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Defining a complex variable

Posted 10 years ago
 In the following two equations with two variables "b" and "c". In my theory if i solve these two equations then i dont get required results, BUT if i explicitly write both variables b and c as complex numbers,i.e,. "b=x1+i y1" and " c= x2+ i y2 " and solve 4 equations like below then i get my required results, but i need to have previous form as it reduces time. CAN ANYONE GUIDE ME ? THANKS eq1 = -2.6496569219504336+ 2.3827335619844803/( 1.29731758306921+ (0.7463275363988051 - b)^2) - ( 0.7684079898461433b)/( 1.29731758306921 + (0.7463275363988051- b)^2) + 2.3827335619844803/( 1.29731758306921+ (0.7463275363988051 - c)^2) - ( 0.7684079898461433c)/( 1.29731758306921 + (0.7463275363988051 - c)^2) eq2 = 0.37580644282399783+ 0.013774982050389344/( 0.21352891577655558+ (0.34484941659029344 - b)^2) + ( 0.1448494165902934b)/( 0.21352891577655558 + (0.34484941659029344- b)^2) + 0.013774982050389344/( 0.21352891577655558+ (0.34484941659029344 - c)^2) + ( 0.1448494165902934c)/( 0.21352891577655558 + (0.34484941659029344 - c)^2)
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Posted 10 years ago
Posted 10 years ago
 In your last reply, In[8] and In[10] represent set of values of b and c?
Posted 10 years ago
 !
Posted 10 years ago
 Once again thanks for your guidance, I have tried your suggestion in my file, but it is not working. I have attached the file, Please see and guide me if i am doing wrong.Thanks Attachments:
Posted 10 years ago
Posted 10 years ago
 Hello Sir,Thanks for your time, Can you please explain as defining first b and c then solving equations for the same b and c has not worked.Thanks
Posted 10 years ago
Posted 10 years ago
 In[1]:= eq1 = -2.6496569219504336 + 2.3827335619844803/(1.29731758306921 + (0.7463275363988051 - b)^2) - (0.7684079898461433 b)/(1.29731758306921 + \ (0.7463275363988051 - b)^2) + 2.3827335619844803/(1.29731758306921 + (0.7463275363988051 - c)^2) - (0.7684079898461433 c)/(1.29731758306921 + \ (0.7463275363988051 - c)^2); In[2]:= eq2 = 0.37580644282399783 + 0.013774982050389344/(0.21352891577655558 + (0.34484941659029344 - b)^2) + (0.1448494165902934 b)/(0.21352891577655558 + \ (0.34484941659029344 - b)^2) + 0.013774982050389344/(0.21352891577655558 + (0.34484941659029344 - c)^2) + (0.1448494165902934 c)/(0.21352891577655558 + \ (0.34484941659029344 - c)^2); In[3]:= NSolve[{eq1 == 0, eq2 == 0}, {b, c}] Out[3]= {{b -> 1.1021 + 0.316679 I, c -> 0.212341 + 0.538527 I}, {b -> 1.10218 - 0.316787 I, c -> 0.212354 - 0.538553 I}, {b -> 0.212358 + 0.538566 I, c -> 1.10216 + 0.316634 I}, {b -> 0.212382 - 0.538559 I, c -> 1.10227 - 0.316857 I}, {b -> 0.0481784 + 0.785998 I, c -> 0.0482403 - 0.78599 I}, {b -> 0.0482052 - 0.785916 I, c -> 0.0481275 + 0.785992 I}, {b -> 0.0134349 + 0.479927 I, c -> 0.0133104 - 0.479802 I}, {b -> 0.013324 - 0.479794 I, c -> 0.0134405 + 0.479929 I}} 
Posted 10 years ago
 Ok Sir,Can you please guide me in the case of 2 equations and 4 equations case as discussed in my last mail. As "I have no problem in solving the 2 equations with 2 variables, but by solving, the results are not favorable, but if i write the variables b and c in equations in the form of "b=x1+i y1 " and "c=x2+i y2 " and choosing 4 equations involving variables x1,y1, x2 and y2, then Using NSolve[{eq1, eq2, eq3, eq4}, {x1, y1, x2, y2}, Reals]. I get my favoured results.How can i get same results.For better understanding, please see the attached file Attachments:
Posted 10 years ago
 The time required by NSolve increases exponentially with the complexity of the problem. For 10 equations and 10 variables, it would be better to use FindRoot.
Posted 10 years ago
 if i use either NSolve[{eq1, eq2}, {b, c}]or NSolve[{eq1, eq2}, {a1, b, c}]Values of both variables b and c are same in both cases. You can see in the attached file.Also sir, if you see my question, it was little different, I have no problem in solving the 2 equations with 2 variables, but by solving, the results are not favorable, but if i write the variables b and c in equations in the form of "b=x1+i y1 " and "c=x2+i y2 " and making 4 equations which is easy in my theory, with 4 variables, then Using NSolve[{eq1, eq2, eq3, eq4}, {x1, y1, x2, y2}, Reals].I get my required results. But this way is not good for me as when i go for higher dimensions, i.e 10 equations and 10 variables , then i face problem that my system stucks and stop running oblem Attachments:
Posted 10 years ago
 First question: saw b and c, assumed there was an a. It's better to put Mathematica code into a code sample, as it's easier to read. Second question: NSolve used that equation to give an answer to a problem with an infinite number of solutions.
Posted 10 years ago
 Thanks for your Answer,Can you please explain, why you have taken " a" as this is not included in equationsIn[4] := NSolve[{eq1, eq2}, {a, b, c}]And can you please explain the following (171802 a)/178835 - (113492 b)/ 178835 - (121484 c)/178835 == 1.
Posted 10 years ago
 NSolve says your equations have an infinite number of solutions. In[1]:= eq1 = -2.6496569219504336 + 2.3827335619844803/(1.29731758306921 + (0.7463275363988051 - b)^2) - (0.7684079898461433 b)/(1.29731758306921 + \ (0.7463275363988051 - b)^2) + 2.3827335619844803/(1.29731758306921 + (0.7463275363988051 - c)^2) - (0.7684079898461433 c)/(1.29731758306921 + \ (0.7463275363988051 - c)^2); In[2]:= eq2 = 0.37580644282399783 + 0.013774982050389344/(0.21352891577655558 + (0.34484941659029344 - b)^2) + (0.1448494165902934 b)/(0.21352891577655558 + \ (0.34484941659029344 - b)^2) + 0.013774982050389344/(0.21352891577655558 + (0.34484941659029344 - c)^2) + (0.1448494165902934 c)/(0.21352891577655558 + \ (0.34484941659029344 - c)^2); In[4]:= NSolve[{eq1, eq2}, {a, b, c}] During evaluation of In[4]:= NSolve::infsolns: Infinite solution set has dimension at least 1. Returning intersection of solutions with (171802 a)/178835-(113492 b)/178835-(121484 c)/178835 == 1. >> Out[4]= {{a -> 1.91912 - 0.59005 I, b -> 1.10211 - 0.316715 I, c -> 0.212318 - 0.538566 I}, {a -> 1.91912 + 0.59005 I, b -> 1.10211 + 0.316715 I, c -> 0.212318 + 0.538566 I}, {a -> 1.10666 + 0.0365591 I, b -> 0.0480565 - 0.785901 I, c -> 0.0480565 + 0.785901 I}, {a -> 1.10666 - 0.0365591 I, b -> 0.0480565 + 0.785901 I, c -> 0.0480565 - 0.785901 I}, {a -> 1.05913 + 0.0223244 I, b -> 0.0133044 - 0.479901 I, c -> 0.0133044 + 0.479901 I}, {a -> 1.05913 - 0.0223244 I, b -> 0.0133044 + 0.479901 I, c -> 0.0133044 - 0.479901 I}, {a -> 1.96051 - 0.579729 I, b -> 0.212318 - 0.538566 I, c -> 1.10211 - 0.316715 I}, {a -> 1.96051 + 0.579729 I, b -> 0.212318 + 0.538566 I, c -> 1.10211 + 0.316715 I}} 
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