pset_5sol

# Here we used the trig identities cosu cosv 2 cosu

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Unformatted text preview: series, we’d like an even extension of f (x) = cos(νx). Since f (x) is already even, we have that f (x) = cos(νx) on −π ≤ x ≤ π and so for ∞ f (x) = a0 + n=1 an cos(nx)dx 8 we have that since cos is even, a0 = −π 1 cos(νx)dx 2π π 1π = cos(νx)dx π0 π 1 = sin(νx) 0 νπ sin(νπ ) = , νπ and an = = = = = = = = = = = 1π cos(νx) cos(nx)dx π −π 2π cos(νx) cos(nx)dx π0 π π 1 cos((ν − n)x)dx + cos((ν + n)x)dx π0 0 π π 1 1 1 sin((ν − n)x) + sin((ν + n)x) 0 0 π ν −n ν+n 1 1 1 sin((ν − n)π ) + sin((ν + n)π ) π ν −n ν +n 1 ν sin((ν − n)π ) + n sin((ν − n)π ) + ν sin((ν + n)π ) − n sin((ν + n)π ) π (ν − n)(ν + n) 2 ν [sin((ν − n)π ) + sin((ν + n)π )] − n[sin((ν + n)π ) − sin((ν − n)π )] π 2(ν − n)(ν + n) 2 ν sin(νπ ) cos(nπ ) − n cos(νπ ) sin(nπ ) π (ν − n)(ν + n) 2 ν sin(νπ ) cos(nπ ) π (ν − n)(ν + n) 1 1 2 ν sin(νπ )(−1)n + π 2ν (ν − n) 2ν (ν + n) sin(νπ ) 1 1 (−1)n + π ν −n ν+n 1 for n ≥ 1. Here we used the trig identities cos(u) cos(v ) = 2 [cos(u − v )+cos(u + v )], sin(u) cos(v ) = 1 [sin(u + v )+sin(u − v )] and cos(u) sin(v ) = 1 [sin(u + v ) − sin(u − 2 2 9 v )]. Thus, the cosine series is given by 1 1 sin(νπ ) sin(νπ ) cos(nx) + (−1)n + f (x) = νπ π ν −n ν+n n=1 = sin(νπ ) 1 1 1 cos(nx) . + (−1)n + π ν n=1 ν−n ν +n ∞ ∞ SOLUTION (b) : If we let x = 0, then f (0) = cos(0) = 1 and cos(nx) = 1 for all n ≥ 1 From the above representation we have 1 π sin(νπ ) =...
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## This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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