It follows from this formula that ei ei 2 cos ei ei

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Unformatted text preview: s((n − m)πt/L) = cos(nπt/L) cos(mπt/L) + sin(nπt/L) sin(mπt/L). Subtracting the first of these from the second and dividing by two yields sin(nπt/L) sin(mπt/L) = and hence 1 (cos((n − m)πt/L) − cos((n + m)πt/L), 2 L sin(nπt/L) sin(mπt/L)dt = 0 1 2 L (cos((n − m)πt/L) − cos((n + m)πt/L)dt. 0 If n and m are positive integers, the integral on the right vanishes unless n = m, in which case the right-hand side becomes 1 2 L dt = 0 L . 2 Hence 2 L π sin(nπt/L) sin(mπt/L)dt = 0 1, for m = n, 0, for m = n. (3.12) Therefore, just as in the previous section, we can evaluate the coefficients of the Fourier sine series of a function f ∈ V0 by simply projecting f onto each element of the orthonormal basis. When we do this, we find that f (t) = b1 sin(πt/L) + b2 sin(2πt/L) + . . . + sin(nπt/L) + . . . , 73 where bn = f , sin(nπt/L) = L 2 L f (t) sin(nπt/L)dt. (3.13) 0 We can treat the Fourier cosine series in a similar fashion. In this case, we can show that the fun...
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This document was uploaded on 01/12/2014.

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