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General sum of cosine and sine series

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                       Finite and Infinite Sums of Cosine and Sine
                                                           ╬

                                           Francis J. O’Brien, Jr., Ph.D.
                                            Aquidneck Indian Council
                                                   Newport, RI


     Francis J. O’Brien, Jr.                                                       July 15, 2014



                                      SOURCES & REFERENCES
Boas, Mary L. Mathematical Methods in the Physical Sciences, 3rd ed., 2006. NY: Wiley.

Carr, G.S. Formulas and Theorems in Pure Mathematics, 2nd ed. New York: Chelsea Publishing
Co., 1970.

Euler, L. Summatio Serierum ex Sin. et Cos. Compositarum [Tom XVIII Novi Comm.
Acad. Sci. Petrop. 1774, pp. 24-36].
http://www.math.dartmouth.edu/~euler/docs/translations/E447tr.pdf.

Gradshteyn, I.S. and I.M. Ryzhik (7th Edition). Table of Integrals, Series, and Products. Alan
Jeffrey and Daniel Zwillinger, Editors. NY: Academic Press, 2007.

Spiegel, M. R. Mathematical Handbook of Formulas and Tables. New York: McGraw-Hill,
1968. [Reprinted Spiegel, Murray R., John Liu, and Seymour Lipschutz. Mathematical
Handbook of Formulas and Tables. New York: McGraw-Hill, 1999.]




Introduction
      This paper explains how to calculate elementary sums of cosine and sine series by use of
complex numbers and geometric series1. More complicated calculations are outlined.

           The derived general forms for finite series are:



1   A good introductory treatment is found in Chapter 2 of the Boas text.
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                         Page 2 of 14


          n

          cosa  kb  cosa  b   cosa  2b   cosa  3b     cosa  nb 
         k 1


                                                                nb 
                                                           sin  
                                               n  1        2 
                                    cos a  b                        (b  0)                               (1)
                                               2  sin  b 
                                                                 
                                                                2

          n 1

           cosa  kb cosa  cosa  b  cosa  2b    cosa  n  1b
          k 0


                                                              nb 
                                                         sin  
                                              n  1   2 
                                   cos a  b                      (b  0)                                   (2)
                                              2  sin  b 
                                                               
                                                              2
                  n

                 sin a  kb  sin a  b   sin a  2b   sin a  3b     sin a  nb 
                 k 1



                                                                nb 
                                                           sin  
                                               n  1        2 
                                    sin a  b                       (b  0)                                (3)
                                               2  sin  b 
                                                                 
                                                                2
                      n 1

                       sin a  kb  sin a  sin a  b  sin a  2b    sina  n  1b
                      k 0


                                                             nb 
                                                        sin  
                                             n  1   2 
                                  sin a  b                      (b  0)                                    (4)
                                             2  sin  b 
                                                              
                                                             2

        NOTE: More complicated forms add a term c k for finite and infinite trig. series; they
will be taken up later.

       For example, the sum of n terms for the series, cos   cos 2  cos 3    cos n ,
can be obtained by Formula (1), and sin   sin 3    sin 2n  1 can be obtained by
                                                                    
Formula (3). The parameters for a & b must be chosen for the given series of interest.

NOTE: Carr (Theorems 783-787) provides similar finite and infinite series2 as well as other
important trigonometric series with brief proofs. Complex numbers is used extensively for the
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                     Page 3 of 14


more advanced results. See also Spiegel, Chap.19, for additional trigonometric series.
Gradshteyn and Ryzhik (Sects. 1.3-1.4) may also be consulted for many more series.

NOTE: Euler’s translated 1774 paper from Dartmouth College is one of the earliest primary
sources for series of cosine and sine.



Geometric Progressions and Elementary Complex Numbers

        Geometric Progressions
                 Finite

        Geometric progression for finite sum of terms:

                                     S n  t1  t1r  t1r 2    t1r n 2  t1r n 1
                                          1 r n
                                         t1
                                            1 r
                                          t      t rn
                                         1  1             t1  0, r  1
                                         1 r 1 r

where n is the numbers of terms, t1 is the first term of the series, t n is the last or n th term,
and the common ratio r is the ratio of any two consecutive terms (usually ist and 2nd) in the finite series.

NOTE: Before applying the formula, it is necessary to show that the progression is actually
geometric by determining the terms n, t1 & r. In the series,

                          n

                         cosk   1  cos  cos2  cos3    cosn 
                        k 0


there exists n  1 total terms3. To use the geometric formula with n terms, we are required to
split up the sum as,
                               n
                      1   cosk   1  cos  cos2  cos3    cosn 
                                              
                              k 1




2There appears to be a typo in Theorem 785 on p.177; the increasing c term is misprinted.
3An easier way to compute the number n is by taking the difference between the upper limit and lower limit of the
summation and adding 1: n  0  1  n  1.
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                         Page 4 of 14


Infinite

        Geometric progression for infinite sum of terms:

                                           S   t1  t1r  t1r 2  
                                                t1
                                                   t1  0, r  1
                                               1 r

NOTE: For example, the finite geometric sum,

                                                             n 1
                                  1 1 1   11
                           Sn         
                                  2 4 8   22
                                                               2            n 1
                                1   1  1   1  1  11
                                            
                                2   2  2   2  2  22
                                           2         3              n
                              1 1 1        1
                                      
                              2 2 2        2
                                   n       k             n
                                   1   1
                                 1  
                                    2
                               k 1    2


                                     1
                                            1                    n
                                              4  1 , and t   1  .
This example shows a series with t1  ; r 
                                     2      1
                                              2
                                                  2      n   
                                                              2

                                                                        n
                                                        1
The infinite sum, when n   , is equal to 1, since lim    0 and r < 1.
                                                   n   2 

        Complex Numbers

        Definitions of cosine and sine derived from Euler’s Formula, e  i  cos   i sin  :

                                          e i  e  i 
                                  cos                 
                                                2       
                                                                                                 (5)
                                          e i  e  i 
                                  sin  
                                               2i       
                                                        

                                2 cos  e i  e  i 
                                                       
                                            i     i 
                                                                                                  (6)
                                2i sin   e  e      

 is a real variable.

 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                Page 5 of 14


NOTE: Alternative derivations for the sum of cosine and sine terms use real parts quantities of
Euler’s Formula4. The real part of

                                    ei  Re ei = Recos   i sin  = cos .

The real part of

                                  e i      e i      cos   i sin  
                                        Re       Re                   sin  .
                                    i         i             i         

Similarly, by De Moivre’s Theorem and Law of Exponents, the real parts of cosine and sine are:

                          Re e i   Ree  Recos k  i sin k   cos k
                                       k          ik




                          Re
                             e   i k
                                            Re
                                                  e  Re cos k  i sin k   sin k
                                                    ik
                                                                             
                                   i                i               i            

NOTE: The proofs of Formulas (1) through (4) as well as more difficult series can be obtained
by using Euler’s definitions of cosine and sine in (5) & (6) or by real parts only analysis.



Introduction of Proofs by Example
        To show the logic and alternative approaches for Formulas (1) through (4), consider the
cosine series:
                           n

                           cosk   cos  cos2  cos3    cosn 
                          k 1


Using real parts only, the sum is:

                                                                      
                                             e i  e i 2  e i 3    e i
                                         Re                                  
                                                                                  n


                                            
                                                                                     
                                                                                      
                                                                         
                                        Re e i  e i 2  e i 3    e in

Clearly the series is a geometric progression of n terms with real exponential terms,



4   Boas on page 79 shows an example in optics (light waves) in which the imaginary part of Euler’s Formula,
      
Im e i  sin  , is combined in a geometric series. The imaginary part of a complex number is needed in some
physics and engineering problems.
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                             Page 6 of 14


                                                     
                                             t1  Re e i

                                             r  Re
                                                    e   Ree 
                                                           i 2
                                                                           i

                                                     e   i


                                             r  Re 
                                               n
                                                      e    in



The geometric sum to calculate is:

                                        S n  t1
                                                     r  Re e   e 
                                                   1   n
                                                                 1    i
                                                                                    in


                                                   1 r                    1  e i

The “trick” is to factor 1  r and 1  r n and use definition of sine in (6).



Factoring 1  r ,

                                                    i  i     i 
                                                          
                                         1 e i  e 2  e 2  e 2 
                                                                  
                                                                  
                                                          
                                                                
                                                                                
                                                                   2i sin
                                                                                2

Factoring 1  r n ,

                                                   in  in       in 
                                                         
                                       1 e in  e 2 e 2  e 2 
                                                                      
                                                                      
                                                        
                                                                     
                                                                   n
                                                          2i sin
                                                                    2

Substituting and simplifying, the sum of cos  cos2  cos3    cosn is,
                                            

                                               in              n                n 1   sin n   
                Re e i
                        
                        1 e in
                                             e 2
                                     Re e i 
                                                         2i sin
                                                                  2
                                                                                  i
                                                                             Re e  2  
                                                                                             
                                                                                                     2
                                                                                                         
                                                                                                         
                         1  e i              i       2i sin                            sin    
                                                                                                      
                                              e2
                                                                 2                             2   
                                      n
                                  sin
                       n 1          2
                 cos       
                       2  sin       
                                       2
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                                              Page 7 of 14


NOTE: as a minimum check on calculations, set k  n  1 (the first term of the summation) to
see if the first term of the series results , cos 
                                                 .

NOTE: The above answer results if we set a  0, b  1 in general Formula (1). The values of
a & b are determined by the nth term of the series.

NOTE: The second way to evaluate cos  cos2  cos3    cosn is the standard
                                    
solution, using Euler’s cosine Formula in (5) and geometric series on each term.

                                              n               n  i
                                                  cosk    
                                                                    e              e  
                                                                                           k                i k

                                                            k 1 
                                                                                           
                                                                                                   2
                                             k 1                                                                  
                                                                                                                    

The same answer results as the reader can verify. However, the real parts only approach, when it
can be applied, involves less work typically.

       The analogous series for sine, sin   sin 2  sin 3    sin n , gives an answer
                                          
similar to the case of cosine. Change the cosine term to sine. The standard approach by Euler
calculates:

                                                        n               e i       n
                                                                                             e   k          i k

                                                     sin k    
                                                                  k 1 
                                                                                                     
                                                                                                           2i
                                                    k 1                                                                 
                                                                                                                          



General Formulas
       Derivation of Formula (1) by Real Parts
                   n

                   cosa  kb
                  k 1

                   cosa  b   cosa  2b   cosa  3b     cosa  nb 
                                                                            
                      
                   Re e i a b   e i a  2b     e i a  nb                 
                   Re
                      e     ia
                                  e   ib     ia
                                            e e        i 2 b
                                                                  e e      ia       inb
                                                                                               
Inspection provides:

                                                                
                                                  t1  Re e ia e ib
                                                         e         ia      ib 2
                                                                 Re 
                                                            e                                              ib
                                                  r  Re            e  ia    ib
                                                          e e
                                                  r  Re 
                                                    n
                                                          e            inb




 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                               Page 8 of 14


The required geometric sum is:

                                        S n  t1
                                                     r  Re e
                                                   1      n
                                                                         ia
                                                                               e ib
                                                                                       e 
                                                                                       1     inb


                                                   1 r                                1  e ib

Now factor the terms 1  r and 1  r n , and use definition of sine in (6):

                                                                           inb                 nb   
                       Re e ia e ib
                                          1 einb
                                                                         e 2
                                                          Re e ia e ib  ib
                                                                                         2i sin
                                                                                                  2
                                                                                                        
                                                                                                         
                                            1 e   ib
                                                                          e 2           2i sin b    
                                                                                                       
                                                                          
                                                                                                 2    
                                                      nb 
                                     n 1    sin        
                            i  a b        
                       Re e          2  
                                                        2 
                                                  sin b 
                                                            
                                                        2 
                                                                nb 
                                                           sin  
                                            n  1           2 
                        cos a  b                  
                                            2  sin  b 
                                                                 
                                                                2

The solution by Euler’s formula for cosine in Formula (5) is:

                                    n
                                                          e i a  kb   e  i a  kb  
                                                                n

                                    cosa  kb   
                                                    k 1                 2                  
                                   k 1                                                      



       Derivation of Formula (2) by Real Parts
                n 1

                 cosa  kb cosa  cosa  b  cosa  2b    cosa  n  1b
                k 0


In abbreviated form for this n term series:

                       cosa  cosa  b  cosa  2b    cosa  n  1b 
                               
                        Re e ia  e ia e ib  e ia e i 2b    e ia e i n1b         
                                                                 
                                                          t1  Re e ia
                                                          r  Re 
                                                                e   ib


                                                          r  Re 
                                                           n
                                                                  e      inb




 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                     Page 9 of 14



                                        S n  t1
                                                     r  Re e   e 
                                                   1     n
                                                                 1       ia
                                                                                         inb


                                                    1 r                       1  e ib

Factoring the terms 1  r and 1  r n , and using definition of sine in (6):

                                                                                             nb 
                                                                                        sin  
                                Re e ia
                                            e  Re e
                                          1        inb                     n 1  
                                                                  i  a b 
                                                                           2 
                                                                                  
                                                                                             2 
                                           1  e ib                                          b
                                                                                         sin  
                                                                                             2
                                                                  nb 
                                                             sin  
                                          n  1               2 
                                cos a  b       
                                          2                   b
                                                              sin  
                                                                  2

Solving by Euler’s definition:

                                 n 1                n 1
                                                           e i a  kb   e  i a  kb  
                                     cosa  kb                                        
                                 k 0                k 0                2                 

NOTE: The following sum represents an oscillating cosine series taken from Gradshteyn &
Rhyzik (Formula 1.341.5):

2 n 1
   1k cosx  ky  cosx  cosx  y  cosx  2 y  cosx  3 y     12n 1cosx  2n  1y 
k 0


                                     2n  1  sin ny
                           sin  x        y        .
                                       2     cos y
                                                    2

This series can be evaluated by the real parts only procedure just outlined with three notable
differences:

       1. The number of terms in the geometric sum is 2n

       2. The factoring of 1  r gives a cosine term in the denominator by (6)

       3. When all factoring and simplifying is done, the following expression for the geometric
          sum is seen:




 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                     Page 10 of 14


                                                              2 n 1 
                                                            i x    y
                                                                   2      sinny
                                              Sn    Re  ie         
                                                                              y
                                                                          cos
                                                                              2

    The negative complex exponential term is evaluated by noting for Euler’s Formula, the real
    parts quantities:

                            
                          Re e ix  Recos x  i sin x   cos x
                          Re ix  Recos x  i 2 sin x  Rei cos x  sin x    sin x
                            ie         i                
                          Re ie ix  sin x
                                    

                   2n  1                                      2n  1  sin ny
    Let x be x           y and the solution results, sin  x        y        .
                     2                                           2     cos y
                                                                              2

    Other  series are found in the Gradshteyn & Rhyzik handbook.



NOTE: Formulas (3) and (4) are very similar to (1) & (2). The calculations are set up for the
reader to supply the details.



       Derivation of Formula (3) by Real Parts
             n

             sin a  kb  sin a  b   sin a  2b   sin a  3b     sin a  nb 
            k 1

                  e i a b  e i a  2b      e i a  nb  
             Re                             
                  i                   i                  i      
                  e ia e ib e ia e i 2b         e ia e inb 
             Re                                               
                  i                     i                   i      




 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                                      Page 11 of 14


                                                           e ia e ib
                                                 t1  Re
                                                                i


                                                        e ia e i 2b
                                                 r  Re       i        Re e ib
                                                           ia ib
                                                         e e
                                                              i


                                                 r n  Re e inb

As shown earlier, factor the terms 1  r and 1  r n , and use definition of sine in (6):

                                                                                    inb               nb   
                S n  Re
                              e   ia ib
                                      e         e  Re e
                                               1    inb              ia ib
                                                                           e       e 2
                                                                                    ib
                                                                                                2i sin
                                                                                                         2
                                                                                                               
                                                                                                                
                                      i        1  e ib                   i                   2i sin b    
                                                                                                              
                                                                                   e 2
                                                                                                        2    
                                 n 1                                                  nb 
                        i  a  b               nb                             sin  
                                 2          sin                              
                                                              sin a  b n  1 
                      e                               2                                    2 
                Re                                                              
                                  i             sin b                  2  sin  b 
                                                                                          
                                                     2                                   2

Calculating the sum by the classical approach evaluates:

                                          n                 n
                                                                 e i a  kb   e  i a  kb  
                                          sin a  kb    
                                                                                                   
                                      k 1                 k 1                 2i                 



        Derivation of Formula (4) by Real Parts

        We leave it to the reader to show by real parts:
               n 1

                sin a  kb  sin a  sin a  b  sin a  2b    sina  n  1b
               k 0


                                                                      nb 
                                                                 sin  
                                                      n  1   2 
                                           sin a  b                               (b  0)
                                                      2  sin  b 
                                                                       
                                                                      2

And by Euler’s Formula:
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                     Page 12 of 14




                                                                 i a  kb 
                                 n 1                 n 1
                                                      e                        e  i a  kb  
                                  sina  kb                             2i                
                                 k 0                 k 0                                      



               Example of Formula (3)

       Consider the series,

                             sin   sin 4  sin 7    sin 3n  2 
                                                                           

       The a & b coefficients of the last term sin 3n  2  are determined from general Formula
                                                           
(3). The coefficients of the first two terms, 1 & 4, are equated to the first two terms of Formula
(3), a  b & a  2b, and solved from the system of equations:

                                                 a b 1
                                                 a  2b  4

This provides a  2, b  3, or the coefficients of the last term are 3n  2.

       Then, the solution of the series is:

                    n
                    sin 3k  2  sin   sin 4  sin 7    sin 3n  2
                   k 1
                                                  3n                         3n 
                                             sin                       sin  
                                 n  1        2   sin  3n  1   2 
                    sin  2  3                         2 
                                 2  sin  3                     sin  3 
                                                                              
                                                  2                          2



Advanced Formulas
       Published sources (e.g., Carr, Spiegel) present solutions for series including forms such
as:

                   cos(a )  c cos(a  b)  c 2 cos(a  2b)    c n 1 cosa  n  1b
                   sin( a )  c sin( a  b)  c 2 sin( a  2b)    c n 1 sina  n  1b



 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                                   Page 13 of 14


The terms c k are coefficients which need to be included in the final form. We show how to
evaluate these series by real parts analysis and by the standard approach using (5) and (6).

            Consider the n term series in Carr, Theorem 785:
n 1

c     k
           cos(a  kb)  cos(a )  c cos(a  b)  c 2 cos(a  2b)    c n 1 cosa  n  1.
k 0


This looks like Formula (2) except for the c terms.

            Solving by real parts:


                   c k cos(a  kb)  Reia  cei(a  b)  c 2 e i(a  2b)    c n 1e ia  n 1b
                  n 1
                                        e
                 k 0
                                           
                                      Re e ia  ce ia e ib  c 2 e ia e i 2b    c n 1e ia e i n 1b     
                                                        
                                                t1  Re e ia

                                                r  Re
                                                       ce e  Rece 
                                                                ia ib
                                                                                          ib

                                                         e     ia


                                                r  Re e 
                                                 n
                                                         c      n inb




                                          S n  t1
                                                       r  Re e   c e 
                                                     1      n
                                                                   1         ia
                                                                                          n inb


                                                         1 r                      1  ce ib

NOTE: In Formula (2), we were able to factor 1  r & 1  r n , use the definition of sine in (6), to
obtain a result corresponding to the standard Euler solution. This is not the best approach when
the series includes the c coefficients. As is turns out the best approach is to adjust S n by
multiplying numerator and denominator by what we might call the “complement of the
conjugate” form5 of 1  ceib which is 1  ce  ib :


                                    S n  t1
                                                 r  Re e   c e   ce 
                                               1     n
                                                             1         1
                                                                      ia
                                                                                  n inb                ib


                                                1 r          1  ce     ce 
                                                                       1            ib                ib



Simplifying, and after some algebra:




5   If we define a complex number in the standard fashion,            z  x  iy  r cos   i sin    re i , then the
conjugate form is, z  x  iy  r cos   i sin    re , so that z z  r , a real number. We call 1  re
                                                                       i                        2                           i

the complement of the conjugate form. In the text c is used for r.
 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.
                                                                                                                                     Page 14 of 14



                         Re e ia
                                     c neinb  ceib 
                                    1           1
                                     1  ce ib  ce  ib 
                                                1


                                  e ia  ce i ( a  b)  c n e i ( a  nb)  c n 1e ia  n 1b 
                          Re
                                                             
                                                      1  c ee  ib  c 2
                                                             ib
                                                            
                                                                              
                                                               2c cos b


                             cos a  c cos(a  b)  c n cos(a  nb)  c n 1 cosa  (n  1)b
                         
                                                              1  2c cos b  c 2



NOTE:              If        we          wanted              to          evaluate              the          infinite            term        case,
 

c
k 0
       k
           cos(a  kb)  cos(a)  c cos(a  b)  c 2 cos(a  2b)   , we                            could           use      the      geometric

                       t1
progression, S           . The assumption for S  is that r  1 when n  . In our series,
                      1 r
r  ce ib ; this means we assume c  1 when n is infinite. Alternatively, we can note that when
c  1, any term with an n exponent ( c n and c n 1 ), goes to 0 when n  . In either case,

                                                                                      cos a  c cos(a  b)
                   cos(a )  c cos(a  b)  c 2 cos(a  2b)                                                             c  1
                                                                                        1  2c cos b  c 2

NOTE: The standard procedure is to use Euler’s Formula and geometric progressions on each
term of the summation:

                                  n 1                  n 1
                                                              c k e i a ( k 1) b  c k e  i a ( k 1) b 
                                      c k cos(a  kb)  
                                                        k 0                         2                          
                                  k 0                                                                           

NOTE: The reader can verify that the sine series, sin(a)  c sin(a  b)    c n1 sina  n  1b

will give the answer derived for the cosine series but change all cosine terms to sine terms in the
numerator. The infinite series also changes cosine to sine in numerator.

                                                                    ———

With these example, many elementary cosine and sine series can be solved. Complex numbers
simplifies the calculations to a considerable degree.



 Francis J. O’Brien, Jr., 2014 <> Aquidneck Indian Council<> All rights reserved.

				
DOCUMENT INFO
Description: A tutorial showing how to evaluate common finite and infinite sums of cosine and sine series. The method uses complex numbers and geometric progressions with numerous examples. A number of references are provided. NOTE: typo in last formula on page 14. The exponent should be kb not (k-1)b.