Integration (2 Viewers)

Woteva636 · May 20, 2009

study-freak said:
$LHS=U(n+1) + U(n-1) - \frac{5}{2}U (n) \\\because $We are given that \int_{0}^{\pi}\frac{cosnx}{5-4cosx}dx, \\$ LHS=\int_{0}^{\pi}\frac{cos(n+1)x+cox(n-1)x- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{2coxnxcosx- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{-\frac{1}{2}cosnx(5-4cosx)}{5-4cosx}dx \\=-\frac{1}{2}\int_{0}^{\pi}cosnxdx \\==-\frac{1}{2n}\left [ sinnx \right ]_{0}^{\pi} \\=0 \\=RHS \\Q.E.D. \\Now, U(0) \\=\int_{0}^{\pi}\frac{1}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{2}{5-4cosx}dx \\=\frac{2}{5}[U(1)+U(-1)] \\=\int_{0}^{\pi}\frac{cosx+cos(-x)}{5-4cosx}dx \\=2\int_{0}^{\pi}\frac{cosx}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{4cosx}{5-4cosx}dx \\\therefore \frac{1}{2}\int_{0}^{\pi}\frac{2-4cosx}{5-4cosx}dx=0 \\\frac{1}{2}\int_{0}^{\pi}(1-\frac{3}{5-4cosx})dx=0 \\\frac{1}{2}(\pi-3U(0))=0 \\U(0)=\frac{\pi}{3}$
Took a long time to do this... It's an interesting question though.
Where did you get this question?

Holy shit, i havnt started intergration yet, i'm a bit behind in the course, but now i'm actually dreading it.

+ rep for that answer though

Timothy.Siu · May 20, 2009

Woteva636 said:
Holy shit, i havnt started intergration yet, i'm a bit behind in the course, but now i'm actually dreading it.

+ rep for that answer though

dw, its actually one of the easiest topics in 4unit, although that question is quite hard.
those questions are pretty rare though.

eldore44 · May 20, 2009

Just wondering, in studyfreaks solution, what happened to the 2/5 in the third line of the second part?

Petyo · May 20, 2009

@@ 'checkin..

Timothy.Siu · May 20, 2009

eldore44 said:
Just wondering, in studyfreaks solution, what happened to the 2/5 in the third line of the second part?

lol nice pick up, i have no idea what happenned to it, i guess someone else can explain it

eldore44 · May 20, 2009

study-freak said:
$tan\frac{\pi}{2}=undefined \\\therefore 3tan\frac{\pi}{2}=undefined \\tan^{-1}(3tan\frac{\pi}{2}) \\=tan^{-1}(undefined) \\=\frac{\pi}{2}$

This seemed quite strange to me, however, my calculator agrees that you can do this kind of thing.
However, it gives me an answer of 3pi/2. Seems reasonable enough.

Petyo · May 20, 2009

study-freak said:
$tan\frac{\pi}{2}=undefined \\\therefore 3tan\frac{\pi}{2}=undefined \\tan^{-1}(3tan\frac{\pi}{2}) \\=tan^{-1}(undefined) \\=\frac{\pi}{2}$

Can we really write like this in the exams?

eldore44 · May 20, 2009

Petyo said:
Can we really write like this in the exams?

Definitely not I would say... I think it is mathematical nonsense

and btw if we do the working with tan properly, we get the answer as Pi

study-freak · May 20, 2009

eldore44 said:
Just wondering, in studyfreaks solution, what happened to the 2/5 in the third line of the second part?

Whoops, made a mistake
solving and writing at the same time in LaTex is quite hard lol...
I'll fix it up later

Drongoski · May 20, 2009

study-freak said:
Yea, that works nicely but the question says "then find U(0)" or something similar so I thought I should use the previous part of the question.

"then find U(0)" could (but maybe not) mean, "after this, go and find . . . " but "thence find U(0)" or "hence find U(0)" would definitely require you to use results just established.

Petyo · May 20, 2009

study-freak said:
Whoops, made a mistake
solving and writing at the same time in LaTex is quite hard lol...
I'll fix it up later

I don't think it's a typing mistake though @@

study-freak · May 20, 2009

Petyo said:
I don't think it's a typing mistake though @@

well i was solving at the same time i was typing

study-freak · May 20, 2009

Petyo · May 20, 2009

I'm not used to Latex so I only wrote my working on papers...
Anyway, if the (2/5) from the third or fourth line is supposed to be 'there' then your method will not lead to the answer pi/3
How come you Study-freak and Drongoski both arrived at the same answer? @@

Petyo · May 20, 2009

study-freak said:
$U(0) \\=\frac{2}{5}[U(1)+U(-1)] \\=\frac{2}{5}\int_{0}^{\pi}\frac{cosx+cos(-x)}{5-4cosx}dx \\=\frac{1}{5}\int_{0}^{\pi}\frac{4cosx-5+5}{5-4cosx}dx \\=\frac{1}{5}\int_{0}^{\pi}(-1+\frac{5}{5-4cosx})dx \\=\frac{-\pi}{5}+U(0) \\\frac{\pi}{5}=0$ which is not true... LOL This method doesn't work now.. Can someone explain this?$

Alright that was what happened to me = =!
I think the only possible explanation is that the question itself is wrong!
Is there anyone who has done or seen this question before???

eldore44 · May 20, 2009

I think the same answers were just a complete coincidence.
Using the t method we get the answer as pi. There is no reason not to believe this.
I get the same answer with the other method, I don't know what it is. There doesnt seem to be any problem with the proof or with the method, so we shouldn't be able to blame the question.
However, I'm gonna go ahead and say the question is flawed. : )

study-freak · May 20, 2009

eldore44How did you handle tan(pi/2) when you used the t method?

lolokay · May 20, 2009

n=/=0 for the condition to hold, as you can see from doing the working of the first part

were there any more parts to the question? eg. find U(0), hence find U(1)?

study-freak · May 20, 2009

lolokay said:
n=/=0 for the condition to hold, as you can see from doing the working of the first part

Oh true that...
No wonder people call you the maths version of Einstein in BoS lol

Petyo · May 20, 2009

lolokay said:
n=/=0 for the condition to hold, as you can see from doing the working of the first part

were there any more parts to the question?

No. that was the full question.
btw can you explain a bit more, what does " n=/=0" mean?

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