Relation of complete homogeneous symmetric polynomials and the elementary symmetric polynomials












4












$begingroup$


I was reading about the symmetric polynomials and saw the following relation:




$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.



(Question) I was wondering if this relation still holds for any length $k$, i.e.




$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




P.S. I tried cases and it holds for any case.






abstract-algebra commutative-algebra symmetric-polynomials







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I was reading about the symmetric polynomials and saw the following relation:




    $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




    The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
    SECOND EDITION I. G. MACDONALD, page: 21.



    (Question) I was wondering if this relation still holds for any length $k$, i.e.




    $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




    P.S. I tried cases and it holds for any case.






    abstract-algebra commutative-algebra symmetric-polynomials







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I was reading about the symmetric polynomials and saw the following relation:




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




      The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
      SECOND EDITION I. G. MACDONALD, page: 21.



      (Question) I was wondering if this relation still holds for any length $k$, i.e.




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




      P.S. I tried cases and it holds for any case.






      abstract-algebra commutative-algebra symmetric-polynomials







      share|cite|improve this question











      $endgroup$




      I was reading about the symmetric polynomials and saw the following relation:




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




      The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
      SECOND EDITION I. G. MACDONALD, page: 21.



      (Question) I was wondering if this relation still holds for any length $k$, i.e.




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




      P.S. I tried cases and it holds for any case.






      abstract-algebra commutative-algebra symmetric-polynomials




      abstract-algebra commutative-algebra symmetric-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 11:06







      Max

















      asked Feb 3 at 10:11









      MaxMax

      1,0041319




      1,0041319






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          We have
          $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
          $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
          $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
          If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






          share|cite|improve this answer











          $endgroup$














            Your Answer








            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%2f3098386%2frelation-of-complete-homogeneous-symmetric-polynomials-and-the-elementary-symmet%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









            2












            $begingroup$

            We have
            $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
            $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
            $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
            If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              We have
              $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
              $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
              $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
              If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                We have
                $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
                $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
                $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
                If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






                share|cite|improve this answer











                $endgroup$



                We have
                $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
                $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
                $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
                If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 19:11

























                answered Feb 3 at 18:48









                Matt SamuelMatt Samuel

                39.3k63870




                39.3k63870






























                    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%2f3098386%2frelation-of-complete-homogeneous-symmetric-polynomials-and-the-elementary-symmet%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

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d
                    Read more

                    SQL update select statement

                    Image
                    0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
                    Read more

                    WPF add header to Image with URL pettitions [duplicate]

                    Image
                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:
                    Read more