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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 =
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
2 L dt =
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) + . . . ,
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.
- Winter '14