Returns the canonical representation for the ratio of two polynomials given a derivation.

Specifically, given two polynomials \(a\) and \(d\) and a derivation \(D\), it returns a tuple of polynomials \((p, (b, d_n), (c, d_s))\) such that

\[ \frac{a}{d} = p + \frac{b}{d_n} + \frac{c}{d_s} \]

where each squarefree factor \(n\) of \(d_n\) is normal and each squarefree factor \(s\) of \(d_s\) is special. The degree of \(b\) will be less than that of \(d_n\), and the degree of \(c\) will be less than that of \(d_s\). In the canonical representation of \(\frac{a}{d}\), \(p\) is the polynomial part, \(\frac{b}{d_n}\) is the normal part, and \(\frac{c}{d_s}\) is the special part.

For example, given the derivation \(D = \frac{d}{dx}\), the canonical representation of \(\frac{x^5+x^2+1}{x^3-3x^2+4}\) is \((x^2 + 3x + 9) + \frac{24x^2 - 12x - 35}{x^3 - 3x^2 + 4} + \frac{0}{1}\).

>>> let a = power 5 + power 2 + 1 :: IndexedPolynomial
>>> let d = power 3 - 3 * power 2 + 4 :: IndexedPolynomial
>>> canonical differentiate $ fromPolynomials a d
(x^2 + 3x + 9,Function (24x^2 + (-12)x + (-35)) (x^3 + (-3)x^2 + 4),Function (0) (1))

In comparison, given the derivation \(Dx = x + 1\), the canonical representation is \((x^2+3x+9) + \frac{215x-319}{9x^2-36x+36} + \frac{1}{9x+9}\).

>>> canonical (extend (const 0) (power 1 + 1)) $ fromPolynomials a d
(x^2 + 3x + 9,Function ((215 % 9)x + ((-319) % 9)) (x^2 + (-4)x + 4),Function ((1 % 9)) (x + 1))

Note for readers of Symbolic Integration I: Transcendental Functions

Returns the splitting factorization of a given polynomial with respect to a given derivation. Specifically, with derivation \(D\) and polynomial \(p\), it returns the polynomials \((q_n, q_s)\) such that \(p = q_n q_s\), the squarefree factors of \(q_n\) are normal, and the squarefree factors of \(q_s\) are special.

For example, with derivation \(D = \frac{d}{dx}\) and \(p = x^3 - 3x^2 + 4\), the splitting factorization is \(p = (x^3 - 3x^2 + 4) \times 1\).

>>> let p = power 3 - 3 * power 2 + 4 :: IndexedPolynomial
>>> splitFactor differentiate p
(x^3 + (-3)x^2 + 4,1)

In comparison, with derivation \(Dx = x + 1\), the splitting factorization is \(p = \left( \frac{x^2 - 4x + 4}{9} \right) \times (9x + 9)\).

>>> splitFactor (extend (const 0) (power 1 + 1)) p
((1 % 9)x^2 + ((-4) % 9)x + (4 % 9),9x + 9)

Returns the squarefree factors in the splitting factorization of a given polynomial with respect to a given derivation. Specifically, with derivation \(D\) and polynomial \(p\), it returns a list of squarefree polynomials \((n_i, s_i)\) such that \(n_i\) are normal and \(s_i\) are special, and also that

\[ p = \left( \prod_{i=1}^n n_i^i \right) \left( \prod_{i=1}^n s_i^i \right) \]

For example, with derivation \(D=\frac{d}{dx}\), the squarefree factors in the splitting factorization of \(p=x^3 - 3x^2 + 4\) are \(\left( (x+1)^1 (x-2)^2 \right) \cdot \left( 1^1 \times 1^2 \right)\).

>>> let p = power 3 - 3 * power 2 + 4 :: IndexedPolynomial
>>> splitSquarefreeFactor differentiate p
[(x + 1,1),(x + (-2),1)]

In comparison, with derivation \(Dx = x+1\), the squarefree factors in the splitting factorization are \(\left( 1^1 \times (-\frac{1}{3}x+\frac{2}{3})^2\right) \cdot \left( (x+1)^1 \times (-3)^2 \right)\).

>>> splitSquarefreeFactor (extend (const 0) (power 1 + 1)) p
[(1,x + 1),(((-1) % 3)x + (2 % 3),(-3))]

For a differential field \((F, D)\), \(t\) transcendental over \(F\), and \(w \in F(t)\), returns the unique extension \(\Delta\) of \(D\) such that \(\Delta t = w\) and \((F(t), \Delta)\) is a differential field.

For example, given the derivation \(Dx=0\) on the rational numbers, a differential field \((\mathbb{Q}(t), \Delta)\) can be obtained where the derivation satisfies \(\Delta t = t + 1\).

>>> let w = power 1 + 1 :: IndexedPolynomial
>>> let derivation = extend (const 0) w
>>> derivation 0
0
>>> derivation $ power 1
x + 1
>>> derivation $ power 2 + power 1 + 1
2x^2 + 3x + 1

As another example, given the derivation \(D = \frac{d}{dx}\), we can derive the unique derivation \(\Delta\) such that \(\Delta t = t^2+1\).

>>> let w' = power 2 + 1 :: IndexedPolynomialWith IndexedPolynomial
>>> let derivation' = extend differentiate w'
>>> derivation' 0
0
>>> derivation' $ power 1
[(0,1),(2,1)]
>>> derivation' $ scale (power 2) (power 1)
[(0,x^2),(1,2x),(2,x^2)]

Check whether a given function \(D\) is consistent with being a derivation with the two given values \(a\) and \(b\).

Specifically, it checks that the following are true:

\[ D(a + b) = D(a) + D(b) \]

\[ D(ab) = a D(b) + b D(a) \]

For example,

>>> consistent differentiate (power 1 - 1) (power 3 :: IndexedPolynomial)
True
>>> consistent abs (-1) 1
False