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infinite_product_of_complex_numbers [2015/04/16 13:21]
nikolaj
infinite_product_of_complex_numbers [2016/04/02 14:53]
nikolaj
Line 14: Line 14:
 >can Euler products arise in this way? >can Euler products arise in this way?
  
-And they may arise as determinants infinite dimensional operators, i.e. products of eigenvalues.+And they may arise ans determinants ​of infinite dimensional operators, i.e. products of eigenvalues.
  
 === Theorems === === Theorems ===
Line 28: Line 28:
 Also, from the sum formula with $a_n=\prod_{k=K}^{n-1}b_k$,​ we get Also, from the sum formula with $a_n=\prod_{k=K}^{n-1}b_k$,​ we get
  
-$\prod_{k=K}^\infty b_k=\lim_{n\to\infty}a_n=\prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,​\prod_{k=K}^{n-1}b_k$ ^+$\prod_{k=K}^\infty b_k=\lim_{n\to\infty}a_n=\prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,​\prod_{k=K}^{n-1}b_k$ 
 + 
 +^ $\prod_{k=0}^\infty b_k = \prod_{k=0}^{M-1}b_k + \sum_{n=M}^\infty(b_n-1)\,​\prod_{k=0}^{n-1}b_k$ ^
  
 <​code>​ <​code>​
 b[n_] = 1 + 1/n!; b[n_] = 1 + 1/n!;
 +
 +Ks = 0;
 +Ke = 100;
 +Mid = 4;
 +
 +Product[b[k],​ {k, Ks, Ke}]
 +
 Ks = 0; Ks = 0;
 Ke = 100; Ke = 100;
 Mid = 4; Mid = 4;
  
-Product[b[k],​ {k, Ks, Ke}] // N 
 Product[b[k],​ {k, Ks, Mid - 1}] +  Product[b[k],​ {k, Ks, Mid - 1}] + 
-  Sum[(b[k] - 1) Product[b[k2],​ {k2, Ks, k - 1}], {k, Mid, Ke}] // N+  Sum[(b[k] - 1) Product[b[k2],​ {k2, Ks, k - 1}], {k, Mid, Ke}]
 </​code>​ </​code>​
  
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