Simplify a weighted average with enumerations as weights
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Could you help me simplify the following expression:
$$forall n_0 ge 0, forall m_1,m_2 gt 0, forall i_1 in [[0,n_0m_1]],forall i_2 in [[0,n_0m_2]],$$
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k=0}^{n_0} w_k k}{sum_{k=0}^{n_0} w_k n_0},$$
$$w_k = w(i_1-k,m_1-1,n_0+1) w(i_2-k,m_2-1,n_0+1),$$
$$w(i,n,m) = sum_{k=0}^n (-1)^k binom{n}{k} biggl(binom{n}{i-km}biggr)$$
(with the convention that $0!=1$ and $forall k lt 0, frac{1}{k!} = 0$, and the notations $m_j = n_j +1$ and $bigl(binom{n}{k}bigr) = binom{n+k-1}{k}$)
which gives:
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2} bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) k_0}{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2}bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) n_0}$$
The goal would be to have a more compact expression, especially for the weights of the average ($w(i,n,m)$), to get rid of the sums, and to make it easier to compute the corresponding function: $$forall x_1,x_2 in [0,1], m(x_1,x_2)=lim_{n_0,m_1,m_2 to infty, } m_{n_0,m_1,m_2}(lfloor x_1 n_0m_1 rfloor,lfloor x_2 n_0m_2 rfloor)$$
I tried to use falling and rising factorials but it didn't led me as far as I'd want to.
I also read about Dixon's identity and MacMahon Master theorem but I don't see how to use them.
To add some background, this is my current attempt to answer this question, which I will edit soon to explain how I get to this expression.
combinatorics summation binomial-coefficients recreational-mathematics average
asked Sep 26 '18 at 14:07
CidToriCidTori
536
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add a comment |
$begingroup$
Could you help me simplify the following expression:
$$forall n_0 ge 0, forall m_1,m_2 gt 0, forall i_1 in [[0,n_0m_1]],forall i_2 in [[0,n_0m_2]],$$
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k=0}^{n_0} w_k k}{sum_{k=0}^{n_0} w_k n_0},$$
$$w_k = w(i_1-k,m_1-1,n_0+1) w(i_2-k,m_2-1,n_0+1),$$
$$w(i,n,m) = sum_{k=0}^n (-1)^k binom{n}{k} biggl(binom{n}{i-km}biggr)$$
(with the convention that $0!=1$ and $forall k lt 0, frac{1}{k!} = 0$, and the notations $m_j = n_j +1$ and $bigl(binom{n}{k}bigr) = binom{n+k-1}{k}$)
which gives:
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2} bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) k_0}{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2}bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) n_0}$$
The goal would be to have a more compact expression, especially for the weights of the average ($w(i,n,m)$), to get rid of the sums, and to make it easier to compute the corresponding function: $$forall x_1,x_2 in [0,1], m(x_1,x_2)=lim_{n_0,m_1,m_2 to infty, } m_{n_0,m_1,m_2}(lfloor x_1 n_0m_1 rfloor,lfloor x_2 n_0m_2 rfloor)$$
I tried to use falling and rising factorials but it didn't led me as far as I'd want to.
I also read about Dixon's identity and MacMahon Master theorem but I don't see how to use them.
To add some background, this is my current attempt to answer this question, which I will edit soon to explain how I get to this expression.
combinatorics summation binomial-coefficients recreational-mathematics average
asked Sep 26 '18 at 14:07
CidToriCidTori
536
$endgroup$
$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00
add a comment |
$begingroup$
Could you help me simplify the following expression:
$$forall n_0 ge 0, forall m_1,m_2 gt 0, forall i_1 in [[0,n_0m_1]],forall i_2 in [[0,n_0m_2]],$$
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k=0}^{n_0} w_k k}{sum_{k=0}^{n_0} w_k n_0},$$
$$w_k = w(i_1-k,m_1-1,n_0+1) w(i_2-k,m_2-1,n_0+1),$$
$$w(i,n,m) = sum_{k=0}^n (-1)^k binom{n}{k} biggl(binom{n}{i-km}biggr)$$
(with the convention that $0!=1$ and $forall k lt 0, frac{1}{k!} = 0$, and the notations $m_j = n_j +1$ and $bigl(binom{n}{k}bigr) = binom{n+k-1}{k}$)
which gives:
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2} bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) k_0}{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2}bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) n_0}$$
The goal would be to have a more compact expression, especially for the weights of the average ($w(i,n,m)$), to get rid of the sums, and to make it easier to compute the corresponding function: $$forall x_1,x_2 in [0,1], m(x_1,x_2)=lim_{n_0,m_1,m_2 to infty, } m_{n_0,m_1,m_2}(lfloor x_1 n_0m_1 rfloor,lfloor x_2 n_0m_2 rfloor)$$
I tried to use falling and rising factorials but it didn't led me as far as I'd want to.
I also read about Dixon's identity and MacMahon Master theorem but I don't see how to use them.
To add some background, this is my current attempt to answer this question, which I will edit soon to explain how I get to this expression.
combinatorics summation binomial-coefficients recreational-mathematics average
asked Sep 26 '18 at 14:07
CidToriCidTori
536
$endgroup$
Could you help me simplify the following expression:
$$forall n_0 ge 0, forall m_1,m_2 gt 0, forall i_1 in [[0,n_0m_1]],forall i_2 in [[0,n_0m_2]],$$
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k=0}^{n_0} w_k k}{sum_{k=0}^{n_0} w_k n_0},$$
$$w_k = w(i_1-k,m_1-1,n_0+1) w(i_2-k,m_2-1,n_0+1),$$
$$w(i,n,m) = sum_{k=0}^n (-1)^k binom{n}{k} biggl(binom{n}{i-km}biggr)$$
(with the convention that $0!=1$ and $forall k lt 0, frac{1}{k!} = 0$, and the notations $m_j = n_j +1$ and $bigl(binom{n}{k}bigr) = binom{n+k-1}{k}$)
which gives:
$$m_{n_0,m_1,m_2}(i_1,i_2) = frac{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2} bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) k_0}{sum_{k_0=0}^{n_0} sum_{k_1=0}^{m_1-1} (-1)^{k_1} binom{m_1-1}{k_1} bigl(binom{m_1-1}{i_1-k_0-k_1(n_0+1)}bigr) sum_{k_2=0}^{m_2-1} (-1)^{k_2} binom{m_2-1}{k_2}bigl(binom{m_2-1}{i_2-k_0-k_2(n_0+1)}bigr) n_0}$$
The goal would be to have a more compact expression, especially for the weights of the average ($w(i,n,m)$), to get rid of the sums, and to make it easier to compute the corresponding function: $$forall x_1,x_2 in [0,1], m(x_1,x_2)=lim_{n_0,m_1,m_2 to infty, } m_{n_0,m_1,m_2}(lfloor x_1 n_0m_1 rfloor,lfloor x_2 n_0m_2 rfloor)$$
I tried to use falling and rising factorials but it didn't led me as far as I'd want to.
I also read about Dixon's identity and MacMahon Master theorem but I don't see how to use them.
To add some background, this is my current attempt to answer this question, which I will edit soon to explain how I get to this expression.
combinatorics summation binomial-coefficients recreational-mathematics average
combinatorics summation binomial-coefficients recreational-mathematics average
asked Sep 26 '18 at 14:07
CidToriCidTori
536
asked Sep 26 '18 at 14:07
CidToriCidTori
536
edited Dec 17 '18 at 8:57
CidTori
asked Sep 26 '18 at 14:07
CidToriCidTori
536
asked Sep 26 '18 at 14:07
CidToriCidTori
536
asked Sep 26 '18 at 14:07
CidToriCidTori
536
536
$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00
add a comment |
$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00
$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00
add a comment |
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$begingroup$
Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = frac{i}{n_0}frac{sum w_k}{sum w_k} = frac{i}{n_0}$, or am I missing something?
$endgroup$
– pisoir
Dec 16 '18 at 16:24
$begingroup$
Nope it was a stupid mistake, thanks! I corrected it.
$endgroup$
– CidTori
Dec 16 '18 at 18:00