Finding Sums of Sines and Series

Date: 03/10/2004 at 08:41:40
From: Jeffery
Subject: sin1 +sin2 +sin3 +sin4 +sin5 .............sin90 = ?

Hi.  I am trying to find the sum of sin1 + sin2 + sin3 + ... + sin90. 
I'm also trying to find the sum of 1^n + 2^n + 3^n + 4^n + ... + n^n. 
Can you help me?

Date: 03/10/2004 at 09:19:24
From: Doctor Jerry
Subject: Re: sin1 +sin2 +sin3 +sin4 +sin5 .............sin90 = ?

Hello Jeffery,

I don't think there is a formula for sin(1)+...+sin(90).  One can 
calculate it, but I'm afraid it must be done the long way.

As to the sum of 1^n + 2^n +...+ n^n, I'll give you a URL.  The
subject is somewhat complicated, but there is a definite formula.  See
the beginning of the URL

  http://mathworld.wolfram.com/BernoulliNumber.html 

and formula (35).

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 03/10/2004 at 12:17:00
From: Doctor Douglas
Subject: Re: sin1 +sin2 +sin3 +sin4 +sin5 .............sin90 = ?

Hi Jeffery.

I'd like to supplement Dr. Jerry's answer.  To calculate the sum of
sines explicitly, I agree that there doesn't seem to be a substitute
for doing it the long way.

If you are willing to go with an approximation, however, there is a 
shortcut (although it requires calculus, which you may not have had).  
The sum of sines is approximately equal to the number of terms (90) 
times the AVERAGE value of the sine function from 0 to 90 degrees:

  sin(1) + ... + sin(90) ~= 90 * <sin(x)>   for 0 < x < 90 deg
     = 90 * <Integral[sin(x)dx]/(pi/2)>
     = 90 * (2/pi)

where, if you haven't had calculus, you only need the result that the 
average value of sin(x) from x = 0 to x = 90 degrees is 2/pi.  Then 
your sum of sines is *approximately* equal to

     = 180/pi = 57.29578.

When I evaluate the sum of sines using a calculator, I obtain 
57.79432506, so my estimate has a discrepancy of slightly less than
one percent.  The discrepancy will become less the more terms you use:

   sin(1/2) + sin(1) + sin(3/2) + ... + sin(179/2) + sin(90) 
      = 115.0908318

and this differs from 180*2/pi = 114.59156 by less than one half of
one percent.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 03/14/2004 at 16:44:27
From: Doctor Korsak
Subject: Formulas exist for sin1 +sin2 +sin3 +sin4 +sin5
.............sin90

Hello Jeffery,

I thought you might be pleased to know that there are in fact formulas 
for your sum.  

Your sum is a special case of the first of a more general pair of 
series:

  sin(A) + sin(A+B) + sin(A+2B) + sin(A+3B) .... + sin(A+(n-1)B)

  cos(A) + cos(A+B) + cos(A+2B) + cos(A+3B) .... + cos(A+(n-1)B)

for which there are formulas.  The book where I found the formulas 
states that to prove them you can repetitively apply the formulas:

  sin(A+B) + sin(A-B) = 2sin(A)cos(B)
 
  cos(A+B) + cos(A-B) = 2sin(A)sin(B)

(Formulas and Theorems in Pure Mathematics, George S. Carr, Chelsea, 
1970)  

Another similar formula can be found in a recently re-issued Dover 
publication, Fourier Series by Georgi P. Tolstov, 1970 (original 
translation by Richard A. Silverman, Prentice-Hall, 1962), pp 98-99,
for the series

  sin(x) + sin(2x) + sin(3x) + .... + sin(nx)

The author works out the derivation using the formula 

  2sinAsinB = cos(A-B) - cos(A+B)

after multiplying the above series through by 2sin(x/2).

In case you are interested in working out a formula by yourself, I am 
leaving out here the actual formulas in the above references. 

I don't know if at your age you already studied about complex numbers 
and DeMoivre's formula and whether you would be allowed to use it at 
your class level, but here is another way you can proceed to derive 
the formula you want.

Since you are summing sines of angles using units of degrees, let's 
convert to units of radians.
 
  sin(n degrees) = sin(n*pi/180 radians)

Strictly speaking, the sin(x) function always has x in terms of 
radians, not degrees.  When people write "sine of 90 degrees = 1" they 
are not thinking of the "formal" sin(x) function with x in units of 
radians.

DeMoivre's formula is

    e^(ix) = cos(x) + i*sin(x)

which means that 
                e^(ix) - e^(-ix)                 e^(ix) + e^(-ix)
      sin(x) =  ----------------    and cos(x) = ---------------- .
                       2i                               2i

Let x = e^(i*pi/180), where i = sqrt(-1), and let N = 90.  Then your 
sum becomes

    Sum (x^n - x^(-n)) / (2i)   for n = 1 to N 
or  
    (1/2i) { x*Sum (x^n) - (1/x)*Sum (1/x)^n }   for n = 0 to N - 1 

which is the difference of the sums of two geometric series.

Although x = e^(i*pi/180) is a complex number, upon making the 
required substitutions you will find that the imaginary terms will all 
cancel out.  

Hint 1: x^N = e^( (i*pi/180)*90 ) = i, i^2 = -1 and 1/i = -i

Hint 2: Using hint 1, the above sum can be simplified to

    (1/2) ( 1 + i(x + 1)/(x - 1) ) 

At this stage you will be undoubtedly wondering how this comes out to 
be a real number value!

Hint 3: Look at the quadrilateral formed by the vectors x and 1 in the 
complex plane.  Note that the vector x + 1 is orthogonal to the vector 
x - 1.  

Hint 4: Think about the ratio of two complex numbers x + 1, x - 1 in
polar coordinates, and then consider the magnitude and direction of
the vector for the complex number (x + 1)/(x - 1).    

If you look up the references above, they all give seemingly different 
formulas, and both differ from what I just derived above, but they are 
all equivalent.  The value of your sum to 15 decimal places is 

               57.797233924590296

Another interesting thing to note: a sum like yours often appears in 
practical applications using Fourier series for analysis of signals.  
What your series would correspond to is a signal comprised of 
uniformly spaced, equal amplitude frequency components.  Such a signal 
is commonly referred to as as a "herring bone" signal because of the 
way it appears on a spectrum analyzer display.

Let me know if you want me to send you the formulas from the above 
mentioned books.
 
- Doctor Korsak, The Math Forum
  http://mathforum.org/dr.math/