High order difference operator applied to 1/u

Question

I need a formula for $\Delta^k \frac 1 u$, where $u(x)$ is a strictly positive function, $\Delta^k$ is the difference operator defined recursively as $\Delta^k=\Delta^1 \Delta^{k-1}$ and $\Delta^1 u(x)=u(x+h)-u(x)$, with $h$ fixed. It is trivial to get a recursive formula by expanding the identity $\Delta^k(u\cdot\frac 1 u)=0$ via the discrete Leibnitz' formula. Thus it should be easy to prove the formula by induction. But it is not easy to guess the correct expression.

I suspect the formula is known, although possibly not widely known (like the formula for the $k$-th derivative of $1/u$). Any pointers to existing literature would be appreciated.

EDIT: I suspect the question is not clearly stated, let me elaborate on what I have in mind. Denote by $u_j(x)=u(x+jh)$. Then the formula for $k=1$ is $$ \Delta^1\frac1u=- \frac{\Delta^1u}{uu_1} $$ the formula for $k=2$ is $$ \Delta^2\frac1u= -\frac{\Delta^2u}{uu_2} +\frac{2\Delta^1u_1\Delta^1u}{uu_1u_2} $$ and so on. The pattern is clear and resembles (obviously) that for the derivatives of $1/u$.

Answer 1

You can apply the formula $$\Delta^p f(x)=\sum_{k=0}^p{p\choose k}(-1)^{p-k}f(x+k)$$ to $f(x)=1/u(x)$. Further simplification will need knowledge how $u(x)$ depends on $x$.
For example, if $u(x)=x$ one has $$\Delta^p x^{-1}=\frac{(-1)^p \Gamma (p+1) \Gamma (x) }{\Gamma (p+x+1)}.$$ If $u(x)=x^2$ one has $$\Delta^p x^{-2}=\frac{(-1)^p \Gamma (p+1) \Gamma (x) \bigl[\psi ^{(0)}(x)-\psi ^{(0)}(p+x+1)\bigr]}{\Gamma (p+x+1)},$$ and if $u(x)=x^3$ one has _{$$\Delta^p x^{-3}=\frac{(-1)^p \Gamma (p+1) \Gamma (x) \left[\bigl(\psi ^{(0)}(x)-\psi ^{(0)}(p+x+1)\bigr)^2-\psi ^{(1)}(p+x+1)+\psi ^{(1)}(x)\right]}{2 \Gamma (p+x+1)}.$$} It may be possible to find the general formula for $u(x)=x^p$.