But is it not possible to solve a set of equations which has exponentials in it because i know that to solve it the way you said it will need three more equations. Infact the original equation is in this form but by using a relation I convert it into the form that I wrote so that there are 6 equations and 6 Unknowns.
Thanks for your help Smichr. Nandan. On Jun 27, 7:49 am, smichr <[email protected]> wrote: > On Jun 26, 6:47 pm, nandan jha <[email protected]> wrote: > > > Hello, > > > I am using Mathematica to solve a set of equations and it keeps > > showing a error message "FindRoot::jsing: Encountered a singular > > Jacobian at the point {A1,A2,A3,Ea1,Ea2,Ea3} = > > {-2629.39,758889.,4.12521*10^6,-19229.4,-21903.5,-71770.6}. Try > > I wonder if you might have an ill-posed problem. Even though there are > 6 equations and unknowns, they always appear in a way that makes it > look like there are 9 unknowns: > > Use the following definitions: > a=(8.314*(48 + 273.15)) > b=(8.314*(44 + 273.15)) > c=(8.314*(41 + 273.15)) > sa=.0183 > sb=0.00995 > sc=0.0075 > ca=0.01784 > cb=0.00983 > cc=0.00742 > A1x=A1*exp(E1/x) where x is [a, b, c] > > Then the equation set looks like this: > A1a + A2a + A3a == sa > A1b + A2b + A3b == sb > A1c + A2c + A3c == sc > A1a**2 + A2a**2 + A3a**2 + 2*A1a*A2a + 2*A3a*A2a - 2*A1a*A3a == ca**2 > A1b**2 + A2b**2 + A3b**2 + 2*A1b*A2b + 2*A3b*A2b - 2*A1b*A3b == cb**2 > A1c**2 + A2c**2 + A3c**2 + 2*A1c*A2c + 2*A3c*A2c - 2*A1c*A3c == cc**2 > > You can solve the 1st for A1a > A1a=>s1-A2a-A3a > ...and substitute this into the 4th and solve for A2a > A2a=>(4*A3a*sa + ca**2 - sa**2 - 4*A3a**2)/(4*A3a) > ...but there is no equation to resolve A3a. > > Is there some other relationship between these quantities? > > /c --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sympy?hl=en -~----------~----~----~----~------~----~------~--~---
