@abbeykabir The algorithms are not in packages. They are all accessible via the method option: dsolve(..., numeric, method= ...). They are documented on the help pages ?dsolve,numeric,IVP , ?dsolve,numeric,BVP , and ?dsolve,numeric,classical .

@abbeykabir The algorithms are not in packages. They are all accessible via the method option: dsolve(..., numeric, method= ...). They are documented on the help pages ?dsolve,numeric,IVP , ?dsolve,numeric,BVP , and ?dsolve,numeric,classical .

Under the rules, why is p(2014) = 30? Shouldn't it be 8?

@Markiyan Hirnyk I didn't feel like looking at or printing out a long number. Also, it's nice to be able to assign the answer to a variable.

@Markiyan Hirnyk I didn't feel like looking at or printing out a long number. Also, it's nice to be able to assign the answer to a variable.

@Vladimir K. Here it is. But there is nothing "wrong" in M14. As Markiyan suggests, such results are to be expected when you use simplify(..., symbolic).

Download symbolic.mw

@abbeykabir Dividing by a matrix is equivalent to multiplying by its inverse. Let J(x) represent the Jacobian matrix evaluated at the vector x. Then the Newton iteration becomes

x[i+1]:= x[i] - J(x[1])^(-1)*F(x[i]).

But recall what I called The first rule of numerical linear algebra: Never invert a matrix; always solve a linear system instead. Computing A^(-1)*B is equivalent to solving A*X=B. This is accomplished in Maple with the command LinearAlgebra:-LinearSolve(A,B).

Now let's step back for a moment to the single-variable Newton iteration:

x[i+1]:= x[i] - f(x[i])/f'(x[i]).

How do we know when to stop iterating?

@abbeykabir Dividing by a matrix is equivalent to multiplying by its inverse. Let J(x) represent the Jacobian matrix evaluated at the vector x. Then the Newton iteration becomes

x[i+1]:= x[i] - J(x[1])^(-1)*F(x[i]).

But recall what I called The first rule of numerical linear algebra: Never invert a matrix; always solve a linear system instead. Computing A^(-1)*B is equivalent to solving A*X=B. This is accomplished in Maple with the command LinearAlgebra:-LinearSolve(A,B).

Now let's step back for a moment to the single-variable Newton iteration:

x[i+1]:= x[i] - f(x[i])/f'(x[i]).

How do we know when to stop iterating?

Are the entries rational functions? How many parameters? Are you looking for a particular eigenvalue?

Try LinearAlgebra:-CharacteristicPolynomial.

Try setting one parameter to 0 or 1.

@mvcr15 Wikipedia has good information. Let's start with Euler's method for a single equation. That's the easiest case. You should be able to find a discussion of it in any serious calculus textbook. Or see the Wikipedia article "Euler method".

@abbeykabir Okay, good. Yes, the Jacobian matrix is the derivative of a multivariate function. The single-variable Newton's method divides by the derivative. In the multivariate case the derivative is a matrix. How can we divide by a matrix?

@abbeykabir Okay, good. Yes, the Jacobian matrix is the derivative of a multivariate function. The single-variable Newton's method divides by the derivative. In the multivariate case the derivative is a matrix. How can we divide by a matrix?

Do you need the methods to solve a single first-order differential equation, or do you need it for a system of equations?

@abbeykabir Ok, that's good. Now let's generalize that to the multivariate case, where x is an n-vector, and f is a vector-valued function on n-vectors (i.e., f: R^n -> R^n). In particluar, what corresponds to the derivative in the multivariable case?

@abbeykabir Ok, that's good. Now let's generalize that to the multivariate case, where x is an n-vector, and f is a vector-valued function on n-vectors (i.e., f: R^n -> R^n). In particluar, what corresponds to the derivative in the multivariable case?