pset_5sol

That is f x n1 bn sinnxdx 4 where bn 1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 > 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...
View Full Document

Ask a homework question - tutors are online