# Tag Info

1

Does this do what you want? polyIter[{a_, b_, c_}, x_] := Module[{q, n}, q = PolynomialQuotient[a, b, x]; n = Exponent[q, x]; {b, a - b #, #} &@If[n < 1, q, (Coefficient[q, x^n] x^n)] // Simplify ] myPolyContinuedFraction[p_, q_, x_] := Last /@ Rest@ NestWhileList[polyIter[#, x] &, {p, q, 0}, Length@CoefficientList[#[[2]], x] > 0 &...

0

The workflow provided in the OP seems to apply to ordinary differential equations up to order 4. The following workflow is more general, but it will require that parameters and independent variables are expressed as monomials. First, set up some test ODE's and PDE's: (*Test ODE's and PDE's*) odeeqn = ϵ y''[x] y[x] + x y'[x] + ϵ^2 y[x] == 0; pdeeqn = U/τ D[u[...

6

As I noted in a comment, you could perform the Thiele continued fraction expansion of a rational function, which is of a slightly different form from the one requested in the OP; this might nevertheless be useful in other contexts. One method for generating this is Viscovatov's algorithm, which can be implemented in Mathematica like so: viscovatov[cof_?...

2

The complete solution space to a system of ODEs can have a complicated structure, including singular components (e.g. Clairaut equations) and multiple branches. Let us call a solution $y_{\bf C}$ a locally general solution to an ODE system with smooth coefficients of dimension $n$ at $(x,{\bf C})=(x_0,{\bf C}_0)$ to a system of ODEs if the mapping that maps ...

2

Assuming you have in mind linear differential equation of n-th order, you have to check two things: That sol is indeed a solution. That you did already. That there are exactly n homogeneous solutions which are linearly independent. The key here is linear independence. In order to detect the linear dependence between functions you can calculate wronskian So,...

1

According to Encyclopedia of Mathematics, in order to prove that sol from eq = y''[x] + 4 y[x] == 7;sol = DSolveValue[{eq}, y[x], x] is a general solution of eq one should prove that sol is a solution of eq for any C[1] and C[2] (done in the question) and any Cauchy problem has a (unique) solution, more exactly, ForAll[{x0, d1, d2}, Exists[{C[1], C[2]}, ...

1

This should find the denominator (factors with a negative power) to factor out in an expression of the form A + B/C + D/E + .... The DeleteDuplicates would delete factors of the denominator with the same base but different exponents, but you would really want to select the correct one, which DeleteDuplicates is not guaranteed to do. It does not matter in ...

1

You cannot get your desired form by using Simplify. Indeed, these are your expressions: expr1 = a + (1 + b^2 + b (2 - (4 d^2 e)/(1 + d (-1 + e))))^(1 - n/2); expr2 = ((a)*(1 + d (-1 + e))^(1 - n/2) + (((1 + b^2)*(1 + d (-1 + e)) + b (2 (1 + d (-1 + e)) - 4 d^2 e))^(1 - n/2)))/(1 + d (-1 + e))^(1 - n/2); Let us count their leaves: ...

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