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













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          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




















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