gfun[istranscendental] test transcendence of a solution to a linear differential equation with initial conditions

Calling Sequence

istranscendental(deq, y(z))

Parameters

deq - a set containing a linear differential equation with polynomial coefficients and initial conditions specifying a unique solution of it;

Description

Given a linear differential equation together with initial conditions specifying a unique solution of it, the istranscendent command determines whether that solution is transcendent or not. It returns either true, followed by an explanation, in which case the solution is transcendental, or FAIL, meaning that it was not able to prove that the solution is transcendental.
This command relies on the command minimizediffeq of the gfun package.
This command is part of the gfun package, so it can be used in the form istranscendental(..) only after executing the command with(gfun). However, it can always be accessed through the long form of the command by using gfun[istranscendental](..).

Examples

> with(gfun):

> f:=exp(z):

> deq:=holexprtodiffeq(f,y(z));

\[\{\frac{d}{d z}y \! \left(z \right)-y \! \left(z \right), y \! \left(0\right) = 1\}\]

> istranscendental(deq,y(z));

\[\mathit{true},\quad \text{``irregular singularity at infinity''}\]

> deq:={(20*z^6+12*z^5)*y(z)+(4*z^7+z^2+3*z-9)*diff(y(z),z)+(z^3-3*z^2)*diff(diff(y(z)
,z),z), y(0) = 1};

\[\{\left(20 z^{6}+12 z^{5}\right) y \! \left(z \right)+\left(4 z^{7}+z^{2}+3 z -9\right) \left(\frac{d}{d z}y \! \left(z \right)\right)+\left(z^{3}-3 z^{2}\right) \left(\frac{d^{2}}{d z^{2}}y \! \left(z \right)\right), y \! \left(0\right) = 1\}\]

> istranscendental(deq,y(z));

\[\mathit{true},\quad \text{``irregular singularity at infinity''}\]

> deq:={z^2*(1+z)*diff(diff(diff(y(z),z),z),z)-z*(2*z^2+2*z-1)*diff(diff(y(z),z),z)+(-
z^2-4*z-1)*diff(y(z),z), y(0) = 0, (D@@2)(y)(0) = 1/4};

\[\left\{z^{2} \left(1+z \right) \left(\frac{d^{3}}{d z^{3}}y \! \left(z \right)\right)-z \left(2 z^{2}+2 z -1\right) \left(\frac{d^{2}}{d z^{2}}y \! \left(z \right)\right)+\left(-z^{2}-4 z -1\right) \left(\frac{d}{d z}y \! \left(z \right)\right), y \! \left(0\right) = 0, D^{\left(2\right)}\! \left(y \right)\! \left(0\right) = {\frac{1}{4}}\right\}\]

> istranscendental(deq,y(z));

\[\mathit{true},\quad \text{``root of multiplicity 2 of the indicial equation ==\textgreater ln at 0''}\]

> f:=sqrt(1-z);

\[\sqrt{1-z}\]

> deq:=holexprtodiffeq(f,y(z));

\[\{\left(2 z -2\right) \left(\frac{d}{d z}y \! \left(z \right)\right)-y \! \left(z \right), y \! \left(0\right) = 1\}\]

> istranscendental(deq,y(z));

\[\mathit{FAIL}\]

A tricky hypergeometric function:

> f:=hypergeom([1/6,5/6],[7/6],z):

> deq:=holexprtodiffeq(f,y(z));

\[\{\left(36 z^{2}-36 z \right) \left(\frac{d^{2}}{d z^{2}}y \! \left(z \right)\right)+\left(72 z -42\right) \left(\frac{d}{d z}y \! \left(z \right)\right)+5 y \! \left(z \right), y \! \left(0\right) = 1\}\]

> istranscendental(deq,y(z));