Can Mathlab solve this

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rob
rob 2013-9-3
评论： Walter Roberson 2022-9-17
Can Mathlab solve this
x1^2 +2.x1 - 2.x2^2 -5.x2 =5
2.x1^2 -3.x1 +x2^2 +3.x2 =19

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回答（4 个）

Thomas
Thomas 2013-9-3
编辑：Thomas 2013-9-3
Yes, look in the symbolic math toolbox http://www.mathworks.com/help/symbolic/solve.html
Go to the bottom of the page for examples
Shashank Prasanna
You can solve a system of nonlinear equations using FSOLVE:
http://www.mathworks.com/help/optim/ug/fsolve.html
This will yield numerical solutions for x1 and x2
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rob
rob 2013-9-3
fsolve is all numerical not algabraic
Walter Roberson
Walter Roberson 2013-9-3
Correct, fsolve() is numeric not algebraic. However can you really make use of the algebraic solutions? For example one of the four solutions to the above system has x1 be
-349/140 + (1/5040) * (4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (1/90720) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344)/(112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) * 6^(1/2) * 36^(1/2) * 2^(1/2) * (((810* (112706532 + 2940 * 1239703701^(1/2))^(1/3) + 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) + 62862) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 1764 * (112706532 + 2940 * 1239703701^(1/2))^(1/3)) * 18^(1/3) * ((112706532 + 2940 * 1239703701^(1/2))^(1/3) / (810 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 125724))^(1/2) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) * 6^(1/2) / (9392211 + 245 * 1239703701^(1/2)))^(1/2) + (1/30240) * (1620 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 2 * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) * 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) + 125724 * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 3528 * (112706532 + 2940 * 1239703701^(1/2))^(1/3)) * 6^(1/2) * ((112706532 + 2940 * 1239703701^(1/2))^(1/3) / (810 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 125724))^(1/2) * 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) / (9392211 + 245 * 1239703701^(1/2))
Would your work seriously be affected if all those 112706532 where 112706533 instead? (That would make a difference in the 6th decimal place.)

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Roger Stafford
Roger Stafford 2013-9-3
It is useful to know how to solve such equations by hand rather than always depending on matlab. The trick is to eliminate either the x1^2 term or the x2^2 term by combining the equations appropriately. If we double the second equation and then add the equations, we get
5*x1^2-4*x1+x2 = 43
which can be solved for x2
x2 = -5*x1^2+4*x1+43
You can then substitute this value of x2 into either one of the original equations and get a fourth degree polynomial equation in x1. The four roots of this can be obtained with matlab's 'roots' program (we need matlab after all) and then corresponding values of x2 from these with the above equation.
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rob
rob 2013-9-4
I dont understand exactly I think lineair dependence is solvable like a +b +c =3 2a +2b - c =5 -a +b -3c = 9
but i was investigating only where the a b and c have a quadratic So a thousend by a thousend has a thousend unkowns of a and a^2 for all i know this is unsolvable and only in fsolve with numerical math.
Walter Roberson
Walter Roberson 2013-9-4
编辑：Walter Roberson 2013-9-4
Suppose you had
a = b^2 + d
b = c^2
c = d^2
then a = d^8 + d, and that has no closed-form solution for d in terms of a. Therefore the generalized 3 x 3 or larger is not always resolvable to algebraic solutions. However, if the forms of the equations are constrained, so that one was not working with the generalized form, then it might be possible to find algebraic solutions; that would vary with the exact constraints.

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Edwin
Edwin 2022-9-17
solve('6/(1-x^2) =5/(1+x) - 3/(1-x)')
Check for incorrect argument data type or missing argument in call to function 'solve'.
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Walter Roberson
Walter Roberson 2022-9-17
Ran in:
syms x
solve(6/(1-x^2) == 5/(1+x) - 3/(1-x))
ans = 
Historically, solve() used to support character vectors like you show, but that changed around R2017b or so.

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