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Is this induction proof correct?
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[QUOTE="johann1301, post: 4848601, member: 353710"] [h2]Homework Statement [/h2] The sequence {x[SUB]n[/SUB]} is given by the recurrence relation x[SUB]n[/SUB] = cos(x[SUB]n-1[/SUB])sin(x[SUB]n-2[/SUB]) for n ≥ 2 and x[SUB]0[/SUB]=2 and x[SUB]1[/SUB]=1,4. Show by induction that 0 ≤ x[SUB]n[/SUB] ≤ 1 for all integers n ≥ 2. [h2]The Attempt at a Solution[/h2] We formulate a statement: P[SUB]n[/SUB]: 0 ≤ x[SUB]n[/SUB] = cos(x[SUB]n-1[/SUB])sin(x[SUB]n-2[/SUB]) ≤ 1 for n ≥ 2 --------------------------------------------------------------- We assume that P[SUB]n[/SUB] is true for n=k, where k ≥ 2: A[SUB]k[/SUB]: 0 ≤ x[SUB]k[/SUB] = cos(x[SUB]k-1[/SUB])sin(x[SUB]k-2[/SUB]) ≤ 1 for k ≥ 2 We then set n=k+1: x[SUB]k+1[/SUB] = cos(x[SUB]k[/SUB])sin(x[SUB]k-1[/SUB]) We know from the assumption A[SUB]k[/SUB] that: 0 ≤ x[SUB]k[/SUB] ≤ 1 This implies that: 0,54 ≤ cos(x[SUB]k[/SUB]) ≤ 1 which again implies 0 < cos(x[SUB]k[/SUB]) < 1 We then have to show: 0 ≤ sin(x[SUB]k-1[/SUB]) ≤ 1 We know that sine to any angle is always equal to or less then 1. We therefore have to prove: 0 ≤ sin(x[SUB]k-1[/SUB]) Since we assumed that 0 ≤ x[SUB]k[/SUB] ≤ 1, this must be true for x[SUB]k-1[/SUB] as well. This means that if the assumption A[SUB]k[/SUB] is true when n=k, its true when n=k+1 also. The last thing is to prove the base case: x[SUB]2[/SUB] = cos(1,4)sin(2) ≈ 0,155 0 ≤ 0,155 ≤ 1 P[SUB]n[/SUB] is thus true. Is this correct argumentation? [/QUOTE]
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Is this induction proof correct?
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