Solution of non-linear set of equations

  • Thread starter Thread starter mtak0114
  • Start date Start date
  • Tags Tags
    Non-linear Set
mtak0114
Messages
46
Reaction score
0
Hi

Consider a set of n non-linear equations and I have m unknowns and n<m, so I don't have enough equations to solve the problem. All the unknowns are parametrized by say t

A possible solution to this that I considered was to take the derivative w.r.t. t and obtain another set of n equations, if 2n > m then I have more equations than unknowns and I these can be solved.

Now will this work?

cheers

M
 
Last edited:
Physics news on Phys.org
If you consider the derivatives as well, you get 2n equation is 2m unknowns. Since the derivatives are unknown as well.
 
Sorry I was not clear the equations are really differential equations but I don't want a solution for all t just for 1 so I'm treating each derivative as an unknown. So I have
f_i(x_i(t),x_i&#039;(t),x_i&#039;&#039;(t)...) = 0 it is possible to have 2n>m+1, if not I just take extra derivatives until it is. Will this work or have I 'cheated'?


cheers

M
 
Maybe if you more specific in the problem you are actually trying solve we could help more. But you can't magically solve an under-determined system by taking derivatives.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top