Homework_Set_5_Solution

# Homework_Set_5_Solution - 2005 MA 1A HOMEWORK 5 SOLUTIONS...

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Unformatted text preview: 2005 MA 1A HOMEWORK 5 SOLUTIONS (1) Define for n = 0 , 1 , . . ., I n = integraldisplay π sin n x dx (a) Prove that I n +2 = ( n + 1) I n- ( n + 1) I n +2 Proof. To integrate I n +2 by parts let u = sin n +1 x , dv = sin x dx . Then du = ( n +1) cos x sin n x dx , v =- cos x . So I n +2 = integraldisplay π sin n +2 x dx =- cos x sin n +1 x bracketrightbig π- integraldisplay π- ( n + 1) cos 2 x sin n x dx =- 1 ·- 1 · 0 + ( n + 1) integraldisplay π (1- sin 2 x ) sin n x dx = ( n + 1) integraldisplay π sin n x dx- ( n + 1) integraldisplay π sin n +2 x dx = ( n + 1) I n- ( n + 1) I n +2 . square (b) Using (a), prove that lim n →∞ I n +2 I n = 1 Proof. By part (a), for n = 0 , 1 , . . . , I n +2 I n = n +2 n +1 , so lim n →∞ I n +2 I n = lim n →∞ n + 1 n + 2 = 1 square (c) Prove that I n +1 ≤ I n and, using (b), then prove lim n →∞ I n +1 I n = 1 Proof. We show that for all x ∈ [0 , 1] and all n = 0 , 1 , . . ., sin n +1 x ≤ sin n x . For the case n = 0, notice that for all x ∈ [0 , 1], 0 ≤ sin x ≤ 1, and sin x = 1, so sin x ≤ sin x . For all x ∈ [0 , 1], ≤ sin x ≤ 1, so sin n x ≥ 0 and 0 ≤ sin n +1 x ≤ sin n x . So, by the monotonicity property of integrals, I n +1 = integraltext π sin n +1 x dx ≤ integraltext π sin n x dx = I n . So we have I n +2 ≤ I n +1 ≤ I n for n = 0 , 1 , . . . . Since I n is positive, we have I n +2 I n ≤ I n +1 I n ≤ 1, so by squeezing limits, we have 1 = lim n →∞ I n +2 I n ≤ lim n →∞ I n +1 I n ≤ 1 so lim n →∞ I n +1...
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## This note was uploaded on 03/30/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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Homework_Set_5_Solution - 2005 MA 1A HOMEWORK 5 SOLUTIONS...

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