Cool Integration Problem (2 Viewers)

[Fus Ro Dah]

Here is a neat result.

Prove that for all natural numbers N,

 

[Carrotsticks]

Re: HSC 2012 Marathon

Here is a neat result.

Prove that for all natural numbers N,

Haha although I do appreciate your enthusiasm, remember that this thread is for MX1. Your question is more suitable for MX2 students. There are a few ways of doing it though (I'm saying this because many MX2 students lurk here).

Method #1: Make the recursive formula for the integral of cos^n(x) using IBP and derive the double factorial expression which split into 2 cases. Replace n with 2n so it forces us to take the even case, then simplify the double factorial and the expression above follows.

Method #2: Use Complex Numbers with the whole cos(x) = 1/2 (z+1/z), raise both sides to the power of n, then use the Binomial Expansion. Integrate the entire expression over pi/2 to 0 and a big chunk of it will cancel out, leaving the desired expression.

 

[RealiseNothing]

http://community.boredofstudies.org/showthread.php?t=284850

I already proved this in this thread, but replace cos with sin (it's the same result anyway as the two integrals are equal).

 

[Carrotsticks]

http://community.boredofstudies.org/showthread.php?t=284850

I already proved this in this thread, but replace cos with sin (it's the same result anyway as the two integrals are equal).

Yep, this is Method #1 that I had. Try doing it Method 2.

 

[RealiseNothing]

Yep, this is Method #1 that I had. Try doing it Method 2.

I'll give it a go tomorrow afternoon since my 4U test will be over (ie I'd rather get some sleep tonight).

 

[Godmode]

Wasn't this a past paper question?

If I remember what i did the first time i tried this question, for carrotsticks' method #2, you go and prove:

i. z^n + z^-n = 2cos(ntheta) by de moivre's theorem & some manipulation

ii. use this to then expand (2cos(theta))^2n = some binomial mess which i won't even attempt to type

iii. and to prove that the integral in question is true, you sub in the result of ii. and all of the cos((2n + k)theta) bits integrate into something like sin((2n + k)theta)/(2n + k), which all cancel out anyway because sin(2k pi/2) = 0 for some integer k

hence the remaining term of the expansion will be the RHS of the intergal

 

[Carrotsticks]

Wasn't this a past paper question?

If I remember what i did the first time i tried this question, for carrotsticks' method #2, you go and prove:

i. z^n + z^-n = 2cos(ntheta) by de moivre's theorem & some manipulation

ii. use this to then expand (2cos(theta))^2n = some binomial mess which i won't even attempt to type

iii. and to prove that the integral in question is true, you sub in the result of ii. and all of the cos((2n + k)theta) bits integrate into something like sin((2n + k)theta)/(2n + k), which all cancel out anyway because sin(2k pi/2) = 0 for some integer k

hence the remaining term of the expansion will be the RHS of the intergal

It may very well be a past HSC question, problems like these are very commonly asked. Proving Trigonometric Series by means of Complex Numbers and expressions.

Easily the best way to prove various Trig series due to the nature of De Moivre's Theorem transferring an exponent to a constant multiple of the variable.

 

[Godmode]

Ha! I found it. It was Q7 b from the '09 paper.

 

[Carrotsticks]

Nice find! I had no idea.

 

[Godmode]

It helps having a past paper book within arm's reach lol.

Now, what happens to the integral in question for negative numbers?

Although de moivre's theorem extends to negatives, I'm assuming the binomial expansion is radically different (if it exists)

 

[Carrotsticks]

It helps having a past paper book within arm's reach lol.

Now, what happens to the integral in question for negative numbers?

Although de moivre's theorem extends to negatives, the binomial expansion is radically different (assuming it exists)

Did you mean for negative limits? I don't fully understand what you're saying.

And yes Binomial Expansions are defined for negative terms, but that is venturing out of the HSC Syllabus.

 

[Godmode]

Sorry, I meant for replacing the exponent of the cosine in the integral, with a negative.

I was thinking about it because for part i. of the solution, as using a negative would result in cos(-2ntheta), which could be rewritten as cos(2ntheta)

I've heard about using the gamma function to do stuff like negative factorials and the like, but it looks too daunting for now; I am more worried about getting to that stage of mathematics first.

 

[seanieg89]

It is not particularly hard to integrate this for negative n (we are just finding the primitive of an even power of sec), although between the two limits given here such an integral will not converge.

 

[Godmode]

Ahh, of course. I forgot about turning it into a sec-based problem, but I am unfamiliar with the relationship between integration and convergence/divergence. I was too busy trying to think of some way I could express this new integral in terms of the original.

I wish this damn HSC would hurry up and finish so I can go learn this all in uni. This level of Series, Sequences and Integration is in 1st year uni, isn't it?

 

[seanieg89]

Yep first year. Its nothing fancy, just sec blows up close to pi/2, and does so fast enough for the integral to not exist. You can check this using the definition of an improper integral.

If you are interested in getting a head start on the early uni stuff check out Spivak's "Calculus"...pretty much the best source to learn calculus from properly.

 

[Godmode]

Thanks man, I'll have a look for it.

 

[Carrotsticks]

Ahh, of course. I forgot about turning it into a sec-based problem, but I am unfamiliar with the relationship between integration and convergence/divergence. I was too busy trying to think of some way I could express this new integral in terms of the original.

I wish this damn HSC would hurry up and finish so I can go learn this all in uni. This level of Series, Sequences and Integration is in 1st year uni, isn't it?

It's also done in second year in Analysis.

You remind me of myself when I was in Year 12. I love Infinite Series and cool integrals.

 

[Godmode]

I can't be the only one who has/had their teachers tell them to shut up in class because they were rambling on and on about tertiary-level maths.

 

[Shadowdude]

I can't be the only one who has/had their teachers tell them to shut up in class because they were rambling on and on about tertiary-level maths.

we have those people at uni too

and tbh, we don't really like them because they sound stuck-up, pompous and arrogant


So... just a heads up. Like, for example, I was in a car today with some friends and we got on the subject of "him" and it was basically 15 minutes of mocking pretty much everything about him because of that sort of thing - going on about higher levels of maths and whatnot.

 

[Godmode]

Fortunately, I have improved somewhat in that regard... I shut up now.

Besides, now I just tell silly stories of my various adventures, people tend to like them more.