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Mysticmin
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#1
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I've just spent several rather ugly hours doing reduction formulae for my maths homework, and a question is still bloody irritating me, so all help is appreciated:

1) Given that

In = Integral (from1 to 0) of 1/[(1+x^2)^n]dx, show that for n>2

2(n-1)In = 2^(1-n) + (2n-3)In-1
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Mysticmin
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I'll give you rep... Plz? I'll honour you as the best mathematician on UKL if you can answer this...
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G4ry
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You have my sympathy, Maths A level was not fun. Ask rayhaydenuk, he's usually very good at answering maths questions.
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Mysticmin
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(Original post by G4ry)
You have my sympathy, Maths A level was not fun. Ask rayhaydenuk, he's usually very good at answering maths questions.
A-level maths is good, A-level further maths is not, especially when your teacher's so scared she won't finish the syllabus that she's rushing through one topic a lesson. cheers, will ask rayhaydenuk.
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Nylex
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I'd help, but I have never done reduction formulae (we haven't even covered them here).
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Mysticmin
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Anyone here actually doing a maths degree?!?! there must be someone... Where are all those aspiring Imperial maths graduates?
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G4ry
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(Original post by Mysticmin)
Anyone here actually doing a maths degree?!?! there must be someone... Where are all those aspiring Imperial maths graduates?
i believe bigcnee is doing a maths degree at Bath, if he's online is another matter.
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TheWolf
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(Original post by G4ry)
i believe bigcnee is doing a maths degree at Bath, if he's online is another matter.
i would love to help but my pure maths is at p2 level
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Mysticmin
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(Original post by G4ry)
i believe bigcnee is doing a maths degree at Bath, if he's online is another matter.
right, will see if i can find him, merci.
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G4ry
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(Original post by Mysticmin)
right, will see if i can find him, merci.
Send him and rayhaydenuk a PM, i'm sure they'd be happy to help when they come online.
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PhilC
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In=S 1/(1+xx)^n

In=uv-Sv du/dx dx

dv/dx=1 means v=x
u=1/(1+xx)^n means du/dx= -2nx/(1+xx)^n+1

follows that

In=x/(1+xx)^n + 2nS xx / (1+xx)^n+1

In= 2^-n + 2nS (1+xx)/(1+xx)^n+1 -2nS 1/(1+xx)^n+1

In= 2^-n +2nIn - 2nIn+1

Means

(2n)In+1 = 2^-n + (2n-1)In

Change n for n-1

2(n-1)In= 2^(1-n) + (2n-3)In-1


QED
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Mysticmin
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(Original post by PhilC)
In=S 1/(1+xx)^n

In=uv-Sv du/dx dx

dv/dx=1 means v=x
u=1/(1+xx)^n means du/dx= -2nx/(1+xx)^n+1

follows that

In=x/(1+xx)^n + 2nS xx / (1+xx)^n+1

In= 2^-n + 2nS (1+xx)/(1+xx)^n+1 -2nS 1/(1+xx)^n+1

In= 2^-n +2nIn - 2nIn+1

Means

(2n)In+1 = 2^-n + (2n-1)In

Change n for n-1

2(n-1)In= 2^(1-n) + (2n-3)In-1


QED
Thank you...now how do you give someone rep? I can't find a button to click.
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TheWolf
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(Original post by Mysticmin)
Thank you...now how do you give someone rep? I can't find a button to click.
next to the edit button of this post
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hungryhound
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I just did this problem, tried to post, got server too busy, then someone else posted...


ohwell

my solution was slightly different, dunno if you want it...

PS. hello again Gary

JM
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Mysticmin
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(Original post by hungryhound)
I just did this problem, tried to post, got server too busy, then someone else posted...


ohwell

my solution was slightly different, dunno if you want it...

PS. hello again Gary

JM
Post it anyway? I hate not being able to do maths, had a mini tantrum before posting this.
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G4ry
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(Original post by hungryhound)
I just did this problem, tried to post, got server too busy, then someone else posted...


ohwell

my solution was slightly different, dunno if you want it...

PS. hello again Gary

JM
Hi, is this a permanent thing or a fleeting visit? I'll pay your forum a visit.
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hungryhound
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(Original post by G4ry)
Hi, is this a permanent thing or a fleeting visit? I'll pay your forum a visit.
I think it's more of a fleeting visit, no time left to do anything any more

As for that solution - it works easier if you transform using the substitution x=tan u and changing the 1 and 0 limits to pi/4 and 0. You should get In = integral cos^2n-2 u du . then do the integration by parts. let one part be sec^2 u and the other be cos^2n u. After you do that, transform using n=n-1 and it should give you the correct solution

JM
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