LChrisman/Computing Derivatives for MetaLog

This page outlines an algorithm for computing the [math]\displaystyle{ i^{th} }[/math] derivative of a Keelin MetaLog.

From Eq(6) in the Baucells, Chrisman, Keelin paper draft:

[math]\displaystyle{ M^{(i)}(y) = \mu^{(i)}(y) + s^{(i)} \ell(y) + \sum_{j=0}^{i-1} s^{(j)}(y) \ell^{(i-j)}(y) }[/math]

thus in

[math]\displaystyle{ y^i (1-y)^i M^{(i)}(y) = y^i (1-y)^i \mu^{(i)}(y) + y^i (1-y)^i s^{(i)} \ell(y) + y^i (1-y)^i \sum_{j=0}^{i-1} s^{(j)}(y) \ell^{(i-j)}(y) }[/math]

the first and third term are polynomials, and the middle term is a polynomial times a logit. Hence, this can always be written in the form:

[math]\displaystyle{ y^i (1-y)^i M^{(i)}(y) = poly_1(y) + poly_2(y) \ell(y) }[/math]

The polynomials for the first term is easy, as is [math]\displaystyle{ poly_2(y) }[/math] Hence, we just need to obtain [math]\displaystyle{ T(y) }[/math] in the form of a polynomial, where [math]\displaystyle{ T(y) = y^i (1-y)^i \sum_{j=0}^{i-1} s^{(j)}(y) \ell^{(i-j)}(y) }[/math]

Change of variable

[math]\displaystyle{ u = y-0.5 }[/math]

[math]\displaystyle{ T(u) = T(y-0.5) }[/math]

Algorithm

The algorithm uses arithmetic operations on a polynomial-coefficients data type. Here [math]\displaystyle{ T(u) }[/math], [math]\displaystyle{ s^{(j)} }[/math], [math]\displaystyle{ R(u) }[/math] and [math]\displaystyle{ N(u) }[/math] are all polynomials. The order of the derivative, [math]\displaystyle{ i }[/math] is given.

1. For each [math]\displaystyle{ j=0 \mbox{ to } i }[/math]
   • Compute [math]\displaystyle{ C_j = \left( \begin{array}{c}i \\ j\end{array}\right) ( i-j-1)! }[/math]
   • Compute [math]\displaystyle{ s^{(j)}(u) }[/math]
   • Compute [math]\displaystyle{ R(u) = (0.5+u)^j (0.5-u)^j }[/math]
   • Compute [math]\displaystyle{ N(u) = (0.5+u)^{i-j} - (-1)^{i-j} ( 0.5-u)^{i-j} }[/math]
   • Compute [math]\displaystyle{ T_j(u) = C_j \cdot s^{(j)}(u) \cdot R(u) \cdot N(u) }[/math]
2. [math]\displaystyle{ T(u) = \sum_{j=0}^i T_j(u) }[/math]