Difference between using "vpasolve" in these 2 codes?

Question

Eman S am 7 Mai 2018

0
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Direkter Link zu dieser Frage

https://de.mathworks.com/matlabcentral/answers/399549-difference-between-using-vpasolve-in-these-2-codes

Bearbeitet: Eman S am 2 Feb. 2020

In the following code, I had defined f(x) using syms and to solve this equation, I used vpasolve(f).

syms  f(x)
alpha=5;
mio=0.6;
B=2;
if alpha==5
      if mio==0.6
            if B==2 
                bast=(x/0.5).^(alpha*(mio-1));
                mqam_part1=3*B*((sqrt(3)/2).^(alpha*mio));
                mqam_part2=((0.5*sqrt(3)).^(-alpha))+((1.5*sqrt(3)).^(-alpha));
                mqam_part3=3*(((x/0.5)).^(alpha*mio));
                mqam_part4=((sqrt(3)-(x/0.5)).^-alpha);
                mqam_part5=((2*sqrt(3))-(x/0.5)).^-alpha;
                mqam_part6=mqam_part4+mqam_part5;
                mqam_part7=2*((3-(x/0.5)).^-alpha);
                mqam_part8=6*(((x/0.5)).^(alpha*mio));
                mqam_part9=6*B*(2.^-alpha);
                eqn_LHS=bast/(mqam_part1+mqam_part2)+(mqam_part3*(mqam_part6+mqam_part7));
                eqn_RHS=B/((mqam_part8*mqam_part6)+mqam_part9);
                f(x)=eqn_LHS-eqn_RHS;
                sol_positive = vpasolve(f);
            end
     end
end

After running this code, it has the following output:

sol_positive =
                                       -0.088528537330827491688440727135052
                                        0.089168332029743739883884412606126
                                         0.33826394036632543530763273039833
                                          1.1599250419465014133554207319948
                                          1.2989629568831597716493684806851
                                          1.6240554937821142318780347702386
                                          4.1380104774956983606108108853334
36340277619978535323175586579368 - 0.17885283113851906156582968732142i
2992644818621369190151455557302 - 0.31440379416295109163597261773161i
4897230890912773483095146949368 - 1.3150774672581101958523122516833i
6282605813254473735848321984622 - 0.083684308295293467728903801421506i
6955614120462881985070978123707 - 0.34019756525211237400051800951627i
8366220103150973884193959600945 - 2.0075730328576518221908039773448i
96894287293815413579933338616546 - 1.0029719645739903595813767103315i
8366220103150973884193959600945 + 2.0075730328576518221908039773448i
2992644818621369190151455557302 + 0.31440379416295109163597261773161i
4897230890912773483095146949368 + 1.3150774672581101958523122516833i
36340277619978535323175586579368 + 0.17885283113851906156582968732142i
96894287293815413579933338616546 + 1.0029719645739903595813767103315i
6282605813254473735848321984622 + 0.083684308295293467728903801421506i
1475584718883716605696238712257 + 0.23111120566119037717218445710823i
6955614120462881985070978123707 + 0.34019756525211237400051800951627i
67755561933436425842885030639685 + 0.89740492763981823171766701235696i
35356088917929933509194226805741 + 0.46392937489127504001635787211895i
1475584718883716605696238712257 - 0.23111120566119037717218445710823i
35356088917929933509194226805741 - 0.46392937489127504001635787211895i
67755561933436425842885030639685 - 0.89740492763981823171766701235696i

But, in the following code,I had defined the parameters of the equation *B,x,mio using syms and to solve this equation, I used vpasolve(eqn1,x,[0 Inf]).

syms alpha mio B x
alpha=5;
mio=0.6;
B=2;
if alpha==5
      if mio==0.6
            if B==2 
                bast=(x/0.5).^(alpha*(mio-1));
                mqam_part1=3*B*((sqrt(3)/2).^(alpha*mio));
                mqam_part2=((0.5*sqrt(3)).^(-alpha))+((1.5*sqrt(3)).^(-alpha));
                mqam_part3=3*(((x/0.5)).^(alpha*mio));
                mqam_part4=((sqrt(3)-(x/0.5)).^-alpha);
                mqam_part5=((2*sqrt(3))-(x/0.5)).^-alpha;
                mqam_part6=mqam_part4+mqam_part5;
                mqam_part7=2*((3-(x/0.5)).^-alpha);
                mqam_part8=6*(((x/0.5)).^(alpha*mio));
                mqam_part9=6*B*(2.^-alpha);
                eqn_LHS=bast/(mqam_part1+mqam_part2)+(mqam_part3*(mqam_part6+mqam_part7));
                eqn_RHS=B/((mqam_part8*mqam_part6)+mqam_part9);
                eqn1=eqn_LHS==eqn_RHS;
                sol_positive = vpasolve(eqn1,x,[0 Inf]);
            end
     end
end

After running this code, it has the following output:

sol_positive =
089168332029743739883884412606126
33826394036632543530763273039833
1599250419465014133554207319948
2989629568831597716493684806851
6240554937821142318780347702386
1380104774956983606108108853334

So, my question is: what is the difference between using vpasolve in the 2 codes and how does it work to give these outputs??

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Image Analyst am 1 Feb. 2020

In MATLAB Online öffnen

Original question

What is the difference between using "vpasolve" in the following 2 codes?

In the following code, I had defined f(x) using syms and to solve this equation, I used vpasolve(f).

syms  f(x)
alpha=5;
mio=0.6;
B=2;
if alpha==5
      if mio==0.6
            if B==2 
                bast=(x/0.5).^(alpha*(mio-1));
                mqam_part1=3*B*((sqrt(3)/2).^(alpha*mio));
                mqam_part2=((0.5*sqrt(3)).^(-alpha))+((1.5*sqrt(3)).^(-alpha));
                mqam_part3=3*(((x/0.5)).^(alpha*mio));
                mqam_part4=((sqrt(3)-(x/0.5)).^-alpha);
                mqam_part5=((2*sqrt(3))-(x/0.5)).^-alpha;
                mqam_part6=mqam_part4+mqam_part5;
                mqam_part7=2*((3-(x/0.5)).^-alpha);
                mqam_part8=6*(((x/0.5)).^(alpha*mio));
                mqam_part9=6*B*(2.^-alpha);
                eqn_LHS=bast/(mqam_part1+mqam_part2)+(mqam_part3*(mqam_part6+mqam_part7));
                eqn_RHS=B/((mqam_part8*mqam_part6)+mqam_part9);
                f(x)=eqn_LHS-eqn_RHS;
                sol_positive = vpasolve(f);
            end
     end
end

After running this code, it has the following output:

sol_positive =
                                       -0.088528537330827491688440727135052
                                        0.089168332029743739883884412606126
                                         0.33826394036632543530763273039833
                                          1.1599250419465014133554207319948
                                          1.2989629568831597716493684806851
                                          1.6240554937821142318780347702386
                                          4.1380104774956983606108108853334
36340277619978535323175586579368 - 0.17885283113851906156582968732142i
2992644818621369190151455557302 - 0.31440379416295109163597261773161i
4897230890912773483095146949368 - 1.3150774672581101958523122516833i
6282605813254473735848321984622 - 0.083684308295293467728903801421506i
6955614120462881985070978123707 - 0.34019756525211237400051800951627i
8366220103150973884193959600945 - 2.0075730328576518221908039773448i
96894287293815413579933338616546 - 1.0029719645739903595813767103315i
8366220103150973884193959600945 + 2.0075730328576518221908039773448i
2992644818621369190151455557302 + 0.31440379416295109163597261773161i
4897230890912773483095146949368 + 1.3150774672581101958523122516833i
36340277619978535323175586579368 + 0.17885283113851906156582968732142i
96894287293815413579933338616546 + 1.0029719645739903595813767103315i
6282605813254473735848321984622 + 0.083684308295293467728903801421506i
1475584718883716605696238712257 + 0.23111120566119037717218445710823i
6955614120462881985070978123707 + 0.34019756525211237400051800951627i
67755561933436425842885030639685 + 0.89740492763981823171766701235696i
35356088917929933509194226805741 + 0.46392937489127504001635787211895i
1475584718883716605696238712257 - 0.23111120566119037717218445710823i
35356088917929933509194226805741 - 0.46392937489127504001635787211895i
67755561933436425842885030639685 - 0.89740492763981823171766701235696i

But, in the following code,I had defined the parameters of the equation *B,x,mio using syms and to solve this equation, I used vpasolve(eqn1,x,[0 Inf]).

syms alpha mio B x
alpha=5;
mio=0.6;
B=2;
if alpha==5
      if mio==0.6
            if B==2 
                bast=(x/0.5).^(alpha*(mio-1));
                mqam_part1=3*B*((sqrt(3)/2).^(alpha*mio));
                mqam_part2=((0.5*sqrt(3)).^(-alpha))+((1.5*sqrt(3)).^(-alpha));
                mqam_part3=3*(((x/0.5)).^(alpha*mio));
                mqam_part4=((sqrt(3)-(x/0.5)).^-alpha);
                mqam_part5=((2*sqrt(3))-(x/0.5)).^-alpha;
                mqam_part6=mqam_part4+mqam_part5;
                mqam_part7=2*((3-(x/0.5)).^-alpha);
                mqam_part8=6*(((x/0.5)).^(alpha*mio));
                mqam_part9=6*B*(2.^-alpha);
                eqn_LHS=bast/(mqam_part1+mqam_part2)+(mqam_part3*(mqam_part6+mqam_part7));
                eqn_RHS=B/((mqam_part8*mqam_part6)+mqam_part9);
                eqn1=eqn_LHS==eqn_RHS;
                sol_positive = vpasolve(eqn1,x,[0 Inf]);
            end
     end
end

After running this code, it has the following output:

sol_positive =
089168332029743739883884412606126
33826394036632543530763273039833
1599250419465014133554207319948
2989629568831597716493684806851
6240554937821142318780347702386
1380104774956983606108108853334

So, my question is: what is the difference between using vpasolve in the 2 codes and how does it work to give these outputs??

Eman S am 2 Feb. 2020

@Image Analyst

It will cause confusion to the reader to have comment has the same content of the question. So, please delete this comment. For the original question, It would be the same as it. I only shorten the title of the question.Don't worry, it wouldn't be edited or deleted.

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Answer 1

Walter Roberson am 7 Mai 2018

1
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Direkter Link zu dieser Antwort

https://de.mathworks.com/matlabcentral/answers/399549-difference-between-using-vpasolve-in-these-2-codes#answer_319084

The fact that you got multiple outputs tells us that the equations define a polynomial. In the case of a polynomial, vpasolve returns all of solutions that meet the defined constraints. In the first case there are no constraints so it returns all of solutions whether real or complex valued. In the second case you included a range, which is equivalent to having added in(x, 0, inf), or (x >= 0 & x <= inf), which acts to eliminate the negative and complex valued solutions.