y+z)^2+1)+z/((x+z)^2+1)+z/((y+z)^2+1)+z/(z^2+1):

eqn1 = eqn2 and eqn2 = eqn3:

I want to proof x=y forever,if I choose eqn1=eqn2 with their first order condition and not ues the equation eqn3.


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question...

beidouxing 10
January 18 2015
0 0

@Axel Vogt  although i find f(x,y)=f(y,x),it can't implie that x==y.And i want to prove x==y independ on z

@Axel Vogt I don't realize the firs...

beidouxing 10
January 18 2015
0 0

@Axel Vogt I don't realize the first you have said.Because i can't find the solutions of  eqn1 and eqn2 ,so i substitute the x,y,z  for tan(a),tan(b),tan(r).And  want to have a try. And I will try the others what you say .Thanks a lot

@Kitonum        ...

beidouxing 10
January 11 2015
0 0

@Kitonum 

 

 

 

eqn1 := (3*y/(y^2+1)^(5/2)+(3*(x+y))/(1+(x+y)^2)^(5/2)+(3*(y+z))/(1+(y+z)^2)^(5/2)+(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-(3*(x+y))/(1+(x+y)^2)^(5/2)-(3*(x+z))/(1+(x+z)^2)^(5/2)-(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2)):

eqn2 := x/(x^2+1)+x/(1+(x+y)^2)+x/(1+(x+z)^2)+x/(1+(x+y+z)^2)-y/(y^2+1)-y/(1+(x+y)^2)-y/(1+(y+z)^2)-y/(1+(x+y+z)^2):

This is my system,and i want to solve the equations .I expect the result x==y orx==z,y==z.


Download 0112.mw

I want the exact analytical solution.can you help me?thanks a lot.

the equation...

beidouxing 10
January 10 2015
0 0


eqn1 := (3*y/(y^2+1)^(5/2)+(3*(x+y))/(1+(x+y)^2)^(5/2)+(3*(y+z))/(1+(y+z)^2)^(5/2)+(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-(3*(x+y))/(1+(x+y)^2)^(5/2)-(3*(x+z))/(1+(x+z)^2)^(5/2)-(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2));

(3*y/(y^2+1)^(5/2)+3*(x+y)/(1+(x+y)^2)^(5/2)+3*(y+z)/(1+(y+z)^2)^(5/2)+3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-3*(x+y)/(1+(x+y)^2)^(5/2)-3*(x+z)/(1+(x+z)^2)^(5/2)-3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))

 (1)

simplify(subs(x = y, eqn1));

0

 (2)

So x-y is a solution for the eqn1=0,so the eqn1 can be divided by the x-y

eqn1/(x-y);

((3*y/(y^2+1)^(5/2)+3*(x+y)/(1+(x+y)^2)^(5/2)+3*(y+z)/(1+(y+z)^2)^(5/2)+3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-3*(x+y)/(1+(x+y)^2)^(5/2)-3*(x+z)/(1+(x+z)^2)^(5/2)-3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2)))/(x-y)

 (3)

``


Download 0111.mw

eqn1 := (3*y/(y^2+1)^(5/2)+(3*(x+y))/(1+(x+y)^2)^(5/2)+(3*(y+z))/(1+(y+z)^2)^(5/2)+(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-(3*(x+y))/(1+(x+y)^2)^(5/2)-(3*(x+z))/(1+(x+z)^2)^(5/2)-(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2));

(3*y/(y^2+1)^(5/2)+3*(x+y)/(1+(x+y)^2)^(5/2)+3*(y+z)/(1+(y+z)^2)^(5/2)+3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-3*(x+y)/(1+(x+y)^2)^(5/2)-3*(x+z)/(1+(x+z)^2)^(5/2)-3*(x+y+z)/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))

 (1)

simplify(subs(x = y, eqn1));

0

 (2)

So x-y is a solution for the eqn1=0,so the eqn1 can be divided by the x-y

eqn1/(x-y);

 

  

``


Download 0111.mw

learning...

beidouxing 10
December 18 2014
0 0

@acer 

I'm a newer of maple,just know a little of it,would you give some suggestions of learning it.thanks a lot

goal...

beidouxing 10
December 18 2014
0 0

@acer 

My purpose is that,the c[1]and c[2] have the same maximize,when x,y,z  satisfy the first order conditon, In fact ,my goal is that x=z,y=z under the condition what i have said above.

the file i upload is the method of my problem,i have lost in this problem for a long time.

I suspect what i say is explicit?

no other condition...

beidouxing 10
December 18 2014
0 0

@acer 

I use the signal{},just for a simplify a good expression after  the eliminate,the{}substitute for().And there is no other conditions of x,y,z.there is no methods for my problem???

@acer  this is my purpose:...

beidouxing 10
December 16 2014
0 0

@acer 


eqn1 := (1/8)*x/((x+y+z)^2+1)+(1/8)*x/((x+y)^2+1)+(1/8)*x/((x+z)^2+1)+(1/8)*x/(x^2+1)-(1/8)*y/((x+y+z)^2+1)-(1/8)*y/((x+y)^2+1)-(1/8)*y/((y+z)^2+1)-(1/8)*y/(y^2+1):

eqn2 := (1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*y+2*z)/(y^2+2*y*z+z^2+1)^(5/2)-3*y/(y^2+1)^(5/2))-(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*x+2*z)/(x^2+2*x*z+z^2+1)^(5/2)-3*x/(x^2+1)^(5/2)):

solve({eqn1, eqn2}, {x, y})this is my purpose:

``

It's my fault,this file is right??

Download problem.mw

@Markiyan Hirnyk  thank you ...

beidouxing 10
December 16 2014
0 0

@Markiyan Hirnyk 

thank you 

help DirectSearch:-SolveEquations...

beidouxing 10
December 16 2014
0 0

@Markiyan Hirnyk 

the maple 18 of me can't evaluate the command of DirectSearch:-SolveEquations,can you tell me the reason???thanks

@acer     ...

beidouxing 10
December 16 2014
0 0

@acer 

 

 

@acer  is it true??...

beidouxing 10
December 16 2014
0 0

@acer 

eqn2:=(1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*y+2*z)/(y^2+2*y*z+z^2+1)^(5/2)-3*y/(y^2+1)^(5/2))-(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*x+2*z)/(x^2+2*x*z+z^2+1)^(5/2)-3*x/(x^2+1)^(5/2));

eqn1 := (1/8)*x/((x+y+z)^2+1)+(1/8)*x/((x+y)^2+1)+(1/8)*x/((x+z)^2+1)+(1/8)*x/(x^2+1)-(1/8)*y/((x+y+z)^2+1)-(1/8)*y/((x+y)^2+1)-(1/8)*y/((y+z)^2+1)-(1/8)*y/(y^2+1)

is it true??

@acer eqn1 := (1/8)*x/((x+y+z)^2+1)+(1/8...

beidouxing 10
December 16 2014
0 0

@acer

eqn1 := (1/8)*x/((x+y+z)^2+1)+(1/8)*x/((x+y)^2+1)+(1/8)*x/((x+z)^2+1)+(1/8)*x/(x^2+1)-(1/8)*y/((x+y+z)^2+1)-(1/8)*y/((x+y)^2+1)-(1/8)*y/((y+z)^2+1)-(1/8)*y/(y^2+1);

eqn2:=(1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*y+2*z)/(y^2+2*y*z+z^2+1)^(5/2)-3*y/(y^2+1)^(5/2))-(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)(-(3/2)*(2*x+2*y+2*z)/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1)^(5/2)-(3/2)*(2*x+2*y)/(x^2+2*x*y+y^2+1)^(5/2)-(3/2)*(2*x+2*z)/(x^2+2*x*z+z^2+1)^(5/2)-3*x/(x^2+1)^(5/2));

There is no problem of the functions' 2D Math.Because I can't upload my flie,i will upload my file if i can.but from this two equations i want to prove {x=z,y=z}.could you help me ??and give me your programming.thanks

thanks a lot...

beidouxing 10
December 15 2014
0 0

@Markiyan Hirnyk 

thanks for your help,

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