Solving non linear equations where equations are equals

I wonder and tried to google it, but I am not sure what to google it, how to solve non linear equations where equations are equal between each other. I am able to write a specific algorithm for two equations but not dynamically for N equations. I will show the example of three (how my equations approximately looks like):

C1, C2, C3, X - are unknows, but in the end I do not need to know result of X.

It can be interpreted like this (last equation C1 + C2 + C3 = 1 is not included here):

Please, don't try to solve this, I am not sure if these equations have results. I just randomly typed coefficients. But this is how my equations can looks like. Only with different coefficients. I tried to calculated it with only two unknows and I have got quadratic equation in the end so with three unknowns there will be cubic in the end. With N unknows there will be polynomial equation with degree N. Also, I have to say, result do not have to be with 100% accuracy. I am not sure if its help somehow or not.

I found on google that maybe using iterative method could help. I look at few iterative methods but I am still not sure how to use it on this kind of problem. I also found, that non linear equation can by linearize. Maybe that would be a option but I am not sure how to do it here.

• Here is a book on iterative methods for nonlinear equations: google.com/books/edition/… Commented Mar 25, 2022 at 0:13

2 Answers

From $$C_1 + C_2 + C_3 = 1$$, you get $$C_3 = 1 - C_1 - C_2$$

Substitute that in your equations, you get

$$\dfrac{ 4 C_1 + 8 C_2 + 16 (1 - C_1 - C_2) }{2 C_1} = \dfrac{ -12 C_1 - 8 C_2 + 16} {2 C_1} = \dfrac{ - 6 C_1 - 4 C_2 + 8} {C_1}$$

and

$$\dfrac{ 9 C_1 + 27 C_2 + 81(1 - C_1 - C_2) }{3 C_2} = \dfrac{ - 24 C_1 - 18 C_2 + 27 }{C_2}$$

and

$$\dfrac{16 C_1 + 64 C_2 + 256 (1 -C_1 - C_2) }{4 C_3} = \dfrac{ -60 C_1 - 48 C_2 + 64 }{C_3}$$

Since these expressions are equal as you have in your question, we can cross multiply to get

$$( - 6C_1 - 4C_2 +8 ) C_2 = (-24 C_1 - 18 C_2 + 27) C_1 \hspace{25pt}(1)$$

and

$$(- 6 C_1 - 4 C_2+8) (1 -C_1 - C_2) = (-60 C_1 - 48 C_2 + 64 ) C_1 \hspace{25pt}(2)$$

Equations (1) and (2) are two quadratic equations in $$C_1$$ and $$C_2$$ and can be solved using the method outlined in the solution of this problem

• I really appreciate your answer and thank you for that but I am really sorry and I don't wanna sounds rude, but, is this way will gonna work for N equations as I mentioned in my question? System of three was just example. In my case, they can be 16 equations or more. Not really infinite, but a pretty large amount. Commented Mar 24, 2022 at 23:52
• No. It doesn't work for $N$ equations. It is for this specific case, where you can cancel $X$ and also where the unknowns $C_1, C_2, C_3$ are related linearly. Commented Mar 25, 2022 at 0:01
• If you wanna solve $N$ equations, then your best bet is to have a good initial guess of the solution, and to use the multi-variate Newton-Raphson numerical method. This method is very versatile, and very fast, and is applicable to a wide range of problems. Commented Mar 25, 2022 at 0:04
• OOoooo, sounds good, I will check that out, THANK you very much for now, I will try it <3 Commented Mar 25, 2022 at 0:05
• Please, have a look at the solution I just posted. Commented Mar 30, 2022 at 12:00

A much more direct way is to consider that this kind of system is composed of an eigensystem condition + (the last equation which is plainly a normalizing condition.

Let us take your example written under the form of a slightly simplified system

$$\begin{cases}2C_1+4C_2+8C_3&=&C_1\\ 3 C_1+9C_2+27C_3&=&C_2\\ 4C_1+16C_2+64C_3&=&C_3\end{cases}$$

giving the matrix-vector eigen-equation (of the form $$MV=XV$$)

$$\underbrace{\begin{pmatrix}2&4&8\\ 3&9&27\\ 4&16&64\end{pmatrix}}_M\underbrace{\begin{pmatrix}C_1\\ C_2\\ C_3\end{pmatrix}}_V=X \underbrace{\begin{pmatrix}C_1\\ C_2\\ C_3\end{pmatrix}}_V$$

where there are only 3 possibilities for $$X$$ and $$(C_1,C_2,C_3)$$ to be chosen among the eigenvalues and associated eigenvectors with normalizing condition $$C_1+C_2+C_3=1$$ giving (for example using the matlab program below):

$$X = 71.5723 \ or \ X=3.2194 \ or \ X=0.2083$$

associated resp. to normalized vectors:

$$\begin{pmatrix}C_1\\ C_2\\ C_3\end{pmatrix}= \begin{pmatrix}0.0888\\ 0.2776\\ 0.6335\end{pmatrix} \ \ or \ \ \begin{pmatrix}0.6198\\ 0.5713\\ -0.1912\end{pmatrix} \ \ or \ \ \begin{pmatrix}2.2101\\ -1.4302\\ 0.2201\end{pmatrix}$$

Matlab program:

 M=[2, 4,  8
3, 9, 27
4,16, 64];
eig(M), % list of eigenvalues
[P,~]=eig(M); % P is a matrix whose columns are eigenvectors
for k=1:3
V=P(:,k);V=V/sum(V), % normalization of eigenvectors
end;

• Nice approach !! [+1] Commented Mar 30, 2022 at 12:09
• @Calm down and have some tea You shouldn't have erased your previous answer ! Commented Mar 30, 2022 at 14:03
• Why do you say that? Commented Mar 30, 2022 at 14:08
• I say that because I feel confused that it looks a consequence of my own answering. Commented Mar 30, 2022 at 14:20
• Yes. You're right. After seeing your answer, I realized that my answer is a bad answer to this question that should be solved in the way you did. Commented Mar 30, 2022 at 15:05