Can Mathlab solve this

Can Mathlab solve this

x1^2 +2.x1 - 2.x2^2 -5.x2 =5

2.x1^2 -3.x1 +x2^2 +3.x2 =19

[Walter Roberson]

You know that has four solutions, right?

[Walter Roberson]

Please read the guide to tags and retag this question.

http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags

Answer 1

[Thomas]

Yes, look in the symbolic math toolbox http://www.mathworks.com/help/symbolic/solve.html

Go to the bottom of the page for examples

Answer 2

[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

[Walter Roberson]

Note: fsolve() will only find one solution at a time.

[rob]

fsolve is all numerical not algabraic

[Walter Roberson]

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.)

Answer 3

[Roger Stafford]

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.

[Walter Roberson]

roots() will give numeric answers in decimal form that approximate the solutions.

MATLAB's symbolic toolbox will not expand symbolic solutions to quartics by default but it can be made to do it.

[rob]

Is this working for a 3 by 3 or 4 by 4 to THus x1 x2 x3 x4

[Walter Roberson]

Possibly, under some restrictions on the combinations of coefficients. In the general case, if you have any power 5 or higher then algebraic solutions can usually not be found. With lower powers it would depend in part on the cross-product terms as to whether there is an algebraic solution or not.

[rob]

And for a 1000 by 1000 What i am asking is really is there a algabraic way to solve it

[Walter Roberson]

You have not defined "it" sufficiently for an answer.

In your example, each variable appears only in the first or second degree, and never in the third degree or higher, and never any fractional or negative exponent. The constant multipliers and the value equated to are all rational and integer in your example. Your example also has no cross-products, no occurrence of x1^n * x2^m for n, m > 0. Effectively, each of your entries is the sum of exactly two quadratic equations with integer coefficients.

in the 3 x 3 and larger versions, which (if any) of those conditions is modified? Is each entry in the 1000 x 1000 still the sum of exactly two quadratic equations in integer coefficients?

[rob]

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]

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.

Answer 4

[Edwin]

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'.

[Walter Roberson]

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.