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Mathematics Dictionary
Dr. K. G. Shih
Series of csc(z)^2


  • Q01 | - Reference
  • Q02 | - Prove csc(z)^2 = Sum[1/((z - k*pi)^2))
  • Q03 | - Verify (pi)^2 = 8*(1/(1^2) + 1/(3^2) + 1/(5^2) + ....
  • Q04 | - Verify (pi)^2 = 9*(Sum[1/(2*n-1)*2] - Sum[1/(6*n-3)^2)
  • Q05 | - Discussion
  • Q06 | - Series of pi
  • Q07 | -
  • Q08 | -
  • Q09 | -
  • Q10 | -
    • Answers


    Q01. feference : Series of csc(z)^2

    Reference 
       Handbook of Mathematical Functions
   Milton Abramowitz and Irene A. Stogun
   Dover Publications, Inc. New York

   Series of csc(z) : Formula 4.3.68, p 75
   Series of csc(z) : Formula 4.3.92, p 75
    Go to Begin

    Q02. Prove csc(z)^2 = Sum[1/(z - k*pi)^2]

    Conditions 
       1. k from -infinte to +infinte
   2. z is not equal 0 or n*pi where n is integer
    Methods 
       1. Taylor's expansion

      It is simple for functions e^x, sin(x), cos(x) sinh(x) and cosh(x).
      For other functions, this method is too complicated. Also we can
      not get the partial fraction form for series of csc(z)^2

   2. Induction method

      It is also not easy.
      We can only use numerical method to prove this seris is true

      [Example 1] Now we use z = pi/2 and csc(z) = 1 to get a series of (pi)^2

      For k = 0, 1, 2, 3, 4, ...
           RHS = 1/(pi/2)^2 + 1/(-pi/2)^2 + 1/(-3*pi/2)^2 + ....
               = (4/(pi^2))*(1/1 + 1/1 + 1/(3^2) + 1/(5^2) + .....)
      For k = -1, -2, -3, -4, ......
           RHS = (4/(pi^2))*(1/(3^2) + 1/(5^2) + 1/(7^5) + .....)

      Hence for k = -infinte to +infinite
           RHS = (8/(pi^2))*(1/1 + 1/(3^2) + 1/(5^2) + ......)

      and we get pi^2 = 8*(1 + 1/(3^2) + 1/(5^2) + 1/(7^2) + ......)

      Is this series true ? we will prove it in Q03.

      [Example 2] Use z = pi/6 and csc(z) = 2 to get another series in Q04
    Go to Begin

    Q03. Verify (pi)^2 = 8*(1/(1^2) + 1/(3^2) + 1/(5^2) + ....

    This series is based on csc(z)^2 = Sum[1/(z - k*pi)^2] when z = pi/2 
       (pi)^2 = 8*Sum[1/(2*n - 1)^2)]
   when n = 40000, pi = Sqr(8*Sum[1/(2*n - 1)^2)])
                      = 3.14159257401265

   This calculation is given in option 5 of NC.exe

   Since pi = 4*Atn(1)
            = 3.14159265358979

   Hence this series is true, but it is convergent very slow. 
   The example shows that the result is accurate to 6 decimal for 40000 terms used.
    Go to Begin

    Q04. Verify (pi)^2 = 9*(Sum[1/(2*n-1)*2] - Sum[1/(6*n-3)^2)])

    This series is based on csc(z)^2 = Sum[1/[(z - k*pi)^2] when z = pi/6 
       k = 0, (pi/6 - 0*pi)^2 = ( 1*pi/6)^2
   k = 1, (pi/6 - 1*pi)^2 = ( 5*pi/6)^2
   k = 2, (pi/6 - 2*pi)^2 = (11*pi/6)^2
   k = 3, (pi/6 - 3*pi)^2 = (17*pi/6)^2
   k = 4, (pi/6 - 4*pi)^2 = (23*pi/6)^2
   k = 5, 
   .....
   k = -1, (pi/6 + 1*pi)^2 = ( 7*pi/6)^2
   k = -2, (pi/6 + 2*pi)^2 = (13*pi/6)^2
   k = -3, (pi/6 + 3*pi)^2 = (19*pi/6)^2
   k = -4, (pi/6 + 1*pi)^2 = (25*pi/6)^2
   k = -5,
   ......
   Since csc(pi/6) = 2
   Hence 2 = 1/( 1*pi/6)^2 + 1/( 5*pi/6)^2 + 1/( 7*pi/6)^2 + 1/(11*pi/6)^2
           + 1/(13*pi/6)^2 + 1/(17*pi/6)^2 + 1/(19*pi/6)^2 + 1/(23*pi/6)^2
           + .....
   The series including 1, 5, 7, 11, 13, 17, 19, 23, 25, .....
   The sequence missing 5, 9, 15, 21, ....
   Missing terms = 1/(3*pi/6)^2 + 1/(9*pi/6)^2 + 1/(15*pi/6)^2, ...
                 = Sum[1/((3 + (n - 1)*6)*pi/6)^2]
   Hence (2)^2 = (36/(pi)^2)*(Sum[1/(2*n-1)^2] - Sum[1/(6*n -3)^2])
               = (36/(pi)^2)*(Sum[1/(2*n-1)^2] - Sum[1/(6*n-3)^2])
   Hence (pi)^2 = 9*(Sum[1/(2*n-1)^2] - Sum[1/(6*n-3)^2])
    Series of Pi 
       (pi)^2 = 9*(1 + 1/(5)^2 + 1/(7)^2 + 1/(11)^2 + 1/(13)^2 + 1/(17)^2 + 1/(19)^2
          + 1/(23)^2 + 1/(25)^2 + ......
    Verification
      (pi)^2 = 9*(Sum[1/(2*n - 1)^2)] - Sum[1/(6*n - 3)^2] when n = 40000, pi = Sqr(9*Sum[1/(2*n - 1)^2) - 1/(6*n - 3)^2]) = 3.14159257401265 This calculation is given in option 5b of NC.exe Since pi = 4*Atn(1) = 3.14159265358979 Hence this series is true, but it is convergent very slow. The example shows that the result is accurate to 6 decimal for 40000 terms used.
    Go to Begin

    Q05. Discussion

    Questions : How to get this series of csc(z)^2 ? 
       1. This formula has been derived long time ago. It is hard to locate the
      paper in literature.

   2. we can get series of csc(x)^2 using d/dx(cot(x)) = csc(x)^2 and the
      the series of cot(x). But it will be different from this expression.

   3. Euler used sine series and general solution of sin(x) = 0 to prove
      that (pi)^2 = 6*(1 + 1/(2^2) + 1/(3^2) + 1/(3^2) + ....).

      The method can be found in keyword "series of Pi" in my web site
      URL : www.b192907.com; By keywords; S; Series of Pi.

      Use d/dx(cot(x)) = csc(x)^2, series of cot(x) and gereral solution of
      csc(x) for x = pi/2. But the result is different.

   4. It is my hope that someone can provide some information about the proof
      of this series.
    Go to Begin

    Q06. Series of pi

    Series In Q03 and Q04 
    
   1. (pi)^2 = 8*(1 + 1/(3)^2 + 1/(5)^2 + 1/(7)^2 + 1/(9)^2 + 1/(11)^2
             + 1/(13)^2 + 1/(15)^2 + 1/(17)^2 + ......}

   2. (pi)^2 = 9*(1 + 1/(5)^2 + 1/(7)^2 + 1/(11)^2 + 1/(13)^2 + 1/(17)^2
             + 1/(19)^2 + 1/(23)^2 + 1/(25)^2 + ......)

   Note : Compare two series, we see that series 2 missing the following terms
          1/(3)^2, 1/(9)^2, 1/(15)^2, 1/(21)^2, 1/(27)^2, ....
          That the numerator has common difference 6
    Go to Begin

    Q07.

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    Q08. Answer
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    Q09. Answer
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    Q10. Answer
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