nth derivative recursive formula

Question

I'm trying to find a recursive series representation of the $n$-th derivative of the following function.

$D^{(1)}_{a, b}(x) = b\sqrt{\frac{p}{q}}D_{a + p - 1, b - 1}(x) - a\sqrt{\frac{q}{p}}D_{a - 1, b + q - 1}(x)$

I think it's pretty clear that the solution can be rearranged in the form $\sum_{k = 0}^{n}C_{a, b}(x, k)D_{a + (n - k)p - n, b + kq - n}(x)$, but as for finding a closed form for $C_{a, b}(x, k)$ I've made little to no progress.

For $k = 0$ I believe $C_{a, b}(x, k) = \frac{b!}{(b - n)!}(\frac{p}{q})^{\frac{n}{2}}$, and $k = n$ I believe $C_{a, b}(x, k) = (-1)^n\frac{a!}{(a - n)!}(\frac{q}{p})^{\frac{n}{2}}$ but this is by little more than inspection.

If it helps, I've noticed $D_{a, b}(x) = \cos^a(x)\sin^b(x)$ when $p = q = 2$.

Any and all help is greatly appreciated. Thanks in advance!

Answer 1

To find a recursive formula for the $n$-th derivative of the given function, we need to understand the provided recurrence relation and then derive the $n$-th derivative in a systematic way.

The given recurrence relation is:

$ D(1)_{a,b}(x) = b\sqrt{pq} D_{a+p-1,b-1}(x) - a\sqrt{qp} D_{a-1,b+q-1}(x) $

We want to express the $n$-th derivative $D^{(n)}_{a,b}(x)$ in terms of a sum of derivatives of the form $D_{a + (n-k)p - n, b + kq - n}(x)$.

Step 1: Define the Series Representation

Let's assume a series representation for the $n$-th derivative:

$ D^{(n)}_{a,b}(x) = \sum_{k=0}^n C_{a,b}(x,k) D_{a+(n-k)p-n, b+kq-n}(x) $

Step 2: Base Cases

Let's first identify the base cases to find some initial values for $C_{a,b}(x,k)$.

1. For $k = 0$: $ C_{a,b}(x, 0) = b! (b-n)! \left(\sqrt{pq}\right)^n 2 $

2. For $k = n$: $ C_{a,b}(x, n) = (-1)^n a! (a-n)! \left(\sqrt{qp}\right)^n 2 $

Step 3: Recurrence Relation for $C_{a,b}(x,k)$

The recurrence relation for the coefficients $C_{a,b}(x,k)$ can be obtained by differentiating the given recurrence relation recursively and matching terms.

$ D^{(n)}_{a,b}(x) = b\sqrt{pq} D^{(n-1)}_{a+p-1,b-1}(x) - a\sqrt{qp} D^{(n-1)}_{a-1,b+q-1}(x) $

Using the series representation for $D^{(n-1)}_{a+p-1,b-1}(x)$ and $D^{(n-1)}_{a-1,b+q-1}(x)$:

$ D^{(n-1)}_{a+p-1,b-1}(x) = \sum_{k=0}^{n-1} C_{a+p-1,b-1}(x,k) D_{a+(n-1-k)p+p-n, b+(k-1)q-n}(x) $

$ D^{(n-1)}_{a-1,b+q-1}(x) = \sum_{k=0}^{n-1} C_{a-1,b+q-1}(x,k) D_{a+(n-1-k)p-1-n, b+kq+q-1-n}(x) $

Matching terms in the recurrence relation gives us:

$ C_{a,b}(x,k) = b\sqrt{pq} \cdot C_{a+p-1,b-1}(x,k-1) - a\sqrt{qp} \cdot C_{a-1,b+q-1}(x,k-1) $

Step 4: General Formula for $C_{a,b}(x,k)$

The general formula for $C_{a,b}(x,k)$ can be expressed recursively as:

$ C_{a,b}(x,k) = b\sqrt{pq} \cdot C_{a+p-1,b-1}(x,k-1) - a\sqrt{qp} \cdot C_{a-1,b+q-1}(x,k-1) $

Summary

The $n$-th derivative $D^{(n)}_{a,b}(x)$ can be represented as:

$ D^{(n)}_{a,b}(x) = \sum_{k=0}^n C_{a,b}(x,k) D_{a+(n-k)p-n, b+kq-n}(x) $

where the coefficients $C_{a,b}(x,k)$ follow the recurrence relation:

$ C_{a,b}(x,k) = b\sqrt{pq} \cdot C_{a+p-1,b-1}(x,k-1) - a\sqrt{qp} \cdot C_{a-1,b+q-1}(x,k-1) $

with initial values for the base cases $k=0$ and $k=n$ provided as:

$ C_{a,b}(x, 0) = b! (b-n)! \left(\sqrt{pq}\right)^n 2 $ $ C_{a,b}(x, n) = (-1)^n a! (a-n)! \left(\sqrt{qp}\right)^n 2 $

This recursive approach should help in finding the coefficients needed for the $n$-th derivative of the function.