How did we come up with this other form for this sum?












1












$begingroup$


Could someone please tell how is it possible to change this sum :



$${x_{3}(m)=frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}} $$



to :



$$ frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l)bigg[sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}}bigg ]$$



?






sequences-and-series







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Could someone please tell how is it possible to change this sum :



    $${x_{3}(m)=frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}} $$



    to :



    $$ frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l)bigg[sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}}bigg ]$$



    ?






    sequences-and-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Could someone please tell how is it possible to change this sum :



      $${x_{3}(m)=frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}} $$



      to :



      $$ frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l)bigg[sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}}bigg ]$$



      ?






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      Could someone please tell how is it possible to change this sum :



      $${x_{3}(m)=frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}} $$



      to :



      $$ frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l)bigg[sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}}bigg ]$$



      ?






      sequences-and-series




      sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 13:12







      Hilbert

















      asked Dec 3 '18 at 10:12









      HilbertHilbert

      1479




      1479






















          1 Answer
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          $begingroup$

          begin{align}&frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} text{, distributive law}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}sum_{l=0}^{N-1}sum_{k=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l) sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}} \
          end{align}



          Notice that we are dealing with finite sum, the order of summation doesn't matter.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:14








          • 1




            $begingroup$
            For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
            $endgroup$
            – Siong Thye Goh
            Dec 3 '18 at 16:19










          • $begingroup$
            Thank you now I can see through it !
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:22











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          $begingroup$

          begin{align}&frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} text{, distributive law}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}sum_{l=0}^{N-1}sum_{k=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l) sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}} \
          end{align}



          Notice that we are dealing with finite sum, the order of summation doesn't matter.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:14








          • 1




            $begingroup$
            For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
            $endgroup$
            – Siong Thye Goh
            Dec 3 '18 at 16:19










          • $begingroup$
            Thank you now I can see through it !
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:22
















          2












          $begingroup$

          begin{align}&frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} text{, distributive law}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}sum_{l=0}^{N-1}sum_{k=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l) sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}} \
          end{align}



          Notice that we are dealing with finite sum, the order of summation doesn't matter.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:14








          • 1




            $begingroup$
            For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
            $endgroup$
            – Siong Thye Goh
            Dec 3 '18 at 16:19










          • $begingroup$
            Thank you now I can see through it !
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:22














          2












          2








          2





          $begingroup$

          begin{align}&frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} text{, distributive law}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}sum_{l=0}^{N-1}sum_{k=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l) sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}} \
          end{align}



          Notice that we are dealing with finite sum, the order of summation doesn't matter.






          share|cite|improve this answer









          $endgroup$



          begin{align}&frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} text{, distributive law}\
          &=frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}sum_{l=0}^{N-1}sum_{k=0}^{N-1}x_{1}(n)x_{2}(l) e^{j2pi kfrac{m-n-l}{N}} \
          &=frac{1}{N}sum_{n=0}^{N-1}x_{1}(n)sum_{l=0}^{N-1}x_{2}(l) sum_{k=0}^{N-1}e^{j2pi kfrac{m-n-l}{N}} \
          end{align}



          Notice that we are dealing with finite sum, the order of summation doesn't matter.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 13:35









          Siong Thye GohSiong Thye Goh

          100k1465117




          100k1465117












          • $begingroup$
            can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:14








          • 1




            $begingroup$
            For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
            $endgroup$
            – Siong Thye Goh
            Dec 3 '18 at 16:19










          • $begingroup$
            Thank you now I can see through it !
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:22


















          • $begingroup$
            can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:14








          • 1




            $begingroup$
            For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
            $endgroup$
            – Siong Thye Goh
            Dec 3 '18 at 16:19










          • $begingroup$
            Thank you now I can see through it !
            $endgroup$
            – Hilbert
            Dec 3 '18 at 16:22
















          $begingroup$
          can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
          $endgroup$
          – Hilbert
          Dec 3 '18 at 16:14






          $begingroup$
          can you please specify how did you use the distributive law in going from :$\ frac{1}{N}sum_{k=0}^{N-1}bigg[sum_{n=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}bigg ]bigg[sum_{l=0}^{N-1}x_{2}(l)e^{-j2pi kfrac{l}{N}}bigg ] e^{j2pi kfrac{m}{N}} $ to : $\ frac{1}{N}sum_{k=0}^{N-1}sum_{n=0}^{N-1}sum_{l=0}^{N-1}x_{1}(n)e^{-j2pi kfrac{n}{N}}x_{2}(l)e^{-j2pi kfrac{l}{N}} e^{j2pi kfrac{m}{N}} \$ Because I think this is really the key step .
          $endgroup$
          – Hilbert
          Dec 3 '18 at 16:14






          1




          1




          $begingroup$
          For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
          $endgroup$
          – Siong Thye Goh
          Dec 3 '18 at 16:19




          $begingroup$
          For example $(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+cd+cf$. Product of sum is the sum of the product.
          $endgroup$
          – Siong Thye Goh
          Dec 3 '18 at 16:19












          $begingroup$
          Thank you now I can see through it !
          $endgroup$
          – Hilbert
          Dec 3 '18 at 16:22




          $begingroup$
          Thank you now I can see through it !
          $endgroup$
          – Hilbert
          Dec 3 '18 at 16:22


















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