Learn How to Solve sin(nx)/n for Infinity with Expert Guidance"

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SUMMARY

The forum discussion focuses on solving the infinite series sum of sin(nx)/n as n approaches infinity. The established results indicate that the sum equals (1/2)*(π - x) for 0 < x < π and -(1/2)*(π + x) for -π < x < 0. Participants emphasize the importance of using Fourier coefficients and the relationship between sine and cosine functions in this context. The discussion also highlights the utility of the logarithmic series in approaching the problem.

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  • Understanding of Fourier series and coefficients
  • Familiarity with complex analysis, specifically Euler's formula
  • Knowledge of convergence of series
  • Basic calculus, particularly limits and integrals
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  • Study Fourier series and their applications in solving differential equations
  • Learn about Euler's formula and its implications in complex analysis
  • Explore the properties of convergence in infinite series
  • Investigate the logarithmic series and its applications in mathematical analysis
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sum (sin(nx)/n)
where n goes from 1 to infinity.
The result should be (1/2)*(pi-x) when 0(x(pi
and -(1/2)*(pi+x) when -pi(x(0

I need someone to show me how to attack this problem.
Thanks
 
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sin(nx)/n = RE[e^inx/n]
(Further Hint : log series)

-- AI
 
Correction to TenaliRaman. sin is imaginary part, cos is real part.
 
I would try computing the Fourier coefficents of x.
 
mathman said:
Correction to TenaliRaman. sin is imaginary part, cos is real part.
OOPS!

-- AI
 

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