Fourier Series Expansion of f(x) cos(x) on [0, π]

In this article, we will explore the Fourier series expansion of the function ( f(x) cos(x) ) on the interval ([0, pi]). We will utilize the Fourier series formula for a function defined on an interval and apply trigonometric identities to find the series coefficients.

Fourier Series Formula

The Fourier series of a function ( f(x) ) defined on an interval ( [a, b] ) is given by:

[ f(x) sim frac{a_0}{2} sum_{n1}^{infty} left( a_n cosleft(frac{2pi nx}{L}right) b_n sinleft(frac{2pi nx}{L}right) right) ]

where ( L frac{b-a}{2} ). In this specific case, ( a 0 ) and ( b pi ), hence, ( L frac{pi}{2} ).

Calculating the Coefficients

The coefficients ( a_0, a_n, ) and ( b_n ) are defined as follows:

Calculating ( a_0 )

The constant term ( a_0 ) is given by:

[ a_0 frac{1}{L} int_{0}^{pi} f(x) , dx frac{2}{pi} int_{0}^{pi} cos(x) , dx ]

Evaluating the integral, we get:

[ a_0 frac{2}{pi} cdot left[ sin(x) right]_{0}^{pi} frac{2}{pi} (0 - 0) 0 ]

Calculating ( a_n )

The cosine coefficients ( a_n ) are given by:

[ a_n frac{1}{L} int_{0}^{pi} f(x) cosleft(frac{2pi n x}{L}right) , dx frac{2}{pi} int_{0}^{pi} cos(x) cos(frac{2nx}{pi}) , dx ]

Using trigonometric product-to-sum identities, we have:

[ cos(A) cos(B) frac{1}{2} left( cos(A-B) cos(A B) right) ]

Substituting ( A 1 ) and ( B frac{2nx}{pi} ):

[ a_n frac{2}{pi} cdot frac{1}{2} left( int_{0}^{pi} cos(1 - frac{2nx}{pi}) , dx int_{0}^{pi} cos(1 frac{2nx}{pi}) , dx right) ]

Evaluating the integrals:

[ a_n frac{1}{pi} left( left[ -frac{pi}{2n} sin(1 - frac{2nx}{pi}) right]_{0}^{pi} left[ -frac{pi}{2n} sin(1 frac{2nx}{pi}) right]_{0}^{pi} right) 0 ]

Hence, ( a_n 0 ) for all ( n ).

Calculating ( b_n )

The sine coefficients ( b_n ) are given by:

[ b_n frac{1}{L} int_{0}^{pi} f(x) sinleft(frac{2pi n x}{L}right) , dx frac{2}{pi} int_{0}^{pi} cos(x) sin(frac{2nx}{pi}) , dx ]

Using the product-to-sum identities:

[ cos(x) sinleft(frac{2nx}{pi}right) frac{1}{2} left( sinleft(x frac{2nx}{pi}right) - sinleft(x - frac{2nx}{pi}right) right) ]

Substituting into the integral:

[ b_n frac{2}{pi} cdot frac{1}{2} left( int_{0}^{pi} sinleft(1 frac{2nx}{pi}right) , dx - int_{0}^{pi} sinleft(1 - frac{2nx}{pi}right) , dx right) ]

Evaluating the integrals:

[ b_n frac{1}{pi} left( left[ -frac{pi}{2n 1} cosleft(1 frac{2nx}{pi}right) right]_{0}^{pi} - left[ -frac{pi}{2n-1} cosleft(1 - frac{2nx}{pi}right) right]_{0}^{pi} right) ]

Hence, ( b_n 0 ) for all ( n ).

Conclusion

With all coefficients ( a_0, a_n, ) and ( b_n ) being zero, the Fourier series expansion of ( f(x) cos(x) ) on the interval ([0, pi]) simplifies to:

[ f(x) sim 0 ]

Therefore, the Fourier series of ( f(x) cos(x) ) on ([0, pi] ) converges to 0 at all points in that interval.