Solving a set of nonlinear quadratic equations

In summary, the conversation discusses solving a system of non-linear quadratic equations expressed in terms of column vectors and a scalar. The goal is to solve for the column vector ϒ, with a proposed algorithm and code to be helpful. The main issue is that the equation is not written correctly and there are concerns about the validity of the equation. The paper referenced also raises doubts about the accuracy of the equation.
  • #1
Tilfani
12
0
I would like to solve this system, which is a sets of non linear quadratic equations, the system needed to be solved can be expressed in general as follow:

ϒϒ'C – ϒα = B

Where ϒ=(ϒ1,ϒ2,...ϒn)’ is a column vector and ϒ’ its transpose

C=(c1,c2,…,cn)’ and B=(b1,b2,…bn)’ are a columns vector

And α is a reel scalar

I would like to solve for ϒ, with approximatively about 30

Can someone propose me an algorihm/method to solve this system.
also a code to do it wil be very useful.
Bests
 
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  • #2
Tilfani said:
I would like to solve this system, which is a sets of non linear quadratic equations, the system needed to be solved can be expressed in general as follow:

ϒ’ϒC – ϒα = B

Where ϒ=(ϒ1,ϒ2,...ϒn)’ is a column vector and ϒ’ its transpose

C=(c1,c2,…,cn)’ and B=(b1,b2,…bn)’ are a columns vector

And α is a reel scalar

I would like to solve for ϒ, with approximatively about 30

Can someone propose me an algorihm/method to solve this system.
also a code to do it wil be very useful.
Bests
Y' Y is a scalar, ##\vec Y \cdot \vec Y = |\vec Y|^2##. If you know the value of ##\vec Y \cdot \vec Y## (but don't know the components of ##\vec Y##), you can rewrite the equation above as a system of equations:
##\alpha Y_1 = \vec Y \cdot \vec Y - b_1##
##\alpha Y_2 = \vec Y \cdot \vec Y - b_2##
.
.
.
##\alpha Y_n = \vec Y \cdot \vec Y - b_n##

Divide both sides by ##\alpha## to get
##Y_1 = 1/\alpha (\vec Y \cdot \vec Y - b_1)##
##Y_2 = 1/\alpha (\vec Y \cdot \vec Y - b_2)##
.
.
.
##Y_n = 1/\alpha (\vec Y \cdot \vec Y - b_n)##
 
  • #4
so we're dealing with real scalars here.
- - - -

So your 'equation' is:

ϒϒ'C – ϒα = B
or

ϒϒ'C = ϒα + B

so you have a real symmetric rank one matrix on the Left Hand side (LHS).

The issue is that every possible c you can choose on the LHS gets mapped to zero or is an eigenvector (i.e. ϒ) or a linear combination of the two aforementioned things. So let's hope that B is either a scaled version of ϒ or else the zero vector. If your B is the zero vector, it should be pretty easy. Otherwise you have problems.

More issues: For starters, why write ϒα + B on the Right hand side... why not just write
##\propto ϒ##
 
  • #5
StoneTemplePython said:
so we're dealing with real scalars here.
- - - -

So your 'equation' is:

ϒϒ'C – ϒα = B
or

ϒϒ'C = ϒα + B

so you have a real symmetric rank one matrix on the Left Hand side (LHS).

The issue is that every possible c you can choose on the LHS gets mapped to zero or is an eigenvector (i.e. ϒ) or a linear combination of the two aforementioned things. So let's hope that B is either a scaled version of ϒ or else the zero vector. If your B is the zero vector, it should be pretty easy. Otherwise you have problems.

More issues: For starters, why write ϒα + B on the Right hand side... why not just write
##\propto ϒ##
B is nonzero column, there is a way to solve that?
 
  • #6
Tilfani said:
B is nonzero column, there is a way to solve that?

then ##B \propto ϒ## or this is not an equation
 
  • #7
StoneTemplePython said:
then ##B \propto ϒ## or this is not an equation
B is a constant.
 
  • #8
Tilfani said:
B is a constant.

I don't know what this means. Your original post, and a quick dimensional check say B is a a column vector.

What I am trying to tell you is your original post is analogous to

## 2 = 3##

or

## 2 = 3 +x##
for real ## x \geq 0##

this is not an equation. It is just wrong.
 
  • #9
Yes i mean B is a constant column vector. Do you think that is wrong to?
 
  • #10
You're not hearing me. It is one of the 3 options

option a)
## B = \mathbf 0##

option b)
##B \propto ϒ##

option c)
this is not an equation. It is just wrong.
- - - - -
I have nothing more to say on the matter. Good luck.
 
  • #11
Ok, please look at eq (2 19) page 8 on this link, this paper, maybe some thing wrong.
Portfolio Theory: Origins, Markowitz and CAPM Based Selection - Springer
PDFhttps://www.springer.com › document
 
  • #12
Tilfani said:
Ok, please look at eq (2 19) page 8 on this link, this paper, maybe some thing wrong.
Portfolio Theory: Origins, Markowitz and CAPM Based Selection - Springer
PDFhttps://www.springer.com › document
Please provide the actual link to the document. The link you show is just to the Springer site.
 
  • #13
Mark44 said:
Please provide the actual link to the document. The link you show is just to the Springer site.
Please find enclosed the document. Go to page 7 to see the original problem, the resolution of lagrangian (which may be wrong) lead to equation posted which is (2 19)
 

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Related to Solving a set of nonlinear quadratic equations

1. How do you solve a set of nonlinear quadratic equations?

To solve a set of nonlinear quadratic equations, you can use various methods such as substitution, elimination, or graphing. The key is to manipulate the equations in such a way that one variable is eliminated, leaving you with a linear equation that can be solved.

2. What is the difference between linear and nonlinear quadratic equations?

Linear equations have a degree of 1, meaning the highest exponent is 1. Nonlinear quadratic equations have a degree of 2, meaning the highest exponent is 2. This makes them more complex to solve, as they often have multiple solutions.

3. Can all nonlinear quadratic equations be solved?

Yes, all nonlinear quadratic equations can be solved using various methods. However, some equations may have complex solutions or may require the use of advanced techniques such as the quadratic formula or completing the square.

4. Why is it important to solve nonlinear quadratic equations?

Solving nonlinear quadratic equations is important in many fields of science and mathematics. They are often used to model real-world situations, such as projectile motion or the growth of a population. By solving these equations, we can better understand and predict the behavior of these systems.

5. Are there any shortcuts or tricks to solving nonlinear quadratic equations?

While there are no shortcuts or tricks that apply to all nonlinear quadratic equations, there are some strategies that can make the process easier. It is important to practice and become familiar with different methods such as factoring, completing the square, and using the quadratic formula. Additionally, understanding the properties of quadratic functions can also help in solving these equations.

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