reduction formulae (1 Viewer)

[freaking_out]

For those who have got a copy of cambridge/arnold, this question is from pg.170 q.20.


if I_n =I{0->1} x(1-x³)ⁿ dx for n>=0 show that

I_n =3n/(3n+2)I_n-1 for n>=1. Hence find an expression for I_n in terms of n for n>=0.


i wish i knew how to type subscripts :mad1:

EDIT: Use and with <> brackets

 

[McLake]

I'm not sure how to do the first part, but the second part is a recursive deffinition, so it will be like a geometric progression, but with * instead of +.

[3n/(3n + 2)]*[(3n - 3)/(3n - 1)]*[(3n - 6)/(3n - 7)]* ..... * I₀

 

[marsenal]

I'll get you started. By parts let u=(1-x³)ⁿ and dv=x
In the intergral part you then get {-3nx⁴(1-x³)^n-1}/2 which actually looks like it's getting worse. However what you need to do is split the x⁴ into x{(1-x³)-1} and you should be able to get it from here.

 

[freaking_out]

split the x4 into x{(1-x3)-1} and you should be able to get it from here

hey, how the hell do u meant to figure that out in the exam.

anyways, thanks Mclake for telling me how to type in _subscripts.

 

[ToRnaDo]

Hey... guys. Im a new member!!

By parts let u=(1-x3)n and dv=x. U will come up with the result
In= 3n/2 I{o-->1} x4(1-x3)^(n-1) (*)

In other hand
In =I{0->1} x(1-x3)n = I{0-->1}(x - x^4)(1-x^3)^n
=In-1 - I{0-->1}x4(1-x3)^(n-1) (**)
U sub I{o-->1} x4(1-x3)n-1 = 2In/3n from (*) to (**) u will get the answer