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please solve this (1 Viewer)
- Thread starter ..:''ooo
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http://www.geocities.com/fourunitmaths/sgs2004.pdf
8 bi) At the start of this you're establishing an inequality under the condition that 0 < x < 1 .... I'm not too hot at inequalities but from this I believe it follows that:
a) 0 < x<sup>n</sup> < 1
b) 0 < (1 - x)<sup>n</sup> < 1
c) 0 < sinπx < 1 ... you can establish this pretty quickly by considering the graph.
If you combine all these you should have little trouble setting up the inequality. For the next part you know that the inequality holds true when 0 < x < 1 which means you can integrate the inequality between 0 and 1 and it will still hold true.
8 bii) The start of this is pretty abstract but if you take d<sup>2</sup>/dx<sup>2</sup>.F<sub>n</sub>(x) then you end up with:
b<sup>n</sup>(π<sup>2n</sup>f<sub>n</sub><sup>(2)</sup>(x) - π<sup>2n-2</sup>f<sub>n</sub><sup>(4)</sup>(x) + ... + (-1)π<sup>2</sup><sup>n-1</sup>f<sub>n</sub><sup>(2n)</sup>(x))
If you look at this you'll see that it is similar to F<sub>n</sub>(x) but the signs are around the wrong way, there is an extra π<sup>2</sup> term and the undifferentiated f<sub>n</sub>(x) is missing. Move things about and add stuff in to get an F<sub>n</sub>(x) term out of it.
Have a play around with it yourself for a bit and see if it works out for you. If it throws up brick walls then don't hesitate to give us a holler.
8 bi) At the start of this you're establishing an inequality under the condition that 0 < x < 1 .... I'm not too hot at inequalities but from this I believe it follows that:
a) 0 < x<sup>n</sup> < 1
b) 0 < (1 - x)<sup>n</sup> < 1
c) 0 < sinπx < 1 ... you can establish this pretty quickly by considering the graph.
If you combine all these you should have little trouble setting up the inequality. For the next part you know that the inequality holds true when 0 < x < 1 which means you can integrate the inequality between 0 and 1 and it will still hold true.
8 bii) The start of this is pretty abstract but if you take d<sup>2</sup>/dx<sup>2</sup>.F<sub>n</sub>(x) then you end up with:
b<sup>n</sup>(π<sup>2n</sup>f<sub>n</sub><sup>(2)</sup>(x) - π<sup>2n-2</sup>f<sub>n</sub><sup>(4)</sup>(x) + ... + (-1)π<sup>2</sup><sup>n-1</sup>f<sub>n</sub><sup>(2n)</sup>(x))
If you look at this you'll see that it is similar to F<sub>n</sub>(x) but the signs are around the wrong way, there is an extra π<sup>2</sup> term and the undifferentiated f<sub>n</sub>(x) is missing. Move things about and add stuff in to get an F<sub>n</sub>(x) term out of it.
Have a play around with it yourself for a bit and see if it works out for you. If it throws up brick walls then don't hesitate to give us a holler.
I don't want to post up my entire solution which is bulky at best. I'd like to point out that this question uses a similar idea to the 2003 HSC q8, as well as Bill Pender's Harder 3u inservice from several years ago. With that said, let's do the question. 
i)a)
Start with the fact that x^n, (1-x)^n and sin(pix) are all between 0 and 1 and work from there perhaps you will note that the product of these 3 expressions must also be between 0 and 1 !
i)b)
Use the result of i)a) and integrate between 0 and 1 with respect to x
ii)a)
Start with LHS and differentiate Fn(x) twice. You will notice you get a Fn(2n+2)(x) term! This is a problem! Until you remember how Fn(x) was originally defined! It is of degreen 2n! Hence, its 2n+2th derivative is... 0!
ii)b)
Just expand using product rule... stuff will cancel You will need to use the result of ii)a) to evaluate the F''n(x) term.
ii)c)
Take the result of ii)b) and integrate both sides between 0 and 1. Easy!!
ii)d)
Fn(0) and Fn(1) are integers [assumed] and hence that expression you evaluated in ii)c) is also an integer! But with some maniupalation using the earlier parts, you can show that as n approaches infinity, your middle expression ni the three-way inequality is sandwiched to 0! (ie. 0 < STUFF < 0). Hence "STUFF" cannot be an integer! Thus our assumptions breakdown, and we deduce pi^2 CANNOT be written in the form a/b and it is irrational!
i)a)
Start with the fact that x^n, (1-x)^n and sin(pix) are all between 0 and 1 and work from there
perhaps you will note that the product of these 3 expressions must also be between 0 and 1 !i)b)
Use the result of i)a) and integrate between 0 and 1 with respect to x
ii)a)
Start with LHS and differentiate Fn(x) twice. You will notice you get a Fn(2n+2)(x) term! This is a problem! Until you remember how Fn(x) was originally defined! It is of degreen 2n! Hence, its 2n+2th derivative is... 0!
ii)b)
Just expand using product rule... stuff will cancel
You will need to use the result of ii)a) to evaluate the F''n(x) term.ii)c)
Take the result of ii)b) and integrate both sides between 0 and 1. Easy!!
ii)d)
Fn(0) and Fn(1) are integers [assumed] and hence that expression you evaluated in ii)c) is also an integer! But with some maniupalation using the earlier parts, you can show that as n approaches infinity, your middle expression ni the three-way inequality is sandwiched to 0! (ie. 0 < STUFF < 0). Hence "STUFF" cannot be an integer! Thus our assumptions breakdown, and we deduce pi^2 CANNOT be written in the form a/b and it is irrational!