Simplify the sum $sum_{i_1+i_2+…+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}cdot…cdot x_{m}^{i_m}}$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 23:21







Y.L

















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


















  • $begingroup$
    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










1 Answer
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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:



$$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$






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    0












    $begingroup$

    $$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$



    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:



    $$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$



      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:



      $$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$



        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:



        $$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$






        share|cite|improve this answer









        $endgroup$



        $$m, z, i_1, ldots, i_m in mathbb{N}, quad m > 0$$



        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:



        $$left(sumlimits_{k=1}^m x_iright)^z = sumlimits_{sumlimits_{k=1}^m i_k=z}frac{z!}{prodlimits_{l=1}^m i_l!}prodlimits_{l=1}^m x_l^{i_l} = sumlimits_{i_1 + ldots + i_m = z}frac{z!}{i_1!cdot ldots cdot i_m!}x_1^{i_1}cdotldotscdot x_m^{i_m}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 23:59









        Snake707Snake707

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