The infinite series C and S are defined C = a*cos(x) + a^2*cos(2x) + a^3*cos(3x) + ..., and S = a*sin(x) + a^2*sin(2x) + a^3*sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = a*sin(x)/(1 - 2a*cos(x) + a^2), and find C.
C + iS = (acos(x) + a^2cos(2x) + a^3cos(3x) + ...) + i( asin(x) + a^2sin(2x) + a^3sin(3x) + ...)
= a(cos(x) + isin(x)) + a^2(cos(2x) + isin(2x)) + a^3(cos(3x) + i*sin(3x)) + ...
= ae^ix + a^2e^i2x + a^3*e^i3x + ...
This is a geometric sequence, with common ratio ae^ix. Using the formula for infinite sum, (S = 1/(a0 - r), where r is the common ratio and a0 is the first term, i.e. ae^ix here) we get after some simplification:
C + iS = (acos(x) - a^2)/(1 - 2acos(x) + a^2) + i(asin(x)/(1 - 2acos(x) + a^2)). Then clearly here C is the real part and S is the imaginary part, so S = (asin(x)/(1 - 2acos(x) + a^2)) as needed and C = (acos(x) - a^2)/(1 - 2acos(x) + a^2).
SF
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