2004 HSC paper Qu 8. b) (1 Viewer)

[roxy_chick]

Hey, I am stuck on all of this....does anyone have the answers for it? thanks....

 

[Necros87]

answers:

i) 1/(n+1)
ii) (-1)^n/(2n-1)

ps these solutions are a page, what part exactly

 

[Sepulchres]

http://www.boredofstudies.org/cours...nternational_School,_J._Hansen,_F._Haggar.pdf

 

[香港!]

Hey, I am stuck on all of this....does anyone have the answers for it? thanks....

Is that the integration one?
I_n=int. tan^n x dx {0 to pi\4}
i) I_n+I_(n+2)=int.[ tan^n x dx+ tan^(n+2) x] dx {both 0 to pi\4}
=int. tan^n x(1+tan²x) dx
=int. tan^n x sec²x dx
=int. tan^n x d(tanx) {0 to pi\4}
=[tan^(n+1) x \ (n+1)]
apply the terminals to get 1\(n+1)

 

[roxy_chick]

lol I meant the proofs for them...I suck at these type of questions so all of them...

 

[香港!]

ahh solutions!

 

[roxy_chick]

Is that the integration one?
I_n=int. tan^n x dx {0 to pi\4}
i) I_n+I_(n+2)=int.[ tan^n x dx+ tan^(n+2) x] dx {both 0 to pi\4}
=int. tan^n x(1+tan²x) dx
=int. tan^n x sec²x dx
=int. tan^n x d(tanx) {0 to pi\4}
=[tan^(n+1) x \ (n+1)]
apply the terminals to get 1\(n+1)

yep, thanks

 

[underthebridge]

I couldn't be bothered with all the working, so I'll tell you the main thing to do for each part:
i) Work out In+2 using integration by parts and then take the In across to the other side.

ii) Use the given thing for Jn to work out Jn - Jn-1 and then use the result u proved in i), only the values for n are a little different.

iii) with the sigma notation part, replace the thing after the sigma sign with Jn - Jn-1 since it was proved in part ii), and then work out the sum of all those things, and u will find that almost all the terms cancel out. What u are left with is Jo and Jm. Work out Jo and it equals pi on 4.

iv) Just a normal substitution integral

v)In is greater than 0 because between x= 0 and 1 from iv), In is always positive or equal to zero. Using part i), In = 1/(n+1) - In+2. since In+2 is always positive (like In), In<=1/(n+1).

From the given formula for In, between x = 0 and pi on 4, tanx is less than one, thus for tanx to the power of n, as n gets larger, every value for tanx to the power of n gets smaller and smaller and thus as n approaches infinity, In and also I2n approahces 0. From the formula provided of Jn, since I2n is approaching zero, Jn also approaches zero.

 

[roxy_chick]

I couldn't be bothered with all the working, so I'll tell you the main thing to do for each part:
i) Work out In+2 using integration by parts and then take the In across to the other side.

ii) Use the given thing for Jn to work out Jn - Jn-1 and then use the result u proved in i), only the values for n are a little different.

iii) with the sigma notation part, replace the thing after the sigma sign with Jn - Jn-1 since it was proved in part ii), and then work out the sum of all those things, and u will find that almost all the terms cancel out. What u are left with is Jo and Jm. Work out Jo and it equals pi on 4.

iv) Just a normal substitution integral

v)In is greater than 0 because between x= 0 and 1 from iv), In is always positive or equal to zero. Using part i), In = 1/(n+1) - In+2. since In+2 is always positive (like In), In<=1/(n+1).

From the given formula for In, between x = 0 and pi on 4, tanx is less than one, thus for tanx to the power of n, as n gets larger, every value for tanx to the power of n gets smaller and smaller and thus as n approaches infinity, In and also I2n approahces 0. From the formula provided of Jn, since I2n is approaching zero, Jn also approaches zero.

Thanks heaps!!!

 

[tim_dawborn]

nice answers