pset_5sol

# That is f x n1 bn sinnxdx 4 where bn 1

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Unformatted text preview: n? Explain this rate of decrease. SOLUTION : For a sine series, we will need the odd extension of f (x) to (−π, π ). For the odd extension, we want −f (−x) on (−π, 0). Then the fourier series of this extended function gives the values f (x) on (0, π ). That is, ∞ f (x) = n=1 bn sin(nx)dx 4 where bn = = = = = 1 π 1 π 1 π 1 π 2 π π f (x) sin(nx)dx −π π 0 f (x) sin(nx)dx − 0 π f (−x) sin(nx)dx −π −π f (x) sin(nx)dx + 0 π 0 π f (−x) sin(nx)dx f (x) sin(nx)dx 0 f (x) sin(nx)dx + 0 π f (x) sin(nx)dx, 0 where the fourth line follows from a change of variables x = −x =⇒ dx = −dx and noting that sin(−nx) = − sin(nx). So, we can compute bn by considering ﬁrst bn = 2 π π π cos(x) sin(nx)dx − 0 0 sin(nx)dx + 2 π π x sin(nx)dx . 0 From double integration by parts, we have that for n = 1, π π cos(x) sin(x)dx 0 π = =⇒ = sin(x) sin(x) 0. π 0 − 0 sin(x) cos(x)dx 2 0 cos(x) sin(x)dx For n &gt; 1, the ﬁrst integral is π π cos(x) sin(nx)dx 0 = = = sin(nx) sin(x) π 0 −n 0 π 0 sin(x) cos(nx)dx π −n − cos(nx) cos(x) −n cos(nπ ) + 1 − n −n 0 π co...
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