blackfriday said:
yep the guy who used to ask dumbass complex numbers questions is back...
In= int (1,0) {x(1-x^3)^n }dx for n>=0
show that In=(3n/3n+2)In-1 for n>=0
hence find an expression for In in terms of n for n>=0
thanks
It's basically one of those questions where you have to cheat. When I write this, just assume that all integral signs and integrated brackets are between 0 and 1.
I<sub>n</sub> = ∫ x(1 - x<sup>3</sup>)<sup>n</sup> dx
use int. by parts where ∫ u'v = uv - ∫ uv' , where u' = x and v=(1 - x<sup>3</sup>)<sup>n</sup>
hence u= 1/2.x<sup>2</sup> and v'= -3nx<sup>2</sup>(1-x<sup>3</sup>)<sup>n-1</sup>
I<sub>n</sub> = uv - ∫ uv' = [1/2.x<sup>2</sup>x(1 - x<sup>3</sup>)<sup>n</sup> ] - ∫ 1/2.x<sup>2</sup>.-3nx<sup>2</sup>(1-x<sup>3</sup>)<sup>n-1</sup> dx
= 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>x<sup>3</sup> dx
Here you cheat by adding and subtracting 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx
I<sub>n</sub> = 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>x<sup>3</sup> dx + 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx - 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx
= 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>(x<sup>3</sup> - 1) dx + 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx
= 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx - 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>(1 - x<sup>3</sup>) dx
= 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n-1</sup>dx - 3n/2 ∫ x(1-x<sup>3</sup>)<sup>n</sup>dx
I<sub>n</sub>= 3n/2.I<sub>n-1</sub> - 3n/2.I<sub>n</sub>
(1+3n/2)I<sub>n</sub> = 3n/2.I<sub>n-1</sub>
I<sub>n</sub> = 3n/(2+3n)I<sub>n-1</sub>
Voila! Thank god for the cut and paste function. If you do a whole bunch of reccurance method questions you'll find that a lot of the harder ones require you to use that "cheating" method in order to factorise and make up a power. I hope that helps you out.
EDIT: For the latter part of the question, try writing out the first few terms and see if you can work out how to notate the entire series.
3n/(3n+2) x 3(n-1)/(3n-1) x 3(n-2)/(3n-4) x 3(n-3)/(3n-7) x....... and remember how to use factorial notation and indice laws.
