Why Is the Coefficient Doubled in the Fourier Sine Series Calculation?

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SUMMARY

The discussion centers on the calculation of the Fourier sine series coefficient, specifically why the coefficient is defined as ##b_n = \frac{2}{\pi} \int_{0}^{\pi} (...)dx## instead of ##b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx##. The rationale is that the Fourier series must represent an odd function, necessitating the use of the odd extension of ##\sin^2(x)## and the application of half-range formulas. This approach ensures that the sine Fourier expansion accurately captures the characteristics of the function being analyzed.

PREREQUISITES
  • Understanding of Fourier series and sine expansions
  • Familiarity with odd and even functions in mathematical analysis
  • Knowledge of integral calculus, particularly definite integrals
  • Experience with half-range Fourier series techniques
NEXT STEPS
  • Study the derivation of half-range Fourier sine series
  • Explore the properties of odd and even functions in Fourier analysis
  • Learn about the application of Fourier series in signal processing
  • Investigate the relationship between Fourier series and eigenfunction expansions
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Students and educators in mathematics, particularly those focusing on Fourier analysis, calculus, and mathematical modeling. This discussion is beneficial for anyone seeking to deepen their understanding of Fourier sine series and their applications in representing functions.

Nikitin
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Homework Statement


https://wiki.math.ntnu.no/_media/tma4120/2013h/tma4120_h11.pdf

Check out the solution to problem 4b)

My question is: Why do they set ##b_n = \frac{2}{\pi} \int_{0}^{\pi}(...)dx## instead of ##b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx##?

Ie, why did they multiply the integral with 2? Did they find the Fourier integral to the odd expansion of ##sin(x)^2##? Because that's the only way for sin(x)^2 to be defined by solely a sine Fourier expansion?

EDIT: Ooops, posted in wrong section. Please move to calculus and beyond homework forum! sorry!
 
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Nikitin said:

Homework Statement


https://wiki.math.ntnu.no/_media/tma4120/2013h/tma4120_h11.pdf

Check out the solution to problem 4b)

My question is: Why do they set ##b_n = \frac{2}{\pi} \int_{0}^{\pi}(...)dx## instead of ##b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx##?

Ie, why did they multiply the integral with 2? Did they find the Fourier integral to the odd expansion of ##sin(x)^2##? Because that's the only way for sin(x)^2 to be defined by solely a sine Fourier expansion?

Yes, that's exactly it. Since they only have sine functions for the eigenfunction expansion, the FS must represent an odd function so they take the odd extension of ##\sin^2(x)## and use the half range formulas.
 
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