Using Linear Algebra to solve systems of non-linear equations

In summary: Linear Algebra can solve for a,b,d, and e if a solution exists. It doesn't do anything regarding x and y.
  • #1
Edwin
162
0
I did some number crunching and found the following:

Given n equations in n unknowns:

a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f

If there is a solution to these equations, you can use substitution to transform these equations into a set of linear equations and solve using linear algebra.

Let f(x) = u and g(y) = v

Then

a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f

which becomes

a*u + b*v = c
d*u + e*v = f

Which can be solved for u and v using linear algebra.

x and y can then be solved for by solving the corresponding equations

u = f(x), and v = g(y) thus making it possible to solve systems of non-linear equations of the form above using linear algebra.

Any thoughts?

Edwin G. Schasteen
 
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  • #2
You aren't using Linear Algebra to solve for x and y.

Linear Algebra can solve for a,b,d, and e if a solution exists. It doesn't do anything regarding x and y.

You can not solve u and v through that at all.

Let f(x) = x^10 + x^9 + x + 1 = u, and g(y) = y^11 + y^10 + y + 1 = v.

According to you, we can solve for u and v.

a*u + b*v = c
d*u + e*v = f

...then for x and y.

Using my simple polynomials, solve for x and y, which you said you can do. To make it easier, let a=1, b=0, and c=0.

a*u + b*v = c
u + 0 = 0
f(x) = 0
x^10 + x^9 + x + 1 = 0

Hmmm... this isn't helping much.

How does Linear Algebra solve that?
 
Last edited:
  • #3
Edwin said:
I did some number crunching and found the following:
Given n equations in n unknowns:
a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f
If there is a solution to these equations, you can use substitution to transform these equations into a set of linear equations and solve using linear algebra.
Let f(x) = u and g(y) = v
Then
a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f
which becomes
a*u + b*v = c
d*u + e*v = f
Which can be solved for u and v using linear algebra.
x and y can then be solved for by solving the corresponding equations
u = f(x), and v = g(y) thus making it possible to solve systems of non-linear equations of the form above using linear algebra.
Any thoughts?
Edwin G. Schasteen

What you are saying is certainly true but I don't see anything new. If I had, for example, the two equations
x2+ y2= 1 and
x2- y2= 1 then I would add the two equations to get 2x2= 2 so x= 1 or -1. Putting that back into either equation I would have 1+ y2= 1 or 1- y2= 1 both of which give y2= 0 or y= 0. The two solutions to the system of equations are x= 1, y= 0 and x= -1, y= 0.

Of course, you are assuming that you have n LINEAR equations in n functions of single variables. Most "non-linear" equations are more complicated than that: for example x2- 2xy+ y2= 1 and x2+ ey= 1 or sin(x)+ cos(x)= 1.
 
  • #4
Maybe I misread your question.

Is u and v constants?
 
  • #5
Linear Algebra can solve for a,b,d, and e if a solution exists. It doesn't do anything regarding x and y.

You can not solve u and v through that at all.
Let me try explaining it a different way:

Suppose the system of equations

au + bv = c
du + ev = f

has a unique solution (u,v) = (m,n). (As is typical with 2 equations and 2 unknowns)


Then, if you were asked to solve the system of equations

a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f

for x and y.

The previous work allows you to immediately reduce this problem to solving

f(x) = m
g(y) = n
 
  • #6
Hurkyl said:
Let me try explaining it a different way:
Suppose the system of equations
au + bv = c
du + ev = f
has a unique solution (u,v) = (m,n). (As is typical with 2 equations and 2 unknowns)
Then, if you were asked to solve the system of equations
a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f
for x and y.
The previous work allows you to immediately reduce this problem to solving
f(x) = m
g(y) = n

Yes, but that doesn't really help at all.
 
  • #7
Really? I think solving

x^10 + x^9 + x + 1 = 7
y^11 + y^10 + y + 1 = 3

looks much much easier than trying to solve

5y^11 + 3x^10 + 5y^10 + 3x^9 + 3x + 5y + 8 = 36
y^11 - x^10 + y^10 - x^9 - x + y = -4

Don't you?


And how else were you planning on solving the equation

cos t + 3 sin s = 2
4 cos t - 2 sin s = -1

?
 
  • #8
Thanks for your input guys!


Hurkyl, I hadn't thought of using this to solve systems of transcendental equations. That's an interesting thought!

Your thoughts regarding simplification of the systems of algebraic equations of the same form into more managable equations of a single variable were exactly what I was aiming at. Given a system of non-linear equations of the form below, you can reduce the equations to a system of non-linear equations of a single variable for each of the variables just as you've shown below. One can then use any variety of numerical methods to approximate sollutions to the given equations.

What do you think?

Inquisitively,

Edwin
 
  • #9
It's what I would expect to be taught in any intermediate algebra course.
The crucial point is that "non-linear equations of the form below" are not general non-linear equations. Your form is restricted to "linear equations of non-linear functions of a single variable". Obviously one can use linear equation methods to reduce to separate non-linear equations.
 
  • #10
That's correct.

Best Regards,

Edwin
 

1. What is Linear Algebra?

Linear Algebra is a branch of mathematics that deals with linear equations and their representations in vector spaces. It involves the study of matrices, vectors, and systems of linear equations, and is used to solve a wide range of mathematical problems.

2. How can Linear Algebra be used to solve systems of non-linear equations?

Linear Algebra techniques can be applied to solve systems of non-linear equations by converting the non-linear equations into a system of linear equations. This can be done by linearizing the equations using methods such as Taylor series expansion or Jacobian matrices. The resulting linear system can then be solved using standard techniques such as Gaussian elimination or matrix inversion.

3. What are the benefits of using Linear Algebra to solve systems of non-linear equations?

Using Linear Algebra to solve systems of non-linear equations allows for the use of efficient and well-developed methods, such as matrix operations and vector spaces, to find solutions. It also provides a systematic approach to solving complex equations and can be applied to a wide range of problems in mathematics, engineering, and science.

4. Are there any limitations to using Linear Algebra to solve systems of non-linear equations?

While Linear Algebra can be a powerful tool for solving systems of non-linear equations, it does have some limitations. It may not always be possible to linearize a non-linear system, and even when it is possible, the resulting linear system may be too complex to solve efficiently. In these cases, other numerical methods may be more suitable.

5. How is Linear Algebra used in real-world applications to solve systems of non-linear equations?

Linear Algebra is used extensively in various fields such as physics, engineering, economics, and computer graphics to solve systems of non-linear equations. For example, it is used in modeling and simulating physical systems, optimizing production processes, and creating 3D animations. It is also essential in machine learning and data analysis, where non-linear systems can be represented and solved using linear models and techniques.

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