Simplify the sum $sum_{i_1+i_2+…+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot…cdot x_{m}^{i_m}}$
$begingroup$
I was wondering is it possible to simplify the following sum:
$$sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot...cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose the sum?
For $m=2$ it is simple. Just could not find it for $m>2$.
Maybe it has to do with multinomial theorem, but how?
Would appreciate your help.
real-analysis calculus sequences-and-series combinatorics summation
asked Jan 8 at 23:14
Y.LY.L
767
$endgroup$
add a comment |
$begingroup$
I was wondering is it possible to simplify the following sum:
$$sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot...cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose the sum?
For $m=2$ it is simple. Just could not find it for $m>2$.
Maybe it has to do with multinomial theorem, but how?
Would appreciate your help.
real-analysis calculus sequences-and-series combinatorics summation
asked Jan 8 at 23:14
Y.LY.L
767
$endgroup$
$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
1
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13
add a comment |
$begingroup$
I was wondering is it possible to simplify the following sum:
$$sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot...cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose the sum?
For $m=2$ it is simple. Just could not find it for $m>2$.
Maybe it has to do with multinomial theorem, but how?
Would appreciate your help.
real-analysis calculus sequences-and-series combinatorics summation
asked Jan 8 at 23:14
Y.LY.L
767
$endgroup$
I was wondering is it possible to simplify the following sum:
$$sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot...cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose the sum?
For $m=2$ it is simple. Just could not find it for $m>2$.
Maybe it has to do with multinomial theorem, but how?
Would appreciate your help.
real-analysis calculus sequences-and-series combinatorics summation
real-analysis calculus sequences-and-series combinatorics summation
asked Jan 8 at 23:14
Y.LY.L
767
asked Jan 8 at 23:14
Y.LY.L
767
edited Jan 8 at 23:21
Y.L
asked Jan 8 at 23:14
Y.LY.L
767
asked Jan 8 at 23:14
Y.LY.L
767
asked Jan 8 at 23:14
Y.LY.L
767
767
$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
1
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13
add a comment |
$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
1
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13
$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
1
1
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$
answered Jan 8 at 23:59
Snake707Snake707
4537
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add a comment |
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$begingroup$
$$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$
answered Jan 8 at 23:59
Snake707Snake707
4537
$endgroup$
add a comment |
$begingroup$
$$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$
answered Jan 8 at 23:59
Snake707Snake707
4537
$endgroup$
add a comment |
$begingroup$
$$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$
answered Jan 8 at 23:59
Snake707Snake707
4537
$endgroup$
$$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$
answered Jan 8 at 23:59
Snake707Snake707
4537
answered Jan 8 at 23:59
Snake707Snake707
4537
answered Jan 8 at 23:59
Snake707Snake707
4537
answered Jan 8 at 23:59
Snake707Snake707
4537
4537
add a comment |
add a comment |
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$begingroup$
What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula?
$endgroup$
– Hagen von Eitzen
Jan 8 at 23:33
$begingroup$
if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+cdots+x_m)^m$
$endgroup$
– Shrey Joshi
Jan 8 at 23:53
$begingroup$
@Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once.
$endgroup$
– Snake707
Jan 9 at 0:00
1
$begingroup$
Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables.
$endgroup$
– Mike Earnest
Jan 9 at 0:31
$begingroup$
Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :)
$endgroup$
– Y.L
Jan 9 at 2:13