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 $1/u$. But I have several guesses for the exact formula and knowing it would save me a lot of time.

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 $1/u$.

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.

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 $1/u$. But I have several guesses for the exact formula and knowing it would save me a lot of time.