Root of Non-Linear Eqn (Numerical Method)

Apteronotus
Messages
201
Reaction score
0
Hi,

I have a non-linear function F: \Re^{3}\rightarrow\Re. I would like to find the roots of this equation numerically, since an explicit formula cannot be derived.

As far as I am aware Newton's method can only be utilized when the domain and the range of the function are of the same degree. (i.e. F: \Re^{n}\rightarrow\Re^{n})

Is there a method that can used for the caseF: \Re^{n}\rightarrow\Re^{m} with m \neq n?

Thanks in advance,
 
Physics news on Phys.org
So you have something like f(x,y,z)= 0 to solve? I see no reason why Newton's method could not be generalized to this. Choose some starting point (x_0, y_0, z_0). The tangent (hyper-)plane to u= f(x,y,z) at that point is u= \nabla f(x_0,y_0,z_0)\cdot(x- x_0,y-y_0,z- z_0)+ f(x_0, y_0, z_0). Set that equal to 0 and solve for (x, y, z) as the next approximation.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top