Finding Sums of Sines and SeriesDate: 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/ |
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