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

697




697












  • $begingroup$
    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
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066850%2fsimplify-the-sum-sum-i-1i-2-i-m-zx-1i-1x-2i-2-cdot-cdot-x%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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

        3737




        3737






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066850%2fsimplify-the-sum-sum-i-1i-2-i-m-zx-1i-1x-2i-2-cdot-cdot-x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            File:DeusFollowingSea.jpg

            Image
            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
            Read more