Is this an empty product?












1












$begingroup$


Assume that we have:
$$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
that can be written as:



$$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$



If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.



It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write



$$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$



Is this wrong? I am not familiar with this notation and it may be very trivial.






algebra-precalculus products







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Assume that we have:
    $$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
    that can be written as:



    $$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$



    If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.



    It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write



    $$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$



    Is this wrong? I am not familiar with this notation and it may be very trivial.






    algebra-precalculus products







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Assume that we have:
      $$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
      that can be written as:



      $$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$



      If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.



      It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write



      $$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$



      Is this wrong? I am not familiar with this notation and it may be very trivial.






      algebra-precalculus products







      share|cite|improve this question











      $endgroup$




      Assume that we have:
      $$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
      that can be written as:



      $$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$



      If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.



      It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write



      $$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$



      Is this wrong? I am not familiar with this notation and it may be very trivial.






      algebra-precalculus products




      algebra-precalculus products






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 11:13









      dallonsi

      1187




      1187










      asked Dec 7 '18 at 10:58









      RScrlliRScrlli

      649114




      649114






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as



          "The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"



          Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Using the same logic if $k=2$ we also have an empty product?
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:18










          • $begingroup$
            as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:28





















          1












          $begingroup$

          no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
          in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:21










          • $begingroup$
            you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:27













          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%2f3029771%2fis-this-an-empty-product%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as



          "The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"



          Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Using the same logic if $k=2$ we also have an empty product?
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:18










          • $begingroup$
            as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:28


















          1












          $begingroup$

          Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as



          "The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"



          Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Using the same logic if $k=2$ we also have an empty product?
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:18










          • $begingroup$
            as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:28
















          1












          1








          1





          $begingroup$

          Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as



          "The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"



          Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.






          share|cite|improve this answer









          $endgroup$



          Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as



          "The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"



          Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 11:15









          dallonsidallonsi

          1187




          1187












          • $begingroup$
            Using the same logic if $k=2$ we also have an empty product?
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:18










          • $begingroup$
            as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:28




















          • $begingroup$
            Using the same logic if $k=2$ we also have an empty product?
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:18










          • $begingroup$
            as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:28


















          $begingroup$
          Using the same logic if $k=2$ we also have an empty product?
          $endgroup$
          – RScrlli
          Dec 7 '18 at 11:18




          $begingroup$
          Using the same logic if $k=2$ we also have an empty product?
          $endgroup$
          – RScrlli
          Dec 7 '18 at 11:18












          $begingroup$
          as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
          $endgroup$
          – Enkidu
          Dec 7 '18 at 11:28






          $begingroup$
          as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
          $endgroup$
          – Enkidu
          Dec 7 '18 at 11:28













          1












          $begingroup$

          no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
          in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:21










          • $begingroup$
            you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:27


















          1












          $begingroup$

          no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
          in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:21










          • $begingroup$
            you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:27
















          1












          1








          1





          $begingroup$

          no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
          in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$






          share|cite|improve this answer











          $endgroup$



          no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
          in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 11:25

























          answered Dec 7 '18 at 11:09









          EnkiduEnkidu

          1,39419




          1,39419








          • 1




            $begingroup$
            makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:21










          • $begingroup$
            you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:27
















          • 1




            $begingroup$
            makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
            $endgroup$
            – RScrlli
            Dec 7 '18 at 11:21










          • $begingroup$
            you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
            $endgroup$
            – Enkidu
            Dec 7 '18 at 11:27










          1




          1




          $begingroup$
          makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
          $endgroup$
          – RScrlli
          Dec 7 '18 at 11:21




          $begingroup$
          makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
          $endgroup$
          – RScrlli
          Dec 7 '18 at 11:21












          $begingroup$
          you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
          $endgroup$
          – Enkidu
          Dec 7 '18 at 11:27






          $begingroup$
          you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
          $endgroup$
          – Enkidu
          Dec 7 '18 at 11:27




















          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%2f3029771%2fis-this-an-empty-product%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

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
          Read more