Converting a Product to a Sum












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How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?










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  • $begingroup$
    If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
    $endgroup$
    – AlgebraicallyClosed
    Jan 30 at 22:12










  • $begingroup$
    @AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
    $endgroup$
    – Yves Daoust
    Jan 30 at 22:15
















1












$begingroup$


How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?










share|cite|improve this question











$endgroup$












  • $begingroup$
    If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
    $endgroup$
    – AlgebraicallyClosed
    Jan 30 at 22:12










  • $begingroup$
    @AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
    $endgroup$
    – Yves Daoust
    Jan 30 at 22:15














1












1








1


0



$begingroup$


How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?










share|cite|improve this question











$endgroup$




How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?







algebra-precalculus products






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edited Jan 30 at 22:14









Kevin Long

3,58121431




3,58121431










asked Jan 30 at 22:03









A. LavieA. Lavie

274




274












  • $begingroup$
    If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
    $endgroup$
    – AlgebraicallyClosed
    Jan 30 at 22:12










  • $begingroup$
    @AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
    $endgroup$
    – Yves Daoust
    Jan 30 at 22:15


















  • $begingroup$
    If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
    $endgroup$
    – AlgebraicallyClosed
    Jan 30 at 22:12










  • $begingroup$
    @AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
    $endgroup$
    – Yves Daoust
    Jan 30 at 22:15
















$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12




$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12












$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15




$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$



Now the expression



$$prod_{i=1}^k(x^i-1)$$



yields irregular polynomials.



$$x-1$$



$$x^3-x^2-x+1$$



$$x^6-x^5-x^4+x^2+x-1$$



$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$



$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$



$$cdots$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes... is there a general expression (a sum) which I can write for each value of k?
    $endgroup$
    – A. Lavie
    Jan 30 at 22:47










  • $begingroup$
    $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
    $endgroup$
    – Thies Heidecke
    Jan 30 at 22:57










  • $begingroup$
    @A.Lavie: there isn't.
    $endgroup$
    – Yves Daoust
    Jan 31 at 8:03



















0












$begingroup$

The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$



    Now the expression



    $$prod_{i=1}^k(x^i-1)$$



    yields irregular polynomials.



    $$x-1$$



    $$x^3-x^2-x+1$$



    $$x^6-x^5-x^4+x^2+x-1$$



    $$x^{10}-x^9-x^8+2x^5-x^2-x+1$$



    $$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$



    $$cdots$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes... is there a general expression (a sum) which I can write for each value of k?
      $endgroup$
      – A. Lavie
      Jan 30 at 22:47










    • $begingroup$
      $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
      $endgroup$
      – Thies Heidecke
      Jan 30 at 22:57










    • $begingroup$
      @A.Lavie: there isn't.
      $endgroup$
      – Yves Daoust
      Jan 31 at 8:03
















    1












    $begingroup$

    $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$



    Now the expression



    $$prod_{i=1}^k(x^i-1)$$



    yields irregular polynomials.



    $$x-1$$



    $$x^3-x^2-x+1$$



    $$x^6-x^5-x^4+x^2+x-1$$



    $$x^{10}-x^9-x^8+2x^5-x^2-x+1$$



    $$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$



    $$cdots$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes... is there a general expression (a sum) which I can write for each value of k?
      $endgroup$
      – A. Lavie
      Jan 30 at 22:47










    • $begingroup$
      $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
      $endgroup$
      – Thies Heidecke
      Jan 30 at 22:57










    • $begingroup$
      @A.Lavie: there isn't.
      $endgroup$
      – Yves Daoust
      Jan 31 at 8:03














    1












    1








    1





    $begingroup$

    $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$



    Now the expression



    $$prod_{i=1}^k(x^i-1)$$



    yields irregular polynomials.



    $$x-1$$



    $$x^3-x^2-x+1$$



    $$x^6-x^5-x^4+x^2+x-1$$



    $$x^{10}-x^9-x^8+2x^5-x^2-x+1$$



    $$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$



    $$cdots$$






    share|cite|improve this answer











    $endgroup$



    $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$



    Now the expression



    $$prod_{i=1}^k(x^i-1)$$



    yields irregular polynomials.



    $$x-1$$



    $$x^3-x^2-x+1$$



    $$x^6-x^5-x^4+x^2+x-1$$



    $$x^{10}-x^9-x^8+2x^5-x^2-x+1$$



    $$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$



    $$cdots$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 30 at 22:27

























    answered Jan 30 at 22:22









    Yves DaoustYves Daoust

    132k676230




    132k676230












    • $begingroup$
      Yes... is there a general expression (a sum) which I can write for each value of k?
      $endgroup$
      – A. Lavie
      Jan 30 at 22:47










    • $begingroup$
      $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
      $endgroup$
      – Thies Heidecke
      Jan 30 at 22:57










    • $begingroup$
      @A.Lavie: there isn't.
      $endgroup$
      – Yves Daoust
      Jan 31 at 8:03


















    • $begingroup$
      Yes... is there a general expression (a sum) which I can write for each value of k?
      $endgroup$
      – A. Lavie
      Jan 30 at 22:47










    • $begingroup$
      $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
      $endgroup$
      – Thies Heidecke
      Jan 30 at 22:57










    • $begingroup$
      @A.Lavie: there isn't.
      $endgroup$
      – Yves Daoust
      Jan 31 at 8:03
















    $begingroup$
    Yes... is there a general expression (a sum) which I can write for each value of k?
    $endgroup$
    – A. Lavie
    Jan 30 at 22:47




    $begingroup$
    Yes... is there a general expression (a sum) which I can write for each value of k?
    $endgroup$
    – A. Lavie
    Jan 30 at 22:47












    $begingroup$
    $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
    $endgroup$
    – Thies Heidecke
    Jan 30 at 22:57




    $begingroup$
    $prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
    $endgroup$
    – Thies Heidecke
    Jan 30 at 22:57












    $begingroup$
    @A.Lavie: there isn't.
    $endgroup$
    – Yves Daoust
    Jan 31 at 8:03




    $begingroup$
    @A.Lavie: there isn't.
    $endgroup$
    – Yves Daoust
    Jan 31 at 8:03











    0












    $begingroup$

    The sums are just geometric series which you can sum, getting
    $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
    The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
    $$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
    but I don't think that is helping.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The sums are just geometric series which you can sum, getting
      $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
      The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
      $$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
      but I don't think that is helping.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The sums are just geometric series which you can sum, getting
        $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
        The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
        $$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
        but I don't think that is helping.






        share|cite|improve this answer









        $endgroup$



        The sums are just geometric series which you can sum, getting
        $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
        The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
        $$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
        but I don't think that is helping.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 22:17









        Ross MillikanRoss Millikan

        301k24200375




        301k24200375






























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