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Fixed the description of the contour.
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Emre
Emre
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A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{n!}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$$$ \frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $n$##

I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.

To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{n!}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $n$##

I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.

To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $n$##

I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.

To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

Made the formula more pleasing to the eye. Made the connection to Cauchy's formula clearer.
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Emre
Emre
  • 833
  • 6
  • 14

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{1}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^\ell} = d^{-\ell}\prod_{j=1}^{\ell-1}\left(\frac{a}{j}-d\right),\quad a,d,\ell \in \mathbb{Z}_{\ge 1}. $$$$ \frac{n!}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $\ell$$n$##

I verified this for thousands of particular values of $(a,d,\ell)$$(a,d,n)$ and proved it for all $(a,d)$ with small $\ell$$n$.

To get a proof for small $\ell$$n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $\ell$$n$ and I don't think a proof for all $\ell$$n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,\ell)$$(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,\ell)$$(a,d,n)$.

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{1}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^\ell} = d^{-\ell}\prod_{j=1}^{\ell-1}\left(\frac{a}{j}-d\right),\quad a,d,\ell \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $\ell$##

I verified this for thousands of particular values of $(a,d,\ell)$ and proved it for all $(a,d)$ with small $\ell$.

To get a proof for small $\ell$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $\ell$ and I don't think a proof for all $\ell$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,\ell)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,\ell)$.

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{n!}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $n$##

I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.

To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

Source Link
Emre
Emre
  • 833
  • 6
  • 14

A natural residue formula

A residue formula

I have strong evindence to believe that the following identity holds: $$ \frac{1}{2\pi i}\oint_{|z-1|<\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^\ell} = d^{-\ell}\prod_{j=1}^{\ell-1}\left(\frac{a}{j}-d\right),\quad a,d,\ell \in \mathbb{Z}_{\ge 1}. $$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

##Verification for a finite sample and proof for small $\ell$##

I verified this for thousands of particular values of $(a,d,\ell)$ and proved it for all $(a,d)$ with small $\ell$.

To get a proof for small $\ell$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $\ell$ and I don't think a proof for all $\ell$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,\ell)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,\ell)$.