Force a sum to be calculated












1












$begingroup$


I want to force this sum to be calculated to 1



Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]


Only DiscretePlot3D gives the correct result showing all point to 1






summation







share|improve this question









$endgroup$












  • $begingroup$
    For what values of $n$ and $m$?
    $endgroup$
    – MarcoB
    5 hours ago
















1












$begingroup$


I want to force this sum to be calculated to 1



Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]


Only DiscretePlot3D gives the correct result showing all point to 1






summation







share|improve this question









$endgroup$












  • $begingroup$
    For what values of $n$ and $m$?
    $endgroup$
    – MarcoB
    5 hours ago














1












1








1





$begingroup$


I want to force this sum to be calculated to 1



Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]


Only DiscretePlot3D gives the correct result showing all point to 1






summation







share|improve this question









$endgroup$




I want to force this sum to be calculated to 1



Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]


Only DiscretePlot3D gives the correct result showing all point to 1






summation




summation






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 5 hours ago









NitraNitra

676




676












  • $begingroup$
    For what values of $n$ and $m$?
    $endgroup$
    – MarcoB
    5 hours ago


















  • $begingroup$
    For what values of $n$ and $m$?
    $endgroup$
    – MarcoB
    5 hours ago
















$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago




$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago










1 Answer
1






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oldest

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2












$begingroup$

The following code might do what you want:



s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;

s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};

s2 = s1 /. {Sin[x_] :> Sin[Factor@x], 

            Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};

s3 = Simplify[s2, m [Element] Integers]


which evaluates to 1. The step to s1 was easy, but I had to do lots of experimentation to find the steps to s2 and s3. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)] assumes that m is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0 but perhaps others can produce them.






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    $begingroup$

    The following code might do what you want:



    s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
    
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
    
s2 = s1 /. {Sin[x_] :> Sin[Factor@x], 
    
            Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
    
s3 = Simplify[s2, m [Element] Integers]


    which evaluates to 1. The step to s1 was easy, but I had to do lots of experimentation to find the steps to s2 and s3. In particular the rule
    Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)] assumes that m is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0 but perhaps others can produce them.






    share|improve this answer











    $endgroup$


















      2












      $begingroup$

      The following code might do what you want:



      s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
      
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
      
s2 = s1 /. {Sin[x_] :> Sin[Factor@x], 
      
            Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
      
s3 = Simplify[s2, m [Element] Integers]


      which evaluates to 1. The step to s1 was easy, but I had to do lots of experimentation to find the steps to s2 and s3. In particular the rule
      Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)] assumes that m is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0 but perhaps others can produce them.






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        The following code might do what you want:



        s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
        
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
        
s2 = s1 /. {Sin[x_] :> Sin[Factor@x], 
        
            Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
        
s3 = Simplify[s2, m [Element] Integers]


        which evaluates to 1. The step to s1 was easy, but I had to do lots of experimentation to find the steps to s2 and s3. In particular the rule
        Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)] assumes that m is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0 but perhaps others can produce them.






        share|improve this answer











        $endgroup$



        The following code might do what you want:



        s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
        
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
        
s2 = s1 /. {Sin[x_] :> Sin[Factor@x], 
        
            Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
        
s3 = Simplify[s2, m [Element] Integers]


        which evaluates to 1. The step to s1 was easy, but I had to do lots of experimentation to find the steps to s2 and s3. In particular the rule
        Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)] assumes that m is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0 but perhaps others can produce them.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 41 mins ago

























        answered 1 hour ago









        SomosSomos

        1,07019




        1,07019






























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