The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)
Concept:
In the interval (-l, l) Fourier series is defined as:-
\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos \frac{{n\pi x}}{l} + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{l}\)
Where, \({a_0} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)dx\)
\({a_n} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)\cos \frac{{n\pi x}}{l}dx\)
\({b_n} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)\frac{{\sin n\pi x}}{l}dx\)
Euler Definition: In the interval (-π, π)
\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin x\)
Where,
\({a_o} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)dx\)
\({a_n} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)\cos nx\;dx\)
\({b_n} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)\sin nx\;dx\)
Calculation:
Given,
f(x) = (x – x2)
\(\therefore {a_0} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)dx = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi \left( {x - {x^2}} \right)dx = \frac{1}{\pi }\left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_{ - \pi }^\pi \)
= - 6.5797Important Questions from Fourier Series
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I. Energy
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