One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. This polynomial is considered to have two roots, both equal to 3. To understand what is meant by multiplicity, take, for example. If has degree, then it is well known that there are roots, once one takes into account multiplicity. The largest exponent of appearing in is called the degree of. Partial Fraction Decomposition CalculatorĪbout solving equations A value is said to be a root of a polynomial if.Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator ![]() Here are some examples illustrating how to formulate queries. To avoid ambiguous queries, make sure to use parentheses where necessary. It also factors polynomials, plots polynomial solution sets and inequalities and more.Įnter your queries using plain English. X = np.linalg.solve(A, np.ones((2 * (n - 1),)))įor M(2) when n=3, which is not what you came up with.Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. 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. For a fixed integer n, I have a set of 2(n-1) simultaneous equations as follows.
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