<TABLE class=forumline cellSpacing=0 cellPadding=0 width="100%" border=0><TBODY><TR><TD class=row1 style="PADDING-RIGHT: 3px; PADDING-LEFT: 3px; PADDING-BOTTOM: 3px; PADDING-TOP: 3px" vAlign=top width="100%" height="100%">I have a big exam soon, can someone help me on these challenge questions please ?
For n = 0, 1, . . ., let In = S(Intergral) (pi/2)(upper bound) and 0 (Lower Bound) sin^n(x) dx.
(a) Show that I0 > I1 > I2 > · · · > 0.
(b) Use integration by parts to show that
In+2 = ((n + 1)/(n + 2))(In) if n is greater or equal to 0.
(c) Use part (b) and the fact that In+2 < In+1 < In, by part (a), to show that ((In+1)/(In)) ->1 as n -> infinity.
(d) Use (b) and induction to show that for all n greater or equal to 0,
I2n =(((2n)!(pi))/((2^(2n))((n!)^(2))(2))) and I2n+1 =((2^(2n))((n!)^2)/((2n + 1)(2n)!))
(e) Using the formula below, for some constant C,
n! = Ce^(-n)n^(n+1/2)an
where an -> 1 as n -> infinity. Use (c) and (d) to show that C = root(2pi)
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For n = 0, 1, . . ., let In = S(Intergral) (pi/2)(upper bound) and 0 (Lower Bound) sin^n(x) dx.
(a) Show that I0 > I1 > I2 > · · · > 0.
(b) Use integration by parts to show that
In+2 = ((n + 1)/(n + 2))(In) if n is greater or equal to 0.
(c) Use part (b) and the fact that In+2 < In+1 < In, by part (a), to show that ((In+1)/(In)) ->1 as n -> infinity.
(d) Use (b) and induction to show that for all n greater or equal to 0,
I2n =(((2n)!(pi))/((2^(2n))((n!)^(2))(2))) and I2n+1 =((2^(2n))((n!)^2)/((2n + 1)(2n)!))
(e) Using the formula below, for some constant C,
n! = Ce^(-n)n^(n+1/2)an
where an -> 1 as n -> infinity. Use (c) and (d) to show that C = root(2pi)
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