This is messy, but solves your problem, barring a very probable mistake transcribing the coefficients: from _future_ import division from sympy import Integer, Symbol, Eq, solve It's not fast, but it allows me to copy the RHS of your equations exactly, limiting the thinking I need to do (always a plus), and gives fractional answers. Here's an entirely different approach, using sympy. Return spsolve(coeff_mat.tocsr(), const).reshape((-1,1)) Solution using scipy.sparse: from scipy.sparse import spdiags, lil_matrix, vstack, hstack Implementation using numpy: import numpy as np This is not reflected in the spreadsheet above but has been implemented in the code below: the row (= eq.) has been derived:Īs Jaime suggested, multiplying by n improves the code. The first column show from which of the above given eqs. The sought after solution vector x is here light green and used to label the columns. The coefficient matrix A is light blue, the constant right side is orange. Here's a snapshot for the example case n=50 showing how you can derive the coefficient matrix and understand the block structure. Then I build up the A-coefficient matrix from 4 block matrices. Here I ordered the equations such that x is N_max.,N_0,M_max.,M_1. The linear system to solve is of the shape A dot x = const 1-vector. This gives the solution in the order N_max.,N_0,M_max.,M_1. Updated: added implementation using scipy.sparse Posted follow up to Using scipy sparse matrices to solve system of equations. Update The answers are great but they use dense solvers where the system of equations is sparse. If I want to plug in n=50, say, how can you set up this system of simultaneous equations in python so that can solve them? This is because the equations become M(1) = 1+(2/2)*N(0) For example, when n=2 the answer should be M(1) = 4, I believe. I actually only want the value of M(n-1). I have been using Maple but I would like to set these up and solve them in python now, maybe using (or any other better method). Notice also that p is just a constant integer in every equation so the whole system is linear. Please see the TI-83 Plus and TI-84 Plus Family guidebooks for additional information.For a fixed integer n, I have a set of 2(n-1) simultaneous equations as follows. Please Note: In addition the Polynomial Root Finder and Simultaneous Equation Solver App can be downloaded onto the TI-84 Plus CE, TI-84 Plus C Silver Edition and TI-84 Plus and TI-83 Plus calculators to provide this feature. Therefore the solution is x = 1.75 and y = -3.375 Press on that option, which then pastes the function onto your home screen.ĩ) Press, which pastes Matrix A into the rref command.ġ0) Close the parenthesis by pressing. To find the reduced row echelon form using the rref( function:ħ) Scroll to "MATH" by pressing the right arrow key one time.Ĩ) Scroll down until you see "rref(", which is the function for reduced row echelon form. To solve your system of equations, you'll now need to use the rref function. Your entries should be 3, -2, 12, 6, 4 and -3.ĥ) Press to exit out of the editing screen. Ĥ) Input the matrix entries, pressing enter after each value. To do so, please follow the steps below:ġ) Press to bring up the Matrix Menu.ģ) Input the dimensions of your matrix by pressing. Solution: First, begin by inputting the matrix. Simultaneous equations can be solved by entering the coefficients of the equations in a matrix, and then using the rref() function on your matrix. How can I solve simultaneous equations on the TI-83 Plus and TI-84 Plus family of graphing calculators? Solution 34599: Solving Simultaneous Equations on the TI-83 Plus and TI-84 Plus Family of Graphing Calculators.
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