Chem Differential Eq HW Solutions Fall 2011 28

Chem Differential Eq HW Solutions Fall 2011 28 - sin kx cos...

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28 Chapter 2 Fourier Series Solutions to Exercises 2.10 5. The cosine part converges uniformly for all x , by the Weierstrass M -test. The sine part converges for all x by Theorem 2(b). Hence the given series converges for all x . 9. (a) If lim k →∞ sin kx = 0, then lim k →∞ sin 2 kx =0 lim k →∞ (1 - cos 2 kx )=0 lim k →∞ cos 2 kx =1 ( * ) . Also, if lim k →∞ sin kx = 0, then lim k →∞ sin( k +1) x ) = 0. But sin( k +1) x = sin kx cos x + cos kx sin x ,so 0 = lim k →∞ (
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Unformatted text preview: sin kx cos x + cos kx sin x ) lim k cos kx sin x = 0 lim k cos kx = 0 or sin x = 0 . By (*), cos kx does not tend to 0, so sin x = 0, implying that x = m . Consequently, if x = m , then lim k sin kx is not 0 and the series k =1 sin kx does not converge by the n th term test, which proves (b)....
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