Sympy can't solve this double angular summation

[ctr26]

Hi all,

    v     v                         
   ____  ____                       
   ╲     ╲                          
    ╲     ╲                         
     ╲     ╲  ⎛       ⎛2⋅π⋅(i - j)⎞⎞
2⋅   ╱     ╱  ⎜1 - cos⎜───────────⎟⎟
    ╱     ╱   ⎝       ⎝     v     ⎠⎠
   ╱     ╱                          
   ‾‾‾‾  ‾‾‾‾                       
  j = 1 i = 1    

In wolfram alpha it comes out correctly as =v

sqrt(sum_i=1^v (sum_j=1^v 2*((((1)*(1))-((1)*(1))cos(2*pi/v*((j-1)-(i-1)))))))

In sympy it outright gives up.

sp.sympify('2*Sum(1 - cos(2*pi*(i - j)/v), (i, 1, v))').doit()

Any thoughts?

Thanks

Craig

[oscarbenjamin]

The code you've shown doesn't match the depicted formula so I'm not sure exactly what sum you want to compute.

I'm not sure what approach could be used for sums like this in general. The only thing I can think is rewriting to exp. If you declare v as a positive integer then rewrite as exp then you get a complicated result:

In [16]: v = symbols('v', integer=True, positive=True)

In [17]: i, j = symbols('i, j')

In [18]: s = 2*Sum(1 - cos(2*pi*(i - j)/v), (i, 1, v), (j, 1, v))

In [19]: s
Out[19]: 
    v     v                         
   ____  ____                       
   ╲     ╲                          
    ╲     ╲                         
     ╲     ╲  ⎛       ⎛2⋅π⋅(i - j)⎞⎞
2⋅   ╱     ╱  ⎜1 - cos⎜───────────⎟⎟
    ╱     ╱   ⎝       ⎝     v     ⎠⎠
   ╱     ╱                          
   ‾‾‾‾  ‾‾‾‾                       
  j = 1 i = 1                       

In [20]: s.doit()
Out[20]: 
    v     v                         
   ____  ____                       
   ╲     ╲                          
    ╲     ╲                         
     ╲     ╲  ⎛       ⎛2⋅π⋅(i - j)⎞⎞
2⋅   ╱     ╱  ⎜1 - cos⎜───────────⎟⎟
    ╱     ╱   ⎝       ⎝     v     ⎠⎠
   ╱     ╱                          
   ‾‾‾‾  ‾‾‾‾                       
  j = 1 i = 1                       

In [21]: s.rewrite(exp)
Out[21]: 
    v     v                                            
  _____ _____                                          
  ╲     ╲                                              
   ╲     ╲    ⎛   2⋅ⅈ⋅π⋅(i - j)        -2⋅ⅈ⋅π⋅(i - j) ⎞
    ╲     ╲   ⎜   ─────────────        ───────────────⎟
     ╲     ╲  ⎜         v                     v       ⎟
2⋅   ╱     ╱  ⎜  ℯ                    ℯ               ⎟
    ╱     ╱   ⎜- ────────────── + 1 - ────────────────⎟
   ╱     ╱    ⎝        2                     2        ⎠
  ╱     ╱                                              
  ‾‾‾‾‾ ‾‾‾‾‾                                          
  j = 1 i = 1                                          

In [22]: s.rewrite(exp).doit()
Out[22]: 
                              ⎛⎧                              2⋅ⅈ⋅π    ⎞   ⎛⎧                      
                              ⎜⎪                              ─────    ⎟   ⎜⎪                      
                              ⎜⎪                                v      ⎟   ⎜⎪                      
                              ⎜⎪           v             for ℯ      = 1⎟   ⎜⎪              v       
       ⎛⎧        -2⋅ⅈ⋅π     ⎞ ⎜⎪                                       ⎟   ⎜⎪                      
       ⎜⎪        ───────    ⎟ ⎜⎪ 2⋅ⅈ⋅π    2⋅ⅈ⋅π⋅(v + 1)                ⎟   ⎜⎪   -2⋅ⅈ⋅π⋅(v + 1)     
   2   ⎜⎪           v       ⎟ ⎜⎪ ─────    ─────────────                ⎟   ⎜⎪   ───────────────    
2⋅v  - ⎜⎨v  for ℯ        = 1⎟⋅⎜⎨   v            v                      ⎟ - ⎜⎨          v           
       ⎜⎪                   ⎟ ⎜⎪ℯ      - ℯ                             ⎟   ⎜⎪- ℯ                + ℯ
       ⎜⎪0     otherwise    ⎟ ⎜⎪───────────────────────    otherwise   ⎟   ⎜⎪──────────────────────
       ⎝⎩                   ⎠ ⎜⎪            2⋅ⅈ⋅π                      ⎟   ⎜⎪              -2⋅ⅈ⋅π  
                              ⎜⎪            ─────                      ⎟   ⎜⎪              ─────── 
                              ⎜⎪              v                        ⎟   ⎜⎪                 v    
                              ⎜⎪       1 - ℯ                           ⎟   ⎜⎪         1 - ℯ        
                              ⎝⎩                                       ⎠   ⎝⎩                      

              -2⋅ⅈ⋅π     ⎞                     
              ───────    ⎟                     
                 v       ⎟                     
         for ℯ        = 1⎟                     
                         ⎟ ⎛⎧        2⋅ⅈ⋅π    ⎞
-2⋅ⅈ⋅π                   ⎟ ⎜⎪        ─────    ⎟
───────                  ⎟ ⎜⎪          v      ⎟
   v                     ⎟⋅⎜⎨v  for ℯ      = 1⎟
                         ⎟ ⎜⎪                 ⎟
───────     otherwise    ⎟ ⎜⎪0    otherwise   ⎟
                         ⎟ ⎝⎩                 ⎠
                         ⎟                     
                         ⎟                     
                         ⎟                     
                         ⎠                     

In [23]: simplify(_)
Out[23]: 
⎧           -2⋅ⅈ⋅π         2⋅ⅈ⋅π    
⎪           ───────        ─────    
⎪              v             v      
⎨ 0    for ℯ        = 1 ∧ ℯ      = 1
⎪                                   
⎪   2                               
⎩2⋅v             otherwise          

I think that the problem is that we need v to be greater than 1 (the formula doesn't hold when v=1) but it isn't easy to express that requirement in SymPy. The Piecewise above does represent that though (if you substitute v=1 you'll get 0).

[ctr26]

I should've specified that the assumptions are that v is a positive integer, but in practise it's larger than 2 and probably 3; and that v is the the number of vertices of a 2D polygon. Rewrite exp looks like the way though I don't understand how mathematica and wolfram know to try that, unless they just have more summation identities?